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Patrice Tauvel Rupert W. T. Yu

Lie Algebras and Algebraic Groups With 44 Figures

123

Patrice Tauvel Rupert W. T. Yu Département de Mathématiques Université de Poitiers Boulevard Marie et Pierre Curie, Téléport 2 – BP 30179 86962 Futuroscope Chasseneuil cedex, France e-mail: [email protected] [email protected]

Library of Congress Control Number: 2005922400

Mathematics Subject Classification (2000): 17-01, 17-02, 17Bxx, 20Gxx ISSN 1439-7382 ISBN-10 3-540-24170-1 Springer Berlin Heidelberg New York ISBN-13 978-3-540-24170-6 Springer Berlin Heidelberg New York This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable for prosecution under the German Copyright Law. Springer is a part of Springer Science+Business Media springeronline.com © Springer-Verlag Berlin Heidelberg 2005 Printed in Germany The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Typesetting: by the authors and TechBooks using a Springer LATEX macro package Cover design: Erich Kirchner, Heidelberg Printed on acid-free paper

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Preface

The theory of groups and Lie algebras is interesting for many reasons. In the mathematical viewpoint, it employs at the same time algebra, analysis and geometry. On the other hand, it intervenes in other areas of science, in particular in diﬀerent branches of physics and chemistry. It is an active domain of current research. One of the diﬃculties that graduate students or mathematicians interested in the theory come across, is the fact that the theory has very much advanced, and consequently, they need to read a vast amount of books and articles before they could tackle interesting problems. One of the goals we wish to achieve with this book is to assemble in a single volume the basis of the algebraic aspects of the theory of groups and Lie algebras. More precisely, we have presented the foundation of the study of ﬁnite-dimensional Lie algebras over an algebraically closed ﬁeld of characteristic zero. Here, the geometrical aspect is fundamental, and consequently, we need to use the notion of algebraic groups. One of the main diﬀerences between this book and many other books on the subject is that we give complete proofs for the relationships between algebraic groups and Lie algebras, instead of admitting them. We have also given the proofs of certain results on commutative algebra and algebraic geometry that we needed so as to make this book as selfcontained as possible. We believe that in this way, the book can be useful for both graduate students and mathematicians working in this area. Let us give a brief description of the material treated in this book. As we have stated earlier, our goal is to study Lie algebras over an algebraically closed ﬁeld of characteristic zero. This allows us to avoid, in considering questions concerning algebraic geometry, the notion of separability, which simpliﬁes considerably our presentation. In fact, under certain conditions of separability, the correspondence between Lie algebras and algebraic groups described in chapter 24 has a very nice generalization when the algebraically closed base ﬁeld has prime characteristic.

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Chapters 1 to 9 treat basic results on topology, commutative algebra and sheaves of functions that are required in the rest of the book. In chapter 10, we recall some standard results on Jordan decompositions and the theory of abstract groups and group actions. Here, the base ﬁeld is assumed to be algebraically closed in order to obtain a Jordan decomposition. Chapters 11 to 17 give an introduction to the theory of algebraic geometry which we shall encounter continually in the chapters which follow. We have selected only the notions that we require in this book. The reader should by no means consider these chapters as a thorough introduction to the theory of algebraic geometry. Chapters 18 and 36 are dedicated to root systems which are fundamental to the study of semisimple Lie algebras. We introduce Lie algebras in chapter 19. In this chapter, we prove important results on the structure of Lie algebras such as Engel’s theorem, Lie’s theorem and Cartan’s criterion on solvability. In chapter 20, we deﬁne the notions of semisimple and reductive Lie algebras. In addition to characterizing these Lie algebras, we discover in this chapter how the structure of semisimple Lie algebras can be related to root systems. The general theory of algebraic groups is studied in chapters 21 to 28. The relations between Lie algebras and algebraic groups, which are fundamental to us, are established in chapters 23 and 24. Chapter 29 presents applications of these relations to tackle the systematic study of Lie algebras. The reader will observe that the geometrical aspects have an important part in this study. In particular, the orbits of points under the action of an algebraic group plays a central role. Chapter 30 gives a short introduction of the theory of representations of semisimple Lie algebras which we need in order to prove Chevalley’s theorem on invariants in chapter 31. We deﬁne in chapter 32 S-triples which are essential to the study of semisimple Lie algebras. Another fundamental notion, treated in chapters 33 to 35, is the notion of nilpotent orbits in semisimple Lie algebras. We introduce symmetric Lie algebras in chapter 37, and semisimple symmetric Lie algebras in chapter 38. In these chapters, we give generalizations of certain results of chapters 32 to 35. In addition to presenting the essential classical results of the theory, some of the results we have included in the ﬁnal chapters are recent, and some are yet to be published. At the end of each chapter, the reader may ﬁnd a list of relevant references, and in some cases, remarks concerning the contents of the chapter. There are many approaches to reading this book. We need not read this book linearly. A reader familiar with the theory of commutative algebra may skip chapters 2 to 8, and consider these chapters for references only. Let us also point out that chapters 18, 19 and 20 constitute a short introduction to

Preface

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the theory of ﬁnite-dimensional Lie algebras and the structure of semisimple Lie algebras. We wish to thank our colleagues A. Bouaziz and H. Sabourin of the university of Poitiers with whom we had many useful discussion during the preparation of this book.

Poitiers, January 2005

Patrice Tauvel Rupert W.T. Yu

Contents

1

Results on topological spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Irreducible sets and spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Noetherian spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Constructible sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Gluing topological spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 4 5 6 8

2

Rings and modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Prime and maximal ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Rings of fractions and localization . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Localizations of modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Radical of an ideal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Local rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Noetherian rings and modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9 Module of diﬀerentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11 11 12 13 17 18 19 21 24 25

3

Integral extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Integral dependence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Integrally closed domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Extensions of prime ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

31 31 33 35

4

Factorial rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Unique factorization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Principal ideal domains and Euclidean domains . . . . . . . . . . . . . 4.4 Polynomials and factorial rings . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Symmetric polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Resultant and discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

39 39 41 43 45 48 50

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Field extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Algebraic and transcendental elements . . . . . . . . . . . . . . . . . . . . . 5.3 Algebraic extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Transcendence basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Norm and trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Theorem of the primitive element . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 Going Down Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8 Fields and derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.9 Conductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

55 55 56 56 58 60 62 64 67 70

6

Finitely generated algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Noether’s Normalization Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Krull’s Principal Ideal Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Maximal ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Zariski topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

75 75 76 81 82 84

7

Gradings and ﬁltrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Graded rings and graded modules . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Graded submodules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Filtrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Grading associated to a ﬁltration . . . . . . . . . . . . . . . . . . . . . . . . .

87 87 88 90 91 92

8

Inductive limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 8.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 8.2 Inductive systems of maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 8.3 Inductive systems of magmas, groups and rings . . . . . . . . . . . . . 98 8.4 An example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 8.5 Inductive systems of algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

9

Sheaves of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 9.1 Sheaves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 9.2 Morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 9.3 Sheaf associated to a presheaf . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 9.4 Gluing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 9.5 Ringed space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

10 Jordan decomposition and some basic results on groups . . . 113 10.1 Jordan decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 10.2 Generalities on groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 10.3 Commutators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 10.4 Solvable groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 10.5 Nilpotent groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

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10.6 Group actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 10.7 Generalities on representations . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 10.8 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 11 Algebraic sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 11.1 Aﬃne algebraic sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 11.2 Zariski topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 11.3 Regular functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 11.4 Morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 11.5 Examples of morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 11.6 Abstract algebraic sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 11.7 Principal open subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 11.8 Products of algebraic sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 12 Prevarieties and varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 12.1 Structure sheaf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 12.2 Algebraic prevarieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 12.3 Morphisms of prevarieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 12.4 Products of prevarieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 12.5 Algebraic varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 12.6 Gluing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 12.7 Rational functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 12.8 Local rings of a variety . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 13 Projective varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 13.1 Projective spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 13.2 Projective spaces and varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 13.3 Cones and projective varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 13.4 Complete varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 13.5 Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 13.6 Grassmannian variety . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180 14 Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 14.1 Dimension of varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 14.2 Dimension and the number of equations . . . . . . . . . . . . . . . . . . . 185 14.3 System of parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 14.4 Counterexamples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 15 Morphisms and dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 15.1 Criterion of aﬃneness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 15.2 Aﬃne morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 15.3 Finite morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 15.4 Factorization and applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 15.5 Dimension of ﬁbres of a morphism . . . . . . . . . . . . . . . . . . . . . . . . 199 15.6 An example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203

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16 Tangent spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 16.1 A ﬁrst approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 16.2 Zariski tangent space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 16.3 Diﬀerential of a morphism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 16.4 Some lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 16.5 Smooth points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 17 Normal varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 17.1 Normal varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 17.2 Normalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 17.3 Products of normal varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 17.4 Properties of normal varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 18 Root systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 18.1 Reﬂections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 18.2 Root systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 18.3 Root systems and bilinear forms . . . . . . . . . . . . . . . . . . . . . . . . . . 238 18.4 Passage to the ﬁeld of real numbers . . . . . . . . . . . . . . . . . . . . . . . 239 18.5 Relations between two roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 18.6 Examples of root systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 18.7 Base of a root system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244 18.8 Weyl chambers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 18.9 Highest root . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 18.10 Closed subsets of roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 18.11 Weights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 18.12 Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 18.13 Dynkin diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256 18.14 Classiﬁcation of root systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 19 Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 19.1 Generalities on Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 19.2 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279 19.3 Nilpotent Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 19.4 Solvable Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286 19.5 Radical and the largest nilpotent ideal . . . . . . . . . . . . . . . . . . . . . 289 19.6 Nilpotent radical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 19.7 Regular linear forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 19.8 Cartan subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294 20 Semisimple and reductive Lie algebras . . . . . . . . . . . . . . . . . . . . . 299 20.1 Semisimple Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299 20.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301 20.3 Semisimplicity of representations . . . . . . . . . . . . . . . . . . . . . . . . . . 302 20.4 Semisimple and nilpotent elements . . . . . . . . . . . . . . . . . . . . . . . . 305 20.5 Reductive Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307

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20.6 Results on the structure of semisimple Lie algebras . . . . . . . . . . 310 20.7 Subalgebras of semisimple Lie algebras . . . . . . . . . . . . . . . . . . . . 313 20.8 Parabolic subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 21 Algebraic groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 21.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 21.2 Subgroups and morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 21.3 Connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322 21.4 Actions of an algebraic group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 21.5 Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326 21.6 Group closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327 22 Aﬃne algebraic groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 22.1 Translations of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 22.2 Jordan decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333 22.3 Unipotent groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 22.4 Characters and weights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338 22.5 Tori and diagonalizable groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 340 22.6 Groups of dimension one . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 23 Lie algebra of an algebraic group . . . . . . . . . . . . . . . . . . . . . . . . . . 347 23.1 An associative algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 23.2 Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348 23.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352 23.4 Computing diﬀerentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354 23.5 Adjoint representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 23.6 Jordan decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362 24 Correspondence between groups and Lie algebras . . . . . . . . . . 365 24.1 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 24.2 An algebraic subgroup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 24.3 Invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368 24.4 Functorial properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372 24.5 Algebraic Lie subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375 24.6 A particular case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380 24.7 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383 24.8 Algebraic adjoint group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383 25 Homogeneous spaces and quotients . . . . . . . . . . . . . . . . . . . . . . . . 387 25.1 Homogeneous spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387 25.2 Some remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389 25.3 Geometric quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391 25.4 Quotient by a subgroup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393 25.5 The case of ﬁnite groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397

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26 Solvable groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401 26.1 Conjugacy classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401 26.2 Actions of diagonalizable groups . . . . . . . . . . . . . . . . . . . . . . . . . . 405 26.3 Fixed points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406 26.4 Properties of solvable groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407 26.5 Structure of solvable groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 27 Reductive groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413 27.1 Radical and unipotent radical . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413 27.2 Semisimple and reductive groups . . . . . . . . . . . . . . . . . . . . . . . . . . 415 27.3 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 27.4 Finiteness properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420 27.5 Algebraic quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422 27.6 Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424 28 Borel subgroups, parabolic subgroups, Cartan subgroups . . 429 28.1 Borel subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429 28.2 Theorems of density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432 28.3 Centralizers and tori . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434 28.4 Properties of parabolic subgroups . . . . . . . . . . . . . . . . . . . . . . . . . 435 28.5 Cartan subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437 29 Cartan subalgebras, Borel subalgebras and parabolic subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441 29.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441 29.2 Cartan subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443 29.3 Applications to semisimple Lie algebras . . . . . . . . . . . . . . . . . . . . 446 29.4 Borel subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 29.5 Properties of parabolic subalgebras . . . . . . . . . . . . . . . . . . . . . . . . 450 29.6 More on reductive Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . 453 29.7 Other applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454 29.8 Maximal subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456 30 Representations of semisimple Lie algebras . . . . . . . . . . . . . . . . 459 30.1 Enveloping algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459 30.2 Weights and primitive elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 461 30.3 Finite-dimensional modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463 30.4 Verma modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464 30.5 Results on existence and uniqueness . . . . . . . . . . . . . . . . . . . . . . . 467 30.6 A property of the Weyl group . . . . . . . . . . . . . . . . . . . . . . . . . . . . 469 31 Symmetric invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471 31.1 Invariants of ﬁnite groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471 31.2 Invariant polynomial functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475 31.3 A free module . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478

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32 S-triples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481 32.1 Jacobson-Morosov Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481 32.2 Some lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484 32.3 Conjugation of S-triples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487 32.4 Characteristic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488 32.5 Regular and principal elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 489 33 Polarizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493 33.1 Deﬁnition of polarizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493 33.2 Polarizations in the semisimple case . . . . . . . . . . . . . . . . . . . . . . . 494 33.3 A non-polarizable element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497 33.4 Polarizable elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 499 33.5 Richardson’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502 34 Results on orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507 34.1 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507 34.2 Some lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 508 34.3 Generalities on orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509 34.4 Minimal nilpotent orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511 34.5 Subregular nilpotent orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513 34.6 Dimension of nilpotent orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 517 34.7 Prehomogeneous spaces of parabolic type . . . . . . . . . . . . . . . . . . 518 35 Centralizers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521 35.1 Distinguished elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521 35.2 Distinguished parabolic subalgebras . . . . . . . . . . . . . . . . . . . . . . . 523 35.3 Double centralizers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525 35.4 Normalizers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 528 35.5 A semisimple Lie subalgebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 530 35.6 Centralizers and regular elements . . . . . . . . . . . . . . . . . . . . . . . . . 533 36 σ-root systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537 36.1 Deﬁnition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537 36.2 Restricted root systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 539 36.3 Restriction of a root . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544 37 Symmetric Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549 37.1 Primary subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549 37.2 Deﬁnition of symmetric Lie algebras . . . . . . . . . . . . . . . . . . . . . . . 553 37.3 Natural subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554 37.4 Cartan subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555 37.5 The case of reductive Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . 557 37.6 Linear forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 559

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38 Semisimple symmetric Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . 561 38.1 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 561 38.2 Iwasawa decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 562 38.3 Coroots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565 38.4 Centralizers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 568 38.5 S-triples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 570 38.6 Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573 38.7 Symmetric invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579 38.8 Double centralizers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584 38.9 Normalizers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 588 38.10 Distinguished elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589 39 Sheets of Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593 39.1 Jordan classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593 39.2 Topology of Jordan classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 596 39.3 Sheets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 601 39.4 Dixmier sheets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603 39.5 Jordan classes in the symmetric case . . . . . . . . . . . . . . . . . . . . . . 605 39.6 Sheets in the symmetric case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608 40 Index and linear forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 611 40.1 Stable linear forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 611 40.2 Index of a representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 615 40.3 Some useful inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 616 40.4 Index and semi-direct products . . . . . . . . . . . . . . . . . . . . . . . . . . . 618 40.5 Heisenberg algebras in semisimple Lie algebras . . . . . . . . . . . . . 621 40.6 Index of Lie subalgebras of Borel subalgebras . . . . . . . . . . . . . . . 625 40.7 Seaweed Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 629 40.8 An upper bound for the index . . . . . . . . . . . . . . . . . . . . . . . . . . . . 630 40.9 Cases where the bound is exact . . . . . . . . . . . . . . . . . . . . . . . . . . . 635 40.10 On the index of parabolic subalgebras . . . . . . . . . . . . . . . . . . . . . 638 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 641 List of notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 645 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 647

1 Results on topological spaces

In this chapter, we treat some basic notions of topology such as irreducible and constructible sets, dimension of a topological space, Noetherian space, which are fundamental in algebraic geometry.

1.1 Irreducible sets and spaces 1.1.1 Deﬁnition. A topological space X is said to be irreducible if any ﬁnite intersection of non-empty open subsets is non-empty. 1.1.2 It follows from the deﬁnition that an irreducible topological space is not empty. 1.1.3 Proposition. Let X be a non-empty topological space. Then the following conditions are equivalent: (i) X is irreducible. (ii) X is not the ﬁnite union of distinct proper closed subsets. (iii) X is not the union of two proper closed subsets. (iv) Any non-empty open subset of X is dense in X. (v) Any open subset of X is connected. Proof. The implications (i) ⇒ (ii) ⇒ (iii) are clear and (iii) ⇒ (iv) follows from the fact that a subset in X is dense if and only if it meets all nonempty open subsets. Now if U is a non-connected non-empty open subset, then U = U1 ∪ U2 where U1 , U2 are non-empty open subsets and U1 ∩ U2 = ∅. Thus (iv) ⇒ (v). The same argument gives (v) ⇒ (i). Remark. If X is irreducible, then it is connected. The converse is not true. 1.1.4 In the rest of this chapter, X is a topological space. A subset of X is called irreducible if it is non-empty and irreducible as a topological space. From the above deﬁnitions, the following result is clear.

2

1 Results on topological spaces

Proposition. Let A be a non-empty subset of X. Then the following conditions are equivalent: (i) A is irreducible. (ii) Let F1 , . . . , Fn be closed subsets of X such that A is contained in the union of the Fi ’s, then there exists j ∈ {1, . . . , n} such that A ⊂ Fj . (iii) Let U , V be open subsets of X such that U ∩ A and V ∩ A are nonempty, then U ∩ V ∩ A = ∅. 1.1.5 Proposition. Let A, B be subsets of X. (i) A is irreducible if and only if its closure A is irreducible. (ii) If A is irreducible and A ⊂ B ⊂ A, then B is irreducible. Proof. For any open subset U , we have U ∩ A = ∅ if and only if U ∩ A = ∅. So (i) and (ii) follow. 1.1.6 Proposition. (i) If X is irreducible, then any non-empty open subset of X is also irreducible. (ii) Let (Ui )i∈I be a covering of X by open subsets such that Ui ∩ Uj = ∅ for all i, j ∈ I. If all the Ui ’s are irreducible, then X is irreducible. Proof. (i) Let U , V be non-empty open subsets of X such that V ⊂ U . If X is irreducible, then V is dense in U . Thus U is irreducible. (ii) Let V be a non-empty open subset of X. There exists k ∈ I such that V ∩ Uk = ∅. Since Ui ∩ Uk = ∅ for all i ∈ I and V ∩ Uk is dense in Uk , V ∩ Ui ∩ Uk = ∅. Hence V ∩ Ui = ∅ for all i. It follows that V ∩ Ui is dense in Ui for all i ∈ I, so V is dense in X. 1.1.7 Proposition. Let Y be a topological space and f : X → Y a continuous map. (i) If A ⊂ X is irreducible, then f (A) is irreducible in Y . (ii) Suppose that Y is irreducible, f is an open map and that f −1 (y) is irreducible for all y ∈ Y . Then X is irreducible. Proof. (i) Let U, V be open subsets of Y such that U ∩ f (A) and V ∩ f (A) are non-empty. Then f −1 (U ) and f −1 (V ) are open subsets whose intersection with A is non-empty. It follows that f −1 (U ∩ V ) = f −1 (U ) ∩ f −1 (V ) meets A. Therefore U ∩ V meets f (A) and assertion (i) follows. (ii) Let U, V be non-empty open subsets of X. Since f is open and Y is irreducible, f (U ) meets f (V ) at some point y. Further, f −1 (y) is irreducible, therefore the open subsets U ∩ f −1 (y) and V ∩ f −1 (y) of f −1 (y) have nonempty intersection. Hence U ∩ V = ∅. 1.1.8 Remark. A map f : X → Y is called dominant if f (X) is dense in Y . It follows from 1.1.5 and 1.1.7 that if X is irreducible and f is continuous and dominant, then Y is irreducible.

1.1 Irreducible sets and spaces

3

1.1.9 Deﬁnition. A maximal irreducible subset of X will be called an irreducible component of X. 1.1.10 Proposition. (i) All irreducible components of X are closed. (ii) Any irreducible subset of X is contained in an irreducible component and X is the union of its irreducible components. Proof. (i) This is clear from 1.1.5. (ii) Let (Ai )i∈I be a family of irreducible subsets of X totally ordered by inclusion and A = i∈I Ai . Let U, V be open subsets which meet A. Since the family is totally ordered, there exist k ∈ I such that U ∩ Ak = ∅ and V ∩ Ak = ∅. Thus U ∩ V ∩ Ak = ∅ because Ak is irreducible. Now the ﬁrst assertion follows from Zorn’s lemma. The second assertion follows from the fact that any singleton in X is irreducible. 1.1.11 Corollary. Any connected component of X is a union of irreducible components of X. Proof. Any irreducible subset of X is connected, so it is contained in a connected component of X. Remark. Two distinct irreducible components of X can have a non-empty intersection. 1.1.12 Proposition. Let U ⊂ X be an open subset. The map Z → Z ∩ U is a bijection between the set of irreducible closed subsets of X meeting U and the set of irreducible subsets of U , closed in U . The reciprocal map is V → V . In particular, C → C ∩ U induces a bijection between the set of irreducible components of X meeting U and the set of irreducible components of U . Proof. Let Z be an irreducible closed subset of X meeting U . Since U ∩ Z is open in Z, it is irreducible by 1.1.6 and dense in Z. Therefore Z = Z ∩ U . Conversely, if V is an irreducible closed subset of U , then V is irreducible by 1.1.5 and V = U ∩ V . 1.1.13 Recall that A ⊂ X is nowhere dense in X if the interior of A is empty. Let us suppose that X has only a ﬁnite number of distinct irreducible n, denote by Ui = X \ ( j=i Xj ) the set components X1 , . . . , Xn . For 1 i of elements of X which are not in j=i Xj . Lemma. The set Ui is open in X and dense in Xi . The disjoint union of the Ui ’s is dense in X. For i = j, Xi ∩ Xj is nowhere dense in X. Proof. It is obviousthat Ui is open in X and is contained in Xi . If Ui is empty, then Xi ⊂ j=i Xj . Since Xi is irreducible, there exists k = i such that Xi = Xk which is impossible by our hypothesis. Thus Ui is dense in Xi since it is a non-empty open subset. The second assertion is now obvious.

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1 Results on topological spaces

Let i = j and U an open subset contained in Xi ∩ Xj . So U is contained in the union of the Xk ’s, k = i. Hence U ∩ Ui = ∅. In the same manner, we have U ∩ Uj = ∅. Thus U = ∅ since Ui ∪ Uj is dense in Xi ∪ Xj . 1.1.14 Lemma. Using the hypotheses of 1.1.13, suppose that X is a subset of a topological space Y . Then X1 , . . . , Xn are the distinct irreducible components of X. Proof. By 1.1.5, the Xi ’s are irreducible. Clearly X = X1 ∪ · · · ∪ Xn . If A is an irreducible subset of X, then it is contained in one of the Xi ’s. Finally, if Xi ⊂ Xj , then Xi ⊂ X ∩ Xj = Xj since Xj is closed in X. This contradicts our hypothesis and so the lemma follows.

1.2 Dimension 1.2.1 Deﬁnition. Let E be a set. A chain of subsets of length n of E is a strictly increasing sequence of subsets E0 ⊂ E1 ⊂ · · · ⊂ En . 1.2.2 Deﬁnition. The dimension of a non-empty topological space X is the supremum of the lengths of chains consisting of irreducible closed subsets. We shall denote it by dim X which is in N ∪ {+∞}. 1.2.3 Proposition. Let us assume that X is non-empty. (i) Let Y be a non-empty subset of X, then dim Y dim X. r (ii) Let X1 , . . . , Xr be closed subsets of X such that X = i=1 Xi , then dim X = max{dim Xi ; 1 i r}. r (iii) Let X1 , . . . , Xr be open subsets of X such that X = i=1 Xi , then dim X = max{dim Xi ; 1 i r}. (iv) Suppose that X is irreducible and dim X is ﬁnite. If Y is a closed subset such that dim Y = dim X, then X = Y . Proof. (i) Let Y0 ⊂ Y1 ⊂ · · · ⊂ Yn be a chain of irreducible closed subsets in Y . The closure Yi of Yi in X is irreducible by 1.1.5 and since Yi = Yi ∩ Y , Yi is strictly contained in Yi+1 . Thus dim Y dim X. (ii) Let Z0 ⊂ Z1 ⊂ · · · ⊂ Zn be a chain of irreducible closed subsets in X. Since Zn is irreducible and Zn = (Zn ∩ X1 ) ∪ · · · ∪ (Zn ∩ Xr ), there exists k such that Zn ⊂ Xk . It follows that dim Xk n. But dim Xk dim X by (i), so (ii) follows. (iii) Let Z0 ⊂ Z1 ⊂ · · · ⊂ Zn be a chain of irreducible closed subsets in X. There exists k such that Xk ∩ Z0 = ∅. It follows that Xk ∩ Zi is a non-empty open irreducible subset in Zk for all 1 i r. By 1.1.3 (iv), Xk ∩ Zi = Zi and Xk ∩ Z0 ⊂ · · · ⊂ Xk ∩ Zn is a chain of irreducible closed subsets in Xk . Hence dim X dim Xk and (iii) follows from (i). (iv) Let dim Y = n and Y0 ⊂ Y1 ⊂ · · · ⊂ Yn be a chain of irreducible closed subsets in Y . If Y = X, then Y0 ⊂ Y1 ⊂ · · · ⊂ Yn ⊂ X is a chain of irreducible closed subsets in X. So dim X > dim Y which contradicts our hypothesis.

1.3 Noetherian spaces

5

1.2.4 Deﬁnition. Let X be non-empty and ﬁnite-dimensional and Y a non-empty subset of X. We deﬁne the codimension of Y in X, denoted by codimX Y , to be dim X − dim Y .

1.3 Noetherian spaces 1.3.1 Proposition. The following conditions are equivalent: (i) Any decreasing sequence of closed subsets of X is stationary. (ii) Any ascending sequence of open subsets of X is stationary. (iii) Any non-empty family of closed subsets of X has a minimal element. (iv) Any non-empty family of open subsets of X has a maximal element. If X satisﬁes the above conditions, then we shall say that X is Noetherian. Proof. This is straightforward. Remark. A Noetherian topological space can be inﬁnite-dimensional. 1.3.2 Proposition. (i) Any subset of a Noetherian space is Noetherian. (ii) Let (Xi )1ip be a ﬁnite covering of X such that the Xi ’s are Noetherian. Then X is also Noetherian. Proof. (i) Let Y be a subset of X and (Fn )n be a decreasing sequence of closed subsets of Y . Then (Fn )n is a decreasing sequence of closed subsets of X. Therefore there exists p such that Fp = Fn for all n p. This implies that Fn = Y ∩ Fn = Y ∩ Fp = Fp , and (i) follows. (ii) Let (Fn )n be a decreasing sequence of closed subsets of X. Then for 1 i p, (Xi ∩ Fn )n is a decreasing sequence of closed subsets of Xi . Hence there exists ni such that Xi ∩ Fni = Xi ∩ Fn for n ni . Set q = max(n1 , . . . , np ), then Fn = Fq for n q. 1.3.3 Proposition. The following conditions are equivalent: (i) X is Noetherian. (ii) Any open subset of X is quasi-compact. Proof. (i) ⇒ (ii) By 1.3.2, it suﬃces to show that X is quasi-compact. Let (Ui )i be an open covering of X. The set of all ﬁnite unions of the Ui ’s has a maximal element U since X is Noetherian. It is clear that U = X. (ii) ⇒ (i) Let (Un )n be an ascending sequence of open subsets of X. The union U of the Un ’s is open and therefore it is quasi-compact. Since (Un )n is an open covering of U , it is clear that the sequence (Un )n is stationary. 1.3.4 The following result is called the principle of transﬁnite induction. Lemma. Let E be a well-ordered set, that is an ordered set such that any non-empty subset has a minimal element. Let F be a subset of E such that: if a ∈ E satisﬁes {x ∈ E; x < a} ⊂ F , then a ∈ F . Then F = E. Proof. Suppose that F = E. Let b be a minimal element in E \ F . Then {x ∈ E; x < b} ⊂ F . Thus b ∈ F which contradicts our hypothesis on b.

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1 Results on topological spaces

1.3.5 Proposition. Let X be Noetherian. Then the set of irreducible components of X is ﬁnite. Proof. Let E be the set of all closed subsets of X ordered by inclusion, and F be the set of all ﬁnite unions of irreducible closed subsets of X. Now let Y ∈ E be such that any proper closed subset of Y is in F . If Y is irreducible, then Y ∈ F . If Y is not irreducible, then Y is the union of two distinct proper closed subsets Y1 , Y2 . But Y1 , Y2 ∈ F , so Y ∈ F . Thus we can apply 1.3.4, and we have E = F . Thus X is the ﬁnite union of irreducible closed subsets X1 , . . . , Xn . Further, we can assume that they are non-comparable. Let Z be an irreducible component of X. Then Z = (Z ∩ X1 ) ∪ · · · ∪ (Z ∩ Xn ) which implies that there exists k such that Z = Z ∩ Xk . Thus Z = Xk . This shows that {X1 , . . . , Xn } is the set of irreducible components of X. 1.3.6 Proposition. Let X be non-empty Noetherian, and X1 , . . . , Xn its irreducible components. For 1 i n, set: Xj . Yi = Xi \ (Xi ∩ Xj ) = X \ j=i

j=i

(i) An open subset U is dense in X if and only if U ∩Xi = ∅ for 1 i n. (ii) For 1 i n, Yi is open in X. Further, Y0 = Y1 ∪ · · · ∪ Yn is open dense in X and the irreducible (connected) components of Y0 are Y1 , . . . , Yn . Proof. Clearly, Yi is open in X and it is non-empty. So it is dense in Xi . Thus Y0 is open dense in X. Further, the Yi ’s are pairwise disjoint. So (ii) follows from 1.1.11. Now if U is open dense in X, then U ∩ Xi ⊃ U ∩ Yi = ∅. Conversely, if an open subset V meets each Xi , then V ∩ Xi is dense in Xi and Xi ⊂ V . Hence V is dense.

1.4 Constructible sets 1.4.1 Proposition. Let Y be a subset of X. The following conditions are equivalent: (i) There exists a family (Ui )i∈I of open subsets of X, whose union contains Y , such that Y ∩ Ui is closed in Ui for all i ∈ I. (ii) Any point of Y admits an open neighbourhood U in X such that U ∩Y is closed in U . (iii) Y is open in Y . (iv) Y is the intersection of an open subset and a closed subset of X. If Y satisﬁes these conditions, then it is called a locally closed subset of X. Proof. Only (ii) ⇒ (iii) needs attention. Let x ∈ Y , and U be an open neighbourhood of x in X such that U ∩ Y is closed in U . Then U ∩ Y = U ∩ Y , which shows that in Y , x is in the interior of Y .

1.4 Constructible sets

7

1.4.2 Let F be a set of subsets of X such that (i) Any open subset of X is in F. (ii) If Z ∈ F, then X \ Z ∈ F. (iii) Any ﬁnite union of elements of F is in F. For example, the set of subsets of X satisﬁes these conditions. The intersection C of all such sets satisﬁes these conditions. An element of C is called a constructible subset of X. 1.4.3 Proposition. The set C is the set of ﬁnite unions of locally closed subsets of X. Proof. Denote by C0 the set of ﬁnite unions of locally closed subsets of X. It is clear that C0 ⊂ C. Let C = (U1 ∩ F1 ) ∪ · · · ∪ (Un ∩ Fn ) ∈ C0 where the Ui ’s (resp. Fi ’s) are open (resp. closed) subsets of X. Then it is clear that X \ C is the union of subsets of the form E1 ∩ · · · ∩ En where for each j, there exists i such that Ej is either X \ Ui or X \ Fi . So the proposition follows. 1.4.4 Proposition. Let Y be a subset of X. (i) If C be a constructible subset of X, then C ∩Y is a constructible subset of Y . (ii) Suppose that Y is closed in X. If C is a constructible subset of Y , then C is a constructible subset of X. Proof. Part (i) is obvious and (ii) follows from the fact that if F is closed in Y and U is open in Y , then F is closed in X and U is locally closed in X. 1.4.5 Lemma. Let C be a constructible subset of X whose closure is irreducible. Then C contains an open subset of C. Proof. Let C = (U1 ∩ F1 ) ∪ · · · ∪ (Un ∩ Fn ) ∈ C0 where the Ui ’s (resp. Fi ’s) are open (resp. closed) subsets of X. We can assume further n that Ui ∩ Fi = ∅ for 1 i n. Since C is irreducible and contained in i=1 Fi , there exists j such that C ⊂ Fj . It follows that Uj ∩ Fj = Uj ∩ C and Uj ∩ Fj is a non-empty subset of C, open in C. 1.4.6 Proposition. Let X be Noetherian. Then any constructible subset Y contains an open and dense subset of Y . Proof. Let Y = Y1 ∪ · · · ∪ Yn be the decomposition of Y into irreducible components. Since irreducible components are closed, they are constructible in Y . By 1.1.14, the Yi ’s are the irreducible components of Y . For 1 i n, set: Ti = Yi \ (Yi ∩ Yj ). j=i

We saw in 1.3.6 that the Ti ’s are non-empty and open in Y . By 1.4.5, there exists a subset Ui in Yi which is open and dense in Yi . Let Vi = Ui ∩ Ti . Then

8

1 Results on topological spaces

Vi is open and dense in Yi . But the Vj ’s are pairwise disjoint, so Vi is in fact open in Y . Now it follows from 1.3.6 (i) that V = V1 ∪ · · · ∪ Vn is a subset of Y which is open and dense in Y .

1.5 Gluing topological spaces 1.5.1 Let (Xi )i∈I be a family of sets and X be the sum of the Xi ’s. For i ∈ I, we denote by θi : Xi → X the canonical injection and we shall identify Xi with its image via θi . Suppose that for all i, j ∈ I, there are subsets Xij ⊂ Xi , Xji ⊂ Xj and a map hji : Xij → Xji such that a) For all i ∈ I, Xii = Xi and hii = idXi is the identity map. b) For all i, j, k ∈ I and x ∈ Xij ∩ Xik , we have hji (x) ∈ Xjk and: hki (x) = hkj (hji (x)). By applying b) to the triples (i, j, i) and (j, i, j), we remark that hij is bijective with reciprocal map hji . Let us deﬁne a binary relation R on X by xRy if and only if there exist i, j ∈ I such that x ∈ Xij , y ∈ Xji , y = hji (x). The preceding remark implies that R is reﬂexive and symmetric. Now suppose that x, y, z ∈ X, xRy and yRz. There exist i, j, k ∈ I such that x ∈ Xij , y ∈ Xji ∩ Xjk , z ∈ Xkj , y = hji (x) , z = hkj (y). By b), we have x = hij (y) ∈ Xik and: z = hkj (y) = hkj (hji (x)) = hki (x). Thus R is transitive and it deﬁnes an equivalence relation on X. We shall say that the set of equivalence classes X/R is obtained by gluing the Xi ’s along the Xij ’s via the bijections hji . Let us denote by ϕ : X → X/R the canonical surjection. Note that we have the following: (i) Each equivalence class has at most one element in each of the Xi ’s. (ii) For i, j ∈ I, Xij is the set of x ∈ Xi such that there exists a y ∈ Xj with xRy. We deduce from these remarks that for i ∈ I, ϕ ◦ θi induces a bijection from Xi to ϕ ◦ θi (Xi ). 1.5.2 Conserving the hypotheses of 1.5.1, let us suppose further that the following properties are satisﬁed: c) Each Xi is a topological space with topology Ti . d) For i, j ∈ I, Xij is open in Xi and hji is a homeomorphism from Xij to Xji .

1.5 Gluing topological spaces

9

We can equip X with the topology S, the sum of the topologies Ti ’s (this is the ﬁnest topology on X for which the injections θi are continuous). Thus a subset A ⊂ X is open (resp. closed) if and only if each θi−1 (A) is open (resp. closed) in Xi . In particular, each θi (Xi ) is open and closed in X. Let T be the topology on X/R, quotient of the topology S. This is the ﬁnest topology for which the maps ϕ ◦ θi are continuous. Therefore a subset B in X/R is open (resp. closed) if and only if B = ϕ(C) where C is an open (resp. closed) subset of X, saturated with respect to R. The topological space X/R is said to be obtained by gluing the Xi ’s along the Xij ’s via the maps hji . 1.5.3 Proposition. Let us suppose that the hypotheses a), b), c), d) of 1.5.1 and 1.5.2 are satisﬁed. Then each ϕ(Xi ) is open in X/R. Further, the restriction of ϕ to Xi is a homeomorphism from Xi onto ϕ(Xi ). Proof. Set Yi = ϕ(Xi ) , Zi =

θj (Xij ).

j∈I

Then Zi is a subset of X saturated with respect to R, and we have Yi = ϕ(Zi ). Since Zi ∩ Xj = Xij and Zi is open in X by our hypotheses, Yi is open in X/R. Let Ai be an open subset of Xi . Then each Ai ∩ Xij is open in Xij . Consequently, each hji (Ai ∩ Xij ) is open in Xji , and therefore open in Xj . It follows that j∈I θj ◦ hji (Ai ∩ Xij ) is open in X. So ϕ(Ai ) is open in X/R, and hence open in Yi . Now the proposition follows. 1.5.4 Let us conserve the hypotheses and notations of 1.5.1 and 1.5.2. To summarize, we have constructed a topological space Y = X/R, and for each i ∈ I, a homeomorphism ψi from Xi onto an open subset ψi (Xi ) in Y such that Y = i∈I ψi (Xi ), and such that ψi (Xij ) = ψj (Xji ) = ψi (Xi ) ∩ ψj (Xj ) , hji (x) = ψj−1 (ψi (x)) for all i, j ∈ I and x ∈ Xij .

References • [8], [26].

2 Rings and modules

Unless otherwise stated, all algebras over a commutative ring considered in this book are associative. We recall here basic concepts of the theory of commutative rings which will allow us to ﬁx vocabulary and notations for the following chapters. Notions such as local rings, ﬁeld of fractions and module of diﬀerentials will often appear in algebraic geometry. All rings considered in this chapter are commutative. Let A be a ring.

2.1 Ideals 2.1.1 Let us ﬁx some notations. • If B is a ring, we denote by Hom(A, B) the set of ring homomorphisms from A to B. • An element x ∈ A is called a zero divisor if there exists y ∈ A \ {0} such that xy = 0. A ring A is called an integral domain if A = {0} and if A has no zero divisors other than 0. • An element x ∈ A is called nilpotent if there exists n ∈ N∗ such that n x = 0. The set of nilpotent elements of A is an ideal called the nilradical of A. We say that A is reduced if its nilradical is {0}. 2.1.2 Proposition. (Chinese Remainder Theorem) Let a1 , . . . , an be ideals of A such that ai + aj = A for all i = j. Let x1 , . . . , xn ∈ A. Then there exists x ∈ A such that x − xi ∈ ai for 1 i n. Proof. First, suppose that n = 2. Then there exist a1 ∈ a1 , a2 ∈ a2 such that a1 + a2 = 1. Hence x = x2 a1 + x1 a2 satisﬁes the conclusion of the proposition. Suppose now that n 3. For i 2, there exist ai ∈ a1 and bi ∈ ai such that ai + bi = 1. The product i2 (ai + bi ) is equal to 1 and it is contained in a1 + a2 · · · an . Thus a1 + a2 · · · an = A and from the previous paragraph, there exists y1 ∈ A such that y1 − 1 ∈ a1 and y1 ∈ a2 · · · an . So y1 − 1 ∈ a1 and y1 ∈ ai for all i 2.

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Proceeding in the same manner for an index j 2, we see that there exists yj ∈ A such that yj − 1 ∈ aj and yj ∈ ai for all i = j. We verify easily that x = x1 y1 + · · · + xn yn satisﬁes the conclusion of the proposition. 2.1.3 Corollary. Let a1 , . . . , an be ideals of A such that ai + aj = A for all i = j. Set b = a1 ∩ · · · ∩ an . (i) We have b = a1 · · · an . (ii) The homomorphism f : A → (A/a1 ) × · · · × (A/an ) induced by the canonical surjections on each factor is surjective and its kernel is b. In particular f induces an isomorphism between A/b and (A/a1 ) × · · · × (A/an ). Proof. (i) From the proof of 2.1.2, a1 + a2 · · · an = A. Therefore by induction, it suﬃces to consider the case n = 2. Clearly, a1 a2 ⊂ a1 ∩ a2 . Let a1 ∈ a1 , a2 ∈ a2 be such that a1 + a2 = 1. If x ∈ a1 ∩ a2 , then x = xa1 + xa2 ∈ a1 a2 . (ii) It is clear that b is the kernel of f . The surjectivity of f follows from 2.1.2 and the last assertion is now obvious.

2.2 Prime and maximal ideals 2.2.1 An ideal m of A is called maximal if it is maximal in the set of proper ideals of A. This is equivalent to the condition that A/m is a ﬁeld. Any ideal of A is contained in a maximal ideal and we shall denote by Spm(A) the set of maximal ideals of A. 2.2.2 An ideal p of A is called prime if it satisﬁes one of the following equivalent conditions: (i) p = A and if x, y ∈ A \ p, then xy ∈ A \ p. (ii) The ring A/p is an integral domain. A maximal ideal is a prime ideal. Let us denote by Spec(A) the set of prime ideals of A. 2.2.3 A chain of prime ideals of length n of A is a sequence p0 ⊂ p1 ⊂ · · · ⊂ p n of pairwise distinct elements of Spec(A). The (Krull) dimension of A, denoted by dim A, is the supremum of the lengths of chains of prime ideals. It is an element of N∪{+∞}. Thus dim A = 0 if and only if prime ideals of A are maximal; this is the case for a commutative ﬁeld. 2.2.4 Proposition. Let B = A1 ×· · ·×An be a product of rings. Any prime (resp. maximal) ideal of B is of the form A1 ×· · ·×Ai−1 ×ai ×Ai+1 ×· · ·×An , where 1 i n and ai is a prime (resp. maximal) ideal of Ai . Proof. If C and D are rings, then C × D is not an integral domain since (0, 1)(1, 0) = (0, 0). The result now follows easily.

2.3 Rings of fractions and localization

13

2.2.5 Proposition. (i) Let a1 , . . . , an be ideals of A and p a prime ideal of A such that a1 · · · an ⊂ p. Then p contains one of the ai ’s. (ii) Let a be an ideal of A and p1 , . . . , pr be prime ideals of A such that a ⊂ p1 ∪ · · · ∪ pr . Then a is contained in one of the pi ’s. Proof. (i) Suppose that ai ⊂ p for 1 i n. There exists ai ∈ ai \ p. But then a = a1 · · · an ∈ a1 · · · an \ p, which contradicts our hypotheses. (ii) We can assume that pi ⊂ pj for i = j. Let us proceed by induction on r. The result is obvious when r = 1. If a ⊂ p1 ∪ · · · ∪ pr−1 , then we obtain our result by induction. So let us suppose that there exists a ∈ a ∩ pr such that a ∈ p1 ∪ · · · ∪ pr−1 . Suppose that there exists y ∈ ap1 · · · pr−1 \ pr . Then a + y ∈ a and a + y ∈ p1 ∪ · · · ∪ pr−1 . It follows that a + y ∈ pr and so y ∈ pr which gives us a contradiction. So we have ap1 · · · pr−1 ⊂ pr . By our assumption, for i < r, we have pi ⊂ pr . So part (i) implies that a ⊂ pr . 2.2.6 Corollary. Let m ∈ Spm(A) and n ∈ N∗ . Then the only prime ideal containing mn is m. Proof. If p ∈ Spec(A) contains mn , then by 2.2.5 (i), m ⊂ p. Hence m = p.

2.2.7 A prime ideal in A is said to be minimal if it is minimal by inclusion in Spec(A). Proposition. If A = {0}, then the set of minimal prime ideals of A is non-empty. Proof. Since A = {0}, Spec(A) is non-empty. We shall endow Spec(A) with the partial order given by reverse inclusion. Let (pi )i∈I be a totally ordered family of prime ideals. Set r = i∈I pi . Let a, b ∈ A be such that ab ∈ r and a ∈ r. So there exists j ∈ I such that a ∈ pj . Now for all k ∈ I such that pk ⊂ pj , a ∈ pk and since pk is prime, b ∈ pk . Consequently b ∈ r and the result follows from Zorn’s lemma.

2.3 Rings of fractions and localization 2.3.1 Deﬁnition. A subset S of A is said to be multiplicative if 1 ∈ S and it is closed under multiplication. 2.3.2 Lemma. Let S be a multiplicative subset of A not containing 0 and a an ideal of A such that a ∩ S = ∅. Then there exists p ∈ Spec(A) such that a ⊂ p and p ∩ S = ∅. Proof. Let A be the set, partially ordered by inclusion, of ideals b of A such that a ⊂ b and b ∩ S = ∅. Since a ∈ A, A is non-empty and it is clearly inductive. By Zorn’s lemma, A has a maximal element p. Suppose that there exist a, b ∈ A \ p such that ab ∈ p. Then the ideals Aa + p and Ab + p contain strictly p. By the maximality of p, there exist

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s ∈ S ∩ (Aa + p) and t ∈ S ∩ (Ab + p). It follows that st ∈ S ∩ p which is a contradiction. We have therefore shown that p is a prime ideal. 2.3.3 Let S be a multiplicative subset of A. We deﬁne the binary relation R on S × A as follows: (s, a)R(t, b) if and only if there exists u ∈ S such that u(sb − ta) = 0. This deﬁnes an equivalence relation on S × A. We shall endow S × A with the following operations: (s, a) + (t, b) = (st, sb + ta) , (s, a).(t, b) = (st, ab). These operations are compatible with R and we see clearly that they induce a ring structure on (S × A)/R. We call the ring (S × A)/R, denoted by S −1 A, the ring of fractions of A by S. We shall denote the class of (s, a) by a/s or a . s The map i : A → S −1 A deﬁned by a → a/1 is a ring homomorphism. We shall call i the canonical homomorphism. We have the following results: • The canonical homomorphism i is injective if and only if S contains no zero divisor. • We have S −1 A = {0} if and only if 0 ∈ S. This is the case if S contains a nilpotent element. • Let s ∈ S. The element i(s) = s/1 is invertible in S −1 A. Its inverse is 1/s. • If p ∈ Spec(A), then S = A \ p is a multiplicative subset of A. The ring of fractions S −1 A, denoted by Ap , is called the localization of A at p. In particular, Ap = {0}. • Suppose that A = {0} and let S be the set of non zero divisors of A. Then S is a multiplicative subset of A. The ring S −1 A, denoted by Fract(A), is called the full ring of fractions of A. Further, if A is an integral domain, then Fract(A) is a ﬁeld which is called the quotient ﬁeld or ﬁeld of fractions of A. 2.3.4 Theorem. Let A, B be rings, S a multiplicative subset of A, i : A → S −1 A the canonical homomorphism and f ∈ Hom(A, B). Suppose further that any element of f (S) is invertible in B. Then there exists a unique homomorphism g : S −1 A → B such that g ◦ i = f . Proof. Let p : S × A → S −1 A be the map sending (s, a) to a/s and h : S × A → B be the map deﬁned by (s, a) → f (a)f (s)−1 . We verify easily that h(s, a) = h(t, b) if a/s = b/t. It follows therefore that there is a unique map g : S −1 A → B such that g ◦ p = h. A direct computation show that g is a ring homomorphism and g ◦ i = f . The uniqueness of g is immediate. 2.3.5 Corollary. Let S, T be multiplicative subsets of A such that S ⊂ T , iS : A → S −1 A and iT : A → T −1 A the corresponding canonical homomorphisms. There exists a unique homomorphism f : S −1 A → T −1 A such that f ◦ iS = iT . Further, if T contains no zero divisors, then f is injective.

2.3 Rings of fractions and localization

15

Remark. If A is an integral domain and S a multiplicative subset of A which does not contain 0, then S −1 A embeds canonically into the quotient ﬁeld Fract(A). 2.3.6 Corollary. Suppose that A is a subring of a ring B and f : A → B the canonical injection. Let S be a multiplicative subset of A (hence of B), and i : A → S −1 A, j : B → S −1 B the canonical homomorphisms. There exists a unique homomorphism g : S −1 A → S −1 B such that g ◦ i = j ◦ f . The homomorphism g is injective and we can therefore identify canonically S −1 A as a subring of S −1 B. Proof. The existence and uniqueness of g follows from 2.3.4. Let (s, a) ∈ S ×A be such that g(a/s) = 0. Then 0 = g((a/1).(1/s)) = g ◦ i(a)[g ◦ i(s)]−1 ⇒ g ◦ i(a) = 0 ⇒ f ◦ j(a) = 0 ⇒ j(a) = 0. It follows that there exists t ∈ S such that ta = 0. Hence (s, a)R(t, 0) and a/s = 0. 2.3.7 Let S be a multiplicative subset of A, i : A → S −1 A be the canonical homomorphism and J (A) (resp. J (S −1 A)) the set of ideals of A (resp. S −1 A). Denote by J (A) the set of ideals a of A which veriﬁes the condition: (1)

s ∈ S , a ∈ A , sa ∈ a ⇒ a ∈ a.

For a ∈ J (A), let S −1 a be the set of elements a/s of S −1 A such that a ∈ a, s ∈ S. We verify easily that S −1 a ∈ J (S −1 A) and thus we have a map ψ : J (A) → J (S −1 A) sending a to S −1 a. The proof of the following result is straightforward. Proposition. (i) If a ∈ J (A), then S −1 a is the ideal of S −1 A generated by i(a). Any ideal of S −1 A is of the form S −1 b for some b ∈ J (A). (ii) If n is the nilradical of A, then S −1 n is the nilradical of S −1 A. (iii) For a, b ∈ J (A), we have: S −1 (a + b) = S −1 a + S −1 b, S −1 (ab) = (S −1 a)(S −1 b), S −1 (a ∩ b) = S −1 a ∩ S −1 b. (iv) The map ψ induces a bijection between J (A) and J (S −1 A), whose reciprocal map is given by b → i−1 (b). 2.3.8 Corollary. The map ψ induces a bijection between the set of prime ideals of A not meeting S and the set of prime ideals of S −1 A. Proof. A prime ideal p of A which does not meet S satisﬁes the condition (1) of 2.3.7. So it suﬃces to show that S −1 p ∈ Spec S −1 A. Let a, b ∈ A, s, t ∈ S be such that (a/s).(b/t) ∈ S −1 p. There exists u ∈ S such that uab ∈ p. Since u ∈ p, ab ∈ p. So either a ∈ p or b ∈ p. Thus either a/s ∈ S −1 p or b/t ∈ S −1 p.

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2.3.9 Let A be a subring of a ring B and a be an ideal of A. Then an ideal b of B is said to lie above a if b ∩ A = a. Let S be a multiplicative subset of A. Let us conserve the notations i, j, f, g of 2.3.6 and let J (S −1 A), J (S −1 B), J (A), J (B) be as in 2.3.7. Proposition. Let p ∈ Spec A be such that S ∩ p = ∅. (i) The map b → S −1 b induces a surjection from the set E of ideals of B lying above p and the set F of ideals of S −1 B lying above S −1 p, and the map b → j −1 (b) is a bijection between F and E ∩ J (B). (ii) The map q → S −1 q induces a bijection between the set of prime ideals in B lying above p and the set of prime ideals of S −1 B lying above S −1 p. Proof. By 2.3.8, we have S −1 p ∈ Spec S −1 A and i−1 (S −1 p) = p. Let b ∈ J (S −1 B) be such that b ∩ S −1 A = S −1 p. Then we have: p = i−1 (S −1 p) = i−1 (b ∩ S −1 A) = A ∩ j −1 (b ). By 2.3.7, S −1 (j −1 (b )) = b and therefore the image of E via the map b → S −1 b contains F. Let b ∈ E, a ∈ A and s ∈ S. The following conditions are equivalent: (i) g(a/s) ∈ S −1 b. (ii) f (a)/f (s) ∈ S −1 b. (iii) There exists t ∈ S such that f (t)f (a) ∈ b. (iv) There exists t ∈ S such that ta ∈ p. (v) a/s ∈ S −1 p. We deduce that S −1 b ∩ S −1 A = S −1 p. Thus the image of E via the map b → S −1 b is F. The rest of the proposition follows from 2.3.7 and 2.3.8. 2.3.10 Proposition. Let S be a multiplicative subset of A and a ∈ J (A) be such that a ∩ S = ∅. Denote by π : A → A/a the canonical projection. Then π(S) is a multiplicative subset of A/a, and the rings S −1 A/S −1 a and π(S)−1 (A/a) are isomorphic. Proof. It is clear that T = π(S) is a multiplicative subset of B = A/a, and that 0 ∈ T since a ∩ S = ∅. Let iS : A → S −1 A and iT : B → T −1 B be the canonical homomorphisms. If s ∈ S, then iT ◦ π(s) is invertible in T −1 B. By 2.3.4, there exists a unique homomorphism f : S −1 A → T −1 B such that f ◦ iS = iT ◦ π. For a ∈ A, s ∈ S, we have: π(a)/π(s) = (π(a)/π(1)).(π(1)/π(s)) = (iT ◦ π(a))(iT ◦ π(s))−1 = (f ◦ iS (a))(f ◦ iS (s))−1 = f ((a/1).(1/s)) = f (a/s). Thus f is surjective. Now suppose that f (a/s) = 0. The preceding computation implies that π(a)/π(s) = 0, and so there exists t ∈ S such that π(ta) = π(a)π(t) = 0. Thus ta ∈ a and it follows that a/s = (ta)/(ts) ∈ S −1 a.

2.4 Localizations of modules

17

2.3.11 Deﬁnition. Let p be a prime ideal of A. We deﬁne the height of p, denoted by ht p, to be the dimension of the ring Ap . 2.3.12 Proposition. The height of a prime ideal p is the supremum of chains p0 ⊂ p1 ⊂ · · · ⊂ pr of prime ideals of A such that pr = p. Proof. This is immediate from 2.3.7 (iii) and 2.3.8.

2.4 Localizations of modules 2.4.1 In this section, S will denote a multiplicative subset of A. Let E be an A-module. We deﬁne the binary relation S on E × S by: for x, y ∈ E, s, t ∈ S, we have (x, s)S(y, t) if and only if there exists u ∈ S such that u(sy − tx) = 0. It is easy to verify that S deﬁnes an equivalence relation on E × S. Let us denote by S −1 E the set of equivalence classes and by x/s the class of (x, s). The proof of the following is similar to the one for rings of fractions. Theorem. Let E be an A-module. (i) Deﬁne a structure of S −1 A-module on S −1 E by x/s + y/t = (tx + sy)/st , (a/s)(y/t) = (ay)/(st) where x, y ∈ E, s, t ∈ S and a ∈ A. The map j : E → S −1 E , x → x/1 is a homomorphism of A-modules, called the canonical homomorphism of E to S −1 E. (ii) Let F be a S −1 A-module and f : E → F be a homomorphism of A-modules. Then there exists a unique homomorphism of S −1 A-modules g : S −1 E → F such that f = g ◦ j. 2.4.2 Proposition. Let f : E → F be a homomorphism of A-modules. There exists a unique homomorphism of S −1 A-modules g : S −1 E → S −1 F such that g(x/1) = f (x)/1 for all x ∈ E. Proof. Let jE : E → S −1 E and jF : F → S −1 F be the canonical homomorphisms. By 2.4.1, there exists a S −1 A-module homomorphism g : S −1 E → S −1 F such that jF ◦ f = g ◦ jE . So g(x/1) = f (x)/1 for all x ∈ E. The uniqueness is clear. 2.4.3 Proposition. Let F be a submodule of E. The homomorphism S −1 F → S −1 E induced by the canonical injection F → E in 2.4.2 is injective, and the S −1 A-modules S −1 (E/F ) and S −1 E/S −1 F are isomorphic. Proof. Let x/s be in the kernel of the homomorphism S −1 F → S −1 E. There exists t ∈ S such that tx = 0 which in turn implies that x/s = 0 in S −1 F .

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The map f : E/F → S −1 E/S −1 F deﬁned by sending x+F to x/1+S −1 F is well-deﬁned and we verify easily that it is a homomorphism of A-modules. By 2.4.1, there exists a unique homomorphism of S −1 A-modules g : S −1 (E/F ) → S −1 E/S −1 F , (x + F )/s → x/s + S −1 F. By construction, g is surjective. Let x ∈ E, s ∈ S be such that x/s ∈ S −1 F . Then there exists t ∈ S such that tx ∈ F . So we have t(x + F ) = 0 in E/F and hence (x + F )/s = 0 in S −1 (E/F ). Thus g is injective. 2.4.4 Proposition. Let E be an A-module, jE : E → S −1 E the canonical homomorphism and a an ideal of A. We have (S −1 a)(S −1 E) = a(S −1 E) = (S −1 a)jE (E). Proof. Let a ∈ a, s, t ∈ S and x ∈ E. We have: (a/s).(x/t) = (a/1).(x/st) = (a/st)(x/1). So the proposition follows. 2.4.5 Remarks. 1) Proposition 2.4.3 justiﬁes the notation S −1 a of 2.3.7. 2) Let M be an A-module, then we see easily that S −1 M and S −1 A ⊗A M are isomorphic S −1 A-modules where we consider S −1 A as an A-algebra via the canonical map A → S −1 A.

2.5 Radical of an ideal 2.5.1 Deﬁnition. Let a be an ideal of A. We deﬁne the radical of a, √ denoted by a, to be the√set of elements a ∈ A for which there exists n ∈ N∗ such that an ∈ a. If a = a, then we say that a is a radical ideal. 2.5.2 Let a be an ideal of A, then by√using the formula of binomial expansion, we obtain easily that its radical a is an ideal of A. Note that the nilradical of A is the radical of the ideal {0}. 2.5.3 Proposition. Let a, b be ideals of A. √ (i) We have a = A if and only if a = A. Further, √ √ √ √ √ √ √ √ √ √ a = a, a + b = a + b, ab = a b = a ∩ b. √ √ ∗ (ii) If a is ﬁnitely generated, then ( a)n ⊂ a for √ nsome n ∈ N . ∗ (iii) If p is a prime ideal of A and n ∈ N , then p = p. Proof.√ (i) Let last equality.√From√ab ⊂ a ∩ b ⊂ a, we deduced √ us prove√the √ ∗ that ab ⊂ a ∩ b ⊂ a ∩ b. Let a ∈ a ∩ b, then there √ exist n, p ∈ N n p n+p such that a ∈ a and a ∈ b.√Hence a ∈ ab and so a ∈ ab as required. Aap . There exists m ∈ N∗ such (ii) Let us suppose that a = Aa1 + · · · + √ m that ai ∈ a for all 1 i p. It√follows that ( a)n ⊂ a if n > mp. √ √ √ (iii) We have p ⊂ p and pn = p by (i). Now if a ∈ p, then there ∗ n exists n ∈ N such that a ∈ p. Since p is prime, a ∈ p.

2.6 Local rings

19

2.5.4 Proposition. Let a be√an ideal of A. (i) The nilradical √ of A/a is a/a. (ii) The ring A/ a is reduced. (iii) If A/a is reduced, then a contains the nilradical of A. √ a∈A Proof. (i) It is clear that the nilradical of A/a is contained in a/a. Let √ and n ∈ N∗ be such that an ∈ a. Then (a + a)n ⊂ a and hence a + a ∈ a/a. (ii) This is immediate from (i) and 2.5.3. (iii) If A/a is reduced, then the kernel of the canonical surjection A → A/a contains the nilradical of A. √ 2.5.5 Proposition. Let a be an ideal of A. Then a is the intersection of prime ideals of A containing a. In particular, the nilradical of A is the intersection of prime ideals of A. Proof. By 2.5.4, we can replace A by A/a and assume that a = {0}. Denote by q the intersection of prime ideals of A and by n the nilradical of A. Clearly, n ⊂ q. Now suppose that there exists an element a ∈ q which is not nilpotent. Then S = {an ; n ∈ N} is a multiplicative subset of A which does not contain 0. It follows from 2.3.2 that there exists p ∈ Spec A such that p ∩ S = ∅. So a ∈ p which contradicts the fact that a ∈ q.

2.6 Local rings 2.6.1 Deﬁnition. The intersection of maximal ideals of A, denoted by rad A, is an ideal called the Jacobson radical of A. 2.6.2 Proposition. Let a ∈ A. Then a ∈ rad A if and only if 1 − ab is invertible for all b ∈ A. Proof. If a ∈ rad A, then ab ∈ rad A. Therefore 1 − ab is not contained in any maximal ideal. So 1 − ab is invertible. If a ∈ rad A, then there exists a maximal ideal m which does not contain a. So m + Aa = A and there exists b ∈ A, c ∈ m such that c + ab = 1. It follows that 1 − ab = c is not invertible. 2.6.3 Proposition. The following conditions are equivalent: (i) The ring A has a unique maximal ideal. (ii) The set A \ rad A is the set of invertible elements of A. (iii) There exists a proper ideal a of A such that any element of A \ a is invertible. Proof. (i) ⇒ (ii) If m is the unique maximal ideal of A, then m = rad A. So (ii) follows. (ii) ⇒ (iii) Take a = rad A. (iii) ⇒ (i) Suppose that a satisﬁes the hypotheses of (iii). If a ∈ a, then 1 − ab is invertible for all b ∈ A. So by 2.6.2, a ⊂ rad A. Now let b be an ideal of A containing a. If b = a, then there exists b ∈ b \ a, which is invertible. This implies that b = A. So a = rad A is the unique maximal ideal of A.

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2.6.4 Deﬁnition. A ring A satisfying the conditions of 2.6.3 is called a local ring. If A is a local ring with maximal ideal mA , we call A/mA the residual ﬁeld of A. A homomorphism u : A → B of local rings is said to be local if u(mA ) ⊂ mB . 2.6.5 Remark. Let A be a local ring with maximal ideal m. Then the rings A and Am are canonically isomorphic by 2.3.3 and 2.6.3. 2.6.6 Proposition. Let p ∈ Spec A. Then Ap is a local ring with maximal ideal pAp , and the residual ﬁeld is canonically isomorphic to the ﬁeld of fractions of A/p. Proof. Since Ap \pAp consists of invertible elements, the ﬁrst assertion follows by 2.6.3. Let f : A → A/p be the canonical projection. Then f (A \ p) is the set A/p \ {0}. So the other assertions follow from 2.3.4 and 2.3.10. 2.6.7 Proposition. (Nakayama’s Lemma) Let a be an ideal of A and E be a ﬁnitely generated A-module such that aE = E. Then there exists f ∈ 1 + a such that f E = 0. Proof. Denote by Mr (A) the ring of r-by-r matrices with coeﬃcients in A, and by Ir the identity matrix in Mr (A). Let {e1 , . . . , er } be a system of generators of E. Our hypothesis says that we can write for 1 i r: ei = ai1 e1 + · · · + air er where aij ∈ a. Set U = [aij ] ∈ Mr (A) and Λ = Ir − U . Then ⎛ ⎞ ⎛ ⎞ e1 0 ⎜ .. ⎟ ⎜ .. ⎟ Λ⎝ . ⎠ = ⎝ . ⎠. er

0

= (det Λ)Ir . If Λ is the transpose of the matrix of cofactors of Λ, then ΛΛ This implies that (det Λ)ej = 0 for 1 j r. Since det Λ is of the form 1 + f with f ∈ a, our result follows. 2.6.8 Corollary. Let a ⊂ rad A be an ideal of A, E a ﬁnitely generated A-module and F a submodule of E. (i) If aE = E, then E = {0}. (ii) If E = aE + F , then E = F . Proof. (i) By 2.6.7, there exists f ∈ a such that (1 + f )E = {0}. But a is contained in rad A, so 1 + f is invertible and hence E = {0}. (ii) Since E/F = a(E/F ), the result follows from (i). 2.6.9 Corollary. Let A be a local ring with maximal ideal m, E a ﬁnitely generated A-module. Let e1 , . . . , en ∈ E be such that their images in the A/mmodule E/mE form a system of generators. Then e1 , . . . , en generate the Amodule E.

2.7 Noetherian rings and modules

21

Proof. Let F be the submodule of E generated by e1 , . . . , en . We have E = mE + F . Since m = rad A by 2.6.3, the result follows from 2.6.8. 2.6.10 Let p ∈ Spec A, S = A \ p and E an A-module. We shall denote by Ep the Ap -module S −1 E. Proposition. Let m be a maximal ideal of A and E an A-module. Suppose that there exists an ideal a of A satisfying the following conditions: (i) The only maximal ideal containing a is m. (ii) We have aE = {0}. Then the canonical homomorphism E → Em is bijective. Proof. Condition (i) implies that A/a is local with maximal ideal m/a. Condition (ii) allows us to consider E as an A/a-module. If s ∈ A \ m, then s + a is invertible in A/a. Thus the map µs : E → E deﬁned by x → sx is bijective. If x, y ∈ E satisfy x/1 = y/1, then there exists s ∈ A \ m such that s(x − y) = 0 = µs (x − y). So x = y. Now let y ∈ E and s ∈ A \ m. There exists x ∈ E such that µs (x) = sx = y and so x/1 = y/s. 2.6.11 Corollary. Let m be a maximal ideal of A, E an A-module and k ∈ N be such that mk E = {0}. Then the canonical homomorphism E → Em is bijective. Proof. This is clear if k = 0. For k > 0, it follows from 2.2.6 and 2.6.10. 2.6.12 Proposition. Let m be a maximal ideal of A, E an A-module and k ∈ N. The canonical homomorphism E → Em /mk Em is surjective. Its kernel is mk E and it induces an isomorphism between E/mk E and Em /mk Em . Proof. By 2.6.10, the canonical homomorphism E/mk E → (E/mk E)m is bijective. Further, we can identify canonically (E/mk E)m and Em /(mk E)m by 2.4.3, and (mk E)m = mk Em by 2.4.4. Thus the map E/mk E → Em /mk Em given by x + mk E → x/1 + mk Em is an isomorphism.

2.7 Noetherian rings and modules 2.7.1 Proposition. For any A-module E, the following conditions are equivalent: (i) Any submodule of E is ﬁnitely generated. (ii) Any ascending sequence of submodules of E is stationary. (iii) Any non-empty family of submodules of E has a maximal element. Proof. This is straightforward. 2.7.2 Deﬁnition. An A-module is called Noetherian if it satisﬁes the conditions of 2.7.1. The ring A is said to be Noetherian if A itself is a Noetherian A-module.

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2.7.3 following (i) E (ii) F

Proposition. Let E be an A-module and F a submodule of E. The conditions are equivalent: is Noetherian. and E/F are Noetherian.

Proof. Again this is straightforward. 2.7.4 Corollary. (i) Any ﬁnite direct sum of Noetherian modules is Noetherian. (ii) Let F, G be submodules of an A-module E such that E = F + G. If F and G are Noetherian, then E is Noetherian. Proof. This is a direct consequence of 2.7.3.

2.7.5 Proposition. Let us suppose that A is Noetherian. For any Amodule E, the following conditions are equivalent: (i) E is Noetherian. (ii) E is ﬁnitely generated. Proof. The implication (i) ⇒ (ii) is clear. Suppose that (ii) is veriﬁed, and let x ∈ E. The submodule Ax is isomorphic to a quotient of A, so it is Noetherian by 2.7.3. Now by applying 2.7.4 (ii), any ﬁnitely generated submodule of E is Noetherian. 2.7.6 Proposition. Let a be an ideal of a Noetherian ring A and S a multiplicative subset of A. (i) The nilradical of A is a nilpotent ideal. (ii) The rings A/a and S −1 A are Noetherian. Proof. Assertion (i) follows from 2.5.3. We deduce from 2.7.3 that A/a is Noetherian because the A/a-submodules of A/a are the A-submodules of A/a. Finally S −1 A is Noetherian by 2.3.7 (iii) and (iv). 2.7.7 Theorem. Let a be a radical ideal of a Noetherian ring A. There exist p1 , . . . , pn ∈ Spec A such that: a = p1 ∩ · · · ∩ pn . If we assume that pi ⊂ pj for i = j, then the preceding decomposition is unique up to a permutation. Proof. Let F be the set of proper radical ideals of A which are not ﬁnite intersections of prime ideals. Let us suppose that F is not empty. By 2.7.1, there is a maximal element in F, say r. Since r is not prime, there exist a, b ∈ A \ r such that ab ∈ r, and the ideals a = r + Aa, b = r + Ab contain strictly r. We claim that a and b are proper. If

2.7 Noetherian rings and modules

23

a = A, then 1 = xa + r for some x ∈ A and r ∈ r. But this would imply that b = xab + rb ∈ r, which is impossible. √ It follows that there√exist prime ideals p1 , . . . , pn , q1 , . . . , qm such that a = p1 ∩ · ·√· ∩ pn√and b = q1 ∩ · · · ∩ qn . Set s = a ∩ b. We have r ⊂ s and if u ∈ s, there exists n ∈ N∗ such that un ∈ a ∩ b. This implies in turn that u2n ∈ ab ⊂ r, and so u ∈ r. Hence r = s which is absurd since s ∈ F. Now suppose that p1 , . . . , pn , q ∈ Spec A be such that p1 ∩ · · · ∩ pn ⊂ q. Then p1 · · · pn ⊂ q, and by 2.2.5 (i), there exists i such that pi ⊂ q. Now the uniqueness of the decomposition follows easily. 2.7.8 Corollary. Let A be a Noetherian ring which is not reduced to {0}. (i) The set of minimal prime ideals is ﬁnite. (ii) Suppose that A is an integral domain. Then any proper non-zero ideal a of A contains a non-zero ﬁnite product of prime ideals. Proof. (i) Let n be the nilradical of A. By 2.7.7, there exist p1 , . . . , pr ∈ Spec A such that n = p1 ∩ · · · ∩ pr . If q is a minimal prime ideal, then n ⊂ q by 2.5.5. So by 2.2.5 (i), there exists i such that q = pi . Thus any minimal prime ideal is one of the pi ’s.√ . . , qs ∈ Spec A are non-zero. (ii) We have a = q1 ∩ · · · ∩ qs where q1 , . √ Since A is Noetherian, by 2.5.3 (ii), for n large, ( a)n ⊂ a. So (q1 · · · qs )n ⊂ a. The result now follows since A is an integral domain. 2.7.9 Theorem. Let A be a Noetherian ring and n ∈ N∗ . The polynomial ring over A in n variables is a Noetherian ring. Proof. By induction on n, it suﬃces to prove that A[X] is Noetherian. Further, we can suppose that A is not reduced to {0}. Suppose that there exists an ideal a of A which is not ﬁnitely generated. Then it is non-zero and let P1 ∈ a \ {0} be such that its degree is minimal among the non-zero elements of a. Since a = A[X]P1 , there exists P2 ∈ a \ A[X]P1 such that its degree is minimal among the elements of a \ A[X]P1 . Repeating this process, we obtain a sequence of elements (Pn )n1 of polynomials in a such that, for any n ∈ N∗ , we have: Pn+1 ∈ a \ (A[X]P1 + · · · + A[X]Pn ), deg Pn+1 deg Q for all Q ∈ a \ (A[X]P1 + · · · + A[X]Pn ). In particular, deg Pn+1 deg Pn for all n ∈ N∗ . Let an be the leading coeﬃcient of Pn and set bn = Aa1 + · · · + Aan . Then (bn )n1 is an ascending sequence of ideals of A. Since A is Noetherian, there exists n ∈ N∗ such that an+1 ∈ bn . So we have an+1 = λ1 a1 + · · · + λn an for some λ1 , . . . , λn ∈ A. Let Q = Pn+1 −

n i=1

λi X deg Pn+1 −deg Pi Pi .

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2 Rings and modules

We have deg Q < deg Pn+1 and Q ∈ a \ (A[X]P1 + · · · + A[X]Pn ). Thus we have obtained a contradiction.

2.8 Derivations 2.8.1 In the rest of this chapter, k will denote a commutative ring and A will denote a commutative k-algebra. Deﬁnition. Let M be an A-module. A derivation of A into M is a map δ : A → M such that for all a, b ∈ A: δ(a + b) = δ(a) + δ(b) , δ(ab) = aδ(b) + bδ(a). Denote by Der(A, M ) the set of such derivations. If A = M , we shall write Der A for Der(A, A), and an element of Der A is called a derivation of A. 2.8.2 Proposition. Let δ ∈ Der(A, M ), x ∈ A and n ∈ N∗ . Then: (i) δ(1) = 0. (ii) δ(xn ) = nxn−1 δ(x). (iii) If x is invertible, then δ(x−1 ) = −x−2 δ(x). Proof. Since δ(1) = δ(1.1) = δ(1) + δ(1), part (i) follows. Part (ii) is a simple induction on n and part (iii) follows directly from (i). 2.8.3 Observe that if δ ∈ Der(A, M ), then the set B of elements a ∈ A such that δ(x) = 0, is a subring of A and δ is B-linear. Conversely, if B is a subring of A such that δ is B-linear, then δ(b) = 0 for all b ∈ B. 2.8.4 Deﬁnition. A derivation of A over k into M is a derivation δ of A into M such that δ is k-linear. The set of such derivations will be denoted by Derk (A, M ). If A = M , then we write Derk A for Derk (A, A) and an element of Derk A is called a derivation of A over k. 2.8.5 Remarks. 1) The partial derivations of the ring k[T1 , . . . , Tn ] are k-derivations of k[T1 , . . . , Tn ]. 2) Let δ ∈ Derk (A, M ) and a ∈ A. Then the map A → M deﬁned by b → aδ(b) is a k-derivation of A with values in M . This induces a natural structure of A-module on Derk (A, M ). 3) If B is a commutative A-algebra and δ ∈ Der(A, B), then for b ∈ B, the map A → B given by a → bδ(a) is a derivation of A into B. So Der(A, B) admits a natural structure of B-module. 4) Let B be a commutative k-algebra, u : A → B be a homomorphism of k-algebras and N a B-module. We can consider N as an A-module via u. If δ ∈ Derk (B, N ), then we verify easily that δ ◦ u ∈ Derk (A, N ). The map C(u) : Derk (B, N ) → Derk (A, N ), δ → δ◦u is a homomorphism of A-modules. We claim that ker C(u) = DerA (B, N ), that is, the following sequence of Amodules is exact: (2)

0 → DerA (B, N ) → Derk (B, N ) → Derk (A, N ).

2.9 Module of diﬀerentials

25

Clearly DerA (B, N ) ⊂ ker C(u) since δ(u(a)) = u(a)δ(1) = 0. Conversely, if δ ∈ ker C(u), then δ(u(a)b) = u(a)δ(b). So ker C(u) = DerA (B, N ). 2.8.6 Proposition. Let A be an integral domain, F the quotient ﬁeld of A and N a F -vector space (considered also as an A-module). Then any derivation of A over k into N can be extended in a unique way to a derivation of F over k into N . Proof. Let δ ∈ Derk (A, N ) and a, b ∈ A, c, d ∈ A \ {0} be such that ad = bc. Then aδ(d) + dδ(a) = bδ(c) + cδ(b). It follows that: d2 bδ(a) − d2 aδ(b) = bd2 δ(a) − bcdδ(b) = b2 dδ(c) + bcdδ(b) − abdδ(d) − bcdδ(b) = b2 dδ(c) − b2 cδ(d). Deﬁne ∆ : F → N as follows, if α = a/b, then: ∆(α) = δ(a)/b − aδ(b)/b2 . The preceding computation implies that ∆ is well-deﬁned and we verify easily that ∆ ∈ Derk (F, N ). Now if ∆ ∈ Derk (F, N ) extends δ to F , then by applying 2.8.2 (iii): ∆ (a/b) = a∆ (1/b) + ∆ (a)/b = ∆ (a)/b − a∆ (b)/b2 = δ(a)/b − aδ(b)/b2 . So ∆ = ∆ and we have uniqueness.

2.9 Module of diﬀerentials 2.9.1 Recall that A is a commutative k-algebra. If M, N are A-modules, then we shall write HomA (M, N ) the set of homomorphisms of A-modules from M to N . By a universal derivation for A over k , we mean a pair (Ω, d) verifying the following properties: (i) Ω is an A-module and d ∈ Derk (A, Ω). (ii) For any A-module M and any δ ∈ Derk (A, M ), there exists a unique f ∈ HomA (Ω, M ) such that f ◦ d = δ. Lemma. Let (Ω, d) and (Γ, D) be universal derivations for A over k. (i) There exists a unique λ ∈ HomA (Ω, Γ ) such that D = λ ◦ d. Further, λ is an isomorphism. (ii) Let M be an A-module. If f ∈ HomA (Ω, M ), then f ◦ d ∈ Derk (A, M ) and the map ψ : HomA (Ω, M ) → Derk (A, M ) , f → f ◦ d is an isomorphism of A-modules.

26

2 Rings and modules

Proof. (i) By deﬁnition, there exist λ ∈ HomA (Ω, Γ ), µ ∈ HomA (Γ, Ω) such that D = λ ◦ d and d = µ ◦ D. So d = µ ◦ λ ◦ d and D = λ ◦ µ ◦ D. By the uniqueness property, µ ◦ λ = idΩ and λ ◦ µ = idΓ . (ii) That f ◦ d ∈ Derk (A, M ) and ψ is a surjective homomorphism of Amodules are straightforward veriﬁcations. The injectivity of ψ follows from the uniqueness property. 2.9.2 The tensor product A ⊗k A has a natural structure of (A, A)bimodule given by a.(x ⊗ y).b = (ax) ⊗ (by) for all a, b, x, y ∈ A. Since A is commutative, we can identify the above structure of (A, A)-bimodule with the structure of A ⊗k A-module induced by the ring structure on A ⊗k A, namely: a.(x ⊗ y).b = (a ⊗ b)(x ⊗ y) Thus A ⊗k A is a left (resp. right) A-module via the ring homomorphism A → A ⊗k A, a → a ⊗ 1 (resp. a → 1 ⊗ a). Denote by m : A ⊗k A → A the k-linear map m(x ⊗ y) = xy. The kernel J of m is an ideal of A ⊗k A and the ring A ⊗k A/J is isomorphic to A. Let δA : A → A ⊗k A be the map given by x → x ⊗ 1 − 1 ⊗ x. Clearly δA (x) ∈ J for any x ∈ A. Lemma. As a left (or right) A-module, J is generated by the image of δA . Proof. Let x1 , . . . , xr , y1 , . . . , yr ∈ A. Then r x i=1 i yi = 0. Thus the lemma follows since r

i=1

xi ⊗ yi = −

r

r

xi δA (yi ) =

i=1

i=1 r

xi ⊗ yi ∈ J implies that δA (xi )yi .

i=1

2.9.3 Set Ωk (A) = J /J 2 . It is clear that Ωk (A) is an A ⊗k A-module. For a, b, x ∈ A, we have: abδA (x) − bδA (x)a = bδA (a)δA (x) ∈ J 2 . In view of 2.9.2, the structures of left and right A-module on Ωk (A) induced by the one on A ⊗k A are the same. We can therefore talk about the A-module ahler) diﬀerentials of A over k. Ωk (A), that we call the module of (K¨ If x ∈ A, denote by dA/k (x) (or dx if there is no confusion), the image of δA (x) in Ωk (A). We say that dx is the diﬀerential of x. By 2.9.2, the set of dx, x ∈ A, is a system of generators of the A-module Ωk (A). Theorem. (i) We have dA/k ∈ Derk (A, Ωk (A)). (ii) The pair (Ωk (A), dA/k ) is a universal derivation for A over k.

2.9 Module of diﬀerentials

27

Proof. (i) Let a, b ∈ A. We have: δA (ab) = aδA (b) + δA (a)b = aδA (b) + bδA (a) + α with α ∈ J 2 . So dA/k ∈ Derk (A, Ωk (A)). (ii) Let M be an A-module and δ ∈ Derk (A, M ). Deﬁne ϕ : A ⊗k A → M , a ⊗ b → bδ(a). If a, x ∈ A, then ϕ(δA (x)) = δ(x) − xδ(1) = δ(x), ϕ(aδA (x)) = δ(ax) − xδ(a) = aδ(x). So ϕ|J ∈ HomA (J , M ). On the other hand: ϕ(δA (a)δA (x)) = δ(ax) − aδ(x) − xδ(a) = 0. We deduce that there exists f ∈ HomA (Ωk (A), M ) such that δ = f ◦ dA/k . Uniqueness follows from 2.9.2. 2.9.4 Example. Let A = k[(Ti )i∈I ] be the ring of polynomials over k in the variables Ti , i ∈ I. Denote dA/k by d. We shall show that Ωk (A) is a free A-module with basis (dTi )i∈I . Let P ∈ A. Since d ∈ Derk (A, Ωk (A)), we have: dP =

∂P dTi . i∈I ∂Ti

It follows that (dTi )i∈I is a set of generators of Ωk (A). Now let M be a free A-module with basis (ei )i∈I . Let δ ∈ Derk (A, M ) be deﬁned as follows: for P ∈ A: ∂P ei . δP = i∈I ∂Ti Let f ∈ HomA (Ωk (A), M ) be such that δ = f ◦ d. Then ei = δ(Ti ) = f (dTi ) for all i ∈ I. So the family (dTi )i∈I is linearly independent over A. 2.9.5 Let u : A → B be a homomorphism of commutative k-algebras. 1) By 2.9.1 and 2.9.3, for any A-module M , there exists an isomorphism of A-modules: ϕk,A : HomA (Ωk (A), M ) → Derk (A, M ) , f → f ◦ dA/k . 2) We have dB/k ◦ u ∈ Derk (A, Ωk (B)). So there exists a unique A-module homomorphism Ω(u) : Ωk (A) → Ωk (B) such that Ω(u) ◦ dA/k = dB/k ◦ u. 3) Let iu : Ωk (A) → Ωk (A) ⊗A B be the canonical homomorphism α → α ⊗ 1. Since Ωk (B) is a B-module and Ωk (A) ⊗A B is a B-module via the right action, there exists a B-linear map Ω0 (u) : Ωk (A) ⊗A B → Ωk (B)

28

2 Rings and modules

such that Ω(u) = Ω0 (u) ◦ iu . 4) Note that dB/A is a derivation over A, so it is a derivation over k. It follows that there exists a unique B-linear map Ωu such that dB/A = Ωu ◦dB/k . Let N be a B-module. From the preceding discussion, we have the following commutative diagram of B-modules: Hom(Ωu ,idN )

(3)

HomB (ΩB (A), N ) −−−−−−−−−→ HomB (Ωk (B), N ) ⏐ ⏐ ⏐ϕk,B ⏐ ϕA,B −−−−→

DerA (B, N )

Derk (B, N )

ju

where ju is the canonical injection and Hom(Ωu , idN ) is the map h → h ◦ Ωu . Proposition. The following sequence of B-linear maps is exact: (4)

Ω0 (u)

Ω

u Ωk (A) ⊗A B −−−−→ Ωk (B) −−−− → ΩA (B) −−−−→ 0.

Proof. It suﬃces to prove that, for any B-module N , the following sequence is exact: h

u u 0 −→ HomB (ΩA (B), N ) −→ HomB (Ωk (B), N ) −→ HomB (Ωk (A) ⊗A B, N )

where hu = Hom(Ωu , idN ) and u = Hom(Ω0 (u), idN ). We have the following commutative diagram: Hom(Ω0 (u),idN )

(5)

HomB (Ωk (B), N ) −−−−−−−−−−→ HomB (Ωk (A) ⊗A B, N ) ⏐ ⏐ ⏐ϕk,A ◦su ϕk,B ⏐ Derk (B, N )

−−−−→

Derk (A, N )

C(u)

where C(u) is as deﬁned in 2.8.5, and su is the canonical isomorphism: su = Hom(iu , idN ) : HomB (Ωk (A) ⊗A B, N ) → HomA (Ωk (A), N ). The vertical arrows of diagram (5) are isomorphisms, and therefore, in view of the diagrams (3) and (5), the result follows from the fact that the sequence (2) in 2.8.5 is exact. 2.9.6 Proposition. Suppose that A is an integral domain. Let B be its ﬁeld of fractions and u : A → B the canonical injection. Then Ω0 (u) is an isomorphism from Ωk (A) ⊗A B to Ωk (B). Proof. Since the vertical arrows of diagram (5) are isomorphisms, it suﬃces to prove that the map C(u) : Derk (B, N ) → Derk (A, N ) is bijective for any B-vector space N . So the proposition follows from 2.8.6.

2.9 Module of diﬀerentials

29

2.9.7 Corollary. Let K be a commutative ﬁeld and E = K((Ti )i∈I ) the ﬁeld of fractions of the ring of polynomials in the variables (Ti )i∈I . Then the family (dTi )i∈I is a basis for the E-vector space ΩK (E). Proof. This is an immediate consequence of 2.9.4 and 2.9.6. 2.9.8 Let kn [T] = k[T1 , . . . , Tn ], P1 , . . . , Pm ∈ kn [T], a the ideal generated by P1 , . . . , Pn and A = kn [T]/a. For 1 i n, let ti be the image of Ti in A. Set t = (t1 , . . . , tn ). Recall that the A-module Ωk (A) is generated by dA/k (ti ), 1 i n. Let (e1 , . . . , en ) be the canonical basis of An and u : An → Ωk (A) the surjective homomorphism given by u(ei ) = dA/k (ti ) for 1 i n. Denote by M the submodule of An generated by ∂Pj ∂Pj (t)e1 + · · · + (t)en ∂T1 ∂Tn where 1 j m. The following lemma extends the result of 2.9.4. Lemma. The kernel of u is M . In other terms, the A-modules Ωk (A) and An /M are isomorphic. Proof. Let α : kn [T] → A and β : An → An /M be the canonical surjections. Since dA/k ∈ Derk (A, Ωk (A)), if Q ∈ kn [T], we have: dA/k (Q(t)) =

∂Q ∂Q (t)dA/k (t1 ) + · · · + (t)dA/k (tn ). ∂T1 ∂Tn

In particular, by taking Q to be one of the Pi ’s, we obtain that M ⊂ ker u. So there exists a unique µ ∈ HomA (An /M, Ωk (A)) such that µ ◦ β = u. Let θ : kn [T] → An be the k-linear map deﬁned by: θ(Q) =

∂Q ∂Q (t)e1 + · · · + (t)en . ∂T1 ∂Tn

We have a ⊂ ker β ◦ θ, and hence there exists a unique ϕ ∈ HomA (A, An /M ) such that ϕ ◦ α = β ◦ θ. It is obvious that ϕ ∈ Derk (A, An /M ). It follows from 2.9.1 and 2.9.5 that there exists a unique ψ ∈ HomA (Ωk (A), An /M ) such that ϕ = ψ ◦dA/k . We obtain that ψ(dA/k (ti )) = ψ ◦dA/k ◦α(Ti ) = β ◦θ(Ti ) = β(ei ) for 1 i n. Since µ◦β(ei ) = dA/k (ti ) and the β(ei )’s generate the A-module An /M , the lemma follows.

References • [8], [10], [11], [52], [56].

3 Integral extensions

In this chapter, we study the notion of integral extensions. In particular, we prove the Going Up Theorem concerning the extension of prime ideals. All rings considered in this chapter are assumed to be commutative. Let A be a subring of a ring B. For x1 , . . . , xn ∈ B, we denote by A[x1 , . . . , xn ] the subring of B generated by A and x1 , . . . , xn .

3.1 Integral dependence 3.1.1 For n ∈ N, denote by Mn (A) the A-algebra of n-by-n matrices with coeﬃcients in A. If M ∈ Mn (A), then we write χM (X) its characteristic polynomial. So χM (X) = det(XIn − M ) where In is the identity matrix. 3.1.2 Deﬁnition. An element x ∈ B is said to be integral over A if there exist n ∈ N∗ and a0 , a1 , . . . , an−1 ∈ A such that: xn + an−1 xn−1 + · · · + a1 x + a0 = 0. Such a relation is called an integral equation or equation of integral dependence for x over A. 3.1.3 Let C be a commutative A-algebra, f : A → C the canonical Aalgebra homomorphism. If an element c ∈ C is integral over f (A), then we shall say that c is integral over A. 3.1.4 Theorem. Let A be a subring of B and x ∈ B. Then the following conditions are equivalent: (i) x is integral over A. (ii) A[x] is a ﬁnitely generated A-module. (iii) There exists a subring C of B, containing A[x] such that C is a ﬁnitely generated A-module. Proof. (i) ⇒ (ii) Let xn +an−1 xn−1 +· · ·+a1 x+a0 = 0 be an integral equation for x over A. Then it is clear that A[x] is generated by 1, x, . . . , xn−1 . (ii) ⇒ (iii) This is obvious.

32

3 Integral extensions

(iii) ⇒ (i) Let {y1 , . . . , yn } be a system of generators of the A-module C. For 1 i n, there exist aij ∈ A, 1 j n, such that: xyi = ai1 y1 + · · · + ain yn . Set M = [aij ] ∈ Mn (A) and for k ∈ N∗ , let M k = [aijk ]. Then we have: xk yi = ai1k y1 + · · · + aink yn . Now χM (M ) = 0 and so χM (x)yi = 0 for 1 i n. Thus χM (x)u = 0 for any u ∈ C. By setting u = 1, we obtain an integral equation for x over A. 3.1.5 Corollary. (i) Let x1 , . . . , xn ∈ B. Suppose that x1 is integral over A and that for 2 i n, xi is integral over A[x1 , . . . xi−1 ]. Then A[x1 , . . . , xn ] is a ﬁnitely generated A-module. (ii) The set of elements of B integral over A is a subring of B which contains A. Proof. (i) Let us prove this by induction on n. For n = 1, the result follows from 3.1.4. Suppose that n 2, then by induction hypothesis, D = A[x1 , . . . , xn−1 ] is a ﬁnitely generated A-module. Let {y1 , . . . , yr } be a system of generators of the A-module D. Now xn is integral over D, so D[xn ] is a ﬁnitely generated D-module. Let {z1 , . . . , zs } be a system of generators of the D-module D[xn ]. Then clearly, {yi zj ; 1 i r, 1 j s} is a system of generators of the A-module A[x1 , . . . , xn ]. (ii) Let x, y ∈ B be integral over A. Since xy, x + y ∈ A[x, y], the result follows from (i) and 3.1.4. 3.1.6 Deﬁnition. Let A be a subring of B. The set C of elements of B integral over A is called the integral closure of A in B. By 3.1.5, C is a subring of B which contains A. If C = B, we shall say that B is integral over A. If C = A, we shall say that A is integrally closed in B. 3.1.7 Proposition. Let A and D be subrings of a ring B. (i) Suppose that A ⊂ D. If D is integral over A, and B is integral over D, then B is integral over A. (ii) If C is the integral closure of A in B, then C is integrally closed in B. Proof. Let x ∈ B and xn + dn−1 xn−1 + · · · + d1 x + d0 = 0 be an integral equation of x over D. By 3.1.5, the A-module A[d0 , . . . , dn−1 , x] is ﬁnitely generated. So x is integral over A. This proves (i), and part (ii) follows. 3.1.8 Proposition. Let b be an ideal of B, a = b∩A and S a multiplicative subset of A. Let us suppose that B is integral over A. Then: (i) B/b is integral over A/a. (ii) S −1 B is integral over S −1 A.

3.2 Integrally closed domains

33

Proof. (i) Let us denote x + b by x, and identify A/a as a subring of B/b. If xn + an−1 xn−1 + · · · + a1 x + a0 = 0 is an integral equation of x over A, then xn + an−1 xn−1 + · · · + a1 x + a0 = 0 is an integral equation of x over A/a. (ii) By 2.3.6, we identify S −1 A as a subring of S −1 B. Let x/s ∈ S −1 B. If n x + an−1 xn−1 + · · · + a1 x + a0 = 0 is an integral equation of x over A, then (x/s)n + (an−1 /s)(x/s)n−1 + · · · + (a1 /sn−1 )(x/s) + a0 /sn = 0 is an integral equation of x/s over S −1 A. 3.1.9 Proposition. Let S be a multiplicative subset of A and C the integral closure of A in B. Then the integral closure of S −1 A in S −1 B is S −1 C. In particular, if A is integrally closed in B, then S −1 A is integrally closed in S −1 B. Proof. By 3.1.8, the integral closure of S −1 A in S −1 B contains S −1 C. Now, if x/s ∈ S −1 B is integral over S −1 A, then there exist a0 , . . . , an−1 ∈ A, t0 , . . . , tn−1 ∈ S such that xn /sn + (an−1 /tn−1 )(xn−1 /sn−1 ) + · · · + (a1 /t1 )(x/s) + a0 /t0 = 0. Set b = t0 · · · tn−1 x and u = (st0 · · · tn−1 )n , then (bn + sn−1 an−1 bn−1 + · · · + s1 a1 b + s0 a0 )/u = 0 where s0 , . . . , sn−1 ∈ S. This implies that there exists v ∈ S such that: v(bn + sn−1 an−1 bn−1 + · · · + s1 a1 b + s0 a0 ) = 0. From this, we deduce that vb = vt0 · · · tn−1 x ∈ C and so x/s ∈ S −1 C.

3.2 Integrally closed domains 3.2.1 Lemma. Let E be an A-module. Then the following conditions are equivalent: (i) The module E is {0}. (ii) For any prime ideal p of A, Ep = {0}. (iii) For any maximal ideal m of A, Em = {0}. Proof. The implications (i) ⇒ (ii) ⇒ (iii) are obvious. Suppose now that (iii) is satisﬁed. If there exists x ∈ E \ {0}, then a = {a ∈ A; ax = 0} is a proper ideal of A. Let m be a maximal ideal containing a. Since Em = {0}, we have x/1 = 0, or equivalently, there exists s ∈ A \ m such that sx = 0 which implies that s ∈ a ⊂ m. We have therefore a contradiction. 3.2.2 Lemma. Let A be an integral domain. Then Am . A= m∈Spm A

34

3 Integral extensions

Proof. For any m ∈ Spm A, Am is a subring of its quotient ﬁeld K by 2.3.5. The intersection B of the Am ’s is therefore a subring of K containing A. Let m ∈ Spm A, S = A \ m. We have Am ⊂ S −1 B ⊂ Am . So Am = S −1 B. It follows from 2.4.3 that S −1 (B/A) = {0}. So by 3.2.1, B/A = {0}. 3.2.3 Deﬁnition. (i) Let A be an integral domain and K its quotient ﬁeld. By the integral closure of A, we mean the integral closure of A in K. (ii) We say that A is integrally closed domain if A is an integral domain, and it is equal to its integral closure. 3.2.4 Lemma. Let B be an integral domain and (A i )i∈I a family of subrings which are integrally closed domains. Then A = i∈I Ai is an integrally closed domain. Proof. Let K, H, Ki be the ﬁeld of fractions of A, B, Ai respectively. We have K ⊂ Ki ⊂ H. If x ∈ K is integral over A, then it is integral over Ai for all i ∈ I. Thus x ∈ Ai for all i ∈ I. 3.2.5 Proposition. Let A be an integral domain. The following conditions are equivalent: (i) The ring A is an integrally closed domain. (ii) For any p ∈ Spec A, Ap is an integrally closed domain. (iii) For any m ∈ Spm A, Am is an integrally closed domain. Proof. We have (i) ⇒ (ii) by 3.1.9. The implication (ii) ⇒ (iii) is obvious. Finally, (iii) ⇒ (i) follows from 3.2.2 and 3.2.4. 3.2.6 Lemma. Let A be an integrally closed domain, K its quotient ﬁeld and P, Q ∈ K[X] monic polynomials such that P Q ∈ A[X]. Then P, Q ∈ A[X]. Proof. Let L be an algebraic closure of K. Then P (X) = (X − α1 ) · · · (X − αm ) , Q(X) = (X − β1 ) · · · (X − βn ), with α1 , . . . , αm , β1 , . . . , βn ∈ L. The αi ’s and the βj ’s are integral over A because they are roots of P Q ∈ A[X]. Since the coeﬃcients of P and Q are polynomials in the αi ’s and the βi ’s with coeﬃcients in A, it follows from 3.1.5 that they are integral over A. Hence P, Q ∈ A[X]. 3.2.7 Lemma. If K is a commutative ﬁeld, then K[X] is an integrally closed domain. Proof. Let P, Q ∈ K[X] \ {0} be relatively prime and let us suppose that P/Q is integral over K[X]. Let (P/Q)n + Fn−1 (P/Q)n−1 + · · · + F0 = 0 be an integral equation of P/Q over K[X]. Suppose further that n is minimal. So F0 = 0, and we have:

3.3 Extensions of prime ideals

35

P n + Fn−1 P n−1 Q + · · · + F0 Qn = 0. It follows that Q divides P n and so Q is invertible in K[X] because P, Q are relatively prime. Thus Q ∈ K \ {0} which implies that P/Q ∈ K[X]. 3.2.8 Theorem. Let A be an integrally closed domain, then A[X] is also an integrally closed domain. Proof. Denote by K and L, the quotient ﬁeld of A and A[X] respectively. Let α ∈ L be integral over A[X]. Then it is integral over K[X]. So by 3.2.7, α ∈ K[X]. Let Q = T m + pm−1 T m−1 + · · · + p0 ∈ A[X][T ] be such that Q(α) is an integral equation of α over A[X]. Let s ∈ N be such that deg α < s and set β = α − X s . Then −β is monic and β is a root of Q(T + X s ) = T m + qm−1 T m−1 + · · · + q0 where q0 , . . . , qm−1 ∈ A[X]. Now q0 = Q(X s ) = X sm + pm−1 X s(m−1) + · · · + p0 , and therefore q0 is monic for s large. But β is a root of Q(T + X s ), so: q0 = −β(β m−1 + qm−1 β m−2 + · · · + q1 ) = −βγ, where γ ∈ A[X]. Thus γ is monic in K[X] for s large. Since q0 ∈ A[X], it follows from 3.2.6 that −β ∈ A[X]. Hence α ∈ A[X].

3.3 Extensions of prime ideals 3.3.1 Proposition. Let A be a subring of an integral domain B such that B is integral over A. Then A is a ﬁeld if and only if B is a ﬁeld. Proof. Suppose that A is a ﬁeld. Let b ∈ B\{0} and bn +an−1 bn−1 +· · ·+a0 = 0 an integral equation of b over A. Since B is an integral domain, we can suppose that a0 = 0. Then in Fract(B): n−1 n−2 b−1 = −(a−1 + a−1 + · · · a−1 0 b 0 an−1 b 0 a1 ) ∈ B.

Now let us suppose that B is a ﬁeld. Let a ∈ A \ {0}, so a−1 ∈ B. Let a + an−1 a−(n−1) + · · · + a0 = 0 be an integral equation of a−1 over A. Then a−1 = −(an−1 + an−2 a + · · · + a0 an−1 ) ∈ A. −n

3.3.2 Corollary. Let A be a subring of a ring B such that B is integral over A and q, q ∈ Spec B. (i) We have q ∈ Spm B if and only if p = q ∩ A ∈ Spm A. (ii) Suppose that q ⊂ q . Then A ∩ q = A ∩ q if and only if q = q . Proof. (i) By 3.1.8, B/q is integral over A/p. So (i) follows from 3.3.1. (ii) Suppose that p = A ∩ q = A ∩ q . Set S = A \ p. Since S ∩ q = S ∩ q = ∅, by 2.3.8, r = S −1 q and r = S −1 q are prime ideals of S −1 B such that r ⊂ r . Since p ⊂ q ⊂ q , we have S −1 p ⊂ r ⊂ r . Hence S −1 p ⊂ r ∩ Ap ⊂ r ∩ Ap . But Ap is a local ring with maximal ideal S −1 p, so S −1 p = r ∩ Ap = r ∩ Ap . It follows from (i) that r, r ∈ Spm S −1 B. Since r ⊂ r , we have equality and we conclude by 2.3.7 that q = q .

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3 Integral extensions

3.3.3 Theorem. (Going Up Theorem) Let A be a subring of a ring B such that B is integral over A. (i) If p ∈ Spec A, then there exists q ∈ Spec B lying above p. (ii) Let p1 , p2 ∈ Spec A and q1 ∈ Spec B lying above p1 . Suppose that p1 ⊂ p2 . Then there exists q2 ∈ Spec B lying above p2 such that q1 ⊂ q2 . Proof. (i) Let S = A \ p, i : A → Ap , j : B → S −1 B the canonical homomorphisms, λ : A → B and µ : Ap → S −1 B the canonical injections. By 2.3.6, we have j ◦ λ = µ ◦ i. By 3.1.8, S −1 B is integral over Ap . Let n ∈ Spm S −1 B, then by 3.3.2, n ∩ Ap ∈ Spm Ap . Since Ap is a local ring with maximal ideal S −1 p, we have S −1 p = n ∩ Ap . So q = j −1 (n) ∈ Spec B and q ∩ A = i−1 (S −1 p) = p. (ii) By 3.1.8, B = B/q1 is integral over A = A/p1 . Denote by p : A → A and q : B → B the canonical surjections, λ : A → B and µ : A → B the canonical injections. We have q ◦ λ = µ ◦ p. Now p(p2 ) ∈ Spec A . It follows from (i) that there exists q2 ∈ Spec B such that A ∩ q2 = p(p2 ). So q2 = q −1 (q2 ) ∈ Spec B satisﬁes q1 ⊂ q2 and A ∩ q2 = p−1 (A ∩ q2 ) = p2 . 3.3.4 Theorem. Let A be a subring of a ring B such that B is integral over A. Then the following conditions are equivalent: (i) The dimension of A is ﬁnite. (ii) The dimension of B is ﬁnite. Further, if these conditions are satisﬁed, then dim A = dim B. Proof. By 3.3.3, any chain of length n of prime ideals of A gives a chain of length n of prime ideals of B. So dim A dim B. Conversely, by using 3.3.2 (ii), we obtain that a chain of length m of prime ideals of B gives a chain of length m of prime ideals of A, so dim B dim A. 3.3.5 Let A be a subring of a ring B. We say that B is a ﬁnite A-algebra if B is a ﬁnitely generated A-module. By 3.1.4, a ﬁnite A-algebra is integral over A. Conversely, by 3.1.5, a ﬁnitely generated A-algebra B, integral over A is a ﬁnite A-algebra. In general, let C be an A-algebra and f : A → C the ring homomorphism inducing the A-algebra structure on C. We say that C is a ﬁnite A-algebra if C is a ﬁnite f (A)-algebra. 3.3.6 Lemma. Let K be a subring of A. Suppose that K is a ﬁeld and A is integral over K. (i) We have Spm A = Spec A. (ii) If A is a ﬁnite K-algebra, then Spec A is a ﬁnite set. Proof. (i) By 3.3.4, dim A = 0. So Spec A = Spm A. (ii) Since Spec A = Spm A, any prime ideal is a minimal prime ideal. If A is a ﬁnite K-algebra, then A is Noetherian by 2.7.5. So the result follows from 2.7.8 (i).

3.3 Extensions of prime ideals

37

3.3.7 Proposition. Let A be a subring of a ring B and p ∈ Spec A. If B is a ﬁnite A-algebra, then the set of prime ideals lying above p is ﬁnite. Proof. Let S = A \ p. In view of 2.3.9, we can replace A, B, p by S −1 A, S −1 B (which is a ﬁnite S −1 A-algebra) and S −1 p. So A is now a local ring with maximal ideal p. Next, replace A and B by A/pA and B/pB so that we can apply 3.3.6 since A is now a ﬁeld and B a ﬁnite A-algebra.

References • [10], [11], [52], [56].

4 Factorial rings

All rings considered in this chapter are commutative. For a ring A, we shall denote by U (A) the group of invertible elements of A.

4.1 Generalities 4.1.1 Let A be a ring. If a ∈ A, then we shall denote the ideal Aa by (a). An ideal of this form is called principal. Let us denote by PA the set of ∗ the set of non-zero principal ideals of A. These principal ideals of A, and PA sets are partially ordered by inclusion. 4.1.2 Let us suppose that A is an integral domain. Let a, b ∈ A. Then we say that a divides b or b is a multiple of a if there exists c ∈ A such that ac = b. We shall write a | b (resp. a b) if a divides (resp. does not divide) b. It is clear that a | b ⇔ (b) ⊂ (a). Note that 0 divides b if and only if b = 0. Further, since A is an integral domain, c is unique when a = 0. We shall call c the quotient of b by a. 4.1.3 Proposition. Let A be an integral domain. The following conditions are equivalent for a, b ∈ A: (i) There exists u ∈ U (A) such that b = au. (ii) a | b and b | a. (iii) (a) = (b). When these conditions are veriﬁed, we say that a and b are equivalent, and we shall write a ∼ b. Proof. This is straightforward. 4.1.4 Deﬁnition. An element a in an integral domain A is called irreducible if the following conditions are satisﬁed: (i) a ∈ U (A). (ii) If b, c ∈ A are such that a = bc, then b ∈ U (A) or c ∈ U (A).

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4 Factorial rings

4.1.5 Proposition. Let a be a non-zero element in an integral domain A. If (a) is a prime ideal, then a is an irreducible element of A. Proof. Since (a) is a prime ideal, (a) = A, so a ∈ U (A). Now let b, c ∈ A be such that a = bc. Then b ∈ (a) or c ∈ (a), i.e. a | b or a | c. By 4.1.3, this implies that either c ∈ U (A) or b ∈ U (A). 4.1.6 Deﬁnition. Let a = (ai )i∈I be a non-empty family of elements of an integral domain A. Denote by U the family (Aai )i∈I in PA . (i) If U has a least upper bound d in PA , then a generator δ of d is called a greatest common divisor, abbreviated to gcd of a. We shall write: (1)

δ = gcd((ai )i∈I ) =i∈I ai .

(ii) If U has a greatest lower bound d in PA , then a generator µ of m is called a least common multiple, abbreviated to lcm of a. We shall write: (2)

µ = lcm((ai )i∈I ) =i∈I ai .

4.1.7 Remarks. 1) The notations in (1) and (2) are not rigorous since δ (resp. µ) is not unique. However, the ideals d and m are unique. So by 4.1.3, a generator of d (resp. m) is unique up to multiplication by an element in U (A). 2) By convention, when I is empty, we set d = {0}, δ = 0, m = A and µ = 1. 4.1.8 Proposition. Let a = (ai )i∈I , δ, µ be elements of an integral domain A. (i) δ is the gcd of the family a if and only if δ divides each ai , and if any common divisor of the ai ’s is a divisor of δ. (ii) µ is the lcm of the family a if and only if µ is a multiple of each ai , and if any common multiple of the ai ’s is a multiple of δ. Proof. We can assume that I is non-empty. If δ = gcd(a), then (ai ) ⊂ (δ), thus δ divides ai for all i ∈ I. Now if r is a common divisor of the ai ’s, then (ai ) ⊂ (r). It follows from the deﬁnition of δ that (δ) ⊂ (r), which implies that r divides δ. Conversely, if δ divides each ai , and if any common divisor r of the ai ’s is a divisor of δ, then (ai ) ⊂ (δ) for all i ∈ I and (δ) ⊂ (r). It follows that δ = gcd(a). This proves (i). The proof of part (ii) is analogue. 4.1.9 Deﬁnition. Let a = (ai )i∈I be elements of an integral domain A. (i) We say that the ai ’s are relatively prime if gcd(a) = 1. (ii) We say that the ai ’s are pairwise relatively prime if gcd(ai , aj ) = 1 for all i, j ∈ I, i = j. 4.1.10 Proposition. Let a, b be non-zero elements in an integral domain A. If a is irreducible, then a, b are relatively prime if and only if a does not divide b.

4.2 Unique factorization

41

Proof. If a divides b, then a is a greatest common divisor of a and b. So a, b are not relatively prime since a ∈ U (A). Conversely, if a does not divide b, then any common divisor d of a and b must be invertible since it divides a. It follows that a and b are relatively prime.

4.2 Unique factorization 4.2.1 Deﬁnition. A commutative ring A is called factorial (or a unique factorization domain) if it is an integral domain and if it veriﬁes the following conditions: (i) For any a ∈ A \ {0}, there exist m ∈ N, p1 , . . . , pm irreducible elements in A, and u ∈ U (A) such that: (3)

a = up1 · · · pm .

(ii) Let m, n ∈ N, p1 , . . . , pm , q1 , . . . , qn irreducible elements in A and u, v ∈ U (A) be such that up1 · · · pm = vq1 · · · qn . Then m = n and there exists a permutation σ of the set {1, . . . , m} such that qi ∼ pσ(i) for all i. We shall express this condition by saying that the decomposition (3) is essentially unique. 4.2.2 Deﬁnition. A subset I of an integral domain A is called a complete system of irreducible elements of A if (i) Any element of I is irreducible. (ii) If q ∈ A is irreducible, then there exists a unique p ∈ I such that p ∼ q. 4.2.3 By 4.2.2, we have another deﬁnition of a factorial ring: Deﬁnition. Let A be an integral domain and I ⊂ A a complete system of irreducible elements of A. We say that A is factorial if for any a ∈ A non-zero, we can associate a unique pair (ua , ν(a)) where ua ∈ U (A) and ν(a) : I → N, p → νp (a), which is zero almost everywhere such that: (4)

a = ua

pνp (a) .

p∈I

We call (4) the decomposition of a into irreducible elements (relative to I), and we call νp (a), the p-adic valuation of a. 4.2.4 Deﬁnition. Let A be an integral domain. (i) We say that A satisﬁes Euclid’s Lemma if for a, b, p ∈ A with p irreducible, we have p | ab ⇒ p | a or p | b. (ii) We say that A satisﬁes Gauss’ Lemma if for a, b, c ∈ A, we have:

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4 Factorial rings

a | bc and a b = 1 ⇒ a | c. (iii) Let I be a complete system of irreducible elements of A. We say that A satisﬁes the condition (D) for I if for any a ∈ A non-zero, there exists a pair (ua , ν(a)) where ua ∈ U (A) and ν(a) : I → N, p → νp (a), such that a decomposes as in (4). Remark. It is clear that if A satisﬁes condition (D) for I, then it satisﬁes condition (D) for any other complete system of irreducible elements of A. We shall therefore state condition (D) without specifying I. 4.2.5 Theorem. Let A be an integral domain satisfying condition (D). Then the following conditions are equivalent: (i) A is factorial. (ii) A satisﬁes Euclid’s Lemma. (iii) An element p ∈ A \ {0} is irreducible if and only if the ideal (p) is prime. (iv) A satisﬁes Gauss’ Lemma. Proof. Let us ﬁx a complete system I of irreducible elements of A. (i) ⇒ (iv) Let a, b, c ∈ A be such that a | bc and a b = 1. If a, b or c is 0, then clearly, a | c. So let us assume that a, b, c are non-zero. Let d ∈ A be such that ad = bc. Since A is factorial, we have νp (a) + νp (d) = νp (b) + νp (c) for all p ∈ I. But a b = 1, so either νp (a) or νp (b) is zero. It follows in both cases that νp (a) νp (c), and so a | c. (iv) ⇒ (ii) This is clear by 4.1.9. (ii) ⇔ (iii) This is straightforward from Euclid’s Lemma and 4.1.7. (ii) ⇒ (i) Let a ∈ A \ {0} and ωp (a) νp (a) p =v p a=u p∈I

p∈I

be two decompositions of a into irreducible elements. Write a = pνp (a) q = pωp (a) r where q p = 1 = r p. Then by Euclid’s Lemma, we must have νp (a) = ωp (a). Now by induction, we have that νp (a) = ωp (a) for all p ∈ I and u = v. 4.2.6 Proposition. Let A be a factorial ring and I a complete system of irreducible elements of A. Let a, b be non-zero elements of A. Then a divides b if and only if νp (a) νp (b) for all p ∈ I. Proof. This is immediate by 4.2.5. 4.2.7 Theorem. Let A be a factorial ring. Any family a = (ai )i∈I of elements of A has a greatest common divisor and a least common multiple.

4.3 Principal ideal domains and Euclidean domains

43

Proof. We can assume that I = ∅ and ai = 0 for all i ∈ I. Let I be a complete system of irreducible elements of A. Set δ = p∈I pνp where for p ∈ I, νp = inf{νp (ai ), i ∈ I}. Setµ = 0 if there exists q ∈ I such that νq (ai ) is not bounded, otherwise, set µ = p∈I pωp where ωp = sup{νp (ai ), i ∈ I}. By 4.1.7 and 4.2.6, it is clear that δ = gcd(a) and µ = lcm(a). 4.2.8 Proposition. Let a, b be non-zero elements of a factorial ring A. Then (a b)(a b) ∼ ab. Proof. This is an immediate consequence of the proof of 4.2.7. 4.2.9 Theorem. A factorial ring A is an integrally closed domain. Proof. Let x ∈ Fract(A) \ {0} be integral over A, and a0 , · · · , an−1 ∈ A be such that xn + an−1 xn−1 + · · · + a0 = 0. Write x = p/q where p, q ∈ A and p q = 1. Then pn + an−1 qpn−1 + · · · + a0 q n = 0, which implies that p | q n . So q ∈ U (A) and x ∈ A.

4.3 Principal ideal domains and Euclidean domains 4.3.1 Deﬁnition. A commutative ring is called a principal ideal domain if it is an integral domain and if all ideals of A are principal. 4.3.2 Remarks. 1) A principal ideal domain is Noetherian. 2) Examples of principal ideal domains are Z and K[X], where K is a commutative ﬁeld. 4.3.3 Proposition. Let A be a principal ideal domain. Suppose that A is not a ﬁeld and a an ideal of A. Then the following conditions are equivalent: (i) a is maximal. (ii) a is prime and non-trivial. (iii) There exists an irreducible element p ∈ A such that a = (p). Proof. This is straightforward from the deﬁnitions and 4.1.5. 4.3.4 Corollary. (i) Let p be an irreducible element of a principal ideal domain A, then A/(p) is a ﬁeld. (ii) The ring A[X] is a principal ideal domain if and only if A is a ﬁeld. Proof. Part (i) is clear from 4.3.3 and the “if” part of (ii) is also clear. Now if A[X] is a principal ideal domain, then clearly A must be an integral domain. Hence U (A[X]) = U (A). It follows that X is an irreducible element of A[X]. By part (i), A[X]/(X) A is a ﬁeld. 4.3.5 Lemma. A Noetherian ring A satisﬁes condition (D).

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4 Factorial rings

Proof. Let E be the set of elements a ∈ A\(U (A)∪{0}) which are not products of irreducible elements. Suppose that E = ∅, and let a1 ∈ E. In particular, a1 is not irreducible. So a1 = a2 a2 where a2 , a2 are both non-equivalent to a1 . Further, since a1 ∈ E, one of them must be in E, say a2 . Now, by repeating the argument with a2 , we obtain a strictly increasing sequence (an )n of ideals of A which contradicts the hypothesis that A is Noetherian. 4.3.6 Theorem. A principal ideal domain is factorial. Proof. Let A be a principal ideal domain. By 4.3.2 and 4.3.5, A satisﬁes condition (D). Now the result follows from 4.3.3 and 4.2.5. 4.3.7 Let A be a principal ideal domain and P the set of ideals of A. If a = (ai )i∈I is a family of elements in A, and δ = gcd(a), µ = lcm(a), then: (ai ) , (µ) = (ai ). (δ) = i∈I

i∈I

We obtain in particular the following well-known result: Theorem. (Bezout’s Theorem) Let a, b be elements of a principal ideal domain A and δ = gcd(a, b). Then there exist x, y ∈ A such that δ = ax + by. 4.3.8 Deﬁnition. A commutative integral domain A is called an Euclidean domain if there exists a map ν : A \ {0} → N such that: (i) For all x, y ∈ A \ {0} such that x | y, we have ν(x) ν(y). (ii) If a, b ∈ A \ {0}, there exist q, r ∈ A such that a = bq + r with r = 0 or ν(r) < ν(b). 4.3.9 The principal ideal domains Z and K[X] are examples of Euclidean domains. In fact, we obtain easily (as for Z) that: Theorem. An Euclidean domain is a principal ideal domain. 4.3.10 Theorem. Let A be an integral domain, S a multiplicative set of A and B = S −1 A. (i) If A is factorial, then so is B. (ii) If A is a principal ideal domain, then so is B. (iii) If A is an Euclidean domain, then so is B. Proof. Clearly, the statements are trivial if B is a ﬁeld. So let us assume that B is not a ﬁeld. (i) Let I be a complete system of irreducible elements of A, and IS the set of elements of I which do not divide any element of S. Since B is not a ﬁeld, IS = ∅. Let q ∈ IS . Let a, b ∈ A, s, t ∈ S be such that q = (a/s)(b/t) in B. Thus qst = ab and the factoriality of A implies that q divides a or b. Let us suppose that q | a and a = qc where c ∈ A. Then (c/s)(b/t) = 1 and so b/t ∈ U (B). Thus q is irreducible in B.

4.4 Polynomials and factorial rings

45

Further, it is easy to see that non-zero elements of B are products of elements of IS and elements of U (B). Thus IS is a complete system of irreducibles in B. Finally, to show that B is factorial, we shall show that (ii) of 4.2.5 is satisﬁed. Let q ∈ IS , b, c ∈ A , s, t ∈ S be such that q/1 divides (b/s)(c/t). We have (qd)/u = (bc)/(st) for some d ∈ A, u ∈ S. Thus qdst = bcu. Further, if qr = u for some r ∈ A, then (q/1)(r/u) = 1 which is absurd since q/1 is irreducible in B. It follows that q divides bc. By Euclid’s Lemma, we have q | b or q | c. It follows that q/1 divides b/s or c/t. Hence B is factorial by 4.2.5. (ii) Let b be an ideal of B and A ∩ b = a. There exists a ∈ A such that a = Aa. So by 2.3.7, b = Ba. (iii) Let ν : A \ {0} → N be the map which makes A an Euclidean domain. By the proof of part (i), B is factorial and IS is a complete system of irreducible elements of A. It follows that for any x ∈ B \ {0}, there exist unique ux ∈ U (B) and ax , a product of irreducibles in IS , such that x = ax ux . Deﬁne νS : B \ {0} → N by νS (x) = ν(ax ). This is a well-deﬁned map. On veriﬁes easily that νS (x) = 1 if x ∈ U (B) and νS (x) νS (y) if x | y. Finally, if x, y ∈ B \ {0} and x = ux ax , y = uy ay as above. Then there exist q, r ∈ A such that ax = qay + r with r = 0 or ν(r) < ν(ay ). It follows that x = (qux /uy )y + ux r with ux r = 0 or νS (ux r) = νS (r) ν(r) < ν(ay ) = νS (y).

4.4 Polynomials and factorial rings 4.4.1 Let A be a factorial ring. The content of a polynomial P ∈ A[X]\{0} is deﬁned to be any greatest common multiple of the coeﬃcients of P , and we shall denote it by c(P ). Of course, c(P ) is unique up to a multiple of an element of U (A). A polynomial P is called primitive if c(P ) ∈ U (A). Clearly, P = c(P )P1 where P1 ∈ A[X] is primitive. 4.4.2 Lemma. Let A be a factorial ring and P, Q ∈ A[X] \ {0}. Then c(P Q) ∼ c(P )c(Q). Proof. We can assume that P, Q are primitive. If c(P Q) ∈ U (A), then there exists an irreducible element p of A which divides c(P Q). Let B = A/(p) and consider the algebra morphism ψ : A[X] → B[X] induced from the canonical surjection A → B. Then ψ(P )ψ(Q) = ψ(P Q) = 0. Since B is an integral domain by 4.2.5, we have that either ψ(P ) = 0 or ψ(Q) = 0. This implies that p divides c(P ) or c(Q) which is absurd. 4.4.3 Proposition. Let K be the ﬁeld of fractions of a factorial ring A and P ∈ A[X]. (i) If P ∈ A, then P is irreducible in A[X] if and only if P is irreducible in A. (ii) If P ∈ A, then P is irreducible in A[X] if and only if P is primitive in A[X] and irreducible in K[X].

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4 Factorial rings

Proof. (i) If P = QR with Q, R ∈ A[X], then deg Q = deg R = 0. So Q, R belong to A. Since U (A) = U (A[X]), the result is clear. (ii) If P is irreducible in A[X], then clearly, it is primitive. Now if P = QR where Q, R ∈ K[X], then there exist primitive polynomials Q , R ∈ A[X] and elements a, b, c, d ∈ A such that Q = ab−1 Q , R = cd−1 Q , a b = 1 = c d. Hence bdP = acQ R and so (bd) = (ac) by 4.4.2. It follows that P = uQ R where u ∈ U (A). Since P is irreducible in A[X], either Q or R is in U (A). This implies that P is irreducible in K[X]. Conversely, suppose that P is primitive in A[X] and irreducible in K[X]. If P = QR with Q, R ∈ A[X] \ {0}, then the irreducibility of P in K[X] implies that either Q or R is in K. Let us suppose that Q ∈ K, then Q ∈ A \ {0}. Now Q | c(P ). Thus Q ∈ U (A) and we conclude that P is irreducible in A[X]. 4.4.4 Corollary. Let the notations be as in 4.4.3. If P ∈ A and P is irreducible in A[X], then P is irreducible in K[X]. 4.4.5 Theorem. Let A be a factorial ring, and X = (Xi )i∈I be a nonempty family of indeterminates over A. The ring A[X] is factorial. Proof. If P ∈ A[X], then there exists a ﬁnite subset J of I such that P ∈ A[(Xj )j∈J ]. Thus it suﬃces to prove the theorem for I ﬁnite. Further, by induction, we are reduced to prove that A[X] is factorial. Let P ∈ A[X] and n = deg P 0. We shall ﬁrst prove by induction on n that P is a product of irreducible elements of A[X]. Since A is factorial, the result is clear for n = 0 by part (i) of 4.4.3. Let us suppose that n 1 and P is not irreducible. By 4.4.1, we can assume further that c(P ) = 1. By part (ii) of 4.4.3, P = QR where 1 deg Q, deg R < n. Applying our induction hypothesis on Q and R, we obtain that P is a product of irreducible elements. We shall show that A[X] satisﬁes Euclid’s Lemma, which, by 4.2.5, would imply that A[X] is factorial. Let P, Q, R ∈ A[X] \ {0} be such that P is irreducible and P | QR. If P ∈ A, then P | c(QR). Hence P divides c(Q) or c(R) since A is factorial. So P divides Q or R as required. If P ∈ A, then P is irreducible in K[X] by 4.4.4 and c(P ) = 1. Now, K[X] is factorial, so P divides Q or R in K[X]. Let us suppose that P F = Q where F ∈ K[X]. There exist a, b ∈ A with a b = 1 such that a−1 bF = G ∈ A[X] is primitive. It follows that aP G = bQ and so a = bc(Q). Thus b ∈ U (A) and P | Q in A[X]. 4.4.6 Remarks. 1) Let K be a (commutative) ﬁeld. The ring K[X, Y ] is a factorial ring which is not a principal ideal domain. In fact, Bezout’s identity is not veriﬁed here, for X Y = 1 and (X) + (Y ) = K[X, Y ].

4.4 Polynomials and factorial rings

47

2) If I is an inﬁnite set, then A[X] is a factorial ring which is not Noetherian. 4.4.7 Theorem. (Eisenstein’s Criterion) Let A be a factorial ring with ﬁeld of fractions K. Let P = an X n + an−1 X n−1 + · · · + a0 ∈ A[X] where an and a0 are non-zero. Suppose that there exists an irreducible element p in A such that (i) p | ai , for 0 i n − 1, (ii) an is not divisible by p, (iii) a0 is not divisible by p2 . Then P is irreducible in K[X]. Further, if c(P ) = 1, then P is irreducible in A[X]. Proof. Let us write P = c(P )(an X n + an−1 X n−1 + · · · + a0 ). By (ii), we have p c(P ) = 1, so the polynomial an X n + an−1 X n−1 + · · · + a0 satisﬁes conditions (i),(ii) and (iii). So by 4.4.3, we are reduced to the case where P is primitive. If P is not irreducible in K[X], then by 4.4.3, we have P = QR with Q = bq X q + bq−1 X q−1 + · · · + b0 , R = cr X r + cr−1 X r−1 + · · · + c0 ∈ A[X], where bq and cr are non-zero, q, r ∈ N∗ and q + r = n. Since a0 = b0 c0 , conditions (i) and (iii) imply that p divides only one of b0 , c0 . Let us suppose that p divides b0 and that c0 is not divisible by p. Let s = min{i such that p bi } > 0. This is well-deﬁned since c(P ) = 1, so c(Q) ∈ U (A) by 4.4.2. We have s q n − 1. But as = b0 cs + b1 cs−1 + · · · + bs c0 where we set cj = 0 if j > r. So condition (i) implies that p | bs which is absurd. 4.4.8 Let a be an ideal of A, B = A/a and ϕ : A → B the canonical surjection. Denote also by ϕ the induced surjection A[X] → B[X]. The image of P ∈ A[X] under ϕ is called the reduction of P modulo a. 4.4.9 Theorem. Let K be the ﬁeld of fractions of a factorial ring A, p a prime ideal of A, B = A/p, L = Fract(B), ϕ : A[X] → B[X] the reduction map as above. Let P ∈ A[X] be such that deg P = deg ϕ(P ). If ϕ(P ) is irreducible in B[X] or L[X], then P is irreducible in K[X]. Proof. Suppose that P = QR where Q, R ∈ A[X] with: Q = bq X q + bq−1 X q−1 + · · · + b0 , R = cr X r + cr−1 X r−1 + · · · + c0 where q + r = deg P . We have bq cr ∈ p since deg P = deg ϕ(P ). Further p is prime, so bq , cr ∈ p which implies that deg Q = deg ϕ(Q) and deg R = deg ϕ(R). Now ϕ(P ) = ϕ(Q)ϕ(R) is irreducible in B[X] or L[X]. We have therefore that either ϕ(Q) or ϕ(R) has degree 0. Thus either deg Q or deg R is zero. By 4.4.3, P is irreducible in K[X].

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4.5 Symmetric polynomials 4.5.1 In this section, A will be a ring and K a commutative ﬁeld. We shall ﬁx an integer n > 0 and write A[X] for A[X1 , . . . , Xn ]. For q ∈ N, we denote by Pq the A-submodule of A[X] of homogeneous polynomials of degree q. We shall also denote by Sn the symmetric group of {1, . . . , n}. 4.5.2 Let σ ∈ Sn . For P ∈ A[X], we deﬁne the polynomial P σ as follows: P σ (X1 , . . . , Xn ) = P (Xσ(1) , . . . , Xσ(n) ). Clearly, P σ is homogeneous of degree q if and only if P is homogeneous of degree q. Further, the map θσ : A[X] → A[X] , P → P σ is an automorphism of the algebra A[X]. We have θσ ◦ θτ = θστ for σ, τ ∈ Sn . Deﬁnition. A polynomial P ∈ A[X] is called symmetric if P σ = P for all σ ∈ S. We denote by A[X]sym the subalgebra of A[X] consisting of symmetric polynomials. 4.5.3 For 1 p n, set Ep =

1i1 <···

Xi1 · · · Xip .

This is a symmetric homogeneous polynomial of degree p. We call this polynomial the elementary symmetric polynomial of degree p in X1 , . . . , Xn . We set E0 = 1 and Ep = 0 if p > n. We see easily that, in A[X1 , · · · , Xn , U, V ], we have (5)

n

n

(U + V Xi ) =

i=1

U n−k V k Ek .

k=0

Thus, in particular, in A[X1 , . . . , Xn , T ]: n

(6)

(1 + T Xi ) =

i=1

(7)

n i=1

(T − Xi ) =

n

Ek T k .

k=0

n

(−1)n−k En−k T k .

k=0

4.5.4 Lemma. Let P ∈ A[X]sym be such that P (X1 , . . . , Xn−1 , 0) = 0. Then P is divisible by En . Proof. We proceed by induction on n. For n = 1, the result is clear. So let n 2.

4.5 Symmetric polynomials

49

Since P (X1 , . . . , Xn−1 , 0) = 0, we have P = P1 Xn + · · · + Pk Xnk where P1 , . . . , Pk ∈ A[X1 , . . . , Xn−1 ]. Further since P is symmetric, the Pj ’s are symmetric in X1 , . . . , Xn−1 . We have P (X1 , . . . , Xn−2 , 0, Xn ) = 0, since P τ = P where τ denotes the transposition (n − 1, n). It follows that Pj (X1 , . . . , Xn−2 , 0) = 0 for all j. Thus the Pj ’s are divisible by X1 · · · Xn−1 . Hence En = X1 · · · Xn divides P as required. 4.5.5 Let ν = (ν1 , . . . , νn ) ∈ Nn and λ ∈ A \ {0}. We deﬁne the weight π(λX1ν1 · · · Xnνn ) of the monomial λX1ν1 · · · Xnνn to be: π(λX1ν1 · · · Xnνn ) = ν1 + 2ν2 + · · · + nνn . The weight π(P ) of a non-zero polynomial P is the maximum of the weights associated to the non-zero terms in P . We set π(0) = +∞. 4.5.6 Lemma. Let P be a symmetric polynomial in A[X] of degree d. There exists Q ∈ A[Y1 , . . . , Yn ] such that π(Q) d and P (X1 , . . . , Xn ) = Q(E1 , . . . , En ). Proof. We can suppose that n 2 and d 1. Let e1 , . . . , en−1 be elementary symmetric polynomials in X1 , . . . , Xn−1 . Set B = A[X1 , . . . , Xn−1 ] and for 1 j n − 1, we have: ej = Ej (X1 , . . . , Xn−1 , 0). We shall proceed by induction on n and d. Since P (X1 , . . . , Xn−1 , 0) ∈ B sym , we have by our induction hypothesis that there exists Q1 ∈ B such that: π(Q1 ) d and P (X1 , . . . , Xn−1 , 0) = Q1 (e1 , . . . , en−1 ). Clearly, deg Q1 (E1 , . . . , En−1 ) d and the polynomial P1 (X1 , . . . , Xn ) = P (X1 , . . . , Xn ) − Q1 (E1 , . . . , En−1 ) has degree d, and P1 (X1 , . . . , Xn−1 , 0) = 0. It follows from 4.5.4 that En divides P1 . Let P2 ∈ A[X] be such that P1 = En P2 . Then deg P2 < d and P2 ∈ A[X]sym . By induction hypothesis, there exists Q2 ∈ A[X] such that: π(Q2 ) n − d and P2 (X1 , . . . , Xn ) = Q2 (E1 , . . . , En ) and hence P (X1 , . . . , Xn ) = Q1 (E1 , . . . , En−1 ) + En Q2 (E1 , . . . , En ). Finally, Q1 (Y1 , . . . , Yn−1 ) + Yn Q2 (Y1 , . . . , Yn ) has weight d.

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4 Factorial rings

4.5.7 Theorem. (i) The algebra A[X]sym is generated by the elementary symmetric polynomials in X1 , . . . , Xn . (ii) The map P → P (E1 , . . . , En ) deﬁnes an isomorphism from the polynomial ring A[T1 , . . . , Tn ] to A[X]sym . Proof. Part (i) is a corollary of 4.5.5. The map in part (ii) deﬁnes clearly a morphism, and part (i) implies that it is surjective. Let us prove the injectivity of the map by induction on n. The case n = 1 is trivial. Let n 2 and suppose that the map is not injective. Choose P ∈ A[T1 , . . . , Tn ] \ A of minimal degree such that P (E1 , . . . , En ) = 0. Write P = P0 + P1 Tn + · · · + Pk Tnk with Pj ∈ A[T1 , . . . , Tn−1 ]. Since P (e1 , . . . , en−1 , 0) = P0 (e1 , . . . , en−1 ) = 0 where the ei ’s are the elementary symmetric polynomials in X1 , . . . , Xn−1 , we have P0 = 0 by induction. Thus P = Tn Q for some Q ∈ A[T1 , . . . , Tn ]. But deg(Q) < deg(P ) and Q(E1 , . . . , En ) = 0 which contradicts the minimality of deg(P ).

4.6 Resultant and discriminant 4.6.1 Let A be a ring and P, Q ∈ A[X] \ A of degree m and n respectively. Let us write: P = a0 + a1 X + · · · + am X m , Q = b0 + b1 X + · · · + bn X n . We deﬁne the resultant of P and Q, denoted by res(P, Q), to be the determinant of the following m + n by m + n matrix: a a1 . . . an−1 an . . . am−1 am 0 ... 0 0 0 . . . an−2 an−1 . . . am−2 am−1 am ... 0 a0 0 0 0 . . . an−3 an−2 . . . am−3 am−2 am−1 . . . ..................................................................... 0 0 ... a0 a1 ... ... ... ... ... am b0 b1 . . . bn−1 bn ... 0 0 ... ... 0 0 . . . bn−2 bn−1 . . . 0 0 ... ... 0 b0 0 ... ... ... ... 0 0 . . . bn−3 bn−2 . . . ..................................................................... 0 0 ... 0 0 ... b0 b1 b2 ... bn In the ﬁrst n rows, we have the coeﬃcients of P , while in the last m rows, we have the coeﬃcients of Q. Let us suppose that A is an integral domain. For λ, µ ∈ A \ {0}, we have: res(λP, µQ) = λn µm res(P, Q) , res(Q, P ) = (−1)mn res(P, Q). Observe also that res(P, Q) is a homogeneous polynomial in the ai ’s and the bj ’s of degree n (resp. m) in the ai ’s (resp. in the bj ’s). 4.6.2 Lemma. If A is factorial, then the following conditions are equivalent:

4.6 Resultant and discriminant

51

(i) P and Q has a non-constant common factor. (ii) There exist F, G ∈ A[X] \ {0} such that P G = QF , deg(F ) < deg(P ) , deg(G) < deg(Q). Proof. Let R ∈ A[X] be a non constant polynomial dividing P and Q. Then there exist F, G ∈ A[X] such that P = F R and Q = GR. Clearly, F and G satisﬁes (ii). So we have (i) ⇒ (ii). Conversely, let F, G be as in (ii). Since deg G < deg Q, there exists an irreducible divisor R of Q and n ∈ N∗ such that Rn divides Q, but Rn does not divide G. It follows that R divides P . 4.6.3 Theorem. Let A be factorial, then the following conditions are equivalent: (i) P and Q has a non-constant common factor. (ii) res(P, Q) = 0. Proof. By 4.6.2, (i) is equivalent to the existence of non zero polynomials F = α0 + α1 X + · · · + αm−1 X m−1 and G = β0 + β1 X + · · · + βn−1 X n−1 such that P G = QF . This means that the following homogeneous system in x0 , . . . , xm−1 , y0 , . . . , yn−1 ⎧ a0 y0 = b0 x0 ⎪ ⎪ ⎪ ⎨ a1 y0 + a0 y1 = b1 x0 + b0 x1 (8) .. .. . .. ⎪ . . ⎪ ⎪ ⎩ am yn−1 = bn xm−1 has a non trivial solution in A. This is equivalent to the existence of a non trivial solution in Fract(A). Now the determinant of (8) is ± res(P, Q), so we have proved our result. 4.6.4 Corollary. Let A be a factorial subring of a factorial ring B, and P, Q ∈ A[X]. If P, Q has a non constant common factor in B[X], then P, Q has a non constant common factor in A[X]. 4.6.5 Corollary. Let A be factorial. There exist F, G ∈ A[X] such that: deg(F ) < deg(Q) , deg(G) < deg(P ) , res(P, Q) = P F + QG. Proof. If res(P, Q) = 0, then the assertion follows from 4.6.2 and 4.6.3. So let us suppose that res(P, Q) = 0. Set X = t (1, X, . . . , X m+n−1 ) ∈ Mm+n,1 (A[X]), Y = (P, XP, . . . , X n−1 P, Q, XQ, . . . , X m−1 Q) ∈ Mn+m,1 (A[X]). t

Let M ∈ Mm+n (A) be the matrix in 4.6.1 deﬁning res(P, Q). We have Y = M X. Let M be the transpose of the matrix of cofactors of M . Then,

52

4 Factorial rings

M Y = M M X = res(P, Q)X.

(9)

Let (c0 , . . . , cn−1 , d0 , . . . , dm−1 ) be the ﬁrst row of M and set : F = c0 + c1 X + · · · + cn−1 X n−1 , G = d0 + d1 X + · · · + dm−1 X m−1 . It follows from (9) that P F + QG = res(P, Q).

4.6.6 Theorem. Let α1 , . . . , αm , β1 , . . . , βn be elements of a factorial ring A, a0 , b0 ∈ A \ {0} and : P = a0 (X − α1 ) · · · (X − αn ) , Q = b0 (X − β1 ) · · · (X − βn ). We have : res(P, Q) = an0 bm 0

n m

(αi − βj ) = an0

i=1 j=1

m

Q(αi ) = (−1)mn bm 0

i=1

n

P (βj ).

j=1

Proof. Let T1 , . . . , Tm , U1 , . . . , Un , V0 , W0 , X be indeterminates over A. Consider the following polynomials in B = A[T1 , . . . , Tm , U1 , . . . , Un ] : f = V0 (X − T1 ) · · · (X − Tm ) = V0 X n + · · · + Vm−1 X + Vm , g = W0 (X − U1 ) · · · (X − Un ) = W0 X n + · · · + Wn−1 X + Wn . With the notations of 4.5.3, we have : Vi = (−1)i V0 Ei (T1 , . . . , Tm ) , Wj = (−1)j W0 Ej (U1 , . . . , Un ). By 4.5.6, we can also view f and g as polynomials over A in the indeterminates V0 , V1 , . . . , Vm , W0 , W1 , . . . , Wn , X. Set : S = V0n W0m

n m

(Ti − Uj ).

i=1 j=1

Clearly, we have : S = V0n

m i=1

g(Ti ) = (−1)mn W0m

n

f (Uj ).

j=1

Let R = res(f, g). Since V0 divides the Vi ’s, and W0 divides the Wj ’s, it follows from the deﬁnition of R that R = V0n W0m H, where H ∈ B. On the other hand, by 4.6.3, it is clear that R is divisible by Ti − Uj . Hence, S divides R. Since R and S are both homogeneous of degree m (resp. n) in Vi (resp. Wj ), we deduce that there exists a ∈ A such that R = aS. Finally, since the coeﬃcient of R and S in the monomial V0n W0m is equal to 1, we have R = S. Our theorem follows immediately.

4.6 Resultant and discriminant

53

4.6.7 Deﬁnition. Let P ∈ K[X] be a polynomial of degree n 2 whose dominant coeﬃcient is an . We deﬁne the discriminant of P , denoted by dis(P ), to be the scalar: dis(P ) = (−1)n(n−1)/2 a−1 n res(P, P ).

4.6.8 Proposition. Let P ∈ K[X] be a polynomial of degree n 2 whose dominant coeﬃcient is an . Suppose further that P is split over K and let α1 , . . . , αn be the roots (not necessarily distinct) of P . Then dis(P ) = an2n−2 (αi − αj )2 . i

Proof. This is a direct consequence of 4.6.6 and the deﬁnition of the discriminant. 4.6.9 Proposition. Let P ∈ K[X] be a polynomial of degree n 2. (i) P and P are coprime if and only if dis(P ) = 0. (ii) Suppose that the characteristic of K is zero and that P is split over K. Then all the roots of P are simple if and only if dis(P ) = 0. Proof. These are direct consequences of 4.6.3 and 4.6.8.

References • [11], [52], [56].

5 Field extensions

We study certain properties of ﬁeld extensions in this chapter. In particular, we prove the well-known Going Down Theorem. All ﬁelds considered here are assumed to be commutative. Let K be a ﬁeld.

5.1 Extensions 5.1.1 An extension of K is a pair (L, i) where L is a ﬁeld and i : K → L is a ring homomorphism. It follows that i is injective and when there is no confusion, we shall identify K with its image i(K). Thus L is an extension of K if K is a subﬁeld of L, and we shall write K ⊂ L. We deﬁne the degree of an extension K ⊂ L to be the dimension of L (ﬁnite or not) as a vector space over K. We shall denote it by [L : K]. An extension of ﬁnite degree is called ﬁnite. 5.1.2 Let K ⊂ L be a ﬁeld extension and S a subset of L. Denote by K[S] (resp. K(S)) the subring (resp. subﬁeld) of L generated by K and S. Clearly, K(S) is the quotient ﬁeld of K[S]. If S = {x1 , . . . , xn }, we shall write K[x1 , . . . , xn ] and K(x1 , . . . , xn ) for K[S] and K(S). An extension K ⊂ L is ﬁnitely generated if there is a ﬁnite subset S in L such that L = K(S). If there exists x ∈ L such that L = K(x), then the extension is called cyclic. 5.1.3 The following propositions are straightforward. Proposition. Let K ⊂ L be a ﬁeld extension, S, T subsets of L and F the set of ﬁnite subsets of S. (i) We have K(S ∪ T ) = K(S)(T ) = K(T )(S). (ii) K(S) is the union of K(U ) for U ∈ F. 5.1.4 Proposition. Let K ⊂ L ⊂ M be ﬁeld extensions. (i) If (ai )i∈I is a K-basis for L and (bj )j∈J is a L-basis for M , then (ai bj )i∈I,j∈J is a K-basis for M . (ii) The extension K ⊂ M is ﬁnite if and only if the extensions K ⊂ L and L ⊂ M are ﬁnite. Further, if K ⊂ M is ﬁnite, then

56

5 Field extensions

[M : K] = [M : L][L : K].

5.2 Algebraic and transcendental elements 5.2.1 Let A be a commutative ring containing K as a subring and S = {x1 , . . . , xn } a ﬁnite subset of A. We say that S is algebraically independent over K if P (x1 , . . . , xn ) = 0 for all P ∈ K[X1 , . . . , Xn ] \ {0}. A subset T of A is said to be algebraically independent over K if any ﬁnite subset of T is algebraically independent over K. 5.2.2 Let K ⊂ L be a ﬁeld extension and x ∈ L. We deﬁne a K-algebra homomorphism : ϕx : K[X] → L , P → P (x). If ker ϕx = 0, then x is algebraically independent over K, and we say that x is transcendental over K. In particular, ϕx induces an isomorphism between K[X] and K[x] which extends to an isomorphism between K(X) and K(x). Thus the extension K ⊂ K(x) is inﬁnite. If ker ϕx = 0, then x is algebraically dependent over K, and we say that x is algebraic over K. In this case, ker ϕx is a prime ideal and by 4.1.4, there exists a unique irreducible unitary polynomial Px such that ker ϕx = (Px ). We shall call Px , the minimal polynomial of x over K. The homomorphism ϕx then induces an isomorphism between K[X]/(Px ) and K[x] = K(x). Thus deg Px = [K(x) : K] and {1, x, . . . , xdeg Px −1 } is a K-basis of K(x). 5.2.3 Proposition. Let K ⊂ L be a ﬁeld extension, S a subset of L algebraically independent over K and x ∈ L \ S. The following conditions are equivalent : (i) The element x is algebraic over K(S). (ii) The set S ∪ {x} is algebraically dependent over K. Proof. This is straightforward.

5.3 Algebraic extensions 5.3.1 Deﬁnition. An extension K ⊂ L is algebraic if any element of L is algebraic over K. An extension which is not algebraic will be called transcendental. 5.3.2 Clearly, by 5.1.4 and 5.2.2, we have : Proposition. (i) Let K ⊂ L be a ﬁeld extension. An element x ∈ L is algebraic over K if and only if [K(x) : K] is ﬁnite. (ii) A ﬁnite extension is algebraic. 5.3.3 Proposition. Let K ⊂ L be a ﬁeld extension and S a subset of L. Suppose that all the elements of S are algebraic over K. Then : (i) The extension K ⊂ K(S) is algebraic. (ii) If S is ﬁnite, then the extension K ⊂ K(S) is ﬁnite.

5.3 Algebraic extensions

57

Proof. By 5.1.3 and 5.3.2, it suﬃces to prove (ii). Part (ii) is clear if S is empty. Let us therefore suppose that S is not empty and proceed by induction on the cardinality of S. Let x ∈ S and set T = S \{s}. We have K(S) = K(T )(x) and [K(S) : K] = [K(T )(x) : K(T )][K(T ) : K] by 5.1.4. By induction, [K(T ) : K] is ﬁnite, and x, being algebraic over K, is also algebraic over K(T ). Hence [K(T )(x) : K(T )] is ﬁnite and we are done. 5.3.4 Proposition. Let K ⊂ L ⊂ M be ﬁeld extensions. The following conditions are equivalent : (i) The extension K ⊂ M is algebraic. (ii) The extensions K ⊂ L and L ⊂ M are algebraic. Proof. The implication (i) ⇒ (ii) is trivial. Conversely, suppose that (ii) is satisﬁed. Let x ∈ M . There exist elements a0 , . . . , an−1 ∈ L such that xn + an−1 xn−1 + · · · + a0 = 0. We have by 5.3.3 that K ⊂ K(a0 , . . . , an−1 ) is ﬁnite. On the other hand, x is algebraic over K(a0 , . . . , an−1 ), so K(a0 , . . . , an−1 ) ⊂ K(a0 , . . . , an−1 )(x) is also ﬁnite. Hence K ⊂ K(x) is ﬁnite and part (i) follows. 5.3.5 Let K ⊂ L be a ﬁeld extension. By 5.3.3 and 5.3.4, the set of elements of L algebraic over K is a subﬁeld M containing K. We shall call M , the algebraic closure of K in L. If M = K, then we shall say that K is algebraically closed in L. 5.3.6 Lemma. Let K ⊂ L ⊂ M be ﬁeld extensions and S a subset of M . (i) If K ⊂ L is algebraic, then K(S) ⊂ L(S) is also algebraic. (ii) If S is a ﬁnite algebraically independent subset over L, then [L : K] [L(S) : K(S)]. Proof. (i) Let x ∈ L(S), then there exist x1 , . . . , xn ∈ L such that x ∈ K(S)(x1 , . . . , xn ). Since the xi ’s are algebraic over K, x is algebraic over K(S) by 5.3.3. (ii) Let S = {x1 , . . . , xr } and B a K-basis of L. We shall show that B is linearly independent over K(S). Let b1 , . . . , bn be pairwise distinct elements of B and f1 , . . . , fn ∈ K(S) be such that f1 b1 + · · · + fn bn = 0. By multiplying by a suitable element of K(S), we can assume that fi ∈ K[S] for all i. We can therefore write for each i: n fi = λα,i xα , λα,i bi xα = 0 α∈Nr

α∈Nr

i=1

where the λα,i ’s are in K. By the algebraic independence of S over L, we have that n λα,i bi = 0 i=1

for all α ∈ N . Finally, B is a K-basis, so all the λα,i ’s are 0, and the result follows. r

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5.4 Transcendence basis 5.4.1 Deﬁnition. Let K ⊂ L be a ﬁeld extension. A subset S of L is called a transcendence basis of L over K if : (i) S is algebraically independent over K. (ii) The extension K(S) ⊂ L is algebraic. Example. Let L = K(X1 , . . . , Xn ) be the quotient ﬁeld of the polynomial ring K[X1 , . . . , Xn ]. Then {X1 , . . . , Xn } is a transcendence basis of L over K. 5.4.2 Deﬁnition. Let K ⊂ L be a ﬁeld extension. A subset B of L is called a purely transcendence basis of L over K if B is a transcendence basis of L over K and L = K(B). If L has a purely transcendence basis, then the extension K ⊂ L is called a purely transcendental extension. 5.4.3 Theorem. Let K ⊂ L be a ﬁeld extension and S, T subsets of L verifying the following conditions : (i) L is algebraic over K(T ). (ii) S is algebraically independent over K. Then there exists a transcendence basis B of L over K such that S ⊂ B ⊂ S ∪ T. Proof. By Zorn’s lemma, there is a maximal (by inclusion) element B among the set of subsets in S ∪ T algebraically independent over K which contain S. Clearly, by 5.2.3, this implies that K(B) ⊂ K(S ∪ T ) is algebraic. On the other hand, by 5.3.3, K(T ) ⊂ K(S ∪ T ) ⊂ L are algebraic extensions. Hence K(B) ⊂ L is algebraic by 5.3.3 and the result now follows. 5.4.4 Corollary. Let K ⊂ L be a ﬁeld extension. There exists a transcendence basis of L over K. Proof. This follows from 5.4.3 by taking S = ∅ and T = L.

5.4.5 Proposition. Let K ⊂ L be a ﬁeld extension and S a subset of L. (i) If S is a transcendence basis of L over K, then it is maximal among subsets of L algebraically independent over K. (ii) If L is algebraic over K(S), then any maximal subset of S algebraically independent over K is a transcendence basis of L over K. Proof. These are easy consequences of 5.4.3.

5.4.6 Theorem. Let K ⊂ L be a ﬁeld extension and B, B two transcendence bases of L over K. Then B and B are equipotent. Proof. It suﬃces to prove that card B card B . First, suppose that B is ﬁnite of cardinal n. If n = 0, then K ⊂ L is algebraic and we have B = ∅. So let n 1 and proceed by induction on n.

5.4 Transcendence basis

59

Let x ∈ B , then applying 5.4.3 with S = {x} and T = B, there exists a proper subset C of B such that C ∪ {x} is a transcendence basis of L over K. Let K1 = K(x) and C = B \ {x}. Clearly, C, C are algebraically independent subsets of L over K1 . It follows that they are transcendence bases of L over K1 . By induction, card C = card C and hence card B card B as required. Let us now consider the case where B is inﬁnite. Since any x in B is algebraic over K(B ), there exists a ﬁnite subset Sx of B such that x is algebraic over K(Sx ). Denote by S the union of the Sx for x ∈ B. We have S ⊂ B and since B is inﬁnite, card S card B. But K(B) ⊂ K(S) ⊂ L are algebraic, it follows from 5.4.5 that S = B and hence card B card B as required. 5.4.7 Deﬁnition. Let K ⊂ L be a ﬁeld extension. We deﬁne the transcendence degree of L over K, denoted by tr degK L, to be the cardinality of any transcendence basis of L over K. Remark. Let L = K(x1 , . . . , xn ) be a ﬁnitely generated extension over K. Then by 5.4.3, tr degK L n. 5.4.8 From the preceding discussion, we obtain easily the following proposition : Proposition. Let K ⊂ L be an extension such that tr degK L = n. (i) Let S ⊂ L be such that K(S) ⊂ L is algebraic. Then card S n, and if card S = n, then S is a transcendence basis of L over K. (ii) Any subset of L algebraically independent over K has at most n elements. If it has n elements, then it is a transcendence basis of L over K. (iii) Suppose that L = K(x1 , . . . , xm ). Then m n, and if m = n, then K ⊂ L is a purely transcendental extension, and x1 , . . . , xm is a purely transcendence basis of L over K. 5.4.9 Theorem. Let K ⊂ L ⊂ M be ﬁeld extensions, S a transcendence basis of L over K and T a transcendence basis of M over L. Then S ∩ T = ∅ and S ∪ T is a transcendence basis of M over K. Proof. Since T is algebraically independent over L, it is also algebraically independent over K(S). Hence S∩T = ∅ and S∪T is algebraically independent over K. Now, K(S) ⊂ L is algebraic, so L(T ) is algebraic over K(S ∪ T ). This implies in turn that M is algebraic over K(S ∪ T ) since M is algebraic over L(T ). This ﬁnishes the proof. 5.4.10 Corollary. Let K ⊂ L ⊂ M be ﬁeld extensions. Then : tr degK M = tr degK L + tr degL M.

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5.4.11 Lemma. Let K ⊂ M be a ﬁnitely generated ﬁeld extension and L the algebraic closure of K in M . Then K ⊂ L is a ﬁnite extension. Proof. The extension L ⊂ M , being also ﬁnitely generated, has a ﬁnite transcendence basis B. Since K ⊂ L is algebraic, K(B) ⊂ L(B) is algebraic by 5.3.6. Now, L(B) ⊂ M is algebraic, so 5.3.4 implies that K(B) ⊂ M is also algebraic. Being also ﬁnitely generated, the extension K(B) ⊂ M is ﬁnite by 5.3.3. Finally, by 5.3.6, [L : K] [L(B) : K(B)] [M : K(B)], and so the result follows. 5.4.12 Proposition. Let K ⊂ L ⊂ M be ﬁeld extensions. If K ⊂ M is ﬁnitely generated, then so is K ⊂ L. Proof. By 5.4.3, any transcendence basis B of L over K is contained in a transcendence basis of M over K. So the hypothesis implies that B is ﬁnite. By 5.4.11, [L : K(B)] is ﬁnite since K(B) ⊂ L is algebraic. It follows that if S is a K(B)-basis of L, then S ∪ B is a ﬁnite system of generators for the extension K ⊂ L.

5.5 Norm and trace 5.5.1 Let K ⊂ L be a ﬁnite extension of degree n. For x ∈ L, the map ux : L → L, y → xy is a homomorphism of vector spaces over K. Set : NormL/K (x) = det ux , TrL/K (x) = tr ux . We call NormL/K (x) (resp. TrL/K (x)) the norm (resp. trace) of x in L relative to K. We have clearly that for λ ∈ K and x, y ∈ L : TrL/K (λx + y) = λ TrL/K (x) + TrL/K (y) , TrL/K (λ) = nλ, NormL/K (λxy) = λn NormL/K (x) NormL/K (y) , NormL/K (λ) = λn . These equalities implies that NormL/K (x) = 0 if and only if x = 0, and if x = 0, then [NormL/K (x)]−1 = NormL/K (x−1 ). 5.5.2 Proposition. Let K ⊂ L be a ﬁnite extension of degree n, x ∈ L and X r + ar−1 X r−1 + · · · + a0 the minimal polynomial of x over K. (i) s = n/r is an integer. (ii) We have TrL/K (x) = −sar−1 and NormL/K (x) = [(−1)r a0 ]s . Proof. Part (i) is clear since n/r = [L : K(x)]. Now let (b1 , . . . , bs ) be a K(x)-basis of L, then (xi bj )0i

5.5 Norm and trace

61

so part (ii) follows. 5.5.3 Let K ⊂ L be a ﬁnite extension of degree n. The map ϕ : L × L → K , (x, y) → TrL/K (xy) is a symmetric K-bilinear form on L. If B = (a1 , . . . , an ) is a K-basis of L, the determinant of (ϕ(ai , aj ))i,j is called the discriminant with respect to the basis B of L. The discriminant is non-zero exactly when ϕ is non-degenerated. Indeed, this is the case when the characteristic of K is zero since for any x = 0, ϕ(x, x−1 ) = n. 5.5.4 Proposition. Let A be an integral domain, K its quotient ﬁeld, K ⊂ L a ﬁeld extension, x ∈ L algebraic over K and P (X) = X r + ar−1 X r−1 + · · · + a0 the minimal polynomial of x over K. Then the following conditions are equivalent : (i) x is integral over A. (ii) a0 , . . . , ar−1 are integral over A. In particular, if x is integral over A and the extension K ⊂ L is ﬁnite, then NormL/K (x) and TrL/K (x) are integral over A. Proof. (ii) ⇒ (i) This is clear by 3.1.5 and 3.1.7. (i) ⇒ (ii) Let Ω be an algebraically closed ﬁeld containing L and x1 , . . . , xr the roots of P in Ω. Since x is integral over A, there exists a polynomial Q(X) = X n + bn−1 X n−1 + · · · + b0 ∈ A[X] such that Q(x) = 0. Then P divides Q in K[X] and x1 , . . . , xr are roots of Q. Thus x1 , . . . , xr are integral over A. Since ±ai is the i-th elementary symmetric polynomial in x1 , . . . , xr , the ai ’s are integral over A by 3.1.5. The last statement follows therefore from 5.5.2. 5.5.5 Theorem. Let A be an integral domain, K its quotient ﬁeld, K ⊂ L a ﬁeld extension, B the integral closure of A in L and F the algebraic closure of K in L. (i) The ring B is an integrally closed domain. (ii) We have F = Fract(B). Proof. Clearly, B ⊂ F and so Fract(B) ⊂ F . Conversely let x ∈ F . There exists P (X) = X r + ar−1 X r−1 + · · · + a0 ∈ where bi , ci ∈ A for all i. K[X] such that P (x) = 0. Let us write ai = bi c−1 i Set d = c0 · · · cr−1 ∈ A, we have that ai d ∈ A for all i. It follows that dx is a root of X r + ar−1 dX r−1 + ar−2 d2 X r−2 + · · · + a0 dr ∈ A[X]. This implies that dx ∈ B. Thus F ⊂ Fract(B), which completes the proof of part (ii). By 3.1.7, B is integrally closed in L, hence it is integrally closed in F . So (i) follows from (ii).

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5.5.6 Theorem. Let A be an integrally closed domain, K = Fract(A), K ⊂ L a ﬁnite extension of degree n and B the integral closure of A in L. If the characteristic of K is zero, then there exists a K-basis (v1 , . . . , vn ) of L such that B is a submodule of the free A-module Av1 ⊕ · · · ⊕ Avn . Proof. Let (u1 , . . . , un ) be a K-basis of L. By 5.5.5, we can suppose that u1 , . . . , un ∈ B. Since the characteristic of K is zero, the bilinear form ϕ (5.5.3) is nondegenerated. Thus there exists a basis (v1 , . . . , vn ) of L over K verifying ϕ(ui , vj ) = δij for all 1 i, j n. Clearly the sum Av1 ⊕ · · · ⊕ Avn is direct. Let z = a1 v1 + · · · + an vn ∈ B where a1 , . . . , an ∈ K. Since ai = TrL/K (zui ), we have ai ∈ A by 5.5.4. Hence B ⊂ Av1 ⊕ · · · ⊕ Avn as required. 5.5.7 Proposition. Let notations be as in 5.5.6. If A is Noetherian, then so is B. Proof. By 5.5.6, B is a Noetherian A-module. So it is also a Noetherian Bmodule.

5.6 Theorem of the primitive element 5.6.1 Let E, F be ﬁeld extensions over K, a K-homomorphism from E to F is a K-algebra homomorphism, i.e. a ring homomorphism leaving invariant the elements of K. Clearly, such a homomorphism is injective. A bijective K-homomorphism will be called a K-isomorphism and a bijective K-homomorphism of E onto itself is called a K-automorphism. 5.6.2 Proposition. Let Ω be an algebraically closed ﬁeld containing K, x, y ∈ Ω algebraic over K. The following conditions are equivalent: (i) x and y have the same minimal polynomial over K. (ii) There exists a K-isomorphism σ : K(x) → K(y) such that σ(x) = y. (iii) There exists a K-homomorphism σ : K(x) → Ω such that σ(x) = y. If these conditions are veriﬁed, we shall say that x and y are conjugate over K. Proof. We have (i) ⇒ (ii) by 5.2.2, and (ii) ⇒ (iii) is obvious. Finally if P is the minimal polynomial of x over K. Then P (σ(x)) = 0. Since P is irreducible over K, we have (iii) ⇒ (i). 5.6.3 Lemma. Let Ω be an algebraically closed ﬁeld containing a ﬁeld K of characteristic zero, and P ∈ K[T ] \ {0} be an irreducible polynomial of degree n. Then P has exactly n distinct roots in Ω. Proof. We can suppose that n 1 and that P is unitary. If P has a multiple root α ∈ Ω, then P (α) = 0. Since P is irreducible, it is the minimal polynomial of α over K. It follows that P divides P , and therefore P = 0. This is absurd since P ∈ K and the characteristic of K is zero.

5.6 Theorem of the primitive element

63

5.6.4 Let σ : K → L be a ring homomorphism. This induces a ring homomorphism K[X] → L[X] given by : K[X] P = an X n + · · · + a0 → P σ = σ(an )X n + · · · + σ(a0 ) ∈ L[X]. Lemma. Let K ⊂ L be a ﬁeld extension, x ∈ L algebraic over K, Ω an algebraically closed ﬁeld and σ a homomorphism from K to Ω. There exists a homomorphism τ : K(x) → Ω extending σ. Proof. Let P be the minimal polynomial of x over K. Then P σ ∈ Ω[X] has non zero degree. It has a root, say y, in Ω. Now if Q(x) = 0, then P divides Q. Thus P σ divides Qσ and therefore σ follows that the map K(x) = K[x] → Ω deﬁned by sending Q (y) = 0. It n n i i a x to i i=0 i=0 σ(ai )y is well-deﬁned. Clearly, this is a homomorphism extending σ. 5.6.5 Theorem. Let K ⊂ L be a ﬁnite extension of degree n and σ a homomorphism from K to an algebraically closed ﬁeld Ω. Then : (i) There exist k extensions of σ to L where 1 k n. (ii) If the characteristic of K is zero, then k = n. Proof. We have L = K(x1 , . . . , xp ) = K[x1 , . . . , xp ] where the xi ’s are algebraic over K. We proceed by induction on p. If p = 1, then the theorem follows from 5.6.3 and the proof of 5.6.4. Suppose that p > 1. Let F = K[x1 , . . . , xp−1 ], then by induction, σ admits r extensions σ1 , . . . , σr to F where 1 r [F : K], and r = [F : K] if the characteristic of K is zero. Again by induction, each σi admits si extensions to L where 1 si [L : F ], and si = [L : F ] if the characteristic of K is zero. The theorem follows immediately. 5.6.6 Corollary. Let K ⊂ Ω be a ﬁeld extension where Ω is an algebraically closed ﬁeld, and K ⊂ L a ﬁnite extension of degree n. The cardinality k of the set of K-homomorphism from L to Ω veriﬁes 1 k n. Further, if the characteristic of K is zero, then k = n. 5.6.7 Theorem. (Theorem of the Primitive Element) Let K ⊂ L be a ﬁnite extension of characteristic zero ﬁelds. Then there exists ξ ∈ L such that L = K(ξ). Proof. Clearly, it suﬃces to prove the theorem in the case where L = K(x, y). Let n = [L : K] and Ω an algebraically closed ﬁeld containing K. Let {σ1 , . . . , σn } be the set of K-homomorphisms from L to Ω (5.6.6). For 1 i = j n, set: Aij = {a ∈ K ; σi (x) + aσi (y) = σj (x) + aσj (y)}. Clearly, Aij contains at most one element. Since the characteristic of K is zero, it is inﬁnite. So there exists a ∈ K not contained in any of the Aij ’s. It follows that σ1 , . . . , σn induce distinct K-homomorphisms from K(x + ay) to Ω. By 5.6.6, we have [K(x + ay) : K] n. Hence L = K(x + ay).

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Remark. Let K ⊂ L be a ﬁnite extension. An element ξ ∈ L is called primitive if L = K(ξ). 5.6.8 Corollary. Let K ⊂ L be a ﬁnite extension of characteristic zero ﬁelds. Then the set of ﬁelds E verifying K ⊂ E ⊂ L is ﬁnite. Proof. By 5.6.7, there exists x ∈ L such that L = K(x). Let E be a subﬁeld of L containing K. The minimal polynomial Q of x over E is a unitary divisor in L[X] of the minimal polynomial P of x over K. It suﬃces therefore to show that E is completely determined by Q because the set of unitary divisors of P in L[X] is ﬁnite. Let E0 be the ﬁeld generated by K and the coeﬃcients of Q. Since E0 ⊂ E, Q is also irreducible in E0 [X]. So Q is the minimal polynomial of x over E0 . But E0 (x) = E(x) = L, so deg Q = [L : E0 ] = [L : E]. Thus E = E0 and E is completely determined by Q. 5.6.9 Theorem. Let Ω, Ω be two algebraic closures of K. There exists a K-isomorphism from Ω to Ω . More precisely, if K ⊂ L ⊂ Ω are ﬁeld extensions and θ : L → Ω is a K-homomorphism, then there exists a Kisomorphism from Ω to Ω extending θ. Proof. By Zorn’s Lemma, there exists a ﬁeld E verifying L ⊂ E ⊂ Ω, maximal such that there exists a K-homomorphism σ : E → Ω extending θ. Let x ∈ Ω. Since x is algebraic over K, it is algebraic over E, so by 5.6.5, σ extends to a K-homomorphism ψ : E(x) → Ω . It follows from the maximality of E that x ∈ E. Hence E = Ω. Finally σ(Ω), being isomorphic to Ω, is an algebraically closed subﬁeld of Ω containing K. We deduce that σ(Ω) = Ω .

5.7 Going Down Theorem 5.7.1 Let K ⊂ E ⊂ F be ﬁeld extensions. Denote by HomK (E, F ) the set of K-homomorphisms from E to F , and Gal(E/K) the group of Kautomorphisms of E. 5.7.2 Lemma. Let K ⊂ E ⊂ Ω be ﬁeld extensions. Suppose that Ω is algebraically closed and E is algebraic over K. If σ ∈ HomK (E, Ω) veriﬁes σ(E) ⊂ E, then σ(E) = E. Proof. Let x ∈ E, P ∈ K[X] its minimal polynomial over K, and x1 , . . . , xn the (distinct) roots of P contained in E. Since σ is injective and σ(xi ) is again a root of P contained in E for all i, it follows that σ is a permutation on the set {x1 , . . . , xn }. Hence x ∈ σ(E). 5.7.3 Deﬁnition. A ﬁeld extension K ⊂ E is called normal if it is algebraic and if any irreducible polynomial in K[X] admitting a root in E is split over E.

5.7 Going Down Theorem

65

5.7.4 Proposition. Let K ⊂ E ⊂ Ω be ﬁeld extensions. Suppose that Ω is algebraically closed and E is algebraic over K. The following conditions are equivalent: (i) The extension K ⊂ E is normal. (ii) For all x ∈ E, the conjugates of x in Ω over K are in E. (iii) For all σ ∈ HomK (E, Ω), we have σ(E) ⊂ E. (iv) For all σ ∈ HomK (E, Ω), we have σ(E) = E. Proof. The equivalence (i) ⇔ (ii) is clear. Further we have (iii) ⇔ (iv) by 5.7.2. Now let x ∈ E, since σ(x) and x share the same minimal polynomial, we have (ii) ⇒ (iii). Conversely, if y ∈ Ω is conjugate to x, then by 5.6.2, there exists a K-homomorphism τ : K(x) → K(y) verifying τ (x) = y. We have seen in 5.6.9 that τ can be extended to a K-homomorphism σ : E → Ω. So y = τ (x) = σ(x) ∈ E and (iii) ⇒ (ii). 5.7.5 Proposition. Let K ⊂ E ⊂ Ω be ﬁeld extensions. Suppose that Ω is algebraically closed and K ⊂ E is a ﬁnite extension. Let F be the subﬁeld of Ω generated by the σ(E) where σ runs over the set HomK (E, Ω). Then K ⊂ F is a ﬁnite normal extension. Further, if K ⊂ L is a normal extension verifying E ⊂ L ⊂ Ω, then F ⊂ L. Proof. By 5.6.6, the set HomK (E, Ω) is ﬁnite. Since K ⊂ σ(E) is a ﬁnite extension for any σ, K ⊂ F is a ﬁnitely generated algebraic extension. It follows from 5.4.11 that K ⊂ F is a ﬁnite extension. Since τ (σ(E)) ⊂ F for all τ ∈ HomK (F, Ω), σ ∈ HomK (E, Ω), normality follows from 5.7.4. The last assertion is a direct consequence of 5.7.4 since any element σ of HomK (E, Ω) can be extended to an element τ ∈ HomK (L, Ω) (5.6.9), so σ(E) = τ (E) ⊂ τ (L) ⊂ L. 5.7.6 Let G be a subgroup of the group of automorphisms of a ﬁeld E. We shall denote the subﬁeld of G-ﬁxed points by E G . Lemma. Let K be a ﬁeld of characteristic zero, K ⊂ E a ﬁnite normal extension and G = Gal(E/K). Then K = E G . Proof. Clearly, K ⊂ E G . Now suppose that there exists x ∈ E G \K. Then the minimal polynomial of x over K is at least of degree two. Since the characteristic of K is zero and K ⊂ E is normal, there exists y ∈ E, x = y, sharing the same minimal polynomial. By 5.6.2, 5.6.9 and 5.7.4, there exists σ ∈ G such that σ(x) = y which is absurd since x ∈ E G . 5.7.7 Proposition. Let A be an integrally closed domain of characteristic zero, K = Fract A, K ⊂ E a ﬁnite normal extension and C the integral closure of A in E. If q, q ∈ Spec C veriﬁes A ∩ q = A ∩ q , then there exists σ ∈ Gal(E/K) such that σ(q) = q .

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Proof. Let Gal(E/K) = {σ1 , . . . , σn } and qi = σi (q). It is clear that σi (C) = C, so qi ∈ Spec C. Suppose that q = qi for all 1 i n. Then by 3.3.2 and 2.2.5, q ⊂ q1 ∪ · · · ∪ qn . Let x ∈ q \ q1 ∪ · · · ∪ qn and y = σ1 (x) · · · σn (x). We have y ∈ E G = K (5.7.5), and since y ∈ C and A is an integrally closed domain, y ∈ A. It follows that y ∈ A ∩ q and by the choice of x, we have y ∈ A ∩ q which contradicts the hypothesis. 5.7.8 Theorem. (Going Down Theorem) Let A be an integrally closed domain of characteristic zero, K = Fract A, and B an integral extension of A whose ﬁeld of fractions E is a ﬁnite extension of K. Let p, p ∈ Spec A be such that p ⊂ p. If q ∈ Spec B veriﬁes p = A ∩ q, then there exists q ∈ Spec B such that p = A ∩ q and q ⊂ q. In particular, ht p = ht q. Proof. Let F be a ﬁnite normal extension of K containing E (5.7.5) and denote by C the integral closure of A (so of B also) in F . By 3.3.3, there exist r, r ∈ Spec C verifying r ⊂ r and A∩r = p, A∩r = p . Let s ∈ Spec C be such that B ∩ s = q. Then A ∩ s = p = A ∩ r, so by 5.7.7, there exists σ ∈ Gal(F/K) such that σ(r) = s. Since A ∩ σ(r ) = p , we have that q = B ∩ σ(r ) veriﬁes q ⊂ q and p = A ∩ q . 5.7.9 Let A be an integrally closed domain of characteristic zero, K = Fract A, K ⊂ L a ﬁnite extension of degree n, B a subring of L integral over A such that L = Fract B. By 5.6.7, there exists ξ = α/β ∈ L, where α, β ∈ B, such that L = K(ξ). Suppose that β ∈ A and let γ0 , . . . , γm−1 ∈ A be such that γ0 = 0 and β m + γm−1 β m−1 +· · ·+γ0 = 0. It follows that γ0 ξ = α(β m−1 +γm−1 β m−2 +· · ·+γ1 ). Since K(γ0 ξ) = K(ξ), there exists b ∈ B such that L = K(b). Let P (T ) = T n + an−1 T n−1 + · · · + a0 be the minimal polynomial of b over K. By 5.5.4, we have a0 , . . . , an−1 ∈ A. Let Ω be an algebraic closure of L and b1 , . . . , bn the conjugates of b in Ω over K. Recall from 5.6.3 that the bi ’s are the n distinct roots of P in Ω. Set ∆= (bi − bj ) = 0 i

whose square is the discriminant d of P . Recall from 4.6.7 that d ∈ A. Lemma. Let us suppose that d is invertible in A. Then: (i) B is a free A-module and {1, b, . . . , bn−1 } is a basis of B over A. (ii) B is the integral closure of A in L. Proof. The set {1, b, . . . , bn−1 } is a basis of L over K. Let x = a0 + a1 b + · · · + an−1 bn−1 ∈ L, a0 , . . . , an−1 ∈ K, be integral over A. Then the conjugates of x in Ω over K are: for 1 i n . xi = a0 + a1 bi + · · · + an−1 bn−1 i

5.8 Fields and derivations

67

This forms a linear system in the ai ’s whose determinant is ∆. It follows that ai = ∆i /∆ where ∆i ∈ L is integral over A since the xi ’s are integral over A. On the other hand, ai = ∆i ∆/d, and d is invertible in A, so the ai ’s are integral over A. Since A is an integrally closed domain, ai ∈ A for all i. Hence x ∈ B and the assertions follow. 5.7.10 Let us conserve the notations and hypotheses of 5.7.9. We suppose further that d is invertible in A. Let m be a maximal ideal of A, k = A/m and for x ∈ A, denote by x its class in k. Let S(T ) = T n + an−1 T n−1 + · · · + a0 ∈ k[T ]. The discriminant of d of S is non-zero because d is invertible in A. Let us suppose that S splits over k and let γ1 , . . . , γn be the distinct roots of S. For 1 i n, set Qi (T ) =

T − γj ∈ k[T ]. j=i γi − γj

Clearly, Qi (γi ) = 1 and Qi (γj ) = 0 if i = j. If R ∈ k[T ] veriﬁes deg R n − 1, then (1)

R(T ) = R(γ1 )Q1 (T ) + · · · + R(γn )Qn (T ).

Let θ denote the class of b in B/mB = B ⊗A k and ei = Qi (θ). Then since {1, θ, . . . , θn−1 } is a basis of B/mB over k by 5.7.9, (1) implies that {e1 , . . . , en } is also a basis of B/mB over k. If i = j, then Qi Qj is divisible by S, so ei ej = 0. On the other hand, Q2i = SM + R where R ∈ k[T ] and deg R n − 1. By (1), we obtain that R = Qi and therefore, ei ei = ei . Thus the ring B/mB is the direct sum of ﬁelds kei = B/ni where ni is the inverse image of j=i kej . It follows that n1 , . . . , nn are the n maximal ideals of B lying above m.

5.8 Fields and derivations 5.8.1 We shall use the notations of 2.8 and 2.9. Further let K be a ﬁeld of characteristic zero. Lemma. Let K ⊂ L be a ﬁnite extension. Then any K-derivation of L in a L-vector space N is zero. In particular, ΩK (L) = {0}. Proof. By 5.6.7, there exists x ∈ L algebraic over K such that L = K(x). Let P (T ) be the minimal polynomial of x over K. Let δ ∈ DerK (L, N ). If Q ∈ K[T ], then δ(Q(x)) = Q (x)δ(x). In particular, 0 = δ(P (x)) = P (x)δ(x). By 5.6.3, P (x) = 0. So δ(x) = 0 and the lemma follows since any element of L can be written as Q(x) for some Q ∈ K[T ]. 5.8.2 Lemma. Let K ⊂ E ⊂ F be ﬁeld extensions such that [F : E] is ﬁnite, and N a F -vector space. Any element of DerK (E, N ) extends uniquely to an element of DerK (F, N ).

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5 Field extensions

Proof. By 5.6.7, there exists x ∈ F such that F = E(x). Let P be the minimal polynomial of x over E and δ ∈ DerK (E, N ). For Q(T ) = n an T n ∈ E[T ], we deﬁne Qδ (x) = n xn δ(an ) ∈ N . Suppose that ∆ ∈ DerK (F, N ) extends δ, then for Q ∈ E[T ], we have: (2)

∆(Q(x)) = Qδ (x) + Q (x)∆(x). In particular, since P (x) = 0, u = ∆(x) is the unique solution in N of 0 = P δ (x) + P (x)u.

(3)

So uniqueness follows. Now, let u = −P (x)−1 P δ (x) ∈ N be the unique solution of (3). We claim that the map ∆ : F → N given by (2) and ∆(x) = u is a well-deﬁned map extending δ. This is just an easy consequence of the following equality: (QR)δ (x) + (QR) (x)u = R(x)(Qδ (x) + Q (x)u) + Q(x)(Rδ (x) + R (x)u) for all Q, R ∈ E[T ]. 5.8.3 Let us conserve the notations of 5.8.2. By 2.8.5 and 5.8.2, we obtain an exact sequence of E-vector spaces: 0 → DerE (F, N ) → DerK (F, N ) → DerK (E, N ) → 0. By 2.9.5, this induces an exact sequence of F -vector spaces: α

β

0 → ΩK (E) ⊗E F → ΩK (F ) → ΩE (F ) → 0 where α(dE/K (x) ⊗ 1) = dF/K (x) and β(dF/K (y)) = dF/E (y) for all x ∈ E, y ∈ F . It follows therefore from 5.8.1 that the F -vector spaces ΩK (E) ⊗E F and ΩK (F ) are isomorphic. Now if K ⊂ E is purely transcendental and tr degK (E) = n, then 2.9.7 implies that the dimension of the E-vector space ΩK (E) is n. So, we have obtained: Theorem. Let K ⊂ F be ﬁnitely generated. Then the dimension of the F -vector space ΩK (F ) is equal to the transcendence degree of F over K. 5.8.4 Proposition. Let u : A → B be a homomorphism of ﬁnitely generated commutative K-algebras which are integral domains. Denote by Ω(u) : ΩK (A) → ΩK (B) the homomorphism of A-modules deﬁned in 2.9.5. Suppose that ΩK (A) (resp. ΩK (B)) is a free A-module (resp. free B-module). Then the following conditions are equivalent: (i) u is injective. (ii) Ω(u) is injective.

5.8 Fields and derivations

69

Proof. (ii) ⇒ (i) In the notations of 2.9.5, Ω(u) is the unique A-module homomorphism such that the following diagram commutes: A ⏐ ⏐ dA/K

u

−−−−→

B ⏐ ⏐d B/K

Ω(u)

ΩK (A) −−−−→ ΩK (B) Let a ∈ ker u, then dA/K (a) = 0. For any x ∈ A, we have ax ∈ ker u, and therefore 0 = dA/K (ax) = adA/K (x). But the elements dA/K (x), x ∈ A, generate the A-module ΩK (A). So a = 0 because ΩK (A) is a free A-module. (i) ⇒ (ii) Let E (resp. F ) be the ﬁeld of fractions of A (resp. B). Since u is injective, we may assume that A ⊂ B and E ⊂ F . By 5.8.3, we have dimF ΩK (F ) = tr degK F , dimF ΩE (F ) = tr degE F , dimF ΩK (E) ⊗E F = dimE ΩK (E) = tr degK E. It follows from a dimension count that the following sequence (2.9.5), induced by the canonical injection v : E → F , is exact: 0 → ΩK (E) ⊗E F → ΩK (F ) → ΩE (F ) → 0. We deduce therefore from 2.9.5 that Ω(v) : ΩK (E) → ΩK (F ) is injective. Let ιA : A → E and ιB : B → F denote the canonical injections. By 2.9.5, 2.9.6, the uniqueness of Ω(ιA ) and the hypothesis that ΩK (A) is a free A-module, we obtain that Ω(ιA ) is the composition : ΩK (A) → ΩK (A) ⊗A E → ΩK (E). In particular, Ω(ιA ) is injective. Now we have the following commutative diagram: u

A −−−−→ ⏐ ⏐ ιA

B ⏐ ⏐ιB

v

E −−−−→ F which induces, from the uniqueness of Ω(ιB ◦ u) = Ω(v ◦ ιA ), the following commutative diagram: Ω(u)

ΩK (A) −−−−→ ΩK (B) ⏐ ⏐ ⏐Ω(ι ) ⏐ Ω(ιA ) B Ω(v)

ΩK (E) −−−−→ ΩK (F ) Since Ω(v) and Ω(ιA ) are injective, we deduce that Ω(u) is also injective.

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5 Field extensions

5.9 Conductor 5.9.1 In this section, A will be a subring of a ring B. The annihilator of the A-module B/A is an ideal of A called the conductor of B in A. It is easy to verify that this is the largest ideal of B contained in A. Proposition. Let S be a multiplicative subset of A, f the conductor of B in A and f1 the conductor of S −1 B in S −1 A. Then S −1 f ⊂ f1 . Furthermore, we have S −1 f = f1 if B is a ﬁnite A-algebra. Proof. Since fB ⊂ A, we have (S −1 f)(S −1 B) ⊂ S −1 A. So S −1 f ⊂ f1 . Now let us suppose that {x1 , . . . , xn } is a system of generators of the Amodule B. Let a/s ∈ f1 where a ∈ A, s ∈ S. Then for each i, (a/s)(xi /1) ∈ S −1 A, i.e. there exists ti ∈ S such that ti axi ∈ A. Set t = t1 · · · tn , then clearly taxi ∈ A. Hence ta ∈ f and a/s ∈ S −1 f. 5.9.2 Corollary. Let f be the conductor of B in A, p ∈ Spec A be such that f ⊂ p, and S = A \ p. If B is a ﬁnite A-algebra, then Ap = S −1 B. Proof. By 5.9.1, S −1 f is the conductor of S −1 B in Ap . Since f ⊂ p, there exists s ∈ S ∩ f. It follows that 1 = s/s ∈ S −1 f. Hence S −1 f = Ap and Ap = S −1 B by the deﬁnition of the conductor. 5.9.3 Proposition. Let A be an integral domain, B be the integral closure of A, f the conductor of B in A and S a multiplicative subset of A not containing 0. (i) A suﬃcient condition for S −1 A to be an integrally closed domain is that S ∩f = ∅. If B is a ﬁnite A-algebra, then this is also a necessary condition. (ii) Suppose that B is a ﬁnite A-algebra. Let p ∈ Spec A, then Ap is not an integrally closed domain if and only if f ⊂ p. Proof. By 3.1.9, S −1 B is the integral closure of S −1 A. Since S −1 f = S −1 A if and only if S ∩ f = ∅, part (i) follows from 5.9.1 and 5.9.2. Finally, part (ii) is just a special case of (i). 5.9.4 Proposition. Let B be an integral domain, A integrally closed in B and x ∈ B such that B is integral over A[x]. Suppose that there exists a unitary polynomial P ∈ A[T ] such that P (x) is contained in the conductor of B in A[x]. Then B = A[x]. Proof. Let b ∈ B. Then bP (x) = Q(x) ∈ A[x] for some Q ∈ A[T ]. Since P is unitary, we have Q = F P + R with F, R ∈ A[T ] and deg R < deg P . Let y = b − F (x), thus yP (x) = R(x). If y = 0, then clearly b = F (x) ∈ A[x]. If y = 0, then P (x) = R(x)/y in Fract(B). Since deg R < deg P , we deduce that x is integral over A[1/y]. On the other hand, y is integral over A[x], and hence over A[x, 1/y]. It follows from 3.1.7 that A[x, y, 1/y] is integral over A[1/y]. So y is integral over A[1/y], and there exist a0 , . . . , am−1 ∈ A[1/y] such that

5.9 Conductor

71

y m + am−1 y m−1 + · · · + a0 = 0. Clearly, this implies that y is integral over A. Since A is integrally closed in B, we have y ∈ A, and therefore b = y + F (x) ∈ A[x]. 5.9.5 Lemma. Let B be an integral domain and x ∈ B an element transcendental over Fract A such that B is a ﬁnite A[x]-algebra. For any maximal ideal n of B, there exists q ∈ Spec B such that q ⊂ n , q = n , n ∩ A = q ∩ A. Proof. Let A (resp. B ) be the integral closure of A in Fract A (resp. B in Fract B). It is clear that B is integral over A [x] and x is transcendental over Fract A = Fract A. Let n be a prime ideal of B lying above n. By 3.3.2, n ∈ Spm B . Set m = n ∩ A and r = n ∩ A [x]. Note that r ∈ Spm A [x] by 3.3.2, and so m A [x] ⊂ r , m A [x] = r since A [x]/m A [x] (A /m )[T ] is not a ﬁeld. Since A [x] is an integrally closed domain by 3.2.8 and m A [x] ∈ Spec A [x], we obtain by 5.7.8 that there exists q ∈ Spec B verifying q ⊂ n , q = n and q ∩ A [x] = m A [x]. Hence q ∩ A = m . Now q = q ∩ B is not a maximal ideal of B by 3.3.2, so we have q = n and q ∩ A = n ∩ A. 5.9.6 Lemma. Let B be an integral domain, A integrally closed in B and x ∈ B such that B is a ﬁnite A[x]-algebra. Suppose that there exists a maximal ideal n of B which is minimal in the set of q ∈ Spec B such that q∩A = n∩A. (i) For all q ∈ Spec B verifying q ⊂ n, the class of x modulo q is algebraic over Fract(A/(q ∩ A)). (ii) The conductor f of B in A[x] is not contained in n. Proof. (i) Denote by x1 the class of x modulo q. The ring B/q is a ﬁnite (A/(q ∩ A))[x1 ]-algebra and n/q ∈ Spm(B/q) is minimal in the set of prime ideals in B/q lying above (n ∩ A)/(q ∩ A). So x1 is algebraic over A/(q ∩ A) by 5.9.5. (ii) Suppose that f ⊂ n. Let q ∈ Spec B be minimal verifying f ⊂ q ⊂ n, and denote p = q ∩ A. Recall from 2.6.6 that since Ap is a local ring with maximal ideal pAp , its residue ﬁeld K is isomorphic to Fract(A/p). Further the ideal qp in Bp is prime and veriﬁes qp ∩ Ap = pAp (2.3.7). By part (i), the class of x modulo q is algebraic over K. Since Bp is integral over Ap by 3.1.8, all elements of Bq /qp are algebraic over K. It follows therefore that Bq /qp is a ﬁeld, and so qp ∈ Spm Bq . Further Ap is integrally closed in Bp by 3.1.9, fp is the conductor of Bp in Ap by 5.9.1, and qp is minimal among the prime ideals of Bp containing fp . So by replacing A, B by Ap , Bp , we can assume that q = n. Under this assumption, n/f is a minimal prime ideal of B/f and (n/f)n/f is the minimal prime ideal of (B/f)n/f . So (n/f)n/f is the nilradical of (B/f)n/f .

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5 Field extensions

Since the class of x modulo n is algebraic over K, there exist y ∈ A \ p ⊂ B \n and P ∈ K[T ] unitary such that yP (x) ∈ n. By the preceding paragraph, the image in (B/f)n/f of the class of yP (x) in B/f is nilpotent. So there exist z ∈ B \ n and k ∈ N∗ such that zy k P (x)k ∈ f, hence P (x)k belongs to the conductor of A[x][zy k B] in A[x]. Now by applying 5.9.4 to A, A[x] and A[x][zy k B], we obtain that zy k B ⊂ A[x]. But this implies that zy k ∈ f ⊂ n which is absurd since y, z ∈ n and n is prime. 5.9.7 Lemma. Let B be an integral domain and A a local ring with maximal ideal m. Let us suppose that: (i) The ring A is integrally closed in B and there exists n ∈ Spm B lying above m. (ii) There exists x ∈ B such that B = A[x] and the class of x modulo n is algebraic over K = A/m. Then B = A. Proof. By (ii), there exists P ∈ A[T ] \ A unitary such that P (x) ∈ mA[x]. Set y = 1 + P (x) and θ, the image of y in C = A[y]/mA[y]. If θ is not invertible in C, then θ is contained in a maximal ideal of C. But A[x]/mA[x] is integral over C, so the image y of y in A[x]/mA[x] is also contained in a maximal ideal by 3.3.2 and 3.3.3. This is absurd since y = 1. Thus we have shown that θ is invertible in C. Next, let X m + αm−1 X m−1 + · · · + α0 be the minimal polynomial of θ over K. Since θ is invertible in C, α0 = 0. For 0 i m − 1, let ai ∈ A be a representative of αi , then there exist n ∈ N and u0 , . . . , un ∈ m such that: y m + am−1 y m−1 + · · · + a0 = un y n + un−1 y n−1 + · · · + u0 . Thus there exists z ∈ A[y] such that yz = a0 − u0 ∈ m. Since A is a local ring, this implies that yz is invertible in A. Hence y is invertible in A[y] ⊂ B, i.e. there exist b1 , . . . , br ∈ A such that br y r + · · · + b1 y = 1. It follows that 1/y is integral over A and (i) implies that 1/y ∈ A. Now if 1/y ∈ m, then 1 ∈ ym ⊂ n which is absurd. So 1/y ∈ A \ m is invertible in A. Thus y ∈ A, which implies that P (x) ∈ A. So the hypotheses on P implies that x is integral over A. Hence x ∈ A and B = A. 5.9.8 Proposition. Let B be an integral domain, A integrally closed in B and x ∈ B such that B is a ﬁnite A[x]-algebra. Let m ∈ Spm(A) and n be a maximal ideal of B which is minimal among the prime ideals of B lying above m. Then Am = Bm = Bn . Proof. By 3.1.9, Am is integrally closed in Bm and Bm is a ﬁnite Am [x]-algebra. Further, nm is a maximal ideal of Bm which is minimal among the prime ideals of Bm lying above mAm (2.3.7 and 3.3.2). We are therefore reduced to the case where A is a local ring with maximal ideal m and B = Bm .

5.9 Conductor

73

By 5.9.6, n does not contain the conductor of B in A[x]. It follows from 5.9.2 that B = Br = A[x]r where r = n ∩ A[x]. By 3.3.2, r ∈ Spm A[x], so n = rA[x]r . Our hypothesis on n implies via 3.3.3 that r is minimal among the prime ideals of A[x] lying above m. Now, 5.9.6 and 5.9.7 imply that A = A[x], and so r = m. Hence Am = B = Bm = Bn as required.

References • [9],[19], [22], [26], [52], [56].

6 Finitely generated algebras

We prove three fundamental results in this chapter : Noether’s Normalization Theorem, Krull’s Principal Ideal Theorem, and Hilbert’s Nullstellensatz. We introduce also the Zariski topology on the set of maximal ideals of a ring. All ﬁelds and rings considered are commutative. Let us denote by K a ﬁeld. If A is a ring, then we shall denote by Specm (A) the set of minimal prime ideals of A.

6.1 Dimension 6.1.1 Let A be a ﬁnitely generated K-algebra and {x1 , . . . , xn } a system of generators of A. Since A is isomorphic to a quotient of the polynomial ring K[T1 , . . . , Tn ], A is Noetherian by 2.7.9 and 2.7.3. Suppose that A is an integral domain and let F be its quotient ﬁeld. Then by 5.4.3, we can extract from {x1 , . . . , xn } a transcendence basis of F over K, and hence tr degK F n. In particular, there exists a transcendence basis of F over K contained in A. 6.1.2 Proposition. Let A be a ﬁnitely generated K-algebra and p, q be distinct prime ideals of A such that p ⊂ q. Then tr degK Fract(A/q) < tr degK Fract(A/p). Proof. Replace A by A/p, we are reduced to the case where A is an integral domain, p = {0} and q = {0}. Let π : A → A/q be the canonical surjection. Let y1 , . . . , yr ∈ A/q be a transcendence basis of Fract(A/q) over K. For each i, ﬁx xi ∈ A such that π(xi ) = yi . Clearly, the xi ’s are algebraically independent over K. It follows from 6.1.1 that: r = tr degK Fract(A/q) tr degK Fract A. If equality holds, then π induces an isomorphism from K[x1 , . . . , xr ] onto K[y1 , . . . , yr ]. Further for any a ∈ q, a is algebraic over K(x1 , . . . , xr ). So there exist P0 , . . . , Pn ∈ K[X1 , . . . , Xr ] such that P0 (x1 , . . . , xr ) = 0 and

76

6 Finitely generated algebras

P0 (x1 , . . . , xr ) + P1 (x1 , . . . , xr )a + · · · + Pn (x1 , . . . , xr )an = 0. Applying π to this equality, we obtain that P0 (y1 , . . . , yr ) = 0 which contra dicts the algebraic independence of the yi ’s. 6.1.3 Corollary. Let A be a ﬁnitely generated K-algebra. Then: dim A

max p∈Spec(A)

tr degK Fract(A/p) =

Proof. This is immediate from 6.1.2.

max p∈Specm (A)

tr degK Fract(A/p).

6.1.4 Corollary. We have dim K[X1 , . . . , Xn ] = tr degK K(X1 , . . . , Xn ) = n. Proof. Let A = K[X1 , . . . , Xn ]. Since X1 , . . . , Xn form a transcendence basis of K(X1 , . . . , Xn ), we have dim A tr degK K(X1 , . . . , Xn ) = n. On the other hand, let pi be the ideal of A generated by X1 , . . . , Xi . Clearly A/pi is isomorphic to K[Xi+1 , . . . , Xn ]. So we have a chain of prime ideals {0} ⊂ p1 ⊂ · · · ⊂ pn in A. So dim A n. 6.1.5 Corollary. Let pi , 1 i n be as in the proof of 6.1.4. Then the height of pi is equal to i. Proof. The chain {0} ⊂ p1 ⊂ · · · ⊂ pi implies that ht pi i. Now the result follows from 6.1.2 and 6.1.4.

6.2 Noether’s Normalization Theorem 6.2.1 Theorem. Let n ∈ N∗ , A = K[X1 , . . . , Xn ] and a a proper ideal of A. Then there exist x1 , . . . , xn ∈ A algebraically independent over K verifying the following conditions: (i) The ring A is integral over B = K[x1 , . . . , xn ]. (ii) If a is non trivial, then there exists s ∈ {1, . . . , n} such that the ideal B ∩ a of B is generated by x1 , . . . , xs . Proof. If a = {0}, then it suﬃces to take xi = Xi for all i. So let us suppose that a = {0}. We shall proceed by induction on n. For n = 1, there exists x ∈ A nonconstant such that a = Ax. Let x1 = x, then (i) and (ii) are satisﬁed. Let us suppose that n > 1 and x1 ∈ a \ {0}. Then x1 ∈ K and we can write aλ X1i1 · · · Xnin x1 = P (X1 , . . . , Xn ) = λ=(i1 ,...,in )∈Nn

∗

where aλ ∈ K. Let k ∈ N be such that k > i1 + · · · + in for all λ = (i1 , . . . , in ) such that aλ = 0. For λ = (i1 , . . . , in ), write l(λ) = i1 + i2 k + · · · + in k n−1 . Then by the choice of k and the uniqueness of the development of an integer

6.2 Noether’s Normalization Theorem

77

in base k, we have that for λ = (i1 , . . . , in ), λ = (j1 , . . . , jn ) such that aλ = 0, aλ = 0, l(λ) = l(λ ) implies that λ = λ . So there exists a unique µ ∈ Nn verifying aµ = 0 such that l(µ) is maximal. i−1 for 2 i n. Then Set ti = Xi − X1k n−1

0 = P (X1 , t2 + X1k , . . . , tn + X1k ) − x1 l(µ) = aµ X1 + Qj (x1 , t2 , . . . , tn )X1j j

where the Qj ’s are polynomials. It follows that X1 is integral over the algebra i−1 for 2 i n, Xi is integral over C = K[x1 , t2 , . . . , tn ]. Since Xi = ti + X1k C by 3.1.7. Thus A is integral over C. Further, by 5.3.3, the extension Fract(C) ⊂ Fract(A) is algebraic and tr degK Fract(A) = n. Therefore by 5.4.3, x1 , t2 , . . . , tn are algebraically independent over K. In particular, C is isomorphic to A, so it is factorial and an integrally closed domain. Let y ∈ A be such that yx1 ∈ C, then y ∈ Fract(C) and since C is an integrally closed domain and y is integral over C, y ∈ C. Hence C∩Ax1 = Cx1 . Suppose ﬁrst that a ∩ K[t2 , . . . , tn ] = {0}. Let S = K[t2 , . . . , tn ] \ {0}. The ideal generated by x1 in S −1 C = K(t2 , . . . , tn )[x1 ] is maximal. So S −1 a = S −1 Cx1 . It follows that if y ∈ a ∩ C, then there exists s ∈ S such that sy ∈ Cx1 . Since x1 , t2 , . . . , tn are algebraically independent, Cx1 ∈ Spec(C). Hence y ∈ Cx1 , and a ∩ C = Cx1 . So we obtain the result by setting x2 = t2 , . . . , xn = tn . Suppose now that a ∩ K[t2 , . . . , tn ] = {0}. Let D = K[t2 , . . . , tn ]. By induction, there exist x2 , . . . , xn ∈ D algebraically independent over K, such that D is integral over E = K[x2 , . . . , xn ] and a ∩ E = Ex2 + · · · Exk for some k with 2 k n. By 3.1.7, A is integral over B = K[x1 , x2 , . . . , xn ], and so x1 , x2 , . . . , xn are algebraically independent over K. Since x1 ∈ a, it follows that a ∩ B = Bx1 + a ∩ E = Bx1 + Bx2 + · · · + Bxk . 6.2.2 Theorem. (Noether’s Normalization Theorem) Let A be a ﬁnitely generated K-algebra and a1 ⊂ a2 ⊂ · · · ⊂ ap a chain of ideals of A such that ap = A. There exist x1 , . . . , xn ∈ A verifying the following conditions: (i) x1 , . . . , xn are algebraically independent over K, A is integral over B = K[x1 , . . . , xn ] and dim A = n. (ii) There exists a chain of integers 0 h(1) · · · h(p) such that for 1 k p, B ∩ ak is {0} if h(k) = 0 and it is the ideal of B generated by x1 , . . . , xh(k) if h(k) > 0. Proof. First of all, observe that if x1 , . . . , xn ∈ A are algebraically independent over K and A is integral over K[x1 , . . . , xn ], then by 6.1.4 and 3.3.4, we have dim A = n. If A is integral over K, then the conditions are void, i.e. n = 0 and B = K. So let us suppose that A is not integral over K.

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Let us ﬁrst establish the theorem for A = K[X1 , . . . , Xn ], the ring of polynomials over K. If p = 1, then the theorem is just a consequence of 6.2.1. So let us suppose that p 2. By induction on p, there exist t1 , . . . , tn ∈ A verifying conditions (i) and (ii) with respect to the chain a1 ⊂ · · · ⊂ ap−1 . Set r = h(p − 1), B = K[t1 , . . . , tn ] and C = K[tr+1 , . . . , tn ]. Applying 6.2.1 to C and the ideal C ∩ ap of C, we obtain elements xr+1 , . . . , xn ∈ C algebraically independent over K such that C is integral over D = K[xr+1 , . . . , xn ], and there is an integer h(p) such that D ∩ ap is generated by xr+1 , . . . , xh(p) . Let xi = ti for 1 i r and E = K[x1 , . . . , xn ]. Then by construction, Fract(B) is algebraic over Fract(E). Thus x1 , . . . , xn are algebraically independent over K. Further, by 3.1.7, A is integral over E. So we have (i). For 1 i p, E ∩ ai contains obviously the ideal Ex1 + · · · + Exh(i) of E. If i p − 1, then E ∩ ai = E ∩ (Bx1 + · · · + Bxh(i) ). But by 5.7.8 and 6.1.5, E ∩ ai and Ex1 + · · · + Exh(i) are prime ideals of E of height h(i). So they must be equal. Let z ∈ E ∩ ap . We can write aλ xi11 · · · xirr z= λ=(i1 ,...,ir )

where aλ ∈ D. Clearly aλ xi11 · · · xirr ∈ ap−1 ⊂ ap if λ = λ0 = (0, . . . , 0). It follows that aλ0 ∈ D ∩ ap . This implies that E ∩ ap ⊂ Ex1 + · · · + Exh(p) , and since clearly x1 , . . . , xh(p) ∈ E ∩ ap , we have condition (ii). Now let us suppose that A is a ﬁnitely generated K-algebra. Then there exists a surjective ring homomorphism π : A = K[X1 , . . . , Xm ] → A for some m ∈ N. Let a0 = ker π and ai = π −1 (ai ) for 1 i p. Since A is a polynomial ring, there exist x1 , . . . , xm ∈ A verifying conditions (i) and (ii) with respect to the chain a0 ⊂ · · · ⊂ ap . By 3.1.8, A is integral over the K-algebra generated by xj = π(xj ), j > h(0). Thus s > h(0) since A is not integral over K. To prove the theorem, it suﬃces therefore to show that xh(0)+1 , . . . , xm are algebraically independent over K. Let t = h(0) + 1. If P (xt , . . . , xm ) = 0 for some non zero polynomial over K, then P (xt , . . . , xm ) ∈ a0 ∩ K[x1 , . . . , xm ] = Bx1 + · · · + Bxh(0) where B = K[x1 , . . . , xm ]. But this contradicts the fact that x1 , . . . , xm are algebraically independent. Hence xt , . . . , xm are algebraically independent over K and the theorem follows. 6.2.3 Corollary. Let A be a ﬁnitely generated K-algebra. We have dim A =

max p∈Spec(A)

tr degK Fract(A/p) =

max p∈Specm (A)

tr degK Fract(A/p).

In particular, if A is an integral domain, then dim A = tr degK Fract A.

6.2 Noether’s Normalization Theorem

79

Proof. By 6.2.2, A is integral over a polynomial ring B contained in A, and dim A = dim B. By 3.1.8, A/p is integral over B/(B ∩ p) for any p ∈ Spec(A). So tr degK Fract(A/p) = tr degK Fract(B/(B ∩ p)) tr degK Fract B. Thus the result follows from 6.1.3 and 6.1.4. Finally, if A is an integral domain, then {0} is the unique minimal prime ideal of A. 6.2.4 Corollary. Let A be a ﬁnitely generated K-algebra such that dim A = 0. If A is an integral domain, then A is a ﬁeld and the extension K ⊂ A is ﬁnite. Proof. If A is an integral domain and dim A = 0, then {0} is a maximal ideal. So A is a ﬁeld. By 6.2.3, the extension K ⊂ A is algebraic. So it is ﬁnite since A is a ﬁnitely generated K-algebra. 6.2.5 Corollary. Let A be a ﬁnitely generated K-algebra. Then we have dim A[X] = 1 + dim A. Proof. By 6.2.2, A is integral over a polynomial ring B contained in A. So A[X] is integral over B[X]. Hence the result follows from 6.2.2. 6.2.6 Proposition. Let A be a ﬁnitely generated K-algebra which is an integral domain. (i) For any p ∈ Spec(A), we have: ht p + dim(A/p) = dim A. (ii) Let p, q ∈ Spec(A) be such that p ⊂ q. Then: ht q = ht p + ht(q/p). Proof. Part (i) is clear for p = {0}. So let us suppose that p = {0}. By 6.2.2, there exist x1 , . . . , xn ∈ A such that B = K[x1 , . . . , xn ] is isomorphic to a polynomial ring, A is integral over B and B ∩ p = Bx1 + · · · + Bxr where 1 r n. By 5.7.8 and 6.1.5, ht p = ht(B ∩ p) = r. On the other hand, B/(B ∩ p) is isomorphic to K[xr+1 , . . . , xn ], so its dimension is n − r. Since A is integral over B, A/p is integral over B/(B ∩ p) (3.1.8). This proves part (i). Finally, since dim A = ht p + dim(A/p) = ht q + dim(A/q), dim(A/p) = ht(q/p) + dim(A/q), part (ii) follows from part (i). 6.2.7 Deﬁnition. A chain p0 ⊂ · · · ⊂ pn of prime ideals of length n of a ring A is called maximal if it satisﬁes the following conditions: (i) p0 ∈ Specm (A) and pn ∈ Spm(A).

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(ii) For any integer i ∈ {0, 1, . . . , n − 1}, the set of q ∈ Spec A verifying pi ⊂ q ⊂ pi+1 is {pi , pi+1 }. 6.2.8 Proposition. Let A be a ﬁnitely generated K-algebra which is an integral domain. Then all maximal chains of prime ideals of A has length dim A. Proof. Let {0} = p0 ⊂ · · · ⊂ pn be a maximal chain of prime ideals of A. For 1 i n, the height of pi /pi−1 ∈ Spec(A/pi−1 ) is 1. Hence by 6.2.6, dim(A/pi−1 ) − dim(A/pi ) = 1. On the other hand, dim(A/pn ) = 0 since, A/pn is a ﬁeld. It follows that dim A = dim(A/p0 ) = n. 6.2.9 Proposition. Let B be a ﬁnitely generated K-algebra which is an integral domain, A a ﬁnitely generated K-subalgebra of B. There exist a nonzero element t ∈ A and a subalgebra C = A[y1 , . . . , yn ] of B verifying the following conditions: (i) The elements y1 , . . . , yn are algebraically independent over K. (ii) The subalgebra B[1/t] of Fract(B) is integral over C[1/t]. Proof. Let L = Fract(A) and z1 , . . . , zr ∈ B be such that B = A[z1 , . . . , zr ]. By applying 6.2.2 to the L-algebra L[z1 , . . . , zr ], we can ﬁnd x1 , . . . , xn ∈ B algebraically independent over L such that L[z1 , . . . , zr ] is integral over L[x1 , . . . , xn ]. Let m i −1 Pij (x1 , . . . , xn )zij = 0 zimi + j=0

be an integral equation for zi over L[x1 , . . . , xn ], where Pij ∈ L[T1 , . . . , Tn ]. Let k = max1jmi −1 {deg Pij } and t ∈ A non-zero such that tPij ∈ A[T1 , . . . , Tn ] for all j. Then there exist Qij ∈ A[T1 , . . . , Tn ] such that tk+1 Pij (x1 , . . . , xn ) = Qij (tx1 , . . . , txn ) and so tk+1 zimi +

m i −1 j=0

Qij (tx1 , . . . , txn )zij = 0.

Set yi = txi . Clearly, the yi ’s are algebraically independent over L, hence over K. The above equality implies that zi is integral over A[y1 , . . . , yn ][1/t]. Hence B[1/t] is integral over C[1/t]. 6.2.10 Proposition. Let A be a ﬁnitely generated K-algebra which is an integral domain, F = Fract(A), F ⊂ L a ﬁnite ﬁeld extension and B the integral closure of A in L. Suppose that the characteristic of K is zero. Then B is a ﬁnitely generated A-module (so B is a ﬁnitely generated K-algebra which is an integral domain). Proof. By 6.2.2, A contains a polynomial ring C over which A is integral. Clearly, B is also the integral closure of C in L. Since Fract(C) ⊂ F is a ﬁnite extension, we can therefore assume that A is a polynomial ring. The result then follows from 4.2.9 and 5.5.6.

6.3 Krull’s Principal Ideal Theorem

81

6.3 Krull’s Principal Ideal Theorem 6.3.1 Lemma. Let C ⊂ A be integral domains, F = Fract(C) and K = Fract(A). Suppose that C is an integrally closed domain,√A is integral over C and F ⊂ K is a ﬁnite extension.√Let x ∈√A be such that Ax ∈ Spec A. Then y = NormK/F (x) ∈ C and C ∩ Ax = Cy. Proof. Let P = X m + cm−1 X m−1 + · · · + c0 be the minimal polynomial of x over F . By 5.5.4, the ci ’s are integral over C. So P ∈ C[X] √ and since ∈ Ax. It follows therefore from 5.5.2 that y ∈ C ∩ Ax. Hence P (x) = 0, c 0 √ √ Cy ⊂ C ∩ Ax. √ n ∈ N∗ and a ∈ A Conversely, if u ∈ C ∩ Ax, √ then there exist n n[K:F ] = NormK/F (un ) = such that u = ax. Hence u ∈ Cy because u NormK/F (a) NormK/F (x) ∈ Cy. 6.3.2 The following theorem is also called Krull’s Hauptidealsatz. Theorem. (Krull’s Principal Ideal Theorem) Let A be a ﬁnitely generated K-algebra, x ∈ A non invertible and p ∈ Spec A minimal such that x ∈ p. Then ht p 1 and equality holds if x is not a zero divisor. Proof. Note that such a p exists by applying 2.2.7 to A/Ax. Let us suppose that ht p = r 1. Let q0 ⊂ · · · ⊂ qr = p be a maximal chain of prime ideals. It suﬃces therefore to show that ht p/q0 = 1 in A/q0 . We are therefore reduced to the case where A is an integral domain and x = 0. By 2.7.8, the set of prime ideals containing x has a ﬁnite number of minimal elements, say p, p2 , . . . , ps . Let u ∈ (p2 ∩ · · · ∩ ps ) \ p, S = {un ; u ∈ N} and B = S −1 A. Since B is a ﬁnitely generated K-algebra and Fract A = Fract B, we have dim A = dim B and ht p = ht(Bp) by 6.2.3 and 2.3.9. Further, Bp is now the unique prime ideal of B minimal among those containing x. So we can suppose that p is the √ unique minimal element of the set of prime ideals containing x. By 2.7.7, Ax = p. By 6.2.2, A contains a polynomial ring C over which A is integral. Let F = Fract(C), K = Fract(A) and y = NormK/F (x). Then 6.3.1 implies that √ Cy. Now C is √ factorial, so there exists z ∈ C irreducible such that C ∩ p = √ Cy = Cz. So ht p = ht Cy = 1 by 5.7.8. 6.3.3 Corollary. Let A be a ﬁnitely generated K-algebra and a a proper ideal of A generated by x1 , . . . , xn . If p is minimal among the prime ideals of A containing a, then ht p n. Proof. Let us proceed by induction on n. We have treated the case n = 1 in 6.3.2. So let us suppose that n 2. By the same argument as in the proof of 6.3.2, we are reduced to the case where A is an integral domain. Now the set E of q ∈ Spec A such that Ax1 + · · · + Axn−1 ⊂ q ⊂ p is not empty. Let q be minimal in E. By our induction hypothesis, ht q n − 1. Further, p/q is minimal among the prime ideals of A/q containing the class of xn modulo q. So the result follows from 6.2.6 and 6.3.2.

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6.3.4 Corollary. Any maximal ideal of A = K[X1 , . . . , Xn ], the polynomial ring in n indeterminates, can be generated by n elements. Proof. Let m ∈ Spm(A). Then by 6.1.4 et 6.2.6, ht m = n. Hence 6.3.3 implies that any system of generators of m has at least n elements. If n = 1, the result is obvious since K[X1 ] is a principal ideal domain. Let us suppose that n 2 and proceed by induction on n. If m ∩ K[X1 ] = {0}, then let S = K[X1 ] \ {0} and S −1 m is a maximal ideal of K(X1 )[X2 , . . . , Xn ]. Therefore ht S −1 m = n − 1. This is not possible since ht S −1 m = ht m = n. Thus m ∩ K[X1 ] is a (non-zero) maximal ideal K[X1 ]x where x ∈ K[X1 ] is irreducible. Clearly, x is also irreducible in A and consequently, Ax is a prime ideal of A. Let y2 , . . . , yn be the images of X2 , . . . , Xn in A/Ax. We have A/Ax = L[y2 , . . . , yn ] where L = K[X1 ]/(K[X1 ]x) and by 6.2.6, dim A/Ax = n − 1. It follows from 6.2.3 that the yi ’s are algebraically independent over L. By the induction hypothesis, the maximal ideal m/Ax of A/Ax is generated by n − 1 elements. Consequently, if we choose representatives of these n − 1 elements in A, then these n − 1 representatives together with x form a system of generators for m.

6.4 Maximal ideals 6.4.1 The following result is a direct consequence of 6.2.4. Proposition. Let A be a ﬁnitely generated K-algebra and m ∈ Spm(A). The extension K ⊂ A/m is ﬁnite. Moreover, if K is algebraically closed, then the natural map K → A → A/m is an isomorphism of K onto A/m. 6.4.2 Theorem. (Hilbert’s Nullstellensatz) Let a be a proper ideal of a √ ﬁnitely generated K-algebra A. Then a is the intersection of maximal ideals of A containing a. Proof. By replacing A by A/a, it suﬃces to prove that the nilradical n of A is the intersection r of maximal ideals of A. Clearly, n ⊂ r (see also 2.5.5). Let a ∈ r. If a ∈ n, then S = {an ; n ∈ N} does not contain 0. The Kalgebra B = S −1 A is again ﬁnitely generated (just add 1/a to a system of generators of A). Let i : A → B be the canonical map (see 2.3.3), m ∈ Spm(B) and i−1 (m) = p ∈ Spec(A) (see 2.3.8). By 6.4.1, K ⊂ B/m is ﬁnite. So we deduce, via the canonical injections K → A/p → B/m, that dim A/p = 0. So p ∈ Spm(A) by 6.2.4. But a ∈ m, so a ∈ p which is absurd since a ∈ r. 6.4.3 Proposition. Let A = K[X1 , . . . , Xn ] be the polynomial ring in n indeterminates, and a an ideal of A. Suppose that K is an algebraically closed ﬁeld. The following conditions are equivalent: n (i) There exist a1 , . . . , an ∈ K such that a = i=1 A(Xi − ai ). (ii) We have a ∈ Spm(A).

6.4 Maximal ideals

83

Proof. (i) ⇒ (ii) This is clear since A/a is isomorphic to K. (ii) ⇒ (i) Let a ∈ Spm A. Then 6.2.6 implies that K ⊂ A/a is ﬁnite and since K is algebraically closed, we have K = A/a. Let ai be the image of Xi in A/a = K. Then m = A(X1 − a1 ) + · · · + A(Xn − an ) ⊂ a. But m ∈ Spm(A), so m = a. 6.4.4 Corollary. Let A, B be ﬁnitely generated K-algebras. Suppose that K is algebraically closed. (i) Let m ∈ Spm(A). Then A = K.1 ⊕ m. (ii) Let ρ : A → B be a K-algebra homomorphism. If m ∈ Spm(B), then ρ−1 (m) ∈ Spm(A). Proof. Part (i) is clear since 1 ∈ m and A/m = K by 6.4.1. (ii) Since ρ(1) = 1 and B/m = K by part (i), ρ−1 (m) = A and the induced map A/ρ−1 (m) → B/m is an isomorphism. So ρ−1 (m) ∈ Spm(A). 6.4.5 For any K-algebras A, B, let us denote by Homalg (A, B) the set of K-algebra homomorphisms from A to B. Given a ﬁnitely generated K-algebra A, it is clear from 6.4.4 that there is a canonical bijection: Homalg (A, K) → Spm(A) , χ → ker χ whose inverse is given by the projection A = K.1 ⊕ m → K = A/m for m ∈ Spm A. Now if B is another ﬁnitely generated K-algebra and ϕ ∈ Homalg (A, B), then the map ϕ∗ : Homalg (B, K) → Homalg (A, K) , χ → χ ◦ ϕ induces a map Spm(ϕ) : Spm(B) → Spm(A) via the canonical bijections above. More precisely, if n ∈ Spm(B), then Spm(ϕ)(n) = ϕ−1 (n). 6.4.6 We can view A as an algebra of functions on Spm(A) as follows: let f ∈ A and m ∈ Spm(A), then we deﬁne f (m) to be the class of f in A/m = K. Note that if ϕ ∈ Homalg (A, B), then for n ∈ Spm(B) and f ∈ A, we have: (f ◦ Spm(ϕ))(n) = f (ϕ−1 (n)) = ϕ(f )(n) since the homomorphism A/ϕ−1 (n) → B/n induced by ϕ is just the identity map from K to itself.

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6.5 Zariski topology 6.5.1 Let A be a ﬁnitely generated K-algebra. For any subset E of A, we shall denote by V(E) the set of maximal ideals of A which contain E. When E = {f1 , . . . , fn }, we shall also write V(f1 , . . . , fn ) for V(E). 6.5.2 Remark. Suppose that K is algebraically closed. Then considering A as functions as in 6.4.6, V(E) can be identiﬁed as the set of common zeros of the elements in E. Note also that if a is an ideal of A generated by E, then V(a) = V(E). 6.5.3 Proposition. Let A be a ﬁnitely generated K-algebra, a, b and (ai )i∈I ideals of A.√We have: (i) V(a) = V( a) and if a ⊂ b, then V(b) ⊂ V(a). (ii) V(a) = ∅ if and only if a = A. √ (iii) V(a) if and only if a ⊂ 0. = Spm(A) V(ai ) and V(ab) = V(a ∩ b) = V(a) ∪ V(b). (iv) V( ai ) = i∈I

i∈I

Proof. The ﬁrst point of (i) and part (iii) are consequences of 6.4.2, while the second point of (i), part (ii) and the ﬁrst point of (iv) are obvious. If m ∈ Spm(A), then m ⊃ ab if and only if m contains either a or b. So V(ab) = V(a) ∪ V(b). The other equalities follow from part (i) and 2.5.3. 6.5.4 In view of 6.5.3, we deﬁne a topology, called Zariski topology on Spm(A), by letting closed subsets to be the V(a)’s where a is an ideal of A. Observe that by part (i) of 6.5.3, we can restrict ourselves to radical ideals. By convention, when we consider Spm(A) as a topological space, we assume that the underlying topology is the Zariski topology. Let a be an ideal of A and B = A/a. Then the map given by m → m/a is a bijection between the set V(a) and Spm B. Clearly, via this identiﬁcation, the Zariski topology of Spm(B) is the one induced by the Zariski topology of Spm(A). Suppose that K is algebraically closed and ϕ : A → B is the canonical surjection. Then the map Spm(ϕ) of 6.4.5 is the canonical injection V(a) → Spm A. 6.5.5 Let A be a ﬁnitely generated K-algebra. For M ⊂ Spm(A), we denote by I(M ) the set of elements a ∈ A such that a ∈ m for all m ∈ M . Clearly, m. I(M ) = m∈M

Proposition. Let a, b be√ideals of A and M ⊂ Spm A. (i) We have I(V(a)) = a. √ √ (ii)√We have √ V(a) ⊂ V(b) (resp. V(a) = V(b)) if and only if b ⊂ a (resp. b = a). (iii) The Zariski closure of M in Spm(A) is V(I(M )).

6.5 Zariski topology

85

Proof. Part (i) is proved in 6.4.2 and part (ii) is a direct consequence of 6.5.3 (i) and the fact that the map M → I(M ) reverses inclusions. To prove part (iii), let a be a radical ideal of A such that M ⊂ V(a). Then a = I(V(a)) ⊂ I(M ) by (i). It follows that V(I(M )) ⊂ V(a). On the other hand, it is clear that M ⊂ V(I(M )). Hence we have proved (iii). 6.5.6 Proposition. Let A be a ﬁnitely generated K-algebra. (i) Spm(A) is a Noetherian topological space whose points are closed. (ii) Let a be a radical ideal of A. Then a ∈ Spec(A) if and only if V(a) is irreducible. Proof. (i) Since V(m) = {m} for all m ∈ Spm(A), points are closed. Let (Fp )p be a decreasing sequence of closed subsets of Spm(A) where Fp = V(ap ), ap a radical ideal of A. Then by 6.5.5, (ap )p is an increasing sequence of radical ideals of A. Since A is Noetherian (6.1.1), this sequence is stationary. It follows that (Fp )p is also stationary. (ii) Suppose that V(a) is irreducible and let f, g ∈ A be such that f g ∈ a. We have V(a) ⊂ V(f g) = V(f ) ∪ V(g). √ Hence V(a) is contained in either V(f ) or V(g). Now 6.5.5 implies that f ∈ a = a or g ∈ a. So a ∈ Spec(A). Conversely, suppose that a ∈ Spec(A). Let b, c be radical ideals of A such √ that V(a) ⊂ V(b) ∪ V(c). So bc ⊂ a = a by 6.5.3 and 6.5.5. Since a is prime, a contains either b or c. Hence V(a) is contained in either V(b) or V(c). 6.5.7 Corollary. Let a be an ideal of A and p1 , . . . , pn be the minimal elements of the set of prime ideals of A containing a. Then the set of irreducible components of V(a) is the ﬁnite set {V(pi ) ; 1 i n}. In particular, the irreducible components of Spm(A) are the closed sets associated to minimal prime ideals of A. So Spm(A) is irreducible if and only √ if 0 ∈ Spec(A). Furthermore, if A is reduced, then Spm(A) is irreducible if and only if A is an integral domain. √ Proof. By 2.7.7 and 2.7.8, we have a = p1 ∩ · · · ∩ pn , and so 6.5.3 implies that V(a) = V(p1 ) ∪ · · · ∪ V(pn ). The result follows from 6.5.6 and 6.5.5. 6.5.8 Proposition. Suppose that K is algebraically closed. Let A, B be ﬁnitely generated K-algebras and ϕ ∈ Homalg (B, A). (i) The map Spm(ϕ) : Spm(A) → Spm(B) deﬁned in 6.4.5 is continuous. (ii) If b is an ideal of B, then Spm(ϕ)−1 (V(b)) = V(Aϕ(b)). (iii) For any ideal a of A, the Zariski closure of Spm(ϕ)(V(a)) is the set V(ϕ−1 (a)). Proof. It suﬃces to prove (ii) and (iii). Let m ∈ Spm(A), then m ∈ V(Aϕ(b)) ⇔ Aϕ(b) ⊂ m ⇔ ϕ(b) ⊂ m ⇔ b ⊂ ϕ−1 (m) = Spm(ϕ)(m) ⇔ Spm(ϕ)(m) ∈ V(b).

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6 Finitely generated algebras

This proves (ii). Let a be an ideal of A. By 6.5.5, the Zariski closure of Spm(ϕ)(V(a)) is V(I(ϕ(V(a)))), the set of maximal ideals of B containing the intersection of the ϕ−1 (m)’s where m ∈ V(a). Now √ ϕ−1 (m) = ϕ−1 m = ϕ−1 ( a) = ϕ−1 (a). m∈V(a)

m∈V(a)

Hence (iii) follows from 6.5.3. 6.5.9 Corollary. Let us keep the notations and hypotheses of 6.5.8. Suppose further that A and B are reduced. Then ϕ is injective if and only if Spm(ϕ) is dominant. Proof. By 6.5.8 (iii), Spm(ϕ) Spm(A) = V(ϕ−1 ({0})). Note that {0} is rad √ ical in the reduced ring A. So the equality ϕ−1 ( a) = ϕ−1 (a) implies that ϕ−1 ({0}) is radical in B. Since B is also reduced, it follows from 6.5.3 and 6.5.5 that V(ϕ−1 ({0})) = Spm(B) if and only if ϕ−1 ({0}) = {0}.

References • [9], [26], [52], [56].

7 Gradings and ﬁltrations

In this chapter, we study the notions of gradings and ﬁltrations on rings and modules. These notions are useful in many areas of mathematics. We shall use them in particular in Chapters 13, 30 and 31.

7.1 Graded rings and graded modules 7.1.1 Deﬁnition. Let G be an abelian group. A grading on G is a family (Gn )n∈Z of subgroups of G such that G is the direct sum of the Gn ’s. A group equipped with a grading is called a graded group. 7.1.2 Let G be a graded group. An element x ∈ G is called homogeneous of degree n if x ∈ Gn . If x is non-zero and homogeneous, then x is contained the degree of x. in a unique Gn and we shall denote by n = deg(x) Any y ∈ G can be written uniquely in the form n yn where yn ∈ Gn . We call yn the homogeneous component of degree n of y. When there is no confusion, we shall denote the grading of a graded group G by (Gn )n∈Z . 7.1.3 Deﬁnition. Let A be a ring. A grading (An )n∈Z on A is said to be compatible with the ring structure if An Am ⊂ An+m for all m, n ∈ Z. A ring equipped with a compatible grading is called a graded ring. 7.1.4 Deﬁnition. Let A be a graded ring and M a left A-module. A grading (Mn )n∈Z on M is said to be compatible with the A-module structure if Am Mn ⊂ Mm+n for all m, n ∈ Z. A left A-module equipped with a compatible grading is called a graded left A-module. 7.1.5 Proposition. Let A be a graded ring. Then A0 is a subring of A. Proof. Clearly, A0 A0 ⊂ A0 . Now let (en )n∈Z be the set of homogeneous components of 1. For any x ∈ Ap , we have x = x.1 = xen = 1.x = en x. n∈Z

n∈Z

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7 Gradings and ﬁltrations

By comparing the homogeneous components of degree p, we obtain that x = xe0 = e0 x. Since any element of A is the sum of its homogeneous components, the equality x = xe0 = e0 x is valid for all x ∈ A. Hence 1 = e0 ∈ A0 . 7.1.6 It follows from 7.1.5 that if M is a graded left A-module, then each Mn is a left A0 -module. We deﬁne graded right A-modules similarly. All modules considered in the rest of this chapter are left modules. 7.1.7 Deﬁnition. (i) Let A and B be graded rings. A ring homomorphism f : A → B is called graded if f (An ) ⊂ Bn for all n ∈ Z. (ii) Let A be a graded ring and M, N graded A-modules. A homomorphism u : M → N of A-modules is called graded of degree p ∈ Z if u(Mn ) ⊂ Nn+p for all n ∈ Z. A homomorphism u : M → N of A-modules is called graded if it is graded of degree p for some p ∈ Z. Note that if u is non-zero, then p is uniquely determined by u.

7.2 Graded submodules 7.2.1 Proposition. Let M be a graded A-module and N a submodule of M . The following conditions are equivalent: (i) N is a graded submodule, i.e. N = n∈Z (N ∩ Mn ). (ii) The homogeneous components of any element of N are in N . (iii) N is generated by homogeneous elements. Proof. This is straightforward. 7.2.2 Corollary. Let N be a graded submodule of a graded A-module M . (i) If (xi )i∈I is a system of generators of N , then the homogeneous components of the xi ’s form a system of homogeneous generators of N . (ii) If N is ﬁnitely generated, then it has a ﬁnite system of homogeneous generators. 7.2.3 Let N be a graded submodule of a graded A-module M , then one veriﬁes easily that M/N = n∈Z (N + Mn )/N is a graded (quotient) module. 7.2.4 Let A be a graded ring. A graded submodule of A is called a graded (left) ideal. If a is a bilateral ideal of A, then the preceding paragraph says that the quotient A/a is again a graded ring. 7.2.5 In the rest of this section, A is a commutative graded ring. We call a graded A-module (Mn )n∈Z positively graded if Mn = {0} for all n < 0. 7.2.6 Proposition. Let a be a graded ideal of A. Then a is prime if and only if given any homogeneous elements x, y ∈ A such that xy ∈ a, we have x ∈ a or y ∈ a.

7.2 Graded submodules

89

Proof. The condition is obviously necessary. Let us prove that it is suﬃcient. Let a, b ∈ A be such that ab ∈ a. Write a = ai1 + · · · + air , b = bj1 + · · · + bjs where i1 < · · · < ir , j1 < · · · < js and the aik ’s and bjl ’s are homogeneous of degree ik and jl . Now we can assume that none of the bjl ’s belong to a. Since ab ∈ a and a is graded, the homogeneous component of degree ir js of ab, which is air bjs , belongs to a. It follows that air ∈ a, and so (a − air )b ∈ a. The result follows by induction. √ 7.2.7 Proposition. Let a be a graded ideal of A. Then a is graded. √ Proof. Let x ∈ a. Write x = xi1 + · · · + xir where i1 < · · · < ir and xik is homogeneous of degree ik . There exists n ∈ N∗ such that xn ∈ a. Since xnir is and a is graded, xnir ∈ a. So the homogeneous component of degree nir of xn √ √ √ xir ∈ a and x − xir ∈ a. By induction, xik ∈ a for k = 1, . . . , r. 7.2.8 Theorem. Let A be positively graded. The following are equivalent: (i) A is Noetherian. (ii) A0 is Noetherian and A is a ﬁnitely generated A0 -algebra. Proof. (i) ⇒ (ii) Set A+ = n∈N∗ An . This is a graded ideal of A and A0 A/A+ . So A0 is Noetherian. Further, A+ is ﬁnitely generated, so by 7.2.2, it is generated by a ﬁnite number of homogeneous elements xi1 , . . . , xir where xik ∈ Aik ⊂ A+ . Denote by B the A0 -subalgebra of A generated by the xik ’s. Clearly A0 ⊂ B. Now assume that Ak ⊂ B for k = 0, . . . , n. Let y ∈ An+1 . Then we can ﬁnd a1 ∈ An+1−i1 , . . ., ar ∈ An+1−ir such that y = a1 xi1 + · · · + ar xir . By our assumption, ak ∈ B for all k = 1, . . . , r. Hence An+1 ⊂ B, and by induction, we have A = B. (ii) ⇒ (i) Since A is a ﬁnitely generated A0 -algebra, A is isomorphic to a quotient of a polynomial algebra A0 [X1 , . . . , Xs ]. The result follows by 2.7.3 and 2.7.9. 7.2.9 Let K be a commutative ﬁeld and A = K[T1 , . . . , Tn ] together with its natural grading. The following result is a graded version of 2.2.5. Proposition. Let p1 , . . . , pr be graded prime ideals of A. If a is a graded ideal of A verifying a ⊂ pi for 1 i r, then there exists a homogeneous element a ∈ a such that a ∈ p1 ∪ · · · ∪ pr . Proof. If a = A, then a = 1 works. So let us assume that a = A. We can further assume that pi ⊂ pj for i = j. We shall proceed by induction on r. The case r = 1 is obvious. So let r 2. For 1 i r, by induction, there exists a homogeneous element ai ∈ a such that ai ∈ j=i pj = Ei . If there exists i such that a ∩ pi ⊂ Ei , then ai ∈ p1 ∪ · · · ∪ pr . If a ∩ pi ⊂ Ei for 1 i r, then we may take ai ∈ a ∩ pi , and by replacing the ai ’s by some suitable powers, we can assume that deg(a1 ) =

90

7 Gradings and ﬁltrations

deg(a2 · · · ar ). Then a = a1 + a2 · · · ar is a homogeneous element of a and clearly a ∈ p1 ∪ · · · ∪ pr .

7.3 Applications 7.3.1 In this section, A is a commutative ring. 7.3.2 Theorem. (Artin-Rees Theorem) Let A be Noetherian, a an ideal of A, M a ﬁnitely generated A-module et N a submodule of M . There exists q ∈ N∗ such that, for all n q, we have N ∩ (an M ) = an−q (N ∩ aq (M )). Proof. Set B=

n0

an , S =

an M , T =

n0

(N ∩ an M ).

n0

One veriﬁes easily that B is a (positively) graded ring, and S is a (positively) graded B-module. Now B is generated by a as an A-algebra. Since A is Noetherian, B is a ﬁnitely generated A-algebra. It follows from 7.2.8 that B is Noetherian. Hence S is ﬁnitely generated, and T is a ﬁnitely generated graded submodule of S. For q ∈ N, set Fq =

q i=0

(N ∩ ai M ) , Gq = Fq ⊕

ai (N ∩ aq M ) .

i1

Since ai (N ∩ aq M ) ⊂ N ∩ aq+i M , (Gq )q is an increasing sequence of submodules of T . Further, the union of the Gq ’s is T . So there exists q such that T = Gq . Therefore, for n q, we have N ∩ an M = an−q (N ∩ aq M ). 7.3.3 Lemma. Let a be an ideal of A and N a ﬁnitely generated A-module. Then N = aN if and only if there exists a ∈ a such that (1 − a)N = 0. Proof. Let x1 , . . . , xs be a system of generators of N . If N = aN , then for 1 i s, there exist aij ∈ a such that xi = ai1 x1 + · · · + ais xs . The determinant of [δij − aij ]1i,js is of the form 1 − a for some a ∈ a. But for 1 i s, (1 − a)xi = 0. Hence (1 − a)N = {0}. The converse is obvious. 7.3.4 Theorem. Let A be Noetherian, a an ideal of A and M a ﬁnitely generated A-module. Then x ∈ n0 an M if and only if there exists a ∈ a such that ax = x. In particular, we have n0 an M = {0} if and only if for any a ∈ a \ {0}, x = 0 is the unique solution in M of ax = x. Proof. This is a consequence of 7.3.2 and 7.3.3 with N = n0 an M .

7.4 Filtrations

91

7.3.5 Corollary. Let A be a Noetherian integral domain and a an ideal of A distinct from A. Then n0 an = {0}. 7.3.6 Corollary. Let A be Noetherian and b an ideal of A. Then b is contained in the Jacobson radical rad A of A if and only if (a + bn ) = a n0

for any ideal a of A. Further, if b ⊂ rad A, then (N + bn M ) = N n0

for any submodule N of a ﬁnitely generated A-module M . Proof. Let b ⊂ rad A, M a ﬁnitely generated A-module and N a submodule of M . One checks easily that bn (M/N ) = (N + bn M )/N for all n ∈ N. By 7.3.4, if x ∈ n0 bn (M/N ), then there exists b ∈ b such that (1 − b)x = 0. But 1 − b is invertible since b ⊂ rad A. So x = 0 and N is the intersection of the (N + bn M )’s, for n 0. In particular, a is the intersection of the a + bn , for n 0. Conversely, if b ⊂ rad A, then b ⊂ m forsome m ∈ Spm(A). Hence bn ⊂ m, and bn + m = A for all n ∈ N. Thus A = n0 (m + bn ) = m. 7.3.7 Corollary. Let A be Noetherian and a an ideal of A such that a ⊂ rad A. Then, for any ﬁnitely generated A-module M : n a M = {0}. n0

7.4 Filtrations 7.4.1 Deﬁnition. A ﬁltration on a group G is an increasing sequence (Gn )n∈Z of subgroups of G such that G = n∈Z Gn . 7.4.2 Deﬁnition. A ﬁltration on a ring A is a ﬁltration (An )n∈Z of the additive group A which is compatible with the ring structure, i.e.: (1) An Am ⊂ Am+n for all m, n ∈ Z. (2) 1 ∈ A0 . 7.4.3 Deﬁnition. Let (An )n∈Z be a ﬁltration on a ring A, M an Amodule. A ﬁltration (Mn )n∈Z is said to be compatible (with the ﬁltration on A) if Am Mn ⊂ Mm+n for all m, n ∈ Z. 7.4.4 Let (Gn )n∈Z be a ﬁltration on G, and H a subgroup of G. Then (H ∩ Gn )n∈Z is called the induced ﬁltration on H. Similarly, if H is normal, then (HGn /H)n∈Z is a ﬁltration on G/H. Let (An )n∈Z be a ﬁltration of a ring A. Then the above procedure provides ﬁltrations on any subring of A and any quotient of A by a bilateral ideal a.

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7 Gradings and ﬁltrations

It is easy to check that the same procedure works for any submodule N of an A-module M with a compatible ﬁltration. 7.4.5 Let (An )n∈Z be a ﬁltration on the ring A and M, N two A-modules with compatible ﬁltrations. A homomorphism u : M → N is said to be compatible if u(Mn ) ⊂ Nn for all n ∈ Z.

7.5 Grading associated to a ﬁltration 7.5.1 Let (Gn )n∈Z be a ﬁltration on an additive group G. For n ∈ Z, let grn (G) = Gn /Gn−1 and grn (G) gr(G) = n∈Z

is a graded commutative group. We call gr(G) the graded group associated to the ﬁltration. 7.5.2 Let (An )n∈Z be a ﬁltration on a ring A, and (Mn )n∈Z a compatible ﬁltration on an A-module M . Let p, q ∈ Z, we deﬁne: (1) grp (A) × grq (M ) → grp+q (M ) , (a + Ap−1 , m + Mq−1 ) → am + Mp+q−1 . One veriﬁes easily that this is a well-deﬁned Z-bilinear map. This induces a Z-bilinear map: (2)

gr(A) × gr(M ) → gr(M ).

In particular, when M = A, we obtain a ring structure on gr(A) via this map. We call gr(A) the graded ring associated to A. It follows that gr(M ) is a graded gr(A)-module called the graded module associated to M . 7.5.3 Let (An )n∈Z be a ﬁltration on a ring A, and M , N be A-modules with compatible ﬁltrations. If u : M → N is a compatible homomorphism, then it is easy to see that u induces a homomorphism gr(u) of the associated graded modules given by gr(u)(m + Mp−1 ) = u(m) + Np−1 for m + Mp−1 ∈ grp (M ). Further, if v : N → P is another compatible homomorphism of A-modules with compatible ﬁltrations, then gr(v ◦ u) = gr(v) ◦ gr(u). In the rest of this section, let (An )n∈Z be a ﬁltration on a ring A and let M be an A-module with a compatible ﬁltration (Mn )n∈Z . Proposition. Let N be a submodule of M . Then we have the following exact sequence of graded gr(A)-modules: gr(j)

gr(p)

0 −→ gr(N ) −→ gr(M ) −→ gr(M/N ) −→ 0 where j : N → M is the canonical inclusion, p : M → M/N is the canonical surjection, and we endow N with the induced ﬁltration. Proof. The homomorphisms j and p are clearly compatible and the exactness of the sequence is a straightforward veriﬁcation.

7.5 Grading associated to a ﬁltration

93

7.5.4 Let N = n∈Z Mn . If N = {0}, then we say that the ﬁltration (on M ) is separated. Let x ∈ M , we deﬁne the order of x, denoted by ν(x) or νM (x), to be: ν(x) = +∞ if x ∈ N , ν(x) = p if x ∈ Mp , x ∈ Mp−1 . 7.5.5 Proposition. Assume that gr(A) has no zero divisors other than 0, then ν(ab) = ν(a) + ν(b) for all a, b ∈ A. Proof. Since n∈Z An is a bilateral ideal of A, the equality is clear if a or b belongs to n∈Z An . Now if ν(a) = r and ν(b) = s where r, s ∈ Z, then clearly ab ∈ Ar+s . Since gr(A) has no zero divisors other than 0, ab ∈ Ar+s−1 . So ν(ab) = ν(a) + ν(b). 7.5.6 Corollary. If gr(A) has no zero divisors other than 0, then neither has A/ n∈Z An . Proof. By 7.5.4, we have ν(ab) = +∞ for any a, b ∈ A \ n∈Z An . Hence ab ∈ n∈Z An . 7.5.7 Proposition. Let u : M → N be a compatible homomorphism of A-modules with compatible ﬁltrations. If the ﬁltration on M is separated and gr(u) is injective, then u is also injective. Proof. Note that, in general, we have u( n∈Z Mn ) ⊂ n∈Z Nn . Let us assume that the ﬁltration on M is separated and gr(u) is injective. The injectivity of gr(u) implies that νM (x) = νN (u(x)) for any x ∈ M . So if x ∈ ker u, then νM (x) = +∞. Since the ﬁltration on M is separated, u is injective. 7.5.8 If there exists n ∈ Z such that Mn = {0}, then Mp = {0} for all p n, and we say that the ﬁltration on M is discrete. Note that a discrete ﬁltration is separated. Proposition. Let u : M → N be a compatible homomorphism of Amodules with compatible ﬁltrations. If the ﬁltration on N is discrete and gr(u) is surjective, then u is also surjective. Proof. Let us assume that the ﬁltration on N is discrete and gr(u) is surjective. There exists q ∈ Z such that Nq = {0}. This implies that u(Mn ) = Nn for n q. Now let us assume that u(Mn ) = Nn for n p. Since gr(u) is surjective, Np+1 = u(Mp+1 )+Np = u(Mp+1 )+u(Mp ) = u(Mp+1 ). It follows by induction that u(Mn ) = Nn for all n ∈ Z. Hence the surjectivity of u. 7.5.9 In the rest of this section, all the ﬁltrations considered are discrete. Let x ∈ M and n = ν(x). We shall denote the image of x in grn (M ) by σM (x) or simply σ(x), and we shall call σ(x) the principal symbol of x. This induces a map σM : M → gr(M ).

94

7 Gradings and ﬁltrations

7.5.10 Proposition. Let (xi )i∈I be a family of elements of M . (i) If (σ(xi ))i∈I is a system of generators of the gr(A)-module gr(M ), then (xi )i∈I generates the A-module M . (ii) If gr(M ) is a free gr(A)-module and (σ(xi ))i∈I is a gr(A)-basis of gr(M ), then M is a free A-module and (xi )i∈I is a A-basis of M . Proof. (i) Let q ∈ Z be such that Aq = {0} and Mq = {0}. Let x ∈ Mn . We shall prove by induction on n that x is contained in the A-submodule generated by (xi )i∈I . This is true for n q. So we may assume that n > q and ν(x) = n. There exist non-zero elements α1 , . . . , αr ∈ gr(A) and indices i1 , . . . , ir ∈ I such that σ(x) = α1 σ(x1 ) + · · · + αr σ(xr ). Set ni = ν(xi ), then we can ﬁnd a1 , . . . , ar ∈ A such that σ(aj ) = αj and ν(aj ) = n − nj for all 1 j r. It follows that: x−

r

aj xj ∈ Mn−1 .

j=1

By our induction hypothesis, x is contained in the A-submodule generated by (xi )i∈I . (ii) Let N be a free A-module with basis (yi )i∈I . Deﬁne a (discrete) ﬁltration on N as follows: for n ∈ Z, Aj yi Nn = j+ni n

where ni = ν(xi ) for i ∈ I. One veriﬁes easily that this is a compatible ﬁltration and the map u : N → M deﬁned by sending yi to xi is a compatible homomorphism of A-modules. The associated map gr(u) is clearly bijective and it follows from 7.5.7 and 7.5.8 that u is an isomorphism. Hence M is free and (xi )i∈I is a basis of M . 7.5.11 Corollary. If gr(M ) is a ﬁnitely generated (resp. Noetherian) gr(A)-module, then M is a ﬁnitely generated (resp. Noetherian) A-module. Proof. These are direct consequences of 7.2.2, 7.5.3 and 7.5.10. 7.5.12 Corollary. If gr(A) is a left Noetherian ring, then so is A.

References • [10], [52], [56].

8 Inductive limits

8.1 Generalities 8.1.1 Recall that a preorder on a set I is a binary relation which is reﬂexive and transitive. Let be a preorder on I, then it is ﬁltered or directed if for all i, j ∈ I, there exists k ∈ I such that i k and j k. A subset J ⊂ I is coﬁnal if for all i ∈ I, there exists j ∈ J such that i j. A set with a ﬁltered preorder is called a ﬁltered set or directed set. 8.1.2 Remarks. 1) A preorder on a set I is ﬁltered if and only if any non-empty ﬁnite subset of I has an upper bound. 2) Let X be a topological space and x ∈ X. Then the set of neighbourhoods of x is a ﬁltered set via the preorder U V if and only if V ⊂ U . 8.1.3 In the rest of this chapter, I is a ﬁltered set with preorder . 8.1.4 Deﬁnition. An inductive system of sets indexed by I is a family of sets (Ei )i∈I and maps (fji )i,j∈I,ij verifying: for i, j, k ∈ I such that i j k, (i) fji is a map from Ei to Ej . (ii) fii is the identity map on Ei . (iii) fkj ◦ fji = fki . When there is no confusion, we shall denote an inductive system of sets indexed by I simply by (Ei , fji ). 8.1.5 Let (Ei , fji ) be an inductive system of sets indexed by I, and F be the sum of the Ei ’s. If x ∈ F , we deﬁne θ(x) to be the unique index i ∈ I such that x ∈ Ei . Deﬁne a binary relation R on F as follows: for x, y ∈ F , let i = θ(x) and j = θ(y), then xRy ⇔ there exists k ∈ I such that k i, k j and fki (x) = fkj (y). One veriﬁes easily that R is an equivalence relation and the quotient lim (E −→ i , fji ) = F/R is called the inductive limit of the inductive system. To

96

8 Inductive limits

simply notations, we shall write −→ lim Ei for −→ lim(Ei , fji ) when there is no confusion. Note that −→ lim(Ei , fji ) is non-empty if at least one of the Ei ’s is non-empty. For i ∈ I, we write fi for the restriction to Ei of the canonical surjection F → −→ lim Ei . In particular, one veriﬁes easily that for i, j ∈ I, (1)

fi = fj ◦ fji whenever i j.

8.1.6 Examples. 1) Let G be a set and consider the inductive system (Ei , fji ) where Ei = G for all i ∈ I and if i j, then fji is the identity map on G. The maps fi are identical for all i ∈ I, and therefore there is a canonical bijection from G to −→ lim Ei . 2) Let A, B be sets and (Vi )i∈I a family of subsets of A such that Vj ⊂ Vi if i j. Let Ei be the set of maps from Vi to B and deﬁne, for i, j ∈ I, i j, fji : Ei → Ej by fji (u) = u|Vj . Then (Ei , fji ) is an inductive system. We call lim Ei the set of germs of functions from Vi to B. This generalizes the case −→ where (Vi )i∈I is the set of neighbourhoods of a point of a topological space A. 8.1.7 Proposition. Let (Ei , fji ) be an inductive system of sets, E = lim −→ Ei and for i ∈ I, fi : Ei → E the canonical map. (i) Let x1 , . . . , xn ∈ E. There exist i ∈ I and y1 , . . . , yn ∈ Ei such that xk = fi (yk ) for 1 k n. (ii) Let i ∈ I and y1 , . . . , yn ∈ Ei be such that fi (y1 ) = · · · = fi (yn ). There exists j ∈ I such that i j and fji (y1 ) = · · · = fji (yn ). Proof. (i) For 1 k n, there exist ik ∈ I and zk ∈ Eik such that xk = fik (zk ). Since I is a directed set, there exists i ∈ I such that i ik for 1 k n. Set yk = fi,ik (zk ). Then (1) implies that xk = fik (zk ) = fi ◦ fi,ik (zk ) = fi (yk ). (ii) For 1 h, k n, there exists αh,k ∈ I such that αh,k i and fαh,k ,i (yh ) = fαh,k ,i (yk ). Let j ∈ I be such that j αh,k for all 1 h, k n. Then fji (yh ) = fji (yk ) since fji = fj,αh,k ◦ fαh,k ,i .

8.2 Inductive systems of maps 8.2.1 Let (Ei , fji ) be an inductive system of sets indexed by I. Set E = lim −→ Ei and for i ∈ I, fi : Ei → E the canonical map. Let G be a set. Proposition. Let (ui )i∈I be a family of maps verifying ui : Ei → G, and (2)

uj ◦ fji = ui if i j. (i) There exists a unique map u : E → G such that

(3)

ui = u ◦ fi for all i ∈ I.

(ii) The map u is surjective if and only if G is the union of the ui (Ei )’s. (iii) The map u is injective if and only if for any i ∈ I and x, y ∈ Ei , ui (x) = ui (y) implies that fji (x) = fji (y) for some j i.

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97

Proof. (i) Let F be the disjoint union of the Ei ’s and f : F → E the canonical surjection. There is a unique map v : F → G such that v|Ei = ui for i ∈ I. Now (2) says that v is compatible with the equivalence relation R. So there is a unique map u : E → G verifying v = u ◦ f . Part (ii) is obvious and part (iii) is just a consequence of 8.1.7. 8.2.2 Remarks. 1) When u is bijective, we shall also call G the inductive limit of (Ei , fji ). 2) If all the fji ’s are injective, then so are the fi ’s. In this case, Ei can be identiﬁed with fi (Ei ), and E can be viewed as the union of the Ei ’s. Conversely, let (Fi )i∈I be a family of subsets of a set F such that F = i∈I Fi and Fi ⊂ Fj if i j. For i j, denote by fji : Fi → Fj the canonical injection. Then F is the inductive limit of the inductive system (Fi , fji ), and for i ∈ I, the canonical map fi : Fi → F is just the canonical injection. 8.2.3 Proposition. Let (Ei , fji ) and (Fi , gji ) be inductive systems of sets indexed by I. Let (ui : Ei → Fi )i∈I be a family of maps such that for all i, j ∈ I verifying i j, the following diagram is commutative: u

i Ei −−−− → ⏐ ⏐ fji

Fi ⏐ ⏐gji

Ej −−−−→ Fj uj

Let E = −→ lim Ei , F = −→ lim Fi and for i ∈ I, let fi and gi be the corresponding canonical maps. There exists a unique map u : E → F such that for any i ∈ I, the following diagram is commutative: ui → Fi Ei −−−− ⏐ ⏐ ⏐gi ⏐ fi E −−−−→ F u

Further, u is injective (resp. surjective) if all the ui ’s are injective (resp. surjective). Proof. The family (gi ◦ ui )i∈I satisﬁes the hypothesis of 8.2.1, so there exists a unique map u : E → F such that u ◦ fi = gi ◦ ui . If the ui ’s are surjective, then gi ◦ ui (Ei ) = u ◦ fi (Ei ) = u( fi (Ei )) = u(E). F = i∈I

i∈I

i∈I

If the ui ’s are injective, then for any i ∈ I, x, y ∈ Ei such that u(fi (x)) = u(fi (y)), we have gi (ui (x)) = gi (ui (y)). Now by 8.1.7, there exists j i such that gji (ui (x)) = gji (ui (y)). Thus uj (fji (x)) = uj (fji (y)), and since uj is injective, fji (x) = fji (y). So u is injective (8.2.1).

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8.2.4 A family (ui : Ei → Fi )i∈I of maps verifying the hypothesis of 8.2.3 shall be called an inductive system of maps from (Ei , fji ) to (Fi , gji ). The map u is called the inductive limit of the family (ui )i∈I and we shall denote it by −→ lim ui when there is no confusion. 8.2.5 Corollary. Let (ui : Ei → Fi )i∈I be an inductive system of maps from (Ei , fji ) to (Fi , gji ), and (vi : Fi → Gi )i∈I an inductive system of maps from (Fi , gji ) to (Gi , hji ). Then (vi ◦ ui )i∈I is an inductive system of maps from (Ei , fji ) to (Gi , hji ) and (4)

lim(vi ◦ ui ) = (lim −→ vi ) ◦ (lim −→ ui ). −→

Proof. This is a straightforward veriﬁcation. 8.2.6 Let J be a directed subset of I, i.e. J is ﬁltered with respect to the preorder induced from I. Then the subfamily of sets (Ei )i∈J and the subfamily of maps (fji )ij,i,j∈J form an inductive system of sets indexed by J, denoted by (Ei , fji )J . We call (Ei , fji )J the inductive subsystem relative to J. lim(Ei , fji )J . For i ∈ I, denote by fi the Now let E = −→ lim Ei and EJ = −→ canonical map. Then (fi )i∈J is an inductive system of maps from (Ei , fji )J to (Ei , fji ), and g = −→ lim(fi )i∈J : EJ → E. lim(fi )i∈K : EK → Let K ⊂ J be another directed subset of I, and g = −→ lim(fi )i∈K : EK → E, then by 8.2.5, we have EJ , g = −→ (5)

g = g ◦ g .

8.2.7 Proposition. Let J be a coﬁnal subset of I. The map g = −→ lim(fi )J : EJ → E is bijective. Proof. Let i ∈ J and x, y ∈ Ei be such that fi (x) = fj (y). By 8.1.7, there exists j ∈ J verifying j i and fji (x) = fji (y). By 8.2.1(iii), g is injective. For i ∈ J, let hi : Ei → EJ be the canonical map. If x ∈ E, then there exist i ∈ I and y ∈ Ei such that x = fi (y). If j ∈ J is such that j i, then x = fj ◦ fji (y) = g(hj ◦ fji (y)). It follows that g is surjective. 8.2.8 Remark. The previous statement says that in order to determine the inductive limit, it suﬃces to determine the inductive limit of the inductive subsystem relative to a coﬁnal subset of I.

8.3 Inductive systems of magmas, groups and rings 8.3.1 Deﬁnition. Let (Ei )i∈I be a family of magmas (resp. semigroups, monoids, groups, rings). An inductive system (Ei , fji ) indexed by I is called an inductive system of magmas (resp. semigroups, monoids, groups, rings) if the fji ’s are homomorphisms of magmas (resp. semigroups, monoids, groups, rings).

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99

8.3.2 Let us denote E = −→ lim Ei and fi : Ei → E the canonical maps. Proposition. There exists a unique structure of magma (resp. semigroups, monoids, groups, rings) on E such that for i ∈ I, fi is a homomorphism of magmas (resp. semigroups, monoids, groups, rings). Further, if the Ei ’s are commutative, then so is E. Proof. Let x, y ∈ E, i ∈ I and xi , yi ∈ Ei be such that x = fi (xi ), y = fi (yi ). Uniqueness follows from the fact that if the fi ’s are homomorphisms of magmas, fi (xi yi ) is the composition of x and y for any law of composition on E. Let us prove its existence. Suppose that aj , bj ∈ Ej verify fj (aj ) = x and fj (bj ) = y. By deﬁnition, there exists k ∈ I, k i, k j such that fkj (aj ) = fki (xi ) and fkj (bj ) = fki (yi ). Since the fji ’s are homomorphisms, it follows that: fi (xi yi ) = fk (fki (xi yi )) = fk (fkj (aj bj )) = fj (aj bj ). Hence xy = fi (xi yi ) deﬁnes a structure of magma on E and it is clear that with respect to this law of composition, the fi ’s are homomorphisms of magma. The proofs for semigroups, monoids, groups, rings and commutativity are similar. 8.3.3 Proposition. Let (Ei , fji ) be an inductive system of magmas (resp. semigroups, monoids, groups, rings), E = −→ lim Ei and fi : Ei → E the canonical maps. Let F be a magma (resp. semigroup, monoid, group, ring) and ui : Ei → F be homomorphisms verifying ui = uj ◦ fji whenever i j. There exists a unique homomorphism u : E → F such that ui = u ◦ fi . Proof. By 8.2.1, there exists a unique map u : E → F such that ui = u ◦ fi . Therefore, we only have to check that u is a homomorphism. Let x, y ∈ E, i ∈ I and xi , yi ∈ Ei be such that x = fi (xi ), y = fi (yi ). Then u(xy) = u(fi (xi yi )) = ui (xi yi ) = ui (xi )ui (yi ) = u(x)u(y). So the result follows. 8.3.4 Remark. As in 8.2.3, if (Ei , fji ) and (Fi , gji ) are inductive systems of magmas (resp. semigroups, monoids, groups, rings), and (ui : Ei → Fi )i∈I is lim Ei → an inductive system of maps which are homomorphisms, then −→ lim ui : −→ is a homomorphism. lim F i −→ 8.3.5 Proposition. Let (Ai , fji ) be an inductive system of rings and A = lim −→ Ai . (i) If the Ai ’s are non-zero, then A is non-zero. (ii) If the Ai ’s are integral domains, then A is an integral domain. (iii) If the Ai ’s are ﬁelds, then A is a ﬁeld.

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Proof. (i) Let 0i , 1i (resp. 0, 1) be the zero and the identity in Ai (resp. A). There exists i ∈ I such that fi (0i ) = 0 and fi (1i ) = 1. If 0 = 1, then there exists j i such that fji (0i ) = fji (1i ). This implies that 0i = 1i which contradicts the fact that Ai is non-zero. (ii) Let x, y ∈ A be such that xy = 0. There exist i ∈ I, xi , yi ∈ Ai such that fi (xi ) = x, fi (yi ) = y. It follows that fi (xi yi ) = 0i and there exists j i such that fji (xi yi ) = fji (xi )fji (yi ) = 0j . Since Aj is an integral domain, fji (xi ) = 0j or fji (yi ) = 0j . But this implies that fj (fji (xi )) = x = 0 or fj (fji (yi )) = y = 0. So we are done. (iii) Let x ∈ A \ {0}, i ∈ I and xi ∈ Ai with x = fi (xi ). Since the Ai ’s are ﬁelds, xi = 0i and fi (x−1 i ) is the inverse of x. So A is a ﬁeld.

8.4 An example 8.4.1 Let A be a commutative ring, S be the set of multiplicative subsets S of A such that 1 ∈ S, 0 ∈ S. For S ∈ S, let iS : A → S −1 A denote the canonical homomorphism and U (S −1 A) the set of invertible elements in S −1 A. Let S, T, V ∈ S verify S ⊂ T ⊂ V . Note that iT (S) ⊂ U (T −1 A). By 2.3.5, there exists a unique homomorphism fT S : S −1 A → T −1 A such that fT S ◦ iS = iT . Since (fV T ◦ fT S ) ◦ iS = iV = fV S ◦ iS , it follows that: fV T ◦ fT S = fV S . 8.4.2 Let us ﬁx p ∈ Spec(A) and let S1 be the set of S ∈ S such that S ∩ p = ∅. Let S, T ∈ S1 and ST = {st; s ∈ S, t ∈ T }. Since p is prime, S, T ⊂ ST ∈ S1 . We conclude that S1 is a directed set, and (S −1 A, fT S )S∈S1 is an inductive system of commutative rings. Let B = −→ lim S −1 A and fS : S −1 A → B the canonical maps. We claim that B is isomorphic to Ap . Let P = A \ p, then P ∈ S and Ap = P −1 A. From 8.4.1, we obtain that the family (fP S : S −1 A → Ap )S∈S1 satisﬁes the hypothesis of 8.2.1. It follows that there is a unique ring homomorphism u : B → Ap such that fP S = u ◦ fS for all S ∈ S1 . Now P is the biggest element in S1 and fP P is just the identity map on Ap . So u is surjective. Finally u is injective by 8.2.1(iii). Hence B is isomorphic to Ap . 8.4.3 Let S2 be the set of S ∈ S not containing any zero divisors of A. Then a similar argument shows that S2 is a directed set and the limit of the inductive system (S −1 A, fT S )S∈S2 is isomorphic to the full ring of fractions of A.

8.5 Inductive systems of algebras 8.5.1 Let A be a ring and (Ei , fji ) be an inductive system of abelian groups with E = −→ lim Ei , and fi the canonical maps.

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101

Let us suppose further that for i ∈ I, Ei is an A-module and whenever i j, fji is a homomorphism of A-modules. Then a simple veriﬁcation shows that E admits a structure of A-module given by: for any a ∈ A, x ∈ E, i ∈ I, xi ∈ Ei verifying fi (xi ) = x, we set ax = fi (axi ). Note that if (Fi , gji ) is another such inductive system of abelian groups, and (ui : Ei → Fi )i∈I is an inductive system of homomorphisms of A-modules, lim Ei → −→ lim Fi is a homomorphism of Athen we verify readily that −→ lim ui : −→ modules. 8.5.2 Let us conserve the notations and hypotheses of 8.5.1. Suppose further that A is a commutative ring, the Ei ’s are A-algebras and the fji ’s are A-algebra homomorphisms. Then the ring structure and the A-module structure on E make E into an A-algebra and the fi ’s are A-algebra homomorphisms. Again, E is associative (resp. commutative) if the Ei ’s are associative (resp. commutative), and if (Fi , gji ) is another such inductive system and (ui : Ei → Fi )i∈I is an inductive lim ui : system of maps such that ui is a homomorphism of A-algebras, then −→ → lim F is a homomorphism of A-algebras. lim E i i −→ −→

References • [8], [9], [34], [52].

9 Sheaves of functions

The notion of a sheaf is essential in many areas in mathematics, and most particularly in algebraic geometry. Indeed, the theory of sheaves plays an important part in the foundation of modern algebraic geometry. We describe in this chapter some basic facts on sheaves of functions.

9.1 Sheaves 9.1.1 In this chapter, X will be a topological space and E a set. We shall denote the set of open subsets of X by Ω X or Ω, and by ΩxX or Ωx the set of open neighbourhoods in X of an element x ∈ X. Let us ﬁx the partial order on Ω and Ωx given by: U V if and only if V ⊂ U. 9.1.2 Deﬁnition. A presheaf of functions on X with values in E is a map F which associates to each U ∈ Ω, a set F(U ) of maps from U to E, such that whenever U, V ∈ Ω verify V ⊂ U , f |V ∈ F(V ) for all f ∈ F(U ). This last condition says that the restriction map, denoted by rVF U or rV U , from F(U ) to F(V ) is well-deﬁned. In the sequel, by a presheaf on X, we shall mean a presheaf of functions on X with values in E. 9.1.3 Remarks. 1) A presheaf on X is an inductive system of sets indexed by Ω. 2) Let F be a presheaf on X. If all the F(U )’s are groups (resp. rings, etc...) and all the rV U ’s are homomorphisms of groups (resp. rings, etc...), we shall say that F is a presheaf of groups (resp. rings, etc...) on X. 3) Let F be a presheaf on X. We shall also use the notation Γ (U, F) for F(U ). The elements of Γ (U, F) are called sections of F over U , and those of Γ (X, F) are called global sections. 9.1.4 Let F be a presheaf on X and x ∈ X. Then (F(U ), rV U )U ∈Ωx is an inductive system of sets indexed by Ωx . Let us denote the inductive limit of

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this system by Fx , and we call Fx the ﬁbre of F at x. The elements of Fx are called the germs of sections of F at x. Further, if F is a presheaf of groups (resp. ring, etc...), then by 8.3.2, Fx is a group (resp. ring, etc...). For U ∈ Ωx , let rU,x : U → Fx be the canonical map. We call rU,x (f ) the germ of f at x. Now let U, V ∈ Ωx , s ∈ Fx , f ∈ F(U ) and g ∈ F(V ). Then s = rU,x (f ) = rV,x (g) if and only if there exists W ∈ Ωx such that W ⊂ U ∩V and rW U (f ) = rW V (g). It follows that f (x) is the same for any f having s as germ at x. We shall call this the value of the germ s at x, and we shall denote it by s(x). 9.1.5 Deﬁnition. A presheaf F on X is a sheaf if it satisﬁes the following condition: if (Ui )i∈I is a family of open subsets of X, U their union and f : U → E a map such that f |Ui ∈ F(Ui ), then f ∈ F(U ). 9.1.6 Examples. 1) Let G be an abelian group and F the presheaf of abelian groups on X deﬁned by F(U ) = Hom(U, G) for all U ∈ Ω, here Hom(U, G) denotes the set of maps from U to G. One veriﬁes easily that F is a sheaf of abelian groups on X. 2) For U ∈ Ω, let F(U ) = C(U, R), the ring of continuous real-valued functions on X. Then F is a sheaf of rings on X. 3) Let X = C and for U ∈ Ω, let F(U ) be the ring of holomorphic functions on U . Then we obtain the sheaf of holomorphic functions on C. 4) Let X = R and for U ∈ Ω, let F(U ) be the ring of bounded continuous real-valued functions on X. It is a presheaf, but not a sheaf. In fact, the identity function is bounded on Un =] − n, n[, but it is not bounded on R.

9.2 Morphisms 9.2.1 Let A, B be sets (resp. groups, rings, etc...), we denote by Hom(A, B) the set of maps (resp. homomorphisms of groups, homomorphisms of rings, etc...) from A to B. 9.2.2 Deﬁnition. Let F and G be presheaves on X. A morphism ϕ : F → G is a family of maps (ϕ(U ) : F(U ) → G(U ))U ∈Ω such that whenever U, V ∈ Ω verify V ⊂ U , the following diagram is commutative: ϕ(U )

(1)

F(U ) −−−−→ G(U ) ⏐ ⏐ ⏐G F ⏐ rV rV U U F(V ) −−−−→ G(V ) ϕ(V )

A morphism of sheaves is a morphism of the underlying presheaves. The morphism ϕ is an isomorphism if ϕ(U ) is an isomorphism for all U ∈ Ω. 9.2.3 Remarks. 1) We shall denote the set of morphisms of (pre)sheaves from F to G by Hom(F, G). It is clear that the composition of two morphisms is again a morphism.

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105

Let us denote by idF the morphism where idF (U ) = idF (U ) for all U ∈ Ω. It follows that given ϕ ∈ Hom(F, G), ϕ is an isomorphism if and only if there exists ψ ∈ Hom(G, F) such that ϕ ◦ ψ = idG and ψ ◦ ϕ = idF . 2) Let x ∈ X and ϕ ∈ Hom(F, G). Recall from 8.2.3 that there is a unique homomorphism ϕx : Fx → Gx such that for all U ∈ Ωx , the following diagram is commutative: ϕ(U )

(2)

F(U ) −−−−→ G(U ) ⏐ ⏐ ⏐rG F ⏐ rU,x U,x Fx

−−−−→ ϕx

Gx

We call ϕx the morphism induced by ϕ. 9.2.4 Theorem. Let ϕ : F → G be a morphism of presheaves on X. The following conditions are equivalent: (i) For all U ∈ Ω, ϕ(U ) is injective. (ii) For all x ∈ X, ϕx is injective. When these conditions are satisﬁed, we say that ϕ is injective. Proof. The implication (i) ⇒ (ii) was proved in 8.2.3. Let us prove (ii) ⇒ (i). Let U ∈ Ω and f, g ∈ F(U ) be such that ϕ(U )(f ) = ϕ(U )(g). Then by (2) F F (f ) = rU,x (g). So there is an open and the injectivity of ϕx , we have that rU,x neighbourhood Vx ∈ Ωx of x such that Vx ⊂ U and f |Vx = g|Vx . Since the Vx ’s, x ∈ U form a covering of U , we deduce that f = g. 9.2.5 Theorem. Let ϕ : F → G be a morphism of sheaves on X. The following conditions are equivalent: (i) For all U ∈ Ω and g ∈ G(U ), there exists an open covering (Ui )i∈I of U and elements fi ∈ F(Ui ) such that ϕ(Ui )(fi ) = g|Ui for all i ∈ I. (ii) For all x ∈ X, ϕx is surjective. When these conditions are satisﬁed, we say that ϕ is surjective. Proof. (i) ⇒ (ii) Let x ∈ X, θ ∈ Gx , U ∈ Ω and g ∈ G(U ) be such that G (g). By our hypotheses, there exist an open covering (Ui )i∈I of U and θ = rU,x elements fi ∈ F(Ui ) such that ϕ(Ui )(fi ) = g|Ui for all i ∈ I. Fix k ∈ I such that x ∈ Uk , then G G G G G rU,x (g) = rU ◦ rU (g) = rU ◦ ϕ(Uk )(fk ) = ϕx (rU (fk )). k ,x k ,U k ,x k ,x G (fk ) ∈ Fx , ϕx is surjective. Since rU k ,x (ii) ⇒ (i) Let U ∈ Ω and g ∈ G(U ). For x ∈ U , there exists ρ ∈ Fx such G (g). Let Ux ∈ Ωx and f x ∈ F(Ux ) be such that Ux ⊂ U and that ϕx (ρ) = rU,x F x ρ = rUx ,x (f ). Then: G F G (g) = ϕx ◦ rU (f x ) = rU ◦ ϕ(Ux )(f x ). rU,x x ,x x ,x

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This implies that there exists an open neighbourhood Vx ∈ Ωx such that Vx ⊂ Ux and g|Vx = ϕ(Ux )(f x )|Vx . Hence g|Vx = rVGx ,Ux ◦ ϕ(Ux )(f x ) = ϕ(Vx ) ◦ rVFx ,Ux (f x ). It follows that the family (Vx , rVFx ,Ux (f x ))x∈U satisﬁes (i).

9.2.6 Remark. Let ϕ : F → G be a morphism of sheaves on X. If ϕ(U ) is surjective for all U ∈ Ω, then by 8.2.3, ϕx is surjective for all x ∈ X. However, the converse is not true. For example, let X = C \ {0} and for U ∈ Ω, let F(U ) be the set of holomorphic functions on U . Consider the morphism of sheaves ϕ : F → F given by g → g . Then ϕ(X) is not surjective since there is no holomorphic function on X such that h (z) = 1/z for all z ∈ X. On the other hand, let U ∈ Ωx and f ∈ F(U ). For any point z ∈ U , ﬁx an open ball Bz of X with centre z, contained in U . Since Bz is simply connected, there exists gz ∈ F(Bz ) such that gz = f |Bz . Hence condition (i) of 9.2.5 is satisﬁed, and ϕx is surjective. 9.2.7 Theorem. Let ϕ : F → G be a morphism of sheaves on X. Then ϕ is an isomorphism if and only if ϕx is bijective for all x ∈ X. Proof. By 9.2.4 and 9.2.6, we only have to prove that if ϕx is bijective for all x ∈ X, then ϕ(U ) is surjective for all U ∈ Ω. Let U ∈ Ω and g ∈ G(U ). By 9.2.5, for any x ∈ U , there exists Ux ∈ Ωx and f x ∈ F(Ux ) such that ϕ(Ux )(f x ) = g|Ux . It follows that for x, y ∈ U , we have: ϕ(Ux ∩ Uy )(f x |Ux ∩Uy ) = g|Ux ∩Uy = ϕ(Ux ∩ Uy )(f y |Ux ∩Uy ). The injectivity of ϕ implies that f x |Ux ∩Uy = f y |Ux ∩Uy . Since F is a sheaf, there exists f ∈ F(U ) such that f |Ux = f x for all x ∈ U . It follows that ϕ(U )(f ) = g since ϕ(U )(f )|Ux = g|Ux for all x ∈ U . 9.2.8 Lemma. Let F be a sheaf on X, U ∈ Ω and f, g ∈ F(U ) Then f = g if and only if rU,x (f ) = rU,x (g) for all x ∈ U . Proof. It is clear that f = g implies that rU,x (f ) = rU,x (g) for all x ∈ U . Conversely, if rU,x (f ) = rU,x (g) for all x ∈ U , then for any x ∈ U , there exists Vx ∈ Ωx such that Vx ⊂ U and f |Vx = g|Vx . So f = g.

9.3 Sheaf associated to a presheaf 9.3.1 Let F be a presheaf on X and U ∈ Ω. Denote by F + (U ) the set of maps f : U → E verifying: for any x ∈ U , there exist V ∈ Ωx and g ∈ F(V ) such that V ⊂ U and f |V = g. We see immediately that F + is a sheaf on X that we shall call the sheaf associated to the presheaf F. If F is a presheaf of groups (resp. rings, etc...),

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then the above deﬁnition implies that F + is a sheaf of groups (resp. rings, etc...). There is a canonical injective morphism of presheaves η F or η : F → F + given by f → f . If F is a sheaf, then F = F + since given U ∈ Ω, f ∈ F + , x ∈ U , there exist Vx ∈ Ωx , Vx ⊂ U and g x ∈ F(Vx ) such that f |Vx = g x . Since the Vx ’s is a covering of U , f ∈ F(U ). 9.3.2 Theorem. Let F be a presheaf on X. (i) For all x ∈ X, the map ηx : Fx → Fx+ is bijective. (ii) If ϕ : F → G is a morphism of presheaves, then there is a unique morphism of sheaves ϕ+ : F + → G + such that ϕ+ ◦ η F = η G ◦ ϕ. (iii) If ϕ : F → G is a morphism of presheaves and G is a sheaf, then there is a unique morphism of sheaves ψ : F + → G such that ψ ◦ η F = ϕ. Proof. (i) Let x ∈ X and fx ∈ Fx+ . There exist U ∈ Ωx , f ∈ F + (U ) such that F+ rU,x (f ) = fx . By deﬁnition, there exist V ∈ Ωx , V ⊂ U and g ∈ F(V ) such F (g)) = fx . Hence ηx is surjective. The injectivity of that f |V = g. So ηx (rV,x ηx is a consequence of 9.2.4 and 9.3.1. (ii) Uniqueness follows from (i) and 9.2.8. Now let U ∈ Ω and f ∈ F + (U ). There exist an open covering (Ui )i∈I of U and elements fi ∈ F(Ui ) such that f |Ui = fi . Set gi = ϕ(Ui )(fi ). Since fi |Ui ∩Uj = fj |Ui ∩Uj , we have gi |Ui ∩Uj = gj |Ui ∩Uj . It follows that there exists g ∈ G + (U ) such that g|Ui = gi . It is easy to see that g does not depend on the choice of the covering (Ui )i∈I and of the elements fi . Now set ϕ+ (U )(f ) = g. Then it is clear that ϕ+ is a morphism of sheaves from F + to G + verifying ϕ+ ◦ η F = η G ◦ ϕ. (iii) This is a direct consequence of (ii) since G + = G. 9.3.3 Remarks. 1) Assume that F is not a sheaf. Then ηx is bijective for all x ∈ X (9.3.2), but η is not an isomorphism. Thus 9.2.7 is false for morphisms of presheaves. 2) Let F be the presheaf of 9.1.6 (4), then F + is the sheaf of 9.1.6 (2) with X = R. 3) Let F, G be presheaves on X, ϕ ∈ Hom(F, G). If ϕx is bijective for all + x ∈ X, then so is ϕ+ x (9.3.2). Consequently, ϕ is an isomorphism (9.2.7). 9.3.4 Let F be a presheaf and G a sheaf on X. The results in 9.3.2 induce a map: ξ : Hom(F, G) → Hom(F + , G), ϕ → ϕ+ . Proposition. The map ξ is bijective. Proof. Let ϕ, ψ ∈ Hom(F, G) be such that ξ(ϕ) = ξ(ψ). Then for U ∈ Ω and f ∈ F(U ), we have ϕ+ (U ) ◦ η F (U )(f ) = ψ + (U ) ◦ η F (U )(f ). Hence ϕ(U )(f ) = ψ(U )(f ) and ξ is injective. Now if µ ∈ Hom(F + , G). Let ν(U )(f ) = µ(U ) ◦ η F (U )(f ) for U ∈ Ω and f ∈ F(U ). Then ν ∈ Hom(F, G) and ν + = µ. So ξ is surjective.

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9.3.5 Proposition. Let F be a presheaf, H a sheaf on X and ρ : F → H a morphism of presheaves. Suppose that for any morphism of presheaves ϕ : F → G, there exists a unique morphism of sheaves ψ : H → G such that ϕ = ψ ◦ ρ. Then F + and H are isomorphic. Proof. There exist ψ ∈ Hom(H, F + ) and µ ∈ Hom(F + , H) such that η F = ψ ◦ ρ and ρ = µ ◦ η F . It follows that η F = ψ ◦ µ ◦ η F and ρ = µ ◦ ψ ◦ ρ. By the uniqueness property of our hypothesis, we obtain that ψ ◦ µ = idF + and µ ◦ ψ = idH . 9.3.6 Proposition. Let ϕ : G → F be a morphism of presheaves on X. Suppose that F is a sheaf and that X has a base of open subsets B such that ϕ(U ) is bijective for all U ∈ B. Then F and G + are isomorphic. Proof. For x ∈ X, let Wx = Ωx ∩ B. Then Wx is a coﬁnal subset of Ωx , it follows from 8.2.7 that: lim G(U ) = Gx . lim F(U ) = Fx , −→ −→

U ∈Wx

U ∈Wx

Our hypothesis and 8.2.3 imply that ϕx is bijective. Hence ϕ+ x is bijective by 9.3.2 and the result follows from 9.2.7. 9.3.7 Let Y ∈ Ω X and F a presheaf on X. The restriction of F to Y , denoted by F |Y is a presheaf on Y endowed with the induced topology. More precisely, F |Y (U ) = F(U ) for U ∈ Ω Y . If F is a sheaf, then F |Y is also a sheaf. Further, if Z ∈ Ω Y , then F |Z = (F |Y )|Z . Let ϕ : F → G be a morphism of presheaves on X. Then we can deﬁne the restriction ϕ|Y : F |Y → G|Y in an obvious way. Clearly, ϕ|Y is a morphism of presheaves and it is an isomorphism if ϕ is an isomorphism. 9.3.8 In general, let Y be any subset of X endowed with the induced topology. Let F be a sheaf on X. We deﬁne, for V ∈ Ω Y , H(V ) to be the set of maps f : V → E such that there exist U ∈ Ω X and g ∈ F(U ) verifying V = Y ∩ U and f = g|V . It is easy to verify that H is a presheaf on Y . However, it is not a sheaf in general. The sheaf H+ associated to H is then deﬁned as follows: for V ∈ Ω Y , + H (V ) is the set of maps f : V → E such that for all x ∈ V , there exist Ux ∈ ΩxX and g ∈ F(Ux ) verifying f |V ∩Ux = g|V ∩Ux . We shall call H+ the restriction of F to Y , and we shall denote it by F |Y . Again, it is easy to see that if Y is open, then this coincides with the deﬁnition in 9.3.7, and that F |Z = (F |Y )|Z if Z ⊂ Y ⊂ X.

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9.4 Gluing 9.4.1 Proposition. Let F be a presheaf, G a sheaf on X, (Xi )i∈I an open covering of X and for i ∈ I, ϕi : F |Xi → G|Xi a morphism of presheaves such that ϕi |Xi ∩Xj = ϕj |Xi ∩Xj for all i, j ∈ I. There exists a unique morphism ϕ : F → G such that ϕ|Xi = ϕi for all i ∈ I. We say that ϕ is obtained by gluing the ϕi ’s. Proof. Let U ∈ Ω, f ∈ F(U ) and for i, j ∈ I, let Ui = U ∩ Xi , fi = f |Ui and fij = f |Ui ∩Uj . If ϕ ∈ Hom(F, G) is such that ϕ|Xi = ϕi for all i, then G ◦ ϕ(U )(f ) = ϕi (Ui )(fi ). rU i ,U

Thus ϕ is unique if it exists. Let us prove the existence of ϕ. By our hypotheses, we have: G ◦ ϕi (Ui )(fi ) = ϕi (Ui ∩ Uj )(fij ) = ϕj (Ui ∩ Uj )(fij ) rU i ∩Uj ,Ui G ◦ ϕj (Uj )(fj ). = rU i ∩Uj ,Uj G Since G is a sheaf, there exists g ∈ G(U ) such that rU (g) = ϕi (Ui )(fi ) for i ,U all i. If we deﬁne ϕ(U )(f ) = g, then we verify easily that ϕ : F → G is a morphism such that ϕ|Xi = ϕi for all i.

9.4.2 Proposition. Let (Xi )i∈I be an open covering of X and for i ∈ I, Fi a presheaf on Xi . Suppose that the following conditions are satisﬁed: (i) There is an isomorphism of sheaves εij : Fi |Xi ∩Xj → Fj |Xi ∩Xj for all i, j ∈ I. (ii) For all i, j, k ∈ I, we have εii = idFi , εjk |Xi ∩Xj ∩Xk ◦ εij |Xi ∩Xj ∩Xk = εik |Xi ∩Xj ∩Xk . Then there exist a sheaf F on X and isomorphisms εi : F |Xi → Fi such that εj |Xi ∩Xj = εij ◦ εi |Xi ∩Xj . Further, the εi ’s and F are unique in the following sense: if F and (εi )i∈I verify the same conclusion, then there is a unique isomorphism ε : F → F such that εi ◦ ε|Xi = εi for all i ∈ I. We say that F is obtained by gluing the sheaves Fi . Proof. Let U ∈ Ω, and for i, j ∈ I, let Ui = U ∩Xi , Uij = U ∩Xi ∩Xj . Set F(U ) to be the set of families (fi ) ∈ i∈I Fi (Ui ) verifying εij (Uij )(fi |Uij ) = fj |Uij for all i, j ∈ I. For V ∈ Ω and V ⊂ U , deﬁne rVF U : F → G by rVF U ((fi )) = (rVFii,Ui (fi )) where Vi = Ui ∩V . Then we verify easily that F is a sheaf on X. Finally, deﬁne the morphisms εi : F |Xi → Fi by εi (U )((fj )) = fi . Then a straightforward veriﬁcation proves the proposition.

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9.5 Ringed space 9.5.1 Lemma. Let B be a base of open subsets of X. For U ∈ B, let L(U ) be a set of maps from U to E. Further, suppose that the following conditions are veriﬁed: (i) If U, V ∈ B are such that V ⊂ U , then f |V ∈ L(V ) for all f ∈ L(U ). (ii) If (Ui )i∈I is a family of open subsets in B whose union U is in B, and f : U → E is a map verifying f |Ui ∈ L(Ui ) for all i, then f ∈ L(U ). Then there exists a unique sheaf F on X such that F(U ) = L(U ) for all U ∈ B. Proof. Any U ∈ Ω is the union of V ∈ B such that V ⊂ U . Let M(U ) be the set of maps f : U → E verifying f |V ∈ L(V ) for all V ∈ B contained in U . It is then clear that if F exists, then F(U ) = M(U ) for all U ∈ Ω. So we only need to show that M is a sheaf. Clearly, M is a presheaf. Let (Ui )i∈I be a family of open subsets of X whose union is U and the elements fi ∈ M(Ui ) be such that fi |Ui ∩Uj = fj |Ui ∩Uj for all i, j ∈ I. Then there exists a unique map f : U → E such that f |Ui = fi for all i ∈ I. Now if V ∈ B is contained in U , then for any W ∈ B contained in V ∩ Ui , f |W ∈ L(W ). Thus f |V ∩Ui ∈ M(V ∩ Ui ). Since V is the union of the set of W ∈ B such that W ⊂ V ∩ Ui for some i ∈ I, we obtain by (ii) that f |V ∈ L(V ). We have therefore proved that M is a sheaf. 9.5.2 Deﬁnition. (i) A Cartan space relative to E is a pair (X, F) where X is a topological space and F a sheaf of functions on X with values in E. (ii) A morphism of Cartan spaces from (X, F) to (Y, G) is a continuous map u : X → Y such that for all V ∈ Ω Y and g ∈ G(V ), the map deﬁned by x → g ◦ u(x) belongs to F(u−1 (V )). 9.5.3 Proposition. Let (X, F) and (Y, G) be Cartan spaces, B and C bases of open subsets for X and Y respectively. The following conditions are equivalent for a map u : X → Y : (i) u is a morphism of Cartan spaces. (ii) For x ∈ X, V ∈ C ∩ Ωu(x) , there exists an open subset U ∈ B ∩ Ωx such that U ⊂ u−1 (V ) and for all g ∈ G(V ), the map (g ◦ u)|U ∈ F(U ). Proof. Obviously, (i) ⇒ (ii). Conversely, suppose that (ii) is veriﬁed. Then u is continuous. Let W ∈ Ω Y , g ∈ G(W ) and x ∈ u−1 (W ). There exist an open neighbourhood Ux ∈ B of x and an open neighbourhood Vu(x) ∈ C of u(x) such that: Ux ⊂ u−1 (W ) , u(Ux ) ⊂ Vu(x) ⊂ W and g ◦ u|Ux ∈ F(Ux ). Since the union of the Ux ’s, with x ∈ u−1 (W ), is u−1 (W ), we have that g ◦ u ∈ F(u−1 (W )).

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9.5.4 Let (X, F) be a Cartan space and Y ⊂ X, then we have seen in 9.3.8 that (Y, F |Y ) is again a Cartan space and the canonical injection j : Y → X is a morphism of Cartan spaces. If u : (X, F) → (Z, G) is a morphism of Cartan spaces, then u ◦ j is again a morphism of Cartan spaces. This is just the restriction of u to (Y, F |Y ). Now let u : (Z, G) → (X, F) be a morphism of Cartan spaces such that u(Z) ⊂ Y . Let us denote by ν : Z → Y the induced map of topological spaces. We claim that ν is a morphism of Cartan spaces. Let V ∈ Ω Y , f ∈ F|Y (V ) and U ∈ Ω X be such that V = U ∩Y . Since there is an open covering (Ui )i∈I of V in X and gi ∈ F(Ui ) such that f |V ∩Ui = gi |V ∩Ui , we can suppose (by replacing Ui by U ∩ Ui if necessary) that the unions of the Ui ∩ Y is equal to V . Now gi ◦ u ∈ G(u−1 (Ui )) = G(u−1 (Ui ∩ Y )). On the other hand, if x is an element of u−1 (Ui ) = u−1 (Ui ∩ Y ), then gi ◦ u(x) = f ◦ u(x). Since G is a sheaf, we obtain that f ◦ u ∈ G(u−1 (V )). So we have proved our claim. 9.5.5 Let (X, F) be a Cartan space and (Yi )i∈I an open covering of X. Suppose that for each i ∈ I, we have a morphism of Cartan spaces ui : (Yi , F |Yi ) → (Z, G) and the ui ’s verify ui |Yi ∩Yj = uj |Yi ∩Yj for all i, j ∈ I. then there exists a unique map u : X → Z such that u|Yi = ui for all i ∈ I. Now if Bi is a base of open subsets of Yi , then their union is a base of open subsets of X. It follows from the deﬁnitions and 9.5.3 that u is a morphism of Cartan spaces. 9.5.6 As a consequence of the preceding remarks, we have: Proposition. Let (X, F), (Y, G) be Cartan spaces, u : X → Y a map, (Vi )i∈I an open covering of Y and for each i ∈ I, (Uij )j∈J is a family of open subsets of X whose union is u−1 (Vi ). For (i, j) ∈ I × J, let uij : Uij → Vi be the map induced by u. Then u is a morphism of Cartan spaces if and only if for all (i, j) ∈ I × J, the map uij is a morphism of Cartan spaces from (Uij , F |Uij ) to (Vi , G|Vi ). 9.5.7 Let us suppose that E is a commutative ring. A Cartan space (X, F) is a ringed space relative to E if for all U ∈ Ω, F(U ) is a subring of the ring of E-valued functions on U . In particular, F is a sheaf of commutative rings on X. Let (X, F), (Y, G) be ringed spaces and u a morphism from (X, F) to (Y, G) as Cartan spaces. Then for any V ∈ Ω Y , the map G(V ) → F(u−1 (V )), g → g ◦ u, is a ring homomorphism. 9.5.8 Remark. Let us suppose that E is a commutative ring and B a base of open subsets of X. Then a Cartan space (X, F) is a ringed space if and only if F(U ) is a subring of the ring of E-valued functions on U for all U ∈ Ω X (9.5.1).

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References and comments • [5], [26], [34], [37]. For our purposes, we have restricted ourselves to sheaf of functions on a topological space. For a more thorough treatment of the theory of sheaves, the reader may refer to [15], [34] and [37].

10 Jordan decomposition and some basic results on groups

The patient reader will soon be rewarded. After this chapter, we shall begin to study algebraic geometry, Lie algebras and algebraic groups. A large part of this chapter covers basic material on the theory of abstract groups. In this chapter, k will denote an algebraically closed (commutative) ﬁeld.

10.1 Jordan decomposition 10.1.1 Let V be a ﬁnite-dimensional k-vector space. We shall denote by End(V ) (resp. GL(V )) the set of endomorphisms (resp. invertible endomorphisms) of V . An element x ∈ End(V ) is said to be: • semisimple if V has a basis consisting of eigenvectors for x. • nilpotent if xn = 0 for some n ∈ N∗ . • unipotent if x − idV is nilpotent. 10.1.2 Theorem. Let x ∈ End(V ). There exists a unique pair (xs , xn ) of endomorphisms of V such that: (i) xs is semisimple, xn is nilpotent and x = xs + xn . (ii) xs ◦ xn = xn ◦ xs . Furthermore, there exist P, Q ∈ k[T ] such that P (0) = Q(0) = 0 and P (x) = xs , Q(x) = xn . We say that x = xs + xn is the additive Jordan decomposition of x and xs (resp. xn ) is the semisimple (resp. nilpotent) component of x. Proof. Let χx (T ) = (T −λ1 )m1 · · · (T −λr )mr be the characteristic polynomial of x where λ1 , . . . , λr are pairwise distinct eigenvalues of x. Let ai be the ideal of k[T ] generated by (T − λi )mi and Vi = ker(x − λi idV )mi . The Vi ’s are x-stable and V = V1 ⊕ · · · ⊕ Vr . By 2.1.2, there exists P ∈ k[T ] such that P (T ) − λi ∈ ai for all 1 i r and P (T ) ∈ a0 = T k[T ] (this last condition is redundant if 0 ∈ {λ1 , . . . , λr }). Set xs = P (x). Then xs (vi ) = λi vi for vi ∈ Vi . So xs is semisimple. Now let xn = x − xs = Q(x) where Q(T ) = T − P (T ). Then xn is nilpotent and the pair (xs , xn ) veriﬁes conditions (i) and (ii).

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If (ys , yn ) is another such pair, then ys , yn commute with x, and hence they commute with xs , xn . It follows that xs − ys = yn − xn is both semisimple and nilpotent. Thus xs = ys and xn = yn . 10.1.3 Corollary. (i) Let x, y ∈ End(V ) be such that x ◦ y = y ◦ x. Then y commutes with xs and xn . Further, (x + y)s = xs + ys and (x + y)n = xn + yn . (ii) Let x ∈ End(V ) and W a x-stable subspace of V . Then W is xs -stable, xn -stable and x|W = xs |W + xn |W is the Jordan decomposition of x|W . (iii) Let x, W be as in (ii) and z, zs , zn be endomorphisms of V /W induced by x, xs , xn . Then z = zs + zn is the Jordan decomposition of z. 10.1.4 Theorem. Let x ∈ GL(V ). There exists a unique pair (xs , xu ) of elements of GL(V ) such that: (i) xs is semisimple, xu is unipotent and x = xs ◦ xu . (ii) xs ◦ xu = xu ◦ xs . We say that x = xs ◦ xu is the multiplicative Jordan decomposition of x and xu is the unipotent component of x. Proof. Let x = xs + xn be the additive Jordan decomposition of x. Since x ∈ GL(V ) and the eigenvalues of x and xs are identical, xs ∈ GL(V ). Hence −1 x = xs ◦(idV +x−1 s ◦xn ). Let xu = idV +xs ◦xn , then clearly (xs , xu ) satisﬁes conditions (i) and (ii). Finally, the proof of uniqueness is as in 10.1.2. 10.1.5 We have also analogues of the results of 10.1.3 for the multiplicative Jordan decomposition. 10.1.6 Let E be a k-vector space (not necessarily ﬁnite-dimensional). If x ∈ End(E) and v ∈ E, we set Ex (v) to be the subspace generated by xn (v), n ∈ N. This is clearly a x-stable subspace of E. Proposition. Let x ∈ End(E). The following conditions are equivalent: (i) The space E is the union of a family of ﬁnite-dimensional x-stable subspaces. (ii) For any v ∈ E, Ex (v) is ﬁnite-dimensional. If these conditions are satisﬁed, we say that x is locally ﬁnite. Proof. Clearly E is the union of the Ex (v), v ∈ E, so (ii) ⇒ (i). Now given (i), any element v is contained in some ﬁnite-dimensional x-stable subspace V . Hence Ex (v) ⊂ V , and dim Ex (v) is ﬁnite. 10.1.7 Remark. Let x be a diagonalisable endomorphism of E and (vi )i∈I a basis of eigenvectors. Let v ∈ E, then there exists a ﬁnite subset J ⊂ I such that v belongs to VJ , the span of the vj ’s, j ∈ J. Since VJ is clearly x-stable, it follows that x is locally ﬁnite. 10.1.8 Proposition. Let x ∈ End(E). The following conditions are equivalent: (i) x is diagonalisable.

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(ii) x is locally ﬁnite, and the restriction of x to any ﬁnite-dimensional x-stable subspace is semisimple. If these conditions are satisﬁed, we say that x is semisimple. Proof. (i) ⇒ (ii) This is clear by 10.1.7 since any ﬁnite-dimensional x-stable subspace is contained in some VJ where J is a ﬁnite subset of I. (ii) ⇒ (i) Since x is locally ﬁnite, V is the union of a family (Vα )α∈A of ﬁnite-dimensional x-stable subspaces. The restriction of x to Vα is semisimple and therefore there exists a basis Bα of eigenvectors for Vα . The union B of the Bα ’s, α ∈ A, is a family of eigenvectors spanning V . We can therefore extract a basis, from B, of eigenvectors for x. 10.1.9 Proposition. Let x ∈ End(E) (resp. x ∈ GL(E)). The following conditions are equivalent: (i) E is the union of a family (Vα )α∈A of ﬁnite-dimensional x-stable subspaces such that x|Vα is nilpotent (resp. unipotent) for all α ∈ A. (ii) x is locally ﬁnite, and the restriction of x to any ﬁnite-dimensional x-stable subspace is nilpotent (resp. unipotent). If these conditions are satisﬁed, we say that x is locally nilpotent (resp. locally unipotent). Proof. Clearly, (ii) ⇒ (i). Conversely, suppose that (i) is veriﬁed. Let V be a ﬁnite-dimensional x-stable subspace and (v1 , . . . , vn ) be a basis of V . Then x|Ex (vi ) is nilpotent (resp. unipotent) for 1 i n. It follows clearly that x|V is nilpotent (resp. unipotent). 10.1.10 Theorem. Let x be a locally ﬁnite endomorphism of E. There exists a unique pair (xs , xn ) of elements of End(E) such that: (i) xs is semisimple, xn is locally nilpotent and x = xs + xn . (ii) xs ◦ xn = xn ◦ xs . Any x-stable subspace is xs -stable and xn -stable. Moreover, if V is a ﬁnitedimensional x-stable subspace of E, then x|V = xs |V + xn |V is the Jordan decomposition (10.1.2). We shall call x = xs + xn the additive Jordan decomposition of x. Proof. First of all, the restriction of x to any ﬁnite-dimensional x-stable subspace has a Jordan decomposition. Further, given two ﬁnite-dimensional xstable subspaces V and W of E, by 10.1.3 (ii) and the uniqueness of the Jordan decomposition, we have (x|V )s |V ∩W = (x|W )s |V ∩W , (x|V )n |V ∩W = (x|W )n |V ∩W . Since x is locally ﬁnite, it follows that there exists a unique endomorphism xs (resp. xn ) of E such that xs |V = (x|V )s (resp. xn |V = (x|V )n ) for all ﬁnitedimensional x-stable subspace V of E. It is clear by construction that the pair (xs , xn ) satisﬁes conditions (i) and (ii), and that any x-stable subspace of E is xs -stable and xn -stable.

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10.1.11 We prove in a similar way the following: Theorem. Let x be a locally ﬁnite invertible endomorphism of E. There exists a unique pair (xs , xu ) of elements of GL(E) such that: (i) xs is semisimple, xu is locally unipotent and x = xs ◦ xu . (ii) xs ◦ xu = xu ◦ xs . Any x-stable subspace is xs -stable and xu -stable. Moreover, if V is a ﬁnitedimensional x-stable subspace of E, then x|V = xs |V ◦ xu |V is the Jordan decomposition (10.1.4). We shall call x = xs ◦ xu the multiplicative Jordan decomposition of x. 10.1.12 Lemma. Let E, F be vector spaces, x ∈ End(E), y ∈ End(F ) and z : E → F a surjective linear map. Suppose further that x is locally ﬁnite and z ◦ x = y ◦ z. Then y is locally ﬁnite. Furthermore: (i) We have z ◦ xs = ys ◦ z and z ◦ xn = yn ◦ z. (ii) If x, y are automorphisms, then z ◦ xs = ys ◦ z and z ◦ xu = yu ◦ z. Proof. Since z ◦ R(x) = R(y) ◦ z for any R ∈ k[T ], it is clear from the surjectivity of z that y is locally ﬁnite and we can assume that E and F are ﬁnite-dimensional. Let P, Q ∈ k[T ] be such that xs = P (x) and xn = Q(x). Let (ei )i∈I be a basis of E consisting of eigenvectors of xs . Then (z(ei ))i∈I is a family of eigenvectors of P (y) and it follows from the surjectivity of z that we can extract from this family a basis of F . Hence P (y) is semisimple. Finally Q(y) is clearly nilpotent and y ◦ z = (P (y) + Q(y)) ◦ z, again the surjectivity of z implies that y = P (y) + Q(y). It is now clear that ys = P (y) and yn = Q(y). The proof of part (ii) is similar. 10.1.13 Let E, F be vector spaces, x ∈ End(E), y ∈ End(F ). We denote by x⊗y the endomorphism of E ⊗k F deﬁned as follows: for v ∈ E and w ∈ F : x ⊗ y(v ⊗ w) = x(v) ⊗ y(w). Lemma. Let x, y be locally ﬁnite automorphisms of E and F respectively. Then x ⊗ y is a locally ﬁnite automorphism of E ⊗k F and (x ⊗ y)s = xs ⊗ ys , (x ⊗ y)u = xu ⊗ yu . Proof. It is clear that x ⊗ y is locally ﬁnite and x ⊗ y = (xs ⊗ ys ) ◦ (xu ⊗ yu ) = (xu ⊗ yu ) ◦ (xs ⊗ ys ). Let (ei )i∈I (resp. (fj )j∈J ) be a basis of E (resp. F ) consisting of eigenvectors of xs (resp. ys ). Then clearly (ei ⊗ fj )(i,j)∈I×J is a basis of E ⊗k F consisting of eigenvectors of xs ⊗ ys . So xs ⊗ ys is semisimple. Finally, we have xu ⊗ yu − idE⊗k F = (xu − idE ) ⊗ yu + idE ⊗(yu − idF ). Since (xu − idE ) ⊗ yu commutes with idE ⊗(yu − idF ), it follows that xu ⊗ yu is locally unipotent.

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10.1.14 Proposition. Let x be a locally ﬁnite endomorphism of a kalgebra A. (i) If x is a k-algebra automorphism, then so are xs and xu . (ii) If x is a derivation of A, then so are xs and xn . Proof. Let B be the image of the linear map q : A ⊗k A → A given by the multiplication of the k-algebra A. (i) If x is a k-algebra automorphism, then q ◦ (x ⊗ x) = x ◦ q. By 10.1.13, x ⊗ x is locally ﬁnite and since q is surjective onto B, we have: q ◦ (xs ⊗ xs ) = q ◦ (x ⊗ x)s = xs ◦ q , q ◦ (xu ⊗ xu ) = q ◦ (x ⊗ x)u = xu ◦ q by 10.1.12 and 10.1.13. Finally, if A has an identity element e, then x(e) = e implies that the line spanned by e is x-stable, hence it is also xs - and xu -stable. Now, xu is unipotent, so xu (e) = e and it follows that xs (e) = e. Hence, xs and xu are k-algebra automorphisms of A. (ii) If x is a derivation, then q ◦ (ι ⊗ x + x ⊗ ι) = x ◦ q where ι denotes the identity map on A. Now we see easily that ι ⊗ x + x ⊗ ι is locally ﬁnite and that (ι ⊗ x + x ⊗ ι)s = ι ⊗ xs + xs ⊗ ι and (ι ⊗ x + x ⊗ ι)n = ι ⊗ xn + xn ⊗ ι. It follows from 10.1.12 that: q ◦ (ι ⊗ xs + xs ⊗ ι) = xs ◦ q , q ◦ (ι ⊗ xn + xn ⊗ ι) = xn ◦ q. Consequently, xs and xn are derivations of A.

10.2 Generalities on groups 10.2.1 In general, we shall present group laws multiplicatively. Let G be a group. Denote by µG : G × G → G, (α, β) → αβ the group multiplication and ιG : G → G, α → α−1 , the inverse. When there is no confusion, we shall simply write µ, ι for µG , ιG . Let U , V be subsets of G, then we set U V = {αβ ; α ∈ U , β ∈ V }. and U −1 = {α−1 ; α ∈ U }. Given a subgroup H of G, recall that NG (H) = {α ∈ G; αHα−1 = H} is a subgroup of G called the normalizer of H in G. If K ⊂ NG (H), then we say that K normalizes H. If NG (H) = G, then H is a normal subgroup of G. Similarly CG (H) = {α ∈ G; αβ = βα for all β ∈ H} is a subgroup of G called the centralizer of H in G. If K ⊂ CG (H), then we say that K centralizes H. 10.2.2 Let us denote the group of automorphisms of G by Aut(G). A subgroup H of G is called characteristic if σ(H) = H for all σ ∈ Aut(G). Given α ∈ G, the map iα : G → G given by β → αβα−1 , is an automorphism of G called an inner automorphism of G. The set Int(G) of inner automorphisms of G is a subgroup of Aut(G). Note that a subgroup of G is normal if and only if it is stable under all inner automorphisms of G.

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10.2.3 Let H be a subgroup of G. We deﬁne the quotient G/H (resp. H\G) to be the set of equivalence classes, called left (resp. right) cosets, of the equivalence relation Rl (resp. Rr ) deﬁned by αRl β ⇔ α−1 β ∈ H (resp. αRr β ⇔ βα−1 ∈ H). Note that a left (resp. right) coset is of the form αH (resp. Hα) for some α ∈ G. It follows that if H is normal, then G/H = H\G and we have an induced group structure on G/H such that the projection G → G/H is a group homomorphism. In general, we have a bijection between G/H and H\G given by αH → Hα−1 . We deﬁne the index of H in G to be the cardinal, denoted by (G : H), of the set G/H (or H\G). Proposition. Let H ⊃ K be subgroups of G. (i) We have card(G) = (G : H) card(H). (ii) We have (G : K) = (G : H)(H : K). 10.2.4 Let H, K be groups and σ : K → Aut(H) a group homomorphism. We deﬁne a law on H × K by: (α, β)(α , β ) = (ασ(β)(α ), ββ ) for α, α ∈ H, β, β ∈ K. We check easily that there is a group law on H × K, denoted by H σ K or simply H K, that we call the semi-direct product of K by H with respect to σ. If eH , eK denote the identity elements of H and K respectively, then (eH , eK ) is the identity element of H K and (α, β)−1 = (σ(β −1 )(α−1 ), β −1 ). Proposition. The maps iH : H → H σ K, α → (α, eK ), iK : K → H σ K, β → (eH , β) and pK : H σ K → K, (α, β) → β are group homomorphisms. Further we have the following exact sequence of group homomorphisms i

pK

H H σ K −→ K −→ 1. 1 −→ H −→

10.2.5 Suppose that H, K are subgroups of G such that K normalizes H and K ∩ H = {eG }. Let σ : K → Aut(H), be the group homomorphism given by σ(β)(α) = βαβ −1 . Then we have: Proposition. The map (α, β) → αβ is an isomorphism of H σ K onto the subgroup HK of G.

10.3 Commutators 10.3.1 Let α, β ∈ G. We deﬁne the commutator of α and β to be (α, β) = αβα−1 β −1 . Clearly, (α, β)−1 = (β, α) and γ(α, β)γ −1 = (γαγ −1 , γβγ −1 ) for γ ∈ G. If H, K are subgroups of G, we shall denote by (H, K) the subgroup generated by the commutators (α, β) where α ∈ H, β ∈ K. The group (H, K) is normal (resp. characteristic) if H and K are normal (resp. characteristic). In

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particular, the subgroup (G, G) is a characteristic subgroup of G called the derived subgroup or the group of commutators. We shall denote (G, G) also by D(G). It is easy to see that: Proposition. A subgroup H of G contains D(G) if and only if G/H is abelian and H is normal. 10.3.2 Proposition. If the index of the centre Z of G is ﬁnite, then D(G) is ﬁnite. Proof. Let n = (G : Z) and a1 , . . . , an be representatives of G/Z. Since (ap , aq ) = (ap z, aq z ) for any z, z ∈ Z, the set C of commutators is ﬁnite and card(C) n2 . Let c1 , . . . , cr be the (distinct) elements of C. Since (G : Z) = n, (α, β)n ∈ Z for all α, β ∈ G. This implies that (α, β)n = −1 β (α, β)n β = (β −1 αβ, β)n , and hence: (α, β)n+1 = (β −1 αβ, β)n (α, β) = (β −1 αβ, β)n−1 (β −1 αβ, β)(α, β) = (β −1 αβ, β)n−1 (β −1 αβ, β 2 ). Thus (α, β)n+1 can be written as a product of n commutators. −1 On the other hand, ci cj = cj (c−1 j ci cj ) and cj ci cj = ck for some k. Now let x = ci1 · · · cis ∈ D(G), 1 i1 , . . . , is r, be such that s is minimal mr 1 for this property. Then the preceding paragraph says that x = cm 1 · · · cr n+1 where m1 + · · · + mr = s. Since ci is a product of n commutators, it follows from the minimality of s that 0 m1 , . . . , mr n. Thus card(D(G)) 2 (n + 1)r (n + 1)n . 10.3.3 Proposition. Let H, K be subgroups of G such that H normalizes K. If the set S = {(α, β); α ∈ H, β ∈ K} is ﬁnite, then so is (H, K). Proof. We can assume that G = HK and so K is normal in G. For β, β1 ∈ H and γ, γ1 ∈ K, we have (γ1 β1 ββ1−1 γ1−1 , γ) = (γ1 , β1 ββ1−1 )(β1 ββ1−1 , γγ1 ). Thus the set T of commutators (αβα−1 , γ), where α ∈ G, β ∈ H and γ ∈ K, is ﬁnite and is contained in (H, K). Moreover, T is stable under inner automorphisms. Consequently, (H, K) is normal in G since S ⊂ T . Since an inner automorphism deﬁnes a permutation of the set T , this induces a homomorphism from G to the group of permutations of T whose kernel L is exactly CG ((H, K)). This implies that L is normal and (G : L) is ﬁnite. Now L ∩ (H, K) is central of ﬁnite index in (H, K), so by 10.3.2, D((H, K)) is ﬁnite. Since D((H, K)) is normal in G, we can replace G by G/D((H, K)) and assume that (H, K) is abelian. The set U = {(α, β); α ∈ H, β ∈ (H, K)} ⊂ S is ﬁnite and generates the subgroup (H, (H, K)). Given α ∈ H, β ∈ (H, K), we have αβα−1 ∈ (H, K) and

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(α, β)2 = (α, β)(αβα−1 )β −1 = (αβα−1 )(α, β)β −1 = (α, β 2 ) ∈ U. Since (H, (H, K)) is abelian and generated by U , it is ﬁnite. Further, (H, (H, K)) is normal in G, therefore we can replace G by G/(H, (H, K)) and assume that H centralizes (H, K). Given α ∈ H, β ∈ K, it follows that (α, β)2 = (α, β)α−1 (α, β)α = α−1 (α2 , β)α = (α2 , β) ∈ S. Since S generates (H, K) and (H, K) is abelian, we obtain that (H, K) is ﬁnite.

10.4 Solvable groups 10.4.1 Deﬁnition. The derived series of G is the sequence (Dn (G))n0 of subgroups of G deﬁned by D0 (G) = G and Dn (G) = (Dn−1 (G), Dn−1 (G)) for n 1. A group G is solvable if there exists an integer n such that Dn (G) = {eG }. 10.4.2 Proposition. (i) Let f : G → H be a homomorphism of groups. We have f (Dn (G)) ⊂ Dn (H) for all n ∈ N. Furthermore, if f is surjective, then f (Dn (G)) = Dn (H). (ii) For any n ∈ N, Dn (G) is a characteristic subgroup of G. Proof. Part (ii) is clear and part (i) is a simple induction. 10.4.3 Deﬁnition. A sequence G = G0 ⊃ G1 ⊃ · · · ⊃ Gm of subgroups of G is called a normal chain if Gi+1 is normal in Gi for 0 i m − 1. 10.4.4 Proposition. The following conditions are equivalent for a group G: (i) G is solvable. (ii) There is a sequence G = G0 ⊃ G1 ⊃ · · · ⊃ Gm = {eG } of normal subgroups of G such that Gi /Gi+1 is abelian for 0 i m − 1. (iii) There is a normal chain G = G0 ⊃ G1 ⊃ · · · ⊃ Gm = {eG } of subgroups of G such that Gi /Gi+1 is abelian for 0 i m − 1. Proof. For (i) ⇒ (ii), take Gi = Di (G), while (ii) ⇒ (iii) is trivial. Finally, given (iii), D(Gi ) ⊂ Gi+1 . By induction, we obtain Di (G) ⊂ Gi for all i. 10.4.5 Proposition. Let H be a normal subgroup of G. Then G is solvable if and only if H and G/H are solvable. Proof. Clearly if G is solvable, H and G/H are solvable. Conversely suppose that H and G/H are solvable. Since the canonical surjection π : G → G/H is a homomorphism of groups, by 10.4.2 (i) there exists p ∈ N such that π(Dp (G)) = {eG/H }, i.e. Dp (G) ⊂ H. Now there exists q ∈ N such that Dp+q (G) ⊂ Dq (H) = {eG }. Thus G is solvable.

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10.4.6 Corollary. Let H, K be normal subgroups of G. If H, K are solvable, then HK is also solvable. Proof. Since HK/H is isomorphic to K/(H ∩ K), it is solvable. Now apply 10.4.5 to HK and H.

10.5 Nilpotent groups 10.5.1 Deﬁnition. The central descending series of G is the sequence (C n (G))n1 of subgroups of G deﬁned by C 1 (G) = G and for n 1, C n+1 (G) = (G, C n (G)). A group G is nilpotent if there exists an integer n such that C n (G) = {eG }. 10.5.2 Remarks. 1) An abelian group is clearly nilpotent. 2) We verify easily that (C p (G), C q (G)) ⊂ C p+q (G), p, q ∈ N∗ . Thus n n D (G) ⊂ C 2 (G), n ∈ N∗ . So a nilpotent group is solvable. 3) Let G be nilpotent and n be minimal such that C n+1 (G) = {eG }. Then n C (G) is contained in the centre of G. In particular, the centre of G is nontrivial if G is non-trivial. 4) Let f : G → H be a homomorphism of groups. As in 10.4.2, we have by induction that f (C n (G)) ⊂ C n (H) for all n ∈ N∗ , and equality holds if f is surjective. It follows that C n (G) is a characteristic subgroup of G for n ∈ N∗ . Further, the quotient of a nilpotent group is nilpotent. 10.5.3 Proposition. Let H be a subgroup of G contained in the centre of G. Then G is nilpotent if and only if G/H is nilpotent. Proof. We have already remarked (10.5.2) that the quotient of a nilpotent group is nilpotent. Conversely, if G/H is nilpotent, then by 10.5.2 there exists n ∈ N∗ such that C n (G) ⊂ H ⊂ Z(G). Thus C n+1 (G) = {eG }. 10.5.4 Proposition. A group G is nilpotent if and only if there exists a sequence G = G1 ⊃ G2 ⊃ · · · ⊃ Gn+1 = {eG } of subgroups of G such that (G, Gk ) ⊂ Gk+1 for 1 k n. Proof. If G is nilpotent, then take Gk = C k (G). Conversely, we verify easily by induction that C k (G) ⊂ Gk for all k. 10.5.5 Proposition. Let G be a nilpotent group, H a subgroup and K a normal subgroup of G. (i) There exists a normal chain G = H1 ⊃ H2 ⊃ · · · ⊃ Hn+1 = H of subgroups of G such that Hk /Hk+1 is abelian for 1 k n. (ii) There exists a sequence K = K1 ⊃ K2 ⊃ · · · ⊃ Km+1 = {eG } of subgroups of G such that (G, Ki ) ⊂ Ki+1 for 1 i m. Proof. Set Gk = C k (G) and n minimal such that Gn+1 = {eG }. (i) Let Hk = Gk H = HGk since Gk is normal in G. If α ∈ H, β ∈ Gk , γ ∈ Gk+1 , then

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αHk+1 α−1 = (αHα−1 )(αGk+1 α−1 ) ⊂ Hk+1 βαγβ −1 = (β, αγ)αγ ∈ (Gk , G)HGk+1 ⊂ Gk+1 HGk+1 = Hk+1 . Thus Hk+1 is normal in Hk . Finally, Gk /Gk+1 is abelian and the canonical homomorphism Gk /Gk+1 → Hk /Hk+1 is surjective. So Hk /Hk+1 is abelian. (ii) It suﬃces to take Ki = K ∩ Gi . 10.5.6 Corollary. Let G be a nilpotent group. (i) Let H = G be a subgroup of G, then NG (H) = H and there is a normal subgroup L containing H, L = G and such that G/L is abelian. (ii) Let K = {eG } be a normal subgroup of G. Then K ∩ Z(G) = {eG }. Proof. (i) Let (Hk ) be a sequence of subgroups verifying condition (i) of 10.5.5. If k is maximal such that Hk = H, then Hk ⊂ NG (H) and H = NG (H). If i is minimal such that Hi = G, then L = Hi is a normal subgroup of G containing H and G/L is abelian as required. (ii) Let (Ki ) be a sequence of subgroups verifying 10.5.5 (ii). If j is maximal such that Kj = {eG }, then K ∩ Z(G) ⊃ Kj ∩ Z(G) = Kj = {eG }.

10.6 Group actions 10.6.1 We shall denote the group of permutations of a set E by S(E). If we have a group homomorphism θ from G to S(E), then we say that G acts on E. For simplicity, we shall often write α.x for θ(α)(x) if α ∈ G and x ∈ E. The set G.x = {α.x; α ∈ G} is called the G-orbit of x, and Gx = {α ∈ G; α.x = x} the stabilizer (or isotropy group) of x. We verify easily that Gx is a subgroup of G. 10.6.2 1) Let G acts on E, x, y ∈ E and α ∈ G be such that α.x = y. Then Gy = αGx α−1 . 2) We shall denote by E G the set of ﬁxed points of E. Thus x ∈ E G if and only if Gx = G if and only if G.x = {x}. 3) Let k ∈ N∗ . An action of G on E is called k-transitive if given x1 , . . . , xk , y1 , . . . , yk where xi = xj and yi = yj if i = j, there exists α ∈ G such that α.xi = yi , 1 i k. When k = 1, we say that G acts transitively on E and that E is a G-homogeneous space. 4) Let E , E be subsets of E. We deﬁne the transporter TranG (E , E ) of E to E to be the set of α ∈ G such that α(E ) ⊂ E . 5) Let G acts on E and F . A map f : E → F is called G-equivariant if f (α.x) = α.f (x) for all x ∈ E and α ∈ G. 10.6.3 Theorem. Let G acts on E and x ∈ E. There is a bijection between G.x and G/Gx . Thus card(G.x) = (G : Gx ). Proof. The map f : G → G.x deﬁned by α → α.x is surjective and we have f −1 (α.x) = αGx .

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10.6.4 Examples. 1) Left multiplication is an action of G on G called the left regular action. The right regular action is given by α.β = βα−1 . 2) The homomorphism G → Aut(G) induced by inner automorphisms is an action called conjugation. The stabilizer of α is just the centralizer CG (α). Two elements α, β are conjugate in G if they are in the same G-orbit. 3) Let H be a subgroup of G. Then G acts naturally on G/H via left multiplication: α.(βH) = αβH. 10.6.5 Let V be a k-vector space, ϕ : G → GL(V ) a group homomorphism. Then G acts on V and V is called a G-module and (V, ϕ) (or ϕ) a representation of G in V . Let (V, ϕ) be a representation of G and V ∗ be the dual of V . Then there is a natural action of G on V ∗ as follows: let f ∈ V ∗ , v ∈ V and α ∈ G, then (α.f )(v) = f (α−1 .v). We call this the contragredient or dual representation of ϕ. If (W, ψ) is another representation of G, then Homk (V, W ) also has a natural G-module structure as follows: let f ∈ Homk (V, W ), v ∈ V , α ∈ G, then (α.f )(v) = ψ(α)f (ϕ(α−1 )(v)). A homomorphism f ∈ Homk (V, W ) is called a G-module homomorphism (or G-homomorphism) if ψ(α)f (v) = f (ϕ(α)(v)) for all α ∈ G. The set of homomorphisms of G-modules will be denoted by HomG (V, W ). We shall also write EndG (V ) for HomG (V, V ). Clearly, HomG (V, W ) = Homk (V, W )G . We can also put a representation on the tensor product V ⊗k W as follows: let v ∈ V , w ∈ W , α ∈ G, then α.(v ⊗ w) = (ϕ(α)(v)) ⊗ (ψ(α)(w)). We can extend this to the n-th tensor product Tn (V ), and consequently to n n V , the n-th alternating product. S (V ), the n-th symmetric product and

10.7 Generalities on representations 10.7.1 Deﬁnition. A G-module E is simple if E = {0} and the only submodules of E are {0} and E. 10.7.2 Proposition. (Schur’s Lemma) (i) If E, F are non-isomorphic simple G-modules, then HomG (E, F ) = {0}. (ii) If E is a simple G-module, then EndG (E) is a ﬁeld. Furthermore, if E is ﬁnite-dimensional, then EndG (E) = k idE . Proof. Since the image and the kernel of a homomorphism of G-modules are G-modules, a homomorphism of simple G-modules is either zero or an isomorphism. So part (i) follows and EndG (E) is a ﬁeld which contains clearly k idE .

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Now, if E is ﬁnite-dimensional, then any u ∈ EndG (E) has an eigenvalue λ (k being algebraically closed). This implies that ker(u − λ idE ) is a non-trivial submodule of E. Hence u = λ idE . 10.7.3 Lemma. Let E be the sum of a family (Ei )i∈I of simple G-modules, and F a submodule of E. There exists J ⊂ I such that E = F ⊕ i∈J Ei . Proof. Let F be the set of subsets K of I such that the sum F + i∈K Ei is direct. Together with the partial order given by inclusion, F is an inductive system. Let J be a maximal element in F. Let V = F ⊕ i∈J Ei . For any k ∈ I, the sum Ek + V is not direct, so Ek ∩ V = {0} and therefore Ek ⊂ V since Ek is simple. Hence E = V . 10.7.4 Proposition. The following conditions are equivalent for a Gmodule E: (i) E is a sum of simple submodules. (ii) E is a direct sum of simple submodules. (iii) Any submodule F of E is a direct summand, i.e. E = F ⊕ F for some submodule F . If these conditions are satisﬁed, we say that E is semisimple or completely reducible. Proof. We may assume that E = {0}. (i) ⇒ (ii) Suppose that E is the sum of a family (Ei )i∈I of simple submodules. Let F be the set (partially ordered by inclusion) of subsets J of I inductive system If K is such that the sum j∈J Ej is direct. Then F is an a maximal element of F, then clearly we have E = k∈K Ek . (ii) ⇒ (iii) This is obvious by 10.7.3. (iii) ⇒ (i) Let F be a non trivial submodule of E, x ∈ F \ {0} and Vx the submodule of F generated by x. Let F be the set (partially ordered by inclusion) of submodules of Vx which are distinct from Vx . Then it is easy to check that F is an inductive system. If M is a maximal element of F, then Vx /M is clearly a simple module. Let M be a submodule of E such that E = M ⊕ M . Then Vx is the direct sum of M and M ∩ Vx because M ⊂ Vx . Since the submodule M ∩ Vx is isomorphic to Vx /M , it is simple. We have therefore proved that F contains a simple submodule. Now let S be the sum of all simple submodules of E and T a submodule such that E = S ⊕ T . So T has no simple submodule. If T = {0}, then the preceding paragraph says that T contains a simple submodule. So S ∩T = {0} which is absurd. Thus E = S. 10.7.5 Proposition. Let E be the sum of a family (Ei )i∈I of simple Gmodules. (i) Any submodule (resp. quotient) F of E is semisimple. Furthermore, there exists J ⊂ I such that F is isomorphic to i∈J Ei . (ii) Any simple submodule of E is isomorphic to some Ei , i ∈ I.

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Proof. By 10.7.4, there exists a submodule F of E such that E = F ⊕ F . Then by 10.7.3, there exists J ⊂ I such that E = F ⊕ ( i∈J Ei ). Since F is isomorphic to E/F , we have part (i). Part (ii) a just a special case of (i).

10.7.6 Proposition. Any representation of a ﬁnite group is semisimple. Proof. Let (E, ρ) be a G-module, F a submodule of E and p ∈ End(E) a projector with respect to F and n = card(G). Set: q=

1 ρ(α)−1 ◦ p ◦ ρ(α). n α∈G

We have q(E) ⊂ F and q(x) = x for x ∈ F . Now for β ∈ G, ρ(β) ◦ q =

1 ρ(αβ −1 )−1 ◦ p ◦ ρ(αβ −1 ) ◦ ρ(β) = q ◦ ρ(β). n α∈G

Thus q is a G-module homomorphism and E = F ⊕ ker q. 10.7.7 Let V be a k-vector space of dimension n.

d Lemma. Let W be a subspace of V of dimension d, y ∈ W \ {0} and d d V such that V = ky ⊕ U . Then V = W ⊕ W where a subspace U ⊂ d−1 W }. W = {x ∈ V ; x ∧ z ∈ U for all z ∈ d−1 d Proof. Let {u1 , . . . , ud } be a basis of W and ϕ ∈ ( V )∗ be such that ϕ(y) = 1 and ϕ|U = 0. Then W = {x ∈ V ; ϕ(x ∧ ui ) = 0 for 1 i d}. It follows that dim W n − d. So it suﬃces to prove that W ∩ W = {0}. Let x ∈ W \{0} and {x, x2 , . . . , xd } be a basis of W . There exists λ ∈ k\{0} d−1 W , we such that x ∧ (x2 ∧ · · · ∧ xd ) = λy ∈ U . Since x2 ∧ · · · ∧ xd ∈ deduce that x ∈ W . 10.7.8 Lemma. Let G be a subgroup of GL(V ) such that V is a semisimple G-module. Let χ : G → k ∗ be a homomorphism of groups, U = {x ∈ V ; α(x) = χ(α)x for all α ∈ G}, and W the subspace generated by α(x)−χ(α)(x), where α ∈ G and x ∈ V . (i) The subspaces U and W are G-submodules. Furthermore, U (resp. W ) is the unique G-submodule such that V = U ⊕ W . (ii) Let H be a subgroup of GL(V ) containing G as a normal subgroup. If χ extends to a homomorphism of H into k ∗ , then U and W are H-stable. Proof. (i) The subspace U is clearly G-stable. On the other hand, β(α(x) − χ(α)(x)) = (βαβ −1 )(β(x)) − χ(βαβ −1 )(β(x)) ∈ W, so W is also G-stable. Now V is semisimple, there exist submodules U , W such that V = U ⊕ U = W ⊕ W .

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If x ∈ W , then α(x) − χ(α)x ∈ W ∩ W = {0}. Hence W ⊂ U . On the other hand, if x ∈ V , there exist y ∈ U , z ∈ U such that x = y +z. For α ∈ G, α(x) − χ(α)(x) = α(z) − χ(α)(z) ∈ U . Hence W ⊂ U . Since U ∩ W = {0}, part (i) follows. (ii) The proof is analogue to the one for (i). 10.7.9 Let V be a semisimple G-module and W the subspace of V generated by α.x − x where α ∈ G, x ∈ V . By 10.7.8, W is a direct summand of V and V G is the unique submodule such that V = V G ⊕ W . The projection of V onto V G corresponding to this direct sum decomposition, denoted by pV , is called the Reynolds operator associated to V .

10.8 Examples 10.8.1 For n, p ∈ N∗ , we denote by Mn,p (k) the space of n by p matrices with coeﬃcients in k, and we write Mn (k) = Mn,n (k). Let In denote the identity matrix and (Eij )1i,jn the canonical basis of Mn (k). Denote by GLn (k), SLn (k), Dn (k), Tn (k) and Un (k) respectively the group of invertible matrices, the group of invertible matrices of determinant 1, the group of invertible diagonal matrices, the group of invertible upper triangular matrices and the group of upper triangular unipotent matrices. We shall denote by diag(λ1 , . . . , λn ) the matrix λ1 E11 + · · · + λn Enn . 10.8.2 It is clear that any matrix A ∈ GLn (k) can be written uniquely as a product diag(1, . . . , 1, det A)B where B ∈ SLn (k). It follows that GLn (k) is isomorphic to the semidirect product SLk (n)H where H = {diag(1, . . . , 1, λ); λ ∈ k∗ } k∗ . Further, we can verify easily that Tn (k) = Un (k) Dn (k). 10.8.3 Let S1 = {1}, and for n 2, let Sn = {Bij (λ) = In + λEij ; 1 i = j n , λ ∈ k} ⊂ SLn (k). A straightforward computation gives: for i = j, Bij (λ)−1 = Bij (−λ) , Bij (λ)Bij (µ) = Bij (λ + µ). 10.8.4 Proposition. Let λ ∈ k ∗ and A ∈ Mn (k). If A commutes with Bij (λ) for all i = j, then A ∈ kIn . Proof. If A commutes with the Bij (λ)’s, then AEij = Eij A for all i = j. Thus, a1i E1j + · · · + ani Enj = aj1 Ei1 + · · · + ajn Ein . So aii = ajj and aki = 0 = ajl for k = i and l = j.

10.8 Examples

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10.8.5 Corollary. (i) The centre of GLn (k) is k ∗ In . (ii) The centre of SLn (k) is k ∗ In ∩ SLn (k) = {µIn ; µn = 1}. 10.8.6 Theorem. (i) The group SLn (k) is generated by Sn . (ii) The group Un (k) is generated by Sn ∩ Un (k). Proof. We shall prove (i) by induction on n. The result is obvious for n = 1. Let us suppose that n 2, and let A = [aij ] ∈ SLn (k). If a1j = 0 for all j 2, then a11 = 0. The matrix A = AB12 (1) = [aij ] veriﬁes a12 = 0. So we can assume that there exists j 2 such that a1j = 0. 1 − a11 ) = (aij ) veriﬁes a11 = 1. Hence Now A = AB1p ( a1p 1 0 Bn1 (−an1 ) · · · B21 (−a21 )A B12 (−a12 ) · · · B1n (−a1n ) = 0C where C ∈ SLn−1 (k). By induction, A is a product of elements of Sn (recall that Bij (λ)−1 = Bij (−λ)). (ii) The proof of part (i) applies since given A = [aij ] ∈ Un (k), a11 = 1 and aij = 0 if i > j. 10.8.7 Since k is algebraically closed, k \ {0, 1, −1} is non-empty. We have by straightforward computations the following result. Lemma. Let λ ∈ k. (i) If n 3 and i, j, k ∈ {1, . . . , n} are distinct, then Bij (λ) = (Bik (λ), Bkj (1)). (ii) Let n = 2, µ ∈ k \ {0, −1, 1} and ν = λ(µ2 − 1)−1 . Then, B12 (λ) = (diag(µ, µ−1 ), B12 (ν)) , B21 (λ) = (B21 (ν)−1 , diag(µ−1 , µ)). 10.8.8 Theorem. The derived subgroup of GLn (k) is SLn (k) and SLn (k) is equal to its derived subgroup. Proof. By 10.8.7, any element of Sn is a commutator of elements of SLn (k). Since Sn generates SLn (k) by 10.8.6, and D(SLn (k)) ⊂ D(GLn (k)) ⊂ SLn (k), the result follows. 10.8.9 Since (diag(µ1 , . . . , µn ), Bij (λ)) = Bij (λ−

λµi ) for i = j, we obtain µj

via 10.8.6 (i): Proposition. The derived subgroup of Tn (k) is Un (k). 10.8.10 Let T+ n (k) denote the subspace of strictly upper triangular ma+ r trices. Then Un (k) = In + T+ n (k). For r 1, set Un,r (k) = In + (Tn (k)) .

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Clearly, Un,n (k) = {In }. Further, it is easy to verify that Un,r (k) are normal subgroups of Un (k) and for r, s ∈ N∗ , (Un,r (k), Un,s (k)) ⊂ Un,r+s (k). Hence, by 10.4.6, 10.5.4 and 10.8.2, we have that: Theorem. The group Un (k) (resp. Tn (k)) is nilpotent (resp. solvable). 10.8.11 Lemma. (Burnside’s Theorem) Let E be a ﬁnite-dimensional kvector space and A a subalgebra of End(E). If the only A-stable subspaces of E are {0} and E, then A = End(E). Proof. Let n = dim E. The result being obvious for n 1. We can assume that n 2. Note that by our hypothesis, Ax = E for all x ∈ E \ {0}. Let u ∈ A \ {0} be of minimal rank r. We claim that r = 1. If r > 1, there exist x, y ∈ E such that u(x) and u(y) are linearly independent. Since Au(x) = E, there exists v ∈ A such that v ◦ u(x) = y. Since u ◦ v ◦ u(x) and u(x) are linearly independent, u ◦ v ◦ u and u are linearly independent. Let F = u(E) which is u ◦ v-stable. Since k is algebraically closed, u ◦ v has an eigenvalue λ and an eigenvector z = u(t) for some t ∈ E \ ker u. So t ∈ ker(u ◦ v ◦ u − λu) and ker u is strictly contained in ker(u ◦ v ◦ u − λu). We deduce that u ◦ v ◦ u − λu is an element of A of rank strictly less than r. But u ◦ v ◦ u − λu = 0 because u ◦ v ◦ u and u are linearly independent. This contradicts the minimality of r. Thus r = 1 and we have proved our claim. Let u0 ∈ A and u ∈ End(E) be both of rank 1. There exist y0 , y ∈ E \ {0} and ϕ0 , ϕ ∈ E ∗ such that u0 (x) = ϕ0 (x)y0 and u(x) = ϕ(x)y for x ∈ E. Let v be the endomorphism of E deﬁned by v(x) = ϕ(x)y0 . The subspace L = {x ∈ E; ϕ0 (w(x)) = 0 for all w ∈ A} is A-stable and L ⊂ ker ϕ0 = E. It follows from our hypothesis that L = {0} and {ϕ0 ◦ w; w ∈ A} = E ∗ . This implies that there exists w ∈ A such that ϕ = ϕ0 ◦ w, and hence v = u0 ◦ w ∈ A. Since Ay0 = E, there exists s ∈ A such that y = s(y0 ). We obtain that u = s ◦ v ∈ A since v ∈ A. We have therefore proved that the algebra A contains all endomorphisms of E of rank 1. It is now clear that A = End(E). 10.8.12 Deﬁnition. Let E be a ﬁnite-dimensional k-vector space and G a subgroup of GL(E). We say that G is unipotent if all the elements of G are unipotent. 10.8.13 Theorem. Let E be a ﬁnite-dimensional k-vector space and G a unipotent subgroup of GL(E). There exists x ∈ E \ {0} such that α(x) = x for all α ∈ G. Proof. Let us proceed by induction on n = dim E. The case n = 1 is trivial. So let us suppose that n 2. Furthermore, if F is a G-stable subspace of E of dimension < n. We can apply the induction hypothesis on F . We can therefore assume that the only G-stable subspaces of E are {0} and E.

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By Burnside’s Theorem (10.8.11), the subalgebra generated by G is End(E). Now, given u, v ∈ G, tr((idE −u) ◦ v) = tr(v) − tr(u ◦ v) = n − n = 0. It follows that tr((idE −u) ◦ w) = 0 for all w ∈ End(E). Hence u = idE and so G = {idE }, which implies that n = 1. Contradiction. 10.8.14 Let E be a k-vector space of dimension n and for u ∈ End(E), we shall denote the matrix of u with respect to a basis E of E by Mat(u, E). Corollary. Let G be a unipotent subgroup of GL(E). (i) There exists a basis E of E such that Mat(u, E) ∈ Un (k) for all u ∈ G. (ii) The group G is nilpotent. Proof. By applying 10.8.13, part (i) is immediate by induction on n. Now 10.8.10 implies (ii).

References • [5], [8], [10], [23], [40], [52], [56], [78].

11 Algebraic sets

In this chapter, we deﬁne algebraic sets, regular functions on them and morphisms between them. These notions will be generalized in the next chapters. From now on, k denotes a commutative algebraically closed ﬁeld of characteristic zero. Let n ∈ N∗ .

11.1 Aﬃne algebraic sets 11.1.1 We call kn the aﬃne space of dimension n (aﬃne line if n = 1 and aﬃne plane if n = 2). Since k is algebraically closed, k is inﬁnite and therefore an element P ∈ k[T] = k[T1 , . . . , Tn ] can be identiﬁed as a k-valued polynomial function on kn sending t = (t1 , . . . tn ) to P (t) = P (t1 , . . . , tn ). 11.1.2 Deﬁnition. Let A ⊂ k[T]. Denote by V(A) the set of points t ∈ kn such that P (t) = 0 for all P ∈ A. A subset X of kn is called an aﬃne algebraic set if there exists A ⊂ k[T] such that X = V(A). 11.1.3 Let A ⊂ k[T]. We verify easily that if a is the ideal of k[T] generated by A, then V(A) = V(a). Since k[T] is Noetherian, a is ﬁnitely generated. Therefore, there exist P1 , . . . , Pn ∈ k[T] such that V(A) = V({P1 , . . . , Pn }) (denoted simply by V(P1 , . . . , Pn )). A subset of the form V(P ), where P ∈ k[T], is called a hypersurface of kn . 11.1.4 Proposition. Let a, b, (ai )i∈I be ideals of k[T]. Then: n = ∅, and if a ⊂ b, V(b) ⊂ V(a). (i) V(0) = k , V(k[T]) (ii) V( i∈I ai ) = i∈I V(ai ) and V(ab) = V(a ∩ b) = V(a) ∪ V(b). Proof. This is analogue to the proof of 6.5.3. 11.1.5 Let m ∈ Spm k[T]. By 6.4.3, there exists a = (a1 , . . . , an ) ∈ kn such that m = (T1 − a1 )k[T] + · · · + (Tn − an )k[T]. We deduce that V(m) = {a}. It follows from 11.1.4 that:

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Proposition. We have V(a) = ∅ if and only if a = k[T]. 11.1.6 For a subset M ⊂ kn , we denote by I(M ) the set of polynomials P ∈ k[T] such that P (a) = 0 for all a ∈ M . It is clear that I(M ) is a radical ideal of k[T]. We shall call I(M ), the ideal of M . Proposition. (i) We have I(∅) = k[T], and if M ⊂ N ⊂ kn , then I(M ) ⊃ I(N ). (ii) If M is an aﬃne algebraic set, then V(I(M )) = M . (iii) The map sending an aﬃne algebraic set√to its ideal is injective. (iv) If a is an ideal of k[T], then I(V(a)) = a. (v) We have I(kn ) = {0}. Proof. Part (i) is obvious. Clearly M ⊂ V(I(M )). Conversely, if M = V(a) for some ideal a ⊂ k[T], then a ⊂ I(M ). Hence V(I(M )) ⊂ V(a) = M (11.1.4). Now (iii) follows immediately from (ii). Since a ⊂ I(V(a)), part (iv) is clear for a = k[T]. Furthermore, for any m ∈ Spm k[T], V(m) = ∅ (11.1.5). It follows that m ⊂ V(m) = k[T]. Hence I(V(m)) = m. √ Now suppose that a = k[T]. Since I(V(a)) is radical, a ⊂ I(V(a)). Conversely, if m is a maximal√ideal containing a, then I(V(a)) ⊂ I(V(m)) = m intersection of maximal ideals containing by 11.1.4 and part (i). Since a is the √ a by 6.4.2, it follows that I(V(a)) ⊂ a. Finally, part (v) follows from (iv) and 11.1.4. 11.1.7 Corollary. The map a → V(a) induces a bijection between the set of radical ideals of k[T] and the set of aﬃne algebraic subsets of kn .

11.2 Zariski topology 11.2.1 By 11.1.4, we have a topology, called the Zariski topology, on kn where the closed subsets are aﬃne algebraic subsets of kn (or the V(a)’s by 11.1.7). In general, this topology is not Hausdorﬀ. Indeed, closed subsets V(a) of k are ﬁnite, thus two non-empty open subsets of k have non-empty intersection. Let a be a radical ideal of k[T] and V = V(a) an aﬃne algebraic subset of kn . The topology induced by the Zariski topology of kn shall also be called the Zariski topology on V . In view of 11.1.4, the closed subsets of V are the V(b)’s where b is an ideal of k[T] containing a. 11.2.2 Proposition. Let M be a subset of kn . The closure of M , with respect to the Zariski topology, is V(I(M )). Proof. Clearly, M ⊂ V(I(M )). Now if a is a radical ideal of k[T] such that M ⊂ V(a), then a = I(V(a)) ⊂ I(M ) by 11.1.6. Thus V(I(M )) ⊂ V(a). 11.2.3 Proposition. (i) Endowed with the Zariski topology, kn is a Noetherian space whose points are closed.

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(ii) Let a be a radical ideal of k[T]. Then V(a) is irreducible if and only if a is prime. Proof. The proof is analogous to the one for 6.5.6. 11.2.4 Corollary. For any ideal a of k[T], the set of irreducible components of V(a) is ﬁnite and it is the set of V(pi )’s where the pi ’s, 1 i r, are the minimal prime ideals of k[T] containing a. Proof. These are direct consequences of 1.3.5 and 11.2.3.

11.3 Regular functions 11.3.1 Deﬁnition. Let V be an aﬃne algebraic subset of kn . A function from V to k is a regular function if it is the restriction to V of a polynomial function on kn . 11.3.2 For any aﬃne algebraic subset V ⊂ kn , we denote by A(V ) the n algebra of regular functions on V . Note that √ A(k ) = k[T]. Let a = I(V ). Then V = V(a) and a = a (11.1.6). Thus the restrictions to V of P, Q ∈ k[T] are equal if and only if P − Q ∈ a. It follows that the algebras A(V ) and k[T]/a are isomorphic. In particular, A(V ) is ﬁnitely generated and reduced. 11.3.3 Proposition. Let V ⊂ kn be an aﬃne algebraic subset. (i) The algebra A(V ) separates the points of V , i.e. if a, b are distinct points of V , then there exists f ∈ A(V ) such that f (a) = f (b). (ii) Elements of A(V ) are continuous maps with respect to the Zariski topology. Proof. (i) Let a = (a1 , . . . , an ), b = (b1 , . . . , bn ) be distinct points in V . Without loss of generality, we can assume that a1 = b1 . The restriction to V of the coordinate function kn → k, (x1 , . . . , xn ) → x1 clearly separates a and b. (ii) For λ ∈ k and ϕ ∈ A(V ), we have ϕ−1 (λ) = V(I(V )) ∩ V(P − λ) where P is an element of k[T] whose restriction to V is ϕ. Thus ϕ−1 (λ) is closed. Since a closed subset of k is either k or a ﬁnite set, ϕ is continuous. 11.3.4 Let V ⊂ kn be an aﬃne algebraic subset. Recall from 11.3.2 that A(V ) is a reduced and ﬁnitely generated k-algebra. By 6.4.5, the map Homalg (A(V ), k) → Spm(A(V )) , χ → ker χ is a bijection. There is a canonical bijection from Spm(A(V )) to V . Namely, given m ∈ Spm(A(V )), there exists m ∈ Spm(k[T]) such that m ⊃ I(V ) and m /I(V ) = m. By 6.4.3, there exist a1 , . . . , an ∈ k such that

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m = (T1 − a1 )k[T] + · · · + (Tn − an )k[T]. Now m depends only on m and since a = (a1 , . . . , an ) ∈ V(m ), a ∈ V(I(V )) = V . So we have a well-deﬁned map from Spm(A(V )) to V . Conversely, given a = (a1 , . . . , an ) ∈ V , I(V ) ⊂ I({a}) = m where m denotes the maximal ideal of k[T] generated by the Ti − ai , 1 i n. Then a corresponds to the maximal ideal m = m /I(V ) of A(V ). Let us consider the composition of the above bijective maps: θ

ψ

V −→ Spm(A(V )) −→ Homalg (A(V ), k). We verify easily that for a ∈ V , ψ ◦ θ(a) is the homomorphism deﬁned by the evaluation at the point a, namely f → f (a), that we shall denote by χa .

11.4 Morphisms 11.4.1 Deﬁnition. Let V ⊂ kn and W ⊂ km be aﬃne algebraic subsets. A map u : V → W is called a morphism (of aﬃne algebraic sets) if for any g ∈ A(W ), g ◦ u ∈ A(V ). 11.4.2 We shall denote by Mor(V, W ) the set of morphisms from V to W . Clearly, the composition of morphisms is again a morphism. An isomorphism of V to W is a bijective morphism whose inverse is a morphism. The function k → k, t → t, is regular on k. It follows that A(V ) is the set of morphisms from V to k. For u ∈ Mor(V, W ), we shall denote by A(u) : A(W ) → A(V ) the map g → g ◦ u. We call A(u) the comorphism of u. If v ∈ Mor(W, Z), then we have A(v ◦ u) = A(u) ◦ A(v). 11.4.3 Proposition. Let V ⊂ kn and W ⊂ km be aﬃne algebraic subsets. For any map u : V → W , denote by ui : V → k, 1 i m, the coordinate maps of u. The following conditions are equivalent: (i) u ∈ Mor(V, W ). (ii) u1 , . . . , um ∈ A(V ). Proof. (i) ⇒ (ii) Let pi : km → k denote the i-th canonical projection. Then ui = pi ◦ u = A(u)(pi ) and since pi |W ∈ A(W ), we have (ii). (ii) ⇒ (i) Let P1 , . . . , Pm ∈ k[T] be such that the restriction of Pi to V is ui . Let g ∈ A(W ) and Q ∈ k[X1 , . . . , Xm ] be such that its restriction to W is g. For any t ∈ V , we have g ◦ u(t) = Q(P1 (t), . . . , Pm (t)). Thus g ◦ u is the restriction of a polynomial function on kn . Hence g ◦ u ∈ A(V ). 11.4.4 Theorem. Let V ⊂ kn , W ⊂ km be aﬃne algebraic subsets. The map u → A(u) is a bijection between Mor(V, W ) and Homalg (A(W ), A(V )). Proof. For ϕ ∈ Homalg (A(W ), A(V )), we deﬁne hϕ : Homalg (A(V ), k) → Homalg (A(W ), k) , χ → χ ◦ ϕ.

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There exists a unique map m(ϕ) : V → W such that the following diagram is commutative: V ⏐ ⏐

m(ϕ)

−−−−→

W ⏐ ⏐

Homalg (A(V ), k) −−−−→ Homalg (A(W ), k) hϕ

where the vertical maps are the isomorphisms x → χx and y → χy of 11.3.4. For x ∈ V , we have therefore: χm(ϕ)(x) = χx ◦ ϕ.

(1)

It follows that for x ∈ V and g ∈ A(W ), g(m(ϕ)(x)) = χm(ϕ)(x) (g) = χx ◦ ϕ(g) = χx (ϕ(g)) = ϕ(g)(x). We deduce that: g ◦ m(ϕ) = ϕ(g). In particular, g ◦ m(ϕ) ∈ A(V ) which shows that m(ϕ) ∈ Mor(V, W ). Furthermore, we have shown that A(m(ϕ)) = ϕ. Now given u ∈ Mor(V, W ), it is clear that A(u) ∈ Homalg (A(W ), A(V )). Taking ϕ = A(u) in the previous argument, we have: χm(A(u))(x) (g) = (A(u)(g))(x) = g(u(x)) = χu(x) (g). Hence m(A(u)) = u. We have therefore proved that the maps ϕ → m(ϕ) and u → A(u) are mutually inverse. 11.4.5 Corollary. Let u ∈ Mor(V, W ). Then u is an isomorphism if and only if the comorphism A(u) of u is an isomorphism of k-algebras. Proof. If u is an isomorphism and v ∈ Mor(W, V ) is such that u ◦ v = idW , v ◦ u = idV , then A(u) ◦ A(v) = idA(V ) and A(v) ◦ A(u) = idA(W ) . This implies that A(u) is an isomorphism of k-algebras. Conversely, if ϕ ∈ Homalg (A(V ), A(W )) veriﬁes A(u) ◦ ϕ = idA(V ) and ϕ ◦ A(u) = idA(W ) , then m(ϕ) ◦ u = idV and u ◦ m(ϕ) = idW . Hence u is an isomorphism.

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11.5 Examples of morphisms 11.5.1 It follows from 11.4.4 that Mor(kn , km ) is in bijection with the set of k-algebra homomorphisms ϕ : A(km ) → A(kn ). Let A(kn ) = k[T1 , . . . , Tn ] and A(km ) = k[X1 , . . . , Xm ]. Then ϕ is uniquely determined by m polynomials P1 , . . . , Pm ∈ k[T1 , . . . , Tn ] verifying: ϕ(Xi ) = Pi (T1 , . . . , Tn ) for 1 i m. By formula (1) of 11.4.4, the morphism u = m(ϕ) (hence A(u) = ϕ) is deﬁned by: χu(x) = χx ◦ ϕ for x = (x1 , . . . , xn ) ∈ kn . Let u(x) = (y1 , . . . , ym ). Then for 1 i m, yi = χu(x) (Xi ) = χx (ϕ(Xi )) = Pi (x1 , . . . , xn ). Thus (2)

u(x) = (P1 (x1 , . . . , xn ), . . . , Pm (x1 , . . . , xn )). 11.5.2 Let us suppose that n m in 11.5.1, then the map u : kn → km , (x1 , . . . , xn ) → (x1 , . . . , xn , 0, . . . , 0)

is a morphism. Its comorphism A(u) is given by: A(u)(Xi ) = Ti if 1 i n , A(u)(Xi ) = 0 if n + 1 i m. 11.5.3 Let us suppose that m n in 11.5.1, then the map v : kn → km , (x1 , . . . , xn ) → (x1 , . . . , xm ) is a morphism. For 1 i m, we have: A(v)(Xi ) = Ti . 11.5.4 Let V ⊂ kn be an aﬃne algebraic subset and a = I(V ) ⊂ A(kn ). By the deﬁnition of A(V ), the canonical injection j : V → kn is a morphism and A(j) is the canonical surjection A(kn ) → A(V ) = A(kn )/a. 11.5.5 Let u ∈ Mor(kn , km ) be given by formula (2) of 11.5.1. Denote by v : kn → kn × km = kn+m the map given by: x = (x1 , . . . , xn ) → (x, u(x)) = (x1 , . . . , xn , P1 (x), . . . , Pm (x)).

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137

The image G of v is called the graph of u. Note that y = (y1 , . . . , yn+m ) ∈ G if and only if y is a zero of the polynomials Tn+i − Pi (T1 , . . . , Tn ) for 1 i m. It follows that G is an aﬃne algebraic set. Further, if we consider v as a map from kn to G, then v ∈ Mor(kn , G) and v is bijective. Now let p : kn+m → kn be the canonical projection (11.5.3) and let j : G → kn+m be the canonical injection. Then by 11.5.3 and 11.5.4, p and j are morphisms (hence p ◦ j is a morphism). But v ◦ (p ◦ j) = idG , (p ◦ j) ◦ v = idkn . Consequently, v is an isomorphism between kn and G. 11.5.6 Again, let u, P1 , . . . , Pm be as in 11.5.1. The ideal I(u(kn )) is the set of polynomials Q ∈ k[X1 , . . . , Xm ] such that for all x ∈ kn , Q(P1 (x), . . . , Pm (x)) = 0. Since k is inﬁnite, I(u(kn )) is the set of polynomials Q ∈ k[X1 , . . . , Xm ] such that Q(P1 (T1 , . . . , Tn ), . . . , Pm (T1 , . . . , Tn )) = 0. The set W = V(I(u(kn ))) is the smallest aﬃne algebraic subset containing u(kn ). Hence W is the Zariski closure of u(kn ). Denote by v : kn → W the morphism induced by u, and j : W → km the canonical injection. We have u = j ◦ v, so A(u) = A(v) ◦ A(j) (by 11.5.4, A(j) is the canonical surjection A(kn ) → A(W )). On the other hand, A(u)(Xi ) = Pi for 1 i m (11.5.1). Let g ∈ A(W ) be such that A(v)(g) = 0. For all x ∈ kn , we have g(v(x)) = 0. But by construction, u(kn ) is dense in W , hence g = 0. We deduced therefore that A(W ) is isomorphic to the subalgebra of A(kn ) generated by P1 , . . . , Pm . Note that in general, u(kn ) = W (i.e. the image of a morphism is not necessarily an aﬃne algebraic set). For example, let u : k2 → k2 , (t1 , t2 ) → (t1 t2 , t2 ). We see easily that the only polynomial Q ∈ k[X1 , X2 ] such that Q(T1 T2 , T2 ) = 0 is zero. Thus the Zariski closure of u(k2 ) is k2 . But clearly, (1, 0) ∈ u(k2 ) and therefore u(k2 ) is not an aﬃne algebraic set. 11.5.7 Let us consider the following morphism u : k → k2 , t → (t2 , t3 ). The image C of u is contained in V(T13 − T22 ). Conversely, given (x1 , x2 ) ∈ V(T13 − T22 ), we have (x1 , x2 ) = u(0) if x1 = 0 and (x1 , x2 ) = u(x2 /x1 ) if x1 = 0. It follows that C = V(T13 −T22 ) and A(C) is

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isomorphic to the k-algebra k[T 2 , T 3 ] (11.5.6). Note however that this algebra, which is an integral domain, is not isomorphic to k[T ], since T ∈ k[T 2 , T 3 ] and T is integral over k[T 2 , T 3 ]. This is an example of a bijective morphism which is not an isomorphism.

11.6 Abstract algebraic sets 11.6.1 Proposition. Any ﬁnitely generated reduced k-algebra A is isomorphic to the algebra of regular functions over some aﬃne algebraic set. Proof. Let s1 , . . . , sn be generators for A. The surjection θ : k[T1 , . . . , Tn ] → A, Ti → si , induces an isomorphism between A and k[T1 , . . . , Tn ]/ ker θ. Since A is reduced, ker θ is a radical ideal. Hence I(V(ker θ)) = ker θ and A is isomorphic to A(V(ker θ)). 11.6.2 Remark. Let A be a ﬁnitely generated reduced k-algebra and V , W be aﬃne algebraic sets such that A is isomorphic to A(V ) and A(W ). Then by 11.4.5, V and W are isomorphic. 11.6.3 Let V ⊂ kn be an aﬃne algebraic set, a = I(V ) ⊂ A(kn ). To avoid confusion between the notations of 6.5.1 and 11.1.3, for b an ideal of A(V ), we shall denote in this section by V (b), the set of maximal ideals of A(V ) containing b. By 11.3.4, there is a canonical bijection: γV : V → Homalg (A(V ), k) , a → χa . On the other hand, we have another canonical bijection (6.4.5): δV : Homalg (A(V ), k) → Spm(A(V )) , χ → ker χ. Thus εV = δV ◦ γV is a bijection between V and Spm(A(V )). It is clear that if b is an ideal of A(kn ) containing a, then εV induces a bijection between V(b) and V (b/a). It follows that εV is a homeomorphism with respect to the Zariski topology (6.5.4 and 11.2.1). Let W ⊂ km be another aﬃne algebraic set. Then by 6.4.5 and 11.4.4, we have for ϕ ∈ Homalg (A(W ), A(V )), the following commutative diagram: V ⏐ ⏐ γV

m(ϕ)

−−−−→

W ⏐ ⏐γW

hϕ

Homalg (A(V ), k) −−−−→ Homalg (A(W ), k) ⏐ ⏐ ⏐δ ⏐ δV W Spm(A(V ))

Spm(ϕ)

−−−−−→

Spm(A(W ))

11.6 Abstract algebraic sets

139

11.6.4 The preceding discussion allows us to consider the category of aﬃne algebraic sets as a subcategory of a bigger (but equivalent) category, namely (abstract) algebraic sets which are not necessarily subsets of kn . The objects of this new category are the triples (X, A, βX ) where: (i) X is a set. (ii) A is a ﬁnitely generated reduced k-algebra. (iii) βX is a bijection from Spm(A) to X. Morphisms of (X, A, βX ) to (Y, B, βY ) are pairs (u, ϕ) where u : X → Y and ϕ ∈ Homalg (B, A) are such that the following diagram is commutative: Spm(ϕ)

Spm(A) −−−−−→ Spm(B) ⏐ ⏐ ⏐β ⏐ βX Y X

u

−−−−→

Y

The composition of morphisms (v, ψ) and (u, ϕ) is (v ◦ u, ϕ ◦ ψ). By abuse of notations, we shall write X and u for (X, A, βX ) and (u, ϕ); we shall also call ϕ the comorphism of u. 11.6.5 To identify the category of aﬃne algebraic sets as a subcategory of the category of algebraic sets, it suﬃces to identify an aﬃne algebraic set V with the triple (V, A(V ), ε−1 V ) (11.6.3), and a morphism u ∈ Mor(V, W ) with the pair (u, A(u)). By 11.4.4, 11.6.1 and 11.6.3, this category is equivalent to the category of aﬃne algebraic sets. Similarly, we can identify a ﬁnitely generated reduced k-algebra A with the triple (Spm(A), A, idSpm(A) ), and a homomorphism ϕ with the pair (Spm(ϕ), ϕ). Again this induces an equivalence between the category of ﬁnitely generated reduced k-algebras and the category of algebraic sets. We deﬁne regular functions on an algebraic set (X, A, βX ) via transport −1 (x) is a maximal ideal of A. For f ∈ A, we of structure. If x ∈ X, βX −1 set f (x) = f (βX (x)) as deﬁned in 6.4.6. Then regular functions on X are morphisms from X to k, and we can identify them as elements of A. For any morphism u : X → Y , we deﬁne A(u) : B → A as in 11.4.2, by g → g ◦ u. 11.6.6 Let (X, A, βX ) be an algebraic set. We deﬁne the Zariski topology on X to be the topology induced, via the bijection βX , by the Zariski topology on (Spm(A), A, idSpm(A) ) (6.5.4). In the case where A = A(V ) for some aﬃne algebraic set V , the Zariski topology of Spm(A) is the one obtained from the Zariski topology on V via the canonical bijection from Spm(A(V )) to V (11.6.3). 11.6.7 Let A be a ﬁnitely generated reduced k-algebra, a a radical ideal of A and βa the bijection Spm(A/a) → V(a) which is the inverse of the canonical bijection m → m/a. It is clear that (V(a), A/a, βa ) is an algebraic set, that we shall denote by V(a) or Spm(A/a). The Zariski topology on this set is induced by the one on Spm(A). The canonical injection V(a) → Spm(A) is a

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morphism, whose comorphism is A → A/a. Thus regular functions on V(a) can be viewed as restrictions of regular functions on Spm(A). 11.6.8 Proposition. Let A, B be ﬁnitely generated reduced k-algebras and ϕ ∈ Homalg (B, A). The following conditions are equivalent: (i) There exist a closed subset V ⊂ Spm(B) and an isomorphism v : Spm(A) → V such that Spm(ϕ) = j ◦ v where j : V → Spm(B) is the canonical injection. (ii) ϕ is surjective. Proof. (i) ⇒ (ii) Let b be a radical ideal of B such that V = V(b), ψ : B → B/b the canonical surjection and θ ∈ Homalg (B/b, A) such that ϕ = θ ◦ ψ. It follows that j = Spm(ψ), v = Spm(θ) and θ is an isomorphism by 11.4.5 and our hypotheses. Hence ϕ is surjective. (ii) ⇒ (i) Let b = ker ϕ, θ : B/b → A the induced isomorphism and ψ : B → B/b the canonical surjection. The ideal b is radical and by 11.4.5, Spm(θ) : Spm(A) → Spm(B/b) is an isomorphism.

11.7 Principal open subsets 11.7.1 Let A be a ﬁnitely generated reduced k-algebra and f ∈ A. We set D(f ) = Spm(A) \ V(Af ). Thus D(f ) is the set of maximal ideals m which do not contain f (or equivalently f (m) = 0). This is an open subset of Spm(A) and an open subset of this form is called a principal open subset of Spm(A). Note that: D(0) = ∅ , D(A) = Spm(A) , D(f g) = D(f ) ∩ D(g). If ϕ ∈ Homalg (B, A) and g ∈ B, then by 6.5.8, we have: Spm(ϕ)−1 (D(g)) = D(ϕ(g)). 11.7.2 Proposition. Principal open subsets of Spm(A) form a base for the Zariski topology on Spm(A). Proof. Let V be a closed subset of Spm(A). If x ∈ V , then there exists f ∈ I(V ) such that f (x) = 0. Thus x ∈ D(f ) and D(f ) ⊂ Spm(A) \ V . 11.7.3 Let A be a ﬁnitely generated reduced k-algebra and f ∈ A. Denote by Af the set of maps from D(f ) to k of the form (3)

x →

g(x) f (x)n

where g ∈ A and n ∈ N. It is clear that Af is a ﬁnitely generated k-algebra. Proposition. (i) An element g of Af is identically zero on D(f ) if and only if f g = 0. (ii) The algebra Af is reduced. (iii) Let S = {f n ; n ∈ N}. The algebras Af and S −1 A are isomorphic.

11.7 Principal open subsets

141

Proof. Part (i) is obvious. Now if (g(x)/f (x)n )m = 0 for all x ∈ D(f ). Then f g = 0 and g(x)/f (x)n = 0 for all x ∈ D(f ). Hence Af is reduced. Let ε : A → Af be the restriction map g → g|D(f ) . If g ∈ S, then ε(g) is invertible. By 2.3.4, there exists a unique homomorphism θ : S −1 A → Af such that θ ◦ i = ε where i : A → S −1 A is the canonical homomorphism. It is clear that θ is surjective. On the other hand, by part (i), θ is injective. Thus θ is an isomorphism. 11.7.4 We shall prove that a principal open subset D(f ) of Spm(A) can be considered canonically as an algebraic set whose topology is the one induced by the topology of Spm(A). Theorem. There exists a canonical bijection β : Spm(Af ) → D(f ) which is a homeomorphism when D(f ) is endowed with the topology induced from the Zariski topology of Spm(A). Proof. Let ε : A → Af , g → g|D(f ) , and γ = Spm(ε) : Spm(Af ) → Spm(A). Set F = ε(f ). If m ∈ Spm(Af ), then F ∈ m and f ∈ ε−1 (m). This implies that the image of γ is contained in D(f ). For m ∈ D(f ), set I(m) = {ε(g)/F n ; g ∈ m, n ∈ N}. The set I(m) is an ideal of Af . If I(m) = Af , then there exist g ∈ m and n ∈ N such that ε(g)/F n = 1. So g/f n = 1 or equivalently (g − f n )/f n = 0. So (g − f n )f = 0 by 11.7.3. Thus gf = f n+1 , hence f n+1 ∈ m and f ∈ m. Contradiction. So I(m) = Af . There exists n ∈ Spm(Af ) such that n ⊃ I(m). It follows that m ⊂ ε−1 (n), so m = ε−1 (n) and I(m) = n. Thus the image of γ is D(f ). Since I(γ(n)) = n for n ∈ Spm(Af ), γ is injective and therefore it induces a bijection β from Spm(Af ) onto D(f ). As γ is continuous (6.5.8), to obtain the result, it suﬃces to prove that β maps open sets to open sets. By 11.7.2, we are reduced to prove that γ(D(ε(g)/F n )) = γ(D(ε(g))) is open for all g ∈ A. Let n ∈ Spm(Af ) be such that ε(g) ∈ n. Then g ∈ ε−1 (n) and γ(D(ε(g))) is contained in D(f ) ∩ D(g). Conversely, let m ∈ Spm(A) be such that f ∈ m and g ∈ m. There exists n ∈ Spm(Af ) such that β(n) = m. Thus m = ε−1 (n). Now there exist h ∈ A, θ ∈ m such that 1 = hg + θ. So 1 = ε(h)ε(g) + ε(θ). Since ε(θ) ∈ n, ε(g) ∈ n. Thus γ(D(ε(g))) = D(f ) ∩ D(g), and we have obtained our result. 11.7.5 Remarks. 1) If A is an integral domain and f = 0, then Af can be identiﬁed with subring A[1/f ] of Fract(A) (since f g = 0 implies that g = 0, the homomorphism A[1/f ] → Af , g/f n → ε(g)/ε(f )n is injective). 2) Let j : D(f ) → Spm(A) be the canonical injection. The diagram

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11 Algebraic sets γ

Spm(Af ) −−−−→ Spm(A) ⏐ ⏐ ⏐id ⏐ β Spm(A) D(f )

j

−−−−→ Spm(A)

is commutative. Consequently, j is a morphism whose comorphism is ε. 3) Note that regular functions on D(f ) are not necessarily restrictions of regular functions on Spm(A). For example, let A = k[T ], then D(T ) = k \ {0} and x → 1/x is a regular function on D(f ) which is not the restriction of a polynomial. 4) The map δ : Mn (k) → k, M → det M is regular and D(δ) = GLn (k). Thus GLn (k) can be considered as an algebraic set. Left multiplication or right multiplication by A ∈ GLn (k), and the map sending M to M −1 are clearly isomorphisms of the algebraic set GLn (k).

11.8 Products of algebraic sets 11.8.1 Let V ⊂ km and W ⊂ kn be aﬃne algebraic sets. Let us identify A(km ) (resp. A(kn )) as the subring k[T1 , . . . , Tm ] (resp. k[Tm+1 , . . . , Tn+m ]) of k[T1 , . . . , Tn+m ]. Let a = I(V ), b = I(W ) and c the ideal in k[T1 , . . . , Tn+m ] generated by a ∪ b. The aﬃne algebraic set V(c) is deﬁned by the equations: P (x1 , . . . , xm ) = 0 , Q(xm+1 , . . . , xn+m ) = 0 for all P ∈ a and Q ∈ b. Thus √ V(c) = V × W , A(V × W ) = k[T1 , . . . , Tn+m ]/ c. Since we have the canonical isomorphisms A(V ) ⊗k A(W ) (k[T1 , . . . , Tm ]/a) ⊗k (k[Tm+1 , . . . , Tm+n ]/b), k[T1 , . . . , Tn+m ] k[T1 , . . . , Tm ] ⊗k k[Tm+1 , . . . , Tm+n ], the ideal c can be written as: c = a ⊗k k[Tm+1 , . . . , Tm+n ] + k[T1 , . . . , Tm ] ⊗k b. Hence A(V ) ⊗k A(W ) can be identiﬁed canonically with k[T1 , . . . , Tm+n ]/c. It follows that there is a canonical surjective homomorphism: ϕ : A(V ) ⊗k A(W ) → A(V × W ) where for f ∈ A(V ), g ∈ A(W ), (x, y) ∈ V × W , ϕ(f ⊗ g)(x, y) = f (x)g(y). Proposition. The homomorphism ϕ is an isomorphism.

11.8 Products of algebraic sets

143

Proof. It suﬃces to prove that ϕ is injective. Let h = f1 ⊗ g1 + · · · + fr ⊗ gr be a non-zero element in ker ϕ. We may assume that h is written with r minimal. It follows that the fi ’s are linearly independent. Now, given any y ∈ W , f1 (x)g1 (y) + · · · + fr (x)gr (y) = 0 for all x ∈ V . Since the fi ’s are linearly independent, gi (y) = 0 for all y ∈ W , i = 1, . . . , r. It follows that the gi ’s are zero, which implies that h = 0. Contradiction. 11.8.2 Corollary. Let A, B be ﬁnitely generated reduced k-algebras, then A ⊗k B is a ﬁnitely generated reduced k-algebra. Proof. It is clear by 11.6.1 and 11.8.1. 11.8.3 From now on, we shall identify A(V × W ) and A(V ) ⊗k A(W ) via ϕ. Denote by pr1 : V × W → V and pr2 : V × W → W the canonical projections. These are morphisms of aﬃne algebraic sets. For (x, y) ∈ V × W , f ∈ A(V ), we have that f ◦ pr1 (x, y) = f (x). It follows that the comorphisms of pr1 and pr2 are given by: A(pr1 ) : A(V ) → A(V ) ⊗k A(W ) , f → f ⊗ 1, A(pr2 ) : A(W ) → A(V ) ⊗k A(W ) , g → 1 ⊗ g. Proposition. Let U, V, W be aﬃne algebraic sets and v ∈ Mor(U, V ), w ∈ Mor(U, W ). The map u : U → V × W , z → (v(z), w(z)) is the unique morphism such that v = pr1 ◦ u and w = pr2 ◦ u. Proof. Clearly u is the unique map having the required properties. Now let f ∈ A(V ), g ∈ A(W ) and z ∈ U . We have: (f ⊗ g)(u(z)) = (f ⊗ g)(v(z), w(z)) = (f ◦ v(z))(g ◦ w(z)). Since f ◦ v, g ◦ w ∈ A(U ), u is a morphism.

11.8.4 Proposition. Let V, W, V , W be aﬃne algebraic sets. For any v ∈ Mor(V, V ) and w ∈ Mor(W, W ), the map u : V × W → V × W , (x, y) → (v(x), w(y)) is a morphism.

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11 Algebraic sets

Proof. Note that u(x, y) = (v ◦ pr1 (x, y), w ◦ pr2 (x, y)). So the result follows from 11.8.3. 11.8.5 Proposition. Let V, W be aﬃne algebraic sets. The canonical projections pr1 , pr2 of V × W onto V and W are open maps. Proof. It suﬃces to show that the image of a principal open subset is open. Let h = f1 ⊗ g1 + · · · + fn ⊗ gn ∈ A(V × W ), and x0 ∈ U = pr1 (D(h)). Then there exists y0 ∈ W such that (x0 , y0 ) ∈ D(h). Set u = g1 (y0 )f1 + · · · + gn (y0 )fn ∈ A(V ). Then x0 ∈ D(u) and D(u) × {y0 } ⊂ D(h). Hence D(u) ⊂ U and U is an open subset of V . The proof for pr2 is analogue. 11.8.6 Proposition. Let V, W be aﬃne algebraic sets. Then V × W is irreducible if and only if V and W are irreducible. Proof. Since the canonical projections are morphisms, V and W are irreducible if V × W is irreducible. Conversely, suppose that V and W are irreducible. The projections pr1 , pr2 of V × W onto V and W are continuous open maps (6.5.8 and 11.8.5). For x ∈ V , pr−1 1 (x) = {x} × W is clearly irreducible. It follows therefore from 1.1.7 that V × W is irreducible. 11.8.7 Corollary. Let A, B be ﬁnitely generated k-algebras. If A and B are integral domains, then A ⊗k B is also an integral domain. 11.8.8 Proposition. Let V, W be aﬃne algebraic sets, V1 (resp. W1 ) a closed subset of V (resp. W ). Then V1 × W1 is a closed subset of V × W . Proof. Let a ⊂ A(V ), b ⊂ A(W ) be ideals such that A(V1 ) = A(V )/a and A(W1 ) = A(W )/b. Then A(V1 × W1 ) = A(V1 ) ⊗k A(W1 ) = A(V × W )/c where c = a ⊗k A(W ) + A(V ) ⊗k b. Hence V1 × W1 = V(c).

11.8.9 Remarks. 1) The Zariski topology on V × W is in general strictly ﬁner than the product of the Zariski topologies on V and W . The open subsets D(f ⊗ g) = D(f ) × D(g), f ∈ A(V ) g ∈ A(W ), form a base with respect to the product topology. However, there exists Zariskiopen subsets of V × W which does not contain any non-empty subsets of this form. For example, if V = W = k, the non-empty subsets D(f ⊗ g) are the complements of ﬁnite unions of subsets of the form {x} × W and V × {y}. Let U = k2 \ V(T1 − T2 ). Since k is inﬁnite, none of the preceding subsets contain U. 2) With the notations of example 4 of 11.7.5, the multiplication GLn (k) × GLn (k) → GLn (k) , (M, N ) → M N is a morphism of algebraic sets.

11.8 Products of algebraic sets

145

References and comments • [5], [19], [22], [26], [37], [40], [78]. All the results in this chapter are classic and can be found in the books cited above.

12 Prevarieties and varieties

In this chapter, we deﬁne an algebraic variety via the theory of sheaves. Local behaviour of an algebraic variety is considered in the last section.

12.1 Structure sheaf 12.1.1 Let A be a ﬁnitely generated reduced k-algebra and X = Spm(A) be endowed with the Zariski topology. For f ∈ A, we shall conserve the notations Af , V(f ) and D(f ) of chapter 11. Theorem. There exists a unique sheaf of functions OX on X, with values in k such that for any f ∈ A, we have: Γ (D(f ), OX ) = Af . We call OX the structure sheaf of the algebraic set X. Proof. By 11.7.2, the D(f )’s, f ∈ A, form a base B for the topology on X. It suﬃces therefore to verify that B satisﬁes the hypotheses of 9.5.1. Let f, g ∈ A be such that D(f ) ⊂ D(g). Then V(g) ⊂ V(f ), hence by 11.1.6: Af = I(V(f )) ⊂ I(V(g)) = Ag. Thus there exist n ∈ N∗ and h ∈ A such that f n = gh. If u/g i ∈ Ag , then its restriction to D(f ) is (uhi )/(g i hi ) = ug i /f ni , which belongs to Af . It follows therefore that if D(f ) = D(g), then Af = Ag . Now let f and (fi )i∈I be non zero elements of A such that D(fi ) is a covering of D(f ). Then V(f ) is the intersection of the V(fi )’s, so V(f ) = V(a), where a is the ideal generated by the fi ’s. Since A is Noetherian, we may assume that I = {1, . . . , r}. Let g : D(f ) → k be a function such that g|D(fi ) ∈ Afi for 1 i r. There exist a1 , . . . , ar ∈ A and n ∈ N∗ such that g(x) = ai (x)/fi (x)n for all x ∈ D(fi ).

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12 Prevarieties and varieties

If x ∈ D(fi )∩D(fj ) = D(fi fj ), then ai (x)fj (x)n = aj (x)fi (x)n . By 11.7.3, it follows that fi fj (ai fjn − aj fin ) = 0. We have V(f1n , . . . , frn ) = V(f1 , . . . , fr ) ⊂ V(f ). So 11.1.6 implies that there exist m ∈ N∗ and b1 , . . . , br ∈ A such that f m = b1 f1n+1 + · · · + br frn+1 . Hence r r ai fi f m = ai bj fi fjn+1 = fin+1 aj bj fj . j=1

j=1

Set a=

r

aj bj fj .

j=1

For 1 i r, we have therefore ai fi f m = afin+1 which implies that if x ∈ D(fi ), a(x) ai (x) = . f (x)m fi (x)n Thus for x ∈ D(f ), g(x) =

a(x) , f (x)m

and g ∈ Af . 12.1.2 By 9.5.8 and 12.1.1, (X, OX ) is a ringed space. Note that OX determines A since Γ (X, OX ) = A. 12.1.3 Proposition. Let X = Spm(A), Y = Spm(B) where A, B are ﬁnitely generated reduced k-algebras. The set of morphisms of the algebraic set X to the algebraic set Y and the set of morphisms of the ringed space (X, OX ) to the ringed space (Y, OY ) are identical. Proof. Let u : (X, OX ) → (Y, OY ) be a morphism of ringed spaces. For any g ∈ B = Γ (Y, OY ), g ◦u ∈ A = Γ (X, OX ). Hence u is a morphism of algebraic sets. Conversely, let u : X → Y be a morphism of algebraic sets and A(u) its comorphism. For g ∈ B, we have u−1 (D(g)) = D(A(u)(g)) (11.7.1). On the other hand, for f = h/g n ∈ Γ (D(g), OY ) where h ∈ B, n ∈ N∗ , we have for x ∈ D(A(u)(g)), (f ◦ u)(x) =

(A(u)(h))(x) h(u(x)) = . g(u(x))n ((A(u)(g))(x))n

Thus f → f ◦ u is a homomorphism (of k-algebras) from Γ (D(g), OY ) = Bg to Γ (u−1 (D(g)), OX ) = AA(u)(g) . Hence by 9.5.3, u is a morphism of ringed spaces. 12.1.4 We can deﬁne by transport of structure, the structure sheaf of an (abstract) algebraic set (11.6.4). Proposition 12.1.3 is again valid in this setting.

12.2 Algebraic prevarieties

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Deﬁnition. An aﬃne algebraic variety over k, or simply an aﬃne variety is a ringed space (X, OX ) where X is an algebraic set. A morphism of aﬃne varieties is a morphism of the underlying ringed spaces.

12.2 Algebraic prevarieties 12.2.1 Deﬁnition. (i) An (algebraic) prevariety over k is a ringed space (X, OX ) relative to k verifying the following conditions: a) X is a Noetherian topological space. b) There is an open covering (Ui )i∈I of X such that each induced ringed space (Ui , OX |Ui ) is an aﬃne algebraic variety over k. (ii) An open subset U of an algebraic prevariety (X, OX ) is called an aﬃne open subset if the induced ringed space (U, OX |U ) is an aﬃne variety. Remarks. 1) An aﬃne variety is a prevariety. 2) When there is no confusion, we shall simply say that X is a prevariety, instead of (X, OX ). 3) A prevariety is quasicompact (1.3.3). Thus there exists a ﬁnite covering of X by open aﬃne subsets. 12.2.2 We deﬁne the category of algebraic prevarieties over k by letting morphisms of prevarieties to be morphisms of ringed spaces. By 9.5.3, we have the following result: Proposition. Let X, Y be prevarieties and u : X → Y a map. Then u is a morphism of prevarieties if and only if for all open aﬃne subset V of Y and x ∈ u−1 (V ), there exists an open aﬃne subset U of X containing x, and contained in u−1 (V ), such that the restriction of u to U is a morphism of algebraic sets from U to V . 12.2.3 It follows from the previous discussion that: Proposition. Let us conserve the notations of 12.2.2. The map u is an isomorphism of prevarieties if and only if u is bijective and for all x ∈ X, there exists an aﬃne open subset U of X containing x, such that u induces an isomorphism from U to an aﬃne open subset of Y . 12.2.4 Remark. When u satisﬁes the condition of 12.2.3 without being bijective, we say that u is a local isomorphism from X to Y . 12.2.5 Proposition. Let X be a prevariety. (i) The topological space X admits a base consisting of aﬃne open subsets. (ii) All points of X are closed. (iii) If U is open in X, then the ringed space (U, OX |U ) is a prevariety. Proof. Let (Ui )i∈I be a covering of X by aﬃne open subsets. (i) Each Ui has a base consisting of aﬃne open subsets (11.7.2 and 11.7.4). So X also has a base consisting of aﬃne open subsets. (ii) A subset F of X is closed if and only if F ∩ Ui is closed for all i ∈ I. Thus the assertion follows from 11.2.3.

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(iii) By 1.3.2, U is Noetherian. If B is a base for X consisting of aﬃne open subsets, then those open subsets in B contained in U form a base for the topology of U . The result now follows. 12.2.6 Proposition. Let (X, OX ) be a prevariety and Y a locally closed subset of X. Then the ringed space (Y, OX |Y ) is a prevariety. Proof. By 1.3.2, Y is Noetherian. We may further assume by 9.5.4 and 12.2.5 (iii) that Y is closed in X. Let (Ui )i∈I be a covering of X by aﬃne open subsets. To obtain the result, it suﬃces to prove that for all i ∈ I, the ringed space (Y ∩ Ui , OX |Y ∩Ui ) is an aﬃne variety. It follows that we may take X = Spm(A) to be an aﬃne variety and Y = V(a) where a is a radical ideal of A. Let us denote ϕ : A → A/a the canonical surjection and j = Spm(ϕ) : Y → X the canonical injection (11.5.4). Let V be an open subset of Y , and G(V ) be the set of maps h : V → k such that there exist an open subset U ⊂ X and g ∈ OX (U ) verifying V = Y ∩ U and g|V = h. The sheaf associated to the presheaf G is OX |Y (9.3.8). Let f ∈ A. We have: D(ϕ(f )) = j −1 (D(f )) = Y ∩ D(f ). It follows from 12.1.1 that OY (D(ϕ(f ))) = (A/a)ϕ(f ) where OY denotes the structure sheaf of Y . Thus if h ∈ OY (D(ϕ(f ))), then there exist g ∈ A and n ∈ N such that h(x) = g(x)/f (x)n , x ∈ Y ∩ D(f ). So the map y → g(y)/f (y)n on D(f ) belongs to OX (D(f )), consequently we obtained that OY (D(ϕ(f ))) ⊂ G(D(ϕ(f ))). Conversely, given h ∈ G(D(ϕ(f ))). There exist an open subset U ⊂ X, and g ∈ OX (U ) such that U ∩ Y = D(ϕ(f )) = Y ∩ D(f ) and g|Y ∩D(f ) = h. Cover U by principal open subsets Ui = D(gi ), gi ∈ A. Then the D(ϕ(gi ))’s cover D(ϕ(f )) and since g|Ui ∈ OX (Ui ) = Agi , ϕ(g)|D(ϕ(gi )) belongs to (A/a)ϕ(gi ) = OY (D(ϕ(gi ))). As OY is a sheaf, we deduce that ϕ(g)|D(ϕ(f )) ∈ OY (D(ϕ(f ))). Hence h ∈ OY (ϕ(f )). The D(ϕ(f ))’s form a base for the topology on Y . It follows therefore (using for example 9.3.6) that OY = OX |Y . 12.2.7 Let (X, OX ) be a prevariety and Y a locally closed subset of X. We call the prevariety (Y, OX |Y ) a sub-prevariety of (X, OX ). From now on, when we considered a locally closed subset of X as a prevariety, it will be as a sub-prevariety of X. 12.2.8 Let u : (X, OX ) → (Y, OY ) be a morphism of prevarieties, Z a locally closed subset of Y containing u(X). Then by 9.5.4, u induces a morphism of prevarieties from (X, OX ) to (Z, OY |Z ). 12.2.9 Let X, Y be prevarieties, Z the sum of the topological spaces X and Y . We identify X and Y as two disjoint open subsets of Z. By 9.5.1, we can deﬁne a structure of prevariety on Z by setting OZ (U ) = OX (U ) and

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OZ (V ) = OY (V ) for open subsets U ⊂ X and V ⊂ Y . This prevariety is called the sum of X and Y , and we shall denote it by X Y . If X and Y are aﬃne varieties, then so is X Y , and we have: A(X Y ) = A(X) × A(Y ) , X = V({0} × A(Y )) , Y = V(A(X) × {0}).

12.3 Morphisms of prevarieties 12.3.1 Let (X, OX ) be a prevariety. An element in Γ (X, OX ) is called a regular function on X. Let f be a regular function on X. If U is an aﬃne open subset of X, then the restriction of f to U is a regular function on U . By 12.2.2, we deduce that f is a morphism from X to the aﬃne line k. 12.3.2 Let X, Y be prevarieties and y0 ∈ Y . Then the map u : X → Y sending any x ∈ X to y0 is a morphism (this is clear by 11.4.3 since we may assume that X and Y are aﬃne varieties by 12.2.2). 12.3.3 Let u : (X, OX ) → (Y, OY ) be a morphism of prevarieties. Then we can associated to u the map Γ (u) : Γ (Y, OY ) → Γ (X, OX ) , g → g ◦ u which is a homomorphism of k-algebras. But contrary to 11.4.4, Γ (u) does not in general determine u. However, Proposition. Let X be a prevariety and Y an aﬃne variety. The map u → Γ (u) is a bijection between the set of morphisms of X to Y and the set of k-algebra homomorphisms from Γ (Y, OY ) to Γ (X, OX ). Proof. Let (Ui )i∈I be an covering of X by aﬃne open subsets. 1) Given u : X → Y a morphism of prevarieties and i ∈ I. The map Γ (Y, OY ) → Γ (Ui , OX ) , g → g ◦ u|Ui is a homomorphism of k-algebras. Clearly this is A(ui ) where ui : Ui → Y is the morphism induced by u. If v : X → Y is a morphism verifying g ◦ u = g ◦ v for all g ∈ Γ (Y, OY ), then ui = vi for all i ∈ I (11.4.4). Hence u = v since the Ui ’s cover X. 2) Let ϕ ∈ Homalg (Γ (Y, OY ), Γ (X, OX )) and ϕi the composition ϕ

Γ (Y, OY ) −−−−→ Γ (X, OX ) −−−−→ Γ (Ui , OX ) where the second map is the restriction homomorphism. By 11.4.4, we have ϕi = A(ui ), where ui ∈ Mor(Ui , Y ). To obtain ϕ = A(u), it suﬃces to show that ui |Ui ∩Uj = uj |Ui ∩Uj . Since aﬃne open subsets form a base for the topology on X, it suﬃces to show that ui |V = uj |V for any aﬃne open subset V ⊂ Ui ∩ Uj . But these two morphisms have the same comorphism

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Γ (Y, OY ) −−−−→ Γ (X, OX ) −−−−→ Γ (V, OX ) where the second map is the restriction homomorphism. So they are identical by 11.4.4. 12.3.4 Corollary. Let (X, OX ) be a prevariety, U an open subset of X and A = Γ (U, OX ). The following conditions are equivalent: (i) U is an aﬃne open subset. (ii) The algebra A is ﬁnitely generated and the morphism u : U → Spm(A), associated to the identity map of A (see 12.3.3), is an isomorphism. Proof. Clearly A is reduced. We have (i) ⇒ (ii) by 11.4.5. Conversely, given (ii), Spm(A) is an aﬃne variety. So by 12.3.3, the morphism u is unique such that Γ (u) = idA . Since u is an isomorphism, (i) follows. 12.3.5 Let us give an example of an open subset which is not aﬃne. Let X = k2 = Spm(k[T1 , T2 ]) and U = X \ {(0, 0)}. We claim that U is not aﬃne. Set U1 = D(T1 ) and U2 = D(T2 ), then U = U1 ∪ U2 . Let f ∈ Γ (U, OX ). Then g = f |U1 ∈ Γ (U1 , OX ), so by 12.1.1, it is of the form P (x1 , x2 ) g(x1 , x2 ) = xm 1 where P ∈ k[T1 , T2 ] and m ∈ N. We may assume that T1 does not divide P . If (x1 , x2 ) ∈ U1 , then xm 1 f (x1 , x2 ) − P (x1 , x2 ) = 0. Now h = T1m f − P deﬁnes a regular function on U . Thus h−1 (0) = F is a closed subset of U containing U1 . But U1 is dense in X (so in U ), we deduce that h is identically zero on U . In particular, m = 0 for otherwise P (0, x2 ) = 0 for all x2 = 0 which contradicts the assumption that T1 does not divide P . We have therefore proved that f is the restriction to U of a polynomial, which is unique since k is inﬁnite. It follows that Γ (U, OX ) = Γ (X, OX ). Now let ϕ : Γ (X, OX ) → Γ (U, OX ) be the restriction homomorphism which is the identity map from the preceding discussion. Let u : U → X be the morphism such that Γ (u) = ϕ (12.3.3). If x ∈ U and g ∈ Γ (X, OX ), we have g(x) = g(u(x)). Thus x = u(x) (11.3.3) and u is the canonical injection. So u is not bijective and 12.3.4 says that U is not aﬃne.

12.4 Products of prevarieties 12.4.1 Let X, Y be prevarieties over k, U ⊂ X and V ⊂ Y aﬃne open subsets. We have seen in 11.8 that U × V has a canonical structure of aﬃne variety for which the elements of A(U × V ) are functions of the form (x, y) → f1 (x)g1 (y) + · · · + fn (x)gn (y)

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where n ∈ N∗ , f1 , . . . , fn ∈ A(U ) and g1 , . . . , gn ∈ A(V ). For h ∈ A(U × V ), we set DU,V (h) = {z ∈ U × V ; h(z) = 0 } ⊂ X × Y. The DU,V (h)’s, h ∈ A(U × V ), form a base for the Zariski topology in U × V (11.7.2). Denote by C the set of DU,V (h)’s for U an open aﬃne subset of X, V an open aﬃne subset of Y and h ∈ A(U × V ). Let DU,V (h), DU ,V (h ) ∈ C with h(x, y) = α1 (x)β1 (y) + · · · + αm (x)βm (y) h (x , y ) = γ1 (x )δ1 (y ) + · · · + γn (x )δn (y ) for (x, y) ∈ U × V and (x , y ) ∈ U × V where α1 , . . . , αm ∈ A(U ), β1 , . . . , βm ∈ A(V ), γ1 , . . . , γn ∈ A(U ) and δ1 , . . . , δn ∈ A(V ). Let (x0 , y0 ) ∈ DU,V (h) ∩ DU ,V (h ). There exist aﬃne open subsets U ⊂ U ∩ U and V ⊂ V ∩ V such that (x0 , y0 ) ∈ U × V . Denote by ai , ci (resp. bi , di ) the restrictions to U (resp. V ) of αi , γi (resp. βi , δi ). Then ai , ci ∈ A(U ) and bi , di ∈ A(V ). Thus the maps deﬁned, for (x, y) ∈ U × V , by s(x, y) = a1 (x)b1 (y) + · · · + am (x)bm (y) t(x, y) = c1 (x)d1 (y) + · · · + cn (x)dn (y) belong to A(U × V ). On the other hand: (x0 , y0 ) ∈ DU ,V (st) , DU ,V (st) ⊂ DU,V (h) ∩ DU ,V (h ). It follows that C is a base for a topology T on X × Y . Let us endow X × Y with this topology. By construction, if U ⊂ X and V ⊂ Y are aﬃne open subsets, then T induces on U × V the Zariski topology on U × V . Aﬃne open subsets in the product of aﬃne open subsets U × V form a base for the Zariski topology on U × V . Consequently, the set B of aﬃne open subsets of a product U × V of aﬃne open subsets is a base for the topology T on X × Y . Since X (resp. Y ) is Noetherian, it is the union of a ﬁnite number of aﬃne open subsets U1 , . . . , Um (resp. V1 , . . . , Vn ). So X × Y is the ﬁnite union of the Noetherian subsets Ui × Vj . Hence X × Y is Noetherian (1.3.2). For W ∈ B, set F(W ) = A(W ). Suppose that F veriﬁes the conditions of 9.5.1. Then there exists a sheaf of functions OX×Y on X × Y such that Γ (W, OX×Y ) = F(W ) for all W ∈ B. The ringed space (X ×Y, OX×Y ) is therefore an algebraic prevariety over k that we shall call the product of the prevarieties X and Y . From now on, when we consider X × Y as a prevariety, it will be with respect to this structure. 12.4.2 Proposition. Let X, Y be prevarieties, X1 ⊂ X and Y1 ⊂ Y . If X1 and Y1 are open (resp. closed, locally closed) in X and Y , then X1 × Y1 is open (resp. closed, locally closed) in X × Y . Furthermore, the subvariety X1 × Y1 of X × Y is the product of the sub-prevarieties X1 and Y1 of X and Y.

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Proof. If X1 and Y1 are open, then the result is clear by the deﬁnition of the topology on X × Y . Next, a locally closed subset is open in its closure, so we are reduced to the case where X1 and Y1 are closed. We have to show that if U and V are aﬃne open subsets of X and Y , then (U × V ) ∩ (X1 × Y1 ) = (U ∩ X1 ) × (V ∩ Y1 ) is closed in U × V , and as an aﬃne variety, it is the product of U ∩ X1 and of V ∩ Y1 . But this is clear by 11.8.8. 12.4.3 Proposition. Let X, X , Y, Y , Z be prevarieties. (i) The canonical projections pr1 : X × Y → X and pr2 : X × Y → Y are morphisms. (ii) If u : Z → X and v : Z → Y are morphisms, then the map w : Z → X × Y , z → (u(z), v(z)) is also a morphism. (iii) If u : X → X and v : Y → Y are morphisms, then t : X × Y → X × Y , (x , y ) → (u(x ), v(y )) is also a morphism. (iv) If x0 ∈ X (resp. y0 ∈ Y ), then the map y → (x0 , y) (resp. x → (x, y0 )) is an isomorphism from Y (resp. X) to the closed sub-prevariety {x0 } × Y (resp. X × {y0 }) of X × Y . Proof. (i) Any point (x, y) ∈ X × Y has an open neighbourhood of the form U × V where U, V are aﬃne open subsets of X and Y . The restrictions of pr1 and pr2 are the canonical projections of this product. So the result is a consequence of 11.8.3. (ii) Let z ∈ Z. There exist an aﬃne open neighbourhood U (resp. V ) of u(z) (resp. v(z)) in X (resp. Y ), and an aﬃne open neighbourhood W of z in Z such that u(W ) ⊂ U and v(W ) ⊂ V . So (ii) follows from 11.8.3. (iii) The argument is analogue to the one used in 11.8.4. (iv) The map Y → {x0 } × Y , y → (x0 , y), is clearly bijective. Taking aﬃne open neighbourhoods, we are reduced to the case where X = Spm(A) and Y = Spm(B) are aﬃne varieties. Let m ∈ Spm(A) be such that x = V(m). Since A/m is isomorphic to k, the canonical homomorphism B → (A/m) ⊗k B is an isomorphism. Hence we have (iv). 12.4.4 Remark. Let X, Y, Z be prevarieties. By 12.4.3, the bijections X × Y → Y × X , (x, y) → (y, x), (X × Y ) × Z → X × (Y × Z) , ((x, y), z) → (x, (y, z)) are isomorphisms of prevarieties. We deduce in particular the deﬁnition of the product X1 × · · · × Xn for a ﬁnite number of prevarieties.

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12.4.5 Proposition. Let X and Y be prevarieties. (i) If X and Y are non-empty, then the topology on X (resp. on Y ) is the quotient topology of the one on X × Y with respect to the relation pr1 (z) = pr1 (z ) (resp. pr2 (z) = pr2 (z )). (ii) If X and Y are irreducible, then X × Y is also irreducible. Proof. (i) Let F ⊂ X be such that F × Y is closed in X × Y . If y ∈ Y , we have (F × Y ) ∩ (X × {y}) = F × {y}. So F × {y} is closed in X × {y}. By 12.4.3 (iv), F is closed in X. So part (i) follows. (ii) Let us write X = U1 ∪ · · · ∪ Um , Y = V1 ∪ · · · ∪ Vn where the Ui ’s and the Vj ’s are non-empty aﬃne open subsets. Then X × Y = Ui × Vj . i,j

Since X, Y are irreducible, (Ui × Vj ) ∩ (Uk × Vl ) = ∅ for 1 i, k m and 1 j, l n. By 1.1.6, it suﬃces to prove that Ui × Vj is irreducible. So we are reduced to the case where X and Y are aﬃne varieties, and the result follows from 11.8.6.

12.5 Algebraic varieties 12.5.1 The diagonal of a set X, denoted by ∆X , is the subset of X × X deﬁned by ∆X = {(x, x) ; x ∈ X}. Proposition. (i) Let X be an aﬃne variety, then ∆X is closed in X × X. (ii) Let X be a prevariety, then ∆X is locally closed in X × X. Proof. (i) Since X is aﬃne, it is closed in an aﬃne space kn . So X × X is closed in kn × kn (11.8.8). Now, ∆X = (X × X) ∩ ∆kn . So it suﬃces to show that ∆kn is closed in kn × kn . But this is clear since it is the aﬃne algebraic subset deﬁned by the polynomials xi − xn+i = 0 for 1 i n. (ii) Let z ∈ ∆X , there is an aﬃne open subset U ⊂ X such that U × U is an open neighbourhood of z in X × X. Now (ii) follows from (i) since ∆X ∩ (U × U ) = ∆U . 12.5.2 Corollary. Let X, Y be prevarieties (resp. aﬃne varieties) and u : X → Y a morphism. The graph Gu of u is locally closed (resp. closed) in X × Y . Furthermore, the graph morphism of u X → X × Y , x → (x, u(x)) is an isomorphism onto the sub-prevariety Gu of X × Y .

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Proof. Set v : X × Y → Y × Y , (x, y) → (u(x), y). By 12.4.3 (iii), v is a morphism and Gu = v −1 (∆Y ). So the ﬁrst part follows from 12.5.1 and the second by observing that pr1 |Gu is the inverse of the graph morphism. 12.5.3 Deﬁnition. An algebraic prevariety X over k is called an (algebraic) variety over k if ∆X is closed in X × X. 12.5.4 Proposition. (i) Any sub-prevariety of a variety is a variety. (ii) The product of two varieties is a variety. (iii) Let X be a prevariety, Y a variety and u : X → Y a morphism. Then the graph of u is closed in X × Y . (iv) Let X be a variety, n ∈ N∗ and ∆nX = {(x, . . . , x) ∈ X n ; x ∈ X }. Then ∆nX is closed in X n . Proof. Part (i) follows from the fact that if Y is a sub-prevariety of a variety X, then ∆Y = (Y × Y ) ∩ ∆X . Let X, Y be varieties, then we can identify ∆X×Y ⊂ (X × Y ) × (X × Y ) with ∆X × ∆Y ⊂ (X × X) × (Y × Y ) (12.4.4). So part (ii) follows from 12.4.2. Part (iii) is proved as in 12.5.2, and part (iv) follows from (ii) and (iii) since ∆nX is the graph of the morphism X → X n−1 , x → (x, . . . , x). 12.5.5 Remarks. 1) An aﬃne variety is a variety. 2) A sub-prevariety of a variety X shall be called a subvariety of X. 3) A variety is quasi-aﬃne if it is isomorphic to a subvariety of an aﬃne variety. We have seen in 12.3.5 that a quasi-aﬃne variety is not necessary aﬃne. 12.5.6 Theorem. Let X be a prevariety. The following conditions are equivalent: (i) X is a variety. (ii) For all morphisms u, v : Y → X of prevarieties, the set of y ∈ Y verifying u(y) = v(y) is closed in Y . (iii) For all aﬃne open subsets U, V of X, U ∩ V is an open aﬃne subset and the algebra A(U ∩ V ) is generated by the functions x → f (x)g(x), for x ∈ U ∩ V , where f ∈ A(U ) and g ∈ A(V ). Proof. (i) ⇒ (ii) Let Y , u, v as in (ii), set: w : Y → X × X , y → (u(y), v(y)). Then {y ∈ Y ; u(y) = v(y)} = w−1 (∆X ), and (ii) follows since X is a variety and w is a morphism (12.4.3). (ii) ⇒ (i) This is clear by taking Y = X × X, u = pr1 and v = pr2 . (iii) ⇔ (i) The subsets U ×V , where U, V are aﬃne open subsets of X, form an open covering of X ×X. So ∆X is closed if and only if ∆X ∩(U ×V ) is closed

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in U ×V for all aﬃne open subsets U, V of X. Now the map u : U ∩V → U ×V , x → (x, x), induces an isomorphism of the sub-prevariety U ∩ V onto the subprevariety ∆X ∩ (U ∩ V ) of the aﬃne variety U × V (12.5.2). If we have (i), then the open subset U ∩ V is aﬃne since (U × V ) ∩ ∆X is closed in U ∩ V . Conversely, suppose that the open subset U ∩ V is aﬃne. Denote by j the canonical injection from (U × V ) ∩ ∆X into U × V . Then w

j

U ∩ V −−−−→ (U × V ) ∩ ∆X −−−−→ U × V, where w denotes the isomorphism induced by u. By 11.6.8, (U × V ) ∩ ∆X is a closed subset of U × V if and only if the comorphism A(U ×V ) → A(U ∩V ) of u is surjective. Hence we have obtained the equivalence of (i) and (iii). 12.5.7 Remark. Let (Ui )i∈I be a covering of a prevariety X by aﬃne open subsets. If for all i, j ∈ I, the open subsets Ui and Uj verify condition (iii) of 12.5.6, then the preceding proof says that X is a variety. 12.5.8 Corollary. Let X be a variety, Y a prevariety, Z a subset of Y and u, v be morphisms from Y to X such that u(y) = v(y) for all y ∈ Z. Then u(y) = v(y) for all y ∈ Z. Proof. This is clear by 12.5.6 (ii). 12.5.9 Proposition. Let X be a prevariety such that any two points in X are contained in some aﬃne open subset. Then X is a variety. Proof. Let x, y be two distinct points of X and U an aﬃne open subset of X containing x and y. The set (U × U ) ∩ ∆X is closed in U × U and its complement Ω in U ×U is open in U ×U , so in X ×X as well. Since (x, y) ∈ Ω and Ω ∩ ∆X = ∅, ∆X is closed in X × X. 12.5.10 Proposition. Let u : X → Y be a morphism of prevarieties. Suppose that the following conditions are satisﬁed: (i) Y is a variety. (ii) There exists an open covering (Vi )i∈I of Y such that each of the open subsets Ui = u−1 (Vi ) is a variety. Then X is a variety. Proof. Since (Ui )i∈I is an open covering of X, it suﬃces to show that the set (Ui × Uj ) ∩ ∆X is closed in Ui × Uj . Let v : X × X → Y × Y , (x, x ) → (u(x), u(x )). Then (Ui × Uj ) ∩ ∆X = [v −1 ((Vi × Vj ) ∩ ∆Y )] ∩ [(Ui × Ui ) ∩ ∆X ]. By condition (i), (Vi × Vj ) ∩ ∆Y is closed in Vi × Vj . So v −1 ((Vi × Vj ) ∩ ∆Y ) is closed in Ui ∩ Uj . Now condition (ii) implies that (Ui × Ui ) ∩ ∆X is closed in Ui × Ui . Hence (Ui × Uj ) ∩ ∆X is closed in Ui × Uj as required.

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12.5.11 Proposition. Let X, Y be irreducible varieties. A morphism u : X → Y which is a local isomorphism is injective. Thus it is an isomorphism from X onto an open subvariety of Y . Proof. Let x, x ∈ X be distinct points such that u(x) = u(x ) = y. There exists, by our hypothesis, an open subset U (resp. U ) of X containing x (resp. x ) such that u|U (resp. u|U ) is an isomorphism from U (resp. U ) onto an open subset V (resp. V ) of Y . Let g, g be the inverse isomorphisms. Then g, g induce morphisms h, h from the open subset V ∩ V of Y to X. Since X is irreducible, U ∩ U is non-empty, and so u(U ∩ U ) is a non-empty open subset of Y contained in V ∩ V . Hence u(U ∩ U ) is dense in V ∩ V since Y is irreducible. We deduce that h and h are identical on V ∩ V which contradicts the fact that h(y) = h (y).

12.6 Gluing 12.6.1 Let X1 , . . . , Xn be prevarieties. Assume that for 1 i, j n, there exist open subsets Xij of Xi , Xji of Xj , and an isomorphism hji : Xij → Xji of prevarieties, such that a) Xii = Xi and hii = idXi for 1 i n. b) hij and hji are mutually inverse isomorphisms for 1 i, j n. c) For 1 i, j, k n, hji |Xij ∩Xik induces an isomorphism of Xij ∩ Xik onto Xji ∩ Xjk , and hki = hkj ◦ hji are identical on Xij ∩ Xik . 12.6.2 With the above assumption, there exist by 1.5.4 a topological space X, and for 1 i n, a homeomorphism ψi from Xi to an open subset ψi (Xi ) of X such that: 1) X = ψ1 (X1 ) ∪ · · · ∪ ψn (Xn ). 2) ψi (Xij ) = ψj (Xji ) = ψi (Xi ) ∩ ψj (Xj ) for 1 i, j n. 3) hji (x) = ψj−1 (ψi (x)) for 1 i, j n and x ∈ Xij . We shall endow X with a structure of prevariety. The sheaf OXi on Xi transports via ψi into a sheaf Fi on ψi (Xi ). Thus if U is an open subset of ψi (Xi ), Fi (U ) is the set of functions z → g(ψi−1 (z)), with g ∈ Γ (ψi−1 (U ), OXi ). We claim that there exists a unique sheaf OX on X such that OX (U ) = Fi (U ) if 1 i n and U an open subset of ψi (Xi ). For 1 i n, let Bi be a base of open subsets of ψi (Xi ). The union B of the Bi ’s is a base of open subsets of X. In view of 9.5.1, it suﬃces to prove that Fi and Fj are identical when restricted to ψi (Xi ) ∩ ψj (Xj ). Let W be an open subset of ψi (Xi ) ∩ ψj (Xj ). For x ∈ W , we have ψj−1 (x) = hji (ψi−1 (x)). But the elements of Fj (W ) are the functions g ◦ ψj−1 , with g ∈ Γ (ψj−1 (W ), OXj ). So, by the hypothesis, the functions g ◦ hji are in Γ (ψi−1 (W ), OXi ) (since the hji ’s are isomorphisms of prevarieties). So we have proved our claim (in fact, we have glued the Fi ’s in the sense of 9.4.2). Since X is a ﬁnite union of Noetherian open subsets, it is a Noetherian topological space. Further, if we take Bi a base of aﬃne open subsets of

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ψi (Xi ), then B is a base of aﬃne open subsets of X. This implies that X is a prevariety. With respect to this structure of prevariety on X, it is clear that each ψi (Xi ) is an open sub-prevariety of X and each ψi is an isomorphism of prevarieties. 12.6.3 With the above notations, we say that the prevariety X is obtained by gluing the prevarieties Xi along the Xi using the isomorphisms hji . In general, we identify via ψi , the prevarieties Xi and ψi (Xi ). 12.6.4 Using the preceding construction, we give below examples of prevarieties which are not varieties. Let X1 = X2 = k and denote by O1 , O2 the point 0 in X1 and X2 respectively. Let U1 = X1 \ {O1 }, U2 = X2 \ {O2 } and glue X1 and X2 along U1 and U2 via the identity map U1 → U2 . The prevariety X thus obtained is the union X1 ∪ {O2 } = X2 ∪ {O1 } and X1 , X2 are aﬃne open subsets of X. Now the open subset X1 ∩ X2 is just U1 (or U2 ), so it is aﬃne. Moreover, X \ (X1 ∩ X2 ) consists of the two distinct points O1 and O2 . The intersection (X1 × X2 ) ∩ ∆X can be identiﬁed with ∆k \ {(0, 0)} where we identify X1 × X2 with k2 . It follows that ∆X is not closed in X × X, and so X is not a variety. Note that if x, y ∈ X \ {O1 } (resp. X \ {O2 }), then x, y are contained in the aﬃne open subset X1 (resp. X2 ). Since X is not a variety, it follows from 12.5.9 that no aﬃne open subset of X contains simultaneously O1 and O2 . The preceding discussion says that any aﬃne open subset of X is contained in X1 or X2 . Hence the intersection of two aﬃne open subsets of X is again an aﬃne open subset. Compare this with condition (iii) of 12.5.6. 12.6.5 Let X1 = X2 = k2 , O1 , O2 the point (0, 0) of X1 and X2 respectively, and X the prevariety obtained by gluing X1 and X2 along X1 \ {O1 } and X2 \ {O2 } via the identity map X1 \ {O1 } → X2 \ {O2 }. Again we may identify X1 and X2 as aﬃne open subsets of X, and identify X1 ∩ X2 with k2 \ {(0, 0)}, which is not aﬃne (12.3.5). So by 12.5.6, X is not a variety.

12.7 Rational functions 12.7.1 In this section, X shall be an irreducible algebraic variety. Lemma. Let U, V be non-empty open subsets of X, u : U → Y and v : V → Y morphisms to a variety Y . If there exists an open subset W in U ∩ V such that u|W = v|W , then u|U ∩V = v|U ∩V . Proof. Since X is irreducible, W is dense in X and so in U ∩ V . The result follows therefore from 12.5.8. 12.7.2 Let U be a non-empty open subset of X and u : U → Y a morphism to a variety Y . By 12.7.1, there exists a largest open subset U0 containing U

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such that u extends to a morphism u0 : U0 → Y . Namely, U0 is the union of all the open subsets containing U where u extends to a morphism to Y . When U0 = U , we say that u is a rational map from X to Y and that U is the domain of deﬁnition of the rational map u. So a morphism of X to Y is a rational map whose domain of deﬁnition is X, and by 12.7.1, any morphism from a non-empty open subset of X to Y extends uniquely to a rational map from X to Y . A rational map from X to k is called a rational function on X. Such a function is therefore the maximal extension of a regular function on a nonempty open subset of X. There is another way to deﬁne rational maps: Let E be the set of pairs (U, u), where U is a non-empty open subset of X and u a morphism from U to Y . Deﬁne the following relation on E (U, u) ∼ (V, v) ⇔ u|U ∩V = v|U ∩V . This is an equivalence relation, and a rational map is simply an equivalence class. There is a third way to deﬁne rational maps. Order E by setting (U, u) (V, v) ⇔ V ⊂ U and u|V = v. Then we have an inductive system, and the set of rational maps is the set of inductive limits of this system. 12.7.3 Example. Let f, g ∈ Γ (X, OX ) with g = 0. The set U of x ∈ X such that g(x) = 0 is a non-empty open subset. The map x → f (x)/g(x) is regular on U since it is regular on all aﬃne open subsets contained in U (11.7.5 and 12.1.1). It follows that the preceding map extends uniquely to a rational function on X. In general, we do not obtain all the rational functions in this way. 12.7.4 From now on, we shall denote by R(X) the set of rational functions on X. We claim that R(X) has a canonical structure of commutative ﬁeld. Firstly, the interpretation of rational functions in terms of inductive limit and the results of 8.3 show that R(X) has a canonical structure of commutative ring. Let f ∈ R(X) \ {0}. The set U of points in X where f is non-zero is a non-empty open subset, and the function x → 1/f (x) is regular on U (12.7.3). Denote by 1/f the unique rational function such that its restriction to U is x → 1/f (x). Then it is obvious that 1/f is the inverse of f in the ring R(X). Thus we have proved our claim. 12.7.5 Remark. Let f ∈ R(X) and x ∈ X. If f is not deﬁned at x, it may happen that g = 1/f is deﬁned at x. Then g(x) = 0 for otherwise f = 1/g would be deﬁned in a neighbourhood of x. We shall say that x is a pole of f . For example, the rational function x2 /x1 on k2 is not deﬁned at the points (0, x2 ). We see that such a point (0, x2 ) is a pole if and only if x2 = 0.

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12.7.6 Theorem. Let X be an irreducible variety. For all aﬃne open subset U of X, the ring Γ (U, OX ) is an integral domain and the canonical map Γ (U, OX ) → R(X), sending a regular function on U to the unique rational function extending it, is injective and extends to an isomorphism from the quotient ﬁeld of Γ (U, OX ) onto R(X). Proof. Since X is irreducible, so is U . Hence Γ (U, OX ) is an integral domain (11.2.3 and 11.3.2). The injectivity of the map Γ (U, OX ) → R(X) is a consequence of 12.7.1. To ﬁnish our proof, we only have to show that the extension of this map to the quotient ﬁeld of Γ (U, OX ) is surjective. Let h ∈ R(X). There exists an aﬃne open subset D(g) ⊂ U , where g ∈ Γ (U, OX ), such that h|D(g) ∈ Γ (D(g), OX ). So h|D(g) is of the form x → f (x)/g(x)m , with f ∈ Γ (U, OX ). If f, g ∈ R(X) are the images of f and g, then h = f/ g m . Thus we have obtained our result. 12.7.7 In view of 12.7.6, we can identify R(X) with the quotient ﬁeld of Γ (U, OX ), for an aﬃne open subset U of X. In particular, if X = Spm(A) is an irreducible aﬃne variety, then R(X) can be identiﬁed with the quotient ﬁeld of A. 12.7.8 Remark. It follows from 12.7.6 that when X is an irreducible variety, R(X) is a ﬁnitely generated extension of k. Conversely, let K be a ﬁnitely generated extension of k and y1 , . . . , yn a system of generators of K, then K is the quotient ﬁeld of the ﬁnitely generated k-algebra A = k[y1 , . . . , yn ] which is also an integral domain. It follows that there exists an irreducible aﬃne variety X ⊂ kn such that A(X) is isomorphic to A. Thus K is isomorphic to R(X). An irreducible variety Y such that R(Y ) is isomorphic to K is called a model of K. The preceding paragraph says that there is an aﬃne model of K. 12.7.9 Proposition. Let X = Spm(A) be an irreducible variety such that A is a factorial ring, and h ∈ R(X). There exist f, g ∈ A verifying the following conditions: (i) The domain of deﬁnition of h is D(g). (ii) For all x ∈ D(g), we have h(x) = f (x)/g(x). Proof. By 12.7.6, there exist f, g ∈ A, g = 0, relatively prime, such that h = f /g (and f, g are unique up to a multiple of an invertible element). Denote by U the domain of deﬁnition of h, we have D(g) ⊂ U . If h is deﬁned on a point x ∈ D(g), then h = f1 /g1 where f1 , g1 ∈ A with g1 (x) = 0. So there exists u ∈ A such that f1 = uf and g1 = ug. This is absurd since g(x) = 0. 12.7.10 Remark. We can not drop the hypothesis that A is factorial. In general, the domain of deﬁnition of h can contain strictly D(g) for any g ∈ A such that h = f /g with f ∈ A.

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12.7.11 Let X, Y be irreducible varieties, u : X → Y a dominant morphism and V ⊂ Y an aﬃne open subset. There exists an aﬃne open subset U of X such that U ⊂ u−1 (V ). Since u is dominant, u(U ) is dense in V . By 6.5.9, the comorphism ϕ : Γ (V, OY ) → Γ (U, OX ) of the restriction U → V of u is injective. Hence by 12.7.6, ϕ can be extended to an injective homomorphism of the ﬁelds θ : R(Y ) → R(X). Let U ⊂ X and V ⊂ Y be aﬃne open subsets such that U ⊂ u−1 (V ). The comorphism ϕ : Γ (V , OY ) → Γ (U , OX ) extends to the same homomorphism θ. In fact, replacing V (resp. U ) by an aﬃne open subset in V ∩ V (resp. U ∩ U ) if necessary, we may assume that V ⊂ V and U ⊂ U . It is then clear that ϕ extends ϕ. When there is no confusion, we shall say that the ﬁeld homomorphism θ : R(Y ) → R(X) is the comorphism associated to u. 12.7.12 Deﬁnition. A dominant morphism u : X → Y of irreducible varieties is called birational if the comorphism θ : R(Y ) → R(X) associated to u is an isomorphism. 12.7.13 Remarks. 1) By 11.5.6, the morphism u : k2 → k2 , (x1 , x2 ) → (x1 x2 , x2 ) is dominant. It is birational since k(T1 , T2 ) = k(T1 T2 , T2 ). However, it is neither surjective (11.5.6) nor injective (since u(x1 , 0) = (0, 0) for all x1 ∈ k). 2) Recall from 11.5.7 the morphism u : k → k2 , t → (t2 , t3 ), whose image C is the algebraic set deﬁned by x31 − x22 = 0. Recall that the morphism v : k → C is bijective, bicontinuous but is not an isomorphism. However, it is a birational morphism since k(T ) = k(T 2 , T 3 ). 3) So a birational morphism can be neither injective nor surjective. If it is bijective and bicontinuous, it does not need to be an isomorphism. Note that a bijective and bicontinuous morphism is not necessarily birational.

12.8 Local rings of a variety 12.8.1 Let (X, OX ) be a variety and x ∈ X. We denote by OX,x , or Ox , the ﬁbre at x of the sheaf OX (9.1.4). If U is an open subset of X containing x and f ∈ Γ (U, OX ), we denote by fx the image of f in OX,x . We deﬁned in 9.1.4 the value of a germ at x. By 8.3, OX,x has a natural structure of commutative ring. 12.8.2 Remarks. 1) If U is an open subset of X containing x, then OX,x = OU,x (8.2.7).

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2) Suppose that X is irreducible, then any regular function on an open neighbourhood of x extends uniquely to a rational function on X deﬁned at x. Moreover, two such functions having the same germ at x extend to the same rational function. Thus OX,x can be identiﬁed as a subring of R(X) of rational functions deﬁned at x. In particular, since X is irreducible, the ring OX,x is an integral domain, and if U is an open subset of X, then Γ (U, OX ) = OX,x . x∈U

12.8.3 Proposition. Let (X, OX ) be a variety and x ∈ X. (i) The ring OX,x is local and its maximal ideal mX,x consists of germs of regular functions which are zero at x. The ﬁelds OX,x /mX,x and k are isomorphic. (ii) Let U be an aﬃne open subset of X containing x, A = Γ (U, OX ) and n the maximal ideal of A corresponding to x. The rings An and OX,x are isomorphic. In particular, there is a bijection between the set of prime ideal of OX,x and the set of closed irreducible subsets of U containing x. (iii) The ring OX,x is Noetherian. Proof. (i) Let ρ : OX,x → k be the map sending a germ to its value at x. This is a surjective ring homomorphism. Thus mX,x = ker ρ is a maximal ideal and the ﬁelds OX,x /mX,x and k are isomorphic. If s ∈ OX,x \ mX,x and (U, f ) a representative of s (9.1.4) where U is an aﬃne open subset of X containing x, then x ∈ D(f ) = {y ∈ U ; f (y) = 0}. Since f is invertible in Γ (D(f ), OU ) = Γ (D(f ), OX ), s is invertible in OX,x . So OX,x is local. (ii) Let ϕ : A → OX,x be the homomorphism sending f ∈ A to its germ at x. If f (x) = 0, that is f ∈ n, then ϕ(f ) is invertible by (i). By 2.3.4, ϕ extends uniquely to a homomorphism ψ : An → OX,x . Let s ∈ OX,x . There exist a principal open subset D(f ) of U containing x and a function g ∈ Γ (D(f ), OX ) such that (D(f ), g) is a representative of s. Since x ∈ D(f ), we have f ∈ A \ n. Moreover, if y ∈ D(f ), then g(y) = h(y)/f (y)n , for some n ∈ N, h ∈ A. Thus ψ(h/f n ) = s, and ψ is surjective. Next, if g ∈ ker ψ, there exist h ∈ A, f ∈ A \ n such that g = h/f . Then ϕ(h) = 0. Thus there exists t ∈ A \ n verifying h|D(t) = 0, and hence ht = 0. So g = 0. We have therefore proved that ψ is an isomorphism, and the second part follows from 2.3.8 and 11.2.3. (iii) This is clear by (ii), 1.3.2 and 2.7.6. 12.8.4 Remark. In general, OX,x is not a ﬁnitely generated k-algebra. Take X = k and x = 0, then OX,x is isomorphic to the subring of k(T ) consisting of rational functions of the form P/Q with Q(x) = 0. Since a nonzero polynomial in one variable has a ﬁnite number of zeros, it is clear that OX,x is not ﬁnitely generated as a k-algebra.

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12.8.5 Proposition. Let (X, OX ) be a variety and x ∈ X Then the following conditions are equivalent: (i) x belongs to a unique irreducible component of X. (ii) The ring OX,x is an integral domain. Proof. Let Y1 , . . . , Yn be the irreducible components of X. (i) ⇒ (ii) Let U = Y2 ∪ · · · ∪ Yn and V = X \ U . By 1.1.13, V ⊂ Y1 is open in X, so it is irreducible. So if x ∈ V , then OX,x = OV,x , and OV,x is an integral domain (12.8.2). (ii) ⇒ (i) Replacing X by an aﬃne open subset containing x, we may assume that X = Spm(A) is aﬃne. Then Yi = V(pi ) where p1 , . . . , pn are the minimal prime ideals of A. Suppose that x ∈ Y1 ∩ Y2 . Again, there is a non-empty aﬃne open subset U1 ⊂ Y1 (resp. U2 ⊂ Y2 ) such that U1 ∩ Yi = ∅ if i 2 (resp. U2 ∩ Yi = ∅ if i = 2). So we can ﬁnd f1 , f2 ∈ A \ {0} such that D(f1 ) ⊂ U1 and D(f2 ) ⊂ U2 . Thus f1 f2 = 0. Now Y1 and Y2 are irreducible, so U1 and U2 are dense in Y1 and Y2 , which implies that the germs of f1 , f2 at x are non-zero. Hence OX,x is not an integral domain. 12.8.6 Let u : X → Y be a morphism of varieties and x ∈ X. If V is an open neighbourhood of u(x) and g ∈ Γ (V, OY ), then g ◦ u is a regular function in the open subset u−1 (V ). Moreover, if g1 , g2 are equal in a neighbourhood of u(x), then g1 ◦ u and g2 ◦ u are equal in a neighbourhood of x. We obtain therefore a map ux : OY,u(x) → OX,x which is clearly a k-algebra homomorphism. We can also verify easily that ux is a local homomorphism. Let U be an aﬃne open subset of X containing x. Any element of OX,x is of the form fx /hx (notations of 12.8.1), where f, h ∈ A(U ) and h(x) = 0. It follows that ux is completely determined by its values on the germs of functions which belong to A(U ). 12.8.7 Lemma. Let X, Y be varieties, x ∈ X, y ∈ Y and U = Spm(A) (resp. V = Spm(B)) an aﬃne open subset of X (resp. Y ) containing x (resp. y). Suppose further that there is a local homomorphism ϕ : OY,y → OX,x . (i) There exist an aﬃne open neighbourhood U0 ⊂ U of x, an aﬃne open neighbourhood V0 ⊂ V of y and a morphism u0 : U0 → V0 such that u0 (x) = y and ϕ is the local homomorphism (u0 )x . (ii) Let U1 ⊂ U (resp. V1 ⊂ V ) be an aﬃne open neighbourhood of x (resp. y), u1 : U1 → V1 a morphism such that u1 (x) = y and ϕ is the local homomorphism (u1 )x . Then u0 and u1 coincide in a neighbourhood of x. Proof. (i) For b ∈ B, denote by iy (b) the canonical image of b in OY,y . Let {b1 , . . . , bm } be a system of generators of B and α : k[T1 , . . . , Tm ] → B the homomorphism deﬁned by α(Ti ) = bi for 1 i m. Let β = ϕ ◦ iy ◦ α,

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a = ker β, {Q1 , . . . , Qr } a system of generators of a, T0 a new indeterminate, and P1 , . . . , Pr homogeneous polynomials of degree h, such that for 1 j r, Qj (T1 /T0 , . . . , Tm /T0 ) = Pj (T0 , T1 , . . . , Tm )/T0h . Each ϕ(iy (bj )) is the germ of a regular function at x which can be written as aj /s where aj , s ∈ A and s(x) = 0. Since the Qj ’s are in a, the regular functions Qj (a1 /s, . . . , am /s) = Pj (s, a1 , . . . , am )/sh are identically zero in a neighbourhood of x. We deduce therefore that there exists t ∈ A verifying t(x) = 0 and Pj (ts, ta1 , . . . , tam ) = 0 in A for 1 j r. Let ρ : k[T1 , . . . , Tm ] → Ats be the homomorphism such that ρ(Ti ) = (tai )/(ts), 1 i m. We have ρ(a) = {0}, so ρ(ker α) = {0}. Thus there exists a homomorphism γ0 : B → Ats such that ρ = γ0 ◦ α; so γ0 (bi ) = (tai )/(ts) for 1 i m. Finally, let u0 = Spm(γ0 ) : U0 = D(st) → V be the associated morphism. If ix : Ats = Γ (D(ts), OX ) → OX,x is the canonical homomorphism, then ix ◦ γ0 = ϕ ◦ iy . It follows that, using the notations of 12.8.3, −1 γ0−1 (I({x})) = γ0−1 (i−1 x (mX,x )) = iy (mY,y ) = I({y}).

Hence ϕ is the local homomorphism (u0 )x . (ii) Replace U0 , U1 (resp. V0 , V1 ) by an aﬃne open subset contained in U0 ∩U1 (resp. V0 ∩V1 ), we may assume that U = U0 = U1 (resp. V = V0 = V1 ). Let γ0 = A(u0 ) and γ1 = A(u1 ). For 1 i m, γ0 (bi ) and γ1 (bi ) are regular functions in a neighbourhood of x having the same germ at x. So there exists s ∈ A verifying s(x) = 0 such that the restriction of γ0 (bi ) and γ1 (bi ) to D(s) are equal. We deduce therefore that u0 |D(s) = u1 |D(s) . 12.8.8 Proposition. Let X, Y be varieties, x ∈ X and y ∈ Y . Suppose further that there is a k-algebra isomorphism ϕ : OY,y → OX,x . (i) There exist an open neighbourhood U0 ⊂ X of x, an open neighbourhood V0 ⊂ Y of y, and an isomorphism u : U0 → V0 such that u(x) = y. (ii) Let U1 ⊂ X (resp. V1 ⊂ Y ) be an open neighbourhood of x (resp. y) and u1 : U1 → V1 an isomorphism. Suppose that the isomorphisms ux and (u1 )x coincide. Then there exists an open neighbourhood U ⊂ U0 ∩ U1 of x such that u|U = u1 |U . Proof. By 12.8.7, it suﬃces to prove (i). Let ψ be the inverse isomorphism of ϕ. Again by 12.8.7, we can ﬁnd aﬃne open neighbourhoods U0 of x, V0 of y and a morphism v from V0 to an open subset of X such that ϕ and ψ are the local homomorphisms ux and vy induced by u and v. Replacing U0 by an aﬃne open subset, we may assume that u(U0 ) ⊂ V0 . Then v ◦ u is a morphism from U0 to an open neighbourhood of x such that (v ◦ u)x is the identity map of OX,x . By 12.8.7, (v ◦ u)(z) = z for all z in an open neighbourhood of x. The same arguments show that there exist aﬃne open neighbourhoods U1 of x, V1 of y such that u|U1 is a morphism from U1 to V1 , v|V1 is a morphism from V1 to U1 and (u ◦ v)(z) = z for all z ∈ V1 . Thus our result is proved.

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Remark. We see from 12.8.8 that the local ring OX,x determines a neighbourhood of x in X up to an isomorphism.

References • [5], [19], [22], [26], [37], [40], [78].

13 Projective varieties

We attach to a projective space a natural structure of an irreducible algebraic variety. More generally, we deﬁne the notion of projective varieties, and discuss properties of these varieties. In particular, we introduce complete varieties. Certain properties of complete varieties are analogue to those of compact topological spaces.

13.1 Projective spaces 13.1.1 Let n be a non-zero positive integer. We deﬁne an equivalence relation R on kn+1 \ {0} by xRy if and only if y = λx for some λ ∈ k \ {0}. The quotient space (kn+1 \ {0})/R is called the projective space of dimension n, and we shall denote it by Pn (k) or Pn . By deﬁnition, Pn is the set of 1-dimensional vector subspaces of kn+1 . Let z ∈ Pn , its homogeneous coordinates are the coordinates x0 , x1 , . . . , xn of a point x in kn+1 whose class is z. Homogeneous coordinates are unique up to multiplication by a non-zero scalar. We shall denote by π : kn+1 \ {0} → Pn the canonical surjection. 13.1.2 We shall call a subset C in kn+1 \ {0} a cone if λx ∈ C for all x ∈ C and for all λ ∈ k \ {0}, or equivalently, C = π −1 (π(C)). The union of {0} and a cone C in kn+1 \ {0} shall be called the aﬃne cone over π(C) in kn+1 . Lemma. Let C be a non-empty closed cone in kn+1 \ {0}. Then C ∪ {0} is the closure C of C in kn+1 . Proof. Let D be a 1-dimensional subspace of kn+1 . Then 0 ∈ D \ {0}. Since C is non-empty, C ∪ {0} ⊂ C. Now C is closed, so C = C ∩ (kn+1 \ {0}). Hence C = C ∪ {0}.

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13.1.3 By 13.1.2, a closed cone in kn+1 \{0} is the intersection of kn+1 \{0} and a closed aﬃne cone in kn+1 . 13.1.4 Let C be a closed aﬃne cone in kn+1 and a = I(C) the associated radical ideal in k[T0 , . . . , Tn ]. Let P ∈ a. Since C is a cone, the condition P (x0 , . . . , xn ) = 0 implies that P (λx0 , . . . , λxn ) = 0 for all λ ∈ k. It follows that all the homogeneous components of P are in a, and so a is a graded ideal. Conversely, given a graded ideal a = k[T0 , . . . , Tn ], the set V(a) is an aﬃne cone in kn+1 . The only maximal graded ideal in k[T0 , . . . , Tn ] is the ideal m0 generated by T0 , . . . , Tn . It follows that for any graded ideal a = k[T0 , . . . , Tn ], √ V(a) = {0} ⇔ a = m0 . By 7.2.7, we see that the map a → V(a) induces a bijection between the set of graded radical ideals of k[T0 , . . . , Tn ] which do not contain m0 and the set of closed aﬃne cones of kn+1 which are not reduced to {0}.

13.2 Projective spaces and varieties 13.2.1 For 0 i n, let us denote by Hi0 (resp. Hi1 ) the hyperplane (resp. aﬃne hyperplane) of kn+1 deﬁned by xi = 0 (resp. xi = 1). Let Ui be the set of 1-dimensional subspaces of kn+1 which are not contained in Hi0 . Thus Ui is the set of points z in Pn whose homogeneous coordinates (x0 , . . . , xn ) verify xi = 0. Now any 1-dimensional subspace in Ui intersects Hi1 at a unique point and this induces a bijection ϕi between Ui and Hi1 . Namely, ϕi sends the point z of Ui ⊂ Pn with homogeneous coordinates (x0 , . . . , xn ) to the point x0 xi−1 xi+1 xn ,..., , 1, ,..., . xi xi xi xi It follows clearly that: Pn = U0 ∪ U1 ∪ · · · ∪ Un . 13.2.2 For i, j ∈ {0, 1, . . . , n}, i = j, Ui ∩ Uj is the set of points whose homogeneous coordinates (x0 , . . . , xn ) verify xi = 0 and xj = 0. Such a point corresponds to the points: x0 x0 xi−1 xi+1 xn xj−1 xj+1 xn ,..., , 1, ,..., ,..., , 1, ,..., and xi xi xi xi xj xj xj xj in the hyperplanes Hi1 and Hj1 respectively. 1 1 1 = Hi1 \ Hj0 and Hji = Hj1 \ Hi0 . We deﬁne the bijection pji : Hij → Set Hij 1 Hji by: x0 xi−1 1 xi+1 xn (1) (x0 , . . . , xi−1 , 1, xi+1 , . . . , xn ) → ,..., , , ,..., . xj xj xj xj xj

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1 Let z ∈ Ui ∩ Uj ∩ Ul , then ϕi (z) is contained in Hij ∩ Hil1 . We verify readily 1 ∩ Hil1 : that ϕj (z) = pji (ϕi (z)) and ϕl (z) = pli (ϕi (z)). Hence, for any x ∈ Hij

(2)

pli (x) = plj (pji (x)).

Let Hii1 = Hi1 and pii = idHi1 . Then we have shown that Pn can be considered as the set obtained by gluing the sets Hi1 , for 0 i n, along the 1 ’s via the bijections pji (1.5.1). Hij 13.2.3 Using the preceding consideration, we can deﬁne a structure of algebraic prevariety on Pn . 1 is an open subvariety of Hi1 The set Hi1 is an aﬃne variety, and Hij 1 1 n+1 (Hij = D(prj ) ∩ Hi where prj : k → k denotes the canonical projection (x0 , . . . , xn ) → xj ). Since regular functions on Hi1 are restrictions of polynomial functions on n+1 , it follows from 11.7.5, 12.1.1 that for i = j, the algebra of regular k 1 is generated by the functions: functions on Hij (x0 , . . . , xn ) →

xl , xj

for 0 l n. We conclude that pji is an isomorphism between the aﬃne 1 1 and Hji . varieties Hij Now equation (2) above is condition c) of 12.6.1. Hence we can endow Pn with a structure of algebraic prevariety over k. From now on, when we consider the projective space as a prevariety, we shall mean the above structure. The main point of the preceding discussion is that the maps ϕi of 13.2.1 allow us to identify Ui with Hi1 , and that we use these maps to deﬁne a structure of prevariety on Pn where the Ui ’s are aﬃne open subsets. 13.2.4 For 0 i n, the map ψi : Hi1 → kn (x0 , . . . , xi−1 , 1, xi+1 , . . . , xn ) → (x0 , . . . , xi−1 , xi+1 , . . . , xn ) is an isomorphism of varieties. Let z ∈ Ui , with homogeneous coordinates (x0 , . . . , xn ). The map x0 xi−1 xi+1 xn ψi ◦ ϕi : Ui → kn , z → ,..., , ,..., xi xi xi xi is an isomorphism of Ui onto kn . Thus regular functions on Ui are of the form: x0 xi−1 xi+1 xn z → P ,..., , ,..., xi xi xi xi where P ∈ k[T0 , . . . , Ti−1 , Ti+1 , . . . , Tn ]. Or equivalently, they are of the form: z →

R(x0 , . . . , xn ) xm i

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where m ∈ N and R ∈ k[T0 , . . . , Tn ] is homogeneous of degree m. xl The algebra A(Ui ) is therefore generated by the , where 0 l n. xi The intersection Ui ∩ Uj , i = j, corresponds via the isomorphism ψi ◦ ϕi to the aﬃne open subset D(Tj ) of kn . From this, we deduce that A(Ui ∩ Uj ) is generated by the functions xi 1 xl for 0 l n and = xi xj (xj /xi ) on Ui ∩ Uj . 13.2.5 For 0 i n, denote by θi : D(Ti ) → Hi1 × (k \ {0}), the map x0 xi−1 xi+1 xn (x0 , . . . , xn ) → ,..., , 1, ,..., , xi . xi xi xi xi By 12.1.1 and 12.4.3 (ii), θi is a bijective morphism. Its inverse ((y0 , . . . , yi−1 , 1, yi+1 , . . . , yn ), yi ) → (y0 yi , . . . , yi−1 yi , yi , yi+1 yi , . . . , yn yi ) is clearly a morphism by the deﬁnition of regular functions on D(Ti ) and Hi1 × (k \ {0}). Hence θi is an isomorphism. 13.2.6 Proposition. The projective space Pn is an irreducible variety. Proof. That Pn is a variety results from 13.2.4 and 12.5.7. Now Ui is irre ducible and Ui ∩ Uj = ∅ for all i, j. Hence Pn is irreducible (1.1.6). 13.2.7 Proposition. The map π is a morphism of varieties. Proof. Since kn+1 \ {0} = D(T0 ) ∪ D(T1 ) ∪ · · · ∪ D(Tn ), it suﬃces to prove that the restriction of π to D(Ti ), 0 i n, is a morphism (9.5.5). Now the image of D(Ti ) via π is the aﬃne open subset Ui . Hence by 9.5.4 and 12.2.8, it suﬃces to show that the map ρi : D(Ti ) → Ui induced by the restriction of π, is a morphism. But we have seen in 13.2.4 that regular functions on Ui are of the form: R(x0 , . . . , xn ) z → xm i where z ∈ Ui with homogeneous coordinates (x0 , . . . , xn ), R ∈ k[T0 , . . . , Tn ] homogeneous of degree m and m ∈ N. So it is clear that if f ∈ A(Ui ), then f ◦ ρi ∈ A(D(Ti )). The result now follows.

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13.3 Cones and projective varieties 13.3.1 Deﬁnition. (i) Any algebraic variety isomorphic to a closed subvariety of the projective space Pn is called a projective variety. (ii) An algebraic variety isomorphic to a (not necessarily closed) subvariety of Pn is called a quasi-projective variety. Remark. An aﬃne or quasi-aﬃne variety is a quasi-projective variety. 13.3.2 Proposition. Two points in a projective variety X are contained in an aﬃne open subset. Proof. A closed subvariety of an aﬃne variety is aﬃne, so we may assume that X = Pn , n 1. Let x, y ∈ Pn . There exists a hyperplane in kn+1 which does not contain the subspaces {0}∪π −1 (x) and {0}∪π −1 (y). By applying a linear transformation, we may further assume that this hyperplane is Hi0 . Hence x, y belong to Ui . 13.3.3 Proposition. The map C → π(C) is a bijection between the set of closed cones in kn+1 \ {0} and the set of closed subsets of Pn . Furthermore, for 0 i n, C ∩π −1 (Ui ) is isomorphic to the product (π(C)∩Ui )×(k\{0}). Proof. If F is a closed subset of Pn , then 13.2.7 says that π −1 (F ) is a closed cone in kn+1 \ {0}. Conversely, given a closed cone C in kn+1 \ {0}, we have by identifying Ui and Hi1 (13.2.1): θi (C ∪ D(Ti )) = (π(C) ∩ Ui ) × (k \ {0}) where θi , Ti are as in 13.2.5. Since C ∩ D(Ti ) is closed in D(Ti ) = π −1 (Ui ) and θi is an isomorphism of D(Ti ) onto Ui × (k \ {0}) (13.2.5), π(C) ∩ Ui is closed in Ui by 12.4.5 (i). Hence π(C) is closed in Pn . The last statement follows from 9.5.4 and 12.2.8. 13.3.4 Corollary. (i) The topology on Pn is the quotient topology of the topology on kn+1 \ {0} with respect to the equivalence relation π(x) = π(x ). (ii) The map π is open. Proof. (i) By 13.3.3, closed subsets of Pn are images of closed subsets saturated by the equivalence relation π(x) = π(x ). Hence we have our result. (ii) Let U be an open subset of kn+1 \ {0} and V = λU = π −1 (π(U )). λ∈k\{0}

Since x → λx is an automorphism of the variety kn+1 \ {0}, it follows that V is open in kn+1 \ {0}. So (i) implies that π is open.

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13.3.5 It follows from 13.1.4 and 13.3.3 that any closed subvariety of Pn is deﬁned by a system of equations: (3)

Pi (x0 , . . . , xn ) = 0

where (x0 , . . . , xn ) are homogeneous coordinates of its points and the Pi ’s are homogeneous polynomials in k[T0 , . . . , Tn ]. A closed subvariety deﬁned by a single non-constant homogeneous polynomial (resp. of degree 1) shall be called a hypersurface (resp. hyperplane) of Pn . Thus a hyperplane of Pn is the image π(H \ {0}) where H is a hyperplane in kn+1 . For example, the complement Hi of Ui is the hyperplane π(Hi0 \ {0}). 13.3.6 Let C be a closed cone in kn+1 \ {0} and X = π(C). By 13.1.2 and 13.1.4, C = V(a) \ {0} for some graded radical ideal a of k[T0 , . . . , Tn ]. We shall say that X is deﬁned by the graded ideal a. The intersection of X with Ui , identiﬁed as kn as in 13.2.4, is a subvariety V(ai ) of kn where ai is the ideal of k[T0 , . . . , Ti−1 , Ti+1 , . . . , Tn ] generated by the polynomials P (T0 , . . . , Ti−1 , 1, Ti+1 , . . . , Tn ), P ∈ a. Let Q ∈ k[T0 , . . . , Ti−1 , Ti+1 , . . . , Tn ] be a polynomial of degree r. Then T0 Ti−1 Ti+1 Tn r ,..., , ,..., P (T0 , . . . , Tn ) = Ti Q Ti Ti Ti Ti is a homogeneous polynomial in k[T0 , . . . , Tn ]. So we have homogenized Q. Now let Z = V(b) be a closed subvariety of Ui , where b is an ideal of k[T0 , . . . , Ti−1 , Ti+1 , . . . , Tn ]. The closure Z of Z in Pn is equal to π(C) for some closed cone C = V(a)\{0} where a is a graded ideal of k[T0 , . . . , Tn ]. It is easy to see that we can take a to be the graded ideal obtained by homogenizing the elements of b. We can also verify easily that Z = Z ∩ Ui . Conversely, let X be a closed irreducible subvariety of Pn such that X is not contained in the complement Hi of Ui . Then X ∩ Ui is a non-empty open subset of X. So X = X ∩ Ui . 13.3.7 Let A = k[T0 , . . . , Tn ] and Q ∈ A a homogeneous polynomial of degree d > 0. Denote by D+ (Q) the set of points of Pn whose homogeneous coordinates (x0 , . . . , xn ) verify Q(x0 , . . . , xn ) = 0 (this is well-deﬁned since Q is homogeneous). Thus D+ (Q), being the complement of a hypersurface, is open in Pn . Lemma. Any non-empty open subset U of Pn is of the form U = D+ (Q1 ) ∪ · · · ∪ D+ (Qr ) where r ∈ N∗ and Q1 , . . . , Qr ∈ A are homogeneous and non-constant. Proof. The closed subset V = Pn \ U is deﬁned by a graded ideal a of A (13.3.6). By 7.2.2, there exist homogeneous elements Q1 , . . . , Qr ∈ A such that a = AQ1 + · · · + AQr . Clearly, we have U = D+ (Q1 ) ∪ · · · ∪ D+ (Qr ).

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13.3.8 Proposition. Let u be a map from Pn to a prevariety X. The following conditions are equivalent: (i) u is a morphism. (ii) u ◦ π is a morphism from kn+1 \ {0} to X. Proof. (i) ⇒ (ii) This is clear by 13.2.7. (ii) ⇒ (i) Suppose that u ◦ π is a morphism. To show that u is a morphism, it suﬃces to prove that u|Ui is a morphism for 0 i n. For Q ∈ k[T0 , . . . , Tn ], we set: D (Q) = {x ∈ kn+1 \ {0} ; Q(x) = 0}. In view of 11.7.2, the D (Q)’s form a base for the topology on kn+1 \ {0}. If x ∈ π −1 (Ui ), then our hypothesis says that there exist an aﬃne open subset V of X containing u(π(x)) and Q ∈ k[T0 , . . . , Tn ] verifying: 1) x ∈ D (Q) and u(π(D (Q))) ⊂ V . 2) For all regular function f on V , the function y → f (u(π(y))) is regular on D (Ti Q) = D (Q) ∩ π −1 (Ui ). Let f ∈ Γ (V, OX ) and g the function y → f (u(π(y))) deﬁned on D (Ti Q). Then there exist r ∈ N and a polynomial R ∈ k[T0 , . . . , Tn ] such that for y = (y0 , . . . , yn ) ∈ D (Ti Q), (4)

g(y) =

R(y0 , . . . , yn ) . yir Qr (y0 , . . . , yn )

Let λ ∈ k \ {0}. There exists Sλ ∈ k[T0 , . . . , Tn ] such that y yn 0 ,..., Sλ (y0 , . . . , yn ) = Q λ λ for all (y0 , . . . , yn ) ∈ kn+1 \ {0}. Set Wλ = λD (Ti Q). A point y = (y0 , . . . , yn ) belongs to Wλ if and only if yi = 0 and Sλ (y) = 0. Hence, Wλ = D(Ti Sλ ). Let y hλ : Wλ → k , y → g . λ There exists Rλ ∈ k[T0 , . . . , Tn ] such that if y ∈ Wλ , then: hλ (y) =

Rλ (y0 , . . . , yn ) . yir Sλr (y0 , . . . , yn )

We deduce that hλ is a regular function on Wλ . If x ∈ Wλ ∩ Wµ , then x = λy = µz for some y, z ∈ D (Ti Q). Thus π(z) = π(y), and from the deﬁnition of g, we obtain that hλ (x) = hµ (x). It follows that g can be extended to a regular function on the open subset Wλ = π −1 (π(D (Q)) ∩ Ui ), U= λ∈k\{0}

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such that (λy) = (y) for all y ∈ U and λ ∈ k \ {0}. By 13.3.4 (ii), π(D (Q)) is open in Pn , so π(D (Q)) ∩ Ui is also open in Pn . By 13.3.7, there exist homogeneous polynomials Q1 , . . . , Qr ∈ k[T0 , . . . , Tn ] such that π(D (Q)) ∩ Ui = D+ (Q1 ) ∪ · · · ∪ D+ (Qr ). Replacing Q by one of the Qi ’s, we can reduced to the case where Q is homogeneous. Then the function g on D (Ti Q) is invariant by the automorphisms y → λy. This implies that the polynomial R in (4) is homogeneous. From the deﬁnition of regular functions on Ui (13.2.4), we conclude that the restriction of u to π(D (Q)) is a morphism. Hence we have our result. 13.3.9 Corollary. A subvariety X of Pn is irreducible if and only if π −1 (X) is an irreducible subvariety of kn+1 \ {0}. Proof. Since π(π −1 (X)) = X, X is irreducible if π −1 (X) is irreducible. Conversely, suppose that X is irreducible. If x ∈ Pn , then π −1 (x), being isomorphic to k \ {0}, is irreducible. It follows from 1.1.7, 13.2.7 and 13.3.4 (ii) that π −1 (X) is irreducible. 13.3.10 Let m, n ∈ N, E = (kn+1 \ {0}) × (km+1 \ {0}) and π : kn+1 \ {0} → Pn , π : km+1 \ {0} → Pm the canonical morphisms. Deﬁne p : E → Pn × Pm , (x, x ) → (π(x), π (x )). The map p is a morphism by 12.4.3. We show as in the previous paragraphs that: • The map p is open. • A subset F in Pn × Pm is irreducible if and only if p−1 (F ) is irreducible in E. • Closed subsets of Pn × Pm are the subsets F such that p−1 (F ) is closed in E and invariant by the maps (x, x ) → (λx, λ x ) where λ, λ ∈ k \ {0}. 13.3.11 Let u : Pn → Pm be a morphism of varieties. By 13.3.8, v = u ◦ π is a morphism of kn+1 \ {0} to Pm . Denote by z0 , . . . , zm the homogeneous coordinates of points in Pm , Vj the aﬃne open subset of Pm deﬁned by zj = 0. For 0 i n and x ∈ π −1 (Ui ), the point v(x) belongs to some Vj . The open subset π −1 (Ui ) ∩ v −1 (Vj ) is then a neighbourhood of x, invariant by the automorphisms y → λy, for λ ∈ k \ {0}. It follows from 13.3.4 (i) and 13.3.7 that it contains a neighbourhood Wx of x in kn+1 \ {0} deﬁned by Q(y0 , . . . , yn ) = 0, where Q is a homogeneous polynomial. Let us identify canonically Vj with the set of points (x0 , . . . , xm ) ∈ km+1 \ {0} such that xj = 1 (13.2.1). The coordinates of v(y), for y ∈ Wx , are regular

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functions, invariant by the automorphisms y → λy. Replacing Q by a suitable power, we may write P0 (y) Pj−1 (y) Pj+1 (y) Pm (y) ,..., , 1, ,..., v(y) = Q(y) Q(y) Q(y) Q(y) where the Pl ’s and Q are homogeneous polynomials of the same degree. Set Pj = Q, then if y ∈ Wx , (5)

u(π(y)) = π (w(y))

with (6)

w(y) = (P0 (y), . . . , Pm (y)).

We may further assume that P0 , . . . , Pm are relatively prime. Proceed in the same manner with another point x . Since kn+1 \ {0} is irreducible, Wx ∩ Wx is non-empty and open. It follows that, multiply the Pl ’s by the non-zero scalar if necessary, we may suppose that (5) is veriﬁed in Wx with the same polynomials Pl . We deduce therefore that (5) is veriﬁed for all y ∈ kn+1 \ {0}. Further, by construction, P0 , . . . , Pn have no common zero in kn+1 \ {0}. Conversely, let P0 , . . . , Pn be homogeneous polynomials of the same degree having no common zero in kn+1 \ {0}. By 13.3.8, formulas (5) and (6) deﬁne a morphism from Pn to Pm . 13.3.12 Proposition. The only regular functions on Pn , that is the elements of Γ (Pn , OPn ), are the constant functions. Proof. Such a function is a morphism from Pn to k ⊂ P1 . If z0 , . . . , zn are homogeneous coordinates of z ∈ Pn , then 13.3.11 says that there exist homogeneous polynomials P, Q of the same degree such that f (z) =

P (z0 , . . . , zn ) Q(z0 , . . . , zn )

for any z ∈ Pn and Q(z0 , . . . , zn ) = 0 if (z0 , . . . , zn ) = (0, . . . , 0). Since Q is homogeneous in at least two indeterminates, it is constant. Hence P is also constant. 13.3.13 Let us conserve the notations of 13.3.11, and identify kn with U0 and km with V0 . Let u : kn → km be a morphism. Under what condition can we extend u to a morphism from Pn to Pm ? Note that if such an extension exists, it is unique by 12.5.6 and 13.2.6. Now we have from 11.5.1 that u(x1 , . . . , xn ) = (Q1 (x1 , . . . , xn ), . . . Qm (x1 , . . . , xn )) where Q1 , . . . , Qm are polynomials. Let r = max{deg Qi ; 1 i m }. Then the extension, if it exists, is necessary the morphism:

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(x0 , x1 , . . . , xn ) →

xr0 , xr0 Q1

x1 xn ,..., x0 x0

, . . . , xr0 Qm

x1 xn ,..., x0 x0

.

We deduce that the condition for the extension of u is that the degree r homogeneous components of the Qi ’s are not simultaneously zero on kn \ {0}.

13.4 Complete varieties 13.4.1 Deﬁnition. An algebraic variety X is said to be complete if for all algebraic variety Y , the projection pr2 : X × Y → Y is a closed morphism. 13.4.2 The subset {(x, y) ∈ k × k ; xy = 1} ⊂ k × k is closed, but its image under pr2 is k \ {0}, which is not closed in k. So the aﬃne line k is not complete. 13.4.3 Proposition. (i) A closed subvariety of a complete variety is complete. (ii) The product of two complete varieties is complete. (iii) Let X be a complete variety and u a morphism from X to a variety Y . Then u(X) is a closed and complete subvariety in Y . Proof. (i) Let X be a complete variety, Y a closed subvariety of X and Z a variety. We know from 12.4.5 and 12.5.4 that Y × Z is a closed subvariety of X × Z. Thus closed subsets of Y × Z are closed in X × Z. It follows that the restriction to Y × Z of pr2 : X × Z → Z is closed. (ii) Let X, Y be complete varieties and Z a variety. Then X ×Y is a variety (12.5.4). If F is a closed subset of (X × Y ) × Z = X × (Y × Z), its projection F on Y × Z is closed, and the projection F of F on Z is also closed. Since F is the projection of F on Z. We are done. (iii) Let Gu be the graph of u. Since Y is a variety, Gu is closed in X × Y (12.5.4). It follows that u(X), the projection of Gu on Y is closed in Y . Let Z be a variety, F a closed subset in u(X) × Z and F the projection of F on Z. Consider the following projections pr13 : X × Y × Z → X × Z , pr23 : X × Y × Z → Y × Z and set F = pr−1 23 (F ) ∩ (Gu × Z). Since u(X) is closed in Y and F is closed in u(X) × Z, F is closed in Y × Z. This implies that F is closed in Gu × Z. But the restriction of pr13 to Gu × Z is an isomorphism to X × Z (12.5.2). Hence pr13 (F ) is closed in X × Z. On the other hand, F is the projection of pr13 (F ) on Z. So F is closed in Z because X is complete. 13.4.4 Corollary. A complete aﬃne variety is a ﬁnite set. Proof. It suﬃces to prove that if X is a non-empty, irreducible and complete aﬃne variety, then X is a singleton, or equivalently, A(X) = k. Let f ∈ A(X), then by 13.4.3, f (X) is a closed irreducible and complete subvariety of k. Since k is not complete and proper closed subsets of k are ﬁnite subsets, f (X) is a singleton, and the result follows.

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13.4.5 Theorem. A projective variety is complete. Proof. By 13.4.3 (i), it suﬃces to prove that Pn is complete. Let Y be a variety, p : Pn × Y → Y the canonical projection and Z a closed subset of Pn × Y . We need to show that p(Z) is closed. a) It suﬃces to show that for all aﬃne open subset V of Y , V ∩ p(Z) is closed in V , that is p(Z ∩ p−1 (V )) is closed in V . Now Z ∩ p−1 (V ) is closed in Pn × V , so we are reduced to the case where Y = Spm(A) is aﬃne. Set B = A[T0 , . . . , Tn ]. b) Using the notations of 13.2.1, Pn × Y is the union of the aﬃne open subsets Wi = Ui × Y , 0 i n. The algebra Ai = A(Wi ) is given by: T0 T0 Tn Tn ,..., ,..., ⊗k A = A . Ai = k Ti Ti Ti Ti Denote by a the graded ideal of B generated by the homogeneous polynomials Q such that, for 0 i n, Tn T0 ,..., ∈ I(Z ∩ Wi ). Q Ti Ti Let Br (resp. ar ) be the A-module of homogeneous polynomials of degree r in B (resp. a). c) For i ∈ {0, . . . , n} and g ∈ I(Z ∩ Wi ), we shall prove that there exist integers k, r such that Tik g ∈ ar . Since g is a polynomial in Tl /Ti , 0 l n, for a suﬃciently large m, Tim g is a homogeneous polynomial of degree m in T0 , . . . , Tn . On the other hand, for any j, (Tim /Tjm )g is zero on Z ∩ Wi ∩ Wj , and (Tim+1 /Tjm+1 )g is zero on (Z ∩ Wj ) \ Wi . Since this holds for any j, we have Tim+1 g ∈ am+1 . d) Let y0 ∈ Y \ p(Z) and m = I({y0 }) ∈ Spm A. To ﬁnish our proof, we need to show that there exists f ∈ A verifying y0 ∈ D(f ) and D(f )∩p(Z) = ∅. Now Z ∩ Wi and Ui × {y0 } are disjoint closed subsets of the aﬃne open set Wi , and I(Ui × {y0 }) = mAi , it follows from 11.1.4 and 11.1.6 that: Ai = mAi + I(Z ∩ Wi ). Thus there exist gi ∈ I(Z ∩ Wi ), αij ∈ m and gij ∈ Ai such that for all 0 i n: 1 = gi + αij gij . j

Paragraph c) says that by multiplying the above equality by Tim with m suitably large, we obtain Tim = hi + αij hij j

where hi ∈ am and hij ∈ Bm . This implies that for N suitably large, any monomial of degree N in T0 , . . . , Tn belongs to aN + mBN , that is, BN = aN + mBN .

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Now apply 2.6.7 to the ﬁnitely generated A-module BN /aN , there exists f ∈ A \ m such that f BN ⊂ aN . Consequently, f (y0 ) = 0 and f.TiN ∈ aN for all i. Hence f is zero on p(Z) as required. 13.4.6 Remarks. 1) Let X, Y be complete varieties and u : X → Y a morphism. Contrary to the case of compact spaces, u can be a bicontinuous bijection without being an isomorphism. For example, let u : P1 → P2 be the map deﬁned by: (x, y) → (x3 , xy 2 , y 3 ). This is a morphism by 13.3.11. Its image Y is a closed subvariety of P2 (13.4.3 (iii) and 13.4.5). Denote by v : P1 → Y the induced morphism. We check easily that v is bijective. Again by using 13.4.3 and 13.4.5, we see that v is bicontinuous. But the image of the aﬃne open subset U0 of P1 is the algebraic set {(x, y) ∈ k2 ; x3 − y 2 = 0}. So by 11.5.7, v is not an isomorphism. 2) There are complete varieties which are not projective varieties.

13.5 Products 13.5.1 Let z ∈ Pn (resp. z ∈ Pm ) with homogeneous coordinates (x0 , . . . , xn ) (resp. (x0 , . . . , xm )). Denote by s(z, z ) ∈ Pnm+n+m the point with homogeneous coordinates the (n + 1)(m + 1) scalars y(i,j) where y(i,j) = xi xj for 0 i n, 0 j m. Thus we have a map s : Pn × Pm → Pnm+n+m . We shall show that s is a morphism, called the Segre embedding. More precisely, we have the following result: Proposition. The map s is a morphism. Its image S is a closed subvariety of Pnm+n+m and s induces an isomorphism of Pn × Pm onto S. Proof. Let Ui (resp. Uj ) be the aﬃne open subset of Pn (resp. Pm ) deﬁned by xi = 0 (resp. xj = 0) and let Wij be the aﬃne open subset of Pnm+n+m deﬁned by y(i,j) = 0. The restriction of s to Ui × Uj (identiﬁed as kn × km ) is clearly a morphism to Wij (identiﬁed as knm+n+m ). Hence u is a morphism. Now Pn × Pm is complete (13.4.3 and 13.4.5), so S is a closed subvariety of Pnm+n+m (13.4.3). From the deﬁnition of s, it is straightforward to check that: • s−1 (Wij ) = Ui × Uj . • The restriction of s to each Ui × Uj is injective. So u is injective. • Let t : Wij → Ui × Uj be the map deﬁned by sending a point with homogeneous coordinates y(l,h) to the point (z, z ) where z (resp. z ) has homogeneous coordinates the yl,j ’s for 0 l n (resp. the yi,h ’s for 0 h m). Then t is a morphism such that t(s(z, z )) = (z, z ) if (z, z ) ∈ Ui × Uj . Hence we have our result. 13.5.2 Proposition. The product of two projective varieties is projective.

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179

Proof. This is clear from 12.4.2 and 13.5.1. 13.5.3 Let d ∈ N∗ and I the set of α = (α(0), . . . , α(n)) ∈ Nn+1 such that |α| = α(0) + · · · + α(n) = d. n+d The cardinality of I is equal to . d Let J be a set containing I and N its cardinality. By 13.3.13, we can deﬁne a morphism v : Pn → PN −1 by sending z ∈ Pn with homogeneous coordinates (x0 , . . . , xn ) to the point of PN −1 with homogeneous coordinates given by the following system: α(0) α(n) if α ∈ I, xα = x0 · · · xn Ph (x0 , . . . , xn ) if h ∈ J \ I, where the Ph ’s are homogeneous polynomials of degree d. Since Pn is complete, the image V of v is a closed subvariety of PN −1 . Proposition. The morphism v induces an isomorphism of Pn onto the closed subvariety V of PN −1 . Proof. For 0 i n, denote by βi = (β(0), . . . , β(n)) ∈ I the elements such that β(i) = d and β(j) = 0 if j = i. Let Wi be the aﬃne open subset of PN −1 consisting of points whose homogeneous coordinate yβi is non-zero. If Ui is the aﬃne open subset of Pn deﬁned by xi = 0, then z ∈ Ui ⇒ v(z) ∈ Wi , z ∈ Ui ⇒ v(z) ∈ Wi . We deduce that v −1 (Wi ) = Ui . Now let w : Wi → Pn be the map sending a point with homogeneous coordinates yh , h ∈ J, to the point of Pn with homogeneous coordinates yγj , 0 j n, where γi = βi and if i = j, γj ∈ I is such that: γj (i) = d − 1 , γj (j) = 1 , γj (h) = 0 if h ∈ {i, j}. The map w is a morphism from Wi to Ui and we verify easily that w(v(z)) = z if z ∈ Ui . Hence the result follows. 13.5.4 Remarks. 1) In the case where J = I, the morphism v is called the Veronese embedding of degree d. 2) Let X be a hypersurface of Pn deﬁned by the homogeneous polynomial P of degree d > 0. Using the notations of 13.5.3, we have aα T α P = α∈L

where L ⊂ I and aα ∈ k \ {0} for α ∈ L. If v is the Veronese embedding of degree d, then v(X) is contained in the hyperplane of PN −1 deﬁned by

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aα Yα = 0.

α∈L

This is one of the reason why the Veronese embedding is useful: it allows us to reduce a problem concerning a hypersurface to a problem concerning a hyperplane. 13.5.5 The following result completes the one in 13.3.7. Proposition. Let Q ∈ k[T0 , . . . , Tn ] be a homogeneous polynomial of degree d > 0. The subset D+ (Q) of Pn is an aﬃne open subset. n+d Proof. Let N = , v : Pn → PN be deﬁned as in 13.5.3 by d v(z) = ((xα )α∈I , Q(x0 , . . . , xn )), and V = v(Pn ) which is closed by 13.5.3. Let DN +1 be the aﬃne open subsets of PN consisting of points such that the last homogeneous coordinate is non-zero. Since V ∩ DN +1 is closed in DN +1 , it is an aﬃne open subset of V . But V ∩ DN +1 = v(D+ (Q)). Hence the result follows since v is an isomorphism of Pn onto V .

13.6 Grassmannian variety

r n 13.6.1 Let Er denote the k-vector space (k ). If (e1 , . . . , en ) is the n canonical basis of k , then the canonical basis of Er is, by deﬁnition, the n N= vectors r eH = eii ∧ ei2 ∧ · · · ∧ eir

for any subset H of {1, . . . , n} consisting of r elements i1 < · · · < ir . We have E0 = k, E1 = kn and Er = {0} if r > n. r 13.6.2 Recall that if V is a subspace of k, then V can be identiﬁed canonically as a subspace of Er . r-dimensional subspaces of kn . If V ∈ Sn,r , then Let r Sn,r be the setof r V = 1, and so V is a line in Er , or a point in the projective space dim P(Er ) = PN −1 . Thus we have a map ψ : Sn,r → PN −1 . Denote by Gn,r the image of ψ. Note that ψ is injective. Indeed, if V, W ∈ Sn,r and (v1 , . . . , vn ) a basis of kn such that (v1 , . . . , vr ) (resp. (vs , . . . , vs+r−1 )) is a basis of V (resp. W ). Then ψ(V ) = ψ(W ) implies that v1 ∧ · · · ∧ vr and vs ∧ · · · ∧ vs+r−1 are proportional. Hence s = 1 (13.6.1) and V = W . 13.6.3 Proposition. The set Gn,r is closed in PN −1 . Proof. Using the notations of 13.6.1, let WH be the aﬃne open subsets of PN −1 deﬁned by eH = 0. Then the N subsets WH cover Er . It suﬃces therefore to prove that Gn,r ∩ WH is closed in WH for all H.

13.6 Grassmannian variety

181

Take for example H = {1, . . . , r}, and write W = WH . Let F0 (resp. F0 ) be the subspace generated by e1 , . . . , er (resp. er+1 , . . . , en ) and p the projection on F0 with respect to the direct sum kn = F0 ⊕ F0 . If (v1 , . . . , vr ) is a basis of V ∈ Sn,r , then writing vi = p(vi ) + wi , wi ∈ F0 , we have ψ(V ) ∈ W ⇔ p(v1 ) ∧ · · · ∧ p(vr ) = 0, which is equivalent to the condition that p induces an isomorphism of the vector spaces V and F0 . So if ψ(V ) ∈ W , then V has a unique basis (v1 , . . . , vr ) of the form vi = ei +

(7)

n

aij ej = ei + ui

j=r+1

for 1 i r. Thus v1 ∧ · · · ∧ vr = e1 ∧ · · · ∧ er +

(e1 ∧ · · · ∧ ui ∧ · · · ∧ er ) + w,

1jr

where w is a linear combination of vectors eH such that H contains at least 2 elements in {r + 1, . . . , n}. Moreover, it is clear that the coeﬃcient of such a vector eH is a polynomial PH in the aij ’s, and this polynomial does not depend on V . Finally, we have e1 ∧ · · · ∧ ui ∧ · · · ∧ er = εaij (e1 ∧ · · · ∧ ei−1 ∧ ei+1 ∧ · · · ∧ er ∧ ej ) j>r

where ε ∈ {−1, 1}. Hence the aij ’s determine completely the decomposition of v1 ∧ · · · ∧ vr in the basis (eH )H . Conversely, given r(n − r) scalars aij and deﬁne the vi as in (7), the subspace V generated by the vectors vi veriﬁes V ∈ Sn,r and ψ(V ) ∈ W . Identifying W with kN −1 , the preceding arguments show that W ∩ Gn,r is isomorphic to the graph of the morphism kr(n−r) → kN −r(n−r)−1 , (aij ) → (PH (aij )). Hence W ∩ Gn,r is closed in W .

13.6.4 Deﬁnition. The closed subvariety Gn,r of PN −1 is called the Grassmannian variety of r-dimensional subspaces of kn . 13.6.5 Remark. Let UH = WH ∩ Gn,r . The proof of 13.6.3 shows that UH is an aﬃne open subset of Gn,r , which is isomorphic to kr(n−r) . 13.6.6 Proposition. The Grassmannian variety Gn,r is irreducible. Proof. Being isomorphic to kr(n−r) , the aﬃne open subsets UH are irreducible. Since the UH ’s cover Gn,r , it suﬃces to show that UH ∩ UJ = ∅ for H = J. Let

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H ∩ J = {α1 , . . . , αs }, H \ (H ∩ J) = {β1 , . . . , βr−s } , J \ (H ∩ J) = {γ1 , . . . , γr−s }, V = keα1 ⊕ · · · ⊕ keαs ⊕ k(eβ1 + eγ1 ) ⊕ · · · ⊕ k(eβr−s + eγr−s ). Then it is clear from the proof of 13.6.3 that ψ(V ) ∈ UH ∩ UJ .

13.6.7 A ﬂag in kn is a chain of subspaces of kn V1 ⊂ V2 ⊂ · · · ⊂ Vn = kn such that dim Vi = i for 1 i n. Denote by Fn the set of ﬂags of kn . Then Fn can be identiﬁed naturally with a subset of G = Gn,1 × Gn,2 × · · · × Gn,n . By 13.5.2, G has a canonical structure of projective variety. Using methods analogue to those in 13.6.3, we can prove that Fn is closed in G. Thus Fn has a structure of projective variety. We shall call Fn , the ﬂag variety of kn .

References and comments • [5], [19], [22], [26], [37], [40], [78]. Being projective, Grassmannian varieties are complete varieties. This fact will prove to be very useful later on.

14 Dimension

The ﬁrst section of this chapter gives an algebraic interpretation of the dimension of a variety as a topological space. We then use this to establish certain results concerning the dimension of algebraic varieties. All the algebraic varieties considered in this chapter are deﬁned over k.

14.1 Dimension of varieties 14.1.1 Recall that the dimension of a topological space and the dimension of an algebra were deﬁned in chapter 1 and chapter 2 respectively. 14.1.2 Proposition. Let X be an aﬃne variety. (i) We have: dim X = dim A(X). (ii) If X is irreducible, then: dim X = tr degk Fract(A(X)). Proof. (i) The map p → V(p) from the set of prime ideals to the set of irreducible subsets of X is an inclusion-reversing bijection (11.1.4, 11.1.7 and 11.2.3). It follows from the deﬁnition of dim X (1.2.2) and the deﬁnition of dim A(X) (2.2.3) that dim X = dim A(X). (ii) Recall from 11.2.3 that if X is an aﬃne variety, then the algebra A(X) of regular functions on X is an integral domain if and only if X is irreducible. The result follows therefore from part (i) and 6.2.3. 14.1.3 Theorem. Let X be an irreducible variety and U a non-empty open subset of X. Then dim X = dim U is ﬁnite and: dim X = tr degk R(X).

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Proof. Let us ﬁrst suppose that X is aﬃne. Let U, V be non-empty aﬃne open subsets of X, then U ∩ V = ∅ because X is irreducible. So there exists f ∈ A(X) such that D(f ) = ∅ and D(f ) ⊂ U ∩ V (11.7.2). It follows from 1.2.3 that: dim D(f ) dim U dim X , dim D(f ) dim V dim X. Hence by remark 1 of 11.7.5, Fract(A(X)) = Fract(A(D(f ))), and 14.1.2 (ii) implies that dim D(f ) = dim U = dim V = dim X. Now any non-empty open subset W of X contains an aﬃne open subset U of X. Using the preceding arguments and 1.2.3, we have dim W = dim X. Let us now suppose that X is not aﬃne. The preceding arguments show that all the aﬃne open subsets of X have the same dimension. Since X is the union of a ﬁnite number of aﬃne open subsets, dim X = dim U for any aﬃne open subset U of X (1.2.3 (iii)). Finally any non-empty open subset W of X contains a non-empty aﬃne open subset. Hence dim W = dim X (1.2.3 (i)). The equality dim X = tr degk R(X) follows immediately from 12.7.6 and 14.1.2. 14.1.4 Corollary. Let X be a non-empty algebraic variety. Then its dimension is ﬁnite and if X1 , . . . , Xn are the irreducible components of X, we have: dim X = max{dim Xi ; 1 i n }. Proof. This follows from 1.2.3 (i), 1.3.5 and 14.1.3. 14.1.5 Examples. 1) A variety of dimension zero is a ﬁnite set (since an irreducible variety of dimension zero is clearly a singleton). 2) We have dim kn = n (6.1.4 and 14.1.2). 3) The projective space Pn is irreducible (13.2.6) and contains an open subset isomorphic to kn (13.2.4). Hence dim Pn = n. 4) The Grassmannian variety Gn,r is irreducible (13.6.6) and contains an open subset isomorphic to kr(n−r) (13.6.5). Hence dim Gn,r = r(n − r). 14.1.6 Proposition. Let Y be a non-empty subvariety of a variety X. (i) We have dim Y dim X. (ii) If X is irreducible and Y is closed, then dim Y = dim X if and only if Y = X. (iii) We have dim Y = dim Y . (iv) If X is irreducible, then dim Y = dim X if and only if Y is open. Proof. Parts (i) and (ii) have been proved in 1.2.3. (iii) Let Y1 , . . . , Yn be the irreducible components of Y . Then by 1.1.14, Y1 , . . . , Yn are the irreducible components of Y . Since Yi is closed in Y (1.1.5),

14.2 Dimension and the number of equations

185

it is locally closed in X. So Yi is open in Yi (1.4.1), and dim Yi = dim Yi (14.1.3). The result follows therefore from 14.1.4. (iv) Let us write Y = F ∩ U where U (resp. F ) is an open (resp. closed) subset of X. By 1.2.3, dim Y dim F dim X. It is now clear by part (ii) and 14.1.3 that dim X = dim Y if and only if Y = U is open. 14.1.7 Proposition. Let X, Y be irreducible varieties. Then: dim(X × Y ) = dim X + dim Y. Proof. By 12.4.5, X ×Y is irreducible. Further, if U ⊂ X and V ⊂ Y are aﬃne open subsets, then U × V is an aﬃne open subset of X × Y . We may therefore assume that X = Spm(A) and Y = Spm(B) are aﬃne. Let m = dim X and n = dim Y . By 6.2.2 and 14.1.2, A (resp. B) contains a subalgebra A0 (resp. B0 ) isomorphic to k[T1 , . . . , Tm ] (resp. k[Tm+1 , . . . , Tm+n ]), and A (resp. B) is generated over A0 (resp. B0 ) by a ﬁnite number of elements f1 , . . . , fr (resp. g1 , . . . , gs ) algebraic over E0 (resp. F0 ), the quotient ﬁeld of A0 (resp. B0 ). Since A(X × Y ) = A ⊗k B, it contains the subalgebra A0 ⊗ B0 k[T1 , . . . , Tn+m ] whose quotient ﬁeld L0 has transcendence degree n + m over k. Now, the quotient ﬁeld L of A ⊗k B is an extension of L0 generated by f1 , . . . , fr , g1 , . . . , gs . So the degree of transcendence of L over k is n + m. 14.1.8 Let us use the notations π : kn+1 \ {0} → Pn and Ui , 0 i n, of 13.1.1 and 13.2.1. Corollary. If X is a closed and irreducible subvariety of Pn , then dim π −1 (X) = 1 + dim X. Proof. If X ∩Ui = ∅, then by 13.3.3, π −1 (X ∩Ui ) is isomorphic to the product (X ∩ Ui ) × (k \ {0}). Thus the result follows from 14.1.7. 14.1.9 Remarks. 1) Let Y be a closed and irreducible subvariety of an irreducible variety X. The codimension of Y in X, denoted by codimX (Y ) or simply codim(Y ) when there is no confusion, is deﬁned to be dim X − dim Y . Hence codimX (Y ) = 0 if and only if Y = X (14.1.6). 2) Let X1 , . . . , Xn be the irreducible components of a variety X. Then by 14.1.6 and 1.1.14, we have codimXi (Xi ∩ Xj ) 1 whenever i = j.

14.2 Dimension and the number of equations 14.2.1 Deﬁnition. A non-empty variety is called of pure dimension or equidimensional if all of its irreducible components has the same dimension. We shall call an irreducible variety of dimension 1 (resp. 2) an algebraic curve (resp. algebraic surface) over k.

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14.2.2 Let X be a variety. As in the aﬃne case, if f ∈ Γ (X, OX ), we denote by V(f ) the closed subvariety of X consisting of elements x ∈ X such that f (x) = 0. If f1 , . . . , fr ∈ Γ (X, OX ), we set: V(f1 , . . . , fr ) = V(f1 ) ∩ · · · ∩ V(fr ). Lemma. We have V(f ) = ∅ if and only if f is invertible in Γ (X, OX ). Proof. If f is invertible, then clearly V(f ) = ∅. Conversely, if V(f ) = ∅, then for any aﬃne open subset U ⊂ X, f |U ∈ Γ (U, OX (U )) veriﬁes V(f |U ) = ∅. Thus f |U is invertible with inverse, say gU . If V is another aﬃne open subset, then gU |U ∩V = gV |U ∩V since it is the inverse of f |U ∩V . So there exists an element g ∈ Γ (X, OX ) such that g|U = gU for any aﬃne open subset U . Hence f g = 1. 14.2.3 The following theorem, also called Krull’s Principal Ideal Theorem (or Hauptidealsatz), is a geometrical translation of 6.3.2. Theorem. Let X be an irreducible variety and f be a non-zero and noninvertible element of Γ (X, OX ). Then V(f ) is a non-empty subvariety of X of pure dimension dim X − 1. Proof. By 14.2.2, V(f ) = ∅. Let Y1 , . . . , Yr be the irreducible components of V(f ). There exists y ∈ Y1 such that y ∈ Y2 ∪ · · · ∪ Yr (1.1.13), and since Y2 ∪ · · · ∪ Yr is closed, there exists an aﬃne open subset U ⊂ X such that y ∈ U and U ∩ V(f ) = U ∩ Y1 is irreducible. By 14.1.3, dim U ∩ Y1 = dim Y1 and dim U = dim X. So we may assume that X = Spm(A) is √ aﬃne with A an integral domain, f ∈ A is non-zero and non-invertible, and Af is a prime ideal. The result follows immediately from 6.2.6 and 6.3.2. 14.2.4 Remark. If X is not irreducible, then V(f ) is not necessary of pure dimension (even if X is of pure dimension). For example, let X = V(T1 T2 ) ⊂ k2 and f the restriction to X of the map (t1 , t2 ) → t1 (t1 + t2 + 1), then we check easily that V(f ) is the disjoint union of a line and a point. 14.2.5 Corollary. Let X be a closed irreducible subvariety of Pn such that dim X 1. Given any homogeneous polynomial P ∈ k[T0 , . . . , Tn ] \ k, the hypersurface V(P ) deﬁned by P (x0 , . . . , xn ) = 0 has a non-empty intersection with X. If V(P ) does not contain X, then X ∩ V(P ) is a subvariety of pure dimension dim X − 1. Proof. Let C = π −1 (X) where π is as in 13.1.1, and C the closure of C in kn+1 . Then dim C 2 (14.1.8) and 0 ∈ C is a zero of P , so the restriction of P to C is non-invertible. Now let C be the zero set of P in C; it is an aﬃne cone and a variety of pure dimension with dim C dim X 1 (14.2.3). It follows that C ∩ (kn+1 \ {0}) = ∅, which implies that π(C ) = V(P ) ∩ X = ∅. The last statement is a direct consequence of 13.3.3.

14.3 System of parameters

187

14.2.6 Proposition. Let X be an irreducible variety and f1 , . . . , fr be elements of Γ (X, OX ). Then for any irreducible component Y of V(f1 , . . . , fr ), we have codimX (Y ) r. Proof. Induction on r. The case r = 1 is just 14.2.3. So let r > 1. There exists an irreducible component Y of V(f1 , . . . , fr−1 ) such that Y ⊂ Y . So: Y ⊂ Y ∩ V(fr ) ⊂ V(f1 , . . . , fr ). It follows that Y is an irreducible component of Y ∩ V(fr ). By induction hypothesis, codimX (Y ) r − 1. Now if fr |Y = 0, then Y = Y . Otherwise, dim Y = dim Y − 1 by 14.2.3. In both cases, we have codimX (Y ) r. 14.2.7 Remark. In 14.2.6, irreducible components of V(f1 , . . . , fr ) do not necessary have the same dimension, and it is possible that they all have codimension strictly less than r. 14.2.8 Proposition. Let X be a closed and irreducible subvariety of Pn and P1 , . . . , Pr ∈ k[T0 , . . . , Tn ] be non-constant homogeneous polynomials. We denote by V(P1 , . . . , Pr ) the set of points of X whose homogeneous coordinates are zeros of P1 , . . . , Pr . (i) Any irreducible component Y of V(P1 , . . . , Pr ) veriﬁes codimX Y r. (ii) If dim X r, then V(P1 , . . . , Pr ) is non-empty. Proof. Part (i) follows from 14.2.6 applied to π −1 (X), while (ii) follows from 14.2.5 and a simple induction. 14.2.9 Corollary. If m < n, then the only morphisms from Pn to Pm are the constant maps. Proof. By 14.2.8 (ii), if P1 , . . . , Pr ∈ k[T0 , . . . , Tn ] are non-constant homogeneous polynomials, with r < n, then V(P1 , . . . , Pr ) = ∅. So the result is a consequence of the description of morphisms given in 13.3.11.

14.3 System of parameters 14.3.1 Proposition. Let X be an irreducible variety and Y be a closed irreducible subvariety of codimension 1 in X. (i) There exist an aﬃne open subset U ⊂ X and f ∈ Γ (U, OX ) \ {0} such that Y ∩ U = ∅ and f |Y ∩U = 0. (ii) For any open subset V ⊂ X and any f ∈ Γ (V, OX ) \ {0} verifying Y ∩ V = ∅ and f |Y ∩V = 0, Y ∩ V is an irreducible component of V(f ). Proof. (i) Being irreducible, Y is non-empty, so there exists an aﬃne open subset U = Spm(A) of X such that Y ∩ U = ∅. Let a be the radical ideal of A such that Y ∩ U = V(a). Then a = {0} by 14.1.6 (iv). Thus any f ∈ a \ {0} works.

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(ii) The irreducibility of Y implies that Y ∩ V is irreducible for any open subset V such that Y ∩ V = ∅. So the open subset Y ∩ V of Y is contained in an irreducible component W of V(f ). By 14.1.6, we have: dim V = dim X , dim(Y ∩ V ) = dim Y = dim X − 1. So we obtain by 14.2.3 that: dim V > dim W dim(Y ∩ V ) = dim V − 1. This implies that dim W = dim(Y ∩ V ), and since W and Y ∩ V are closed irreducible subsets of V , we have W = Y ∩ V (14.1.6). 14.3.2 Corollary. Let X be an irreducible variety, F the set of closed irreducible subvarieties of X which are distinct from X, and F∗ the set of maximal elements of F. If Y ∈ F, then Y ∈ F∗ if and only if codimX (Y ) = 1. Proof. If codimX (Y ) = 1, then Y ∈ F∗ by 14.1.6. Conversely, let Y ∈ F∗ . For any aﬃne open subset U = Spm(A) of X, Y ∩ U is a closed irreducible subset of U , distinct from U . Thus Y ∩ U = V(p) for some p ∈ Spec A. It follows that Y ∩ U ⊂ V(f ) for any f ∈ p \ {0}, and so Y ∩ U is contained in an irreducible component Z of V(f ), and dim Z = dim X − 1. By 1.1.12, Y = Y ∩ U and so Y ⊂ Z. Since Z is closed in U , it is locally closed in X. So Z is open in Z and dim Z = dim Z (14.1.6). So the result follows from the maximality of Y . 14.3.3 Using the notations of 14.3.1, we can ask if for any x ∈ Y , there is an open subset U of X containing x and f ∈ Γ (X, OX ) such that the variety Y ∩ U is exactly V(f ). In general, the answer is no. However: Proposition. Let X be an irreducible aﬃne variety such that A(X) is factorial. If Y is a closed subvariety of X of pure dimension dim X − 1, then there exists f ∈ A(X) such that Y = V(f ). Proof. Let Y1 , . . . , Yr be the irreducible components of Y and pi = I(Yi ) ∈ Spec(A(X)), 1 i r. By 11.2.4, the pi ’s are minimal among the prime ideals of A(X). Hence pi = A(X)fi , where fi is an irreducible element of A(X). Finally, we obtain that Y = V(f ) where f = f1 · · · fr . 14.3.4 Corollary. Any closed subvariety Y of Pn of pure dimension n − 1 is deﬁned by a single equation P (x0 , . . . , xn ) = 0 where P ∈ k[T0 , . . . , Tn ] is a non-constant homogeneous polynomial. In other words, Y is a hypersurface. Proof. Let Y1 , . . . , Yr be the irreducible components of Y and Ci = π −1 (Yi ), C = π −1 (Y ). By 14.1.8, 13.3.3 and 13.3.9, the Ci ’s are closed irreducible cones of dimension n in kn+1 \ {0}. The closure Ci of Ci in kn+1 is again irreducible of dimension n. We see easily that the Ci ’s are the irreducible components of C. Thus C is of pure dimension n in kn+1 . Since it is a closed aﬃne cone, there exists a non-constant homogeneous polynomial P ∈ k[T0 , . . . , Tn ] such that C = V(P ) (14.3.3). Hence our result.

14.3 System of parameters

189

14.3.5 Proposition. Let X = Spm(A) be an irreducible aﬃne variety and Y a closed irreducible subvariety of codimension r 1 in X. For any integer s between 1 and r, there exist f1 , . . . , fs ∈ A such that: (i) Y ⊂ V(f1 , . . . , fs ). (ii) The irreducible components of V(f1 , . . . , fs ) are of codimension s. Consequently, there exist f1 , . . . , fr ∈ A such that Y is an irreducible component of V(f1 , . . . , fr ). We say that the fi ’s form a system of parameters of Y. Proof. Let us proceed by induction on s. Since Y = X, I(Y ) = {0}. If s = 1, any f1 ∈ I(Y ) \ {0} works by 14.2.3. Let us suppose that s > 1. The induction hypothesis says that there exist f1 , . . . , fs−1 ∈ A such that Z = V(f1 , . . . , fs−1 ) is a subvariety of pure dimension containing Y , and its irreducible components Z1 , . . . , Zn have codimension s − 1 in X. Since s − 1 < r, Zi ⊂ Y , so I(Y ) ⊂ I(Zi ). Hence I(Y ) ⊂ I(Z1 ) ∪ · · · ∪ I(Zn ) (2.2.5). Let fs ∈ I(Y ) \ (I(Z1 ) ∪ · · · ∪ I(Zn )). We have Y ⊂ V(f1 , . . . , fs ). If W is an irreducible component of V(f1 , . . . , fs ), then by 14.2.6, we have codimX (W ) s. But W ⊂ V(f1 , . . . , fs−1 ), so W ⊂ Zi for some i and W = Zi since fs ∈ I(Zi ). We conclude by 14.1.6 that codimX (W ) s. Hence codimX (W ) = s. 14.3.6 Corollary. Let A be a ﬁnitely generated k-algebra which is an integral domain, p ∈ Spec(A), n = dim A, and r = dim A/p. Then there exists a chain {0} = p0 ⊂ p1 ⊂ · · · ⊂ pn−r = p ⊂ pn−r+1 ⊂ · · · ⊂ pn of prime ideals of A, containing p, of length n. Proof. The existence of the pk ’s, k n − r, is just the deﬁnition of dim A/p. Consider A as the algebra of regular functions on an irreducible aﬃne variety X, p corresponds to a closed irreducible subvariety of X of dimension r. So the result follows by applying 14.3.5. 14.3.7 Corollary. Let X be a closed irreducible subvariety of Pn . For any closed irreducible subvariety Y of codimension r in X, there exist r nonconstant homogeneous polynomials P1 , . . . , Pr ∈ k[T0 , . . . , Tn ] such that Y is an irreducible component of X ∩ V(P1 , . . . , Pr ). Proof. We prove, as in 14.3.5, that there exist non-constant homogeneous polynomials P1 , . . . , Ps , 1 s r, such that Y ⊂ V(P1 , . . . , Ps ) and X ∩ V(P1 , . . . , Ps ) is a subvariety of pure dimension. If Z1 , . . . , Zm are the irreducible components of X ∩ V(P1 , . . . , Ps ), then the ideals I(π −1 (Zj )) and I(π −1 (Y )) are graded. We can ﬁnish the proof as in 14.3.5 by using 7.2.9.

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14.4 Counterexamples 14.4.1 Let us return to the point mentioned in 12.7.10. Denote by s : P1 × P1 → P3 the Segre embedding (13.5.1). Thus the points z, z ∈ P1 with homogeneous coordinates (x0 , x1 ) and (x0 , x1 ) are sent to the point s(z, z ) of P3 with homogeneous coordinates: (x0 x0 , x0 x1 , x1 x0 , x1 x1 ). It is easy to check that the image C of s is the set of points of P3 whose homogeneous coordinates y0 , y1 , y2 , y3 verify y0 y3 − y1 y2 = 0. By 13.5.1, C is irreducible. Let π : k4 \ {0} → P3 be the canonical surjection (13.1.1), and X the aﬃne cone in k4 deﬁned by x1 x4 − x2 x3 = 0. Then π(X \ {0}) = C. It follows from 13.1.2 and 13.3.9 that X is an irreducible subvariety of k4 . On the open subset D(x2 ) ∩ D(x4 ) = D(x2 x4 ) of X, the functions x1 /x2 and x3 /x4 are identical. They are therefore the restriction of a rational function h on X whose domain of deﬁnition U contains D(x2 ) and D(x4 ). Suppose that h ∈ A(X). Then x2 h is regular and is zero on the points (1, 0, γ, 0) of X. But x2 h and x1 are identical on D(x2 x4 ), so the irreducibility of X implies that x2 h and x1 are identical on X. Contradiction. Now suppose that there exists g ∈ A(X) non-invertible such that U = D(g). The complement V(g) of U in X is contained in the plane P ⊂ X of k4 deﬁned by the equations x2 = 0 and x4 = 0. Since P is irreducible, we obtain by 14.1.6 and 14.2.3 that P = V(g). So the restriction of g to the plane P ⊂ X deﬁned by x1 = 0 and x3 = 0 is regular and (0, 0, 0, 0) is the only zero. This is absurd in view of 14.2.3. The preceding argument says that U is contained strictly in D(g) for all functions f, g ∈ A(X) such that h = f /g. As we have seen in 12.7.10, this proves that A(X) is not factorial. 14.4.2 Let us conserve the notations of 14.4.1 and set x = (0, 0, 0, 0). If U is an open subset of X and f ∈ Γ (U, OX ), let VU (f ) = {y ∈ U ; f (y) = 0}. Let us suppose that there exist an open subset U of X containing x, and f ∈ Γ (U, OX ) such that VU (f ) = P ∩ U . Let W = P ∩ U . Then W contains x and is an open subset of P . So dim W = 2. Let g = f |W . Then g ∈ Γ (W, OX |P ) and VW (g) = VU (f ) ∩ P . But VU (f ) ∩ P = {x}, so its dimension is zero. This contradicts 14.2.3. Thus we have proved that there does not exist any open subset U of X containing x and any f ∈ Γ (U, OX ) verifying VU (f ) = P ∩ U (see 14.3.3).

References • [5], [19], [22], [26], [37], [40], [78].

15 Morphisms and dimension

We study properties of morphisms of algebraic varieties in this chapter. The notions of an aﬃne morphism and a ﬁnite morphism are introduced. We also consider the relation between dimension and morphism. In particular, the dimension of a ﬁbre of a morphism is considered.

15.1 Criterion of aﬃneness 15.1.1 Let X be a variety and U an open subset of X, we shall denote often Γ (U, OX ) by OX (U ). Furthermore, if U is aﬃne, then we shall also write A(U ) for OX (U ). If f ∈ OX (U ), we set: DU (f ) = {x ∈ U ; f (x) = 0}. 15.1.2 Lemma. Let X be a variety and f ∈ OX (X). Then we have DX (f ) = X if and only if f is invertible in OX (X). Proof. This is just a reformulation of 14.2.2. 15.1.3 Lemma. Let X be a variety, A = OX (X) and f ∈ A. (i) The restriction to DX (f ) of f is invertible in OX (DX (f )). (ii) The identity map A → A induces an isomorphism of Af onto OX (DX (f )). Proof. Let Y = DX (f ) and OY = OX |Y (9.3.7 and 9.3.8). By 12.2.6 and 12.5.4, (Y, OY ) is a variety. If g = f |Y , then DY (g) = Y and so g is invertible in OX (Y ) by 15.1.2. It follows from 2.3.4 that there is a unique ϕ ∈ Homalg (Af , OX (Y )) such that ϕ(h/f n ) = (h|Y )/(f n |Y ). We shall prove that ϕ is an isomorphism. Let (Ui )i∈I be a covering of X by aﬃne open subsets. Since DX (f ) ∩ Ui = DUi (f |Ui ), we deduce that OX (Y ∩ Ui ) = A(Ui )f |Ui (12.1.1).

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Let h/f n ∈ ker ϕ. Then (h|Y )/(f n |Y ) = 0, so (h|Y ∩Ui )/(f n |Y ∩Ui ) = 0. Thus (f |Ui )(h|Ui ) = 0 (11.7.3), which implies that f h = 0. Hence h/f n = 0, and ϕ is injective. Let g ∈ OX (Y ). Then g|Y ∩Ui ∈ OX (Y ∩ Ui ) = A(Ui )f |Ui . So there exist ni ∈ N and hi ∈ A(Ui ) such that (f ni |Y ∩Ui )(g|Y ∩Ui ) = hi |Y ∩Ui . Let m be maximal among the ni ’s and i = (f m−ni |Ui ).hi ∈ A(Ui ). We have ( i −

j )|Y ∩Ui ∩Uj = 0. Since Ui ∩Uj is an aﬃne open subset (12.5.6), (f |Ui ∩Uj ).(( i −

j )|Ui ∩Uj ) = 0 (11.7.3), and so (f i )|Ui ∩Uj = (f j )|Ui ∩Uj . We deduce therefore that there exists ∈ A such that |Ui = (f |Ui ) i for all i. It follows that (f m+1 |Y )g = |Y , and so ϕ is surjective. 15.1.4 Let X, Y be varieties with Y aﬃne. Recall that given a morphism v : X → Y , we have the associated k-algebra homomorphism: Γ (u) : OY (Y ) → OX (X) , g → g ◦ v. By 12.3.3, the map v → Γ (v) is a bijection between Mor(X, Y ) and Homalg (OY (Y ), OX (X)). Suppose further that A = OX (X) is a ﬁnitely generated k-algebra, then the identity map A → A = OX (X) induces a morphism u : X → Spm(A). Proposition. The following conditions are equivalent: (i) X is an aﬃne variety. (ii) The k-algebra A is ﬁnitely generated and u is an isomorphism. (iii) The k-algebra A is ﬁnitely generated and u is a homeomorphism. Proof. We have seen in 12.3.4 that (i) ⇒ (ii), and (ii) ⇒ (iii) is clear. Suppose that we have (iii). Set Y = Spm(A). Since the DY (f )’s, f ∈ A, form a base of open subsets of Y , the subsets DX (f ) = u−1 (DY (f )) form a base of open subsets of X. Moreover, if we denote uf : DX (f ) → DY (f ) the morphism induced by u, it follows from 15.1.3 and the bijectivity of u that Γ (uf ) : OY (DY (f )) = Af → OX (DX (f )) is an isomorphism. Hence u is an isomorphism (12.2.3) and X is aﬃne.

15.1.5 Theorem. Let X be a variety, A = OX (X) and f1 , . . . , fr ∈ A be such that X = DX (f1 ) ∪ · · · ∪ DX (fr ) and each DX (fi ), 1 i r, is an aﬃne variety. The following conditions are equivalent: (i) X is an aﬃne variety. (ii) A = Af1 + · · · + Afr . Proof. (i) ⇒ (ii) If X is aﬃne, the identity map idA induces an isomorphism u : X → Y = Spm(A) (15.1.4). Since u−1 (DY (f )) = DX (f ), we have Spm(A) = DY (f1 ) ∪ · · · ∪ DY (fr ) , V(Af1 + · · · + Afr ) = ∅.

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193

Hence A = Af1 + · · · + Afr as required (6.5.3). (ii) ⇒ (i) Suppose that A = Af1 + · · · + Afr . First, we shall show that A is a ﬁnitely generated k-algebra. By our hypotheses and 15.1.3, Afi , being isomorphic to OX (DX (fi )), is ﬁnitely generated. Let {si,1 /fini , . . . , si,ri /fini } be a system of generators of Afi where si,k ∈ A. Now there exist 1 , . . . , r ∈ A such that: (1)

1 = 1 f1 + · · · + r fr .

Denote by B the subalgebra of A generated by the fi ’s, the i ’s and the si,k ’s. Let g ∈ A, then g/1 ∈ Afi . So for each i, there exists mi ∈ N such that fimi g = hi ∈ B. Suppose that mi 1. Then by taking a suitable power of (1), there exist t1 , · · · , tr ∈ B such that: 1 = t1 f1m1 + · · · + tr frmr . Hence g = t1 h1 + · · · + tr hr ∈ B, and B = A is ﬁnitely generated. Finally, the identity map idA induces an isomorphism ϕi : Afi → OX (DX (fi )) by 15.1.3, which in turn induces an isomorphism (since DX (fi ) is aﬃne) vi : DX (fi ) → Spm(Afi ). Moreover, vi (x) = u(x) for x ∈ DX (fi ) and u−1 (DY (fi )) = DX (fi ). As vi is a homeomorphism, so is u. Hence X is aﬃne (15.1.4).

15.2 Aﬃne morphisms 15.2.1 Deﬁnition. Let X, Y be varieties. A morphism u : X → Y is called aﬃne if there is a covering (Vi )i∈I of Y by aﬃne open subsets such that u−1 (Vi ) is an aﬃne open subset of X for all i ∈ I. 15.2.2 Lemma. Let X, Y be aﬃne varieties and u : X → Y a morphism. Then for all aﬃne open subset V in Y , u−1 (V ) is an aﬃne open subset. In particular, u is aﬃne. Proof. Since Y is aﬃne, there exist g1 , . . . , gn ∈ A(Y ) such that V = DY (g1 )∪ · · · ∪ DY (gn ). Let fi = gi ◦ u ∈ A(X) and v : u−1 (V ) → V the morphism induced by u. We have u−1 (DY (gi )) = DX (fi ), DY (gi ) = DV (gi |V ) and so u−1 (DY (gi )) = Du−1 (V ) (hi ) where hi = Γ (v)(gi |V ) (here, the notations are as in 15.1.4). Since V is aﬃne, there exist, by 15.1.5, t1 , . . . , tn ∈ OY (V ) such that 1 = t1 (g1 |V ) + · · · + tn (gn |V ). Hence 1 = Γ (v)(t1 )h1 + · · · + Γ (v)(tn )hn . Since X is aﬃne, so is DX (fi ) = Du−1 (V ) (hi ). It follows from 15.1.5 that u−1 (V ) is aﬃne.

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15.2.3 Proposition. The following conditions are equivalent for a morphism u : X → Y of varieties: (i) u is aﬃne. (ii) For any aﬃne open subset V of Y , u−1 (V ) is an aﬃne open subset. Proof. Clearly, (ii) implies (i). Conversely, let u be aﬃne and (Vi )i∈I a covering of Y by aﬃne open subsets such that u−1 (Vi ) is aﬃne for all i ∈ I. Given any aﬃne open subset V of Y and v : u−1 (V ) → V the morphism induced by u, we have ni DV (gij ), V ∩ Vi = j=1

where gij ∈ OY (V ). We have u−1 (DV (gij )) = Du−1 (V ) (hij ) where hij = Γ (v)(gij ). Applying 15.2.2 to the morphism u−1 (Vi ) → Vi , we see that u−1 (Dv (gij )) is aﬃne. Since V is aﬃne, there exist, by 15.1.5, tij ∈ OY (V ) verifying 1= tij gij . i,j

Hence 1=

Γ (v)(tij )hij .

i,j

Thus u−1 (V ) is aﬃne (15.1.5).

15.3 Finite morphisms 15.3.1 Deﬁnition. A morphism u : X → Y of varieties is called ﬁnite if there is a covering (Vi )i∈I of Y by aﬃne open subsets such that (i) u−1 (Vi ) is an aﬃne open subset of X for all i ∈ I. (ii) For all i ∈ I, A(u−1 (Vi )) is, via the comorphism A(Vi ) → A(u−1 (Vi )) of the morphism u−1 (Vi ) → Vi induced by u, a ﬁnite A(Vi )-algebra. Remarks. 1) In 15.3.1, we do not exclude the possibility that u−1 (Vi ) may be empty. 2) A ﬁnite morphism is aﬃne. So the inverse image of an aﬃne open subset under a ﬁnite morphism is aﬃne (15.2.3). 15.3.2 Proposition. Let X = Spm(A), Y = Spm(B) be aﬃne varieties, ϕ ∈ Homalg (B, A) and u = Spm(ϕ) : X → Y the morphism associated to ϕ. (i) If A is (via ϕ) a ﬁnite B-algebra, then u is ﬁnite, and for any open subset V of Y , the morphism u−1 (V ) → V , induced by u, is also ﬁnite. (ii) Conversely, if u is ﬁnite, then A is (via ϕ) a ﬁnite B-algebra. Proof. (i) The ﬁrst part is clear. As for the second, it suﬃces to prove the result in the case where V = D(g), g ∈ B. Then u−1 (V ) = D(h) where h = ϕ(g). Let f1 , . . . , fm ∈ A be a system of generators of A viewed as a B-module. If f ∈ A, then there exist t1 , . . . , tm ∈ B such that:

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195

f = ϕ(t1 )f1 + · · · + ϕ(tm )fm . Since regular functions on D(h) are of the form x → f (x)/hn (x), they can be written as: n

n

x → [t1 (u(x))/ (g(u(x))) ]f1 (x) + · · · + [tm (u(x))/ (g(u(x))) ]fm (x). Thus the restriction to D(h) of the fi ’s generates A(D(h)) as a A(D(g))module. (ii) By the second part of (i), we can write Y = D(g1 ) ∪ · · · ∪ D(gm ) with gi ∈ B, and the ring Ahi is a ﬁnitely generated Bgi -module where hi = ϕ(gi ). Denote by (fij /hni i )j a ﬁnite system of generators of Ahi . Multiplying the fij by some power of hi , we may assume that all the ni ’s are equal to the some integer n. Let f ∈ A, and fi its restriction to D(hi ). We have fi = ϕ(tij )fij /hni ⇒ hni f = ϕ(tij )fij . j

j

Since the set of D(gi ) = D(gin ) covers Y , there exist s1 , . . . , sm ∈ B such that n . But 1 = s1 g1n + · · · + sm gm ϕ(si gin )f = ϕ(si tij )fij , j

so we have f=

ϕ(si tij )fij .

i,j

Hence A is a ﬁnite B-algebra. 15.3.3 Corollary. Let u : X → Y be a morphism of varieties. The following conditions are equivalent: (i) u is ﬁnite. (ii) For all aﬃne open subset V ⊂ Y , u−1 (V ) is an aﬃne open subset of X and A(u−1 (V )) is a ﬁnite A(V )-algebra. Proof. This is clear by 15.2.3 and 15.3.2. 15.3.4 Proposition. Let u : X → Y be a ﬁnite morphism of varieties. (i) The map u is closed. (ii) For all y ∈ Y , the set u−1 (y) is ﬁnite. Proof. (i) Let (Vi )i∈I be a covering of Y by aﬃne open subsets. If F is a closed subset of X, we have u(F ) ∩ Vi = u(F ∩ u−1 (Vi )). It suﬃces therefore to show that u(F ) ∩ Vi is closed in Vi for all i. So we are reduced to the case where X = Spm(A) and Y = Spm(B) are aﬃne and u = Spm(ϕ) for some ϕ ∈ Homalg (B, A).

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Let Z = V(a) be a closed subset of X and b = ϕ−1 (a). The points of u(Z) are ϕ−1 (m), with m a maximal ideal of A containing a. Identifying B/b with a subring of A/a via ϕ, A/a is a ﬁnitely generated B/b-module. Now by 3.3.2 and 3.3.3, if n ∈ Spm(B/b), then n = m ∩ (B/b) for some m ∈ Spm(A/a). Let m ∈ Spm(A) and n ∈ Spm(B) be such that m = m/a and n = n/b. Then n = ϕ−1 (m). It follows that u(Z) = V(b), and u(Z) is closed in X. (ii) Again, we may assume that X = Spm(A) and Y = Spm(B) are aﬃne and we only need to consider y ∈ u(X). By (i), u(X) = V(b) where b = ker ϕ. Identify B/b as a subring of A as above. According to 3.3.7, there is a ﬁnite number of maximal ideal of A lying above a maximal ideal of B/b. So the result follows. 15.3.5 Proposition. (i) The composition of ﬁnite morphisms is again ﬁnite. (ii) Let Y be a closed subvariety of X. The canonical injection Y → X is ﬁnite. (iii) Let u : X → Y be a ﬁnite morphism, Z be a closed subvariety of Y containing u(X), j : Z → Y the canonical injection and v : X → Z the morphism such that u = j ◦ v. Then v is ﬁnite. (iv) Let ui : Xi → Yi , i = 1, 2, be ﬁnite morphisms. Then the morphism v : X1 × X2 → Y1 × Y2 deﬁned by (x1 , x2 ) → (u1 (x1 ), u2 (x2 )), is ﬁnite. Proof. (i) This is clear from 15.3.3 and 3.1.7. (ii) Since the intersection of an aﬃne open subset U of X with Y is an aﬃne open subset of Y which is closed in U , we are reduced to the case where X = Spm(A) is aﬃne and Y = V(a) with a a radical ideal of A. The result now follows from the fact that the A-module A/a is generated by 1. (iii) Let z ∈ Z and Vz an aﬃne open neighbourhood of z in Y . Then u−1 (Vz ) = v −1 (Z ∩ Vz ) is an aﬃne open subset and the morphism u−1 (Vz ) → Vz induced by u is ﬁnite (15.3.3). Thus we may assume that X = Spm(A), Y = Spm(B) are aﬃne and Z = Spm(B/b) where b is an ideal of B. Since u is ﬁnite, A is a ﬁnitely generated B-module via A(u) which is just the composition B → B/b → A. Hence A is also a ﬁnitely generated B/b-module. (iv) The aﬃne open subsets V ×W , with V (resp. W ) an aﬃne open subset of Y1 (resp. Y2 ), cover Y1 × Y2 . So we may assume that Xi = Spm(Ai ) and Yi = Spm(Bi ), i = 1, 2, are aﬃne. If Ai is a ﬁnitely generated Bi -module, then it is clear that A1 ⊗k A2 is a ﬁnitely generated B1 ⊗k B2 -module. 15.3.6 Remark. Let Y be a subvariety of X. In general, the canonical injection Y → X is not ﬁnite. For example, the injection k \ {0} → k is not ﬁnite since its image is not closed (15.3.4). 15.3.7 Example. Let us consider the morphism u : k → k2 , t → (t2 , t3 ), in 11.5.7. Recall that its image is C = V(T13 − T22 ), and A(C) = k[T 2 , T 3 ]. Let v : k → C be the morphism induced by u. Since the A(C)-algebra k[T ]

15.4 Factorization and applications

197

is ﬁnite, v is ﬁnite. The canonical injection j : C → k2 is ﬁnite by 15.3.5 (ii), and so by 15.3.5 (i), u = j ◦ v is ﬁnite. 15.3.8 Corollary. Let u : X → Y be a ﬁnite dominant morphism of irreducible varieties. (i) u is surjective and R(X) is a ﬁnite algebraic ﬁeld extension of R(Y ). In particular, dim X = dim Y . (ii) If C ⊂ X is nowhere dense in X, then u(C) is nowhere dense in Y . Proof. (i) Since u(X) is dense in Y and closed (15.3.4), u is surjective. Let V be a non-empty aﬃne open subset of Y . Then A(u−1 (V )) is a ﬁnitely generated A(V )-module. Let x1 , . . . , xn be a system of generators, then each xi is integral over A(V ). This implies that the xi ’s are algebraic over R(Y ) = Fract(A(V )) (12.7.6). Since R(X) = Fract(A(u−1 (V ))), the result follows from 14.1.2. (ii) Since u(C) = u(C), we may assume that C is closed. It suﬃces then to prove that if Z is an irreducible component of C (so Z is closed in X by 1.1.5), then u(Z) is nowhere dense in Y . Since u(Z) is an irreducible closed subset of Y (1.1.7 and 15.3.4), it follows from 15.3.5 that the restriction Z → u(Z) of u is a surjective ﬁnite morphism. According to (i) and 14.1.6, dim u(Z) = dim Z < dim X = dim Y. Using 14.1.6 again, we deduce that u(Z) is nowhere dense in Y . 15.3.9 Corollary. Let u : X → Y be a ﬁnite morphism of varieties. If Y is complete, then X is complete. Proof. Let Z be a variety, p : X × Z → Z and q : Y × Z → Z the canonical projections. The morphism v : X × Z → Y × Z deﬁned by (x, z) → (u(x), z), satisﬁes p = q ◦ v. By 15.3.5 (iv), v is ﬁnite. So v is a closed map (15.3.4), and since q is closed by hypothesis, so is p. Thus X is a complete variety.

15.4 Factorization and applications 15.4.1 Proposition. let X be an irreducible aﬃne variety of dimension n. There exists a surjective ﬁnite morphism u : X → kn . Proof. Let X = Spm(A), where A is a ﬁnitely generated k-algebra which is an integral domain. According to 6.2.2 and 14.1.2, there is a subalgebra B of A, isomorphic to k[T1 , . . . , Tn ], such that A is integral over B. So A is a ﬁnite Balgebra (3.3.5), and the morphism u : X → Spm(B) = kn which corresponds to the canonical injection B → A, is then a ﬁnite morphism (15.3.2). It is dominant by 6.5.9, and so it is surjective (15.3.4). 15.4.2 Proposition. Let u : X → Y be a dominant morphism of irreducible varieties.

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(i) There exist m ∈ N, V ⊂ Y a non-empty aﬃne open subset and a covering (Ui )1ir of u−1 (V ) by aﬃne open subsets such that the restriction ui : Ui → V factors through v

pr

i Ui −→ V × km −→ V,

where pr is the canonical surjection and vi a surjective ﬁnite morphism. (ii) The set u(X) contains a non-empty open subset of Y . Proof. We only need to prove (i). Given W = Spm(B) a non-empty aﬃne open subset of Y , u−1 (W ) is a non-empty open subset of X since u is dominant. So u−1 (W ) is the ﬁnite union of non-empty aﬃne open subsets U1 , . . . , Ur . Again, since u is dominant, the u(Ui )’s are dense in Y . Let Ui = Spm(Ai ) where Ai is a ﬁnitely generated k-algebra which is an integral domain. The morphism B → Ai induced by u is injective (6.5.9). There exist, by 6.2.9, si ∈ B \ {0} and a subalgebra Ci of Ai containing B such that Ci B[T1 , . . . , Tmi ] = B ⊗k k[T1 , . . . , Tmi ] and Ai [1/si ] is a ﬁnitely generated Ci [1/si ]-module. The quotient ﬁeld of Ai [1/si ] is Li = Fract(Ai ) and the quotient ﬁeld of Ci is isomorphic to K(T1 , . . . , Tmi ) = Li where K = Fract(B). It follows that Li and Li have the same transcendence degree over k (3.3.4 and 6.2.3). This degree is equal to dim Ui = dim X (14.1.2). We deduce therefore that all the mi ’s are equal to some integer m. Let s = s1 · · · sr . Then Ci [1/s] = B[1/s] ⊗k k[T1 , . . . , Tm ]. Take V = Spm(B[1/s]) = DY (s) , Ui = Spm(Ai [1/s]) , 1 i r. Then we are done since the morphism vi is the one associated to the injection Ci [1/s] → Ai [1/s]. 15.4.3 Proposition. Let u : X → Y be a morphism of varieties and C a constructible subset of X. Then u(C) is constructible. Proof. Let X1 , . . . , Xr be the irreducible components of X. Since u(C) = u(C ∩ X1 ) ∪ · · · ∪ u(C ∩ Xr ) and C ∩ Xi is constructible in Xi , we may assume that X is irreducible. We shall proceed by induction on the dimension n of X. The case n = 0 is obvious. Since C is Noetherian, it is the ﬁnite union of its irreducible components. They are constructible in X since they are closed in C. Thus we may further assume that C is irreducible. So C is an irreducible closed subvariety of X. Suppose that C = X. Then dim C < X and C is constructible in C. By the induction hypothesis, u(C) is constructible. Suppose that C = X. It suﬃces to show that u(C) is constructible in u(X). So we may assume that Y is irreducible and u is dominant. Since C is constructible and dense in X, C contains a non-empty open subset U of X (1.4.5), and the restriction of u to U is still dominant. By 15.4.2, u(U ) ⊂ u(C) contains a non-empty open subset V of Y . So we have:

15.5 Dimension of ﬁbres of a morphism

199

u(C) = V ∪ u(C ∩ u−1 (Y \ V )). The set u−1 (Y \ V ) is a closed subvariety of X distinct from X since u is dominant and Y \ V is nowhere dense in Y . Moreover, C ∩ u−1 (Y \ V ) is constructible in u−1 (Y \ V ). By the induction hypothesis, u(C ∩ u−1 (Y \ V )) is constructible in Y , and we have ﬁnished our proof.

15.5 Dimension of ﬁbres of a morphism 15.5.1 Let u : X → Y be a morphism of varieties and X , Y subvarieties of X and Y such that u(X ) ⊂ Y . We say that X dominates Y if the morphism X → Y induced by u is dominant. 15.5.2 Lemma. Let u : X → Y be a surjective ﬁnite morphism of irreducible varieties and W an irreducible closed subvariety of Y . There is at least one irreducible component of u−1 (W ) which dominates W , and any such irreducible component has dimension dim W . Proof. Let Z1 , . . . , Zr be the irreducible components of u−1 (W ). Then u(Zi ) is closed in Y (15.3.4) and since u is surjective, W = u(Z1 ) ∪ · · · ∪ u(Zr ). The irreducibility of W implies that W = u(Zi ) for at least one i. Now if W = u(Zi ), then 15.3.5 implies that the morphism Z → W induced by u is ﬁnite, and so dim W = dim Zi (15.3.8). 15.5.3 Theorem. Let u : X → Y be a dominant morphism of irreducible varieties. (i) We have dim X dim Y . (ii) If W is an irreducible closed subvariety of Y and Z an irreducible component of u−1 (W ) which dominates W . Then (2)

dim Z dim W + dim X − dim Y.

(iii) There exists a non-empty open subset V of Y such that for all irreducible closed subvariety W of Y verifying W ∩ V = ∅, there is at least one irreducible component of u−1 (W ) which dominates W , and for such an irreducible component Z, we have (3)

dim Z = dim W + dim X − dim Y.

Proof. Since u is dominant, the comorphism (of ﬁelds) R(Y ) → R(X) is injective (12.7.11). Therefore, dim Y = tr degk R(Y ) tr degk R(X) = dim X. By 15.4.2 (i), there exist an integer m ∈ N, a non-empty open subset V ⊂ Y and a covering (Ui )1ir of u−1 (V ) by aﬃne open subsets such that u|Ui = pr ◦vi where vi : Ui → V × km is a surjective ﬁnite morphism and pr : V × km → V is the canonical surjection. Since vi is ﬁnite and surjective, we have m = dim X − dim Y (15.3.8). Let W be an irreducible subvariety of Y verifying W ∩ V = ∅. Then (W ∩ V ) × km is an irreducible closed subvariety of dimension m + dim W

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in V × km . Applying 15.5.2 to vj , we obtain that there is at least one irreducible component of vi−1 ((V ∩ W ) × km ) which dominates W , and such an irreducible component has dimension m + dim W . Since u−1 (V ∩ W ) is the union of vi−1 ((V ∩ W ) × km ), 1 i r, we have proved that there exists an irreducible component of u−1 (W ) which dominates W . If Z is such an irreducible component, then Z ∩ Ui is dense in Z for some i and Z ∩ Ui dominates (V ∩ W ) × km . Hence dim Z = m + dim W . So we have part (iii). To prove (ii), replace Y by an aﬃne open subset V which meets W , and replace X by u−1 (V ), we may assume that Y is aﬃne. Let s = codimY (W ). According to 14.3.5, there exist f1 , . . . , fs ∈ A(Y ) such that W is an irreducible component of V(f1 , . . . , fs ). Let gi = fi ◦ u ∈ A(X). If Z is an irreducible component of u−1 (W ) dominating W , then Z ⊂ T = V(g1 , . . . , gs ). We claim that Z is an irreducible component of T . Let Z be an irreducible component of T containing Z, then W = u(Z) ⊂ u(Z ) ⊂ V(f1 , . . . , fs ). Since W is an irreducible component of the latter, we have u(Z ) ⊂ W and so Z ⊂ Z ⊂ u−1 (W ). Since Z is an irreducible component of u−1 (W ), we have Z = Z as claimed. From our claim, we deduce that codimX (Z) s = codimY (W ) (14.2.6). Hence we have (2). 15.5.4 Corollary. Let u : X → Y be a dominant morphism of irreducible varieties. (i) If y ∈ u(X), each irreducible component of u−1 (y) has dimension at least dim X − dim Y . (ii) There exists a non-empty open subset of V ⊂ u(X) such that for any y ∈ V , each irreducible component of u−1 (y) has dimension dim X − dim Y . 15.5.5 Corollary. Let u : X → Y be a morphism of varieties. (i) If dim u−1 (y) r for all y ∈ Y , then dim X r + dim Y . (ii) If u is dominant and dim u−1 (y) = r for all y ∈ u(X), then dim X = r + dim Y . Proof. (i) Let Y1 , . . . , Yn be the irreducible components of Y and for 1 i n, (Xij )j be the irreducible components of u−1 (Yi ). Let z ∈ u(Xij ). Applying 15.5.4 to the restriction Xij → u(Xij ), we have dim Xij dim u(Xij ) + dim u−1 (z) r + dim Y. Since X is the union of the Xij ’s, the result follows. (ii) Let us conserve the notations above andlet Yi be an irreducible component of Y of dimension dim Y . Since Yi \ ( j=i Yj ) is open and nonempty in Y (1.3.6), the restriction u−1 (Yi ) → Yi is dominant. Removing certain Xij ’s if necessary, we may assume that all the Xij ’s dominate Yi . If dim Xij < r + dim Y for all j, then applying 15.5.4 (ii) to the restriction Xij → Yi , we can ﬁnd points y such that dim u−1 (y) < r which contradicts our hypothesis. Hence we have dim X = r + dim Y .

15.5 Dimension of ﬁbres of a morphism

201

15.5.6 Corollary. Let u : X → Y be a dominant birational morphism of irreducible varieties. There exists a non-empty open subset V of Y such that the restriction u−1 (V ) → V of u is an isomorphism. Proof. By our hypothesis, dim X = dim Y . Replace Y by an aﬃne open subset W and X by u−1 (W ), we are reduced to the case where Y = Spm(B) is aﬃne. Let U = Spm(A) be a non-empty aﬃne open subset of X (so u(U ) is dense in Y ) and W = u(X \ U ). If T is an irreducible component of X \ U , then by 14.1.6, dim T < dim X = dim Y . Thus dim Z < dim Y for any irreducible component Z of W (15.5.3). It follows from 14.1.6 that there exists g ∈ B such that W ∩ DY (g) = ∅, and so u−1 (DY (g)) ⊂ U . If f = g ◦ u ∈ A, then u−1 (DY (g)) = DU (f ). Replacing Y by DY (g) and X by DU (f ), we may suppose that X = Spm(A) and Y = Spm(B) are aﬃne where A, B are integral domains, and the comorphism ϕ : B → A extends to an isomorphism of ﬁelds Fract(B) → Fract(A). It follows that if a1 , . . . , am ∈ A are such that A = k[a1 , . . . , am ], then there exist b1 , . . . , bm ∈ B and s ∈ B \ {0} such that ai = bi /s for 1 i m. So A[1/s] and B[1/s] are isomorphic. Thus V = DY (s) works. 15.5.7 Let E be a topological space and f : E → N a map. We say that f is upper semi-continuous if {x ∈ E; f (x) n} is closed in E for all n ∈ N. Theorem. Let u : X → Y be a morphism of irreducible varieties. For x ∈ X, set e(x) to be the maximum dimension of any irreducible component of dim u−1 (u(x)) containing x. Then the map e : X → N is upper semicontinuous. Proof. Replacing Y by u(X), we may assume that u is dominant. Set r = dim X − dim Y 0 (15.5.3), and for n ∈ N, Sn (u) = {x ∈ X ; e(x) n}. We proceed by induction on p = dim Y . The case p = 0 is obvious since Sn (u) is the union of irreducible components of X of dimension at least n. If n r, then Sn (u) = X (15.5.4 (i)). So suppose that n > r. By 15.5.4, there is a non-empty open subset V of Y such that Sn (u) ⊂ X \ u−1 (V ). Let W1 , . . . , Ws be the irreducible components of Y \ V . We have dim Wj < dim Y (14.1.6). For 1 j s, let (Zij )j be the irreducible components of u−1 (Wi ). Denote by vij : Zij → Wi the restriction of u. By the induction hypothesis, the Sn (vij )’s are closed in Zij , and so in X. But if x ∈ X \u−1 (V ), any irreducible components of u−1 (u(x)) containing x is −1 (vij (x)). contained in one of the Zij ’s, so it is an irreducible component of vij Thus Sn (u) is the union of the Sn (vij ), and so Sn (u) is closed in X. 15.5.8 Corollary. Let X, Y be varieties. The canonical projections pr1 : X × Y → X, pr2 : X × Y → Y are open.

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Proof. Let X1 , . . . , Xm (resp. Y1 , . . . , Yn ) be the irreducible components of X (resp. Y ). If U is open in X × Y , then pr1 (U ) is open if pr1 ((Xi × Y ) ∩ U ) = Xi ∩ pr1 (U ) is open in Xi for all i. Moreover, pr1 (U ∩ (Xi × Y )) = pr1 (U ∩ (Xi × Y1 )) ∪ · · · ∪ pr1 (U ∩ (Xi × Yn )). So we may assume that X, Y are irreducible. Let U be an open subset of X × Y and F = X \ pr1 (U ). If x ∈ F and −1 z ∈ pr−1 1 (x), then pr1 (x) = {x} × Y is the unique irreducible component of −1 pr1 (pr1 (z)) ∩ ((X × Y ) \ U ) containing z. Its dimension is dim Y . Now if x ∈ F , U ∩ pr−1 1 (x) is not empty, so the irreducible components of (x) are all of dimension strictly less than dim Y (14.1.6). ((X × Y ) \ U ) ∩ pr−1 1 (F ) is the set of points z of the variety (X × Y ) \ U Hence F × Y = pr−1 1 such that e(z) dim Y (with respect to the restriction of pr1 ). By 15.5.7, F × Y is closed, and so F is closed (12.4.5). 15.5.9 Examples. 1) Let X ⊂ k3 be the cone deﬁned by the equation x1 x3 − (x21 + x22 ) = 0 and u : X → k2 = Y the projection (x1 , x2 , x3 ) → (x1 , x2 ). The morphism u is dominant and birational since k(T1 , T2 , (T12 + T22 )T1−1 ) = k(T1 , T2 ). Clearly V = D(T1 ) ⊂ Y satisﬁes the conclusion of 15.5.6. Denote by W the line x1 = 0, which is just the complement of V in k2 . The set u−1 (W ) is the line deﬁned by x1 = 0 and x2 = 0 in k3 . Thus u(u−1 (W )) = {(0, 0)}, and no irreducible components of u−1 (W ) dominates W . Moreover dim u−1 (W ) = 1 > dim X − dim Y = 0. Since u−1 (W ) = u−1 (0, 0), we see that, in 15.5.4, we can have dim u−1 (y) > dim X − dim Y . 2) Let us consider the following morphism v : X = k2 → k3 , (t1 , t2 ) → (t21 − 1, t1 (t21 − 1), t2 ). The image Y of v is the cylinder x22 − x21 − x31 = 0. Let u : X → Y be the morphism induced by v. It is a ﬁnite morphism. The image W of the morphism w : k → Y , t → (t2 − 1, t(t2 − 1), t), is irreducible of dimension 1. We verify easily that the diagonal ∆ of k and the point (1, −1) are the irreducible components of u−1 (W ). Clearly the irreducible components consisting of the point (1, −1) does not dominate W since dim W = 1. Note that u(k2 \ ∆) = (Y \ W ) ∪ {(0, 0, 1)}. Thus the image of the open set 2 k \ ∆ is not open. So a surjective ﬁnite morphism is not necessarily open.

15.6 An example

203

15.6 An example 15.6.1 For p, q ∈ N∗ , let Mp,q be the set of p by q matrices with coeﬃcients in k and we write Mp for Mp,p . Endow Mp,q with its natural structure of irreducible aﬃne algebraic variety of dimension pq. Then the map Mr,p × Mp,q × Mq,s → Mr,s , (A, B, C) → ABC is a morphism. 15.6.2 The group GLn (k) is a principal open subset of Mn (11.7.5), so it is an irreducible aﬃne variety of dimension n2 . The map GLn (k) → GLn (k), A → A−1 , is an automorphism of the variety GLn (k). 15.6.3 Let r ∈ {0, 1, . . . , min(p, q)}. Denote by Cr = {A ∈ Mp,q ; rk(A) r} , Cr = {A ∈ Mp,q ; rk(A) = r}. In other words, A ∈ Cr if and only if all the (r + 1)-minors of A are zero. It follows that Cr is closed in Mp,q . Similarly, if A ∈ Cr , then A ∈ Cr if and only if at least one of the r-minors of A is non-zero. Thus Cr is open in Cr . Let Ir ∈ Mr be the identity matrix and Ir 0 ∈ Mp,q . J= 0 0 It is well-known that the image of the morphism w : Mp × Mq → Mp,q , (P, Q) → P JQ, is Cr . So Cr is irreducible in Mp,q . Let u : Mp × Mq → Cr be the morphism induced by w. We have u(GLp (k) × GLq (k)) = Cr . Let v : GLp (k) × GLq (k) → Cr be the morphism induced by u. Let A ∈ Cr and (U, V ) ∈ GLp (k) × GLq (k) be such that A = U JV . If (P, Q) ∈ GLp (k) × GLq (k), then (P, Q) ∈ v −1 (A) ⇔ (U −1 P, QV −1 ) ∈ v −1 (J). We deduce that all the ﬁbres of v are isomorphic. Writing AB XY P = , Q= CD Z T where A, X ∈ Mr , we have P JQ =

AX AY CX CY

.

Hence (P, Q) ∈ v −1 (J) ⇔ A ∈ GLr (k) , X = A−1 , C = 0 , Y = 0.

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It is now clear that the map GLr (k) × GLp−r (k) × GLq−r (k) × Mr,p−r × Mq−r,r → v −1 (J) −1 AD A 0 (A, B, C, D, E) → , 0 B E C is an isomorphism. We deduce from the above discussion that for any A ∈ Cr , v −1 (A) is irreducible and: dim v −1 (A) = r2 + (p − r)2 + (q − r)2 + r(p − r) + r(q − r). Since Cr is open in the irreducible variety Cr , we have dim Cr = dim Cr = r(p + q − r).

References and comments • [5], [19], [22], [26], [37], [40], [78]. Many results of this chapter are due to Chevalley, for example: 15.4.2, 15.4.3, 15.5.7.

16 Tangent spaces

The reader may have come across the notion of a tangent space in diﬀerential geometry. We deﬁne here the analogous algebraic object, the Zariski tangent space, for an algebraic variety. We shall see later that the Zariski tangent space appears in relating algebraic groups and Lie algebras.

16.1 A ﬁrst approach 16.1.1 Given P ∈ k[T1 , . . . , Tn ] = k[T] and x ∈ kn , set: Dx (P ) =

∂P ∂P (x)T1 + · · · + (x)Tn . ∂T1 ∂Tn

The map P → Dx (P ) is k-linear, and if P, Q ∈ k[T], then (1)

Dx (P Q) = P (x)Dx (Q) + Q(x)Dx (P ).

16.1.2 Let X be a closed subvariety of kn , a = I(X) the associated radical ideal in k[T] and x = (x1 , . . . , xn ) ∈ X. Denote by nx = I({x}) the maximal ideal of A = A(X) = k[T]/a associated to x. 16.1.3 Let ax be the ideal of k[T] generated by the Dx (P ), P ∈ a. Then by (1), if (Pi )i∈I is a system of generators of a, then (Dx (Pi ))i∈I is a system of generators of ax . We call the tangent space of X at x, denoted by Tanx (X), the vector subspace V(ax ) of kn . If r is the rank of the set of linear forms Dx (P ), P ∈ a, then the dimension of Tanx (X) is n − r. Let {Q1 , . . . , Qr } be a basis of the k-vector space spanned by the Dx (P ), P ∈ a. There exist indices i1 , . . . , in−r such that {Q1 , . . . , Qr , Ti1 , . . . , Tin−r } is a basis of the space of homogeneous polynomials of degree 1. Clearly we can identify k[T]/ax with k[Ti1 , . . . , Tin−r ]. It follows that ax is prime and the algebra of regular functions on Tanx (X) is k[T]/ax .

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16.1.4 Now let L ⊂ kn be a line containing x and v = (v1 , . . . , vn ) ∈ kn be such that L is the set of points of the form x + tv, t ∈ k. Let (Pi )1is be a system of generators of a. Then the points in L ∩ X are obtained by solving the following system of equations in t: (2)

P1 (x + tv) = · · · = Ps (x + tv) = 0. For 1 i n, we have Pi (x + tv) = t

n ∂P i (x)vj + t2 Qi (t) j=1 ∂Tj

where Qi ∈ k[T ]. Thus, t = 0 is a “multiple root” of (2) if and only if, ∂Pi ∂Pi (x)v1 + · · · + (x)vn = 0 ∂T1 ∂Tn for 1 i s. Hence t = 0 is a multiple root of (2) if and only if v ∈ Tanx (X). 16.1.5 Let P, Q ∈ k[T] be such that P − Q ∈ a. By the deﬁnition of Tanx (X), the restriction of Dx (P ) − Dx (Q) to Tanx (X) is zero. So we can deﬁne a linear map Dx : A → (Tanx (X))∗ , the dual space of Tanx (X), by setting Dx (f ) = Dx (F )|Tanx (X) where f ∈ A and F ∈ k[T] is a representative of f . The map Dx is obviously surjective. Furthermore, since A = k ⊕ nx and Dx (λ) = 0 for any λ ∈ k, Dx induces a linear surjection dx : nx → (Tanx (X))∗ . Proposition. The kernel of dx is n2x . Thus dx induces an isomorphism nx /n2x → (Tanx (X))∗ of k-vector spaces. Proof. Let Nx = k[T](T1 − x1 ) + · · · + k[T](Tn − xn ) be the maximal ideal of k[T] associated to x. Then a ⊂ Nx and nx = Nx /a. It is clear from (1) that n2x ⊂ ker dx . Conversely, let f ∈ ker dx and F ∈ Nx be a representative of f . Since I(Tanx (X)) = ax (16.1.3), there exist F1 , . . . , Fs ∈ a and λ1 , . . . , λs ∈ k such that Dx (F ) = λ1 Dx (F1 ) + · · · + λs Dx (Fs ). Set G = F − λ1 F1 − · · · − λs Fs , then Dx (G) = 0. So ∂G ∂G (x) = · · · = (x) = 0. ∂T1 ∂Tn Since G(x) = 0, it follows from Taylor’s formula that G is contained in the ideal generated by (Ti − xi )(Tj − xj ), 1 i, j n, which is N2x . Finally, G is also a representative of f because F1 , . . . , Fs ∈ a. Hence f ∈ n2x .

16.2 Zariski tangent space

207

16.1.6 It follows from the preceding discussion that we can identify Tanx (X) with (nx /n2x )∗ . We now give another interpretation of Tanx (X). Let ϕ : A → OX,x be the map sending f ∈ A to its germ at x, and mX,x the maximal ideal of OX,x . We have seen in 12.8.3 that ϕ induces a local isomorphism ψ : Anx → OX,x . By 2.6.12, the vector spaces nx /n2x and nx Anx /n2x An are isomorphic. So we have proved: Proposition. The vector spaces Tanx (X) and (mX,x /m2X,x )∗ are isomorphic.

16.2 Zariski tangent space 16.2.1 Using the discussion in 16.1, we can deﬁne the tangent space at a point of an arbitrary variety as follows: Deﬁnition. Let X be a variety and x ∈ X. We deﬁne the Zariski tangent space at x, denoted by Tx (X), the vector space (mX,x /m2X,x )∗ . 16.2.2 Let us denote by ρ : mX,x → mX,x /m2X,x the canonical surjection. If s ∈ OX,x , we denote by s(x) its value at x. Recall that OX,x = k ⊕ mX,x . Let us consider k as an OX,x -module via the homomorphism deﬁned by the evaluation at x, that is s.λ = s(x)λ for all s ∈ OX,x , λ ∈ k. A k-derivation of OX,x in k (2.8.1) is then a k-linear map δ : OX,x → k verifying δ(st) = s(x)δ(t) + t(x)δ(x) where s, t ∈ OX,x . Such a derivation shall be called a point derivation of OX,x , and we denote by Derxk (OX,x , k) the set of these derivations. In the same way, if U is an aﬃne open subset containing x, we can consider k as an A(U )-module by setting f.λ = f (x)λ, for f ∈ A(U ) and λ ∈ k. We shall also call a k-derivation of A(U ) to k, a point derivation of A(U ), that is, a k-linear map ∆ : A(U ) → k verifying for f, g ∈ A(U ), ∆(f g) = f (x)∆(g) + g(x)∆(f ). We shall denote by Derxk (A(U ), k) the set of these derivations. 16.2.3 Let λ ∈ Tx (X) and deﬁne a map Lλ : OX,x → k by Lλ |k = 0 , Lλ |mX,x = λ ◦ ρ. If s ∈ OX,x , then s − s(x) ∈ mX,x . We deduce easily that if s, t ∈ OX,x , then (3)

Lλ (st) = s(x)Lλ (t) + t(x)Lλ (s).

Thus Lλ ∈ Derxk (OX,x , k). Conversely, let δ ∈ Derxk (OX,x , k). If s is the germ at x of the constant function 1, then δ(s) = δ(ss) = s(x)δ(s) + s(x)δ(s) = 2δ(s).

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Hence δ(s) = 0. Further, if s, t ∈ OX,x verify s(x) = t(x) = 0, then δ(st) = 0. Thus δ|m2X,x = 0. It follows that there is a linear form λ on mX,x /m2X,x such that δ|mX,x = λ ◦ ρ = Lλ . Hence δ = Lλ . We have therefore identiﬁed the tangent space Tx (X) of X at x with the vector space Derxk (OX,x , k). An element of Derxk (OX,x , k) shall be called a tangent vector of X at x. 16.2.4 Let U be an aﬃne open subset of X containing x and let δ ∈ Derxk (OX,x , k). For f ∈ A(U ), set ∆U δ (f ) = δ(fx ), where fx denotes the germ ∈ Derxk (A(U ), k). of f at x. We verify easily that ∆U δ x Conversely, let ∆ ∈ Derk (A(U ), k). If g ∈ A(U ) is such that its germ at x is 0, then by 11.7.3, there exists f ∈ A(U ) verifying f (x) = 0 and f g = 0. So 0 = ∆(f g) = f (x)∆(g) + g(x)∆(f ) = f (x)∆(g). Hence ∆(g) = 0. Now let s ∈ OX,x , there exists f ∈ A(U ) such that f (x) = 0 and s is the germ of some function g/f on D(f ). Let us assume that s is also the germ of a function g1 /f1 with f1 (x) = 0. Then gf1 − g1 f is identically zero in a neighbourhood of x. From the preceding paragraph, we have ∆(gf1 −g1 f ) = 0, which implies that g(x)∆(f1 ) − f (x)∆(g1 ) = g1 (x)∆(f ) − f1 (x)∆(g). A simple computation gives: f12 (x)[f (x)∆(g) − g(x)∆(f )] = f 2 (x)[f1 (x)∆(g1 ) − g1 (x)∆(f1 )]. From this, we deduce that if h = g/f is deﬁned on D(f ) where f, g ∈ A(U ) and f (x) = 0, we can deﬁne a linear form δ on OX,x by setting: δ(hx ) = f −2 (x)[f (x)∆(g) − g(x)∆(f )] where hx is the germ of h at x. We now verify easily that δ ∈ Derxk (OX,x , k) and ∆U δ = ∆. We have therefore shown that the tangent space Tx (X) can be identiﬁed with Derxk (A(U ), k). In particular, if U is an aﬃne open subset of X containing x, then we can identify Tx (X) with Tx (U ). 16.2.5 We shall give yet another interpretation of Tx (X). Let U be an aﬃne open subset of X containing x. Consider k as an A(U )-module as in 16.2.2, and denote it by kx . Let ΩU = Ωk (A(U )) be the module of diﬀerentials of A(U ) over k (see 2.9.3). Recall from 2.9.5 that the map f → f ◦ dA(U )/k is a k-linear isomorphism from HomA(U ) (ΩU , kx ) to Derk (A(U ), kx ) = Derxk (A(U ), k). Let nx = I({x}) be the maximal ideal of A(U ) associated to x, and set

16.3 Diﬀerential of a morphism

209

ΩUx = ΩU /nx ΩU . Since A(U ) = k ⊕ nx as vector spaces, we can identify the A(U )-module kx with A(U )/nx via the canonical surjection A(U ) → A(U )/nx . So the vector spaces ΩUx and kx ⊗A(U ) ΩU are isomorphic. On the other hand, it is easy to verify that the map ΩU → kx ⊗A(U ) ΩU , α → 1 ⊗ α, induces a k-linear bijection Ψ : Homk (kx ⊗A(U ) ΩU , k) → HomA(U ) (ΩU , kx ), deﬁned by (Ψ (u))(α) = u(1 ⊗ α) for all u ∈ Homk (kx ⊗A(U ) ΩU , k) and α ∈ ΩU . It follows from the preceding paragraphs that there is a k-linear isomorphism from Derxk (A(U ), k) to Homk (ΩUx , k). So by 16.2.4, we may also identify the tangent space Tx (X) with the dual of ΩUx .

16.3 Diﬀerential of a morphism 16.3.1 Let X, Y be varieties, u ∈ Mor(X, Y ) and x ∈ X. We saw in 12.8.6 that u induces a local morphism ux : OY,u(x) → OX,x . Since ux (mY,u(x) ) is contained in mX,x , we have ux (m2Y,u(x) ) ⊂ m2X,x . It follows that ux induces a map ux : mY,u(x) /m2Y,u(x) → mX,x /m2X,x . We shall call the dual map Tx (u) : Tx (X) → Tu(x) (Y ) of ux , the diﬀerential of the morphism u at x. It is also sometimes denoted by dux . To avoid confusions with the elements of Ωk (A(U )), we shall not use the notation dux in this chapter. A straightforward veriﬁcation shows that if v : Y → Z is a morphism, then: Tx (v ◦ u) = Tu(x) (v) ◦ Tx (u). 16.3.2 Let us conserve the notations of 16.3.1. Let y = u(x), and θx : Tx (X) → Derxk (OX,x , k) , θy : Ty (Y ) → Deryk (OY,y , k) be the isomorphisms deﬁned in 16.2.3. Denote by Tx (u) : Derxk (OX,x , k) → Deryk (OY,y , k) the unique linear map such that Tx (u) ◦ θx = θy ◦ Tx (u). If λ ∈ Tx (X) and µ = Tx (u)(λ), then θx (λ)|k = 0 , θx (λ)|mX,x = λ ◦ ρx , θy (µ)|k = 0 , θy (µ)|mY,y = µ ◦ ρy where ρx : mX,x → mX,x /m2X,x and ρy : mY,y → mY,y /m2Y,y denote the canonical surjections. Since µ = λ ◦ ux , we have θy (µ)|mY,y = λ ◦ ux ◦ ρy = λ ◦ ρx ◦ (ux |mY,y ). We deduce therefore that Tx (u) is the map: Derxk (OX,x , k) → Deryk (OY,y , k) , δ → δ ◦ ux .

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16.3.3 Let V be an aﬃne open neighbourhood of y = u(x) in Y and U an aﬃne open neighbourhood of x in X such that u(U ) ⊂ V . Denote by αx : Derxk (OX,x , k) → Derxk (A(U ), k) , αy : Deryk (OY,y , k) → Deryk (A(V ), k) the isomorphisms in 16.2.4. Let Tx (u) : Derxk (A(U ), k) → Deryk (A(V ), k) be the unique linear map such that αy ◦ Tx (u) = Tx (u) ◦ αx . It follows from 16.2.4 and 16.3.2 that if δ ∈ Derxk (OX,x , k) and g ∈ A(V ), (αy ◦ Tx (u)(δ))(g) = (αy ◦ δ ◦ ux )(g) = δ ◦ ux (gy ) = δ((g ◦ u)x ) = αx (δ)(g ◦ u). where gy is the germ of g at y. It follows that if A(u) is the comorphism of u, then Tx (u) : Derxk (A(U ), k) → Deryk (A(V ), k) , ∆ → ∆ ◦ A(u). Remark. When there is no confusion, we shall also denote Tx (u) and by Tx (u).

Tx (u)

16.3.4 Let us take X = km and Y = kn . Then A(X) = k[T1 , . . . , Tm ], A(Y ) = k[S1 , . . . , Sn ], Tanx (X) = km , Tany (Y ) = kn . Let P1 , . . . , Pn ∈ A(X) be such that u(x) = (P1 (x), . . . , Pn (x)) for all x ∈ X. We have the following bijective linear maps: βx : Derxk (A(X), k) → Tanx (X) = km , ∆ → (∆(T1 ), . . . , ∆(Tm )) , βy : Deryk (A(Y ), k) → Tany (Y ) = kn , ∆ → (∆(S1 ), . . . , ∆(Sn )). Let Fx (u) : km → kn be the linear map such that Fx (u) ◦ βx = βy ◦ Tx (u). If Q ∈ A(Y ) and ∆ ∈ Derxk (A(X), k), then 16.3.3 implies that: m ∂(Q ◦ u) (Tx (u)(∆)) (Q) = ∆(Q ◦ u) = (x)∆(Ti ) ∂Ti i=1 m n ∂Q ∂Pj (u(x)) (x) ∆(Ti ) = ∂Ti j i=1 j=1 ∂S n ∂Q m ∂P j (y) (x)∆(Ti ) . = j=1 ∂Sj i=1 ∂Ti

Hence Fx (u) is the linear map k

m

m m ∂P ∂P1 n → k , (v1 , . . . , vm ) → (x)vi , . . . (x)vi . i=1 ∂Ti i=1 ∂Ti n

Let Mx (u) = [aij ] ∈ Mn,m (k) be the matrix of Fx (u) with respect to the canonical bases of km and kn . Then for 1 i m, 1 j n, we have: aij =

∂Pj (x). ∂Ti

16.3 Diﬀerential of a morphism

211

16.3.5 Now assume that X is a closed subvariety of Y = kn . Let j : X → Y be the canonical injection, a = I(X) ⊂ A(Y ) = k[T1 , . . . , Tn ] and, for 1 i n, let ti be the image of Ti in A(X) = A(Y )/a. Let v = (v1 , . . . , vn ) ∈ Y . For P ∈ a and v ∈ Tanx (X), we saw in 16.1 that ∂P ∂P (x)v1 + · · · + (x)vn = 0. ∂T1 ∂Tn We can then deﬁne an element ∆v ∈ Derxk (A(X), k) by setting for f ∈ A(X) and P any representative of f in A(Y ), ∆v (f ) =

∂P ∂P (x)v1 + · · · + (x)vn . ∂T1 ∂Tn

Conversely, given ∆ ∈ Derxk (A(X), k). We have, for P ∈ A(Y ), ∆(P (t1 , . . . , tn )) =

∂P ∂P (x)∆(t1 ) + · · · + (x)∆(tn ). ∂T1 ∂Tn

In particular, if P ∈ a, then ∆(P (t1 , . . . , tn )) = 0. It follows that we have a bijective linear map γx : Derxk (A(X), k) → Tanx (X) , ∆ → (∆(t1 ), . . . , ∆(tn )). Let Gx (j) : Tanx (X) → Tany (Y ) be the unique linear map such that Gx (j) ◦ γx = βy ◦ Tx (j) where βy is as in 16.3.4. It follows immediately from the preceding discussion that Gx (j) is the canonical injection from Tanx (X) to kn = Tany (Y ). 16.3.6 Now let X and Y be closed subvarieties of km and kn respectively and denote by j1 : X → km and j2 : Y → kn the canonical injections. There exist P1 , . . . , Pn ∈ k[T1 , . . . , Tm ] (not necessarily unique in general) such that u(z) = (P1 (z), . . . , Pn (z)) for all z ∈ X. Let v : km → kn be the morphism deﬁned by z → (P1 (z), . . . , Pn (z)). Then we have j2 ◦ u = v ◦ j1 . From this, we obtain the following commutative diagram: Hx (u)

Tanx (X) ⏐ γx ⏐

−−−−→

Tany (Y ) ⏐γy ⏐

Derxk (A(X), k) ⏐ ⏐ Tx (j1 )

−−−−→

Tx (u)

Deryk (A(Y ), k) ⏐ ⏐T (j ) y 2

Tx (v)

Derxk (k[T1 , . . . , Tm ], k) −−−−→ Deryk (k[S1 , . . . , Sn ], k) where Hx (u) is the unique linear map such that Hx (u) ◦ γx = γy ◦ Tx (u). Consequently, Hx (u) is the linear map

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Tanx (X) → Tany (Y ) , (v1 , . . . , vn ) →

n ∂P n ∂P 1 n (x)vi , . . . , (x)vi . i=1 ∂Ti i=1 ∂Ti

In particular, if : kn → km is the linear map whose matrix [aij ] with re∂Pi (x), then spect to the canonical bases of km and kn is deﬁned by aij = ∂Tj

(Tanx (X)) ⊂ Tany (Y ) and (v) = Hx (u)(v) for all v ∈ Tanx (X). 16.3.7 Proposition. Let X be a subvariety of a variety Y , j : X → Y the canonical injection and x ∈ X. The map Tx (j) : Tx (X) → Tx (Y ) is injective. Proof. The set X is locally closed in Y (12.2.7), so it is the intersection of a closed subset F and an open subset U of Y . Let j1 : X → U , j2 : U → Y be the canonical injections. Since j = j2 ◦ j1 , we have Tx (j) = Tx (j2 ) ◦ Tx (j1 ). By 16.2.4 and 16.3.3, Tx (j2 ) is an isomorphism. So we are reduced to the case where X is closed in Y . We may assume that Y = Spm(A) is aﬃne and X = Spm(A/a) where a is a radical ideal of A. By 16.3.3, if ϕ : A → A/a is the canonical surjection, then Tx (j) is the map Derxk (A/a, k) → Derxk (A, k), ∆ → ∆ ◦ ϕ. It is now clear that Tx (j) is injective. 16.3.8 Let U, V be aﬃne varieties and (x, y) ∈ U × V . Recall from 11.8.1 that the map associating f ⊗ g ∈ A(U ) ⊗k A(V ) to the function (u, v) → f (u)g(v) on U × V is an isomorphism from A(U ) ⊗k A(V ) to A(U × V ). For D ∈ Derxk (A(U ), k) and D ∈ Deryk (A(V ), k), we deﬁne a map θD,D : A(U × V ) → k by setting, for f ∈ A(U ), g ∈ A(V ), θD,D (f ⊗ g) = D(f )g(y) + f (x)D (g). (x,y)

We verify easily that θD,D ∈ Derk (A(U × V ), k). So we have a linear map (x,y) θ : Derxk (A(U ), k)×Deryk (A(V ), k) → Derk (A(U ×V ), k), (D, D ) → θD,D . Since θD,D (f ⊗ 1) = D(f ) and θD,D (1 ⊗ g) = D (g), the map θ is injective. (x,y) Now let ∆ ∈ Derk (A(U × V ), k). For f ∈ A(U ) and g ∈ A(V ), we set: D(f ) = ∆(f ⊗ 1) , D (g) = ∆(1 ⊗ g). Then D ∈ Derxk (A(U ), k), D ∈ Deryk (A(V ), k) and ∆ = θD,D . We have therefore proved that θ is an isomorphism. Proposition. Let X, Y be varieties and (x, y) ∈ X × Y . Denote by j : X × {y} → X × Y , j : {x} × Y → X × Y the canonical injections. Then the map T(x,y) (X × {y}) × T(x,y) ({x} × Y ) → T(x,y) (X × Y ) (D, D ) → T(x,y) (j)(D) + T(x,y) (j )(D ) is a bijection. In particular, Tx (X)×Ty (Y ) and T(x,y) (X ×Y ) are isomorphic as k-vector spaces.

16.4 Some lemmas

213

Proof. By 16.2.4, we may assume that X and Y are aﬃne. Let us identify X × {y} with X and {x} × Y with Y . The comorphism of j (resp. j ) is the map sending f ⊗ g to the function u → f (u)g(y) on X (resp. v → f (x)g(v) on Y ). It follows from 16.3.3 that T(x,y) (j)(D) = θD,0 and T(x,y) (j )(D ) = θ0,D . So the result follows.

16.4 Some lemmas 16.4.1 In this section, k is a commutative integral domain and K its quotient ﬁeld. For λ ∈ k \ {0}, set Sλ = {λn ; n ∈ N} and kλ = Sλ−1 k. Since k is an integral domain, we may identify Sλ−1 k with the subring of K consisting of elements of the form λ−n a with n ∈ N and a ∈ k. For any k-module M , denote by Mλ the kλ -module kλ ⊗k M , which we may identify with Sλ−1 M (2.4.1). Let m, p ∈ N∗ and r ∈ {1, . . . , min(m, p)}. We deﬁne the following matrix: Ir 0 Jr = ∈ Mm,p (k). 0 0 16.4.2 Let Λ = [αij ] ∈ Mm,p (k) and (e1 , . . . , ep ) be the canonical basis of k p . Denote by Mk (Λ) or M(Λ), the quotient of k p by the submodule generated by the elements εi = αil e1 + · · · + αip ep , 1 i m. Lemma. (i) Let P ∈ GLm (k), Q ∈ GLp (k). Then M(P Λ) = M(Λ), and the k-modules M(ΛQ) and M(Λ) are isomorphic. (ii) There exist λ ∈ k \ {0}, P ∈ GLm (kλ ) and Q ∈ GLp (kλ ) such that Λ = P Jr Q, where r is the rank of the matrix Λ considered as an element of Mm,p (K). Proof. (i) Let P = [pij ], P Λ = [αij ] and for 1 i m, set εi = αi1 e1 + · · · + αip ep .

Denote by L (resp. L ) the submodule of k p generated by the εi (resp. εi ). Since εi = pi1 ε1 + · · · + pim εm for 1 i m, L ⊂ L. Now the same argument applied to P −1 implies that L ⊂ L . Therefore L = L and M(P Λ) = M(Λ). ] and for 1 i m and 1 j p, set: Next let Q = [qij ], ΛQ = [αij e1 + · · · + αip ep , ej = qj1 e1 + · · · + qjp ep . εi = αi1

So (e1 , . . . , ep ) is a basis of k p and for 1 i m, εi = αi1 e1 + · · · + αip ep . Let u be the automorphism of k p verifying u(ej ) = ej for 1 j p. Then u induces an isomorphism from M(Λ) to M(ΛQ). (ii) It is well-known that there exist P ∈ GLm (K) and Q ∈ GLp (K) such that Λ = P Jr Q. To obtain the result, it suﬃces to take λ such that the coeﬃcients of P , Q, P −1 , Q−1 belong to kλ .

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16 Tangent spaces

16.4.3 Lemma. Let us conserve the notations of 16.4.2. (i) There exists λ ∈ k \ {0} such that M(Λ)λ is a free kλ -module of rank p − r. (ii) We can choose λ such that p − r images of the elements 1 ⊗ ei of kλ ⊗k k p form a basis of the kλ -module M(Λ)λ . Proof. Part (i) is clear by 16.4.2. As for part (ii), for 1 i p, denote by νi the canonical image of 1 ⊗ ei in M(Λ)λ . Let (f1 , . . . , fp−r ) be a basis of the kλ -module M(Λ)λ . Since the νi ’s generate M(Λ)λ , we may assume that there exists M = [βij ] ∈ Mp−r (kλ ) verifying det M = 0 and νj = β1j f1 + · · · + βp−r,j fp−r for 1 j p − r. Modify λ if necessary, we may assume that the coeﬃcients of the matrix M −1 ∈ GLp−r (K) are in kλ . So part (ii) follows. 16.4.4 Let X be an irreducible aﬃne variety and x ∈ X. The ring k = A(X) is an integral domain and recall from 12.7.6 that we can identify its quotient ﬁeld K with the ﬁeld of rational functions R(X) on X. Let Λ = [αij ] ∈ Mm,p (k) be as in 16.4.3, r its rank considered as an element of Mm,p (K) and let Λ(x) ∈ Mm,p (k) be the matrix [αij (x)]. Consider k as a k-module, denoted by kx , as in 16.2.2. Set Mk (Λ)x = Mk (Λ)/nx Mk (Λ). Then Mk (Λ)x = kx ⊗k Mk (Λ), and it is clear that we may identify Mk (Λ)x with Mk (Λ(x)) as k-vector spaces. Lemma. Let E = {x ∈ X; dimk Mk (Λ)x = p − r}. (i) E is a non-empty open subset of X. (ii) We have dimK K ⊗k Mk (Λ) = p − r. (iii) If x ∈ E, then there exists f ∈ k = A(X) verifying f (x) = 0, and Mk (Λ)f is a free kf -module of rank p − r. Proof. (i) Let Ls (resp. Ls (x)) be the set of square submatrices of size s in Λ (resp. Λ(x)). Then det Θ = 0 if Θ ∈ Ls with s > r, and there exists Θ ∈ Lr such that det Θ = 0. Similarly, since det Θ ∈ k, we have det A = 0 if A ∈ Ls (x) with s > r, and the set F of x ∈ X for which there exists A ∈ Lr (x) verifying det A = 0, is a non-empty open subset of X. Since we can identify Mk (Λ)x with Mk (Λ(x)), we have E = F. So we have proved part (i). (ii) This is clear because the rank of Λ in Mm,p (K) is r. (iii) Let x ∈ E. There exists A ∈ Lr (x) such that det A = 0. Reindexing if necessary, we may assume that A = [αij (x)]1i,jr and denote by Θ the matrix [αij ]1i,jr . We have Θ ∈ GLr (kf ) where f = det Θ. Let (e1 , . . . , ep ) be the canonical basis of k p and for 1 j p, νj the image of 1 ⊗ ej in Mk (Λ)f . Then for 1 i m, αi1 ν1 + · · · + αip νp = 0, or equivalently

16.5 Smooth points

215

αi1 ν1 + · · · + αir νr = −(αi,r+1 νr+1 + · · · + αip νp ). It follows from our choice of A that ν1 , . . . , νr are kf -linear combinations of νr+1 , . . . , νp . So by (ii), the elements 1 ⊗ νr+1 , . . . , 1 ⊗ νp form a basis of the K-vector space K ⊗kf Mk (Λ)f . Consequently νr+1 , . . . , νp are linearly independent over kf . Thus Mk (Λ)f is a free kf -module of rank p − r.

16.5 Smooth points 16.5.1 Proposition. Let X be a variety and x ∈ X. The dimension of Tx (X) is greater than or equal to the dimension of any irreducible component of X containing x. Proof. By 16.2.4 and 16.3.7, we may assume that X is irreducible and aﬃne. Let X = Spm(A) and nx = I({x}). Since the k-vector spaces nx /n2x and mX,x /m2X,x are isomorphic, it suﬃces to show that dimk nx /n2x dim X. Let f1 , . . . , fr ∈ nx be such that their images in nx /n2x form a basis of the k-vector space nx /n2x . By 2.6.9, we have nx = Af1 + · · · + Afr . Consider the morphism u : X → kr , y → (f1 (y), . . . , fr (y)). If u(y) = 0, then nx ⊂ ny and so nx = ny , which in turn implies that x = y. For y ∈ X, denote by e(y) the maximum dimension of an irreducible component of u−1 (u(y)) containing y. By 15.5.7, the set of y ∈ X verifying e(y) 1 is closed in X, so the set U of y ∈ X such that e(y) = 0 is open in X. From the previous paragraph, we have x ∈ U . Let v : U → kr denote the restriction of u to U . Then dim v −1 (z) = 0 for all z ∈ v(U ). Thus dim X = dim U dim kr = r (15.5.4 (i)) which proves our result. 16.5.2 Deﬁnition. Let X be an irreducible variety of dimension n. A point x ∈ X is called smooth (or simple) if dim Tx (X) = n. The variety X is called smooth (or non-singular) if all its points are smooth. 16.5.3 Remarks. 1) When X is not irreducible, let dimx (X) denote the maximum dimension of irreducible components of X containing x. Then x is a smooth point of X if Tx (X) = dimx X. Similarly, the variety X is smooth if all its points are smooth. 2) We saw in 16.1 that the aﬃne space kn is smooth. Similarly the projective space Pn is also smooth, because any point of Pn is contained in an open aﬃne neighbourhood isomorphic to kn . By using the same argument, we see that the Grassmannian varieties Gn,r are smooth. 3) If x is a smooth point in an irreducible variety X, then we can show that the ring OX,x is factorial, and hence an integrally closed domain. 16.5.4 Proposition. Let X = Spm(A) be an irreducible aﬃne variety, x ∈ X a smooth point, nx the maximal ideal of A associated to x, and f1 , . . . , fn ∈ nx be such that their images in nx /n2x form a basis of the kvector space nx /n2x . Then f1 , . . . , fn are algebraically independent over k, and

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so they form a transcendence basis of the ﬁeld of rational functions R(X) over k. Proof. Consider the following morphism: u : X → kn , y → (f1 (y), . . . , fn (y)). Suppose that there exists P ∈ k[T1 , . . . , Tn ] \ {0} such that P (f1 , . . . , fn ) = 0, then u(X) ⊂ V(P ) and so dim u(X) n − 1 (14.2.3). Now proceed as in the proof of 16.5.1, we deduce that dim X n − 1 which is absurd. So the fi ’s are algebraically independent over k. The last part follows from 12.7.6 and 5.4.8. 16.5.5 Let X be an irreducible variety, x ∈ X and U an aﬃne open subset of X containing x. Recall that ΩU = Ωk (A(U )) is the module of diﬀerentials of A(U ) over k. Theorem. Let X be an irreducible variety of dimension n. (i) Let x be a smooth point of X. There is an aﬃne open neighbourhood U of x such that ΩU is a free A(U )-module of rank n. (ii) The set of smooth points in X is a non-empty open subset of X. Proof. Let U be an aﬃne open subset of X. By 12.7.6, we may identify the quotient ﬁeld of A(U ) with R(X). Further, the R(X)-vector spaces R(X)⊗A(U ) ΩU and Ωk (R(X)) are isomorphic (2.9.6). Since the characteristic of k is zero, it follows from 5.8.3 and 14.1.2 that dimR(X) Ωk (R(X)) = tr degk R(X) = n. So we may assume that X = Spm(A) is aﬃne with A = k[T1 , . . . , Tp ]/a where a is a radicalidealof k[T1 , . . . , Tp ]. Let {P1 , . . . , Pm } be a system of generators ∂Pi ∈ Mm,p (A(X)) and r the rank of Λ considered as an for a, Λ = ∂Tj ij element of Mm,p (R(X)). Using the notations of 16.4.2, the A(X)-modules ΩX and MA(X) (Λ) are isomorphic (2.9.8). In view of 16.4.4, we obtain that n = p − r. By 16.2.5 and 16.4.4, the k-vector spaces Tx (X) and MA(X) (Λ)x have the same dimension. So part (ii) follows by 16.4.4. Finally, let x ∈ X be a smooth point. By 16.4.4, there exists f ∈ A(X) such that f (x) = 0 and MA(X) (Λ)f is a free A(X)f -module of rank n. If U = D(f ), then U is an aﬃne open subset of X containing f such that A(U ) = A(X)f . So we have part (i). 16.5.6 Remark. If X is equidimensional and if X is the disjoint union of its irreducible components, then 16.5.5 implies that the set of smooth points points of X is a dense open subset of X.

16.5 Smooth points

217

16.5.7 Let u : X → Y be a morphism of irreducible aﬃne varieties, x ∈ X and y = u(x). Recall that any A(X)-module can be considered as an A(Y )module via the comorphism A(u) of u. Denote by kx (resp. ky ) the A(X)-module (resp. A(Y )-module) such that f.λ = f (x)λ (resp. g.λ = g(y)λ) for all f ∈ A(X), g ∈ A(Y ) and λ ∈ k. Observe that kx and ky are isomorphic as A(Y )-modules. Denote dA(X)/k and dA(Y )/k by dX and dY . We saw in 2.9.5 that there exists a unique homomorphism θ(u) : ΩY → ΩX of A(Y )-modules such that θ(u) ◦ dY = dX ◦ A(u). Recall from 16.3.3 that Tx (u) can be identiﬁed with the map Derk (A(X), kx ) → Derk (A(Y ), ky ) , ∆ → ∆ ◦ A(u). Now, using the following canonical isomorphisms (2.9.5): HomA(X) (ΩX , kx ) → Derk (A(X), kx ), HomA(Y ) (ΩY , ky ) → Derk (A(Y ), ky ), we can identify Tx (u) with the map HomA(X) (ΩX , kx ) → HomA(Y ) (ΩY , ky ) , ϕ → ϕ ◦ θ(u). Finally, with the identiﬁcations in 16.2.5, Tx (u) can be identiﬁed with the x , k) → Homk (ΩYy , k), ϕ → ψ, where map from Homk (ΩX ψ(λ1 dY (g1 ) + · · · + λs dY (gs )) = ϕ(λ1 dX (g1 ◦ u) + · · · + λs dX (gs ◦ u)) for λ1 , . . . , λs ∈ k and g1 , . . . , gs ∈ A(Y ). Theorem. Let u : X → Y be a morphism of irreducible varieties. (i) Suppose that there exists a smooth point x ∈ X such that y = u(x) is a smooth point in Y and Tx (u) : Tx (X) → Ty (Y ) is surjective. Then u is dominant. (ii) Suppose that u is dominant. There exists a non-empty open subset U of X such that for all x ∈ U , u(x) is a smooth point in Y and Tx (u) is a surjection from Tx (X) onto Tu(x) (Y ). Proof. In view of 16.5.5, we are reduced to the case where X and Y are aﬃne, smooth and irreducible, and ΩX (resp. ΩY ) is a free A(X)-module (resp. A(Y )-module) of rank m = dim X (resp. n = dim Y ). We have the following homomorphism of free A(X)-modules: ϕ(u) : A(X) ⊗A(Y ) ΩY → ΩX , 1 ⊗ α → θ(u)(α). Let us ﬁx bases for these free modules and let Λ = [aij ] ∈ Mm,n (A(X)) be the matrix of ϕ(u) with respect to these bases. For x ∈ X, set Λ(x) = [aij (x)] ∈ Mm,n (k). With the notations of 16.2.5, let x ϕ(u)x : ΩYy = kx ⊗A(X) (A(X) ⊗A(Y ) ΩY ) → kx ⊗A(X) ΩX = ΩX

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be the homomorphism induced by ϕ(u) by scalar extension. In suitable bases, Λ(x) is the matrix of ϕ(u)x . The preceding description shows that ϕ(u)x is the dual map of Tx (u). Thus ϕ(u)x and Tx (u) have the same rank. (i) Suppose that Tx (u) is surjective. Then ϕ(u)x is injective, so the rank of Λ(x) is n. As in the proof of 16.4.4, we deduce that the rank of Λ (over R(X)) is at least n. But the rank of Λ is at most n, so we have equality. It follows that ϕ(u) is injective, which implies in turn that θ(u) is injective. Finally, since ΩX (resp. ΩY ) is a free A(X)-module (resp. A(Y )-module), A(u) is injective (5.8.4). Hence u is dominant (6.5.9). (ii) If u is dominant, then A(u) is injective. By 5.8.4, ϕ(u) is injective. Hence the rank of Λ (over R(X)) is n. There exists therefore a non-empty open subset U of X such that for x ∈ U , the rank of Λ(x) is n. It follows from the discussion above that Tx (u) is surjective for all x ∈ U .

References • [5], [19], [22], [26], [37], [40], [78].

17 Normal varieties

17.1 Normal varieties 17.1.1 Let us begin by a remark which will be useful later on. Let X be a prevariety and U, V aﬃne open subsets of X such that U ∩ V = ∅. Any x ∈ U ∩ V has a fundamental system of neighbourhoods consisting of subsets D(s) such that D(s) ⊂ U ∩ V and s ∈ A(U ). For each of these subsets D(s), there exists t ∈ A(V ) such that D(t) is a neighbourhood of x in D(s). Denote by u : A(V ) → A(U )s = A(D(s)) the comorphism of the canonical injection D(s) → V . We have D(t) = D(u(t)), and u(t) is of the form r/sn for some r ∈ A(U ). Thus D(u(t)) = D(sr) where sr ∈ A(U ). Hence open subsets contained in U ∩ V which are principal in both U and V , form a base for the topology on U ∩ V . 17.1.2 Deﬁnition. A point x in a variety X is said to be normal if the local ring OX,x of X at x is an integrally closed domain. The variety X is normal if all its points are normal. 17.1.3 Remarks. 1) If X is irreducible, then it is normal if and only if it admits an open covering consisting of normal subvarieties. 2) If x is a normal point of X, then it is contained in a unique irreducible component of X (12.8.5). We deduce that if X is normal, then irreducible components of X are pairwise disjoint. So we may reduce the study of normal varieties to irreducible normal varieties. 17.1.4 Proposition. Let X = Spm(A) be an irreducible aﬃne variety. Then X is normal if and only if A is an integrally closed domain. Proof. Suppose that X is normal. By 12.8.2, we have OX,x . A= x∈X

It follows from 3.2.4 that A is an integrally closed domain.

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Suppose that A is an integrally closed domain. For p ∈ Spec(A), the ring Ap is an integrally closed domain (3.2.5). So X is normal since OX,x is isomorphic to Anx , where nx = I({x}) is the maximal ideal of A associated to x (12.8.3). 17.1.5 Corollary. The set of normal points of an irreducible variety X contains a non-empty open subset of X. Proof. We may assume that X = Spm(A) is aﬃne. By 6.2.2, there exist t1 , . . . , tn ∈ A algebraically independent over k such that A is integral over C = k[t1 , . . . , tn ]. Let B be the integral closure of A. Then B is integral over C (3.1.7), and by 5.5.6, B is a ﬁnitely generated C-module, hence B is a ﬁnitely generated A-module. Let Y = Spm(B) and u : Y → X be the morphism induced by the injection A → B. The morphism u is dominant (6.5.9) and birational since Fract(A) = Fract(B). It follows from 15.5.6 that there exists a non-empty open subset U of X such that the restriction u−1 (U ) → U of u is an isomorphism. By 17.1.4, Y is normal, and the result follows. 17.1.6 Example. The aﬃne space kn is normal (4.2.9 and 17.1.4). Since Pn has an open covering by subsets isomorphic to kn , it is normal. In the same way, the Grassmannian varieties Gn,r are also normal. 17.1.7 Proposition. Let X, Y be irreducible varieties and u : X → Y a surjective ﬁnite morphism. Assume further that Y is normal. Then for all x ∈ X and for all irreducible subvariety W of Y containing u(x), there exists an irreducible component of u−1 (W ) containing x which dominates W . Proof. Since u is ﬁnite, there is an aﬃne open subset V containing u(x) such that U = u−1 (V ) is an aﬃne open neighbourhood of x. We may therefore assume that X = Spm(A), Y = Spm(B) are aﬃne where B ⊂ A are integral domains, B is an integrally closed domain and A is a ﬁnitely generated Bmodule. Let p = I(W ), m = I({x}), n = I({u(x)}). We have n = B ∩ m and p ⊂ n. By 5.7.8, there exists q ∈ Spec(A) such that q ⊂ m and p = B ∩ q. So x ∈ V(q) and u(V(q)) = W . Hence there is an irreducible component of u−1 (W ) containing V(q) and it dominates W . 17.1.8 Corollary. Let u : X → Y be a dominant morphism of irreducible varieties. There exists a non-empty open subset V of Y such that: (i) The variety V is normal. (ii) For all y ∈ V , x ∈ u−1 (y) and irreducible subvariety W of Y containing y, there exists an irreducible component of u−1 (W ) containing x which dominates W . Proof. By 15.4.2, there exist a non-empty aﬃne open subset V of Y , m ∈ N, and a covering (Ui )i∈I of u−1 (V ) by aﬃne open subsets such that the

17.2 Normalization v

221

pr

restriction vi : Ui → V of u factors through Ui →i V × km → V , where pr denotes the canonical projection. Note that vi is a surjective ﬁnite morphism. We may assume that V is normal (17.1.5), and x belongs to one of the Ui ’s. We are reduced to the case where X is one of the Ui ’s, so X is aﬃne. Since A(V × km ) = A(V ) ⊗k k[T1 , . . . , Tm ] = A(V )[T1 , . . . , Tm ], it follows from 3.2.8 that A(V × km ) is an integrally closed domain. So V × km is normal (17.1.4). Now vi is a surjective ﬁnite morphism, so the result follows from 17.1.7.

17.2 Normalization 17.2.1 Theorem. Let X be an irreducible variety and K a ﬁnite ﬁeld extension of R(X), the ﬁeld of rational functions on X. There exist a normal irreducible variety Y such that R(Y ) is isomorphic to K, and a surjective ﬁnite morphism u : Y → X such that there is a covering (Ui )i∈I of X by aﬃne open subsets verifying u−1 (Ui ) is aﬃne and A(u−1 (Ui )) is the integral closure of A(Ui ) in R(Y ). Furthermore, if Y1 is an irreducible variety and u1 : Y1 → X is a surjective ﬁnite morphism having the same properties, then there is a unique isomorphism v : Y → Y1 such that u = u1 ◦ v. Proof. Let us prove the existence of Y . First, suppose that X = Spm(A) is aﬃne. Let B be the integral closure of A in K, and Y = Spm(B). Then B is a ﬁnitely generated A-module (6.2.10). Take u : Y → X to be the morphism induced by the canonical injection A → B. The result follows since K is the quotient ﬁeld of B (5.5.5). Now let us treat the general case. Let (Ui )i∈I be a covering of X by nonempty aﬃne open subsets. For i ∈ I, let Vi = Spm(Bi ) where Bi is the integral closure of Ai = A(Ui ) in K. Denote by θi : Ai → Bi the canonical injection and ui = Spm(θi ). Let i, j ∈ I. By 12.5.6, Ui ∩ Uj is a non-empty aﬃne open subset, and the algebra Aij = A(Ui ∩ Uj ) is generated by the functions x → f (x)g(x) where x ∈ Ui ∩ Uj , f ∈ Ai and g ∈ Aj . Let Bij be the integral closure of Aij in K, θij : Aij → Bij the canonical injection and uij = Spm(θij ). Let U ⊂ Ui ∩ Uj be a principal open subset of both Ui and Uj (17.1.1). We have U = D(t) = D(si ) = D(sj ) with t ∈ Aij , si ∈ Ai and sj ∈ Aj . Observe that if s ∈ Ai \ {0}, then the ring of regular functions on u−1 i (D(s)) = D(θi (s)) is equal to Bi [1/θi (s)], the integral closure in K of the ring (Ai )s = Ai [1/s] of regular functions on D(s) (3.1.9). The same applies for Aij , Bij , θij . −1 It follows from the above observation that the rings A(u−1 i (U )), A(uj (U )) −1 and A(uij (U )) are all equal to Bij [1/θij (t)], which is the integral closure in K of Aij [1/t]. Since the set of open subsets in Ui ∩ Uj which are principal in both Ui and Uj form a base of the topology on Ui ∩ Uj (17.1.1), it follows −1 from 9.5.1 and 12.1.1 that the open subsets u−1 i (Ui ∩ Uj ) and uj (Ui ∩ Uj ) can be identiﬁed canonically.

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Let Y be the prevariety deﬁned by gluing the Vi ’s along the intersections Vi ∩ Vj = Spm(Bij ) via these canonical identiﬁcations. Then the map u : Y → X, which is ui on Vi , is a morphism. Since X is a variety and each Vi = u−1 (Ui ) is a variety, Y is also a variety (12.5.10). Furthermore, the Vi ’s are irreducible and the Vi ∩Vj ’s are non-empty, so Y is irreducible (1.1.6). Let us now prove that Y is unique up to an isomorphism. Let Y1 be another such variety. Let (Uj )j∈J be a covering of X by aﬃne open subsets. The open subsets contained in Ui ∩ Uj which are principal in both Ui and Uj form a base of the topology of Ui ∩ Uj (17.1.1). Using the same arguments, we have our result. 17.2.2 The normal variety Y in 17.2.1 is called the normalization of X in K. When K = R(X), we shall simply say that Y is the normalization of X; in this case, the morphism u is birational. 17.2.3 Corollary. Let X be an irreducible complete variety. If Y is the normalization of X in a ﬁnite extension of R(X), then Y is complete. Proof. Since the morphism Y → X is ﬁnite (17.2.1), the result is a consequence of 15.3.9. 17.2.4 Proposition. Let X, Y be irreducible varieties, u : X → Y a surjective ﬁnite morphism and n = [R(X) : R(Y )]. Suppose that Y is normal. (i) For all y ∈ Y , the cardinality of u−1 (y) is at most n. (ii) There exists a non-empty open subset V of Y such that u−1 (V ) is isomorphic to the normalization of V in R(X), and the cardinality of u−1 (y) is n for all y ∈ V . Proof. (i) Since u is ﬁnite, u−1 (y) is a ﬁnite set for all y ∈ Y (15.3.4). Further, if V is an aﬃne open subset of Y , u−1 (V ) is an aﬃne open subset of X, and A(u−1 (V )) is a ﬁnite A(V )-algebra (15.3.3). So we are reduced to the case where X = Spm(A) and Y = Spm(B) are aﬃne. Let y ∈ Y and u−1 (y) = {x1 , . . . , xm }. There exists a ∈ A(X) such that the a(xi ) s, 1 i m, are pairwise distinct. Let P (T ) = T p + αp−1 T p−1 + · · · + α0 the minimal polynomial of a over R(Y ). We have deg(P ) n, and by 5.5.4, αp−1 , . . . , α0 are integral over B. Hence αp−1 , . . . , α0 ∈ B. Let Q(T ) = T p + αp−1 (y)T p−1 + · · · + α0 (y). Then a(x1 ), . . . , a(xm ) are roots of Q. Hence m n. (ii) Again, we may assume that X = Spm(A) and Y = Spm(B) are aﬃne. By 5.6.7, there exists a ∈ R(X) such that R(X) = R(Y )(a). We may assume

17.3 Products of normal varieties

223

further that a ∈ A (proceed as in 5.7.9). The degree of the minimal polynomial P of a over R(Y ) is n, and as above, P (T ) = T p + αp−1 T p−1 + · · · + α0 where αp−1 , . . . , α0 ∈ B. Let d ∈ B be the discriminant of P . Then d = 0 since the characteristic of k is zero. Replace Y by D(d) and X by u−1 (D(d)), we may assume that d is invertible in B. In view of 5.7.9, A is the integral closure of B in R(X) and {1, a, . . . , an−1 } is a basis of the B-module A. Further, by 5.7.10, the cardinality of u−1 (y) is n for all y ∈ Y . By 17.1.5, there exists a non-empty open subset U of X which is a normal variety. The set u(X \U ) is closed and nowhere dense in Y (15.3.4 and 15.3.8). So there exists an aﬃne open subset V of Y such that u−1 (V ) is an aﬃne open subset contained in U . Thus the aﬃne variety u−1 (V ) is normal; its ﬁeld of rational functions is R(X). Finally, A(u−1 (V )) is the integral closure of A(V ) in R(X). Hence we have our result. 17.2.5 Example. Let X = {(x1 , x2 ) ∈ k2 ; x31 − x22 = 0}. We saw in 11.5.7 that X is the image of the morphism u : k → k2 , t → (t2 , t3 ). Thus X is irreducible. However, A(X) = k[T 2 , T 3 ] is not an integrally closed domain (11.5.7). So X is not normal. The set Y = X \ {(0, 0)} is the principal open subset DX (T 2 ) of X, and A(Y ) = k[T, T −1 ]. It follows from 3.2.5 that Y is normal. Thus (0, 0) is the only point in X which is not normal. The morphism v : k → X, t → u(t), is bijective, bicontinuous, birational (12.7.13) and ﬁnite (15.3.7). The integral closure of k[T 2 , T 3 ] in k(T ) is k[T ]. Thus k is the normalization of X.

17.3 Products of normal varieties 17.3.1 Lemma. Let X = Spm(A) be an irreducible aﬃne variety and E a ﬁnite-dimensional k-vector subspace of A. Denote by F the subspace of A generated by the elements of A which, in R(X), can be represented in the form uv −1 , with u ∈ E and v ∈ E \ {0}. Then F is ﬁnite-dimensional. Proof. Let us ﬁx a ﬁnite normal extension L of Fract(A) (5.7.5). There exist t1 , . . . , tn ∈ A algebraically independent over k such that A is integral over B = k[t1 , . . . , tn ] (6.2.2). The integral closure B of B in L is also the integral closure of A in L (3.1.7). By 5.5.6, B is a ﬁnitely generated B-module, so it is a ﬁnitely generated k-algebra. Replace X by Spm(B ), we may assume that A is the integral closure of B in L. Denote by K = Fract(B). The extension K ⊂ L is ﬁnite and normal. Let d be its degree, which is also the order of the Galois group Gal(L/K) (5.6.6 and 5.7.4). Let Gal(L/K) = {σ1 , . . . , σd } with σ1 = idL . If u ∈ L, then σ1 (u) · · · σd (u) ∈ K. Replace E by the sum of the σi (E)’s, we may assume that E is Gal(L/K)stable.

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Let E1 be the k-vector space spanned by the products of k elements of E, with k d. Clearly E1 is ﬁnite-dimensional, and the same goes for E1 ∩ K. Since any element of E1 is integral over the integrally closed domain B, we deduce that E1 ∩ K ⊂ B. Denote by Br of elements in B[t1 , . . . , tn ] of degree at most r in the ti ’s. Then there exists r ∈ N such that E1 ∩ K ⊂ Br . Now let w ∈ A, u, v ∈ E be such that w = uv −1 . Then: w = uσ2 (v) · · · σd (v)/(σ1 (v) · · · σd (v)). We deduce that w = u (v )−1 where u ∈ A and v ∈ Br \ {0}. There exists a basis (z1 , . . . , zd ) of L over K such that A ⊂ Bz1 ⊕ · · · ⊕Bzd (5.5.6). It follows that there exists s ∈ N such that E1 ⊂ Bs z1 ⊕ · · · ⊕ Bs zd . Let us write u = p1 z 1 + · · · + p d z d , w = q 1 z 1 + · · · + q d z d where p1 , q1 , . . . , pd , qd ∈ B. Then qi v = pi for 1 i d, and we have pi ∈ Bs , for 1 i d, because u = uσ2 (v) · · · σd (v) ∈ E1 . Thus qi ∈ Bs , and hence F is ﬁnite-dimensional. 17.3.2 Proposition. Let X, Y be irreducible normal varieties. Then the variety X × Y is normal. Proof. We may assume that X = Spm(A) and Y = Spm(B) are aﬃne. Let f ∈ R(X × Y ) be integral over A ⊗k B and D its domain of deﬁnition. Then f can be represented in the form g/h where g = u1 ⊗ v1 + · · · + un ⊗ vn , h = u1 ⊗ w1 + · · · + un ⊗ wn with u1 , . . . , un ∈ A linearly independent and v1 , w1 , . . . , vn , wn ∈ B. For y ∈ Y , deﬁne gy , hy ∈ A by: gy = v1 (y)u1 + · · · + vn (y)un , hy = w1 (y)u1 + · · · + wn (y)un . The set V of elements y ∈ Y such that wi (y) is non-zero for some i is a nonempty open subset of Y . If y ∈ V , then hy is a non-zero element of A. So there exists x ∈ X such that (x, y) ∈ DX×Y (h) ⊂ D and fy = gy /hy ∈ R(X) is a regular function on DX (hy ). Denote by F the set of maps from V to k, G the subset of F consisting of maps u such that there exist a non-empty open subset W ⊂ V and v ∈ B verifying u|W = v|W . Since Y is irreducible, it is easy to see that G is a subspace of F. Let y ∈ V . Since f is integral over A⊗k B, we verify easily that fy is integral over A. Since X is normal, there exists a function ty ∈ A which represents fy . On the other hand, fy is the quotient of two elements of the vector subspace E of A generated by the ui ’s. By 17.3.1, there exist s1 , . . . , sm ∈ A linearly independent, such that ty = ρ1 (y)s1 + · · · + ρm (y)sm where ρ1 , . . . , ρm ∈ F. In particular, for all x ∈ X such that (x, y) ∈ D, ty (x) = f (x, y).

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225

A similar argument shows that there is a non-empty open subset U of X verifying the following conditions: for all x ∈ U , there exist y ∈ Y such that (x, y) ∈ D, and tx ∈ B such that tx (y) = f (x, y) for all (x, y) ∈ D. For x ∈ U , denote by Vx the set of y ∈ V such that (x, y) ∈ D. It is a non-empty open subset of V . Let x ∈ U . For y ∈ Vx , we have ρ1 (y)s1 (x) + · · · + ρm (y)sm (x) = ty (x) = f (x, y) = tx (y). So the restrictions of tx and s1 (x)ρ1 + · · · + sm (x)ρm to Vx are identical. It follows that s1 (x)ρ1 + · · · + sm (x)ρm ∈ G. Hence, if λ is a linear form on F which is zero on G, then s1 (x)λ(ρ1 ) + · · · + sm (x)λ(ρm ) = 0. This being true for all x in the dense open subset U of X, we have: λ(ρ1 )s1 + · · · + λ(ρm )sm = 0. But the si ’s are linearly independent, so λ(ρi ) = 0 for 1 i m. Hence ρi ∈ G, 1 i m. Thus there exist t1 , . . . , tm ∈ B such that f can be represented in the form s1 ⊗ t1 + · · · + sm ⊗ tm ∈ A ⊗k B. So our result follows.

17.4 Properties of normal varieties 17.4.1 Lemma. Let A ⊂ B be ﬁnitely generated k-algebras, X = Spm(A), Y = Spm(B), u : Y → X the morphism associated to the canonical injection A → B, n ∈ Spm(B), m = n ∩ A, and y ∈ Y the point associated to n. The following conditions are equivalent: (i) n is minimal among the prime ideals q ∈ Spec(B) such that q∩A = m. (ii) The point y is isolated in the ﬁbre u−1 (u(y)) of the morphism u. Proof. The algebra A/m is a subalgebra of B/n, which is isomorphic to k (6.4.1). So m ∈ √ Spm(A). Thus any proper ideal b of B containing m veriﬁes b ∩ A = m. Let Bm = q1 ∩ · · · ∩ qs where q1 , . . . , qs ∈ Spec(B) with qi ⊂ qj whenever i = j. Then q1 , . . . , qs are exactly the minimal prime ideal lying above m. Also, Z = u−1 (u(y)) = V(q1 ) ∪ · · · ∪ V(qs ) is the decomposition of Z into irreducible components. If part (i) is true, then we may take n = q1 , and Z is the disjoint union of {y} and V(q2 ) ∪ · · · ∪ V(qs ). Hence {y} is open in Z, which proves (ii). Conversely, if y is isolated in Z, then it is an irreducible component of Z. In view of the above discussion, we have (i). 17.4.2 Proposition. Let A ⊂ C ⊂ B be ﬁnitely generated k-algebras verifying: they are integral domains, B is integral over C, A is integrally closed in B and Fract(B) is a ﬁnite extension of Fract(A). Let m ∈ Spm(A). Suppose that there exists n ∈ Spm(B) minimal in the set of q ∈ Spec(B) verifying q∩A = m. Then the canonical homomorphism Am → Bn is bijective.

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Proof. From our hypotheses, C is a ﬁnitely generated A-algebra, so C = A[u1 , . . . , um ], where u1 , . . . , um ∈ C. The case m = 1 is exactly 5.9.8. Let us proceed by induction on m. Suppose that m > 1. Let B be the integral closure of A[u1 , . . . , um−1 ] in B. It is ﬁnitely generated (3.3.5) and integrally closed in B. We have C ⊂ B [um ] ⊂ B, so B is integral over B [um ] and Fract(B) is a ﬁnite extension of Fract(B ). The ideal n = n ∩ B ∈ Spm(B ) and if q ∈ Spec(B) veriﬁes q ⊂ n and q ∩ B = n , then q ∩ A = n ∩ A = n ∩ A = m, so it follows from our hypothesis on n that q = n and n is minimal among the prime ideals of B lying above n . We may therefore apply 5.9.8 to B ⊂ B [um ] ⊂ B, and we obtain that the homomorphism Bn → Bn is bijective. Denote by u : Spm(B) → Spm(A) , v : Spm(B ) → Spm(A) , w : Spm(B) → Spm(B ) the morphisms induced by the canonical injections. Note that u = v ◦ w. By 17.4.1, the point y corresponding to n is isolated in u−1 (u(y)). It follows that y is also isolated in w−1 (w(y)). Since the homomorphism Bn → Bn is bijective, it follows from 12.8.8 that w induces an isomorphism from a neighbourhood of y in Spm(B) onto a neighbourhood of w(y) in Spm(B ). Hence w(y) is isolated in v −1 (v(w(y))). Applying 17.4.1, we deduce that n is minimal among the prime ideals of B lying above m. By induction on A ⊂ A[u1 , . . . , um−1 ] ⊂ B , the homomorphism Am → Bn is bijective. So we are done. 17.4.3 Let u : X → Y be a morphism of irreducible varieties. Suppose that there exist x ∈ X and an open neighbourhood U of x in X such that u induces an isomorphism from U onto an open subset V of Y . We may assume that U is aﬃne. We see in particular that u is dominant and birational. Further U ∩ u−1 (u(x)) = {x}, and so x is isolated in u−1 (u(x)). In this situation, we have the following more precise statement. Theorem. (Zariski’s Main Theorem) Let u : X → Y be a dominant morphism of irreducible varieties. Suppose that there exists x ∈ X such that x is isolated in the u−1 (u(x)). (i) We have dim X = dim Y , so R(X) is a ﬁnite extension of R(Y ). (ii) There is an open neighbourhood V of u(x) in Y and an open neighbourhood U of x in u−1 (V ) such that the restriction of u factorizes through: j

U −→ V −→ V v

where v is ﬁnite and surjective, and j is an isomorphism from U onto an open subset of V . (iii) If X is normal, we may suppose that V = Y and V is the normalization of Y in R(X). Proof. Since {x} is an irreducible component of u−1 (u(x)), we have dim X = dim Y by 15.5.3 and 15.5.4. In particular, R(X) is a ﬁnite extension of R(Y ).

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To prove the other parts, we may assume that X = Spm(B), Y = Spm(A) are aﬃne, and A ⊂ B. Let A (resp. A ) be the integral closure of A in B (resp. Fract(B)). By 6.2.10, A is a ﬁnitely generated A-algebra and since A is Noetherian, A is a ﬁnitely generated A-module. Thus A is a ﬁnitely generated k-algebra which is also an integral domain. Hence Y = Spm(A ) is irreducible and the morphism v : Y → Y , associated to the canonical injection A → A , is ﬁnite (15.3.2). It follows from the injections A → A → B that u factorizes through: v w X −→ Y −→ Y. Since w−1 (w(x)) ⊂ u−1 (u(x)), the point x is isolated in w−1 (w(x)). So by 17.4.1, n = I({x}) is minimal among the prime ideals q of B verifying q∩A = n ∩ A = n = I({w(x)}). Applying 17.4.2 to A ⊂ B ⊂ B, we obtain that the canonical homomorphism An → Bn is bijective. So w induces an isomorphism from a neighbourhood of x in X onto a neighbourhood of w(x) of Y (12.8.8). So we have proved (ii). Let Z be the normalization of Y in R(X). There exist a surjective ﬁnite morphism s : Z → Y and a covering (Vi )i∈I of Y by aﬃne open subsets such that s−1 (Vi ) is an aﬃne open subset and A(s−1 (Vi )) is the integral closure of A(Vi ) in R(Z) R(X). Suppose that X is normal. Replacing Y by a Vi containing u(x) and U by an aﬃne open neighbourhood U of x in X such that U ⊂ u−1 (Vi ), we are reduced to the case where Y = Spm(A), X = Spm(B) with B is an integrally closed domain. Then A (as deﬁned above) is isomorphic to A(Z). So (iii) follows. 17.4.4 Corollary. Let u : X → Y be a dominant morphism of irreducible varieties. The following conditions are equivalent: (i) X is normal and for all x ∈ X, x is isolated in u−1 (u(x)). (ii) The extension R(Y ) ⊂ R(X) is ﬁnite and u factorizes through: j

X −→ Y −→ Y v

where Y is the normalization of Y in R(X), v is a surjective ﬁnite morphism as in 17.2.1, and j is an isomorphism from X onto an open subset of Y . If these conditions are satisﬁed, there exists an open subset V of Y such that V is normal, and for all y ∈ V , card u−1 (y) = [R(X) : R(Y )]. Proof. The fact that (ii) implies (i) is clear. Furthermore, the last part follows from 17.1.5 and 17.2.4. Let us suppose that (i) is true. Let Y be the normalization of Y in R(X), v : Y → Y be as in 17.2.1 and x ∈ X. By 17.4.3, there exists an isomorphism jU from an aﬃne open neighbourhood U of x in X onto an open subset of Y such that u|U = v ◦ jU . Let U be another aﬃne open neighbourhood of x and jU an isomorphism from U to an open subset of Y such that u |U = v ◦ jU . Then there exists an

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aﬃne open neighbourhood U ⊂ U ∩U of x such that u(U ) is contained in an aﬃne open subset V of Y . Let A = A(V ), B = A(U ) with B an integrally closed domain, and A the integral closure of A in R(X). The restrictions jU |U and jU |U correspond both to the injection A → B. So they are identical. We deduce that there exists a morphism j : X → Y such that j|U = jU for all aﬃne open subset U with the above properties. Thus j is a local isomorphism, and (ii) follows by 12.5.11. 17.4.5 Corollary. Let u : X → Y be a birational morphism of irreducible varieties. Suppose that Y is normal and for all x ∈ X, x is isolated in u−1 (u(x)). Then (i) u is an isomorphism from X onto an open subset of Y . (ii) If X is complete, then u is an isomorphism from X to Y . Proof. Let v : X → X be the normalization of X (17.2.1). The morphism u◦v has the same properties as u. Since Y is normal and R(X ) = R(X) = R(Y ), it follows from 17.4.4 that u ◦ v induces an isomorphism w from X onto an open subset of Y . So w−1 ◦u is the reciprocal isomorphism of v. Hence X = X and (i) follows. If X is complete, then u(X) is closed in Y (13.4.3), so u(X) = Y . 17.4.6 Corollary. Let u : X → Y be a bijective birational morphism of irreducible varieties. If Y is normal, then u is an isomorphism. 17.4.7 Corollary. Let u : X → Y be a morphism of irreducible normal varieties. Suppose that there exists n ∈ N∗ such that card u−1 (y) = n for all y ∈ Y . Then u is a ﬁnite morphism and u factorizes through: j

X −→ Y −→ Y v

where v : Y → Y is the normalization of Y in R(X) and j is an isomorphism from X to Y . Proof. Note that our hypothesis implies that u(X) = Y and that for any x ∈ X, x is isolated in u−1 (u(x)) (since it is a ﬁnite set). So by 17.4.4, u j

v

factorizes through X → Y → Y , where v is a surjective ﬁnite morphism and j is an isomorphism from X onto an open subset of Y . So to obtain the result, we only need to show that j(X) = Y . Suppose that there exists y ∈ Y \ j(X). Then v −1 (v(y )) is a ﬁnite set (15.3.4). Since j −1 (v −1 (v(y ))) = u−1 (v(y )), card v −1 (v(y )) n. If card v −1 (v(y )) = n, then y ∈ j(X). So card v −1 (v(y )) n + 1. On the other hand, Y \ j(X) is closed and nowhere dense in Y . By 15.3.4 and 15.3.8, v(Y \ j(X)) is also closed and nowhere dense in Y . Let V = Y \v(Y \j(X)). Then V is open. But v −1 (V ) ⊂ j(X) and card v −1 (y) = n if y ∈ V . It follows from 17.2.4 that card v −1 (v(y )) n. Contradiction.

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229

17.4.8 Corollary. Let u : X → Y be a bijective morphism of irreducible varieties. If Y is normal, then u is an isomorphism. Proof. Let v : X → X be the normalization of X (17.2.1), and w = u ◦ v. Since v is ﬁnite and u is bijective, w−1 (y) is ﬁnite for all y ∈ Y (15.3.4). By 17.4.4, R(X) = R(X ) is a ﬁnite extension of R(Y ). Let V be a non-empty open subset of Y such that u−1 (V ) is normal (17.1.5), and v1 : v −1 (u−1 (V )) → u−1 (V ) the morphism induced by v. Then v1 is ﬁnite and surjective (15.3.3), and in view of 17.2.4, card v1−1 (z) = [R(X ) : R(X)] = 1 for all z ∈ u−1 (V ). Thus card w−1 (y) = 1 for all y ∈ V . The morphism u ◦ v1 : v −1 (u−1 (V )) → V is surjective and ﬁnite (17.4.7), so by applying 17.2.4, we obtain that R(X) = R(Y ). Hence u is birational, and the result follows from 17.4.6. 17.4.9 Corollary. Let u : X → Y be a dominant morphism of irreducible varieties. Suppose that Y is normal. If x ∈ X is isolated in u−1 (u(x)), then for all neighbourhood U of x in X, u(U ) is a neighbourhood of u(x) in Y . In particular, if x is isolated in u−1 (u(x)) for all x ∈ X, then u is open. Proof. Let v : X → X be the normalization of X (17.2.1), w = u ◦ v and x ∈ X be isolated in u−1 (u(x)). Then any x ∈ X verifying v(x ) = x is isolated in w−1 (w(x )). If U is a neighbourhood of x in X, then v −1 (U ) is a neighbourhood of x in X . So it suﬃces to prove that w(v −1 (U )) is a neighbourhood of u(x) in Y . Therefore, we may assume that X is normal. By 17.4.3, there exists an open neighbourhood V of u(x) in Y and an open neighbourhood U of x in X such that the restriction u|U : U → V factorizes j

v

through U → V → V , where V is the normalization of V in R(X) and j is an isomorphism from U onto an open subset of V . So we can further reduce to the case where X = Y is the normalization of Y in a ﬁnite extension of R(X). Let W be an open neighbourhood of x in X. Then u(W ) = Y \ T where T = {y ∈ Y ; u−1 (y) ⊂ X \ W }. To obtain the result, it suﬃces to prove that u(x) ∈ T . Suppose that u(x) ∈ T . Then u(x) is contained in the closure of an irreducible component T0 of T . By 17.1.7, there exists an irreducible component Z of u−1 (T0 ) containing x verifying u(Z) = T0 . If S ⊂ Z is an open subset containing x, then u(S) contains a non-empty open subset of T0 (15.4.2), hence u(S) ∩ T0 = ∅. It follows that S ∩ u−1 (T0 ) = ∅, so x ∈ u−1 (T0 ) ⊂ X \ W which contradicts the fact that W is a neighbourhood of x in X. 17.4.10 Theorem. Let u : X → Y be a dominant morphism of irreducible varieties and r = dim X − dim Y . Suppose that Y is normal. If x ∈ X is such that all the irreducible components of u−1 (u(x)) containing x has dimension r, then the image via u of a neighbourhood of x in X is a neighbourhood of u(x) in Y . Proof. Let U ⊂ X (resp. V ⊂ Y ) be an aﬃne open neighbourhood of x (resp. u(x)) such that U ⊂ u−1 (V ). Since the intersection of U and an irreducible

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component of u−1 (u(x)) containing x is dense in the irreducible component, the restriction U → V satisﬁes the same hypotheses. So we may assume that X = Spm(B) and Y = Spm(A) are aﬃne, where A ⊂ B are integral domains and A is an integrally closed domain. Also, we only need to prove that u(X) is a neighbourhood of u(x) in Y . Replacing X by a smaller aﬃne open subsets, we may assume further that all the irreducible components of Z = u−1 (u(x)) have dimension r. √ Denote by m = I({u(x)}) ∈ Spm(A). We have Z = V(Bm), and if a = Bm, then A(Z) = B/a. Since all the irreducible components of Z have the same dimension r, it follows from 6.2.2 and 6.2.3 that there exist elements s1 , . . . , sr ∈ B/a algebraically independent over k such that B/a is a ﬁnitely generated C-module, where C = k[s1 , . . . , sr ]. For 1 i r, ﬁx ti ∈ B such that its class in B/a is si . Denote by ϕ : A[T1 , . . . , Tr ] → B the A-module homomorphism such that ϕ(Ti ) = ti . Since ϕ the canonical injection A → B factorizes through A → A[T1 , . . . , Tr ] → B, u factorizes through: pr v X −→ Y × kr −→ Y. If W = pr−1 (u(x)) = {u(x)} × kr , then Z = v −1 (W ). Since W = Spm(C), the restriction Z → W of v is ﬁnite and surjective. It follows that any point y of u−1 (u(x)) is isolated in u−1 (u(y)). Let S = {x ∈ X; x is isolated in v −1 (v(x ))}. The set S contains Z and is open in X (15.5.7). Since Y × kr is normal (17.1.6 and 17.3.2), the restriction of v to S is open (17.4.9). But pr is also open (15.5.8), we deduce therefore that u(S) is open in Y , and hence u(X) is a neighbourhood of u(x) in Y . 17.4.11 Corollary. Let u : X → Y be a dominant morphism of irreducible varieties. Suppose that Y is normal and for all y ∈ u(X), all the irreducible components of u−1 (y) have dimension dim X − dim Y . Then u is open. 17.4.12 Proposition. Let u : X → Y be a dominant morphism of irreducible varieties. If f is a regular function which is constant on the ﬁbres u−1 (y), y ∈ Y , then f , considered as an element of R(X), belongs to R(Y ). Proof. Let v : X → Y × k be the morphism x → (u(x), f (x)) and p : Y × k → Y the canonical projection. Our hypothesis implies that the restriction of p to v(X) is injective. Let W ⊂ v(X) be such that W is open and dense in v(X) (15.4.2). Denote by w the restriction p to W , then w is injective and p(W ) = Y . Replacing W by a smaller open subset if necessary, we may assume that W is normal (17.1.5). The set u(W ) contains a non-empty open subset V of Y (15.4.2) that we may also assume to be normal. Replacing W by w−1 (W ), we may further assume that w : W → V is bijective. By 17.4.8, w induces an isomorphism from W onto V . The restriction q to W of the canonical projection Y × k → k is a regular function on W , and q ◦ (w|W )−1 is regular on V . For x ∈ u−1 (W ), we have q ◦ (w|W )−1 (u(x)) = f (x). So we are done.

17.4 Properties of normal varieties

231

17.4.13 Proposition. Let x be a normal point of an irreducible variety X and f be a rational function on X whose domain of deﬁnition is D. Suppose that x ∈ D, and denote by E the domain of deﬁnition of f = 1/f . There exists a subvariety Y of X containing x verifying Y ∩ E = ∅ and such that f (y) = 0 for all y ∈ Y ∩ E. Proof. Recall from 12.8.3 that A = OX,x is Noetherian. By our hypothesis, f ∈ A. Let a = {h ∈ A; f h ∈ A}. We observe that a is an ideal of A contained in the maximal ideal mX,x of A. Let p1 , . . . , pr be the (pairwise distinct) prime ideals minimal among the prime ideals containing a. By 2.5.3 and 2.7.7, for some n ∈ N∗ , we have pn1 pn2 · · · pnr ⊂ a. It follows that pi Ap1 = Ap1 for i 2 since p1 Ap1 is the unique maximal ideal of the local ring Ap1 . Let k ∈ N be minimal such that f \ Ap1 , so pk1 f ⊂ Ap1 . Our hypothesis implies that k > 0. Let g ∈ pk−1 1 p1 g ⊂ Ap1 . Since Ap1 is an integrally closed domain (3.2.5), g is not integral over Ap1 . If p1 g ⊂ p1 Ap1 , then multiplication by g induces a homomorphism of the ﬁnitely generated Ap1 -module p1 Ap1 . Proceed as in the proof of 3.1.4, we would obtain that g in integral over Ap1 . Contradiction. So we deduce that p1 Ap1 g ⊂ p1 Ap1 . Thus p1 g contains an invertible element of Ap1 , hence 1/g ∈ p1 Ap1 and (1/g)Ap1 = p1 Ap1 . Let h = f /g k , then h ∈ f pk1 Ap1 ⊂ Ap1 . If h ∈ p1 Ap1 , then f /g k−1 ∈ Ap1 which contradicts the minimality of k. So h is invertible in Ap1 and f = 1/f = h−1 (1/g k ) ∈ p1 Ap1 . Let 1 , . . . , s ∈ A be generators of the ideal p1 and U an aﬃne open neighbourhood of x contained in the domain of deﬁnition of the i ’s. Let Y = {y ∈ U ; i (y) = 0 for all 1 i s}. Since p1 ⊂ mX,x , x ∈ Y . But f ∈ p1 Ap1 , so f (z) = 0 for all z ∈ Y ∩ E. 17.4.14 Proposition. Let u : X → Y be a ﬁnite morphism of irreducible varieties and x ∈ X a simple point such that u−1 (u(x)) = {x}. Suppose that the map Tx (u) : Tx (X) → Tu(x) (Y ) is injective. Then there exists an open neighbourhood V of y = u(x) in Y such that the restriction of u to u−1 (V ) is an isomorphism onto a closed subvariety of V . Proof. We may assume that X = Spm(B) and Y = Spm(A) are aﬃne where B is a ﬁnite A-algebra. Let ϕ be the comorphism of u and m, n the maximal ideals of A and B corresponding to y and x respectively. By our hypothesis, the dual map, m/m2 → n/n2 , of Tx (u) is surjective, and so a system of generators of m gives a system of generators of n/n2 . Thus n = Bϕ(m) + n2 . It follows that n2 ⊂ Bϕ(m) + n3 , so n = Bϕ(m) + n3 . By induction, we obtain that n = Bϕ(m) + nk for all k ∈ N∗ . By 17.4.1, n is the unique prime ideal of B containing Bϕ(m). Thus n/Bϕ(m) is the nilradical of B/Bϕ(m). Since B/Bϕ(m) is Noetherian, it follows from 2.5.2 and 2.5.3 that nk ⊂ Bϕ(m) for some k ∈ N∗ . We deduce therefore that n = Bϕ(m), and so Bϕ(m) + ϕ(A) = B because B = n ⊕ k. The

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A-module B/ϕ(A) is ﬁnitely generated and veriﬁes ϕ(m)(B/ϕ(A)) = B/ϕ(A). So by 2.6.7, there exists f ∈ A \ m such that ϕ(f )B = ϕ(A). Hence the homomorphism ϕf : Af → Bϕ(f ) is surjective. By 11.6.8, u is an isomorphism from the open subset D(ϕ(f )) of X onto a closed subvariety of D(f ) in Y .

References and comments • [5], [19], [22], [26], [37], [40], [78]. The result stated and proved in 17.4.11 is very useful for showing that a morphism is an open map.

18 Root systems

In this chapter, we are concerned with the theory of root systems. This simple geometric notion will play an amazingly important role in the structure theory of semisimple Lie algebras. Throughout this chapter, k is a ﬁeld of characteristic zero and V is a kvector space of dimension l. If u ∈ End(V ), we shall denote by rk(u) the rank of u.

18.1 Reﬂections 18.1.1 Deﬁnition. We say that s ∈ End(V ) is a reﬂection if s2 = idV and rk(idV −s) = 1. 18.1.2 Let s be a reﬂection of V and denote by Vs+ (resp. Vs− ) the eigenspace of s with eigenvalue 1 (resp. −1). Then V = Vs+ ⊕ Vs− , dim Vs− = 1 , det s = −1. 18.1.3 Lemma. The following conditions are equivalent for s ∈ End(V ): (i) s is a reﬂection. (ii) There exist a ∈ V and a∗ ∈ V ∗ such that a∗ (a) = 2 and s(x) = x − a∗ (x)a for all x ∈ V . Proof. The implication (ii) ⇒ (i) is obvious. Let us prove that (i) ⇒ (ii). Since rk(idV −s) = 1, there exist a ∈ V and a∗ ∈ V ∗ such that x − s(x) = a∗ (x)a for all x ∈ V . Finally, s2 = idV implies that a∗ (a) = 2. 18.1.4 If a ∈ V and a∗ ∈ V are such that a∗ (a) = 2, then we shall denote by sa,a∗ the reﬂection of V deﬁned by: sa,a∗ (x) = x − a∗ (x)a. We can also deﬁne a reﬂection sa∗ ,a of V ∗ by: sa∗ ,a (x∗ ) = x∗ − x∗ (a)a∗ . We have sa∗ ,a = t (sa,a∗ ) = t (sa,a∗ )−1 .

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18.1.5 Proposition. Let s be a reﬂection of V , u ∈ End(V ) and W ⊂ V a subspace. (i) We have u ◦ s = s ◦ u if and only if Vs+ and Vs− are u-stable. (ii) We have s(W ) ⊂ W if and only if Vs− ⊂ W or W ⊂ Vs+ . Proof. Part (i) is clear. Let us prove part (ii). If W ⊂ Vs+ , then s(W ) ⊂ W . Now if Vs− ⊂ W , then x − s(x) ∈ Vs− ⊂ W for all x ∈ W . Hence s(W ) ⊂ W . Conversely, suppose that s(W ) ⊂ W and W ⊂ Vs+ . Then there exists x ∈ W such that y = x − s(x) = 0. It follows that y ∈ Vs− ∩ W , hence Vs− ⊂ W since dim Vs− = 1. 18.1.6 Proposition. Let R be a ﬁnite subset of V which generates V , s, t reﬂections of V and α ∈ R \ {0} verifying s(α) = t(α) = −α and s(R) = t(R) = R. Then s = t. Proof. Let G ⊂ GL(V ) be the stabilizer of R. Since R generates V , G can be identiﬁed with a subgroup of the symmetric group on R, so G is ﬁnite. By 18.1.4, we can write s = sα,α∗ and t = sα,β ∗ . A simple induction gives: (s ◦ t)n (x) = x + n (β ∗ (x) − α∗ (x)) α. Taking n to be the order of s ◦ t in G, we see that α∗ = β ∗ . So s = t. 18.1.7 Let B be a bilinear form on V . An endomorphism u of V is an orthogonal transformation (with respect to B) if B(u(x), u(y)) = B(x, y) for all x, y ∈ V . Suppose that B is symmetric or antisymmetric. Then U, W ⊂ V are orthogonal if B(U, W ) = {0}. A vector x ∈ V is isotropic if B(x, x) = 0. A subspace W of V is non-degenerate if W ∩ W ⊥ = {0} where W ⊥ is the set of x ∈ V such that B(x, y) = 0 for all y ∈ W . Proposition. Let B be a non-degenerate symmetric bilinear form on V . (i) A reﬂection s of V is an orthogonal transformation if and only if Vs+ and Vs− are orthogonal. In this case, Vs+ and Vs− are non-degenerate. (ii) Let H be a non-degenerate hyperplane of V . Then there exists a unique reﬂection s of V such that s is an orthogonal transformation and s|H = idH . If α ∈ H ⊥ \ {0} , then H ⊥ = kα is non-degenerate and: s(x) = x − 2

B(x, α) α. B(α, α)

We say that s is the reﬂection orthogonal to the hyperplane H. Proof. (i) Let x ∈ Vs+ , y ∈ Vs− . If s is an orthogonal transformation, then B(x, y) = B(s(x), s(y)) = −B(x, y).

18.2 Root systems

235

Hence Vs+ and Vs− are orthogonal. These eigenspaces are non-degenerate since B is non-degenerate and V = Vs+ ⊕ Vs− . The converse is immediate. (ii) By our assumptions, kα is non-degenerate and V = H ⊕ kα. Clearly, the reﬂection s described above veriﬁes Vs+ = H and Vs− = kα. So part (i) says that s is an orthogonal transformation. Now let t be another such orthogonal transformation, then Vt+ = H and − Vt = kα by (i). Let α∗ ∈ V ∗ be such that α∗ (α) = 2 and t = sα,α∗ . Then H = ker α∗ and there exists λ ∈ k such that α∗ (x) = λB(x, α) for all x ∈ V . It follows that 2 = α∗ (α) = λB(α, α). So s = t. 18.1.8 Proposition. Let G be a subgroup of GL(V ). Suppose that V is a simple G-module and that G contains a reﬂection s. (i) The centralizer of G in End(V ) is k idV . (ii) Let B be a G-invariant non-zero bilinear form on V . Then B is nondegenerate, symmetric or antisymmetric. Any bilinear form G-invariant on V is proportional to B. Proof. (i) Let D = (s − idV )(V ). Then dim D = 1. By 18.1.5, if u ∈ End(V ) commutes with s, then u(D) ⊂ D. It follows that there exists λ ∈ k such that (u − λ idV )(D) = 0. Hence ker(u − λ idV ) = V because V is a simple G-module. (ii) Let M = {x ∈ V ; B(x, V ) = {0}} and N = {x ∈ V ; B(V, x) = {0}}. Then M , N are G-submodules of V since B is G-invariant. Hence M = N = {0} and B is non-degenerate. Let b be a G-invariant bilinear form on V . Since B is non-degenerate, there exists u ∈ End(V ) such that b(x, y) = B(u(x), y) for all x, y ∈ V . If α ∈ G and x, y ∈ V , then: B(u◦α(x), y) = b(α(x), y) = b(x, α−1 (y)) = B(u(x), α−1 (y)) = B(α◦u(x), y). So u commutes with G. By (i), u = λ idV for some λ ∈ k. Take, in particular, the bilinear form b(x, y) = B(y, x), then B(y, x) = b(x, y) = λB(x, y) = λb(y, x) = λ2 B(y, x). Hence λ2 = 1 and B is symmetric or antisymmetric.

18.2 Root systems 18.2.1 Deﬁnition. A subset R ⊂ V is called a root system in V if the following conditions are satisﬁed: (R1) R is a ﬁnite subset of V which spans V , and 0 ∈ R. (R2) For all α ∈ R, there exists α∨ ∈ V ∗ such that α∨ (α) = 2 and sα,α∨ (R) = R. (R3) For all α ∈ R, α∨ (R) ⊂ Z. Further, if for all α ∈ R, the only elements of R proportional to α are α and −α, we say that R is reduced.

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18 Root systems

18.2.2 From now on, let R be a root system in V . The elements of R are called the roots of R, the dimension of V is the rank of R that we shall denote by rk R. Let α ∈ R. By 18.1.6, there exists a unique α∨ satisfying (R2). It follows that (−α)∨ = −α∨ . So we may write sα for sα,α∨ and R∨ = {α∨ ; α ∈ R}. By (R1), the set of u ∈ GL(V ) verifying u(R) ⊂ R is a ﬁnite group, denoted by A(R). The subgroup W (R) of A(R) generated by the sα ’s is called the Weyl group of R. Let α ∈ R. We have −α = sα (α) ∈ R by (R2), and so −R = R. Hence − idV ∈ A(R). However, − idV is not necessarily contained in W (R). For α, β ∈ R, set: aα,β = β ∨ (α). From the deﬁnition of a root system, we have: ⎧ ⎨ aα,β ∈ Z , aα,α = 2, a−α,β = aα,−β = −aα,β , (1) ⎩ sβ (α) = α − aα,β β. 18.2.3 Let V = V1 ⊕ · · · ⊕ Vn . For 1 i n, let Ri be a root system in Vi . Identify V ∗ with the direct sum of the Vi∗ ’s. If α ∈ Ri , α∨ can be identiﬁed with an element of V ∗ which is zero on the Vj for j = i. Thus the union R of the Ri ’s is a root system in V that we shall call the direct sum of the Ri ’s. We have R∨ = R1∨ ∪ · · · ∪ Rn∨ . If α ∈ Ri and j = i, then Vj ⊂ ker α∨ and kα ⊂ Vi . We deduce that sα |Vj = idVj and sα (Vi ) ⊂ Vi by 18.1.5. The group W (R) is the product of the groups W (Ri ). Deﬁnition. A root system is irreducible if it is not the direct sum of two non-empty root systems. 18.2.4 Proposition. A root system R is the direct sum of a ﬁnite family (Ri )i∈I of irreducible root systems, which is unique up to a permutation of indices. We call the Ri ’s, the irreducible components of R. Proof. If R is non-empty and non-irreducible, then it is the direct sum of two non-empty root systems. By induction on the cardinality of R, we obtain that R is the direct sum of a ﬁnite family (Ri )i∈I of irreducible root systems. To prove the uniqueness, it suﬃces to show that if R is the direct sum of R and R , then Ri ⊂ R or Ri ⊂ R . Let Ri = Ri ∩ R , Ri = Ri ∩ R and V , V , Vi , Vi the vector subspaces spanned by R , R , Ri , Ri . If α ∈ Ri , then sα (R ) ⊂ R , sα (Ri ) ⊂ Ri (18.2.3), so Ri is a root system in Vi . Similarly, Ri is a root system in Vi . It follows easily that Ri is the direct sum of Ri and Ri . But Ri is irreducible, so Ri = ∅ or Ri = ∅. So we are done. 18.2.5 Lemma. Let U be a subspace of V and W the subspace of V spanned by R ∩ U . Then R ∩ U is a root system in W .

18.2 Root systems

237

∨ ∨ Proof. If α ∈ R ∩ U , then αW = α∨ |W . For β, γ ∈ R ∩ U , we have βW (β) = 2 ∨ ∨ (γ) = γ − β (γ)β ∈ U . So sβ,β ∨ (γ) ∈ R ∩ U by (R2). The other and sβ,βW W properties for a root system are clear.

18.2.6 Proposition. Let V = V1 ⊕· · ·⊕Vn and Ri = R∩Vi for 1 i n. The following conditions are equivalent: (i) The subspaces V1 , . . . , Vn are W (R)-stable. (ii) We have R ⊂ V1 ∪ · · · ∪ Vn . (iii) For 1 i n, Ri is a root system in Vi and R is the direct sum of the Ri ’s. Proof. (i) ⇒ (ii) Let α ∈ R. If sα (Vi ) ⊂ Vi , then kα ⊂ Vi or Vi ⊂ ker α∨ (18.1.5). Since V is the direct sum of the Vi ’s, Vj ⊂ ker α∨ for some j. So α ∈ Vj . (ii) ⇒ (iii) This is clear by 18.2.5. (iii) ⇒ (i) This follows from 18.2.3. 18.2.7 Corollary. The root system R is irreducible if and only if V is a simple W (R)-module. Proof. This follows from 10.7.6 and 18.2.6. 18.2.8 Proposition. If R is a root system in V , then R∨ is a root system in V ∗ . Furthermore, α∨∨ = α for all α ∈ R. Proof. Let us ﬁrst show that R∨ spans V ∗ . We may assume by 18.2.3 and 18.2.4 that R is irreducible. The result is clear if rk R = 1. Suppose that rk R 2. If there exists x ∈ V \ {0} such that α∨ (x) = 0 for all α ∈ R, then the subspace kx is W (R)-stable. But this is absurd by 18.2.7. So R∨ spans V ∗. Now, let α, β ∈ R, γ = sα (β) and θ = β ∨ − β ∨ (α)α∨ . Then θ(γ) = 2 , sγ,θ = sα ◦ sβ ◦ sα . Hence sγ,θ (R) = R, and so θ = γ ∨ = sα∨ ,α (β ∨ ) by 18.1.6. Thus sα∨ ,α (R∨ ) = R∨ . It follows that condition (R2) is satisﬁed and α∨∨ = α. Finally, we see readily that (R3) is also satisﬁed. 18.2.9 We call R∨ the dual root system of R. The map R → R∨ , α → α∨ , is a bijection, that we shall call canonical. We observe that aα,β = aβ ∨ ,α∨ for α, β ∈ R. Since sα∨ ,α = t (sα,α∨ )−1 , the map W (R) → W (R∨ ) , w → t w−1 is an isomorphism of groups. So we may identify W (R) and W (R∨ ) via this isomorphism and consider that W (R) acts on V and V ∗ . Let α ∈ R and g ∈ A(R). Set β = g(α), θ = t g −1 (α∨ ). Then

238

18 Root systems

θ(β) = 2 , sβ,θ = g ◦ sα ◦ g −1 . Hence sβ,θ (R) = R and θ = β ∨ . Hence t g(R∨ ) = R∨ and the map g → t g −1 is an isomorphism from A(R) onto A(R∨ ). So we may consider that A(R) acts on V ∗ . Note that the previous discussion shows that W (R) is a normal subgroup of A(R).

18.3 Root systems and bilinear forms 18.3.1 Proposition. For x, y ∈ V , the map V × V → k deﬁned by: ∨ (x, y) → (x|y) = α (x)α∨ (y) α∈R

is a non-degenerate A(R)-invariant symmetric bilinear form on V . Proof. If g ∈ A(R), we saw in 18.2.9 that t g(R∨ ) = R∨ . Thus t (g(x)|g(y)) = g(α∨ )(x)t g(α∨ )(y) = (x|y). α∈R

So the bilinear form (.|.) is A(R)-invariant. Clearly it is symmetric. For α, β ∈ R, we have α∨ (β) ∈ Z and β ∨ (β) = 2. So (β|β) ∈ N and (β|β) 4. So this bilinear form is non-zero. Now let R1 , . . . , Rn be the irreducible components of R and Vi the subspace spanned by Ri . Then by 18.2.3, for i = j, α∨ (x)α∨ (y) = 0 for all x ∈ Vi , y ∈ Vj and α ∈ R. Thus Vi and Vj are orthogonal with respect to (.|.). We are therefore reduced to the case where R is irreducible and the proposition follows from 18.1.8 and 18.2.7. 18.3.2 Proposition. Let R1 , . . . , Rn be the irreducible components of R, Vi the subspace spanned by Ri and B a W (R)-invariant symmetric bilinear form on V . (i) The restriction of B to Vi is proportional to the one deﬁned in 18.3.1. (ii) The Vi ’s are pairwise orthogonal with respect to B. Proof. (i) By 18.2.7, for 1 i n, Vi is a simple W (Ri )-module. So (i) follows from 18.1.8 and 18.3.1. (ii) Let Ui be the subspace of Vi spanned by w(x) − x, w ∈ W (Ri ) and x ∈ Vi . If w, w ∈ W (Ri ) and x ∈ Vi , then: w (w(x)) − w (x) = [w (w(x)) − w(x)] + [w(x) − x] − [w (x) − x]. So Ui is a W (Ri )-module. Further, sα (α) − α = −2α = 0 if α ∈ Ri . So Ui = {0}. We deduce from 18.2.7 that Ui = Vi . Now let x ∈ Vi , y ∈ Vj with i = j. We saw in 18.2.3 that w|Vj = idVj for all w ∈ W (Ri ). It follows that: B(w(x), y) = B(w(x), w(y)) = B(x, y). So w(x) − x is orthogonal to y. Thus Vi and Vj are orthogonal with respect to B.

18.4 Passage to the ﬁeld of real numbers

239

18.3.3 Let (.|.) be a non-degenerate W (R)-invariant bilinear form on V . For α, β ∈ R, it follows from 18.1.7 and 18.3.2 that aβ,α = 2

(α|β) . (α|α)

Hence: (α|β) = 0 ⇒

(β|β) aβ,α , aα,β = 0 ⇔ aβ,α = 0 ⇔ (α|β) = 0. = aα,β (α|α)

If x ∈ V , then: sβ ◦ sα (x) − sα ◦ sβ (x) = aα,β α∨ (x)β − aβ,α β ∨ (x)α. Consequently, if α and β are not proportional, then aα,β = 0 ⇔ aβ,α = 0 ⇔ sα ◦ sβ = sβ ◦ sα .

18.4 Passage to the ﬁeld of real numbers 18.4.1 Let R be a root system in V . Denote by VQ (resp. VQ∗ ) the Q-vector subspace spanned in V (resp. V ∗ ) by R (resp. R∨ ). Let K be an extension of k. We can identify canonically V as a subset of K ⊗k V , and V ∗ as a subset of K ⊗k V ∗ = (K ⊗k V )∗ . Then R is a root system in K ⊗k V and the α∨ ’s are the same. 18.4.2 Proposition. (i) R is a root system in VQ . (ii) The vector space V (resp. V ∗ ) identiﬁes canonically with k⊗Q VQ (resp. k ⊗Q VQ∗ ). The canonical bilinear form on V × V ∗ allows us to identify VQ∗ with (VQ )∗ and VQ with (VQ∗ )∗ . ∨ on VQ . It is Proof. (i) By (R3), α∨ (VQ ) ⊂ Q, so α∨ deﬁnes a linear form αQ ∨ now easy to verify that R is a root system in VQ , and αQ is the linear form associated to α in (VQ )∗ . (ii) Let i : k ⊗Q VQ → V be the canonical homomorphism and t i : V ∗ → k ⊗Q (VQ )∗ its dual. The map i is surjective since R spans V . Further, if α ∈ R, ∨ ∨ . By (i) and 18.2.8, the αQ ’s span (VQ )∗ , and so t i is then t i(α∨ ) = 1 ⊗ αQ t surjective. Hence i and i are isomorphisms. They allow us to identify VQ∗ with (VQ )∗ and VQ with (VQ∗ )∗ .

18.4.3 By 18.4.1 and 18.4.2, we can consider R as a root system in VQ and in VR = R ⊗Q VQ , and the corresponding Weyl groups are canonically isomorphic. Let us use the notations of 18.3.1. If x, y ∈ VQ , we have (x|y) ∈ Q. Denote again by (.|.) the canonical extension of this bilinear form to VR . Then for z ∈ VR : ∨ [α (z)]2 0. (z|z) = α∈R

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18 Root systems

By 18.3.1, equality holds if and only if z = 0. Thus (.|.) is a W (R)-invariant inner product on VR . Conversely, let us ﬁx a W (R)-invariant inner product on VR . We can consider the length of a root and the angle between two roots. By 18.3.2, this angle does not depend on the choice of the inner product. Similarly, if two roots belong to the same irreducible component, then the ratio of their lengths does not depend on the choice of the inner product.

18.5 Relations between two roots 18.5.1 We shall assume in the rest of this chapter that k = R. Let us ﬁx a root system R in V and a W (R)-invariant inner product (.|.) on V . If α, β ∈ R, denote by θαβ ∈ [0, π[ the angle between α and β, ωαβ the order of sα ◦ sβ and α the length of α. 18.5.2 If α, β ∈ R, then 18.3.3 implies that: aα,β aβ,α = 4 cos2 θα,β 4. Furthermore, aα,β aβ,α ∈ Z. Assuming α β, we see easily that the only possibilities are listed in the following table: θαβ β2 /α2 π undetermined ωαβ 2 π 1 ωαβ 3 2π 1 ωαβ 3 π 2 ωαβ 4 3π 2 ωαβ 4 π 3 ωαβ 6 5π 3 ωαβ 6

aα,β

aβ,α

0

0

1

1

−1

−1

1

2

−1

−2

1

3

−1

−3

2

2

0

1

β=α

−2

−2

π

1

β = −α

1

4

0

4

β = 2α

−1

−4

π

4

β = −2α

=2 =3 =3 =4 =4 =6 =6

18.5 Relations between two roots

241

18.5.3 We deduce from the table above the following proposition. Proposition. Let α and β be roots. ! 1" (i) If β = λα with λ ∈ k, then λ ∈ ± 1, ±2, ± . 2 (ii) If β ∈ kα and α β, then aα,β ∈ {−1, 0, 1}. (iii) If aα,β > 0, that is (α|β) > 0, then α − β ∈ R unless α = β. (iv) If aα,β < 0, that is (α|β) < 0, then α + β ∈ R unless α = −β. (v) If α − β and α + β do not belong to R ∪ {0}, then (α|β) = 0. 1 α ∈ R. 2 2) It may happen that α + β ∈ R and (α|β) = 0. Also, if α − β and α + β do not belong to R ∪ {0}, we say that α and β are strongly orthogonal. Remarks. 1) A root in R is reduced if

18.5.4 Proposition. Let α, β be roots which are not proportional. Set Iα,β = {j ∈ Z; β + jα ∈ R} and Sα,β = {β + jα; j ∈ Iα,β }. (i) The set Iα,β is an interval [−p, q] of Z containing 0. (ii) The set Sα,β is stable under sα and sα (β + qα) = β − pα , sα (β − pα) = β + qα. We say that Sα,β is the α-string through β, and that β − pα is the source, β + qα is the target and p + q the length. (iii) We have aβ,α = p − q. Proof. Let q = max Iα,β , −p = min Iα,β . Since β ∈ R, 0 ∈ Iα,β . Suppose that Iα,β = [−p, q]. Then there exist −p r, s q such that r, s ∈ Iα,β , r < s − 1 , r + i ∈ Iα,β if 1 i s − r − 1. It follows from 18.5.3 that (α|β + sα) 0 and (α|β + rα) 0. But this is absurd since: (α|β + sα) − (α|β + rα) = (s − r)α2 > 0. So Iα,β = [−p, q]. Next, we have sα (β + jα) = β − (aβ,α + j)α. Since sα (R) = R, sα (Sα,β ) = Sα,β . So (ii) follows from the fact that j → −aβ,α − j is an order-reversing bijection from Iα,β to itself. Finally, since sα (β + qα) = β − pα = β − (aβ,α + q)α, part (iii) follows from (ii). 18.5.5 Corollary. Let α, β be roots that are not proportional, S the αstring through β and γ its source. The length of S is −aγ,α and it is equal to 0, 1, 2 or 3. Proof. Note that S = Sα,γ , so the ﬁrst part follows from 18.5.4 (iii) and the second follows from the table in 18.5.2.

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18.5.6 Proposition. Let α, β be roots which are not proportional and p, q be the integers such that Iα,β = [−p, q]. If α + β ∈ R, then β + α2 p+1 . = β2 q Proof. Let γ = β−pα be the source of the α-string through β. Since α+β ∈ R, the length l = p + q of Sα,γ = Sα,β is in {1, 2, 3} (18.5.5). The results follows since by using 18.5.2 and 18.5.4, the only possibilities are listed in the following table: l p q aβ,α 1 2

β=γ β=γ

0 1 −1 β + α2 = β2 0 2 −2 2 β + α2 = β2

2 3 3

β =γ+α β=γ β =γ+α

1 1 0 β + α2 = 2 β2 0 3 −3 3 β + α2 = β2 1 2 −1 β + α2 = β2

3 β = γ + 2α 2 1

1

β + α2 = 3β2

Let us take for example the case p = 0 and q = 2, then by 18.5.2, aα,β = −1. Now β + α2 = β2 + α2 + 2(β|α). Replacing 2(β|α) by β2 aα,β , we obtain β + α2 = α2 . So we have 2β + α2 = β2 by replacing 2(β|α) by α2 aβ,α in the above equality. 18.5.7 Proposition. Assume that R is irreducible. Let α, β ∈ R, then there exists w ∈ W (R) such that w(β) = α if and only if α = β. Proof. If α = β, then since R is irreducible, there exists w ∈ W (R) such that α and w(β) are not orthogonal. So we may assume that (α|β) = 0. Replacing β by sβ (β) if necessary, we may further assume that (α|β) > 0. By 18.5.2, either α = β or aα,β = aβ,α = 1. In the latter, sα ◦ sβ ◦ sα (β) = sα ◦ sβ (β − α) = sα (−β − α + β) = α. So we are done. The converse is clear since the inner product is W (R)invariant. 18.5.8 Proposition. Assume that R is irreducible and reduced. (i) For α, β ∈ R, we have: # β2 1 1 , ∈ 1, 2, 3, . α2 2 3 (ii) The set {α; α ∈ R} contains at most two elements.

18.6 Examples of root systems

243

Proof. (i) In view of 18.2.7, the subspace spanned by w(β), w ∈ W (R), is V . Thus there exists w ∈ W (R) such that (α|w(β)) = 0. Since w(β) = β, the result follows from 18.5.2. (ii) This is a simple consequence of part (i). 18.5.9 Assume that R is irreducible and reduced. If {α; α ∈ R} = {λ, µ} where λ < µ, then a root of length λ (resp. µ) is called a short (resp. long) root. If all the roots have the same length, then, by convention, they are called long.

18.6 Examples of root systems 18.6.1 In view of the results of the previous sections, we give some examples of root systems of small rank over the real numbers. The reader may verify easily that the examples below are indeed roots systems. • Rank 1. A1 −α

α

Clearly, this is the only irreducible reduced root system of rank 1. • Rank 2. A1 × A1 β −α

α −β

The roots α and β are orthogonal, and this root system is not irreducible. A2 β

β+α

−α

α

−β−α

−β

Here, all the roots have the same length, and the angle between two adjacent roots is π/3. This is an irreducible reduced root system of rank 2.

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18 Root systems

B2 β+α

β

−α

β+2α

α

−β−2α

−β−α

−α

√ This is again an irreducible reduced root system of rank 2. We have β = 2α, and the angle between two adjacent roots is π/4. G2 2β+3α

β

β+α

β+2α

−α

−β−3α

β+3α

α

−β−2α

−β−α

−β

−2β−3α

√ In this root system, β = 3α, and the angle between two adjacent roots is π/6. This rank 2 root system is also irreducible and reduced.

18.7 Base of a root system 18.7.1 In this rest of this chapter, all root systems considered are assumed to be reduced. 18.7.2 We deﬁne a lexicographic order on V to be an order obtained in the following way: choose a R-basis (e1 , . . . , el ) of V , and write λ1 e1 + · · · + λl el ≺ µ1 e1 + · · · + µl el if for some integer n, we have λ1 = µ1 , . . ., λn = µn , and λn+1 < µn+1 . Clearly, this is a total order on V which is compatible with addition.

18.7 Base of a root system

245

18.7.3 Let ≺ be a lexicographic order on V . Set: R+ = {α ∈ R; α 0} , R− = {α ∈ R; α ≺ 0}. An element of R+ (resp. R− ) is called a positive (resp. negative) root. Denote by B the set of elements of R+ which is not the sum of two positive roots. The set B is called a base of R. 18.7.4 Theorem. Let R be a root system, ≺ a lexicographic order on V , R+ the set of positive roots and B the base of R with respect to this order. Then: (i) Any positive root is the sum of roots in B. (ii) If α, β ∈ B are distinct, then aα,β 0 and α − β ∈ R. (iii) If B = {β1 , . . . , βq }, then (β1 , . . . , βq ) is a basis of V . (iv) If α ∈ R+ \ B, then there exists β ∈ B such that aα,β > 0 and α − β ∈ R+ . (v) If α ∈ R+ , β ∈ B and α = β, then sβ (α) ∈ R+ . In particular, sβ permutes the elements of R+ \ {β}. Proof. (i) Let α1 ≺ · · · ≺ αn be the elements in R+ . It is clear that α1 ∈ B. Suppose that α1 , . . . , αi−1 can all be expressed as a sum of elements of B. If αi ∈ B. then αi = αr + αs where αr , αs ∈ R and r, s < i. We deduce therefore that αi is also a sum of elements of B. So by induction, we have (i). (ii) If α − β ∈ R, then β − α ∈ R. So either α − β or β − α is positive. This is impossible since α, β ∈ B and α = (α − β) + β, β = (β − α) + α. Thus α − β ∈ R, and aα,β 0 by 18.5.3 (iii). (iii) Let λ1 , . . . , λq ∈ R+ be such that, reindexing if necessary, (2)

v = λ1 β1 + · · · + λp βp = λp+1 βp+1 + · · · + λq βq . Then by part (ii), (v|v) = (λ1 β1 + · · · + λp βp |λp+1 βp+1 + · · · + λq βq ) 0.

Hence v = 0. Since βi 0, it follows that λi = 0 for all i. So the elements of B are linearly independent, and part (i) implies that B is a basis of V . (iv) By (i), there exist β1 , . . . , βp ∈ B pairwise distinct, and n1 , . . . , np ∈ N∗ such that α = n1 β1 + · · · + np βp . Since (α|α) > 0, it follows that (βi |α) > 0 for some i. Then 18.5.3 says that α − βi and βi − α are roots. Since βi ∈ B, this forces α − βi to be positive. (v) Let us write α = n1 β1 + · · · + np βp as above. Since α = β, there exists r such that βr = β. It follows that the coeﬃcient of βr in sβ (α) = α − aβ,α β is nr . But by (i) and (iii), the coeﬃcients of a root in the basis B are either all positive or all negative. So sβ (α) ∈ R+ . 18.7.5 Remarks. 1) An element of B is called a simple root of R with respect to B.

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18 Root systems

2) In general, let V be a k-vector space and consider R as a root system in VR . The same proof (18.7.4) shows that B is a basis of V , VQ and VR . 18.7.6 Corollary. Set: ρ=

1 α. 2 α∈R+

Then sβ (ρ) = ρ − β and β ∨ (ρ) = 1 for all β ∈ B. Proof. This is a simple consequence of 18.7.4 (v).

18.7.7 Proposition. Let ≺ be a lexicographic order on V and α1 , . . . , αn positive roots such that α = α1 + · · · + αn ∈ R+ . Then there is a permutation σ of {1, 2, . . . , n} such that ασ(1) + · · · + ασ(i) ∈ R+ for 1 i n. Proof. Since (α|α) > 0, there exists j such that (α|αj ) > 0. If α = αj , then n = 1 and the result is obvious. Otherwise by 18.5.3, α − αj ∈ R+ , and the result follows by induction on n. 18.7.8 Proposition. Let ≺ be a lexicographic order on V , B be a base of R and WB the subgroup of W (R) generated by sβ , β ∈ B. (i) We have R = {w(β); w ∈ WB , β ∈ B}. (ii) We have WB = W (R). Proof. Let S = {w(β); w ∈ WB , β ∈ B} ⊂ R. (i) Since sβ (β) = −β, it suﬃces to show that R+ ⊂ S. Let α1 ≺ · · · ≺ αn be the elements of R+ . Since α1 ∈ B and B ⊂ S, α1 ∈ S. Let us suppose that α1 , . . . , αi−1 ∈ S. If αi ∈ B, then αi ∈ S. Otherwise, by 18.7.4 (iv), there exists β ∈ B such that aα,β > 0. Thus δ = sβ (α) = α − aα,β β ≺ α. Further δ ∈ R+ (18.7.4 (v)). So δ ∈ S and hence α = sβ (δ) ∈ S. Now (i) follows by induction. (ii) It suﬃces to prove that sα ∈ WB for all α ∈ R. By part (i), there exist w ∈ WB and β ∈ B such that α = w(β). Hence sα = w ◦ sβ ◦ w−1 ∈ WB . 18.7.9 Proposition. (i) Let B be a base of R, B ⊂ B, V the subspace spanned by B and R = R ∩ V . Then B is a base for the root system R . (ii) Let W be a subspace of V , R = R ∩ W and B a base of R . Then there exists a base of R containing B . Proof. (i) We may assume that B = ∅. By 18.2.5, R is a root system in V . = V ∩ R+ . Then it is clear Let B = (β1 , . . . , βl ), B = (β1 , . . . , βp ) and R+ that with respect to the lexicographic order associated to B (resp. B ) on V ) is the set of positive roots. Hence B is the base of (resp. V ), R+ (resp. R+ R with respect to this order. (ii) Let B = (β1 , . . . , βp ) and complete to a basis (β1 , . . . , βl ) of V . The lexicographic order on V associated to this basis deﬁnes a base B of R which clearly contains B .

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247

18.8 Weyl chambers 18.8.1 We shall endow V with the usual topology deﬁned for example by the W (R)-invariant inner product. Let E = (e1 , . . . , el ) be a basis of V and λ1 , . . . , λl ∈ V ∗ such that for any x ∈ V , x = λ1 (x)e1 + · · · + λl (x)el . The set {x ∈ V ; λi (x) > 0, 1 i l} will be called the open simplicial cone associated to the basis E. 18.8.2 Let B = {β1 , . . . , βl } be a base of R and si = sβi for 1 i l. For w ∈ W (R), by 18.7.8 (ii), there exist 1 i1 , . . . , in l such that w = si1 ◦ · · · ◦ sin . Such an expression is called a reduced decomposition if n is minimal, and we say that n is the length of w that we shall denote by (w). By convention, (idV ) = 0. 18.8.3 Lemma. Let B be a base of R. (i) Let β1 , . . . , βp ∈ B not necessary distinct. If s1 ◦ · · · ◦ sp−1 (βp ) ∈ R− , si = sβi , then s1 ◦ · · · ◦ sp = s1 ◦ · · · ◦ sq−1 ◦ sq+1 ◦ · · · ◦ sp−1 for some q ∈ {1, . . . , p − 1}. (ii) Let w ∈ W (R) and β1 , . . . , βp ∈ B such that w = s1 ◦ · · · ◦ sp , si = sβi . If p = (w), then w(βp ) ∈ R− . (iii) Let w ∈ W (R) and n(w) the number of positive roots α such that w(α) ∈ R− . Then n(w) = (w). Proof. (i) Let αp−1 = βp and αi = si+1 ◦ · · · ◦ sp−1 (βp ), 0 i p − 2. We have α0 ∈ R− and αp−1 ∈ R+ . Let q be the smallest integer such that αq ∈ R+ . Then sq (αq ) = αq−1 ∈ R− . So αq = βq by 18.7.4 (v). Since sw(α) = w◦sα ◦w−1 for w ∈ W (R) and α ∈ R, we have: sq = (sq+1 ◦ · · · ◦ sp−1 ) ◦ sp ◦ (sp−1 ◦ · · · ◦ sq+1 ). So part (i) follows. (ii) We have w(βp ) = −s1 ◦ · · · ◦ sp−1 (βp ). If s1 ◦ · · · ◦ sp−1 (βp ) ∈ R− , then p > (w) by part (i). So we have proved (ii). (iii) Proceed by induction on (w). The case (w) = 0 is obvious. Let w = s1 ◦ · · · ◦ sp be a reduced decomposition of w where the notations are as above. By (ii), w(βp ) ∈ R− . So it follows from 18.7.4 (v) that n(w ◦ sp ) = n(w) − 1. But it is clear that (w ◦ sp ) = (w) − 1, so we have by induction that n(w) = (w). 18.8.4 For α ∈ R, deﬁne Pα = ker α∨ = {x ∈ V ; (x|α) = 0}. Let P be the union of Pα , α ∈ R. It is a closed subset of V , and the connected components of P = V \ P are open subsets of V , called the Weyl chambers (or chambers) of R. By 18.3.2, they do not depend on the choice of the W (R)-invariant inner product on V . Since the inner product is W (R)-invariant, W (R) acts on P and P , and W (R) permutes the chambers of R.

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18.8.5 Lemma. The group W (R) acts transitively on the set of Weyl chambers of R. Proof. Let C1 , C2 be chambers in R, x1 ∈ C1 and x2 ∈ C2 . Let w ∈ W (R) be such that x1 − w(x2 ) = inf{x1 − w (x2 ); w ∈ W (R)}. Set I = {tx1 + (1 − t)w(x2 ); t ∈ [0, 1]}. Suppose that I ∩ Pα = ∅ for some α ∈ R, then since x1 , w(x2 ) ∈ Pα , there exists t0 ∈]0, 1[ such that (t0 x1 + (1 − t0 )w(x2 )|α) = 0. Thus x1 − sα ◦ w(x2 )2 = x1 − w(x2 )2 − < x1 − w(x2 )2 .

4(1 − t0 ) (w(x2 )|α)2 t0 α2

But this contradicts the assumption on w, and since I is connected, I is contained in a chamber. Hence x1 , w(x2 ) ∈ C1 . Since w permutes the chambers and C1 ∩ w(C2 ) = ∅, we have C1 = w(C2 ). 18.8.6 Proposition. Let B = {β1 , . . . , βl } be a base of R, (β1 , . . . , βl ) the basis of V dual to B with respect to the inner product (.|.), and l # xi βi ; x1 , · · · , xl > 0 . C(B) = {x ∈ V ; (x|βi ) > 0, 1 i l} = i=1

Then C(B) is a Weyl chamber of R. Proof. It is clear from 18.7.4 that C(B) ⊂ P . Since C(B) is convex, it is connected. So C(B) is contained in a chamber, say C, of R. Suppose that there exists x ∈ C \ C(B). Then (x|β) < 0 for some β ∈ B. Set C1 = {x ∈ C; (x|β) > 0} and C2 = {x ∈ C; (x|β) < 0}. Then C is the disjoint union of the open subsets C1 and C2 . Now x ∈ C2 and C(B) ⊂ C1 . This implies that C is not connected. Contradiction. 18.8.7 Theorem. Let R be a root system in V . (i) Any chamber of R is of the form C(B) for some base B of R. In particular, any chamber of R is an open simplicial cone. (ii) The map B → C(B) induces a bijection between the set of bases of R and the set of chambers of R. (iii) The group W (R) acts simply transitively on the set of bases of R and on the set of chambers of R. Proof. (i) If w ∈ W (R) and B a base of R, then clearly w(B) is another base of R. So (i) follows from 18.8.5 and 18.8.6. (ii) The map B → C(B) is surjective by (i). Since C(B) is uniquely determined by R+ βi , 1 i l, (notations of 18.8.6), the map is injective. (iii) By 18.8.5 and (ii), W (R) acts transitively on the set of bases of R. Let B be a base and w ∈ W (R) such that w(B) = B. Then 18.8.3 (ii) forces w to be idV . Hence the action is simply transitive on the set of the bases and, by (ii), on the set of chambers as well.

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249

18.8.8 Remarks. 1) For a ﬁxed base B of R, the chamber C(B) is called the fundamental Weyl chamber. 2) If C is a Weyl chamber, denote by B = B(C) the corresponding base of R, R+ = R+ (C) and R− = R− (C). For β ∈ B, the hyperplane Pβ is called a wall of the chamber C. 3) Suppose that R is the direct sum of root systems R1 , . . . , Rn . Let Ci be a chamber of Ri and Bi the corresponding base of Ri , then C = C1 × · · · × Cn is a chamber of R corresponding to the base B = B1 ∪ · · · ∪ Bn . By 18.8.7 (iii), all the chambers and bases of R are obtained in this way. 18.8.9 Proposition. Let B = {β1 , . . . , βl } be a base of R and C = C(B). (i) For all x ∈ V , there exists w ∈ W (R) such that w(x) ∈ C. (ii) If x, y ∈ C and w ∈ W (R) verify w(x) = y, then x = y. (iii) Any W (R)-orbit of V intersects C at a unique point. Proof. Let us deﬁne the following partial order on V : x y ⇔ y − x ∈ R+ β1 + · · · + R+ βl . (i) Let w be maximal with respect to in the set {w (x); w ∈ W (R)}. If w(x) ∈ C, then there exists β ∈ B such that β ∨ (w(x)) = (β|w(x)) < 0. But sβ ◦ w(x) − w(x) = −β ∨ (w(x))β, so w(x) ≺ sβ ◦ w(x) which contradicts the maximality of w. So w(x) ∈ C. (ii) We shall show by induction on (w) that w is the product of reﬂections sβ , β ∈ B such that sβ (x) = x. The case (w) = 0 is trivial. So let us suppose that (w) > 0. Then by 18.7.8 (ii) and 18.8.3 (ii), there exists β ∈ B such that w(β) ∈ R− and (w ◦ sβ ) = (w) − 1. Since x, y ∈ C, 0 (β|x) = (β|w−1 (y)) = (w(β)|y) 0. Hence (β|x) = 0. Thus sβ (x) = x and w ◦ sβ (x) = y. The result follows by induction. Thus x = y. (iii) This is clear from (i) and (ii). 18.8.10 Remark. The partial order introduced in the proof of 18.8.9 is called the partial order associated to B (or to C). 18.8.11 Remark. The notions of bases and Weyl chambers can easily be extended to a non-reduced root system R. In particular, R has a base, and W (R) acts simply transitively on the set of bases and the set of Weyl chambers.

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18.9 Highest root 18.9.1 Lemma. Assume that R is irreducible. Let B be a base of R. Then B is not the (disjoint) union of two orthogonal subsets. Proof. Suppose that B is the disjoint union of (non-empty) subsets B1 and B2 such that (β1 |β2 ) = 0 for β1 ∈ B1 , β2 ∈ B2 . By 18.7.8 (ii), the subspace of V spanned by B1 is proper and W (R)-stable. This contradicts 18.2.7. 18.9.2 Theorem. Assume that R is irreducible. Let B = {β1 , . . . , βl } be a base of R, C = C(B) the corresponding chamber. (i) There exists a root α = n1 β1 + · · · + nl βl such that for any root α = p1 β1 + · · · + pl βl , we have ni pi , 1 i l. (ii) We have α ∈ C. (iii) We have α α for all α ∈ R. In particular, α is a long root. α}, then aα,α ∈ {0, 1}. (iv) If α ∈ R+ \ { Proof. Let α = n1 β1 + · · · + nl βl be a maximal element of R with respect to the partial order associated to B. Clearly, α ∈ R+ . Let I = {1, . . . , l} , I1 = {i ∈ I; ni > 0} , I2 = {i ∈ I; ni = 0} = I \ I1 . We claim that I2 = ∅, because otherwise, for each i ∈ I, (α|βi ) 0 (18.7.4), and so by 18.9.1, (βi |βj ) = 0 for some i ∈ I2 and j ∈ I1 . It follows that (α|βi ) < 0 and hence α ≺ α + βi ∈ R+ (18.5.3), which contradicts the maximality of α. So we have our claim. Using 18.5.3 again, we deduce that (α|βi ) 0 for all i ∈ I, and so (α|βi ) > 0 for some i ∈ I. Now, if α is another maximal element of R with respect to , then (α|α ) > 0. So if α = α , then α − α ∈ R (18.5.3), and therefore α α or α α which again contradicts the maximality of α and α . Thus we have proved (i), and part (ii) follows from 18.5.3. Let α ∈ R. To prove (iii), we may assume that α = α and α ∈ C (18.8.9). α| α − α) 0 and Since 0 ≺ α − α, (x| α − α) 0 for all x ∈ C. In particular, ( (α| α − α) 0. Hence ( α| α) (α| α) (α|α). α}, then α α and ( α|α) 0 since α ∈ C. So Finally, if α ∈ R+ \ { (iv) is a consequence of 18.5.2. 18.9.3 Assume that R is irreducible. The root α in 18.9.2 is called the highest root of R with respect to the base B (or chamber C).

18.10 Closed subsets of roots 18.10.1 Deﬁnition. Let P be a subset of R. (i) We say that P is symmetric if P = −P . (ii) We say that P is closed if given α, β ∈ P verifying α + β ∈ R, then α + β ∈ P. (iii) We say that P is parabolic if P is closed and R = P ∪ (−P ). 18.10.2 Proposition. Let P ⊂ R be a closed subset such that P ∩(−P ) = ∅. There exists a chamber C of R such that P ⊂ R+ (C).

18.10 Closed subsets of roots

251

Proof. 1) Let n ∈ N∗ , α1 , . . . , αn ∈ P and α = α1 + · · · + αn . We claim that α = 0. This is obvious if n = 1. Let us proceed by induction on n. If n 2 and α = 0, then (α1 |α2 +· · ·+αn ) < 0. So there exists some j such that (α1 |αj ) < 0. Since P is closed, it follows from 18.5.3 that α1 + αj ∈ P . Thus α = 0 is the sum of n − 1 elements of P , which contradicts the induction hypothesis. So we have proved our claim. 2) We claim there exists α ∈ P such that (α|β) 0 for all β ∈ P . For otherwise, there exist α1 , α2 ∈ P such that (α1 |α2 ) < 0, then α1 + α2 ∈ P , and there exists α3 ∈ P such that (α1 + α2 |α3 ) < 0, and so α1 + α2 + α3 ∈ P . Repeating this process, we obtain a sequence (αi )i1 of elements of P such that α1 + · · · + αi ∈ P for all i ∈ N∗ . It follows that there exists i < j such that α1 + · · · + αi = α1 + · · · + αj . Hence αi+1 + · · · + αj = 0 which contradicts 1). 3) Finally, let us prove that there is a basis E of V such that the elements of P are positive with respect to the lexicographic order associated to E. The case l = 1 is trivial. Let us proceed by induction on l and assume that l 2. From 2), there exists v1 ∈ V \ {0}, such that (v1 |α) 0 for all α ∈ P . Let H be the hyperplane orthogonal to v1 and V1 the subspace of V spanned by R ∩ H. It follows from 18.2.5 that P ∩ H is a closed subset of R ∩ H. By our induction hypothesis, there exists a basis (v2 , . . . , vl ) of H such that the elements of P ∩H are positive with respect to the associated lexicographic order. It is then clear that the elements of P are positive with respect to the lexicographic order associated to the basis (v1 , . . . , vl ) of V . 18.10.3 Corollary. The following conditions are equivalent for a subset P of R. (i) There exists a chamber C of R such that P = R+ (C). (ii) P is parabolic and P ∩ (−P ) = ∅. If these conditions are veriﬁed, the chamber C such that P = R+ (C) is unique. Proof. The equivalence is clear from 18.10.2. If P = R+ (C), then C ∨ is the set of ϕ ∈ V ∗ verifying ϕ(α) > 0 for all α ∈ P . Hence the uniqueness of C. 18.10.4 Lemma. Let C be a chamber of R, B = B(C), R+ = R+ (C), P a closed subset of R containing R+ , Σ = B ∩ (−P ) and Q the set of roots which is the sum of elements of −Σ. Then P = R+ ∪ Q. Proof. Since P is closed, Q ⊂ P (18.7.7). So R+ ∪ Q ⊂ P . Now let α ∈ P \R+ . Then α is the sum of p elements of −B. We shall show by induction on p that α ∈ Q. The case p = 1 is obvious. So let us assume that p 2. By 18.7.7, there exist β ∈ −B, γ ∈ R− such that α = β + γ, and γ is the sum of p−1 elements of −B. Since R+ ⊂ P , −γ, −β ∈ P . So β = (−γ)+α ∈ P and γ = (−β) + α ∈ P , because P is closed. Thus β ∈ −Σ, and γ ∈ Q by our induction hypothesis. Again since P is closed, α = β + γ ∈ Q.

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18.10.5 Proposition. The following conditions are equivalent for a subset P of R. (i) P is parabolic. (ii) P is closed and there exists a chamber C of R such that R+ (C) ⊂ P . (iii) There exists a chamber C of R and a subset Σ of B(C) such that P is union of R+ (C) and the set of roots which can be expressed as a sum of elements of −Σ. Proof. (i) ⇒ (ii) Let C be a chamber such that the cardinal of P ∩ R+ (C) is maximal. Set B = B(C), R+ = R+ (C). Suppose that there exists β ∈ B \ P . So −β ∈ P and −β = sβ (β) ∈ where C = sβ (C) and R+ = R+ (C ). Hence −β ∈ P ∩ R+ . sβ (R+ ) = R+ If α ∈ R+ \ {β}, then sβ (α) ∈ R+ (18.7.4 (v)). So α = sβ ◦ sβ (α) ∈ . Hence P ∩ R+ ⊂ P ∩ R+ . sβ (R+ ) = R+ which contradicts our choice We deduce that {−β} ∪ (P ∩ R+ ) ⊂ P ∩ R+ of C. Hence B ⊂ P and R+ ⊂ P . (ii) ⇒ (iii) This is clear by 18.10.4. (iii) ⇒ (i) We only need to show that P is closed. Let Σ = {β1 , . . . , βp } and B = B(C) = Σ ∪ {βp+1 , . . . , βl }. Let α, β ∈ P be such that α + β ∈ R. Since R+ (C) ⊂ P , we may assume that α + β = −n1 β1 − · · · − nl βl ∈ R− (C), where n1 , . . . , nl ∈ N. Since P = R+ (C) ∪ Q, ni = 0 for p + 1 i l. Hence α + β ∈ Q ⊂ P . 18.10.6 Proposition. Let P be a subset of R, Γ (resp. V ) the subgroup (resp. subspace) of V generated (resp. spanned) by P . Then the following conditions are equivalent: (i) Γ ∩ R = P . (ii) P is closed and symmetric. (iii) P is closed and is a root system in V . If these conditions are veriﬁed, then the map α → α∨ |V is a bijection from P onto P ∨ . Proof. (i) ⇒ (ii) This is clear. (ii) ⇒ (iii) With the notations of 18.2.1, P veriﬁes (R1) in V . Let α, β ∈ P . If β = ±α, then sα (β) = −β ∈ P . If β = ±α, then for 0 j aβ,α , β − jα ∈ R. Since P is symmetric and closed, sα (β) = β − aβ,α α ∈ P . Thus sα (P ) ⊂ P , and (R2) is veriﬁed. Finally, it is clear that (R3) is veriﬁed. So P is a root system V and last part of the proposition follows also. (iii) ⇒ (i) Clearly, P ⊂ Γ ∩ R. Conversely, if α ∈ Γ ∩ R, then α = α1 +· · ·+αn where αi ∈ P , since P = −P . From 0 < (α|α) = (α|α1 +· · ·+αn ), we deduce that (α|αi ) > 0 for some i. If α = αi , then α ∈ P . Otherwise, α − αi ∈ R (18.5.3) and α − αi ∈ Γ . By induction on n, α − αi ∈ P and since P is closed, α = (α − αi ) + αi ∈ P .

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253

18.11 Weights 18.11.1 Let us use the notations of 18.5.1 and 18.8.1. We deﬁne an isomorphism ϕ : V → V ∗ , ϕ(x)(y) = (x|y) for all x, y ∈ V . Denoting (ϕ(x)|ϕ(y)) = (x|y), we obtain a W (R∨ )-invariant inner product on V ∗ . For α ∈ R, we have: 2α∨ ϕ(α) = ∨ ∨ . (α |α ) Thus ϕ(Pα ) = Pα∨ (18.8.4). From this, we deduce that the image of a chamber of R via ϕ is a chamber of R∨ . 18.11.2 Let C be a chamber of R, B = B(C) = {β1 , . . . , βl }, C ∨ = ϕ(C) and B = B(C ∨ ) = {γ1 , . . . , γl }, S(C) = {x ∈ V ; (x|y) 0 for all y ∈ C} and S(C ∨ ) = {λ ∈ V ∗ ; (λ|µ) 0 for all µ ∈ C ∨ }. By 18.8.6, we have: S(C) = R+ β1 + · · · + R+ βl , S(C ∨ ) = R+ γ1 + · · · + R+ γl . It is then clear that S(C ∨ ) = R+ β1∨ + · · · + R+ βl∨ , and we deduce that B = B ∨ = {β1∨ , . . . , βl∨ }. In particular, B ∨ is a base of R∨ . 18.11.3 Let α = n1 β1 + · · · + nl βl ∈ R. Then: l 2βi∨ 2α∨ = ϕ(α) = . n i (α∨ |α∨ ) (βi∨ |βi∨ ) i=1

Hence α∨ =

l

ni βi 2 α−2 βi∨ .

i=1

By 18.11.2, ni βi 2 α−2 ∈ Z for 1 i l. 18.11.4 Denote by Q(R) the sublattice of V spanned by R. By 18.7.4, it is a free abelian subgroup of rank l in V . Any base of R is a basis of Q(R). In the same way, Q(R∨ ) is a sublattice of rank l in V ∗ . The lattice Q(R) is called the root lattice of R and its elements are called radical weights. 18.11.5 Proposition. Let P (R) be the set of x ∈ V such that α∨ (x) ∈ Z for all α ∈ R. We call P (R) the weight lattice of R and its elements are called weights. (i) P (R) is a sublattice of rank l in V . (ii) We have Q(R) ⊂ P (R) ⊂ VQ . (iii) If B ∨ is a base of R∨ , the dual basis of B ∨ in V is a basis of P (R). Proof. If x ∈ V , the following conditions are equivalent: (i) a∗ (x) ∈ Z for all a∗ ∈ Q(R∨ ). (ii) β ∨ (x) ∈ Z for all β ∨ ∈ B ∨ . (iii) The coordinates of x in the dual basis of B ∨ are integers. Consequently, we have parts (i) and (iii). By (R3), we have Q(R) ⊂ P (R).

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Let B = {β1 , . . . , βl } be a base of R, B ∨ = {β1∨ , . . . , βl∨ } as in 18.11.2 and {α1 , . . . , αl } the dual basis of B ∨ . For 1 p, q l, write αp =

l

λpj βj , and δpq =

i=1

l

λpj βq∨ (βj ).

j=1

This is a Cramer’s system with coeﬃcients in Q, so P (R) ⊂ VQ . 18.11.6 Let C be a chamber in R, B = B(C), B ∨ = {β1∨ , . . . , βl∨ } the base of R∨ as in 18.11.2. The dual basis { 1 , . . . , l } of B ∨ is a basis of P (R), and they are called the fundamental weights of R with respect to the base B (or chamber C). We shall conserve this notation for fundamental weights. If x ∈ V , then x ∈ C if and only if β ∨ (x) > 0 for all β ∈ B. Hence C = R∗+

1

+ · · · + R∗+

l

, C = R+

1

+ · · · + R+

l.

18.11.7 Proposition. Let B = (β1 , . . . , βl ) be a base of R, 1 , . . . , l the corresponding fundamental weights and C = C(B). For x ∈ V , the following conditions are equivalent: (i) β ∨ (x) ∈ N for all β ∈ B. (ii) x = n1 1 + · · · + nl l where n1 , . . . , nl ∈ N. (iii) x ∈ C ∩ P (R). (iv) For all w ∈ W (R), x − w(x) ∈ Q(R) and x − w(x) 0 in the partial order associated to B. If x ∈ V veriﬁes these conditions, then x is called a dominant weight of R with respect to B. Proof. The implications (i) ⇒ (ii) ⇒ (iii) are obvious. (iii) ⇒ (iv) Let w ∈ W (R) be such that w(x) is maximal in X = {w (x); w ∈ W (R)} with respect to the partial order associated to B. We saw in the proof of 18.8.9 (i) that w(x) ∈ C. So 18.8.9 (iii) implies that w(x) = x. Thus x is the unique maximal element of X. Since x ∈ P (R), (iv) follows from 18.7.8 (ii). (iv) ⇒ (i) This is clear since x − sβ (x) = β ∨ (x)β for β ∈ B. 18.11.8 Proposition. Let B be a base of R, C = C(B) and the partial order associated to B. If x ∈ V , the following conditions are equivalent: (i) x ∈ C. (ii) sβ (x) x for all β ∈ B. (iii) w(x) x for all w ∈ W (R). Proof. Since x − sβ (x) = β ∨ (x)β, we have (i) ⇔ (ii). Finally, (iii) ⇒ (ii) is obvious and (i) ⇒ (iii) is similar to the implication (iii) ⇒ (iv) of 18.11.7. 18.11.9 Corollary. With the hypotheses of 18.11.8, we have x ∈ C if and only if w(x) ≺ x for all w ∈ W (R) \ {idV }.

18.12 Graphs

255

Proof. The “if” part is clear by 18.11.8 and the deﬁnition of C. Conversely, if x ∈ C, then w(x) x for all w ∈ W (R). If w(x) = x and w = idV , then it follows from the proof of 18.8.9 (ii) that w is the product of reﬂections sβ , β ∈ B, verifying sβ (x) = x. Thus β ∨ (x) = 0 which is absurd since x ∈ C. 18.11.10 Proposition. With the notations of 18.11.6, let ρ be the half sum of the positive roots in R. Then: (i) ρ = 1 + · · · + l ∈ C. (ii) We have β2 = 2(ρ|β) for all β ∈ B. Proof. This is clear since β ∨ (ρ) = 1 = 2(ρ|β)/(β|β) for β ∈ B (18.7.6).

18.11.11 With the notations of 18.2.2 and 18.11.6, if 1 i l, then: βi =

l

αij

j.

j=1

Then A = [αij ] ∈ GLl (k). On the other hand, δij = βi∨ (

j)

=2

(βi | j ) . (βi |βi )

Thus αij = βj∨ (βi ) = aβi ,βj , and the matrix [aβi ,βj ] is invertible.

18.12 Graphs 18.12.1 Deﬁnition. A graph is a pair (S, A) where S is a ﬁnite set and A is a subset of P(S) (the set of subsets of S) whose elements are subsets of cardinal 2. A normed graph is a pair (Γ, f ) with the following properties: (a) Γ = (S, A) is a graph. (b) Let E be the set of pairs (i, j) such that {i, j} ∈ A. Then f is a map from E to R such that f (i, j)f (j, i) = 1 for all (i, j) ∈ E. 18.12.2 Let Γ = (S, A) be a graph. An element of S is called a vertex and an element of A is called an edge. Two vertices x, y are linked if {x, y} is an edge. A vertex is called a terminal vertex (resp. a ramiﬁcation point) if it is linked to at most one vertex (resp. at least three vertices). In general, we represent a graph by a picture consisting of points corresponding to the elements of S and two points are linked by a line if and only if the corresponding vertices are linked. 18.12.3 Let Γ = (S, A) and Γ = (S , A ) be graphs. An isomorphism of Γ onto Γ is a bijection from S to S which transports A to A . If S ⊂ S and A ⊂ A, then we say that Γ is a subgraph of Γ . Furthermore, if A = A ∩ P(S ), then we say that Γ is a full subgraph of Γ .

256

18 Root systems

18.12.4 Let Γ = (S, A) be a graph. If a, b ∈ S, then a path from a to b is a sequence C = (x0 , . . . , xn ) of elements of S verifying: (i) x0 = a and xn = b. (ii) For 0 i n − 1, {xi , xi+1 } ∈ A. The integer n is the length of the path C. The path C is injective if the xi ’s are pairwise distinct. Thus a path from a to b of minimal length is injective. We deﬁne an equivalence relation ∼ on S as follows: a ∼ b if and only if there is a path from a to b. The equivalence classes are called the connected components. A graph is connected if it has at most one connected component. Thus in a connected graph, there is always a path between two vertices. Let Γ1 , . . . , Γn be the connected components of Γ , considered as full subgraphs of Γ . Then each Γi is connected and for i = j, there is no path between a vertex of Γi and a vertex of Γj . A cycle in Γ is a path C = (x0 , . . . , xn ) of length n 2 verifying x0 = xn and C = (x0 , . . . , xn−1 ) is an injective path. A graph is a forest if Γ has no cycles. A connected forest is called a tree. For n ∈ N∗ , we denote by An the graph with vertices {1, 2, . . . , n} and edges are the subsets {i, j} such that i − j = ±1: 1

2

3

n -1

n

A graph which is isomorphic to An+1 is called a chain of length n 0. 18.12.5 Proposition. Let Γ be a non-empty graph. Then Γ is a chain if and only if Γ is a tree without ramiﬁcation points. Proof. It is clear that if Γ is chain, then it is a tree without ramiﬁcation points. Conversely, suppose that Γ is a tree without ramiﬁcation points. Let C = (x0 , . . . , xn ) be an injective path of maximal length in Γ . Suppose that a is a vertex which is distinct from the xi ’s. Since Γ is connected, there is a path C from a to one of the xi ’s. Replacing a by a vertex in C , we may assume that a and xj are linked. If j = 0 (resp. n), then (a, x0 , . . . , xn ) (resp. (x0 , . . . , xn , a)) is an injective path of length n + 1. But this contradicts our choice of C. If 0 < j < n, then xj is a ramiﬁcation point which is absurd. Hence {x0 , . . . , xn } is the set of vertices of Γ and since Γ is a tree without ramiﬁcation points, Γ is necessarily a chain.

18.13 Dynkin diagrams 18.13.1 Deﬁnition. Let R, R be root systems in V and V . We say that R and R are isomorphic if there exists a bijective linear map F : V → V such that F (R) = R and for all α, β ∈ R, we have: (F (α)|F (β)) (α|β) = . (F (β)|F (β)) (β|β)

18.13 Dynkin diagrams

257

18.13.2 Observe that the notion of isomorphism in 18.13.1 does not depend on the choice of invariant inner products (18.3.2). Suppose that R and R are isomorphic and F : V → V the corresponding bijective linear map. Then the map w → F ◦ w ◦ F −1 is an isomorphism of the groups W (R) and W (R ). In particular, let R1 , . . . , Rm be the irreducible components of R. Then given λ1 , . . . , λm ∈ R∗ , the direct sum R of the root systems λi Ri is isomorphic to R. 18.13.3 Deﬁnition. Let B = (β1 , . . . , βl ) be a base of R. The matrix [aβi ,βj ]1i,jl is called the Cartan matrix of R with respect to B. 18.13.4 It follows from 18.8.7 that, up to a permutation, the Cartan matrix of R does not depend on the choice of the base B. Clearly, isomorphic root systems have the same Cartan matrix, up to a permutation. The converse is given in the following proposition. Proposition. Let R, R be root systems in V, V , B, B bases of R, R . Suppose that there is a bijection f : B → B transforming the Cartan matrix of R to the Cartan matrix of R . Then R and R are isomorphic via an isomorphism F : V → V which extends f . In particular, the Cartan matrix of a root system determines the root system up to an isomorphism. Proof. Let B = (β1 , . . . , βl ), B = (β1 , . . . , βl ) where βi = f (βi ). Since B, B are basis of V and V , it is clear that f extends uniquely to an isomorphism F : V → V . Let α, β ∈ B, α = F (α), β = F (β). Then aβ,α = aβ ,α and: sF (α) (F (β)) = β − aβ ,α α = F (β − aβ,α α) = F (sα (β)). Since B is a basis of V , we deduce that sF (α) ◦ F = F ◦ sα for α ∈ B. Hence 18.7.8 implies that the map w → F ◦ w ◦ F −1 is an isomorphism from W (R) onto W (R ), sending sα to sF (α) if α ∈ B. Let β ∈ R. By 18.7.8, β = w(α) for some α ∈ B and w ∈ W (R). It follows that F (β) = (F ◦ w ◦ F −1 )(F (α)). Hence F (R) = R . On the other hand, since sw(α) = w ◦ sα ◦ w−1 , we have: F ◦sβ = F ◦w◦F −1 ◦F ◦sα ◦w−1 = (F ◦w◦F −1 )◦sF (α) ◦F ◦w−1 = (F ◦w◦F −1 )◦sF (α) ◦(F ◦w◦F −1 )−1 ◦F = sF (w(α)) ◦F = sF (β) ◦F. Finally, for α, β ∈ R, we have: sF (α) F (β) = F (β − aF (β),F (α) α) , F ◦sα (β) = F (β − aβ,α α). Hence aF (β),F (α) = aβ,α .

258

18 Root systems

18.13.5 We associate to R a normed graph Γ (R) = (X, f ), called the Dynkin diagram as follows: Fix a base B = (β1 , . . . , βl ) of R, and for 1 i, j l, let nij = aβi ,βj . The vertices of the graph X is {1, . . . , l} (or {β1 , . . . , βl }) and {i, j} is an edge if and only if (βi |βj ) = 0 (or equivalently nij = 0 or nji = 0). If {i, j} is an edge, we deﬁne: βi 2 nij = . f (i, j) = nji βj 2 It is clear that f (i, j)f (j, i) = 1. Let θij be the angle between βi and βj . Then the only possible cases, up to a permutation of i and j, are listed below: π 1) i and j are not linked ; nij = nji = 0, θij = . 2 2π . 2) f (i, j) = f (j, i) = 1 ; nij = nji = −1, βi = βj , θij = 3 √ 1 3π . 3) f (i, j) = 2, f (j, i) = ; nij = −2, nji = −1, βi = 2βj , θij = 2 4 √ 1 5π . 4) f (i, j) = 3, f (j, i) = ; nij = −3, nji = −1, βi = 3βj , θij = 3 6 Thus the Dynkin diagram determines R up to an isomorphism (18.13.4). 18.13.6 In practice, we represent Γ (R) = (X, f ) by a picture with points and lines as follows: the points are the vertices of X, and we draw nij nji lines from i to j. Further, if f (i, j) > 1 and θij = π/2 (cases 3) and 4) above), then we place the symbol > on the lines joining i to j, that is from the long root to the short root: i

j

f(i,j) = 2

i

j

f(i,j) = 3

18.13.7 Proposition. Let R be root system in V and Γ (R) its Dynkin diagram. Then: (i) R is irreducible if and only if Γ (R) is connected. (ii) Γ (R) is a forest, and if R is irreducible, then Γ (R) is a tree. Proof. (i) Suppose that R is the direct sum of two non-empty root systems R1 and R2 . Let B1 , B2 be bases of R1 , R2 . Then B = B1 ∪ B2 is a base of R and Γ (R1 ), Γ (R2 ) are full subgraphs of Γ (R). Further there is no edge joining a vertex of Γ (R1 ) to a vertex of Γ (R2 ). So Γ (R) is not connected. Conversely, suppose that Γ (R) is not connected, then R has a base of the form B = B1 ∪ B2 where B1 , B2 are non-empty orthogonal subsets. Let Vi be the subspace spanned by Bi . If α ∈ B1 , then the orthogonal of α is the direct sum of V2 and a hyperplane in V1 . It follows that sα (Vi ) = Vi for i = 1, 2. The same result applies for α ∈ B2 . So V1 and V2 are W (R)-stable (18.7.8), and R is not irreducible (18.2.7).

18.14 Classiﬁcation of root systems

259

(ii) Let B be a base of R. If Γ (R) is not a forest, then there exist β1 , . . . , βn ∈ B pairwise distinct such that (β1 , . . . , βn ) is a cycle. Let γi = βi /βi . If {i, j} is an edge, then by 18.13.5, (γi |γj ) −1/2. Hence: $ $2 n $ $ $ γi $ = n + 2 (γi |γj ) n + 2[(γ1 |γ2 ) + · · · + (γn−1 |γn ) + (γn |γ1 )] $ $ i=1

i

n − n = 0. This is absurd. So Γ (R) is a forest, and (i) implies that Γ (R) is a tree.

18.13.8 Let B = (β1 , . . . , βl ) be a base of R. If α = n1 β1 + · · · + nl βl ∈ R, we set H(α) to be the set of βi ∈ B such that ni = 0. Proposition. (i) If α ∈ R, then H(α) is a connected subset of B (considered as the set of vertices in the Dynkin diagram of R). (ii) If H is a non-empty connected subset of B, then β∈H β ∈ R. Proof. (i) We may assume that α ∈ R+ and we shall proceed by induction on the cardinality n of H(α). The case n = 1 being trivial, we shall assume that n 2. By 18.7.4, there exists β ∈ B such that α − β ∈ R. Let r be maximal such that γ = α − rβ ∈ R. So r 1 and the β-string through γ is of the form {γ, . . . , γ + qβ}, with q r. Thus aγ,β = −q −r −1 (18.5.4), and so (γ|β) = 0. It follows that there is an edge joining β and an element of H(γ). Since H(α) = H(γ) ∪ {β}, it follows from the induction hypothesis that H(α) is connected. (ii) Let α = β∈H β. We shall prove by induction on the cardinality n of H(α) that α ∈ R. The case n = 1 being trivial, we may assume that n 2. Let (γ0 , γ1 , . . . , γm ) be an injective path of maximal length in H. Since H is a tree (18.13.7) and the length of the path is maximal, γ0 is a terminal vertex in H. Let M = H \ {γ0 }, then M is connected. By the induction hypothesis, α − γ0 ∈ R. Further, γ0 and γ1 are linked, so (α − γ0 |γ0 ) < 0 (18.7.4). Hence α ∈ R (18.5.3).

18.14 Classiﬁcation of root systems 18.14.1 Theorem. Let R be an irreducible root system in V . Then the Dynkin diagram of R is isomorphic to one of the following : Al

l vertices, l 1

Bl

l vertices, l 2

Cl

l vertices, l 3

Dl

l vertices, l 4

260

18 Root systems

E6

E7

E8

F4 G2 Moreover, these diagrams are pairwise non-isomorphic. Proof. We can check easily that these diagrams are pairwise non-isomorphic. So let us prove the ﬁrst part. Let B = (β1 , . . . , βl ) be a base of R and Γ be the Dynkin diagram of R. Set γi = βi /βi . Recall from the proof of 18.13.7 that (γi |γj ) −1/2 if {i, j} is an edge, and (γi |γj ) = 0 otherwise. Let f (i, j) be as deﬁned in 18.13.5. 1) Let us suppose that there exist vertices i, j of Γ such that f (i, j) = 3. Since Γ is connected (18.13.7), if Γ is not G2 , then there exists a vertex k distinct from i and j such that √ 3 1 , (γi |γk ) 0 , (γj |γk ) − . (γi |γj ) = − 2 2 It follows that

√ √ √ 1 3 3γi + 2γj + γk 2 3 + 4 + 1 − 2 3 × 2 − 2 × 2 × = 0. 2 2

This is absurd. So Γ G2 . Let us assume from now on that f (i, j) = 3 for all i, j. 2) Let us suppose that there exist vertices i, j, k, l of Γ such that f (i, j) = 2 and f (k, l) = 2. Again, since Γ is connected, renumbering the vertices if necessary, we may assume that Γ contains the following subdiagram : 1

2

3

n 2

n 1

n

1 with n 3. So, (γ1 |γ2 ) = (γn−1 |γn ) = − √ . Hence : 2 $ $2 $ 1 $ $ √ γ1 + γ2 + · · · + γn−1 + √1 γn $ 1 + 1 +(n−2)−4 √1 √1 −2(n−3) 1 = 0. $ 2 2 2 $ 2 2 2 2

18.14 Classiﬁcation of root systems

261

Again, this is absurd. So we may further assume that Γ has at most one edge {i, j} such that f (i, j) = 2. 3) Let us suppose that Γ has an edge {i, j} such that f (i, j) = 2, and a ramiﬁcation point. Then Γ contains a subdiagram of the form : n 1

2

3

n 2

n 3

n 1

1 1 with n 4. So (γ1 |γ2 ) = − √ and (γi |γj ) = − for the other edges in this 2 2 subdiagram. Consequently, we have $ $2 $ γ1 $ $ √ + γ2 + · · · + γn−2 + γn−1 + γn $ 1 + (n − 3) + 1 + 1 − 2 √1 √1 $ 2 2 2 $ 2 4 4 2 2 11 11 1 −2 = 0. −2(n − 4) − 2 2 22 22 This is absurd. 4) Suppose that Γ has an edge {i, j} such that f (i, j) = 2. By the previous considerations, Γ is a chain. If Γ is not isomorphic to Bl or Cl or F4 , then it contains a subdiagram of the form : 1

2

3

4

5

It follows that : $ $√ √ √ √ $ 2γ1 + 2 2γ2 + 3γ3 + 2γ4 + γ5 $2 2 + 8 + 9 + 4 + 1 − 2 2.2 2 1 2 √ 1 1 1 −2.2 2.3 √ − 2.3.2 − 2.2 = 0. 2 2 2 Again, this is absurd. So Γ is isomorphic to either Bl , Cl or F4 . We may therefore assume in the rest of the proof that f (i, j) = 1 for all edges {i, j} of Γ . 5) If Γ has no ramiﬁcation point, then it is isomorphic to Al . 6) Let us suppose that Γ has two distinct ramiﬁcation points. Then Γ contains a subdiagram of the form : 1

n 1 3

2

4

n 3

n 2 n

We obtain therefore that : $γ γ2 γn $ 1 γn−1 1 $2 $ 1 + γ3 + · · · + γn−2 + + $ 4 + (n − 4) − 2(n − 5) $ + 2 2 2 2 4 2 11 11 11 11 −2 −2 −2 −2 22 22 22 22 = 0.

262

18 Root systems

This is absurd. 7) We are left with the case where Γ has a unique ramiﬁcation point. In particular, Γ are at least 4 vertices, and it is of the form :

If Γ contains a subdiagram of the form : 1

4

5

6

7

3

2

then 1 γ1 + 2γ2 + 3γ3 + 2γ4 + γ5 + 2γ6 + γ7 2 24 − 2 (2 + 6 + 6 + 6 + 2 + 2) = 0. 2 This is absurd. If Γ contains a subdiagram of the form : 1

2

3

4

5

6

7

8

then we can check easily that γ1 + 2γ2 + 3γ3 + 4γ4 + 3γ5 + 2γ6 + γ7 + 2γ8 2 0, which is absurd. If Γ contains a subdiagram of the form : 1

2

3

4

5

6

7

8

9

then we can check easily that γ1 + 2γ2 + 3γ3 + 4γ4 + 5γ5 + 6γ6 + 4γ7 + 2γ8 + 3γ9 2 0, which is absurd. It is now easy to see that Γ is isomorphic to either Dl , E6 , E7 or E8 .

18.14.2 In the following tables, we give explicit descriptions of irreducible root systems. The irreducible root system R is considered to be a subset of some Rn where n is not necessarily l, the rank of R. Let B = (β1 , . . . , βl ) be a base of R and (ε1 , . . . , εn ) denote the canonical basis of Rn .

18.14 Classiﬁcation of root systems

263

The vertices of the Dynkin diagram correspond to the elements of B. We add a vertex which corresponds to − α, where α is the highest root of R with respect to B and concerning the edges to this vertex, we use the convention of 18.13.5 and 18.13.6. This new graph is called the extended Dynkin diagram of R. If α = n1 β1 + · · · + nl βl ∈ R, we set: |α| = n1 + · · · + nl . We call |α| the height of α, and we denote it also by ht(α). We ﬁx an order on the roots: let α, β ∈ R+ , then α ≺ β if either |α| < |β| or, |α| = |β| and α is smaller than β in the lexicographic order associated to (βl , βl−1 , . . . , β1 ). For the graphs of type E6 , E7 , E8 , F4 , G2 , we give an explicit list of positive roots in increasing order.

References • [13], [29], [39], [41], [80].

264

18 Root systems

Root system of type Al (l 1) • V is the hyperplane of Rl+1 consisting of points such that the sum of its coordinates is zero. • Roots: εi − εj (1 i, j l + 1, i = j). • Number of roots: n = l(l + 1). • Base: β1 = ε1 − ε2 , β2 = ε2 − ε3 , . . . , βl = εl − εl+1 . • Highest root: α = β1 + · · · + βl . • Cartan matrix: ⎞ ⎛ 2 −1 0 0 · · · 0 0 ⎜ −1 2 −1 0 · · · 0 0 ⎟ ⎟ ⎜ ⎜ 0 −1 2 −1 · · · 0 0 ⎟ ⎟ ⎜ ⎜ 0 0 −1 2 · · · 0 0 ⎟ ⎟ ⎜ ⎜ ··· ··· ··· ··· ··· ··· ···⎟ ⎟ ⎜ ⎝ 0 0 0 0 · · · 2 −1 ⎠ 0 0 0 0 · · · −1 2 • Extended Dynkin diagram (l 2):

1

2

l-1

l

• Order of W (R): (l + 1)!. • Positive roots: εi − εj = βi + · · · + βj−1

(1 i < j l + 1).

• Fundamental weights: i

i (ε1 + · · · + εl+1 ) (1 i l) = (ε1 + · · · + εi ) − l+1 1 [(l − i + 1)β1 + 2(l − i + 1)β2 + · · · + (i − 1)(l − i + 1)βi−1 = l+1 +i(l − i + 1)βi + i(l − i)βi+1 + · · · + iβl ].

18.14 Classiﬁcation of root systems

Root system of type Bl (l 2) • • • • • •

V = Rl . Roots: ±εi (1 i l), ±εi ± εj (1 i < j l). Number of roots: n = 2l2 . Base: β1 = ε1 − ε2 , β2 = ε2 − ε3 , . . . , βl−1 = εl−1 − εl , βl = εl . Highest root: α = β1 + 2β2 + 2β3 + · · · + 2βl . Cartan matrix: ⎞ ⎛ 2 −1 0 0 · · · 0 0 ⎜ −1 2 −1 0 · · · 0 0 ⎟ ⎟ ⎜ ⎜ 0 −1 2 −1 · · · 0 0 ⎟ ⎟ ⎜ ⎜ 0 0 −1 2 · · · 0 0 ⎟ ⎟ ⎜ ⎜ ··· ··· ··· ··· ··· ··· ···⎟ ⎟ ⎜ ⎝ 0 0 0 0 · · · 2 −2 ⎠ 0 0 0 0 · · · −1 2

• Extended Dynkin diagram: l = 2: 2

1

l3: 1 2

3

l-1

l

• Order of W (R): 2l .l!. • Positive roots: ⎧ (1 i l) ⎨ εi = βi + · · · + βl (1 i < j l) εi − εj = βi + · · · + βj−1 ⎩ εi + εj = βi + · · · + βj−1 + 2(βj + · · · + βl ) (1 i < j l). • Fundamental weights: i

l

= ε1 + · · · + ε i (1 i < l) = β1 + 2β2 + · · · + (i − 1)βi−1 + i[βi + · · · + βl ] 1 1 = (ε1 + · · · + εl ) = (β1 + 2β2 + · · · + lβl ). 2 2

265

266

18 Root systems

Root system of type Cl (l 3) • • • • • •

V = Rl . Roots: ±2εi (1 i l), ±εi ± εj (1 i < j l). Number of roots: n = 2l2 . Base: β1 = ε1 − ε2 , β2 = ε2 − ε3 , . . . , βl−1 = εl−1 − εl , βl = 2εl . Highest root: α = 2β1 + 2β2 + · · · + 2βl−1 + βl . Cartan matrix: ⎞ ⎛ 2 −1 0 0 . . . 0 0 ⎜ −1 2 −1 0 . . . 0 0 ⎟ ⎟ ⎜ ⎜ 0 −1 2 −1 . . . 0 0 ⎟ ⎟ ⎜ ⎜ 0 0 −1 2 . . . 0 0 ⎟ ⎟ ⎜ ⎜ ··· ··· ··· ··· ··· ··· ···⎟ ⎟ ⎜ ⎝ 0 0 0 0 . . . 2 −1 ⎠ 0 0 0 0 . . . −2 2

• Extended Dynkin diagram: 1

2

l-2

l-1

l

• Order of W (R): 2l .l!. • Positive roots: ⎧ (1 i l) ⎨ 2εi = 2(βi + · · · + βl−1 ) + βl (1 i < j l) εi − εj = βi + · · · + βj−1 ⎩ εi + εj = βi + · · · + βj−1 + 2(βj + · · · + βl−1 ) + βl (1 i < j l). • Fundamental weights: i

= ε1 + · · · + ε i

(1 i l)

1 = β1 + 2β2 + (i − 1)βi−1 + i[βi + βi+1 + · · · + βl−1 + βl ]. 2

18.14 Classiﬁcation of root systems

267

Root system of type Dl (l 4) • • • • • •

V = Rl . Roots: ±εi ± εj (1 i < j l). Number of roots: n = 2l(l − 1). Base: β1 = ε1 − ε2 , β2 = ε2 − ε3 , . . . , βl−1 = εl−1 − εl , βl = εl−1 + εl . Highest root: α = β1 + 2β2 + · · · + 2βl−2 + βl−1 + βl . Cartan matrix: ⎞ ⎛ 2 −1 . . . 0 0 0 0 ⎜ −1 2 . . . 0 0 0 0 ⎟ ⎟ ⎜ ⎜ ··· ··· ··· ··· ··· ··· ···⎟ ⎟ ⎜ ⎜ 0 0 . . . 2 −1 0 0 ⎟ ⎟ ⎜ ⎜ 0 0 . . . −1 2 −1 −1 ⎟ ⎟ ⎜ ⎝ 0 0 . . . 0 −1 2 0 ⎠ 0 0 . . . 0 −1 0 2

• Extended Dynkin diagram (l 4): 1

l- 1 2

3

l-3

l-2

l

• Order of W (R): 2l−1 .l!. • Positive roots:

⎧ (1 i < j l) ⎨ εi − εj = βi + · · · + βj−1 (1 i < l) εi + εl = βi + · · · + βl−2 + βl ⎩ εi + εj = βi + · · · + βj−1 + 2(βj + · · · + βl−2 ) + βl−1 + βl (1 i < j < l). • Fundamental weights: i

l−1

l

= ε1 + · · · + ε i (1 i l − 2) = β1 + 2β2 + · · · + (i − 1)βi−1 1 +i[βi + βi+1 + · · · + βl−2 ] + i[βl−1 + βl ] 2 1 = (ε1 + ε2 + · · · + εl−2 + εl−1 − εl ) 2 1 1 1 = [β1 + 2β2 + · · · + (l − 2)βl−2 ] + lβl−1 + (l − 2)βl 2 4 4 1 = (ε1 + ε2 + · · · + εl−2 + εl−1 + εl ) 2 1 1 1 = [β1 + 2β2 + · · · + (l − 2)βl−2 ] + (l − 2)βl−1 + lβl . 2 4 4

268

18 Root systems

Root system of type E6 • V is the subspace of R8 consisting of points whose coordinates (ξi ) verify ξ6 = ξ7 = −ξ8 . • Roots: 5 1 ±εi ± εj (1 i < j 5), ± (−1)ν(i) εi , ε8 − ε7 − ε6 + 2 i=1 where ν(i) are positive integers such that ν(1) + · · · + ν(5) is even. • Number of roots: n = 72. • Base: 1 (ε1 − ε2 − ε3 − ε4 − ε5 − ε6 − ε7 + ε8 ), 2 β2 = ε1 + ε2 , β3 = ε2 − ε1 , β4 = ε3 − ε2 , β5 = ε4 − ε3 , β6 = ε5 − ε4 . β1 =

• Highest root: α = β1 + 2β2 + 2β3 + 3β4 + 2β5 + β6 . • Cartan matrix: ⎞ ⎛ 2 0 −1 0 0 0 ⎜ 0 2 0 −1 0 0 ⎟ ⎟ ⎜ ⎜ −1 0 2 −1 0 0 ⎟ ⎟ ⎜ ⎜ 0 −1 −1 2 −1 0 ⎟ ⎟ ⎜ ⎝ 0 0 0 −1 2 −1 ⎠ 0 0 0 0 −1 2 • Extended Dynkin diagram: 1

3

4

5

6

2

• Order of W (R): 27 .34 .5. • Positive roots: ⎧ j 5) ⎨ ±εi + εj (1 i < 5 5 1 (−1)ν(i) εi with ν(i) even. ε8 − ε7 − ε6 + ⎩ 2 i=1 i=1 Let us represent the root aβ1 + bβ2 + cβ3 + dβ4 + eβ5 + f β6 by the positive roots are:

a c d e f b

. Then

18.14 Classiﬁcation of root systems 1 0 0 0 0 1 1 0 1 1 1 1 0 1 1 1

0 0 0 0 0 0 1 1

0 0 0 1 0 0 1 0 1 1 1 0 1 1 2 1

0 0 1 0 1 0 1 0

0 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 2 1

0 0 1 1 1 0 1 1 2 1

0 0 1 0 1 1 1 0 0 0 1 1 0 1 2 1 1 2 2 1

0 0 0 0 1 1 1 1 2 1

0 0 0 0 0 1 1 1 0 1 1 0 1 2 2 1 1 2 3 1

1 0 0 0 1 1 1 0 2 1

0 0 0 0 0 0 1 1 1 1 1 1 1 1 2 1 1 2 3 2

0 1 1 0 1 0 1 1

1 1 0 0 0 1 1 0 0 1 2 1 0 1 2 1

0 0 1 0 1 0 2 1

2 1

• Fundamental weights: 1

2

3

4

2 (ε8 − ε7 − ε6 ) 3 1 = [4β1 + 3β2 + 5β3 + 6β4 + 4β5 + 2β6 ] 3 1 = (ε1 + ε2 + ε3 + ε4 + ε5 − ε6 − ε7 + ε8 ) 2 = β1 + 2β2 + 2β3 + 3β4 + 2β5 + β6 =

5 1 (ε8 − ε7 − ε6 ) + (−ε1 + ε2 + ε3 + ε4 + ε5 ) 6 2 1 = [5β1 + 6β2 + 10β3 + 12β4 + 8β5 + 4β6 ] 3 = ε3 + ε4 + ε5 − ε6 − ε7 + ε8 =

= 2β1 + 3β2 + 4β3 + 6β4 + 4β5 + 2β6 5

6

2 (ε8 − ε7 − ε6 ) + ε4 + ε5 3 1 = [4β1 + 6β2 + 8β3 + 12β4 + 10β5 + 5β6 ] 3 1 = (ε8 − ε7 − ε6 ) + ε5 3 1 = [2β1 + 3β2 + 4β3 + 6β4 + 5β5 + 4β6 ]. 3 =

0 0 1 1 0 0 1 0 1 1 1 0 1 2 2 1

269 0 0 1 1 1 1 1 1

270

18 Root systems

Root system of type E7 • V is the hyperplane of R8 orthogonal to ε7 + ε8 . • Roots: 6 1 ±(ε7 − ε8 ), ±εi ± εj (1 i < j 6), ± (−1)ν(i) εi , ε7 − ε8 + 2 i=1 where ν(i) are positive integers such that ν(1) + · · · + ν(6) is odd. • Number of roots: n = 126. • Base: 1 (ε1 − ε2 − ε3 − ε4 − ε5 − ε6 − ε7 + ε8 ), β2 = ε1 + ε2 , 2 β3 = ε2 − ε1 , β4 = ε3 − ε2 , β5 = ε4 − ε3 , β6 = ε5 − ε4 , β7 = ε6 − ε5 . β1 =

• Highest root: α = 2β1 + 2β2 + 3β3 + 4β4 + 3β5 + 2β6 + β7 . • Cartan matrix: ⎞ ⎛ 2 0 −1 0 0 0 0 ⎜ 0 2 0 −1 0 0 0 ⎟ ⎟ ⎜ ⎜ −1 0 2 −1 0 0 0 ⎟ ⎟ ⎜ ⎜ 0 −1 −1 2 −1 0 0 ⎟ ⎟ ⎜ ⎜ 0 0 0 −1 2 −1 0 ⎟ ⎟ ⎜ ⎝ 0 0 0 0 −1 2 −1 ⎠ 0 0 0 0 0 −1 2 • Extended Dynkin diagram: 1

3

4

5

6

7

2

• Order of W (R): 210 .34 .5.7. • Positive roots: ⎧ ε 8 − ε7 ⎪ ⎪ ⎨ ±ε + ε (1 i < j 6) i j 6 6 1 ⎪ ⎪ (−1)ν(i) εi with ν(i) odd. ⎩ ε8 − ε7 + 2 i=1 i=1 Representing the root aβ1 + bβ2 + cβ3 + dβ4 + eβ5 + f β6 + gβ7 by the positive roots are:

a c d e f g , b

18.14 Classiﬁcation of root systems 1 0 0 0 1 1 0 0 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 1 2 1 1 2 3 1 1 2 3 1

0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 1 1 1 2 1 0 2 1 0 2 2 1

0 0 0 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 2 2 1 1 1 2 1 1 2 2 1 1 2 3 2

0 0 0 0 0 0 1 0 0 1 1 0 1 1 1 1 0 0 1 1 1 2 1 1 2 2 1

0 1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 1 2 1 0 1 2 1 1 1 2 1 1 2 3 1

0 0 0 0 0 0 1 0 0 1 1 0 1 1 1 1 1 0 2 1 1 2 2 1 3 2 1

0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 1 1 2 1 0 1 2 1 1 2 2 1 1 2 3 2 1 2 3 2

0 0 0 1 0 0 1 1 0 1 1 1 1 0 0 2 1 0 2 1 0 2 1 0 3 2 1

0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 1 2 3 1 1 2 4 2

1 0 0 1 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 1 1 2 1 1 3 2 1

0 0 0 0 0 0 0 0 1 1 1 1 0 1 2 1 0 1 2 1 0 1 2 1 1 1 2 1 1 2 2 1 1 3 4 2

0 1 0 0 1 1 0 0 0 1 0 0 1 1 0 1 1 1 2 1 1 2 2 1 3 2 1

• Fundamental weights: 1

= ε8 − ε 7 = 2β1 + 2β2 + 3β3 + 4β4 + 3β5 + 2β6 + β7

2

=

3

4

5

6

7

1 (ε1 + ε2 + ε3 + ε4 + ε5 + ε6 − 2ε7 + 2ε8 ) 2 1 = [4β1 + 7β2 + 8β3 + 12β4 + 9β5 + 6β6 + 3β7 ] 2 1 = (−ε1 + ε2 + ε3 + ε4 + ε5 + ε6 − 3ε7 + 3ε8 ) 2 = 3β1 + 4β2 + 6β3 + 8β4 + 6β5 + 4β6 + 2β7 = ε3 + ε4 + ε5 + ε6 − 2ε7 + 2ε8 = 4β1 + 6β2 + 8β3 + 12β4 + 9β5 + 6β6 + 3β7 1 (2ε4 + 2ε5 + 2ε6 − 3ε7 + 3ε8 ) 2 1 = [6β1 + 9β2 + 12β3 + 18β4 + 15β5 + 10β6 + 5β7 ] 2 = ε5 + ε 6 − ε 7 + ε 8 = 2β1 + 3β2 + 4β3 + 6β4 + 5β5 + 4β6 + 2β7 =

1 (2ε6 − ε7 + ε8 ) 2 1 = [2β1 + 3β2 + 4β3 + 6β4 + 5β5 + 4β6 + 3β7 ]. 2

=

0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 2 2 1 0 1 2 1 1 2 3 2 2 3 4 2

271 0 0 1 0 0 0 1 0 0 1 1 0 1 1 1 1 1 0 2 2 1 2 1 1 3 2 1

272

18 Root systems

Root system of type E8 • V = R8 . • Roots: ±εi ± εj

(1 i < j),

8 1 (−1)ν(i) εi , 2 i=1

where ν(i) are positive integers such that ν(1) + · · · + ν(8) is even. • Number of roots: n = 240. • Base : 1 (ε1 − ε2 − ε3 − ε4 − ε5 − ε6 − ε7 + ε8 ), β2 = ε1 + ε2 , β3 = ε2 − ε1 , 2 β4 = ε3 − ε2 , β5 = ε4 − ε3 , β6 = ε5 − ε4 , β7 = ε6 − ε5 , β8 = ε7 − ε6 .

β1 =

• Highest root: α = 2β1 + 3β2 + 4β3 + 6β4 + 5β5 + 4β6 + 3β7 + 2β8 . • Cartan matrix: ⎞ ⎛ 2 0 −1 0 0 0 0 0 ⎜ 0 2 0 −1 0 0 0 0 ⎟ ⎟ ⎜ ⎜ −1 0 2 −1 0 0 0 0 ⎟ ⎟ ⎜ ⎜ 0 −1 −1 2 −1 0 0 0 ⎟ ⎟ ⎜ ⎜ 0 0 0 −1 2 −1 0 0 ⎟ ⎟ ⎜ ⎜ 0 0 0 0 −1 2 −1 0 ⎟ ⎟ ⎜ ⎝ 0 0 0 0 0 −1 2 −1 ⎠ 0 0 0 0 0 0 −1 2 • Extended Dynkin diagram: 1

3

5

4

6

7

8

2

• Order of W (R): 214 .35 .52 .7. • Positive roots: ⎧ j 8) ⎨ ±εi + εj (1 i < 7 7 1 ν(i) (−1) εi with ν(i) even. ε8 + ⎩ 2 i=1 i=1 Representing the root aβ1 + bβ2 + cβ3 + dβ4 + eβ5 + f β6 + gβ7 + hβ8 by a c d e f g h , the positives roots are: b 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1

18.14 Classiﬁcation of root systems 0 1 1 0 0 1 1 1 0 1 2 1 1 1 2 1 0 1 1 0 1 1 1 0 1 1 1 1 1 1 2 1 1 1 2 1 1 1 2 1 1 1 2 1 1 2 3 2 1 2 3 2 2 3 4 2 1 2 4 2 2 3 4 2 2 4 5 3

1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 1 0 0 0 1 1 1 0 1 0 0 0 1 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 0 1 1 1 1 1 1 1 0 1 2 1 1 1 1 1 0 1 2 1 2 1 1 1 0 1 2 1 2 2 1 1 0 1 2 1 2 2 2 1 1 2 3 2 3 2 1 0 1 2 3 2 2 2 2 1 1 2 3 1 3 2 1 0 1 3 4 2 3 3 2 1 2 3 4 2 4 3 2 1 1 3 5 2 4 3 2 1 2 4 6 3

1 1 0 0 0 0 0 0 1 1 0 0 0 1 1 0 1 1 0 0 0 1 1 1 1 1 0 0 0 1 2 1 1 0 0 0 1 1 2 1 1 1 1 1 1 2 2 1 1 1 1 1 1 2 2 1 2 1 1 1 1 2 3 1 2 2 1 1 1 2 3 2 2 2 2 1 1 2 3 2 2 2 1 0 1 2 3 1 2 2 1 1 1 2 3 1 3 2 2 1 1 3 4 2 3 2 1 1 1 2 4 2 3 2 2 1 1 3 4 2 4 3 2 1 2 3 5 2 4 3 2 1 2 4 6 3

1 1 1 0 0 0 0 0 1 1 0 0 0 0 1 0 1 1 0 0 0 0 1 1 1 1 0 0 1 1 1 0 1 1 0 0 0 1 2 1 1 1 0 0 1 1 2 1 2 1 0 0 1 2 2 1 2 1 0 0 1 2 2 1 2 1 0 0 1 2 3 1 2 1 1 0 1 2 3 1 3 2 1 0 1 2 3 2 3 2 1 1 1 2 3 1 3 2 1 0 1 2 4 2 3 2 2 1 1 2 3 2 3 3 2 1 1 2 4 2 4 3 2 1 1 3 5 3 5 3 2 1 2 4 6 3

0 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 0 0 1 1 0 1 1 1 0 0 1 1 1 2 1 0 0 1 1 1 1 2 1 0 0 1 1 2 1 1 1 1 0 1 1 2 1 2 1 1 0 1 1 2 1 2 1 1 0 1 2 2 1 2 2 1 0 1 2 3 1 2 1 1 1 1 2 3 1 2 2 2 1 1 2 4 2 3 2 1 1 1 2 3 2 3 3 2 1 2 3 4 2 4 3 2 1 2 3 4 2 4 3 2 1 2 3 5 3 5 4 2 1 2 4 6 3

0 0 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 0 1 1 1 0 0 0 1 1 1 1 1 0 0 1 2 1 1 1 1 0 0 1 2 1 2 1 1 0 0 1 2 1 2 2 1 0 1 2 2 1 2 2 1 0 1 2 2 1 2 1 1 1 1 2 2 1 2 2 1 1 1 2 2 1 3 2 1 0 1 2 3 2 3 2 2 1 1 2 3 1 3 2 1 1 1 3 4 2 3 3 2 1 1 3 4 2 4 3 2 1 2 4 5 2 5 4 3 1 2 4 6 3

1 0 0 0 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0 2 1 1 0 2 2 1 0 1 1 1 1 2 1 1 1 2 2 1 1 2 2 2 1 3 2 1 1 3 3 2 1 3 2 2 1 4 3 2 1 4 3 2 1 5 4 3 2

• Fundamental weights: 1

2

3

4

5

6

7

8

= 2ε8 = 4β1 + 5β2 + 7β3 + 10β4 + 8β5 + 6β6 + 4β7 + 2β8 1 = (ε1 + ε2 + ε3 + ε4 + ε5 + ε6 + ε7 + 5ε8 ) 2 = 5β1 + 8β2 + 10β3 + 15β4 + 12β5 + 9β6 + 6β7 + 3β8 1 = (−ε1 + ε2 + ε3 + ε4 + ε5 + ε6 + ε7 + 7ε8 ) 2 = 7β1 + 10β2 + 14β3 + 20β4 + 16β5 + 12β6 + 8β7 + 4β8 = ε3 + ε4 + ε5 + ε6 + ε7 + 5ε8 = 10β1 + 15β2 + 20β3 + 30β4 + 24β5 + 18β6 + 12β7 + 6β8 = ε4 + ε5 + ε6 + ε7 + 4ε8 = 8β1 + 12β2 + 16β3 + 24β4 + 20β5 + 15β6 + 10β7 + 5β8 = ε5 + ε6 + ε7 + 3ε8 = 6β1 + 9β2 + 12β3 + 18β4 + 15β5 + 12β6 + 8β7 + 4β8 = ε6 + ε7 + 2ε8 = 4β1 + 6β2 + 8β3 + 12β4 + 10β5 + 8β6 + 6β7 + 3β8 = ε7 + ε 8 = 2β1 + 3β2 + 4β3 + 6β4 + 5β5 + 4β6 + 3β7 + 2β8 .

273

274

18 Root systems

Root system of type F4 • V = R4 . • Roots: ±εi

(1 i 4), ±εi ± εj

(1 i < j 4),

1 (±ε1 ± ε2 ± ε3 ± ε4 ). 2

• Number of roots: n = 48. • Base: β1 = ε2 − ε3 , β2 = ε3 − ε4 , β3 = ε4 , β4 =

1 (ε1 − ε2 − ε3 − ε4 ). 2

• Highest root: α = 2β1 + 3β2 + 4β3 + 2β4 . • Cartan matrix: ⎛ ⎞ 2 −1 0 0 ⎜ −1 2 −2 0 ⎟ ⎜ ⎟ ⎝ 0 −1 2 −1 ⎠ 0 0 −1 2 • Extended Dynkin diagram: 1

3

2

4

• Order of W (R) : 27 .32 . • Positive roots: ⎧ (1 i 4) ⎪ ⎨ εi (1 i < j 4) εi ± ε j 1 ⎪ ⎩ (ε1 ± ε2 ± ε3 ± ε4 ). 2 Denote by

a b c d

the root aβ1 + bβ2 + cβ3 + dβ4 , then the positive roots are:

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

1 1 0 0

0 1 1 0

0 0 1 1

1 1 1 0

0 1 2 0

0 1 1 1

1 1 2 0

1 1 1 1

0 1 2 1

1 2 2 0

1 1 2 1

0 1 2 2

1 2 2 1

1 1 2 2

1 2 3 1

1 2 2 2

1 2 3 2

1 2 4 2

1 3 4 2

2 3 4 2

• Fundamental weights: 1 2 3 4

= ε1 + ε2 = 2β1 + 3β2 + 4β3 + 2β4 = 2ε1 + ε2 + ε3 = 3β1 + 6β2 + 8β3 + 4β4 1 = (3ε1 + ε2 + ε3 + ε3 ) = 2β1 + 4β2 + 6β3 + 3β4 2 = ε1 = β1 + 2β2 + 3β3 + 2β4 .

18.14 Classiﬁcation of root systems

Root system of type G2 • V is the hyperplane of R3 deﬁned by ξ1 + ξ2 + ξ3 = 0. • Roots: ±(ε1 − ε2 ), ±(ε1 − ε3 ), ±(ε2 − ε3 ), ±(2ε1 − ε2 − ε3 ), ±(2ε2 − ε1 − ε3 ), ±(2ε3 − ε1 − ε2 ). • • • •

Number of roots: n = 12. Base: β1 = ε1 − ε2 , β2 = −2ε1 + ε2 + ε3 . Highest root: α = 3β1 + 2β2 . Cartan matrix: 2 −1 −3 2

• Extended Dynkin diagram: 1

2

• Order of W (R): 12. • Positive roots: β1 , β2 , β1 + β2 , 2β1 + β2 , 3β1 + β2 , 3β1 + 2β2 . • Fundamental weights: 1 2

= −ε2 + ε3 = 2β1 + β2 = −ε1 − ε2 + 2ε3 = 3β1 + 2β2 .

275

19 Lie algebras

The aim of this chapter is to give a detailed account of the general theory of Lie algebras. The representation theory of sl2 (k) in section 19.2 will be used very often in the rest of the book. Results such as Engel’s theorem, Lie’s theorem and Cartan’s Criterion are the backbone of the subject. The notions of regular linear forms and Cartan subalgebras are introduced in the ﬁnal sections.

19.1 Generalities on Lie algebras 19.1.1 Let g be a k-vector space. A bilinear map g×g → g, (x, y) → [x, y], is a Lie bracket on g if for all x, y, z ∈ g, we have • [x, y] = −[y, x] (antisymmetry), • [x, [y, z]] = [[x, y], z] + [y, [x, z]] (Jacobi identity). Endowed with a Lie bracket, g is called a Lie algebra. Let g be a Lie algebra. A subspace h ⊂ g is a Lie subalgebra (resp. an ideal) if [x, y] ∈ h for all x, y ∈ h. (resp. [x, h] ⊂ h for all x ∈ g). When h is an ideal of g, the quotient g/h inherits a natural structure of Lie algebra given by: [x + h, y + h] = [x, y] + h. Examples. Let us give some examples of Lie algebras. Let A be an associative k-algebra. Then, A has a Lie algebra structure given by: for x, y ∈ A, [x, y] = xy − yx. So if V is a k-vector space of dimension n, then End(V ) (resp. Mn (k)) endowed with the above Lie bracket is a Lie algebra that we shall denote by gl(V ) (resp. gln (k)). The following subsets of gln (k) are clearly Lie subalgebras: • sln (k) (resp. sl(V )): the set of matrices in gln (k) (resp. endomorphisms in gl(V )) whose trace is zero. • tn (k) (resp. nn (k)): the set of upper triangular matrices (resp. strictly upper triangular matrices) in gln (k). • dn (k) the set of diagonal matrices in gln (k).

278

19 Lie algebras

Observe that sl(V ) is an ideal of gl(V ). Let h, h be subsets of g, then we denote by [h, h ] the subspace spanned by the elements [x, y] where x ∈ h, y ∈ h . When h and h are ideals of g, then clearly [h, h ] is also an ideal of g. Let g, g be Lie algebras endowed with Lie brackets [, ] and [, ] , then a map u : g → g is a morphism of Lie algebras if u([x, y]) = [u(x), u(y)] for all x, y ∈ g. So we may talk about isomorphisms and automorphisms. A linear map d : g → g is called a derivation of g if for all x, y ∈ g, d([x, y]) = [d(x), y] + [x, d(y)]. We denote by Der g the set of derivations of g. Note that Der g, endowed with the following Lie bracket, [d, d ] = d ◦ d − d ◦ d, for d, d ∈ Der g, is a Lie algebra. Let x ∈ g. The Jacobi identity implies that the linear map y → [x, y], denoted adg x or ad x, is a derivation of g, called an inner derivation of g. So it induces a linear map g → Der(g) which is called the adjoint representation of g. Convention. In this chapter, we shall only consider ﬁnite-dimensional Lie algebras and g will denote a ﬁnite-dimensional Lie algebra. Set: ker(ad x)n , g• (x) = (ad x)n (g). g0 (x) = n0

n0

0

We call g (x) the nilspace of ad x. We have: [g0 (x), g0 (x)] ⊂ g0 (x) , [g0 (x), g• (x)] ⊂ g• (x) , g = g0 (x) ⊕ g• (x). Let n = dim g and χx the characteristic polynomial of ad x. Then χx (T ) = an (x)T n + an−1 (x)T n−1 + · · · + a1 (x)T + a0 (x) where ai are polynomial functions on g with an = 1 and a0 = 0 (since [x, x] = 0). If p is the smallest integer such that ap (x) = 0, then by decomposing g into characteristic subspaces of ad x, we obtain that dim g0 (x) = p. Deﬁne rk(g) = min{l; al = 0} to be the rank of the Lie algebra g. If x ∈ g veriﬁes dim g0 (x) = rk(g), then we say that x is generic in g. Denote by ggen the set of generic elements of g. It is clear from its deﬁnition that ggen is a dense open subset of g. 19.1.2 Lemma. If h is a Lie subalgebra of g, then h ∩ ggen ⊂ hgen . Proof. For x ∈ h, denote by u(x) the endomorphism of g/h induced by adg x. Let χu(x) be the characteristic polynomial of u(x) and χx,h the characteristic polynomial of adh x. Then χx = χu(x) χx,h and the result follows easily.

19.2 Representations

279

19.1.3 Let p, q be subsets of g. The set of elements of q which commute with all the elements of p, denoted by cq (p), is called the centralizer of p in q. The set cg (g), denoted by z(g), is called the centre of g. If g = z(g), then we say that g is abelian (or commutative). If x ∈ g, we shall write qx for cq ({x}). The set of x ∈ q such that [x, p] ⊂ p, denoted by nq (p), is called the normalizer of p in q. If p is a Lie subalgebra, then so is ng (p), and p is an ideal of ng (p). 19.1.4 Let us denote by Aut g the set of automorphisms of g, N the set of elements x ∈ g such that ad x is a nilpotent endomorphism of g. For x ∈ N , deﬁne γx ∈ GL(g) as follows: γx =

∞ 1 (ad x)n . n=0 n!

Then γx ∈ Aut g. Such an automorphism is called an elementary automorphism of g. The subgroup Aute g of Aut g generated by elementary automorphisms is called the group of elementary automorphisms of g. Observe that if x ∈ N , α ∈ Aut g, then α(x) ∈ N and ad α(x) = α ◦ (ad x) ◦ α−1 . So γα(x) = α ◦ γx ◦ α−1 , and Aute g is a normal subgroup of Aut g.

19.2 Representations 19.2.1 A representation of g, or a g-module, is a pair (V, σ), where V is k-vector space and σ : g → gl(V ) is a Lie algebra homomorphism. If V is ﬁnite-dimensional, then we say that the representation is ﬁnite-dimensional. For x ∈ g, v ∈ V , we shall denote σ(x)(v) by x.v. We set V g = {v ∈ V ; x.v = 0 for all x ∈ g}. We deﬁne the notions of submodule, quotient module and direct sum in the obvious way. The g-module (V, σ) is said to be simple if V = {0} and the only submodules of V are V and {0}. If V is the direct sum of simple submodules, then we say that V is semisimple, or completely reducible. The analogues of 10.7.2 and 10.7.5 are true for g-modules. Let V be a g-module. A chain 0 = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V of submodules of V is called a Jordan-H¨ older series if the modules Vi /Vi−1 are simple for 1 i n. Let (V, σ) be a representation of g, then we deﬁne its dual, or contragredient, representation (V ∗ , π) as follows: for x ∈ g, v ∈ V and f ∈ V ∗ , (π(x)(f ))(v) = −f (σ(x)(v)). The adjoint representation is indeed a representation of g. Its dual representation is called the coadjoint representation of g. If h is a Lie subalgebra of g, then the map h → gl(g/h) induced by x → ad x is called the adjoint representation of h in g/h. A representation (V, σ) is said to be faithful if σ is injective. We shall admit the following result.

280

19 Lie algebras

Theorem. (Ado’s Theorem) Any ﬁnite-dimensional Lie algebra g has a faithful ﬁnite-dimensional representation. 19.2.2 Let (V, σ) be a ﬁnite-dimensional representation of g. We deﬁne a symmetric bilinear form βσ on g associated to σ as follows: for x, y ∈ g, βσ (x, y) = tr(σ(x)σ(y)). The symmetric bilinear form associated to the adjoint representation is called the Killing form of g, denoted by Lg . If a is an ideal of g, then the restriction of Lg to a × a is the Killing form of a, and the orthogonal of a with respect to βσ is an ideal of g. Observe that if x, y, z ∈ g, then βσ ([x, y], z) + βσ (y, [x, z]) = 0. If x ∈ g, denote by fx the linear form on g such that fx (y) = Lg (x, y) for all y ∈ g. It follows that the map x → fx is a homomorphism of g-modules, called the Killing homomorphism of g, from the adjoint representation of g to the coadjoint representation of g. 19.2.3 Lemma. Let V be a ﬁnite-dimensional vector space and f an endomorphism of V . Then f is nilpotent if and only if tr(f n ) = 0 for all n ∈ N∗ . Proof. This is an elementary result from linear algebra. 19.2.4 In this rest of this section, g is the Lie algebra sl2 (k). Let 01 1 0 00 e= , h= , f= . 00 0 −1 10 Then (e, h, f ) is a basis of g and [e, f ] = h , [h, e] = 2e , [h, f ] = −2f. Let r ∈ N and for 1 i r, let µi = i(r − i + 1) and denote by (kr+1 , σr ) the representation of g given by: ⎞ ⎛ r 0 0 ... 0 ⎜0 r − 2 0 ... 0 ⎟ ⎟ ⎜ ⎟ ⎜ σr (h) = ⎜ 0 0 r − 4 . . . 0 ⎟ , ⎜ .. .. .. . . .. ⎟ ⎝. . . . ⎠ . 0 ⎛

0 µ1 ⎜0 0 ⎜ ⎜ σr (e) = ⎜ ... ... ⎜ ⎝0 0 0 0

0 ... µ2 . . . .. . . . . 0 ... 0 ...

0

0

. . . −r

⎞

⎛

00 ⎟ ⎜1 0 ⎟ ⎜ ⎟ ⎜. . ⎟ , σr (f ) = ⎜ .. .. ⎟ ⎜ ⎝0 0 µr ⎠ 00 0 0 0 .. .

Lemma. The representation σr is simple.

⎞ ... 0 0 0 ... 0 0 0⎟ ⎟ . . .. .. .. ⎟ . . . . .⎟ ⎟ ... 1 0 0⎠ ... 0 1 0

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281

Proof. Let (ε1 , . . . , εr+1 ) be the canonical basis of kr+1 . Then for any non zero vector v ∈ kr+1 , there exist n ∈ N and λ ∈ k \ {0} such that σr (f )n (v) = λεr+1 . So any non-trivial submodule contains εr+1 , and the result follows since kεi = kσr (e)r−i+1 (εr+1 ). 19.2.5 Let (V, σ) be a ﬁnite-dimensional representation of g where dim V = 0. Set H = σ(h), E = σ(e) and F = σ(f ). By a simple induction, we have the following identities for n ∈ N, m ∈ N∗ : [H, E] = 2nE n , [H, F n ] = −2nF n , (1) [F, E m ] = −m(H − (m − 1) idV ) ◦ E m−1 Thus E and F are nilpotent (19.2.3). Let q = min{n ∈ N; E q+1 = 0} and w ∈ V such that v = E q (w) = 0. Set v0 = v and for n ∈ N∗ , vn = F n (v). Then by using (1), we obtain: 0 = [F, E q+1 ](w) ⇒ H(v0 ) = qv0 . Hence, again by (1), we deduce that: (2)

H(vn ) = (q − 2n)vn . Further, a simple induction shows that:

(3)

E(vn ) = n(q − n + 1)vn−1 .

It follows that if vn = 0, it is an eigenvector for H with eigenvalue (q −2n). Let s be such that vs = 0 and vn = 0 for n > s, and let W be the subspace spanned by v0 , . . . , vs . Then W is a g-submodule of V and (v0 , . . . , vs ) is a basis of W by (2). Denote by HW , EW , FW the restrictions of H, E, F to W . Since [EW , FW ] = HW , 0 = tr(HW ) =

s

(q − 2n) = (s + 1)(q − s).

n=0

Hence s = q and the matrices of HW , EW , FW in the basis (v0 , . . . , vq ) are the ones in 19.2.4 with r = q. We deduce that the representations σ|W and σq are equivalent. So we have obtained the following result: Theorem. Let r ∈ N and (V, σ) a simple representation of sl2 (k) of dimension r + 1. Then σ is equivalent to σr . Moreover, the eigenvalues of σ(h) are −r, −r + 2, . . . , r − 2, r, and if v ∈ V \ {0} veriﬁes σ(e)(v) = 0 (resp. σ(f )(v) = 0), then v is a σ(h)-eigenvector of eigenvalue r (resp. −r). 19.2.6 Theorem. Any ﬁnite-dimensional representation of g = sl2 (k) is semisimple.

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Proof. Let (V, σ) be a ﬁnite-dimensional representation of g. If dim V = 0, there is nothing to prove. So we may assume that dim V 1. By 19.2.5, there is a simple submodule W of dimension q + 1 in W . We shall prove that there is a submodules U of V such that V = V ⊕ W . Then the result would follow by induction on the dimension of V . Denote by η the dual representation of σ. Let w ∈ V and (v0 , . . . , vq ) be the basis of W as in 19.2.5. Let g ∈ V ∗ be such that g(v0 ) = 1. Then (η(E)q (g)) (w) = g((−E)q (w)) = (−1)q . So f = η(E)q (g) = 0 and q is the smallest integer such that η(E)q+1 = 0. By 19.2.5, the subspace M of V ∗ spanned by fn = η(F )n (f ), 0 n q, is a simple submodule of dimension q + 1. Using (3), we verify easily that: fn (vq−n ) = (−1)q−n (q!)2 , fn (vp ) = 0 , if n + p > q.

(4)

Let U be the orthogonal of M in V . Then dim V = dim U + dim W and U is a g-submodule of V . Further using (4) we deduce that U ∩ W = {0}. So V is the direct sum of the submodules U and W . 19.2.7 Corollary. Let (V, σ) be a ﬁnite-dimensional representation of sl2 (k). (i) There exist r1 , . . . , rn ∈ N such that σ is equivalent to the direct sum of the representations σri , 1 i n. (ii) We have V = σ(e)(V ) ⊕ ker σ(f ) = σ(f )(V ) ⊕ ker σ(e). (iii) σ(h) is semisimple with integer eigenvalues. (iv) If V is non-trivial, then σ is irreducible if and only if the eigenvalues of σ(h) are simple and they are either all even or all odd. Proof. Part (i) follows from 19.2.6 and 19.2.5. The rest is clear for σr . The general case follows from (i).

19.3 Nilpotent Lie algebras 19.3.1 An ideal of g is said to be characteristic if it is invariant under all the derivations of g. We deﬁne by induction the following decreasing chains of characteristic ideals of g: C 1 (g) = g , C 2 (g) = [g, g] , . . . , C i+1 (g) = [g, C i (g)] , . . . D (g) = g , D1 (g) = [g, g] , . . . , Di+1 (g) = [Di (g), Di (g)] , . . . 0

We call (C i (g))i1 the central descending series of g, and (Di (g))i0 the derived series of g. In particular, D(g) = D1 (g) is called the derived ideal of g. If u : g → h is a Lie algebra homomorphism, then u(C i (g)) ⊂ C i (h) and u(Di (g)) ⊂ Di (h) for all i. Further, these inclusions are equalities when u is surjective.

19.3 Nilpotent Lie algebras

283

19.3.2 Proposition. The following conditions are equivalent: (i) There exists an integer i such that C i (g) = {0}. (ii) There exists an integer j such that ad x1 ◦ ad x2 ◦ · · · ◦ ad xj = 0 for all x1 , . . . , xj ∈ g. (iii) There exists a chain g = g1 ⊃ g2 ⊃ · · · ⊃ gn = {0} of ideals of g such that [g, gi ] ⊂ gi+1 for i < n. If these conditions are veriﬁed, we say that g is nilpotent. Proof. The equivalence (i) ⇔ (ii) is straightforward by the deﬁnition of C i (g). Next, we have (i) ⇒ (iii) by taking gi = C i (g). Finally, to show that (iii) ⇒ (i), it suﬃces to show that C i (g) ⊂ gi , which is a simple induction. 19.3.3 Proposition. (i) If g is nilpotent, then any subalgebra (resp. quotient) of g is nilpotent. (ii) Let a be a subalgebra of g which is contained in the centre of g. Then g is nilpotent if and only if g/a is nilpotent. Proof. Part (i) is obvious. If g/a is nilpotent, then there exists k ∈ N such that C k (g/a) = {0}, or C k (g) ⊂ a. Since a is in the centre of g, C k+1 (g) = {0}. The converse follows from part (i). 19.3.4 Proposition. Let g be nilpotent. (i) If g = {0}, then z(g) = {0}. (ii) If h is a subalgebra of g distinct from g, then ng (h) = h. (iii) The Killing form of g is zero. Proof. (i) The last non-zero C i (g) is central in g. (ii) Let j = max{i; C i (g) + h = h}. Then C j (g) + h ⊂ ng (h). (iii) By 19.3.2, ad x ◦ ad y is a nilpotent endomorphism for all x, y ∈ g. 19.3.5 Lemma. Let V be a vector space of dimension n > 0. (i) If x ∈ End(V ) is nilpotent, ad x is a nilpotent endomorphism of gl(V ). More precisely, if xp = 0, then (ad x)2p−1 = 0. (ii) If g is a Lie subalgebra of gl(V ) consisting of nilpotent endomorphisms, then V g = {0}. Proof. Part (i) is clear from the fact that if y ∈ gl(V ), then (ad x)n (y) is the sum of terms of the form ±xi yxj with i + j = n. To prove (ii), we proceed by induction on dim g = p. The result is obvious when p 1, so let us assume that p > 1. Let a be a non-zero Lie subalgebra of g of dimension q < p. For x ∈ a, the endomorphism σ(x) induced by adg x on g/a is nilpotent by (i). So by induction, there exists y ∈ g \ a such that [y, a] ⊂ a. Consequently, a is an ideal of a ⊕ ky. By iterating this process, we see that g contains an ideal h of codimension 1. By the induction hypothesis, W = V h = {0}. Now for y ∈ g \ h, z ∈ h and w ∈ W , we have,

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z ◦ y(w) = y ◦ z(w) + [y, z](w) = 0, since [y, z] ∈ h. Thus y(W ) ⊂ W . Since y is nilpotent, y(v) = 0 for some v ∈ W \ {0}. Hence V g = {0}. 19.3.6 Theorem. (Engel’s Theorem) Let (V, σ) be a representation of g of dimension n > 0. Assume that σ(g) is a nilpotent endomorphism for all g ∈ g. Then there is a ﬂag {0} = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V of V such that σ(g)Vi ⊂ Vi−1 for 1 i n. Proof. We may assume that g ⊂ gl(V ). The result is clear when n = 1. So let us proceed by induction on n. By 19.3.5, there exists v ∈ V g which is nonzero. Let ϕ : V → V /kv = W be the canonical surjection. Then by applying the induction hypothesis to W , we have a ﬂag (Wi )0in−1 of W with the required properties. Set V0 = {0} and Vi = ϕ−1 (Wi−1 ) for 1 i n, then the ﬂag (Vi )0in clearly satisﬁes σ(g)Vi ⊂ Vi−1 for 1 i n. 19.3.7 Corollary. A Lie algebra g is nilpotent if and only if for all x ∈ g, ad x is a nilpotent endomorphism of g. 19.3.8 Let V be a ﬁnite-dimensional vector space and x ∈ End(V ). For α ∈ k, set: ker(x − α idV )n . Vα (x) = ker(x − α idV ) , V α (x) = n0

The subspace V α (x) is called the generalized eigenspace of x relative to α, and V 0 (x) is the nilspace of x. Clearly Vα (x) ⊂ V α (x) and α V (x). Vα (x) = {0} ⇔ V α (x) = {0} , V = α∈k

Lemma. Let x, y ∈ End(V ). Then (ad x)n (y) = 0 for some n ∈ N if and only if y(V α (x)) ⊂ V α (x) for all α ∈ k. Proof. Suppose that (ad x)n (y) = 0 for some n ∈ N. Since (ad x)(y) = (ad(x − α idV ))(y), we only need to show that y(V 0 (x)) is contained in V 0 (x). This is trivial when n = 0. Let us proceed by induction on n. Set z = [x, y], then (ad x)n−1 (z) = 0. By the induction hypothesis, z(V 0 (x)) ⊂ V 0 (x). For q ∈ N∗ , we have the following identity: xq y = yxq +

q i=1

xq−i (xy − yx)x−i−1 = yxq +

q

xq−i zxi−1 .

i=1

Since V is ﬁnite-dimensional, there exists r ∈ N∗ such that xr (V 0 (x)) = {0}. It follows that x2r−1 y(V 0 (x)) = {0} and therefore y(V 0 (x)) ⊂ V 0 (x).

19.3 Nilpotent Lie algebras

285

Conversely, suppose that y(V α (x)) ⊂ V α (x) for all α ∈ k. For α ∈ k, there exists rα ∈ N∗ such that (x − α idV )rα |V α (x) = 0. This implies that (ad(x − α idV ))2rα −1 (y)|V α = 0 (19.3.5). But (ad x)(y) = (ad(x − α idV ))(y), so (ad x)2rα −1 (y)|V α = 0. Now V is ﬁnite-dimensional and V is the sum of the generalized eigenspaces V α (x), so (ad x)n (y) = 0 for some n ∈ N. 19.3.9 Let g be a Lie algebra, P the set of maps from g to k and (V, σ) a ﬁnite-dimensional representation of g. For λ ∈ P , set: λ(x) Vλ (g) = Vλ(x) (σ(x)) , V λ (g) = V (σ(x)). x∈g

x∈g

The sum of V λ (g), λ ∈ P , is clearly direct. Proposition. Assume that g is nilpotent. (i) For all λ ∈ P , V λ (g) is a submodule of V . (ii) If V λ (g) = {0}, then Vλ (g) = {0}, and λ is a linear form on g which is zero on [g, g]. (iii) We have: λ V (g). V = λ∈P

If λ ∈ P , there exists a basis of V λ (g) such that for any x ∈ g, the matrix A(x) = [aij (x)] of σ(x)|V λ (g) with respect to this basis is strictly upper triangular, with aii (x) = λ(x) for all i. Proof. (i) Since g is nilpotent, it follows from 19.3.8 (i) that V λ(x) (σ(x)) is a submodule. (ii) If W = V λ (g) = {0}, then by (i), W is a submodule. If x ∈ g, there exists r ∈ N∗ such that (σ(x)|W − λ(x) idW )r = 0. It follows that λ(x) is the unique eigenvalue of σ(x)|W . Hence (dim W )λ(x) = tr σ(x)|W , and therefore λ is a linear form on g which is zero on [g, g]. It is then clear that the map g → End(W ), x → σ(x)|W − λ(x) idW , is a representation of g. The result now follows from Engel’s theorem (19.3.6). (iii) We shall prove part (iii) by induction on dim V = r. The case r = 0 is trivial. So let us assume that r 1. If for all x ∈ g, λ(x) is the unique eigenvalue of σ(x), then V = V λ (g). From the proof of (ii), x → σ(x) − λ(x) idV is a representation of g in V , so by applying Engel’s theorem (19.3.6) to this representation, we have the required basis. Now suppose that there exists x ∈ g such that σ(x) has two distinct eigenvalues. Since V = α∈k V α (x), dim V α (x) < r and the V α (x)’s are submodules of V (19.3.8), the result follows by applying the induction hypothesis.

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19.4 Solvable Lie algebras 19.4.1 Proposition. The following conditions are equivalent: (i) There exists k ∈ N such that Dk (g) = {0}. (ii) There exists a chain g = g0 ⊃ g1 ⊃ · · · ⊃ gn = {0} of ideals of g such that [gi , gi ] ⊂ gi+1 for 0 i n − 1. If these conditions are veriﬁed, we say that g is solvable. Proof. The proof is analogue to the one for 19.3.2. 19.4.2 Proposition. (i) A nilpotent Lie algebra is solvable. (ii) Subalgebras and quotients of a solvable Lie algebra are solvable. (iii) Let a be an ideal of g. Then g is solvable if and only if a and g/a are solvable. Proof. Parts (i) and (ii) are straightforward. Now if a and g/a are solvable, then Di (g) ⊂ a for some integer i and Di+j (g) ⊂ Dj (a) = {0} for some integer j. So g is solvable. The converse follows from (ii). 19.4.3 Lemma. Let (V, σ) be a ﬁnite-dimensional representation of g, a an ideal of g, λ ∈ a∗ and W = {v ∈ V ; σ(x)(v) = λ(x)(v) for all x ∈ a}. Then W is a submodule of V . Proof. We may assume that g ⊂ gl(V ) and W is non-zero. For v0 ∈ W \ {0} and x ∈ g, set vn = xn v0 , n ∈ N∗ , and p the smallest integer such that v0 , . . . , vp are linearly independent. Denote by U the subspace spanned by v0 , . . . , vp , then it is clear that x(U ) ⊂ U . Now if y ∈ a, then: y(v1 ) = yx(v0 ) = xy(v0 ) + [y, x](v0 ) ∈ λ(y)v1 + kv0 y(v2 ) = yx(v1 ) = xy(v1 ) + [y, x](v1 ) ∈ λ(y)v1 + kx(v0 ) + kv1 + kv0 ⊂ λ(y)v2 + kv1 + kv0 . If 0 i p, then we have by induction that: y(vi ) ∈ λ(y)vi + kvi−1 + · · · + kv0 . It follows that y(U ) ⊂ U and tr(y|U ) = (p + 1)λ(y). Since [x, y] ∈ a, we have that (p + 1)λ([x, y]) = 0. Hence λ([x, y]) = 0. Finally, yx(v0 ) = xy(v0 ) + [y, x](v0 ) = λ(y)x(v0 ) + λ([y, x])(v0 ) = λ(y)x(v0 ). So x(v0 ) ∈ W .

19.4.4 Theorem. (Lie’s Theorem) Let (V, σ) be a ﬁnite-dimensional representation of a solvable Lie algebra g. Then there exists a ﬂag {0} = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V of V such that σ(g)(Vi ) ⊂ Vi for 0 i n.

19.4 Solvable Lie algebras

287

Proof. We may assume that g ⊂ gl(V ) and V = {0}. We shall prove by induction on dim g = p that there exists µ ∈ g∗ and v ∈ V \ {0} such that x(v) = µ(x)v for all x ∈ g. We can then conclude as in the proof of 19.3.6. The case p 1 is obvious, so let us assume that p 2. Since D(g) = g, any codimension 1 subspace of g containing D(g) is an ideal of g. Let a be such an ideal. By induction, there exist λ ∈ a∗ and v ∈ V \ {0} such that y(v) = λ(y)v for all y ∈ a. Let W be as in 19.4.3, then W = {0} and g(W ) ⊂ W . Let x ∈ g \ a, then since k is algebraically closed, x|W has an eigenvector w. It follows that z(w) ∈ kw for all z ∈ g. So we are done. 19.4.5 Corollary. (i) Any ﬁnite-dimensional simple representation of a solvable Lie algebra is 1-dimensional. (ii) If g is solvable, then there exists a ﬂag {0} ⊂ g0 ⊂ · · · ⊂ gn = g consisting of ideals of g. (iii) If g is solvable, then any element of ad[g, g] is nilpotent. (iv) A Lie algebra g is solvable if and only if D(g) is nilpotent. Proof. Parts (i) and (ii) are consequences of 19.4.4, while part (iii) follows from (ii). Finally, (iv) follows from (iii) and 19.4.2 (iii). 19.4.6 Corollary. Let (V, σ) be a representation of dimension n > 0 of a nilpotent Lie algebra g such that ker σ(x) = {0} for all x ∈ g. Then there exists v ∈ V \ {0} such that σ(x)(v) = 0 for all x ∈ g. Proof. We shall prove the result by induction on n. The case n = 1 is obvious, so we may assume that n > 1. By 19.4.4, there is a basis (v1 , . . . , vn ) of V such that for all x ∈ g, the matrix of σ(x) in this basis is upper triangular with diagonal entries λ1 (x), . . . , λn (x) where λ1 , . . . , λn ∈ g∗ . Since ker σ(x) = {0}, λ1 · · · λn = 0. If λ1 · · · λn−1 = 0, then by applying the induction hypothesis on the restriction of σ to W = kv1 + · · · + kvn−1 , we have our result. So let us assume that λ1 · · · λn−1 = 0. Then λn = 0 since the ring of polynomial functions on g is an integral domain. Let x ∈ g be such that λ1 (x) · · · λn−1 (x) = 0. We have σ(x)(v) = 0 for some v ∈ V \ {0}. By our choice of x, v ∈ W , and σ(g)(v) ∈ W since λn = 0. Deﬁne f : g → W by f (y) = σ(y)(v). Then f ([x, y]) = σ(x)(f (y)) for all y ∈ g because σ(x)(v) = 0. It follows that the restriction of σ(x) to f (g) is nilpotent. If f (g) = {0}, then σ(x)(w) = 0 for some w ∈ f (g) \ {0}. But this is absurd since λ1 (x) · · · λn−1 (x) = 0. Thus f (g) = {0} and σ(g)(v) = {0}. 19.4.7 Lemma. Let V be a ﬁnite-dimensional vector space and g ⊂ gl(V ) a Lie subalgebra such that tr(xy) = 0 for all x, y ∈ g. Then any element of D(g) is a nilpotent endomorphism of V . Proof. By 19.4.4 and 19.4.5, it suﬃces to prove that g is solvable. We shall prove this by induction on dim g = r. The case r = 0 is trivial. So let r > 0.

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If D(g) = g, then by the induction hypothesis, D(g) is solvable and therefore g is solvable. So let us assume that D(g) = g and we shall show that this case is not possible. Let h be a maximal (proper) Lie subalgebra of g. Then dim h 1 and h is solvable by induction. Applying 19.4.4 to the adjoint representation of h in g/h, we deduce from the maximality of h that there exist x ∈ g \ h and λ ∈ h∗ verifying g = h ⊕ kx and [y, x] − λ(y)x ∈ h for all y ∈ h. 1) Let (W, σ) be a simple ﬁnite-dimensional representation of g and W0 a simple h-submodule of W . Then by 19.4.5, W0 = kw0 for some w0 ∈ W \ {0}. Let µ ∈ h∗ be such that σ(y)(w0 ) = µ(y)w0 for all y ∈ h. Set W−1 = {0} and for n ∈ N, wn = σ(x)n (w0 ) and Wn = kw0 + · · · + kwn . If y ∈ h, then: σ(y)wi+1 = σ(y)σ(x)wi = σ([y, x])wi + σ(x)σ(y)wi = σ([y, x] − λ(y)x)wi + σ(x)(σ(y)wi ) + λ(y)σ(x)wi , and (σ(y) − µ(y) − (i + 1)λ(y))wi+1 is equal to σ([y, x] − λ(y)x)wi + σ(x)(σ(y) − µ(y) − iλ(y))wi . It follows easily by induction that for all i ∈ N and y ∈ h, σ(h)(Wi ) ⊂ Wi and σ(y).wi − (µ(y) + iλ(y))wi ∈ Wi−1 . Since W is a simple g-module, W = Wq where q is the largest integer such that w0 , . . . , wq are linearly independent. The equality in the previous paragraph implies that 1 tr(σ(y)) = (q + 1)µ(y) + q(q + 1)λ(y). 2 Now g = D(g), so tr(σ(y)) = 0 and we have 2µ(y) = −qλ(y), q q q 2 σ(y).wi − (i − )λ(y)wi ∈ Wi−1 , tr(σ(y)2 ) = λ(y)2 . j− 2 2 j=0 2) If {0} = V0 ⊂ · · · ⊂ Vp = V is a Jordan-H¨ older series of the g-module V , and q(j) = dim(Vj+1 /Vj ) − 1, then, using the preceding computation, we obtain that % 2 & q(j) p−1 q(j) 0 = tr(y 2 ) = λ(y)2 . i− 2 j=0 i=0 Since D(g) = g = h, λ = 0 and so q(j) = 0 for all j. Again, the fact that D(g) = g implies that (Vj+1 /Vj )g = Vj+1 /Vj . Hence g consists of nilpotent endomorphisms of V . By 19.3.6, g is nilpotent. But this contradicts the assumption that D(g) = g and dim g > 0. 19.4.8 Theorem. (Cartan’s Criterion) Let (V, σ) be a ﬁnite-dimensional representation of g and βσ the bilinear form associated to σ. The following conditions are equivalent: (i) σ(g) is a solvable Lie algebra. (ii) We have βσ (D(g), D(g)) = {0}.

19.5 Radical and the largest nilpotent ideal

289

Proof. We may assume that g ⊂ gl(V ). Then clearly (i) ⇒ (ii) by 19.4.5. Conversely by 19.3.6 and 19.4.7, D2 (g) is nilpotent. So D(g) and g are solvable. Hence (ii) ⇒ (i).

19.5 Radical and the largest nilpotent ideal 19.5.1 Lemma. Let a be an ideal of g and (V, σ) a ﬁnite-dimensional representation of g. (i) Assume that σ is simple and all the elements of σ(a) are nilpotent. Then σ(a) = {0}. older series of V . (ii) Let {0} = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V be a Jordan-H¨ Then all the elements of σ(a) are nilpotent if and only if σ(a)(Vi ) ⊂ Vi−1 for 1 i n. Proof. Part (i) follows from 19.3.5 and 19.4.3, while (ii) follows from (i). 19.5.2 Lemma. Let V be a ﬁnite-dimensional vector space, g a Lie subalgebra of gl(V ) and a an abelian ideal of g. If V is a simple g-module, then a ∩ D(g) = {0}. Proof. 1) Let S be the associative subalgebra of End(V ) generated by a. Let b ⊂ a be an ideal of g satisfying tr(bs) = 0 for all b ∈ b and s ∈ S. In particular, tr(bn ) = 0 for all b ∈ b and n ∈ N∗ , so b is nilpotent (19.2.3). By 19.5.1, b = {0}. 2) Since a is abelian, we have for x ∈ g, a ∈ a and s ∈ S that tr([x, a]s) = tr(xas − axs) = tr(xas − xsa) = tr(xas − xas) = 0. It follows from point 1) that [g, a] = {0}, and so a ⊂ z(g). Hence xs = sx for all x ∈ g and s ∈ S. 3) Finally, we have tr([x, y]s) = tr(xys − yxs) = tr(xys − xsy) = tr (x(ys − sy)) = 0 for x, y ∈ g and s ∈ S. So point 1) implies that a ∩ D(g) = {0}.

19.5.3 Proposition. There is a unique maximal solvable ideal in g. It is called the radical of g, and is denoted by rad g. Proof. Let a, b be solvable ideals of g. Then a + b is an ideal of g which is solvable (19.4.2) since (a + b)/b and a/a ∩ b are isomorphic. So there is a unique maximal solvable ideal in g. 19.5.4 Proposition. The radical of g is the smallest ideal a of g such that rad(g/a) = {0}.

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Proof. Let r = rad g and p : g → g/r be the canonical surjection. Then o = p−1 (rad(g/r)) is an ideal of g such that o/r is solvable. So o is solvable by 19.4.2 and o = r. Let a be an ideal of g and ϕ : g → g/a the canonical projection. If rad(g/a) = {0}, then ϕ(r) = {0}. Thus a ⊃ r. 19.5.5 Proposition. Let (V, σ) be a ﬁnite-dimensional representation of older series of V , βσ the bilinear g, {0} = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V a Jordan-H¨ form on g associated to σ and σi the representation of g on Vi /Vi−1 induced by σ, for 1 i n. (i) There is a unique maximal ideal n among the ideals a of g such that the elements of σ(a) are nilpotent. (ii) We have n = ker σ1 ∩ · · · ∩ ker σn and βσ (n, g) = {0}. We call n the largest nilpotent ideal of σ. Proof. By 19.5.1, for an ideal a of g, σ(x) is nilpotent for all x ∈ a if and only if a ⊂ ker σ1 ∩ · · · ∩ ker σn . So we have (i) and the ﬁrst part of (ii). Let x ∈ g, y ∈ n, then σ(x)σ(y)(Vi ) ⊂ σ(x)(Vi−1 ) ⊂ Vi−1 , for 1 i n. So σ(x)σ(y) is nilpotent and βσ (n, g) = {0}. 19.5.6 Lemma. Let (V, σ) be a ﬁnite-dimensional representation of g. (i) If σ is simple, then σ(rad g ∩ D(g)) = {0}. (ii) If n is the largest nilpotent ideal of σ, then rad g ∩ D(g) ⊂ n. Proof. (i) Let r = rad g and i be the smallest integer such that σ(Di+1 (r)) = {0}. Then the image under σ of a = Di (r) is an abelian ideal of σ(g). By Schur’s Lemma, σ(a) ⊂ k idV , and so σ(D(g) ∩ Di (r)) = {0}. If i > 0, then Di (r) ⊂ D(g) and so σ(a) = {0}. This is impossible by the minimality of i. So i = 0 and σ(r ∩ D(g)) = {0}. Part (ii) follows clearly from (i) and 19.5.5. 19.5.7 Let us conserve the notations of 19.5.5. In general, the fact that σ(x) is nilpotent does not imply that x ∈ n. Nevertheless, we have the following result: Proposition. Let (V, σ) be a ﬁnite-dimensional representation of g and n the largest nilpotent ideal of σ. If x ∈ rad g and σ(x) is nilpotent, then x ∈ n. older Proof. Let r = rad g and {0} = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V be a Jordan-H¨ series of the r-module V . By 19.4.5, Wi = Vi /Vi−1 is 1-dimensional, and the representation of r on Wi is given by a linear from λi ∈ r∗ which is zero on D(r). The set of elements x ∈ r such that σ(x) is nilpotent is therefore the ideal a = ker λ1 ∩ · · · ∩ ker λn of r. By 19.5.6, [g, a] ⊂ r ∩ D(g) ⊂ r ∩ n ⊂ a. Thus a is an ideal of g and the result follows from 19.5.5. 19.5.8 Proposition. Let L be the Killing form of g and n the largest nilpotent ideal of the adjoint representation of g. Then n is the largest nilpotent ideal of g, and L(g, n) = {0}.

19.6 Nilpotent radical

291

Proof. Let a be an ideal of g and x ∈ a. Then adg x is nilpotent if and only if ada x is nilpotent. Thus a is nilpotent if and only if adg x is nilpotent for all x ∈ a (19.3.7). Since any nilpotent ideal is contained in the radical of g, the result follows from 19.5.7 and 19.5.5.

19.6 Nilpotent radical 19.6.1 It follows from 19.5.5 that the intersection of the kernels of simple ﬁnite-dimensional representations of g is equal to the intersection of the largest nilpotent ideals of ﬁnite-dimensional representations of g. This is an ideal of g that we shall call the nilpotent radical of g. In view of 19.5.8, the nilpotent radical of g is contained in the radical of g. 19.6.2 Theorem. Let r = rad g and s the nilpotent radical of g. Then s = r ∩ D(g). In particular, when g is solvable, then s = D(g). Proof. Any λ ∈ g∗ verifying λ(D(g)) = {0} determines a representation of g of dimension 1. So s ⊂ ker λ. Since s ⊂ r, we have s ⊂ r ∩ D(g). The reverse inclusion follows from 19.5.6. 19.6.3 Corollary. Let r = rad g. (i) Let (V, σ) be a ﬁnite-dimensional representation of g and β the bilinear form on g associated to σ. Then β(r, D(g)) = {0}. (ii) r is the orthogonal of D(g) with respect to the Killing form L of g. (iii) r is a characteristic ideal of g. (iv) If a is an ideal of g, then rad a = a ∩ r. Proof. (i) Let n be the largest nilpotent ideal of g and s the nilpotent radical of g. By 19.6.2 and 19.6.1, [g, r] ⊂ s ⊂ n. So the result follows from 19.5.5 and 19.2.2. (ii) Let u be the orthogonal of D(g) with respect to L. Then u is an ideal of g containing r (19.2.2). Since the Killing form of u is L|u×u , adg u is solvable (19.4.8). But adg u is isomorphic to u/z(g), so u is solvable, and u = r. (iii) Let x, y ∈ g and d ∈ Der g. Since ad d(x) = [d, ad x], we verify easily that L(d(x), y) + L(x, d(y)) = 0. So the orthogonal of a characteristic ideal with respect to L is characteristic, and the result follows from (ii). (iv) Clearly, a ∩ r ⊂ rad a. On the other hand, by applying (iii) to a, we obtain that rad a is a solvable ideal of g. So rad a ⊂ r. 19.6.4 Proposition. Let r = rad g and n the largest nilpotent ideal of g. (i) We have d(r) ⊂ n for all derivation d of g. (ii) n is a characteristic ideal of g. Proof. It suﬃces to prove (i). We deﬁne a Lie algebra h = kv ⊕ g extending the Lie bracket of g by deﬁning [v, x] = d(x) for all x ∈ g. Then g is an ideal of h and by 19.6.3 (iii), r ⊂ rad h, and so d(r) = [v, r] ⊂ [h, h] ∩ rad h = o. By 19.6.2, o is a nilpotent ideal of h. So d(r) is contained in the nilpotent ideal o ∩ g of g. Hence d(r) ⊂ n (19.5.8).

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19.7 Regular linear forms 19.7.1 Let n = dim g, d ∈ {0, 1, . . . , n} and Gr(g, d) the Grassmannian variety of d-dimensional vector subspaces of g. Recall from 13.6 that given B = (e1 , . . . , en ) a basis of g, the set UB of elements o ∈ Gr(g, d) verifying o ∩ (ked+1 + · · · + ken ) = {0} is open in Gr(g, d). Further, o ∈ UB if and only if o has a unique basis (v1 , . . . , vd ) of the form vi = ei +

n

aij ej .

j=d+1

The map o → (a1,d+1 , . . . , a1,n , . . . , ad,d+1 , . . . , ad,n ) is an isomorphism of UB onto kd(n−d) . 19.7.2 Lemma. (i) The set of d-dimensional Lie subalgebras of g is a closed subset of Gr(g, d). (ii) The set of d-dimensional solvable (resp. nilpotent, abelian) Lie subalgebras of g is a closed subset of Gr(g, d). (iii) The set of (o, f ) ∈ Gr(g, d) × g∗ such that f ([g, o]) = {0} (resp. f ([o, o]) = {0}) is a closed subset of Gr(g, d) × g∗ . Proof. Let us prove (i), and (ii) for solvable Lie subalgebras. The proofs of the rest are similar. We shall use the notations of 19.7.1. An element o ∈ Gr(g, d) is a Lie subalgebra if and only if [vi , vj ] ∈ o for 1 i, j d. These conditions correspond to the vanishing of determinants in the apq ’s. Since the open subsets UB , for B a basis of g, form an aﬃne open covering of Gr(g, d), part (i) follows. Further, a d-dimensional Lie subalgebra o is solvable if and only if Dd (o) = {0}. By expressing this condition in the basis (v1 , . . . , vd ), we see that it is equivalent to the vanishing of certain polynomials in the apq ’s. 19.7.3 For f ∈ g∗ , deﬁne Φf to be the alternating bilinear form on g given by: Φf (x, y) = f ([x, y]). For a subset a of g, denote by a(f ) the orthogonal of a with respect to Φf . If a is an ideal of g, then it is easy to check that a(f ) is a Lie subalgebra of g. In particular, the kernel g(f ) of Φf is a Lie subalgebra of g. Since Φf induces a non-degenerate alternating bilinear form on g/g(f ) , dim g − dim g(f ) is an even integer. We deﬁne the index of g to be the integer χ(g) = min{dim g(f ) ; f ∈ g∗ }. We say that f ∈ g∗ is regular if dim g(f ) = χ(g), and we denote by g∗reg the set of regular elements of g∗ . 19.7.4 Lemma. Let u be a subspace of g. (i) The set of elements f ∈ g∗ verifying u ∩ g(f ) = {0} is closed in g∗ . (ii) Let f ∈ u∗ and g ∈ g∗ be such that g|u = f and dim g(g) dim g(h) for all h ∈ g∗ verifying h|u = f . Then [g(g) , g(g) ] ⊂ u.

19.7 Regular linear forms

293

Proof. (i) Let E = (e1 , . . . , en ) be a basis of g such that (e1 , . . . , ep ) is a basis of u, and (e∗1 , . . . , e∗n ) the basis of g∗ dual to g. For 1 i, j n, set [ei , ej ] =

n

γijk ek .

k=1

For f = λ1 e∗1 + · · · + λn e∗n , we have: u = µ1 e1 + · · · + µp ep ∈ u ∩ g(f ) ⇔ 0 = f ([u, ej ]) =

p n

µl γljk λk for all j.

l=1 k=1

( ' n So u∩g(f ) = {0} is equivalent to the condition that the matrix k=1 γljk λk , 1 l p and 1 j n, has rank strictly less than p. So we are done. (ii) Fix a subspace w such that g is the direct sum of g(g) and w. Let h ∈ g∗ be such that h|u = 0. If t ∈ k, the assumptions on g imply that g = w ⊕ g(g+th) ⇔ w ∩ g(g+th) = {0}. By (i), this deﬁnes an open subset K of k containing 0. Let (e1 , . . . , en ) be a basis of g such that (e1 , · · · , ep ) is a basis of w. For x = λ1 e1 + · · · + λn en ∈ g and t ∈ K, write x = x1 (t) + x2 (t) where x1 (t) = µ1 e1 + · · · + µp ep ∈ w and x2 (t) ∈ g(g+th) . The coeﬃcients µi are obtained via the following equalities: p i=1

µi (g + th)([ei , ej ]) =

n

λk (g + th)([ek , ej ]) , 1 j n.

k=1

' ( The set K of t ∈ K such that the matrix (g + th)([ei , ej ]) , 1 i p, 1 j n, has rank p, is open and it contains 0. It follows that the maps x → x1 (t) and x → x2 (t) are rational on K . Let x, y ∈ g(g) , and x2 (t), y2 (t) their projections in g(g+th) with respect to the direct sum g(g+th) ⊕ w. Then for t ∈ K , we have: (g + th)([x2 (t), y2 (t)]) = 0. Taking the derivatives on both sides at t = 0, we obtain: h([x, y]) + g([x, y2 (0)]) + g([x2 (0), y]) = 0. Hence h([x, y]) = 0. This is true for all h verifying h|u = 0, so we have [x, y] ∈ u. 19.7.5 Proposition. (i) The set g∗reg is open and dense in g∗ . (ii) If f ∈ g∗reg , then the Lie algebra g(f ) is abelian. (iii) For all f ∈ g∗ , g(f ) contains an abelian Lie subalgebra of dimension χ(g).

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Proof. (i) We have f ∈ g∗reg if and only if g(f ) ∩ u = {0} for all subspaces of dimension dim g − χ(g). So the result follows from 19.7.4 (i). (ii) Just take u = {0} in 19.7.4 (ii). (iii) Let χ(g) = d, p : Gr(g, d) × g∗ → g∗ be the canonical surjection and F the set of elements (u, f ) ∈ Gr(g, d) × g∗ such that f ([g, u]) = {0} and [u, u] = {0}. Since p is a closed map (13.4.5 and 13.6.3), it follows from 19.7.2 (iii) that p(F) is closed. But (ii) says that g∗reg ⊂ p(F). So part (i) implies that p(F) = g∗ and we are done. 19.7.6 Fix an integer d 0 and let g∗d = {f ∈ g∗ ; dim g(f ) = d}. The set is locally closed in g∗ , so it is a subvariety of g∗ . Let ϕ : g∗d → Gr(g, d) be the map deﬁned by f → g(f ) .

g∗d

Proposition. The map ϕ is a morphism of varieties. Proof. Let h ∈ g∗d , E = (e1 , . . . , en ) be a basis of g such that (e1 , . . . , ed ) is a basis of g(h) and (e∗1 , . . . , e∗n ) the basis!of g∗ dual to E. Denote " by V the subspace spanned by ed+1 , . . . , en , u = f ∈ g∗d ; V ∩ g(f ) = {0} , and U = {p ∈ Gr(g, d); p ∩ V = {0}}. By 19.7.1 and 19.7.4, u and U are open subsets of g∗d and Gr(g, d). Note that ϕ−1 (U ) = u. Let f = λ1 e∗1 + · · · + λn e∗n ∈ u. For 1 i n and d + 1 j n, set αi,j = f ([ei , ej ]). Since g(f ) ∩ V = {0}, the condition 0 = f ([ei , νd+1 ed+1 + · · · + νn en ]) = νd+1 αi,d+1 + · · · + νn αi,n , 1 i n implies that νd+1 = · · · = νn = 0. So the rank of the matrix [αi,j ] is n − d. Now, for 1 k d, there is a unique (vd+1,k , . . . , vn,k ) ∈ kn−d such that 0 = f ([ei , ek + νd+1,k ed+1 + · · · + νn,k en ]) , 1 i n or equivalently, νd+1,k αi,d+1 + · · · + νn,k αi,n = (λ1 e∗1 + · · · + λn e∗n )([ek , ei ]) , 1 i n. It follows from the formulas for solving a Cramer’s system that there exists an open subset o of u such that the maps f → νj,k , d + 1 j n, 1 k d, are regular on o. So the result follows from the remarks in 19.7.1.

19.8 Cartan subalgebras 19.8.1 For x ∈ g, α ∈ k, we shall denote by gα (x) and gα (x) the subspaces gα (ad x) and gα (ad x) (19.3.8). Let h be a Lie subalgebra of g. Consider g as a h-module via the adjoint representation. If λ ∈ h∗ \ {0} is such that gλ (h) = {0} (19.3.9), then we say that λ is a root of g relative to h. The set of roots, denoted by R(g, h), is called the root system of g relative to h. 19.8.2 For x, y, z ∈ g, n ∈ N and λ, µ ∈ k, we have:

19.8 Cartan subalgebras

(ad x − λ − µ)n ([y, z]) =

295

n n [(ad x − λ)i (y), (ad x − µ)n−i (z)]. i i=0

particular, g0 (x) is a We deduce therefore that [gλ (x), gµ (x)] ⊂ gλ+µ (x). In λ Lie subalgebra of g. Recall also from 19.3.8 that g = λ∈k g (x) and so if y ∈ gλ (x) with λ = 0, then ad y is nilpotent. 19.8.3 Deﬁnition. A nilpotent Lie subalgebra h of g verifying h = ng (h) is called a Cartan subalgebra of g. 19.8.4 Lemma. A Cartan subalgebra h of g is a maximal nilpotent Lie subalgebra. Proof. If t is a nilpotent Lie subalgebra containing h, then by applying 19.3.6 to the adjoint representation of h on t/h, we have h = t since h = ng (h). 19.8.5 Lemma. Let x ∈ g and h a Lie subalgebra of g. (i) If g0 (x) ⊂ h, then h = ng (h). (ii) Suppose that there exists an element x ∈ h such that h ⊂ g0 (x) and dim g0 (x) dim g0 (y) for all y ∈ h. Then g0 (x) ⊂ g0 (y) for all y ∈ h. (iii) If x ∈ ggen , then g0 (x) is the unique Cartan subalgebra of g containing x. In particular, g has a Cartan subalgebra. (iv) Let t be a Cartan subalgebra of g. There exists an element x ∈ g such that t = g0 (x). Proof. (i) Since g0 (x) ⊂ h, ad x induces a bijective endomorphism u of ng (h)/h. But x ∈ h implies that [x, ng (h)] ⊂ h, so u = 0 and h = ng (h). (ii) Let t = g0 (x) and y ∈ h. For λ ∈ k, denote by uλ and vλ the endomorphisms of t and g/t induced by ad(x + λy). Let χλ , χλ and χλ be respectively the characteristic polynomial of ad(x + λy), uλ and vλ . Then χλ = χλ χλ , and if n = dim g and r = dim t, then: χλ (T ) = T r +

r−1 i=0

ai (λ)T i , χλ (T ) = T n−r +

n−r−1

bj (λ)T j ,

j=0

where ai and bj are polynomial functions on k. Since t = g0 (x), b0 (0) = 0. So the set E of λ ∈ k such that b0 (λ) = 0 is open dense in k; so k \ E is a ﬁnite set. Hence g0 (x + λy) ⊂ t if λ ∈ E, and by our choice of x, this implies that g0 (x + λy) = t if λ ∈ E. It follows that χλ (T ) = T r if λ ∈ E, and thus for all λ ∈ k. This means that t ⊂ g0 (x + λy). So we have our result by replacing y by y − x and by taking λ = 1. (iii) If x ∈ ggen , then the conditions of (ii) are veriﬁed for h = g0 (x). So 0 g (x) ⊂ g0 (y) for all y ∈ h. Hence g0 (x) is nilpotent (19.3.7) and g0 (x) is a Cartan subalgebra by (i). Now if t is a Cartan subalgebra containing x, then t ⊂ g0 (x) since t is nilpotent. So t = g0 (x) (19.8.4). (iv) For y ∈ t, denote by σ(y) the endomorphism of g/t induced by ad y. Since t = ng (t), 0 is the only element v ∈ g/t such that σ(y)(v) = 0 for all

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y ∈ t. But t is nilpotent, so by 19.4.6, there exists x ∈ t such that σ(x) is bijective. Consequently g0 (x) = t. 19.8.6 Theorem. Let h be a nilpotent Lie subalgebra of g. If λ ∈ h∗ , let by gλ . us denote gλ (h) simply λ λ µ λ+µ . In particular, g0 is a (i) We have g = λ∈h∗ g , and [g , g ] ⊂ g Lie subalgebra of g containing h. (ii) h is a Cartan subalgebra if and only if h = g0 . (iii) If g0 is nilpotent, then g0 is a Cartan subalgebra of g. (iv) Let λ, µ ∈ h∗ , then L(gλ , gµ ) = {0} if λ = −µ. In particular, L(h, gλ ) = {0} if λ = 0. (v) Let x, y ∈ h. Then: (dim gλ )λ(x)λ(y). L(x, y) = λ∈h∗

Proof. (i) This follows from 19.3.9 and 19.8.2. (ii) Applying 19.3.6 to the adjoint representation of h on g0 /h, we obtain that if g0 = h, then ng (h) = h. So h is not a Cartan subalgebra. Conversely, since [h, ng (h)] ⊂ h and h is nilpotent, h ⊂ ng (h) ⊂ g0 . So if 0 g = h, h is a Cartan subalgebra. (iii) If t = g0 is nilpotent, then t ⊂ g0 (t). Since h ⊂ t, we deduce that 0 g (t) ⊂ t. Hence g0 (t) = t and by (ii), t is a Cartan subalgebra. (iv) Let λ, µ, ν ∈ h∗ , x ∈ gλ and y ∈ gµ . By (i), we have: (ad x◦ ad y)(gν ) ⊂ gλ+µ+ν . If λ + µ = 0, then ad x◦ ad y is nilpotent, hence L(x, y) = 0. Since h ⊂ g0 , we are done. (v) Let x, y ∈ h. There is a basis of gλ such that the matrix of ad x◦ ad y|gλ with respect to this basis is upper triangular with a unique eigenvalue λ(x)λ(y) (19.3.9). The result is now clear. 19.8.7 Corollary. Assume that the Killing form L on g is non-degenerate. Let h be a Cartan subalgebra R = R(g, h). of g and α . g (i) We have g = h ⊕ α∈R (ii) If α ∈ R, then −α ∈ R and L|gα ×g−α is non-degenerate. (iii) The restriction of L to h × h is non-degenerate. (iv) The elements of R span the vector space h∗ . (v) The Lie algebra h is abelian. Proof. Parts (i), (ii) and (iii) are immediate consequences of 19.8.6. Let x ∈ h be such that α(x) = 0 for all α ∈ R. By 19.8.6 (v), L(x, h) = {0}. Now (iii) implies that x = 0. So we have proved (iv). Finally let x, y ∈ h. If α ∈ R, then [x, gα ] ⊂ gα and [y, gα ] ⊂ gα . Since (ad[x, y])|gα = [ad x|gα , ad y|gα ],

19.8 Cartan subalgebras

297

we see that the trace of (ad[x, y])|gα is zero. But by 19.3.9, this is equal to (dim gα )α([x, y]), so α([x, y]) = 0. Thus [x, y] = 0 by (iv).

References and comments • [12], [14], [29], [38], [39], [42], [43], [80]. For a proof of Ado’s theorem (19.2.1), see [12] or [29]. The notion of a Lie algebra extends to any base ﬁeld k (although there is a slight ambiguity of antisymmetry when the characteristic of k is 2). Some of the results in this chapter can be extended. It is a good exercise for the reader to work out to what extent the results in this chapter remain valid. For example, Lie’s theorem fails in general if k has positive characteristic. Lie algebras over the real numbers are fundamental in the theory of Lie groups. For more details, the reader may refer to [24], [59], [84] and [87].

20 Semisimple and reductive Lie algebras

In this chapter, we study the structure of semisimple and reductive Lie algebras. We introduce certain important objects related to such Lie algebras such as Weyl groups, Borel subalgebras and parabolic subalgebras. Let g be a ﬁnite-dimensional Lie algebra over k and L its Killing form.

20.1 Semisimple Lie algebras 20.1.1 Deﬁnition. (i) We say that g is simple if dim g > 1 and the only ideals of g are {0} and g. (ii) A Lie algebra g is semisimple if {0} is the only abelian ideal of g. 20.1.2 Theorem. The following conditions are equivalent: (i) g is semisimple. (ii) rad g = {0}. (iii) L is non-degenerate. Proof. (i) ⇒ (ii) Since r = rad g is an ideal, Dr (r) are ideals of g for all r 0. Moreover r is solvable, so if r = {0}, then Dr (r) is non-zero and commutative for some r. (ii) ⇒ (iii) The kernel a of L is an ideal of g and L|a×a is the Killing form of a. So L|a×a = 0 and a is solvable (19.4.8). Hence a ⊂ r = {0}. (iii) ⇒ (i) This follows from 19.5.8. 20.1.3 Corollary. (i) g is semisimple if and only if the Killing homomorphism of g is an isomorphism. (ii) A product of a ﬁnite number of semisimple Lie algebras is semisimple. (iii) The Lie algebra g/ rad g is semisimple. 20.1.4 Lemma. Let a be an ideal of g and b the orthogonal of a with respect to L. Suppose that {0} is the only solvable ideal of g contained in a. Then g = a ⊕ b and the Lie algebras g and a × b are isomorphic.

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20 Semisimple and reductive Lie algebras

Proof. Since b is an ideal, so is c = b ∩ a. The Killing form on c is zero, so it is solvable (19.4.8) and we deduce that c = {0}. So the result follows since dim g = dim a + dim b. 20.1.5 Proposition. Assume that g is semisimple. (i) The centre of g is trivial. In particular, the adjoint representation of g is faithful. (ii) We have g = [g, g]. (iii) All derivations of g are inner. (iv) For all ﬁnite-dimensional representation (V, σ) of g, σ(g) ⊂ sl(V ). Moreover, if σ is faithful, then the associated bilinear form βσ on g is nondegenerate. Proof. Part (i) is clear and (ii) follows from 19.6.3 and 20.1.2. For (iii), let u = ad g and d = Der g. We have u g, and u is an ideal of d. It follows from 20.1.4 that d = u × v where v is the orthogonal of u in d with respect to the Killing form. Since [d, ad x] = ad d(x) for all d ∈ d and x ∈ g, it is now clear that v = {0}. The ﬁrst part of (iv) is a direct consequence of (ii). Let a be the orthogonal of g with respect to βσ . By Cartan’s Criterion, σ(a) is solvable and so a = {0} if σ is faithful. 20.1.6 Proposition. Let a be an ideal of a semisimple Lie algebra g and b the orthogonal of a in g with respect to the Killing form L. (i) a and g/a are semisimple. (ii) We have g = a ⊕ b and the Lie algebras g and a × b are isomorphic. Proof. These are direct consequences of 19.6.3 (iv), 20.1.2 and 20.1.4. 20.1.7 Theorem. The following conditions are equivalent: (i) g is semisimple. (ii) g is a direct product of simple Lie algebras. Proof. A simple Lie algebra is semisimple since its radical is {0}. So by 20.1.6, the implication (i) ⇒ (ii) is a simple induction on dim g. Conversely, if g is a direct product of simple Lie algebras, then its Killing form is clearly non-degenerate (since simple Lie algebras are semisimple). So g is semisimple by 20.1.2. 20.1.8 Proposition. Suppose that g = a1 × · · · × an where the ai ’s are simple Lie algebras. Then any ideal of g is a direct product of certain ai ’s. In particular, the ai ’s are the minimal ideals of g. We call the ai ’s the minimal components of g.

20.2 Examples

301

Proof. Let a be an ideal of g. By permuting the indices, we may assume that a ∩ ai = {0} if and only if 1 i p for some integer p. Then by the simplicity of the ai ’s, a ∩ ai = ai for 1 i p and [a, ai ] ⊂ a ∩ ai = {0} for i > p. It follows that a ⊃ a1 × · · · × ap and [a, ap+1 × · · · × an ] = {0}. Thus a ∩ (ap+1 × · · · × an ) = {0} since it is in the centre of ap+1 × · · · × an which is semisimple. As a ⊃ a1 × · · · × ap , we have a = a1 × · · · × ap .

20.2 Examples 20.2.1 Lemma. Let g be a Lie subalgebra of gl(V ) where V is a ﬁnitedimensional vector space. Assume that z(g) = {0} and that V is a simple g-module. Then g is semisimple. Proof. By 19.5.6, an abelian ideal a in g veriﬁes [a, g] ⊂ a ∩ D(g) = {0}. So a ⊂ z(g) = {0}. 20.2.2 Theorem. Let V be a ﬁnite-dimensional vector space and a the set of scalar endomorphisms of V . (i) We have z(gl(V )) = a and z(sl(V )) = {0}. (ii) gl(V ) a × sl(V ). (iii) The Lie algebra sl(V ) is semisimple. Proof. Part (i) is just 10.8.4 and (ii) is clear. Part (ii) implies that any sl(V )submodule of V is a gl(V )-submodule. So (iii) follows from (i) and 20.2.1. 20.2.3 Theorem. Let V be a n-dimensional vector space, β a nondegenerate symmetric or alternating bilinear form on V and g the Lie subalgebra of gl(V ) consisting of elements f ∈ gl(V ) such that β(f (v), w) + β(v, f (w)) = 0 for all v, w ∈ V . Then g is semisimple except when β is symmetric and n = 2. Proof. For a subspace W of V and we denote by W ⊥ its orthogonal with respect to β. If n 1, then g = {0} which is semisimple. If n = 2 and β is symmetric, then by choosing a suitable basis of V , we see that dim g = 1. Thus g is not semisimple. If n = 2 and β is alternating, then by choosing a suitable basis of V , we see that g sl2 (k) which is semisimple by 20.2.2. Suppose that n 3. Let W be a g-submodule of V distinct from V and {0} and u, v ∈ V , w ∈ W be such that β(u, w) = 0, β(v, w) = 0. Deﬁne f ∈ End(V ) as follows: for x ∈ V ,

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f (x) = β(x, u)v − β(v, x)u. We verify easily that f ∈ g and since f (w) = −β(v, w)u, we have u ∈ f (W ) ⊂ W . We conclude that (kw)⊥ ⊂ W and therefore dim W n − 1. But V = W , so W = (kw)⊥ . This is true for all w ∈ W \ {0}. It follows that W = W ⊥ and 2(n − 1) = dim W + dim W ⊥ = n. But this is impossible for n 3. We have therefore proved that V is a simple g-module. In view of 20.2.1, we are left to show that z(g) = {0}. By Schur’s Lemma, z(g) ⊂ k idV and it is clear from the hypotheses on β and g that g∩k idV = {0}. So we are done.

20.3 Semisimplicity of representations 20.3.1 Lemma. If dim g = 1, then g has a 2-dimensional representation which is not semisimple. Proof. Let g = kx and (u, v) a basis of a 2-dimensional vector space. Deﬁne σ : g → gl(V ) by setting σ(λx)(u) = λv and σ(λx)(v) = 0. Then it is clear that (V, σ) is a representation of g in which kv is the unique submodule of dimension 1. In particular, V is not semisimple. 20.3.2 Lemma. Assume that g is semisimple. Let (V, σ) be a simple ﬁnitedimensional faithful representation of g and β the bilinear form on g associated to σ. (i) β is non-degenerate. (ii) Suppose that dim g = n = 0. Let (x1 , . . . , xn ) be a basis of g and (y1 , . . . , yn ) the basis of g dual to (x1 , . . . , xn ) with respect to β. Then: c = σ(x1 )σ(y1 ) + · · · + σ(xn )σ(yn ) =

n idV . dim V

Proof. Since σ is faithful, we may assume that σ(g) = g ⊂ gl(V ). (i) If a is the kernel of β, then a is a solvable ideal (19.4.8). So a = {0}. (ii) For z ∈ g and 1 i n, we have: [z, xi ] =

n j=1

λij xj , [z, yi ] =

n

µij yj .

j=1

So λij = β([z, xi ], yj ) = −β(xi , [z, yj ]) = −µji . It follows easily that [z, c] = 0. Since tr(c) = β(x1 , y1 ) + · · · + β(xn , yn ) = n, part (ii) follows from Schur’s Lemma. 20.3.3 Theorem. (Whitehead’s First Lemma) Let g be a semisimple Lie algebra, (V, σ) a ﬁnite-dimensional representation of g and f : g → V a linear map. The following conditions are equivalent: (i) f ([x, y]) = σ(x)f (y) − σ(y)f (x) for all x, y ∈ g. (ii) There exists v ∈ V such that f (x) = σ(x)(v) for all x ∈ g.

20.3 Semisimplicity of representations

303

Proof. The implication (ii) ⇒ (i) is straightforward. Let us prove (i) ⇒ (ii). Suppose that (i) is veriﬁed and we may assume that g = {0}. First, suppose that σ is simple. Let ker σ = a. By 20.1.7 and 20.1.8, g a × b where a, b are ideals which are semisimple. So a = [a, a] and by our hypothesis on f , f (a) = {0}. Let (x1 , . . . , xn ) be a basis of b and (y1 , . . . , yn ) the basis of b dual to (x1 , . . . , xn ) with respect to the bilinear form on b associated to the restriction n idV . Set of σ to b. By 20.3.2, c = σ(x1 )σ(y1 ) + · · · + σ(x1 )σ(y1 ) = dim V v = c−1 (σ(x1 )f (y1 ) + · · · + σ(xn )f (yn )). Let z ∈ g. We have by the proof of 20.3.2 that for 1 i n, [z, xi ] =

n

λij xj , [z, yi ] =

j=1

n

−λji yj .

j=1

Since f satisﬁes (i), we obtain that c ◦ f (z) = = =

n i=1 n i=1 n

σ(xi )σ(yi )f (z) = σ(xi )f ([yi , z]) +

n

σ(xi )f ([yi , z]) +

n

i=1 n

i=1 n

i=1

i=1

σ([xi , z])f (yi ) +

σ(xi )σ(z)f (yi )

σ(z)σ(xi )f (yi )

σ(z)σ(xi )f (yi ) = c ◦ σ(z)(v).

i=1

So σ(z)(v) = f (z) for all z ∈ g as required. In the general case, we proceed by induction on dim V . From the previous paragraph, we may assume that V contains a submodule W distinct from V and {0}. Let π : V → V /W be the canonical surjection. Then by the induction hypothesis, there exists v ∈ V such that (π ◦ f )(x) = π ◦ σ(x)(v) for all x ∈ g. Let θ(x) = f (x) − σ(x)(v). Then θ(x) ∈ W and θ ∈ Hom(g, W ) satisﬁes (i). By the induction hypothesis, there exists w ∈ W such that θ(x) = σ(x)(w) for all x ∈ g. So f (x) = σ(x)(v + w) for all x ∈ g. 20.3.4 Theorem. (Weyl’s Theorem) Let g be a Lie algebra. The following conditions are equivalent: (i) All ﬁnite-dimensional representations of g is semisimple. (ii) g is semisimple. Proof. We may assume that g = {0}. (i) ⇒ (ii) If (i) is veriﬁed, then the adjoint representation of g is semisimple. If a is an abelian ideal of g, then g = a⊕b where b is an ideal of g. So a ⊂ z(g). If z(g) = {0}, then g has an ideal of codimension 1. But 20.3.1 would imply that g has a non-semisimple representation of dimension 2. Contradiction. (ii) ⇒ (i) Let (V, σ) be a ﬁnite-dimensional representation of g, U a nontrivial submodule of V , π : V → V /U the canonical surjection and (V /U, τ )

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the representation of g induced by σ. Denote by M the set of linear maps from V /U to V and set N = {φ ∈ M ; φ(V /U ) ⊂ U }. The map λ : g → End(M ), deﬁned by setting for x ∈ g and φ ∈ M : λ(x)(φ) = σ(x) ◦ φ − φ ◦ τ (x). then deﬁnes a representation of g in which N is a submodule. Let us ﬁx φ0 ∈ M such that π ◦ φ0 = idV /U . The linear map f : g → M , x → λ(x)(φ0 ) veriﬁes f (g) ⊂ N since for any w ∈ V /U , π (f (x)(w)) = π◦σ(x)◦φ0 (w) − π◦φ0 ◦τ (x)(w) = τ (x)◦π◦φ0 (w) − τ (x)(w) = τ (x)(w) − τ (x)(w) = 0. So by 20.3.3, condition (i) of 20.3.3 is valid for the representation (N, µ) induced by λ. Thus there exists ψ0 ∈ N such that f (x) = µ(x)(ψ0 ) for all x ∈ g. Set θ0 = φ0 − ψ0 , then λ(x)(θ0 ) = 0 for all x ∈ g. Hence π ◦ θ0 = idV /U and θ0 ◦ τ (x) = σ(x) ◦ θ0 for all x ∈ g. It follows that V is the direct sum of the submodules U and θ0 (V /U ). So σ is semisimple. 20.3.5 Theorem. (Levi-Malcev Theorem) Let g be a Lie algebra and r = rad g. There exists a Lie subalgebra s of g such that g = s ⊕ r. Moreover, s is semisimple and it is called a Levi subalgebra of g. Proof. Observe that such a Lie algebra is isomorphic to g/r, so it is semisimple by 20.1.3. We shall prove by induction on dim r that such a Lie subalgebra exists. The case r = {0} is obvious, so we may assume that r = {0}. 1) Suppose that there is an ideal a of g contained in r such that a = {0} and a = r. Then rad g/a = r/a and by our induction hypothesis, there is a Lie subalgebra b of g such that b ∩ r = a , g = b + r. Since b/a g/r is semisimple, rad b ⊂ a and so rad b = a. By induction, there is a Lie subalgebra s of g such that b = s ⊕ a. It follows from the deﬁnition of b that g = s ⊕ r. 2) We are left with the case where r is commutative. In this case, z(g) = {0} or r since [g, r] is an ideal of g contained in r. When z(g) = r, the adjoint representation of g is semisimple since ad g g/r is semisimple. Thus there is an ideal s of g such that g = s ⊕ r. When z(g) = {0}, g is isomorphic to the Lie subalgebra ad g in End(g). Let (End(g), σ) be the representation of g deﬁned by σ(x)(u) = [ad x, u] for x ∈ g and u ∈ End(g), and M = {u ∈ End(g); u(g) ⊂ r and u|r = λu idr for some λu ∈ k}. Denote by N the kernel of the linear form u → λu on M . Since r is abelian, R = adg r ⊂ N . Also, we have σ(g)(M ) ⊂ N , and σ(g)(R) ⊂ R. A straightforward computation shows that for x ∈ r and u ∈ M , we have σ(x)(u) = −λu ad x. So σ(r)(M ) ⊂ R. It follows that the representation

20.4 Semisimple and nilpotent elements

305

(M/R, θ) of g induced by σ is semisimple since θ(r) = {0} (20.3.4). Moreover, θ(g)(M/R) ⊂ N/R, so there exists v ∈ M such that λv = −1 and for all x ∈ g, σ(x)(v) ∈ R. Since z(g) = {0}, we have a well-deﬁned linear map α : g → r, given by ad α(x) = σ(x)(v). We have α|r = idr because σ(x)(u) = −λu ad x for x ∈ r and u ∈ M . It follows that g = s ⊕ r where s = ker α = {x ∈ g; [ad x, v] = 0} is clearly a Lie subalgebra of g. 20.3.6 Corollary. Let r = rad g and n the nilpotent radical of g. Then n = r ∩ D(g) = [g, r]. Proof. Let s be a Levi subalgebra of g. Then we have D(g) = [s, s] + [s, r] + [r, r] = s + [s, r] + [r, r] = s ⊕ [g, r]. So n = r ∩ D(g) = [g, r] (19.6.2).

20.4 Semisimple and nilpotent elements 20.4.1 Lemma. Let V be a ﬁnite-dimensional vector space and x an endomorphism of V . (i) If x is nilpotent (resp. semisimple), then so is ad x. (ii) Let x = xs + xn be the Jordan decomposition of x. Then ad xs + ad xn is the Jordan decomposition of ad x. Proof. Part (ii) is clear from (i). In view of 19.3.5, we only need to prove (i) when x is semisimple. Let (v1 , . . . , vn ) be a basis of eigenvectors of V for x, then it is easy to check that the endomorphisms zij ∈ End(V ) deﬁned by zij (vi ) = vj , zij (vk ) = 0 if k = i, form a basis of eigenvectors of End(V ) for ad x. 20.4.2 Proposition. Let V be a ﬁnite-dimensional vector space and g a semisimple Lie subalgebra of gl(V ). Then g contains the semisimple and nilpotent components of its elements. Proof. Let V be the set of g-submodules of V , and for W ∈ V, we deﬁne gW = {x ∈ gl(V ); x(W ) ⊂ W, tr(x|W ) = 0}. Since g = [g, g], g ⊂ gW . Let t = ngl(V ) (g) ∩ gW . W ∈V

Let x ∈ t, X = adgl(V ) x, S = adgl(V ) xs and N = adgl(V ) xn . Then the Jordan decomposition of X is S +N (20.4.1). So xs , xn ∈ t because S, N (resp. xs , xn ) are polynomials in X (resp. x) with no constant term. We verify easily that t is a Lie subalgebra of gl(V ) and g is an ideal of t. So t = g × a for some ideal a of t (20.1.4). Moreover [a, g] ⊂ a ∩ g = {0}. Take W to be a minimal element of V \ {0}. Then W is simple. For a ∈ a, tr(a|W ) = 0, so Schur’s Lemma says that a|W = 0. Since V is the sum of minimal elements of V \ {0}, we deduce that a = {0}. Hence t = g and xs , xn ∈ g.

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20.4.3 Corollary. With the hypotheses of 20.4.2, an element x ∈ g is semisimple (resp. nilpotent) if and only if adg x is semisimple (resp. nilpotent). Proof. Since (g, ad) is a g-submodule of (g, adgl(V ) ), 20.4.1 says that if x is semisimple (resp. nilpotent), then so is adg x. Conversely, if adg x is semisimple (resp. nilpotent), then since the adjoint representation is faithful, it follows by 20.4.2 that x = xs (resp. x = xn ). 20.4.4 Theorem. Let g be semisimple and x ∈ g. Then the following conditions are equivalent: (i) adg x is semisimple (resp. nilpotent). (ii) There is a ﬁnite-dimensional faithful representation (V, σ) of g such that σ(x) is semisimple (resp. nilpotent). (iii) For all ﬁnite-dimensional representation (V, σ) of g, σ(x) is semisimple (resp. nilpotent). Proof. We have (iii) ⇒ (ii) since the adjoint representation is faithful. Also (ii) ⇒ (i) follows from 20.4.3. Let us prove (i) ⇒ (iii). Let (V, σ) be a ﬁnite-dimensional representation of g, h = g/ ker σ and (V, π) the representation of h obtained from σ. Let y be the image of x in h. If adg x is semisimple (resp. nilpotent), then so is adh y. Thus π(y) = σ(x) is semisimple (resp. nilpotent) by 20.4.3. 20.4.5 Let g be semisimple. An element x ∈ g is called semisimple (resp. nilpotent) if it satisﬁes the conditions of 20.4.4. In view of 20.4.2, an element x ∈ g can be written uniquely in the form s + n where s is semisimple, n is nilpotent and [s, n] = 0. We say that s (resp. n) is the semisimple component (resp. nilpotent component) of x. By 20.4.2, we have the following result: Corollary. Let h be a semisimple Lie subalgebra of a semisimple Lie algebra g, x ∈ h and s, n the semisimple and nilpotent components of x in g. Then s, n ∈ h. 20.4.6 Proposition. Let g be semisimple. Then g is generated, as a Lie algebra, by its nilpotent elements. Proof. Let h be a Cartan subalgebra of g and R the root system of g relative to h. By 19.8.6 and 19.8.7, h is abelian and [h, gα ] ⊂ gα for all α ∈ R. Since g = h ⊕ ( α∈R gα ), we have: g = [g, g] ⊂

α∈R

gα +

[gα , gβ ].

α,β∈R

But all the elements of gα are nilpotent (19.8.2). So the result follows.

20.5 Reductive Lie algebras

307

20.5 Reductive Lie algebras 20.5.1 Deﬁnition. (i) A Lie algebra g is said to be reductive if its adjoint representation is semisimple. (ii) A Lie subalgebra h of g is reductive in g if the adjoint representation of h in g is semisimple. 20.5.2 Remark. If h is reductive in g, then h is reductive since h is a h-submodule of g. 20.5.3 Lemma. An abelian Lie algebra h has a ﬁnite-dimensional representation such that the associated bilinear form is non-degenerate. Proof. Let n = dim h, then the representation on kn deﬁned by identifying h with the set of diagonal matrices works. 20.5.4 Proposition. Let r = rad g and n the nilpotent radical of g. The following conditions are equivalent: (i) g is reductive. (ii) g = s × a where s is semisimple and a is abelian. (iii) g has a ﬁnite-dimensional representation such that the associated bilinear form is non-degenerate. (iv) g has a faithful semisimple representation of ﬁnite dimension. (v) n = {0}. (vi) r is the centre of g. Proof. (i) ⇒ (ii) If g is reductive, then g is the direct sum of its minimal (non-trivial) ideals a1 , . . . , an . It follows that ai is simple or dim ai = 1 for all i. So we have (ii). (ii) ⇒ (iii) By 20.5.3, a has a ﬁnite-dimensional representation (V, σ) such that the associated bilinear form on a is non-degenerate. Also the bilinear form on s associated to the adjoint representation (s, ads ) of s is non-degenerate. So the bilinear form on g associated to the representation (s ⊕ V, ads ⊕σ) of g is non-degenerate. (iii) ⇒ (iv) In the notations of 19.5.5 (ii), take the direct sum of the σi ’s. (iv) ⇒ (v) This is clear by the deﬁnition of n and 19.5.1. (v) ⇒ (vi) Since z(g) ⊂ r and n = [g, r] (20.3.6), the result is clear. (vi) ⇒ (i) If r = z(g), then ad g is a semisimple Lie subalgebra of gl(g) (20.3.5). So the adjoint representation is semisimple (20.3.4). 20.5.5 Corollary. A Lie algebra g is reductive if and only if D(g) is semisimple. In particular, if g is reductive, then g D(g) × z(g). Proof. If D(g) is semisimple, then g = D(g) × a for some ideal a of g (20.1.4). Since [a, g] ⊂ a ∩ D(g) = {0}, a = z(g). So the result follows from 20.5.4.

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Remark. By 20.2.2, gl(V ) is a reductive Lie algebra. 20.5.6 Let (V, σ) and (W, π) be representations of g. We deﬁne the representation (V ⊗k W, σ ⊗ π) of g by setting for x ∈ g: (σ ⊗ π)(x) = σ(x) ⊗ idW + idV ⊗π(x). Corollary. If σ and π are semisimple of ﬁnite dimension, then σ ⊗ π is semisimple. Proof. We may assume that σ and π are simple and since the nilpotent radical n of g is contained in ker σ ∩ ker π (19.5.1), replacing g by g/n, we may assume further that n = {0}, so g is reductive (20.5.4). Let g = s×a be as in 20.5.4 (ii). By Schur’s Lemma, σ(a) ⊂ k idV and π(a) ⊂ k idW . So σ⊗π(a) ⊂ k idV ⊗ idW . Since σ ⊗ π|s is semisimple (20.3.4), the result follows. 20.5.7 Proposition. Assume that g is reductive. Let x ∈ D(g), y ∈ z(g) and z = x + y. The following conditions are equivalent: (i) x is semisimple in D(g). (ii) There exists a faithful semisimple ﬁnite-dimensional representation (V, σ) of g such that σ(z) is semisimple. (iii) For all semisimple ﬁnite-dimensional representation (V, σ) of g, σ(z) is semisimple. Proof. By Schur’s Lemma, σ(y) is a scalar multiple of idV for any simple ﬁnite-dimensional representation (V, σ) of g. In particular, σ(y) is semisimple. Since y ∈ z(g), we may assume that y = 0. Thus we have (i) ⇒ (iii) by 20.4.4, (iii) ⇒ (ii) by 20.5.4 (iv), and ﬁnally (ii) ⇒ (i) by 20.4.4. 20.5.8 Proposition. With the notations of 20.5.7, the following conditions are equivalent: (i) x is nilpotent in D(g) and y = 0. (ii) There exists a faithful semisimple ﬁnite-dimensional representation (V, σ) of g such that σ(z) is nilpotent. (iii) For all semisimple ﬁnite-dimensional representation (V, σ) of g, σ(z) is nilpotent. Proof. We have (i) ⇒ (iii) by 20.4.4 and (iii) ⇒ (ii) by 20.5.4 (iv). So let us prove (ii) ⇒ (i). Let (V, σ) be as in (ii). Let s, n be the semisimple and nilpotent components of x in D(g). By 20.4.4 and 20.5.7, σ(n) is nilpotent and σ(s + y) is semisimple. Furthermore, they commute and their sum σ(z) is nilpotent. So σ(s + y) = 0. Since σ is faithful and D(g) ∩ z(g) = {0}, we deduce that y = s = 0, and hence x = n is nilpotent.

20.5 Reductive Lie algebras

309

20.5.9 Assume that g is reductive. An element z ∈ g is called semisimple (resp. nilpotent) if it veriﬁes the conditions of 20.5.7 (resp. 20.5.8). We deﬁne as in 20.4.5 the semisimple and nilpotent components of z. We shall denote by Sg (resp. Ng ) the set of semisimple (resp. nilpotent) elements of g. 20.5.10 Theorem. Let r = rad g, (V, σ) a ﬁnite-dimensional representation of g, g = σ(g) and r = σ(r). The following conditions are equivalent: (i) σ is semisimple. (ii) g is reductive and the elements of its centre are semisimple. (iii) The elements of r are semisimple. (iv) The restriction of σ to r is semisimple. Proof. (i) ⇒ (ii) This is clear from 20.5.4 (iv) and 20.5.7. (ii) ⇒ (iii) This is obvious. (iii) ⇒ (iv) If (iii) is veriﬁed, then by 20.3.6, [g , r ] = {0}. So r is abelian, and since its elements are semisimple, (iv) follows. (iv) ⇒ (i) Using the notations of 19.3.9, we have V = λ∈r∗ Vλ (r) by 19.4.5 (ii). Since [g , r ] = {0} (20.3.6), each Vλ (r) is a g-module. So (i) follows from 20.3.4. 20.5.11 Lemma. We have V = V g ⊕ σ(g)(V ) for all semisimple representation (V, σ) of g. Proof. Clearly, we may assume that V is simple. Since σ(g)(V ) is a submodule of V , the result follows. 20.5.12 Lemma. Let t be a Lie subalgebra of a semisimple Lie algebra g verifying: a) L|t×t is non-degenerate. b) If x ∈ t, then its semisimple and nilpotent components are in t. Then t is reductive in g. Proof. Condition a) and 20.5.4 (iii) imply that t is reductive. Let x ∈ z(t) and s, n the semisimple and nilpotent components of x. Then condition b) and 10.1.2 imply that s, n ∈ z(t). It follows that for all y ∈ t, (ad n) ◦ (ad y) is nilpotent (since [y, n] = 0) and hence L(n, y) = 0. We deduce from condition a) that n = 0 and so all the elements of z(t) are semisimple. The result now follows from 20.5.10 (ii). 20.5.13 Proposition. Assume that g is semisimple. Let t be a Lie subalgebra of g which is reductive in g, and c = cg (t). Then: (i) c veriﬁes the conditions of 20.5.12, and so it is reductive in g. (ii) g = c ⊕ [t, g] and [t, g] is the orthogonal of c with respect to L. Proof. By 10.1.2 and 20.4.5, c contains the semisimple and nilpotent components of its elements. For (x, y, z) ∈ t × c × g, we have:

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20 Semisimple and reductive Lie algebras

L([x, z], y) = L(z, [y, x]) = L(z, 0) = 0. Thus c and [t, g] are orthogonal. So by 20.5.11, we have (ii). Finally since L is non-degenerate, it is non-degenerate on c × c. So (i) follows. 20.5.14 Proposition. Assume that g is semisimple. Let (s, n) ∈ Sg × Ng be such that [s, n] = 0. Then n ∈ [gs , gs ]. Proof. By 20.5.13, gs is reductive in g. So the result follows by applying 20.5.8 to the adjoint representation of gs in g. 20.5.15 Deﬁnition. Let g be semisimple. A Lie subalgebra of g is called a torus if all its elements are semisimple. 20.5.16 Lemma. Assume that g is semisimple. Let t be a torus of g. (i) t is abelian and reductive in g. (ii) If t is a maximal torus, then cg (t) = t. Proof. (i) Since rad t consists of semisimple elements, t is reductive (20.5.10) and so D(t) is semisimple. Thus D(t) = {0} (20.4.6), and t is abelian. (ii) By 20.5.13, c = cg (t) is reductive in g, so c is the direct product of its centre z and D(c). Since t is maximal, z = t and D(c) contains no non-zero semisimple elements. Hence D(c) = {0}.

20.6 Results on the structure of semisimple Lie algebras 20.6.1 In this section, g is a semisimple Lie algebra and h is a Cartan subalgebra of g. We set R = R(g, h) and gλ = gλ (h). If x, y ∈ g, we shall write x, y for L(x, y). Let λ ∈ h∗ . By 19.8.7 (iii), there exists a unique hλ ∈ h such that λ(h) = hλ , h for all h ∈ h. The map h∗ → h, λ → hλ , is a linear isomorphism. For λ, µ ∈ h∗ , deﬁne λ, µ = hλ , hµ . Thus we obtain a non-degenerate symmetric bilinear form on h∗ . Note that we have: λ, µ = hλ , hµ = λ(hµ ) = µ(hλ ). It follows also from 19.8.7 that hα , α ∈ R, span h∗ . 20.6.2 Proposition. Let t be a Lie subalgebra of g. Then t is a Cartan subalgebra of g if and only if t is maximal torus of g. Proof. Suppose that t is a Cartan subalgebra. Let h ∈ t and h = s + n its Jordan decomposition. By 10.1.2, s, n ∈ ng (t) = t. So 19.8.7 (iii) and 20.5.12 imply that t is reductive in g. Hence t is abelian. By 20.5.10 (ii), t is a torus of g and it is maximal since ng (t) = t. Conversely, suppose that t is a maximal torus in g. By 20.5.16 (i), t is abelian and there is a subspace m of g such that g = t ⊕ m and [t, m] ⊂ m. It follows that ng (t) = cg (t) = t by 20.5.16 (ii).

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311

20.6.3 Corollary. (i) Any generic element of g is semisimple, and any semisimple element is contained in a Cartan subalgebra of g. So Sg is the union of Cartan subalgebras of g. (ii) If h ∈ h, λ ∈ h∗ and x ∈ gλ , then [h, x] = λ(h)x. Proof. It is immediate from 19.8.5 and 20.6.2. 20.6.4 Proposition. Let α ∈ R. (i) If x ∈ gα , y ∈ g−α and h ∈ h, then: h, [x, y] = α(h)x, y = h, hα x, y , [x, y] = x, yhα . (ii) The vector space [gα , g−α ] is khα . (iii) We have α(hα ) = 0. Proof. (i) By 20.6.3, [h, x] = α(h)x, so h, [x, y] = [h, x], y = α(h)x, y = h, hα x, y = h, x, yhα . Since [x, y] ∈ g0 = h and L|h×h is non-degenerate, we have [x, y] = x, yhα . (ii) This follows from (i) since gα , g−α = {0} (19.8.7). (iii) Let (x, y) ∈ gα × g−α be such that x, y = 1, so [x, y] = hα . If α(hα ) = 0, then a = khα + kx + ky is a nilpotent Lie algebra. Applying 19.4.4 to the adjoint representation of a in g, we see that if z ∈ [a, a], then adg z is nilpotent. Thus, hα is nilpotent. By 20.6.2, we obtain that hα = 0. Contradiction. 20.6.5 In view of 20.6.4 (iii), for α ∈ R, we may deﬁne: Hα = Then: α(Hα ) = 2 , Hα , Hα =

2hα . hα , hα

4 , Hα = Hα , Hα hα . hα , hα 2

Let Xα ∈ gα \ {0}. There exists X−α ∈ g−α verifying Xα , X−α = 0. By 20.6.4, [Xα , X−α ] = Xα , X−α hα . Choose X−α so that [Xα , X−α ] = Hα . Then we obtain: [Hα , Xα ] = 2Xα , [Hα , X−α ] = −2X−α , [Xα , X−α ] = Hα . So the Lie algebra sα = kHα + kXα + kX−α is isomorphic to sl2 (k). 20.6.6 Proposition. (i) For α, β ∈ R, we have dim gα = 1 and β(Hα ) is an integer. (ii) If x, y ∈ h, then x, y = α∈R α(x)α(y).

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Proof. (i) Suppose that dim gα > 1. If y ∈ g−α \ {0}, there exists Xα ∈ gα \ {0} such that Xα , y = 0. Let X−α and sα be as in 20.6.5. By 20.6.4, [Xα , y] = Xα , yhα = 0. So it follows from 19.2.5 and 19.2.7 that y is a linear combination of eigenvectors of ad Hα whose eigenvalues are in N. But [Hα , y] = −α(Hα )y = −2y. Contradiction. Hence dim gα = 1. By 20.6.3, [Hα , x] = β(Hα )x for α, β ∈ R, x ∈ gβ . So β(Hα ) ∈ Z (19.2.7 and 20.6.5). (ii) This is clear from part (i) and 19.8.6. 20.6.7 For α, β ∈ R, set aβα = β(Hα ). The integers aβα are called the Cartan integers of (g, h). We have aαα = 2 and: aβα = hβ , Hα = 2

β, α Hβ , Hα hβ , hα =2 =2 . hα , hα α, α Hα , Hα

By 20.6.6, Hα , Hβ ∈ Z. It follows that the ratio between hα and Hα is rational, and that hα , hβ ∈ Q. 20.6.8 Theorem. Let α, β ∈ R. (i) There exist p, q ∈ N such that [−p, q] is the set of integers t such that β + tα ∈ R ∪ {0}. Moreover, we have aβα = p − q. (ii) We have β − aβα α ∈ R. (iii) If β − α ∈ R ∪ {0}, then aβα 0, p = 0 and q = −aβα . (iv) The only elements of R proportional to α are α and −α. (v) If α + β ∈ R, then [gα , gβ ] = gα+β . Proof. (i) Set u = t∈Z gβ+tα , s = gα + g−α + kHα sl2 (k). Since [s, u] ⊂ u, u is a s-module. The set of eigenvalues for Hα in u is {(β +tα)(Hα ) = aβα +2t, t ∈ Z and β + tα ∈ R ∪ {0}}. But by 19.2.5, 19.2.6 and 19.2.7, it is also of the form {−r, −r + 2, . . . , r − 2, r} for some r ∈ N. Hence the set of t ∈ Z such that β + tα ∈ R ∪ {0} is [−p, q] for some p, q ∈ Z. Since β ∈ R, 0 ∈ [−p, q], so p, q ∈ N. Finally, p − q = aβα since r = 2q + aβα and −r = −2p + aβα . (ii) By (i), −p q − p = −aβα q, so β − aβα α ∈ R ∪ {0}. If β = aβα α, then by 20.6.7, we have: aβα = 2

α, α β, α = 2aβα = 2aβα . α, α α, α

So aβα = 0, which implies that β = 0. Contradiction. (iii) This is clear from (i). (iv) Let λ ∈ k be such that γ = λα ∈ R. Then γ(Hα ) = 2λ ∈ Z. Exchanging the roles of γ and α, we deduce that 2λ−1 ∈ Z. Thus λ belongs to {−2, −1, −1/2, 1/2, 1, 2}. Since R = −R, we need only to prove that if α ∈ R, then 2α ∈ R. Suppose that 2α ∈ R and y ∈ g2α \ {0}. Then from the preceding discussion, 3α ∈ R and so [Xα , y] = 0. Since [X−α , y] ∈ gα = kXα , we have

20.7 Subalgebras of semisimple Lie algebras

313

[Xα , [X−α , y]] = 0 and 4y = [Hα , y] = [[Xα , X−α ], y] = 0. Thus y = 0. Contradiction. (v) If α + β ∈ R, then q 1. So [Xα , gβ ] = 0 (19.2.5, 19.2.6 and 19.2.7). Since dim gα+β = 1 (20.6.6), we are done. 20.6.9 For α ∈ R, let sα ∈ End(h∗ ) be deﬁned by: sα (λ) = λ − λ(Hα )α = λ − 2

λ, α α. α, α

Thus: s2α = idh∗ , sα (λ), sα (µ) = λ, µ. If β ∈ R, then sα (β) = β − aβα α ∈ R. It follows that sα (R) = R. The subgroup W (g, h) of GL(h∗ ) generated by the sα ’s, α ∈ R, is called the Weyl group relative to (g, h). Since w(R) = R for all w ∈ W (g, h) and R spans h∗ (19.8.7), W (g, h) is a ﬁnite group. 20.6.10 From the discussion in the preceding paragraphs, we deduce that R is a reduced root system in h∗ (18.2.1). Moreover, the set of Hα ’s form a dual root system of R in h. Let QHα , h∗Q = Qα , hR = R ⊗Q hQ , h∗R = R ⊗Q h∗Q hQ = α∈R

α∈R

Then we may identify h with k⊗Q hQ , and h∗ with k⊗Q h∗Q . The Weyl chambers are deﬁned in hR and h∗R . The group W (g, h) acts on h∗ and h∗Q . It acts also on h et hQ (18.2.9), so on hR and h∗R . The form ., . takes on rational values on hQ and h∗Q , and deﬁnes a nondegenerate positive-deﬁnite form on them. It can be extended to an inner product on hR and h∗R . Finally these forms are all W (g, h)-invariant.

20.7 Subalgebras of semisimple Lie algebras 20.7.1 In this section, g is a semisimple Lie algebra, h is a Cartan subalgebra of g. We denote R(g, h) by R and W (g, h) by W . For α ∈ R, we denote hα = kHα = [gα , g−α ]. For a subset P of R, we set: α gP = g , hP = hα . α∈P

α∈P

If P, Q ⊂ R, then it is easy to check that: (1)

[h, gP ] = gP , [gP , gQ ] = hP ∩(−Q) + g(P +Q)∩R . We deduce immediately the following result from these equalities.

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Lemma. Let h be a subspace of h and P ⊂ R. The following conditions are equivalent: (i) h + gP is a Lie algebra of g. (ii) P is closed subset of R and hP ∩(−P ) ⊂ h . 20.7.2 Proposition. The following conditions are equivalent for a Lie subalgebra a of g: (i) [h, a] ⊂ a. (ii) There exist a closed subset P ⊂ R and a subspace h of h such that hP ∩(−P ) ⊂ h and a = h + gP . Proof. The implication (ii) ⇒ (i) is straightforward. Suppose that (i) is veriﬁed. Let P = {α ∈ R; gα ⊂ a} and h = a ∩ h. By 19.3.9, we have a = h + α∈P gα . So (ii) follows from 20.7.1. 20.7.3 Lemma. Let a be a Lie subalgebra of g such that [h, a] ⊂ a, h the subspace of h and P ⊂ R such that a = h + gP . Let h be a subspace of h and Q ⊂ R verifying h ⊂ h and Q ⊂ P . Then the following conditions are equivalent: (i) h + gQ is an ideal of a. (ii) We have (P + Q) ∩ R ⊂ Q and hP ∩(−Q) ⊂ h ⊂ α∈P \Q ker α. Proof. We have [h + gP , h + gQ ] = hP ∩(−Q) + [h , gQ ] + [h , gP ] + g(P +Q)∩R . Thus h + gQ is an ideal of a if and only if hP ∩(−Q) ⊂ h , [h , gP ] ⊂ gQ , g(P +Q)∩R ⊂ gQ . So the result is clear. 20.7.4 Proposition. Let a be a Lie subalgebra of g such that [h, a] ⊂ a, h the subspace of h and P ⊂ R such that a = h + gP . Set: Q = {α ∈ P ; −α ∈ / P } , t = {x ∈ h ; α(x) = 0, α ∈ P ∩ (−P )}. (i) The radical of a is t + gQ . (ii) gQ is a nilpotent ideal of a. Proof. Let α ∈ P , β ∈ Q be such that α + β ∈ R. Since P is closed, α + β ∈ P . However, α + β ∈ −P , for otherwise −β = −(α + β) + α ∈ P , this contradicts the fact that β ∈ Q. So α + β ∈ Q, and (P + Q) ∩ R ⊂ Q. Since P ∩ (−Q) = ∅ (so Q ∩ (−Q) = ∅), it follows from 18.10.2 and 20.7.3 that gQ is a nilpotent ideal of a. We have, from the deﬁnition of t, that [t, gP ] ⊂ gQ . So t + gQ is an ideal of a and it is solvable since gQ is nilpotent. Let r = rad a. Then by 19.6.3, [h, r] ⊂ r, and hence r = h + gS where h ⊂ h and S ⊂ P . To prove (i), it suﬃces to prove that S ⊂ Q and h ⊂ t. If α ∈ S is such that −α ∈ P , then hα = [gα , g−α ] ⊂ r. This implies in turn that g−α = [hα , g−α ] ⊂ r. So r contains a Lie subalgebra isomorphic to sl2 (k). This contradicts the fact that r is solvable. Hence S ⊂ Q. Next, if x ∈ h and α ∈ P ∩ (−P ), then [x, gα ] ⊂ gα ∩ r = {0}. So α(x) = 0 and h ⊂ t.

20.7 Subalgebras of semisimple Lie algebras

315

20.7.5 Proposition. Let a be a Lie subalgebra of g such that [h, a] ⊂ a, h the subspace of h and P ⊂ R such that a = h + gP . (i) All the elements of a are nilpotent if and only if h = {0}. If this is the case, then P ∩ (−P ) = ∅ and a is nilpotent. (ii) a is solvable if and only if P ∩ (−P ) = ∅. If this is the case, then [a, a] = gS where S = ((P + P ) ∩ R) ∪ {α ∈ P ; α(h ) = {0}}. (iii) a is reductive in g if and only if P = −P . (iv) a is semisimple if and only if P = −P and h = hP . If this is the case, hP is a Cartan subalgebra of a and the root system of (a, hP ) is {α|hP ; α ∈ P }. Proof. (i) If all the elements of a are nilpotent, then a is nilpotent (19.3.7) and h = {0} (20.6.2). Conversely, if h = {0}, then P ∩ (−P ) = ∅ (20.7.1). It follows from 18.10.2 that all the elements of gP are nilpotent. (ii) If P ∩(−P ) = ∅, then gP is a Lie subalgebra (20.7.1) which is nilpotent by (i). Since [a, a] = [h , gP ] + g(P +P )∩R ⊂ gP , we see that a is solvable and we have the description [a, a] = gS as required. Conversely, if α ∈ P ∩ (−P ), then a contains the Lie subalgebra hα + gα + g−α which is isomorphic to sl2 (k). So a is not solvable. (iii) In the notations of 20.7.4, since rad a = t + gQ , we have from 20.7.4 (i) that a is reductive in g ⇔ adg x is semisimple for all x ∈ t + gQ ⇔ Q = ∅ ⇔ P = −P. (iv) If a is semisimple, P = −P by (iii). Since a = [a, a] = hP + gP , we have h = hP . Conversely, if P = −P and h = hP , then a is reductive by (iii) and its radical is {0} by 20.7.4. So a is semisimple. Finally, suppose that a is semisimple. Since hP is ad h-stable, so is n = na (hP ). It follows that n = hP + gS where S ⊂ P . For α ∈ S, we have gα = [hα , gα ] ⊂ [hP , gα ] ⊂ hP which is absurd. So S = ∅ and hP = n is a Cartan subalgebra of a. The remaining statements now follow. 20.7.6 Proposition. Let a be a Lie subalgebra of g containing h. (i) We have a = ng (a). (ii) a contains the semisimple and nilpotent components of its elements. Proof. (i) Write a = h + gP with P ⊂ R. Then ng (a) = h + gQ where Q ⊃ P . Since gQ = [h, gQ ] ⊂ a, we have Q = P . (ii) Let x ∈ a, and s, n its semisimple and nilpotent components. Then [s, a] ⊂ a and [n, a] ⊂ a (10.1.2). Hence s, n ∈ a by (i). 20.7.7 Proposition. Let P be a closed subset of R and a = h + gP . (i) a is solvable if and only if P ∩ (−P ) = ∅. If this is the case, then [a, a] = gP . (ii) a is reductive if and only if P = −P .

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Proof. Part (i) follows from 20.7.5 (ii) and if P = −P , then a is reductive by 20.7.5 (iii). Finally, if a is reductive, then [a, a] = h + gP , where h ⊂ h, is semisimple. So P = −P by 20.7.5 (iv).

20.8 Parabolic subalgebras 20.8.1 Let us conserve the notations of 20.7.1. Let B be a base of R and denote by R+ (resp. R− ) the corresponding set of positive (resp. negative) roots. Set: α α n+ = g , n− = g , b+ = h ⊕ n+ , b− = h ⊕ n− . α∈R+

α∈R−

By 20.7.5, n+ and n− (resp. b+ and b− ) are nilpotent (resp. solvable) Lie subalgebras of g. The decomposition: (2)

g = n− ⊕ h ⊕ n+

is called the triangular decomposition of g relative to B. The Lie subalgebra b+ is called the Borel subalgebra of g relative to h and B. In general, a Lie subalgebra b of g is called a Borel subalgebra of (g, h) if there exists a base B of R such that b is the Borel subalgebra of g relative to h and B . 20.8.2 Proposition. (i) The Lie algebra n+ (resp. n− ) is generated by gα , α ∈ B (resp. α ∈ −B). (ii) The Lie algebras b+ and b− are maximal solvable Lie subalgebras of g; moreover, ng (b± ) = b± and z(b± ) = {0}. Proof. (i) Let α ∈ R+ . By 18.7.7, there exist α1 , . . . , αn ∈ B such that α = α1 + · · · + αn and α1 + · · · + αi ∈ R for 1 i n. So the result follows from 20.6.8 (v). (ii) We have ng (b± ) = b± by 20.7.6. If t is a Lie subalgebra of g containing b+ , then it contains h. So by 20.7.2, t = h + gP for some closed subset P of R containing R+ . If P contains a negative root α, then t contains the Lie subalgebra gα + g−α + hα sl2 (k), so t is not solvable. Hence b+ is a maximal solvable Lie subalgebra. The same argument applies for b− . Finally, if X = H + α∈R+ λα Xα ∈ z(b+ ), then [h, X] = {0} implies that X = H. On the other hand [X, n+ ] = {0}, so X = 0. Again the same argument applies for b− . 20.8.3 Proposition. Let l = dim h. (i) For (h, x) ∈ h × n+ , the characteristic polynomial of adg (h + x) is: Xl X − α(h) . α∈R

(ii) The largest nilpotent ideal, the nilpotent radical and the set of nilpotent elements of b+ are all equal to n+ . Furthermore, we have:

20.8 Parabolic subalgebras

[n+ , n+ ] =

317

gα .

α∈R+ \B ∗ α Proof. (i) Let us ﬁx a lexicographic order on hβR deﬁning B. For y ∈ g , we have: [h + x, y] = α(h)y + z with z ∈ β>α g . So in a suitable basis, the matrix of ad(h + x) is triangular with diagonal entries 0 (l times) and α(h), α ∈ R. (ii) This is clear by (i), 19.6.2, 18.7.7 and 20.6.8 (v).

20.8.4 Proposition. Let b = h + gP be a Lie subalgebra of g containing h. Then the following conditions are equivalent: (i) b is a Borel subalgebra of (g, h). (ii) There exists a chamber C of R such that P = R+ (C). (iii) R is the disjoint union of −P and P . Proof. The subset P is closed by 20.7.1. So the equivalence (ii) ⇔ (iii) follows from 18.10.3. If b is a Borel subalgebra of (g, h), then P = R+ , and so (i) ⇒ (iii) ⇒ (ii). Finally if (ii) is veriﬁed, then it is clear from the deﬁnition of a Borel subalgebra that b is a Borel subalgebra of (g, h). Hence (ii) ⇒ (i). 20.8.5 Proposition. Let p = h + gP be a Lie subalgebra of g containing h. Then the following conditions are equivalent: (i) p contains a Borel subalgebra of (g, h). (ii) There exists a chamber C of R such that P ⊃ R+ (C). (iii) P is parabolic. If these conditions are veriﬁed, we say that p is a parabolic subalgebra of (g, h). Proof. We have (i) ⇔ (ii) by 20.8.4 and (ii) ⇔ (iii) by 18.10.5. 20.8.6 Theorem. Let p = h + gP be a parabolic subalgebra of (g, h), Q = P \ (−P ) and s = h + gP ∩(−P ) . Then (i) p = s ⊕ gQ . (ii) s is reductive in g. (iii) gQ is the largest nilpotent ideal and the nilpotent radical of p. It is also the orthogonal of p with respect to the Killing form of g. (iv) z(p) = {0}. (v) p = ng (p). (vi) rad p = t ⊕ gQ , where t = {x ∈ h; α(x) = 0 for all α ∈ P ∩ (−P )}. Proof. Part (i) is clear and (ii), (v), (vi) follow from 20.7.5 (iii), 20.7.6 and 20.7.4 respectively. (iii) By 20.7.4 (ii), gQ is a nilpotent ideal of p. If n is the largest nilpotent ideal of p, then gQ ⊂ n ⊂ h + gQ (20.7.4). If x ∈ h ∩ n, then adp x is nilpotent (19.5.8), so α(x) = 0 for all α ∈ P , and x = 0 since P is parabolic. Thus n = gQ . Since [h, gQ ] = gQ , the nilpotent radical of p contains gQ (20.3.6),

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and so it is equal to gQ . Finally, part (i) and 19.8.6 imply that gQ is the orthogonal of p with respect to the Killing form of g. (iv) Let z = x + y ∈ z(p) where x ∈ h, y ∈ α∈P gα . Since [z, h] = {0}, y = 0. Now [z, gα ] = {0} for all α ∈ P . But P is parabolic, so z = 0. 20.8.7 Let h ∈ h \ {0}, e, f ∈ g verifying the relations of 19.2.4. If p ∈ Z, denote by gp the ad h-eigenspace of eigenvalue p. By 19.2.5, 19.2.6 and 19.2.7, we have: g = p∈Z gp . Set: p= g p , s = g0 , n = gp . p0

p>0

Then h ⊂ s and p = h ⊕ gP where P = {α ∈ R; α(h) 0}. It is easy to check that P contains a base of R. So p is a parabolic subalgebra of (g, h) whose nilpotent radical is n and s = h ⊕ gP ∩(−P ) is reductive in g. 20.8.8 Let B be a base of R and b the Borel subalgebra of g relative to B. For J ⊂ B, denote by Q(J) the set of roots which can be written as a sum of elements of −J. Set: P (J) = R+ ∪ Q(J) , pJ = h ⊕ gP (J) . It follows from 18.10.5 that pJ is a parabolic subalgebra of g containing b, and any parabolic subalgebra of g containing b is obtained in this way. It is clear that if J, J ⊂ B, then pJ ⊂ pJ if and only if J ⊂ J , and pJ = pJ if J = J . Thus the parabolic subalgebras of g containing b, distinct from g, and maximal with these properties are of the form pK , where K = B \ {β} for some β ∈ B.

References • [14], [29], [38], [39], [42], [43], [80].

21 Algebraic groups

In this chapter, we present some general properties of algebraic groups. All the varieties considered in this chapter are algebraic varieties over k.

21.1 Generalities 21.1.1 Let G be a group. Denote by µG : G × G → G, (α, β) → αβ, the group multiplication and ιG : G → G, α → α−1 , the inverse. As in 10.2.1, we set, for U, V ⊂ G, U V = {αβ; α ∈ U, β ∈ V } and U −1 = {α−1 ; α ∈ U }. Deﬁnition. Let G be a group endowed with a structure of algebraic variety. If the group multiplication µG and the inverse ιG are morphisms of algebraic varieties, then we say that G is an algebraic group. Furthermore, if G is an aﬃne algebraic variety, we say that G is an aﬃne algebraic group. Remark. Note that the topology on G × G here is the Zariski topology, and not the usual product topology. So an algebraic group is not a topological group. 21.1.2 Let G, G be algebraic groups. If α ∈ G, then translations β → αβ, β → βα, and conjugation, β → αβα−1 , are isomorphisms of the algebraic variety G. A map G → G is a morphism of algebraic groups if it is a homomorphism of groups and a morphism of algebraic varieties. We deﬁne in an obvious way the notion of isomorphisms and automorphisms of algebraic groups. 21.1.3 Example. 1) The additive group Ga = k and the multiplicative group Gm = k \ {0} are algebraic groups. 2) For n ∈ N∗ , GLn (k) is an algebraic group (11.7.5 and 11.8.9). 3) Any closed subgroup of an algebraic group is an algebraic group. Thus SLn (k) and the subgroup of GLn (k) consisting of diagonal (resp. upper triangular, lower triangular) matrices are algebraic groups. 4) Let n ∈ N∗ . Denote by Jn = [aij ] ∈ GLn (k) the matrix where aij = 1 if i + j = n + 1 and aij = 0 otherwise. Set:

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21 Algebraic groups

# 0 Jn 0 Jn A= . −Jn 0 −Jn 0 ⎧ ⎛ ⎞ ⎞⎫ ⎛ 1 0 0 ⎬ 1 0 0 ⎨ O2n+1 (k) = A ∈ GL2n+1 (k) ; t A ⎝ 0 0 Jn ⎠ A = ⎝ 0 0 Jn ⎠ . ⎭ ⎩ 0 Jn 0 0 Jn 0 # 0 Jn 0 Jn O2n (k) = A ∈ GL2n (k) ; t A A= . Jn 0 Jn 0

Sp2n (k) =

A ∈ GL2n (k) ; t A

Since these subgroups are closed in the corresponding GLp (k), they are algebraic groups. 5) Let A be a k-algebra of ﬁnite dimension (not necessarily associative). Then the group of automorphisms of A, denoted by Aut A, is an algebraic group. 6) If G, G are algebraic groups, then their direct product G × G endowed with the Zariski product topology is an algebraic group. 21.1.4 Lemma. Let G be an algebraic group. Then there is a unique irreducible component of G which contains the identity element e. Proof. Let X1 , X2 be two irreducible components of G which contain e, and X = X1 X2 = µ(X1 × X2 ). Since the Xi ’s are irreducible, so is X1 × X2 , and hence X is irreducible. Thus X ⊂ X1 or X ⊂ X2 . On the other hand e ∈ Xi , i = 1, 2, it follows that Xi ⊂ X, i = 1, 2. Hence X = X1 = X2 and we are done. 21.1.5 We shall denote by G◦ the unique irreducible component of G containing e, and we shall call G◦ the identity component of G. 21.1.6 Theorem. Let G be an algebraic group. (i) The identity component G◦ is a closed normal subgroup of G of ﬁnite index. The cosets of G modulo G◦ are at the same time the irreducible components and the connected components of G. In particular, G is a pure variety. (ii) If H is a closed subgroup of ﬁnite index in G, then H contains G◦ . Proof. Since µG (G◦ × G◦ ) = G◦ G◦ and ιG (G◦ ) = (G◦ )−1 are irreducible subsets of G containing e, they are contained in G◦ (21.1.4). So G◦ is a subgroup of G. Similarly, for α ∈ G, αG◦ α−1 is an irreducible subset of G◦ containing e, so αG◦ α−1 ⊂ G◦ and G◦ is normal. Since translations are isomorphisms of the algebraic variety G (21.1.2), the coset αG◦ is the unique irreducible component of G containing α. So part (i) follows from the fact that G is the disjoint union of the cosets of G modulo G◦ . (ii) Let H be such a subgroup. Let H1 , . . . , Hn be the left cosets of G modulo H having non-empty intersection with G◦ . Then G◦ is the disjoint union of the closed subsets H1 ∩ G◦ , . . . , Hn ∩ G◦ . Since G0 is irreducible, we obtain that n = 1 and G◦ ⊂ H.

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21.1.7 Remarks. 1) By 21.1.6, an algebraic group is irreducible if and only if it is connected. We shall use the term connected algebraic groups instead of irreducible algebraic group. 2) Let G be a connected algebraic group. If α, β ∈ G, we have β = (βα−1 )α. Since the map γ → (βα−1 )γ is an automorphism of the variety G, we see that the point β is smooth if and only if α is smooth. Since the set of smooth points is non-empty (16.5.5), we deduce that G is a smooth variety. We obtain in the same way that G is a normal variety (17.1.5). 21.1.8 In the rest of this chapter, G will denote an algebraic group.

21.2 Subgroups and morphisms 21.2.1 Lemma. Let U, V be dense open subsets of G. Then G = U V . Proof. Let α ∈ G. By 21.1.2, αV −1 is a dense open subset of G. So it follows that αV −1 ∩ U = ∅ and α ∈ U V . 21.2.2 Proposition. Let H be a subgroup of G. (i) H is a subgroup of G. (ii) If H contains a non-empty open subset of H, then H = H. In particular, if H is constructible, then H is closed. Proof. (i) Let α ∈ H. Since H = αH ⊂ αH and αH is closed by 21.1.2, H ⊂ αH. Hence α−1 H ⊂ H and HH ⊂ H. If β ∈ H, then Hβ ⊂ H. It follows −1 that Hβ = Hβ ⊂ H. Hence H H ⊂ H. Similarly, by 21.1.2, H = H −1 = H. So H is a subgroup of G. (ii) Let U be an open subset of H contained in H. If α ∈ H and β ∈ U , then α ∈ αβ −1 U ⊂ H and αβ −1 U is an open subset of H. Thus H is open in H. It follows that H is a constructible subset of G and therefore H contains a dense open subset of H (1.4.6). In view of 21.2.1, we have H = HH = H. 21.2.3 Corollary. If H, K are closed subgroups of G such that K normalizes H, then HK is a closed subgroup of G. Proof. By our hypothesis, HK is a subgroup of G. On the other hand, it is equal to µG (H × K). Since µ is a morphism of varieties, HK is constructible (15.4.3). So the result follows from 21.2.2 (ii). 21.2.4 Proposition. Let u : G → H be a morphism of algebraic groups. (i) ker u and im u are closed subgroups of G and H respectively. (ii) u(G◦ ) = u(G)◦ . (iii) dim G = dim(ker u) + dim(im u).

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Proof. (i) Since u is a homomorphism of groups, ker u and im u are subgroups of G and H respectively. Since u is morphism of varieties, ker u is closed and im u is constructible (15.4.3). So the result follows from 21.2.2. (ii) By (i), u(G◦ ) is a closed connected subgroup of im u. Since G◦ has ﬁnite index in G (21.1.6), u(G◦ ) has ﬁnite index in im u. Thus, again by 21.1.6, we obtain that u(G◦ ) = u(G)◦ . (iii) Since the ﬁbres of the morphism G → u(G) induced by u are the cosets of G modulo ker u, they all have dimension dim(ker u). So (iii) follows from 15.5.5 (ii). 21.2.5 Let A ∈ Mn (k) be a nilpotent matrix and p ∈ N be such that Ap+1 = 0. For t ∈ k, we set: ϕA (t) = In +

t2 t tp A + A2 + · · · + Ap . 1! 2! p!

Then this deﬁnes a morphism of algebraic groups ϕA : Ga → GLn (k). It follows from the previous results that ϕA (Ga ) is a closed connected subgroup of GLn (k). 21.2.6 Proposition. Let u : G → H be a morphism of algebraic groups. Then the following conditions are equivalent: (i) u is an isomorphism of groups. (ii) u is an isomorphism of algebraic groups. Proof. The implication (ii) ⇒ (i) is obvious. Suppose that u is an isomorphism of groups. By 21.2.4 (ii), u(G◦ ) = H ◦ . Since G◦ and H ◦ are normal varieties, the result is just a consequence of 17.4.8 and 21.1.6.

21.3 Connectedness 21.3.1 Theorem. Let (Xi )i∈I be a family of irreducible varieties and for i ∈ I, ui : Xi → G a morphism. Assume that e ∈ Yi = ui (Xi ) for all i ∈ I. Denote by H the subgroup of G generated by the Yi ’s and K the smallest closed subgroup of G containing the Yi ’s. (i) The closed subgroup K is connected and is equal to H. (ii) There exist n ∈ N, i1 , . . . , in ∈ I and ε1 , . . . , εn ∈ {−1, 1} such that: H = Yiε11 · · · Yiεnn . Proof. If i ∈ I, the map Xi → G, α → ui (α)−1 , is again a morphism. So replacing I if necessary, we may assume that for any i ∈ I, Yi−1 = Yj for some j ∈ I. Observe that we have clearly K = H. For m ∈ N and λ = (i1 , . . . , im ) ∈ I m , set Yλ = Yi1 · · · Yim . Since Yλ is the image of the morphism Xi1 × · · · × Xim → G, (α1 , . . . , αm ) → α1 · · · αm , it follows that Yλ and Yλ are irreducible.

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Let µ = (j1 , . . . , jn ) ∈ I n , n ∈ N. Set (λ, µ) = (i1 , . . . , im , j1 , . . . , jn ) ∈ . We claim that Yλ Yµ ⊂ Y(λ,µ) . Let us prove our claim. For α ∈ Yλ , the I translation morphism β → αβ maps Yµ into Y(λ,µ) . It follows that Yλ Yµ ⊂ Y(λ,µ) . Now for β ∈ Yµ , we have Yλ β = Yλ β ⊂ Y(λ,µ) . Hence Yλ Yµ ⊂ Y(λ,µ) as claimed. Let F = {Yλ ; n ∈ N, λ ∈ I n }. Since the topological space G is Noetherian, F has a maximal element, say Yµ . Let n ∈ N and λ ∈ I n . Since e ∈ Yλ , it follows from the claim above that: m+n

Yµ ⊂ Yµ Yλ ⊂ Y(λ,µ) . Since Yµ is maximal, we have Yµ = Yµ Yλ = Y(µ,λ) . Taking λ = µ, we deduce that Yµ is stable under multiplication. Similarly, −1 taking λ such that Yλ = Yµ−1 , we see that Yµ ⊂ Yµ . So Yµ is a subgroup of G which is closed and connected by deﬁnition. Clearly K = Yµ . Furthermore, being the image of a morphism Yµ is a constructible subset of G (15.4.3) and therefore by 1.4.6 and 21.2.1, Yµ = Yµ Yµ = Y(µ,µ) . Hence H ⊂ H = K = Yµ = Y(µ,µ) ⊂ H. 21.3.2 Let (Gi )i∈I be a family of subgroups of G, and ji : Gi → G the canonical injections. In order to apply 21.3.1 in this situation, the Gi ’s need to have a structure of algebraic variety and the ji ’s need to be morphisms. Corollary. Let (Gi )i∈I be a family of closed connected subgroups of G and H the subgroup of G generated by the Gi ’s. Then H is a closed connected subgroup of G. 21.3.3 Corollary. With the notations of 10.8.1, the groups SLn (k), Dn (k), Tn (k) and Un (k) are closed connected subgroups of GLn (k). Proof. These are clearly closed subgroups of GLn (k). The group Dn (k) is connected because it is isomorphic to (Gm )n . The connectedness of SLn (k) and Un (k) follows from 10.8.6 and 21.3.2. Finally, Tn (k) is connected because it is isomorphic to Dn (k) × Un (k). 21.3.4 Let g be a ﬁnite-dimensional Lie algebra over k. By 21.2.5 and 21.3.2, we have: Corollary. The group Aute g is a connected algebraic group which is a closed normal subgroup of Aut g. 21.3.5 Let H, K be closed subgroups of G. It is not true in general that the subgroup (H, K) is closed in G. However, we have the following result: Proposition. Let H, K be closed subgroups of G.

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(i) If either H or K is connected, then (H, K) is a closed connected subgroup of G. (ii) Assume that H normalizes K. Then the (H, K) is a closed subgroup of G. In particular, (G, G) is a closed subgroup of G. Proof. (i) Assume that H is connected. For β ∈ K, the map uβ : H → G, α → αβα−1 β −1 , is a morphism of varieties. Since (H, K) is the subgroup of G generated by uβ (H), β ∈ K, the result follows from 21.3.1. (ii) We may assume that G = HK (21.2.3). Then K and (H, K) are normal subgroups of G. By (i), (H ◦ , K) and (H, K ◦ ) are closed connected subgroups of G. Let L be the subgroup of G generated by α(H ◦ , K)α−1 and β(H, K ◦ )β −1 , α, β ∈ G. Since (H, K) is normal, we have L ⊂ (H, K). Moreover, L is closed (21.3.2). So we are done if L has ﬁnite index in (H, K). Let θ : G → G/L be the canonical surjection. Then θ(H ◦ ) centralizes θ(K), and θ(K ◦ ) centralizes θ(H). Since H ◦ (resp. K ◦ ) has ﬁnite index in H (resp. K), it follows that S = {(γ, δ); γ ∈ θ(H), δ ∈ θ(K)} is a ﬁnite set. So (H, K)/L is ﬁnite by 10.3.3. 21.3.6 Corollary. If G is connected, then Dn (G) and C n+1 (G) are closed connected subgroups of G for all n ∈ N. 21.3.7 Proposition. Let α ∈ G and H a closed subgroup of G. (i) The centralizers CG (α) of α in G and CG (H) of H is G are closed subgroups of G. (ii) The set {β ∈ G; βHβ −1 ⊂ H} is the closed subgroup NG (H) of G. Proof. (i) Let u, v denote the translations β → αβ and β → βα respectively. Then CG (α) = {β ∈ G; u(β) = v(β)}. So it is closed by 12.5.8. Since CG (H) is the intersection of CG (β), β ∈ H, it is also closed. (ii) For β ∈ G, the map iβ : G → G, α → βαβ −1 , is an automorphism of the algebraic group G. So Kβ = iβ (H) is a closed subgroup of G of dimension dim H and iβ (H ◦ ) = (Kβ )◦ . On the other hand, (H : H ◦ ) = (Kβ : (Kβ )◦ ). If βHβ −1 ⊂ H, then Kβ ⊂ H and Kβ◦ is a closed connected subgroup of H containing e of dimension dim H ◦ . So (Kβ )◦ = H ◦ . It follows that H = Kβ , and so NG (H) = {β ∈ G; βHβ −1 ⊂ H}. Let β ∈ H. The map uβ : G → G, α → αβα−1 , is a morphism because it is the composition of G → G × G , α → (αβ, α−1 ) and G × G → G , (α, γ) → αγ. −1 So u−1 ⊂ H} is the β (H) is closed in G. Since NG (H) = {β ∈ G; βHβ −1 intersection of uβ (H), β ∈ H, it is closed in G.

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325

21.4 Actions of an algebraic group 21.4.1 Deﬁnition. Let G acts on a variety X. We say that G acts rationally on X, or X is a G-variety if the map G × X → X , (α, x) → α.x is a morphism of algebraic varieties. So, if X is a G-variety, the map x → α.x is an automorphism of X for all α ∈ G. 21.4.2 Proposition. Let X be a G-variety. (i) If Y and Z are subsets of X with Z closed in X, then TranG (Y, Z) is a closed subset of G. (ii) For all x ∈ X, Gx is a closed subgroup of G. (iii) If α ∈ G, then the set of ﬁxed points of α is closed in X. Thus X G is closed in X. (iv) If G is connected, then G stabilizes each irreducible component of X. Proof. (i) If x ∈ X, the map ux : G → X, α → α.x is a morphism. So u−1 x (Z) is closed in G. Since TranG (Y, Z) is the intersection of u−1 y (Z), y ∈ Y , it is closed in G. (ii) This is clear by (i) since Gx = TranG ({x}, {x}). (iii) The diagonal ∆ of X in X × X is closed (12.5.3). For α ∈ G, the map u : X → X × X, x → (x, α.x), is a morphism. The set of ﬁxed points of α is u−1 (∆), so it is closed in X. (iv) Let G be connected, Y an irreducible component of X and H the stabilizer of Y in H. By (i), H is closed in G and since the elements of G permutes the irreducible components of X, we obtain that (G : H) is ﬁnite. Hence G = G◦ ⊂ H by 21.1.6. 21.4.3 Proposition. Let X be a G-variety, x ∈ X and Y = G.x. (i) The set Y is locally closed in X, so it is a subvariety of X. (ii) If G is connected, then Y is irreducible and smooth. (iii) We have dim(G.x) = dim G − dim Gx . Proof. It is clear that Y is stable under the action of G. (i) The map ux : G → X, α → α.x is a morphism and Y = im ux . So Y is constructible (15.4.3) and it contains a dense open subset of Y (1.4.6). Since G acts transitively on Y and Y is stable under the action of G, we deduce that Y is open in Y . (ii) Since Y = im ux , Y is irreducible when G is connected. It is smooth since G acts transitively on Y . (iii) Let vx : G → Y be the morphism induced by ux . Then the ﬁbres of vx are left cosets of G modulo Gx . So the results follows from 15.5.5 (ii). 21.4.4 Proposition. Let X and Y be G-varieties. Assume that X is a G-homogeneous space.

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(i) The G◦ -orbits of X are open and closed in X. They all have the same dimension and they are the pairwise disjoint irreducible components of X. In particular, X is a pure variety. (ii) For n ∈ N, {y ∈ Y ; dim(G.y) n} is a closed subset of Y . Proof. (i) Let x ∈ X. For α ∈ G, the right coset G◦ α deﬁnes a G◦ -orbit, G◦ α.x of X. Since G acts transitively on X, and (G : G◦ ) is ﬁnite, there exist elements α1 , . . . , αn ∈ G such that X = (G◦ α1 .x) ∪ · · · ∪ (G◦ αn .x) and G◦ αi .x ∩ G◦ αj .x = ∅ if i = j. Let Xi = G◦ αi .x. So the Xi ’s are exactly the G◦ -orbits in X. Suppose for example that dim X1 dim Xi for 2 i n and Y = X1 \ X1 is non-empty. Since X1 is irreducible (21.4.3) and Y is closed in X1 , dim Y < dim X1 = dim X1 . But this is absurd because Y , being G◦ -stable, is the union of certain Xi ’s with i 2. So X1 is closed in X. Now Xi = G◦ αi .x = αi G◦ .x = (αi α1−1 ).X1 , thus Xi is closed in X for all i. The result is now clear. (ii) By (i), G.y is a pure variety whose irreducible components are G◦ orbits. So we may assume that G is connected. By 21.4.2 (iv), we may further assume that X is irreducible. Let u : G × X → X × X be the morphism sending (α, x) to (x, α.x). Then u−1 (u(α, x)) = (αGx , x) for all α ∈ G and x ∈ X. By 21.1.4 and 21.4.2 (ii), (α(Gx )◦ , x) is the unique irreducible component of (αGx , x) containing (α, x). Its dimension is dim Gx . So the result is a consequence of 21.4.3 (iii) and 15.5.7. 21.4.5 Proposition. Let X be a G-variety, x ∈ X and Y = G.x. (i) The set Y is the union of Y and orbits of dimension strictly less than dim Y . (ii) Orbits of minimal dimension are closed. In particular, X contains a closed orbit. Proof. Let Y1 , . . . , Yn be the irreducible components of Y . They are G◦ -orbits by 21.4.4. By 1.1.14, Y1 , . . . , Yn are the irreducible components of Y . Since Y \ Y ⊂ (Y1 \ Y1 ) ∪ · · · ∪ (Yn \ Yn ), we have dim(Y \ Y ) < dim Y . So we have (i) since Y \ Y is G-stable. Finally (ii) follows easily from (i).

21.5 Modules 21.5.1 Let V be a k-vector space of dimension n. By identifying V with kn via a basis B of V , we can transport the Zariski topology of kn on V . It is clear that the topology obtained on V does not depend on the choice of B. Hence we have a notion of Zariski topology on V , and in a similar way, on End(V ) and GL(V ). 21.5.2 Let (V, ρ) be a ﬁnite-dimensional G-module. We say that V is a rational G-module if the map ρ : G → GL(V ) is a morphism of algebraic groups. Lemma. The following conditions are equivalent:

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(i) V is a rational G-module. (ii) The map θ : G×V → V , (α, x) → ρ(α)(x), is a morphism of varieties. (iii) For all x ∈ V , the map θx : G → V , α → ρ(α)(x), is a morphism of varieties. Proof. Since θ is the composition of G × V → End(V ) × V , (α, x) → (ρ(α), x), End(V ) × V → V , (u, x) → u(x), where the second map is a morphism of varieties, we have (i) ⇒ (ii). The implication (ii) ⇒ (iii) is obvious. Let B = (x1 , . . . , xn ) be a basis of V and (ϕ1 , . . . , ϕn ) the basis of V ∗ dual to B. The map V → k, v → ϕi (v), is a morphism of varieties for all i. So if (iii) is veriﬁed, then the map G → Mn (k) sending α ∈ G to the matrix ρ(α) with respect to B is a morphism of varieties. Hence (iii) ⇒ (i). 21.5.3 Remark. Let V be a ﬁnite-dimensional rational G-module, then the map θ in 21.5.2 (ii) makes V into a G-variety. So we may apply the results of 21.4 to rational G-modules. 21.5.4 Deﬁnition. Let V be a G-module. We say that V is locally ﬁnite if for all x ∈ V , there exists a ﬁnite-dimensional G-submodule Vx of V containing x. If V is a locally ﬁnite G-module, then V is a rational G-module if for all x ∈ V , we can choose Vx to be a rational G-module. 21.5.5 Proposition. Let (V, ρ) be a rational G-module and W a submodule of V . Then W is a rational G-module. Proof. For x ∈ W , let Vx be a rational G-submodule of ﬁnite dimension containing x. By 21.5.2, for y ∈ Vx , the map G → V , α → ρ(α)(y), is a morphism of varieties. If y ∈ W ∩Vx , then the map G → W ∩Vx , α → ρ(α)(y), is again a morphism of varieties. It follows from 21.5.2 that W ∩ Vx is a ﬁnitedimensional rational G-submodule containing x.

21.6 Group closure 21.6.1 Let H be a subgroup of G. In this section, we study the properties of the map H → H. Recall from 21.2.2 that H is a subgroup of G. Lemma. (i) If H is commutative, then so is H. (ii) If H is normal in G, then so is H. Proof. (i) For β ∈ G, denote by uβ and vβ the morphisms from G to itself deﬁned by uβ (α) = αβ and vβ (α) = βα. Then if β ∈ H, then uβ and vβ are identical on H, so they are identical on H (12.5.8). This implies in turn that for β ∈ H, uβ and vβ are identical on H, hence on H. Thus H is commutative. (ii) For β ∈ G, the morphism of varieties ϕβ : G → G, α → βαβ −1 , veriﬁes ϕβ (H) ⊂ H, and so ϕβ (H) ⊂ H.

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21.6.2 Let P ⊂ G. The intersection of all the closed subgroups of G containing P , denoted by A(P ), is called the group closure of P in G. Clearly, A(P ) is the smallest closed subgroup of G containing P . If H is a subgroup of G, then A(H) = H. So A(P ) can also be identiﬁed as the closure of the subgroup of G generated by P . 21.6.3 Lemma. Let G, H be algebraic groups, P, Q ⊂ G, R ⊂ H and u : G → H a morphism of algebraic groups. (i) If P is a dense subset of Q, then A(P ) = A(Q). (ii) If P normalizes (resp. centralizes) Q, then A(P ) normalizes (resp. centralizes) A(Q). (iii) We have A(P × R) = A(P ) × A(R). (iv) We have u(A(P )) = A(u(P )). Proof. (i) Since Q ⊂ P ⊂ A(P ), A(Q) ⊂ A(P ). The reverse inclusion is obvious. (ii) If x ∈ G normalizes Q, then it normalizes the subgroup generated by Q, and hence A(Q) (same proof as in 21.6.1 (ii)). Thus P ⊂ NG (A(Q)). Since NG (A(Q)) is closed in G (21.3.7), A(P ) ⊂ NG (A(Q)). (iii) Since A(P ) × A(R) is a closed subgroup of G × H containing P × R, it contains A(P × R). Conversely, A(P × R) contains P × {eH } and {eG } × Q, so it contains A(P ) × {eH } and {eG } × A(Q). So A(P × R) ⊃ A(P ) × A(R). (iv) Since the subgroup u(A(P )) is closed in H (21.2.4) and it contains u(P ), we have A(u(P )) ⊂ u(A(P )). Moreover, P ⊂ u−1 (A(u(P ))) ⇒ A(P ) ⊂ u−1 (A(u(P ))) ⇒ u(A(P )) ⊂ A(u(P )). So we have proved our result. 21.6.4 Proposition. Let H, K be subgroups of G. Then the groups (H, K) and (H, K) have the same closure in G. Proof. The map u : G × G → G , (α, β) → αβα−1 β −1 , is a morphism of varieties. Since H × K is dense in H × K, u(H × K) is dense in u(H ×K). So the result follows from 21.6.3 (i) since A(u(H ×K)) = (H, K) and A(u(H × K)) = (H, K). 21.6.5 Proposition. Let H, K be subgroups of G. If H normalizes K, then H normalizes K and (H, K) = (H, K). Proof. By 21.6.3 (ii), H ⊂ NG (K), so HK is a closed subgroup of G (21.2.3). We may therefore assume that G = HK. Then K is normal in G and (H, K) is closed in G (21.3.5). So the result follows from 21.6.4.

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21.6.6 Corollary. Let H, K be subgroups of G. (i) For all n ∈ N, we have: Dn (H) = Dn (H) , C n+1 (H) = C n+1 (H). (ii) Assume that K is a normal subgroup of H. If H/K is abelian (resp. nilpotent, solvable), then so is H/K. Proof. By 21.6.5, (i) is a simple induction on n. If H/K is solvable, then Dn (H) ⊂ K for some n ∈ N. So (i) implies that Dn (H) = Dn (H) ⊂ K. Thus H/K is solvable. The proofs of the other cases are similar. 21.6.7 Corollary. The following conditions are equivalent: (i) G is solvable. (ii) There is a sequence G = G0 ⊃ G1 ⊃ · · · ⊃ Gn+1 = {e} of closed subgroups of G such that (Gi , Gi ) ⊂ Gi+1 for 0 i n. Proof. In view of 21.3.6 and 21.6.6 (i), we have (i) ⇒ (ii) by taking Gi = Di (G). Conversely, given (ii), a simple induction shows that Di (G) ⊂ Gi . So we have (i). 21.6.8 Corollary. The following conditions are equivalent: (i) G is nilpotent. (ii) There is a sequence G = G0 ⊃ G1 ⊃ · · · ⊃ Gn+1 = {e} of closed subgroups of G such that (G, Gi ) ⊂ Gi+1 for 0 i n. Proof. This is analogue to the proof of 21.6.7.

References • [5], [23], [38], [40], [78].

22 Aﬃne algebraic groups

In the rest of this book, all the algebraic groups that we are going to come across are aﬃne. We shall therefore assume from now on that by an algebraic group, we mean an aﬃne algebraic group. In fact, we shall see that under this assumption, an algebraic group can be identiﬁed with a closed subgroup of a general linear group. By convention, G will always denote an (aﬃne) algebraic group.

22.1 Translations of functions 22.1.1 Let G be an algebraic group, X an aﬃne G-variety and π : G × X → X , (α, x) → α.x the morphism deﬁning the action of G on X. The comorphism of π A(π) : A(X) → A(G × X) = A(G) ⊗k A(X) sends f ∈ A(X) to the function F on G × X deﬁned by F (α, x) = f (α.x). From this, we obtain that if α ∈ G and f ∈ A(X), then the function τα f on X deﬁned by (τα f )(x) = f (α−1 .x) belongs to A(X). We verify easily that this deﬁnes a group homomorphism τ : G → GL(A(X)), α → τα . We shall endow A(X) with this G-module structure. Proposition. Let E be a ﬁnite-dimensional subspace of A(X). (i) There exists a ﬁnite-dimensional G-submodule of A(X) containing E. (ii) The subspace E is a G-submodule of A(X) if and only if A(π)(E) is contained in A(G) ⊗k E. Proof. (i) We may assume that dim E = 1. So E is spanned by an element f ∈ A(X). Let

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22 Aﬃne algebraic groups

A(π)(f ) = u1 ⊗ v1 + · · · + un ⊗ vn where u1 , . . . , un ∈ A(G) and v1 , . . . , vn ∈ A(X). For α ∈ G and x ∈ X, (τα f )(x) = A(π)(f )(α−1 , x) = u1 (α−1 )v1 (x) + · · · + un (α−1 )vn (x). Thus the G-submodule generated by f is contained in the subspace spanned by v1 , . . . , vn . (ii) Let (v1 , . . . , vn ) be a basis of E. Suppose that A(π)(E) ⊂ A(G) ⊗k E. If f ∈ E, then there exist u1 , . . . , un ∈ A(G), such that A(π)(f ) = u1 ⊗ v1 + · · · + un ⊗ vn . Then for α ∈ G, we have as seen in the proof of (i) that τα f = u1 (α−1 )v1 + · · · + un (α−1 )vn . So E is a G-submodule of A(X). Suppose that E is a G-submodule of A(X). For f ∈ E, there exist w1 , . . . , wm ∈ A(X), u1 , . . . , un+m ∈ A(G) such that the elements v1 , . . . , vn , w1 , . . . , wm are linearly independent and A(π)(f ) = u1 ⊗ v1 + · · · + un ⊗ vn +

m

un+i ⊗ wi .

i=1

It follows that for α ∈ G, τα f = u1 (α−1 )v1 + · · · + un (α−1 )vn +

m

un+1 (α−1 )wi .

i=1

Since E is a G-submodule, un+i (α−1 ) = 0 for all α ∈ G and 1 i m. So un+i = 0 for 1 i m and we have A(π)(f ) ∈ A(G) ⊗k E as required. 22.1.2 Proposition. Let G be an algebraic group and X an aﬃne Gvariety. Then the G-module A(X) is rational. Proof. Let f ∈ A(X). By 22.1.1, f is contained in a ﬁnite-dimensional Gsubmodule E of A(X). Let (v1 , . . . , vn ) be a basis of E. Then as we have seen in the proof of 22.1.1, there exist u1 , . . . , un ∈ A(G), such that τα f = u1 (α−1 )v1 + · · · + un (α−1 )vn for all α ∈ G. Since the maps G → k, α → ui (α−1 ), are morphisms of varieties, the map G → E, α → τα f , is also a morphism of varieties. So E is a rational G-module (21.5.2). 22.1.3 Proposition. Let G be an algebraic group and X an aﬃne Gvariety. Then there exist a ﬁnite-dimensional rational G-module V and a Gequivariant morphism u : X → V such that u is an isomorphism from X onto a G-stable closed subvariety of V .

22.2 Jordan decomposition

333

Proof. Let v1 , . . . , vn be generators of the algebra A(X). By 22.1.2, there is a ﬁnite-dimensional rational G-submodule W of A(X) containing all the vi ’s. The dual W ∗ of W is also a rational G-module. Denote by ϕ : X → W ∗ the map deﬁned as follows: ϕ(x)(w) = w(x) for all w ∈ W , x ∈ X. Then it is clear that ϕ is a G-equivariant morphism. Since the vi ’s generate A(X), it is easy to see that A(ϕ) is surjective. So we can conclude by 11.6.8. 22.1.4 Let G be an aﬃne algebraic group. The group G acts on itself by left and right translations. We obtain from the preceding discussion two G-module structures λ, ρ : G → GL(A(G)) on A(G) deﬁned as follows: for α, β ∈ G, u ∈ A(G), (λα u)(β) = u(α−1 β) , (ρα u)(β) = u(βα). We call λα (resp. ρα ) the left translation (resp. right translation) of functions. 22.1.5 Theorem. Let G be an aﬃne algebraic group. There exists n ∈ N such that G is isomorphic (as an algebraic group) to a closed subgroup of GLn (k). Proof. Applying 22.1.3 to the representation of G on itself by left translations, we obtain that there exist a rational G-module (V, ρ) and a G-equivariant morphism u : G → V such that u is an isomorphism from G onto a Gstable closed subvariety of V . Let α ∈ G be such that ρ(α) = idV . Since u is G-equivariant, we have u(αβ) = ρ(α)u(β) = u(β). So α = e because u is injective. We conclude that ρ is injective and therefore G is isomorphic as an algebraic group to the closed subgroup ρ(G) of GL(V ) (21.2.6). 22.1.6 Lemma. Let G be an aﬃne algebraic group, H a closed subgroup, and a = I(H) the deﬁning ideal of H in A(G). Then H = {α ∈ G; λα (a) ⊂ a} = {α ∈ G; ρα (a) ⊂ a}. Proof. Let α, β ∈ H, u ∈ a. Since α−1 β ∈ H, we have u(α−1 β) = 0, so λα u, ρβ u ∈ a. Conversely, if λα (a) ⊂ a, then u(α−1 ) = (λα u)(e) = 0 for all u ∈ a. So −1 α ∈ H and hence α ∈ H. The proof for ρα is analogue.

22.2 Jordan decomposition 22.2.1 Let us consider A(G) as a G-module by right translation. Then A(G) is rational (22.1.2) and by 10.1.11, ρα decomposes into (ρα )s (ρα )u with (ρα )s semisimple, (ρα )u locally unipotent and (ρα )s (ρα )u = (ρα )u (ρα )s . Theorem. If α ∈ G, there exists a unique pair (αs , αu ) of elements of G such that: (ρα )s = ραs , (ρα )u = ραu , α = αs αu = αu αs . The decomposition α = αs αu is called the Jordan decomposition of α and αs (resp. αu ) is called the semisimple (resp. unipotent) component of α. If α = αs (resp. α = αu ), then we say that α is semisimple (resp. unipotent).

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Proof. Since ρα is an automorphism of the k-algebra A(G), so are (ρα )s and (ρα )u (10.1.14). Thus the maps f → ((ρα )s f ) (e) , f → ((ρα )u f ) (e) from A(G) → k are k-algebra homomorphisms. By 11.3.4, there exist αs , αu ∈ G such that ((ρα )s f ) (e) = f (αs ) and ((ρα )u f ) (e) = f (αu ) for all f ∈ A(G). Let β ∈ G. Since ρα and λβ commute, (ρα )s and λβ commute also. It follows that ((ρα )s f ) (β) = λβ −1 ((ρα )s f ) (e) = (ρα )s (λβ −1 f ) (e) = (λβ −1 f )(αs ) = f (βαs ). Therefore (ρα )s = ραs and in the same way, we have (ρα )u = ραu . Finally, the map G → GL(A(G)), α → ρα , is injective, so α = αs αu = αu αs since ρα = (ρα )s (ρα )u = (ρα )u (ρα )s . 22.2.2 Remark. As in 22.2.1, we have (λα )s = λαs and (λα )u = λαu . 22.2.3 Proposition. Let G and H be algebraic groups. (i) If θ : G → H is a morphism of algebraic groups, then θ(αs ) = θ(α)s and θ(αu ) = θ(α)u for all α ∈ G. (ii) Suppose that G is a closed subgroup of GL(V ) where V is a ﬁnitedimensional k-vector space. Let α ∈ G. Then its semisimple and unipotent components deﬁned in 22.2.1 coincide with those deﬁned in 10.1.4. Proof. (i) Since K = im θ is a closed subgroup of H (21.2.4) and θ factorizes through G → K = im θ → H, it suﬃces to prove the result in the following two cases: 1) G is a closed subgroup of H and θ is the canonical injection. Since A(G) = A(H)/I(G) and I(G) is stable under ρα , α ∈ G (22.1.6), the result follows from 10.1.3 and 10.1.5. 2) θ is surjective. Then A(θ) is injective and A(θ) ◦ ρθ(α) = ρα ◦ A(θ). Thus A(H) can be identiﬁed as a subspace of A(G) stable under ρα , α ∈ G. So we may conclude as in the previous case. (ii) In view of (i), it suﬃces to prove the result in the case where G = GL(V ) and by ﬁxing a basis, we may assume that G = GLn (k). Then A(G) = k[T11 , T12 , . . . , Tnn ]D where D is the determinant (11.7.5). Writing α = [αij ], then for 1 i n, ρα T1i = α1i T11 + · · · + αni T1n . Thus the subspace spanned by T1i , 1 i n, is stable under ρα . Again we may conclude by 10.1.3 and 10.1.5. 22.2.4 Corollary. Let α ∈ G and θ an isomorphism of algebraic groups from G onto a closed subgroup of GL(V ) where V is a ﬁnite-dimensional k-vector space. The following conditions are equivalent:

22.3 Unipotent groups

335

(i) α is semisimple (resp. unipotent). (ii) θ(α) is a semisimple (resp. unipotent) automorphism of V . 22.2.5 Proposition. The set of unipotent elements of G is closed in G. Proof. By 22.1.5 and 22.2.3 (i), it suﬃces to prove the result for G = GL(V ) where V is a ﬁnite-dimensional k-vector space. Since α ∈ GL(V ) is unipotent if and only if (α − idV )dim V = 0, the result follows. Remark. In general, the set of semisimple elements of G is neither open nor closed in G. 22.2.6 Proposition. Let G be commutative and denote by Gs (resp. Gu ) the set of semisimple (resp. unipotent) elements of G. (i) Gs and Gu are closed subgroups of G. (ii) The map π : Gs × Gu → G, (α, β) → αβ, is an isomorphism of algebraic groups. (iii) If G is connected, then Gs and Gu are also connected. Proof. We may assume that G is a closed subgroup of GLn (k) (22.1.5). So we may apply 22.2.4 with V = kn . (i) Since G is commutative, we have (αβ)s = αs βs and (αβ)u = αu βu . So Gs and Gu are subgroups of G. We saw in 22.2.5 that Gu is closed. There exists α ∈ GLn (k) such that αGs α−1 ⊂ Dn (k) because Gs is commutative. It follows that Gs = G ∩ (α−1 Dn (k)α) and it is closed in G since Dn (k) is closed in GLn (k). (ii) The uniqueness of the Jordan decomposition implies that π is a bijective morphism of algebraic groups. So the result follows from 21.2.6. (iii) This follows from (ii) and 11.8.6.

22.3 Unipotent groups 22.3.1 Deﬁnition. Let G be an algebraic group. (i) We say that G is unipotent if all the elements of G are unipotent. (ii) A morphism λ : Ga → G of algebraic groups is called an additive one-parameter subgroup of G. 22.3.2 Proposition. The following conditions are equivalent: (i) G is unipotent. (ii) G is isomorphic, as an algebraic group, to a closed subgroup of Un (k). Proof. (i) ⇒ (ii) Since G is isomorphic to a closed subgroup H of GLn (k) (22.1.5) and the elements of H are unipotent automorphisms by 22.2.4, the result follows from 10.8.14. (ii) ⇒ (i) This is clear by 22.2.3 and 22.2.4.

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22 Aﬃne algebraic groups

22.3.3 Let V be a ﬁnite-dimensional k-vector space. If n is a nilpotent endomorphism of V and u an unipotent automorphism of V , we may deﬁne: exp(n) = en =

∞ 1 ∞ (−1)k+1 nk , log(u) = (u − idV )k . k! k k=0 k=1

Then en is an unipotent automorphism of V and log(u) is a nilpotent endomorphism of V . Furthermore, it is well-known that: exp(log(u)) = u and log(exp(n)) = n. Now let ϕn : Ga → GL(V ) , t → etn . We saw in 21.2.5 that ϕn is a morphism of algebraic groups. It follows that ϕn (Ga ) is a connected closed subgroup of GL(V ). Suppose that n = 0. For t ∈ k, we have log(etn ) = tn. So ϕn induces a bijection between Ga and ϕn (Ga ) whose inverse map is log |ϕn (Ga ) . It follows from the deﬁnition of log that ϕn is an isomorphism of algebraic groups from Ga onto ϕn (Ga ). 22.3.4 Proposition. Let G be a closed subgroup of GL(V ), u an unipotent element of G and n = log(u). Then ϕn (Ga ) ⊂ G. Consequently, ϕn (Ga ) is the smallest closed subgroup of G (and so of GL(V )) containing u. Proof. Let k ∈ Z. Since exp(log(uk )) = uk = exp(k log(u)), we deduce that ϕn (Z) ⊂ G. Since G is closed and Z is dense in Ga , we have by the continuity of ϕn that ϕn (Ga ) = ϕn (Z) ⊂ ϕn (Z) ⊂ G. So the result follows. 22.3.5 Proposition. Let V be a ﬁnite-dimensional k-vector space and λ : Ga → GL(V ) a group homomorphism. The following conditions are equivalent: (i) λ is an additive one-parameter subgroup of GL(V ). (ii) There exists n ∈ End(V ) nilpotent such that λ(t) = etn for all t ∈ k. Proof. The implication (ii) ⇒ (i) is clear. So let us suppose that (i) is veriﬁed. By 22.2.4 and 22.3.3, Ga is unipotent. So λ(Ga ) is a connected closed unipotent subgroup of GL(V ). Let n = log(λ(1)). Then ϕn (Ga ) ⊂ λ(Ga ) (22.3.4) and for k ∈ Z, λ(k) = λ(1)k = ekn . Since Z is dense in the irreducible variety Ga , we obtain that λ = ϕn (12.5.8). 22.3.6 Theorem. Let G be a unipotent algebraic group. (i) G is connected. (ii) If V is a non-zero rational G-module, then V G = {0}. (iii) Let X be an aﬃne G-variety. Then the G-orbits in X are closed.

22.3 Unipotent groups

337

Proof. (i) We may assume that G ⊂ GL(V ) for some ﬁnite-dimensional kvector space V . Let N be the set of log(u), u ∈ G. By 22.3.4, G is generated by ϕn (Ga ), n ∈ N . Thus the result follows from 21.2.5 and 21.3.1. (ii) We may assume that V is ﬁnite-dimensional and the result was proved in 10.8.13. (iii) Suppose that there exists x ∈ X such that Z = G.x is not closed in X. Set Y = Z \ Z. Then Y is a non-empty closed subset of Z (21.4.3). Let a = I(Y ). Then a = {0} and by (ii), there exists f ∈ a \ {0} such that α.f = f for all α ∈ G. Hence f (α.x) = f (x) for all α ∈ G and f is constant on Z, and so on Z. But f is zero on Y , and so f = 0. Contradiction. 22.3.7 Let X be an irreducible aﬃne G-variety. We can extend the Gaction on A(X) (22.1.1) to Fract(A(X)) as follows: for α ∈ G, u, v ∈ A(X) with v = 0, α(uv −1 ) = (α.u)(α.v)−1 . It is clear that Fract(A(X))G is a subﬁeld of Fract(A(X)). Recall also from 12.7.7 that we may identify Fract(A(X)) with R(X). Corollary. Let G be a unipotent algebraic group and X an irreducible aﬃne G-variety. Then R(X)G = Fract A(X)G . Proof. Clearly R(X)G ⊃ Fract A(X)G . Let u be an element of R(X)G . Denote by V the set of s ∈ A(X) such that us ∈ A(X). Then V is a Gsubmodule of A(X). Since u ∈ Fract A(X), V = {0}. So by 22.3.6, there exists v ∈ V G \ {0} ⊂ A(X)G \ {0}. It follows that w = uv ∈ A(X)G and u = wv −1 ∈ Fract(A(X)G ). 22.3.8 Lemma. Let K be a commutative ﬁeld and G a group of automorphisms α of the polynomial ring K[T ] such that α.K ⊂ K. (i) We have K(T )G = Fract(K[T ]G ). (ii) There exists p ∈ K[T ]G such that K(T )G = K G (p). Proof. Let α ∈ G and u = α.T . If deg(u) = 0, then since α.K ⊂ K, we have α.K[T ] ⊂ K which is absurd. Similarly, if deg(u) 2, then deg(α.v) 2 for all v ∈ K[T ] \ K, and so α.K[T ] = K[T ], which is again absurd. We deduce that deg(u) = 1 and hence deg(α.v) = deg(v) for all v ∈ K[T ]. (i) Let f ∈ K(T )G \ Fract K[T ]G . Replacing f by f −1 if necessary, we may assume that f = uv −1 where u, v ∈ K[T ] are relatively prime and deg(u) deg(v) > 0. If α ∈ G, then u(α.v) − v(α.u) = 0. Since u and v are relatively prime, we deduce that there exists χ(α) ∈ K \ {0} such that α.v = χ(α)v and α.u = χ(α)u. Now there exist q, r ∈ K[T ] such that u = qv + r and deg(r) < deg(v). Then χ(α)u = χ(α)(α.q)v + α.r. It follows that α.q = q and α.r = χ(α)r. Moreover f = q + rv −1 and α.(rv −1 ) = rv −1 . So we obtain (i) by induction on deg(u) + deg(v).

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(ii) If K[T ]G ⊂ K, then by (i), we have K(T )G = K G = K G (1). So let us suppose that K[T ]G \ K = ∅. Take p ∈ K[T ]G \ K of minimal degree. For u ∈ K[T ]G \ K, there exist q, r ∈ K[T ] unique such that u = qp + r and deg(r) < deg(p). It follows from our choice of p that r ∈ K G . By induction on deg(u), we obtain that u ∈ K G [p]. Now the result follows from (i). 22.3.9 The following result applies in particular for unipotent groups. Proposition. Let n ∈ N and G a subgroup of Tn (k). Consider the natural action of G on kn , which induces an action of G on k(T1 , . . . , Tn ). Then k(T1 , . . . , Tn )G is a purely transcendental extension of k. Proof. Let K = k(T1 , . . . , Tn−1 ) and T = Tn , then G is a group of automorphisms of K[T ] leaving K stable. So the result follows from 22.3.8 and induction on n.

22.4 Characters and weights 22.4.1 Lemma. Let Γ be a group and X the set of homomorphisms from Γ to k\{0}. Then X is a family of linearly independent elements of the vector space of maps from Γ to k. Proof. Suppose that there exist χ1 , . . . , χn ∈ X pairwise distinct linearly dependent with n minimal with this property. Then χi = 0, n > 1 and there exist λ1 , . . . , λn−1 ∈ k \ {0} such that: χn = λ1 χ1 + · · · + λn−1 χn−1 . Since χ1 = χn , χ1 (β) = χn (β) for some β ∈ Γ . It follows that if α ∈ Γ , then χn (α)χn (β) = λ1 χ1 (α)χ1 (β) + · · · + λn−1 χn−1 (α)χn−1 (β) = (λ1 χ1 (α) + · · · + λn−1 χn−1 (α))χn (β) So λ1 (χ1 (β) − χn (β))χ1 + · · · + λn−1 (χn−1 (β) − χn (β))χn−1 = 0. But this is absurd by our choice of n. 22.4.2 Deﬁnition. Let G be an algebraic group. (i) A morphism χ : G → Gm of algebraic groups is called a (multiplicative) character of G. (ii) A morphism λ : Gm → G is called a multiplicative one-parameter subgroup of G. 22.4.3 Denote by X ∗ (G) the set of characters of G. It is a multiplicative subgroup of the set of maps from G to k and by 22.4.1, the elements of X ∗ (G) are linearly independent in the k-vector space of maps from G to k. If χ ∈ X ∗ (G) and α, β ∈ G, then it is clear that χ(αβα−1 β −1 ) = 1, and hence χ|D(G) = 1. In particular, X ∗ (SLn (k)) = {1} (10.8.8).

22.4 Characters and weights

339

Let H be a closed normal subgroup of G. If χ ∈ X ∗ (H) and α ∈ G, then the map χα : H → Gm , β → χ(αβα−1 ), is a character of H. Thus the map (α, χ) → χα deﬁnes an action of G on X ∗ (H). 22.4.4 Denote by X∗ (G) the set of multiplicative one-parameter subgroups of G. If G is abelian, then X∗ (G) is a multiplicative subgroup of the set of maps from Gm to G. Let λ ∈ X∗ (G) and α ∈ G. Then the map λα : Gm → G, t → αλ(t)α−1 , is again a multiplicative one-parameter subgroup of G. Thus the map (α, λ) → λα deﬁnes an action of G on X∗ (G). 22.4.5 Let (V, π) be a rational G-module. For χ ∈ X ∗ (G), denote by Vχ = {v ∈ V ; π(α)(v) = χ(α)v for all α ∈ G}. Then Vχ is a subspace of V . We say that χ is a weight of G in V if Vχ = {0}. If χ is a weight of G in V , then Vχ is called the weight space of weight χ of G in V , and a non-zero vector of Vχ is called a weight vector of weight χ of G in V . A non-zero vector of V is called a weight vector if it is a weight vector for some χ. Let v ∈ V \ {0}. Then it is easy to see that v is a weight vector if and only if the subspace spanned by π(G)(v) is 1-dimensional. Let H be a closed normal subgroup of G. We saw in 22.4.3 that G acts on X ∗ (H). Let χ ∈ X ∗ (H) and denote by Vχ,H the weight space of weight χ of H in V . Then we see that π(α)(Vχ,H ) ⊂ Vχα ,H . 22.4.6 Lemma. Let (V, π) be a rational G-module. Then the sum Vχ χ∈X ∗ (G)

is direct. In particular, if V is ﬁnite-dimensional, the set of weights of G in V is ﬁnite. Proof. If the sum is not direct, then there exist an integer n 2, pairwise distinct elements χ1 , · · · , χn of X ∗ (G), and vi ∈ Vχi \ {0}, 1 i n, such that v1 + · · · + vn = 0. We may assume that n is minimal with this property. Since χ1 = χ2 , there exists α ∈ G such that χ1 (α) = χ2 (α). Then 0 = π(α)(v1 + · · · + vn ) = χ1 (α)v1 + · · · + χn (α)vn , and hence, (χ2 (α) − χ1 (α))v2 + · · · + (χn (α) − χ1 (α))vn = 0. But this contradicts the minimality of n.

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22.5 Tori and diagonalizable groups 22.5.1 Deﬁnition. Let G be an algebraic group. (i) We say that G is a torus if it is isomorphic, as an algebraic group, to Dn (k) for some n ∈ N. (ii) We say that G is diagonalizable if it is isomorphic, as an algebraic group, to a closed subgroup of Dn (k) for some n ∈ N. Remarks. 1) A torus is a connected diagonalizable algebraic group. 2) It is clear that a diagonalizable algebraic group is commutative, and that all the elements of a diagonalizable group are semisimple. 22.5.2 Let n ∈ N∗ and G = Dn (k). We have: A(G) = k[T1 , T1−1 , . . . , Tn , Tn−1 ] =

(a1 ,...,an )∈Zn

kT1a1 · · · Tnan .

Moreover, we verify easily that invertible elements of A(G) are the elements λT1a1 · · · Tnan with λ ∈ k \ {0} and (a1 , . . . , an ) ∈ Zn . For (a1 , . . . , an ) ∈ Zn , set χ(a1 ,...,an ) = T1a1 · · · Tnan ∈ A(G). Then we verify easily that χ(a1 ,...,an ) ∈ X ∗ (G). Conversely, given χ ∈ X ∗ (G), then χ is invertible in A(G) and χ(1) = 1, so it is of the form χ(a1 ,...,an ) for some (a1 , . . . , an ) ∈ Zn . It follows that the map: Zn → X ∗ (G) , (a1 , . . . , an ) → χ(a1 ,...,an ) , is an isomorphism of groups, and the elements of X ∗ (G) form a basis of A(G). Similarly, for (b1 , . . . , bn ) ∈ Zn , the map λ(b1 ,...,bn ) : Gm → G , t → diag(tb1 , . . . , tbn ), deﬁnes an element of X∗ (G), and so the map Zn → X∗ (G) , (b1 , . . . , bn ) → λ(b1 ,...,bn ) , is an isomorphism of groups. Proposition. Let n ∈ N∗ and G = Dn (k). (i) The groups X ∗ (G) and X∗ (G) are free of rank n and the elements of X ∗ (G) form a basis of A(G). (ii) If χ ∈ X ∗ (G) and λ ∈ X∗ (G), then there exists an integer χ, λ such that χ(λ(t)) = tχ,λ for all t ∈ Gm . Furthermore, the pairing X ∗ (G) × X∗ (G) → Z , (χ, λ) → χ, λ, is bilinear and non-degenerate. (iii) If (V, π) is a rational G-module, then V = Vχ . χ∈X ∗ (G)

In particular, V is the direct sum of G-submodules of dimension 1.

22.5 Tori and diagonalizable groups

341

Proof. Part (i) is just a summary of the preceding discussion, and (ii) follows from: χ(a1 ,...,an ) (λ(b1 ,...,bn ) (t)) = ta1 b1 +···+an bn . Finally for (iii), we may assume that V is ﬁnite-dimensional. Let B be a basis of V . For α ∈ G, denote by [πij (α)] the matrix of π(α) with respect to B. Since πij ∈ A(G), part (i) says that they are linear combinations of characters of G. So we have: χ(α)uχ , π(α) = χ∈X ∗ (G)

where uχ ∈ End(V ). If α, β ∈ G, then: χ(α)χ(β)uχ = π(α)◦π(β) = π(αβ) = χ∈X ∗ (G)

χ,χ ∈X ∗ (G)

χ(α)χ (β)uχ ◦uχ .

Applying 22.4.1 to the characters of G × G, we deduce that uχ ◦uχ = uχ and uχ ◦uχ = 0 if χ = χ . On the other hand: uχ = π(e) = idV . χ∈X ∗ (G)

So if Mχ = uχ (V ), then: V =

χ∈X ∗ (G)

Mχ .

Finally, if v ∈ Mχ , then uχ (v) = v and: π(α)(v) = χ (α)uχ ◦uχ (v) = χ(α)u2χ (v) = χ(α)v. χ ∈X ∗ (G)

So Mχ = Vχ and we are done. 22.5.3 Theorem. The following conditions are equivalent for an algebraic group G. (i) G is diagonalizable. (ii) X ∗ (G) is a ﬁnitely generated abelian group whose elements form a basis of A(G). (iii) Any rational G-module is the direct sum of G-submodules of dimension 1. Proof. (i) ⇒ (ii) Suppose that G is closed subgroup of H = Dn (k). Then A(G) is a quotient of A(H). If χ ∈ X ∗ (H), then its image in A(G) belongs to X ∗ (G). Hence X ∗ (G) is a basis of A(G) by 22.5.2 and 22.4.1. Let (ε1 , . . . , εn ) be the canonical basis of the Z-module Zn and φi the image of χεi in A(G). Then φa1 1 · · · φann , a1 , . . . , an ∈ Z, span A(G), and they are characters of G. It follows that any φ ∈ X ∗ (G) is a linear combination of the φa1 1 · · · φann ’s. By 22.4.1, φ is one of the φa1 1 · · · φann ’s. Thus X ∗ (G) is ﬁnitely generated.

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22 Aﬃne algebraic groups

(ii) ⇒ (iii) The proof is the same as the one for 22.5.2. (iii) ⇒ (i) We may assume that G is a closed subgroup of GL(V ) where V is a ﬁnite-dimensional k-vector space (22.1.5). Then V is the direct sum of G-submodules of dimension 1, and the result is clear. 22.5.4 Corollary. Let G be diagonalizable and H a closed subgroup of G. (i) H is diagonalizable. (ii) There exist χ1 , . . . , χn ∈ X ∗ (G) such that H = (ker χ1 ) ∩ · · · ∩ (ker χn ). (iii) If θ ∈ X ∗ (H), then there exists χ ∈ X ∗ (G) such that χ|H = θ. Proof. (i) This is obvious. ∗ ∗ (ii) Let XH (G) = {χ ∈ X ∗ (G); H ⊂ ker χ}. If χ ∈ XH (G), then χ − 1 belongs to I(H). Let f ∈ I(H) \ {0}. By 22.5.3, there exist λ1 , . . . , λn ∈ k \ {0} and χ1 , . . . , χn ∈ X ∗ (G) pairwise distinct such that f = λ1 χ1 + · · · + λn χn . Reindexing if necessary, we may assume that there exist m0 = 0 < m1 < · · · < mr = n such that χi |H = θj for mj−1 + 1 i mj , where θ1 , . . . , θr are pairwise distinct elements of X ∗ (H). Applying 22.4.1 to the θj ’s, we obtain that λmj−1 +1 + · · · + λmj = 0 for 1 j r. Set: µj =

m j i=mj−1 +1

λi (χ−1 mj χi − 1).

∗ Then f = χm1 µ1 + · · · + χmr µr , and we have χ−1 mj χi ∈ XH (G) if mj−1 + 1 ∗ (G). Since A(G) i mj . Thus the ideal I(H) is generated by χ − 1, χ ∈ XH ∗ is Noetherian, there exist χ1 , . . . , χr ∈ XH (G) such that

I(H) = A(G)(χ1 − 1) + · · · + A(G)(χr − 1). So the result follows. (iii) By 22.5.3, there exist λ1 , . . . , λn ∈ k \ {0} and χ1 , . . . , χn ∈ X ∗ (G) such that θ = λ1 (χ1 |H ) + · · · + λn (χn |H ). In view of 22.4.1, χi |H = θ for all i, and the sum λ1 + · · · + λn = 1. 22.5.5 Lemma. (i) The only closed connected subgroups of Gm are {1} and Gm . (ii) If G is a connected algebraic group, then X ∗ (G) is torsion free. Proof. (i) Let H be a closed connected subgroup of Gm . Since dim H dim Gm = 1 (14.1.6), we have dim H = 0 or 1. But H is connected, so H is either {1} or Gm . (ii) Let χ ∈ X ∗ (G). Then χ(G) is a closed connected subgroup of Gm (21.2.4). By (i), χ(G) = {1} or Gm . In the ﬁrst case, χ = 1. In the second case, χn = 1 for all n.

22.5 Tori and diagonalizable groups

343

22.5.6 The group algebra k[Zn ] of the abelian group Zn is isomorphic to k[T1 , T1−1 , . . . , Tn , Tn−1 ]. It follows from 22.5.3 that if G is diagonalizable, then the group algebra k[X ∗ (G)] is isomorphic to A(G). 22.5.7 Theorem. Let n ∈ N∗ and G an aﬃne algebraic group. The following conditions are equivalent: (i) As an algebraic group, G is isomorphic to Dn (k). (ii) G is a torus of dimension n. (iii) G is a connected diagonalizable algebraic group of dimension n. (iv) G is diagonalizable and X ∗ (G) is isomorphic to Zn . Proof. Clearly, we have (i) ⇒ (ii) ⇒ (iii). (iii) ⇒ (iv) By 22.5.3 and 22.5.5, X ∗ (G) is a ﬁnitely generated torsion free abelian group. So it is of the form Zr for some r ∈ N. In view of 22.5.6, A(G) is isomorphic to k[T1 , T1−1 , . . . , Tr , Tr−1 ]. Thus r = n by 6.1.4, 6.2.3 and 14.1.2. (iv) ⇒ (i) Let χ1 , . . . , χn be a basis of the free abelian group X ∗ (G). The map u : G → Dn (k) , α → diag χ1 (α), . . . , χn (α) is a morphism of algebraic groups. Since that A(G) = k[T1 , T1−1 , . . . , Tn , Tn−1 ] by 22.5.6, A(u) is bijective. Thus u is an isomorphism of algebraic groups (11.4.5). 22.5.8 Lemma. Let A be a commutative ring and f : E → F a surjective homomorphism of A-modules. Suppose that F is a free A-module. Then there is a submodule E of E such that (i) E = E ⊕ ker f . (ii) f induces an isomorphism from E onto F . Proof. Let (yi )i∈I be a basis of F . For i ∈ I, ﬁx xi ∈ E such that f (xi ) = yi . Then (xi )i∈I is a linearly independent subset. So the submodule E = i∈I Axi satisﬁes the required properties. 22.5.9 Theorem. Let G be diagonalizable. Then G◦ is a torus, and there exists a ﬁnite subgroup H (not unique in general) of G such that G is, isomorphic as an algebraic group, to the product H × G◦ . Proof. That G◦ is a torus follows from 22.5.4 and 22.5.7. Let π : G → Dn (k) be a morphism of algebraic groups inducing an isomorphism from G to a closed subgroup of Dn (k). By 22.5.4 (iii), restriction to H induces a surjective homomorphism from Zn X ∗ (Dn (k)) to X ∗ (G◦ ). By 22.5.5, X ∗ (G◦ ) is a free Z-module. Applying 22.5.8 and using the fact that Z is a principal ideal domain (so any submodule of a free module is free), we obtain a basis χ1 , . . . , χn of X ∗ (Dn (k)) such that χ1 , . . . , χr generate the subgroup consisting of characters χ verifying χ|G◦ = 1. The automorphism α → diag(χ1 (α), . . . , χn (α))

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22 Aﬃne algebraic groups

of Dn (k) then sends π(G◦ ) to the set of matrices diag(µ1 , . . . , µn ) verifying µ1 = · · · = µr = 1. Thus Dn (k) = Dr (k) × π(G◦ ). So as abstract groups, π(G) = H × π(G◦ ) where H = π(G) ∩ Dr (k). Since H is isomorphic to G/G◦ , it is ﬁnite (21.1.6). Finally π induces an isomorphism of algebraic groups from G onto H × π(G◦ ) since the restriction of π to each irreducible component of G is an isomorphism. 22.5.10 The following result is called the rigidity of diagonalizable groups. Proposition. Let G, H be diagonalizable algebraic groups, X an irreducible aﬃne variety and u : X × G → H a morphism of varieties. Assume that for all x ∈ V , the map vx : G → H , α → u(x, α), is a morphism of algebraic groups. Then u(x, α) does not depend on x. Proof. Let (θ, x) ∈ X ∗ (H) × X. By our hypotheses, the map α → θ(u(x, α)) is a character of G and the map (x, α) → θ(u(x, α)) is a regular function on X × G. By 22.5.3, we have: fθ,χ (x)χ(α), θ u(x, α) = χ∈X ∗ (G)

where fθ,χ ∈ A(X) for all χ ∈ X ∗ (G). It follows from 22.4.1 that given x ∈ X, there exists χθ,x ∈ X ∗ (G) such that θ(u(x, α)) = χθ,x (α) for all α ∈ G. So fθ,χθ,x (x) = 1 and fθ,χ (x) = 0 if χ ∈ X ∗ (G) \ {χθ,x }. Thus fθ,χ (x) ∈ {0, 1}. For ε ∈ {0, 1}, set Xε = {x ∈ X; fθ,χ (x) = ε}. Since X is the disjoint union of the closed subsets X0 and X1 , we deduce that χθ,x depends only on θ. But 11.3.3 and 22.5.3 say that the elements of X ∗ (H) separate the points of H, so the result follows. 22.5.11 Corollary. Let ◦ H be a closed ◦ diagonalizable subgroup of G. (i) We have NG (H) = CG (H) . (ii) The index of CG (H) in NG (H) is ﬁnite. Proof. The subgroups NG (H) and CG (H) are closed in G by 21.3.7. Applying 22.5.10 to the following map ◦ u : NG (H) × H → H , (β, α) → βαβ −1 , ◦ we obtain that u(β, α) = u(e, α) = α for all α ∈ H and β ∈ NG (H) . So we have proved (i). Finally, (ii) follows from (i) and 21.1.6 (i).

22.6 Groups of dimension one

345

22.6 Groups of dimension one 22.6.1 Lemma. If G is connected and dim G = 1, then G is commutative. Proof. Let us ﬁx β ∈ G and denote by u : G → G the morphism α → αβα−1 . Since u(G) is irreducible (1.1.5 and 1.1.7) and dim G = 1, either u(G) = {β} or u(G) = G. Suppose that u(G) = G. Then u(G) contains a non-empty open subset of G (15.4.3 and 1.4.6). Thus G \ u(G) is a ﬁnite set (14.1.3, 14.1.5 and 14.1.6). We may assume that G is a closed subgroup of GLn (k) (22.1.5). For α ∈ G, denote by χα its characteristic polynomial. There exist f0 , . . . , fn−1 ∈ A(G) such that χα (T ) = T n + fn−1 (α)T n−1 + · · · + f0 (α). Since χβ = χαβα−1 , the morphism G → kn , α → (f0 (α), . . . , fn−1 (α)), is constant on u(G) and hence on G. Thus χα (T ) = χe (T ) = (T − 1)n for all α ∈ G, and G is unipotent. Now u(G)β −1 ⊂ D(G) and so G \ D(G) is ﬁnite. It follows that D(G) = G, so D(G) = G (21.3.6). But this contradicts the fact that G is unipotent (10.8.10 and 22.3.2). So αβα−1 = {β} for all α, β ∈ G and G is commutative. 22.6.2 Theorem. Let G be a connected algebraic group of dimension 1. (i) If G is not unipotent, then G is isomorphic to Gm as algebraic groups. (ii) If G is unipotent, then G is isomorphic to Ga as algebraic groups. Proof. We may assume that G is a closed subgroup of GLn (k). By 22.6.1, G is commutative. So we have G = Gs × Gu . Since dim G = 1 and G is connected, we have either G = Gs or G = Gu . If G = Gs , then since G is commutative, G is diagonalizable (22.2.4). Therefore G is isomorphic to Gm by 22.5.7. If G = Gu , then we may identify G with a closed subgroup of Un (k) (22.3.2). Let α ∈ G \ {e}, then the image of the additive one-parameter subgroup ϕlog(α) (Ga ) is a closed subgroup of G isomorphic to Ga (22.3.4). But dim G = 1, so G is isomorphic to Ga .

References and comments • [5], [23], [38], [40], [78]. Examples of non-aﬃne algebraic groups are abelian varieties. The reader may refer to [53] and [58] for a detailed account of the subject.

23 Lie algebra of an algebraic group

This chapter explains how to associate a Lie algebra to an algebraic group, and studies some basic properties of this association.

23.1 An associative algebra 23.1.1 In this chapter, G will denote an algebraic group with identity element eG or e, µG or µ the group multiplication and ιG or ι the inverse. For α ∈ G, deﬁne χα : A(G) → k, χα (f ) = f (α) to be the evaluation at α. It is clear that χα is an element of the dual A(G)∗ of A(G). 23.1.2 If x, y ∈ A(G)∗ , then x ⊗ y ∈ (A(G) ⊗k A(G))∗ is the element deﬁned as follows: (x ⊗ y)(f ⊗ g) = x(f )y(g) for all f, g ∈ A(G). Denote by x · y = (x ⊗ y) ◦ A(µ). So x · y ∈ A(G)∗ . Lemma. Endowed with the operation (x, y) → x · y, A(G)∗ is an associative k-algebra whose identity element is χe . The map G → A(G)∗ , α → χα , induces an injective homomorphism of G into the group of invertible elements of A(G)∗ . Proof. Let f ∈ A(G) and A(µ)(f ) = s1 ⊗t1 +· · ·+sn ⊗tn where s1 , t1 , . . . , sn , tn ∈ A(G). For α ∈ G, we have: f (α) = f (eα) =

n

si (e)ti (α) = f (αe) =

i=1

n

si (α)ti (e).

i=1

Thus f = s1 (e)t1 + · · · + sn (e)tn = t1 (e)s1 + · · · + tn (e)sn . If x ∈ A(G)∗ , then: (χe · x)(f ) = (χe ⊗ x)◦A(µ)(f ) =

n

n si (e)x(ti ) = x si (e)ti = x(f ).

i=1

i=1

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23 Lie algebra of an algebraic group

It follows that χe · x = x. Similarly, we have x · χe = x. Consequently, we have χα · χβ = χαβ for all α, β ∈ G and the map α → χα is injective. The associativity of the group multiplication implies that: idA(G) ⊗A(µ) ◦A(µ) = A(µ) ⊗ idA(G) ◦A(µ). So for all x, y, z ∈ A(G)∗ : (x ⊗ y ⊗ z)◦I ⊗ A(µ)◦A(µ) = x ⊗ (y ⊗ z)◦A(µ) ◦A(µ) = x · (y · z), (x ⊗ y ⊗ z)◦ A(µ) ⊗ I ◦A(µ) = (x ⊗ y)◦A(µ) ⊗ z ◦A(µ) = (x · y) · z, where I = idA(G) . We have therefore proved the result. 23.1.3 Lemma. Let u : G → H be a morphism of algebraic groups and A (u) the map dual to A(u). Then A (u) ∈ Homalg (A(G)∗ , A(H)∗ ), and A (u)(χα ) = χu(α) for all α ∈ G. Proof. Since u is a morphism of algebraic groups, we have: A(µG )◦A(u) = A(u) ⊗ A(u) ◦A(µH ). So for x, y ∈ A(G)∗ : (x · y)◦A(u) = (x ⊗ y)◦A(µ G )◦A(u) = (x ⊗ y)◦ A(u) ⊗ A(u) ◦A(µH ) = x◦A(u) · y◦A(u) . Thus A (u) is a homomorphism of algebras and the second part follows immediately.

23.2 Lie algebras 23.2.1 For α ∈ G, we shall denote the left and right translation of functions by α by λα and ρα as in 22.1.4. Since αG◦ is the unique irreducible component of G containing α, we have dim G = dim G◦ . It follows from 16.3.7 and the smoothness of points in G that Te (G◦ ) can be identiﬁed with Te (G), which in turn, can be identiﬁed with Derek (A(G), k) (16.2.4). Let Lie(G) be the set of derivations X of A(G) verifying: X◦λα = λα ◦X for all α ∈ G. Such a derivation is said to be left invariant and Lie(G) is a Lie subalgebra of Derk (A(G)). Let X ∈ Lie(G). For f ∈ A(G), set: θ(X)(f ) = X(f )(e) = χe X(f ) . Clearly θ(X) ∈ Derek (A(G), k) and so we have a linear map:

23.2 Lie algebras

349

θ: Lie(G) → Derek (A(G), k) = Te (G) , X → θ(X). We shall prove that θ is bijective. 23.2.2 Let f ∈ A(G) and x ∈ Derek (A(G), k). Deﬁne a function f ∗ x on G as follows: for α ∈ G, set (f ∗ x)(α) = x(λα−1 f ). This is called the right convolution of f by x. Let A(µ)(f ) = s1 ⊗ t1 + · · · + sn ⊗ tn where s1 , t1 , . . . , sn , tn ∈ A(G). For α, β ∈ G, we have A(µ)(f )(α, β) = f (αβ) = s1 (α)t1 (β) + · · · + sn (α)tn (β). So λα−1 f = s1 (α)t1 + · · · + sn (α)tn . Since (f ∗ x)(α) = x(λα−1 f ), it follows that: f ∗ x = x(t1 )s1 + · · · + x(tn )sn . Hence f ∗ x ∈ A(G). So right convolution by x is an endomorphism of A(G). We claim that it is a derivation of A(G). Let f, g ∈ A(G), x ∈ Derek (A(G), k) and α ∈ G. Then: (f g ∗ x)(α) = x λα−1 f λα−1 g = x(λα−1 f )(λα−1 g)(e) + x(λα−1 g)(λα−1 f )(e) = x(λα−1 f )g(α) + x(λα−1 g)f (α) = (f ∗ x)(α)g(α) + (g ∗ x)(α)f (α). Hence (f g) ∗ x = (f ∗ x)g + (g ∗ x)f and we have proved our claim. For β ∈ G. we have: λβ (f ∗ x) (α) = (f ∗ x)(β −1 α) = x(λ α−1 β f ) = x λα−1 (λβ f ) = (λβ f ) ∗ x (α). So λβ (f ∗ x) = (λβ f ) ∗ x and the derivation f → f ∗ x is left invariant. It follows that x → ∗x deﬁnes a map η : Derek (A(G), k) → Lie(G). We claim that η is the inverse of θ. If X ∈ Lie(G), f ∈ A(G) and α ∈ G, then: (η◦θ(X)(f ))(α) = (f ∗ θ(X))(α) =θ(X)(λα−1f ) = X(λα−1 f )(e) = λα−1 (Xf ) (e) = X(f )(α). So η◦θ(X) = X. Similarly, if x ∈ Derek (A(G), k), then θ◦η(x) (f ) = (f ∗ x)(e) = x(λe−1 f ) = x(f ). Hence θ◦η(x) = x. So our claim follows and we have obtained the following result.

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23 Lie algebra of an algebraic group

Proposition. The linear map θ : Lie(G) → Derek (A(G), k) = Te (G) is an isomorphism. In particular, Lie(G) is a Lie algebra over k of dimension dim G. 23.2.3 We can transport the Lie algebra structure on Lie(G) to Te (G) via the isomorphism θ, and the Lie algebra thus obtained, denoted by L(G), shall be called the Lie algebra of G. This Lie algebra structure can be recovered from the structure of associative algebra on A(G)∗ deﬁned in 23.1. Let x, y ∈ L(G) = Te (G), f ∈ A(G) and A(µ)(f ) = s1 ⊗ t1 + · · · + sn ⊗ tn where s1 , t1 , . . . , sn , tn ∈ A(G). Then: (f ∗ x) ∗ y (e) = x(t1 )(s1 ∗ y)(e) + · · · + x(tn )(sn ∗ y)(e) = y(s1 )x(t1 ) + · · · + y(sn )x(tn ) = (y · x)(f ). Hence θ [η(x), η(y)] = x · y − y · x. and the Lie bracket on L(G) is given by: [x, y] = x · y − y · x. 23.2.4 Let u : G → H be a morphism of algebraic groups and denote by du : L(G) → L(H) the diﬀerential due (or Te (u)) of u at e (16.3.1). Proposition. The map du is a Lie algebra homomorphism. Proof. Let x, y, z ∈ L(G). Since du(z) = z◦A(u) (16.3.3) and we saw in the proof of 23.1.3 that: (x · y − y · x)◦A(u) = x◦A(u) · y◦A(u) − y◦A(u) · x◦A(u) , the result follows from 23.2.3. 23.2.5 Let H be a closed subgroup of G. By 16.3.7 and 23.2.4, we may identify L(H) with a Lie subalgebra of L(G). More precisely: Lemma. Let a = I(H). Then: L(H) = {x ∈ L(G); a ∗ x ⊂ a} = {x ∈ L(G); x(a) = {0}}. Proof. Let j : H → G be the canonical injection, then A(j) is just the restriction of an element of A(G) to H and if x ∈ L(H), then dj(x) = x◦A(j). Hence dj(x)(f ) = 0 for all f ∈ a. Let f ∈ a, x ∈ L(H) and α ∈ H. Since λα−1 f ∈ a (22.1.6), we have f ∗ dj(x) (α) = dj(x)(λα−1 f ) = 0. Thus f ∗ dj(x) ∈ a Conversely, let x ∈ L(G) be such that a ∗ x ⊂ a. If f ∈ a, then (f ∗ x)(e) = 0 = x(f ). We deduce that x induces an element y of Derek (A(G)/a, k) = Derek (A(H), k), because A(H) = A(G)/a. Clearly, x = dj(y). Finally if f ∈ a and x ∈ L(G) verify f ∗ x ∈ a, then x(f ) = (f ∗ x)(e) = 0. Conversely, if x(g) = 0 for all g ∈ a, then, since λα−1 g ∈ a for α ∈ H, we have (g ∗ x)(α) = x(λα−1 g) = 0, and so g ∗ x ∈ a.

23.2 Lie algebras

351

23.2.6 A derivation X ∈ Derk (A(G), k) is said to be right invariant if X ◦ ρα = ρα ◦ X for all α ∈ G. We shall denote the set of right invariant derivation of A(G) by Lie (G). It is a Lie subalgebra of Derk (A(G)). As in the case of left invariant derivations, we have a linear map θ : Lie (G) → Te (G) deﬁned as follows: for X ∈ Lie (G) and f ∈ A(G), set θ (X)(f ) = X(f )(e) = χe (X(f )). Also for f ∈ A(G) and x ∈ Te (G), we have the notion of left convolution x ∗ f of f by x deﬁned as follows: for α ∈ G, (x ∗ f )(α) = x(ρα f ). Using the same arguments as in 23.2.2, right convolution deﬁnes a map η which is the inverse of θ . Thus we have, via θ , another Lie algebra structure on Te (G) that we shall denote by L (G). Then we see that the Lie bracket on L (G) is given by: [x, y] = y · x − x · y. So L (G) is the opposite Lie algebra of L(G). The analogue of 23.2.5 for L (H) (with left convolution) remains valid. 23.2.7 Let α ∈ G, x ∈ L(G), and f, s1 , t1 , . . . , sn , tn ∈ A(G) be such that A(µ)(f ) = s1 ⊗ t1 + · · · + sn ⊗ tn . Since (χα · x)(f ) = s1 (α)x(t1 ) + · · · + sn (α)x(tn ), it follows from 23.2.2 that: (χα · x)(f ) = (f ∗ x)(α). Similarly, we have: (x · χα )(f ) = (x ∗ f )(α). Let us evaluate χα · x · χα−1 . For 1 i n, let A(µ)(ti ) = ui1 ⊗ vi1 + · · · + uip ⊗vip where uij , vij ∈ A(G). Then ρα−1 ti = vi1 (α−1 )ui1 +· · ·+vip (α−1 )uip . Thus x(ρα−1 ti ) = x(ui1 )vi1 (α−1 ) + · · · + x(uip )vip (α−1 ) and we deduce that: ' ( (χα · x · χα−1 )(f ) = χα ⊗ (x ⊗ χα−1 )◦A(µ) ◦A(µ)(f ) p n n si (α) x(uij )vij (α−1 ) = si (α)x(ρα−1 ti ). = i=1

j=1

i=1

On the other hand, let iα : G → G, β → αβα−1 . Then: (f ◦iα )(β) = f (αβα−1 ) = s1 (α)t1 (βα−1 ) + · · · + sn (α)tn (βα−1 ). Hence f ◦iα = s1 (α)(ρα−1 t1 ) + · · · + sn (α)(ρα−1 tn ). So we have: (χα · x · χα−1 )(f ) = x(f ◦iα ).

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23 Lie algebra of an algebraic group

23.3 Examples 23.3.1 Let G = Ga and A(G) = k[T ]. The Lie algebras Lie(G) and L(G) are 1-dimensional, so they are abelian. Let X ∈ Derk (A(G)) be the derivation deﬁned by X(P ) = P (T ) for all P ∈ k[T ]. If α ∈ G, then λα ◦ X = X ◦ λα . So X spans Lie(G). It follows that L(G) is spanned by x, where x(P ) = P (0) for P ∈ k[T ]. 23.3.2 Let G = Gm and A(G) = k[T, T −1 ]. Again Lie(G) and L(G) are 1-dimensional. The unique derivation X of A(G) verifying X(T ) = T is left invariant, and therefore it spans Lie(G). It follows that L(G) is spanned by x ∈ Derek (A(G), k) where x(P ) = P (1) for P ∈ A(G). 23.3.3 Let G = GLn (k) and denote by I the identity matrix. Since GLn (k) is an aﬃne open subset of Mn (k), TI (GLn (k)) may be identiﬁed with TI (Mn (k)) (16.2.4), which is Mn (k). Let A(G) = k[T11 , T12 , . . . , Tnn , D−1 ] where D is the polynomial in Tij such that D(M ) = det M for all M ∈ Mn (k). It follows that an element x ∈ L(G) is completely determined by its values on Tij . Denote by ϕ : L(G) → gln (k) the map given by x → [x(Tij )]. It is clear that ϕ is linear and injective. Since L(G) and gln (k) are both of dimension n2 , ϕ is bijective. Let x, y ∈ L(G), X = [xij ] = ϕ(x) and Y = [yij ] = ϕ(y). The formulas for matrix multiplication imply that A(µ)(Tij ) = Ti1 ⊗ T1j + · · · + Tin ⊗ Tnj for 1 i, j n. Thus (x · y)(Tij ) = xi1 y1j + · · · + xin ynj . It follows from 23.2.3 that [x, y] = ϕ−1 ([X, Y ]). So we have proved that ϕ is a Lie algebra isomorphism and we may identify L(G) with gln (k) via ϕ. Now let V be a ﬁnite-dimensional k-vector space, H = GL(V ) and h = L(H). Then the same applies for H. In fact, the map End(V )∗ → A(H), ψ → ψ|H , is injective since H is dense in End(V ). So any x ∈ h identiﬁes with a linear form on End(V )∗ . It follows that given x ∈ h, there is a unique X ∈ End(V ) such that x(g) = g(X) for all g ∈ End(V )∗ . This deﬁnes a linear map ϕ : h → gl(V ) = End(V ) which is injective since A(H) is generated by End(V )∗ and 1/ det. They have the same dimension, and by using the same arguments as above, it is clear that ϕ is a Lie algebra isomorphism. So we may identify h with gl(V ) via ϕ. 23.3.4 Let H be a closed subgroup of G = GLn (k) and a (resp. b) the ideal of H in A(G) (resp. A(Mn (k)) = k[T11 , T12 , . . . , Tnn ]). Then clearly a is the ideal of A(G) generated by b.

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353

Take H = Tn (k) the set of upper triangular matrices in G. Then a is generated by Tij , i > j. If x ∈ L(G) and X = ϕ(x) = [xij ] as in 23.3.3, then by 23.2.2, we have, for 1 i, j n, Tij ∗ x = x1j Ti1 + x2j Ti2 + · · · + xnj Tin . Thus if i > j, then Tij ∗ x ∈ a if and only if xkj = 0 for k i. By 23.2.5, we see that L(Tn (k)) identiﬁes with tn (k), the set of upper triangular matrices in Mn (k). The same arguments show that the Lie algebra of Dn (k) is dn (k), and the Lie algebra of Un (k) is nn (k) (in the notations of 10.8.1 and 19.1.1). 23.3.5 Let G = GLn (k) and A = [aij ] ∈ Mn (k) a nilpotent matrix. Recall from 22.3.3 that the image H of the map u : Ga → G, t → etA , is a connected closed subgroup of G and u induces an isomorphism from Ga onto H. We saw in 23.3.1 that L(Ga ) is spanned by x, where x(P ) = P (0) for P ∈ k[T ]. We deduce that L(H) is spanned by du(x). We have du(x)(Tij ) = x(Tij ◦ u) = aij , so du(x) = ϕ−1 (A) where ϕ is as in 23.3.3. Identifying L(G) with gln (k), we see that L(H) = kA. 23.3.6 Let us use the notations of 23.3.3. We may consider the algebra A = A(Mn (k)) = k[T11 , T12 , . . . , Tnn ] as a subalgebra of A(G). If X ∈ Mn (k) and f ∈ A(G), we denote f ∗ ϕ−1 (X) by f ∗ X (there is no confusion since ϕ is a Lie algebra isomorphism). The map f → f ∗ X is a derivation of A(G) and if X = [xij ], then Tij ∗ X = x1j Ti1 + · · · + xnj Tin . We deduce that f ∗ X ∈ A if f ∈ A. In the same way, if f ∈ A and α ∈ G, then ρα f and λα f belong to A. Proposition. Let H be a closed subgroup of G = GLn (k) and b the ideal of H in A. Then: H = {α ∈ G; ρα (b) = b} = {α ∈ G; λα (b) = b}, L(H) = {X ∈ gln (k); b ∗ X ⊂ b}. Proof. Let S = {Dn ; n ∈ N} and a the ideal of H in A(G). Then a = S −1 b (2.3.7). So a = A(G)b, b = A ∩ a. Let α ∈ G and X ∈ gln (k). For f ∈ A(G) and g ∈ A, we have ρα (f g) = (ρα f )(ρα g) , (f g) ∗ X = (f ∗ X)g + f (g ∗ X). It follows that if ρα (b) ⊂ b and b ∗ X ⊂ b, then ρα (a) ⊂ a and a ∗ X ⊂ a. So 22.1.6 and 23.2.5 imply that α ∈ H and X ∈ L(H). Conversely, given α ∈ H and X ∈ L(H). Since a ∗ X ⊂ a (23.2.5) and A ∗ X ⊂ A, we have b ∗ X ⊂ b. Similarly, we have ρα (a) ⊂ a and ρα−1 (a) ⊂ a, so ρα (a) = a. But ρα (A) = A, we deduce therefore that ρα (b) = b.

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23.4 Computing diﬀerentials 23.4.1 Let G be an algebraic group and u : G → X a morphism of varieties. When there is no confusion, we shall denote due by du. Recall that we may identify L(GLn (k)) with gln (k) via the map x → [x(Tij )] (23.3.3). 23.4.2 Lemma. Let u : G → GLn (k) be a morphism of algebraic groups and uij ∈ A(G), 1 i, j n, be such that u(α) = [uij (α)] for all α ∈ G. If x ∈ L(G), then du(x) = [x(uij )]. Proof. This is clear since du(x)(Tij ) = x(Tij ◦ u) = x(uij ).

23.4.3 Let V be a ﬁnite-dimensional subspace of A(G) such that for all α ∈ G, ρα (V ) ⊂ V , and ρ : G → GL(V ) the morphism of algebraic groups given by α → ρα |V . Proposition. For x ∈ L(G) and f ∈ V , we have: dρ(x)(f ) = f ∗ x. Proof. Let B = (f1 , . . . , fn ) be a basis of V . If α ∈ G and 1 j n, then ρα fj = ρ1j (α)f1 +ρ2j (α)f2 +· · ·+ρnj (α)fn , where ρij ∈ A(G) for 1 i, j n. By 23.4.2, if x ∈ L(G), then dρ(x)(fj ) = x(ρ1j )f1 + · · · + x(ρnj )fn . On the other hand, we have: fj (βα) = (λβ −1 fj )(α) = (ρα fj )(β) = ρ1j (α)f1 (β) + · · · + ρnj (α)fn (β) for α, β ∈ G. Hence: (fj ∗ x)(β) = x(λβ −1 fj ) = x(ρ1j )f1 (β) + · · · + x(ρnj )fn (β) = dρ(x)(fj )(β), and we have proved the required result. 23.4.4 Corollary. Let V be a ﬁnite-dimensional subspace of A(G) verifying ρα (V ) ⊂ V for all α ∈ G. Then V ∗ x ⊂ V for all x ∈ L(G). 23.4.5 In relation with 23.4.3, we give a result concerning the determinant. Let n ∈ N∗ , Sn the symmetric group of {1, . . . , n} and if σ ∈ Sn , denote by ε(σ) the signature of σ. If α ∈ G = GLn (k), denote by D(α) the determinant of α. As an element of A(G), we have: D= ε(σ)T1σ(1) · · · Tnσ(n) . σ∈Sn

For α, β ∈ G and x = [x(Tij )] ∈ L(G) = gln (k), we have λα−1 D = D(α)D because (λα−1 D)(β) = D(αβ) = D(α)D(β). Since x ∈ L(G) = Derek (A(G), k), we see that x(λα−1 D) is equal to D(α)

σ∈Sn

ε(σ)

n i=1

δ1σ(1) · · · δi−1,σ(i−1) x(Tiσ(i) )δi+1,σ(i+1) · · · δn,σ(n) ,

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355

where δij denotes the symbol of Kronecker. Thus (D ∗ x)(α) = D(α) tr(x). Consequently: D ∗ x = tr(x)D. Now H = SLn (k) = {X ∈ Mn (k); D(X) = 1} is a connected closed subgroup of G of dimension n2 − 1 (21.3.3 and 14.2.3). Denote by b the ideal of H in A(Mn (k)). Then D − 1 ∈ b. If x ∈ L(H), then (D − 1) ∗ x = tr(x)D ∈ b (23.3.6). Since b = A(Mn (k)), it follows that tr(x) = 0. Thus L(H) ⊂ sln (k). Since dim L(H) = dim sln (k) = n2 − 1, we deduce that L(H) = sln (k). 23.4.6 Let G, H be algebraic groups, x ∈ L(G) and y ∈ L(H). For f ∈ A(G) and g ∈ A(H), set: θx,y (f ⊗ g) = x(f )g(eH ) + f (eG )y(g). We saw in 16.3.8 that the map θ : L(G) × L(H) → L(G × H), (x, y) → θx,y , is a linear bijection. We claim that it is a Lie algebra isomorphism. Let x, x ∈ L(G), y, y ∈ L(H), z = θx,y , z = θx ,y . Then it is easy to check that: (f ⊗ g) ∗ z = (f ∗ x) ⊗ g + f ⊗ (g ∗ y). It follows that (f ⊗ g) ∗ z ∗ z is equal to: (f ∗ x) ∗ x ⊗ g + (f ∗ x) ⊗ (g ∗ y ) + (f ∗ x ) ⊗ (g ∗ y) + f ⊗ (g ∗ y) ∗ y . This implies that (f ⊗ g) ∗ z ∗ z − (f ⊗ g) ∗ z ∗ z is equal to: (f ∗ x) ∗ x − (f ∗ x ) ∗ x ⊗ g + f ⊗ (g ∗ y) ∗ y − (g ∗ y ) ∗ y . Hence we have proved our claim (23.2.3). We shall identify the Lie algebras L(G) × L(H) and L(G × H) via θ. 23.4.7 Proposition. Let x, y ∈ L(G), then: dµ(x, y) = x + y , dι(x) = −x. Proof. Let f ∈ A(G) and A(µ)(f ) = s1 ⊗t1 +· · ·+sn ⊗tn where si , ti ∈ A(G) pour 1 i n. Set z = θx,y (23.4.6). Then: dµ(z)(f ) = z

n i=1

n x(si )ti (e) + si (e)y(ti ) . si ⊗ ti = i=1

On the other hand, we saw in the proof of 23.1.2 that: f = s1 (e)t1 + · · · + sn (e)tn = t1 (e)s1 + · · · + tn (e)sn . So it is clear that dµ(z)(f ) = x(f ) + y(f ). Next, let π : G → G × G denote themorphism α → α, ι(α) . Then µ◦π(α) = e, so dµ◦dπ = 0. But dπ(x) = x, dι(x) , so we obtain from the formula for dµ that x + dι(x) = 0.

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23.4.8 Corollary. Let f, g ∈ A(G) be such that g(α) = f (α−1 ) for all α ∈ G. If x ∈ L(G), then x(g) = −x(f ). Proof. Note that g = f ◦ ι and f, g are morphisms from G to Ga . So dg = df ◦dι = −df (23.4.7). But df (x)(h) = x(h◦f ) and dg(x)(h) = x(h◦g) for x ∈ L(G) and h ∈ k[T ] = A(Ga ). So the result follows by taking h = T . 23.4.9 Corollary. Let (V, π) be a ﬁnite-dimensional rational representation of G, (V ∗ , σ) its contragredient representation (which is clearly rational), x ∈ L(G), f ∈ V ∗ , and v ∈ V . Then: (dσ(x)f (v) = −f dπ(x)v . Proof. Let (v1 , . . . , vn ) be a basis of V and (ϕ1 , . . . , ϕn ) its dual basis. We have n n πij (α)vi , σ(α)ϕj = σij (α)ϕi , π(α)vj = i=1

i=1

where πij , σij ∈ A(G). Then we check easily that σij (α) = πji (α−1 ). It follows from 23.4.2 and 23.4.8 that for x ∈ L(G), we have: dσ(x)(ϕj ) = −x(πj1 )ϕ1 − · · · − x(πjn )ϕn . Hence dσ(x)(ϕj )(vi ) = −x(πji ) = −ϕj dπ(x)vi . 23.4.10 Proposition. Let u : G → G be a morphism of varieties such that u(e) = e, and set v : G → G, α → u(α)α−1 . Then dv = du − idL(G) . Proof. Let θ : G → G × G be the morphism α → (u(α), α−1 ). Then v = µ ◦ θ. If x ∈ L(G), then by 23.4.7, we have dv(x) = dµ(du(x), −x) = du(x) − x. 23.4.11 Remark. Let V be a ﬁnite-dimensional subspace of A(G) verifying λα (V ) ⊂ V for all α ∈ G. Denote by λ : G → GL(V ) the morphism of algebraic groups α → λα |V . Using the same computations as in 23.4.3 and the result obtained in 23.4.8, we obtain that for x ∈ L(G) and f ∈ V , dλ(x)(f ) = −x ∗ f.

23.4.12 Let π : G → GL(U ) and τ : H → GL(V ) be ﬁnite-dimensional rational representations of the algebraic groups G and H, and S = U ⊕ V , T = U ⊗k V . We deﬁne the rational representations (S, π ⊕ τ ) and (T, π ⊗ τ ) of G × H as follows: for α ∈ G, β ∈ H, u ∈ U and v ∈ V , (π ⊕ τ )(α, β)(u, v) = π(α)(u), τ (β)(v) , (π ⊗ τ )(α, β)(u ⊗ v) = π(α)(u) ⊗ τ (β)(v).

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357

Proposition. Let x ∈ L(G), y ∈ L(H), u ∈ U and v ∈ V . Then: d(π ⊕ τ )(x, y)(u, v) dτ (y)(v) = dπ(x)(u), , d(π ⊗ τ )(x, y)(u ⊗ v) = dπ(x)(u) ⊗ v + u ⊗ dτ (y)(v) . Proof. Let (e1 , . . . , em ) and (f1 , . . . , fn ) be bases of U and V . Then π(α)ej =

m

πij (α)ei , τ (β)fl =

i=1

n

τkl (β)fk ,

k=1

where α ∈ G, β ∈ H, πij ∈ A(G) and τkl ∈ A(H). Let I = {1, . . . , m} × {1, . . . , n} and for (j, l) ∈ I, set g(j,l) = ej ⊗ fl . The g(j,l) ’s, for (j, l) ∈ I, form a basis of U ⊗k V . We have (π ⊗ τ )(α, β)g(j,l) =

(i,k)∈I

(j,l)

(j,l)

θ(i,k) (α, β)g(i,k) , (j,l)

where θ(i,k) ∈ A(G × H) = A(G) ⊗k A(H). So θ(i,k) = πij ⊗ τkl . Let z = (x, y) ∈ L(G × H). By 23.4.6, we have: (j,l)

z(θ(i,k) ) = πij (eG )y(τkl ) + x(πij )τkl (eH ). Hence we obtain the second equality (23.4.2). The proof for the ﬁrst equality is analogue. 23.4.13 Let us conserve the notations of 23.4.12 with H = G. We shall also denote by (π ⊗ σ, U ⊗k V ) the rational representation of G deﬁned by: (π ⊗ τ )(α)(u ⊗ v) = π(α)u ⊗ τ (α)v . It follows from the preceding discussion that for x ∈ L(G): d(π ⊗ τ )(x) = dπ(x) ⊗ idV + idU ⊗dτ (x). 23.4.14 Let (V, π) be a ﬁnite-dimensional rational representation of G. n (π) of G. We deﬁne, as in 10.6.5, the representations Tn (π), Sn (π) and They are also rational and we have for x ∈ L(G), v1 , . . . , vn ∈ V : d Tn (π)(x)(v1 ⊗ · · · ⊗ vn ) =

n

v1 ⊗ · · · ⊗ vi−1 ⊗ dπ(x)vi ⊗ vi+1 ⊗ · · · ⊗ vn ,

i=1

d Sn (π)(x)(v1 · · · vn ) = d

n

(π)(x)(v1 ∧ · · · ∧ vn ) =

n

n

v1 · · · vi−1 dπ(x)vi vi+1 · · · vn ,

i=1

v1 ∧ · · · ∧ vi−1 ∧ dπ(x)vi ∧ vi+1 ∧ · · · ∧ vn .

i=1

23.4.15 Let A be a ﬁnite-dimensional k-algebra (not necessarily associative or with unit). A k-linear map δ : A → A is called a derivation of A if

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for all x, y ∈ A, δ(x.y) = δ(x).y + x.δ(y). Let Der A (resp. Aut A) denote the set of derivations (resp automorphisms) of A. By 21.1.3, Aut A is an algebraic group. Proposition. Let π : G → Aut A be a rational representation of G. Then for x ∈ L(G), dπ(x) ∈ Der A. Proof. Let (e1 , . . . , en ) be a basis of A. There exist γijk ∈ k and πij ∈ A(G) such that: n n γijk ek , π(α)ej = πij (α)ei . ei .ej = i=1

k=1

Since π(α)(ei .ej ) = π(α)ei . π(α)ej ), we have:

n

γijk πlk (α) =

n

γrsl πri (α)πsj (α).

r,s=1

k=1

Denote by δuv the symbol of Kronecker, then: n

γijk x(πlk ) =

k=1

=

γrsl δri x(πsj ) + x(πri )δsj

n r,s=1 n

n

s=1

r=1

γisl x(πsj ) +

γrjl x(πri ).

On the other hand, 23.4.2 implies that: dπ(x)(ei .ej ) =

n n l=1

γijk x(πlk ) el .

k=1

Similarly, dπ(x)ei .ej + ei . dπ(x)ej is equal to: n n l=1

n n γrjl x(πri ) el + γisl x(πsj ) el .

r=1

l=1

s=1

So the result follows. 23.4.16 Proposition. Let g be a semisimple Lie algebra and G the group of elementary automorphisms of g. Then L(G) = g. Proof. Since G ⊂ Aut g, L(G) ⊂ ad g (23.4.15 and 20.1.5) and L(G) is a Lie subalgebra of ad g (16.3.7 and 23.2.4). Finally, if ad x is nilpotent, then ad x ∈ L(G) by 23.3.5. So the result is a consequence of 20.4.6. 23.4.17 Proposition. Let (U, π) and (V, τ ) be ﬁnite-dimensional rational representations of an algebraic group G.Assume that there exists a linear map θ:U → V such that θ π(α)u = τ (α) θ(u) pour (α, u) ∈ G × U . Then we have θ dπ(x)u = dτ (x) θ(u) for x ∈ L(G) and u ∈ U .

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359

Proof. Let U and V be bases of U and V . Denote by A(α) = [πij (α)] (resp. B(α) = [τkl (α)]) the matrix of π(α) (resp. τ (α)) with respect to the basis U (resp. V). If x ∈ L(G), 23.4.2 implies that M (x) = [x(πij )] (resp. N (x) = [x(τkl )]) is the matrix of dπ(x) (resp. dτ (x)) with respect to the basis U (resp. V). Let Λ be the matrix of θ with respect to the bases U and V. Then our hypothesis says that ΛA(α) = B(α)Λ. From this, we deduce immediately that ΛM (x) = N (x)Λ. This proves the result.

23.5 Adjoint representation 23.5.1 In the rest of this chapter, G is an algebraic group and g its Lie algebra. 23.5.2 For α ∈ G, denote by iα the inner automorphism of G deﬁned by β → αβα−1 . Recall that the set Int G of inner automorphisms of G is a subgroup of Aut G. Let AdG (α) or Ad(α) be the diﬀerential of iα . By 16.3.1 and 23.2.4, it is an automorphism of the Lie algebra g. Since iαβ = iα ◦ iβ for α, β ∈ G, we deduce that Ad(αβ) = Ad(α) ◦ Ad(β). Thus the map Ad : G → GL(g) , α → Ad(α) deﬁnes a representation of G, called the adjoint representation of G. By 23.2.7, if α ∈ G and x ∈ g, then: Ad(α)(x) = χα · x · χα−1 . Let u : G → H a morphism of algebraic groups. If α ∈ G, then u◦iα = iu(α) ◦u. It follows that: du◦ AdG (α) = AdH u(α) ◦du. Let α ∈ G, x ∈ g and y = Ad(α)(x). In the notations of 23.2.2, let X, Y ∈ Lie(G) be deﬁned by X = η(x), Y = η(y). It is clear that ρα ◦X◦ρα−1 is a derivation of A(G). Moreover, since left and right translations commute, ρα ◦X◦ρα−1 ∈ Lie(G). Let f ∈ A(G) and β ∈ G. Then: (ρα ◦X◦ρα−1 )(f )(β) = X(ρ α−1 f )(βα) = x λ(βα)−1 (ρα−1 f ) = x (λβ −1 f )◦iα = y(λβ −1 f ) = Y (f )(β). Hence Y = ρα ◦X◦ρα−1 . Let f1 , . . . , fr ∈ A(G), be linearly independent elements which generate A(G) as an algebra. Complete this family to a basis (fn )n∈N of the k-vector space A(G). Since the G-module A(G) is rational (for the representations λ and ρ), we have an,k (α)fn fk ◦iα = n∈N

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23 Lie algebra of an algebraic group

where, for 1 k r, (an,k )n is a family of regular functions on G verifying for α ∈ G, an,k (α) = 0 only if n belongs to a ﬁnite subset (which depends on k and α) of N. Let x ∈ g, y = Ad(α)(x) and xn = x(fn ). For n r + 1, xn is a polynomial in x1 , . . . , xr . Moreover: ank (α)xn . y(fk ) = x(fk ◦iα ) = n∈N

An element z ∈ g is completely determined by z(fk ), 1 k r. It follows that if x ∈ g, then the map α → Ad(α)(x) is a morphism from G to g. By 21.5.2, the adjoint representation of G in g is rational. 23.5.3 Lemma. Let α ∈ G and u : G → G be the morphism of varieties β → αβα−1 β −1 . Then du = Ad(α) − idg . Proof. Apply 23.4.10 with u = iα . 23.5.4 Lemma. Assume that G is a closed subgroup of GLn (k), so we may identify g with a Lie subalgebra of gln (k) (23.2.5 and 23.3.3). If α ∈ G and x ∈ g, then Ad(α)(x) = αxα−1 . Proof. Let α = [αij ], β = [βij ] ∈ G, α−1 = [αij ] and γ = αβα−1 = [γij ]. If x ∈ g, we may identify x with the matrix [x(Tij )]. Now

n n Ad(α)(x)(Tij ) = x(Tij ◦iα ) = x αik αlj Tkl = αik x(Tkl )αlj . k,l=1

k,l=1

On the other hand, we have γij =

n

αik βkl αlj .

k,l=1

So Ad(α)(x) = αxα−1 . 23.5.5 Let us determine the diﬀerential of Ad. Let α, β ∈ G, x, y ∈ g, f ∈ A(G) and A(µ)(f ) = s1 ⊗ t1 + · · · + sn ⊗ tn where s1 , t1 , . . . , sn , tn ∈ A(G). We obtain ρα f = t1 (α)s1 + · · · + tn (α)sn as in 23.2.2. Hence (1)

(y ∗ f )(α) = y(ρα f ) = y(s1 )t1 (α) + · · · + y(sn )tn (α). It follows that:

(2)

x(y ∗ f ) = y(s1 )x(t1 ) + · · · + y(sn )x(tn ) = (y.x)(f ).

Denote Ad by u. If x ∈ g, we saw in 23.3.3 that du(x) may be identiﬁed with an element (x) of End(g) and for all ψ ∈ End(g)∗ ⊂ A GL(g) , we have: ψ (x) = du(x)(ψ).

23.5 Adjoint representation

361

Let f ∈ A GL(g) and y ∈ g. Deﬁne ϑf,y ∈ End(g)∗ as follows: for ϕ ∈ End(g), ϑf,y (ϕ) = ϕ(y) (f ). Then for x, y ∈ g and f ∈ A GL(g) , we have: du(x)(y) (f ) = (x)(y) (f ) = ϑf,y (x) = du(x)(ϑf,y ) = x(ϑf,y ◦u). On the other hand, if α ∈ G, then: (ϑf,y ◦ u)(α) = u(α)(y) (f ) = diα (y) (f ) = y(f ◦iα ). But if A(µ)(f ) = s1 ⊗ t1 + · · · + sn ⊗ tn , we saw in 23.2.7 that: f ◦iα = s1 (α)(ρα−1 t1 ) + · · · + sn (α)(ρα−1 tn ). It follows from 23.2.6 that: −1 y(f ◦iα ) = s1 (α)(y ∗ t1 )(α−1 ) + · · · + sn (α)(y ∗ tn )(α ) = s1 (y ∗ t1 )◦ι + · · · + sn (y ∗ tn )◦ι (α). We deduce therefore that: ϑf,y ◦u = s1 (y ∗ t1 )◦ι + · · · + sn (y ∗ tn )◦ι. In view of 23.4.7, we obtain that: n n x(ϑf,y ◦u) = x(si )(y ∗ ti )(e) − si (e)x(y ∗ ti ). i=1

i=1

It follows from (1), (2) and the deﬁnition of y ∗ ti that: n n x(ϑf,y ◦u) = x(si )y(ti ) − si (e)(y · x)ti . i=1

i=1

Finally 23.1.2 and 23.2.3 imply that: du(x)(y)(f ) = x(ϑf,y ◦u) = (x · y)(f ) − (χe · y · x)(f ) = (x · y)(f ) − (y · x)(f ) = [x, y](f ). Thus we have proved the following result: Theorem. The diﬀerential of the adjoint representation AdG is adg . 23.5.6 Corollary. If H is a closed normal subgroup of G and h is the Lie algebra of H. Then h is an ideal of g. Proof. Let a be the ideal of H in A(G), α ∈ G, f ∈ a and x ∈ h. If H is normal in G, then f ◦ iα ∈ a, so x(f ◦ iα ) = 0 (23.2.5). Thus Ad(α)(x)(f ) = 0, and this implies that Ad(α)(x) ∈ h (23.2.5). Let (x1 , . . . , xn ) be a basis of g such that (x1 , . . . , xp ) is a basis of h. If y ∈ g, then 23.4.2 implies that for 1 j n, we have πij ∈ A(G) such that: n n πij (α)xi , (d Ad)(y)(xj ) = y(πij )xi . Ad(α)(xj ) = i=1

i=1

So Ad(α)(h) ⊂ h implies that πij = 0 if 1 j p and p + 1 i n. Thus (d Ad)(x)(h) ⊂ h and the result follows from 23.5.5.

362

23 Lie algebra of an algebraic group

23.6 Jordan decomposition 23.6.1 Let (V, π) be a representation of g. We say that the representation π is locally ﬁnite if any v ∈ V is contained in a ﬁnite-dimensional g-submodule of V . 23.6.2 Let us consider A(G) as a rational G-module via ρ. If x ∈ g, denote by η(x) the derivation of A(G) deﬁned by η(x)(f ) = f ∗ x for f ∈ A(G) (see 23.2.2). The map g × A(G) → A(G) , (x, f ) → f ∗ x deﬁnes a representation of g which is locally ﬁnite (23.4.4). Theorem. Let x ∈ g. There exists a unique pair (xs , xn ) of elements of g such that: x = xs + xn , [xs , xn ] = 0 , η(x)s = η(xs ) , η(x)n = η(xn ). The element xs (resp. xn ) is called the semisimple component (resp. nilpotent component) of x, and x = xs + xn is called the Jordan decomposition of x. Proof. Let X = η(x). If α ∈ G, then the automorphism λα of A(G) commutes with X. By 10.1.2, λα commutes also with Xs and Xn . On the other hand, Xs and Xn are derivations of A(G) (10.1.14). So Xs , Xn ∈ Lie(G) and we are done since η : g = L(G) → Lie(G) is a Lie algebra isomorphism (23.2.2). 23.6.3 Proposition. Let u : G → H be a morphism of algebraic groups. For all x ∈ g, we have: (du(x))s = du(xs ) , (du(x))n = du(xn ). Proof. Set X = ηG (x) and Y = ηH du(x) . Let f ∈ A(H), then we have du(x)(f ) = x(f ◦u). So if α ∈ G, then: X(f ◦u)(α) = x λα−1 (f ◦u) = x (λu(α)−1 f )◦u = du(x)(λu(α)−1 f ) = Y (f ) u(α) . It follows that X◦A(u) = A(u)◦Y . Let E = A(u) A(H) . Then X(E) ⊂ E. Since A(u) induces a surjection from A(H) onto E, 10.1.12 implies that Xs ◦A(u) = A(u)◦Ys and Xn ◦A(u) = A(u)◦Yn . Thus: du(xs )(f ) = xs (f ◦u) = Xs (f ◦u)(eG ) = Ys (f ) u(eG ) = Ys (f )(eH ) = (du(x))s (f ). It follows that du(xs ) = (du(x))s . We obtain in a similar way du(xn ) = (du(x))n . 23.6.4 Proposition. Let V be a ﬁnite-dimensional k-vector space and G a closed subgroup of GL(V ). If x ∈ g, then the Jordan decomposition of x deﬁned in 23.6.2 coincide with the one deﬁned in 10.1.2.

23.6 Jordan decomposition

363

Proof. By 23.6.3, we may assume that G = GL(V ) and by ﬁxing a basis of V , we are reduced to the case where G = GLn (k). We have A(G) = k[T11 , T12 , . . . , Tnn ]D where D is the determinant. Let x = [xij ] ∈ g. Then as in 23.3.4: Tij ∗ x = x1j Ti1 + x2j Ti2 + · · · + xnj Tin . Thus Wi = kTi1 + · · · + kTin is a subspace of A(G) such that η(x)(Wi ) ⊂ Wi . Moreover, the action of η(x) on Wi may be identiﬁed as the matrix operation of x on kn . We deduce therefore that η(x)|Wi = η(xs )|Wi + η(xn )|Wi is the Jordan decomposition of η(x)|Wi . It follows that if W is the sum of the Wi ’s, for 1 i n, then η(x)|W = η(xs )|W + η(xn )|W is the Jordan decomposition of η(x)|W . By 10.1.10, we obtain that η(xs )|W = η(x)s |W and η(xn )|W = η(x)n |W . But derivations of A(G) which are identical on W are equal. So we are done.

References • [5], [23], [38], [40], [78].

24 Correspondence between groups and Lie algebras

We describe more precisely the correspondence between algebraic groups and their Lie algebras in this chapter. We give functorial properties between them and how certain properties are preserved. We also explain how to attach to a Lie algebra an algebraic group (namely, its algebraic adjoint group).

24.1 Notations 24.1.1 Let G and H be algebraic groups. We say that G and H are isomorphic if they are isomorphic as algebraic groups. A closed subgroup of G will also be called an algebraic subgroup of G. Recall that L(G) denote the Lie algebra of G and Lie(G) the Lie algebra of left invariant derivations of A(G) (23.2.1). We shall denote by µG or µ (resp. ιG or ι) the group multiplication (resp. inverse) of G, and for α ∈ G, iα denotes the inner automorphism of G deﬁned by iα (β) = αβα−1 . 24.1.2 Let g be the Lie algebra of G. Recall that AdG or Ad denotes the adjoint representation of G. If a is a subspace of g, we denote by CG (a) (resp. NG (a)) the set of elements α ∈ G verifying Ad(α)(x) = x (resp. Ad(α)(x) ∈ a) for all x ∈ a. If x ∈ g, we shall also denote CG (kx) by CG (x).

24.2 An algebraic subgroup 24.2.1 In this section, g denote the Lie algebra of an algebraic group G. For α ∈ G, let χα be the linear form on A(G) deﬁned by χα (f ) = f (α). Recall from 23.1.2 that A(G)∗ has a structure of associative algebra whose unit is χe , where e denotes the identity element of G. If x ∈ A(G)∗ and n ∈ N∗ , we shall denote by xn the product in A(G)∗ of n factors of x. By convention, x0 = χe . 24.2.2 Let us ﬁx x ∈ g. Set jx to be the subspace of A(G) consisting of elements f verifying xn (f ) = 0 for all n ∈ N.

366

24 Correspondence between groups and Lie algebras

Denote by ηG (x) or η(x), the left invariant derivation of A(G) deﬁned by η(x)(f ) = f ∗ x for f ∈ A(G) (see 23.2.2). Thus x(f ) = η(x)(f ) (e). Let f ∈ A(G), then A(µ)(f ) = s1 ⊗ t1 + · · · + sp ⊗ tp where si , ti ∈ A(G). We saw in 23.2.2 that η(x)(f ) = x(t1 )s1 + · · · + x(tp )sp . 1) If n ∈ N, then: η n+1 (x)(f ) = η n (x) η(x)(f ) = η n (x) x(t1 )s1 + · · · + x(tp )sp = η n (x)(s1 )x(t1 ) + · · · + η n (x)(sp )x(tp ). We have x(f ) = η(x)(f ) (e). Now suppose that xn (g) = η n (x)(g) (e) for all g ∈ A(G), then: n ) = xn (s1 )x(t xn+1 (f ) = (xn ⊗ x)◦A(µ)(f n1 ) + · · · + x (sp )x(tp ) n = η (x)(s1 ) (e)x(t1 ) + · · · + η (x)(sp ) (e)x(tp ).

It follows that xn+1 (f ) = η n+1 (x)(f ) (e), and so by induction, we have xn (f ) = η n (x)(f ) (e) for all n ∈ N (the case n = 0 is clear) and f ∈ A(G). 2) Let n ∈ N and f ∈ jx . Then: xn η(x)(f ) = η n (x) η(x)(f ) (e) = η n+1 (x)(f ) (e) = 0. Thus η(x)(jx ) ⊂ jx . 3) Let f ∈ jx and g ∈ A(G). Then: χe (f g) = f (e)g(e) = 0 , x(f g) = x(f )g(e) + f (e)x(g) = 0. Set h = η(x)(f ) and = η(x)(g). Suppose that xn (uv) = 0 for all u ∈ jx and v ∈ A(G). Then: xn+1 (f g) = η n (x) η(x)(f g) (e) = η n (x) hg + f (e) = xn (hg) + xn (f ). Hence xn+1 (f g) = 0 since h ∈ jx . We have therefore proved that jx is an ideal of A(G). 4) Let f ∈ A(G) and g = f ◦ι. For α, β ∈ G, we have: f (αβ) = A(µ)(f )(α, β) = s1 (α)t1 (β) + · · · + sp (α)tp (β). It follows that: A(µ)(g)(α, β) = g(αβ) = f (β −1 α−1 ) = A(µ)(f )(β −1 , α−1 ) = s1 (β −1 )t1 (α−1 ) + · · · + sp (β −1 )tp (α−1 ). Hence: A(µ)(g) = (t1 ◦ι) ⊗ (s1 ◦ι) + · · · + (tp ◦ι) ⊗ (sp ◦ι). Now χe (g) = f (e) = χe (f ) and x(g) = −x(f ) (23.4.8). Suppose that we have shown that xn (u◦ι) = (−1)n xn (u) for all u ∈ A(G). Then:

24.2 An algebraic subgroup

367

xn+1 (g) = (xn ⊗ x)◦A(µ)(g) = xn (t1 ◦ι)x(s1 ◦ι) + · · · + xn (tp ◦ι)x(sp ◦ι) = (−1)n+1 [xn (t1 )x(s1 ) + · · · + xn (tp )x(sp )] = (−1)n+1 (x ⊗ xn )A(µ)(f ) = (−1)n+1 xn+1 (f ). Thus xn (g) = (−1)n xn (f ) for all n ∈ N. So if f ∈ jx , then f ◦ι ∈ jx . 5) Let f, g ∈ A(G) \ jx . Denote by p and q the smallest integers such that xp (f ) and xq (g) are non-zero. Since η(x) is a derivation of A(G), we have: xp+q (f g) = η p+q (x)(f g) (e) =

=

m+n=p+q

(p + q)! m η (x)(f ) η n (x)(g) (e) m+n=p+q m!n! (p + q)! m (p + q)! p x (f )xn (g) = x (f )xq (g) = 0. m!n! p!q!

Thus f g ∈ jx , and we have shown that jx is a prime ideal of A(G). 6) Let (x) = V(jx ) be the set of α ∈ G verifying f (α) = 0 for all f ∈ jx . By deﬁnition, e ∈ (x). We shall show that if α ∈ (x) and f ∈ jx , then λα−1 f ∈ jx where λα is the left translation of functions (22.1.4). Since the derivation η(x) is left invariant (23.2.2), we have η n (x)(λα−1 f ) = n λα−1 η (x)(f ) . From this, we deduce that: xn (λα−1 f ) = η n (x)(λα−1 f ) (e) = λα−1 η n (x)(f ) (e) = η n (x)(f ) (α).

a

a

a

It follows from point 2 that η(x)(jx ) ⊂ jx . Hence our result. 7) Points 2, 4 and 6 prove that (x) is an algebraic subgroup of G. Moreover, by 10.1.6 and point 5, the deﬁning ideal of (x) in A(G) is: I (x) = jx = jx .

a

a

a

a

It follows that (x) is connected. 8) By 11.8.1 and the preceding discussions, the ideal Jx of (x) × (x) in the algebra A(G×G) = A(G)⊗k A(G) is equal to Jx = jx ⊗k A(G)+A(G)⊗k jx . Let f ∈ jx . If α, β ∈ (x), then A(µ)(f )(α, β) = f (αβ) = 0. Thus A(µ)(jx ) ⊂ Jx . For p, q ∈ N, denote by Tp,q,x the kernel of the linear form xp ⊗ xq on A(G) ⊗k A(G), and let Tx be the intersection of the Tp,q,x ’s, p, q ∈ N. We have xp+q (f ) = (xp ⊗xq )◦A(µ)(f ), so f ∈ jx implies that A(µ)f ∈ Tx . It follows that Tx ⊂ Jx . Since we have clearly Jx ⊂ Tx , we obtained that Tx = Jx .

a

a

a

24.2.3 Theorem. Let G be an algebraic group and x ∈ L(G). (i) The group (x) is a connected and commutative algebraic subgroup of G such that x ∈ L( (x)). (ii) Any closed subgroup H of G verifying x ∈ L(H) contains (x). Proof. Points 2 and 7 of 24.2.2 imply that I (x) = jx and η(x)(jx ) ⊂ jx . So by 23.2.5, x ∈ L (x) . Let us prove that (x) is abelian.

a

a

a

a a

a

368

24 Correspondence between groups and Lie algebras

Denote by ζ : A(G) ⊗k A(G) → A(G) ⊗k A(G) the linear map deﬁned by ζ(f ⊗ g) = g ⊗ f . If p, q ∈ N and f ∈ A(G), we have: (xp ⊗ xq )◦ζ◦A(µ)(f ) = (xq ⊗xp )◦A(µ)(f ) = xq+p (f ) = (xp ⊗ xq )◦A(µ)(f ). Thus (xp ⊗ xq )◦ ζ◦A(µ) − A(µ) = 0. By point 8 of 24.2.2, we deduce that ζ◦A(µ) − A(µ) A(G) ⊂ Ix . If α, β ∈ (x) and f ∈ A(G), then:

a

f (αβ) − f (βα) = A(µ)(f )(α, β) − A(µ)(f )(β, α) p p si (α)ti (β) − si (β)ti (α) = i=1 i=1 = A(µ) − ζ◦A(µ) (f )(α, β) = 0. So we have αβ = βα. (ii) Let H be a closed subgroup of G verifying x ∈ L(H), and a = I(H) the ideal of H. We have η(x)(a) ⊂ a (23.2.5), so η n (x)(a) ⊂ a and xn (a) = {0}, if n ∈ N by point 1 of 24.2.2. Thus a ⊂ jx , and (x) ⊂ H.

a

24.2.4 Remark. Let V be a ﬁnite-dimensional vector space, G a closed subgroup of GL(V ) and x ∈ L(G). If x is a nilpotent endomorphism, then by 22.3.4 and 23.3.5, the group (x) is the set of etx , t ∈ k.

a

a

24.2.5 Corollary. Let h be a Lie subalgebra of g, and (h) the intersection of all algebraic subgroups of G whose Lie algebra contains h. Then (h) is connected and h ⊂ L( (h)).

a a Proof. It is clear that a(h) is an algebraic subgroup of G, and since L(a(h)) = L(a(h)◦ ), a(h) is connected. By 24.2.3, we have a(x) ⊂ a(h) for all x ∈ h. So it follows from 16.3.7, 23.2.4 and 24.2.3 that x ∈ L(a(x)) ⊂ L(a(h)). Hence h ⊂ L(a(h)). 24.2.6 Remark. Let H be a closed connected subgroup of G, h its Lie algebra. Since (h) ⊂ H, we have L( (h)) ⊂ h. It follows from 24.2.5 that h = L( (h)). Thus dim (h) = dim H = dim h. Since H is connected, we deduce that (h) = H.

a

a

a

a

a

24.3 Invariants 24.3.1 Let g be the Lie algebra of an algebraic group G, (V, π) a rational representation of G of dimension n, and B a basis of V . If u ∈ End(V ), denote of u with respect to B. We by [ϕij (u)] the matrix deﬁne πij ∈ A(G) by πij (α) = ϕij π(α) , and if x ∈ g, set dij (x) = ϕij dπ(x) , 1 i, j n. Let x1 , . . . , xr ∈ g. Then: ϕij dπ(x1 )◦ · · · ◦dπ(xr ) = di,k1 (x1 )dk1 ,k2 (x2 ) · · · dkr−1 ,j (xr ). 1k1 ,...,kr−1 n

∗ Moreover, for x ∈ g and ϕ ∈ End(V ) , we have by 23.3.3 that:

24.3 Invariants

369

ϕ dπ(x) = dπ(x)(ϕ) = x(ϕ◦π). Thus: ϕij dπ(x1 )◦ · · · ◦dπ(xr ) =

1k1 ,...,kr−1 n

x1 (πi,k1 )x2 (πk1 ,k2 ) · · · xr (πkr−1 ,j ).

Hence: ϕij dπ(x1 )◦ · · · ◦dπ(xr ) = (x1 · · · xr )(πij ) = (x1 · · · xr )(ϕij ◦π). So by linearity, we obtain the following result:

∗ Lemma. Let r ∈ N∗ , x1 , . . . , xr ∈ g and ϕ ∈ End(V ) . Then: ϕ dπ(x1 )◦ · · · ◦dπ(xr ) = (x1 · · · xr )(ϕ◦π).

24.3.2 Theorem. Let (V, π) be a ﬁnite-dimensional rational representation of an algebraic group G, E the subspace of End(V ) spanned by π(α)−idV , α ∈ G, and F the subspace of End(V ) spanned by the products dπ(x1 ) ◦ · · · ◦ dπ(xr ) where r ∈ N∗ and x1 , . . . , xr ∈ g = L(G). (i) For x ∈ g, we have dπ(x) ∈ E. (ii) Assume that G is connected. If α ∈ G, then π(α) − idV ∈ F . Proof. (i) Let ϕ ∈ End(V )∗ be such that ϕ|E = 0. Then α → ϕ ◦ π(α) is a constant function on G and so x(ϕ ◦ π) = 0 for all x ∈ g. So by 23.3.3, we have 0 = dπ(x)(ϕ) = ϕ(dπ(x)). Hence dπ(x) ∈ E. (ii) Let x ∈ g, n ∈ N∗ and ϕ ∈ End(V )∗ verifying ϕ|F = 0. We have xn (ϕ ◦ π) = 0 by 24.3.1. So ϕ ◦ π − ϕ(idV ) ∈ jx (notations of 24.2.2). It follows that if α ∈ (x), then ϕ(π(α)−idV ) = 0. Hence π(α)−idV ∈ F for α ∈ (x). If α, β ∈ G, then: π(αβ) − idV = π(α) − idV ◦ π(β) − idV + π(α) − idV + π(β) − idV .

a

a

a

So for n ∈ N∗ , x1 , . . . , xn ∈ g and αi ∈ (xi ), 1 i n, we obtain that π(α1 · · · αn ) − idV ∈ F . Let H be the subgroup of G generated by (x), x ∈ g. It algebraic subgroup of G (21.3.2 and 24.2.3). If α ∈ H, is a connected ) which are zero on F , then ϕ π(α) − idV = 0 for all linear forms on End(V so π(α) − idV ∈ F . If x ∈ g, then x ∈ L (x) ⊂ L(H) (24.2.3). We have therefore g = L(H), hence dim G = dim H. Since G is connected, we have G = H, and the result follows.

a

a

24.3.3 Corollary. Let G be a connected algebraic group, g its Lie algebra and V a ﬁnite-dimensional rational G-module. (i) We have V G = V g . (ii) A subspace W of V is a G-submodule if and only if W is a gsubmodule.

370

24 Correspondence between groups and Lie algebras

24.3.4 Let (E, π) be a rational (not necessarily ﬁnite-dimensional) Gmodule (see 21.5.4) and V, W two ﬁnite-dimensional G-submodules of E. Denote by σ : G → GL(V ), τ : G → GL(W ) and θ : G → GL(V + W ) the corresponding subrepresentations of G. If α ∈ G, then θ(α)|V = σ(α) and θ(α)|W = τ (α). We deduce easily that for x ∈ g, dθ(x)|V = dσ(x) and dθ(x)|W = dτ (x). In particular, dσ(x)|V ∩W = dτ (x)|V ∩W and so there exists a unique (x) ∈ End(E) such that (x)|U = d(π|U )(x) for all ﬁnite-dimensional G-stable subspace U of E. We shall also denote (x) by dπ(x). It is clear that the map g → gl(E), x → dπ(x), is a Lie algebra homomorphism. Thus E is a g-module. We deduce immediately from 24.3.3 the following result: Proposition. Let G be a connected algebraic group, g its Lie algebra and E a rational G-module. (i) We have E G = E g . (ii) A subspace F of E is a G-submodule if and only if F is a g-submodule. 24.3.5 Proposition. Let H, K be algebraic subgroups of G whose Lie algebras are h and k respectively. (i) If H ⊂ K, then h ⊂ k. (ii) Assume that H and K and connected. If h ⊂ k, then H ⊂ K. (iii) We have L(H ∩ K) = h ∩ k. Proof. (i) This is a consequence of 16.3.7 and 23.2.4. (ii) If h ⊂ k, then (h) ⊂ (k). So H ⊂ K by 24.2.6. (iii) We have L(H ∩K) ⊂ h∩k by (i). On the other hand, (24.2.5). So h ∩ k ⊂ L( (h ∩ k)) ⊂ L(H ∩ K).

a a

a

a(h∩k) ⊂ H ∩K

24.3.6 Let (V, π) be a ﬁnite-dimensional rational representation of G, W a subspace of V , v ∈ V and: NG (W ) = {α ∈ G; π(α)(W ) = W } , ng (W ) = {x ∈ g; dπ(x)(W ) ⊂ W }, CG (v) = {α ∈ G; π(α)(v) = v} , cg (v) = {x ∈ g; dπ(x)(v) = 0}, CG (v) , cg (W ) = cg (v). CG (W ) = v∈W

v∈W

If H is an algebraic subgroup of G, set: NH (W ) = H ∩ NG (W ) , CH (v) = H ∩ CG (v). These are clearly algebraic subgroups of G (21.4.2). Proposition. With the above notations, we have: (i) LNG (W) = ng (W ). (ii) LCG (v) = cg (v). (iii) LCG (W ) = cg (W ). (iv) L NH (W ) = h ∩ ng (W ) , L CH (v) = h ∩ cg (v).

24.3 Invariants

371

Proof. (i) Let us ﬁx a basis (v1 , . . . , vn ) of V such that (v1 , . . . , vp ) is a basis of W . For α ∈ G, denote by [πij (α)] the matrix of π(α) with respect to this basis. Then α ∈ NG (W ) if and only if πij (α) = 0 for 1 i p and p + 1 j n. So 23.2.5 and 23.4.2 imply that L(NG (W )) ⊂ ng (W ). For f ∈ V ∗ and v ∈ V , deﬁne ϑf,v ∈ End(V )∗ by ϑf,v (ϕ) = f ϕ(v) if ϕ ∈ End(V ). If n ∈ N∗ and x ∈ g, then by 24.3.1: n f dπ(x) (v) = ϑf,v (dπ(x))n = xn (ϑf,v ◦π). n So x ∈ ng (W ) if and only if f dπ(x) (v) = 0 for all n ∈ N∗ , v ∈ W and f ∈ W ⊥ , the orthogonal of W in V ∗ . Let x ∈ ng (W ). We have therefore ∈ jx (notations of 24.2.2). It follows that if α ∈ (x), ϑf,v ◦π − ϑf,v ◦π(e) then f π(α)(v) = ϑf,v ◦π(α) = ϑf,v ◦π(e) = 0 for all v ∈ W and f ∈ W ⊥ . (x) ⊂ NG (W ). By 24.2.3 and 24.3.5, we have x ∈ L (x) ⊂ Consequently, L NG (W ) . So we have proved (i). (ii) G (v)) ⊂ cg (v), and x ∈ cg (v) if and only if As innpart (i), we have L(C = 0 for all n ∈ N∗ and f ∈ V ∗ . So we can ﬁnish our proof as ϑf,v dπ(x) in (i). p p (iii) Since CG (W ) = i=1 CG (vi ) and cg (W ) = i=1 cg (vi ), the result follows from part (ii) and 24.3.5 (iii). (iv) This follows from (i), (ii) and 24.3.5 (iii).

a

a

a

24.3.7 Let (V, σ) and (W, τ ) be ﬁnite-dimensional rational representations of G. We have already seen that there is a natural structure of rational representation π of G on Hom(V, W ) given by: for f ∈ Hom(V, W ) and v ∈ V , π(α)(f ) (v) = τ (α) f σ(α−1 )(v) . Similarly, let (V, σ ) and (W, τ ) be representations of a Lie algebra g, then there is a natural structure of representation π of g on Hom(V, W ) given by: for x ∈ g, f ∈ Hom(V, W ) and v ∈ V , π (x)(f ) (v) = τ (x) f (v) − f σ (x)(v) . Suppose that g = L(G) and σ = dσ, τ = dτ . Then we deduce easily from 23.4 that π = dπ. Now let A be a ﬁnite-dimensional k-algebra (not necessarily associative or with unit). The algebra structure on A is deﬁned by ν ∈ End(A ⊗k A, A) such that ν(a ⊗ b) = ab for all a, b ∈ A. Denote by τ (resp. τ ) the identity representation of G = GL(A) (resp. g = gl(A)) and let σ = τ ⊗ τ (see 23.4.13), σ = dσ. Construct π and π as above with V = A ⊗k A and W = A. Then we see easily that for α ∈ G (resp. x ∈ g), α ∈ Aut A (resp. x ∈ Der A) if and only if π(α)(ν) = ν (resp. π (x)(ν) = 0). Since π = dπ, it follows from 24.3.6 (ii) that: Proposition. The Lie algebra of Aut A is Der A.

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24 Correspondence between groups and Lie algebras

24.4 Functorial properties 24.4.1 Theorem. Let u : G → H be a morphism of algebraic groups and K the kernel of u. Then L(K) is the kernel of du. Proof. Let a = I(K) be the deﬁning ideal of K in A(G) and b the set √ of elements f ◦ u − f ◦ u(eG ) for f ∈ A(H). Clearly K = V(b) and so a = b. Let x ∈ L(K) and f ∈ A(H). Then by 23.2.5, we have: 0 = x f ◦u − f ◦u(eG ) = x(f ◦u) = du(x)(f ). So x belongs to the kernel of du. Let x ∈ L(G) and f ∈ A(H). Then for α ∈ G, η(x)(f ◦u)(α) = λα−1 η(x)(f◦u) (eG ) = η(x) λα−1 (f ◦u) (eG ) = x λα−1 (f ◦u) = x (λu(α)−1 f )◦u = du(x)(λu(α)−1 f ).

a

So if x ∈ ker(du), then b ⊂ jx = I( (x)) (notations as in 24.2.2). Since jx is prime, we have a ⊂ I( (x)), hence (x) ⊂ K. So by 24.2.3 and 24.3.5, x ∈ L(K).

a

a

24.4.2 Theorem. Let G be an algebraic group, Z(G) its centre, g its Lie algebra and z(g) the centre of g. (i) We have Z(G) ⊂ ker(AdG ) and L ker(AdG ) =z(g). (ii) If G is connected, then Z(G) = ker(AdG ) and L Z(G) = z(g). Proof. (i) Let α ∈ Z(G), then iα = idG . So for x ∈ g and f ∈ A(G), we have: Ad(α)(x)(f ) = diα (x)(f ) = x(f ◦iα ) = x(f ). Hence Ad(α) = idg and α ∈ ker(AdG ). The second part of (i) follows easily from 24.4.1 and 23.5.5. (ii) By 22.1.5, G has a faithful ﬁnite-dimensional rational representation (V, π). So 24.4.1 implies that the representation (V, dπ) of g is also faithful. If α ∈ G, then we have (23.5.2): dπ◦ AdG (α) = AdGL(V ) π(α) ◦dπ. So if α ∈ ker(AdG ) and x ∈ g, then (23.5.4) dπ(x) = π(α)◦dπ(x)◦π(α)−1 , which means that π(α) belongs to the centralizer of dπ(g) in End(V ). If G is connected, then 24.3.2 says that the subalgebra (with unit) of End(V ) generated by dπ(g) is equal to the subalgebra (with unit) in End(V ) generated by π(G). So if α ∈ ker AdG , then π(αβ) = π(βα) for all β ∈ G. Since π is injective, α ∈ Z(G). So by (i), we have Z(G) = ker(AdG ) and L Z(G) = z(g).

24.4 Functorial properties

373

24.4.3 Remark. If G is not connected, then it may happen that Z(G) = ker(AdG ) and L Z(G) = z(g). Let us give an example. Using the notations of example 4 of 21.1.3, take G = O2 (k). We verify easily that G has two connected components G◦ and G1 given by: # # a 0 0 b ◦ ; b ∈ k \ {0} . G = ; a ∈ k \ {0} , G1 = 0 a−1 b−1 0 Let I2 denote the identity matrix of M2 (k), then: Z(G) = {−I2 , I2 } , L Z(G) = {0}. Since G◦ = SL2 (k) ∩ D2 (k), it follows from 23.3.4, 23.4.5 and 24.3.5 (iii) that g = L(G) is the set of matrices diag(u, −u) with u ∈ k. By 23.5.4, we have ker(AdG ) = G◦ . So Z(G) = ker(AdG ) and L Z(G) = z(g). 24.4.4 Theorem. Let G be an algebraic group. (i) If G is commutative, then L(G) is commutative. (ii) If G is connected and L(G) is commutative, then G is commutative. Proof. (i) If G is commutative, then G = ker(AdG ). So L(G) is commutative (24.4.2 (i)). (ii) By 22.1.5, G has a faithful ﬁnite-dimensional rational representation (V, π). If g = L(G) is commutative, then the subalgebra (with unit) of End(V ) generated by dπ(g) is also commutative. If G is connected, then 24.3.2 implies that π(G) is commutative. Since π is injective, we conclude that G is commutative. 24.4.5 Remark. If G is not connected, then the example in 24.4.3 shows that although L(G) is commutative, G is not commutative. 24.4.6 Lemma. Let α ∈ G, H an algebraic subgroup of G and h its Lie algebra. We have: Ad(α)(h) = L(αHα−1 ). Proof. Let a = I(H) (resp. b = I(αHα−1 )) be the ideal of H (resp. αHα−1 ) in A(G). It is clear that b = {f ◦iα−1 ; f ∈ a}. Let x ∈ L(G), y = Ad(α)(x) and f ∈ A(G). Since y(f ) = x(f ◦iα ), we have by 23.2.5 that: y ∈ L(αHα−1 ) ⇔ y(b) = {0} ⇔ x(a) = {0} ⇔ x ∈ h. So we have obtained the result. 24.4.7 Theorem. Let G be an algebraic group, H an algebraic subgroup of G, g = L(G) and h = L(H). (i) If H is normal in G, then h is an ideal of g. (ii) Assume that G and H are connected. If h is an ideal of g, then H is normal in G.

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24 Correspondence between groups and Lie algebras

Proof. We have already seen (i) in 23.5.6. Let us prove (ii). We have ad(g)(h) ⊂ h, so Ad(G)(h) = h (24.3.3). It follows from 24.4.6 that for all α ∈ G, L(αHα−1 ) = h, hence αHα−1 = H (24.3.5). 24.4.8 Theorem. Let G be an algebraic group, H an algebraic subgroup of G, g = L(G) and h = L(H). (i) We have L NG (H) ⊂ ng (h) and L CG (H) ⊂ cg (h). (ii) If H is connected, then: N G (H) = NG (h) , CG (H) = CG (h), L NG (H) = L NG (h) = ng (h) , L CG (H) = L CG (h) = cg (h). Moreover, Proof. The groups NG (H) and CG (H) are closed in G (21.3.7). (H), h is an ideal of L N (H) (24.4.7 (i)), so since H is normal in N G G L NG (H) ⊂ ng (h). Let (V, π) be a ﬁnite-dimensional faithful rational representation of G (22.1.5), A (resp. B) the subalgebra with unit of End(V ) generated by π(H) (resp. dπ(h)). By 24.3.2 (i), we have B ⊂ A. It follows that CG (H) ⊂ CG (h). Therefore, by 24.3.5 (i) and 24.3.6 (iii), we have L CG (H) ⊂ cg (h). If H is connected, then A = B (24.3.2). Thus NG (H) = NG (h) et CG (H) = CG (h) and the rest follows from 24.3.6. 24.4.9 Let H, K be subgroups of G. Recall that (H, K) denotes the subgroup of G generated by αβα−1 β −1 , α ∈ H, β ∈ K. Recall also that the derived subgroup (G, G) of G is closed (21.3.5) and it is connected if G is connected (21.3.6). Proposition. Let H, K be closed subgroups of G, L the closure of (H, K) in G (it is a subgroup by 21.2.2), and g, h, k, l the Lie algebras of G, H, K, L respectively. Let α ∈ H, β ∈ K, x ∈ h, and y ∈ k. Then [x, y] , Ad(α)(y) − y , Ad(β)(x) − x are elements of l. Proof. For α ∈ H, denote by uα : K → L the map β → αβα−1 β −1 . If y ∈ k, then duα (y) = Ad(α)(y) − y (see 23.4.10). Thus Ad(α)(y) − y ∈ l. Similarly, we have Ad(β)(x) − x ∈ l. It follows that for x ∈ h, we have a map vx : K → l given by vx (β) = Ad(β)(x) − x for β ∈ K. If y ∈ k, then dvx (y) = [y, x]. Hence [x, y] ∈ l. 24.4.10 Corollary. Let G be an algebraic group and g its Lie algebra. Then [g, g] ⊂ L (G, G) . 24.4.11 Remark. We shall see later that [g, g] = L (G, G) if G is connected. This is however not true if G is not connected. In the example of 24.4.3, [g, g] = {0}, (G, G) = G◦ and so L (G, G) = g.

24.5 Algebraic Lie subalgebras

375

24.5 Algebraic Lie subalgebras 24.5.1 In this section, G will denote an algebraic group, g its Lie algebra. Recall that if x ∈ g (resp. h is a subspace of g), then (x) (resp. (h)) is the intersection of algebraic subgroups of G whose Lie algebra contains x (resp. h). Moreover, (x) and (h) are connected and (x) is commutative (24.2.3 and 24.2.5). Let m be a subset of g and H an algebraic subgroup of G such that m ⊂ L(H). Then (x) ⊂ H for all x ∈ m, and so H contains the subgroup K generated by the (x)’s, x ∈ m. Since K is a connected algebraic subgroup of G (21.3.2) whose Lie algebra contains m (24.3.5), we deduce that K is the smallest algebraic subgroup of G whose Lie algebra contains m. We shall denote K by (m).

a

a

a

a

a

a a

a

Deﬁnition. A Lie subalgebra h of g is called algebraic if h = L(H) for some algebraic subgroup H of G. 24.5.2 Proposition. Let (hi )i∈I be a family of algebraic Lie subalgebras of g, and Hi an algebraic subgroup of G such that L(Hi ) = hi . Set: h= hi , H = Hi . i∈I

i∈I

Then h is an algebraic Lie subalgebra of g, and h = L(H). Proof. The group H is closed in G and veriﬁes L(H) ⊂ h by 24.3.5. Moreover,

a(h) ⊂ Hi, so a(h) ⊂ H. Thus h ⊂ La(h) ⊂ L(H).

24.5.3 Proposition. Let u : G → H be a morphism of algebraic groups, and h = L(H). (i) The Lie algebra of u(G) is du(g). (ii) If k is an algebraic Lie subalgebra of g, then du(k) is an algebraic Lie subalgebra of h. (iii) If l is an algebraic Lie subalgebra of h, then (du)−1 (l) is an algebraic Lie subalgebra of g. Proof. (i) By 21.2.4, u(G) is a closed subgroup of H, and we may assume that G is connected. The morphism v : G → u(G) induced by u is dominant. Let α, β ∈ G. Denote by w : G → G (resp. w : u(G) → u(G)) the automorphism of varieties deﬁned by γ → (αβ −1 )γ (resp. δ → v(αβ −1 )δ). We have w(β) = α, w (v(β)) = v(α) and v ◦ w = w ◦ v. So dwβ (resp. dwv(β) ) is an isomorphism from Tβ (G) onto Tα (G) (resp. from Tv(β) (u(G)) onto Tv(α) (u(G))) and we have dvα ◦ dwβ = dwv(β) ◦ dvβ . Thus 16.5.7 (ii) implies that du induces a surjection from g onto L(u(G)). (ii) We may assume that k = g. So the result follows from (i). (iii) Applying (i) and 24.5.2, we may assume that u and du are surjective. Let L = (l) and K = u−1 (L). Since ker u ⊂ K, ker du = L(ker u) ⊂ L(K) (24.3.5 and 24.4.1). Applying (i) to the surjective morphism K → L induced

a

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24 Correspondence between groups and Lie algebras

by u, we obtain that du(L(K)) = l. So we have (du)−1 (l) = L(K) + ker du = L(K). 24.5.4 Let m be a non-empty subset of g. Denote by a(m) the intersection of all the algebraic Lie subalgebras of g containing m. By 24.5.2, a(m) is the smallest algebraic Lie subalgebra of g containing m. Clearly, we have a(m) = L( (m)). If m is a Lie subalgebra of g, then we call a(m) the algebraic hull of m in g.

a

Proposition. Let u : G → H be a morphism of algebraic groups and m a non-empty of subset g. Then: (i) du a(m) = a du(m) . du(m) . (ii) u (m) =

a

a

Proof. (i) By 24.5.3, du(a(m)) is an algebraic Lie subalgebra of L(H) containing du(m), so it contains a(du(m)). Conversely, again by 24.5.3, (du)−1 (a(du(m))) is an algebraic Lie subalgebra of g containing m, so it contains a(m). It follows that du(a(m)) ⊂ a(du(m)). (ii) We have L( (du(m))) = a(du(m)) and L(u( (m))) = du(a(m)) (24.5.3). Since the groups u( (m)) and (du(m)) are connected, the result follows from (i) and 24.3.5.

a

a

a

a

24.5.5 Theorem. Let (V, π) be a ﬁnite-dimensional rational representation of G, and E, F vector subspaces of V such that E ⊂ F . Then H = {α ∈ G; (π(α) − idV )(F ) ⊂ E} is an algebraic subgroup of G and h = {x ∈ g; dπ(x)(F ) ⊂ E} is a Lie subalgebra of g verifying h = L(H). Proof. By deﬁnition, H ⊂ NG (F ) and h ⊂ ng (F ). So we may assume that V = F , G = NG (F ) and g = ng (F ) (24.3.6). Let B = (v1 , . . . , vn ) be a basis of V such that (v1 , . . . , vp ) is a basis of E. For α ∈ G and x ∈ g, denote by [πij (α)] (resp. [xij ]) the matrix of π(α) (resp. dπ(x)) with respect to B. Then x ∈ h (resp. α ∈ H) if and only if xij = 0 (resp. πij (α) = δij where δij denotes the Kronecker symbol) for p + 1 i n and 1 j n. We deduce therefore that H is an algebraic subgroup of G, h is a Lie subalgebra of g, and we have L(H) ⊂ h by ∗23.4.2. For f ∈ V ∗ and v ∈ V , deﬁne ϑf,v ∈ End(V ) as in the proof of 24.3.6, and denote by E ⊥ the orthogonal of E in V ∗ . We have: x ∈ h ⇔ xn (ϑf,v ◦π) = 0 for all n ∈ N∗ , (f, v) ∈ E ⊥ × V, α ∈ H ⇔ ϑf,v ◦π(α) = ϑf,v ◦π(e) for all (f, v) ∈ E ⊥ × V. Proceeding as in the proof of 24.3.6, we obtain that h ⊂ L(H).

24.5.6 Corollary. Let m, n be subspaces of g such that m ⊂ n. Then H = {α ∈ G; (Ad(α) − idg )(n) ⊂ m} is an algebraic subgroup of G, and

24.5 Algebraic Lie subalgebras

377

h = {x ∈ g; ad(x)(n) ⊂ m} is an algebraic Lie subalgebra of g verifying h = L(H). 24.5.7 Theorem. Let h be a Lie subalgebra of g. (i) Any ideal of h is an ideal of a(h). (ii) We have [a(h), a(h)] = [h, h]. Proof. (i) Let a be an ideal of h. By 24.5.6, the set of x ∈ g verifying [x, a] ⊂ a is an algebraic Lie subalgebra k of g. Since h ⊂ k, we have a(h) ⊂ k. Thus a is an ideal of a(h). (ii) As in (i), the set of x ∈ g verifying [x, h] ⊂ [h, h] contains a(h). So the set of y ∈ g such that [y, a(h)] ⊂ [h, h] is an algebraic Lie subalgebra g containing h, so it contains a(h). We deduce therefore that [a(h), a(h)] ⊂ [h, h]. Since h ⊂ a(h), we have [a(h), a(h)] ⊃ [h, h]. 24.5.8 Let (Xi )i∈I be a family of irreducible subvarieties of G. Suppose that the following conditions are satisﬁed: (i) e ∈ Xi for all i ∈ I. (ii) If i ∈ I, there exists j ∈ I such that Xi−1 = {α−1 ; α ∈ Xi } = Xj . Let H be the subgroup of G generated by the Xi ’s. By 21.3.1, H is a connected algebraic subgroup of G. Let h be its Lie algebra. Proposition. The vector space h is spanned by the vector subspaces Ad(α)(Te (βi−1 Xi )) of g, for α ∈ H, i ∈ I and βi ∈ Xi . Proof. 1) By 21.3.1, there exist i1 , . . . , in ∈ I such that the morphism u : Xi1 × · · · × Xin → H , (β1 , . . . , βn ) → β1 · · · βn is surjective. To simplify notations let us write Xk for Xik if 1 k n. Let X = X1 × · · · × X n . There exists a = (α1 , . . . , αn ) ∈ X such that the map dua : Ta (X) → Tu(a) (H) is surjective (16.5.7). For 1 i n, set: γi = α1 · · · αi , Yi = αi−1 Xi , Zi = γi Yi γi−1 . Let Y = Y1 × · · · × Yn and Z = Z1 × · · · × Zn . Deﬁne the following maps: s : Y → X , (β1 , . . . , βn ) → (α1 β1 , . . . , αn βn ), t : Y → Z , (β1 , . . . , βn ) → (γ1 β1 γ1−1 , . . . , γn βn γn−1 ), v : Z → H , (β1 , . . . , βn ) → β1 · · · βn , −1 . θ : H → H , β → β u(a) We verify easily that θ◦u◦s = v◦t. Moreover if ε = (e, . . . , e) ∈ Y , then s(ε) = a. The maps s and θ are isomorphisms of varieties and dua is surjective. It follows that the diﬀerential

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24 Correspondence between groups and Lie algebras

d(v◦t)ε : Tε (Y ) → Te (H) = h of v◦t = θ◦u◦s at the point ε is surjective. 2) Let x = (x1 , . . . , xn ) ∈ Tε (Y ), so xi ∈ Te (αi−1 Xi ) for 1 i n. By 23.4.7, we have: d(v◦t)ε (x) =

n k=1

d(iγk )e (xk ) =

n

Ad(γk )(xk ).

k=1

Since γk ∈ H and d(v◦t)ε is surjective, we have proved that h is contained in the sum of the vector subspaces Ad(α)(Te (βi−1 Xi )), for α ∈ H, i ∈ I and βi ∈ Xi . Now, each Ad(α)(Te (βi−1 Xi )) is clearly contained in h, so the result follows. 24.5.9 Theorem. Let (Hi )i∈I be a family of connected algebraic subgroups of G, H the connected algebraic subgroup of G generated by the Hi ’s (see 21.3.2) and h, hi the Lie algebras of H, Hi , i ∈ I respectively. (i) The vector space h is the sum of the vector spaces Ad(α)(hi ), α ∈ H, i ∈ I. (ii) As a Lie algebra, h is generated by hi , i ∈ I. Proof. (i) Since Hi is a group, we have βi−1 Hi = Hi for all βi ∈ Hi and Te (βi−1 Hi ) = hi . So the result follows from 24.5.8. (ii) Let k be the Lie subalgebra of g generated by hi , i ∈ I. Since Hi ⊂ H, hi ⊂ h, and as h is a Lie algebra, we deduce that k ⊂ h. To show that h ⊂ k, it suﬃces, by (i), to establish that Ad(α)(k) ⊂ k for α ∈ H, that is, H ⊂ NG (k). Since NG (k) is a group, we are reduced to prove that Hi ⊂ NG (k) for all i ∈ I. The Hi ’s are connected, so this is equivalent to proving that hi ⊂ ng (k) (24.4.8). But this is obvious since hi ⊂ k. 24.5.10 Corollary. The following conditions are equivalent for a Lie subalgebra h of g: (i) h is an algebraic Lie subalgebra of g. (ii) For all x ∈ h, a(x) ⊂ h. (iii) The vector space h is a sum of algebraic Lie subalgebras of g. (iv) As a Lie algebra, h is generated by algebraic Lie subalgebras of g. Proof. The implications (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) are clear, while (iv) ⇒ (i) follows from 24.5.9. 24.5.11 Let H, K be connected normal algebraic subgroups of G. By 21.3.5, L = (H, K) is a connected algebraic subgroup of G. Let h, k, l be the Lie algebras of H, K, L. Theorem. We have l = [h, k].

24.5 Algebraic Lie subalgebras

379

Proof. 1) We have [h, k] ⊂ l by 24.4.9. 2) For α, β ∈ G, let cα (β) = αβα−1 β −1 . Since (αβα−1 β −1 )−1 = βαβ −1 α−1 = α−1 (αβα−1 )α(αβα−1 )−1 , we deduce that cα (K)−1 = cα−1 (K). By 21.3.1, there exist n ∈ N∗ and α1 , . . . , αn ∈ H such that the map u : K n → L , (γ1 , . . . , γn ) → cα1 (γ1 ) · · · cαn (γn ) is surjective. So by 16.5.7, there exists b = (β1 , . . . , βn ) ∈ K n such that dub : Tb (K n ) → Tu(b) (L) is surjective. 3) Let s : K n → K n and θ : L → L be deﬁned as follows: −1 . s(γ1 , . . . , γn ) = (β1 γ1 , . . . , βn γn ) , θ(γ) = γ u(b) These are isomorphisms of varieties. For 1 k n and γ ∈ K, let: −1 −1 Ck (γ) = cα1 (β1 ) · · · cαk−1 (βk−1 )cαk (βk γ) cαk (βk ) · · · cα1 (β1 ) . Finally, deﬁne t : K n → K n and v : K n → H by: t(γ1 , . . . , γn ) = C1 (γ1 ), . . . , Cn (γn ) , v(γ1 , . . . , γn ) = γ1 · · · γn . We check easily that: v◦t = θ◦u◦s. Denote by ε = (e, . . . , e) ∈ K n . Since θ and s are isomorphisms of varieties, the diﬀerential d(v◦t)ε : Tε (K n ) → Te (L) = l of v◦t at the point ε is surjective. Let δk = cα1 (β1 ) · · · cαk (βk ), 1 k n. Then we obtain easily that: Ck (γ) = iδk iβk αk (γ)iβk (γ −1 ) . n If x = (x1 , . . . , xn ) ∈ Te (K) = Tε (K n ), then 23.4.7 implies that: d(v◦t)ε (x) =

n

Ad(δk ) Ad(βk ) Ad(αk )(xk ) − xk .

k=1

Since [k, [h, k]] ⊂ [h, k] and K is connected, 24.5.5 implies that Ad(γ)([h, k]) ⊂ [h, k] for all γ ∈ K. As δk , βk ∈ K, to obtain the result, it suﬃces therefore to prove that if x ∈ k and α ∈ H, then Ad(α)(x) − x ∈ [h, k]. But this follows again from 24.5.5 and the fact that H is connected. 24.5.12 Theorem. Let h be a Lie subalgebra of g. Then [h, h] is an algebraic Lie subalgebra of g. Furthermore, if H is a connected subgroup of G such that h = L(H), then [h, h] is the Lie algebra of D(H).

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Proof. Since [h, h] = [a(h), a(h)] (24.5.7), we may assume that h is algebraic. Then H = (h) and the result follows from 24.5.11.

a

24.5.13 Corollary. (i) If G is solvable (resp. nilpotent), then g is solvable (resp. nilpotent). (ii) Assume that G is connected. If g is solvable (resp. nilpotent), then G is solvable (resp. nilpotent). Proof. We have L(G) = L(G◦ ) and if G is solvable (resp. nilpotent), then so is G◦ . So we may assume that G is connected. Then 24.5.11 implies that L Dn (G) = Dn (g) and L(C n (G) = C n (g) (notations of 19.3.1). So the result follows.

24.6 A particular case 24.6.1 In this section, G denotes an algebraic group and g its Lie algebra. If x ∈ g, then an element of a(x) is called a replica of x. and nilpotent components. 24.6.2 Let x ∈ g and xs , xn its semisimple 1) We have xs , xn ∈ a(x) = L (x) . Thus a(xs ) ⊂ a(x), a(xn ) ⊂ a(x) (24.5.10), and (xs ) ⊂ (x), (xn ) ⊂ (x) (24.2.3). Note that the Lie algebra a(x) is commutative (24.4.4). 2) Let H ⊂ (x) be the subgroup of G generated by (xs ) and (xn ). Then H is a connected algebraic subgroup of G (21.3.1). Moreover, xs ∈ L (xs ) ⊂ L(H) and xn ∈ L( (xn ) ⊂ L(H). Hence x ∈ L(H), and (x) ⊂ H (24.2.3). It follows that H = (x). Since (x) is commutative, (x) = (xs ) (xn ). Applying 24.5.9 we obtain that a(x) = a(xs ) + a(xn ). 3) Let us reduce to the case where G is a closed subgroup of GL(V ) with V a ﬁnite-dimensional vector space (22.1.5). Suppose that x is semisimple. Using a basis of diagonalization for x, we may assume that G ⊂ GLn (k) and x ∈ dn (k). Since dn (k) = L Dn (k) , (x) ⊂ Dn (k) and a(x) ⊂ dn (k). Thus the elements of (x) (resp. a(x)) are semisimple. Suppose that x is nilpotent. We may assume that x ∈ nn (k). Again we see that (x) ⊂ Un (k) and a(x) ⊂ nn (k). Thus the elements of (x) are unipotent, and those of a(x) are nilpotent. In the general case, the preceding discussion shows that a(xs ) (resp. a(xn )) is the set of semisimple (resp. nilpotent) elements of a(x). Similarly, (xs ) (resp. (xn )) is the set of semisimple (resp. unipotent) elements of (x). As in 22.2.6, we obtain:

a a

a a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

Proposition. Let us conserve the above notations. (i) We have a(x) = a(xs ) ⊕ a(xn ). (ii) The map (xs ) × (xn ) → (x), (α, β) → αβ, is an isomorphism of algebraic groups.

a

a

a

a

24.6.3 Suppose that G is a closed subgroup of GL(V ). To determine (x) (resp. a(x)), it suﬃces to know (xs ) and (xn ) (resp. a(xs ) and a(xn )).

a

a

24.6 A particular case

381

a

For (xn ) and a(xn ), we may use 23.3.5 and 24.2.4. Let us consider xs . For simplicity, let us suppose that x semi-simple. We may assume, via a basis of diagonalization, that G ⊂ GLn (k) and x = diag(s1 , . . . , sn ) where s1 , . . . , sn ∈ k. Denote by: Lx = {(a1 , . . . , an ) ∈ Zn ; a1 s1 + · · · + an sn = 0}.

a

We saw in 24.6.2 that (x) ⊂ Dn (k) and a(x) ⊂ dn (k). Moreover, by 22.5.4 and 22.5.7, (x) is a torus and it is the intersection of the kernel of certain characters of Dn (k). Let H = Dn (k), h = dn (k), A(H) = k[T1 , T1−1 , . . . , Tn , Tn−1 ] and we identify L(H) with h via the map y → [y(Ti )] (see 23.3.3 and 23.3.4). Thus if P ∈ A(H), then: n ∂P x(P ) = si (e). ∂Ti i=1

a

Let of H. By 22.5.2, there exist integers a1 , . . . , an such that χ be a character χ diag(t1 , . . . , tn ) = ta1 1 · · · tann for all t1 , . . . , tn ∈ k \ {0}. If Q ∈ k[T, T −1 ] ∈ A(Gm ), then: n n ∂(Q◦χ) si (e) = si ai Q (1). dχ(x)(Q) = x(Q◦χ) = ∂Ti i=1 i=1 Hence (23.3.2):

d dχ(x) = (a1 s1 + · · · + an sn ) . dT T =1

a

a

By 24.4.1 and the deﬁnition of (x), we deduce that (x) ⊂ ker χ if and only if (a1 , . . . , an ) ∈ Lx . Since (x) is the intersection of the kernel of certain characters of H, we have proved that (x) is the set of matrices diag(t1 , . . . , tn ) such that ta1 1 · · · tann = 1 for all (a1 , . . . , an ) ∈ Lx . On the other hand, Lx is a subgroup of Zn , so it is free and ﬁnitely generated. Applying 24.3.5 (iii) and 24.4.1, we see that a(x) is the set of matrices diag(u1 , . . . , un ) such that a1 u1 + · · · + an un = 0 for all (a1 , . . . , an ) ∈ Lx . To summarize, we have obtained the following result:

a

a

Theorem. Let x = diag(s1 , . . . , sn ) ∈ dn (k) and

a

Lx = {(a1 , . . . , an ) ∈ Zn ; a1 s1 + · · · + an sn = 0}.

The group (x) is the set of matrices diag(t1 , . . . , tn ) ∈ Dn (k) verifying ta1 1 · · · tann = 1 for all (a1 , . . . , an ) ∈ Lx . The Lie algebra a(x) is the set of matrices diag(u1 , . . . , un ) ∈ dn (k) such that a1 u1 + · · · + an un = 0 for all (a1 , . . . , an ) ∈ Lx . 24.6.4 Proposition. Let V be a ﬁnite-dimensional vector space and let x ∈ gl(V ). (i) Any element of (x) can be written as P (x) for some P ∈ k[T ]. (ii) Any replica of x can be written as Q(x) for some Q ∈ k[T ] without constant term.

a

382

24 Correspondence between groups and Lie algebras

a

a

a

Proof. Since (x) = (xs ) (xn ), a(x) = a(xs ) + a(xn ) (24.6.2) and xs , xn are polynomials without constant term in x, it suﬃces to consider on the one hand the case where x is nilpotent, and on the other hand, the case where x is semisimple. If x is nilpotent, then (x) = {etx ; t ∈ k} and a(x) = kx (23.3.5 and 24.2.4). So the result is clear. Suppose that x is semisimple. Denote by s1 , . . . , sp the pairwise distinct eigenvalues of x and V1 , . . . , Vp the corresponding eigenspaces. For i ∈ {1, . . . , p} and v ∈ Vi , set:

a

Gv = {α ∈ GL(V ); α(kv) ⊂ kv} , gv = {y ∈ gl(V ); y(kv) ⊂ kv}. By 24.5.6, Gv is an algebraic subgroup of GL(V ) whose Lie algebra is gv . Since x ∈ gv , we deduce easily that for all α ∈ (x) and y ∈ a(x), V1 , . . . , Vp are eigenspaces of α and y. So let α ∈ (x) and t1 , . . . , tp ∈ k \ {0} be such that α|Vi = ti idVi for 1 i p. There exists P ∈ k[T ] such that ti = P (si ) for 1 i p. Thus α = P (x), and we have proved (i). We see similarly that if y ∈ a(x), then there exists Q ∈ k[T ] such that y = Q(x). If x is invertible, by using the minimal polynomial of x, we see that idV is a polynomial without constant term in x, so we may assume that Q has no constant term. If x is not invertible, then we have, for example, s1 = 0. The set of elements y ∈ gl(V ) verifying y|V1 = 0 is an algebraic Lie subalgebra of gl(V ) containing x (24.5.6), hence y|V1 = 0 for all y ∈ a(x). But y = Q(x), so Q has no constant term.

a

a

24.6.5 Proposition. Let V be a ﬁnite-dimensional vector space and x ∈ gl(V ). Then x is nilpotent if and only if tr(xy) = 0 for all replica y of x. Proof. If x is nilpotent, then so is xp for all p ∈ N∗ , so tr(xp ) = 0. So 24.6.4 implies that tr(xy) = 0 for all replica y of x. Conversely, suppose that tr(xy) = 0 for all replica y of x. Since xy = yx, xn y = yxn and xn y is nilpotent. Consequently, tr(xs y) = 0 for all replica y of x. Since a(xs ) ⊂ a(x), it suﬃces to prove that if x is semisimple and tr(xy) = 0 for all y ∈ a(x), then x = 0. Let (v1 , . . . , vr ) be a basis of V consisting of eigenvectors of x with eigenvalues s1 , . . . , sr , and E the Q-vector subspace of k spanned by the si ’s. For a Q-linear form ϕ on E, deﬁne y ∈ gl(V ) to be the element such that y(vi ) = ϕ(si )vi for 1 i r. It is clear from 24.6.3 that y ∈ a(x). So: tr(xy) = s1 ϕ(s1 ) + · · · + sr ϕ(sr ) = 0. We deduce from this that: 2 2 ϕ(s1 ) + · · · + ϕ(sr ) = 0. Thus ϕ(s1 ) = · · · = ϕ(sr ) = 0. Since the si ’s span E, we have E = {0}.

24.8 Algebraic adjoint group

383

24.7 Examples 24.7.1 In this section, G denotes an algebraic group whose Lie algebra is g. We shall give examples of algebraic Lie subalgebras of g. 24.7.2 Let h be a Lie subalgebra of g. • If h = [h, h], then h is an algebraic Lie subalgebra of g (24.5.12). In particular, if h is semisimple, h is an algebraic Lie subalgebra of g. • Suppose that h = ng (a), where a is a Lie subalgebra of g. Then by 24.5.6, h is algebraic in g. In particular, if h is its own normalizer in g, then it is an algebraic Lie subalgebra of g. • Similarly, if h = cg (a), where a is a subspace of g, then h is an algebraic Lie subalgebra of g. 24.7.3 Lemma. Let h be an ideal of g. Then [g, a(h)] ⊂ h. In particular, a(h) is an ideal of g. Proof. Let k = {x ∈ g; [x, g] ⊂ h}. Then 24.5.6 says that k is algebraic in g. If h is an ideal of g, then h ⊂ k. Hence a(h) ⊂ k. 24.7.4 Proposition. Let h be a Lie subalgebra of g and r the radical of h. The following conditions are equivalent: (i) h is an algebraic Lie subalgebra of g. (ii) r is an algebraic Lie subalgebra of g. Proof. (i) ⇒ (ii) Suppose that h = a(h). To obtain the result, we may assume that g = h. So r is the largest solvable ideal of g. By 24.7.3, a(r) is an ideal of g and it is solvable since [a(r), a(r)] = [r, r] (24.5.7). Hence r = a(r). (ii) ⇒ (i) By 20.3.5, h = s ⊕ r where s is a semisimple Lie subalgebra of g. Since s is algebraic (24.7.2), if r is algebraic, then so is h (24.5.10). 24.7.5 Remark. Let r be the radical of h and s a semisimple Lie subalgebra of g verifying h = s ⊕ r. Then we obtain easily that a(h) = s ⊕ a(r). Furthermore, a(r) is the radical of a(h). 24.7.6 Let h be a Lie subalgebra of g such that x = xn for all x ∈ h. We claim that h is algebraic in g. We may assume that G = GL(V ). By 23.6.4, any x ∈ h is a nilpotent endomorphism. So by 23.3.5, kx is algebraic in g for all x ∈ h. Hence 24.5.10 implies that h is an algebraic Lie subalgebra of g.

24.8 Algebraic adjoint group 24.8.1 Deﬁnition. We deﬁne the algebraic adjoint group of a Lie algebra g to be the smallest algebraic subgroup of GL(g) whose Lie algebra contains adg g.

a

24.8.2 For any Lie subalgebra h of gl(g), denote by (h) the smallest algebraic subgroup of GL(g) whose Lie algebra contains h, and a(h) the smallest algebraic Lie subalgebra of gl(g) containing h.

384

24 Correspondence between groups and Lie algebras

a

The algebraic adjoint group of g is therefore G = (ad g) and we have L(G) = a(ad g). If ad g is an algebraic Lie subalgebra of gl(g), then we say that G is the adjoint group of g, and that g is ad-algebraic. Let Aut g be the group of automorphisms of g. We have already seen that Aut g is an algebraic subgroup of GL(g) whose Lie algebra is Der g (24.3.7). So G ⊂ Aut g. If α ∈ Aut g, then α normalizes ad g, and therefore normalizes G. Thus G is a normal subgroup of Aut g. Applying 22.3.4 and 23.3.5, we see also that G contains the group of elementary automorphisms of g. We have also from 23.4.16 that G = Aute g if g is semisimple. 24.8.3 Let g be a Lie algebra and G its algebraic adjoint group. • Let x ∈ g. If α ∈ (ad x), there exists P ∈ k[T ] such that α = P (ad x) (24.6.4). It follows that any ideal of g is G-stable. Conversely, if a subspace of g is G-stable, then it is also ad g-stable (24.3.3). Thus G-stable subspaces of g are exactly the ideals of g. • The derivation ad x of g extends to a derivation of the symmetric algebra S(g) of g, denoted by adS(g) x. An ideal of S(g) is said to be g-invariant if it is stable under adS(g) x for all x ∈ g. In view of 23.4.14 and 24.3.4, an ideal of S(g) is g-invariant if and only if it is G-stable. By 24.3.4, we also have:

a

S(g)G = S(g)g . 24.8.4 Let us look at an example. Let α ∈ k \ {0} and g the Lie algebra with basis (x, y, z), and Lie bracket deﬁned as follows: [x, y] = y , [x, z] = αz , [y, z] = 0, all other brackets can be deduced from these. Let X = ad x, Y = ad y, Z = ad z. To determine the algebraic adjoint group G of g and L(G), it suﬃces to determine (X), (Y ), (Z), and a(X), a(Y ), a(Z) (24.5.9). We shall identify an element of End(g) with its matrix with respect to the basis (x, y, z). By 23.3.5 and 24.2.4, we have:

a

a

a(Y ) = {etY ; t ∈ k} , a(Z) = {etZ ; t ∈ k} , a(Y ) = kY

a

, a(Z) = kZ.

The eigenvalues of X are 0, 1, α. In the notations of 24.6.3, LX is the set of elements (a1 , a2 , a3 ) ∈ Z3 such that a2 +a3 α = 0. Let us distinguish two cases: r 1) α = ∈ Q where r, s ∈ Z \ {0} are coprime. Then: s LX = {(n, pr, −ps); n, p ∈ Z}.

a

Thus a(X) = kX. Therefore g is ad-algebraic. The group (X) is the set of −ps = 1 for all n, p ∈ Z. So it is the set of matrices diag(t1 , t2 , t3 ) with tn1 tpr 2 t3 s r matrices diag(1, t , t ) where t ∈ k \ {0}. We deduce easily that:

24.8 Algebraic adjoint group

385

⎫ ⎧⎛ ⎞ ⎬ ⎨ 1 0 0 G = ⎝ u ts 0 ⎠ ; u, v ∈ k , t ∈ k \ {0} . ⎭ ⎩ v 0 tr 2) α ∈ Q. Then LX = Z × {0} × {0}. Thus a(X) is 2-dimensional and it is spanned, as a vector space, by X and T = diag(0, 0, 1). We have a(ad g) = kX ⊕ kT ⊕ kY ⊕ kZ, so a(ad g) is of dimension 4. The group (X) is the set of matrices diag(1, t, t ) with tt = 0. We obtain that: ⎧⎛ ⎫ ⎞ ⎨ 100 ⎬ G = ⎝ u t 0 ⎠ ; u, v ∈ k , t, t ∈ k\{0} . ⎩ ⎭ v 0 t

a

In particular, g is not ad-algebraic. 24.8.5 Proposition. Let g be a Lie algebra, G its algebraic adjoint group, h a Lie subalgebra of g, K = (adg h) and k = a(adg h). (i) For x ∈ h, α ∈ K and X ∈ k, we have α(x) ∈ h and X(x) ∈ h. (ii) Let L = {α|h ; α ∈ K} and l = {X|h ; X ∈ k}. Then L is the algebraic adjoint group of h, and l is the smallest algebraic Lie subalgebra of gl(h) containing adh h.

a

a

Proof. (i) Since K is generated by the (adg y), y ∈ h, the fact that K(h) ⊂ h follows from 24.6.4 (i). In the same way, X(h) ⊂ h follows from 24.6.4 (ii). (ii) Let u : K → GL(h) be the map α → α|h . Then is a rational (h, u) du(adg h) by representation of K. We have L = u(K) = u (adg h) = 24.5.4, hence u(K) = (adh h). Thus L is the algebraic adjoint group of h. The second part of (ii) is clear since l = L(u(K)) by 24.5.3.

a

a

a

References and comments • [5], [23], [38], [40], [78]. The presentation of 24.2 given here is inspired by [38]. Most of the notions (like replica) of sections 24.5, 24.6, 24.7 and 24.8 are taken from [23]. The fact that the base ﬁeld is of characteristic zero is essential in establishing the results of this chapter. When the characteristic of the base ﬁeld is positive, the correspondence between algebraic groups and their Lie algebras becomes less “nice”.

25 Homogeneous spaces and quotients

We consider in this chapter actions of an algebraic group on an algebraic variety. The notions of homogeneous space and geometric quotient are studied, and we show that the quotient of an algebraic group by a closed subgroup has a natural structure of algebraic variety. In this chapter, G is an algebraic group.

25.1 Homogeneous spaces 25.1.1 Proposition. Let X be an irreducible G-variety. If X is a Ghomogeneous space, then X is a smooth normal variety. Proof. This follows from 16.5.5 and 17.1.5. 25.1.2 Theorem. Let X, Y be G-varieties, u : X → Y a G-equivariant morphism and r = dim X − dim Y . Suppose further that X and Y are Ghomogeneous spaces. (i) The map u is open. For all y ∈ Y , each irreducible component of u−1 (y) has dimension r. (ii) Let W be a closed irreducible subvariety of Y and Z an irreducible component of u−1 (W ). Then Z dominates W and r = dim Z − dim W . (iii) Let Z be a variety and v : X × Z → Y × Z be the morphism deﬁned by (x, z) → (u(x), z). Then v is open. (iv) The map u is an isomorphism if and only if u is bijective. Proof. In view of 21.4.4, we may reduce easily to the case where G is connected and X, Y are irreducible. Note that since X and Y are G-homogeneous spaces and u is G-equivariant, u is surjective. (i) By 15.5.4, there exists a non-empty open subset V of Y such that for all y ∈ V , all the irreducible components of u−1 (y) have dimension r. Let y ∈ Y . Since G acts transitively on Y , we have α.y ∈ V for some α ∈ G. It follows easily from the G-equivariance of u that if Z1 , . . . , Zs are the irreducible components of u−1 (y), then α.Z1 , . . . , α.Zs are the irreducible components of

388

25 Homogeneous spaces and quotients

u−1 (α.y). Now the map x → α.x is an isomorphism of the variety X, so dim Zi = r for all 1 i s. We deduce from 17.4.11 and 25.1.1 that u is open. (ii) By 17.1.8 and 15.5.3, there exists a non-empty open subset V of Y such that V is normal and for all y ∈ V , x ∈ u−1 (y) and irreducible subvariety Y of Y containing y, there is at least one irreducible component of u−1 (Y ) containing x which dominates Y and for such an irreducible component X , we have r = dim X − dim Y . Let Z be an irreducible component of u−1 (W ), then there exists z ∈ Z which is contained in no other irreducible components of u−1 (W ). Moreover, there exists α ∈ G such that u(α.z) = α.u(z) ∈ V . It is clear that α.Z is an irreducible component of u−1 (α.W ) containing α.z. Applying the preceding result with x = α.z, y = u(α.z), and Y = α.W , we obtain that α.Z dominates α.W and dim(α.Z) − dim(α.W ) = r. So (ii) follows. (iii) We may assume that Z is aﬃne. Moreover, if (iii) is true for Z, then it is also true for any closed subvariety of Z. So we may further assume that Z = kn . By (i), all the ﬁbres of v have dimension r. So the result follows from 17.4.11 since Y × kn is normal (17.3.2 and 25.1.1). (iv) Since Y is normal (25.1.1), the result follows from 17.4.8. 25.1.3 Let X be a G-variety and x ∈ X. The stabilizer Gx of x in G is an algebraic subgroup of G (21.4.2) and the orbit G.x of x is a locally closed subset of X (21.4.3). Further, we have by 21.4.3: dim(G.x) = dim G − dim Gx . Recall (21.4.4) that G.x is pure and its irreducible components are exactly the G◦ -orbits (thus pairwise disjoint) in G.x. Let C be an irreducible component of G.x. Then it is a G◦ -homogeneous space, so it is normal and smooth (25.1.1). Let π : G → G.x, α → α.x, be the orbit morphism. If we consider the G-action on G by left translation, then π is G-equivariant. By 25.1.2, π is open. Note also that if α ∈ G, then π −1 (α.x) = αGx . Proposition. The map dπe : L(G) → Tx (G.x) is surjective and its kernel is L(Gx ). Proof. Let Y be an irreducible component of G.x containing x. We have Tx (G.x) = Tx (Y ) since irreducible components of G.x are pairwise disjoint, and Y is the image of the restriction of π to G◦ . So we may consider π to be a surjective morphism from G◦ onto Y . By 16.5.7, there exists a non-empty open subset U of G◦ such that dπα : Tα (G◦ ) → Tα.x (Y ) is surjective for all α ∈ U . Since G◦ and Y are both G◦ -homogeneous spaces, we deduce that dπe is surjective. Thus dim ker(dπe ) = dim G − dim(G.x) = dim L(Gx ). But it is clear that L(Gx ) ⊂ ker(dπe ). So we have equality.

25.2 Some remarks

389

25.2 Some remarks 25.2.1 Let u : (X, OX ) → (Y, OY ) be a morphism of varieties, U ⊂ X and V ⊂ Y open subsets such that u(U ) ⊂ V . If g ∈ OY (V ), we deﬁne U U uU V (g) : U → k by uV (g)(x) = g ◦ u(x) for all x ∈ U . Clearly uV (g) ∈ OX (U ). 25.2.2 In this section, X denotes a G-variety. If U is an open subset of X, we denote by OX (U )G the set of f ∈ OX (U ) such that f (x) = f (y) if x and y belong to the same G-orbit. Lemma. Let U be an open subset of X, f ∈ OX (U )G and V = α∈G α.U . There exists a unique element g ∈ OX (V )G such that g|U = f . Proof. The uniqueness is obvious. So let us prove the existence of g. For α ∈ G, the map uα : α.U → U , x → α−1 .x is an isomorphism of varieties. So gα : α.U → k, x → f (α−1 .x), is an element of OX (α.U ). For α, β ∈ G and x ∈ α.U ∩ β.U , there exist y, z ∈ U such that x = α.y = β.z. It follows that y, z ∈ G.x, so gα (x) = f (y) = f (z) = gβ (y). By the properties of a sheaf, there exists g ∈ OX (V ) such that g|α.U = gα for all α ∈ G. 25.2.3 Assume that U is an open G-stable subset of X. If α ∈ G, the map α.f : U → U , x → f (α−1 .x), is an element of OX (U ). It is then easy to check that G × OX (U ) → OX (U ) , (α, f ) → α.f is an action of G on OX (U ). The set of ﬁxed points under this action is clearly OX (U )G . 25.2.4 Lemma. Assume that X is irreducible. Let U, V be non-empty open subsets of X such that U ⊂ V , and f ∈ OX (U ), g ∈ OX (V ) verifying g|U = f . Then the following conditions are equivalent: (i) g ∈ OX (V )G . (ii) f ∈ OX (U )G . Proof. (i) ⇒ (ii) This is obvious. (ii) ⇒ (i) Suppose that there exist α ∈ G and x ∈ V such that α.x ∈ V and g(x) = g(α.x). Then x ∈ α−1 .V . Set U = U ∩(α−1 .U ) and V = V ∩(α−1 .V ). Let θ : X → X ×X be the morphism y → (y, α.y). The map h = g⊗1−1⊗g belongs to OX×X (V ×V ), and the restriction of h◦θ to V belongs to OX (V ). We have h ◦ θ(x) = 0 and h ◦ θ|U = 0. But this is absurd since U is dense in V . Hence g ∈ OX (V )G . 25.2.5 Let us assume that X is irreducible. Let f ∈ OX (U ) and h ∈ R(X) the rational function deﬁned by f . Denote by V the domain of deﬁnition of h and g ∈ OX (V ) the function induced by h. If f ∈ OX (U )G , then 25.2.4 says that g ∈ OX (V )G . Hence the open subset V is G-stable (25.2.2).

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25 Homogeneous spaces and quotients

Conversely, given h ∈ R(X), V its domain of deﬁnition, and g ∈ OX (V ) the function induced by h. By 25.2.2, if g ∈ OX (V )G , then V is G-stable and g|U ∈ OX (U )G for all open subsets U in V . Let us denote by R(X)G the set of rational functions h verifying the following condition: if V is the domain of deﬁnition of h, and g ∈ OX (V ) the function induced by h, then g ∈ OX (V )G . We can interpret R(X)G in another way. Namely, let α ∈ G, U an open subset of X and uU α : α.U → U the isomorphism of varieties deﬁned by x → α−1 .x. If f ∈ OX (U ), then f ◦ uU α ∈ OX (α.U ), and if V is an open subset )| of U , then (f |V ) ◦ uVα = (f ◦ uU α α.V . Denote by uα : R(X) → R(X) the inductive limit of the family (uU α )U . Thus we obtain an action of G on R(X). Then R(X)G is the set of invariants of G under this action. 25.2.6 Assume that X is irreducible. We deduce from the preceding discussion that for a non-empty aﬃne open subset U of X, we have Fract(OX (U )G ) ⊂ R(X)G . 25.2.7 Let x ∈ X, Z = G.x, H = Gx and π : G → Z be as in 25.1.3. Endow G with the structure of H-variety via right translation. Let V be a non-empty open subset of Z and U = π −1 (V ). For α ∈ U and β ∈ H, we have αβ ∈ U , so U is H-stable, and OG (U )H is the set of functions h ∈ OG (U ) verifying f (αβ) = f (α) for all (α, β) ∈ U × H. Let g ∈ OZ (V ) and f = πVU (g) ∈ OG (U ). Then for (α, β) ∈ U × H, we have: f (αβ) = g((αβ).x) = g(α.x) = f (α). Thus f ∈ OG (U )H and the map OZ (V ) → OG (U )H , g → πVU (g), is injective. We shall show that it is bijective. Let C1 , . . . , Cn be the irreducible components of Z. By 21.4.4 and 25.1.1, they are open and closed, smooth, normal, pairwise disjoint G◦ -orbits in Z. Let Vi = V ∩ Ci and Ui = π −1 (Vi ), 1 i n. We may assume that Vi = ∅ for 1 i r and Vi = ∅ if i > r. Clearly, the open subsets Ui are pairwise disjoint and H-stable. So to obtain the result, it suﬃces to consider the case where V is irreducible. Assume that V is irreducible. Then V = Ci = G◦ .y for some i and y ∈ Z. Moreover, V is a normal variety by the preceding discussion. Let f ∈ OG (U )H . Set ∆ ⊂ U × k = {(α, f (α)); α ∈ U }. It is closed in U × k (12.5.4). Let u : G × k → Z × k , (α, λ) → (π(α), λ). By 25.1.2 (iii), u is an open morphism. Let ∆ = u(∆) ⊂ V × k. Let (π(β), f (β)) ∈ ∆ where β ∈ U , and (α, λ) ∈ (U × k) \ ∆. If (π(β), f (β)) = u(α, λ), then π(β) = π(α) and λ = f (β). So α ∈ βH and f (α) = f (β) = λ, hence (α, λ) ∈ ∆ which is absurd. It follows that u−1 (∆ ) = ∆ and ∆ is a closed subvariety of V × k. Let G◦ β1 , . . . , G◦ βr be the irreducible components of G having non-empty intersection with U . Set Ui = G◦ βi ∩ U . Then we have π(Ui ) = V because π(G◦ βi ) = G◦ .y.

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391

Let γ ∈ U \U1 , then there exists γ1 ∈ U1 such that γ.x = γ1 .x. So γ ∈ γ1 H, and hence π(γ) = π(γ1 ) and f (γ) = f (γ1 ). It follows that ∆ is the image of the morphism U1 → V × k, β → (π(β), f (β)). Since U1 is open in G◦ β1 , it is irreducible. We obtain therefore that ∆ is also irreducible. Let p : ∆ → V (resp. q : ∆ → k) be the restriction to ∆ of the canonical projection Z × k → Z (resp. Z × k → k). Then p is bijective, and so by 17.4.8, p is an isomorphism. Thus q ◦ p−1 ∈ OZ (V ) and if α ∈ U , then q ◦ p−1 (π(α)) = f (α). Hence f = πVU (q ◦ p−1 ) and we are done.

25.3 Geometric quotients 25.3.1 Deﬁnition. Let X be a G-variety. A geometric quotient of X by G is a pair (Y, π) where Y is a variety and π : X → Y is a morphism verifying the following conditions: (i) π is open, constant on G-orbits and deﬁnes a bijection from the set of G-orbits onto Y . (ii) If V is an open subset of Y and U = π −1 (V ), πVU induces a bijection from OY (V ) onto OX (U )G . 25.3.2 Let G and X be as in 25.3.1. If a geometric quotient (Y, π) of X by G exists, then the ﬁbres of π are the G-orbits, so they are closed. We can further precise condition (i) of 25.3.1. Lemma. Let X be a G-variety and θ : X → Y a surjective morphism of varieties whose ﬁbres are the G-orbits in X. Then the following conditions are equivalent: (i) The map θ is open. (ii) For all G-stable open subset U of X, θ(U ) is an open subset of Y . (iii) For all G-stable closed subset F of X, θ(F ) is a closed subset of Y . In particular, if θ is a closed map, then it is also open. Proof. Clearly, (i) ⇒ (ii). Conversely, if U is an open subset of X, then U = α∈G α.U is a G-stable open subset of X such that θ(U ) = θ(U ). So (ii) ⇒ (i). Let F be a G-stable closed subset of X, then U = X \ F is a G-stable open subset of X. Thus θ(F ) = Y \ θ(U ). So (ii) ⇒ (iii). Conversely, if U is a G-stable open subset of X, then F = X \ U is a G-stable closed subset of X and θ(U ) = Y \ θ(F ). Hence (iii) ⇒ (ii). 25.3.3 In the rest of this section, X is a G-variety. Proposition. Assume that a geometric quotient (Y, π) of X by G exists. Let Z be a variety and u : X → Z a morphism which is constant on G-orbits in X. Then there exists a unique morphism v : Y → Z such that u = v ◦ π. Proof. The hypothesis implies that there is a unique map v : Y → Z such that u = v ◦ π. Let W be an open subset of Z. Then u−1 (W ) = π −1 (v −1 (W ))

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is open in X. Since π is surjective and open, v −1 (W ) = π(u−1 (W )) is open in Y . Thus v is continuous. Now let V = v −1 (W ), U = u−1 (W ) and f ∈ OZ (W ). To obtain the result, we need to prove that f ◦ v ∈ OY (V ). But f ◦ v ◦ π = f ◦ u ∈ OX (U ) and since π is constant on G-orbits, f ◦ v ◦ π ∈ OX (U )G . As πVU induces a bijection from OY (V ) onto OX (U )G , f ◦ v ∈ OY (V ). 25.3.4 Corollary. If (Y, π) and (Y , π ) are geometric quotients of X by G, then there exists a unique isomorphism v : Y → Y such that π = v ◦ π. 25.3.5 In view of 25.3.4, we can talk about, if it exists, the geometric quotient of X by G. The existence of a geometric quotient of X by G requires very strong conditions on the G-orbits as we shall see in the following result. Proposition. Assume that X is irreducible. Let θ : X → Y be a surjective morphism of varieties whose ﬁbres are the G-orbits in X. (i) The G-orbits in X have dimension r = dim X − dim Y . (ii) If Y is normal, then (Y, θ) is the geometric quotient of X by G. Proof. (i) By 15.5.4, dim G.x r for all x ∈ X and there is a non-empty open subset V of Y such that dim G.x = r for all x ∈ θ−1 (V ). Let T = {(α, x, α.x); α ∈ G, x ∈ X} ⊂ G × X × X be the graph of the morphism (α, x) → α.x. It is closed in G × X × X (12.5.4). Let ∆ be the diagonal of X, Z = T ∩ (G × ∆) and p : Z → ∆ the canonical projection. If x ∈ X, then p−1 (x, x) = Gx × {(x, x)}, so all the irreducible components of p−1 (x, x) are of the same dimension. Let C be an irreducible component of Z containing {e} × ∆ and q = p|C . Then q is surjective and if x ∈ X, then q −1 (x, x) is the union of irreducible components of p−1 (x, x). Applying 15.5.4, we obtain that dim Gx dim C − dim ∆ for all x ∈ X and dim Gx = dim C − dim ∆ if x belongs to a non-empty open subset U of X. It follows that for x ∈ X, r dim G.x = dim G − dim Gx dim G − dim C + dim ∆ = s. Since dim G.x = r = s for x ∈ U ∩ θ−1 (V ), we obtain that dim G.x = r for all x ∈ X. (ii) Assume that Y is normal. Since G-orbits are pure varieties (21.4.4), it follows from 17.4.11 and part (i) that θ is open. Let V be a non-empty open subset of Y and U = θ−1 (V ). The open subset U is G-stable and if g ∈ OY (V ), then θVU (g) ∈ OX (U )G . The map g → θVU (g) is therefore an injection from OY (V ) into OX (U )G . We need to prove that it is surjective. Let f ∈ OX (U )G and ∆ = {(x, f (x)); x ∈ U } ⊂ U × k. By 12.5.4, ∆ is closed in U × k. Let u : X × k → Y × k be the morphism (x, λ) → (θ(x), λ).

25.4 Quotient by a subgroup

393

For x ∈ X and λ ∈ k, u−1 (θ(x), λ) = G.x × {λ}. So all the irreducible components of the ﬁbres of u have dimension r. Since Y ×k is normal (17.3.2), it follows from 17.4.11 that u is open. Since X is irreducible, so is ∆ and hence ∆ = u(∆) is also irreducible. Proceeding as in 25.2.7, we obtain that f ∈ θVU (OY (V )). 25.3.6 Proposition. Assume that X is irreducible and that (Y, π) is a geometric quotient of X by G. Then the comorphism ϕ of π is an isomorphism from R(Y ) onto R(X)G . Proof. Since π is dominant, the comorphism ϕ : R(Y ) → R(X) is welldeﬁned. Clearly, ϕ(R(Y )) ⊂ R(X)G . Conversely, let h ∈ R(X)G . Its domain of deﬁnition U is G-stable and V = π(U ) is an open subset such that π −1 (V ) = U . Since ϕ induces a bijection between OY (V ) and OX (U )G , we have h ∈ ϕ(R(Y )).

25.4 Quotient by a subgroup 25.4.1 Let V be a ﬁnite-dimensional k-vector space, W a subspace d of V dimension d > 0, G = GL(V ) and g = gl(V ). The exterior power d W . Denote by π (resp. dπ) the representation of G contains the line L = d V given in 10.6.5 (resp. 23.4.14). (resp. g) on Lemma. Let α ∈ G and x ∈ g. (i) We have α(W ) = W if and only if π(α)(L) = L. (ii) We have x(W ) ⊂ W if and only if dπ(x)(L) ⊂ L. Proof. If α(W ) = W (resp. x(W ) ⊂ W ), then it is clear that π(α)(L) = L (resp. dπ(x)(L) ⊂ L). (i) Let (e1 , . . . , en ) be a basis of V such that (e1 , . . . , ed ) is a basis of W and (el+1 , . . . , el+d ) a basis of α(W ). Clearly L = ke1 ∧ · · · ∧ ed and π(α)(L) = kel+1 ∧ · · · ∧ el+d . Since the elements ei1 ∧ · · · ∧ eid , 1 i1 < · · · < id n, d form a basis of V , it is clear that if π(α)(L) = L, then l = 0 and hence α(W ) = W . n aij ej and v = e1 ∧ · · · ∧ vd . Then: (ii) For 1 i d, let x(ei ) = i=1

dπ(x)(v) = =

d

e1 ∧ j=1 n d

· · · ∧ ej−1 ∧ x(ej ) ∧ ej+1 ∧ · · · ∧ ed

aij e1 ∧ · · · ∧ ej−1 ∧ ei ∧ ej+1 ∧ · · · ∧ ed .

i=1 j=1

If x(W ) ⊂ W , then aij = 0 for some d < i n, 1 j d, and we obtain clearly that dπ(x)(L) ⊂ L.

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25.4.2 Let g be the Lie algebra of G, H a closed subgroup of G and h the Lie algebra of H. Let us consider the action of G on A(G) by right translation and conserve the notation ρα of 22.1.4. We shall also used the right convolution deﬁned in 23.2.2. Lemma. There exists a ﬁnite-dimensional subspace V of A(G), stable by right translations, and a subspace W of V such that: H = {α ∈ G; ρα (W ) = W } , h = {x ∈ g; W ∗ x ⊂ W }. Proof. Let (u1 , . . . , ur ) be a system of generators of the deﬁning ideal a = I(H) in A(G). There exists a ﬁnite-dimensional G-submodule V of A(G) containing u1 , . . . , ur (22.1.1). Set W = V ∩ a. Let α ∈ H. Then ρα (W ) = W (22.1.6). Conversely, if α ∈ G veriﬁes ρα (W ) = W , then ρα (ui ) ∈ a for 1 i r, so ρα (a) ⊂ a. Hence α ∈ H by 22.1.6. The second part is proved in the same manner by using 23.2.5 and 23.4.4. 25.4.3 Theorem. Let g be the Lie algebra of G, H a closed subgroup of G and h its Lie algebra. There exists a rational ﬁnite-dimensional G-module V and a line L in V such that: H = {α ∈ G; π(α)(L) = L} , h = {x ∈ g; dπ(x)(L) ⊂ L}. Proof. This is immediate by 25.4.1 and 25.4.2. 25.4.4 Corollary. Let us conserve the notations of 25.4.3. There exists a quasi-projective G-variety X which is a G-homogeneous space, and an element x ∈ X such that: (i) H is the stabilizer of x in G. (ii) Let θ : G → X be the morphism θ(α) = α.x for α ∈ G. The ﬁbres of θ are the left cosets αH, α ∈ G. Proof. Let V and L be as in 25.4.3, P(V ) the projective space associated to V , ϕ : V \ {0} → P(V ) the canonical surjection and x = ϕ(L). Set α.ϕ(v) = ϕ π(α)(v) for α ∈ G and v ∈ V \{0}. This deﬁnes a rational action of G on P(V ). Denote by X the G-orbit of x in P(V ). By 21.4.3, X is a quasi-projective variety. So the corollary follows from 25.4.3. 25.4.5 Theorem. Let g be the Lie algebra of G, H a closed normal subgroup of G and h its Lie algebra. Then there exists a ﬁnite-dimensional rational G-module (W, ϕ) such that H = ker ϕ and h = ker dϕ. Proof. Let V, π, L be as in 25.4.3 and X ∗ (H) the group of characters of H. For χ ∈ X ∗ (H), denote by Vχ the weight space of weight χ of H in V . Recall from 22.4.5 that since H is normal, π(G) permutes the Vχ ’s. Further L ⊂ Vχ0

25.4 Quotient by a subgroup

395

for some χ0 ∈ X ∗ (H), so we may assume that V is the direct sum of the weight spaces Vχ (22.4.6). Let E be the subspace of gl(V ) consisting of endomorphisms such that

(Vχ ) ⊂ Vχ for all χ. Observe that E is isomorphic to the direct sum of the gl(Vχ )’s. Then for α ∈ G, ∈ E and χ ∈ X ∗ (H), we have π(α) ◦ ◦ π(α−1 )(Vχ ) = π(α) ◦ (Vχα ) ⊂ π(α)(Vχα ) ⊂ Vχ where χα is as deﬁned in 22.4.3. So π(α) ◦ ◦ π(α−1 ) ∈ E and this deﬁnes a −1 map ϕ : G → GL(E) where ϕ(α)( ) = π(α) ◦ ◦ π(α ). It follows from 23.5.4 that ϕ(α) = AdGL(V ) π(α) . So ϕ is a rational representation of G. Since π(α), α ∈ H, acts on each Vχ by scalar multiplication, we obtain that ϕ(α)( ) = for all ∈ E, and so α ∈ ker ϕ. Conversely, if α ∈ ker ϕ, then π(α) ◦ = ◦ π(α) for all ∈ E. It follows that π(α)(Vχ ) ⊂ Vχ for all χ ∈ X ∗ (H) and π(α)|Vχ is central in gl(Vχ ). Hence π(α) acts on Vχ by scalar multiplication. In particular π(α)(L) ⊂ L and so α ∈ H (25.4.3). The fact that h = ker dϕ now follows from 24.4.1. 25.4.6 Let X be a G-variety, x ∈ X, H = Gx and θ : G → G.x the morphism α → α.x. If α ∈ G, then θ−1 (α.x) = αH. Now H acts on G in the following way: for α ∈ G and β ∈ H, β.α = αβ −1 . Then the ﬁbres of π are the H-orbits of G. Proposition. With the above notations, (G.x, θ) is the geometric quotient of G by H. Proof. By 25.1.2, θ is open. So the result is just a reformulation of 25.3.5. 25.4.7 Let H be a closed subgroup of G and π : G → G/H the canonical surjection. By 25.4.4 and 25.4.6, there exists a geometric quotient (X, θ) of G by H (action of H on G as in 25.4.6) and x ∈ X, verifying: • X is a smooth quasi-projective G-variety. • Gx = H. • If α ∈ G, then θ(α) = α.x. Deﬁne ϕ : G/H → X by ϕ(αH) = α.x. This deﬁnes a bijection from G/H onto X. Transporting the topology of X to G/H, (G/H, π) is then a geometric quotient of G by H. Moreover, by 25.3.4, the topology thus obtained on G/H (which makes it into a quasi-projective variety) is the unique topology for which (G/H, π) is a geometric quotient of G by H. Unless otherwise stated, we shall always endow G/H with this structure. 25.4.8 Proposition. Let H, K be closed subgroups of G such that H ⊂ K. Then the canonical map G/H → G/K is a morphism of algebraic varieties. Proof. This is clear by 25.3.3 and 25.4.7.

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25.4.9 For i = 1, 2, let Gi be an algebraic group, Hi a closed subgroup of Gi and πi : Gi → Gi /Hi the canonical surjection. Let G = G1 × G2 , H = H1 × H2 and u : G/H → (G1 /H1 ) × (G2 /H2 ) , (α1 , α2 )H → (π1 (α1 ), π2 (α2 )). Proposition. (i) The map u is an isomorphism of varieties. (ii) Let v : G1 → G2 be a morphism of algebraic groups such that v(αH1 ) ⊂ v(α)H2 for all α ∈ G. Then the map w : G1 /H1 → G2 /H2 such that w ◦ π1 = π2 ◦ v is a morphism of varieties. Proof. (i) The group G acts naturally on G1 /H1 × G2 /H2 and the stabilizer (resp. G-orbit) of (H1 , H2 ) is H (resp. (G1 /H1 ) × (G2 /H2 )). Let θ : G → (G1 /H1 ) × (G2/H2 ) be the morphism (α1 , α2 ) → π1 (α1 ), π2 (α2 ) . Then in view of 25.4.6, (G1 /H1 ) × (G2 /H2 ), θ is a geometric quotient of G by H. So (i) follows from 25.3.4. (ii) Since π2 ◦ v : G1 → G2 /H2 is a morphism, the result follows clearly by 25.3.3. 25.4.10 Proposition. Let H be a closed subgroup of G. The map u : G × (G/H) → G/H , (α, βH) → αβH is a morphism of varieties. Proof. Applying 25.4.9 with G1 = G2 = G, H1 = {e} and H2 = H, we obtain that the varieties (G × G)/({e} × H) and G × (G/H) are isomorphic. Let π : G → G/H be the canonical surjection. Then the morphism v = π ◦ µG : G × G → G/H is constant on the ({e} × H)-orbits in G × G (via the right translation action). So the result follows from 25.3.3. 25.4.11 Theorem. Let H be a closed normal subgroup of G and π : G → G/H the canonical surjection. (i) The group structure of G/H together with its structure of variety deﬁne a structure of aﬃne algebraic group on G/H. The map π is a morphism of algebraic groups. (ii) The map dπ induces an isomorphism from L(G)/L(H) onto L(G/H). Proof. By 25.4.5, there is a ﬁnite-dimensional rational G-module (V, ϕ) such that H = ker ϕ and L(H) = ker dϕ. Let K = ϕ(G), then K is a closed subgroup of GL(V ), so it is aﬃne. Endow GL(V ) with the following G-variety structure: for α ∈ G and β ∈ GL(V ), α.β = ϕ(α) ◦ β. Then K is the G-orbit of idV and GidV = H. Let π : G → G. idV = K be the morphism α → ϕ(α) ◦ idV . By 25.4.6, (K, π) is a geometric quotient of G by H. Thus G/H is an aﬃne variety. Moreover, by construction, the group structure on G/H is the one from the algebraic group K. So (i) follows. By 24.4.7, L(H) is an ideal of L(G), so (ii) follows from 25.1.3.

25.5 The case of ﬁnite groups

397

25.4.12 Let H, K be closed normal subgroups of G such that H ⊂ K. Denote by π : G → G/H the canonical surjection. Proposition. (i) The set π(K) is a closed normal subgroup of G/H, isomorphic to the algebraic group K/H. (ii) The algebraic groups G/K and (G/H)/π(K) are isomorphic. Proof. (i) Clearly, π(K) is a normal subgroup of G/H. It is closed by 21.2.4 and 25.4.11 (or by 25.3.2). Let θ : K → π(K) be the morphism induced by π. By 25.3.3, there is a unique morphism of varieties u : K/H → π(K) such that θ = u ◦ π where π : K → K/H is the canonical surjection. It is clear that u is a bijective group homomorphism. Hence by 21.2.6, u is an isomorphism of algebraic groups. (ii) The kernel of the composition G → G/H → (G/H)/π(K) of canonical surjections is K. By 25.3.3, this deﬁnes a morphism of varieties v : G/K → (G/H)/π(K). The morphism v is a bijective group homomorphism, so the result follows again from 21.2.6. 25.4.13 Let H be a closed subgroup of G. In general, if H is not normal, G/H is not aﬃne. However, we have the following result: Proposition. Let G be a unipotent algebraic group and H a closed subgroup of G. Then the variety G/H is aﬃne. Proof. Let V, L, π be as in 25.4.3. Since H is unipotent, LH = {0} (22.3.6). It follows that LH = L and for x ∈ L \ {0}, the map θ : G → G.x, α → α.x, induces an isomorphism from G/H to G.x (25.4.6). Since G.x is aﬃne (22.3.6), we are done.

25.5 The case of ﬁnite groups 25.5.1 Proposition. Let G be a ﬁnite group consisting of automorphisms of an aﬃne variety X. Then the algebra of invariants A(X)G is ﬁnitely generated. Moreover, if X is irreducible, then R(X)G = Fract(A(X)G ). Proof. Let f1 , . . . , fn be a system of generators for A(X) and α1 , . . . , αs the elements of G with α1 = eG . For 1 i n, set: Pi (T ) = (T − α1 .fi ) · · · (T − αs .fi ). Then G permutes the roots of Pi , so the coeﬃcients of Pi are in A(X)G . Let B be the subalgebra of A(X)G generated by the coeﬃcients of the Pi ’s. Since Pi (fi ) = 0, fi is integral over B, and so A(X) is integral over B (3.1.5). Moreover, A(X) is a ﬁnite B-algebra (3.3.5). Since B is noetherian, A(X)G is a ﬁnitely generated B-module, so the algebra A(X)G is ﬁnitely generated. Suppose that X is irreducible. Let f = uv −1 ∈ R(X)G where u, v ∈ A(X) and: v1 = (α1 .v) · · · (αs .v) , u1 = u(α2 .v) · · · (αs .v). Then v1 ∈ A(X)G and f = u1 v1−1 . Hence u1 ∈ A(X)G and so we have f ∈ Fract(A(X)G ).

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25.5.2 Let G be a ﬁnite group consisting of automorphisms of an irreducible aﬃne variety X. By 25.5.1, there is an aﬃne variety Y such that Y = Spm(A(X)G ). Let u : X → Y be the morphism deﬁned by the canonical injection A(X)G → A(X). Proposition. The pair (Y, u) is the geometric quotient of X by G. Proof. We saw in the proof of 25.5.1 that A(X) is integral over A(X)G . If n ∈ Spm(A(X)G ), then there exists m ∈ Spm(A(X)) such that n = m ∩ A(X)G (3.3.2 and 3.3.3). Thus u is surjective. Given x ∈ X, we have G.x ⊂ u−1 (u(x)). Suppose that there exists y ∈ −1 u (u(x)) such that y ∈ G.x. Since the ﬁnite sets G.x and G.y are disjoint and closed, it follows from 11.3.3 that there exists f ∈ A(X) verifying f |G.x = 1 and f |G.y = 0. Set g = (α1 .f ) · · · (αs .f ) where α1 , . . . , αs are the elements of G. Then g ∈ A(X)G and g|G.x = 1, g|G.y = 0. This is absurd since u(x) = u(y). Thus the ﬁbres of u are the G-orbits. By 15.3.2, u is a ﬁnite morphism, and therefore u is closed (15.3.4). Since the ﬁbres of u are the G-orbits, 25.3.2 implies that u is open. Let V be an open subset of Y and U = u−1 (V ). Then U is G-stable and G G it is clear that uU V (OY (V )) ⊂ OX (U ) . Conversely, let f ∈ OX (U ) . Since G G G X is irreducible, we have OX (U ) ⊂ R(X) = Fract(A(X) ) (12.8.2 and 25.5.1). It follows that there exist g, h ∈ A(X)G = A(Y ) such that f = g/h and h|U = 0. Hence g/h ∈ OY (V ), and uU V is bijective. So we have proved that (Y, u) is the geometric quotient of X by G. 25.5.3 In the case of a linear action, we have the following more precise statement: Proposition. Let V be a ﬁnite-dimensional vector space, G a ﬁnite subgroup of GL(V ) and N the order of G. Then the algebra A(V )G is generated by homogeneous invariants of degree at most N . Proof. Let p : A(V ) → A(V ) denote the map deﬁned by: p(u) =

1 α.u. N α∈G

We have up(v) = p(uv) if u ∈ A(V )G and v ∈ A(V ). Moreover, the image of p is A(V )G and deg(p(u)) deg(u) for all u ∈ A(V ). Let B be the subalgebra of A(V )G generated by invariants of degree at most N , and AN the subspace of A(V ) consisting of polynomial functions of degree at most N −1. We claim that A(V ) = BAN . Since A(V ) is generated by powers of linear forms, to prove our claim, it suﬃces to prove that n ∈ BAN for all ∈ V ∗ and n ∈ N. This is clear if n < N . Now, (T − α. ) = T N + aN −1 T N −1 + · · · + a0 α∈G

25.5 The case of ﬁnite groups

399

where a0 , . . . , aN −1 ∈ B, so N ∈ B + B + · · · + B N −1 . By induction, we obtain that n ∈ BAN if n N . Let f ∈ A(V )G . It follows from our claim that there exist a1 , . . . , ar ∈ B and f1 , . . . , fr ∈ AN such that f = a1 f1 + · · · + ar fr . So: f = p(f ) = a1 p(f1 ) + · · · + ar p(fr ). Since deg(p(fi )) deg(fi ) N − 1, we have p(fi ) ∈ B, and so f ∈ B as required.

References and comments • [5], [21], [38], [40], [78]. More general results concerning quotients of an algebraic variety can be found in [30]. In general, for a G-variety X, the geometric quotient of X by G does not exist. In [72], Rosenlicht proved that there exists a dense open G-stable subvariety Y of X such that the geometric quotient of Y by G exists. Other proofs of this result can be found in [35], [67], [75]. In particular, we deduce from this that tr degk R(X)G = dim X − ρ where ρ is the maximal dimension of G-orbits in X.

26 Solvable groups

Structure of diagonalizable groups and solvable groups are studied in this chapter. We prove in particular two important results : Lie-Kolchin Theorem and Borel’s Fixed Point Theorem. In this chapter, G denotes an algebraic group whose Lie algebra is g.

26.1 Conjugacy classes 26.1.1 Let H be a closed subgroup of G, h its Lie algebra and x ∈ g. We set: CH (x) = {α ∈ H; (Ad α)(x) = x} , ch (x) = {y ∈ h; [x, y] = 0}. The set CH (x) (resp. ch (x)) is a closed subgroup of H (resp. Lie subalgebra of h). We have CH (x) = H ∩ CG (x) and ch (x) = h ∩ cg (x). Moreover, by 23.5.5, 24.3.5 and 24.3.6, L CH (x) = ch (x). For γ ∈ G, set: CH (γ) = {α ∈ H; αγ = γα} , ch (γ) = {x ∈ h; (Ad γ)(x) = x}. Again, CH (γ) (resp. ch (γ)) is a closed subgroup of H (resp. Lie subalgebra of h). More generally, if L is a subset G, we set CH (L) (resp. ch (L)) to be the intersection of the CH (γ)’s, with γ ∈ L. (resp. ch (γ)’s, γ ∈ L). It follows that CH (L) (resp. ch (L)) is a closed subgroup of H (resp. Lie subalgebra of h). Lemma. (i) If γ ∈ G, then L CH (γ) = ch (γ). (ii) If L is a subset of G, then L CH (L) = ch (L). Proof. By 24.5.2, it suﬃces to prove (i). Since CH (γ) = H ∩ CG (γ) and ch (γ) = h ∩ cg (γ), it follows from 24.3.5 (iii) that it suﬃces to prove the result for H = G. Let u, v, w : G → G be the morphisms deﬁned as follows: u(α) = αγα−1 , v(α) = αγα−1 γ −1 , w(α) = αγ, α ∈ G. Then u = w ◦ v and due = dwe ◦ dve = dwe ◦(Ad(γ)−idg ) (23.5.3). Since CG (γ) is the stabilizer of γ under the action

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26 Solvable groups

of G given by conjugation, 25.1.3 implies that ker(due ) = L(CG (γ)). So the result follows because dwe is bijective. 26.1.2 For γ ∈ G and x ∈ g, set: ClH (γ) = {αγα−1 ; α ∈ H} , clH (x) = {(Ad α)(x) ; α ∈ H}. The set ClH (γ) is the H-orbit of γ under the action of H on G by conjugation, so it is locally closed in G (21.4.3). In the same way, clH (x) is locally closed in g. Lemma. If (Ad α)(x) − x ∈ h for all α ∈ H, then [x, h] ⊂ h. Proof. Apply 24.5.5 to the subspaces E = h and F = kx + h of g. 26.1.3 Let us conserve the above notations and let π : H → ClH (γ) be dπe induces a surjection from h onto the morphism α → αγα−1. By 25.1.3, Tγ (ClH (γ)) with kernel L CH (γ) = ch (γ). Suppose further that γ is semisimple and γHγ −1 ⊂ H. Then (Ad γ)(h) ⊂ h and the restriction of Ad γ to h is semisimple. So h = ch (γ) ⊕ m where m is the sum of eigenspaces of (Ad γ)|h whose eigenvalues are not equal to 1. We see, via the automorphism α → αγ −1 of G, that Te (ClH (γ)γ −1 ) = m. 26.1.4 Theorem. Let H be a closed subgroup of G and h its Lie algebra. (i) Let γ be a semisimple element of G verifying γHγ −1 ⊂ H. Then ClH (γ) is closed in G. (ii) Let x be a semisimple element of g verifying (Ad α)(x) − x ∈ h for all α ∈ H. Then clh (γ) is closed in g. Proof. We may assume that G = GL(V ) for some ﬁnite-dimensional vector space V . (i) For α ∈ G, let µα be its minimal polynomial. For α ∈ NG (H), we have (Ad α)(h) ⊂ h and we shall denote by χα the characteristic polynomial of (Ad α)|h . Let W be the set of elements α ∈ NG (H) verifying: µγ (α) = 0 , χα = χγ . Then γ ∈ W , and if α ∈ W , then βαβ −1 ∈ W for all β ∈ H. Moreover, if α ∈ W , then µα divides µγ . It follows that the elements of W are semisimple. Let α ∈ W . Then by 21.4.3 and 26.1.1, we have: dim ClH (α) = dim H − dim CH (α) = dim h − dim ch (α). Since χα = χγ and (Ad α)|h is semisimple, dim ch (α) = dim ch (γ). Thus the H-orbits of the variety W (under conjugation) are all of the same dimension. By 21.4.5, these orbits are closed in W . Since W is clearly closed in NG (H) and NG (H) is closed in G, the result follows.

26.1 Conjugacy classes

403

(ii) Let ng (H) = {y ∈ g; (Ad α)(y) − y ∈ h for all α ∈ H}. We have x ∈ ng (H) and we saw in 26.1.2 that if y ∈ ng (H), then [y, h] ⊂ h. Let us denote by χy the characteristic polynomial of (ad y)|h . Let µx be the minimal polynomial of x as an element of End(V ) and set: w = {y ∈ ng (H); µx (y) = 0 , χy = χx }. Then w contains x, and it is a closed subset of g verifying (Ad α)(w) ⊂ w for all α ∈ H. Using the same argument as in (i), we obtain that the H-orbits of w are all of the same dimension, and so they are closed. 26.1.5 Theorem. Let U be a closed unipotent subgroup of G, γ a semisimple element of G verifying γU γ −1 ⊂ U , and M the set of elements αγα−1 γ −1 , with α ∈ U . (i) The group CU (γ) is closed and connected. The set M is a closed irreducible subset of U . (ii) The map u : M × CU (γ) → U , (α, β) → αβ, is an isomorphism of varieties. (iii) The morphism v : G → G, α → αγα−1 γ −1 , induces an automorphism of the variety M . Proof. Since γU γ −1 ⊂ U , we have M ⊂ U . Moreover, U is connected (22.3.6). 1) Clearly, CU (γ) is closed in U and it is connected (22.3.6). The fact that M is closed follows from 26.1.4 and M is irreducible because M = v(U ). So we have proved (i). 2) Let α ∈ U , β = αγα−1 γ −1 . If β ∈ CU (γ), then γβ = βγ = αγα−1 . Since β is unipotent and γ is semisimple, βγ = γβ is the Jordan decomposition of αγα−1 . But this element is semisimple. So β = e and M ∩ CU (γ) = {e}. 3) Let α, β ∈ U . Then: v(αβ) = αβγβ −1 α−1 γ −1 = α(βγβ −1 γ −1 )α−1 (αγα−1 γ −1 ) = αv(β)α−1 v(α). So if β ∈ CU (γ), then v(β) = e and: v(αβ) = v(α) = v(α)v(β). 4) Using the equalities in point 3, we obtain that if α is central in U , then: v(βα) = v(αβ) = v(β)v(α). 5) Let us prove (ii) in the case where U is commutative. By point 4, the map v induces a homomorphism from the group U to itself whose image is M and kernel is CU (γ). So dim U = dim M + dim CU (γ) (21.2.4) and u is a homomorphism of algebraic groups. The connectedness of U , CU (γ) and M imply therefore that u is surjective. Let (α, β) ∈ M × CU (γ) be such that u(α, β) = αβ = e. Then α = β −1 ∈ M ∩ CU (γ) = {e} by point 2. It follows that α = β = e. The homomorphism u

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26 Solvable groups

is therefore injective, and we have proved that u is an isomorphism of algebraic groups (21.2.6). 6) Let us prove (ii) in the general case. We proceed by induction on the dimension of U . The case dim U = 0 is obvious, so let us assume that dim U > 0. Since U is unipotent, it is nilpotent (22.3.2 and 10.8.14). Its centre Z is a non-trivial closed unipotent subgroup, so it is also connected. If Z = U , then the result follows from point 5. Otherwise, Z = U . Note that γ and U are contained in the normalizer NG (Z) of Z in G. Let π : NG (Z) → NG (Z)/Z be the canonical surjection, G = NG (Z)/Z, U = U/Z and γ = π(γ). Set: M = {α γ (α )−1 (γ )−1 ; α ∈ U } = π(M ) , N = {αγα−1 γ −1 ; α ∈ Z}. Applying the induction hypothesis to (G , U , γ ) and (G, Z, γ), we obtain that the morphisms u : M × CU (γ ) → U and u0 : N × CZ (γ) → Z are isomorphisms of varieties. • Let us show that u is injective. Let α1 , α2 ∈ CU (γ) and β1 , β2 ∈ M be such that β1 α1 = β2 α2 . Let α = α1 α2−1 ∈ CU (γ), then β1 α = β2 . So π(β1 )π(α) = π(β2 ). Since π(α) ∈ CU (γ ) and π(β1 ), π(β2 ) ∈ M , we obtain that π(α) = e, or equivalently α ∈ Z. Let us write β1 = v(δ1 ), β2 = v(δ2 ) where δ1 , δ2 ∈ U . The fact that α ∈ Z ∩ CU (γ) implies that δ2 γδ2−1 = β2 γ = β1 αγ = (δ1 γδ1−1 )α = α(δ1 γδ1−1 ). The elements δ1 γδ1−1 et δ2 γδ2−1 are semisimple and α is unipotent. We deduce therefore that α = e, and so α1 = α2 and β1 = β2 . • Next, we show that v(Z) = Z ∩ M . If α ∈ Z, then γα−1 γ −1 ∈ Z, so v(Z) ⊂ Z ∩ M . Conversely, since u0 is surjective, any α ∈ Z is of the form δβ, where β ∈ CZ (γ) and δ ∈ N = v(Z). Furthermore, if α ∈ M , then α = δ ∈ v(Z) because u is injective. • We claim that CU (γ ) = π CU (γ) . Let α ∈ U be such that π(α)γ = γ π(α). Then v(α) ∈ M ∩ ker π = Z ∩ M = v(Z). Thus v(α) = v(β) for some β ∈ Z. By applying point 4 to β and β −1 α, we have β −1 α ∈ CU (γ). As π(β −1 α) = π(α), we have therefore proved our claim. u is surjective. Since u is surjective, π(M ) = M and • Letus show that π CU (γ) = CU (γ ), we have U = M CU (γ)Z = M ZCU (γ). The surjectivity of u0 implies then that U = M v(Z)CZ (γ)CU (γ). Finally, as M v(Z) = M by point 4, we have U = M CU (γ). • So u is bijective and it follows from 17.4.8 that isomorphism. u is an 7) Since u is surjective, we have M = v(U ) = v M CU (γ) = v(M ) by point 3. On the other hand, if α, β ∈ M verify v(α) = v(β), then β −1 α ∈ CU (γ) and u(α, e) = u(β, β −1 α). It follows from the injectivity of u that α = β. Thus v induces a bijection from M onto itself. But M = ClU (γ)γ −1 , so 25.1.1 implies that M is normal. Therefore (iii) follows from 17.4.8.

26.2 Actions of diagonalizable groups

405

26.2 Actions of diagonalizable groups 26.2.1 In this section, T denotes a diagonalizable algebraic group. Let us suppose that T acts rationally on G and that if α ∈ T , the morphism θα : G → G , β → α.β is an automorphism of the group G. Denote (dθα )e by πα . The map πα is an automorphism of the vector space g, and α → πα deﬁnes a rational representation of T on g. Let GT (resp. gT ) be the set of elements β ∈ G (resp. x ∈ g) verifying θα (β) = β (resp. πα (x) = x) for all α ∈ T . Recall that X ∗ (T ) denotes the set of characters of T . If χ ∈ X ∗ (T ), denote by gχ the set of x ∈ g such that πα (x) = χ(α)x for all α ∈ T . Denote by 1T ∈ X ∗ (T ) the trivial character of T , that is 1T (α) = 1 for all α ∈ T , and we set: ! " Φ(T, G) = χ ∈ X ∗ (T ) \ {1T }; gχ = {0} . Then we have: g = gT ⊕

gχ .

χ∈Φ(T,G)

Let χ ∈ X ∗ (T ) and T χ the kernel of χ. Denote by CG (T χ ) the set of β ∈ G verifying θα (β) = β for all α ∈ T χ . By identifying G and T as subgroups of the semidirect product G θ T , we may always assume that G and T are closed subgroups of an algebraic group. Proposition. (i) We have L(GT ) = gT . (ii) Assume that G is connected and G = GT . Then G is generated (as a group) by the subgroups CG (T χ ), χ ∈ X ∗ (T ). Proof. (i) Identify G and T as closed subgroups of an algebraic group, then apply 26.1.1 (ii). (ii) The condition G = GT implies that Φ(G, T ) = ∅. Denote by G the subgroup generated by the CG (T χ ), χ ∈ Φ(G, T ). Applying (i) with T = T χ , χ we obtain that L(CG (T χ )) = gT . So gT + gχ ⊂ L(G ) if χ ∈ Φ(T, G). Since g is the sum of the gT + gχ , χ ∈ Φ(T, G), it follows that g ⊂ L(G ). So G = G◦ ⊂ G and the result follows. 26.2.2 Let us suppose that T acts rationally on the algebraic groups G and H via group automorphisms of G and H respectively, as in 26.2.1, and denote them by θαG and θαH for α ∈ T . Proposition. Let σ : G → H be a T -equivariant surjective morphism of algebraic groups. Then σ induces a surjective morphism from (GT )◦ onto (H T )◦ . Proof. Since σ◦θαG = θαH ◦σ for all α ∈ T , we have dσ◦dθαG = dθαH ◦dσ for all α ∈ T . The surjectivity of u implies that du is surjective. Thus du induces a surjection from L(G)T onto L(H)T . By 26.2.1 (i), this surjection is the diﬀerential of the restriction GT → H T . So the result follows.

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26 Solvable groups

26.2.3 Proposition. Let T be a torus of G and K a closed subgroup of G such that T ⊂ NG (K), and k = L(K). Then there exists α ∈ T such that: CK (α) = CK (T ) , ck (α) = ck (T ). Proof. We may assume that G ⊂ GL(V ) where V is a ﬁnite-dimensional vector space. Denote by χ1 , . . . , χr the pairwise distinct weights of T in V , and let Vi = Vχi , 1 i r. We have V = V1 ⊕ · · · ⊕ Vr (22.5.2). If M = GL(V1 ) × · · · × GL(Vr ) ⊂ GL(V ) and m = L(M ), then it is clear that CK (T ) = M ∩ K and ck (T ) = m ∩ k. is a non-trivial character of T , its kernel If 1 i, j r and i = j, χi χ−1 j Tij is a subgroup of T distinct from T . Since the ﬁeld k is inﬁnite, we deduce that there exists α ∈ T which is not contained in any of the Tij ’s, i = j. Therefore CK (α) = M ∩ K and ck (α) = m ∩ k.

26.3 Fixed points 26.3.1 Lemma. Let V be a k-vector space of dimension n > 0 and G a connected solvable subgroup of GL(V ). Then there exists v ∈ V \ {0} such that Gv ⊂ kv. Proof. In view of 21.6.6, we may assume that G is an algebraic subgroup of GL(V ). We shall proceed by induction on dim G. If dim G = 1, then G is commutative (22.6.1), and the result is then clear because G = Gs or G = Gu (22.2.6). So let us suppose that dim G 2. Let H = (G, G) be the group of commutators of G, then it is a connected solvable algebraic subgroup of G (21.3.6) verifying dim H < dim G. So by induction and 22.4.5, there exists χ ∈ X ∗ (H) such that Vχ = 0 where Vχ denotes the weight space of weight χ of H in V . Since H is normal in G, we saw in 22.4.5 that the (direct) sum W of the Vχ ’s, χ ∈ X ∗ (H), is G-stable. So we may assume that V = W . Note that the set of weights of H in V is ﬁnite (22.4.6). Let us ﬁx a basis of V consisting of weight vectors of H, and identify GL(V ) with GLn (k) via this basis. Then the elements of H are diagonal matrices. For β ∈ H, the map G → H, α → αβα−1 , is a morphism of varieties whose image is connected since G is connected. But the image is a ﬁnite set since its elements are diagonal matrices having the same eigenvalues as for the diagonal matrix β. We deduce therefore that each Vχ is G-stable. So we may further assume that V = Vχ where χ ∈ X ∗ (H). Now H = (G, G), so det β = 1 for all β ∈ H. But for β ∈ H, we have β = χ(β) idV . Hence the connected subgroup H is ﬁnite. So H = {e} and G is commutative. But the result is clear when G is commutative. So we are done. 26.3.2 Theorem. (Lie-Kolchin Theorem) Let V be a ﬁnite-dimensional kvector space and G a connected solvable subgroup of GL(V ). Then there exists a basis of V with respect to which the elements of G are upper triangular.

26.4 Properties of solvable groups

407

Proof. Let us proceed by induction on dim V . The case dim V 1 is obvious. So let us assume that dim V 2. By 26.3.1, there exists a weight vector v1 of G in V . For α ∈ G, denote by α the automorphism of W = V /kv1 induced by α, and v → v, the canonical surjection V → W . Then clearly the set G of the α , α ∈ G, is a connected solvable subgroup of GL(W ). So by the induction hypothesis, there exists v2 , . . . , vn ∈ V such that (v2 , . . . , v, n ) is a basis of W with respect to which the elements of G are upper triangular. It follows that (v1 , . . . , vn ) is a basis of V with respect to which the elements of G are upper triangular. 26.3.3 Corollary. Let G be a connected solvable algebraic group, then its derived subgroup D(G) = (G, G) is a closed connected unipotent subgroup. Proof. In view of 21.3.6, we only need to show that D(G) is unipotent. We may assume that G is a closed subgroup of some GLn (k) (22.1.5). By 26.3.2, there exists α ∈ GLn (k) such that αGα−1 ⊂ Tn (k) (notations of 10.8.1). Hence αD(G)α−1 ⊂ Un (k) (10.8.9), and D(G) is unipotent. 26.3.4 Theorem. (Borel’s Fixed Point Theorem) Let G be a connected solvable algebraic group and X a non-empty complete G-variety. Then the set X G is not empty. Proof. We shall prove the result by induction on d = dim G. If d = 0, then G = {e} and X = X G . So let us assume that d > 0 and let H = D(G). Since dim H < d, the set Y = X H is a non-empty closed subvariety of X (21.4.2). So Y is complete (13.4.3). Now H is normal in G, so if x ∈ Y , α ∈ G and β ∈ H, then: β.(α.x) = α. (α−1 βα).x = α.x, so Y is G-stable. There exists by 21.4.5 x ∈ Y such that the orbit G.x is closed, and therefore complete. The group Gx contains H, so Gx is normal in G. By 25.4.11 and 25.4.7, the variety G/Gx is aﬃne and it is isomorphic to G.x. But G.x is irreducible since G is connected, it follows therefore from 13.4.4 that G.x = {x}.

26.4 Properties of solvable groups 26.4.1 Let Gs (resp. Gu ) denote the set of semisimple (resp. unipotent) elements of G. Recall from 22.2.5 that Gu is a closed subset of G. 26.4.2 Proposition. Let G be connected and solvable. (i) The set Gu is a closed connected nilpotent subgroup of G. It contains D(G) and so it is normal in G. (ii) The group G/Gu is a torus. Proof. (i) By 22.1.5 and 26.3.2, we may assume that G ⊂ Tn (k) for some n ∈ N∗ . Since the set of unipotent elements of Tn (k) is Un (k), we have

408

26 Solvable groups

Gu = G ∩ Un (k) by 22.2.3. So part (i) follows from 10.8.9, 10.8.10, 22.2.5 and 22.3.6 (i). (ii) Since D(G) ⊂ Gu , G/Gu is abelian. It is connected because it is the image of G under the canonical surjection G → G/Gu . Let β ∈ G/Gu and α ∈ G such that π(α) = β. Denote by αs and αu the semisimple and unipotent components of α. It follows that β = π(αs ). Hence β is semisimple (22.2.3) and the result follows from 22.5.7. 26.4.3 Proposition. Let G be connected and solvable. (i) There exists a chain G = Gp ⊃ Gp−1 ⊃ · · · ⊃ G1 ⊃ G0 = closed connected normal subgroups of G such that dim(Gi /Gi−1 ) = 1 1 i p. (ii) There exists a chain G = Gn ⊃ Gn−1 ⊃ · · · ⊃ G1 ⊃ G0 = closed connected normal subgroups of G such that dim(Gi /Gi−1 ) = 1 1 i n.

Gu of for all {e} of for all

Proof. (i) Let π : G → G/Gu = T be the canonical surjection. By 26.4.2, T is isomorphic to Dp (k) (Gm )p for some p. It is therefore clear that there is a chain T = Tp ⊃ Tp−1 ⊃ · · · ⊃ T1 ⊃ T0 = {e} of closed connected subgroups of T such that Ti /Ti−1 Gm for 1 i p. The subgroups Hi = π −1 (Ti ) of G are closed and normal in G, and so are the subgroups Gi = Hi◦ . Moreover, dim(Hi /Hi−1 ) = dim(Ti /Ti−1 ) = 1 (25.4.12) and (Hi−1 ∩ Gi )/Gi−1 is a ﬁnite set. So dim(Gi /Gi−1 ) 1 and the result follows. (ii) By 26.3.2, we may assume that G ⊂ Tr (k) and Gu ⊂ Ur (k) for some r ∈ N∗ . Let J be the set of pairs (i, j) where i, j ∈ {1, . . . , r} and i < j, and endow J with the following total order: (i, j) < (k, l) if j < l or if j = l and i > k. Let λ1 < · · · < λs be the elements of J in ascending order, and for (i, j) ∈ J, let U(i,j) be the set of matrices [xkl ] ∈ Ur (k) verifying xkl = 0 for (k, l) (i, j). Then we verify easily that Ur (k) = Uλ0 ⊃ Uλ1 ⊃ · · · ⊃ Uλs = {Ir } is a chain of closed normal subgroups of Tr (k). Moreover, we have Uλi / Uλi+1 Ga for 0 i < s. Let Gi = (G ∩ Uλi )◦ , then Gi is a closed normal subgroup of G contained in Gu and we have dim(Gi /Gi+1 ) 1. So the result follows from (i). 26.4.4 Proposition. Let G be connected and solvable. Then the following conditions are equivalent: (i) The set Gs is a subgroup of G. (ii) The group G is nilpotent. If these conditions are veriﬁed, then Gs is a connected subgroup of G contained in the centre of G, and the map θ : Gs × Gu → G, (α, β) → αβ, is an isomorphism of algebraic groups. Proof. Let us denote by π : G → G/Gu the canonical surjection. (i) ⇒ (ii) Suppose that Gs is a subgroup of G. Since π|Gs is injective, it follows from 26.4.2 (ii) that Gs is abelian.

26.5 Structure of solvable groups

409

We may assume that G ⊂ GLn (k) and since Gs is abelian, we may assume further that Gs ⊂ Dn (k). So Gs = G ∩ Dn (k) is closed in G. Thus the restriction of π to Gs is a bijective morphism of algebraic groups, hence an isomorphism (21.2.6). In particular, Gs is connected. Since αβα−1 ∈ Gs for all β ∈ Gs and α ∈ G, we have G = NG (Gs ). So by 22.5.11: G = NG (Gs ) = NG (Gs )◦ = CG (Gs )◦ . Thus Gs is contained in the centre of G. Since G/Gs is unipotent, it is nilpotent (10.8.14), and so G is nilpotent (10.5.3). Note that since Gs is contained in the centre of G, the map θ is a bijective morphism of algebraic groups, and hence it is an isomorphism (21.2.6). (ii) ⇒ (i) Suppose that G is nilpotent. Fix α ∈ Gs and deﬁne the following morphisms of varieties: u : G → G , β → αβα−1 and v : G → G , β → αβα−1 β −1 . Since G is nilpotent, there exists n ∈ N∗ such that v n (β) = e for all β ∈ G. So (dv)n = 0, and dv is a nilpotent endomorphism of g = L(G). As dv = Ad(α) − idg and α ∈ Gs , it is also a semisimple endomorphism of g (22.2.3). Hence Ad(α) = idg . It follows from 24.4.2 and the connectedness of G that α is central in G. It is now clear that Gs is a subgroup of G. 26.4.5 Proposition. Let G be connected and solvable and g its Lie algebra. (i) L(Gu ) is the set of nilpotent elements of g. (ii) If G is nilpotent, then L(Gs ) is the set of semisimple elements of g. Proof. (i) We may assume that G ⊂ Tn (k) and Gu ⊂ Un (k). Then L(Gu ) ⊂ L(Un (k)) = nn (k), so all the elements of L(Gu ) are nilpotent. Let π : G → G/Gu be the canonical surjection. If x is a nilpotent element of g, then since G/Gu is a torus, 23.6.3 implies that dπ(x) = 0, or x ∈ ker dπ. So x ∈ L(ker π) = L(Gu ) by 24.4.1. (ii) Assume that G is nilpotent. Then π induces an isomorphism from Gs onto the torus G/Gu (26.4.2), so all the elements of L(Gs ) are semisimple. By 26.4.4, Gs is contained in the centre of G, so L(Gs ) is in the centre of g (24.4.2). Again by 26.4.4, G and Gs × Gu are isomorphic, so g is isomorphic to the direct product L(Gs ) × L(Gu ). The result follows therefore from (i) and the uniqueness of the Jordan decomposition in g.

26.5 Structure of solvable groups 26.5.1 In this section, G denotes a connected solvable algebraic group whose Lie algebra is g. We saw in 26.4.2 that Gu is normal in G, so its Lie algebra n is an ideal of g (24.4.7) and n is the set of nilpotent elements of g (26.4.5). Moreover, since D(G) ⊂ Gu (26.4.2), we have [g, g] ⊂ n (24.3.5 and 24.5.12).

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Proposition. There exists a Lie subalgebra t verifying the following conditions: (i) t is an algebraic Lie subalgebra of g. (ii) t is abelian and all its elements are semisimple. (iii) As a vector space, g = t ⊕ n. Proof. We may assume that G ⊂ GL(V ) and g ⊂ gl(V ) for some ﬁnitedimensional vector space V . The elements of n are then nilpotent endomorphisms of V (23.6.4). Let A be the set of abelian Lie subalgebras of g whose elements are semisimple endomorphisms of V . Then A = ∅ since {0} ∈ A. Let t be an element of A of maximal dimension. By 23.6.4, the representation t → adg t of t in g is semisimple, and t + n is stable under this representation. So there exists a subspace m such that g = m ⊕ (t + n) and [t, m] ⊂ m. Since [g, g] ⊂ n, we deduce that [t, m] = {0}. Let x ∈ m and xs , xn its semisimple and nilpotent components. By 23.6.4, xs , xn ∈ g and since [x, t] = {0}, we have [xs , t] = {0} (10.1.3 and 23.6.4). It follows that kxs + t is an abelian subalgebra whose elements are semisimple. Our choice of t implies that xs = 0. So all the elements of m are nilpotent. Hence m ⊂ n, which in turn implies that m = {0}. Thus g = t + n and it is clear that t ∩ n = {0}. So we are left to prove that t is algebraic. Let x ∈ t and y a replica of x. Then by 24.6.4 (ii), y is semisimple and [y, t] = {0}. Again by our choice of t, we deduce that y ∈ t. It follows from 24.5.10 that t is an algebraic Lie subalgebra of g. 26.5.2 Let us conserve the hypotheses and notations of 26.5.1 and let T be a connected algebraic subgroup of G verifying L(T ) = t. Proposition. (i) The group T is a torus of dimension dim(G/Gu ). (ii) The map u : T × Gu → G, (α, β) → αβ is an isomorphism of varieties. Proof. (i) We may again assume that G ⊂ GL(V ) and g ⊂ gl(V ). Note that T = (t) (see 24.5.1). If x ∈ t, then any element of (x) can written as P (x) for some P ∈ k[X] (24.6.4) and T is generated by the (x)’s, x ∈ t. It follows that T is abelian and its elements are semisimple. Since T is connected, it is a torus (22.5.7) of dimension dim(G/Gu ) (26.5.1 (iii)). (ii) Since T is a torus, T ∩ Gu = {e}. Let T Gu be the set of products αβ, α ∈ T and β ∈ Gu . Then T Gu is a subgroup of G because Gu is normal in G. It is also closed in G by 21.2.3 and connected since it is the image of the morphism u. Let us consider the following actions of T × Gu on T × Gu and G:

a

a

a

(α, β).(α , β ) = (αα , β β −1 ) , (α, β).γ = αγβ −1 . where α, α ∈ T , β, β ∈ Gu and γ ∈ G. Then u is injective and equivariant with respect to these actions. Moreover, the stabilizer of e in T × Gu is {e}

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411

because T ∩ Gu = {e}. As the (T × Gu )-orbit of e is T Gu , it follows from 21.4.3 that: dim(T Gu ) = dim(T × Gu ) = dim T + dim Gu = dim G. Hence the connectedness of G implies that G = T Gu . So u is a bijective morphism of (T × Gu )-homogeneous spaces, and 25.1.2 says that u is an isomorphism of varieties. 26.5.3 Set:

C ∞ (G) =

C n (G).

n1

We have C ∞ (G) ⊂ D(G) ⊂ Gu , so C ∞ (G) is a normal unipotent subgroup of G. If T is a torus of G, then T ∩ Gu = {e}, so dim T dim(G/Gu ). In particular, if dim T = dim(G/Gu ), then T is a maximal torus. Now 26.5.2 says that such a torus exists and if T is such a torus, then the map T × Gu → G is an isomorphism of varieties. 26.5.4 Lemma. Let T ⊂ G be a torus of dimension dim(G/Gu ). If α is an element of Gs , then there exists β ∈ C ∞ (G) such that βαβ −1 ∈ T . Proof. We have already noted that G = T Gu . Let us proceed by induction on dim G. Suppose that G is nilpotent. Then T = Gs by 26.4.4 and so the result is obvious. Suppose that G is not nilpotent, then C ∞ (G) = {e}. The unipotent group ∞ C (G) is connected and nilpotent (10.8.10 and 21.3.6). The last non-trivial term of the central descending series of C ∞ (G) is connected (21.3.6), so it is contained in the identity component N of the centre of C ∞ (G). It follows that N = {e} and N is a normal subgroup of G. In particular, N ⊂ Gu and N ∩ T = {e}. Let G = G/N and π : G → G the canonical surjection. It is clear that the map π(T ) × (G )u → G , (γ, δ) → γδ, is an isomorphism of varieties. Applying the induction hypothesis on G , we obtain that there exists γ ∈ C ∞ (G ) = π(C ∞ (G)) such that γπ(α)γ −1 ∈ π(T ). Let δ ∈ C ∞ (G) be such that π(δ) = γ. Then by replacing α by δαδ −1 , we may assume that α ∈ T N = N T . Let β ∈ N and γ ∈ T be such that α = βγ. By 26.1.5, there exists δ ∈ N and ε ∈ CN (γ) such that β = (δγδ −1 γ −1 )ε. Then α = (δγδ −1 )ε = ε(δγδ −1 ) since ε commutes with γ and N is abelian. But δγδ −1 is semisimple and ε is unipotent, so α = (δγδ −1 )ε is the Jordan decomposition of α. Hence ε = e and δ −1 γδ = γ ∈ T . 26.5.5 Theorem. Let S, T be maximal tori of a connected solvable algebraic group G. (i) There exists α ∈ C ∞ (G) such that S = αT α−1 . In particular, dim S = dim T = dim(G/Gu ). (ii) The map T × Gu → G, (α, β) → αβ, is an isomorphism of varieties.

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Proof. In view of 26.5.2 and 26.5.3, it suﬃces to show that there exists an element α ∈ C ∞ (G) such that S = αT α−1 where T is a maximal torus of dimension dim(G/Gu ). By 26.2.3, there exists γ ∈ S such that CG (S) = CG (γ). By 26.5.4, we may assume that γ ∈ T . Then T ⊂ CG (γ)◦ = CG (S)◦ . Let C = CG (S)◦ . Any torus of C is a torus of G, and so by 26.5.2 (i) and 26.5.3, T is a torus of C of dimension dim(C/Cu ). It follows by 26.5.4 that if β ∈ S, then δβδ −1 ∈ T for some δ ∈ C. Since S is central in C, we deduce that S ⊂ T . But S is a maximal torus, so S = T . 26.5.6 Proposition. Let G be a connected solvable algebraic group. (i) If T is a maximal torus of G, then any semisimple element of G is conjugate to a unique element of T . (ii) Let S be a (not necessarily closed) subgroup of G consisting of semisimple elements. Then S is contained in a torus. Moreover, the group CG (S) is connected and is equal to NG (S). Proof. (i) Recall that the canonical surjection π : G → G/Gu induces an isomorphism from T onto G/Gu . Let α ∈ Gs . By 26.5.4, α is conjugate to an element of T . Now if β ∈ G, then π(βαβ −1 ) = π(α) since G/Gu is abelian. So the result follows. (ii) The restriction of the canonical surjection π to S is injective, so S is abelian. If α ∈ NG (S) and β ∈ S, then π(βαβ −1 ) = π(α) since G/Gu is abelian. So βαβ −1 = α from the injectivity of π|S . Hence CG (S) = NG (S). Now, S is a closed diagonalizable subgroup of G verifying CG (S) = CG (S) (21.6.3). So we may assume that S is closed in G. Let T be a maximal torus in G. If S is central in G, then CG (S) = G is connected and since any element of S is conjugate to an element of T (26.5.4), we have S ⊂ T . Suppose that S ⊂ Z(G) and s ∈ S \Z(G). Again by 26.5.4, we may assume that s ∈ T . Then T ⊂ CG (s) = T (CG (s))u , which proves that H = CG (s) is connected (22.3.6). Since H = G, we can argue by induction on dim G, that S is conjugated in H to a subgroup of T and that CH (S) = CG (S) is connected. So we have proved (ii).

References and comments • [5], [40], [78], [79]. In this chapter, we have followed closely the presentation of solvable groups given in [5].

27 Reductive groups

Semisimple groups and reductive groups are very useful objects in the study of algebraic groups. In this chapter, we consider these groups and their representations. We prove Hilbert-Nagata Theorem on invariants of rational reductive group actions, and we introduce the notion of algebraic quotient.

27.1 Radical and unipotent radical 27.1.1 In this section, H is an algebraic group and G is a closed normal subgroup of H. Denote by R1 (G) the set of normal solvable subgroups of G and R2 (G) (resp. R(G)) the set of elements of R1 (G) which are closed (resp. closed and connected) in G. Let R1 (G), R2 (G) and R(G) be the subgroups of G generated by R1 (G), R2 (G) and R(G) respectively. These subgroups are normal in G and the subgroup R(G) is called the radical of G. Lemma. (i) We have R1 (G) = R2 (G). The group R1 (G) is the largest normal solvable subgroup of G. It is a closed normal subgroup of H. (ii) The radical R(G) of G is a connected solvable algebraic subgroup of G. It is the largest connected normal solvable subgroup of G. It is normal in H and we have R(G) = R(G◦ ) = R1 (G)◦ . Proof. (i) If H ∈ R1 (G), then H ∈ R2 (G) (21.6.1 and 21.6.6). So it is clear that R1 (G) = R2 (G). If R1 (G) = {e}, then it is solvable. Otherwise, there exists K ∈ R2 (G) such that R1 (G) ⊃ K = {e}. Since the group R1 (G)/K is contained in R1 (G/K), we obtain by induction on dim G that R1 (G)/K is solvable. Hence R1 (G) is solvable (10.4.5). Again by using 21.6.1 and 21.6.6, we see that R1 (G) is closed. We deduce therefore that R1 (G) is the largest normal solvable subgroup of G. If α ∈ H, then αR1 (G)α−1 is again a normal solvable subgroup of G. So it is contained in R1 (G). It follows that R1 (G) is normal in H.

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(ii) Since R(G) ⊂ R1 (G), it is solvable. Also, it is clear that R(G) is normal in G, and it is closed and connected by 21.3.2. So R(G) ⊂ R1 (G)◦ . Now R1 (G) is normal in G, so R1 (G)◦ is also normal in G. Moreover R1 (G)◦ is solvable, so R(G) = R1 (G)◦ . We deduce therefore that R(G) is the largest connected normal solvable subgroup of G. By (i), R1 (G) is normal in H, so R(G) = R1 (G)◦ is also normal in H. Finally, R(G) is connected, so it is contained in G◦ . Therefore R(G) ∈ R(G◦ ). Hence R(G) ⊂ R(G◦ ). Conversely, since G◦ is normal in G, R(G◦ ) is a connected normal solvable algebraic subgroup of G, so R(G◦ ) ⊂ R(G). 27.1.2 Let us conserve the notations of 27.1.1. Let U(G) be the set of normal unipotent algebraic subgroups of G and denote by Ru (G) the subgroup of G generated by U(G). We call Ru (G) the unipotent radical of G. If K ∈ U(G), then K ∈ R(G), so Ru (G) ⊂ R(G). Moreover, by 21.3.2, Ru (G) is a closed connected subgroup of G. Lemma. The group Ru (G) is the set R(G)u of unipotent elements of R(G). It is the largest normal unipotent subgroup of G. It is normal in H and we have Ru (G) = Ru (G◦ ). Proof. If K ∈ U(G), then K ∈ R(G) and K ⊂ R(G)u . So Ru (G) ⊂ R(G)u . Since R(G) is normal in H and R(G)u is a closed normal subgroup of R(G) (26.4.2), it is clear that R(G)u is a normal algebraic subgroup of H. In particular, R(G)u ⊂ Ru (G), and hence R(G)u = Ru (G). Finally, R(G) = R(G◦ ) by 27.1.1, so Ru (G) = R(G)u = R(G◦ )u = Ru (G◦ ). 27.1.3 Proposition. Let g be the Lie algebra of G, r the radical of g and t the nilpotent radical of g. (i) The Lie algebra of R(G) is r. (ii) We have t ⊂ L Ru (G) . Proof. In view of 27.1.1 and 27.1.2, we may assume that G is connected. (i) By 24.4.7 and 24.5.13, L R(G) is a solvable ideal of g, so L R(G) ⊂ r. On the other hand, r is an algebraic Lie subalgebra of g (24.7.4). Let H be the smallest closed subgroup of G such that L(H) = r. Then by 24.4.7 and 24.5.13, H is a connected normal solvable algebraic of G, so H ⊂ R(G). subgroup Hence r = L(H) ⊂ L R(G) . Thus r = L R(G) . (ii) Let x ∈ t. Then ρ(x) is nilpotent for all ﬁnite-dimensional representation ρ of g. It follows that η(x) is locally nilpotent where η is the representation of g in A(G) deﬁned in 23.6.2. Thus x is a nilpotent element of r. Since is the set of nilpotent elements of r (26.4.5 and 27.1.2), we have L Ru (G) x ∈ L Ru (G) . 27.1.4 Remarks. 1) In general, we have t = L R (G) . For example, if u G = Ga , then Ru (G) = G, L Ru (G) = g and t = {0}. 2) The group (G, R(G)) is connected (21.3.5) and its Lie algebra is equal to [g, r] = t (20.3.6 and 24.5.11). So 27.1.3 implies that (G, R(G)) ⊂ Ru (G).

27.2 Semisimple and reductive groups

415

27.2 Semisimple and reductive groups 27.2.1 Deﬁnition. An algebraic group G is said to be semisimple if R(G) = {e}. It is said to be reductive if Ru (G) = {e}. 27.2.2 Proposition. Let G be an algebraic group and g its Lie algebra. (i) The group G is semisimple if and only if g is semisimple. (ii) The group G is reductive if and only if g is reductive and all the elements of the centre of g are semisimple. Proof. Let r be of g, and t the nilpotent radical of g. the radical (i) Since L R(G) = r and R(G) is connected, we have R(G) = {e} if and only if r = {0}. So the result follows from the Levi-Malcev Theorem. (ii) Suppose that G is reductive. Then by 27.1.3, t = {0}. So g is reductive and r is the centre of g. Thus R(G) is the centre of G◦ (24.4.2). But Ru (G) = R(G)u (27.1.2), we deduce therefore from 26.4.5 that all the elements of r are semisimple. Conversely, suppose that g is reductive and all the elements of r = z(g) are semisimple. Since Ru (G) = R(G)u is connected and L Ru (G) is the set of nilpotent elements of r (26.4.5), Ru (G) = {e}. Thus G is reductive. 27.2.3 Remarks and examples. 1) Let G is a connected algebraic group. Recall from 21.3.6 and 24.5.12 that (G, G) is a connected algebraic subgroup of G whose Lie algebra is [g, g]. So if G is semisimple, then G = (G, G) (20.1.5). Similarly, if G is reductive, then (G, G) is semisimple (20.5.5). 2) The product of two semisimple (resp. reductive) algebraic groups is semisimple (resp. reductive). 3) Recall from 20.2.2 that sln (k) is semisimple. So the group SLn (k) is semisimple (23.4.5 and 27.2.2). 4) The Lie algebra g can be reductive while G is not reductive. This is the case for the additive group G = Ga (27.1.4). 27.2.4 Proposition. Let G be a connected algebraic group. Then the following conditions are equivalent: (i) G is semisimple. (ii) The only connected normal abelian algebraic subgroup of G is {e}. Proof. The implication (i) ⇒ (ii) is clear since such a subgroup is contained in R(G) = {e}. Conversely if R(G) = {e}, then the last non-trivial term of the derived series of R(G) is a connected normal abelian algebraic subgroup of G (21.3.6). 27.2.5 Proposition. Let G be a connected reductive algebraic group. (i) Its radical R(G) is a torus, and it is equal to the identity component of Z(G). (ii) The group (G, G) ∩ R(G) is ﬁnite.

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Proof. (i) Since R(G)u = Ru (G) = {e}, we have R(G) = R(G)s . As R(G) is connected, it is a torus (22.5.7). By 27.1.3, L R(G) isthe radical of g, which is equal to z(g) (20.5.4 and 27.2.2). Since L Z(G) = z(g) (24.4.2) and R(G) is connected, we have R(G) = Z(G)◦ (24.3.5). (ii) Let H = (G, G) ∩ R(G). Then H ◦ is a connected normal abelian algebraic subgroup of (G, G). By 27.2.4, H ◦ = {e}. So H is ﬁnite.

27.3 Representations 27.3.1 Let us start by a remark. Let G be an algebraic group, E a rational G-module and F a simple submodule of E. If x ∈ F \ {0}, then the submodule generated by x is clearly F . It follows that F is ﬁnite-dimensional. 27.3.2 Theorem. Let G be an algebraic group, (V, ρ) a ﬁnite-dimensional rational representation of G and σ the restriction of ρ to G◦ . Then the following conditions are equivalent: (i) ρ is semisimple. (ii) σ is semisimple. Proof. (i) ⇒ (ii) By 10.7.4, we may assume that ρ is simple. So V is ﬁnitedimensional and it contains a simple G◦ -submodule W . Now G◦ is normal in G, so ρ(α)(W ) is a simple G◦ -module for all α ∈ G. Set: S= ρ(α)(W ). α∈G

Then S is a non-zero G-submodule of V . Hence S = V and therefore σ is semisimple. (ii) ⇒ (i) Let W be a d-dimensional G-submodule of V . We shall show that V = W ⊕ W for some G-submodule W of V , and the result would follow. Let g be the Lie algebra of G. By 27.2.2, dσ(g) = g is reductive, and the elements of its centre are semisimple. So if (E, θ) is a ﬁnite-dimensional rational representation of σ(G), then p the representation p dθ(g ) is semisimple. V ) (resp. (σp , V )) the representation For an integer p, denote by (ρp , of G (resp. G◦ ) associated to ρ (resp. σ) as deﬁned in 23.4.14. Then by the above discussion and 20.5.10, dσp is semisimple, and so 24.3.3 implies that σp is semisimple. d d W be a non-zero element. Then W = ky and for α ∈ G, Let y ∈ we have ρd (α)(y) = θ(α)y where θ(α) = det(ρ(α)|W ) is a character of G. Let d V ; σd (α)(x) = θ(α)(x) for all α ∈ G◦ } and T be the subspace of S = {x ∈ d d V spanned by σd (α)(x) − θ(α)x, α ∈ G◦ and x ∈ V. d V = S⊕T Applying 10.7.8 to θ, G◦ and G, we obtain the decomposition d V into G-submodules. of We claim that the restriction λ of ρd to S is a semisimple representation of G. If α ∈ G, set λ1 (α) = θ(α)−1 λ(α). Then (λ1 , S) is a representation of G

27.3 Representations

417

whose kernel contains G◦ . Thus λ1 (α) depends only on the class of α modulo G◦ . This induces a representation (λ2 , S) of the ﬁnite group G/G◦ which is semisimple by 10.7.6. It follows that λ1 , and hence λ, is semisimple. We have therefore proved our claim. Since ky is a G-submodule of S, it followsthat S = ky ⊕ U for some d V = ky ⊕ U for some GG-submodule U . This implies, in turn, that submodule U . It follows from 10.7.7 that V = W ⊕ W where W is the set of d−1 W. x ∈ V such that x ∧ z ∈ U for all z ∈ To ﬁnish the proof, it suﬃces therefore to prove that W is a G-submodule. d−1 W . Since ρ(α)(W ) = W , we have Let x ∈ W , α ∈ G and u ∈ d−1 d−1 d−1 ρd−1 (α)( W) = W . So u = ρd−1 (α)(u ) for some u ∈ W. We deduce therefore that ρ(α)(x) ∧ u = ρ(α)(x ∧ u ) ∈ U since U is a G submodule. Hence W is a G-submodule as required. 27.3.3 Theorem. The following conditions are equivalent for an algebraic group G: (i) G is reductive. (ii) Any ﬁnite-dimensional rational representation of G is semisimple. (iii) Any rational representation of G is semisimple. Proof. (i) ⇒ (ii) Suppose that G is reductive. Then g = L(G) is reductive and all the elements of r are semisimple (27.2.2). Let (V, ρ) be a ﬁnite-dimensional rational G-module. By 20.5.10, the representation (V, dρ) of g is semisimple. Therefore, 24.3.3 implies that the restriction of ρ to G◦ is semisimple. So ρ is semisimple by 27.3.2. (ii) ⇒ (i) We may assume that G ⊂ GL(V ) for some ﬁnite-dimensional vector space V . Suppose that G is not reductive, then H = Ru (G) = {idV } and the group H acts non-trivially on V . So V H = V and V H = {0} (22.3.6). Now since H is normal in G, V H is a G-submodule of V . If V is a semisimple G-module, then V = V H ⊕ W for some non-zero G-submodule W . But again by 22.3.6, W H = {0} which implies that V H ∩ W = {0}. Contradiction. (ii) ⇒ (iii) If V is a rational G-module, then it is the sum of ﬁnitedimensional rational G-submodules. So if (ii) is veriﬁed, then V is the sum of simple G-modules. (iii) ⇒ (ii) This is obvious. 27.3.4 Let G be an algebraic group and let us consider the action ρ of G on A(G) via right translations (22.1.4). Assume that G is reductive. The subspace A(G)G is the set of constant functions and we may identify it with k. By 10.7.9, the subspace V of A(G) spanned by ρα f − f , α ∈ G and f ∈ A(G), is the unique submodule of A(G) such that A(G) = k ⊕ V . Let us denote by IG or I the Reynolds operator associated to A(G). Note that I ◦ρα = I for all α ∈ G. It is clear that λα (k) = k if α ∈ G, where λα denotes the left translation of functions (22.1.4). Furthermore, we see immediately that λα (V ) = V . Thus I ◦λα = I.

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27 Reductive groups

27.3.5 In the rest of this section, for an algebraic group G, we denote by G the set of isomorphism classes of ﬁnite-dimensional simple rational G-modules. - then Mω denotes a representative of ω, and M ∗ denotes the dual If ω ∈ G, ω G-module (see 10.6.5). If E is a rational G-module, then we set E(ω) to be the sum of submodules of Eω which are isomorphic to Mω . The G-submodule E(ω) is called the isotypic component of type ω of E. If E and F are ﬁnite-dimensional rational G-modules, we shall endow Hom(E, F ) and E ⊗k F with the G-module structures as deﬁned in 10.6.5. 27.3.6 Theorem. Let G be a reductive algebraic group and E, F rational G-modules. - then any simple submodule of E(ω) is isomorphic to Mω . (i) If ω ∈ G, (ii) We have: E(ω) . E= ω∈G

- then f (E(ω) ) ⊂ F(ω) . (iii) If f ∈ HomG (E, F ) and ω ∈ G, - the map (iv) For all ω ∈ G, θω : HomG (Mω , E) ⊗k Mω → E(ω) , f ⊗ v → f (v) , is an isomorphism of G-modules. Proof. (i) This is clear by 10.7.5 (ii). (ii) Since E is a semisimple G-module (27.3.3), it is equal to the sum of the - and E the sum of the E(ω) ’s where ω ∈ G - \ {θ}. E(ω) ’s (27.3.1). Let θ ∈ G, If E ∩ E(θ) is non-zero, then it contains a simple submodule E . But part (i) and 10.7.5 imply that E is isomorphic to Mθ and Mω for some ω = θ. This is absurd (10.7.2). So the sum of the E(ω) ’s is direct. (iii) In view of 10.7.2, this is obvious. (iv) By (iii), θω is well-deﬁned, and a simple veriﬁcation shows that θω is a homomorphism of G-modules. It is also clear that θω is surjective. Let (Vi )i∈I be a family of submodules of E verifying Vi Mω for all i ∈ I, and: Vi . E(ω) = i∈I

Denote by (pi )i∈I the family of projectors associated to this direct sum decomposition. If f ∈ HomG (Mω , E), then the image of f is contained in E(ω) and f = i∈I pi ◦ f . It follows easily that: HomG (Mω , E) = HomG (Mω , Vi ). i∈I

Let us ﬁx ui ∈ HomG (Mω , Vi ) \ {0}. By 10.7.2, ui is bijective and all the elements of HomG (Mω , Vi ) are proportional to ui . Let f1 , . . . , fn ∈ HomG (Mω , E) and x1 , . . . , xn ∈ Mω be such that

27.3 Representations

419

θω (f1 ⊗ x1 + · · · + fn ⊗ xn ) = 0. For all i ∈ I, we have pi f1 (x1 ) + · · · + fn (xn ) = 0. So we may assume that f1 , . . . , fn ∈ HomG (Mω , Wi ), fk = λk ui , with λi ∈ k for 1 k n. Thus λ1 x1 + · · · + λn xn = 0 since ui is injective. We deduce therefore that f1 ⊗ x1 + · · · + fn ⊗ xn = 0. Hence θω is injective. 27.3.7 Remarks. Let us conserve the hypotheses of 27.3.6. 1) In the proof of 27.3.6 (iv), the cardinality of the set I is called the multiplicity of Mω in E, and we shall denote it by multω (E). The module HomG (Mω , E) has dimension multω (E), and we have the following isomorphism of G-modules: multω (E)Mω . E ω∈G

2) If L is a submodule of E, then L(ω) = L ∩ E(ω) for all ω ∈ G. 27.3.8 Proposition. Let G be a reductive algebraic group acting rationally on a commutative k-algebra A by automorphisms, and let B = AG . (i) If a is a G-stable ideal of A, then the G-modules B/(B∩a) and (A/a)G are isomorphic. (ii) Let (ai )i∈I be a family of G-stable ideals of A. Then: ai = (B ∩ ai ). B∩ i∈I

i∈I

- But a(ω) = Proof. (i) By 27.3.6, a is the direct sum of the a(ω) ’s, for ω ∈ G. a ∩ A(ω) by 27.3.7. We deduce therefore that: A/a = (A(ω) /a(ω) ). ω∈G

It follows that (A/a)(ω) = A(ω) /a(ω) , and we obtain the result by taking ω to be the isomorphism class of the trivial G-module. (ii) We have: ai = (ai )(ω) = (ai )(ω) . i∈I

So

i∈I

ai

(ω)

=

i∈I ω∈G -

i∈I (ai )(ω)

- i∈I ω∈G

- and we conclude as in (i). for all ω ∈ G,

27.3.9 Let E, F be G-modules and consider the action of G × G on E ⊗k F deﬁned in 23.4.12, and the action of G × G on A(G) deﬁned as follows: for α, β, γ ∈ G and f ∈ A(G), (α, β).f (γ) = f (α−1 γβ). Theorem. Let G be a reductive algebraic group and

420

27 Reductive groups

φ:

ω∈G

Mω∗ ⊗k Mω → A(G)

the map sending f ⊗ x ∈ Mω∗ ⊗k Mω to the function α → f (α.x). (i) φ is an isomorphism of (G × G)-modules. (ii) We have multω A(G) = dim Mω for all ω ∈ G. Proof. Let x ∈ Mω , f ∈ Mω∗ , and α, β, γ ∈ G. Then: (α, β).φ(f ⊗ x) (γ) = φ(f ⊗ x)(α−1 γβ) = f (α−1 γβ.x) = (α.f )(γβ.x) = φ (α.f ) ⊗ (β.x) (γ). So φ is a homomorphism of (G × G)-modules. 1) Let us ﬁx f ∈ Mω \ {0}. Then (e, α).(f ⊗ x) = f ⊗ (α.x) for x ∈ Mω . Thus f ⊗k Mω is a simple G-module isomorphic to Mω . It follows immediately that since ρα φ(f ⊗ x) = φ f ⊗ (α.x) , φ induces a homomorphism of the simple G-module f ⊗k Mω into the Gmodule A(G). It is clear that φ(f ⊗k Mω ) is non-zero, so it is isomorphic to Mω and contained in A(G)(ω) . Hence A(G)(ω) = {0} for all ω ∈ G. 2) Now, let V be a simple submodule of A(G) which is isomorphic to Mω and x → ux a G-module isomorphism from Mω onto V . Deﬁne g ∈ Mω∗ by g(x) = ux (e), x ∈ Mω . Then for α ∈ G and x ∈ Mω , we have: φ(g ⊗ x)(α) = g(α.x) = uα.x (e) = (α.ux )(e) = ux (α). We deduce that φ(g ⊗k Mω ) = V . Since φ(Mω∗ ⊗k Mω ) ⊂ A(G)(ω) by point 1, it follows that φ(Mω∗ ⊗k Mω ) = A(G)(ω) . So φ is surjective by 27.3.6 (ii). - and M = Mω . If V is a submodule of M ∗ , its 3) Finally, let ω ∈ G orthogonal in M is a submodule of M . So M ∗ is a simple G-module. We claim that M ∗ ⊗k M is a simple (G × G)-module. Denote by σ and σ the representations of G on M and M ∗ , and θ the associated representation of G × G on M ∗ ⊗k M . By 10.8.11, the subalgebra of End(M ) (resp. End(M ∗ )) generated by σ(G) (resp. σ (G)) is equal to End(M ) (resp. End(M ∗ )). Since End(M ∗ ⊗k M ) is isomorphic to End(M ∗ )⊗k End(M ), we deduce that the subalgebra of End(M ∗ ⊗k M ) generated by θ(G × G) is equal to End(M ∗ ⊗k M ). So we have proved our claim. Since A(G)(ω) = φ(Mω∗ ⊗k Mω ) = {0}, our claim and point 2 imply that φ is bijective. So we have proved (i), and (ii) follows immediately.

27.4 Finiteness properties 27.4.1 In this section, G denotes a reductive algebraic group. Let A be a commutative k-algebra on which G acts rationally by automorphisms. Denote by p the Reynolds operator associated to A. Lemma. For u ∈ AG and v ∈ A, we have p(uv) = up(v). Consequently, the unique complementary submodule ker(p) of AG in A, is an AG -module.

27.4 Finiteness properties

421

Proof. There exist, by 10.7.9, α1 , . . . , αn ∈ G and w1 , . . . , wn ∈ A such that: v = p(v) +

n

(αi .wi − wi ).

i=1

Since αi .(uwi ) = u(αi .wi ), we have: uv = up(v) +

n

αi .(uwi ) − uwi .

i=1

On the other hand, up(v) ∈ AG , so the result follows from 10.7.9. 27.4.2 Let V be a ﬁnite-dimensional rational G-module and A the algebra of polynomial functions on V . Endow A with the G-module structure deﬁned in 22.1.1. Proposition. The algebra AG is ﬁnitely generated. Proof. For n ∈ N, let An be the subspace of A consisting of homogeneous polynomial functions of degree n. Then An is G-stable and so: G AG = An . n∈N

Let I be the ideal of A generated by AG n , n 1. Since A is Noetherian, I is ﬁnitely generated. Let f1 , . . . , fr ∈ AG be a system of generators of I that we may assume to be homogeneous. Now let f ∈ AG be homogeneous of degree non-zero, then: f = f1 g1 + · · · + fr gr where g1 , . . . , gr are homogeneous. Let p denote the Reynolds operator of A. It follows from 27.4.1 that: f = p(f ) = f1 p(g1 ) + · · · + fr p(gr ). But hi = p(gi ) ∈ AG for 1 i r, so if hi ∈ k[f1 , . . . , fr ] for 1 i r, then f ∈ k[f1 , . . . , fr ]. Since deg(hi ) < deg f , we obtain by induction (on the degree of f ) that AG = k[f1 , . . . , fr ]. 27.4.3 Theorem. (Hilbert-Nagata Theorem) Let G be a reductive algebraic group and A a ﬁnitely generated commutative k-algebra on which G acts rationally by automorphisms. (i) The algebra B = AG is ﬁnitely generated, and is a direct summand of the B-module A. (ii) If b is an ideal of B, then B ∩ (Ab) = b. Proof. The fact that B is a direct summand of the B-module A follows from 27.4.1. Let us denote by p the Reynolds operator associated to A.

422

27 Reductive groups

Let b be an ideal of B. Then b ⊂ B∩(Ab). Conversely, if f ∈ B∩(Ab), then there exist f1 , . . . , fn ∈ b and g1 , . . . , gn ∈ A such that f = g1 f1 + · · · + gn fn . By 27.4.1, we have: f = p(f ) = f1 p(g1 ) + · · · + fn p(gn ). Thus f ∈ b. Now let x1 , . . . , xn be a system of generators of A. There exists a ﬁnitedimensional submodule W of A containing x1 , . . . , xn . Denote by V the dual of W . The injection V ∗ = W → A extends to a surjective G-module homomorphism u from A(V ) onto A. Since u◦pA(V ) = p◦u, u induces a surjective homomorphism A(V )G → AG . Hence the result follows from 27.4.2. 27.4.4 Let us conserve the notations and hypotheses of 27.4.3. Let V be a rational G-module. If u ∈ HomG (V, A) and f ∈ AG , then clearly the map x → f u(x) belongs to HomG (V, A). Thus HomG (V, A) has a natural structure of AG -module. Proposition. Let V be a ﬁnite-dimensional rational G-module. Then HomG (V, A) is a ﬁnitely generated AG -module. Proof. We have: G G G A ⊗ A(V ) = A ⊗ A(V )n A ⊗ A(V )n = n0 n0 G A ⊗ A(V )n = AG ⊕ (A ⊗ V ∗ )G ⊕ n2 G A ⊗ A(V )n . = AG ⊕ HomG (V, A) ⊕ n2

G By 27.4.3, A ⊗ A(V ) is a ﬁnitely generated commutative AG -algebra. In G particular, the ideal n1 A ⊗ A(V )n of this AG -algebra is ﬁnitely generated. Since HomG (V, A) = (A ⊗ A(V )1 )G , the result follows. 27.4.5 Corollary. Let us conserve the notations and hypotheses of 27.4.3. - the subspace A(ω) is a ﬁnitely generated AG -module. For all ω ∈ G, Proof. This is clear by 27.3.6 (iv) and 27.4.4.

27.5 Algebraic quotients 27.5.1 In this section, G is a reductive algebraic group and X an aﬃne G-variety. Recall from 22.1.1 that G acts naturally on A(X). By 27.4.3, the reduced algebra A(X)G is ﬁnitely generated. So 11.6.1 implies that A(X)G is the algebra of regular functions of an aﬃne variety that we shall denote by X//G, instead of Spm(A(X)G ). Let π : X → X//G be the morphism induced by the injection A(X)G → A(X). The pair (X//G, π), or X//G, is called the algebraic quotient of X by G.

27.5 Algebraic quotients

423

Proposition. (i) The map π is surjective and constant on G-orbits. (ii) Let Y be an aﬃne G-variety and u : X → Y be a morphism which is constant on the G-orbits of X. Then there exists a unique morphism v : X//G → Y such that u = v ◦ π. (iii) If X is normal, then X//G is also normal. Proof. (i) For α ∈ G, x ∈ X et f ∈ A(X)G = A(X//G), we have: f (x) = A(π)(f )(x) = f π(x) = f (α.x) = f π(α.x) . So π(x) = π(α.x). Let ξ ∈ X//G and a its ideal in A(X)G . Then aA(X) ∩ A(X)G = a by 27.4.3. It follows that aA(X) = A(X). Let b be a maximal ideal of A(X) containing aA(X) and x ∈ X be the point deﬁned by b. Then we have ξ = π(x). (ii) The assumptions imply that A(u) A(Y ) ⊂ A(X)G . Thus A(u) = A(π)◦θ where θ is an algebra homomorphism from A(Y ) to A(X)G . The morphism v = Spm(θ) veriﬁes u = v◦π. Clearly, v is unique with this property. (iii) If f ∈ Fract A(X)G is integral over A(X)G , then f is integral over A(X), so f belongs to A(X) since X is normal. Hence f ∈ A(X)G . 27.5.2 Proposition. Let us conserve the hypotheses and notations of 27.5.1. (i) If Y is a closed G-stable subset of X, then π(Y ) is a closed subset of X//G. Furthermore, π(Y ), π|Y is the algebraic quotient of Y by G. (ii) Let (Yi )i∈I be a family of closed G-stable subsets of X. Then: Yi = π(Yi ). π i∈I

i∈I

Proof. (i) Let j : Y → X be the canonical injection and (Y //G, ω) the algebraic quotient of Y by G. The map A(j) is the surjection f → f |Y . Let us denote by pA(X) , pA(Y ) the corresponding Reynolds operators. Since A(j)◦pA(X) = pA(Y ) ◦A(j), A(j) induces a surjection θ : A(X)G → A(Y )G . The map A(ω) is the canonical injection A(Y )G → A(Y ) and we have A(ω)◦θ = A(j)◦A(π), so Spm(θ)◦ω = π◦j. By 11.6.8, Spm(θ) = i◦v where v is an isomorphism of Y //G onto a closed subvariety Z of X//G, and i : Z → X//G is the canonical injection. So (i) follows. (ii) If a is an ideal of A(X) (resp. A(X)G ), denoteby V(a) (resp. V (a)) the variety of zeros of a in X (resp. X//G). We have i∈I Yi = V i∈I ai . So (i) and 27.3.8 imply that: π Yi = π Yi = V A(X)G ∩ ai = V (A(X)G ∩ ai ) i∈I i∈I i∈I i∈I = V (A(X)G ∩ ai ) = π(Yi ) = π(Yi ). i∈I

i∈I

i∈I

424

27 Reductive groups

27.5.3 Proposition. Let us conserve the hypotheses and notations of 27.5.1. If x ∈ X, then π −1 (π(x)) contains a unique closed G-orbit O and we have: π −1 π(x) = {y ∈ X; O ⊂ G.y}. Proof. The set Y = π −1 π(x) is closed and G-stable, so it contains a closed orbit by 21.4.5. Suppose that O1 and O2 are two distinct closed orbits contained in Y . Then 27.5.2 (ii) implies that π(O1 ) ∩ π(O2 ) is empty, which is absurd. So Y contains a unique closed G-orbit O. If y ∈ Y , then the closed G-stable subset G.y of Y contains a closed orbit (21.4.5), so O ⊂ G.y. Conversely, as π is constant on the closures of orbits, if O ⊂ G.y, then π(x) = π(y). 27.5.4 Remark. In view of 27.5.3, we can interpret X//G as the set of closed G-orbits of X.

27.6 Characters 27.6.1 Deﬁnition. Let G be an algebraic group. (i) A function f ∈ A(G) is said to be central if f (αβα−1 ) = f (β) for all α, β ∈ G. We shall denote by C(G) the subalgebra of A(G) consisting of central functions. (ii) Let (V, π) be a ﬁnite-dimensional rational G-module. The character of V is deﬁned to be the map χV : G → k, α → tr π(α). 27.6.2 If V, W are ﬁnite-dimensional rational G-modules, then: χV ⊕W = χV + χW , χV ⊗W = χV χW . Moreover, it is clear that χV ∈ C(G) and that χV = χW if V and W are - then we denote χM simply by χω . isomorphic G-modules. If ω ∈ G, ω Proposition. Let G be a reductive algebraic group. The space C(G) is spanned by the χω ’s, with ω ∈ G. - and set M = Mω . Let (e1 , . . . , en ) be a basis of M , Proof. Fix ω ∈ G ∗ ∗ (e1 , . . . , en ) the dual basis and φ as deﬁned in 27.3.9. If α ∈ G, then: χω (α) =

n i=1

∗

e∗i (α.ei ) =

n

φ(e∗i ⊗ ei )(α).

i=1

Thus χω ∈ φ(M ⊗k M ). By 27.3.9, to obtain the result, it suﬃces to prove that any central function contained in φ(M ∗ ⊗k M ) is proportional to χω . Let θ : M ∗ ⊗k M → Endk (M ) be the map sending f ⊗ x to the endomorphism y → f (y)x. It is well-known that θ is a linear isomorphism and we verify easily that θ is a G-module homomorphism. Moreover, for α, β ∈ G, f ∈ M ∗ and x ∈ M , we have: φ α.(f ⊗ x) (β) = f (α−1 βαx) = φ(f ⊗ x)(α−1 βα).

27.6 Characters

425

We deduce therefore that if ξ ∈ M ∗ ⊗k M , then the following conditions are equivalent: (i) φ(ξ) ∈ C(G). (ii) α.ξ = ξ for all α ∈ G. (iii) α.θ(ξ) = θ(ξ) for all α ∈ G. (iv) α.θ(ξ)(α−1 .x) = θ(ξ)(x) for all (α, x) ∈ G × M . (v) θ(ξ) ∈ EndG (M ) = k idM (10.7.2). But, we have: idM = θ(e∗1 ⊗ e1 + · · · + e∗n ⊗ en ). Since θ is bijective, the result follows from equivalence of conditions (i) and (v). 27.6.3 Let G be a reductive algebraic group, I the Reynolds operator associated to A(G) and (V, σ) a ﬁnite-dimensional rational G-module. Let (e1 , . . . , en ) be a basis of V and (e∗1 , . . . , e∗n ) the dual basis. Set α.ej =

n

σij (α)ei

i=1

for α ∈ G and 1 j n. With φ in the notations of 27.3.9, we have σij = φ(e∗i ⊗ ej ). If α, β ∈ G, then: n n σlj (αβ)el = σij (β) σli (α)el i=1 l=1 l=1 n n = σli (α)σij (β) el .

(αβ).ej =

n

i=1

l=1

Hence: (1)

ρβ σlj =

n

σij (β)σli , λα−1 σlj =

i=1

n

σli (α)σij .

i=1

Since I ◦ρβ = I ◦λα−1 = I (27.3.4), we have therefore (2)

I(σlj ) =

n

σij (β) I(σli ) =

i=1

n

σli (α) I(σij )

i=1

for all α, β ∈ G. We deduce that: (3)

I(σlj ) =

n

I(σli ) I(σij ).

i=1

Deﬁne q ∈ End(V ) as follows: for 1 j n: q(ej ) =

n i=1

I(σij )ei .

426

27 Reductive groups

If x ∈ V and f ∈ V ∗ , then: f q(x) = I φ(f ⊗ x) .

(4)

Lemma. The map q is the Reynolds operator pV associated to V . Proof. The equalities in (2) imply that q◦σ(α) = σ(α)◦q = q for all α ∈ G and G ∗ (3) implies that q 2 = q. Finally if x ∈ V , for f ∈ V , we have φ(f ⊗x)(α) = f (α.x) = f (x), so I φ(f ⊗ x) = f (x). By (4), we obtain that f q(x) = f (x) for all f ∈ V ∗ , hence q(x) = x. Now the result follows from 10.7.9. 27.6.4 Let us deﬁne an involution f → f ∗ in A(G) and a bilinear form (f, g) → f, g on A(G) as follows: f ∗ (α) = f (α−1 ) , f, g = I(f g ∗ ). If h ∈ A(G) and α ∈ G, then (ρα h−h)∗ = λα h∗ −h∗ . By 10.7.9 and 27.3.4, we deduce that I(f ) = I(f ∗ ) for f ∈ A(G). The form . , . is therefore symmetric. Let us endow A(G) with this bilinear form in the rest of this section. 27.6.5 Let (V, σ) and (W, τ ) be non-isomorphic simple rational G-modules of dimension m and n, E (resp. F) a basis of V (resp. W ). For α ∈ G, denote by S(α) = [σij (α)] and T (α) = [τij (α)] the matrices of σ(α) and τ (α) with respect to the basis E et F. Lemma. (i) For 1 i, j m and 1 k, l n, we have σij , τkl = 0. (ii) For 1 i, j, k, l m, we have mσij , σkl = δil δjk , where δ denotes the Kronecker symbol. Proof. (i) For A ∈ Mm,n (k), set A(α) = S(α)AT (α−1 ) = [θij (α)] and let B = [I(θij )] ∈ Mm,n (k). If α, β ∈ G, then A(αβ) = S(α)A(β)T (α−1 ). It follows as in 27.6.3 that S(α)B = BT (α). So 10.7.2 implies that B = 0. But if A = [aij ], then: θij (α) = σik (α)akl τlj∗ (α). k,l

Hence: I(θij ) =

akl σik , τlj .

k,l

So (i) follows by taking A to be the matrices of the canonical basis of Mm,n (k). (ii) If σ = τ , then 10.7.2 implies that B = µA Im , with µA ∈ k. So I(θij ) = 0 if i = j, and the same argument as in (i) proves the result for i = j. On the other hand, we have tr A(α) = tr(A) for all α ∈ G, so tr(B) = tr(A). It follows that µA = m−1 tr(A). So (ii) follows as in (i) by taking A to be the matrices of the canonical basis of Mm (k).

27.6 Characters

427

27.6.6 Theorem. Let G be a reductive algebraic group. (i) The decomposition A(G) = (Mω∗ ⊗ Mω ) ω∈G

in 27.3.9 is orthogonal. Moreover, the restriction of . , . to each Mω∗ ⊗ Mω is non-degenerate. - form an orthonormal basis of C(G). (ii) The characters χω , ω ∈ G, - then: (iii) If V is a ﬁnite-dimensional rational G-module and ω ∈ G, multω (V ) = χω , χV . (iv) Any ﬁnite-dimensional rational G-module is determined up to an isomorphism by its character. Proof. Let V and W be two simple rational G-modules of ﬁnite dimension, (e1 , . . . , em ) (resp. (ε1 , . . . , εn )) a basis of V (resp. W ), and (e∗1 , . . . , e∗m ) (resp. (ε∗1 , . . . , ε∗n )) the dual basis. Let φ be as in 27.3.9. Suppose that V and W are not isomorphic. Then φ(e∗i ⊗ej ), φ(ε∗k ⊗εl ) = 0 by 27.6.3 and 27.6.5. Hence we have (i). - where ω = θ. We saw in the proof of 27.6.2 that χω ∈ Let ω, θ ∈ G, φ(Mω∗ ⊗ Mω ). So (i) implies that χω , χθ = 0. Moreover: χω , χω = φ(e∗i ⊗ ei ), φ(e∗j ⊗ ej ). i,j

So 27.6.5 says that χω , χω = 1, and (ii) follows from 27.6.2. Part (iii) follows from (ii) and 27.3.7, and (iv) is an immediate consequence of (iii).

References • [5], [38], [40], [51], [77], [78].

28 Borel subgroups, parabolic subgroups, Cartan subgroups

In this chapter, our interest turns to certain fundamental notions of the theory of algebraic groups : Borel subgroups, parabolic subgroups, Cartan subgroups. One of the main result we prove in this chapter is that Borel subgroups (resp. Cartan subgroups) of an algebraic group are conjugate. Let G be a connected algebraic group and g its Lie algebra. Two elements x, y ∈ g are conjugate if there exists α ∈ G such that y = Ad(α)(x).

28.1 Borel subgroups 28.1.1 Deﬁnition. A maximal connected solvable subgroup of G is called a Borel subgroup of G. 28.1.2 Remarks. Let H be a subgroup of G. If H is solvable and connected, then so is H. Thus a Borel subgroup is a maximal connected solvable algebraic subgroup of G. In particular, for reasons of dimension, G has a Borel subgroup. If H is unipotent, then H is also unipotent (10.8.14), so it is nilpotent and connected (10.8.10 and 22.3.6). Thus H is contained in a Borel subgroup of G. In the same way, a torus of G is contained in a Borel subgroup of G. 28.1.3 Theorem. (i) If B is a Borel subgroup, then G/B is projective. (ii) All the Borel subgroups of G are conjugate. Proof. Let P be a Borel subgroup of G of maximal dimension. By 25.4.3, there exists a ﬁnite-dimensional rational representation (V, π) of G and a line L of V such that: P = {α ∈ G; π(α)(L) = L} , p = L(P ) = {x ∈ g; dπ(x)(L) ⊂ L}. Applying 26.3.2 to the action of P on V /L, we obtain that P ﬁxes a ﬂag H0 : L = V1 ⊂ V2 ⊂ · · · ⊂ Vn = V of V . Since G acts rationally on the ﬂag variety F of V , our choice of L implies that P is the stabilizer of H0 in G.

430

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Now let H ∈ F and H the stabilizer of H in G. Then π(H) is solvable. On the other hand, ker π ⊂ P , so ker π is solvable. It follows that H is solvable. Hence dim H dim P . We deduce therefore that the G-orbit of H0 is of minimal dimension. So it is closed (21.4.5) and projective (13.6.7). Since the morphism G/P → G.H0 , αP → α.H0 , is G-equivariant and bijective, it follows from 25.1.2 that it is an isomorphism. Thus G/P is projective. Finally, let B be a Borel subgroup of G. Then B acts on G/P by (β, αP ) → βαP . By 13.4.5 and 26.3.4, there exists α ∈ G such that BαP = αP . So α−1 Bα ⊂ P , or B ⊂ αP α−1 . It follows from the deﬁnition of a Borel subgroup that α−1 Bα = P . 28.1.4 Corollary. Let P be a closed subgroup of G. Then the following conditions are equivalent: (i) The variety G/P is projective. (ii) The variety G/P is complete. (iii) The group P contains a Borel subgroup of G. If these conditions are veriﬁed, then P is called a parabolic subgroup of G. Proof. (i) ⇒ (ii) This is clear by 13.4.5. (ii) ⇒ (iii) Let B be a Borel subgroup of G. Then by 26.3.4, the action of B on G/P , given by (β, αP ) → βαP , has a ﬁxed point, say αP . So BαP = αP and P contains the Borel subgroup α−1 Bα of G. (iii) ⇒ (i) Let B be a Borel subgroup contained in P . Since G/B is projective (28.1.3) and the canonical morphism G/B → G/P is surjective, 13.4.3 implies that G/P is complete. But G/P is quasi-projective (25.4.7), so it is a projective variety. 28.1.5 Remark. Let H be a closed connected subgroup of G. By 28.1.4, the following conditions are equivalent: (i) H is a Borel subgroup of G (ii) H is solvable and G/H is projective. 28.1.6 Corollary. Let B be a Borel subgroup of G. (i) Let u : G → G be an automorphism of the algebraic group G. If u(α) = α for all α ∈ B, then u = idG . (ii) We have Z(G)◦ ⊂ Z(B) ⊂ CG (B) = Z(G). Proof. (i) Let π : G → G/B be the canonical surjection and v : G → G the morphism α → u(α)α−1 . If u(α) = α for all α ∈ B, then v is constant on αB for all α ∈ G. By 25.3.3 and 25.4.7, there exists a morphism w : G/B → G such that w◦π = v. It follows therefore from 13.4.3, 13.4.5 and 28.1.3 that v(G) is closed in G. So v(G) is aﬃne and complete. Moreover, v(G) is connected. Thus v(G) = v(B) = {e} by 13.4.4, and u = idG . (ii) The subgroup Z(G)◦ is connected and solvable. So Z(G)◦ is contained in a Borel subgroup P of G. By 28.1.3, there exists α ∈ G such that αP α−1 = B. So Z(G)◦ ⊂ Z(B) since αZ(G)◦ α−1 = Z(G)◦ . Clearly, Z(G) ∪ Z(B) ⊂

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CG (B). Finally, let α ∈ CG (B) and iα the inner automorphism of G deﬁned by α. Then iα |B = idB , so iα ∈ idG by part (i). Hence α ∈ Z(G). Remark. We shall see in 28.2.3 that Z(B) = Z(G). 28.1.7 Corollary. (i) The set of maximal tori of G is equal to the set of maximal tori of the Borel subgroups of G. Two maximal tori of G are conjugate. (ii) The set of maximal unipotent subgroups of G is equal to the set of maximal unipotent subgroups of the Borel subgroups of G. Two maximal unipotent subgroups of G are conjugate. Proof. (i) Let T be a maximal torus. We observed in 28.1.2 that T is contained in a Borel subgroup B and T is a maximal torus in B. By 26.5.5, any maximal torus of B is conjugate to T . So the result follows from the fact that Borel subgroups are conjugate (28.1.3). (ii) Again, we saw in 28.1.2 that any maximal unipotent subgroup U is contained in a Borel subgroup B and U is a maximal unipotent subgroup in B. So U = Bu by 26.4.2. The result follows again from 28.1.3. 28.1.8 Corollary. Let B be a Borel subgroup of G. (i) If B = Bs , then G is a torus. (ii) If {e} is the only torus of B, then G is unipotent. Proof. (i) By 26.5.5, B = T Bu where T is a maximal torus of B. If B = Bs , then B = T , and T = Z(B) ⊂ Z(G) by 28.1.6. So T is normal in G, and the variety G/T is irreducible, aﬃne and complete (25.4.11 and 28.1.3). So it is reduced to a single point (13.4.4). (ii) If B = Bu , then H = Z(B)◦ is not trivial unless G = {e}, and H ⊂ Z(G) (28.1.6). The group B/H is a unipotent Borel subgroup of G/H. By induction on dim G, we deduce that G/H = B/H. Hence G = B is unipotent. 28.1.9 Corollary. The following conditions are equivalent: (i) G is nilpotent. (ii) G has a nilpotent Borel subgroup. (iii) G has a unique maximal torus. (iv) There exists a maximal torus of G contained in Z(G). Proof. (i) ⇒ (ii) This is obvious. (ii) ⇒ (iii) Let B be a nilpotent Borel subgroup. By 26.4.4, B = T × Bu where T = Bs is a maximal torus of G. By 28.1.6, T ⊂ Z(B) ⊂ Z(G). Since any maximal torus of G is conjugate to T (28.1.7), T is the unique maximal torus of G. (iii) ⇒ (iv) If G has a unique maximal torus T , then T is normal in G (28.1.7). For α ∈ G, the map T → T , β → αβα−1 , is a morphism of the algebraic group T . So by 22.5.10, we have T ⊂ Z(G).

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(iv) ⇒ (i) Let T be a maximal torus of G contained in Z(G). By 26.5.5 and 28.1.7, we have B = T × Gu and T = Bs . So B is nilpotent (26.4.4). The group B/T is then a nilpotent Borel subgroup of G/T . By induction on dim G, we deduce that G/T is nilpotent, and since T ⊂ Z(G), G is nilpotent (10.5.3). 28.1.10 Corollary. (i) If T is a maximal torus of G, then C = CG (T )◦ is nilpotent and C = NG (C)◦ . (ii) If dim G 2, then G is solvable. Proof. (i) By 28.1.7 and 28.1.9, T is the unique maximal torus of C and C is nilpotent. Moreover if α ∈ NG (C), then αT α−1 is a maximal torus of C, so αT α−1 = T . Thus T is normal in NG (C), and so T commutes with NG (C)◦ by 22.5.10. Hence NG (C)◦ ⊂ C and the result follows. (ii) Let B be a Borel subgroup of G. Then B = T Bu where T is a maximal torus of G. If B = G, then dim B 1, so B = T or B = Bu . But 28.1.8 would imply that G = B. Contradiction. 28.1.11 Theorem. Let B be the set of Borel subgroups of G. Then: ◦ ◦ R(G) = B , Ru (G) = Bu B∈B

B∈B

Proof. Let H (resp. K) be the identity component of the intersection of B (resp. Bu ), B ∈ B. Since R(G) is connected, normal and solvable, it is contained in H (28.1.3). On the other hand, H is normal (28.1.3), connected and solvable. So H ⊂ R(G). The equality K = Ru (G) is proved in the same way.

28.2 Theorems of density 28.2.1 Proposition. Let H be a closed subgroup of G and X= αHα−1 . α∈H

(i) X contains a dense open subset of X. (ii) If the variety G/H is complete, then X is closed in G. (iii) Suppose that there exists γ ∈ G which is contained only in a ﬁnite number of conjugates of H. Then H = G. Proof. Denote by pr1 : (G/H) × G → G/H and pr2 : (G/H) × G → G the canonical projections, π : G → G/H the canonical surjection and u, v the morphisms: u : G × G → G × G , (α, β) → (α, αβα−1 ), v : G × G → (G × G)/(H × {e}) = (G/H) × G , (α, β) → (π(α), β).

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Finally, let M = {(α, β) ∈ G × G; β ∈ αHα−1 }, N = v(M ). (i) Since X = pr2 ◦v◦u(G × H), it is a constructible subset of G (15.4.3). So the result follows from 1.4.6. (ii) We have M = u(G × H) and u is an isomorphism of varieties. Thus M is closed in G × G. It follows from 25.3.2 and 25.4.7 that N is closed in (G/H) × G. So if G/H is complete, then X = pr2 (N ) is closed. (iii) Let p = pr1 |N . If α ∈ G, then p−1 π(α) = {π(α)} × (αHα−1 ). We obtain therefore from 15.5.5 that dim N = dim(G/H) + dim H = dim G. Consider the action of G on G/H given by α.π(β) = π(αβ) and denote q = pr2 |N . If α ∈ G, then q −1 (α) = Eα × {α} where Eα = {π(β); β −1 αβ ∈ H} = {π(β); α.π(β) = π(β)}. The set q −1 (γ) is ﬁnite by assumption. Let us consider the morphism N → X = pr2 (N ) induced by q. Then 15.5.4 (i) says that dim X = dim N . Thus dim X = dim G and, since G is connected, we have X = G by 1.2.3. 28.2.2 Theorem. Let B be a Borel subgroup of G, T a maximal torus in G and C = CG (T )◦ . Then: G= αBα−1 , Gu = αBu α−1 , Gs = αT α−1 . α∈G

Furthermore, the set

α∈G

α∈G

αCα−1 contains a dense open subset of G.

α∈G

Proof. By 28.1.10, C is nilpotent. So 26.4.4 implies that C = T ×Cu . Fix α ∈ T such that CG (α) = CG (T ) (26.2.3) and let β ∈ G be such that βαβ −1 ∈ C. Then βαβ −1 ∈ T and CG (T ) ⊂ CG (βαβ −1 ). But dim CG (βαβ −1 ) = dim C, so we deduce that CG (βαβ −1 ) = CG (βαβ −1 )◦ = C. Thus βCβ −1 = C, that is β ∈ NG (C). Since C = NG (C)◦ (28.1.10), we have proved that the set of conjugates of α which are contained in C is ﬁnite. We deduce therefore from 28.2.1 that the union of the conjugates of C contains a dense open subset of G. The group C is connected and nilpotent, so up to conjugation, we may assume that C ⊂ B (28.1.3). So the union D of the set of Borel subgroups contains a dense open subset of G. But G/B is projective (28.1.3), so D is closed in G by 28.2.1. Hence D = G. The other equalities follow now from 26.5.6 and 28.1.7. 28.2.3 Corollary. (i) We have Z(B) = Z(G) for any Borel subgroup B of the group G. (ii) The set Z(G)s is the intersection of all the maximal tori of G. Proof. (i) Let α ∈ Z(G). By 28.2.2, a conjugate of α is in B, so α ∈ B and Z(G) ⊂ Z(B). The result follows therefore from 28.1.6. (ii) Let α ∈ Z(G)s . As in the proof of (i), we have α ∈ B. So by 26.5.6, α is contained in a maximal torus of B. Hence α belongs to the intersection H of

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all the maximal tori of G (28.1.7). Conversely, H is a diagonalizable subgroup of G and clearly H is normal in G. So 22.5.10 implies that H is central in G. Hence H = Z(G)s .

28.3 Centralizers and tori 28.3.1 Recall that for α ∈ G, αs (resp. αu ) denotes the semisimple (resp. unipotent) component of α. Proposition. (i) Let H be a connected solvable subgroup of G. For any α ∈ CG (H), there exists a Borel subgroup of G containing H and α. (ii) Let S be a torus of G and α ∈ CG (S). There exists a torus of G containing S and αs . The group CG (S) is connected. (iii) If α ∈ G, then α ∈ CG (αs )◦ . Proof. (i) The group H is solvable, connected and α ∈ CG (H). We may therefore assume that H is closed. Let B be a Borel subgroup containing α (28.2.2). Consider the action of G on G/B given by: (β, γB) → (βγ)B. The variety X of ﬁxed points of α is non-empty since it contains eB, and it is H-stable. It follows from 26.3.4 that there exists β ∈ G such that βB ∈ X and γ.βB = βB for all γ ∈ H. So βBβ −1 is a Borel subgroup of G containing H and α. (ii) By (i), S and α are contained in a Borel subgroup B of G. We have αs ∈ B and the ﬁrst part follows from 26.5.6. Again by 26.5.6, CB (S) is connected, so α ∈ CG (S)◦ . Hence CG (S) = CG (S)◦ . (iii) Let B be a Borel subgroup containing α. Then it contains also αs , and α ∈ CB (αs ) = CB (αs )◦ (26.5.6). Hence α ∈ CG (αs )◦ . 28.3.2 Proposition. Let S be a torus of G and C = CG (S). (i) If B is a Borel subgroup of G containing S, then C ∩ B is a Borel subgroup of C. (ii) Any Borel subgroup of C is of the form C ∩ B, where B is a Borel subgroup of G containing S. Similar results hold for maximal tori and maximal unipotent subgroups. Proof. If D is a Borel subgroup of C, then S ⊂ Z(C) ⊂ Z(D) (28.2.3). Since all the Borel subgroups of C are conjugate (28.1.3), it suﬃces to prove (i). The group C ∩ B = CB (S) is solvable and connected (26.5.6), so the result would follows from 28.1.4 if the variety C/(C ∩ B) is complete. Observe that SBu is a connected algebraic subgroup of G (21.2.3). Consider the action of C on G/B given by (α, βB) → (αβ)B. Then C ∩ B is the stabilizer of eB = eB in C. So the orbit map C/(C ∩ B) → CeB is a bijective C-equivariant morphism, hence, it is an isomorphism by 25.1.2. It suﬃces therefore to show that CeB is closed in G/B. Let π : G → G/B be the canonical surjection. Then X = CB = π −1 (CeB ) is B-stable (on the right), it follows that X is also B-stable. Since π is open (25.4.7), π sends B-stable

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closed subsets to closed subsets of G/B (25.3.2). So we are reduced to show that X is closed in G. If α ∈ X, then α−1 Sα ⊂ B, and this is still true for α ∈ X by continuity. Let us consider the morphism X × S → B/Bu , (α, β) → α−1 βαBu . It veriﬁes the hypotheses of 22.5.10. It follows that α−1 βα ∈ βBu for α ∈ X and β ∈ S. Thus α−1 Sα is a maximal torus of the group SBu . So 28.1.7 implies that there exists γ ∈ SBu such that α−1 Sα = γ −1 Sγ. Hence αγ −1 ∈ NG (S), and so α ∈ NG (S)B. Let Y = NG (S)B. We obtain therefore that CB = X ⊂ X ⊂ Y . Since Y = π −1 (NG (S)eB ) is the inverse image of a locally closed subset (21.4.3), it is locally closed, and hence it is a subvariety of G. Let us consider the action of C × B on Y given by (α, β) · γ = αγβ −1 . The orbits under this action are of the form CγB, γ ∈ NG (S). By 22.5.11 and 28.3.1, C = NG (S)◦ and so C is normal in NG (S) (21.1.6). It follows that for α ∈ NG (S), αCB = CαB, and for α, β ∈ NG (S), the morphism αCB → βCB, γ → βα−1 γ, is an isomorphism of irreducible subvarieties of G. Consequently, since C has ﬁnite index in NG (S) (21.1.6), it follows from the preceding discussion that Y has a ﬁnite number of C × B-orbits all of the same dimension. In particular, these orbits are closed (21.4.5). Since X is also the closure of X in Y , we have X = X (21.4.5). The proofs for the cases of maximal tori and maximal unipotent subgroups are analogue. 28.3.3 Corollary. Let T be a maximal torus of G. If B is a Borel subgroup of G containing T , then B contains CG (T ). Proof. The group CG (T ) is connected (28.3.1) and nilpotent (28.1.9). So the result follows from 28.3.2.

28.4 Properties of parabolic subgroups 28.4.1 Lemma. Let u : G → H be a surjective morphism of algebraic groups and B = T Bu a Borel subgroup of G where T is a maximal torus of G. We have u(B) = u(T )u(Bu ) and u(B) is a Borel subgroup of H. Moreover, u(T ) is a maximal torus of H and u(Bu ) = u(B)u . In particular, any Borel subgroup of H is the image of a Borel subgroup of G. Proof. The composition G → H → H/u(B) of morphisms induces a surjective morphism G/B → H/u(B). So H/u(B) is a complete variety (28.1.3 and 13.4.3). Thus u(B) is a parabolic subgroup of H (28.1.4). Since u(B) is solvable and connected, it is a Borel subgroup of H. Finally, it is clear that u(B) = u(T )u(Bu ), and we have u(Bu ) = u(B)u by 22.2.3. So u(T ) is a maximal torus of u(B), and hence of H.

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28.4.2 Theorem. Let G be a connected algebraic group. (i) If B is a Borel subgroup of G, then NG (B) = B. (ii) For any parabolic subgroup P of G, we have NG (P ) = P = P ◦ . Proof. (i) Let N = NG (B) and proceed by induction on dim G. The result is clear if dim G 2 by 28.1.10. Let α ∈ N and T a maximal torus of B. Since αT α−1 is a maximal torus of B, there exists β ∈ B such that αT α−1 = βT β −1 (28.1.7). So αβ −1 ∈ NG (T ) and by replacing α by αβ −1 , we may assume that α ∈ NG (T ). Consider the morphism u : T → T , γ → αγα−1 γ −1 . Since T is commutative and α normalizes T , we see that u is an endomorphism of the group T . We have two cases: 1) u is not surjective. Then S = (ker u)◦ is a non-trivial torus of T and α ∈ CG (S). Moreover, α normalizes CG (S) ∩ B which is a Borel subgroup of CG (S) (28.3.2). If CG (S) = G, then α ∈ B by the induction hypothesis. Otherwise if CG (S) = G, then S ⊂ Z(G) = Z(B). So by passing to the quotient G/S, we obtain the result by 28.4.1 and the induction hypothesis. 2) u is surjective. By 25.4.3, there exists a ﬁnite-dimensional rational Gmodule V and v ∈ V \ {0} such that N = {α ∈ G; α.v ∈ kv}. If β ∈ Bu , then β.v = v by 22.3.6. If β ∈ T , then the surjectivity of u implies that β is the commutator of two elements of N , so β.v = v. It follows that β.v = v for all β ∈ B. The map γ → γ.v induces therefore a morphism G/B → V . By 13.4.3, 13.4.4 and 13.4.5, this morphism is constant. So γ.v = v for all γ ∈ G. Hence G = N , and therefore G = N = B by 28.2.2. (ii) Let P be a parabolic subgroup of G and B a Borel subgroup of G contained in P . In particular, B ⊂ P ◦ . Let α ∈ NG (P ). Then αBα−1 is a Borel subgroup of P ◦ , so there exists β ∈ P ◦ such that αBα−1 = βBβ −1 . Now part (i) says that β −1 α ∈ NG (B) = B. So α ∈ βB ⊂ P ◦ . We have therefore proved that NG (P ) = P ◦ = P . 28.4.3 Corollary. Let B be a Borel subgroup of G. (i) We have NG (Bu ) = B. (ii) Any parabolic subgroup of G is conjugate to a unique parabolic subgroup containing B. (iii) If H is a solvable subgroup (not necessarily connected or closed) of G containing B, then H = B. Proof. (i) Let U = Bu and N = NG (U ). Since B normalizes U , it is a Borel subgroup of N ◦ . By 28.2.2, any unipotent element of N ◦ is conjugate in N ◦ to an element of U . As U is normal in N ◦ , it follows that U = Nu . We deduce therefore that the group N ◦ /U is a torus, and so, it is solvable. Thus N ◦ is solvable and B = N ◦ . Now N normalizes N ◦ , so N ⊂ NG (B) = B.

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(ii) Let P, Q be parabolic subgroups containing B verifying Q = αP α−1 with α ∈ G. Then B and αBα−1 are Borel subgroups of Q. So there exists β ∈ Q such that αBα−1 = βBβ −1 (28.1.3). Thus β −1 α ∈ NG (B) = B, and so α ∈ βB ⊂ Q. Hence P = Q. (iii) It suﬃces to show that H = B. But H is a parabolic subgroup. So H is connected 28.4.2, and since it is solvable, we have H = B (28.1.5). Remark. Condition (iii) of 28.4.3 does not say that a maximal solvable subgroup of G is a Borel subgroup of G.

28.5 Cartan subgroups 28.5.1 Deﬁnition. A subgroup of G is called a Cartan subgroup of G if it is equal to the centralizer of a maximal torus of G. 28.5.2 Cartan subgroups of G are connected and nilpotent (28.1.10 and 28.3.1). Two Cartan subgroups are conjugate (28.1.7) and their common dimension is called the rank of G, denoted by rk(G). By 28.2.2, if C is a Cartan subgroup of G, then the union of the conjugates of C contains a dense open subset of G. By 28.1.7 and 28.3.2, the map T → CG (T ) is a bijection from the set of maximal tori of G onto the set of Cartan subgroups of G. We have also, for a maximal torus T , CG (T ) = T × CG (T )u . 28.5.3 Lemma. (i) Let H be a connected nilpotent algebraic group and K a closed subgroup of H distinct from H. Then dim K < dim NH (K). (ii) If C is a Cartan subgroup of G, then it is a maximal connected nilpotent subgroup of G. Proof. (i) Since H = K, H = {e}. The last non-trivial term in the central descending series of H is connected (21.3.6) and central in H. So Z = Z(H)◦ = {e}. If Z ⊂ K, then the connected algebraic subgroup ZK contains strictly K and normalizes K. So dim NH (K) dim ZK > dim K. Suppose that Z ⊂ K. Let π : H → H/Z be the canonical surjection. We have: NH/Z (K/Z) = π(NH (K)) , dim NH (K) = dim Z + dim NH/Z (K/Z). So the result follows by induction on dim H. (ii) Suppose that H is a connected nilpotent subgroup of G containing strictly C. We may assume that H is closed. By (i), C = NH (C)◦ . Let α ∈ NH (C) and write C = CG (T ) where T is a maximal torus of G. Then αT α−1 is a maximal torus of C, and so αT α−1 = T since T is central in C. Thus NH (C)◦ ⊂ NH (T )◦ = CH (T ) = C (28.1.10 and 28.3.1). We have therefore obtained a contradiction.

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28.5.4 Let α ∈ G. By 28.2.2, αs belongs to a maximal torus of G. We deduce therefore that dim CG (αs ) rk(G). We say that α is generic in G if dim CG (αs ) = rk(G). Denote by Ggen the set of generic elements of G. Lemma. Let T be a maximal torus of G and α ∈ T . Then the following conditions are equivalent: (i) α ∈ Ggen . (ii) CG (α)◦ = CG (T ). (iii) χ(α) = 1 for all χ ∈ Φ(T, G) (in the notations of 26.2.1). Proof. The equivalence (i) ⇔ (ii) is clear since CG (T ) is connected (28.3.1), while (ii) ⇔ (iii) follows from 26.1.1 and 26.2.1. 28.5.5 Proposition. Let α be a semisimple element of G. The following conditions are equivalent: (i) α ∈ Ggen . (ii) CG (α)◦ is a Cartan subgroup of G. (iii) CG (α)◦ is a nilpotent subgroup of G. (iv) α belongs to a unique maximal torus of G. (v) α belongs to a ﬁnite number of maximal torus of G. Proof. The equivalence (i) ⇔ (ii) (resp. (ii) ⇔ (iii)) follows from 28.5.4 (resp. 26.4.4 and 28.5.3). Let T be a maximal torus of G containing α (28.2.2). Then T ⊂ CG (α)◦ . If CG (α)◦ is nilpotent, then T is its unique maximal torus (26.4.4). So (iii) ⇒ (iv) ⇒ (v). To ﬁnish the proof, we shall show that (v) ⇒ (ii). Let H = CG (α)◦ and T a maximal torus of G containing α. Since maximal tori of H are conjugate in H, the connected variety H/NH (T ) is ﬁnite, and hence it is reduced to a point. Thus T is normal in H. It follows from 22.5.10 that T is central in H. So H ⊂ CG (T ). But CG (T ) ⊂ CG (α)◦ = H, hence H = CG (T ). 28.5.6 Lemma. Let H be a connected algebraic group which contains a unipotent element α belonging to a unique Cartan subgroup C of H. Then H is nilpotent. Proof. Let T be a maximal torus such that CG (T ) = C, and B a Borel subgroup of H containing T . By 28.3.3, C ⊂ B. Let Bu = Nn ⊃ Nn−1 ⊃ · · · ⊃ N0 = {e} be the central descending series of Bu . We have N0 ⊂ C. Suppose that Ni ⊂ C. Since α ∈ Bu , we have α−1 βαβ −1 ∈ Ni if β ∈ Ni+1 , so βαβ −1 ∈ αNi ⊂ C, and α ∈ β −1 Cβ. Since β −1 Cβ is a Cartan subgroup of H containing α, our hypothesis on α implies that β −1 Cβ = C. We deduce therefore that Ni+1 ⊂ NH (C). Since Ni+1 is connected (21.3.6), we have Ni+1 ⊂ NH (C)◦ = C. We have therefore proved that Bu ⊂ C. Now T ⊂ C and B = T Bu , so we have B = C. Consequently B is nilpotent (28.1.10 and 28.3.1), and so H is nilpotent (28.1.9).

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28.5.7 Theorem. Let G be a connected algebraic group. (i) An element α ∈ G is generic if and only if it belongs to a unique Cartan subgroup of G. (ii) The set Ggen contains a dense open subset of G. Proof. (i) If α is generic, then C = CG (αs )◦ is the unique Cartan subgroup containing αs (28.5.5), and α ∈ C by 28.3.1. Now let D be a Cartan subgroup containing α. Then αs ∈ Ds and D = Ds × Du . Thus D ⊂ CG (Ds )◦ ⊂ CG (αs )◦ , and hence D = C. Conversely, suppose that α belongs to a unique Cartan subgroup C. Since αs ∈ Cs ⊂ Z(C), we deduce that H = CG (αs )◦ contains C. So any Cartan subgroup of H contains αs ∈ Z(H). Since α is contained in a unique Cartan subgroup, it follows that αu = αs−1 α is contained in a unique Cartan subgroup of H. Hence H is nilpotent by 28.5.6, and so H is a Cartan subgroup of G (28.5.5) which implies that αs , and hence α, is generic. (ii) Observe that by 28.5.4 (iii), the set Ggen is non-empty. Let C = CG (T ) = T × Cu be a Cartan subgroup of G and T0 the set of α ∈ T verifying χ(α) = 1 for all χ ∈ Φ(T, G) (notations of 26.2.1). The set C0 = T0 × Cu is dense and open in C and by 28.5.4, C0 = C ∩ Ggen . Since any generic element belong to a Cartan subgroup, we deduce that Ggen is the image of the morphism u : G × C0 → G , (α, γ) → αγα−1 . It follows from the inclusion C0 ⊂ u(G × C0 ) = Ggen that C = C0 ⊂ Ggen . As Ggen is stable by conjugation, it contains the union of the conjugates of C, which contains a dense open subset of G (28.5.2). 28.5.8 Let H be a nilpotent algebraic group. By 26.4.4, T = (H ◦ )s is the unique maximal torus of H ◦ and H ◦ = T × (H ◦ )u . So T ⊂ Z(H ◦ ) and T is normal in H. Lemma. We have T ⊂ Z(H). Proof. Let t = L(T ) and for α ∈ H, let uα and vα be the morphisms of T into itself deﬁned by: uα (γ) = αγα−1 , vα (γ) = αγα−1 γ −1 . Then uα (e) = vα (e) = e, so duα and dvα are endomorphisms of t. Since H/H ◦ is ﬁnite, uα and duα are automorphisms of ﬁnite order. So duα is a semisimple automorphism of t. Similarly, since H is nilpotent, dvα is nilpotent. But dvα = duα − idt , so uα = idT . Hence T ⊂ Z(H). 28.5.9 Theorem. Let G be a connected algebraic group and C a subgroup (not necessarily closed) of G. The following conditions are equivalent: (i) C is a Cartan subgroup of G. (ii) C is a maximal nilpotent subgroup of G, and any subgroup of C of ﬁnite index in C, has ﬁnite index in NG (C). (iii) C is a closed connected nilpotent subgroup of G and C = NG (C)◦ .

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Proof. (i) ⇒ (ii) Let C = CG (T ) be a Cartan subgroup of G where T is a maximal torus of G, and H a nilpotent subgroup containing C. We may assume that H is closed. Then T is a maximal torus of H ◦ . By 28.5.8, we have T ⊂ Z(H), and so H ⊂ CG (T ) = C. Now C is connected, so if K is a subgroup of C of ﬁnite index in C, then K = C. Thus NG (K) ⊂ NG (C). Since NG (C)◦ = C by 28.1.10, C has ﬁnite index in NG (C). The inclusions K ⊂ C ⊂ NG (C) and 10.2.3 show that K has ﬁnite index in NG (C). (i) ⇒ (iii) This follows from 28.1.10, 28.3.1 and 28.5.3. (iii) ⇒ (i) We have C = S × Cu where S = Cs (26.4.4). Let B be a Borel subgroup containing C, T a maximal torus of B containing S, and D the centralizer of S in B. The group D is connected (28.3.1) and B = T Bu , D = T Du . Since S ⊂ Z(D), the group SDu is connected, nilpotent and it contains C. Now C = NG (C)◦ , it follows that SDu = C (28.5.3). The group D is connected and solvable. So (D, D) ⊂ Du ⊂ C (26.4.2), and hence C is normal in D. It follows from the equality C = NG (C)◦ that D = C, so T ⊂ C, and S = T . Thus C = CB (T ) is a Cartan subgroup of B, and hence of G. (ii) ⇒ (i) If C veriﬁes (ii), then C is closed since C is nilpotent. Next, C ◦ has ﬁnite index in C and therefore C ◦ has ﬁnite index in NG (C). Hence C ◦ = NG (C)◦ . It follows from (iii) ⇒ (i) that C ◦ is a Cartan subgroup of G, and (i) ⇒ (ii) shows that C ◦ is maximal nilpotent. So C = C ◦ .

References • [5], [38], [40], [78].

29 Cartan subalgebras, Borel subalgebras and parabolic subalgebras

In this chapter, we apply the results in Chapter 28 on Cartan subgroups, Borel subgroups and parabolic subgroups to their Lie algebras. We generalize the deﬁnitions of Borel subalgebras and parabolic subalgebras to arbitrary Lie algebras, and we establish the relations between the group objects and Lie algebra objects. In particular, we prove analogues of results on conjugacy of such subalgebras.

29.1 Generalities 29.1.1 In this chapter, g is a ﬁnite-dimensional k-Lie algebra and G its algebraic adjoint group (24.8.1). By 24.5.7 and 24.5.12, ad[g, g] = [ad g, ad g] is an algebraic Lie subalgebra of gl(g). Let us denote by G1 the connected algebraic subgroup of G such that L(G1 ) = ad[g, g]. 29.1.2 As in 19.8.1, we let gλ (x) to be the generalized eigenspace of ad x relative to λ ∈ k, and denote by N(g) the set of x ∈ g verifying x ∈ gλ (y) for some y ∈ g and λ ∈ k \ {0}. We saw in 19.8.2 that if x ∈ N(g), then ad x is nilpotent. Let Auts g be the subgroup of Aute g generated by ead x , x ∈ N(g). If h is a Lie subalgebra of g, then N(h) ⊂ N(g). So if we denote by Auts (g, h) the group generated by eadg x , x ∈ N(h), then we may identify {γ|h ; γ ∈ Auts (g, h)} with Auts h. The group Auts g is a connected algebraic subgroup of Aute g (21.2.5 and 21.3.2). Lemma. Let f : g → h be a surjective morphism of Lie algebras and β ∈ Auts h. Then there exists α ∈ Auts g such that β ◦ f = f ◦ α. Proof. Let y ∈ h and x ∈ g be such that y = f (x). Then f (gλ (x)) = hλ (y), so f (N(g)) = N(h). To obtain the result, we may assume that β = ead y where y ∈ N(h). Then y = f (x) for some x ∈ N(g). It is then easy to see that α = ead x veriﬁes the conclusion of the lemma.

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29.1.3 Assume that g is semisimple. Let h be a Cartan subalgebra of g, R = R(g, h) the root system of g relative to h, and W = W (g, h) its Weyl group. Recall from 18.2.9 that W acts on h. Lemma. Let w ∈ W be considered as an automorphism of h. There exists θ ∈ Auts g such that θ|h = w. Proof. We may assume that w = sα , where α ∈ R. Let Xα ∈ gα , X−α ∈ g−α be such that [Xα , X−α ] = Hα . Let: θ = ead Xα ◦e− ad X−α ◦ead Xα . Then θ ∈ Auts g. To obtain the result, it suﬃces to prove that θ(h) = h − α(h)Hα for all h ∈ h. This is clear if α(h) = 0. On the other hand, we have: θ(Hα ) = ead Xα ◦e− ad X−α (Hα −2Xα ) = −Hα +2Xα −2Xα = −Hα = sα (Hα ). So we are done.

29.1.4 Proposition. Let H be an algebraic group, h its Lie algebra and (V, π) a ﬁnite-dimensional rational H-module. Deﬁne θ to be the morphism of varieties: θ : H × V → V , (β, w) → π(β)(w). (i) If (α, v) ∈ H × V and (x, w) ∈ Tα (H) × V , then: dθ(α,v) (x, w) = dπα (x)(v) + π(α)(w). (ii) For all (α, v) ∈ H × V , the linear maps dθ(α,v) and dθ(eH ,v) have the same rank. Proof. (i) Deﬁne the morphisms: p : H → H × V , β → (β, v) and q : V → H × V , w → (α, w). Then θ ◦ p(β) = π(β)(v) and θ ◦ q(w) = π(α)(w) and so: d(θ ◦ p)α (x) = dπα (x)(v) , d(θ ◦ q)v (w) = π(α)(w). On the other hand, we have dpα (x) = (x, 0) and dqv (w) = (0, w). Thus: dπα (x)(v) = d(θ◦p)α (x) = dθ(α,v) ◦dpα (x) = dθ(α,v) (x, 0), π(α)(w) = d(θ◦q)v (w) = dθ(α,v) ◦dqv (w) = dθ(α,v) (0, w). Since dθ(α,v) (x, w) = dθ(α,v) (x, 0) + dθ(α,v) (0, w), the result follows. (ii) Fix α ∈ H and consider the following isomorphisms of varieties: σ : H × V → H × V , (β, w) → (αβ, w) , ρ : V → V, w → π(α)(w). Then θ ◦ σ = ρ ◦ θ. If v ∈ V , then d(θ ◦ σ)(eH ,v) = d(ρ ◦ θ)(eH ,v) , that is, dθ(α,v) ◦ dσ(eH ,v) = π(α) ◦ dθ(eH ,v) . So the result follows.

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29.2 Cartan subalgebras 29.2.1 We shall use the notations of chapter 19. Lemma. Let G1 be as in 29.1.1, h a Lie subalgebra of g and x ∈ h be such that ad x induces an automorphism of g/h. The image of the morphism θ : G1 × h → g , (β, y) → β(y) contains a dense open subset of g. Proof. The hypothesis on x implies that g = h + [x, g]. In particular, we have g = h + [g, g] and so g = h + [x, g] = h + [x, h + [g, g]] = h + [x, [g, g]]. Now 29.1.4 says that dθ(e,x) (u, y) = u(x) + y for u ∈ L(G1 ) and y ∈ h. So the image t of dθ(e,x) is h + L(G1 )(x). But L(G1 ) = ad[g, g], and so t = g. It follows from 16.5.7 that θ is dominant, and so the result follows from 15.4.2. 29.2.2 Lemma. Any Cartan subalgebra h of g contains a generic element of g. Proof. For x ∈ h, let π(x) denote the endomorphism of g/h induced by ad x. Since h = ng (h), 0 is the only element v of g/h verifying π(x)(v) = 0 for all x ∈ h. As h is nilpotent, it follows from 19.4.6 that there exists x0 ∈ h such that π(x0 ) is bijective. By applying 29.2.1 and using the fact that the set of generic elements ggen of g contains a dense open subset of g (19.1.1), we deduce that G1 (h) ∩ ggen = ∅. Hence we obtain h ∩ ggen = ∅ because G1 (ggen ) = ggen . 29.2.3 Theorem. (i) A Lie subalgebra of g is a Cartan subalgebra of g if and only if h = g0 (x) for some x ∈ ggen . (ii) Let h and t be Cartan subalgebras of g. There exists α ∈ Auts g such that t = α(h). In particular, all the Cartan subalgebras of g have dimension the rank of g. Proof. (i) This is clear by 19.8.5 (iii) and 29.2.2. (ii) By (i), h = g0 (x) for some x ∈ ggen . Let λ1 , . . . , λn be the pairwise distinct non-zero eigenvalues of ad x. Then: g = g0 (x) ⊕ gλ1 (x) ⊕ · · · ⊕ gλn (x). If yi ∈ gλi , then yi ∈ N(g), and so ad yi belongs to the Lie algebra of Auts g. Consider the morphism θ : (Auts g) × h → g , (β, z) → β(z). If z ∈ h et yi ∈ gλi , then 29.1.4 implies that: dθ(e,x) (0, y) = y , dθ(e,x) (ad yi , 0) = [yi , x].

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Since ad x induces a bijective endomorphism of gλi , the image of dθ(e,x) is g. The same argument as in 29.2.1 shows that (Auts g)(h) contains a dense open subset of g. So the same is true for (Auts g)(k). We deduce therefore from 26.2.2 that h ∩ ggen ∩ (Auts g)(k) = ∅. So the result follows easily by 19.8.5 (iii). 29.2.4 Proposition. Let f : g → g be a surjective morphism of Lie algebras. (i) The image under f of a Cartan subalgebra h of g is a Cartan subalgebra of g . (ii) Let h be a Cartan subalgebra of g . Then any Cartan subalgebra of −1 f (h ) is a Cartan subalgebra of g. (iii) Any Cartan subalgebra of g is the image under f of a Cartan subalgebra of g. Proof. (i) Let a = ker f and π : g → g/a g be the canonical surjection. If π(x) ∈ ng/a (π(h)), then x ∈ ng (h + a) = h + a (19.8.5 (i) and 29.2.3 (i)). Thus ng/a (π(h)) = π(h), and ng (f (h)) = f (h). Since f (h) is nilpotent, it is a Cartan subalgebra of g . (ii) Let h be a Cartan subalgebra of t = f −1 (h ). By (i), f (h) is a Cartan subalgebra of h , so f (h) = h (19.8.4). If x ∈ ng (h), then f (x) ∈ ng (h ) = h . Thus x ∈ t, and therefore ng (h) = nt (h) = h. As h is nilpotent, it is a Cartan subalgebra of g. (iii) This is direct consequence of (i) and (ii). 29.2.5 Proposition. Let L be a connected algebraic group and l its Lie algebra. A Lie subalgebra of l is a Cartan subalgebra of l if and only if it is the Lie algebra of a Cartan subgroup of L. Proof. Let H be a Cartan subgroup of L and h its Lie algebra. Then H is a connected nilpotent closed subgroup verifying H = NL (H)◦ (28.5.9). So 24.4.8 and 24.5.13 imply that h is a Cartan subalgebra of l. Conversely, let h be a Cartan subalgebra of l. Since h = nl (h), h is an algebraic Lie algebra of l (24.7.2). Let H be a connected algebraic subgroup of L such that L(H) = h. Then H is nilpotent (24.5.13) and H = NL (H)◦ by 24.3.5 since L(NL (H)◦ ) = L(NL (H)) = nl (h) = h (24.4.8). So H is a Cartan subgroup of L by 28.5.9. 29.2.6 Proposition. Let l be the Lie algebra of a connected algebraic group L. (i) An element x ∈ l is semisimple if and only if x belongs to the Lie algebra of a torus of L. (ii) If x ∈ l is semisimple, then x belongs to a Cartan subalgebra of l. Proof. (i) If T is a torus of L, then all the elements of L(T ) are semisimple (26.4.5). Conversely, suppose that x ∈ l is semisimple. Let H = CL (x). Then

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cl (x) = L(H) (24.3.6), so x ∈ L(H). Let T be a maximal torus of H and C = CH (T )◦ . Then L(C) = ch (T ) (26.1.1), so x ∈ L(C). Now C is nilpotent (28.1.10) and so C = T × Cu (26.4.4). It follows therefore from 26.4.5 that x ∈ L(T ). (ii) Let T be a maximal torus of L such that x ∈ L(T ). Then C = CL (T ) is a Cartan subgroup of L containing T . Hence 29.2.5 implies that L(C) is a Cartan subalgebra of l containing x. 29.2.7 Proposition. Let m be the Lie algebra of a connected algebraic group M , a a Lie subalgebra of m and s = a(a). (i) We have rk(s) = rk(a) + dim(s/a). (ii) If k is a Cartan subalgebra of s, then a ∩ k is a Cartan subalgebra of a. (iii) If h is a Cartan subalgebra of a, then k = a(h) is a Cartan subalgebra of s such that h = k ∩ a. Proof. 1) Let k be a Cartan subalgebra of s and h = a ∩ k. Since k is algebraic (29.2.5), we have a(h) ⊂ k. Let x ∈ k ∩ sgen (29.2.2). We have s = k ⊕ l where l is a subspace of s which is (ad x)-stable, and ad x induces an automorphism of l. Thus l = [x, l] ⊂ [s, s], and 24.5.7 implies that s = k + [a, a]. We deduce therefore that a = h + [a, a]. Moreover, a(h) and [a, a] are algebraic, so a(h) + [a, a] is an algebraic Lie subalgebra of m containing a (24.5.10). It follows that s = a(h) + [a, a]. Consequently, if x ∈ k, then x = y + z where y ∈ a(h) and z ∈ [a, a]. Hence z = x − y ∈ k, and so z ∈ k ∩ a = h which in turn implies that x ∈ a(h). We have therefore proved that a(h) = k. 2) Denote by S the connected algebraic subgroup of M such that L(S) = s. Let N = {α ∈ S; Ad(α)(h) = h} and n = L(N ). Then n = ns (h) (24.3.6). But if α ∈ N and p is an algebraic Lie subalgebra of s containing h. Then Ad(α)(p) also contains h. It follows from the deﬁnition of a(h) = k that Ad(α)(k) = k. We deduce therefore that n ⊂ ns (k) = k (24.3.5). Hence n ∩ a ⊂ h, and so h = na (h). As h is nilpotent, it is a Cartan subalgebra of a. We have therefore proved (ii). 3) Now s = k + a, so dim(s/a) = dim(k/h). So (i) follows. 4) Finally, let h be a Cartan subalgebra of a, x ∈ h ∩ agen , and u the endomorphism of s/a induced by ads x. Since a is an ideal of s containing [s, s] = [a, a] (24.5.7), we have u = 0. Thus dim s0 (x) = dim a0 (x) + dim(s/a), and by (i), we deduce that x ∈ sgen . It follows that k = s0 (x) is a Cartan subalgebra of s (19.8.5). Since h = a0 (x), we have h = k ∩ a. So part (iii) follows from point 1.

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29.3 Applications to semisimple Lie algebras 29.3.1 Assume that g is semisimple, and denote by L its Killing form. Since L is non-degenerate, the Killing homomorphism x → fx deﬁned in 19.2.2 is an isomorphism. In the notations of 19.7.3, we have for x, y, z ∈ g: Φfx (y, z) = fx ([y, z]) = L(x, [y, z]). We deduce immediately that: g(fx ) = gx . Thus we say that x ∈ g is regular if fx ∈ g∗reg . Let us denote by greg the set of regular elements of g; it is a dense open subset of g. 29.3.2 Proposition. If g is semisimple, then its rank is equal to its index. Proof. Since ggen and greg are both dense open subsets of g, there exists x ∈ g which is regular and generic. So the result is clear since gx = g0 (x) (19.8.7 (v)). 29.3.3 Theorem. Let g be a semisimple Lie algebra, h1 and h2 two Cartan subalgebras of g. Then the root systems R1 = R(g, h1 ) and R2 = R(g, h2 ) are isomorphic. Proof. By 29.2.3, there exists θ ∈ Aute g such that θ(h1 ) = h2 . Let us deﬁne τ : h∗1