Induced Modules over Group Algebras (North-Holland Mathematics Studies)

INDUCED MODULES OVER GROUP ALGEBRAS

NORTH-HOLLAND MATHEMATICS STUDIES 161 (Continuation of the Notas de Matematica)

Editor: Leopoldo NACHBIN Centro Brasileiro de Pesquisas Fisicas Rio de Janeiro, Brazil and University of Rochester New York, U.S.A.

NORTH-HOLLAND -AMSTERDAM

NEW YORK

OXFORD TOKYO

INDUCED MODULES OVER GROUP ALGEBRAS

Gregory KARPILOVSKY Department of Mathematics California State University Chico, CA, U.S.A.

1990

NORTH-HOLLAND -AMSTERDAM

NEW YORK

OXFORD TOKYO

ELSEVIER SCIENCE PUBLISHERS B.V. Sara Burgerhartstraat 25 P.O. Box 211, 1000 AE Amsterdam, The Netherlands Distributors for the U.S.A. and Canada: ELSEVIER SCIENCE PUBLISHING COMPANY, INC. 655 Avenue of the Americas New York, N.Y. 10010, U.S.A.

Library of Congress Cataloging-in-Publication D a t a K a r p i l o v s k y . G r e g o r y , 1940Induced modules over group algebras Gregory Karpilovsky. p. cm. -- ( N o r t h - H o l l a n d m a t h e m a t i c s s t u d i e s , 161) Includes bibliographical references. ISBN 0-444-88414-9 ( U . S . ) 1 G r o u p a l g e b r a s . 2. M o d u l e s ( A l g e b r a ) I. T i t l e . 1 1 . S e r i e s . P A 1 7 1 . K 3 7 8 1990 512 .24--dC20 @9-49199 f

CIP

ISBN: 0 444 88414 9 ELSEVIER SCIENCE PUBLISHERS B.V., 1990 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior written permission of the publisher, Elsevier Science Publishers B.V./Physical Sciences and Engineering Division, P.0.Box 103, 1000 AC Amsterdam, The Netherlands. Special regulations for readers in the U.S.A. - This publication has been registered with the Copyright Clearance Center lnc. (CCC), Salem, Massachusetts. Information can be obtained from the CCC about conditions under which photocopies of parts of this publication may be made in the U.S.A. All other copyright questions, including photocopying outside of the U.S.A., should be referred to the publisher. No responsibility is assumed by the publisher for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions or ideas contained in the material herein. Printed in the Netherlands

To my wife Helen, who typed it all

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vii

In 1898 Frobenius discovered a construction which, in present terminology, associates with every module of a subgroup the induced module of a group. This construction proved t o be of fundamental importance and is one of the basic tools in the entire theory of group representations. The present monograph is designed for research mathematicia,ns and advanced graduate students and gives a picture of the general theory of induced modules as it exists at present. Much of the material presented has heretofore been available only in research articles. Our approach is not intended to be encyclopedic, but each topic is considered in sufficient depth that the reader may obtain a c1ea.r idea of the major results in the area. A word about notation. As is customary, Theorem 5.4.2 denotes the second result of Section 4 of Chapter 5; however, for simplicity, all references to this result within Chapter 5 itself are designated as Theorem 4.2. The following is a brief description of the content of the book. After establishing algebraic preliminaries (Chapter l),the general facts about induced modules are provided (Chapter 2). Following presentation of some formal properties of induced modules, we examine in detail their annihilators and record a number of applications. Among other results, we prove some classical theorems of Mackey and tie together relative projectivity and induced modules. In Chapter 3 we provide a detailed information on the process of induction from normal subgroups. Among other results, we discover conditions under which induced modules are totally indecomposa.ble or absolutely indecomposable. Special attention is drawn to the study of crossed products over prime rings. The information obtained is then applied to establish a criterion for induced modules to be homogeneous. The chapter ends with the discussion of the Loewy length of induced modules.

viii

Preface

Let H be a subgroup of G, let F be a field and let V be an F H module. It is natural to investigate circumstances under which the induced module V G has projective indecomposable summands. An important contribution to this problem was marked by the appearance of the paper of Robinson (1989) to which Chapter 4 is devoted. Among other results, we prove a useful reciprocity theorem and demonstrate that if V is irreducible, then projective summands of V G (if there are any) have relatively large dimension. The aim of Chapter 5 is twofold: first to present some basic results of Green theory with refinements and extensions achieved in scope of recent developments, and second to provide a number of applications. The chapter ends with a result of Alperin (1986) which relates the Green correspondence with normal subgroups. In Chapter 6 we investigate in detail circumsta.nces under which restriction and induction of irreducible modules are completely reducible modules. As one of the applications, we show that if H is a subgroup of G and F is an arbitrary field of characteristic p > 0, then the following conditions are equivalent: (i) V Gis completely reducible for any irreducible FH-module V and MH is completely reducible for any irreducible FG-module M . (ii) There exists a normal suhgroup N of G such that N H and N has p'-index in G. Chapter 7 provides a detailed account of a distinguished class of induced modules, namely permutation modules. These modules hold much information about the p-modular representations of G, as well as information about the fusion of p-subgroups of G. The main results presented are due to Alperin (1988), Brouk and Robinson (1986), Robinson (1988), Dress (1975), Saksonov (1971) and Brouk (1985). Chapter 8, the final chapter, is based exclusively on an important work of Weiss (1988). A number of applications to the isomorphism problem for group rings is also presented. I would like to express my gratitude to my wonderful wife for the tremendous help and constant encouragement which she has given me in the preparation of this book (as well as in the previous 11 books).

ix

Contents Preface Chapter 1. 1. 2. 3. 4.

5. 6.

Preliminaries Notation and terminology Tensor products Artinian, noetherian and completely reducible modules The radical of modules and rings The Wedderburn- Artin theorem Group algebras and group representations

Chapter 2. 1. 2. 3. 4. 5. 6. 7. 8.

General properties of induced modules . Induced modules, representations and characters Formal properties of induced modules Annihilators of induced modules Clifford’s theorem Dual and contragredient modules Induction, restriction, and outer tensor products Mackey theorems and their applications Counting induced modules and characters 9. Relative trace maps 10. Induction and relative projectivity 11. Unique decompositions 12. Projective covers

Chapter 3. 1. 2. 3.

4. 5. 6. 7. 8. 9.

Induction from normal subgroups Complete noetherian local rings Reduction to G-invariant modules Group-graded algebras and crossed products The endomorphism ring of induced modules Relations between the decomposition of V G and E n d ~ cV( c ) Twisted group algebras over fields Total and absolute indecomposability of induced modules Crossed products over prime rings Homogeneity of induced modules

1 1 4 8 19 29 37

45 45 54 66 74 79 87 89 97 103 107 120 123 131 132 139 141 153 159 167 180 191 208

Contents

X

Frobenius and symmetric algebras Symmetric crossed products EndFc( V') is symmetric Graded modules Induction from irreducible modules and their projective covers Inflated modules over twisted group algebras Induction of absolutely irreducible modules Applications The Loewy length of induced modules

210 220 230 232

Chapter 4. 1. 2. 3.

Projective summands of induced modules The Reynolds ideal Projective summands Applications

273 273 280 295

Chapter 5. 1. 2. 3.

Green theory Vertices and sources The Green correspondence The endomorphism ring of the Green correspondents The Green correspondence and Brauer's induction theorem The Green correspondence and the Brauer lift The Green correspondence and normal subgroups

299 299 309

Chapter 6. 1. 2. 3. 4. 5.

Simple induction and restriction pairs Blocks of algebras Defect groups of blocks Blocks and vertices Simple induction and restriction pairs Complete reducibility of induced modules

34 1 34 1 350 358 365 373

Chapter 7. 1. 2. 3. 4. 5. 6. 7.

Permutation modules Preliminary results Hecke algebras Fusion and permutation modules Complete reducibility of ( 1 ~ ) ~ Induction from Sylow p-subgroups Loewy series for transitive permutation modules Characterizations of p-perrnutaion modules

383 383 392 403 409 416 426 433

10. 11. 12. 13. 14. 15. 16. 17. 18.

4. 5. 6.

238 242 256 26 1 266

318 329 334 337

xi

Contents 8. 9. 10.

Chapter 8. 1. 2. 3. 4. Bibliography Notation Index

The Brauer morphism Scott modules The Brauer morphism and p-permutation modules

438 448 45 1

Permutation lattices Generalized permutation lattices Permutation lattices and normal subgroups Some bimodule isomorphisms Applications

459 459 473 484 489

499 511 516

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1

Chapter 1 Preliminaries Our aim here is to review some basic ring-theoretic results and to fix conventions and notations for the rest of the book. Since we presuppose a familiarity with various elementary ring-theoretic terms, only a brief description of them is presented. Many readers may wish to glance briefly at the contents of this chapter, referring back to the relevant sections when they are needed later.

1.

Notation and terminology

All rings in this book are associative with 1 # 0 and subrings of a ring R are assumed to have the same identity element as R. Each ring homomorphism will be assumed to preserve identity elements. Let R be a ring. An element x in R is called nilpotent if xn = 0 for some integer n 2 1. An ideal J in R is nil if every element of J is nilpotent, while J is nilpotent if there is an integer n 2 1 such that J" = 0, where J" is the product of J with itself n times. An element e in R is an idempotent if e2 = e. Two idempotents u,v of R are orthogonal, if uv = vu = 0. A nonzero idempotent is primitive if it cannot be written as a sum of two nonzero orthogonal idempotents. We denote by Z ( R )the centre of R. An idempotent e is called centrally primitive if e is a primitive idempotent of Z ( R ) . The ring of all n x n matrices over R is denoted M,(R). Let R be a ring. From this we can construct a new ring R", called the opposite ring of R. Both the underlying set and the additive structure

Preliminaries

2

of R" are just those of R. But the multiplication, denoted by o is given bY z ~ =y y z for all z,y E R It is easy to verify that R" is a ring with these operations. Clearly, Z ( R ) = Z(R")and R is commutative if and only if R = R". Assume that A is a ring, R a commutative ring, and p : R -+ Z ( A ) a ring homomorphism. The resulting system ( A ,R, p ) is called an Ralgebra. It will be convenient to suppress the p and speak of A as an R-algebra or as an algebra over R. Thus A is an R-algebra (with respect to some p ) if and only if there is an ideal I of R with R / I isomorphic to a subring of Z ( R ) . Setting r a = p ( r ) a , r E R, a E A, it follows that A is an R-algebra such that r(zy) = z(ry) = ( w ) y

for all

r E R,x,y E A

Conversely, if A is an R-module for which the above equalities hold, then the map R + Z ( A ) , r H r 1 is a ring homomorphism and hence A is an R-algebra. By a homomorphism of R-algebras, we understand a ring homomorphism which is also a homomorphism of R-modules. Let (&), i E I , be a family of rings and let R be the direct product set &rRj. We can define addition and multiplication on R by the rules

-

(G) t (Yi) = ( X i

t Y;)

(Z;>(Yi)

= (Wi)

(z;,y; E R;) . An immediate verification shows that R is a ring ; we shall refer to R as the direct product of the f a m i l y (Ri),i E I . We now record some conventions concerning modules. By an Rmodule, we always understand a unital R-module; the symbols RM or M R will be used to underline the fact that M is a left or right Rmodule, respectively. We usually consider left R-modules, and, in this case, speak simply about R-modules. A nonzero R-module V is indecomposable if 0 and V are the only direct summands of V . A nonzero R-module V is irreducible if 0 and V are the only submodules of V . Let V and W be R-modules. We denote by H o r n ~ ( VW , ) the additive abelian group of all R-homomorphisms V + W . The same notation will be used in case V and W are right R-modules. In case V = W ,

1. Notation and terminology

3

we write E n d ~ ( vinstead ) of Horn~(V, V ) . Note that E n d ~ ( vis) a

ring under the multiplication

Let V be an R-module. A submodule W of V is called maximal if W is a proper submodule (i.e.W # V ) and W is not contained in any proper submodule of V . An R-module V is said to have an R-basis (vi), i E I, if there exist elements o; E V such that each o E V can be written as a finite sum = Crp; with uniquely determined coefficients r; E R. Such modules V are called free R-modules. Given a pair of rings R, S, we say that V is an ( R ,S ) - b i m o d u f e if V is a left R-module and a right S-module, with the actions of R and S on V commuting : ( r v ) s = ?-(‘us)

(7.

E R, v E

v,s E S )

Unless explicitly stated otherwise, all groups are assumed to be multiplicative. The multiplicative group of a field F is denoted by F*. If S is a subset of G, < S > will denote the subgroup of G generated by S, while cG(s) and NG(s) is the centralizer and normalizer of S in G, respectively. As usual, Z(G) denotes the centre of G. Let H be a subgroup of a group G. A subset of G containing just one element from each left coset zH is called a left transversal for H in

G, and right transversals are defined correspondingly. Let g be an element of a finite group G and let p be a prime. We say that g is a p e l e m e n t if the order of g is a power of p ; we say that g is a p’-element (or g is p-regular) if its order is prime to p. Each g E G can be written in a unique way g = gpgpt, where g p is a pelement, gpt is a p’-element, and gP and gpt commute. The elements g, and gpt are called the p and p’-parts of g , respectively. A conjugacy class C of G is called p r e g u f a r (respectively, p-singular) if the order of the elements of C is not divisible by p (respectively, divisible by p ) . Let S be an arbitrary set. An action of G on S is a homomorphism

Preliminaries

4

8 of G into the permutation group of S. When the action is understood, we shall suppress the symbol 8 and instead write s H g s for O(g),g E G, s E S. The stabilizer G(s) of s E S is defined by G(s) = { g E G, 's = s} and the orbit of Gs of s E S is defined by G~ = {'slg

E G}

We say that G acts transitively if Gs = S for some (and hence all) s E S. One readily verifies that G(s) is a subgroup of G and that IGsl = (G : G(s))

Throughout the book, S, and A, denote symmetric and alternating groups of degree n, respectively, while Z n is a cyclic group of order n. 2.

Tensor products

Let R be a ring. Given a right R-module V and a left R-module W, the 23-module V @ R W , called the tensor product of V and W , is defined as follows. Let F be a free 23-module with V x W as a basis. Each element of F can be uniquely written in the form

c

Zij(Vi, Wj)

(Zjj

E

z,

vj

E

v,wj E W )

with finitely many zij distinct from 0. Let T be the subgroup of F generated by all elements of the form-

(21,

w1+ w2)- (v,w1) - (21, w2)

(v,r w ) - ( ' U f , 20) with v,v; E V and w,wi E W . Then V @R W is defined as the factor group FIT. The image of (v,w) under the natural homomorphism F + F / T is denoted by v 8 w. Observe that the Z-module V @ R W consists of all finite sums C vi @ w;with vi E V,wi E W . Assume that

2. Tensor products

5

V is an ( S ,R)-bimodule. Then V

@R

W can be regarded as a (left)

S-module by putting s(v@w) = s v @ w

( W E

v , w E w,s E S )

In particular, if V and W are modules over a commutative ring R, then V @ R W is an R-module. Let R be a ring, V,V' two right R-modules, W,W' two left Rmodules and 4 : V + V',$ : W -+ W' two R-linear maps.Then the map

$@$:V@RW+VI@RW' defined by

(4 0$)(W

€3

= +)

@ $(w)

is a 23-linear map. Moreover, if V,V' are ( S ,R)-bimodules and 4 is also S-linear, then $ 8 $ is S-linear. We next record a number of standard properties of tensor products (see Bourbaki (1974), (1972)).

2.1. Proposition. Let V,V',V" be right R-modules, let W,W',W'' be left R-modules and let

be exact sequences. (i) The sequences

w 3' V ~ wR'5 vN@R w o v@Rw'3v @ R w 'Yv 8 R w"+ o

v'@R

4

of 23 -homomorphisms are exact. (ii) The homomorphism g 0t ; V € 3 W ~ -+ V" @ R W" is surjective and its kernel is equal to I m ( f @ 1 ) + I m ( 1 0s ) . In particular, if V' and W' are submodules of V and W , respectively, and f and s are inclusion maps, then the map

(v/v')@R (w/w') ( v@ R w ) / ( I m ( f@ 1 ) + I m ( 1 €3 s)) (v

+ V ' )@ (wt W')

(W

€3 w)t (h(f 6 1)

+ I m p @ s))

Preliminaries

6

is a Z-isomorphism. 2.2. Proposition. Let ( K ) , i E I , be a family of right R-modules and (Wj),j E J a family of left R-modules. Then the map

is a Z-isomorphism. 2.3. Proposition. Let V be an ( S ,R)-bimodule. Then

as S-modules. 2.4. Proposition. If V and W are too free modules over a commutative ring R and (v;),(wj) are R-bases of V and W , respectively, then (v;8 wj) is an R-basis of V @ R W .

2.5. Proposition. Let R be a subring of a ring S and let V be a free left R-module with el,. . . , en as an R-basis. Then S @R V is a free left S-module with 1 8 el,. .. ,1 @ en as an S-basis. 2.6. Proposition. Let V be a right R-module, W a left R-module, V' a submodule of V and W' a submodule of W . If V' is a direct summand of V and W' is a direct summand of W , then the canonical

homomorphism

v' @ R w' v @ R w 4

is injective and the image of V' @R N .

@R

W' is a direct summand of the

Z-module M

Let R be a ring, V a right R-module and W a left R-module. The module V is said to be flat i f for every injective homomorphism f : W' t W of left R-modules, the homomorphism

7

2. Tensor products

is injective. By a projective module, we understand a direct summand of a free module.

2.7. Proposition.Every projective right R-module is pat. Let V be a right R-module and let A be a left ideal of R. The homomorphism V @R A -i V which sends v 8 a to va is called canonical: its image is the subgroup V A consisting of all finite sums C v ; u ; with v; E V,a; E A. The induced homomorphism V 8~A -iV A is called canonical. The canonical map V @ R R + V is an isomorphism. Note that if V is an ( S ,R)-bimodule, then all canonical maps are homomorphisms of S-modules. 2.8. Proposition. Let V be a right R-module. Then the following properties are equivalent: (i) V is flat. (ii) For any finitely generated submodule W' of a left R-module W , the canonical homomorphism l v @ i : V @ R W' + V @ R W (i is the inclusion map) is injective. d (iii) For every exact sequence of left R-modules W' + W + W" the sequence

*

is exact. (iv) For every finitely generated left ideal A of R, the canonical map V @ R A + V A is an isomorphism. Let V be a flat right R-module and let W' be a submodule of a left R-module W . Then the canonical injection V @ R W' + V @ , RW allows us to identify V @ R W' with its image in V @ R W . In the future, we shall always use this identification. 2.9. Proposition. Let V be a flut right R-module. (i) I f f : X + Y is a homomorphism of left R-modules, then

Preliminaries

8

and

Im(1v 8 f ) = v

OR

(Imf)

(ii) If W',W" are two submodules of a left R-module W , then

3.

Artinian, noetherian a n d completely reducible modules

Throughout this section, R denotes an arbitrary ring. Let V be an R-module. We say that V is artinian (respectively, noetherian) if every descending (respectively, ascending) chain of submodules of V terminates. The ring R itself is called artinian (respectively, noetherian) if the left regular R-module R R is artinian (respectively, noetherian). We call V finitely generated if V = Cy=lRvi for some wl, . . . $0, in V . The module V is said to be finitely cogenerated if for every family{ li E I}of submodules of V with niErK= 0, there exists a finite subset J of I such that

3.1. Proposition.For any R-module V , the following conditions are equivalent: (i) V is artinian. (ii) Every nonempty set of submodules of V has a minimal element. (iii) Every factor-module of V is finitely cogenerated. Proof.(i)=+ (ii): Assume that there is a nonempty set S of submodules of V such that S does not have a minimal element. Then, for any given W E S, the set {W' E SlW' c W } is nonempty. Owing to the axiom of choice, there is a function W L-+ W' with W 3 W' for any W in S. By choosing any W in S, we therefore obtain an infinite descending

chain

w 3 W ' 3 W " > ....

of submodules of V. (ii)+ (iii): It suffices to show that if W is a submodule of V and

3. Artinian, noetherian and completely reducible modules

{Eli E I} is

nj&

a family of submodules of V with n;& = W for some finite subset J of I . Let us put

9

= W , then

S = {nkEKvklICI is finite } By assumption, S has a minimal element, say njE& with J C_ I and J finite. It is obvious that W = njEJv,as required. (iii)+ (i): Consider a descending chain & _> & 2 . .. 2 V , 2 . . . of submodules of V . If W = nn>lVn, then by hypothesis V/W is finitely cogenerated. Thus W = V, for some n 2 1. But then V,+j = V, for all i 2 1, as desired. W 3.2. Corollary,Let V irreducible submodule.

#0

be an artinian module. Then V has an

Proof.Apply Proposition 3.1. for the nonempty set of all nonzero submodules of V . W Turning to noetherian modules, we next record

3.3. Proposition. For any R-module V , the following conditions are equivalent: (i) V is noetherian. (ii) Every nonempty set of submodules of V has a maximal element. (iii) Every submodule of V is finitely generated. Proof.The proof of this result is dual to that of Proposition 3.1. and therefore will be omitted. Proposition. If V is a finitely generated R-module, then every proper submodule of V is contained in a maximal submodule ( in particular, if in addition V # 0 , then V has a maximal submodule ). 3.4.

Proof. By hypothesis, V = Ci=lRv; for some v1,. . . ,v, in V . Denote by W a proper submodule of V and by S the set of all proper submodules of V which contain W . Then W E S and if (Wj) is a chain in S , then U;W; is a submodule. If V = U;W;, then vj E W,, for

Preliminaries

10

some nj, 1 5 j 5 r. Let W, be the largest of the modules W,, , . . . ,W,,. Then W contains vl,,. .,u, and hence V = W,, a contradiction. Thus S is inductive and therefore, by Zorn's lemma, it has a maximal element. This gives us a maximal submodule of V containing W , as desired. W

Let 0 --f U --f V -, W --f 0 be an exact sequence ofR-modules. Then V is artinian (noetherian) i f and only i f both U and W are artinian (noetherian). 3.5. Proposition.

Proof. Suppose that V is artinian. Then, because U and W are isomorphic to a submodule and factor module of V , respectively, it is obvious from the definition that both U and W are artinian. Conversely, assume that U and W are artinian. To show that V is artinian, we may assume that U 2 V and W = V/U. Consider a descending chain

v, 2 vz 2 ... 2 v, 2 ...

of submodules of V . Since V/U is artinian, there is an integer rn such that (i = 1,2, . . .) = vm+; u

+

v, + u

Because U is artinian, there is an integer n 2 rn such that

Taking into account modularity and that V, 2 V,+i, we have for each i = 1,2, ...

+

V, = V , n (v, U ) = V , n (Vn+l =

+U )

K+i + (V, n U )= Vn+i + (K+;n U )

= Vn+i

Hence V is artinian. The proof of the noetherian case is dual. W 3.6. Corollary. Let V = @y=lq.Then V is artinian (noetherian) if and only if each is artinian (noetherian).

P r o o f . The case n = 2 follows from Proposition 3.5. The general case is a straightforward induction on n. '

3. Artinian, noetherian and completely reducible modules

11

3.7. Corollary. Let R be a ring. Then R is artinian (noetherian) if and only if every finitely genemted R-module is artinian (noetherian). Proof. Assume that R is artinian. If V is a finitely generated R-module, then V is a factor module of a free R-module W of finite rank. By Corollary 3.6, W is artinian, hence so is V , by Proposition 3.5. Conversely, if every finitely generated R-module is artinian, then so is R, since R is generated by 1. The proof of the noetherian case is similar . 3.8. Corollary. Let A be an algebra over a commutative ring R such that A is a finitely generated R-module. If R is artinian (noetherian), then A is artinian (noetherian).

Proof. This is a direct consequence of Corollary 3.7. Let V be a nonzero module. A finite chain of n -t 1 submodules of

V

v = vo 3 v, 3 . . . 3 v, = 0

is called a composition series of length n for V provided that each K-I/V, is irreducible, 1 5 i 5 n , i.e. provided each term in the chain is maximal in its predecessor. The irreducible modules

are called the composition factors of the series. 3.9. Proposition. A nonzero module V has a composition series if and only if V is both artinian and noetherian.

Proof. Suppose that V is both artinian and noetherian. Because any factor module of V is artinian, it follows from Corollary 3.2 that there is an ascending chain

0

c v1 c vz c .

of submodules of V where each term in the chain is maximal in its

successor. But V is noetherian, hence the chain terminates at V after

Preliminaries

12

a finitely many terms. Hence

V has a composition series .

Conversely, assume that V has a composition series. We argue by induction on n, where n is the minimum length of all such series. If n = 1, then V is irreducible and there is nothing to prove. Otherwise, if = vo 3 6 3 . . . 3 =0

v

v,

is a composition series of minimal length for V ,then V/V1is irreducible and V1 has a composition series of length n - 1. Now apply Proposition

3.5. Suppose that

v = vo 3 6 3 ... 3 v, = 0 and

v = wo 3 w,3 ... 3 w, = 0 are two composition series for V. We say that these two series are equivalent in case n = rn and there is a permutation r~ of { 1,2,. . . ,n} such that vi/K+l Wo(i)/WC+)+l ( 2 = 1,. . ., n )

V V

3.10. Proposition. (The Jordan-Holder theorem). If a module has a composition series, then every pair of composition series for are equivalent.

Proof. We argue by induction on c ( V ) , where c ( V ) is the minimum length of composition series for V. The case c ( V ) = 1 being obvious, we may assume that c(V)= n > 1 and that any module with a composition series of smaller length has all of its composition series equivalent. Choose a composition series

(1)

v = v o 3 VI 3 . . . 3 v, = 0

of minimal length and fix any other composition series for

(2)

v = wo 3 w13 ... 3 w, = 0

V:

3. Artinian, noetherian and completely reducible modules

13

If & = W1, then by the induction hypothesis and in view of c ( K ) 5 n - 1, (1) and (2) are equivalent. We may thus assume that V1 # W1, in which case V = V, -t Wl since V, is a maximal submodule of V. It follows that

V, n W1 is maximal in both Wl 3.9, V1 n IV1 has a composition series whence

and V1. Owing to Proposition

and hence

are composition series for V, and W l . Since c(Vl) < n, every two composition series are equivalent. Thus the two series

and

v = VO 2 v1 2 xo 3 . .. 3 XI, = 0

are equivalent. It follows that k < n - 1 and c(W1) < n , proving that every two composition series for Wl are equivalent. Hence the two series = 3 3 ... 3 W, = 0

v wo w,

and

v = w o > w l 2x03 ... > x k = o

are equivalent. Because, by (3), V/Vl Z Wl/XO and V/W1 &/XO, the series (1) and (2) are equivalent, thus completing the proof. Assume that V is both artinian and noetherian. Then, by Propositions 3.9 and 3.10, for such a module V we can define its composition len,gth c(V) unambiguously by c(V) = 0 if V = 0, and c(V) = n if V has a composition series of length n. We next examine direct decompositions of modules.

Preliminaries

14

Let R be any ring and let V # 0 be an Rmodule which is either artinian or noetherian. Then V is a finite direct sum of indecomposable submodules. 3.11. Proposition.

For any nonzero R-module M that is not a finite direct Proof. sum of indecomposable submodules, choose a proper decomposition M = M' @ N' where M' is not a finite direct sum of indecomposable submodules. Assume by way of contradiction that V cannot be written as a finite direct sum of indecomposable submodules. Then

v = V' @ w',V' = v"@ W",.. . is a sequence of proper decompositions. Hence there exist infinite chains

W' c W' @ W" c . . . and V 3 V' 3 V'' 3 . . . proving that V is neither artinian nor noetherian. w

3.12. Proposition. Let R R = & @ . . . @ Vn with K # 0 for i = 1,2,. . . ,n, and write 1 = el - . en with e; E K. Then { e l , . . . e,} is a set of pairwise orthogonal idempotents in R and V , = R e ; , 1 5 i 5 n. Conversely, if { e l , . . . , e n } is a set of pairwise orthogonal idempotents

+ +

in R, then

n

R(Ce;) = @y=lRei i=l

Proof. Given r E R , we have r = r . 1 = 1 . r = re1 Hence, in particular,

+ . - + re,.

which shows that e;ej = &jei for all i , j . Moreover, Rei V , and RR = Re1 @ - . @ Re,, which implies that = R e ; for all i. Conversely, suppose that { e l , . . . ,e n } is a set of pairwise orthogonal idempotents and put e = CZ1e;. Then e2 = e and ee; = e;e = e; for R e ; . If E r i e ; = 0 , then multiplication on the all i. Hence R e = C:=t=, right by ej gives r j e j = 0 for all j. Thus R e = @y=lRe;, as asserted.

3. Artinian, noetherian and completely reducible modules

15

3.13. Corollary. Assume that R is artinian or noetherian. Then there exists a set { e l , . . . ,en} of pairwise orthogonal primitive idempotents of R with el - - . en = 1 . Moreover

+ +

where each Re; is an indecomposable R-module. Proof.

Apply Propositions 3.11 and 3.12.

We shall refer to the R-modules R e l , . . ., Re, in Corollary 3.13 as the principal indecomposable R-modules. An R-module V is said to be completely reducible if every submodule of V is a direct summand of V .

3.14. Proposition. Let V be a completely reducible R-module. (i) Every submodule ofV is isomorphic to a homomorphic image of V and every homomorphic image ofV is isomorphic to a submodule of

V. (ii) Every submodule and every homomorphic image ofV is comple t ely reducible, (iii) If V # 0 , then V contains an irreducible submodule. Proof. (i) If W is a submodule of V , then V = W @ W’ for some submodule W’ of V . Then W E V/W’ and V/W W‘, as required. (ii) Let W be a submodule of V and let U = V / W . By (i), it suffices to show that U is completely reducible. Assume that U1 = & / W is a submodule of U . We may choose V2 with V = Vl @ V2. Then

U = Ul@

(vz + W ) / W

and therefore U is completely reducible. (iii) Fix a nonzero 2, in V , By Zorn’s lemma, there is a submodule W of V maximal with respect to the property that TI 4 W . Write V = W @ W’ for some submodule PV’ of V . If W’ is not irreducible, then W‘ = W1@W2 for some nonzero submodules Wl, W2of V . Because

(W @ Wl) n (W @I W 2 )= W

Preliminaries

16

it follows that w $ W @ Wi for i = 1 or i = 2. But this contradicts the maximality of W . Thus W’ is irreducible, as desired.

We are now ready to provide two important characterizations of completely reducible modules.

3.15. Proposition. For any nonzero R-module V , the following conditions are eqivalent: (i) V is completely reducible. (ii) V is a direct sum of irreducible submodules. (iii) V is a sum of irreducible submodules.

Proof. (i)Ij(ii) : Consider the collection of sets of irreducible submodules of V whose sum is direct. By Proposition 3.14 (iii), it is nonempty and, by Zorn’s lemma, there is a maximal element, say {K}, in this collection. Let W = and let V = W @ W’. If W’ # 0, then by Proposition 3.14 (ii), (iii), W‘ contains an irreducible submodule V‘. Hence W V’ = V‘ @ (@V,),contrary to the maximality of {K}. Accordingly, W’ = 0 and V = W , as required, (ii)+ (iii) : Obvious. (iii)+ (i) : Let W be a submodule of V . Owing to Zorn’s lemma, there is a submodule W’ of V maximal with the property that WnW’ = 0. Hence we must have W W’ = W @ W’. We are thus left to show that V = W e W’. Assume by way of contradiction that W $ W’ # V and let v E V be such that w $ W @ W‘. By hypothesis, w = v1 . . . w, where v; E V , and V , is an irreducible submodule of V , 1 5 i 5 n. Accordingly, v j 6 W @ W’ for some j and thus

+

+

+ +

V, n (W $ W’) # V, However,V, is irreducible, so

V, n (W @ W‘) = 0 and hence

W 03 W’+ v, =

w @ W‘$ v,

which shows that W n (W’ @ V,) = 0. Since this contradicts the maximality of W’, the result follows. m

3. Artinian, noetherian and completely reducible modules

17

3.16. Corollary. Let V # 0 be a completely reducible R-module. Then the following conditions are equivalent: (i) V is artinian. (ii) V is noetherian. (iii) V is a finite direct sum of irreducibze submodules. Proof. By Corollary 3.6, if V = @LIK then V is artinian (noetherian), if and only if each V , is artinian (noetherian). Now apply Proposition 3.15 (ii). 3.17. Corollary. If R R is completely reducible, then every Rmodule is completely reducible. Proof. Let V be an R-module and let v E V . Then the map R R --$ Rv, r H r v is a surjective homomorphism of R-modules. Therefore, by Proposition 3.14 (ii), Rv is completely reducible. Because V = CvEVRv,the result follows by applying Proposition 3.15 (iii).

3.18 Lemma. Let V = @K where each Vi is irreducible and let W be an irreducible submodule of V . Then W @I$ where j ranges over those i for which W E V,. Proof. Given a nonzero w E W , we may write w = Cv;with v; E L$ and with finitely many v; # 0. If v; # 0, then the map r w I-+ rvi,r E R, is a nonzero homomorphism from W = Rw to Vl = Rv;.Since W,K are irreducible, W V , as desired.

Let V be a completely reducible R-module. We say that V is homogeneous if it can be written as a sum of isomorphic irreducible submodules. The sum of all irreducible submodules of V which are isomorphic to a given irreducible submodule of V is called a homogeneous component of V .

Let V = @iEzV,, where each V , is irreducible 3.19. Proposition. j E J , are all representatives of the and let J C I be such that the isomorphism classes of V,, i E I . For each j E J , denote b y Wj the sum

K,

Preliminaries

18

of all F with r/: E 6 . Then (i) The Wj, j E J , are all homogeneous components of V . (ii) V = @ j E ~ W j .

Proof. (i) Apply Lemma 3.18. (ii) This follows from the definition of Wj. Let V be any R-module. A submodule W of V is said to be fully invariant in V if it is mapped into itself by all R-endomorphisms of V .

3.20. Proposition. Let V # 0 be a completely reducible Rmodule. A submodule W of V is fully invariant if and only i f W is a sum of certain homogeneous components of V .

Proof. Keeping the notation of Proposition 3.19, fix f E EndR(V). If V , 2 V, then either f ( X ) = 0 or f(V,) Z V , %' vj, since V;: is irreducible. Thus f(Wj) C Wj, so each Wj (and hence the sum of certain Wj) is fully invariant. Conversely, assume that W is fully invariant. It suffices to show that if P is an irreducible submodule of W and Q is an irreducible submodule of V such that Q Z P , then Q C W. If Q = P , this is clear; otherwise P n Q = 0, so P -t Q = P @ Q and thus V = P @ Q @ U for a submodule U of V . Let f : P -+ Q be the given isomorphism and define an endomorphism g of V by

Then W _> g(W) 2 g ( P ) = f ( P )= Q, as asserted. 1 We close by recording the following useful observation. 3.21. Corollary. Let V # 0 be a completely reducible R-module and write V as the direct sum of its homogeneous components: V = @ i ~ W jThen . En&(V) % EndR(Wj)

n

jEJ

4. The radical of modules and rings

19

Proof.

It is clear that any family (fj) with f j E E n d ~ ( W j ) determines a unique endomorphism f of V . The map

is obviously an injective ring homomorphism. Because each Wj is fully invariant (Proposition 3.20), any endomorphism f of V maps each Wj into itself. Setting f, = flWj,j E J , it follows that ( f j ) H f, as desired. 4.

The radical of modules and rings

Throughtout, R denotes an arbitrary ring. Given an R-module V , the radical J ( V ) of V is defined to be the intersection of all maximal submodules of V . If V contains no maximal submodules, then by definition J ( V ) = V . Observe that if V # 0 is finitely generated, then J ( V ) # V by virtue of Proposition 3.4. The Jacobson radical J ( R )of R is defined by

J ( R )= J(RR) Hence, by definition, J ( R ) is the intersection of all maximal left ideals of R. We say that R is semisimple if J ( R ) = 0. 4.1 Proposition. For any nonzero R-module V , the following conditions are equivalent: (i) V is a finite direct sum of irreducible modules. (ii) V is artinian and com.pletely reducible. (iii) V is artinian and J ( V ) = 0.

Proof.

(i)+ (ii): This follows from Corollary 3.6 and Proposition

3.15. (ii)+ (iii): We may clearly assume that V # 0, in which case V is a direct sum of irreducible submodules, by Proposition 5.15. Therefore, by the definition of J ( V ) ,J ( V ) = 0 (iii)+ (i): Owing to Proposition 3.1, V is finitely cogenerated. Since J ( V ) = 0 it follows that f~r=,,K= 0 for some maximal submodules V,,. . . ,V, of V . Thus V is isomorphic to a submodule of (V/K).

ny=,

Preliminaries

20

Hence, by Proposition 3.14 (ii), V is completely reducible. Now apply Corollary 3.16 (iii). 4.2 Corollary.

For any ring R, the following conditions are equiv-

alent:

(i) RR is a finite direct sum of irreducible modules. (ii) RR is completely reducible. (iii) R is semisimple artinian. (iv) Every R-module is completely reducible. Proof. (i)+ (ii): Follows from Proposition 3.15. (ii)+ (iii): Because 1 lies in the sum of finitely many irreducible submodules, it follows that R R is a finite direct sum of irreducible modules. Now apply Proposition 4.1 for V = R R. (iii)+ (iv): By Proposition 4.1 (applied for V = R R),RR is completely reducible. Now apply Corollary 3.17. (iv)+ (i): By assumption, RR is completely reducible. Thus R is semisimple artinian, by the implication (ii)+ (iii). The desired conclusion follows from Proposition 4.1 applied for V = R R. We now proceed to examine J ( V ) in detail.

4.3. Lemma. (i) Let f : W + V be a homomorphism of Rmodules.Then f ( J ( W ) )& J ( V ) with equality i f f is surjective and Ke rf C J ( W ) . (ii) If W is a submodule of an R-module, then J ( W ) C J ( V ) and

J ( V / W )2 ( J ( V )t W ) / W . (iii) If W is a submodule of an R-module V with W 2 J ( V ) , then J ( V / W )= J ( V ) / W . (iv) IfV is an R-module, then J ( V / J ( V ) )= 0 and J ( V ) W for any submodule W ofV for which J ( V / W )= 0 (v) If W is a submodule of an R-module V , then W = J ( V ) if and only ifW J ( V ) and J ( V / W )= 0 . Proof. (i) If V has no maximal submodules, then J ( V )= V and therefore f ( J ( W ) )C J ( V ) .Now assume that M is a maximal submodule of V . Then the map f * : W --+ VIM given by f*(w) = f ( w ) M

+

4. The radical of modules and rings

21

is an R-homomorphism. Since K e r f * is a maximal submodule of W,J ( W ) C K e r f * and so f ( J ( W ) )C M . Hence f ( J ( W ) ) J ( V ) . Suppose that f is surjective and K e r f C J ( W ) . If W has no maximal submodules, then so does V , in which case

f ( J ( W ) )= f ( W )=

v = J(V)

We may thus assume that the set {MiliE I } of all maximal submodules of W is nonempty. By assumption, K e r f Mi for all i E I . If X and Y are the lattices of submodules of W containing K e r f and submodules of V , respectively, then the map U H f ( U ) , U E X , is an isomorphism of X onto Y. Therefore { f ( M i ) l i E I } is the set of all maximal submodules of V and

as asserted.

(ii) and (iii). This follows from (i) applied to the cases where f is the inclusion map and f : V --f V/W is the natural homomorphism. (iv) This follows from (iii) and (ii). (4 Apply (iv>* 4.4.Corollary.

Let R be an arbitrary ring and let I be an ideal of

R. (i) I = J ( R ) if and only if I J ( R ) and J ( R / I ) = 0 (ii) If1 C J ( R ) , then J ( R / I ) = J ( R ) / I . Proof. This is a direct consequence of Lemma 4.3 (v), (iii) and the obvious fact that the radical of the R-module R / I is the same as the Jacobson radical of the ring R / I .

A submodule W of an R-module V is called superfluous if for every submodule X of V , W + X = V implies X = V . 4.5. Proposition. Let V be an R-module. Then J ( V ) is the su'm of all superfluous submodules in V . Futhermore, if V is finitely generated, then J ( V ) is the unique largest superfluous submodule of V .

Preliminaries

22

Proof. Suppose that W is a superfluous submodule of V . To prove that W C J ( V ) ,we may assume that V has a maximal submodule M . If W M , then M + W = V and so M = V , a contradiction. Hence every superfluous submodule of V is contained in J ( V ) . On the other hand, let z E V . If W is a submodule of V with Rz + W = V , then either W = V or there is a maximal submodule M of V such that W & M and z $! M . If z E J ( V ) ,then the latter cannot occur. Thus z E J ( V )implies that Rz is a superfluous submodule of V , proving the first assertion. Suppose that V is finitely generated. By the foregoing, it suffices to show that J ( V )is a superfluous submodule of V . To this end, assume that V = W J ( V ) for a submodule W of V . Owing to lemma 4.3 (ii), J ( V / W )= V / W . Since V/W is finitely generated it follows from Proposition 3.4 that V/W = 0. Hence V = W as required.

+

4.6 Proposition. Let V be an R-module generated b y v l , . . . ,v, and let v E V . Then v E J ( V ) if and only if for all r; E R, 1 5 i 5 n, the elements v; r;v, 1 5 i 5 n, generate V .

+

Proof. Let v E J ( V )and let W be the submodule of V generated by the elements vi + riv, 1 5 i 5 n. Then W + J ( V ) = V so that W = V , by Proposition 4.5. Conversely, suppose that v 6 J ( V ) , Then there is a maximal subW , so for each module W of V such that v $! W . Hence V = Rv i there exists r; E R such that vi E -r;w W , whence v; r;v E W . Thus the elements v; + r;v, 1 5 i 5 n , do not generate V .

+

+

+

Let V be an R-module. The annihilator of V , written ann(V),is defined by

ann(V)= { T

E

RlrV = 0)

It is clear that ann(V) is an ideal and that V may be viewed as an R/ann(V)-module. We say that V is faithful if ann(V)= 0. An ideal I of R is said to be primitive if the ring R / I has a faithful irreducible module. Obviously, 1 is primitive if and only if I is the annihilator of an irreducible R-module.

4. The radical of modules and rings

23

Lemma. (i) An R-module V is irreducible if and only if V E R / X for some maximal left ideal X of R. (ii) Every maximal left ideal of R contains a primitive ideal and every primitive ideal is the intersection of the maximal left ideals containing it. (iii) J ( R ) is the intersection of the annihilators of irreducible Rmodules. In particular, J ( R ) is an ideal of R. 4.7.

Proof. (i) It is obvious that R / X is irreducible. Conversely, suppose that V is irreducible and let 0 # u € V . Then Ru is a nonzero submodule of V and so V = Rv. The map R -+ V,r H ru is therefore a surjective R-homomorphism. Thus R / X E V for some left ideal X of R. Since V is irreducible, X is maximal. (ii) Let X be a maximal left ideal of R. By (i), R / X is an irreducible R-module. Hence the annihilator of R / X is a primitive ideal contained in X . Let I be a primitive ideal in R and let V be an irreducible R-module whose annihilator is I . Given a nonzero IE: in V , put V, = { r E Rlrz = 0). Then V, is a left ideal in R. Since V is irreducible , V = Ra: E R/V, and so V, is a maximal left ideal of R. Because I is the intersection of all V, with 0 # II: E V , (ii) is proved. (iii) Apply (i) and (ii). Let V be an R-module. (i) J(R)V C J ( V ) with equality i f R / J ( R )is artinian. (ii) Assume that R/ J ( R)is artinian. Then V is completely reducible if and only if J(R)V = 0. 4.8. Proposition.

(i) If V has no maximal submodules, then J ( V ) = V and so J(R)V C J ( V ) . Let M be a maximal submodule of V . Then VIM is an irreducible R-module. Because J ( R ) annihilates VIM (Lemma 4.7 (iii)), it follows that J(R)V & M . Thus J(R)V & J ( V ) . Suppose that R / J ( R ) is artinian. Owing to Lemma 4.3 (v), it suffices to verify that J ( V / J ( R ) V )= 0. Because J ( R / J ( R ) )= 0 (Corollary 4.4 (i) and Lemma 4.7 (iii)) and R / J ( R )is artinian, every R / J (R)-module is completely reducible (Corollary 4.2). In particular, the R/J(R)-module V / J ( R ) Vis completely reducible. Hence V / J ( R ) V is a completely reducible R-module and therefore J ( V / J ( R ) V )= 0. Proof.

Preliminaries

24

(ii) If V is completely reducible, then J(R)V = 0 by Lemma 4.7 (iii). Conversely, assume that J(R)V = 0. Then V can be regarded as a module over the semisimple artinian ring R/J(R). Hence, by Corollary 4.2 V is a completely reducible R / J ( R)-module. Thus V is a completely reducible R-module. 4.9. Proposition.

(Nakayama’s lemma.) Let W be a submodule If W J(R)V = V , then W = V .

of a finitely generated R-module V .

+

Proof. Owing to Proposition 4.8 (i), J(R)V C_ J ( V ) and hence W + J ( V ) = V . But, by Proposition 4.5, J ( V ) is a superfluous submodule of V , hence the result. W The rest of this section will be devoted to ring-theoretic properties of J ( R ) . An element x of R is called a left (respectively, right) unit if there exists y E R such that yx = 1 (respectively, xy = 1). By a unit, we understand an element x in R which is both a left and a right unit. Hence x E R is a unit if and only if there exists y E R (denoted by x-l) such that xy = yx = 1 4.10. Proposition. Let x be an element of a ring R. Then x E J ( R ) if and only if for all r E R, 1 - rx is a left unit. In particular

J(R ) contains no nonzero idempotents. Proof. We first note that 1 - T X is a left unit if and only if R( 1 rx) = R. Therefore the first assertion follows from Proposition 4.6 applied to V = R R. Assume that e is an idempotent in J ( R ) . Then, by the above, x(l - e) = 1 for some x E R. Thus e = 1 e = x(1 - e)e = 0, as we wished to show. W 4.11. Proposition. For any ring R, J ( R ) is the unique largest ideal I of R such that 1 - rx is a unit of R for all r E R, x E I .

Proof. Owing to Proposition 4.10, it suffices to show that 1 - x is a unit for all 2 E J ( R ) . Since, by Proposition 4.10, 1 - x is a left unit, we have y ( 1 - x) = 1 for some y E R. Therefore z = 1 - y = -ya:

4. The radical of modules and rings

25

is in J ( R ) and so 1 = y'( 1 - z ) = y'y for some y' E R. Thus y is a unit and 1 - x = y-* is also a unit.

4.12 Corollary. For any ring R, J(R")= J ( R ) and, in particular, J ( R ) is the intersection of all maximal right ideals of R. Proof.

Apply Proposition 4.11.

4.13. Corollary. Let I be a left or right nil ideal of R. Then

I

C J(R).

Proof. Suppose that I is a left nil ideal of R and let x E I . Then, for any r E R,rx E I and so (rx)" = 0 for some n 2 1. Invoking the identity

(1 - y ) ( l

+ y + . . + y"-') *

= (1

+ y + - + yn-')(l * *

- Y) = 1 - y n

we see that 1 - r x is a unit. Thus by Proposition 4.10, I C J ( R ) . If I is a right nil ideal, then the same argument applied to R" yields the

result. W 4.14. Proposition.

For a n y ring R, the following conditions

are equivqlent: (i) R is artinian. (ii) R is noetherian, R / J ( R ) is artinian and J ( R ) is nilpotent. Proof. (i)+ (ii): Put J = J R and consider the chain J 2 J 2 2 . . .. By hypothesis, J k = J"' = . - . for some k 2 1. Setting I = J k we have I 2 = I . Assume by way of contradiction that I # 0. Then there exists a left ideal X in R with I X # 0, for example X = R. Let M be a minimal element in the set of all such X . Then I ( J M ) = ( I J ) M = I M # 0 and, becuase J M C M , we have J M = M . By Proposition 4.9, we are thus left to show that A4 is finitely generated. By assumption, there exists xlinM with I x # 0 and hence I ( R x ) # 0. Since R x C_ M , w e have M = R x , proving that J ( R ) is nilpotent. Since R is artinian, so is R / J ( R )by Proposition 3.5. To prove that R is noetherian, we argue by induction on the nilpotency index n of

Preliminaries

26

J ( R ) . If n = 1, then there is nothing to prove. Consider the exact sequence 0 + J"-1 + R + R/J"-' + 0 Since J ( R / J " - ' ) = J/Jn-' (Corollary 4.4 (ii)), it follows from the induction hypothesis that R/J"-' is noetherian. Hence, by Proposition 3.5, we need only show that Jn-' is noetherian. But J"-' is artinian (as a submodule of an artinian module RR) and J"-' is completely reducible, by Proposition 4.8 (ii). Hence, by Corollary 3.16, J"-l is noetherian, as required. (ii)+(i): We argue by induction on the nilpotency index 7~ of J(R). If n = 1 , there is nothing to prove. By induction hypothesis, R/J(R)"-l is artinian. On the other hand, J(R)"-' is noetherian (as a submodule of a noetherian module RR). Since R / J ( R )is artinian and J(R)" = 0, it follows from Corollary 3.16 and Proposition 4.8 (ii) that J(R)"-' is artinian. Thus, by Proposition 3.5, R is artinian. W 4.15. Corollary. Let R be an artinian ring. Then every finitely generated R-module is both artinian and noetherian. Proof. Apply Proposition 4.14 and Corollary 3.7. W

We next record the following useful observation. 4.16. Proposition,

Let (Ri),i E I , be a family of rings. Then

Proof. An element (r;) E n i E I R is ; a left unit if and only if ri, is a left unit of R; for all i E I . Now apply Proposition 4.10. H

In what follows, we use the standard fact that if el and e2 are idempotents of a ring S , then Sel Sez as left S-modules if and only if elS 2 e2S as right S-modules (see Jacobson (1956)).

s

4.17. Proposition. Let I be a nil ideal of a ring S and, for each E S , let S be the image of s in S = S / I . Then

4. The radical of modules and rings

27

(i) Each idempotent c E S can be lifted to an idempotent e E S , that

e = E. (ii) If e and f are idempotents of S , then e S E fS as right S modules .if and only if & Z $9 as right S-modules, (equivalently, S e E S f as left S-modules aj and only if SZE Sf as left S-modules). (iii) If I = el + €2 + . . . + en is a decomposition of I into orthogonal idempotents in S, then there exist orthogonal idempotents e l , e2,. . . ,en E is

S such that 1 =el+e2+-**+en, Moreover,

~j

e; = E;

is primitive if and only if so is ei.

Proof. (i) Fix an idempotent 6 E S and choose u E S with ii = E . Then u - u2 E I and therefore (u- u2)rn= 0 for some m 2 1. We have

Observe that on the right each term after the first m is divisible by (1 - u ) ~ while , the first rn terms are divisible by urn. Hence if e denotes the sum of the first m terms, then 1 = e + (1 - e)rng,where g is a polynomial in u. Now u ( 1 - u) E I, so

that is = ii = E . Since e ( l - e ) = e ( l - u)"g = 0 , e is a required i dempo tent. (ii) Let e , f be idempotents in S. We first show that e S E f S as right S-modules if and only if there exists a,b E R such that a = e a f , b = fbe,ab=e,

and

ba=

f

(1)

First suppose that 0 : e S ---f fS is an isomorphism that maps e to b and a to f. Then f b = b and O ( e ) e = O ( e ) ,hence be = b, so f b e = b, and similarly eaf = a. Hence e = ( @ - l @ ) ( e )= 0 - ' ( b ) = 0 - ' ( f b ) = O - ' ( f ) b = ab, and similarly f = ba. Conversely, if a and b satisfy (l), then x H bx is a homomorphism from e S to fS with inverse y H ay. Now assume that e S E f S and let a , b E S satisfy (1). Then 2 , b

Preliminaries

28

satisfy the same conditions for the isomorphism ES E fS, so &' E Now assume that d E fS and a, b E S are such that a

G

e a f , b 3 fbe,ab

e , ba

f

fs.

(modl)

and put a1 = e a f , bl = f b e . Then albl = e - z , where z E e l e . Because z is nilpotent, 1 - z has an inverse 1 - z' and so z z' = zz' = z'z. Setting 2" = ez'e, we have z zN = zz" = z"z and it follows that a l b l ( e - z") = e. Now put a2 = a l , b2 = bl(e - z"), then

+

+

a2 3

a , b2

b ( m o d l ) , a2b2 = e

Next write b2a2 = f -y, then y E flf.Because (ha2)' = bzeaz = bza2, we have

f-y=(f

-Y)2=f2-fy-yf+y2=

f -2y+y2

Hence y2 = y and because y is nilpotent, we find that y = 0 and thus bzaz = f. Since a2bz = e and b2a2 = f imply a2 = eu2 f,b2 = f bze, the required assertion follows. (iii) Let e , f be idempotents in S such that ef = f E = 0. We first claim that there exists an idempotent g such that g = f and eg = ge = 0 . To verify the claim, we note that f e is nilpotent and so 1 - f e is a unit. Put

h = (1 - f e ) - ' f ( 1 - f e ) Then h2 = h, he = 0 and

h = f. Now put

g = (1 - e ) h . Then

proving the claim. We now demonstrate that if e l , . . . ,en are idempotents such that Eiej = 0 for all i # j , then there exist idempotents e: such that = ~i and e:es = 0 for all i # j. For n = 1 there is nothing to prove, so we may assume that n > 1 and use induction on n. In other words, we may assume that e;ej = 0 for i # j, i , j > 1. Put e = e2 + e n , then e is an idempotent such that Fli? = EFl = 0. Applying the claim, there exists an idempotent el, such that e: = Cl and eel, = e i e = 0. It follows

4

+

5. The Wedderburn-Artin theorem

29

that e:, e 2 , . . . ,e, are pairwise orthogonal idempotents. By (i) and the above, there exist pairwise orthogonal idempotents e l , e2,. . . ,en such that E; = c i , 1 5 i 5 n. Thus for u = el - - - e, we have U = i, so 1 - u is a nilpotent idempotent, and therefore u = 1. To prove the final assertion, suppose that e E S is a primitive idempotent. Because e # 0, we have e # 0. If 6 = ~t: y for some orthogonal idempotents 2,y E 3, then there exist orthogonal idempotents el, e2 E S with C1 = 1t:,t?2 = y and hence with E = el e2. By (ii), (el e 2 ) S , which is possible only in the case el = 0 or e2 = 0. eS Thus either 5 = 0 or y = 0 and so iii is primitive. Conversely, assume that e is primitive. Then E # 0 and so e # 0. If e = u v, where u , v are orthogonal idempotents, then ?I = U 5 , ii5 = V i i = 0 and so either U = 0 or V = 0. Thus either u = 0 or v = 0, proving that e is primitive. This completes the proof.

+ +

+

+

+

5.

+

+

The Wedderburn-Artin theorem

In this section we provide the ring-theoretic structure of semisimple artinian rings. Throughout, R denotes an arbitrary ring and, for any interger n 2 l , M , ( R ) is the ring of all n x n matrices over R. Let i,j E (1,. . . ,n } and let e;j be the matrix with (i,j)-th entry 1 and 0 elsewhere. The elements eij E M,(R) are called matriz units. As is customary, we shall identify R with its image in M,(R) consisting of all scalar matrices diag(r, T , . . . ,T ) , r E R. As a preliminary to our first result, let us observe that the matrix units satisfy the following properties : ( i ) e;jeks = 0 if j # k and e i j e k , = ei, if j = k. (ii) 1 = ell . . - enn. (iii) The centralizer of { e ; j } in M,(R) is R. (iv) R Z ellM,(R)ell. It turns out that (i)) (ii), and (iii) determine M,(R) up to isomorphism. Namely, we have the following result.

+ +

5.1. Proposition. Let S be any ring containing elements v,j, 1 5 i ,j 5 n , satisfying (i) and (ii) and let R be the centralizer of {vij} in

Preliminaries

30

S . Then the map (aij) H C i , j aijvij

is an isomorphism of rings and R-modules. Furthermore, R Z v l l S q l .

Proof. Note that M,(R) is a free R-module freely generated by the matrix units eij. Hence the map ejj H v;j extends to a homomorphism of R-modules, which clearly coincides with $. Applying (i), (ii), and (iii), we immediately derive that 1c, is also a ring homomorphism. Suppose that Ci,ja;jv;j = 0 and fix k,s E { 1,.. . ,n}. Then, for all t E { I , . ..,n}, 0 = v t k ( x aijvij)vst = aksvtt i,j

and so 0=

c

aksvtt

= a k s ( z vtt) = ak3

t

t

Thus (ajj) = 0 and therefore $ is injective. To show that 1c, is also surjective, fix s E S and for each i,j put

Then, for all

v,t,

we have

and aijvrt

=

VkiSVjkvrt

= VTjSVjt

k

Hence a;j commutes with all vTt and therefore the foregoing, +((Ujj))

=

c i,j

ajjv;j =

c

aij

E R. Moreover, by

v1;svjj

i,j

= (cv;j)s(xvjj) = s

5. The Wedderburn-Artin theorem

31

proving that 1c, is surjective. Finally, taking into account that

the result follows. For any integer n 2 1 and any (left) R-module V , we write V" for the n-th direct power of V . 5.2. Proposition.

Let V be an R-module and let n

Proof. Fix z,j E { 1,. . . ,n } and define

f&, where

vj

f;j

2 1. Then

E E n d ~ ( v "by )

. . . ,W") = (0,. . . ,v j , o , . . . , O )

is in the i-th place. Then clearly n

fijfks

= ajkfis

and

cfi; = i=l

1

+

If E EndR(V") and $ ( v l , . . . ,on) = ($l ( V l ) , * * * ,+n(vn)), then $ obviously centralizes all f i j if and only if t,bl = $2 = . . . = +". Thus the centralizer of {fij) in EndR(V")is identifiable with EndR(V). The desired conclusion is therefore a consequence of Proposition 5.1. Let V be an R-module and let n 2 1. Then the R-module V" can also be viewed as an M,(R)-module in the following natural way. We visualize the elements of V" as column vectors and, for each A E M,(R) and x E V " , define Az as the matrix multiplication. We now are ready to record the following result. 5.3. Proposition. (i) The map W H W" is an isomorphism from the lattice of submodules of V onto the lattice of M,( R)-submodules of V " . (ii) EndM,,(R)( V " ) EndR(V ).

Preliminaries

32

(iii)The map V H V" induces a bijective correspondence between the isomorphism classes of R ans M,(R)-modules. The inverse of this correspondence is given b y W H ell W . f W" obviously preserves orProof. (i) The correspondence W +

der. Thus we need only show that it has an inverse which is also order-preserving . Consider the projection ?rl : V" + V on the first factor and, for any submodule X of V", put g ( X ) = ?rl(X). Then the correspondence X H g ( X ) is order-preserving and clearly ( g f ) ( W = ) W . From the action of M,(R) on V", we see that X can be written in the form X = V;.. for some submodule 6 of V. Thus ( f g ) ( X )= f(6)= V;2 = X , proving (i). (ii) Put S = M,(R) and choose f E Ends(V"). Then, by the nature of action of S on V", f has the same projections, say X j , on all factors. Conversely, any given 1c, E EndR(V) determines an element of Ends(V") whose projections on all factors are equal to +. Thus the map f H X j provides a desired isomorphism. (iii) It suffices to show that ellVn 2 V and (ellW)" W , where V and W are R and M,(R)-modules, respectively. By the definition of

ell 7

ellVn = V x 0 x . . . x 0 Z V

Observe also that

Becuase eklw = 0 if and only if ellw = 0, 1 5 Ic

5 n, w E W , the map

is well defined and is at least an R-isomorphism of W onto (ellW)". Consequently, we need only show that

Since the latter is a consequence of the action of M,(R) on (ellW)", the result is verified.

5. The Wedderburn-Artin theorem

33

For any ideal I of R, put

M,(I) = {(a,j)E M,(R)la;j E I

for all i , j E {1,..., n } }

5.4. Proposition. (i) The map I H M,(I) is a bijection between the sets of ideals of R and M,(R). In particular, R is simple if and only i f so is M,(R). (ii) Mn(R)/Ma(I) Mn(R/I). (iii) If R = I1 @ . . . @ I, is a two-sided decomposition of R, then

M,(R) = Mn(Ii)@ . . . @ Mn(Is) is a two-sided decomposition of M,(R). Furthermore, the ideal I; is indecomposable if and only if so is M,(I;).

Proof. (i) It is clear that M,(I) is an ideal of M,(R) and that the given map is injective. Suppose that J is an ideal of R and let I C_ R consist of all entries of elements in J. Then I is an ideal of R such that

J = M,(I). (ii) The natural homomorphism R + R / I induces a surjective homomorphism M,(R) + M,(R/I) whose kernel is M n ( I ) . (iii) This follows from the facts that Z(M,(R)) = Z(R) and that, for any r E R, M,(R)r = M,(Rr). H 5.5. Proposition.

For any ring R and any positive interger n,

J(Mn(R)) = M"(J(R)) Proof. Let V be an R-module and let V" be an M,(R)-module as in Proposition 5.3. Then ann(V") = M,(ann(V)),by the definition of V". Now apply Proposition 5.3 and Lemma 4.7 (iii). H

Let R be a finite direct product of f i l l matrix 5.6. Corollary. rings over division rings. Then R is semisimple artinian. Proof. That R is semisimple follows from Propositions 4.16 and 5.5. To prove that R is artinian, we may assume that R = M,(D)for some n 2 1 and some division ring D. Now apply Proposition 5.3.

Preliminaries

34

Our next aim is to establish the converse of the above result. As a preliminary, we record the following classical observation.

.

(Schur's lemma). Let V be an irreducible module. 5.7. Lemma Then E n d R ( V ) is a division ring. Proof. Let f : V + V be a nonzero R-homomorphism. Since f(V) is a nonzero submodule of V ,f ( V ) = V . Because f # 0, K e r f # V and so K e r f = 0. Thus f is an isomorphism, as required. Let V be an R-module and let e be an idempotent of R. Then HomR(Re, V ) E eV as additive groups. Similarly, i f V is a right R-module, then H o m R ( e R , V ) V e . 5.8. Lemma.

Proof. If f E H o m R ( R e , V ) , then e f ( e ) = f ( e 2 ) = f ( e ) E eV. Therefore the map f H f ( e ) is a homomorphism from HornR(Re, V ) to e V . Conversely, if v E e V , then the map g , : z e H zv is an Rhomomorphism from R e to V . Because the map H gv is an inverse of f H f ( e ) , the first isomorphism follows. The second isomorphism is proved by a similar argument. W 5.9. Lemma. Let e be an idempotent of a ring R. Then

EndR(Re) E (eRe)" and E n d R ( e R ) E eRe

as rings

In particular,

End&R)

E R"

and E n d R ( R R )

R

Proof. Applying Lemma 5.8 for V = R e , it follows that the map f H f ( e ) is an isomorphism of the additive group of E n d R ( R e ) onto the additive group of eRe. Given f , g E E n d R ( R e ) , write f ( e ) = erle and g ( e ) = er2e for some r1, r2 E R. Then ( f g ) ( e )= f(er2e) = erzerle = (erze)(erle)= g ( e ) f ( e ) proving that f H f ( e ) reverses multiplication. Since e is the identity element of the ring eRe, the above map preserves identity elements.

5. The Wedderburn-Artin theorem

35

This establishes the first isomorphism. The second isomorphism follows from Lemma 5.8 applied for V = eR. W We have now come to the demonstration of the following classical result. 5.10 Theorem.

(Wedderburn-Artin). Let R be a semisimple ar-

tinian ring. (i) There exist only finitely many, say Vi, . . . , V , of nonisomorphic irreducible R-modules and V , 2 Re; for some primitive idempotent e; o f R , 15 i 5 r . (ii) RR =" nl& @ . . .@ n,V, for some positive integers n;, 1 5 i 5 r , where n;T/: is a direct sum ofn; copies of K. (iii) R 2 niz1Mn,(D;),where D;= EndR(x)" S e;Re; is a division ring, 1 5 i 5 r . (iv) The integers n; and r are unique and each D; is determined up to isomorphism.

Proof. (i) and (ii): Owing to Corollary 4.2, RR is a finite direct sum of irreducible modules. Therefore, by Proposition 3.12, there exist idempotents e l , . . . ,en of R such that R R = Rel$. . .$Re, and each Re; is irreducible (in particular, e; is a primitive idempotent of R). We may assume that Rel, .. . , Re,, 1 5 r 5 n , are all nonisomorphic modules of the set { R e l , .. . ,Re,}. If V is any irreducible R-module, then by Lemma 4.7 (i) and Proposition 3.10, V 2 Re; for some i E {1,2,. . . ,r } . This establishes (i) and (ii). (iii) Put Di = EndR(V,)",1 5 i 5 r . Then we have

R"

EndR(RR)

(by Lemma 5.9)

r

!2

nEnd~(n;K)

(by Corollary 3.21)

i=l

n r

Mni(D;o)

(by Proposition 5.2)

i=l

and thus r

Preliminaries

36

(e;Re;)" we have D; Z e;Rei, Because, by Lemma 5.9, EndR(K) proving (iii). (iv) Assume that R z Mk,(Dj), where each Di is a division ring. Then there exist central idempotents u1,. . .,us of R such that

ng=,

Mkj(Di)is a simple ring. Hence Ruj is a direct sum of kj and Ruj copies of an irreducible R-module and Rui, Ruj, i # j , have no common composition factors. Thus r = s and, by renumbering the u j , we have I C3. = n,, 1 5 j 5 r, by Proposition 3.10. Put A = Mkj(Di),D = Di and let V be an irreducible A-module. E n d o ( o D ) . Hence, by Then, by Proposition 5.3(ii), EndA(V) Lemma 5.9, EndA(V) 2 Do. On the other hand, V % V,, hence

This proves that

Di

Dj and the result follows. W

5.11. Corollary. For any ring R, the following conditions are equivalent : (i) R is semisimple artinian. (ii) R is a finite direct product of full matrix rings over division rings. (iii) RR is completely reducible. (iv) Every R-module is completely reducible.

Proof. This is a direct consequence of Corollaries 4.2 and 5.6 together with Theorem 5.10. We close by proving the following classical result. 5.12. Theorem. (Wedderburn.) Let A be a finite-dimensional algebra over a field F and let I be an ideal o f A which has an F-basis consisting of nilpotent elements. Then I is a nilpotent ideal of A .

Proof. We may harmlessly assume that F is algebraically closed. Moreover, because (I J ( A ) ) / J ( A )is an ideal of A / J ( A ) having a

+

6. Group algebras and group representations

37

basis consisting of nilpotent elements, we may assume that J ( A ) = 0. Hence, by Theorem 5.10,

. . ,n,. Let

be the projection of I into M n i ( F ) 1, 5 i 5 r . It obviously suffices to show that 7 r i ( l ) = 0 for all i E ( 1 , . . . , r } . Since 7 r i ( I ) is an ideal of M n i ( F ) ,either 7 r ; ( l ) = 0 or 7 r ; ( l ) = M n i ( F ) . In the latter case, it follows that M n i ( F )has a basis consisting of nilpotent elements. But then the trace of each matrix in M n i ( F )would be zero, a contradiction. Thus 7r;(I) = 0 and the result is established. for some positive integers

6.

121,.

7ri

Group algebras and group representations

Let R be a commutative ring and let G be an arbitrary group. The group algebra RG of G over R is defined to be the free R-module on

the elements of G , with multiplication induced by that in G. More explicitly, RG consists of all formal linear combinations C xg - g, xg E R,g E G, with finitely many xg # 0 subject to for all g E G (i) C xg - g = C yg - g if and only if zg = yg (4 c x g ‘ 9 C Y g Y = C(xg Yg) ‘9 (iii) (C xg . g ) ( C yh - h ) = C zt . t , where t = Cgh=t zg- y h for all r E R (iv) r(C xg - g ) = C ( r x g ). g One easily verifies that these operations define RG as an associative R-algebra with 1 = 1~ l ~where , 1~ and 1~ are identity elements of R and G, respectively. Applying the injective homomorphisms

+

+

we shall in future identify R and G with their images in RG. With

these identifications, the formal sums and products become ordinary sums and products. For this reason, we drop the dot in xg - g. We shall also adopt the convention that RG 2 R H means an isomorphism of R-algebras.

Preliminaries

38

Let x = C x g g E RG. Then the support of x , Supp(x), is defined by

SUPP(4 = ( 9 E Glx,

# 0)

It is obvious that Supp(x)is a finite subset of G that is empty if and only if x = 0. For convenience of reference, we next record some elementary properties of group algebras. 6.1. Proposition.

Let A be an R-algebra and let

f : G -+ U ( A ) be a homomorphism of G into the unit group U ( A ) o f A . Then the map f* : RG -+ A defined b y

is a homomorphism of R-algebras. In particular, i f f is injective and A is R-free with f(G) as a basis, then RG Z A.

Proof. Because RG is R-free freely generated by G , f* is a homomorphism of R-modules. Let

be two elements of RG. Then

as required.

Let R be a commutative ring and let G be a group. Given a nonzero R-module V , by a representation of G on V we understand a homomorphism p : G + AutR(V)

6. Group algebras and group representations

39

We say that p is faithful if Kerp = 1. By Proposition 6.1, to each representation p of G on V there corresponds an RG-module structure on V given by (v E

V,zgE R,g E G)

Conversely, any nonzero RG-module V determines a representation p of G on V given by p(g)v = gv for all g E G,v E V . Thus there is a bijection between the class of all nonzero RG-modules and the class of all representations of G on R-modules. We say that the representation p of G is irreducible (indecomposable, completely reducible) if the corresponding RG-module is irreducible (indecomposable, completely reducible). Let V be an R-module which is R-free of finite rank n. Then

A u ~ R ( V ) GLn(R) where GLn(R) is the unit group of M,(R). Thus to each representation of G on V , there is a matrix representation p* : G + GL,(R). In particular, if R = F is a field, then all finitely generated F-modules are free of finite rank. In this case the study of FG-modules, representations of G on F-modules, and matrix representations of G over F are essentially equivalent. For the rest of this section, all vector spaces are assumed to be finite-dimentional over a ground field and G is a finite group. Let V be a vector space over a field F having a basis v l , . . . ,v, of n elements, and let f E EndF(V). With respect to this basis, f can be represented as an n x n matrix ( a ; j ) over F . By the trace of f , written tr f, we mean n

t r f = tr(a;j)= Caii i=l

If a different basis for V is chosen, then the two different matrices for f are similar, and hence have the same trace. Thus f H t r f is a well defined F-valued function on EndF(V). The map

Preliminaries

40

g) = t r f . trg. Now suppose that V is an FG-module. Then V determines a homomorphism 1c, : F G + EndF(V) of F-algebras. By the character of V, written x v , we understand the map xv : F G + F given by x v ( x ) = tr+(x) for all x E FG. The map x : G --f F defined by x ( g ) = xv(g) is called the character of G a$orded by V. In view of the equality X V E "99) = %XV(9) = xgx(9) is obviously F-linear, but need not satisfy the equality t r ( f

c

c

the characters x v and x determine each other. Suppose that p : G + GL(V) is a representation of G. By the character of p, (or character of G) we understand the map x : G + F given by x ( g ) = trp(g) for all g E G. It is clear from the foregoing discussion that : (i) We may treat the terms character of G and character of G afforded by an FG-module as interchangeable. (ii) The character of an FG-module V is completely determined by the character of G afforded by V. Let V and W be FG-modules. Define the sum xv xw of the characters of V and W , respectively , by the rule

+

+

Then obviously xV@w = xv xw and therefore the sum of two characters is again a character. Note that if charF = p > 0, then the constant function 0 is the character of the direct sum of p copies of any FG-module V . On the other hand, if charF = 0, then identifying Z with its image in F , we obtain

In this case, we say that x(1) is the degree of x. Let p : G 3 GL(V) be a representation of G and let x be the character of p. Then x is a class function; that is its value is cwstant on any conjugacy class of G. Indeed, for all x , y E G, we have

6. Group algebras and group representations

41

If C is a conjugacy class of G, then we put x(C)= ~ Suppose that the map

pi

( z with ) z E

C.

: G + GL(K),i = 1,2,is a representation of G. Then

P*

@'2

:G

+

GL(V1 @F

VZ)

defined by ( P I @ P 2 ) ( 9 ) = /h(g) @ P 2 ( 9 )

is a.gain a representation of G. We shall refer to pi 8 p2 as the (inner) tensor product of pi and p 2 . Note that if X I , x 2 are the characters of p1, p2, respectively, then the product xlx2defined by

is the character of pi @ p2.

We next record the following classical result. 6.2. Proposition. (Maschke's theorem). Let G be a finite group and let F be a field of characteristic p 2 0 . Then F G is semisimple i f and only i f p does not divide the order of G.

Proof. If p > 0 divides [GI, then z = CSECgsatisfies z E Z(FG) and z2 = IGlz = 0. Hence F G z E J ( F G ) # 0. Conversely, assume that p does not divide [GI, and let W be a submodule of an FG-module V . We may write V = W@W'for some F-subspace W'. Let 8 : V -+ W be the projection map. Define 1c, : V + V by

Because for all v E V and y E G,

1c, is an FG-homomorphism. Now assume that v E V . Then, for any z E G,z-'v E W , so O(x-lv) = z-lv. Accordingly, z8z-l~= v and $(v) = v. Setting

Preliminaries

42

W" = Ker$, it follows that W'' is an FG-submodule of V such that W" n W = 0. Finally, let v E V. Then, by the above, v - $(v) E W", so 21 = $(v) (v - $(v)) E W W'. Thus V = W @ W' and the result follows. rn

+

+

Suppose that A is a finite-dimensional algebra over a field F. Let V be a vector space over F, let E be a field extension of F , and form the tensor products AE=E@FA

and

VE=E@FV

If V is an A-module, then the vector space VE becomes an AE-module under a module action

An A-module V is said to be absolutely irreducible if, for every field extension E/F, VE is an irreducible AE-module. We say that F is a splitting field for A if every irreducible A-module is absolutely irreducible. We say that F is a splitting field for G if F is a splitting field for FG. We close by quoting certain standard facts pertaining to group representations ( see Curtis and Reiner (1962) Dornhoff (1971), Huppert and Blackburn (1982)).

.

6.3. Proposition Let F be an arbitrary field and let ~ 1 , ... ,X , be the characters of G aforded b y all nonisomorphic irreducible FGmodules. Then the characters xl,.. . ,xT are linearly independent (as functions from G to F . )

6.4.Proposition. Let F be a splitting field for G of characteristic 0, and let X I , . . . ,xT be all irreducible characters of G over F. For each i E (1, ...,r } , put

ei = Xi(i)/IGl

c

x;(g-l)g

g€G

Then e l , . . . ,e, is a complete system of centrally primitive idempotents of F G .

6. Group algebras and group representations

43

6.5. Proposition. Let F be a field of characteristic 0, and let V and W be arbitrary FG-modules. Then V 2 W i f and only i f xv = X W .

6.6. Proposition. (Orthogonality relations). Let xv and xw be the characters afforded b y the irreducible FG-modules V and W . Then (i) If F then CgECxv(9)XW(9-1)= 0, (ii) If F is a splitting field for F G and charF does not divide /GI, then

v w,

c xv(s)xv(g-l)

= IGI

g€G

(iii) If F is a splitting field for FG with charF not dividing \GI, CI, . . . ,C, conjugacy classes of G and X I , . . . ,xT irreducible characters of G over F , then

where gj E Cj.

In what follows if p = 0, then by definition all elements of G are p'-element s. Let F be an arbitrary field of characteristic p 2 0, let n be the least common multiple of the orders of the p'-elements in G and let E be a primitive n-th root of unity over F . Denote by I, the multiplicative group consisting of those integers r , taken modulo n, for which E I+ E' defines an automorphism of F ( E )over F . Two $-elements a, b E G are called F-conjugate if z-lbz = a' for some z E G and some r E I,. Thus an F-conjugacy class is a union of ordinary conjugacy classes. 6.7. Proposition. (Witt-Berman theorem). Let F be an arbitrary field of characteristic p >_ 0 and let G be a finite group. Then the number of nonisomorphic irreducible FG-modules is equal to the number of F-conjugacy classes of p'-elements in G.

We close by quoting some standard results on absolutely irreducible modules (see Huppert and Blackburn (1982)).

Preliminaries

44

Let A be a finite-dimensional algebra over a field F , and let V be an irreducible A-module. Then the following conditions are equivalent: (i) V is absolutely irreducible. (ii) If E is the algebraic closure of F , then VE is an irreducible AEmodule. (iii) EndA(V) = F , i.e. each A-endomorphism o f V is a left multiplication b y an element of F . 6.8. Proposition.

6.9. Proposition. Let A be a finite-dimensional algebra over a field F . (i) F is a splitting field for A if and only if

for some r 2 1 and some n; 2 1, 1 5 i 5 r . (ii) If F is a splitting field for A, then for any field extension E I F , E is a splitting field for A E . 6.10. Proposition. (Burnside). Let V be an absolutely irreducible A-module, where A is a finite-dimensional algebra over a field F . Then the homomorphism A EndF(V), a H $a, where $a(v) = av, a E A, v E V , is surjective.

45

Chapter 2 General properties of induced modules In this chapter we record a number of basic results concerning induced modules. After presenting some formal properties of these modules, we examine in detail their annihilators and provide a number of applications. Among other results, we prove some classical theorems of Mackey and tie together relative projectivity and induced modules. The chapter ends with a short discussion of projective covers of induced modules.

1.

Induced modules, representations and characters

Throughout this section, R denotes a commutative ring and G a group. Let H be a subgroup of G. Then the group algebra RH can be identified with the subalgebra of RG consisting of all R-linear combinations of the elements h E H . If V is an RG-module, then we shall denote by VH th RH-module obtained by the restriction of algebra; thus as an R-module, VH equals V , but only action of RH is defined on VH. This process will be called restriction and it permits us to go from any RG-module V to a uniquely determined RH-module V'. There is a dual process of induction. Let V be any RH-module. Since we may consider RH as a subalgebra of RG, we can define an RG-module structure on the tensor product

General properties of induced modules

46

This is the induced module and we denote it by V G . 1.1. Lemma. Let H be a subgroup of G. If T is a left (right) transversal for H in G, then RG is a free right (left) RH-module with T as a basis.

Proof. Suppose that T is a left transversal for H in G. Then, for any t E T , t(RH) is the R-linear span of the coset t H . Therefore, for any tl,. . . ,in E T ,t l ( R H ) -.. t,(RH) is the R-linear span of UY="=,t;H.As is apparent from the definition of RG, if X and Y are disjoint subsets of G, their R-linear spans meet at 0. Thus

+ +

proving that RG is a free right RH-module with basis T . A similar argument proves the case where T is a right transversal. 1.2. Corollary. Let H be a subgroup ofG, let T be a left transversal for H in G and let V be an RH-module. (i) VG = @V (direct sum of R-modules) (ii) V G= (l@V)$(CtET,tgH t@V) (direct sum of RH-modules) (iii) The map V -+ t @ V, v H t @ v is an R-isomorphism, while the map V + 1 8 V , v H 1 @I v is an RH-isomorphism. Proof.

(i) By Lemma 1.1, RG = etETt(RH),hence

as required.

(ii) By (i), it suffices to prove that CtET,tgHt@V is an RH-submodule . h E H and t E T - H. Then ht = t'h' for some h' E H of ( V G ) ~Fix and t' E T - H. Since h(t @ V) = ht @ V = t'h' @ V = t' 8 V, (ii) is established. (iii) The map V -+ R H @wV = 1 @ V, v H 1 @ v is obviously an RH-isomorphism. If t @ v = 0 then 18v = 0, hence v = 0. This shows that V -+ t @ V , v H t @ v is an injective R-homomorphism. Since the above map is obviously surjective, the result follows.

1. Induced modules, representations and characters

47

Let V be an RG-module. By the kernel of V , written I<erV, we understand the kernel of the representation of G afforded by V , i.e. I<erV = { g E Glgv = v for all v E V } 1.3. Corollary. Let H be a subgroup of an arbitrary group G and let V be an RH-module. Then

Proof. Let T be a left transversal for H in G. Then obviously

By Corollary 1.2(i), g E Ii'erVG if and only if g ( t @ v) = t 8 v

for all t E T ,v E V

If g t = t'h with t' E T , h E H , then g ( t @ v) = t ' @ hv. Hence, by Corollary 1.2(i), g ( t 8 v) = t 8 v if and only if g t = t h with hv = v. Thus g E I<er(VG)if and only if g E ntETtKerVt-l,as required. The process of induction is closely related to the concept of imprimitivity of a module. We say that an RG-module V is imprimitive if V can be written as a direct sum

of R-modules r/: with 111 a permutation of the set

> 1 such that for each g E G, V;:

g V , is {r/:li E I } . The latter is equivalent to the assumption that G acts on the set { K J iE I } with each g E G sending V, to g v . We refer to { K l i E I } as a system of imprimitivity for V . The following result ties together imprimitive and induced modules. H

1.4. Proposition. (i) Let H be a proper subgroup of G, let V be an RH-module and let T be a left transversal for H in G . Then { t 8 Vlt E T } is a system of imprimitivity for V G and the action of G on it is transitive. (ii) Let V be an imprimitive RG-module with {r/:liE I } as a system

General properties of induced modules

48

of imprimitivity and let G act transitively on

{xli E I ) . If H is the

stabilizer of any W E { K l i E I } , then W is an RH-module such that VS'G

Proof. (i) This follows from Corollary 1.2(i) and the fact that the left multiplication by g E G induces a transitive action of G on the set

{ t 63 Vlt E T } . ( 2 ) Let T be a left transversal for H in G. Then { K l i E I } = { t W l t E T } and V = etETtW (direct sum of R-modules). Hence, by Corollary l.2(i), the map t 63 W -+ t W , t 63 w H t w induces an Risomorphism W G V . Given g E G, t E T , we may write gt = t'h for some t' E T , h E H . Hence f ( g ( t 8 w)) = f ( t ' h @ W ) = f (t' 63 h w ) = t'hw = gtw = g ( t w ) = g f ( t 8 w ) as required.

Let V be an RH-module and let g E G. We define gV to be the R(gHg-')-module whose underlying R-module is V and on which the elements z E gN9-l act according to the rule 2

* 2, = (g-1zg)v

1.5. Lemma. Let H be a subgroup o f G , let V be an RH-module and let g E G. Then (i) gV E g 8 V and hV V for all h f H . (ii) ( g ~ ) ' z vG. (iii) For all z, y E G, "YV "(YV). (iv) V is irreducible (completely reducible, indecomposible) if and only i f so is gV. (v) If R = F is a field, then for any field extension E / F , the E(gHg-')-modules ( g V ) E and g ( V E ) are isomorphic. . -

Proof. (i) The map f : gV --f g @ V , v H g @ v is an Risomorphism, by Corollary 1.2(iii). Given z = ghg-', h E H , we have

1. Induced modules, representations and characters

proving that gV

%g @

49

V . If h E H , then

as required. (ii) Choose a left transversal T for H in G with g = t E 2'. Then the stabilizer of t @ V is tHt-'. Hence, by (i) and Proposition 1.4,

as desired. (iii) Applying (i), we have XY

v = (zy) @ v = z(y @ V ) E "(y @ V )z "("),

as required.

(iv) This follows from the fact that W is an RH-submodule of V if and only if W is an R(gHg-')-submodule of gV. (v) Observe that ( ~ V ) and E ~ ( V Ecoincide ) as vector spaces. Let z E gHg-', X E E and w E V . Then the action of z on X 8 v E g(V') is given by 2

* (A @ w)

= g-*zg(X @ v) = X @ g-lzgw

Bearing in mind that the corresponding action in (gV), is given by z(X @ v) = X @ z

*

2)

= X @ g-'zgv,

the result follows. W 1.6. Lemma. Let H be a subgroup of G, let E f F be a field extension and let V be an FH-module. Then

( V G )Z ~(VE)'

as

EG-modules

Proof. Let { v;/iE I }be an F-basis of V and let T be a left transversal for H in G. Then

General properties of induced modules

50

are E-bases of ( V G )and ~ ( V E ) ~respectively. , Hence the map 6' : ( V G )+ ~ ( V E )defined ~ by B ( 1 €3 (t @ v;)) = t @ ( 1 €3 v;) is an Eisomorphism. Fix g E G, t E T and write g t = t'h for some t' E T , h E H . Then, for z = 1 €3 (t €3 vi), we have

as required. W

Let F be a field and let V , W be FG-modules. Then the vector space V @F W is an FG-module where the action of elements g E G is defined by g(v €3 w)= gv €3 gw and then extended to V € 3 W ~ and FG by F-linearity. Note that if pv, pw are representations of G afforded by V , W respectively and d i m ~ V< 00, dirn~W< co,then V @IF W affords the tensor product Pv 63 Pw.

1.7. Lemma. Let EIF be a f i e l d extension and let V , W be FGmodules. Then

(V @ F W ) E% VE@ E WE as EG-modules Proof. Let {v;li E I } and {wjlj E J } be F-bases of V and W , respectively. Then (1 @ (vi @ wj)li E I , j E J }

and

{(I €3 vi) €3 (1 €3 wj)li E I , j E J }

are E-bases of (V @F W ) Eand VE@ E W E ,respectively. Hence the map

extends to an E-isomorphism of (V @ F W ) Eonto VE@ E WE. It is obvious that for all v E V , w E W ,

1. Induced modules, representations and characters

51

Since for all g E G,

and 9((1@ V i ) 8 (1 63 Wj)) = (g(1 8 v;) 8 g(l €3 Wj)) = (1 @ 9%) €3 (1 8 gwj)

we have +(9(1€3 i.(

€3 Wj))= g+(l C3 (0;€3 Wj)),

as required.

For the rest of this section, G denotes a finite group and all vector spaces over a field F are assumed to be finite-dimensional. 1.8. Proposition. Let H be a subgroup of G, let V be an F H module with F-basis {vl,. . . ,urn} and let p be the matrix representation of H aforded b y V in this basis. If gl,g2,.. . ,gn is a left transversal for H in G , then (91 @ v l , * - - , g l

is an F-basis of V G with respect to which the matrix representation pG aforded b y V G is given b y ~ ~ (=9(Pij(9)) )

(1 L i , j L n>

where p;j(g) = p(grlggj) if gT'ggj E H and p;j(g) = 0 if gr'ggj

9 H.

Proof. That the elements (9;@ vj} constitute an F-basis of V Gis a consequence of Corollary 1.2(i). We may write p(h) = (yij(h)), where hv; = Cj y,;(h)vj. To find the matrix pG(g), fix j E { 1,. . . ,n}and write ggj =gkh(h E H).Then g(gj C3 vt) = ggj 8 vt = gkh C3 vt = gk C3 hvt m

s=l

m s=l

General properties of induced modules

52

Now extend the coefficients -yst to the whole of G by setting yst(z) = 0 for x E G- H. Because for a fixed j there is only one k with gk'ggj E H, we have m n S(9j

63 4 =

y Y s t ( 9 ; 1 g 9 j ) ( 9 i (23 us), s=l i = l

as required.

Let V be an FH-module and let x be the character of H afforded by V . Then the character of G afforded by the induced module V Gis called the character induced from x; we shall use the notation xG for such a character. 1.9. Proposition. Let x be a character of H and let 91,. . . , g n be a left transversal for H in G. Set ~ ( x=) 0 for all x E G - H. (i) X G ( g ) = CZ1 x ( g i l g g i ) and IHIXG(9)= x(z-'gz) (9 E G )

c

XG

(ii)

If charF does not divide \HI, then

XG(d

=

V1-l

c

x(z-19z)

(9 E G )

zEG

(iii) If charF = 0 , then for any given g E G ,

where h l , . . . h , are the representatives for the conjugacy classes of H contained in the conjugacy class Cg of g and, b y convention, X G ( g ) = 0 if H n C, = 0. )

Proof. (i) Let x be afforded by an FH-module V. Then, in the notation of Proposition 1.8, we have trpii (9)= X ( g t l g g i ) and hence i=l

i=l

1. Induced modules, representations and characters

53

as required.

Since the value of xG does not depend on the basis of V G with respect to which xG is computed, it must also be independent of the choice of the left transversal for H in G. Bearing in mind that G can be written as a disjoint union of IHI left transversals for H in G, it follows from (1) that IHIXG(d =

c

x(s-lg4

(2)

XEG

as asserted. (ii) This is a direct consequence of (2). (iii) If H n Cg= 8, then the assertion follows from (i). Assume that H n Cg # 0. For any h in H n Cg,let X h denote the conjugacy class of H containing h. Then

H n cg = x h l u X h , u . - .u x h , for some h l , . . . ,hm in H . Since x is constant on each

xh,,

we have

Now if h f H n Cg, then as s runs over G, s-'gz = h for exactly C c ( g ) values of 2. Applying (2) and (3), we deduce that

as desired. We now apply Proposition 1.9(iii) to exhibit an orthogonality relation proved by Osima (1952). 1.10. Proposition. Let H be a subgroup of G, let F be an algebraically closed field of characteristic 0 and let X I , . . . ,xr be all irreducible characters of H over F . Then, for all g E G, h E H ,

General properties of induced modules

54

Proof. If H n Cg = 8, then g is not conjugate to h. Hence the result follows from Proposition 1.9(iii). We may thus assume that Cgn H # 8. Denote by h l , hz, . . . ,h , the representatives for the conjugacy classses of H contained in Cgsuch that h E { h l , . . . ,h,} whenever g is conjugate to h. Since the number of conjugacy classes of H is equal to r , we may choose the representatives h,+l,. . . , h, for the conjugacy classes of H not contained in Cgwith h E {h,+l, . . . ,h,} whenever h is not conjugate to g . By Proposition 1.6.6(i), we have

2

Xk(hi)Xk(hi')

= IcH(hi)lhij

(4)

k=l

Now h = h, for some s E { 1,. .. ,r } and h is conjugate to g if and only if 1 5 s 5 m. Applying Proposition 1.9(iii), we derive

and, by (4),the latter is equal to IC,(g)l if 1 5 s 5 rn and 0 if s So the proposition is true. 2.

> rn.

Formal properties of induced modules

Throughout this section, R denotes a commutative ring and G an arbitrary group. 2.1. Lemma. (Transitivity of the induction). If L, H are subgroups of G with H L, then for any RH-module V ,

s

(VL)G

Proof. We have

2VG

2. Formal properties of induced modules

55

as required.

2.2. Lemma.

Let V be an RH-module. Then the map

is an injective homomorphism of RH-modules. Furthermore, the image

1 @ v of V in V G is a direct summand

Of(VG)H.

Proof. Since R H is a direct summand of the RH-module RG, RH @ R H V is a direct summand of the RH-module VG.Since the map

is an isomorphism of RH-modules, the result follows. H

We shall refer to the injective RH-homomorphism V + V G ,v H 1@ 2) as the canonical injection. With the aid of this injection, we now provide a universal characterization of V G . Let H be a subgroup of G, let V be an RHmodule and let f : V + V G be the canonical injection. Then, for a n y RG-module W and for any Q E HomRH(V, W H ) ,there exists one and only one $ E HomRG(VG,W ) such that tp = 1C, o f . 2.3. Proposition.

Proof. Since x@v = z ( l @ v ) , x E R G , v E V ,any RG-homomorphisr V G + W is uniquely determined by its restriction to 1 @ V . Thus there exists at most one such +. Consider the map RG x V + W , (2,v) H zcp(v). Then this map is balanced and therefore determines a 12 -homomorphism V G3 W , x @ v H xcp(v), x E R G , v E V . Because

General properties of induced modules

56

for all z,z1 E RG and v E V, 1c, is an RG-homomorphism. Taking into account that $(l 8 v ) = cp(v) for all v E V the result follows. W

Let S be a subring of a ring L and let M be a (left) S-module. Given 1 E L and 1c, E Homs(L,M), define lcp : L + M by for all x E L

(/y)(x) = cp(z1)

It is straightforward to verify that the additive group Homs(L,M) becomes a (left) L-module. Let H be a subgroup of G and let V be an RH-module. Since R H is a subring of RG, we may apply the foregoing construction so that N o m m ( R G , V ) becomes a (left) RG-module. We shall denote this module by GV and refer to GV as a coinduced module. 2.4. Proposition. Let H be a subgroup of an arbitrary group G, let V be an RH-module, and let W be an RG-module. Then (i) HOmRG(VG, % HOmRH( WH) as R-modules (ii)HomRG(W,G V) S HomRH(WH, V) as R-modules

w)

v,

Proof. (i) Given X E Homm(V, W H )we may, by Proposition 2.3, define A* E HomRG(VG, w) by X*(z €3 v) = .(X(v))

The map X

H

w E V,x E R G

A* is obviously R-linear. Moreover, if A * = 0, then X * ( l @ w) = X(v) = o

for all v E V

and thus X = 0. Finally, given 1c, E HomRG(VG,W), define cp : V + W by cp(v) = $(1 €3 v) for all w E V. Then cp E H o ~ R H ( V WH) , and 1c, = cp* as required. ( V), define y(X) E H O ~ R H WH, ( V ) by (ii) For any X E H O ~ R GW,G

2. Formal properties of induced modules

57

Then clearly X H p(X) is an R-linear map. If a E then the map $ ( a ) defined by

[Ilt(a>(w)l(4 = 44

H u ~ R H ( WV H ) ,,

(wE W,sE RG)

belongs to H u ~ R G ( W V ),.~Moreover, ($cp)(X) = X and Hence the map X H $(A) is a required isomorphism. H

((P$J)(CY)

= a.

Let H be a subgroup of an arbitrary group G, let V be an RH-module and let W be an RG-module. (i) If V and W are irreducible, then V is isomorphic to a submodule of WH if and only i f W is isomorphic to a factor module of V G . (ii) If V is irreducible, then for any irreducible factor module X of V", V is isomorphic to a submodule of XH. 2.5. Corollary.

Proof. (i) Note that V is isomorphic to a submodule of W H if and only if H O ~ R H ( W V ,H )# 0. On the other hand, W is isomorphic to a factor module of V Gif and only if H o ~ R ~ (WV) # ~ 0. , Now apply Proposition 2.4 (i). (ii) This is a direct consequence of (i). H We next establish the following result which will allow us to identify, under certain circumstances, induced and coinduced modules.

Let H be a subgroup of G offinite index and let V be an RH-module. Then 2.6. Proposition.

Proof. Let (91, g 2 , . . . ,gn} be a left transversal for H in G. For every $ E" v, put n

d(4)

= csi @ 4(gi') i=l

E VG /'

Then an elementary calculation shows that 8 :G V --+ V G is an Rhomomorphism. To show that 8 is an RG-homomorphism, it therefore suffices to verify that S(g4) = g O ( 4 )

for all g E G, 4 eG V

General properties of induced modules

58

For every g E G, there is an h; E H an a permutation i H r ( i ) of the

index set {1,2, . . . ,n} such that 97'9 = hjg&. It follows that n

n

i=l n

i=l n

=

proving that 8 is an RG-homomorphism. If Cy=lgi 8 $(gi') = 0, then 4(gi1) = 0 for all i E { l , .. . ) n } . Because {gT1,. . . ,g;'} is a right transversal for H in G, given g E G, we have g = hg;' for some h E H and i E (1,. . . ,n}. Hence 4(g) = h4(gT1) = 0 and so q5(Csgg) = C z g 4 ( g ) = 0, proving that 8 is injective. The elements gT',gz',. . . , g i l constitute a basis of a free left RHmodule RG. It follows that for every set {vi E VI1 5 i 5 n } , there is a 4 eG V such that 4(gc') = vi, 1 5 i 5 n. Hence 8 is surjective and the result follows. 2.7. Corollary. Let H be a subgroup of G of finite index, let V be RH-module and let W be an RG-module. Then

Proof. Apply Propositions 2.6 and 2.4(ii). Suppose that U is a submodule of an RH-module V , and let i : V be the inclusion map. Since RG is a free RH-module, RG is flat, so the canonical homomorphism

U

+

is injective. For this reason, from now on we shall identify image in V G .

U G with its

2. Formal properties of induced modules

59

2.8. Lemma.

Let &and V2 be submodules of an RH-module V. (i) &G G VF if and only if V1 G V,. (ii) Vp = V : i f and only if V1 = V,.

(iii) (Vl + V2)G= KG + .v : ( i ~ ) (& n v ~= )hG~n ~ 2 ~ . (w) If V = V, @ V,, then VG = V F @ CG.

Proof. We first observe that (ii) is a consequence of (i), that (v) follows from (iii) and (iv) and that (iv) is a consequence of the fact that RG is a free RH-module. To prove (i), note that V1 C V2 obviously implies YGG Vp. Let T be a left transversal for H in G containing 1. Then each element of VG can be uniquely written in the form CtET taut with finitely many vt E V not equal to 0. Therefore if 01 E V1 is such that l @ v l E V p then v1 E V2. Thus qG VF implies & C VZ,proving (i). To prove (iii), observe that the inclusion VF V? C (V1 V2)G follows from (i). The opposite inclusion being a consequence of the

+

+

equality

t 69 (v1

+

v2)

= t €3 211

+ t €3

02

(v1 E

v1,v2

E V2,t E T )

the result follows. The following theorem contains some important formal properties of induced modules.

Let H be a subgroup of an arbitrary group G and let U A V 3 W be a sequence of homomorphisms of RH-modules. Then the following properties hold: (i) The sequence 2.9. Theorem.

is exact if and only if the corresponding sequence of RG-modules

is exact. (ii) Suppose that the sequence (1) is exact. Then (1) splits if and

General properties of induced modules

60

only if (2) splits. (iii) I f U is a submodule of V , then V G / U G (V/U)". Proof. (i) That exactness of (1) implies that of (2) is a consequence of the fact that RG is a flat RH-module (see Proposition 1.2.8(iii)). Conversely, suppose that (2) is exact. By Proposition 1.2.9(i), I ( e r ( l 8 p ) = (I(erp)G

and

Im(1 €9 A) =

Because I(er(1 €9 p ) = Im(1 €9 A), it follows from Lemma 2.8(ii) that K e r p = ImA. A similar argument shows that Ker(1 €9 A) = 0 implies KerX = 0 and that Im(1 €9 p ) = W Gimplies I m p = W . Hence the sequence (1) must also be exact. (ii) Suppose that 0 + U 5 V 1: W -+ 0 is a split exact sequence and let y : W t V be a splitting homomorphism. Then 1 €9 y : WG+ V G is obviously a splitting homomorphism for the induced sequence. Conversely, assume that the exact sequence

of RG-modules splits. Let T be a left transversal for H in G containing 1. Consider the mappings

WSWG

VG

W H l ( 8 W

CtfT

v

1,

t (8 V t

U1

Then c and T are well defined RH-linear maps. Let $I : W G+ V Gbe a splitting homomorphism for (2) and let w E W . Then, on the one hand, for some vt E +(I €9 w)= t B vt

C

t€T

and, on the other hand,

v

2. Formal properties of induced modules

Hence

20

61

= p ( q ) and therefore

This shows that ~ $ :0W V is a splitting homomorphism for (1) and the required assertion follows. (iii) Apply (i) to the natural exact sequence 0 -+ U -+ V -+ V / U + 0. H

2.10. Corollary. Let H be a subgroup of an arbitrary group G and let V be an RH-module. I f V G is completely reducible, then so is V.

Proof. Let W be a submodule of V and let

0

-+

w

-+

v

-+

v/w+ 0

be the natural exact sequence. The corresponding sequence 0 + WG+

vG (v/w)G0 --f

-+

splits by the hypothesis. Hence, by Theorem 2.9, the sequence 0+

w v + v/w+ 0 -+

splits, proving that V is completely reducible. Let S be a ring and let V be a left S-module. Given a (left or right) ideal I of S,let IV denote the set of all finite sums of the form

Then IV is an additive subgroup of is in fact a submodule of V .

V and if I is a left ideal, then IV

2.11. Proposition. Let H be a subgroup of an arbitrary group G and let X C Y be left ideals of R H .

General properties of induced modules

62

(i) ( Y / X ) GE A . Y / A- X fii) Let I be a right ideal of RG such that I 2 RG . X . If V is an RH-module, then there is a canonical injective map $J : I ( V G )+ ( X V ) G

of additive groups

Proof. All tensor products below are to be taken over R H . (i) Consider the diagram

0 -+ R G @X

+ R G @Y -+

If

R G @ ( Y / X ) -+ 0

1s

0 --* RG * X

--+

RG. Y

-+

RG * Y / R G* X

+

0

where all maps are canonical. Then the diagram is commutative and since RG is a flat RH-module, both rows are exact, and f , g are isomorphisms (see Proposition 1.2.8). This establishes the required assertion. (ii) Denote by X the natural homomorphism V + V / X V . Then the sequence

o -+ R G B X V --+

R G @v ' 9 R G B (V/XV)

is exact. Let x E I , y E RG, and v E V . Then xy E I can be written in the form

o E RG . X, so xy -+

Thus

and hence I(RG @ V ) C I(er(l@ A), as required. For the rest of this section, G denotes a finite group and F an arbitrary field. All FG-modules are assumed to be finite-dimensional

2. Formal properties of induced modules

63

F-spaces.

Let U and V be any FG-modules. Then the intertwinning number for U and V is defined to be

Let U be a completely reducible FG-module, and let V be an irreducible FG-module. We say that a nonnegative interger rn is the multiplicity of V as an irreducible constituent of U if a decomposition of U into irreducible submodules contains exactly rn submodules isomorphic to

V. 2.12. Lemma. Let U and V be completely reducible FG-modules. (i) i(U,V)= i(V,U). (ii) I f V is irreducible and F is a splitting field for FG, then i(U, V ) is equal to the multiplicity o f V as an irreducible constituent of U . (iii) I f F is a splittingfield for F G , then U is irreducible if and only if2(U,U) = 1. Proof. (i) Suppose that U and V are irreducible. If U

V , then

V , then obviously HornFG(U, V ) 2 Horn=~(V, U ) . This proves the case where U and V are irreducible. In the general case, write

If U

where the { U j } and { V k } are irreducible. Because

we have

i ( U 1 e Uz, U3) = i(U1,U3) for any FG-modules U,, Uz, and

U3.

+ i(U2,

Similarly,

u3)

64

General properties of induced modules

Thus

proving (i). (ii) If V is irreducible, then by the above

On the other hand, since F is a splitting field for FG,

proving (ii). (iii) Write U = mlU1 @ - . @ mtUt with U l , . .. ,Ut pairwise nonisomorphic and irreducible. Then, by (2), i(U, Us)= rn, for all s E {1,...,t } . It follows that

i(U,U ) = m:

+ - - + rn:

whence the result.

As a preliminary to the next theorem, we introduce the following definitions. Let A be a finite-dimensional algebra over a field F , and let V be an A-module. By the socle Soc(V) of V , we understand the sum of all irreducible submodules of V. We shall refer to the factor module V / J ( V ) as the head of V . 2.13. Lemma. (i) V / J ( V ) is completely reducible, and for any submodule W such that V/W is completely reducible, J ( V ) W . (ii) Soc(V) is completely reducible, and for any completely reducible submodule W of V, W C Soc(V). (iii) J ( V ) is the intersection of the kernels o j t h e homomorphisms V 4 W , where W ranges ouer all irreducible A-modules. (iv) For any given A-modules V and W , W is isomorphic t o an irreducible constituent of V /J ( V ) if and only if W is isomorphic to an irreducible factor module o j V .

2. Formal properties of induced modules

65

Proof. (i) This is a direct consequence of Lemma 1.4.3 and Proposition 1.4.1. (ii) Apply Proposition 1.3.15. (iii) This is a direct consequence of the definition of J ( V ) . (iv) Apply (i). 2.14. Theorem. (Nakayama reciprocity). Let F be an arbitrary field, let H be a subgroup of G and let V and W be irreducible F H and FG-modules, respectively. If m is the multiplicity of W as an irreducible constituent in the head of V G and n is the multiplicity o f V as an irreducible constituent in the socle of W,, then m - i(W,W )= n i(V,V ) In particular, m = n if F is a splitting field for F G and FH

Proof. We first observe that HomFG(U, W )= 0 for all irreducible FG-modules U with U 9 W . Applying Lemma 2.13(i), (iii), we therefore deduce that

HomFc(VG,W ) Z H O ~ F G ( V ~ / J (WV)~ ) , Z horn^^( W,W )@ - . -@ HomFG(W, W ) (m times) Thus i(VG,W ) = m . i(W,W ) . By Lemma 2.13(ii), the image of any FH-homomorphism V -+ WHis contained in the socle S of WH.Accordingly,

and therefore i(V,W H ) = n - i(V,V ) . Bearing in mind that, by Proposition 2.4(i), i(VG,W ) = i(V,W H )the , result follows. 2.15. Corollary. (Frobenius reciprocity). Let H be a subgroup of G , let F be a n algebraically closed field such that charF does not divide /GI, and let V and W be irreducible FH and FG-modules, respectively. Then the multiplicity of V as an irreducible constituent of W H is equal to the multiplicity of W as an irreducible constituent of VG.

66

General properties of induced modules

Proof. Apply Maschke’s theorem and Theorem 2.14. Let F be a splitting field for F G of characteristic 0 and let Cf(G, F ) be the F-vector space of class functions f : G + F . By defining

we obtain a symmetric nonsingular bilinear form Cf(G, F)x C f(G, F ) -+ F, called the inner product. If X I , . . . ,xr are the irreducible characters of G, then X I , . . . ,x r form an orthogonal basis for Cf(G, F ) . This can be easily seen by applying Proposition 1.6.6. Note also that, for any character x of G over F,

and a; =< x,xi >. Applying Corollary 2.15, we immediately derive 2.16. Corollary. (Frobenius reciprocity for characters). Let F be an algebraically closed field of characteristic 0, let H be a subgroup of G and let x and 0 be irreducible characters ofH and G, respectively. Then

< X G , O >=< x,e, >

3.

Annihilators of induced modules

Throughout this section, G denotes an arbitrary group and R a commutative ring. Unless explicitly stated otherwise, RG-modules are assumed to be left RG-modules. Let H be a subgroup of G and let V be an RH-module. We should like to read properties of RG from VGand conversely, properties of V G from RG. One very valuable tool for obtaining some of this exchange of information is propvided by “annihilators”, defined bellow. Let M be an RG-module. Then the annihilator of M , written ann(M),is defined by

a n n ( M ) = {z E RGlzM = 0)

3. Annihilators of induced modules

67

The annihilator of a right RG-module is defined similarly. Note that in both cases the annihilator is an ideal of RG. Our main result in this section describes ann(VG) exclusively in terms of ann( V ) ;most of the material presented is contained in a work of Kniirr(1979 ). Let H be a subgroup of G. Then the natural projection

R:RG+RH is defined by Y

It is clear that T is an R-linear map; in fact property, namely

R

satisfies a stronger

3.1. Lemma. Let H be a subgroup ofG and let R : RG be the natural projection. (i) R is a homomorphism of left and right RH-modules. (ii) For any left ideal I of RG, I E RG R ( I ) . (iii) For any right ideal I of RG, I R ( I ) - RG.

+

RH

-

Proof. (i) Let T be a right transversal for H in G such that 1 E T . Owing to Lemma 1.1, the map t H 0, 1 H 1, 1 # t E T , extends to a homomorphism R' : RG + RH of left RH-modules. Since 7r = R', we conclude that R is a homomorphism of left RH-modules. A similar argument, by taking T to be a left transversal, shows that R is a homomorphism of right RH-modules. (ii) Let T be a left transversal for H in G. By Lemma 1.1, each a E I can be uniquely written in the form

a =xtat

(at E RH)

tCT

with finitely many nonzero at. For each t f T , we have

t-'a E I

and at = r(t-'a) E w(I)

Hence a E RG . R( I ) , proving (ii). (iii) Choose T to be a right transversal for H in G and apply the argument in (ii).

General properties of induced modules

68

3.2. Lemma. Let N be a normal subgroup of G and let T be a transversal for N in G. Then (i) For any ideal A of R N ,

RG - A = $tET t A - t

(direct sum of RN-modules)

where t A = tAt-' (ii) For any family {A,JiE I } of ideals of R N ,

RG - (fI;czAi) = niczRG * Ai (iii) For any ideals A, B of R N , A RG*B.

c

B if and only if RG - A

c

Proof. (i) Since N d G , t A is an ideal of R N . In particular, t A RN and hence, by Lemma 1.1,

C t A * t= $ t c T t A * t t€T

is a left ideal of RG contained in RG-A. Since for t E N , tA.t = t A = A , the required assertion follows. (ii) By (i), we have RG - A; = e t E tA;~* t and therefore

niEzRG - A; = =

t ) = $tET (nicztAi) t (nicZAi) ' t

niEz(etETtAi @tET

= RG * (niEzAi)

as asserted (iii) By (i), we have

-

Hence, if RG A c RG - B , then A c B. Conversely, if A C B , then certainly RG A C RG - B. If RG - A = RG - B , then by the above, A = B , a contradiction. Thus RG A C RG B , as required. H

-

-

For any additive subgroup A of RG, let I d ( A ) denote the sum of all two sided ideals of RG contained in A. Thus Id(A) is the unique largest ideal of RG contained in A.

3. Annihilators of induced modules

69

3.3. Lemma. Let H be a subgroup of G, let I be an ideal of R H and let a : RG + RH be the natural projection. (i) Id(RG I ) = ann(RG/RG. I ) = ngEGRG(gI) where gI = gig-'. (ii) Id(RG.I) is the unique largest two sided ideal J of RG satisfying

(iii) Id( I RG) = Id( RG - I ) . (iv) If H is a normal subgroup of G, then *

I d ( RG * I ) = RG( n g E G ’ I ) Proof. (i) Because ann(RG/RG.I ) is an ideal of RG, the equality

Id(RG - I ) = ann(RG/RG - I ) is a consequence of the definition of Id(RG.1). Since I d ( R G - I ) RG-I and I d ( R G . I ) is an ideal of RG, we have

Id(RG.I ) Accordingly,

cg( R G .I ) = RG(’I)

Id(RG - I )

c ngEGRG(gI)

But ngEGRG(gI)is obviously an ideal of RG contained in RG I . Thus

as required.

(ii) Let J be an ideal of RG such that a ( J ) C I . Then, by Lemma 3.l(ii),

J~RG*T(J)GRG.I and so J C I d ( R G . I ) . On the other hand, let J be any ideal of RG contained in RG I . Then T ( J ) 2 a(RG - I ) G I , since a is a homomorphism of right RH-modules (Lemma 3.l(i)) and I is a left ideal of RH.

(iii) Applying Lemma S.l(iii) and the argument of (ii), we conclude that I d ( I RG) is also the unique largest two-sided ideal J of RG

-

General properties of induced modules

70

satisfying ? r ( J ) 2 I . Thus I d ( I *RG) = Id(RG * I ) as required. (iv) This is a direct consequence of (i) and Lemma 3.2(ii). H We are now ready to describe the annihilators of induced modules.

3.4. Theorem. Let H be a subgroup of an arbitrary group G , let V be a left RH-module and let I = ann(V). (i) ann(V G )= Id( RG I ) = ng,GRG(gI). (ii) If H d G, then ann(VG)= RG(flseG”). (iii) If V G is completely reducible, then J ( R G ) G RG(gI) for all g E G. The converse is true provided R G / J ( R G ) is artinian.

-

Proof. It is clear that (iii) is a consequence of (i). Hence, by Lemma 3.3(i),(iv), it suffices to verify that ann(VG)= Id(RG * I ) . To this end, let J C RG.1 be an ideal of RG. Since IV = 0, Proposition 2.11(ii) tells us that J ( V c ) = 0. Hence J & ann(VG)and therefore

Id(RG * I )

ann(VG)

To prove the reverse containment, let T be a left transversal for H in G. Then, by Corollary 1.2, V G = @I V . If 5 E ann(VG),say 5 = &Ttzt, xt E RH, with finitely many xt # 0, then for all o E V ,

Hence each Q E I and therefore 5 E RG.1. It follows that ann(VG)G RG I and thus ann(VG) Id(RG I ) , as desired. H

-

We close by proving a number of applications of the above result. As a preliminary observation, we first record 3.5. Lemma. Let H be a subgroup of an arbitrary group G. Then

J(RG)

RG*J ( R H )

if and only if J ( R G ) 5 J ( R H ) * RG

3. Annihilators of induced modules

71

Proof. Apply Lemma 3.3(iii) for I = J ( R H ) . H 3.6. Theorem. Let H be a subgroup of an arbitrary group G and let R be a commutative ring such that R H / J ( R H ) is artinian. Then

the following conditions are equivalent: (i) For a n y irreducible RH-module V , ( V G )is~completely reducible. (ii) J ( R H ) . RG C RG J ( RH ). (iii) J ( R H ) RG = RG J ( RH ) . Moreover, if these conditions are satisfied, then for any given positive integer n, ( R G * J ( R H ) ) " = RG * J ( R H ) " Proof. (i)+(ii): P u t V = R H / J ( R H ) . Since R H / J ( R H ) is semisimple artinian, V is completely reducible, so by assumption ( V G ) ~ is completely reducible; hence J ( R H ) G ann(VG). Now J ( R H ) = ann(V) and thus, by Theorem 3.4(i), ann(V G )= Id( RG - J ( R H ) ) Hence J ( R H ) R G

s I d ( R G - J ( R H ) ) and therefore J ( R H ) - RG

RG-J(RH). (ii)

+ (iii): It follows from (ii) that RG - J ( R H )- R G

C RG

*

J(RH)

whence

RG J ( R H ) *

RG - J ( R H ) * R G G Id(RG - J ( R H ) ) = I d ( J ( R H ) RG), *

by Lemma 3.3(iii). Hence RG. J ( R H ) J ( R H ) - R Gand the required assertion follows. (iii) =$ (i): It is clear from (iii) that R G . J ( R H ) is an ideal of RG. Let V be an irreducible RH-module and let I = ann(V). Then J ( R H ) 5 I and so RG.J(RH)c R G . 1

General properties of induced modules

72

Applying Theorem 3.4(i), we conclude that

J ( R H ) C RG - J ( R H ) C I d ( R G - I ) = a n n ( V G ) Since R H / J ( R H ) is semisimple artinian, the latter implies that (V')H is completely reducible. The final assertion follows by induction on n, using (iii). 3.7. Theorem. Let H be a subgroup o f a group G and let R be a

commutative ring such that both R H / J ( R H ) and R G / J ( R G ) are artinian. Then the following conditions are equivalent: (i) For any irreducible RH-module V , the RG-module V G is completely reducible. (ii) J ( RG) RG J ( R H ) . (ii) J ( R G ) C J ( R H ) - RG. Proof.

That (ii) is equivalent to (iii) is a consequence of Lemma

3.5. (i) =+(ii): Put V = R H / J ( R H ) . Since R H / J ( R H ) is semisimple artinian, V is a completely reducible RH-module, hence by hypothesis V G is completely reducible. Therefore J ( R G ) V G = 0. But, by Proposition 2.ll(i),

VG RG/RG. J ( R H ) and thus J ( R G ) R G - J ( R H ) . (ii) + (i): Let W be any irreducible RH-module and let V be as in (i). Since W is irreducible, W S R H / M for some maximal left ideal A4 of R H . Hence W C V/U where U = M / J ( R H ) . Since W G VG/UG (Theorem 2.9(iii)), we are left to verify that V Gis completely reducible. But J ( R G ) C_ RG J ( R M ) implies that J ( R G ) V G = 0. Since by assumption R G / J ( R G ) is artinian, we conclude that V Gis completely reducible, as required. 3.8. Lemma. Let H be a subgroup of G. If for any irreducible RG-module V , the RH-module VHis completely reducible, then J ( R H )

J ( R G ) . The converse is true under the additional assumption that R H / J ( R H ) is artinian.

3. Annihilators of induced modules

73

Proof. Let I be any maximal left ideal of RG and let V = R G / I . By assumption VHis completely reducible. Hence J ( R H ) V = 0 and so J ( R H ) I . Thus J ( R H ) 5 J ( R G ) , as required. Conversely, suppose J ( R G ) and that R H / J ( R H ) is artinian. Let V be that J ( R H ) any irreducible RG-module. Since J ( R G ) V = 0, we must also have J ( R H ) V = 0. But R H / J ( R H ) is artinian, hence VH is completely

s

reducible.

Let H be a subgroup o f a group G and let R be a commutative ring such that both R H / J ( R H ) and R G / J ( R G ) are artinian. Then the following conditions are equivalent: (i) J ( R G ) = RG' - J ( R H ) . (ii) J ( RG) = J ( R H ) - RG. (iii) For any irreducible RH-module V , V G is completely reducible, and for any irreducible RG-module M , MH is completely reducible. Furthermore, under the stronger assumption that G is finite and R is artinian, each o f t h e above conditions is equivalent to (iv) For any irreducible RG-module M , (MH)' is completely reducible. 3.9. Theorem.

Proof. The equivalence of (i), (ii) and (iii) follows from Theorem 3.7 and Lemma 3.8. Furthermore, it is clear that (iii) implies (iv). Assume that G is finite and R is artinian. Then both RG and R H are artinian and hence so are R H / J ( R H ) and R G / J ( R G ) . Let V be an irreducible RH-module. Then V Gis a finitely generated module over an artinian ring RG, hence V Gis artinian. It follows, from Corollary 1.3.2, that V Ghas an irreducible submodule, say M . Then, by Corollary 2.7,

and therefore there is an exact sequence

By Theorem 2.9, the latter gives rise to the exact sequence

General properties of induced modules

74

Hence (iv) implies that V G is completely reducible. Finally, let M be any irreducible RG-module. Then (iv) implies ( M H )is~ completely reducible, hence MH is completely reducible by Corollary 2.10. 4.

Clifford's theorem

Let N be a normal subgroup of an arbitrary group G and let R be a commutative ring. Given an RN-module V and g E G, let gV be the RN-module whose underlying R-module is V and on which N acts accordingly to the rule

n * v = (g-lng)~

(v E V )

where n * v denotes the module operation in gV and nw the operation in V. Such an RN-module gV is called a conjugate of V. 4.1. Lemma. Let V be an RN-module, let U be an RG-module and let g E G. (i) g 8 V is an RN-submodule of(VC)N and g @ V 2 gV. (ii) V is irreducible (indecomposable, completely reducible) if and only if so is gV. (iii) If W is an RN-submodule of UN, then gW is an RN-submodule of UN a n d g W E

gw.

(iv) (")G E V G . (v) 'V E V and "(YV) "YV for all x, y E G, i.e. G acts on the isomorphism classes of RN-modules by conjunction. Proof. It is clear that g @ V is an RN-submodule of ( V G ) ~ . Hence, by Lemma 1.5, we need only verify (iii). It is obvious that gW is an R-submodule of UN. If w E W and n E N, then n(gw) = g(g-'ng)w E gW.Thus gW is an RN-submodule of U N . Finally, the map gW + gW,w H gw is obviously an RN-isomorphism.

Let

V be an RN-module and let H = (9 E GIV 2 W}

4. Clifford's theorem

75

Then, by Lemma 4.l(v), H is a subgroup of G. We shall refer to H as the inertia group of V ;in case H = G we shall say that V is G-invariant. We are now ready to establish the following classical result. 4.2. Theorem.

(Cliflord (1937)). Let N be a normal subgroup of an arbitrary group G, let R be any commutative ring and let U be an irreducible RG-module. If GIN is infinite, assume that UN is artinian. Then (i) UN is completely redzlcible and, in particular, UN contains as irreducible submodule, say V . (ii) If H is the inertia group of V , then (G : H ) < 0;) and there exists a positive integer e such that UN g e

(@tErtv)

where T is a left transversal for H in G and { t V ( t E T } are all nonisomorphic conjugates of V . (iii) If W is the sum of all submodules of UN isomorphic t o V , then W is an irreducible RH-module such that

W N ~ and ~ V U g WG (Here, of course, eV denotes the direct sum of e copies of V ) .

Proof. If G I N is infinite, then by Corollary 1.3.2, U N contains an irreducible submodule V . Hence UN is completely reducible, since U = CsE~gV and each gV is an irreducible RN-module (Lemma 4.1). We may therefore assume that GIN is finite, say GIN = { g l N , . . . , g k N } . Because U is irreducible,

U = RGu = RN(g1u) +

* * *

+ RN(9kt.d)

for any 0 # u E U . Thus U N is a finitely generated RN-module. Applying Proposition 1.3.4, we conclude that UN has a maximal submodule, say M . We now set

MO

= ngEGgM

Owing to Lemma 4.l(iii), each gM is an RN-submodule of U N . Moreover, if g M M' c U for some RN-submodule M' of U , then

A4 5 g - W '

c 9-'u

General properties of induced modules

76

Thus M = g-'M', so M' = gM and therefore gM is a maximal RNsubmodule of U N . Hence each uN/gM is an irreducible RN-module. If g E G, then g = gin for some i E (1,. . . ,k}, n E N and gM = giM. Therefore k Mo = n;=,g;M

But Mo is an RG-submodule of U with Mo C M # U , hence Mo = 0. It follows that there is an injective homomorphism

and consequently UN is completely reducible, as required. (ii) and (iii) . Because CgEGgV is a nonzero RG-submodule of U , we have

u=cgv g€G

where, by Lemma 4.1(ii), (iii), each gV is an irreducible RN-module isomorphic to g V . If GIN is infinite, then by hypothesis there are only finitely many direct summands gV appearing in U N . But every conjugate gV of V is a direct summand of U , hence there are only finitely many nonisomorphic conjugates of V. This shows that (G : H ) = n < 00 and that if { 1 = 91,. . . ,g,} is a left transversal for H in G, then glV,. . . ,gnV are all nonisomorphic conjugates of V . Let W;be the sum of all submodules of U N isomorphic to giV (hence W = W l ) ,1 5 i 5 n. Then, by Proposition 1.3.19(ii),

Hence, to prove (ii), we are left to verify that each W; contains the same number of irreducible direct summands. Now for each g E G, gU = U and the modules gWi, gWj, i # j , have no composition factor in common. Thus W, = giW which proves (ii). It is clear that W is an RH-module and, by the above,

U = W @ g2W

@

* *.

@ g,W

(direct sum of R-modules)

Hence { W,g2W, . . . ,gnW} is a system of imprimitivity for U on which G acts transitively. Because H is the stabilizer of W , it follows from

4. Clifford’s theorem

77

Proposition 1.4(ii) that U Z W G . Finally, assume that X # 0 is an RH-submodule of W . Then

U‘ = x f3 gzx f3 *

*

f3 gnx# 0

is an RG-submodule of U . Therefore U = U’and so X = W , proving that W is irreducible. This completes the proof of the theorem.

4.3. Corollary. Let N be a normal subgroup of G of finite index and let R be any commutative ring. Then J ( R N ) J ( R G ) .

Proof. Apply Theorem 4.2 and Lemma 3.8. Let N be a normal subgroup of a finite group G, let F be a field and let U be an FG-module, dimFU < 00, such that U = CgEGgVfor some irreducible submodule V of U N . Let H and W denote, respectively, the inertia group of V and the sum of all submodules of UN isomorphic to V . (i) UN Z e ( & T t V ) , where e is a positive integer, T is a left transversal for H in G and {“It E T } are all nonisomorphic conjugates of V . (ii) W is an FH-module such that WN 2 eV and U S W G . (iii) U is irreducible if and only if W is irreducible. 4.4. Proposition.

Proof. Except for the irreducibility of W , properties (ii) and (iii) of Theorem 4.2 were deduced from the fact that U = x g E G g V . Hence the properties (i) and (ii) of the present proposition also hold. By the foregoing, we are left to verify (iii). If U is irreducible) then W is irreducible by Theorem 4.2(iii). Conversely, suppose that the FH-module W is irreducible. Let X be an irreducible submodule of U and let Vo denote an irreducible submodule of X N . Because VO is a n irreducible submodule of U N and UN is the sum of irreducible FN-modules gV, g E G, it follows that V, 2 gV

Z

gV

for some g E G

Now X = CgEGgVO,since X is irreducible. On the other hand,

g-lvo2

9-1

v

0 -N 9-1

(”)

v

General properties of induced modules

78

by virtue of Lemma 1.5. Hence, replacing & by g-lVO, we may assume that V E Vo, in which case H is also the inertia group of Vo, Let Wo be the sum of all submodules of X , isomorphic to Vo. By Theorem 4.2(iii), Wois an FH-module such that X E W:. Now Wo is a nonzero submodule of W , so W = Wo, by the irreducibility of W . Thus, by (ii), we have

x 5% w,"= WG5%u Hence U is irreducible and the result follows.

Let N be a normal subgroup of a finite group G, let E I F be a field extension and let U and V be, respectively, an irreducible FG-module, dimFU < 00, and an irreducible FN-submodule of U N . Let H and W denote, respectively, the inertia group of V and the sum of all submodules of UN isomorphic to V . If VE is an irreducible EN-module, then (i) UN E e ( e t 'E V )~where e is a positive integer, T a left transversal fro H in G and {'V(t E T } are all nonisomorphic conjugates of V . (ii) (UE)N e ( e t 'E (VE ~) ) and {'(VE)JtE T } are all nonisomorphic conjugates of VE. (iii) W is an irreducible FH-module and W E is an EH-module such that ( W E ) N 2 eVE , UE (WE)G 4.5. Corollary.

(iv) U E is irreducible if and only if WE is irreducible. In particular, if V and W are absolutely irreducible, then so is U . Proof. Because U = CgEGgVand

it follows that

By Deuring-Noether theorem (see Huppert and Blackburn (1982)), for any 2, y E G , "V E u V if and only if ("V), E (YV),. Because

5. Dual and contragredient modules

79

g ( V ~2 ) ( ~ V ) E(Lemma 1.5(v)), it follows that H is the inertia group of VE. Bearing in mind that W = & H h V , we also have WE = C h E H h Vand ~ the latter is, of course, the sum of all submodules of (UE)N isomorphic to VE. Applying Theorem 4.2 and Proposition 4.4, we conclude therefore that (i)-(iv) hold except possibly for the value of e in (ii). The correctness of value of e being a consequence of UN Z e ( B L E'rV ) ,the equality (UE)N= ( U N ) Eand the isomorphismg(VE) (gV)E, the result follows. 5.

D u a l a n d contragredient modules

Throughout this section, R is a commutative ring, G a finite group and A is an R-algebra. By convention, an A-module means a left A-module. Let V be an A-module. Then the dual $' of V is a right A-module whose underlying additive group is H o r n ~ ( v , R )with the action of A given by (fa)(.)

= f(m) for all

f E H o r n ~ ( V , R ) , Ea A , v E V

Similarly, if V is a right A-module, then the dual $' of V is an A-module whose underlying additive group is H o r n ~ ( VR , ) with the action of A given by (af)(v) = f ( v a )

for all

f E H o r n ~ ( VR, ) ,a E A, v E V

If V, W are A-modules and f E H o r n ~ ( VW , ) ,then is defined by ( j + ) ( v ) = +(f(v>> for all

vE

v,$E r/i/

f* E H o r n ~ ( l / ic) /, (1)

If V , W are right A-modules and f E HomA(V, W ) ,then f^ E H o ~ A (9 ~ ) , is defined analogously. Let U be an A-module. If V is a submodule of U , define

vL= {$ E fil+(V)= 0) Then V L is obviously a right submodule of the right A-module is also clear that if V, C V, then V' 2

K ' .

fi.

It

General properties of induced modules

80

5.1. Lemma. Let Then

U ,V , W be A-modules such that U = V 63 W .

o=W'@V'

and W L g Q ,

vl z2

w

Proof. It is clear that W' n V' = 0. Given f E 0,we may write f = f1 f 2 where f1, f 2 E 0 satisfy f l ( W )= f2(V) = 0. Thus 0 = W' @ V*. Finally, if f E W', then f H f l V is obviously a n isomorphism of W' onto p. Similarly, if f E V I , then f H flW is a n isomorphism of V* onto

+

w.

5.2. Lemma. Let V and W be A-modules. (i) Iff : V -+ W is a surjective A-homomorphism, then f : W -+ 6' is an injective A-homomorphism. (ii) The map V -+ v H v*, where v*($) = $(v) for all 1c, E P, is an A-homomorphism.

(vr,

Proof. (i) This is a direct consequence of the definition o f f given by (1). and (ii) The given map is obviously additive. If ZI E V , It, E a E A, then

as required. In what follows, given a n A-module V , we write VRfor V regarded as an R-module. Assume that VRis free with a finite R-basis vl, . . . ,v,,. Then P is free with a basis 61,.. . ,6, given by 6;(vj) = &j. We shall refer t o C 1 , . . . ,Gn as the dual basis of 01,. . . ,v,. 5.3. Lemma. Let V be an A-module such that VR is free with a finite R-basis vl,. . . ,v,,, and let G1,.. . ,G, be the dual basis of v1,. . . , v,. If a E A and av; = Cj"= rjivj , with rj; E R, then 6 j a = r$j.

cjn=l

5. Dual and contragredient modules

81

Proof. For all i, k E { 1 , 2 , . . . ,n}, we have

Thus 6;a =

cy=lr;j6j, as required.

5.4. Lemma. Let V and W be A-modules such that VR and WR are finitely generated free R-modules. (i) The map V --$ v H v*, where v*($) = $(v) for all $ E $' is an A-isomorphism.

($'I,

A

(ii) I f f E HornA(V, W ) ,then j ( v * ) = f(v)* for all v E V . Furthermore, i f f is an isomorphism of V onto W , then f is an isomorphism o j w onto

Q.

Proof. (i) By Lemma 5.2(ii), the given map is an A-homomorphism. Let q , . . . ,v, be an R-basis of V and let Gl,. . . ,f i n be a dual basis of V. Then vT(Gj) = Gj(vi) = 6jj which shows that TI;, . . . ,v;t is an R-basis of

c. Thus the given map is an isomorphism.

(ii) For any given v E V and ?j E W , we have

proving the first assertion. Assume that f is an isomorphism of V onto W . If vl, . .. ,v, is an R-basis of V , then maps the dual basis f ( v l ) ,. . . ,f(vn) onto 61,. . . ,Gn. Thus jis an isomorphism of r/i/ onto h

1

h

v. rn

5.5. Lemma. Let V be a right A-module such that V is R-free. Then ann(V) = ann(C/)

+

Proof. Assume that z E ann(V)and let be an arbitrary element of Then, for all v E V , (z$)(v) = vz) = $(O) = 0, so z E ann(P). Conversely, suppose that z E ann(V). Then, for all $ E and all v E v,

c.

$1

$(4 = (.$)(.I

Q

=0

General properties of induced modules

82

Since V is R-free, we conclude that vz = 0 for all v E V . Hence z E ann(V),as required. Turning our attention to induced modules over group algebras, we now prove 5.6. Proposition. Let H be a subgroup of an arbitrary group G and let V be an RH-module. Then

(vGj 2 (QG Proof. Let T be a left transversal for H in G. By Corollary 1.2(j), a typical element of V G can be uniquely written in the form

Ct8vt

E V)

(Vt

tET

with finitely many ut # 0. Because T-l is a right transversal for H in G, each element of (p)G can be uniquely written in the form

c

$t

8 t-l

($t

E

t€T

with finitely many

gt # 0. Consider the map 8 : (P)" -+(vGj

defined by

Then 8 is obviously an R-homomorphism. If 8 (CtET $t 8 t-') = 0, then $t(vt) = 0 for all t E T , vt E V . Hence each $t = 0 and so 8 is injective. If X E (V"], then

v,

hence t9(C At 8 t - l ) = X where Xt(vt) = X(t 8 vt). But each At is in and so 8 is surjective. We are therefore left to verify that for z = $ @ t - l ,

5. Dual and contragredient modules

83

Thus O ( z g ) = O ( z ) g and the result follows. Let V be an RG-module. An element v E V is said to be Ginvariant (or simply an invariant element) if g v = v for all g E G. In what follows Inv(V) denotes the set of invariant elements of V . It is clear that Inv(V) is the largest R-submodule of V on which G acts trivially. 5.7. Lemma. Let V and W be RG-modules. Then the RGmodule H o ~ R ( V W, ) becomes an RG-module under the following action of G:

(sf)(.) = s ( f ( 9 - W

(9 E G, ZI E

v,f E H O ~ R ( Vw) ,

Fu rt h e rmo re, Inv( HomR( V,W ) )= H O ~ R GV,(W ) Proof. The first statement is obvious. To prove the second, note that f E Inv(Horn~(V,W) if and only if f ( v ) = ( g f ) ( v ) = g ( f ( g - ' v ) ) for all v E V , g E G. The last condition is equivalent to f E H o ~ R G ( V W, ) . The special case of the RG-module HornR(V,W ) ,in which W = R and g r = r for all r E R, g E G deserves a separate mention. In this case, HomR(V,R ) = as an R-module and the action of G on is given by

c

( g f ) ( v >= f ( g - l v )

(v E V , g E G,f E

c)

General properties of induced modules

04

We shall denote this module by V*and refer to V' as the contragredient of

v.

5.8. Lemma. Let V be an RG-module which is R-fiee on afinite basis, say vl,. . . , v n and let r : G + G L n ( R ) be the matrix representation of G agorded by V with respect to v1,. . .,vn. Then, with respect to the R-basis $ 1 , . . . ,V;, of V* the matrix representation of G agorded b y V*is given b y g Ht r(g-*) (9 E G) where T ( g - ' ) is the transpose of q g - ' ) .

Proof. Since the left action of g on l;i in V* is the same as the right action of g-' on .L;i in P, the result follows by virtue of Lemma

5.3. Our next aim is to provide an analog of Proposition 5.6 for contragredient modules. This will be achieved as an easy consequence of the following general result.

.

5.9. Proposition Let H be a subgroup of an arbitrary group G, let V be an RG-module and let W be an RH-module. Then (i) om^( W,v)]' HornR( wG,V ) (ii) [ H O ~ ~ ( w)]" V ~ , 2 H O ~ R ( V w") ,

Proof. Let T be a left transversal for H in G. (i) A typical element f E [ H o ~ R ( W VH)]" , can be uniquely written in the form f=Ct@ft

(ft

E H o m ( W 7 VH))

t€T

with finitely many ft # 0. Consider the map

w,V H ) I G + HomR( W G V, )

8 : [HOrnR( defined by

5. Dual and contragredient modules

85

(only finitely many wt # 0). Then 8 is clearly an R-homomorphism. If 6(f) = 0, then O ( f ) ( t 8 w)= t f t ( w ) = 0 for all t E T ,w E W . Hence v) each f t = 0 and therefore 0 is injective. If X E H o r n ~ ( W ~ ,then (&T t 8 w t ) = E t E T tXt(wt),where xt(wt) = x(t 8 wt).Since each At is in horn^( VH),we see that 8(f) = for f = &T t 8 At. Hence 6 is surjective. Fix g E G and, for each t E T , write g t = t'ht with ht E H and t I+ t' a permutation of 2'. Hence, if f = CtETt @ f t ,then g f = CtETt ' 8 htft. Moreover,

w,

and

proving (i). (ii) A typical element f E [ H o r n ~ ( WIG v ~ , can be uniquely written in the form

with finitely many

ft

# 0.

Consider the map

8 : [ H O r n R (V H , W)IG--+ H O r n R (

v,W C )

General properties of induced modules

86

defined by O(f>(v) =

ct 8

ft(t-'v)

tCT

Then 8 is obviously an R-homomorphism. If S(f) = 0, then ft(t-'v) = 0 for all t E T , v E V . Hence f t = 0 for all t E T , so f = 0 and thus 8 is injective. If X E H o r n ~ ( VW"), , then X(v) = & - t8 Xt(t-'v) for some At E H o r n ~ ( vW~ ), with finitely many X t # 0. Hence e(z) = X for z = CtET t 8 A t ) proving that 0 is surjective. Fix g E G and, for each t E T , write g t = t'ht with t H t' a permutation of 2'. Then, for all 'u E V , (9W))(.)

= ge(f)(g-''u) =

c

9t 8 ft(t-lg-lv)

t U

and

proving (ii). 5.10. Corollary. Let H be a subgroup of an arbitrary group G and let W be an RH-module. Then

(W*)"c (W")* Proof. This is a special case of Proposition 5.9(i) in which V = R and G acts trivially on V . 5.11. Corollary. The group ring RG is self-contragredient, that is, (RG)*E RG as left RG-modules. Proof. Let H = 1 and let W be the RH-module R on which H acts trivially. Then W" = RG 8~ R E RG and W* E W.Hence, by Corollary 5.10, (RG)*E (W')" Z W GE RG

87

6. Induction, restriction, and outer tensor products

as required. H

Let V be a finitely generated projective RGmodule. Then V* is also projective. 5.12. Corollary.

Proof. By hypothesis, RG @ . . . @ RG = V @ W for some RGmodule W . Hence, by Corollary 5.11, V* is a direct summand of R G @ . . . @ RG. Thus V* is projective. H 6.

Induction, restriction, and outer tensor products

The aim of this section is twofold: first, to relate the induction and restriction processes, and second, to prove that the operation of forming outer tensor products commutes with the operation of forming induced modules. Throughout, R denotes a commutative ring and G an arbitrary group. Let V and W be RG-modules. Then the R-module V @ R W is an RG-module, where the action of the elements g,g E G is defined by

(v E v , w E W )

g(v@w) =gv@gw

and then extended to V @ R W and RG by R-linearity. We shall refer to V @ R W as the (inner) tensor product of V and W . It is clear that V @ R W S' W @ R V. Suppose now that G1 and G2 are groups and that & and V, are RGl and RG2-modules, respectively. Then the outer tensor product V, # V2 of V, and V2 is an R(G1 x Gz)-module whose underlying R-module is V, @R V2 with the module operation given by ( z , y ) ( v l8 v2) = X v 1 @ yv2

(z

E Gl7y E

G27v1

E

K,v2

E

b)

and extended to R(Gl x G 2 )and V, @ R V2 by R-linearity. Another interpretation of the outer tensor product V, # V2 comes from the isomorphism RGI @ R RG2 R(G1 x G2). Indeed, V, @ R V, is an RG1 @ RG2-module by means of (U

8 b)(v1 @ ~

2 = ) U V ~8 bv2

(a E

RG1, b E RG2, ~1 E V , 7

~2

E V,)

Hence, if we identify RG1 @ RG2 with R(GI x G2), the module above is nothing else but the outer tensor product of V, and V2.

General properties of induced modules

88

6.1. Theorem. Let H be a subgroup ofG, let U be an RH-module and let V be an RG-module. Then

(U€3 vH)G Proof.

= U G €3 v

as RG-modules.

Let T be a left transversal for H in G. Then the map

f :UG x

v + (U€3RVH)G

given by

is balanced. Hence f determines a unique R-homomorphism

v

€3 G (U €3 VH)G €3 u ) €3 2, H t €3 ( u €3 t-lv)

{ UG (t

Similarly, the map ( U ~ V H -,)UG@V, ~ t@(u@t-lv)H (t@u)@vis an R-homomorphism which is inverse to f*. Thus f* is an R-isomorphism. Fix g E G and write gt = tlh with tl E T , h E H . Then

f*(g[(t€3 U ) €3 v]) = f * ( ( t l€3 hu) €3 gv) = t i €3 (hu 8 t;'gh) and

Hence f* is an RG-homomorphism, as required. 6.2. Theorem. Let Hi be a subgroup of a group G;, i = 1,2, and let V, be an RH;-module, i = 1,2. Then

Proof. We may identify R(H1 x H2) with R H I C ~ R R H and ~ R(G1 x G2) with RG1 € 3 RG2. ~ With these identifications, ( a €3 b ) ( q €3 vz) = av1 @ bv2 for all v1 E V1, v2 E V2, a E RH1, b E RH2 and a similar

7. Mackey theorems and their applications

formuls holds for the action of RGl €3 RG2 on result, we have to establish an isomorphism

89

KG1#bGz. To prove the

V = (RGi € 3 RG2) ~ € ~ ‘ R H ~ @ R(V,#v2) H~ (RGi @ R H ~h)# (RG2 @ R H ~VZ) = W of (RG1 €3 RG2)-modules. Let Ti, i = 1,2, be a left transversal for H;

in G;. Then, by Corollary 1.2,

and

kv = @ ( t l , t z ) ~ ~ l x€3~ Vi) z ( t €3 i ( t 2 8 V2)

(direct sum of R-modules)

f “tl €3 t2) €3 (01 €3 0 2 ) ] = (tl €3 01) 0 (t2 €3 v2) is obviously an R-isomorphism, we see that f induces an R-isomorphism f* : V --f W. We are thus left to verify that for any given 91 E GI, 92

E Gz,

To this end, write g;t; = sihi with

s;

E T,, hi E H , i = 1,2. Then

as required. W

7.

Mackey theorems and their applications

Throughout, unless explicitly stated otherwise) G denotes an arbitrary group and R a commutative ring. Our aim is to prove some classical

General properties of induced modules

90

results of Mackey and to provide some applications of them. In particular, we present a criterion for two induced modules to be disjoint and a criterion for the irreducibility of induced modules. The section ends with the proof that if V and W are irreducible RN-modules, N 4 G, then VG S W Gif and only if V Z gW for some g E G. In what follows, we put t &Ht-'nS = (tV)tHt-lnS

7.1. Theorem (Mackey Decomposition (Mackey (1958)). Let H and S be subgroups of G , let T be a full set of double coset representatives for ( S ,H ) in G and let V be an RH-module. Then

Proof. Let {g;Ii E I} be a left transversal for H in G. Then, by Corollary 1.2 and Lemma 1.5(i),

V G= @iefgi8 V

(direct sum of R-modules)

and gi €3 V

"V

(as R(g;Hg;l-modules)

(1)

Put X = {g; €3 Vli E I } . Then G and, in particular S , acts on X . Moreover, 9; €3 V and gj €3 V lie in the same S-orbit if and only if gi and gj belong to the same double (S,H)-coset. For each t E T , let Wt denote the sum of the gj €3 V for which gj E S t H . Then each Wt is an RS-module and

(vG)s= @tETWt Setting I4 to be the restriction of tV to R ( t H t - l to verify that Wt E (K)". Let J C I be such that

n S ) , we are thus left

Then S acts transitively on the set {gj@VIj E J } and under this action the stabilizer o f t €3 V is {S

E Slst 8

v = t Q V } = {S

E Slt-lst E H } = t H t - l n s

7. Mackey theorems and their applications

91

Hence, by (1) and Proposition 1.4(ii), Wt Z' (&)' as required. W

7.2. Theorem. (Mackey Tensor Product Theorem (Mackey (1958)). Let H and S be subgroups of G , let V be an RH-module and let W be an RS-module. Then

where T is a full set of double coset representatives for ( S ,H ) in G.

Proof. Invoking Theorems 6.1 and 7.1, we have

On the other hand, by Theorem 6.1,

The desired conclusion is therefore a consequence of the transitivity of induction. In what follows, given an RG-module V , we write I n v ( V ) for the R-module of G-invariant elements of V .

7.3. Lemma. Let H be a subgroup of G of finite index, let T be a left transversal for H in G with 1 E T and let V be an RH-module. Then, the m a p I n v ( V ) + Inv(VG) &Tt

@ 'U

is an R-isomorphism.

Proof. It is clear that if w is H-invariant, then

&Tt

@

v is G-

invariant. Since the given map is obviously an injective R-homomorphism, we are left to verify that any G-invariant element of V Gis of the form

General properties of induced modules

92

CtETt 8 v for some v E I n v ( V ) . So assume that x E V Gis G-invariant and write

x=Ct@vt

(vt

EV)

t€T

Then the equalities t x = z and hx = x ( h E H ) imply, respectively, vt = v1 and hvl = vl, t E T , as required. W In what follows, for any RG-module V , V' denotes the contragredient of V .

7.4. Lemma. Let V be a finitely generated R-free RG-module and let W be an RG-module. Then

V*8 W

HomR(V, W )

as RG-modules

Proof. Let v l , . . . ,v, be an R-basis of V and let v;, . . . ,vz be the dual basis of V*. Define 1c, : V*8 W + H o r n ~ ( VW , ) by

[+(f €3 41(4= [f(v>lw

(v E

v,w E W,f E V*)

Then 1c, is obviously an R-homomorphism. If 1c, (CZ,vr €3 w;) = 0, then wj

C V8 ~

= {IC,

20;

1

}(~j) =

o

proving that 1c, is injective. If X E Horn~(V, W ) ,then X = II, (CZ, v t €3 X(v;) and thus 1c, is an R-isomorphism. Finally, fix g E G , v E V,w E W and f E V*. Then

7.5. Lemma. Let F be a field, let G be a finite group and let V and W be finitely generated FG-modules. Then

dirnFlnv(V* 8 W ) = i ( V , W )

7. Mackey theorems and their applications

Proof.

93

Applying Lemmas 5.7 and 7.4, we have

as required. W

As an easy application of the above, we now prove

7.6. Theorem. (Mackey Intertwining Number Theorem (Mackey (1958)). Let F be an arbitrary field, let H and S be subgroups of a finite group G and let V and W be finitely generated F H and FS-modules, respectively. Then

where T is a full set of double coset representatives for ( S ,H ) in G.

Proof. We have

where the last equality follows from Lemma 7.5 and the obvious fact that '(V') = ('V)'. In what follows, all modules are assumed to be finitely generated over their ground rings. Let F be a field, let G be a finite group and let V and W be FG-modules. We say that V and W are disjoint if they have no composition factors in common.

General properties of induced modules

94

7.7. Lemma. Let F be a field, let G be a jnite group such that charF does not divide IGl and let V and W be FG-modules. Then V and W are disjoint if and only if i(V,W ) = 0. Proof. By Maschke's theorem, we may write V = CBVj and W = (4) and {Wk} are irreducible FG-modules. Hence

$Wk, where the

i(v,W ) = C i ( 4 ,Wk) j,k

Since y $?! Wk if and only if i ( y ,Wk) = 0 , we see that i(V)W ) = 0 if and only if vj $?! Wk for all j , k. But the latter is equivalent to the disjointness of V and W , hence the result. We are now ready to provide two applications of Theorem 7.6. The first gives a criterion for two induced modules to be disjoint and the second provides a criterion for the irreducibility of induced modules.

7.8. Theorem. Let G be a finite group, let F be a field with charF not dividing IGl and let H and S be subgroups of G. If V and W are F H and FS-modules, respectively, and T is a full set of double coset representatives for ( S ,H ) in G, then V G and W G are disjoint if and only if for all t E T , the F(tHt-' n S)-modules t q H t - l n S and WtHt-lnS are disjoint. Proof. Apply Lemma 7.7 and Theorem 7.6.

7.9. Corollary. Further to the hypotheses and notation of Theorem 7.8, assume that both V Gand W Gare irreducible. Then V G$?! W G i f and only if f o r all t E T , the F(tHt-' fl S)-modules t&Ht-lnS and WtHt-lnS are disjoint. Proof. Since V G and W Gare irreducible, they are disjoint if and only if V G W G .Now apply Theorem 7.8. 7.10. Theorem. Let H be a subgroup of afinite group G, let F be an algebraically closed field with charF not dividing [GI and let V be an FH-module. Denote b y T a full set of double coset representatives for

7. Mackey theorems and their applications

95

( H ,H ) in G. Then V Gis irreducible if and only i f V is irreducible and for all t E T - H , the F(tHt-' n H)-modules t&Ht-'nH and &Ht-'nH are disjoint. Proof.

Applying Theorem 7.6 for W = V and S = H , we have

Also, by Lemma 2.12(iii), V Gis irreducible if and only if i(VG,V G )= 1. Since for t E H , t K H t - l n H S K H t - I n H (Lemma 1.5), it follows that V G is irreducible if and only if

and

i(V,V ) = 1 Now apply Lemmas 2.12(iii) and 7.7. H

A particular case where H 4 G deserves special attention and is recorded in the following: 7.11. Corollary. Let N be a normal subgroup of a finite group G, let F be an algebraically closed field with charF not dividing [GI and let V be an irreducible FN-module. Then V G is irreducible if and only i f V g V for a l l g E G - N . Proof.

Direct consequence of Theorem 7.10.

Applying Theorem 7.10 to one-dimensional representation of H , we obtain

7.12. Corollary. Let H be a subgroup o f a finite group G , let F be an algebraically closed field with charF not dividing IGJand let p be a one-dimensional representation of pi. Then the induced representation pG is irreducible if and only iJ for each g E G - H , there exists x E gHg-' n H such that p ( z ) # ~ ( g - l z g ) .

General properties of induced modules

96

Proof. Let V be an FH-module which affords p . Then, for any g E G, the representation g p of gHg-' given by "(z) = p(g-'zg) is afforded by g V . Since, for any given g E G, the F(gH9-l n H ) modules '&Hg-InH and &Hg-lnH are irreducible, they are disjoint if and only if they are nonisomorphic. In terms of representations, the latter is equivalent to the existence of z E gHg-' nH such that p ( z ) # g p ( z ) = p(z-'gz). The desired conclusion is therefore a consequence of Theorem 7.10. W We close by presenting two results pertaining to an arbitrary group G and an arbitrary commutative ring R. The following observation will clear our path.

7.13. Lemma. Let S be an arbitmy ring, let V be an S-module such that V = @,EIK whem each V , is irreducible, and let W be an irredu'cible submodule of V . (i) W Z I.j. for some j E I . (ii) If all the are nonisomorphic, then W = I.j. for some j E I . (i) Let J C I be such that the 5 ,j E J are all representatives of the isomorphism classes of V,. For each j E J , denote by Wj the sum of all V; with V; E %. By Proposition 1.3.19(i), the Wj,j E J are all homogeneous cornponenets of V . Hence W Wj for some j E J . Since W is irreducible, W C Xl @ - @ X , for some finitely many S-modules X I , . . . ,X,, isomorphic to 5. Thus W E % by the Jordan-Holder theorem. (ii) If all the V; are nonisomorphic, then Wj = 5 and W C Since the result is established. the latter is possible only in the case W = 6 ,

Proof.

- -

K.

W

7.14. Theorem. Let N be a normal subgroup of an arbitrary group G, let R be a commutative ring and let V be an irreducible R N module. If gV 9 V for all g E G - N , then V G is irreducible.

Proof.

Let W be a nonzero submodule of

VG. Then

W N

is a

8. Counting induced modules and characters

97

submodule of ( V G )and ~ (VC)N

= $ t u t €3

v

(direct sum of RN-modules)

where T is a transversal for N in G. Hence ( V G ) N (and therefore W N ) is completely reducible. Let X be an irreducible submodule of W'. Then X is an irreducible submodule of ( V G )and, ~ by hypothesis and Lemma 1.5(i), the t €3 V,t E T , are nonisomorphic. Thus, by Lemma 7.13(iii), X = t o @ V for some to E T . For any g E G , g X Wwhich implies that t €3 V C W for all t E T . Thus Vc = W as we wished to show. W

7.15. Theorem. Let N be a normal subgroup of an arbitrary group G, let R be a commutative ring and let V and W be irreducible RN-modules. Then V GE W G i f and only i f V 2 gW for some g E G. Proof. If V 2 gW, then by Lemma 1.5(ii), V G (gW)GE WG. Conversely, assume that V G W G .If T is a transversal for N in G, then and (W'), = $t& 8 W (V')N = $t& €3 V Since (VG)' 2 (W C )and ~ 1€3 V is an isomorphic copy of V contained in (VG)', V is isomorphic to a submodule, say X of ( W c ) N . Since, by Lemma 7.13(i) and 1.5(i), X is isomorphic to t €3 W E tW for some t E T , the result follows. w 8.

Counting induced modules and characters

Throughout, N denotes a normal subgroup of a finite group G and F an arbitrary field of characteristic p 2 0. Our first aim is to find the number of nonisomorphic FG-modules induced from the irreducible FN-modules. We begin by recording some preliminary information. All FG-modules bellow are assumed to be finitely generated. Let V be an FG-module. We say that V is a permutation module if there exists a basis {vl,. . . ,vn} of V on which G acts as a permutation group.

8.1. Lemma. Let V be a permutation module and let B = {vl,. . . ,v,} be a basis of V on which G acts as a permutation group.

General properties of induced modules

98

(i) If B1,. . . ,B, are G-orbits of B and H; is the stabilizer of any given element in Bi,l 5 i 5 r then

V

E

$i=l(l~i)G and

r = dirnFInv(V)

(ii) v E V'. (iii) IfB' = {ui,.. . ,vk} is another basis o f V on which G acts as a permutation group, then B' and B have the same number of G-orbits. Moreover, ifcharF = 0 , then for each g E G, the number of elements of B fixed b y g is equal to the number of elements of B' fixed by g . Proof. (i) This is a direct consequence of Proposition 1.4 and Lemma 7.3. (ii) Apply Corollary 5.10 and (i). (iii) The first statement follows from (i). To prove the second, assume that churF = 0 and fix g E G. Denote by M and M' the matrices of the linear transformation g with respect to B and B', respectively. Then M and M' are similar and hence t r M = trM'. Because g permutes the elements of B and B', trM and trM' are, respectively, the numbers of elements of B and B' fixed by g. This completes the proof of the lemma. Given an FN-module V and g E G, denote by x and characters of N afforded by V and gV respectively. Thus

"(4= x(9-lx9)

for all

5

g x

the F -

EN

Let n be the least common multiple of the orders of the p'-elements of G and let e be a primitive n-th root of unity over F . Denote by I, the multiplicative group consisting of those integers p , taken modulo n, for which e H e p defines an automorphism of F ( E )over F . Recall that, by definition, two p'-elements a, b E G are F-conjugate if z-lbx = up for some x E G and some p E I,. Owing to Proposition 1.6.7, the number of nonisomorphic irreducible FG-modules is equal to the number of F-conjugacy classes of p'-elements in G. 8.2. Lemma. Let x be a p'-element of N and let T, and I(, be the F-conjugacy classes of G and N, respectively, with x E T and x E K .

8. Counting induced modules and characters

99

Then

Tz = UgEGKg-Izg Proof. Let m be the least common multiple of the orders of the $-elements in N , so that n = mk for some natural number k and S = E~ is a primitive m-th root of unity over F . Assume that s E T,. Then s = g - ' s p g for some g E G and some p E I,. If p = r ( m o d m ) , 0 5 r 5 m - 1, then s = g-ls'g. The automorphism E H cfi of F ( E ) over F induces the automorphism 6 H 6p = 6' of F(6) over F . Hence r E I, and s E Kg-lZg. Since every automorphism of F ( 6 ) over F extends to that of F ( E )over F , it follows that if two elemets of N are F-conjugate in N , then they are F-conjugate in G. This proves the opposite inclusion, as desired. 8.3. Lemma. Let V and W be irreducible FN-modules and let x v and xw be the characters afforded b y V and W , respectively. Then V G W G if and only if x v = gxw for some g E G.

Proof. This is a direct consequence of Theorem 7.15 and Proposition 1.6.3. W Let x be the character of an F-representation p of G. If the $-elements a , b E G are F-conjugate, then x ( u ) = x(b). 8.4. Lemma.

By hypothesis, s-'bs = a' for some x E G and some Proof. r E I,. Let u be the automorphism of F ( E )over F such that a(&)= E'. We may choose a nonsingular matrix M such that p ( u ) = M-'diag(c*',

where k = degp and

~ ( b=) x ( a r ) = as required. W

.. . ,E @ ~ ) M

. . ,cyk are positive integers. Thus + . . . + &*k' - a(&*]+ * . + E a r ) = o(x(a))= x ( a ) ~1,.

Let x be the character of an F-representation of G and let K be an F-conjugacy class. Owing to Lemma 8.4, we may define x ( K ) unambiguously by setting

x(10 = x ( g )

for

g EK

General properties of induced modules

100

Recall that, by Propositions 1.6.7 and 1.6.3, the number of irreducible F-characters of G is equal to the number of F-conjugacy classes of G. We are now ready to prove the following crucial result. 8.5. Theorem. Let xl,. . . ,xm and K1,. . . ,Km be the irreducible F-characters of G and the F-conjugacy classes of p'-elements of G, respectively. Let A be a group acting on {XI, . . . ,xm}and { Kl, . .. ,K m } such that

x;(Kj) = "x;("Kj)

for all

a E A , i , j E { l ) .. . ,rn}

Then the number of A-orbits of { X I , . . . ,Xm} is equal to the number of A-orbits of {KI,. . . ,K,}. Furthermore, if charF = 0 , then for each a E A, the number of elements of {XI,... ,xm}fixed b y a is equal to the number of elements of {Kl, . . . ,K,} fixed by a.

Proof. We denote by V and W the permutation FA-modules corresponding to the action of A on { X I , . . . ,x m ) and { K l , . . . Km} respectively. Let pv and pw be the matrix representations of A afforded by V and W with respect to the bases {xl,.. . ,x,} and {Kl,. . . ,Km}, respectively. Put M = (x;(Kj)), 1 5 i , j 5 rn. By Proposition 1.6.3, the characters X I , . .. ,Xm are F-linearly independent and hence M is nonsingular. Fix a E A and write pv(a) = (a;j),pw(a) = ( b ; j ) . Then, for any # xi. given j E (1,. . .,rn}, a;, = 1 if c x j = xi and a;, = 0 if Similarly, b;j = 1 if "Kj = K; and b;, = 0 if "Kj # K;. Put )

(cij) = pv(a)M

and

(dij) = M p w ( a )

k=l

where the last equality follows from the assumption that x;(Kj) = "xi( " K j ) . On the other hand,

8. Counting induced modules and characters

101

which shows that pv(a)M = M p w ( a ) for all a E A. Since M is nonsingular, we conclude that V 2 W . Thus, by Lemma 8.1(i), the number of A-orbits of { X I , . .. ,x,} is equal to the number of A-orbits of {KI,. . . ,K,}. Finally, assume that charF = 0 and let xv and x w be the characters of V and W . Since V W , we have x v ( a ) = Xw(a) for all a E A. But x v ( a ) is the number of elements of {xl,.. . ,x,} fixed by a , while Xw(a) is the number of elements of { K l , . . . ,K,} fixed by a , hence the result. We shall say that two F-characters a,,f?of N are G-conjugate if /? = ga for some g E G.

Let N be a normal subgroup of G and let F be an arbitrary field of characteristic p 2 0. (i) The number of G conjugacy classes of irreducible F-characters o f N is equal to the number of F-conjugacy classes of p'-elements of G contained in N . (ii) If charF = 0 or charF does not divide IN1 and F is algebraically closed, then for any g E G the number of irreducible F-characters x of N with g x = x is equal to the number of F-conjugacy classes I( of p'-elements of N such that gKg-' = I<. 8.6. Corollary.

Proof. (i) Let { X I , . . . ,xn} and { K l , . . . ,K,} be the sets of irreducible F-characters of N and F-conjugacy classes of p'-elements of N , respectively. Then G acts on { x l , .. . x,} and { K l , . . . ,I<,} by sending, for each g E G , x; to g x ; and I(; to gK; = 9Kig-l. Since, for all g E G , g x i ( W j ) = "i(gKjg-1) = Xj(Kj) it follows from Theorem 8.5 that the number of G-orbits of { X I , . . . ,xm} is equal to the number of G-orbits of { K l , .. . ,Km}. But the former number is equal to the number of G-conjugacy classes of irreducible F-characters of N , while the latter number, by Lemma 8.2, is equal to the number of F-conjugacy classes of p'-elements of G contained in N . Thus (i) is established. (ii) The case charF = 0 follows from (i) and Theorem 8.5. Assume that charF does not divide IN[ and that F is algebraically closed. Then

General properties of induced modules

102

as is well known (see Curtis and Reiner (1981, p.420)) that the number of xi with g x ; = x; is independent of the characteristic of F . Hence we may assume that charF = 0 and the result follows. W

8.7. Corollary. (Karpilovsky (1978)). Let N be a normal subgroup of a finite group G and let F be an arbitrary field of characteristic p 0. Then the number of nonisomorphic FG-modules induced from the irreducible FN-modules is equal to the number of F-conjugacy classes of p'-elements of G contained in N .

>

Proof. By Lemma 8.3, the number of G-conjugacy classes of irreducible F-characters of N is equal to the number of nonisomorphic FG-modules induced from the irreducible 3"-modules. The desired conclusion is therefore a consequence of Corollary 8.6(i). H The groups G satisfying the hypothesis of the theorem below are commonly known as Frobenius groups. 8.8. Theorem. Let F be an algebraically closed field of characteristic p 2 0 let N be a normal subgroup of G with charF not dividing IN[ and assume that CG(z) N for every 1 # x E N . Then (i) If V is a nontrivial irreducible FN-module, then V G is irre-

ducible. (ii) The number of irreducible FG-modules induced from the irreducible FN-moduEes is equal to the number of nonidentity conjugacy classes of G contained in N . Proof. (i) Let x be the character of N afforded by V . Owing to V. Theorem 7.14, it suffices to show that for any g E G - N , gV Hence, by Proposition 1.6.3, we need only show that for any g E G- N , 9% # x. Applying Corollary 8.6(ii), we are thus left to verify that for any nonidentity conjugacy class C of N , gCg-' # C. Let x E C and assume, by way of contradiction, that gxg-l E C. Then 9x9-l = nxn-' for some n E N . Hence g-'n E CG(Z) N and therefore g E N, a contradict ion. (ii) Let r be the number of conjugacy classes of G contained in N . By Corollary 8.7, there exist exactly r , say FG,. . . ,YG,of nonisomor-

9. Relative trace maps

103

phic FG-modules induced from the irreducible FN-modules. By (i), there are precisely r - 1 irreducible modules among the qG,1 5 i 5 r , hence the result.

We close by proving th following result. 8.9. Theorem. (Osima (1952)). Let H be a subgroup of G, let F be an algebraically closed field of characteristic 0 and let xl,.. . ,xT be all distinct irreducible F-characters of H . Then the number of linearly independent F-characters of G among the induced characters xf,.. . ,x,G is equal to the number of conjugacy classes of G which contain an element of H .

Proof. Let 91, g2, . . . ,gs be the representatives for the conjugacy classes of G containing an element of H and let rn be the dimension of the vector space spanned by x?, . . . ,x:. Replacing each gi by its conjugate, if necessary, we may assume that gi E H , 1 5 i 5 s. By Proposition 1.1.9, each x: takes value 0 on each conjugacy class of G which contains no element of H . This shows that rn = rankA, where A is the r x s matrix defined by

(i and j are row and column index, respectively). Setting B = ( x ; ( g J y l ) ) , 1 5 i 5 r , 1 5 j 5 s, it folloews from Proposition 1.1.10 that tAB = diag(n,, 122,. . .,n,) where n; = ICc(gi)l, 1 5 i 5 s, and t A is the transpose of A . Because tAB is nonsingular, the rank of A is equal to s. Thus rn = s and the result follows. 9.

Relative trace maps

Throughout, G denotes an arbitrary group, H a subgroup of G of finite index and R a commutative ring. Given an RG-module V , we write I n v ( V ) for the R-module of G-invariant elements of V . In particular,

General properties of induced modules

104

1nv(V') denotes the R-module of H-invariant elements of the module V. 9.1. Lemma. Let T be a left transversal for H in G and let V be an RG-module. Then the map

Trg : Inv(V')

--+

Inv(V)

given by

is an R-homomorphism which is independent of the choice of T . Proof. Assume that T' is another left transversal for H in G. Then, for each t E T , we may find ht E H such that T' = {th,]tE T } . Hence, for any v E Inw(V')

Ctv = C t h t v = C t'v tCf

t€T

t'ET'

proving that Trg is independent of the choice of T. If g E G , then is another left transversal for H in G, hence

gT

gTrg(v) = C ( g t ) v= Trg(w) tQ

and thus T r s ( v )E Inv(V). Since Trg is obviously an R-homomorphism, the result follows. H In what follows, we refer to Tr$ as the (interior) relative trace map. For the convenience of future reference , we first record some elementary properties of relative trace maps. 9.2. Lemma. Let S

c H be subgrozlps of G offinite index. Then

T r z = Trg 0 T r f . (ii) Tr:(Inv(Vs))c T r z ( I n w V ~ ) . (a)

Proof. (i) Let X and Y be left transversals for H in G and S in H , respectively. Then XY is a left transversal for S in G. Hence, for

9.

Relative trace maps

105

any v E Inv(Vs),

as required.

(ii) By (i), we have

as asserted. H 9.3. Lemma. Let H be a subgroup of G of finite index and let V be an RG-module. Then, for any given g E G, v E Inv(VH) if and only zf gv E Inv( &Hg-l ). Moreover

and, in particular,

Proof. The first statement follows from the fact that, for any h E H , v E V , hv = v if and only if (ghg-')(gv) = gv. Let T be a left transversal for H in G. Because gTg-' is a left transversal for 9Hg-' in G, we have

as required. H 9.4. Lemma. Let H be a subgroup of G offinite index and let V be an RH-module. Then the map T r g : I n v ( ( V G ) H )+ I n v ( V G ) restricts to an R-isomorphism of I n v ( l @ V )onto I n v ( V G ) . In particular, Trg is surjective.

106

General properties of induced modules

Proof. Let T be a left transversal for H in G. Then

The desired conclusion is therefore a consequence of Lemma 7.3, since I n w ( V ) 2 I n v ( l 8 V ) via v H 1 8 v.

Let H be a subgroup ofG offinite index and let an RG-module V be isomorphic to a direct summand o f ( V ~ ) Then ~ , the map T r ; : I ~ ~ ( v H+ ) I~V(V) 9.5. Lemma.

is surjective. Proof. Our assumption ensures that Inv(V ) / T r g ( I n v (V H )is ) isomorphic to a direct summand of Inv((VH)G)/Tr$(Inv( (V H ) ~ ) H Since, ). by Lemma 9.4, the latter is 0, the result follows. 9.6. Lemma. Let H , K be subgroups of a finite group G and let V be an RG-module. (i) For all v E I n u ( v K ) , T r g ( v )= CtETT r g t - l n H ( t v ) , where T as a set of all double coset representatives for ( H ,I() in G. (ii) Trg(Inu(VK)) CtET TTgt-l.H(Inv(V,Kt-lnH)).

Proof. (i) Fix t E T . It is clear that if 91,. . . ,g8 is a left transversal for tKt-l n H in H , then g l t , . . . ,g,t is a complete set of representatives of the left cosets of I( in H t K . Since t v E InvtKt-l(V) InqKt-lnH(V), it follows that

which clearly implies the assertion. (ii) This is a direct consequence of (i).

Of special interest will be the case where the relative trace map is applied to the RG-module V = H o r n ~ ( UW , ) where U and W are

10. Induction and relative projectivity

107

RG-modules. Then VH = H o ~ R ( U H , W Hand, ) by Lemma 5.7,

I ~ ~ ( V= H )H o ~ R H ( ~WHH, ) I n v ( v ) = HomRG(U,W) Thus, by Lemma 9.1,

Trg : HomRH(UH, W H )+ HOmRG(U,

w)

is an R-homomorphism. Note also that, by the definition of Trg and the action o f t E T on f E HomRH(UH, WH) V , we have

T r g ( f )= C t f t - ' t@

9.7. Lemma. Let G be an arbitrary group, let H be a subgroup of finite index and let U , V , and W be RG-modules. (i) For a n y f E HomRH(U, v ) and g E HomRG(V, W ) , W39.f)=9T4(f)

(ii) For any f E HomRG(U, V ) and g E HomRH(V,W), T r Z ( 9 f )= T G ( 9 ) f

(iii) Tr$(EndRH(U)) is an ideal of EndRG(V). Proof. (i) Let T be a left transversal for H in G. Then tgt-' = g for all t E T , hence T r g ( g f )=

C t(9f)t-l = g C t f t - l t€T

=gTrg(f)

t€T

(ii) This is proved by a similar argument. (iii) Apply (i) and (ii) for the case U = V = W . W 10.

Induction and relative projectivity

Throughout, G denotes an arbitrary group and R a commutative ring. Let H be a subgroup of G and let V be an RG-module. We say that V is H-projective if every exact sequence of RG-modules

O+U+W+V+O

General properties of induced modules

108

for which the associated sequence of RH-modules

splits is also a split exact sequence of RG-modules. The RG-module V is said to be H-injective if every exact sequence of RG-modules

for which 0

VH

WH -+

UH

--f

0

is a split sequence of RH-modules, is also a split sequence of RGmodules. Hence, if R is a field, then an RG-module is projective (respectively, injective) if and only if it is l-projective (respectively, l-injective). Our aim is to provide various characterizations of H projective modules and record a number of applications. AS a point of departure, we first establish the following two preliminary results.

10.1. Lemma. Let H be a subgroup ofG offinite index, let T be a lest transversal for H in G containing 1 and let V be an RG-module. Define f : (VH)G

v

by

f

Ct@vt =Ctvt (tET

(Vt

E V)

tET

Then f is a surjective RG-homomorphism such that Kerf is a direct summand of ( ( V H ) ~ ) H .

Proof. It is clear that f is a surjective R-homomorphism. To show that f is an RG-homomorphism, fix g E G. Then, for any t E T , gt = t’h, for some t’ E T and h, E H. Thus

10. Induction and relative projectivity

109

and therefore

)

= gf C t @ v t ET

(t

proving that f is a surjective RG-homomorphism. Let i : VH -+ (VH)' be the canonical injection. It is clear that ~(VH n )K e r f = 0. Because

C t ~ v t - 1 8Ctvt ~ ~ e r j , tET

(tET

we conclude that ((vH)')

= Icerf

i(vH)

as asserted.

10.2. Lemma. Let H be a subgroup o f G offinite index, let T be a left transversal for H in G and let V be an RG-module. The map

f : V -+ (VH)'

defined b y f(v) =

ct

@ t-lv

tET

is an injective RG-homomorphism which is i n ~ e ~ e ~ of~ ethen tchoice ofT and f ( V ) , is a direct summand of ((V","),.

Proof. Assume that T' is another left transversal for H in G. Then, for any given t E T , there exists ht E H and t' E T' such that t = t'ht. Thus

C t @ t-'v = C t'ht 8 htl(t')-'v tET

=

C t' 8 (t')-'v, t'ET'

t6T

proving that f is independent of the choice of T . If g E G , then

f (gv)

=

C t 63 t-'(gv) = 9 C ( g - ' t ) 8 (t-lg)v tET

=

sfW,

tQ

General properties of induced modules

110

proving that f is an RG-homomorphism (since f is clearly an Rhomomorphism). Bearing in mind that

(VH)'

= @tcTt (8 V H ,

f is an injection. Note further that, by Corollary 1.'2@),W' = @teT,tgHt@ V ' is an RH-submodule of ( ( V H ) " ) ~ We . claim that ((VH)')>,

= w'@f(V)H

Indeed, if f ( v ) = CtfTt @t-'v E f ( V )n W', then v = 0. On the other hand, if t' E T n H, then

ct

@ vt -

ct

@ t-lt'vt1 E

W',

tET

tfT

thus completing the proof. H We are now ready to prove the following result essentially due to Higman( 1954). 10.3. Theorem. Let G be an arbitrary group, let R be a commutative ring and let H be a subgroup ofG offinite index. Then, for any

RG-module V , the following conditions are equivalent: (i) V is H-projective. (ii) V is isomorphic to a direct summand of(VH)G. (iii) V is isomorphic to a direct summand of W G ,where W is an R H -module. (iv) E n d R ( V ) is H-projective. (v) T r g : EndRH(VH) 4 EndRG(V) is surjective. (vi) There exists $ E EndRH(VH)) such that T r s ( $ ) = 1. (vii) V is H-injective. Proof. (i)+(ii): By Lemma 10.1, there is an exact sequence of RG-modules

0 u (VH)' v 0 such that the associated sequence of RH-modules splits. By hypothesis, V is H-projective, so K e r f is a direct summand of ( V H ) ~ Thus . ( v H ) ~=

K e r f @ V'

10. Induction and relative projectivity

111

where V' 2 V . (ii) + (iii): Obvious. (iii) + (vi): Let us show first that any RG-module V of the form V = W G ,where W is an RH-module, satisfies (vi). To this end, denote by T a left transversal for H in G containing 1 and define 1c, : V + V bv

It is easily verified that 1c, is an RH-homomorphism. Moreover, for all w E W , we have

(t' €3 w)=

c t$(t-'t'

€3 w)= t' €3 w

(t' E T )

t€T

Thus II, satisfies (vi). Turning to the general case, we may harmlessly assume that

W G=

v Cf3 V'

(direct sum of RG-modules)

Let T : W G --+ V be the projection map and let II, satisfy (vi) with respect to W G . Then T 0 II, induces an RH-homomorphism V -+ V , and for all v E V , we have

since 7~ is an RG-homomorphism. The desired implication follows. (vi)+ (vii): Suppose that V is an RG-submodule of the RG-module U such that UH = VH@ W for some RH-submodule W . Let 7r : U,y + VH be the projection map and let 'p E HornRc(U,V) be defined by 'p = TrE(ll,o T ) , where II, E E ~ R H ( V H satisfies ) (vi). If v E V , then ~ ( v= ) v and hence U = V @ Ker'p.

(vii)+ (i): Assume that V is H-injective. Owing to Lemma 10.2, V is isomorphic to a direct summand of ( V H ) G . Applying implication (iii)+ (vi), there exists II, E E ~ R H ( V such H ) that TT$($)= 1. Now assume that f : U + V is a surjective RG-homomorphism for which

U = Kerf

Cf3

W

(direct sum of RH-modules)

General properties of induced modules

112

The restriction f 1 = f IW is an RH-isomorphism of W onto VH. Set cp = and put 8 = Trz(cp$f). Then 8 E EndRG(U) and

fc'

(f 0 e)(4

=

c

ftcp$ft-'u

t€T

= Ctfcp$t-lu tET

= C t $ t - ' f u = f(u) t€T

proving that U = 8 ( U ) 8(u) =

+ K e r f . If 8(u) E 8 ( U ) n K e r f , then

c

c tcp+

ft-lu =

t ET

tcp+t-'fu = 0

tET

Thus U = O(U) @ K e r f , proving (i). (ii)+ (iv): Since V is isomorphic to a direct summand of ( V H ) it~ , follows that EndR(V) is isomorphic to a direct summand of H o m R ( ( v ~ ) ~ , But, by Proposition 5.9(i),

H o m ~( (V H ) ~V, )

H o m ~VH, ( V H )=~(EndR(V ) H ) ~

hence EndR(V) is H-projective, by the equivalence of (i) and (ii). (iv)+ (v): This is a direct consequence of Lemma 9.5 and the implication (i)+ (ii). (v)+ (vi): Obvious. H We now record a number of consequences of the result above. 10.4. Corollary. Let H be a subgroup of G of finite index n such that n is a unit of R. Then any RG-module is H-projective.

Proof. Let V be an RG-module and let $ : V +(v) = n-lv Then

for all

+ E EndRH(VH) and for all v E V,

(E

t$t-l

1

2)

= 2)

2)

--+

E

V

V be defined by

10. Induction and relative projectivity

113

where T is a left transversal for H in G. This shows that TT$($)= 1 and the result follows by virtue of Theorem 10.3(vi). W 10.5. Corollary. Let H be a subgroup of G offinite index and let V be an RG-module which is R-free offinite rank. Then V is H projective if and only if

is surjective. Proof.

Owing to Lemma 7.4,

Now apply Theorem 10.3. 10.6. Corollary. Let G be an arbitrarygroup, let H be a subgroup of finite index and let V , W be RG-modules. Then, for any given X E H o m R c ( V , W ) ,the following are equivalent: (i) X E Tr$(HomRH(V, W ) ) . f (ii) Assume that U is an H-projective RG-module and that U -+ f WH + 0 splits. W + 0 is an exact sequence for which UH -+ Then there exists ?,h E HomRG(v with = f o $. (iii) Assume that U is an H-injective RG-module such that 0 -+ f f V U is an exact sequence for which 0 -+ VH -+ UH splits. Then there exists q!~E HomRG(U, with = $ o f .

u)

w)

Proof. We will demostrate that (i) is equvalent to (ii). The dual of our argument will easily establish the equivalence of (i) and (iii). (i)+ (ii): Since UH f, WH-+ 0 splits, there exists p E HomRH(W, such that f p = 1. By hypothesis, there exists Xo E HomRH(V, W )such that X = T r $ ( X o ) . Setting $ = Tr$(pXo) E H o w L R G ( V , U we ) , obtain

u)

f$ = T r g ( f&) = T r g ( X o )= A,

(by Lemma 9.6(i))

General properties of induced modules

114

as required.

(ii)+ (i): Since U is H-projective, it follows from Theorem 10.3(vi) that T r $ ( A,) = 1 for some E EndRH( U ) . Hence

Tr%f A,$>

= f T r Z (A,)$ = f$=A,

(by Lemma 9.6(i), (ii))

as required. We close by recording a number of useful properties of H-projective modules and some applications.

10.7. Theorem. Let S 5 H be subgroups of G offinite index and let V be an RG-module. (i) Let V = @y=lK, where each E is an RG-module. Then V is H-projective if and only if each K is H-projective. (ii) IfV is S-projective, then V is also H-projective. (iii) If V is H-projective and VH is S-projective, then V is Sprojective. (iv) For any RH-module V , the RG-module VG is H-projective. (v) If W is an S-projective RH-module, then any RG-module which is a direct summand of W G is S-projective.

Proof. (i) Suppose that V is H-projective. Then, by Theorem 10.3(vi), there exists $ E E ~ ~ R H ( Vsuch H ) that Tr$($) = 1. Let ~ i 1 ,5 i 5 n, be the projection of V onto T/: defined by the decomposition V = @K. Then K ~ O $E E n d R H ( T / : ) H and obviously Tr$(r;o$) = 1. Thus, by Theorem 10.3(vi), each V , is H-projective. Conversely, if each $; E E ~ ~ R H ( Eis )such H that Tr$($i) = 1, then t,b E EndRH(V'&) defined by $lK = $; obviously satisfies T r $ ( $ ) = 1. Thus V is H projective. (ii) Let 0 + U -+ W + V -+ 0 be an exact sequence of RG-modules such that the associated sequence 0 -+ U, t WH + VH + 0 splits. Then the sequence 0 + US -+ WS -+ VS -+ 0 also splits. Because V is S-projective, the sequence 0 t U ---f W + V + 0 splits and therefore V is H-projective. (iii) Owing to Theorem 10.3(ii), we have (VH)G = V' @ X and

10. Induction and relative projectivity

(Vs)H = V" @ Y , where V'

%

115

V and V" 2 VH.It follows that

( vs)G ( v H ) G

@

Y G= v' @ x @ Y G

and thus V is isomorphic to a direct summand of ( V S ) ~Now . apply Theorem 10.3(ii), (iv) This was established in the course of the proof of Theorem 10.3. (v) Owing to Theorem 10.3(ii), ( W S )=~W ' $ X , where W' E W ,so that (Ws)G % W G@ XG.By (iv), (WS)G is S-projective and therefore, by (i) any RG-module which is a direct summand of WGmust be Sprojective. H 10.8. Theorem. Let H be a subgroup of G of finite index n. (i) If V is a projective RG-module, then V H is a projective R H module. The converse is true if n is a unit of R. (ii) If V is an RH-module, then the induced module V G is projective if and only i f V is projective. (iii) If R as a field of characteristic p > 0 and P is a Sylow p subgroup of a finite group G, then an RG-module V is projective if and only i f Vp is projective.

Proof. (i) Let V be a projective RG-module. Then there exists a free RG-module X with X = V @ W for some RG-module W . Since X H = VHfBWHand X H is a free RH-module, we conclude that VHis projective. Conversely, assume that V is an RG-module such that VH is a projective RH-module and that n is a unit of R. Let

be an exact sequence of RG-modules. Then the associated sequence

splits, since VH is projective. By Corollary 10.4, V is H-projective, hence the sequence

O-+U+WtV+O also splits. Thus V is a projective RG-module. (ii) Let T be a left transversal for H in G. Then, by Corollary

General properties of induced modules

116

1.2( ii) ,

(VG),=(l@V)$

c (tET,W

)

t@V

(direct sum of RH-modules)

where, of course, 1 8 V is identifiable with V . Suppose now that VGis projective. Then, by (i), ( V G ) is ~ also projective. Since V S 1 @ V is a direct summand of (VG),, we deduce that V is projective. Conversely, assume that V is projective. If V is a free RH-module, then the RG-isomorphism

implies that VG is a free RG-module. Thus V G is projective. In the general case, we have W 2 V @ U for some free RH-module W and some RH-module U . Then W GS VG$UGand hence V Gis projective. (iii) This is a direct consequence of (i).

A ring S is said to be local if S / J ( S ) is a division ring. In what follows, we write U(S)for the group of units of S. We next record certain properties of local rings, some of which will be required for subsequent investigations. 10.9. Lemma. For any ring S , the following conditions are equivalent: (i) S is local. (ii) S has a unique maximal left ideal.

(iii) J ( S ) = s - U ( S ) .

(iv) The set of nonunits of S is a left ideal.

Proof. (i)+(ii): Let I be a maximal left ideal of S. Then I / J ( S ) is a proper ideal of S / J ( S ) and thus I = J ( S ) . (ii)+(iii): Let J be a unique maximal left ideal of S. Since J ( S ) is the intersection of all maximal left ideals of S , we have J ( S ) = J , and so J is an ideal. The inclusion J ( S ) S - U(S)being obvious, suppose that x E S - U(S).If Sx # S , then Sx lies in a maximal left ideal of S , whence Sx C J ( S ) , so x E J ( S ) . On the other hand,

10. Induction and relative projectivity

117

if S x = S then yx = 1 for some y E S . Clearly y 4 J ( S ) , otherwise 1 = y E J ( S ) . Hence S y = S , so zy = 1 for some z E S and hence z = x. Thus z E U ( S ) ,a contradiction. (iii) (iv): 0bvious. (iv)+(i): Let 5 @ J(S)and let I = S - U ( S ) . If M is a maximal left ideal of S , then M C_ I # S , whence M = I = J ( S ) . Hence x is a unit and thus x J ( S ) is a unit of S / J ( S ) . Consequently, S / J ( S )is a division ring as required.

+

+

10.10. Corollary. Let S be a local ring. Then (i) The only idempotents o f S are 0 and 1. (ii) For any ideal I o f S , S / I is a local ring. Proof. (i) Let e # 1 be an idempotent of S. Then s is a nonunit and hence, by Lemma 10.9, e E J ( S ) . Thus, by Proposition 1.4.10, e = 0. (ii) Let J / I be a maximal left ideal of S / I . Then J is a maximal left ideal of S , hence J = J ( S ) by Lemma 10.9. Thus S / I has a unique maximal left ideal and, by Lemma 10.9, S / I is local. 1

We next provide necessary and sufficient conditions under which group algebras of finite groups are local rings. 10.11. Proposition. Let G # 1 be afinite group and let R be a commutative ring. Then RG is local if and only if R / J ( R ) is a field of prime characteristic p and G is a p-group. Proof. Assume that RG is local and put F = R / J ( R ) . Since both R and FG are homomorphic images of RG, they are local by Corollary lO.lO(ii). Now the augmentation ideal I ( G ) of FG is a maximal left ideal, hence I ( G ) = J ( F G ) is nilpotent. By Maschke’s theorem, we conclude that charF = p > 0 and pl /GI. If G has an element g of prime order q # p , then g - 1 E I < g > is not nilpotent by Maschke’s theorem. This is impossible since I < g >C_ I ( G ) , hence G is a p group. Conversely assume that G is a p-group and R/ J ( R ) is a field of chracteristic p . It suffices to show that R G / J ( R G ) R / J ( R ) . If

General properties of induced modules

118

J ( R ) = 0, then R is a field and I ( G ) has an R-basis { g - 111 # g E G } consisting of nilpotent elements. Hence I ( G ) is nilpotent by Theorem 1.5.12. But then J(RG) = I ( G ) and RG/J(RG)Z R, proving the case J ( R )= 0. Turning to the general case, we first note that J(R)G J ( R G ) by Corollary 4.3. Hence, by Corollary 1.4.4(ii), J ( R G / J ( R ) G )= J ( R G ) / J ( R ) G Since RG/J(R)G2 ( R / J ( R ) ) Git, follows from the special case above that RG/J(RG) R / J ( R ) ,as required. The following preliminary observation will allow us to apply the above result.

10.12. Lemma. Let S be a local ring. Then any finitely generated projective S-module is free. Proof. Let V be a finitely generated projective S-module, and let

o1,. . . ,o, be a minimal generating set for V . Assume that we are given

a relation

We claim that all such s; E J ( S ) . Indeed, if si 6 J ( S ) , then it is a unit in S , and we can solve for the corresponding oi as an S-linear sum of the vis. This, however, is impossible since otherwise the generator o; can be deleted. Thus all s; are in J ( S ) , as claimed. Let W be a free S-module freely generated by wl,. . . ,w, and map W onto V by sending w;to 0;. By the foregoing, the kernel of this map, say K ,is contained in J ( S ) W . By hypothesis, V is projective so that W = IC @ V’ with V’ V . Because K 2 W/V’ it must be finitely generated. Moreover, J ( S ) W = J ( S ) K $ J ( S ) V ’ and hence the inclusion I( C_ J ( S ) W forces I( = J ( S ) K . Thus I( = 0 by Nakayama’s lemma (Proposition 1.4.9) and W 2 V , as required. The following result is essentially due to Dickson (1907).

10. Induction and relative projectivity

119

10.13. Theorem. Let G be a finite group and let P be a Sylow p-subgroup of G. Assume that R is a local ring such that R/ J ( R ) is of characteristic p and that V is a projective RG-module which is R-free of finite rank n. Then [PI divides n.

Proof. By Theorem 10.8(i), Vp is a projective RP-module. Since Vp is finitely generated and, by Proposition 10.11, R P is local, it follows from Lemma 10.12, that Vp is free. Thus Vp is isomorphic to a direct sum of, say r , copies of R P . But each R P is R-free of rank lPl, hence n = IPIr, as required. We close by recording the following result. 10.14. Theorem. Let R be an arbitrary commutative ring and let N be a normal subgroup of G of finite index n. Then the following conditions are equivalent: (i) J ( R G ) = RG J ( R N ) . (ii) n is a unit of R. (iii) For any irreducible RN-module V , V G is completely reducible. *

Proof. The equivalence of (i) and (ii) follows from Theorems 3.9 and 4.2. (ii)=+(iii): Let T be a transversal for N in G. Then V G= etETt@V where, by Lemma 1.5(i),(iv), each t 8 V is an irreducible RN-module. Let Mr be an RG-submodule of V Gand let 0

+

w + V G+ V G / W+ 0

be the natural exact sequence. Then the sequence

0+

w, + ( V " ) ,

+(VG/W)N +0

splits, since ( V " ) , is a complete reducible RN-module. The desired assertion now follows by virtue of Corollary 10.4. (iii)=+(ii): Let I be any maximal ideal of R. It suffices to show that n = n - 1 I . To this end put V = R / I . Then the irreducible R-module V is also an irreducible RN-module under the trivial action of N . Furthermore, we obviously have

V G (R/I)(G/N)

General properties of induced modules

120

Since, by hypothesis, V Gis completely reducible, it follows that ( R / I ) ( G / N ) is semisimple. Thus, by Proposition 1.6.2, n $ I and the result is established. 11.

Unique decompositions

In this section, we shall collect a number of general results which will be used repeatedly in our subsequent investigations. Throughout, unless explicitly stated otherwise, R denotes an arbitrary ring.

11.1. Lemma. Let V be an indecomposable R-module which is both artinian and noetherian. Then every f E E n d R ( V ) is either a unit or nilpotent.

Proof. Let L, and X n be, respectively, the image and the kernel of f",n 2 1. We assert that

to a sufficient large n; if substantiated, it will follow that f is nilpotent, provided f is a nonunit, hence the result. Since V is both artinian and noetherian, there exists n 2 1 such that Ln = L2n and X n = X2n. Then, for any x E V , we may find y E V for which u(x)= u2(y),where u = f". Thus

Finally, given x E L, n X,, we have u(x)= 0 and x = u(y) for some y E V . Then y E X2,, = X,, whence x = u(y) = 0, proving (1). Let R be an arbitrary ring and let V # 0 be an R-module. We say that V is strongly indecomposable if EndR(V) is a local ring. 11.2. Lemma.

If V is strongly indecomposable, then V is inde-

composable. The converse is true if V is both artinian and noetherian. Proof. Assume that V is strongly indecomposable and that V = V' $ W is a direct decomposition of V . If .n : V + V' is the projection

121

11. Unique decompositions

map, then x is an idempotent of the local ring E n d R ( V ) , so x = 0 or T = 1, by Corollary lO.lO(i). Hence either V' = 0 or V' = V , proving that V is indecomposable. Conversely, assume that V is indecomposable and both artinian and noetherian. By Lemma 10.9, it is enough to show that every nonunit f E E = EndR(V) is in J ( E ) . Let g be an arbitrary element of E . Then gf is a nonunit and hence is nilpotent, by Lemma 11.1. Thus 1 - gf is a unit of E , so by Proposition 1.4.10, f E J ( E ) as we wished to show. W The following result is Azumaya's generalization of a classical theorem of Krull and Schmidt. 11.3. Theorem.

Let R be a ring, let V be an R-module and let

where each of the R-modules K, Wj is strongly indecomposable. Then m = n and, after possibly reordering the Wj, 2 W; for all i E { l , .. . , m } .

Proof. We argue by induction on min{m,n}. If m = 1 or n = 1, then the result is obvious since V is indecomposable. Let ei : V 3 K and fj : V + Wj be the projection maps, 1 5 i 5 m, 1 5 j 5 n. Note that el = elfjel and that, by hypothesis, E n d ~ ( 6is) local. Thus e l f j e l is a unit of EndR(V1)for some j . By renumbering the Wj, we may therefore assume that e l f l e l is an automorphism of Vl = elV. Thus flV1 C W1 and the kernel of f1 on V, is 0. Our next aim is to show that

@z2K.

To this end, suppose that II: E flVl n Then el2 = 0 and 5 = f i e l y for some y E V . Consequently, e l f l e l y = e1II: = 0 and therefore e1y = 0, since elfie1 is an automorphism on Vl. Hence II: = f l e l y = 0 , proving that flV, n = 0. Now fix x E V . Then e l z E elV = elflelV and hence e l x =

@z2K

General properties of induced modules

122

elflelw for some w E 5

- flelw E

W . It follows that el(z

-

f i e l w) = 0 and

Hence m

5

= flelw

+ (a: - flelw) E f1V + C V , i=2

proving (2). Since flV, C W1, it follows from (2) that

Wl = flvl @ {Wl n @Z1W But W1 is indecomposable and flV,

# 0, so Wl = fl&

S

Vl and

v = Wl @ ( @ 3 3 Consequently,

@Z2K= V/W1 = @jn=2Wj

and therefore, by induction, m = n and after possible reordering the W j , V , S W j f o r a l l i E { 1 , ...,m}. Let V be an R-module. We say that V has the unique decomposition property if the following two properties hold: (i) V is a finite direct sum of indecomposable modules; (ii) If V = = @YzlWj, where each K, Wj is indecomposable, W; for all then m = n and, after possibly reordering the Wj, i E (1,. . . , m } .

@clK

11.4. Corollary. Let R be a noetherian ring such that each indecomposable R-module is strongly indecomposable. Then any finitely generated nonzero R-module has the unique decomposition property.

Proof. Because R is noetherian, it follows from Corollary 1.3.7 that any finitely generated R-module V # 0 is noetherian. Hence, by Proposition 1.3.11, V is a finite direct sum of indecomposable modules. The desired conclusion is therefore a consequence of Theorem 11.3. 11.5. Corollary. (Krull - Schmidt theorem). Let V # 0 be an Rmodule which is both artinian and noetherian. Then V has the unique decomposition property.

12. Projective covers

123

Proof. By Proposition 1.3.11, V is a finite direct sum of indecomposable modules. Since these indecomposable modules are again artinian and noetherian, it follows from Lemma 11.2 that they are strongly indecomposable. Now apply Theorem 11.3.

11.6. Corollary. Let R be an artinian ring. Then any finitely generated R-module V # 0 has the unique decomposition property. Proof. Because R is artinian, the module V is both artinian and noetherian, by virtue of Corollary 1.4.15. The desired conclusion is therefore a consequence of Corollary 11.5. 12.

Projective covers

Our aim in this section is twofold: first to provide some general information pertaining to projective covers of modules and second to record a result concerning projective covers of induced modules. In what follows, R denotes an arbitrary ring and all modules are assumed to be finitely generated left modules. We know that each Rmodule is a homomorphic image of a projective module. For some modules V a stronger assertion is possible. A homomorphism

f:V-+W of R-modules is said to be essential if for every proper submodule V' of V , f(V') # f ( V ) . We say that an R-module P is a projective cover of V in case P is projective and there is an essential epimorphism

P-tV 12.1. Lemma. (i) A homomorphism f : V + W of R-modules is essential if and only i f K e r f is superfluous. (ii) If K , . . . , V, are superfluous submodules of V , then so is Vi *. t (iii) If P; is a projective cover of V;, then P I @ - .$P, is a projective cover of I/, @ , . . @ V, . (iv) A projective cover of an irreducible module is indecomposable.

v,.

+

General properties of induced modules

124

Proof. (i) Assume that f : V + W is an essential homomorphism. If W is a submodule of V such that V = W K e r f , then f(V) = f(W) and so W cannot be a proper submodule of V . This shows that V = W and hence Kerf is a superfluous submodule of V . Conversely, assume that K e r f is superfluous. If V' is a submodule of V with f(V) = f (V'),then V = V' K e r f . Since K e r f is superfluous, we must have V = V' and therefore f is essential. (ii) Suppose that W is a submodule of V such that

+

+

+ +- + + +- + +

Then V1 (V2 - . V, W ) = V and, since & is superfluous, we - * V, W = V . Now apply induction on n. have V2 (iii) If fi : P; -+ is an essential epimorphism, then

is an epimorphism whose kernel @y=llcerfiis superfluous, by virtue of (ii). Hence, by (i), @ fi is essential, which implies that @Pi is a projective cover of @K. (iv) Suppose that $ : P + V is an essential epimorphism, where V is an irreducible R-module. If P = P' @ P", for some proper submodules P' and P", then $(PI) = $(P") = 0 since 0 is the only proper submodule of V . But then V = $(PI) $(PI') = 0, a contradiction. Hence P is indecomposable, as asserted. W

-+

One of the consequences of the following lemma is that if a module does have a projective cover, then it is unique up to isomorphism.

12.2. Lemma. Assume that an R-module V has a projective cover P and let M be an R-module such that V is a homomorphic image of M . (i) If M is projective, then P is isomorphic to a direct summand of

M. (ii) If M has a projective cover L , then P is isomorphic to a direct summand of L. (iii) If M is another projective cover of V , then it4 % P .

12. Projective covers

125

Proof. (i) Fix an epimorphism cp : M -+ V and an essential epimorphism $ : P + V. Since M is projective, there exists a homomorphism 8 : M + P such that cp = $ 0 8. Because $ is an essential epimorphism and $08 is an epimorphism, 8 is also an epimorphism. But P is projective, hence there is a splitting homomorphism f : P + M and so M = Irn f @ Ker8. Since f is injective, (i) is established. (ii) Assume that L is a projective cover of M and let f : L + M be an epimorphism. Then cpf : L + V is an epimorphism, hence by (i) P is isomorphic to a direct summand of L. (iii) Suppose that M is a projective cover of V and choose 'p in (i) to be essential. It will be demonstrated that 0 is also essential, from which will follow that M = Imf 2 P. Assume that M' is a proper submodule of M and put P' = 6(M'). Then

and so P'

# P.

Hence 8 is essential, as required. 1

From now on, we write P ( V ) for a projective cover of V . Owing to Lemma 12.2(iii), if P ( V ) exists, then it is unique up to isomorphism. 12.3. Theorem. Let R be an artinian ring and let V be an Rmodule. (i) V has a projective cover. (ii) A projective R-module W is a projective cover of V if and only

i f W / J ( R ) Ws V / J ( R ) V . In particular, (a) IfV is completely reducible, then V 2 P ( V ) / J ( R ) P ( V ) . (b) I f U and V are completely reducible R-modules, then U E V if and only i f P ( U )C P ( V ) . (iii) If V is projective and i f V / J ( R ) V= @L1K is a decomposition as a direct sum of irreducible R-modules, then V Z @Y."=,(K)is a decomposition as a direct sum of indecomposable R-modules. (iv) I f V and W are projective R-modules, then V S W i f and only

if V / J ( R ) V2 W / J ( R ) W Proof. (i) Let L be a projective module such that V = L / S for some submodule S of L. For any submodule W of S , let fw : L / W + L / S be

126

General properties of induced modules

the canonical homomorphism. Next choose W to be minimal in S such that fw is essential; such a submodule exists, since fs is essential and S is artinian. We claim that W is a direct summand of L ; if sustained, it will follow that L/W is a projective module and hence that L / W is a projective cover of V . Let W’ be a submodule of L minimai among those whose projection cp : W’ -+ L/W is surjective, and let n : L + L/W be the projection map. Since L is projective, there exists a homomorphism n’: L -+ W’ with ?r = cp o n’,and the minimality of W’ forces ?r’(L)= W’. Denote by L‘ the kernel of n’.The projection f~ factors onto

L/L’+ L/W

-+

L/S

and the two factors are essential. Because L’ is contained in W , the minimality of W implies that W = L’, i.e. that Q is an isomorphism. It follows that L = W @ W’, as required. (ii) Let W be a projective module, let f : W + V be an essential epimorphism and let 71 : V + V / J ( R ) Vbe the projection. Since n is essential, so is nf and thus Kernf E J(R)W (Lemma 12.l(i) and Proposition 1.4.5). On the other hand

and therefore J(R)W

K e r n f . Thus Kernf = J(R)W and so

W / J ( R ) W2 V / J ( R ) V Conversely, assume that W is a projective R-module such that the above isomorphism holds. Let f : V + V / J ( R ) V be the projection map and let g : W + V / J ( R ) Vbe the homomorphism induced by the isomorphism W / J ( R ) WE V / J ( R ) V . Since f is an essential epimorphism and since W is projective, there is an epimorphism g’ : W -+ V with fg‘ = g . Hence

Kerg’ C Kerg = J(R)W is a superfluous submodule of W . Thus W is a projective cover of V , as required.

12. Projective covers

127

(iii) By Propositions 1.4.5 and 1.4.8, J(R)V is a superfluous submodule of V . Since V is projective, it follows that P ( V / J ( R ) V )= V . Invoking Lemma 12.l(iii) and (iv), we deduce that

where each P( E) is indecomposable. (iv) If V E W , then obviously V / J ( R ) VZ W/J(R)W. Conversely, assume that V and W are projective R-modules such that V / J ( R ) VE W / J ( R ) W .By the Krull-Schmidt theorem, we may write

and

W / J ( R ) W= w163 . . * @ W, where each we have

rZ:, Wi is irreducible and

2 Wi, 1 5

i 5 n. Then, by (iii),

as required. H

Assume that R is an artinian ring and let n 2 1 be such that

where the U; are principle indecomposable R-modules. Owing to the Krull-Schmidt theorem, the Ui are uniquely determined up to isomorphism and the order in which they appear. Furthermore, the above decomposition determines a complete set {el,. . . ,e,} of primitive idempotents in R such that Vi = Re;. Except when J ( R ) = 0, the principal indecomposable R-modules form only a small subclass of all indecomposable R-modules. However, it is the class which plays a very important role due to the following result.

Let R be an artinian ring. (i) The following conditions are equivalent: (a) V is a projective cover of an irreducible R-module.

12.4. Theorem.

General properties of induced modules

128

(b) V is a projective indecomposable R-module. (c) V is a principal indecomposable R-module. (ii) If Rel, , ,. ,Re, are all nonisomorphic principal indecomposable R-modules, then R e l / J ( R ) e l , .. .,Re,/J(R)e, are all nonisomorphic irreducible R-modules. Proof. (i) The implications (a)+(b) and (c)+(a) are consequences of Lemma 12.l(iv) and Theorem 12.3(iii), respectively. Now assume that (b) holds. Since V is projective, there exists n 2 1 such that R" = V @I V', where R" denotes a direct sum of n copies of R. Now the indecomposable components of R" are principal indecomposable Rmodules and V is indecomposable. Hence (c) follows by virtue of the Krull-Schmidt theorem. (ii) Invoking (i), we may find irreducible R-modules Vi, . . . ,V m such that Rei = P ( x ) , in which case K E Rei/J(R)ei by Theorem 12.3(ii). Re, and so i = j . If V is any irreducible RIf then Rei module, then by (i), P ( V ) E P ( K ) for some i E (1,. . . ,n } . Thus, by Theorem 12.3(ii), V Z V , as required.

4,

We close by proving the following result.

12.5. Proposition. Let H be a subgroup of a finite group G, let F be a field and let V and W be FH and FG-modules, respectively. Then (i) P(VG)is isomorphic to a direct summand of P ( V ) G . (ii) P ( W H )is isomorphic to a direct summand of P ( W ) H . (iii) If H is normal, then for all g E G , P(") = " ( V ) . In particular, V is G-invariant if and only if P ( V ) is G-invariant.

Proof. (i) By the definition of P ( V ) ,there is an FH-epimorphism P ( V ) --+ V . Hence, by Theorem 2.9(i), there is an FG-epimorphism P(V)" V G . By Theorem 10.8@), P(V)Gis projective. Thus, by Lemma 12.2(i), P ( V G )is isomorphic to a direct summand of P(V)". (ii) Since there is a homomorphism of P ( W ) onto W and hence of P ( W ) H onto W H ,P(WH)is isomorphic to a direct summand of P ( P ( W ) H ) .Because P ( W ) is projective, so is P(W)Hby Theorem 10,8(i). Hence, by Lemma 12.2(i), P ( W H )is isomorphic to a direct

12. Projective covers

129

summand of P ( W ) H . (iii) Let cp : P ( V ) + V be an essential epimorphism. Then cp can also be regarded as an epimorphism " ( V ) + gV. Because W is a submodule of P ( V ) if and only if it is a submodule of " ( V ) , the latter epimorphism is essential. It therefore suffices to prove that if V is projective, then so is gV. So assume that V is projective and write FH @I - @I FH = V @ W for some FH-module W . Then

--

and, since obviously

g(

F H ) S F H , the result follows.

This Page Intentionally Left Blank

131

Chapter 3 Induction from normal subgroups In this chapter we provide a detailed information on the process of induction from normal subgroups. Our first major result (Theorem 2.1) reduces the study of V G to the case where V is G-invariant. Under this assumotion we then show that E ~ ~ R G is ( Va crossed ~) product of GIN over E n d ~ ~ ( vThis ) . will allow us to discover circumstances under which V G is totally indecomposable or absolutely indecomposable. As a consequence of our main results, we derive a classical theorem known as the Green Indecomposability Theorem. Special attention is drawn to the study of crossed products over prime rings. The information obtained is then applied to provide a criterion for V G to be homogeneous. Our next major result asserts that if G is a finite group and S a finite-dimensional semisimple algebra over a field F , then any F-algebra which is a crossed product of G over S is symmetric. As an application, it is shown that if V is a completely reducible FN-module ( N is a normal subgroup of G), then the F-algebra E n d r ; . ~ ( Vis~ ) symmetric. This fact will allow us to derive a number of important facts concerning induction from irreducible modules and their projective covers. The chapter ends with the discussion of the Loewy length of induced modules.

Induction from normal subgroups

132

1.

Complete noetherian local rings

Throughout this section, R denotes a commutative ring. We begin by recording the following piece of information. A metric space ( M , p ) is a set M together with a map

p:MxM-+IR such that for all z, y, z E M , (9 P ( Z , Y) = P(Y, 4 2 0. (ii) p ( z , y ) = 0 if and only if x = y. 6;;) P ( 5 , Z ) I P b , Y ) P ( Y , Z ) . A Cuuchy sequence in M is a sequence {z;} of elements of M such that for any e > 0, there exists a positive integer k such that

+

P(2",4

<E

m,n > k

for all

A sequence {xi} is said to converge to provided

lim;,,p(z,z;)

3:

E M , written limxi =

z,

=0

Assume that I is an ideal of a ring S such that

Then, for any distinct z,y E S, there exists a unique m 5 -y

1 0 such that

E I" - Im+l

where by convension Io = S. We define PI : S x S 3 IR by

It is then easily verified that (S,pl) is a metric space. We refer to S as being complete in the I-udic topology if each Cauchy sequence in (S,P I ) converges. Let { s i } be a sequence in S. The following properties are direct consequences of the definitions: (i) {s;} is Cauchy if and only if for any n such that i,j > m implies s; - sj E I".

> 0, there exists m > 0,

1. Complete noetherian local rings

133

(ii) limsi = s if and only if for any n > 0, there exists m > 0 such that i > m implies s; - s E I". (iii) If limsi = s and lims: = s', then lirn(si si) = s s' and

+-

+

lim(sis:) = ss'. 1.1. Lemma. Assume that R is a noetherian ring and that A is an R-algebra which is finitely generated as R-module. If I is an ideal of R such that I J ( A ) , then fl;=,I"A = 0. In particular, if R is a noetherian local ring with maximal ideal P = J ( R), then n =:, P" = 0.

Proof. Let L be an arbitrary ideal of R and let V be a finitely generated R-module. It is well known (see Bourbaki (1961)) that L W = W for W = n;=,L"V. Applying this fact for V = A , L = I and W = nr=,I"A, we conclude that IW = W and hence that J(A)W = W . Because W is a finitely generated A-module, it follows from Nakayama's lemma that W = 0. So the lemma is true. 1.2. Lemma. Let A be an R-algebra which is finitely generated as R-module. Then J(R)A C J ( A ) .

Proof. Let V be any irreducible A-module. Then V = Av for each nonzero v in V , hence V is a finitely generated R-module. Now J ( R)V is an A-submodule of V , hence is 0 or V . But J ( R ) V # V , otherwise by Nakayama's lemma we find that V = 0. Hence J ( R ) V = 0 and so J ( R ) annihilates every such V . Thus the same is true for J ( R ) A and so J ( R ) A J ( A ) .

c

1.3. Lemma. Let R be a local ring with maximal ideal P and residue class field F = R / P . Suppose that A is an R-algebra which is finitely generated as R-module and put A = A / P A . Then (i) J ( A ) = J ( A ) / P A and A / J ( A ) S A / J ( A ) as F-algebras. (ii) For any finite group G , R G / J ( R G ) F G / J ( F G ) .

c

Proof. (i) By Lemma 1.2, P A J ( A ) and so, by Corollary 1.4.4(ii), J ( A ) = J ( A ) / P A . The latter obviously implies the second assertion.

Induction from normal subgroups

134

(ii) Put A = RG and note that

A = R G / P ( R G ) ( R / P ) G= FG Now apply (i). W

Assume that R is a local noetherian ring with maximal ideal P . Owing to Lemma 1.1, n : y n = 0. We say that R is complete if R is complete in the P-adic topology. 1.4. Lemma. Let R be a complete noetherian local ring with maximal ideal P , and let A be an R-algebra which is finitely generated as R-module. Then, for any ideal J C J ( A ) of A , A is complete in the J-adic topology.

Proof. By Lemma 1.2, P A C J ( A ) ,while by Lemma 1.3(i), J" C_ J(A)" P A for some n 2 1. Thus we may assume that J = PA. Note that, by Lemma 1.1, the intersection of all (PA)" = P"A is zero. Write A = Ro1 . . . Ro,, the o; E A, and let { w ; } be a Cauchy sequence in A. Given an integer n > 0, denote by m(n) the largest integer such that for all i , j 2 n wi - wj E P"(")A

+ +

(if no such m ( n ) exists, set m(n) = n ) . Bearing in mind that {wi} is Cauchy, we have limn-,mm(n)= 00. Let

and write

We then have

135

1. Complete noetherian local rings

where

and t=n

so for any t , {bnt} is a Cauchy sequence in R. Since R is complete, limb,, = b, for some b, E R. Setting w = C;=l btvt, it follows that

thus completing the proof.

We have now accumulated all the information necessary to prove the following result. In what follows, we use the standard fact that if el and e2 are idempotents of a ring S , then Sel Z Se2 as left S-modules if and only if e l S 2 e2S as right S-modules (see Jacobson (1956)). 1.5. Theorem. Let R be a complete noetherian local ring, let A be an R-algebra which is finitely generated as R-module and let J C J ( A ) be an ideal of A . For each x E A, let 5 be the image o f x in A = A / J . (i) Each idempotent E E A can be lifted to an idempotent e E A, that is e = E . Moreover, e is primitive if and only if is primitive. (ii) If e l , e2 are idempotents of A, then Ael E Aez as left A-modules if and only if Ael 2 A Z 2 as left A-modules (equivalently, e l A e 2 A as right A-modules if and only i f e l A E e2A as right A-modules). (iii) If I = E~ e2 t.- . en is a decomposition of I into orthogonal idempotents in A, then there exists orthogonal idempotents e l , . . . ,en E A such that 1 = el . . e, and e; = ci

-+

+

+ +

Proof. (i) Given an idempotent e E A, choose x1 E A with 51 = E . Then nl = x: - x1 E J . Once x ; and n; are chosen, put x;+1 = x ;

+ n; - 2x;n;,

n;+l =

2

- xi+l

(1)

Induction from normal subgroups

136

Then n; E J2iand thus {xi} is a Cauchy sequence. By Lemma 1.4, A is complete in the J-adic topology, so limx; = e for some e E A. Then e2 - e = limn; = o and, moreover, e = 51 = c , so e is a required idempotent of A. Let e be an idempotent in A and let E = €1 6 2 , where €1 and ~2 are orthogonal idempotents in A. It will next be shown that there exist orthogonal idempotents el, e2 in A with

+

To this end, choose a E A such that si = €1 and put z1 = eae. Then 51 = E s i E = €1 so we may consider the Cauchy sequence {z;} given by (1). Since z1 E eAe, each x; E eAe and thus ex; = x; = x;e for all i. Accordingly, setting e; = Zimzi, we deduce that el is an idempotent of A such that E l = €1 and (by taking limits) eel = el = ele. Letting 2 = e2 # 0, 22 = €2: and e1e2 = e2e1 = 0, e2 = e - el, we now have e2 proving (2). If E is not primitive, then so is e, by virtue of (2). The converse being obvious, it follows that e is primitive if and only if so is 2. (ii) Assume that e1,e2 are idempotents in A. Any isomorphism Ael 2 Ae2 of left A-modules carries J e l onto Je2 and induces an isomorphism A E 1 A22 of left A-modules. Conversely, let f : /IE1 --+ Ac!~ be an isomorphism of left A-modules and let 3 = f-'. Because A = Aei @ A ( l - e,), each Ae; is a projective A-module. Thus we can find A-homomorphisms f : Ael -+ Ae2 and g : Ae2 -+ Ael lifting f and g,'respectively. We claim that 6 = gf is an automorphism of Ael,; if sustained, it will follow by symmetry, that fg is an automorphism of Ae2 and hence that f is an isomorphism. therefore (8 - 1)Ael C PAel. Setting It is obvious that 4 lifts p = 1 - 8, we have PkAel C Jkel. Thus 1 p t p2 t . . . is a well defined endomorphism of Ael, and is a two-sided inverse of 1 - p. Because 1 - p = 8, this demonstrates that 8 is an A-automorphism of Ael, as desired. (iii) The case n = 1 being obvious, we argue by induction on n. So assume that n > 1 and that the result holds for n-1. Put 6 = en-l +E,,

sf,

+

1. Complete noetherian local rings

so that 1 = ~1

137

+. . . + en-2 + 6 is an orthogonal decomposition. By the

induction hypothesis, we may find an orthogonal decomposition:

- 1 it follows that Applying (2) to the decomposition ?t = 6 = ~ ~ +c,, e= en for some orthogonal idempotents en-l, en in A such that En-1 = En-1 and = c,. Thus 1 = el+- - .+en is the desired orthogonal decomposition.

+

As an easy consequence of the above result, we derive 1.6. Corollary. Let R be a complete noetherian local ring and let A be an R-algebra which is finitely generated as R-module. Then A is local i f and only if 1 is the only nonzero idempotent of A .

Proof. If A is local, then 0 and 1 are the only idempotents of A , by Corollary 2.10.10(i). Conversely, suppose that 1 is the only nonzero idempotent in A . Let P be the maximal ideal of R and put A = A / P A . By Lemma 1.3(i), A / J ( A ) A / J ( A ) and so it suffices to show that A is local. By Theorem 1.5, I is the only nonzero idempotent of A. Since A is a finite-dimensional algebra over a field, the result follows.

The next lemma will enable us to take full advantage of the results so far obtained.

1.7. Lemma. Let R be a noetherian ring and let V be a finitely generated R-module. Then the R-algebra E n d R ( V ) is a finitely generated R-module.

+ +

Proof. We may write V = Rvl Rv, for some vi E V . Let W be a free R-module with basis wl,. . . ,w, let 'p : W 3 V be the R-homomorphism such that 'p(w;) = v,, and put IC = Ker'p. Since E n d R ( W ) Z M,(R), E ~ ~ R (isWa )finitely generated R-module. Put Eo = {f E E n d R ( W ) J f ( K ) I(}. I f f E Eo, then f induces ) an S E E n d ~ ( vby -

f(.

+ IC) = f(.) + K

a

(z E

W)

Induction from normal subgroups

138

For any given g E E n d ~ ( v write ),

and define g E E n d ~ ( wby )

j=1

If Cj bjwj E I(, then Cj bjvj = 0 and

so

proving that g E Eo. It follows that g H g is a surjective R-homomorphism Eo + EndR(V). Now EndR(W) is a finitely generated R-module and R is noetherian. Hence Eo and EndR(V) are finitely generated Rmodules, as required. W 1.8. Corollary. Let R be a complete noetherian local ring and let A be an R-algebra which is finitely generated as R-module. Then every nonzero finitely generated A-module has the unique decomposition property.

Proof. By Corollary 1.3.8, A is noetherian. Hence, by Corollary 2.11.4, it suffices to verify that any given finitely generated indecomposable A-module V is strongly indecomposable. Because V is inde. to composable, 0 and 1 are the only idempotents of E n d ~ ( v )Owing Lemma 1.7, E n d ~ ( v(and ) hence E n c l ~ ( V )is) a finitely generated Rmodule. The desired conclusion now follows by appealing to Corollary 1.6.

2. Reduction to G-invariant modules

2.

139

Reduction to G-invariant modules

Throughout this section, N denotes a normal subgroup of a finite group G. All modules are assumed to be finitely generated over their ground rings. It will be also assumed that R is a commutative ring of one of the following types: (i) R is artinian. (ii) R is a complete noetherian local ring. These assumptions are made to guarantee that if A is an R-algebra which is finitely generated as R-module, then any nonzero A-module has the unique decomposition property (see Corollaries 2.11.6 and 1.8). Let V be an indecomposable RN-module and let H be the inertia group of V. Our aim is to show that the decomposition of V G is determined uniquely by that of V H .This will allow us to concentrate on the case where V is G-invariant, i.e. gV E V for all g E G. 2.1. Theorem. (Ward (1968) and Willems (1976)). Let V be an indecomposable RN-module, let H be the inertia group o f V and write

where the K are the indecomposable RH-modules. Then (i) V G = KG @ - - .03 Vf, where the yG are indecomposable RGmodules such that qGE yGimplies 2 4. (ii) If, for some i E (1,. . . ,s } , is irreducible, then KG and V are irreducible.

.@KG

Proof. (i) The equality V G= Kc@. . follows from the facts that induction is transitive (Lemma 2.2.1) and preserves direct sums (Lemma 2.2.8). To show that each yGis indecomposable, suppose that

XG=X@Y is is a direct decomposition. We know, from Corollary 2.1.2, that identifiable with a direct summand of ( y G ) ~Hence, . by the KrullSchmidt theorem, we may assume that I$ is a direct summand of X H ,

say

X*=I/:@X'

(2)

Induction from normal subgroups

140

Observe also that ( V H )is~the direct sum of isomorphic RN-modules of the form h 8 V with h E H . Invoking the Krull-Schmidt theorem, we deduce from (1) that

for some positive integer m;. Accordingly, for any g E G, we obtain RN-isomorphisms (9c3 V t ) N

9 €9

(K)N 2

m;(g €9 V )

which in turn implies

(yG)N

= ( @ t c T t €9 v i ) N 2 mi(t €9 V )

(4)

where T is a left transversal for H in G. Restricting (2) to N, it follows from (3) that X N % mzv @ X & But X is an RG-module, so

proving that for each t E T , t €9 V appears as a direct summand of X N with multiplicity at least mi. Bearing in mind that for distinct t l , t 2 E T , tl 8 V and t 2 8 V are nonisomorphic RN-modules, we de~ isomorphic to a direct summand of XN. duce from (4) that ( q ' ) is But X N is a direct summand of (V;.')N, hence by the Krull-Schmidt theorem, X = KG. This shows that KG is indecomposable. Assume by way of contradiction that KG S '3/1 but V;: V,. Because is identifiable with a direct summand of (K')H, we may write

(KG)* = V , 63 q'

and

( y G =) vj~ @ q'

where K' and are RH-modules. Invoking the Krull-Schmid theorem, we see that K is isomorphic to an RH-direct summand of 5'. It follows that ( K ) N is isomorphic to an RN-direct summand of Applying (3)) we therefore deduce that V is isomorphic to an RN-direct summand of ( 5 ' )and ~ that

(y)~.

(FG)N = ( Y ) N @ ( 6 ' ) N 2 mjV 63 ( 6 ' ) N

3. Group-graded algebras and crossed products

141

However, by (4), the multiplicity of V as an indecomposable direct summand of ( y G is ) ~exactly mj. This contradiction proves the required assertion. (ii) Suppose that there exists an i E (1,. . . ,s} such that r/: is irreducible. Then by Clifford's theorem (Theorem 2.4.2), ( K ) N is completely reducible. Thus, by (3), V is irreducible. Let W be an irre(such W exists, by virtue of Proposition ducible factor module of 1.3.4). Then, by Corollary 1.2.5, V , is isomorphic to an RH-submodule of W,. Consequently, is isomorphic to an RN-submodule of W,. Therefore, by (3), W N contains a submodule X Z m;V. Thus, for any g E G, WN contains the RN-submodule gX Z m;(g@I V ) . Let T be a left transversal for H in G. Then the RN-module t @I V , t E T , are irreducible and nonisomorphic and they appear as direct summands of WN with multiplicity at least mi. On the other hand, by (4), the multiplicity of each t @I V as a direct summand of ( y G ) N is exactly m;. Now W = v G / X for some submodule X of V,' and WN 2 ( K G ) ~ / XThus ~.

KG

(v)~

and, by the foregoing, X = 0. The conclusion is that W = irreducible, as desired. H

YG is

2.2. Corollary. Let V be an irreducible RN-module, let H be the inertia group of V and let ( H : N ) be a unit of R. Then V G is comple t ely reducible .

Proof. Invoking Theorem 2.1, we may harmlessly assume that H = G. Now apply Theorem 2.10.14. H 3. Group-graded algebras and crossed products

This section contains a number of general results which will enable us to scrutinize the structure of the endomorphism ring of induced modules. Our exposition is strongly influenced by a work of Dade (1970a). Throughout, A denotes an algebra over a commutative ring R. If X , Y are R-submodules of A , then X Y denotes the R-submodule

Induction from normal subgroups

142

of A consisting of all finite sums (2;E

x,y;E Y )

Let G be a multiplicative group. Then A is called a G-graded algebra if there is a family { A h E GI of R-submodules of A indexed by the elements of G such that the following conditions hold: (direct sum of R-modules). (a) A = @gEGAg for all x , y E G. (b) A A , A,, We shall refer to (a) as a G-grading of A and to A, as the g-component of A. When (b) is replaced by the stronger condition, namely (4 A A y = A,, for all x,y E G we say that A is a strongly G-graded algebra. From now on, A denotes a G-graded algebra over R and U ( A )denotes the unit group of A.

c

3.1. Lemma. (i) A1 is a subalgebra of A with 1 E Al. (ii) For each g E G, A, is an ( A l ,Al)-bimodule under left and right multiplication b y the elements of Al. (iii) A is strongly G-graded if and only if 1 E A,A,-l for all g E G.

Proof. (i) By definition, Al is an R-submodule of A and, by (b), Al is multiplicatively closed. It therefore suffices to show that 1 E Al. Invoking (a), we may write

l=Ca, SEG

where a, E A, for all g E G, and all but finitely many of a, are zero. Now fix some h 6 G and aL E A h . Because G is a group, (a) implies that A = @g€GAgh (1) Owing to (b), the product agai lies in Agh for all g E G. Therefore, aI h = l . a h = I

c

,€G

a g 4

3. Group-graded algebras and crossed products

143

is precisely the expansion of a; in the decomposition (1). But ah already lies in the summand A1.h = A h in (1). Hence all the agai for g # 1 must be zero and ala: must be u i . This proves that a1 acts as a left identity on A h for all h E G. In view of (a), we deduce that al is a left identity for the algebra A. Thus al E Al is the identity element of A, as desired. (ii) This is a direct sequence of (b). (iii) Suppose that A is strongly G-graded. Because 1 E Al, it follows from (a)that 1 E AgAg-1 for all g E G. Conversely, assume that 1 E AgAg-l for all g E G. Invoking (b), we then have = 1 - A,, = AxA,-~Ax,

A,,

A x A x - ~ z= y A,A, for all x , y E G. Hence A,A, = A,, for all z,y E G, as asserted. W

We refer to a unit u of A as being graded if u E A, for some g E G. In this case, we say that g is the degree of u and write g = deg(u) = degA(u) The set of all graded units of A will be denoted by GrU(A). 3.2. Lemma. (2) I f u E G T U ( A )is of degree g , then u-l is of degree 9 - l . (ii) GrU(A) is a subgroup o f U ( A ) and the map deg : GrU(A)-+ G

is a group homomorphism with kernel U(A1). (iii) The map G r U ( A ) + Aut(A1) u H i,

{

where

i&)

= uzu-l

for all

x E A1

Induction from normal subgroups

144

is a homomorphism. (iv) Right multiplication b y any u, E A,

n U ( A ) is an isomorphism

of left Al-modules.

Proof. (i) We first write u-l = CxEGu,with a, E A, and with A,A, A,, for all x E G, it finitely many nonzero a,. Since ua, follows that uu-l = C I E G u a ,is a unique expansion for ua-l in the decomposition

c

A = fBxEGAgz = @XEGAX Now, by Lemma 3.1(i), uu-' = 1 lies in A1 = A,A,-1. Hence the az for # 9-l must be zero, and 1 must equal uag-l. Therefore u-' = a,-' lies in A,-], as asserted. (ii) That GrU(A)is a subgroup of U ( A )follows from ( i ) and the fact that A,A, 5 A,, for all x,y E G. Since A,A, A x y ,the given map is a homomorphism with kernel Al n U ( A ) . Owing to (i), A1 n U ( A ) U(Al) and since the opposite inclusion is trivial, the required assertion is established. (iii) By hypothesis, u E A, for some g E G. Thus, by (i) we have

x

uA1u-l 5 A,AlA,-i

A1

Bearing in mind that GrU(A)is a group, (iii) follows. (iv) Since u, is a unit of A , right multiplication by ug is an Alisomorphism of Al onto Alu,. Because u, e A,, we have Alu, C A,. Owing to (i), we also have

A, = A,u,'u,

C AgAg-lUgC Aiug

The conclusion is that A, = Alu, and the result follows. W

We next introduce a very important class of group-graded algebras, namely crossed products. Owing to Lemma 3.2(ii), the sequence of group homomorphisms

1

+ U(A1)+.

GrU(A)%' G + 1

(2)

3. Group-graded algebras and crossed products

145

is always exact, except possibly at G. We say that a G-graded algebra A is a crossed product of G over A1, written A = A1 * G, provided the sequence (2) is exact. Thus A is a crossed product of G over A1 if and only if for any g E G, there exists 3 E A, n U(A). In the special case, where (2) is an exact splitting sequence, we shall refer to A as a skew group ring of G over Al. Thus A is a skew group ring of G over A1 if and only if for each g E G, there exists E A,nU(A) such that zy = zy for all z,y E G

A G-graded algebra A is called a twisted group ring of G over A1 if for all g E G, there exists g E A, n U(A) such that g centralizes Al. In a particular case, where Al Z(A) and A, n U ( A ) # 8 for all g E G, we shall refer to A as a twisted group algebra of G over Al. Finally, if Al * G is a twisted group algebra and a skew group ring, A1 * G is notihng else but the group algebra of G over Al. Our next aim is to provide a condition under which a strongly Ggraded algebra A is a crossed product of G over Al. Let A be a strongly G-graded algebra whose subalgebra Al is a local ring. Then A is a crossed product of G over A1 . 3.3. Proposition.

Proof. Since A,A, = A,, for all

J(A1) C A1 = A,A,-1 Thus we may find a, E A, and

2,y

E G, we have

for all g E G

E A,-I for which

Because A1 is a local ring, we have U(Al) = Al - J(Al). Hence

is a right inverse to a, in A. The idempotent a, lies in A,-] A, Because 0 # 1 = l2 = (u,u,.-,)2 = ug(ag-lug)ag-l tI

C Al.

Induction from normal subgroups

146

we have a;-lag # 0. Therefore a$)-lag= 1 and thus a;-, is a two sided inverse to a, in A. We conclude that

A,

n U ( A )# 8

for all g E G

which shows that A is a crossed product of G over A l . W We now proceed to describe crossed products more explicitly. Let A be a crossed product of G over Al. A map Q

: G -+

Aut(A1)

is called an automorphism system for A if, for all g E G, there exists a unit g of A in A, with I = 1 such that

for all x E Al

a ( g ) ( s ) = gxg-'

(3)

Now suppose that B is an R-algebra. Given maps Q

: GtAut(B)

(Y

:

GxG+U(B)

we say that (G,B , Q, a ) is a crossed system for G in B if, for all 2,y , z E G, and b E B, the following properties hold:

where ,b = a ( g ) ( b )for all g E G, b E B.

Let a G-graded R-algebra A be a crossed product of G over A l . For a n y g E G, fix a unit g of A in Ag with I = 1, let n : G + Aut(A,), be the corresponding automorphism system given b y (3) and let (Y : G x G U(A1) 3.4. Proposition.

--j

be defined b y +,Y)

= ZY"Y

(7)

3. Group-graded algebras and crossed products

147

Then the following properties hold: (i) A is a strongly G-graded algebra with A, = A1ij = jA1. (iz) ( G ,A1,o, a ) is a crossed system for G over A1 (to which we refer as corresponding to A) (iii) A is a free (left and right) Al-module freely generated b y the elements g , g E G (iv) For all x,y E G and r1, 7-2 E Al,

Conversely, for any R-algebra B and a n y crossed system ( G ,B , 0,a ) for G over B , the free B-module C freely generated b y the eIements g , g E G with multiplication given b y (8) (with r1,r2 E B ) is a Ggraded R-algebra (with C, = Bg for all g E G) which is a crossed product of G over C1 = B and having (G,B, 0,a ) as a corresponding crossed system.

Proof. Assume that u E A, n U ( A ) for some g E G. Owing to Lemma 3.2(i), u-' E A,-] n U ( A ) and hence 1 = uu-l E A,A,-I. Invoking Lemma 3.1(iii), we conclude that A is a strongly G-graded algebra. By Lemma 3.2(iv), A, = A1g and the argument of that lemma applied to left multiplication shows that A, = gA1, proving (i). Let x,y E G and b E Al. Then

"( Yb) = z(yby-')Z-'

= a ( x ,y)ZybZy-lcu(x, y)-' = ~ ( zy ,) " y b a ( x ,y ) - 1 ,

proving (4). For any x , y , z E G, we also have

(.y). -

.(Y.)

__

= a ( x ,y ) z y z = a ( x ,y ) a ( z y ,2)= q " ( y , z ) y z = "cu(y,z)ZyZ= Z a ( y , Z ) a ( x , y Z ) q z ,

proving ( 5 ) . Since = 1, (6) is also true, proving (ii). Property (iii) is a direct consequence of (i). To prove (iv), fix x,y E G and r l , r2 E Al. Then ( r 1 z ) ( r 2 y= ) rl(%r2ii-')(.y) = r1 " r 2 a ( x ,y)-,

Induction from normal subgroups

148

proving (8). To prove the converse, it suffices to show that the multiplication given by (8) is associative. To this end, fix x , y , z E G and rl,r2,r3 E B. Then

( r i z )[(r2Y)(r@)]= ri "[r2'r3a(Y,z ) ] a ( zY, z ) m = rl "r2"( Yr3)" a ( y , z ) a ( z ,y z ) ~ (bY(4)) = rl I r z a ( ~ , yzYr3a(x,y)-1 ) =a(y,z ) a ( z ,y z ) z y ~ (by ( 5 ) ) = r1 "r2a(z,y) "Yr3a(z,y)-la(z,y)a(zy, z ) z y ~ =

[ ( T i2) (

f a Y)] (r3z).

as required.

Returning to our study of G-graded algebras, we now introduce the following important notion. Let A be a G-graded R-algebra. An ideal I of A is called a graded ideal if

I = @ g E G ( I n Ag) Assume that B is another G-graded R-algebra. A homomorphism

f:A+B of R-algebras is said to be graded if f ( A g )C B, for all g E G.

3.5. Lemma. (i) If I is a graded ideal of A, then A / I is a G graded R-algebra by setting ( A / I ) , = ( A , I ) / I for all g E G. (ii) I f f : A -+ B is a graded homomorphism of G-graded Ralgebras, then Kerf is a graded ideal. Moreover, if f is surjective, then there is a graded isomorphism A / K e r f --+ B . (iii) An ideal I of A is graded i f and only i f I is the kernel of a graded homomorphism A -+ B of G-graded R-algebras.

+

3. Group-graded algebras and crossed products

149

as asserted. (ii) First of all, we know that Kerf is an ideal of A. Let z E Kerf and write x = CgEG x,, xg E A,. Then

0 = f (4=

cf

(z,>

and

f ( 4E 4

,EG

Since B = $,&Ideduce ,, that each z, E Kerf and therefore Kerf is a graded ideal. Now assume that f is surjective and let

f* : A / K e r f + B be the induced homomorphism of R-algebras. Then, for all g E G,

proving that f* is a graded isomorphism. (iii) If I is a graded ideal of A , then the natural map A -+ A / I is a graded homomorphism with kernel I . The converse is a consequence of (ii). H An ideal X of A1 is called G-invariant if

A,XA,-i = X

for all g E G

In the special case where A is a crossed product of G over A1 and ij E U ( A ) n A,, we have A,XA,-I = 3Xg-l. Thus in this case X is G-invariant if and only if gxg-1 =

x

for all g E G

We are now in a position to describe all graded ideals of A . 3.6. Theorem. (Dude (1970~)). Let A be a strongly G-graded R - a lg e b ra . (i) If I is a graded ideal of A and X = I n Al, then X is a Ginvariant ideal of A1 such that

I = AX = X A

and

I n A, = A,X = X A , for all g E G

Induction from normal subgroups

150

(ii) For any G-invariant ideal X of Al, I = X A = AX is a graded ideal o f A such that

I n A, = AgX = XA,

for all g E G

(iii) J(A1) is a G-invariant ideal of A1 and hence A - J ( A l ) = J(A1)- A is a graded ideal of A. Proof. (i) By hypothesis, I = e g E ~ where I g I, = I n A, for all g E G. In particular, we must have X = 1 1 . Given z,y E G, we have

A,Iy = A,(I n AY)C A,I n A,Ay In&,

c

T

Thus A,-lIacy I y and therefore

Consequently, AxIy = I,, and, in particular, A,I1 = I, for all g E G. A similar argument shows that IlA, = 1, for all g E G. Given g E G, we then also have

AgIIAg-l= IgAg-1= 11, proving that Finally,

11

is G-invariant.

as required. (ii) By assumption, AgXAg-l = X for all g E G, so multiplication on the right by A, gives A,X = XA, for all g E G. Since A = $,EGA,, we have

I = AX = $,,GA,X

= $,EGXA, = X A ,

proving that I is a graded ideal of A such that I n A, = A,X = X A , for all g E G.

3. Group-graded algebras and crossed products

151

(iii) Owing to (ii), it suffices to show that J(A1) is a G-invariant ideal of A1. Let V be an irreducible Al-module and let L be the annihilator of V . For any given g E G, it will be shown that A,LA,-I is the annihilator of some irreducible Al-module. Because J ( A l ) is the intersection of all such annihilators, the desired assertion will follow. Choose a maximal left ideal I of A1 such that V E A1/I. Since A,A, = Azy, for all 2,y E G and 1 E Al, the map X + A,X sends bijectively the set of all left ideals X of A1 onto the set of left Alsubmodules of A,, with Y 3 A,-IY as the inverse map. These maps clearly preserve inclusion, hence A,I is an Al-submodule of A,. An elementary calculation shows that the annihilator of the irreducible Al-module A,/A,I is precisely A,LA,-l, as required. H We have now come to the main result of this section. 3.7. Theorem. Let A be a strongly G-graded R-algebra. (i) A/A J(A1) is a strongly G-graded R-algebra by setting +

(A/A - J ( A I ) ) ~= (A,

+ A - J(Al))/A - J(A1)

f o r all g E G

Mo re0 v e r,

(A/A J(A1))l

A,/&%)

and, provided G is finite,

(ii) If A is a crossed product (skew group ring, twisted group ring, twisted group algebra) of G over Al, then A/A J(A1) is a crossed product (skew group ring, twisted group ring, twisted group algebra) of G over Al / J(A,). (iiz) If A is a crossed product of G over A1 and A1 = R. 1~ J(A1), then A/A - J(A1) is a twisted group algebra of G over R . ~ A / (- R 1~n

-

+

J(A& Proof. (i) The fact that A / A . J ( A l ) is a G-graded R-algebra with the given g-component is a consequence of Lemma 3.5(i) and Theorem 3.6(ii), (iii). Because A is strongly G-graded, so is A/A J(Al). By

Induction from normal subgroups

152

definition,

and therefore

On the other hand,

and therefore

proving that ( A / A - J ( A l ) )2, Al/J(A,). If G is finite, then A.J(A1)C J ( A ) ,by virtue of Corollary 4.2.3 in Karpilovsky (1987a). (ii) Assume that A is a crossed product of G over A,. Then, for any g E G, there exists a unit g of A in A,. Hence ij A . J(A1) is a unit of A / A - J(A1)in ( A / A J ( A l ) ) ,for all g E G. This proves that A / A J ( A l )is a crossed product of G over A l / J ( A l ) ,by applying (i). Suppose that A is a skew group ring of G over A,. Then, for any g E G, there exists a unit g of A in A, such that

+

5g = zy

But then for all

2,y

for all x , y E G

E G,

proving that A / A J(A1) is a skew group ring of G over A,/J(A,). Now assume that A is a twisted group ring of G over Al. Then, for any g E G, we may choose g E U(A)n A, such that 3 centralizes Al. But then, for each g E G, ij+A.J(A,) centralizes (Al+ A . J ( A ) / A - J ( A , ) ) , proving that A / A - J(A1)is a twisted group ring of G over A,/J(Al). Finally, assume that A is a twisted group algebra of G over A,. Then A1 G Z(A) and hence

4. The endomorphism ring of induced modules

153

Thus A / A - J ( A l ) is a twisted group algebra of G over A l / J ( A , ) , as required. (iii) Our assumption guarantees that

and therefore A / A . J ( A l )is a twisted group algebra of G over AlIJ(A1). Since the homomorphism R . 1~ + A l / J ( A l ) ,r - 1 H r 1 J ( A 1 ) is surjective with kernel R 1~ n J ( A l ) ,the result follows.

- +

4.

The endomorphism ring of induced modules

The proof of most of the results of this section can be found in Dade (1980),(1981), although many of the essential ideas are contained in previous works of Conlon (1964) and Ward (1968). Throughout, N denotes a normal subgroup of a finite group G and R an arbitrary commutative ring. Let V be an RN-module. Recall that V is said to be G-invariant if gV Z V for all g E G. If W is another RN-module, we say that V weakly divides W if there exists a positive integer k and an injective homomorphism f : V -+ kW = W

@

W

(k copies)

such that f ( V ) is a direct summand of kW. Weshall say that V and W are weakly isomorphic if each weakly divides the other. Of course, this is an equivalence relation among RN-modules. Finally, we shall refer to V as being weakly G-invariant if it is weakly isomorphic to each g V , g E G. 4.1. Lemma. Let V be an RN-module. (i) The map

is an injective homomorphism of R-algebras whose image consists of all $ E EndRG(V") for which $(1@ V ) C 1 @ V .

Induction from normal subgroups

154

(ii) Iff : 1 €3 V

€3 V is a n RN-isomorphism, t h e n f can be uniquely extended t o a n RG-isomorphism V G+ V G . +g

Proof. (i) It is plain that the given map is an injective homomorphism of R-algebras. Moreover, for all f E E n d m ( V ) ,l @f sends 1@V into 18V. Suppose that $ E EndRG(VG)is such that $(1€3V) l @ V . Then there exists f E EndRN(V)such that $ ( l @v) = 1 @ f ( v ) for all v E V . Because $ is uniquely determined by its restriction to 1 €3 V , we have $ = 1 €3 f , as required. (ii) Thanks to a universal characterization of induced modules (Proposition 2.2.3) f can be uniquely extended to an RG-homomorphism $ : VG4 V G .Since for any z E G,

it follows that 1c, is surjective. To show that $ is also injective, denote by T a transversal for N in G. Then, any given w E V G ,can be uniquely written in the form

If $(w)= 0, then

c tf

(1€3 vt) = 0

tCl'

Because t f (1 €3 vt) E tg €3 V and Tg is a new transversal for N in G, we have t f (1 €3 vt) = 0. Thus 1 €3 vt = 0 and therefore w = 0, as we wished to show. Let V be an RN-module. We say that V can be extended to an RG-module if there exists an RG-module V * , whose additive group coincides with that of V while the multiplication

* : RG x V* -, V*

in

V*

satisfies z*v=zv

forall

~ E V , ~ E R(1) N

We are now ready to prove the following fundamental result.

4. The endomorphism ring of induced modules

s

155

4.2. Theorem.

Let V be an RN-module, let E = EndRG(VG), = EndRN(V) and, for any g E G , define EgN by

(i) E is a GIN-graded R-algebra with E g as ~ its gN-component, g E G and the identity component of E is identifiable with S . (ii) E is strongly GIN-graded i f and only i f V is weakly G-invariant, while E is a crossed product of G I N over S if and only if V is G -

invariant. (iii) E is a skew group ring of GIN over S if and only if V can be extended to an RG-module. In fact, there is a bijective correspondence between all extensions V* o f V to RG-modules and all splitting homomorphisms y for 1 + U ( S ) + G r U ( E ) 2 G I N + 1, in which V* corresponds to y if and only if 1 @ g * v = y ( g N ) ( g@v)

for all

E V,g E G ( 2 )

-

(iv) If V is weakly G-invariant, then E J ( S ) = J ( S ) .E is a graded ideal of E contained in J ( E ) and such that E I E - J ( S ) is a strongly G/ N-graded R-algebra with ( E / E- J ( S)),, = (E g N +E. J ( S ) ) / E .J ( S ) , g E G . Furthermore, i f V is G-invariant, then E I E . J ( S ) is a crossed product o f G / N over S / J ( S ) . (v) If V is G-invariant and S = R - 1s J ( S ) , then E I E - J ( S ) is a twisted group algebra of G I N over R l s / ( R 1s n J ( S ) ) S / J ( S ) .

+

-

Proof. (i) Observe that if X N = y N , then x @ V = y 8 V which shows that E g N is well-defined. Let T be a transversal for N in G. Then (direct sum of RN-modules) V G= $ t c T t @I V and so if rt : V G+ t @I V is a projection map, then

Now assume that f E E and let gt : 1 @ V + t 8 V be the R N homomorphism that is the restriction of rt 0 f to 1 @ V . Owing to a

Induction from normal subgroups

156

universal characterization of induced modules (Proposition 2.2.3), there exists ft E E such that

It follows that ft E Et-"

whence f = CtETft. Now assume that then have and

&T

vt

and, applying (3) and (4),we have

CtETft= 0 with

= 0. Thus

ft E

Et-IN. For any v E V, we

vt = f t ( 1 c3 v) E t @ v vt

= 0 and f t = 0 for all

t E T , proving that

I f f E E=N,cp E E,N, then

and thus (fcp)(l@ V)

c f(y-'

63 V ) c y - ' f ( l @ V ) E y-'z-l@

v

It follows that f ( o E E z y and ~ therfore E is a GIN-graded R-algebra with E I l as ~ its gN-component. Finally, by Lemma 4.1(i), the identity component of E is identifiable with S, proving (i). (ii) Thanks to Lemma 3.l(iii), E is strongly GIN-graded if and only if for any given g E G, there exists (o; E E g ~y5i, E E g - l ~1, 5 i 5 n = n(g), such that

It will next be shown that ( 5 ) is equivalent to V and gV to be weakly isomorphic, which will imply the first assertion.

4. The endomorphism ring of induced modules

157

If (5) holds, then any map f : l@V + n(g@V),w t-+ ($l(v), . . . ,$ n ( v ) ) is an injective RN-homomorphism. Moreover, the map (wl, . . . ,wn) H ‘pI(w1) * v n ( v n ) is a homomorphism from n(g QD V ) to 1 QD V which is a left inverse to f . This proves that V weakly divides gV. By symmetry, V weakly divides g-’V and thus gV weakly divides V , proving that V and gV are weakly isomorphic. Conversely, suppose that V and gV are weakly isomorphic. Then V weakly divides g 8 V ,so there is a positive integer n and an injective RN-homomorphism $ : 1 @ V + n(g @ V ) such that $(1 @ V ) is a direct summand of n(g 8 V ) . We may write $(z) = ($1(z),. .. ,$ n ( S ) ) , where : 1 @ V + g @ V is an RN-homomorphism, in which case $; can be regarded as an element of E g - l N . Let f : n(g V )-+ 1 8 V be an RN-homomorphism which is a left inverse to $. Then f determines RN-homomorphisms

+-+

+;

‘p;

:g @

v + 1 QD v

(1 5 i 5 n )

(which can be regarded as elements of E g ~such ) that ( 5 ) holds. If V is G-invariant, then E is a crossed product of GIN over S, by virtue of Lemma 4.1(ii). Conversely, suppose that for any g E G there exists fg E U ( E )n E g ~ Then . fg-l : 1 @ V + g @ V is an injective RN-homomorphism such that fg-ifg is an automorphism of 1@ V and hence of g 8 V . Thus fg-1 is an RN-isomorphism, as desired. (iii) Assume that V’ is an extension of V to an RG-module. Then (1) implies that there is a unique RN-isomorphism of g 8 V onto V sending g 8 w into g * w, for all g E G, w E V . This isomorphism obviously depends only on the coset g N . Hence, by Lemma 4.1(ii), there is a unique y(gN) E E g n~U ( E ) such that (2) holds. Now fix z,y E G. Then, for all v E V ,

+N)r(yN)(w @ 4 = 7 ( 4 z 7 ( ” y QD 4 = r(zN)z(lQD Y*4 (by (2))

*

= 7 ( z N ) ( zQD y v) = 1@ z * (y * v) = 1 8 (zy) = 7 ( z y N ) h €3 4

*

2,

(by (2))

which forces r(zN)y(yN) = y(zyN). This demonstrates that y is a splitting homomorphism, the only such homomorphism satisfying (2).

Induction from normal subgroups

158

Conversely, suppose that 7 : G/N + GrU(E) is a homomorphism . any g E G, v E V , let such that for all g E G, r ( g N ) E E s ~ For g * v E V* be defined by (2). Then for any n E N, 1 8 n * o = r ( N ) ( n8 v) = n 8 v = 1 8 nu and therefore n * v = nv. Furthermore, given s , y E G and v E V , we have

Thus V extends to the RG-module V*,the only such module satisfying (2). (iv) Suppose that V is weakly G-invaraint. Then, by (ii), E is strongly GIN-graded, so by Theorem 3.6(iii), E . J ( S ) = J ( S ) - E is a graded ideal of E . By Theorem 3.7(i), this graded ideal is contained in J ( E ) . The rest of the first statement is a consequence of Theorem 3.7(i). The second statement is a consequence of (ii) and Theorem 3.7(ii) . (v) Apply (ii) and Theorem 3.7(iii). 4.3. Corollary. Let V be a strongly indecomposable RN-module.

Then V is G-invaraint if and only if V is weakly G-invariant.

Proof. Because V is strongly indecomposable, S = EndRN(V) is a local ring. Now suppose that V is weakly G-invariant. Then, by Theorem 4.2(ii), E = EndRG(VG)is strongly GIN-graded. Because S is local, it follows from Proposition 3.3 that E is a crossed product of GIN over S. Thus, by Theorem 4.2(ii), V is G-invariant. Conversely, if V is G-invariant, then V is obviously weakly G-invariant, hence the result. Our next aim is to provide a condition under which J ( E ) n S = J ( S ) , where E = EndRc(VG)and S = EndRN(V). The following simple observation will clear our path. 4.4. Lemma. Let A be an R-algebra which is finitely generated

5. Relations between the decompositions of VG and End&VG)

159

as an R-module and let R / J ( R )be artinian. Then

J(A)" C J ( R ) A

for some n

21

Proof. We know, from Lemma 1.2, that J ( R ) AC J ( A ) . Hence, by Corollary 1.4.4(ii), we have

J ( A / J ( R ) A )= J ( A ) / J ( R ) A But A / J ( R ) Ais an R/J(R)-algebra which is finitely generated as an R/J(R)-module. Hence, by Corollary 1.3.8, A / J ( R ) Ais artinian. Invoking Proposition 1.4.14, we therefore deduce that J ( A ) / J ( R ) Ais nilpotent, hence the result. 4.5. Theorem. Let V be a finitely generated G-invariant R N module, let E = EndRG(VG) and let S = EndRN(V). If R is artinian or a noetherian local ring, then

J ( E )n S = J ( S ) Proof. Owing to Theorem 4.2(iv), J ( S ) C J ( E ) n S . Now V is a finitely generated R-module, hence EndR(V) (and therefore S ) is a finitely generated R-module (Lemma 1.7, Propositions 1.4.14 and 1.3.3, Corollary 1.3.7). Similarly, V Gis a finitely generated R-module and hence E is a finitely generated R-module. Thus, by Lemmas 4.4 and 1.2, J ( R ) S C J ( S ) and J(E)" & J ( R ) E for some n 2 1. Moreover, by Theorem 4.2(i), E = $ z c ~ / ~ E with z El = S. Hence J ( R ) E n S = J ( R ) S and

( J ( E )f l S)" C J ( R ) En S = J(R)S C J ( S ) It follows that ( J ( E )n S ) / J ( S ) is a nilpotent ideal of S / J ( S ) and therefore J ( E )n S J ( S ) ,as required. 5.

Relations between the decompositions of V Gand EndRG(VG)

Throughtout this section, N denotes a normal subgroup of a finite group G and R an arbitrary commutative ring. Given an RN-module

160

Induction from normal subgroups

V , it will be convenient to identify EndRN(V) with the subring of E n d ~ c ( Vconsisting ~) of all $ E EndRG(VG)for which $(1 @ V ) 1@ V (see Lemma 4.1(i)), In what follows we shall freely use this identification. For convenience of reference, we first record some general observations which provide a link between a module and its endomorphism ring. Let S be a ring, let V be a left S-module and let E = Ends(V). For any subset L of E and any subset W of V , let LW denote the set of all finite sums

It is clear that LV is a submodule of V . Note also that if I is a principal right ideal of E generated by $, then

We may regard V as a left E-module via 'pv = cp(v),cp E The E @ E V becomes a left E-module via

E,v E V .

Moreover, the canonical map

is an isomorphism of E-modules. 5.1. Lemma. Let S be a ring, V a left S-module and I a right ideal of E = Ends(V) which is a direct summand o j E . Then I = Ee for some idempotent e of E c n d (i) IV = e ( V ) (ii) The map I @ E V + IB

is an isomorphism of E-modules.

5. Relations between the decompositions of V G and End&VG)

161

Proof. The fact that I = E e for some idempotent e of E is a consequence of the right analog of Proposition 1.3.12. Hence (i) follows from (1). Because I is a direct summand of E , we may identify I @ E V with its image in E @IE V, in which case (2) maps I @ E V onto IV. This proves (ii) and hence the result.

5.2. Lemma. Let S be an arbitrary ring, let V be a left S-module and let E = E n d s ( V ) . (i) If V = c&V;. and ei : V 3 6 is the projection map, then el,. . . ,e, are mutually orthogonal idempotents of E with sum 1. Conuersely, if el,. . . ,e, are mutually orthogonal idempotents of E with sum 1, then V = c&V, where = e;(V) and ei : V + K is the projection

map.

(ii) K is indecomposable if and only i f e; is primitive. (iii) K Z 4 as left S-modules if and only if e;E Z ejE as right E-modules. be the Proof. (i) Assume that V = @i=lx and let e; : V + projection map. It is plain that e: = ei, 1 5 i 5 T . If i # j , then (e;ej)(V) = e;(y)= 0 , and thus e;ej = 0. Moreover, we obviously have C:==, e; = 1.

Conversely, assume that el,. . . ,e, are mutually orthogonal idempotentsofEwithsum1. Since1 = e l + - - - + e , , w e h a v e V = V,+-..+V,. If wi E V;., 1 5 i 5 r , are such that el(v1) - e,(v,) = 0, then by taking the image of both sides under e;, we obtain e;(v;)= 0, 1 5 i 5 r , Thus V = $r==,K and obviously e; : V + K is the projection map. (ii) If e; = e f , where e, f are orthogonal idempotents, then K = e(V )$ f ( V ) . Conversely, if K = V,' @ Kii and A; : V -+ p; : V + are projection maps, then & , p i are orthogonal idempotents of E with e; = A; pi. Thus V , is indecomposable if and only if ei is primitive. (iii) Assume that eiE 2 e j E as right E-modules and put I; = e;E, 1 5 i 5 r. Then, by Lemma 5.1, IiV = K and V; % V, as E-modules. as S-modules. Hence Q Z Conversely, suppose that 8 : r/: -, vj is an S-isomorphism and fix v E V. Given cp f e;E, we have Imcp C ei(V) = K, so Im(89) C 4.

+- - +

+

+

v',

Induction from normal subgroups

162

Thus ( ~ Q ) ( v = ) ej(V1) for some ~1 E V, so e j ( b ) ( v ) = ej(V1) = ( b ) ( v ) and therefore

BQ = ej(0cp) E e j E If we now define p : eiE -t ejE by p(cp) = 09, then p is obviously a homomorphism of right E-modules. A similar argument shows that the map X : ejE + eiE defined by A($) = 0-'$ is also an E-homomorphism. Because X is the inverse of p , we deduce that eiE E ejE, as desired. Let V be a free module over a ring S. In case any two bases of V have the same cardinality, we refer to this cardinality as the rank of V. 5.3. Lemma. Let V be a free module over a local ring S. Then any two bases of V have the same cardinality.

Proof. Let {v;li E I} be two bases of V. Denote by V; and tijj the images of vi and wj, respectively, in 3 = V/J(S)V. Then (6ili E I}and {tijjlj E J } are two bases of the S/J(S)-module V . But, by hypothesis, S / J ( S ) is a division ring, hence 111 = lJl, as required. The observation above, will allow us to speak about

"

the rank of

V" whenever V is a free module over a local ring. Let S be a ring and let V be a left S-module. Then V can be regarded as a right Ends(V)'-module by setting VQ

=~ ( v )

for all v E V,cp E Ends(V)'

5.4. Lemma. Let V be a G-invariant RN-module, let S = EndRN(V and let E = EndRG(VG)". For each g E G, choose 'pg E U ( E ) mapping 18 V isomorphically onto g 8 V, put 91 = 1 and define $g : V + V by 'p;l(g 8 V ) = 1 8 $ ~ ~ ( v )Then . V 8 s E is an (RG,E)-bimodule with the actions

5. Relations between the decompositions of V G and End&VG>

163

Proof. Because V is a right S-module and E is a left 5'-module, the tensor product V 8 s E is at least a 2-module. Given s E S, g E G,vE V , we have the following equality in VG:

which implies

We therefore conclude that

Given g E G, we now introduce f : V x E + V 8 s E given by

It is clear that f is R-bilinear. Since, for any s E S,

the m a p v 8 e H g(v 8 e) is an R-endomorphism of V 8 s E . We now claim that, given z,y E G and v E V ,

where s = 'pz'py'p;i E S. Indeed, we have

Induction from normal subgroups

164

proving (4). It follows that X(Y(V

@ el) = z($J&) @ vve) = $W%/(v)) @ cpxvve

= $z($v(v)) @ ~ ' P q = e 4$Jz($Jv(v))>@ vzve = $z&) @ cpxve (by (4)) = (zY>(v@ 4

Because vl = 1 (hence &(TI) = v for all v E V ) ,we also have l ( v @ e ) = v @ e. Hence V 8 s E is a left RG-module. It is clear that V @sE is a right E-module under the given action of E. Moreover, for all g E G,u E V,e,e'E E ,

[g(v8 e)]e' = $Jg(v) cpgee' = g ( v

eel) = g[(v@ e)er],

thus completing the proof. H The following theorem is the main result of this section.

(Cline (1972), Theorem A). Let V be a G-invariant RN-module, let E = End&VG)' and let S = EndRN(V)' be identified with the subring of E consisting of all cp E E with cp(l@V)C l @ V . Then the map 5.5. Theorem.

PsE + v@e

H

e(l@v) vG

is an (RG,E)-bimodule isomorphism, where the (RG, E)-bimodule structure of V 8 s E is given by Lemma 5.4. Proof. The given map, say f, is obviously R-linear. Since for all seG f (g(v @ 4 ) = f ($Jg(v>@ cpge) = (cpge)(l@$ J g W = (vge)(cp;'(g @ 4) = 4s @ 4 = g e ( l @v) = g f ( v @ e ) ,

it follows that f is an RG-homomorphism. Because for all e,e' E E and v E V f((v

e)e') = f ( v eel) = ( e e ' ) ( l @w) = ( e ( I @v))e' = f ( v e)e',

5. Relations between the decompositions of V G and En&c(VG)

165

we deduce that f is also an E-homomorphism and hence is an (RG,E)bi mo dule homomorphism. Let T denote a transversal for N in G. Then

Because pt(l @ V ) = t @ V , it follows from (5) that f is surjective. Owing to Theorem 4.2(ii) and Proposition 3.4(iii), E is a free left Smodule freely generated by all yt, t E T . Thus any element of V 8 s E can be uniquely written in the form

Because

f ( p w t )= Ct Et TP t ( l @ v t ) and cpt( 1@ vt) E t @ V , it follows from ( 5 ) that if CtETvt €3 pt E K e r f , then pt(l @ vt) = 0 for all t E T . Thus all vt = 0 and therefore C t Evt ~€3 (Pt = 0. Hence f is injective and the result follows. As a second major result, we now prove the following theorem in which the RN-module V is not assumed to be finitely generated.

Let V be a strongly indecomposable G-invariant RN-module and let S = EndRN(V) be identified with the subring of E = EndRG(VG) consisting of all (P E E with p(l @ V ) 1 @ V . (i) I f a right ideal I of E is a direct summand of E, then I is a free S-module offinite rank. (ii) E contains a finite complete set of orthogonal primitive idempotents el,. . . ,en such that for Ij = e;E the following properties hold: I;V where each I;V is an indecomposable RG-module (a) V G= such that (IjV)N is a direct sum o f n ; copies o f V , nj = ranks(I;) (V is identified with 1 8 V ) . (b) I;V IjV if and only i f e j E Z ejE. (c) I f V is R-free offinite rank, then I;V is R-free with 5.6. Theorem.

run kR( 1;V ) = ranks (I;)runkR( V )

Induction from normal subgroups

166

Proof. (i) By Theorem 4.2(ii) and Proposition 3.4(iii), E is a finitely generated free S-module. Since I is a direct summand of E , I is a finitely generated projective S-module. But, by hypothesis, S is local, hence I is a free S-module, by Lemma 2.10.12. Finally, by Lemma 5.3, I has a uniquely determined rank which is finite since I is finitely generated. (ii) It is a consequence of (i) that E admits a finite decomposition E = $Z,E; where each E; is indecomposable. Hence E; = e;E for some orthogonal primitive idempotents e; of E with el - e, = 1. By Lemma 5.2, we deduce that

+ +

where each ei(VG)is indecomposable and e;(VG)Z ej(VG)if and only if e;E ejE. We now claim that

e;(VG)= I;V

for all i

(6)

Indeed, since V C V G and e;(VG)= I;VG,we obviously have I;V e;(VG).On the other hand, V Gis the sum of all g 8 V,g E G. Hence, to prove (6), we need only verify that ei(g 8 V

) C IiV

for all

g EG

Fix g E G and w E V . Let cp E E map V isomorphically onto g 8 V . Then g @ w = cp(v1) for some 01 E V . Hence ei(g 8 v) = ( e i c p ) ( Y ) E IiV,

proving (6). All the remaining assertions will now follow provided we show that (IjV)N is a direct sum of nj copies of V , n; = ranlcs(1;). Since Ij is a free right S-module of rank n;, E" o e; is a free (left) 9-module of E" 0 e; under the isomorphism rank n;. Moreover, the image of V of Theorem 5.5 is I;V. Hence

which immediately implies the required assertion by virtue of the fact that V @ s o So V .

6. Twisted group algebras over fields

6.

167

Twisted group algebras over fields

In this section, we shall collect a number of general observations concerning twisted group algebras. The results recorded will play a significant role in our subsequent investigations. Throughout, G denotes a finite group and F* the multiplicative group of a field F . Let Z2(G,F*)be the set of all functions

a:GxG-,F* which satisfy the following identities:

We shall refer to the elements of Z2(G,F') as cocycles. Given a , p E Z2(G7F'), define a@by the rule

It is then obvious that a@is also a cocycle and that Z2(G7F * ) becomes an abelian group. Let t : G -+ F* be such that t ( 1 ) = 1 and let 6t : G x G + F* be defined by W(z, 3)=W ( y ) t ( z y ) - ' We shall refer to S t as a coboundary. It is routine to verify that the set B2(G,F') of all coboundaries constitutes a subgroup of Z 2 ( G ,F*). The factor group

H2(G7F') = Z2(G7F * ) / B 2 ( GF, * ) is said to be the second cohomology group of G over F*. The elements of H2(G,F * ) are called cohomology classes ; any two cocycles contained in the same cohomology class are said to be cohomologous. Given (Y E Z2(G7F * ) ,we shall write &. for the cohomology class containing a. Given a E Z 2 ( G 7 F * )we , write F*G for the vector space over F

Induction from normal subgroups

168

with basis {gig E G} which is in one-to-one correspondence with G. Define multiplication in F"G distributively using

zg = a ( z , y ) w ,

Xa: = a:X

for all z, y E G, X E F. Then F*G is a twisted group algebra of G over F and, conversely, any twisted group algebra arises in this manner. Note that if a ( z , y ) = 1 for all z , y E G, then F a G Z FG. More generally, we have 6.1. Lemma. For any given (Y E Z2(G,F*),the following conditions are equivalent. (i) F"G E F G as F-algebras. (ii) F"G admits an F-algebra homomorphism into F. (iii) a is a coboundary.

Proof. (i)+(ii): This follows from the fact that the augmentation map F G + F is a homomorphism of F-algebras. (ii)+(iii): Let f : F*G + F be a homomorphism of F-algebras. Then, for all x , y E G,

f(a:)f($=

fm)= f ( + , Y ) z y )

= +,Y)f(zy)

and so a = bt, where t : G -+ F*is defined by t ( g ) = f(g). (iii)+(i): Assume that a = St for some t : G --.) F* with t ( 1 ) = 1. Then the map qh : F"G 4 F G , which is the extension of g H t ( g ) g by F-linearity gives the desired isomorphism. H 6.2. Lemma.

(i) Each element of H2(G,F*)has order dividing

IGI-

(ii) If G is a p-group and F a perfect field of characteristic p , then F G for all H2(G,F*) = 1. In particular, b y Lemma 6.1, F*G a E Z2(G,F*). FG (iii) If G is cyclic and F is algebraically closed, then F"G for all a E Z2(G,F*)(in particular, b y Lemma 6.1, H2(G,F') = 1).

Proof. (i) Assume that a E Z2(G,F'). Because 4 3 , Y)+Y,

4 =4 Y ,4 4 x 9Y

4

6. Twisted group algebras over fields

169

Thus a" = S t , as asserted. (ii) By (i), it suffices to show that if QP is a coboundary, then so is a. To this end, assume that t : G 3 F', t ( 1 ) = 1 is such that

Since F is perfect and charF = p , we have

for some p : G + F' with p(1) = 1. Thus a = Sp, as required. (iii) Let g be a generator of G, say of order n. Then g" = X - 1 for some A E F'. Because F is algebraically closed, X-' = p" for some p E F'. Thus (pij)" = 1. Since the elements I, pij, .. . ,(pij)"-' form an F-basis of F"G, we conclude that FOG Z FG.

6.3. Lemma. Let a E Z2(G,P)and let L be a field extension of F . Then the map

z@g

zg

is an isomorphism of L-algebras. Proof. Let {ijlg E G} be an F-basis of F"G with zij = a ( x , y ) Z y for all x , y E G. For each g E G, put ij = 1 8 9 . Then {jig E G) is an L-basis of L @F F"G and, for all x,y E G,

So the lemma is verified.

170

Induction from normal subgroups

We are now ready to prove the following useful result. 6.4. Proposition. Let G be a nonidentity group and let CY E Z2(G,F*).If charF = p > 0 and G is a p-group, then F"G is local.

The converse is true if F is algebraically closed.

Proof. Let charF = p > 0 and let G be a pgroup. If E is the algebraic closure of F, then F"G admits an injective homomorphism into E"G (Lemma 6.3). Since E is perfect of characteristic p and G is a pgroup, E"G Z E G by Lemma 6.2(ii). Hence, by Proposition 2.10.11, E"G is local. Thus F"G must also be local. Conversely, suppose that F is algebraically closed and F"G is local. Assume by way of contradiction that G has an element g of order n > 1 where n # 0 in F. Then, for H =< g >, we have F"H Z FH. Since e = (l/n)Ch,=Hh is a nontrivial idempotent of F H , we obtain the desired contradiction. Another useful observation is recorded in the following proposition. 6.5. Proposition. Let F be a field of characteristic p > 0 , let G be a p-group and let Q E Z2(G,F*). Then F"G/J(F"G) is a finite

purely inseparabze field extension of F.

Proof. Let E be the algebraic closure of F . Then, by Lemmas 6.3 and 6.2(iii)

Since G is a p-group and charF = p , we also have E G / J ( E G ) s E . Thus EaG/J(EaG) E E. Now put I = J(EaG) n F"G. Then I is a nilpotent ideal of F"G with F"G/I an F-subalgebra of E. Thus F"G/I is a finite field extension of F and therefore I = J(F*G). Furthermore, the field extension is purely inseparable since it is generated by the images of the element 3,s E G and these satisfy ijpn E F for some n. H Our next aim is to determine explicitly a basis for the centre Z(F"G) Q E Z2(G,F') and a basis ( Z l s E G}

of FOG. To this end, we now fix

6.

Twisted group algebras over fields

171

of F"G such that

iy = a(x,y)zy

for all x,y E G

An element g E G is said to be a-regular if

Expressed otherwise, g is a-regular if and only if

It is obvious that the identity element of G is a-regular. The following elementary properties of a-regular elements are worthwhile mentioning. 6.6. Lemma. (i) An element g E G is a-regular i f and only i f it is @-regular for a n y cocycle @ cohomologous t o a. (ii) If g E G is a-regular, t h e n so is a n y conjugate of 9.

Proof. (i) Let g be an a-regular element and let p be cohomologous to a, say @(x,y) = a ( x , y ) t ( . ) t ( y ) t ( x y ) - l for some t : G + F*. Then, for x E C G ( g ) ,

proving that g is @-regular. The converse is obvious. (ii) Let y E G and let x E C ~ ( y g y - ' ) . Then y-'xy E C G ( ~and ) therefore y - l z y . g = ji . y-1.y Since y - l x y = X ~ - ' C C for ~ some X = X(x,y) E F*, it follows that y-lzys = g y - l x y

Hence i commutes with ygy-'. Again, since ygy-1 = pyijy-l for some p = p ( g , y ) E F*, we see that i commutes with ygy-1. Thus ygy-l is also a-regular.

Induction from normal subgroups

172

Let Cgbe the conjugacy class of G containing g. We say that Cg is a-regular if g (and hence, by Lemma 6.6(ii), any element of Cg)is aregular. The set of all a-regular elements of G will be denoted by Go. We also say that the cocycle a is standard if it satisfies the following two properties: for all x E G

a ( x , x - ~= ) 1

for all g E G o , x E G

a ( x ,g ) a ( z g ,2 - 1 ) = 1

Expressed otherwise, a is standard if and only if the following two properties hold

___-

xgx

= xgx-1

for all

x E G,g E G"

6.7. Lemma. Let G be a finite group of order n , let F be an algebraically closed field of characteristic p 2 0, and let a E Z 2 ( G ,F*). If p divides n, let P be the largest normal p-subgroup of G. Then, after making a diagonal change of basis, if necessary, the following properties hold: (i) 9-' = ij-' for all g E G (ii) Z g 5 - l = xgx-l for all x E G,g E Go (iii) ~y = ~y for all x , y E P .

Proof. Owing to Lemma 6.2(ii), there exists t : P + F* such that a ( x ,y ) = t(z)t(y)t(xy)-'

for all

2,y

E

P

Let X : G + F* be defined by X(x) = t ( z ) - l if x E P and X(x) = 1 if x $ P . Then, replacing g by X ( g ) g , we may assume that (iii) holds. We now show that (ii) holds for g E P . Indeed, write 5g5-l = pxgx-l for some p E F* and put rn = ]PI. Then, by (iii),

gm whence

= xgx - 1 m = 1

6. Twisted group algebras over fields

173

Since m is a power of p , p = 1 and therefore (ii) holds. Suppose now that g E G" - P and let Cg = { g = g 1 , g 2 , . - . , g r } . Because g is a-regular, ij has r conjugates in I' = {XglX E F*,g E G}. Choose i j arbitrarily, take g 2 , . . . , g r as its conjugates in I' and leave other basis elements unchanged. Then we have conditions (ii) and (iii) holding. Due to (iii), to prove (i), we may assume that g $ P . Consider the elements of G - P and write G - P as a disjoint union

where S1 consists of the a-regular elements. For one of each pair 9,g-l E Sz with g # 9-', leave g unchanged and replace by g-' = a(g,g-')-'g-l. For each g E S2 with g = g-' replace i j by a ( g , g-')-'/Zg. Now consider g E S1 and write Cg = ( 9 = 91,. . . , g p ) . We still have above the choice of g at our disposal. If Cg # Cg-l, we choose ij,g-' asin the case g# 9-'. If Z g 2 - l = gi, then 2ij-lz-l = gi-' = g i ' , by choice of gi,gy'. Finally, let Cg = Cg-1. Thus 9-' = ygy-' for some y E G and 9-' - Y9Y (1)

----'

and so all g; by c ~ ( g , g - ' ) - ' / ~ g ; , we still Replacing g by a(g,g-')-'12ij, have (1) and also 9;' = g i - ' . This completes the proof.

6.8. Lemma. Let a E Z 2 ( G , F * ) ,let (91,. ..,st} be a full set of representatives for the a-regular classes of G and let Ti be a left transversal for Cc(g;)in G . Then the elements CzETi ZgiZ-', 15 i 5 t f o r m an F-basis for Z(F*G). In particular, if a is standard and C; is the conjugacy class of g i , then the elements

form an F-basis for Z(F*G). 7

Proof. Setting sgiz-1 = Zg;Z> z E Ti, 1 5 i 5 t , and = i j if g ----I is not regular, we have zgz - zgz-l for all z E G and all a-regular

Induction from normal subgroups

174

g E G. Thus we may assume that 3 g Z - l = xgz-' for all x E G and all a-regular g E G and we must show that the elements of (2) form an

F-basis for Z(F"G). To this end, put ki = EgECig, 1 5 i _< t . Let y = Eygg E F"G be the elements of F* such that and let

Then y is central if and only if for all x E G

9

9

Now put Suppy = (9 E GI?, # 0) and fix t E Suppy,a E Cc(t). If y is central, then comparing the coefficients of f, we get yt = ytX,,t. Hence = 1 and therefore t is a-regular. Since Zij5-l = zgz-l for all z E G and all a-regular g E G, it follows from (3) that y is central if and only if (i) Suppy consists of a-regular elements (4 r g s = 7gzg"-' = 7z-lgz9- for all z E G or, equivalently, if and only if y is an F-linear combination of k;,1 I: i 5 t. Since the elements Ici are obviously linearly independent, the result follows.

c

c

c

Let A be an F-algebra and define [A,A ] ,the commutator subspace of A , to be the F-linear span of all Lie products [x,y] = zy -yz, z, y E A. The following observation illustrates how the commutator subspace can be brought into argument. 6.9. Lemma. Let A be an algebra over a field of characteristic

P>O cz1 ay"(mod[A,A ] ) (i) ( c E 1 a*)'" (ii) I f s E [ A ,A ] , then s p E [A,A]

(a; E A , n E N)

Proof. (i) By induction argument, it suffices to show that (a

+ b)P" E a'" + bP"(mod[A,A ] )

6. Twisted group algebras over fields

175

for all a, b E A. Suppose first that n = 1. Expanding by the distributive law, we have (a+

b ) P = up

+ bp +

&la2

*"UP

(4)

where the sum is over all products a1 - + - a of p p terms, a; E { a , b } , not all equal to a or b. With each word a1a2 - - up associate its cyclic permutations

-

U1U2

*

- - U p , U2U3

* *

a

UpUl,

* *

,U p U l - - U p - l *

All these products are congruent modulo [A,A ] . This is so since for x = a;,a;, - . . u ; ~y , = a; a;,+,

-

1

a;,ql

- . . a;3-1

we have x - y = [y, 61 E [A,A ] , where y = a;, a,,

-

*

a;J-l and

S = aiJa;j+l

- a;,

Therefore the sum of these cyclic permutations is p x 1 - x p modulo [ A , A ] ,and hence it belongs to [ A , A ] .Applying (4),it follows that

and hence that

(ab - b ~ E) (~~ 6 ) ' - (ba)P(mod[A,A ] ) G [a,(ba)P-'b](rnod[A, A]) Accordingly

The required assertion now follows from (5) and (6) by induction on n. (ii) Apply (6). We next describe the commutator subspace of FaG. 6.10. Lemma. Let a be a standard cocycle and let y = E y Z z E FOG. Then y E [FOG,F"G] if and only if the sum of the coeficients yz over each a-regular class of G is zero.

Induction from normal subgroups

176

Proof. Denote by C1, ,C, the a-regular classes of G and let : F*G -+ F x F x - * x F (s times) be the F-linear mapping given

t,b

by

c

-

7 3 H (Ply ' ' ' P a )

where

Pi

=

c

7x

XECi

xEG

We must show that [F"G,F"G] = Kert+b. Observe that [F"G,F"G] is spanned by all Sy - yii with z,y E G. First assume that zy is not a-regular. Then so is yz since zy and yz are conjugate. But then

zy - yz = a ( x ,y) zy - a ( y , z)yz which shows that [ z , y ] E K e r $ . Suppose next that z y is a-regular. Then, bearing in mind that a is standard, we have

zy - yz = 5 y - z-l(zy)z = a ( z ,y )z y - a@,y)5-1"Yz = Q(Z, y ) ( a - z-'(zy)z) Hence [Z,y] E Kerzl, and therefore [F"G,F"G] KerG. To prove the opposite containment, we first demonstrate L a t i g E G is not an a-regular element, then g E [F*G,F*G].Indeed, let g be a nonregular element so that Zg # 33 for some x E cG(g). Because z commutes with g, we have zg2-l = X i j for some X E F* with X # 1. In view of the identity

g - 235-1 = (33-1 )z- - "35-1)

(7)

we deduce that (1 - X)g, and hence g, belongs to [F*G,F"G]. We are thus left to verify that if xi = CIECi y&i and if CIECi~i = 0, then x; E [ P G ,FaGI. The latter being a consequence of (7) and a vector space identity

if

xy=lX i = 0, the result follows. H As a generalization of Maschke's thpwern) we next prove

6.11. Proposition. Suppose that the characteristic of the field F does not divide the order of G. Then FOG is semisimple.

6. Twisted group algebras over fields

177

Proof. Let W be a submodule of an PG-module V . Since V is a vector space over F , its subspace W has a complement in V, say V = W@W'.Let 8 : V + W be the projection map, and let $ : V + V be defined by

$(v) = IGI-'

C z8s-lv

(v E V )

xEG

Because for all v E V,y E G,

$(yv)

C ZOZ-lijv = 1GI-l CjE6jjZ-1yv 1GI-l C i j Z ~ r ( yz, ) - ~ ~ c Y~ )( Z~ -, ~ i j - ~ i j v 1GI-l C i j Z 8 Z - l ~

= 1GI-l

xEG

=

ZEG

zEG

=

zEG

= Y+(v>,

it follows that $ is an FQG-homomorphism. Now assume that v E W . Then, for any x E G,Z-'v E W , so O(Z-lv) = Z-lv and thus ZOii-'v = v. Consequently, $(v) = v for all v E W . This implies that W is a direct summand of V and the result follows. H The next lemma is a final step in the preparation for the proof of the main result. 6.12. Lemma. Let A be a finite-dimensional algebra over an algebraically closed field of characteristic p > 0 and let S = [A,A].

Then

T = (5 E AlzPmE S

for some integer m 2 1)

is an F-subspace of A and the number of nonisomorphic irreducible A-modules equals dimF(A/T). Proof. Owing to Lemma 6.9(ii), a E [A,A]implies ap E [A,A]. Thus S C 2'. By Lemma 6.9(i), if a , b E T , then for sufficiently large m,

(a

+ b)pm = apm+ bp'"

3

0

(mod S)

Induction from normal subgroups

178

+

which shows that a b E T . Clearly, Xu E T for all X E F and a E 2'. Hence T is an F-subspace of A. Since the Jacobson radical J ( A ) of A is nilpotent, it follows from the definition of T that J ( A ) & T . Now put A = A / J ( A ) and T = T / J ( A ) .If r is the number of nonisomorphic irreducible A-modules, then there exists positive integers n l , . . . ,n, such that P

A 2 IIM,;(F) i=l

+

Now put = (S J ( A ) ) / J ( A ) Then . clearly S is the F-subspace of spanned by all ab - bu ( a ,b E A ) and

T = {z E A l s p "

ES

for some integer

A

rn 2 I}.

Let Si, 2';be the analogous F-subspaces defined for the F-algebra A; = M n i ( F ) .Then P

d i r n ~ ( A / T=) d i m ~ ( A / T=)

C dirn~(A;/T;) i=l

and therefore it suffices to verify that

Let est be the ni x ni-matrix with (s, t)th entry equal to 1 and all other entries 0. Then, for any s # t , est = esjejt - ejtesj E Si

and ess - ett = estets - etsest E S;

It follows that S; contains the n' - 1 linearly independent elements est(l 5 s , t 5 ni,s # t ) and ess - e l l ( s = 2 , . . . ,ni). Because S; C T.j, this shows that dirn~(A;/Ti)= d i m ~ ( A i-) d i m ~ ( T ;5) n: - (ns - 1) = 1 On the other hand, the elements of 5'; are all n; x n;-matrices with trace 0. In particular, ell # S;. But eyy = ell for all m 2 1, so

6. Twisted group algebras over fields

179

4

T,. Thus Ti # A;, so from dirnF(A/T;) 5 1, we deduce that dirn~(A;/T;) = 1, as desired. ell

In what follows, if p = 0 then by definition all elements of G are p'-elements. 6.13. Theorem. (Asano-Osima-Takahasi (1937)). Let G be a finite group, let F be an algebraically closed field of characteristic p 2 0 and let a E Z2(G,F * ) . Then the number of nonisomorphic irreducible F"G-modules is equal to the number ofa-regular classes of p'-elements of G.

Proof. Let r be the number of nonisomorphic irreducible F*Gmodules. If charF = 0 , then FaG is semisimple (Proposition S.ll), hence r = ( Z ( F " G ): F ) This proves the case charF = 0, by applying Lemma 6.8. Now assume that charF = p > 0 and put S = [F*G,F"G] and

T = { x E F"G1xPm E S

for some integer

rn 2 1)

Choose an F-basis a l , . . . , a t of T . We may find the integer m so large that afm E S for each i and pm is at least as large as the order of a Sylow p-subgroup of G. Owing to Lemma 6.9, we have

for all A; E F . Hence x = equivalently

xET

Cxgg E T if and only if xPm E S or,

if and only if

c

x;mXSpmgpmE

s

where, by our choise of rn, each gPm is a $-element and for each g E G, A, E F* satisfies g P m = A:mgpm. Let C1, Cz, . . . ,C, denote the a-regular classes of p'-elements of G, and Put Cz*= (9 E GlgPmE C;} (1 5 i 5 s )

Induction from normal subgroups

180

By Lemma 6.7, we may assume that CY is a standard cocycle. Applying (8) and Lemma 6.10, we therefore deduce that x = C xga E T if and only if /

\

(c

g€C?

pm

= XgXg)

c

(XgXg)Prn

=0

S€CY

for each i E (1,. . . ,s}. Thus xgg E T

if and only if

c

xgXg = 0

(9)

gECf

for all i E (1,. . . ,s}. Now define 1c, : F"G + F x the F-linear map given by

x F (s times ) as

with ,8; = C g E C f zgXg. Then 1c, is clearly surjective and, by (9), T is the kernel of +. Thus, by Lemma 6.12, r = dirnF(F"G/T) = s

and the result follows.

7.

Total and absolute indecomposability of induced modules

The notion of indecomposability of a module can be strengthened in two different ways. The first is absolute indecomposability in the sense of Huppert and the second is absolute indecomposability in the sense of Green. Since we shall use both notions, it will be convenient to rename the stronger one (namely absolute indecomposability in the sense of Huppert) as total indecomposability. Throughout this section, N denotes a normal subgroup of a finite group G, R a commutative local ring and A an R-algebra. Let V be an A-module and let S = EndA(V). We say that V is totally indecomposable if = R * 1s J ( S )

s

+

7.1. Lemma. The following conditions are equivalent: (i) V is totally indecomposable.

7. Total and absolute indecomposability of induced modules

181

(ii) J ( R ) S E J ( S ) and S / J ( S ) is ofR/J(R)-dimension I . (iii) J(R)S C J ( S ) and S / J ( S ) E R / J ( R ) as R/J(R)-algebras. (Note that if R is noetherian and V finitely generated, then J ( R ) S J ( S ) b y Lemmas f.7and 1.2).

+

Proof. (i)+(ii): By hypothesis, S = R 1s J ( S ) ,hence

S / J ( S ) 2 R . i s / ( R . is n J ( S ) )

-

Since R is local, so is its homomorphic image R 1s and hence Re 1s n J ( S ) J ( R . l s ) . Thus, by Corollary 1.4.4(i), R . l s n J ( S ) = J ( R - I s ) . SInce J ( R 1s) = J ( R ) ls, it follows that J ( R ) S C J ( S ) . Thus S / J ( S ) is an algebra over the field R / J ( R ) having dimension 1 since = R 1s -t J ( S ) . (ii)=+(iii): Obvious. (iii)+(i): Obvious.

-

-

s

It follows from Lemma 6.1 that if V is totally indecomposable, then

V is strongly indecomposable and hence indecomposable. Moreover, as a partial converse, we have

7.2. Lemma. Let V be finitely generated, let R be a complete noetherian local ring and let the jield R/ J ( R ) be algebraically closed. If V is indecomposable, then V is totally indecomposable.

Proof. Let V be indecomposable. By the proof of CaroDary V is strongly indecomposable, hence S / J ( S )is a division ring. By Lemma 1.7 and 1.2, J ( R ) S J ( S ) and so S / J ( S )is an R/J(R)-algebra. Furthermore, by Lemma 1.7, S is a finitely generated R-module. Hen= S / J ( S ) is a finite-dimensional algebra over the field R/J(R). Since R / J ( R ) is aIgebraically closed, it follows that S / J ( S ) is of R / J ( R ) dimension 1. Hence, by Lemma 1.7, V is totally indecomposable. W The next observation will justify our concentration on totally indecomposable modules.

7.3. Lemma. Let V be a totally indecomposable G-invariant R N module, let S = EndRN(V) and let E = EndRG(VG). Then

182

Induction from normal subgroups

(i) E - J ( S ) J ( E ) and A = E / E - J ( S ) is a twisted group algebra of G / N over the field R / J ( R ) . (ii) V G is strongly indecomposable if and only if A is local. (iii) V G is totally indecomposable if and only i f A/ J ( A ) is of R/ J ( R ) dimension 1. Proof. (i) This is a direct consequence of Theorem 4.2(iv), (v) and Lemma 7.l(iii). (ii)and (iii) Since E - J ( S ) 2 J ( E ) , we have E / J ( E ) E A / J ( A ) as R/ J ( R)-algebras. Hence E is local (respectively, E / J ( E ) is of R/ J ( R)dimension 1)if and only if A is local (respectively, A / J ( A ) is of R / J ( R ) dimension 1). Invoking some general facts on twisted group algebras, we now record the following consequences of the results so far obtained.

7.4. Theorem. Let V be a totally indecomposable G-invariant RN-module, let the field R / J ( R ) be of prime characteristic p and let GIN be a p-group. Then V G is strongly indecomposable. Proof. Let A be as in Lemma 7.3(i). Then, by Proposition 6.4, A is local. Hence, by Lemma 7.3(ii), V Gis strongly indecomposable.

7.5. Theorem. Let V be a totally indecomposable G-invariant RN-module. If V G is totally indecomposable, then either G = N or c h a r R / J ( R ) = p > 0 and G / N is a p-group. The converse is true if the field R/ J ( R ) is perfect. Proof. We may harmlessly assume that G # N . Let A be as in Lemma 7.3(i), let F = R / J ( R ) and let H = G / N . Then A F"H for some a E Z 2 ( H , F * )and, by Lemma 7.3(iii), V G is totally indecomposable if and only if

dimFF"H/J(F"H) = 1

(1)

Assume that (1) holds. Then, by Lemma 6.1(ii), F"H Z F H and hence F H is local. Since H # 1, we conclude from Proposition 2.10.11 that charF = p > 0 and H is a pgroup.

7. Total and absolute indecomposability of induced modules

183

Conversely, assume that F is perfect, charF = p > 0 and H is a F H . Hence (1) holds and pgroup. Then, by Lemma 6.2(ii), F a H the result follows.

7.6. Theorem. Let R be a complete noetherian local ring, let V be a finitely generated totally indecomposable RN-module and let H be the inertia group of V . If charR/ J ( R ) = p > 0 and H / N is a p-group, then V G is indecomposable.

Proof. Owing to Theorem 2.1, we may assume that G = H , in which case the result follows by virtue of Theorem 7.4. H The result above can be strengthened under the additional assumption on R. This is the content of the following theorem.

7.7. Theorem. Let R be a complete noetherian local ring such that the field R/ J ( R ) is algebraically closed. If V is a finitely generated indecomposable RN-module and H is the inertia group of V , then the following conditions are equivalent: (i) V G is indecomposable. (ii) V G is totally indecomposable. (iii) H = N or c h a r R J J ( R )= p > 0 and HIN is a p-group. Proof. b y Lemma 7.2,(i) and (ii) are equivalent. Moreover, by Theorem 2.1, V G is indecomposable if and only if V H is indecomposable. Hence we may assume that G = H , i.e. that V is G-invariant. By Lemma 7.2, V is totally indecomposable. Moreover, since R is a complete noetherian local ring, V G is indecomposable if and only if V G is strongly indecomposable (see proof of Corollary 1.8). Hence, by Lemma 7.3(ii), V G is indecomposable if and only if the twisted group algebra A of that lemma is local. The desired conclusion is therefore a consequence of Proposition 6.4. Assume that V is a finitely generated totally indecomposable Ginvariant RN-module, where R is a complete noetherian local ring. Put F = R / J ( R ) ,E = EndRG(VG)and S = EndRN(V). Then, by

184

Induction from normal subgroups

Lemma 7.3(i),

E / E . J ( S )E F " ( G / N ) and

for some

CY

E Z 2 ( G / N ,F*)

c

E * J(S) J(E) By Theorem 1.5, any decomposition F"(G/N) = L1@

@ L,

into indecomposable right ideals of F " ( G / N ) can be lifted to a decomposition E = II@*..@I, into indecomposable right ideals of E. We are now ready to prove the following result essentially due to Conlon (1964). 7.8. Theorem. Let V be a finitely generated totally indecomposable G-invariant RN-module, where R is a complete noetherian local ring, Then, in the notation above, the following properties hold: (i) V G = @r=lI;V is a decomposition into indecomposable RGmodules such that

( I i v ) 2~ e;V

with e; = ranks(Ii) = dimFL;

(ii) I;V E IjV as RG-modules if and only i f L; % Lj as right F"(G/N)-modules. In particular, the number of nonisomorphic indecomposable direct summands ofVG is equal to the number of nonisomorphic irreducible Fa(GIN)-modules. (iii) If V is R-free, then each I;V is R-free and rankR(I;V) = (dimFL;)(rankR(V))

(1 5 i 5 n )

Proof. All of the stated properties will follow from Theorems 5.6, 1.5 and 2.12.4 provided we show that ranks(I;) = dimFL;

(1 5 i 2 n )

-

Because F a ( G / N )is identifiable with E / E J ( S ) ,the latter will follow provided we show that for e; = ranks(1;)

+

dimF(1; E J ( S ) ) / E- J ( S ) )= e;

(2)

7. Total and absolute indecomposability of induced modules

185

To prove (2), fix i E { 1,. . . ,n } and put ei = e. Then choose an S-basis { d l , . . . ,d e } of Ij and extend it to an S-basis

(this is always possible because all direct summands of E are free Smodules by Theorem 5.6(i)). Then we have

which implies that

Ij f l E J ( S ) = $;=,d;J(S)

+

-

Thus each element in (I; E - J ( S ) ) / E J ( S ) is a unique S / J ( S ) linear combination of the images of d l , . . . ,d , in ( I ; E J ( S ) / E J ( S ) . Invoking Lemma 7.1, we conclude that (2) holds and the result is established.

+ -

*

7.9. Corollary. Let R be a complete noetherian local ring such that the field F = R / J ( R ) is algebraically closed of characteristicp > 0 . If V is a finitely generated indecomposable RN-module and H is the inertia group of V , then f o r

E = EndRH(VH)

and S = EndRN(V)

the following properties hold: (i) E / E . J ( S ) 2 F " ( H / N ) for some LY E Z 2 ( H / N , F * ) . (ii) The number of nonisomorphic indecomposable direct summands o f V G is equal to the number of a-regular classes ofp'-elements of H / N . Proof. (i) By Lemma 7.2, V is totally indecomposable. Now apply Lemma 7.3(i). (ii) By Theorem 2.1(i), we may assume that H = G, i.e. that V is G-invariant . The desired conclusion now follows by applying Theorems 7.8(ii) and 6.13.

For the rest of this section, we assume that R satisfies the following two conditions:

186

Induction from normal subgroups

(i) R is a complete local ring. (ii) R is a principal ideal domain. (for example, R can be a field, or the ring of P-adic integers, P a prime ideal of an algebraic number field). If R is not a field and R satisfies (i) and (ii), then R is said to be a complete discrete valuation ring. An integral domain S containing R is an extension of R, written SIR, if the following hold: (i) S is a principal ideal domain and a local ring. (ii) S is R-free. (iii) J ( S ) e= J ( R ) S for some integer e 2 1. We refer to e as the ramification index of SIR. In case e = 1, we say that SIR is unramified. Note that if SIR is an extension of R, then S / J ( S ) is a field extension of R / J ( R ) .Of course, if R is a field, then an extension SIR is simply an arbitrary field extension. Also, if R is a field, then all extensions S/R are unramified. We say that the extension SIR is finite if S is a finitely generated R-module. In this case, if R is a complete discrete valuation ring, then so is S. This is a consequence of a classical theorem of Cohen (1946). Let A be an R-algebra and let SIR be an extension of R. Then we Put As = S @ R A It is clear that A s is an S-algebra. Now assume that A is R-free of finite rank and let { q ,. . . ,a , } be an R-basis of A. Then the map

is an injective homomorphism of R-algebras. Identifying A with its image in As, it follows that each element of A s can be uniquely written in the form EKl X;a;, X i E S. If V is an A-module, then V . = S @ R V becomes an As-module under a module action

We say that V is absolutely indecomposable if for every finite extension SIR, Vs is an indecomposable As-module.

7. Total and absolute indecomposability of induced modules

187

7.10. Proposition. Let A be an R-algebra which is R-free of jinite rank, let V , W be finitely generated A-modules and let SIR be an extension of R. Then

If V = W , then the two sides of this are isomorphic S-algebras. Proof. Let { s;} be an R-basis of S . Each element in HornA (V,W ) , can be written uniquely in the form Csj 8 fi with fj E H o ~ A ( V , W ) and with finitely many f; # 0. Consider the map given by

+

HomA(V,w)S

HOmAs(VS,WS)

Then clearly 1c, is an S-homomorphism and, in case V = W , 1c, is also a ring homomorphism. If $(Cs; @I f j ) = 0, then Cs; 8 f;(v) = 0 for all v E V . Thus all f; = 0 and therefore 1c, is injective. Now fix X E H o ~ A ~ ( Ws). V S , For each i, define f; by

Because V is finitely generated, there are only finitely many nonzero fi. Clearly, f; E HornA( V,W ) for each i and $(C s; 8 fi) = A. Thus 1c, is surjective and the result follows. H 7.11. Lemma. Let A be an R-algebm which is R-free of finite rank and let SIR be a finite extension of R. Then, for E = EndA(V), (i) J ( E ) s J ( E s ) . (ii) E s / J ( E ) s ( E / J ( E ) ) sas S-algebras.

c

Proof. (i) By Lemmas 1.7 and 1.2, E is a finitely generated Rmodule with J ( R ) E 5 J ( E ) ,so E / J ( R ) Eis a finite-dimensional algebra over the field R / J ( R ) with J ( E / J ( R ) E )= J ( E ) / J ( R ) E .Hence J(E)" 5 J ( R ) E for some rn >_ 1. If sl,. . . ,,s E S and x1).. . ,x, E J ( E ) ,then we have

188

Induction from normal subgroups

and thus J ( E ) s / J ( R ) E sis a nilpotent ideal of Es/J(R)Es. Hence

and therefore J ( E ) s E J(Es). (ii) We may write S = Rsl @ Then

... @ Rs, for some sl,. . . ,s,

E S.

which implies that

as required.

7.12. Lemma. Let D be a division ring and let F be a subfield of Z ( D ) such that dimFD < 00 and D # F . If Xo E D - F , let f ( X ) be the rnonic irreducible polynomial over F with root Xo) and let K be a field extension of F by a root of f(X). Then K @ F D is not a division ring. Proof. By hypothesis, deg f ( X ) = n > 1 and

Since f ( X ) is reducible over K 8 1 %! K , we may write

and g and h are of degree at least 1. Owing to (3), we have

Since

7. Total and absolute indecomposability of induced modules

189

it follows that K 693 D has zero divisors. 7.13. Proposition. Let A be an R-algebra which is R-free of finite rank, let V be a finitely generated indecomposable A-module and

let E = EndA(V). (i) If V is totally indecomposable, then for a n y finite extension S I R , Vs is totally indecomposable. (ii) I f V is totally indecomposable, then V is absolutely indecomposable. (iii) There exists a finite unmmified extension S I R such that

where each Proof. 7.11,

I$ is a totally indecomposable As-module. (i) By hypothesis E / J ( E ) Z R / J ( R )and so, by Lemma

E s / J ( E ) sz S 6 9 R~/ J ( R )2 S / J ( R ) S Hence

Es/J(&)

=S/J(S),

proving (i) by applying Proposition 7.10. (ii) Direct consequence of (i). (iii) Given a finite unramified extension S I R , it follows that Vs is a direct sum of, say n ( S ) ,nonzero indecomposable As-modules, where 1 5 n ( S ) 5 dirnR,JcRl(V/J(R)V).We may therefore choose S so that n ( S ) is maximum. It is obvious that every component of Vs remains indecomposable when tensored with any finite unramified extension of S. Replacing R by S, we may thus assume that Vs is indecomposable for any finite unramified extension SIR. For any finite unramified extension S I R , put

Then D ( S ) is a division ring and a finite-dimensional S/J(S)-algebra. If T / S is a finite unramified extension, then T / R is a finite unramified extension. We now show that D ( T ) is a homomorphic image of T @s

Induction from normal subgroups

190

D W Owing to Lemma 7.11(i), we have

VT

= ( t l @V S ) €B

..-€B ( t , €3 VS)

Consider the map X : T x EndA,( Vs) + EndA, (V T )given by

Then X obviously induces a well defined S-bilinear map

T x D ( S )+ D ( T ) and a homomorphism

Since by Proposition 7.10,

it follows that

is surjective. Thus D ( T ) is a homomorphic image of

T 63s D ( S ) . Choose a finite unramified extension K / R such that d i m ~ , ~ ( q D ( I < ) is minimum. If D ( K ) # l " / J ( K ) , then there is a monic irreducible polynomial f ( X ) E K / J ( K ) [ X ]of degree at least two having a root in D ( K ) . Choose fO(X) E K [ X ] such that the image of f o ( X ) in K / J ( K ) [ X ]is f ( X ) . Let S = K[b]where b is a root of f o ( X ) . It is easily verified that S / K is a finite unramified extension. By Lemma 7.12, S @K D ( K ) is not a division ring, hence D ( S ) is a proper homomorphic image of S € 3 D~ ( K ) and dirnsp(s)D(S)< d i r n ~ l ~ ( ~ ) D ( K ) , a contradiction. Thus D ( K ) = K / J ( K ) and the result follows. N

Lemma. For any two finite extensions S I / R and SZ/R, there is a finite extension T / R such that T is an extension of both S1 and S2. 7.14.

8. Crossed products over prime rings

191

Proof. This is an easy consequence of basic results in valuation theory. For a self-contained proof refer to Dornhoff (1972, p.328). W We are now ready to prove the following fundamental result.

7.15. Theorem. (Green (1959)) Feit (1962)). Let R be a complete local ring and a principal ideal domain. Let V be a finitely generated absolutely indecomposable R N - m o d u l e , let the field R / J ( R ) be of prime characteristic p and let GIN be a p-group. T h e n V Gis absolutely indecomposable.

Proof. Owing to Proposition 7.13(iii), we may choose a finite extension S I R such that every indecomposable component of (VG)s% ( Vs)" is totally indecomposable. Again, by Proposition 7.13(iii), there is a finite extension T / S such that VT is totally indecomposable. By Proposition 7.13(i), every indecomposable component of ( V G )=~( ( V G ) s ) , is totally indecomposable. But, by Theorem 7.6, (V"), is indecomposable. Hence ( V"), is totally indecomposable. If the theorem is false, then there is a finite extension K / R such that ( V G )is~decomposable. Invoking Lemma 7.14, we may choose a common finite extension L of T and K . By Proposition 7.13(i), (VG), is totally indecomposable. But K L , so (VG)Lis decomposable, a contradiction. 8.

Crossed products over prime rings

Throughout this section, G denotes a finite group, S a ring and S * G a crossed product of G over S. &call, from Proposition 3.4, that if (G, S,0,a ) is a crossed system of S * G, then S * G is a free (left and right) S-module freely generated by the elements ij,g E G, with multiplication given by

~ " s ~ ~ ~ , Y )( z~ Y1 ~ E3 2S, Z , Y E G) where a ( z , y ) E S and Is2 = a(z)(sz) E S. We now introduce some termonology. First of all, recall that an ideal f of S is said to be G-invariant if ( S I ~ : ) ( S ~= Y )s

gig-1 = I

for all g E G

Induction from normal subgroups

192

A G-invariant ideal I of S is said to be G-prime if III2 C I for Ginvariant ideals I; of S imply 11 E I or I2 E I. The ring S is said to be prime if for all nonzero ideals A, B of S we have AB # 0. An ideal P of S is called a prime ideal if SIP is a prime ring. A minimal prime ideal of S is a prime ideal P of S such that P is not properly contained in any other prime ideal. Our aim is to prove some purely ring-theoretic results concerning the structure of S * G, where S is a prime ring. These results will be applied in the next section to investigate the homogeneity of induced modules. 8.1. Lemma. If P is a prime ideal of S * G, then P n S is a G-prime ideal of S . Conversely) if A is a G-prime ideal of S ) then there exists a prime ideal P o f S * G such that P n S = A.

Proof. Observe that if I is an ideal of S * G, then since I is Ginvariant so is I n S. Conversely, if A is a G-invariant ideal of S , then by Theorem 3.6(ii), A . (S * G ) = ( S * G ) A = A * G is a two sided ideal of S * G with (A * G) n S = A. Assume that P is a prime ideal of S * G. Then, by the above, P n S is at least G-invariant. Now suppose that A and B are G-invariant ideals of S with AB P n S. Then (A * G ) ( B* G) = ( S * G)AB(S* G)

c c

( S * G)(P n S)(S * G) p,

c

so the primeness of P ensures that A * G P or B * G P. Thus A P n S or B C_ P n S, proving that P n S is G-prime. Conversely, let A be a G-prime ideal of S. Then A is a G-invariant ideal of S and so A*G is an ideal of S*G with (A*G)nS = A. Applying Zorn’s lemma, we may thus choose an ideal P of S * G maximal with respect to P nS = A. Let I , J be ideals of S* G properly containing P. Then I n S and J n S are G-invariant ideals of S properly containing A. Because A is G-prime, this impiies that

( I nS)(Jn S ) g A

8. Crossed products over prime rings

and hence I J

9 P.

Thus

193

P is prime as required.

In what follows, given a subset T of G, we write S * T for the set of elements of S * G with support in T . 8.2. Lemma. Let I be a nonzero left and right S-submodule of S* G and let T = {gl,g2,. . . , g n } be a subset of G with g1 = 1. Assume

that and I n ( S * T ' ) = O forall T ' c T

Iil(S*T)#O

For each i E { 1,. . . ,n } , define B; b y n

B; = { s E SI them exists p = C s j i j j E I

with s = sj}

j=1

Then (i) Each B; is a nonzem ideal of S . (ii) There exists a natural bijection f; : B1 4 B; such that for all s,t E S,a E B1 (a) f;(sat) = sfj(u)git (b) fl is the identity map. (iii) The elements of I n ( S * T ) are precisely the elements of the form:

c n

Q

=

f;(.)ji

with u E B1

i=l

Proof. Since I is a left and right S-submodule of S*G, it is clear that each B; is an ideal of S. Furthermore, the minimality of T implies that each B; is nonzero. Observe also that for each b; E B;, the minimality of T ensures that there exists a unique ,O = C s j i j j E I with si = b;. This is so since if p and p' are two such elements, then p - p' E I is an element of smaller support and so /? - p' = 0. By the foregoing, for each a E B1, there exists a unique CY = C;=,b& E I with bl = a . We may thus define fi : B1 4 Bi by setting f , ( a ) = b;. By the definition of f;,fl is the identity map and

Induction from normal subgroups

194

Conversely, it is clear that each element of I n (S * 2’) is of this form. Furthermore, each fi is clearly an additive bijection. Finally, let a E B1 and cy be as above and let s , t E S. Then

and since S; = 1, this implies that fj(sat) = sb;”t = s f ; ( a ) ” t

thus completing the proof. H For the rest of this section, S denotes a prime ring. In order to make further progress, we briefly discuss a certain ring of quotients M = Qo(S) which is defined in Martindale (1969). Consider the set X of all left S-module homomorphisms

f:A+S where A ranges over all nonzero ideals of S. We refer to two such homomorphisms f : A + S,g : B 3 S as equivalent if there exists a nonzero ideal C of S with C C A n B and with f(c) = g(c)

for all c E C

This relation is obviously reflexive and symmetric. It is also transitive, because for all nonzero ideals A, B of S, A n B is a nonzero ideal of S. We have thus defined an equivalence relation on the set X . For each f E X , let [f,A] (or f if no confusion can arise) denote the class of the function f above. We can define an arithmetic on the set &o(S) of these classes by

if,A1

[9, BI = [fs,BAI where fg is the composition first f then g . In this way we obtain a ring, called the Martindale ring of quotients of S. *

8. Crossed products over prime rings

195

The following five lemmas provide some basic properties of the ring Qo(S)* 8.3. Lemma. For any f : A + S in X , f = 0 if and only if f = 0. In particular, i f g : B + S is another element o f X , then f = ij

ifand o n l y i f f = g o n A n B . Proof. It is obvious that f = 0 implies f = 0. Conversely, suppose that f = 0. Then there exists a nonzero ideal B of S contained in A such that f ( B )= 0. Note that B A G B and so 0 = f ( B A )= B f ( A ) Because S is prime, it follows that f ( A ) = 0 and therefore f = 0. The second assertion follows from the first and the fact that f = g if and only if f - g = 0.

For every a E S, let

a, E

Homs(S,S) be defined by

a,(z) = sa.

8.4. Lemma. (i) The map S + Qo(S) d efined b y a H a, is an injective ring homomorphism. (ii) Iff : A + S is an element of X , then iirf = f ( a ) , .

+

Proof. (i) If a , b E S, then T, = 1, arb+- = (ab), and a, br = ( a b),. Hence the given map is a ring homomorphism. If a, = 0, then a, = 0, by Lemma 8.3. Thus a = 0 and (i) follows. (ii) For every a E A, Sa G A, hence f is defined on Sa. Therefore, for every x E S ,

+

proving that

- = f ( a ) r *W

~ , =f a,f

Thanks to Lemma 8.4(i), we may identify S with its image in Qo(S). With this identification, Lemma 8.4(ii) tells us that for f : A + S in

x,

af = f (4

for all a E A, f E QO(S)

(1)

Induction from normal subgroups

196

8.5. Lemma. Let M = Qo(S) be as above. (i) If m E M and Am = 0 for some nonzero ideal A of S , then

m=0. (ii) If ml,m2,...,mk E M , then there exists a nonzem ideal A of S with each Am; in S . (iii) M is a prime ring. (iv) If Q is an automorphism of S , then 0 extends uniquely t o an automorphism of M . (v) Z ( M ) = CM(S)and Z ( M ) is a field. Proof. (i) Let m E M with Am = 0. If m = f,then (1) shows that ! ( a ) = 0 for all a E A. Thus f vanishes on an ideal in its domain and therefore m = f = 0. (ii) Let ml,m2,. . . ,mk E M with m; = &. We may harmlessly assume that all the fi are defined on the common domain A. Then, by (11,

-

am; = afi = f;(a) E S

for all a E A. (iii) Let I be a nonzero ideal of M . If 0 # m E I , then by (ii) and (i), 0 # Am E I n S for some nonzero ideal A of S. Hence every nonzero ideal of A4 meets S nontrivially. Because S is prime, so is M . (iv) Let Q be an automorphism of S and let f : A + S. Then the map f":A"+S given by

mu) = f (4"

is clearly a left S-module homomorphism. From this we easily deduce that the map f H f" gives rise to an automorphism of M extending 0. To establish uniqueness of extension, it suffices to verify that if T is an automorphism of M fixing S elementwise, then T = 1. To this end, let m E M and let A be a nonzero ideal of S with Am C S. Then, for all a E A, we have

am = (am). = armT= am7 Thus A(m - m') = 0 and (i) implies that m7 = m. (v) Suppose that m E CM(S) and that m # 0. Because m cen-

8. Crossed products over prime rings

197

tralizes S, it is clear that T = {t E Sltm = 0) is a two-sided ideal of S and so we must have T = 0 by (i). Owing to (ii), we may choose a nonzero ideal A of S with Am C S. Then the map f : A -+ S defined by f(a) = am is an injective homomorphism of left S-modules. If B = f ( A ) ,then B = Am = mA since m E CM(S).Now there exists an inverse map g : B + S so that f g = 1 A . Since B is an ideal of S , ij E M and g is an inverse of f = rn. Moreover, because rn centralizes S, we must have 7n-l E C M ( S )and therefore C M ( S )is at least a division ring. Finally, conjugation by m induces an automorphism of M which is trivial on S. Hence, by (iv), the automorphism must also be trivial on M . Thus m E Z ( M ) and the result follows. H The preceding results indicate that the structure of Qo(S) is very close to that of S. Our next lemma will indicate that Qo(S) is large enough to contain certain needed additional units. 8.6. Lemma. Let CT be an automorphism o f S and let a , b E S be fixed nonzero elements. If for all s E S

as6 = bs'au

then there exists a unit m E M = Qo(S) such that b = am such that m-'sm = so

for all s E S

Proof. Put A = SaS,B = SbS and define the maps f : A + B and g:B+Aby

To prove that f is well-defined, it suffices to verify that C z i a y i = 0 implies Cx;by,O= 0. To this end, suppose that C x i a y i = 0. Then, for all s E S, the formula asb = bs'au yields

198

Induction from normal subgroups

and so C xiby: = 0 since a" # 0 and S is prime. Similarly, if C xiby; = 0, then for all s E S, we have

and we conclude that Cx;ayp-' = 0. Thus both f and g are welldefined and since they are clearly homomorphisms of left S-modules, we have f = m E M and ij E M . Furthermore, fg = 1~and g f = lg, so g = rn-l and m is a unit of M . Note that a,f is defined on S and for all x E S we have

Thus ii,f = b,, or equivalently, am = b. Finally, let c E S. Then gcrf is defined on B and for all xby E B we have

Thus gC,f =

and so rn-lcrn = c", as required.

8.7. Lemma. Let M = &O(S), let c be an automorphism of S and let A,B be nonzero ideals of S . Suppose that f : A + B is an additive bijection which satisfies f ( s a t )= s f ( a ) t " for all s,t E S and a E A . Then m = f is a unit an M , m-lsm = s" for all s E S and f ( a ) = am for all a E A .

Proof. Because f is an additive bijection, its inverse g : B + A is also an additive bijection. Furthermore, for all s , t E S and b E B , g(sbt) = sg(b)t"-' Observe also that f and g are left S-module homomorphisms so that m = f and g are elements of M . Moreover, because g is an inverse cif

8. Crossed products over prime rings

199

f,we have g = m-l. Let q E S. Then g q r f is defined on B and for all b E B , we have

Thus g q r f = q: and this yields rn-lqrn = q' for all q E S. Finally, because m = f we have f(a) = a m for all a E A. Let CT be an automorphism of S. Following Kharchenko (1975), we say that is X-inner if it is induced by conjugation by a unit of M = Qo(S). In other words, these automorphisms arise from those units m E M with m-lSm = S.If ml and m2 are two such units, then clearly so is mlm2. Applying Lemma 8.5(iv), we deduce that the set of all X-inner automorphisms of S is a normal subgroup of A u t ( S ) . Now let S * G be given and let G* be the group of graded units of S * G. Then G* acts on S by conjugation and the elements of U ( S ) surely act as X-inner automorphisms. Taking into account that G ' / U ( S ) 2 G, we conclude that

Ginn = { g E GIs I+

is an X-inner automorphism of

gs

S}

is a normal subgroup of G.

By Lemma 8.5(iv), the automorphism s H g s of S extends to a unique automorphism of M which we denote by the same symbol. We now define M*G to be the free M-module freely generated by (31s E G } and with multiplication given by (az)(@) = a"ba(x, y)zy

for all a , b E M and x , y E G. Here of course a : G x G + U ( S ) is given by

a ( x , y ) = zyzy- 1

The first part of the lemma below demonstrates that M G over M extending S * G.

* G is in fact

a unique crossed product of

Let M = Qo(S) and let F = Z ( M ) . (i) There exists a unique crossed product M * G extending S (ii) If E = C M * G ( M ) , then E C M * G;,,, M * Ginn = M

8.8. Lemma.

* G. @F

E

Induction from normal subgroups

200

and E = FtG;,,, some twisted group algebra of Gin,, over the field F . (iii) If H is a subgroup of Gin,, then

M

*H

=

( E n M * H ) ) and E n ( M * H ) = F t H

(iv) If L is a G-invariant ideal o f E , then L ( M * G ) = ( M * G ) L is an ideal ofM*G. Furthermore, (M*G)L considered as a left M-submodule of M * G is a direct summand of M * G . Also, (M*G)Ln(M*Gi,,) = M L and if L # E , then ( M * G ) L n M = 0. Proof. (i) Owing to Lemma 8.5(iv), the automorphisms of S extend uniquely to those of M . Thus, the above definition of M * G is the only possible extension of S * G. We are therefore left to verify the associativity of the multiplication. Let G* be the group of graded units of S * G. observe that G* acts on S and therefore the uniqueness of extension ensures that we obtain a group action of G* on M . This fact will be used implicitly in the computations below. Given a , b, c E M and x, y, z E G, we have

-

[ ( a i ) ( b i j ) ] ( c .= ~ )mlxyz

and

( a i )[ (bij)(cz)] = m a w for some ml,m2E M . An easy computation shows that ml = a"ba(x, y)"Yca(x, y)-lsl

and m2

= a"b"( YC)SZ

where s1 and s2 are elements of S independent of a , b, and c. Because

the a , b,and c terms in the two expressions are equal. But in the special case when a = b = c = 1 the above products belong to S * G and so s1 = s2. Hence ml = m2 and M * G is associative. (ii) Let y E E and let x E Suppy, say y = rnI .. By Lemma

+-

8. Crossed products over prime rings

8.5(i),(ii), there exists a E S with ay E S Because y commutes with sa, we have asay = aysa

201

* G and with x E S u p p a y .

for all s E S

We deduce therefore that

asamii = amiisa = am"s"aii Since b = am is a nonzero element of S, we see that the identity

asb = bxsxa holds for all s E S. Applying Lemma 8.6, we deduce that s H "s is an X-inner automorphism of S and thus x E Ginn. This proves that E C M * Gin,. For each g E G;,,, choose a unit m, E M inducing the automorphism s H g s on S and put ij = mi's. Then the elements 6 for all g E Gin, form an M-basis for M * G;,,. The elements # , g E Ginn are obviously F-linearly independent and we claim that they form an F-basis for E. To this end, observe that each ij is a unit in M * G which acts by conjugation on M centralizing all of S. Invoking Lemma 8.5(iv), we conclude that j must centralize all of M , i.e. j E E . Suppose that y E E C M * Gin,. Then we can write y = C-y,ij with y, E M and it is clear that each ygij centralizes M . Because ij is a unit of E , we have y, E M r ) E = F and we infer that the elements form an F-basis for E . Furthermore, because F C Z ( E ) we must have M * Gin, = M @= E . To establish the Iast assertion, note that E is an associative Falgebra with basis (519 E Gin,}. Furthermore, for s,y E G;,,, itg E E and itij = m ( z , y ) @ for some m ( z , y ) E M . It follows that E is isomorphic to FtG;,,, some twisted group algebra of Gin, over the field F . (iii) This is a direct consequence of the way the algebra FtG;,, is constructed. (iv) Let L be a G-invariant ideal of E. Then, by definition of E , we have LM = M L and the G-invariance yields L j = gL for all g E G.

Induction from normal subgroups

202

We deduce therefore that L(M * G) = ( M * G)L is an ideal of M

* G.

Let Y be a transversal for Gin, in G and let L' be the F-complement for L in E. Because M * Ginn = M @gF E = M L @ ML', we have

M

*G=CM YEY

L @~

CML'~ YEY

Moreover, since L is G-invariant,

( M * G)L = C

( M * G;,,)Ly

=

Y€Y

C MLij Y €Y

which implies that ( M * G)L is an M-direct summand of M thermore, C y e yMLjj is clearly a direct sum and so

* G.

Fur-

( M * G)L n ( M * Gin,) = M L If L # E , then we can choose L' above to contain the identity element 1. But then CyEY ML'Y contains M and hence ( M * G)L n M = 0, as required. From now on, we shall employ the following notation:

M = &o(S),E= CM*G(M) and F = Z ( M ) We wish to apply the foregoing results in order to establish a bijective correspondence between the prime ideals P of S * G satisfying P n S = 0 and the G-prime ideals of E. Recall that the group G acts on E by the rule: g x = gxg-1 for all x E E , g E G and that, by definition, a G-invariant ideal A of E is G-prime if A1 A2 A for G-invariant ideals A; of E implies that A1 C A or A2 E A. If L is a G-invariant ideal of E , then we put L" = L ( M

* G) n S * G

so that L" is an ideal of S * G by Lemma 8.8(iv). For any ideal I of S * G we put

I d = { 7 E El Ay 5 I

for some nofizero ideal A of S}

8. Crossed products over prime rings

203

We claim that I d is a G-invariant ideal of E . Indeed, assume that y1,y2 E I d with Alyl I , A272 I and let t E E. Owing to Lemma 8.5(ii), there exists a nonzero ideal B of S with B t C S*G. Since y 1 , ~ 2 and t centralize S, we have

AIB(7lt) = (AlTl)(Bt)c I and

BAl(tT1) = (Bt)(Al71)c I

+

Thus 71 y 2 , y l t , t ~ 1E I d so I d is an ideal of E which is clearly Ginvariant, as claimed. Let I be an ideal of S * G. Following Lorenz and Passman (1979), we say that I is S-cancelable if for every x E S * G and any nonzero G-invariant ideal A of S , Ax I implies that x E I . It is obvious that any prime ideal P of S * G satisfying P n S = 0 is S-cancelable.

c

Let L, Ll,and L2 be G-invariant ideals of E . Then (i) L" is S-cancelable. (ii) L;"Ly (L1L,)".

8.9. Lemma.

Proof. (i) Suppose that A z C L" for some x E S * G and some nonzero G-invariant ideal A of S. By definition of L", we have Ax C ( M * G)L. By Lemma 8.8.(iv), ( M * G)L is a left M-module direct summand of M * G and hence

M

* G = ( M * G ) L@ K

for some M-submodule K of M ( M * G ) L ,x 2 E I<, we have

* G.

Writing x =

21

A ( x - X I ) = Ax2 E ( M * G ) L f l I< = 0 Applying Lemma 8.5(i), we see that x = z1 and thus

z E ( M * G)L n S * G = L",

+ x2 with

XI

E

Induction from normal subgroups

204

proving (i). (ii) Owing to Lemma 8.8(iv), we have ( M * G)Li = Li(M * G) and so

LyL; C L i ( M * G)L2(M * G ) = L1L2(M * G ) Moreover, LyLi C S * G and thus

LyLy

c L1L2(M * G ) n S * G = (LiL2)",

as desired. As a final preparatory result, we now record the following technical lemma. 8.10. Lemma. With the notation above, we have (i) For any G-invariant ideal L of E , L = LUd. (ii) If I is an ideal of S * G, then I Id".

c

Proof. (i) Given y E L , it follows from Lemma 8.5(ii) that By C S * G for some nonzero ideal B of S. Then By = L ( M * G ) n S * G = L", so y E LUdand therefore L 5 LUd.Conversely, if y E Lud,then A y C L" for a nonzero ideal A of S. Thus

by Lemma 8.8(iv). Let L' be an F-complement for L in E and write y = y1 t 7 2 with y1 E L and 7 2 E L'. Then

and Lemma 8.5(i) implies that y = y1 E L. (ii) Fix x E I. The case )SuppxJ= 0 being trivial, we show that x E Id" by induction on ISuppxl. Suppose that x # 0 and that the result is established for all elements y E I of smaller support size. Choose T C Suppx minimal with respect to the property that I n S * T # 0. We may assume that 1 E .'2 Indeed, if y E T , then suppxy-' = (Suppx)y

-' 3- Ty-l

8. Crossed products over prime rings

205

Now Ty-* also has this minimal property and 1 E Ty-'. Because it suffices to verify that xy-' E Id", we can replace x by sj-l and T by

Ty-'. If T = {gl = 1,g2,. . . ,g,}, then Lemma 8.2 applies and we employ its notation. There exist nonzero ideals A = B1, B2,. . . ,B, of S and additive bijections fi : A + Bi satisfying f ( s a t ) = sfi(a)git

for all a E A , s , t E S

Thus, by Lemma 8.7, there exist units ml, m2,. . . ,m, of M such that m-lsm = g r s for all s E S and f;(u) = am* for all a E A. Moreover, since fl is the identity function, we have ml = f 1 = 1. Let ,f3 = Crnig; E M * G. Because gls = gjsg;' for all s E S, we see that migi centralizes S. But m;gi then yields an automorphism of M acting trivially on S. Applying Lemma 8.5(iv), we deduce that mig; E E = CM*G(M) and thus p E E. Furthermore, since f i ( a ) = ami for all a E A, it follows from Lemma 8.2(iii) that for all a E A

Thus, by definition, we have ,f3 E I d . Let s be the identity coefficient of x and let a E A . Put 7 = ax - aps. Since Suppp = T C_ Suppx, 1 E T and ml = 1, we see that ISuppyl < ISuppzl. Moreover, y E I since u p E I , so by induction we have 7 E Id". Observe that /3 E I d and ups E S * G so ups E Id" and thus ax E Id" for all a E A. Thus, if

then D is a nonzero G-invariant ideal of S with Dz C_ Id". But, by Lemma 8.9(i), Id" is S-cancelable and therefore x E Id". The desired conclusion now follows by induction. We are now ready to prove the main result of this section. 8.11. Theorem. (Lorenz and Passman (1979)). Let S * G be a crossed product of the finite group G ouer the prime ring S and let E = FtGinn be the centratizer of M = Qo(S) in M * G where F = Z ( M ) .

Induction from normal subgroups

206

Then the map P I-+ Pd yields a bijective correspondence between the prime ideals P of S * G with P n S = 0 and the G-prime ideals of E . The inverse of this map is given b y L H L". Proof. Let P be a prime ideal of S * G with P n S = 0, and let L be a G-prime ideal of E. We must show that (i) Pd is a G-prime ideal of E and P = Pdu. (ii) L" is a prime ideal of S * G with L" n S = 0 and L = Lud. We first demonstrate that P = Pdu. Owing to Lemma 8.10(ii), we have P C Pd". Suppose that z E Pdu. Then z E S * G and z E (A4* G)Pd so

By definition of P d , there exist nonzero ideals D;of S with Di& = G;Di C P . Setting D = r)gEGnr=l gD;, it follows that D is a nonzero Ginvariant ideal of the prime ring S with 6;D P for all i. Furthermore, by Lemma 8.5(ii), there exists a nonzero ideal B of S such that BP; C S * G for all i and by the foregoing argument we may also assume that B is G-invariant. The conclusion is that

( B * G ) z ( D* G) = ( S * G ) B z D ( S* G) 5 C ( S * G)BPiSjD(S * G ) i

But P is prime and P does not contain either B * G or D * G since P n S = 0. Hence z E P , so Pdu 5 P and thus P = Pd". To show that P d is G-prime, suppose that MIA42 C_ P d for some G-invariant ideals MI, M 2of E . Then, by Lemma 8.9(ii),

M,"M," C_

(MlM2)"

2 PdU= P

Since P is prime, this implies Mi" C P for some i and, by Lemma 8.10(i), Mi = MTd 5 P d . Thus Pdis G-prime and (i) is established. Because E is a finite-dimensional algebra over the field F , we see that L is in fact G-maximal. Moreover, it is an immediate consequence

8. Crossed products over prime rings

207

of Lemma 8.8(iv), that L" n S = 0 and, by Lemma 8.10(i), we have L = Lad. Suppose that L" c I for some ideal I of S * G. Then I d 2 LUd= L and I d # L since otherwise Lemma 8.10(ii) would yield I Id" = L". Thus the G-maximality of L implies that I d = E and therefore I n S # 0. To prove that L" is prime, assume that L" C 11, L" C I2 for some ideals I1,12 of S * G. Then, by the above, 1 1 n S # 0 and 1 2 n S # 0, hence 0 # ( I , n S ) ( I zn S ) 1112n S

c

Since L" n S = 0, we deduce that result is established. 4

1112

L". Thus L" is prime and the

As an easy application of the results so far obtained, we finally record

8.12. Theorem. (Lorenz and Passman (1979)). Let S * G be a crossed product of the finite group G over the prime ring S . (i) A prime ideal P of S * G is minimal if and only if P n S = 0 . (ii) There are only jnitely many such minimal prime ideals, say Pl, .. . ,P,. Furthermore, if chars = p is a prime and G is a p-group, then n = 1. (iii) J = n;=, Pi is a unique largest nilpotent ideal of S * G. Proof. Because E = FtG;,, is a finite-dimensional algebra over the field F , there are only finitely many, say L1,. . . , Ln, of G-prime ideals of E . Furthermore, if chars = p is a prime and G is a p-group, then since charF = chars, it follows from Proposition 6.5 that E has a unique prime ideal. Thus in this case E has a unique G-prime ideal and therefore n = 1. Returning to the general case, we obviously have J ( E ) = n;=lLi. If P; = Ly , then by Theorem 8.11, PI, . . . , P, are the unique prime ideals of S * G which are disjoint from S - (0). Put J = n?.=,P;.Then J C Ly so J d C Lyd = L; by Theorem 8.11 and therefore

J d C n;==,Li =J(E)

208

Induction from normal subgroups

Applying Lemma 8.10(ii), we see that J C Jdu J(E)". Because J ( E ) is nilpotent, we deduce, from Lemma 8.9(ii), that both J ( E ) " and J are nilpotent. On the other hand, each Pi surely contains all nilpotent ideals of S * G. Thus J 2 J(E)" and hence J = J(E)" is clearly the largest nilpotent ideal of S * G. Finally, let P be any prime ideal of S * G. Because J is nilpotent, P 2 J and so P 2 P; for some i. Thus the minimal ideals of S * G are the minimal members of the set {PI,. . . ,P,}. But P; 2 Pj implies that L; = pi" 3 - pd3 = Lj and thus, since Lj is G-maximal, we must have i = j . This shows that P I , .. . ,P, are precisely the minimal prime ideals of S * G, thus completing the proof. 9.

Homogeneity of induced modules

Throughout this section, N denotes a normal subgroup of a finite group G and R a commutative ring which is either artinian or a complete noetherian local ring. Let V be a finitely generated indecomposable RN-module. Our aim is to discover circumstances under which the induced module V Gis homogeneous, i.e. V Gis a direct sum of isomorphic indecomposable modules. The following ring- theoretic result will clear our path. 9.1. Theorem. Let p be a prime and let S*G be a crossed product of a finite p-group G over a simple artinian ring S of characteristic p . Then S * G is an artinian ring such that S * G / J ( S * G ) is simple.

Proof. Because S * G is a finitely generated S-module and S is artinian, S * G is certainly artinian. Since S is simple, S is also prime. Applying Theorem 8.12(ii), we deduce that S * G has a unique minimal prime ideal, say P . Again, by Theorem 8.12(iii), P is a unique largest nilpotent ideal of S * G. But S * G is artinian, hence P = J ( S * G). Because P is prime, we deduce that S * G / J ( S * G) is a prime ring. But S * G / J ( S * G) is semisimple artinian, hence a direct product of, say n, simple rings. If n > 1, then there are two nonzero ideals of

9. Homogeneity of induced modules

209

S * G / J ( S * G) whose product is zero, which is impossible. Hence S * G/J(S * G) is simple and the result follows. The following result in the special case where R is a perfect field of characteristic p > 0 was established by Harris (1986). The general case is due t o Karpilovsky (1989).

Let R be a commutative ring which is either artinian or a complete noetherian focal ring and assume that R / J ( R ) is of prime characteristic p . If N is a normal subgroup of a finite group G and V a finitely generated indecomposable RN-module such that H/N is a p-group, where H is the inertia group of V , then 9.2. Theorem.

for some positive integer e and some indecomposable RG-module U . Proof. Owing to Theorem 2.1, we may harmlessly assume that H = G, in which case GIN is a p-group and V is G-invariant. Put

E = E ~ ~ RVG G )( , S = E n d ~ ~ ( vand ) A =E/E J(S) Because V is G-invariant, A is a crossed product of GIN over S / J ( S ) and E J ( S ) C J ( E ) , by Theorem 4.2(iv). Observe also that, by Lemmas 2.11.2 and 1.7, and Corollary 1.6, S is local. Hence S / J ( S ) is a division ring. Furthermore, J ( R ) S C J ( S ) by Lemmas 1.2 and 1.7. Thus S / J ( S ) is of characteristic p and therefore, by Theorem 9.1, A / J ( A ) is simple. Moreover, A / J ( A ) is artinian since it is a finitely generated R/J(R)-module. But

J ( A ) = J ( E ) / E- J ( S ) since E . J ( S ) C J ( E ) . Thus A / J ( A ) E E / J ( E )and therefore E / J ( E ) is simple artinian.

. . ,(p7 be the primitive idempotents of E corresponding to Let yl,. the decomposition of V G into indecomposable direct summands. By Lemma 5.2(iii), it suffices to show that p;E S pjE for all z , j E { 1 , . . . , T } . Let pi be the image of (pi in E = E / J ( E ) . By Proposition 1.4.17 and Theorem 1.5, it suffices to verify that &; E p j for

Induction from normal subgroups

210

all i , j . But each (pi is a primitive idempotent of the simple artinian algebra El hence the result. W

Remark. Assume that R is the field of two elements, G the symmetric group of degree 3 and N the alternating group of degree 3. Then there exists a unique (up to isomorphism) irreducible RN-module of dimension 2. It is well known (see Huppert and Blackburn (1982, Ex.70)) that V G2 2U where U is the unique irreducible RG-module of dimension 2. Consequently, the positive integer e of Theorem 9.2 need not be equal to 1. 10.

F’robenius and symmetric algebras

Our intent here is to establish a basic knowledge about Frobenius and symmetric algebras. All the information recorded will be applied in the next section. Throughout this section, we set A* = l $ o r n ~ ( A , F )where , A denotes a finite-dimensional algebra over a field F . The left (right) Amodule means the regular left (right) A-module. 10.1. Lemma. Let V be a finite-dimensional space over a field F and let f : V x V + F be a bilinear form. Then the following conditions

are equivalent: (i) f (z, V ) = 0 implies z = 0. (ii) f (V,z) = 0 implies x = 0. Proof. Let el, e 2 , . . . ,en be a basis of V and let b;j = f (ei, e j ) . Then it is clear from the bilinearity that f ( z , V ) = 0 if and only if f(z,e i ) = 0 , 1 _< i 5 n. Similarly, f ( V ,z) = 0 if and only if f ( e ; ,z) = 0, 1 5 i 5 n. Now write z = C A j e j with A j E F . Then

Hence f ( x , V ) = 0 if and only if (A1, A 2 , . . . , A,) system of homogeneous linear equations

is a solution of the

211

10. Frobenius and symmetric algebras

Similarly, f (V, z) = 0 if and only if the A's satisfy bilA1

+

+ + binA,

= 0,

15 i 5 n

(2)

We know from linear algebra that (1) or (2) has only a zero solution if and only if det(b;j)# 0. Thus the lemma is proved. We refer to a bilinear form f : V x V + F as being nonsingular if it satisfies the equivalent conditions of Lemma 10.1. Let f : A x A 4 F be a bilinear form. Then f is said to be associative if f ( z y , z ) = f (5,y z )

for all z, y , z E A

while f is called symmetric if f(Z,Y)

=f ( Y 4 )

for all 3 , Y E A

10.2. Corollary. Let $ E A* and let f : A x A 3 F be defined by f(z,y) = $(zy). Then f is an associative bilinear form on A which is symmetric if and only if $ ( z y ) = $(yz) for all z, y E A . Furthermore, the following conditions are equivalent: (i) KerlC, contains no nonzero right ideals of A. (ii) Ker$ contains no nonzero left ideals of A . (iii) f is nonsingular.

Proof. It is obvious that f is a bilinear form on A. Given z, y, z E A , we have f ( z y 7 z ) = Itt((zy)z) = Itt(+Z))

=f

(Z7 Y Z ) ,

proving that f is associative. By definition, f(z, y) = f ( y , 5) if and only if $ ( z y ) = $(yz). Hence f is symmetric if and only if $ ( z y ) = $(yz) for all z , y E A . Note that f ( z , A ) = 0 if and only if $ ( z A ) = 0, while f ( A , z ) = 0 if and only if $ ( A x ) = 0. The desired conclusion is therefore a consequence of Lemma 10.1. 10.3. Lemma. The following conditions are equivalent: (i) There exists $ E A* such that Ker$ contains no nonzero right

Induction from normal subgroups

212

ideals of A. (ii) There exists $ E A" such that Ker$ contains no nonzero left ideals of A. (iii) There exists a bilinear form f : A x A + F which is both associative and nonsingular.

Proof. (i)+(ii): Apply Corollary 10.2. (ii)+(iii): Apply Corollary 10.2. (iii)+(i): Let ll, : A + F be defined by $(z) = f(z,1). It is obvious that ll, E A*. Let a E A be such that $ ( a A ) = 0. Then f ( a b , 1) = 0 for all b E A. But f(ab, 1) = f ( a , b), so the nonsingularity of f implies that a = 0. H We say that A is a Frobenius algebra if it satisfies the equivalent conditions of Lemma 10.3.

The following conditions are equivalent: (i) There exists $ E A* such that Kerll, contains no nonzero right ideals of A and ll,(xy) = +(yz) for all z , y E A . (ii) There exists E A* such that ker$ contains no nonzero left ideals of A and $ ( x y ) = ll,(yz)for all z, y E A . (iii) There exists a bilinear form f : A x A * F which is associative, nonsingular and symmetric. 10.4. Lemma.

+

Proof. (i) +(ii): Apply Corollary 10.2. (ii) +(iii): Apply Corollary 10.2. (iii) +(i): Define II, : A + F by $(z) = f ( z , l ) . Then, by Lemma 10.3, $ E A* and ICerII, contains no nonzero right ideals of A. Furthermore, since f is symmetric, we have $(ZY)

= f b y , 1) = f (z, Y ) = f ( Y , 2) = f ( y z ,1) = $ ( Y 4

as required. We say that A is a symmetric algebra if it satisfies the equivalent conditions of Lemma 10.4. In order to avoid exessive verbosity, it will be convenient to say that ( A , $ ) is a Frobenius (or symmetric) algebra whenever $ satisfies conditions of Lemma 10.3(or Lemma 10.4).

10. Frobenius and symmetric algebras

213

As a preliminary to the next observation, note that A* is an ( A ,A)bimodule via

for all z , a E A and f E A*. Lemma. (i) Suppose that ( A , $ ) is a Frobenius algebra. Then the map f : A + A* given b y f ( a ) ( z )= $ ( z a ) for all z, a E A is an isomorphism of left A-modules. Moreover, if ( A , $ ) is a symmetric algebra, then f is an isomorphism of (A,A)-bimodules. (ii) Suppose that f : A + A* is an isomorphism of left A-modules and set $ ( a ) = f ( l ) ( a ) for all a E A. Then (A,$) is a Frobenius algebra. Moreover, i f f is an isomorphism of (A,A)-bimodules, then ( A , + ) is a symmetric algebra. 10.5.

Proof. (i) It is clear that for all a E A , f ( a ) E A*, and that f (a: t y ) = f (a:) f ( y ) for all a:, y E A. Given r, z E A, we have

+

proving that f is a homomorphism of left A-modules. By the definition of +, f is injective, and since dirnFA = dimFA*, f is a bijection. If ( A , $ ) is a symmetric algebra, then for all r,z E A, we have

as required.

(iii) It is obvious that 1c, E A*. Suppose that $(aA) = 0. Then, for all x E A , 0 = f ( l > ( a a : )= f ( z ) ( a ) and hence a = 0. Thus ( A , $ ) is a Frobenius algebra. Assume that f is an isomorphism of (A, A)-bimodules. Then, for all x , y E A , we have

as desired. W

Induction fkom normal subgroups

214

10.6. Corollary. The following conditions are equivalent: (i) A is a Frobenius (respectively, symmetric) algebra. (ii) A E A* as left A-modules (respectively, A E A* as ( A , A ) -

bimodules). Proof. Direct consequence of Lemma 10.5. W

n:==,

10.7. Lemma. Let Al,Az,. . . ,A, be F-algebras. Then Ai is a Frobenius (respectively, symmetric) algebra i f and only i f each A; is a Frobenius (respectively, symmetric) algebra.

Proof. Assume that $; E A* is such that (A;,$;)is a Frobenius A; and define : A + F by algebra, 1 5 i 5 n. Set A =

n?=,

+

Then obviously $ E A*. Put a = ( a l , . . . , a , ) and assume that +(Aa)= 0. Let z = ( 2 1 , . . . ,z,) E A be such that zj = 0 for j # i. Then $ ( x u ) = $;(zja;) = 0 for all zi E A;. Hence each ai = 0 and therefore a = 0, proving that ( A ,$) is a Frobenius algebra. Conversely, if ( A ,$) is a Frobenius algebra, put $; = $ o X i where X i : Ai + A is the canonical injection. Then (Ai,$;) is obviously a Frobenius algebra. If each ( A ,$i) is a symmetric algebra, then for a = ( a l , . . . ,an),b = ( b l , . . . ,b,) we have

proving that ( A ,$) is a symmetric algebra. Conversely, if ( A ,+) is a symmetric algebra, then for ai, b; E A, we have

Thus each (Ai,$;)is symmetric and the result follows.

10.8. Lemma. If A is a Frobenius (symmetric) algebra, then for a n y positive integer n, M,(A) is a Frobenius (symmetric) algebra.

10. Frobenius and symmetric algebras

215

Proof. Let $ E A* be such that ( A , $ ) is a Frobenius algebra. Define the map X : Mn( A ) + F by

Then obviously X E Mn(A)*.Assume that

(ajj) E

Mn(A) is such that

We must show that ( a i j ) = 0. To this end, fix i , j E (1,. . . ,n} and denote by e;j the matrix with (i,j)-th entry 1 and 0 elsewhere. Then, for any a E A , X[(a;j)ae;j]= $(uj;u) = O Since ( A , $ ) is a Frobenius algebra, we deduce that aj; = 0. Thus ( a ; j ) = 0 and therefore M n ( A )is a Frobenius algebra. If ( A ,$) is a symmetric algebra, then n

n

/ n

n

n

\

n

n

n

proving that M,(A) is also symmetric. W 10.9. Proposition. symmetric.

I f A is a semisimple algebra, then A is

Proof. By Wedderburn’s theorem, A is isomorphic to a finite direct product of full matrix rings over division rings. Hence, by Lemmas 10.7 and 10.8, we may assume that A is a division algebra over F . Then, for any nonzero $ in A*, ( A , $ ) is a Frobenius algebra. Denote by [ A , A ]the additive subgroup of A generated by all elements of the form ab - ba, with a , b E A . It is clear that [A,A] is an F-subspace of A . To prove that ( A ,$) is a symmetric algebra, it suffices to verify that $ ( [ A ,A ] )= 0 for some nonzero $ in A*. The latter will follow provided

Induction from normal subgroups

216

we show that [A,A] # A. This last statement being independent of the ground field, we may assume that F is the centre of A . Let E / F be a field extension such that AE E M n ( E ) for some n 2 1. Because each matrix in [ M , ( E ) , M , ( E ) ]has trace 0 , we have [AE,AE] # A E . But [AE,A E ]= [A,A ] E ,so [ A ,A] # A , as required. 10.10. Proposition. If A is a Frobenius dgebra and G is a finite group, then any crossed product F-algebra A * G is Frobenius.

Proof. Let $ E A* be such that ( A , $ ) is a Frobenius algebra. Define the map X : A * G + F by

Then it is clear that X is an F-linear map. Let x = C x g g E A * G be such that X ( x y ) = 0 for all y E A * G. For any a E A , g E G, we have zag-l = C x t i a g - 1 = c x t t atg ---I tEG

t€G

Thus

0 = X(zag-') = $ ( x g g a )

for all a E A

and so $ ( x g A ) = 0 . Hence each x g = 0 and therefore x = 0, as we wished to show. Again, let A be a finite-dimensional algebra over a field F . For each subset X of A, we denote by I ( X ) and r ( X ) the left and right annihilators of X defined by

I ( X ) = { a E AlaX = 0 } , r ( X ) = { a E AlXa = 0} It is clear that l ( X ) and r ( X ) are left and right ideals of A , respectively. Moreover, if X is a left (respectively, right) ideal of A , then I ( X ) (respectively, r ( X ) )is a two-sided ideal of A. Suppose that ( A ,$) is a Frobenius algebra and that X is a subset of A. Define the subsets I X and X I of A by

' X = { a E AI$(aX) = 0 } X I = {a €Al$(Xa)=O}

217

10. F'robenius and symmetric algebras

It is clear that if X is a subspace of A, then so are ' X and X I . 10.11. Theorem. Let ( A , $ ) be a Frobenius algebra and let X be an F-subspace of A. (i) ('X)* = ' ( X ' ) = X and dimFX' = dirnF'X = dzrnFA -

dimFX. (ii) If X is a left (respectively, right) ideal of A, then r ( X ) = X I (respectively, l ( X ) = ' X ) . In particular, for every idempotent e of A (Ae)' = (1 - e ) A A/(Ae)' E eA as right A-modules d i m F e A = dimFAe

(iii) If X is a left ideal of A, then dimFX

+ d i m F r ( X ) = dimFA

and I ( r ( X ) )= x

If X is a right ideal of A, then dimFX

+ dimFl(X) = dimFA

and r ( l ( x ) )= x

(iv) The mapping X H r ( X ) is a duality of the lattice of left ideals of A onto the lattice of right ideals of A, that is X H r ( X ) is a bijection of the set of left ideals of A onto the set of right ideals of A such that

for all left ideals X l , X 2 of A . Similarly, X H i ( X ) is a duality of the lattice of right ideals of A onto the lattice of left ideals of A. (v) If ( A ,$) is symmetric, then r ( X ) = l ( X )for any two-sided ideal X ofA.

Proof. (i) Let { a l , .. . ,a , } be an F-basis of A such that { a l ,. . . ,a,} is an F-basis of X ( m 5 n ) . The map

Induction from normal subgroups

218

is an isomorphism of F-spaces such that the image of X' consists of all ( A 1 , . . . ,A,) with A1 = . . . = A, = 0. Hence dimFX' = dimFA dimFX and, by a similar argument, d i m F ' X = dimFA - dimFX. We conclude that d i r n F ( 'X)' = dirnFX Because ('X)' 2 X , it follows that ('X)' = X , as desired. The equality ' ( X ' ) = X is proved similarly. (ii) Suppose that X is a right ideal of A. Then a E ' X if and only if $ ( a x ) = 0 or if and only if $ ( a X A ) = 0. SInce the latter is equivalent to aX = 0, we have 1 ( X ) = ' X . The equality r ( X ) = X' for a left ideal X of A follows by the same argument. Aez = 0 ++ ex = 0 ++ Given z E A, we have z E (Ae)' z E (1 - e)A, so (Ae)' = (1 - e)A and A/(Ae)' G eA. Finally, dimFAe = dimFeA, by applying (i). (iii) This follows directly from (i) and (ii). (iv) That the given map is a bijection follows from (iii). Clearly we have

Applying (iii), it follows that

The corresponding assertion for left ideals follows similarly. (v) Assume that ( A , $ ) is a symmetric algebra. Then I X = X' for any subset X of A. The desired conclusion follows from (ii). Let V be a left A-module. Then the socZe of V , written SocV, is defined as the sum of irreducible submodules of V . The socle of a right A-module is defined similarly. The Zeft and right socles of A are defined as the socles of A A and AA, respectively. In case the left and right socles of A coincide, we shall write SocA to denote both of them. 10.12. Theorem. Suppose that A is a Frobenius aZgebra.

(i) The left and right socles of A coincide. Moreover, dimFSocA =

10. Fkobenius and symmetric algebras

219

dirnF(A/J(A)). (ii) For each primitive idempotent e E A, (SocA)e is the unique irreducible submodule of Ae. In particular, Soc(Ae) = (SocA)e. (iii) If A as symmetric, then for any primitive idempotent e E A, Soc(Ae) E A e / J ( A ) e Proof. Setting J = J ( A ) , it follows that r ( J ) and l ( J ) are the left and right socles of A, respectively. Let e be a primitive idempotent of A. Then J e is the unique maximal submodule of Ae and A e / J e is irreducible, by Theorem 2.12.4. By Theorem 10.11, the map X H X' is an inclusion-reversing bijection between the set of left ideals X of A contained in Ae and the set of right ideals of A containing (Ae)'. We between thus obtain an inclusion-reversing bijection X H X'/(Ae)' the set of submodules of A e and the set of submodules of A/(Ae)*. Now (Ae)' = (1 - e ) A and A/(Ae)' 2 eA by Theorem 10.11. Thus we have an inclusion-reversing bijection X H eX' between the set of submodules X of A e and submodules of eA. In particular, e(Je)' is the unique irreducible submodule of eA. Further, for y E A, we have y E ( J e ) l H Jey =

o w ey E r ( J )

which implies that e(Je)* = e r ( J ) . Thus e r ( J ) is the unique irreducible submodule of eA. Because J annihilates all irreducible A-modules, we have e r ( J ) J = 0. Thus r ( J ) J is annihilated by every primitive idempotent e E A. The latter implies r ( J ) J = 0 and so r ( J ) l ( J ) . The reverse inclusion holds by symmetry. The moreover assertion being a consequence of Theorem 10.11, (i) is established. The same argument (reversing left and right) also proves (ii). To prove (iii), let $ E A* be such that ( A ,$) is a symmetric algebra. By the foregoing, r ( J ) e is the unique irreducible submodule of Ae and J r ( J ) e = 0. Thus every f E H o m A ( A e , r ( J ) e ) induces a map X E HornA(Ae/Je,r(J)e). Hence we need only show that HornA( Ae, r ( J ) e ) # 0. Since there is an isomorphism of additive groups HornA(Ae,r ( J ) e ) e r ( J ) e given by f H f ( e ) , we are left to verify that e r ( J ) e # 0. Assume by way of contradiction that e r ( J ) e = 0. Then

Induction from normal subgroups

220

and so r ( J ) e = 0, a contradiction. This completes the proof of the theorem. 11.

Symmetric crossed products

Our aim is to prove that if S is a finite-dimensional semisimple algebra over a field F and G a finite group, then any F-algebra which is a crossed product of G over S is symmetric. Let S be an arbitrary ring, let G be a finite group and let S * G be a crossed product of G over S. As usual, for any s E S,g E G, we put

Then z H g z , z E Z(S), provides an action of G on Z ( S ) . Note that if S is an algebra over a field F, then S * G is an F-algebra if and only if G acts trivially on F . We now fix the following data: E is a G-invariant field contained in Z ( S ) I( = EG = { a E Elga = a for all g E G} F is a subfield of IC such that K / F is a finite field extension and dimES C 00 t = trE/K E + I(. Note that

t ( E )= I(

(1)

and t(ga)= t(u)

for all a E E,g E G

(2)

(see Lang (1984, Chapter VIII)). Let II, E H o ~ E ( SE,) . We say that 1c, is G-invariant if cp( gs) = gcp(s)

for all g E G, s E S

11.1. Lemma. I f ( S ,cp) is a Frobenius E-algebra, 0 and f : S * G + F is defined b y

then (i) ( S * G , f ) is a Frobenius F-algebra.

#X

(3) E Homf;(K,F )

11. Symmetric crossed products

221

(ii) If (S,cp) is a symmetric E-algebra and cp is G-invariant, then ( S * G , f) is a symmetric F-algebra. Proof. (i) If 1c, = X 0 t cp, then 1c, E H o r n ~ ( Sand, ) by (l),( A o t ) ( E ) = F . Assume that +(US) = 0 for some a E S. If cp(aS) # 0, then cp(aS) = E and $(US) = ( A 0 t ) ( E )= F = 0, a contradiction. Thus cp(aS) = 0 and therefore a = 0, proving that (S,T)) is a Frobenius F-algebra. The desired assertion now follows from Proposition 10.10. (ii) Assume the additional hypotheses of (ii) and let z = C zgg, y= C ygg be two elements of S * G , zg,yg E S. Then 0

c

f (4 = Wcp(

zggyg-lg-')))

g€G

=

-

c c Nt(cp(sY,-ls-'4>>

A(t(cp(~,(sv,-ls-'>>>>

SEG

=

s€G

-

since jyg-1g-l E S and cp(ab) = cp(h) for all a , b E S. Hence

f (4 =

c Wcp [ c c

g€G

=

g(yS-lF%s)]

-

wgcp(Yg-1g-lzgg)))

(bY(3))

gEG

=

A(t(cp(Yr19-'%g)))

g€G

=

(bY(2))

-

Wcp(CY g - 1 g - l z g 3 ) )

= f ( Y 4

as required.

11.2. Lemma. Let R be a commutative ring and let a G-graded R-algebra A be a crossed product of G over Al. Let e l , . . . ,en be orthogonal idempotents of Z ( A ) n A1 with 1 = el + - - . e,. Then

+

A = $y.=le;A is a direct sum decomposition of A into ideals e;A where each e;A is a G-graded R-algebra which is a crossed product of G over e;Al such that

Induction from normal subgroups

222

(i) e; is the identity element of e;A. (ii) ( e ; A ) , = e;A, for all g E G . (iiif e i j E U(e;A)n ((eiA),) for all g E G . (iu) U ( A )= @ L 1 U ( e ; A ) .

Proof. The proof is straightforward and so will be omitted. In order to make further progress, we need to introduce reduced characteristic polynomials. Let A be a finite-dimensional algebra over a field F . Then A is said to be separable if for every field extension E of F , AE = E @ F A is a semisimple E-algebra. For each a E A , let char.pol.AIF(a)be the characteristic polynomial of the F-linear map x H ax, x E A . The trace map TA/Fand norm map N A I Fare then defined by

where m = dimFA. Now assume that A is a separable F-algebra. Then there exists a field extension E / F and an isomorphism

of E-algebras. For each a E A , let

We now define the reduced characteristic polynomial of a E A by S

red.char.pol.A,F(a) =

char.pol.cpi(a), i=l

where char.pol.cp;(a) is the characteristic polynomial of the matrix cpi(a). It can be shown (see Curtis and Reiner (1981, p.159)) that this reduced characteristic polynomial is independent of the choice of E and of the isomorphism f . Further, its coefficients all lie in the ground field

11. Symmetric crossed products

223

F. Now put m = E n ; and write

for a E A. Then trA/F is called the reduced trace, and nrA/F is called the reduced norm. It is easily verified that trA/F : A -+ F is an F-linear map and

trA/F(ab)= t T A / F ( b U ) for all a , b E A

(4)

We now quote the following two standard facts. 11.3. Proposition. Let A be a separable F-algebra. Then the reduced trace trA/F gives a symmetric bilinear associative nonsingular form, namely

(a,b) ++ Proof.

~ T A I F ( ~ ~ (a,b ) E

A)

See Curtis and Reiner (1981, p.165).

11.4. Proposition. Let A be a semisimple F-algebra. Then A is separable if and only if the centres of the division algebras associated with the Wedderburn components of A are separable field extensions of F. Proof.

See Curtis and Reiner (1981, p.145).

Let A, B be isomorphic rings and let u : A -+ B be a ring isomorphism. Let F be a field contained in Z ( A ) such that A is a finitedimensional separable F-algebra. If L = a(F),then L is a field contained in Z ( B ) ,B is a finite-dimensional separable L-algebra and

Let a E A and let

where n

1 1 is an integer and A; E F for all i E ( 1 , . . . ,n } .

Induction from normal subgroups

224

11.5. Lemma.

With the notation above,

red.char.pol.slL(a(a))= X "

+ Q ( X ~ ) X " - ' + + a ( X , - l ) ~+ a(X,) * *

Proof. Let F and L be algebraic closures of F and L , respectively. It is well known (see Lang (1984, VII, Theorem 2.8) that Q : F + L can be extended to a field isomorphism i? : F + 1. Thus there is a ring isomorphism CY

:F 8 F

A

+L 8 L

B

such that cY(X 8 a ) = e(X) 8 a ( a )

for all X E F and all a E A. Because P 8~ A is a finite-dimensional semisimple F-algebra, there exists an F-algebra isomorphism

for some positive integer rn and some positive integers r; for all 1 5 i rn. The isomorphism 8 : F + 1 induces the ring isomorphism m

M

i=l

r=l

Setting 7 = p o p o a-', we obtain an 1-algebra isomorphism m

Also let

5

11. Symmetric crossed products

225

m

red.char.pol.B,L(a(a)) =

char.pol.$i(~(u)) i=l

We have

~i[char.pol.(cpi(a))] = char.pol.$;(a(a))

(1 5 i 5 rn)

thus completing the proof. W We have now come to the demonstration for which this section has been developed. 11.6. Theorem. (Harris (1989)). Let G be afinite group and let S be a finite-dimensional semisimpte algebra over a field F. Then any F-algebra A which is a crossed product ofG over S is symmetric.

Proof. Let el, . . . ,en be pairwise orthogonal primitive idempotents of Z(S) with 1 = el ... en. Setting Bi = eiS, we then have S = B1 @ - - - @ B, where each Bi is a simple F-algebra. The map ei I-+ gei = jeig-l, g E G, provides an action of G on {el,. .. ,en}. If f 1 , . . . ,f k , k 5 n, are the G-orbit sums, then f l , . .. ,f k are mutually orthogonal idempotents of Z ( A ) n S. Since a finite direct product of symmetric F-algebras is a symmetric F-algebra (Lemma 10.7), it follows from Lemma 11.2 that it suffices to assume that G acts transitively on { e l , . . . ,en}. Put e = el, f = 1 - e , B = B1 = eS, E = Z ( B ) and denote by H the stabilizer of e. It is obvious that E is a field, F E F e = eF 5 E , AH = {C shhlsh E S, h E H } is an F-algebra which is a crossed product of H over S, e , f are orthogonal idempotents of Z ( A H )f l S with 1 = e -t f . By Lemma 11.2, eAH is a crossed product of H over

+ +

Induction from normal subgroups

226

B. For each h E H , put k = eh. Then k is a graded unit of eAH of degree h and the map p H kpk-', p E E , h E H , provides an action of H as a group of automorphisms of E. Let I< = E H be the H-fixed subfield of E , so that F E2 Fe = eF C I<. Put t = t r E / K : E + K and fix 0 # X E H o ~ F ( K , F )Note . that B is a finite-dimensional simple E-algebra, hence the reduced trace ~ T B I EE H o r n ~ ( BE, ) is defined by Proposition 11.4. Furthermore, by (4), t r B ~ E ( a b= ) trB/E(ba)

for all a,b E B

and, by Proposition 11.3, the kernel of t r B / E contains no nonzero right ideals of B. Also t r B / E is H-invariant by Lemma 11.5. Define f : eAH + F as in Lemma 11.1 with t r B / E playing the role of cp. Then, by Lemma 11.1 (ii), ( e A H , f ) is a symmetric F-algebra. Hence, for any graded unit u of AH and any b E B ,

f(eubeu-') = f(b)

(5)

Choose gl,. . . ,gn in G such that e; = g;e;Z-l. Then B; = zBl~-' for a l l i E { l , . . . , n } . Definecp:S=@Z1B; + E by : if y = Cr=lyi E S for yi E Bi, put

c f(G-'!/;G) n

cp(d =

i=l

Then obviously cp E H o ~ F ( SF ,) , cp(ab) = cp(ba) for all a , b E S and Kercp contains no nonzero right ideals of S. Applying Lemma l.l(ii), with F playing the role of E , we are left to verify that cp( g x ) = cp(x) for all x E S,g E G. The latter will follow provided we show that for any fixed j E (1,. . .,n}, g E G, and z E Bj, cp( g z ) = ~ ( 2 ) . We have gBj3-l = ijijjB1G-'ij-' = B k for some k E { 1,. . . , n}. Hence ijE = u a where u is a graded unit of AH. SInce z-'zz E B1 = B , we have u z z g j u = e u z z s - eu

-'

and hence, by ( 5 ) ) f(uz-'z~u-')= f ( j 7 ' Z i j j ) . cp(

gz)

But then

= cp(gzg-') = f(g&Mj-li&) = f(ug;'Zijju-') = f(g371zgi) =

44,

227

11. Symmetric crossed products

thus completing the proof. H

11.7. Corollary. Let G be a finite group and let S be a finitedimensional algebra over a field F . If an F-algebra A is a crossed product of G over S, then A / A - J ( S ) is a symmetric F-algebra.

-

Proof. By Theorem 3.7(ii), A / A J ( S ) is a crossed product of G over S / J ( S ) . Since S / J ( S ) is a finite-dimensional semisimple Falgebra, the result follows by virtue of Theorem 11.6. H

To provide another application of Theorem 11.6, we need the following simple observation. 11.8. Lemma. Let R be a commutative ring and let A be a G graded R-algebra. If N is a normal subgroup of G , then A can be regarded as an GIN-graded R-algebra by means of

AgN = @xEgNAz Moreover, if A is a crossed product of G over Al, then A can also be regarded as z crossed product of G I N over

AN = @xENAx Proof. Let T be a transversal for N in G. It is clear that

A = @tETAtN Given t l , tz E T , we have

AtlNAtzN =

NAr) (@y€t2NAy) @xEtlN,yEtzNAxy= @zEtitzNAz = AtItzN (@x€tt

Assume that A is a crossed product of G over A l . Then, for all g E G, there exists a unit g of A in Ag in which case j E A g ~as, required. H 11.9. Corollary. Let F * G be a crossed product of a finite group G over a field F . If N is a normal subgroup of G , then

F

* G / ( F * G ) ( J ( F* N ) )

Induction from normal subgroups

228

is a symmetric FG-algebra, where FG is the fixed field of G. Proof. We first note that A = F * G is an Fc-algebra. Since G is finite, F/FGis a finite Galois extension. In particular, F is a finitedimensional algebra over the field FG. Hence S = F * N is a finite dimensional algebra over FG.By Lemma 11.8, A can be regarded as a crossed product of GIN over S. Thus, by Corollary 11.7, A / A . J ( S ) is a symmetric FG-algebra, as required. H The following result for ordinary group algebras is due to Harris (1989).

11.10. Corollary. Let F"G be a twisted group algebra o f G over

F. Then FaG/FaG- J(F"N)

is a symmetric F-algebra. Proof. Since G acts trivially on F, we have F = FG. Now apply Corollary 11.9. R As a preliminary to our final result, we next record

11.11. Lemma. Let n be a positive integer and let A = F[X]/(X"). Then A is a symmetric F-algebra.

-+

Proof. Put s = X ( X " ) . Then { l,s,s2,. . . ,sn-l} is an F-basis of the commutative F-algebra A. Since any ideal of F[X] containing ( X " ) is of the form ( X ' ) , 0 5 i 5 n , ( X o ) = F [ X ] , it follows that are all nonzero ideals of A. The map A , A s , ... ,

given by

is obviously F-linear. Since for all i E {0,1,. , . ,n - l}, si # Ker+,

11. Symmetric crossed products

229

it follows that kerll, contains no nonzero ideals of A. Thus ( A , $ ) is a symmetric algebra, as required. Assume that an F-algebra A is a crossed product of G over an Falgebra S. It is natural to consider the following problem. Problem. What are necessary and sufficient conditions for A to be a symmetric F-algebra? An important sufficient condition is given by Theorem 11.6. The following example illustrates that the above problem in full generality is extremely complicated.

11.12. Example. (Dude). There exists a finite-dimensional algebra A over a field F such that A is a crossed product of a group G over an F-algebra S such that fi) A is a symmetric F-algebra.. (ii) S is not a symmetric F-algebra. Proof. Put B = F [ X ] / ( X 2 )so , that B = F [ x ]with x = X + ( X 2 ) and B is a symmetric F-algebra by Lemma 11.11. Let A = M z ( B ) and let eij, 1 2 i,j 2 2, be the usual matrix “units)) of A. Then A is a symmetric F-algebra, by virtue of Lemma 10.7. The elements eij, xejj, 1 5 z,j 5 2 constitute an F-basis of A and xejj = eijx, Let G =< g

x2 = o

(1 5 i , j

5 2)

> be a cyclic group of order 2 and put

A, = Felz

+ Fe21 + F s e l l + Fxez2

Then one immediately verifies that A becomes a G-graded algebra. Since e21 e12 E A, and

+

it follows that A is a crossed product of G over S. By the foregoing, we are left to verify that S is not a symmetric F-algebra. To this end, observe that J ( S ) = Fze21 Fxe12 and S =

+

Induction from normal subgroups

230

Sell @ Sezz (direct sum of S-modules). Put P = Sell. Then P = Fell Fzezl and J ( P ) = J ( S ) P = Fse21. Thus P is a projective indecomposable S-module such that SocP = J ( P ) . Since ellJ(P)= 0 and e,l(P/J(P))# 0, we have

+

SOCP = J ( P ) y P / J ( P ) Thus S is not a symmetric F-algebra, by virtue of Theorem 10.12(iii). The proof is therefore complete. 12.

EndFG(VG)is symmetric

All modules below are assumed to be finitely generated over their ground fields. Throughout, N denotes a normal subgroup of a finite group G, F an arbitrary field and V an FN-module. Our aim is to prove that if V is completely reducible, then the F-algebra EndFG(VG) is symmetric. We begin by recording the following two elementary results. 12.1. Lemma. Let V be an irreducible FN-module and let H be the inertia group o j V . Then any FH-homomorphism 8 : V H + V H extends to a unique FG-homomorphism 0' : V" t V G and the map EndFH( V H ) + EndFc( V") 6 H 8' is an isomorphism of F-algebras.

Proof. Let gl,gZ,. . . ,gs be a transversal for N in H and gl,. . . ,gk a transversal for N in G ( k 2 s). Then

O'(9i 8 v) = g i q 1 €3 v)

(v E V )

Then it is immediate that 8' is a unique element of EndFG(v") extending 6. Thus the map 0 H 8' is an injective homomorphism of

is symmetric

12. E + G ( V ~ )

231

F-algebras. Note that ( V H )is~the sum of all submodules of ( V G ) , isomorphic to V . Since V is assumed to be irreducible, we conclude that for any t,b E EndFG(VG),we have @ ( V H ) V H .This shows that the given map is surjective and the result follows.

12.2. Lemma. Let V be a completely reducible FN-module. Then there exist a positive integer s , irreducible nonconjugate FN-modules V,, . . . ,V , and positive integers n1, n2,. . . ,n, such that

n 8

EndFG( V G )

Mni(EndFH,(qHi)

i= 1

where H, is the inertia group of K .

Proof. Write V = f&m;V; where each is irreducible and mi 2 1 is an integer. We may assume that V,,. .. ,V , ( s 5 k) are such that Vl, . . . ,V, are mutually nonconjugate and each Q , j E (1,. . . ,k} is conjugate to some K, i E ( 1,. . . ,s} (hence yGg KG). It follows that there exist positive integers n l , 1 2 2 , . . . ,n, such that

vG2 $:=ln;KG

($tlniK)"

Thus we may harmlessly assume that V = $:=lniV;. Let T be a transversal for N in G containing 1. Assume that i , j E (1,. . . ,s} and i # j. Then, by Proposition 2.2.4(i), HOmFG

( n j & ) G ) HomFN ( n i K , ( n j q G ) N )

Thus HomFG

( ( n i K ) G(njl$)G) , g ninj ( $ t , g H O m F N ( V , , t C3 4) = 0,

since V , and t 8 V, are irreducible and nonisomorphic FN-modules for all t E G. This fact immediately

EndFG(VG)

g

N

N

Induction from normal subgroups

232

where the last isomorphism follows from Lemma 12.1. It is now an easy matter to prove the following important result. 12.3. Theorem. (Harris (1989)). Let F be an arbitraryfield and let V be a finitely generated completely reducible FN-module. Then the F-algebm EndFc(VG) is symmetric.

Proof. Owing to Lemmas 12.2, 10.8 and 10.7, we may harmlessly assume that V is irreducible and G-invariant. Then, by Theorem 4.2(ii), the F-algebra EndFc(VG) is a crossed product of GIN over S = EndFN(V). But V is irreducible, hence S is a division ring by Schur's lemma. Since V is finitely generated, S is of finite F-dimension. Now apply Theorem 11.6. Another application of Theorem 11.6 is given by 12.4. Theorem. Let F be an arbitrary field and let V be a finitely generated G-invariant FN-module. Then the F-algebra

is symmetric.

Proof. Put E = EndFc(VG) and S = End&V). Since V is Ginvariant it follows from Theorem 4.2(iv) that E / E - J ( S ) is a crossed product of G I N over S / J ( S ) . Bearing in mind that S / J ( S ) is a finitedimensional semisimple F-algebra, the result follows by applying Theorem 11.6. m 13.

Graded modules

Throughout this section, G denotes a group, R a commutative ring and A a G-graded R-algebra. While we shall have occasion to use right modules, any unspecified module will be understood to be left and unitary. All the information recorded here will be utilized in the next section.

13. Graded modules

233

We say that an A-module M is G-graded (or simply graded) provided there exists a family

P g I 9 E GI of Al-submodules of M indexed by G such that the following conditions hold:

M = $gEGMg

(direct sum of Al-modules)

AxMyG Mzy

for all z,y E G

(1) (2)

The above definition certainly implies that the regular module A A is graded (with M, = A, for all g E G ) . We refer to M, as the 9 component of M . By the g-component of rn E M , we understand the m,. unique rn, E M, defined by rn = CgEG A submodule N of a graded module M is said to be a graded submodule if for all rn E N and all g E G, rn, E N . Expressed otherwise, N is graded if

N = @gEG(N fl Mg) Thus if N is a graded submodule of M , then N is a graded module with Ng = N fl Mg for all g E G. A graded left A-submodule 1 of A is called a graded left ideal of A. We call (1) a G-grading of M and refer to M, as the g-component of M . When (2) can be replaced by the stronger condition

AxMy = Mzy

for all z,y E G

(3)

we say that M is a strongly graded A-module. Let M and N be two graded A-modules. An A-homomorphism

is called graded if

f(M,)

Ng

for all g E G

(4)

In case f is an isomorphism, we say that M and N are isomorphic as graded A-modules.

234

Induction from normal subgroups

13.1. Lemma. Let f : M + N be a graded homomorphism of two graded A-modules M and N . Then (i) I - e rf is a graded submodule of M . (ii) f ( M ) is a graded submodule of N . (iii) M / K e r f E f ( M ) as graded A-modules, where the g-component of M / K e r f is defined b y

+

( M l K e r f ) , = (Mg K e r f ) / K e r f

(9 E G )

Proof. For any given rn E M , f ( m )= f(Erng)= cf(rng), where each f ( r n g ) is in N g . This shows that f(m),= f ( r n g )

for all g E G , m E M

(5)

If rn E K e r f , then f ( r n ) = 0 and so, by ( 5 ) , f ( r n g ) = 0. Thus m, E K e r f , proving (i). Property (ii) is a direct consequence of (5). To prove (iii), let f’ : M / K e r f -+ f ( M ) be the isomorphism induced by f. Then

for all rn E M . Thus f * is a graded isomorphism, as required. Let V be any Al-module. Then the tensor product

is naturally an A-module with

for all z, y E A and o E V . We shall refer to V Aas the induced module. Bearing in mind that

A = @gEGAg it follows from Lemma 3.l(ii) that we may identify the Al-module A, 6 3 V~ with ~ its image in V A ,for any g E G. With this identification, we have the following direct decomposition of Al-modules:

13. Graded modules

235

Setting

( V A ) ,= A,

@Al

V

for all g E G

it follows from (6) and the inclusion

A,A, C A,,

for all s , y E G

that V A is a graded A-module. Let M be a graded A-module. For each g E G, put

Mg = M and define

for all s E G. Then Mg is obviously a graded A-module which differs from M only by grading. We shall refer to Mg as a conjugate of M . 13.2. Theorem.

(Dude (1980)). The following properties of a G-graded algebra A are equivalent to each other: (i) A is strongly G-graded. (ii) Every graded A-module is strongly graded. (iii) For any graded A-module M , the natuml map

is a graded isomorphism. (iv) For any graded A-module M , there exists an Al-module V such that M s VA as graded A-modules.

Proof. (i)

=+

(ii): Let M be a graded A-module. Because A is strongly graded, A,A, = A,, for all z , y E G , so by Lemma 3.1(i), we have Mxy= AIM,, = A,A,-1 MxyC A x M y C M,,

Induction from normal subgroups

236

where we have used (2) twice. Thus A,M, = ME,, as required. (ii) =+ (iii): Fix g E G. Then

so $ is a graded epimorphism. The kernel N of $ is a graded A-module (Lemma 13.l(i)) whose g-component N, is given by

Ng = N n (A,

@ A ~MI)

By assumption, N is strongly graded so that for all g E G, Ng = A,N1. Now N1 is the kernel of the natural isomorphism of A1 8~~Ml onto Ml, so Nl = 0. Thus N = 0 and therefore $ is a graded isomorphism. (iii) + (iv): Put V = MI and apply (iii). (iv) + (i): Invoking Lemma 3.1(i), (ii), we obtain for all g E G

A,Al = A, Hence, for any Al-module V ,

Applying the hypothesis, we conclude therefore that

for any graded A-module M . The regular A-module A is a graded A-module with g-component As for all g E G. Letting M to be the conjugate module A" for some z E G, we define

A,A, = Ag(A")l = (A"), = A,, by (7). Thus A is strongly G-graded, as asserted.

To provide a useful application of the above result, we need to record the following simple observation. 13.3. Lemma. If V is projective A-module.

a

projective Al-module, then V A is

a

13. Graded modules

237

Proof. By hypothesis, A1 .$A1 = V$ 6 for some A1-module V,. Since A @ I A ~A1 2 A, it follows that A @ - * @ A VA @ FA. Thus V Ais a direct summand of a free A-module and therefore is projective, as required. $ a

+

13.4. Corollary. Let A be a strongly G-graded algebra such that Al is semisimple artinian. Then every graded A-module is projective.

Proof. Let M be a graded A-module. Then, by Theorem 13.2(iii), there is a graded isomorphism M + M t = A @ I A ~ MI. Since A1 is semisimple artinian, it follows from Corollary 1.5.11 that every Almodule is projective. Hence Ml is a projective Al-module. Thus, by Lemma 13.3, A4 is projective. As a preliminary to our final result, let us recall the following piece of information. Let A be a strongly G-graded R-algebra and let I be a G-invariant ideal of Al. Then, by Theorem 3.6, I A = A I is a graded ideal of A. Hence, by Lemma 3.5(i), A/IA is a G-graded R-algebra with

( A I I A ) , = (A,

+ IA)/IA

for all g E G

13.5. Proposition. Let A be a strongly G-graded R-algebra, let I be a G-invariant ideal ofA1 and let V be a G-graded A-module. Then, for any n 2 0, I"V/I"+lV is a G-graded A/IA-module (here I' = A1).

Proof. Since I is G-invariant, so is I" for all n 3 1. Hence each I"V is a submodule of V. Since I A annihilates I"V/I"+'V, it follows that I"V/I"+'V can be regarded as an AIIA-module. Because V is G-graded, there exists a family {Vglg f G} of Alsubmodules of V such that

for all z, y E G. Setting

+

(I"V/I"+'V), = (I"T/, I"+lV)/I"t'V

(g E G)

Induction from normal subgroups

238

we immediately deduce that I"V/I"+'V is a direct sum of all (I"V/In+lV)g, g E G. Finally, for any z,y E G, we have

+ +

(A / I A), (I" V/In+'V ) , = (A,I"V, In+' V)/I"+lV = (I"A,V, I"+'V)/I"+'V (PV,, -I- I"+'V)/I"+'V = (I"V/In+lV),y,

c

thus completing the proof. 14.

Induction from irreducible modules and their projective covers

Throughout this section, N denotes a normal subgroup of a finite group G and F an arbitrary field. All vector space algebras have finite dimension over the stipulated field. If n is a positive integer and V is a module, then nV denotes the direct sum of n copies of V . Let P ( V ) be a projective cover of an irreducible FN-module V . Our aim is to tie together the decompositions of V Gand P(V)" and to describe indecomposable direct summands. Among other properties, it will be demonstrated that if W is an indecomposable direct summand of V G ,then Soc(W) and W / J ( W )are isomorphic irreducible FG-mod ules. Let S be a ring and let M be a (left) S-module. For any subset X of S, we put A ~ ~ M (=X{)m E M l X m = 0} and

X M = (C2imjIxi E X,mi E M } If X is a right ideal of S , then AnnM(X) is a submodule of M . For convenience of reference, we first record the following elementary observat ion. 14.1. Lemma.

L e t S be a ring, let M be a n S - m o d u l e a n d let

14. Induction from irreducible modules and their projective covers 239

be a direct sum decomposition of M . Then, for any left ideal I of S and for any subset X o f S we have (i) I M and IMj are submodules o f M and I M = $j,jIMj. (ii) M / I M 2 $jEj(Mj/IMj) as S-modules. (iii) A n n ~ ( x=) $jE~AnnM3 (X).

Proof. The proof is straightforward and so will be omitted. H 14.2. Lemma. Let H be a subgroup ofG containing N and let W be an FH-module. Then (i) w ~ / J ( F N ) W E (~w / J ( F N ) w ) ~ . (ii) [ Annw ( J (F N ) ) I G2 A n n w (~J (F N ) ). Proof. (i) Since J(FN)FG = F G . J ( F N ) , J ( F N ) W is obviously a submodule of W . Consider the natural exact sequence 0 + J ( F N ) W -%

w -5W / J ( F N ) W -+ 0

By Theorem 2.2.9(i), the corresponding sequence of FG-modules 0 + ( J ( F N ) W ) Gl% W GlY ( W / J ( F N ) W ) " -+o is exact. Since

the desired conclusion follows. (ii) Since g J ( F N ) = J ( F N ) g for all g E G, A n n w ( J ( F N ) )is a submodule of W . Consider the natural exact sequence

0

+ A n n v ( J ( F N ) )5 W

5W/Annv(J(FN))

-+

0

As above, the corresponding sequence of FG-modules

is exact. Since fm(1 8 a ) = A ~ ~ ~ G ( J ( F(ii) N is ) )established. , H

240

Induction from normal subgroups

14.3. Lemma. Let V be an N-projective FG-module and let n be a positive integer such that J ( F N ) " V = 0 . Then in the chain

V = J(FN)OV 2 J ( F N ) V 2

* * .

2 J(FN)'+'V 2 J ( F N ) n V = 0

of FG-submodules of V , each factor

v;. = J ( F N ) ' v / J ( F N ) ' + ' v

(0

5 i 5 n - 1)

is a projective module over the symmetric F-algebra F G / F G . J ( F N ) . Proof. Since V is N-projective, it follows from Theorem 2.10.3(ii) Thus, by Lemma that V is isomorphic to a direct summand of 14.1, V;. is a direct summand of W; = J(FN>'(VN)"/J(FN)'+'(V~)~. By Corollary 11.10, F G / F G - J ( F N ) is a symmetric F-algebra. It therefore suffices to verify that W; is a projective F G / F G J ( F N ) module. We know, from Lemma 11.8, that F G is a crossed product of GIN over F N , where

-

( F G ) g=~ gFN

for all g E G

Hence, by Theorem 3.7(ii), F G / F G . J ( F N ) is a crossed product of GIN over F N / J ( F N ) . Now F N / J ( F N ) is semisimple artinian, hence by Corollary 13.4, it suffices to verify that W; is a (GIN)-graded F G / F G . J(FN)-module. Since the latter is a consequence of the fact that (VN)' is a GIN-graded FG-module and Proposition 13.5, the result is established.

We have now come to the demonstration for which this section has been developed.

(Harris (1989)). Let N be a normal subgroup of a finite group G , let F be an arbitrary field and let V be an irreducible FN-module. Let V1,V2,. . . , V, be all nonisomorphic irreducible FG-modules for which V is a direct summand of (K)', let r; be the multiplicity of as a composition factor of VG/J ( V G ) , and let W; = P ( V ; . ) / J ( F N ) P ( V , )Then . 14.4. Theorem.

14. Induction from irreducible modules and their projective covers 241

and each W; satisfies the following properties: (i) W; is indecomposable and W; E Wj implies i = j . (ii) W; 2 A n n p ( v ) ( J ( F N ) ) . (iii) Wj/J(Wj) socwj 2 K.

=

Proof. By Theorem 2.12.3, P ( V ) / J ( F N ) P ( VS ) V . Hence, by Lemma 14.2(i), P ( V ) G / J ( F N ) P ( V )s GVG (1) and therefore P(V)"/J(P(V)G)2 VG/J(VG) (2) Now, by Theorem 2.2.14, V1,.. . ,V, are all nonisomorphic composition factors of V G / J ( V G )hence ,

v ~ / J ( 2v er=lr;K ~)

(3)

Because P ( V ) Gis projective and, by (2) and (3),

P ( v ) ~ / J ( P ( v ) ~ey=lrjx, ) it follows from Theorem 2.12.3(iii) that

P(V)"

2

@y="=,jP(v,)

(4)

Applying (1) and Lemma 14.1(ii), we also deduce that

V G 2 @r=lrr;W;

(5)

Now SocP(V)= A n n p ( v ) ( J ( F N )E ) V by Theorem 10.12(iii). Hence, by Lemma 14.2(ii) applied to W = P ( V ) ,we have G

V G2 [Annp(v)( J (F N ) ) ] 2 A n n p p q ~( J (F N ) ) Thus, by (4) and Lemma 14.1(iii),

V G $:=lr;Annp(v)( J (F N ) )

(6)

Now fix i E { 1,. . . ,n } . Then

Wa/J(Wi)

=

P(K)/J(P(K)) 2 v, E Soc(P(v;:)) =

SOC ( A n n P ( v ) ( J ( F N ) ) )

(7)

Induction from normal subgroups

242

Hence W; and Annp(v,)(J ( F N ) ) are indecomposable FG-modules. Note that P ( v ) is a projective (hence N-projective) FG-module. Therefore, by Lemma 14.3, W;is a projective module over the symmetit follows ric F-algebra F G / F G - J ( F N ) . Since, by ( 7 ) , W;/J(Wi) from Theorem 10.12(iii) and (7) that

x

By ( 5 ) , (6) and the Krull-Schmidt theorem, A n n p ( ~ l ( J ( F i Vis) )a projective module over F G / F G J ( F N ) . Hence, by (8) and Theorem 10.12(iii), Wi AnnP(v,)(J(FN)),

-

=

thus completing the proof. H

(Harris (1989)). Let N be a normal subgroup of a finite group G, let F be an arbitrary field and let W be an irreducible FG-module. Then W is N-projective if and only if 14.5. Corollary.

J ( F N ) P ( W )= J ( F G ) P ( W )

(9)

Proof. Assume that W is N-projective. Then, by Theorem 2.10.3(ii), W is isomorphic to a direct summand of (WN)'. Hence there exists an irreducible FN-module V such that V is a direct summand of WN and W is a direct summand of VG. Since W is a direct summand of V G , we have W Z W; for some i, where W; is as in Theorem 14.4. Now W; is irreducible, hence W E V , (by Theorem 14.4(iii)) and J ( F G ) P ( V , )= J ( F N ) P ( V , ) ,proving (9). Conversely, assume that (9) holds. Let V be an irreducible direct summand of W N . By Theorem 14.4, P ( W ) / J ( F N ) P ( W )is a direct summand of V G . Since by (9), P ( W ) / J ( F N ) P ( W ) W , it follows that W is a direct summand of VG.Therefore W is a direct summand ~ so W is N-projective, by applying Theorem 2.10.3(ii). of ( W N )and

15. Inflated modules over twisted group algebras Throughout this section, N denotes a normal subgroup of a finite group G and F an arbitrary field. Given an absolutely irreducible FN-module

15. Inflated modules over twisted group algebras

243

V , our future aim is to provide a detailed description of the decomposition of VG.Since twisted group algebras and projective representations will play the main role in the ensuing discusion, it will be convenient to relegate all the necessary background into a separate section. Let V be a finite-dimensional vector space over F . A mapping p : G -+

GL(V)

is called a projective representation of G over F if there exists a mapping a : G x G t F* such that

p(1) = 1v

(2)

Thus an ordinary representation is a projective representation with a ( x , y ) = 1 for all z,y E G. If we identify G L ( V ) with G L ( n , F ) , n = d i r n ~ Vthe , resulting map is called a projective matrix representation of G over F . As in the case of ordinary representations, we shall treat the terms “projective representation” and “projective matrix representation” as interchangeable. Applying the associativity of the multiplication in G, it follows from (1) and (2) that a E Z2(G,F*). To stress the dependence of p on V and 0, we shall often refer to p as an a-representation on the space V . Two projective representations p; : G t GL(V;),i= 1,2, are said to be linearly equivalent if there exists a vector space isomorphism f : t V . such that

for all g E G. The projective representation p on the space V is called irreducible if 0 and V are the only subspaces which are sent into themselves by all the transformations p(g),g E G. An a-representation p on the space V is called completely reducible if for any subspace W invariant under all transformations p ( g ) , g f G, there exists another such subspace W’ with V = W @I W‘. The following result shows that the study of a-representations is equivalent to the study of F”G-modules.

244

Induction from normal subgroups

15.1. Lemma. There is a bijective correspondence between arepresentations of G and F"G-modules. This correspondence preserves sums and maps bijectively linearly equivalent (irreducible, completely reducible) representations into isomorphic (irreducible, completely reducible) modules.

Proof. Let p be an a-representation of G on the space V . We may clearly define a homomorphism f : F"G + E n d ~ ( vby ) setting f ( i j ) = p(g) and extending by linearity. Thus V becomes an F"Gmodule by setting

Conversely, given an F"G-module V, and hence a homomorphism f : FOG + E n d ~ ( v )put , p ( g ) = f ( i j ) . Then p ( g ) E GL(V) since i j is a unit of F"G. Furthermore, p(1) = 1 and for all 2,y E G ,

so that p is an a-representation on V. This sets up a bijective corre-

spondence between a-representations and PG-modules. A subspace W of V is invariant under all p ( g ) if and only if W is an F"G-submodule of V . Consequently, the correspondence preserves sums and maps bijectively irreducible (completely reducible) representations into irreducible (completely reducible) modules. We next note that an F-isomorphism f : V, + V, of F"G-modules is an F"G-isomorphism if and only if Jf(v) = f(ijv) for all g E G, E V , Suppose now that p; : G + GL(V,),i = 1 , 2 are two arepresentations. Then p1 is linearly equivalent to p2 if and only if there is an F-isomorphism f : V i + & such that p2(g)f = f p l ( g ) for all g E G. The latter is equivalent to p z ( g ) f ( v ) = f p l ( g ) v or to gf(v) = f ( g v ) , for all g E G , v E V,. Thus two a-representations are linearly equivalent if and only if the corresponding modules are isomorphic. W Given a E Z 2 ( G / N ,F * ) ,we denote by infa the element of Z2(G,P ) defined by (infa)(z, Y )=a(xK

Y W

15. Inflated modules over twisted group algebras

for all x , y E G. Then the map a

H

H 2 ( G / N ,F')

245

inf Q induces a homomorphism

+ H2(G,F*)

called the inflation map. Observe that since (infa)(z,y)= 1 for all x , y E N , we may embed F N in FpG by n H fi for all n E N , where p = i n f a or p = (infa)-'.

Theorem. (Cliford (1937)). Let V be an absolutely irreducible G-invariant FN-module. Then there exists w = WG(V)E Z 2 ( G / N ,F') such that for 7 = infw, V can be extended to an FrGmodule, i.e. there exists an FYG-module W such that WN E V . Furthermore, if w is a coboundary, then V can be extended to an FGmodule. 15.2.

Proof. Let r be the representation of N afforded by V . To prove the first statement, it suffices (by Lemma 15.1) to exhibit -y E Z 2 ( G , F * ) ,whose values are constant on the cosets of N and a 7representation p of G such that p(n) = r ( n )for all n E N . By assumption, for each g E G , there exists a nonsingular matrix L ( g ) such that r(9-'n9) = L ( g ) - ' W L ( g )

We may, of course, assume that L(1) = 1. Let T be a transversal for N in G containing 1 and put

Then, for all g E G , n E N ,

From Schur's lemma, we readily verify that p(x)p(y)differs from p(xy) only by a scalar

Induction from normal subgroups

246

Assume that w is a coboundary, say

w ( z N ,y N ) = X(sN)t(yN)X(syN)-' for some X : G/N + F* with X(N) = 1. Then, setting p*(g) = X(gN)-'p(g), it is immediate that p* is an ordinary representation of G such that p*(n) = p ( n ) = r ( n )for all n E N . From now on, we shall refer to 2u = wG(V) as the obstruction cocycle

of

v.

We now return to our study of F"G-modules. Assume that p ; : G --+ GL(V,)

(i = 1,2)

is an a;-representation. Consider the map pi €3 p2 : G

--t

GL(Vi 8 V2)

given by €3 P2)(9) = Pl(S) €3 P 2 b ) for all g E G. One immediately verifies that p1 8 p2 is an q a 2 representation. We shall refer to p1 €3 p2 as the (inner) tensor product of the projective representations p1 and p2. It will convenient to reformulate the above in terms of modules. Let V and W be F"G and FPG-modules, respectively. Then the vector space V @ F W is an F"PG-module, where the action of elements g, g E G, is defined by (P1

g(v@w)=gv€3gw

(v E v,wE W )

15. Inflated modules over twisted group algebras

247

and then extended to V@FW and FaPG by F-linearity. It is clear that if pv and pw are cy and P-representations of G afforded by V and W , respectively, then pv @ pw is afforded by the F"PG-module V € 3 W ~ .

15.3. Lemma. Let V and W be FaG and FPG-modules, respectively, and let EIF be a field extension. Then

(V @ F W ) E2 VE@E WE

as EaPG-modules

Let ( ~ 1 , . . . ,v,} and ( ~ 1 , . . . ,w,} be F-bases of V and Proof. W , respectively. Then (1 @ (v; @ w j ) ) and ((1 €3 v ; ) 8 (1 8 wj)}, 1 5 i 5 n, 1 5 j I m, are E-bases of (V 8~W ) Eand V , @ E WE, respectively. Consequently, the map

extends to an E-isomorphism of (V 8~ W ) Eonto VE@ E WE. It is obvious that for all v E V , w E W,$(l @ (v @ w)) = (1@ v) €3 (1 @ w). Since, for all g E G,

and

We now come to the main topic of this section. Let a E Z2(G/N, F*), where F is an arbitrary field and GIN acts trivially on F*. Given an P(G/N)-module V , we can form an F'"f"G-module inf(V)whose underlying space is V and on which elements g , g E G, act according to the rule: gv = g N v (v E V ) We shall refer to z n f ( V ) as being inflated from V .

248

Induction from normal subgroups

15.4. Lemma. Let CY E Z2(G/N,F*)and let V be an F"(G/N)module. (i) V is irreducible (absolutely irreducible, indecomposable, totally indecomposable) if and only if inf(V) is irreducible (absolutely irreducible, indecomposable, totally indecomposable). (ii) For any F"(G/N)-modules Vl and %,& E v2 if and only if inf( &) Z inf (V2). (iii) If W is an F'"f*G-module such that fiw = w for all n E N, w E W , then W = inf(U) for some F*(G/N)-module U. (iv) IfE/F as a field extension, then inf(VE) (infv),.

Proof. The proof is straightforward and so will be omitted. Let H be a subgroup of G. Then, for any CY E Z2(G,F*),the restriction of CY to H x H is an element of Z2(H,F*).To prevent our expressions from becoming too cumbersome, we shall use the same symbol for an element of Z2(G,F*) and its restriction to Z2(H,F*).With this agreement, we will identify F*H with the subalgebra of F*G consisting of all F-linear combinations of the elements &, h € H . Given an F*G-module V, we denote by VH the F"H-module obtained by the restriction of algebra; thus as an F-module, V H equals V, but only the action of F*H is defined on VH. Given an F"H-module W , we form the F"G-module

and refer to W Gas the induced module. 15.5. Lemma. F"S-module. Then

Let S be a subgroup of Q = G/N and let Vbe an inf(VQ) z inj(V)'

Proof. Let H be the subgroup of G with S = H/N. If {gl,. . . ,gn} is a left transversal for H in G, then {glN, . . . ,gnN} is a left transversal for H / N in GIN. Consequently, the map

15. Inflated modules over twisted group algebras

249

and

Consequently, putting X = a ( g N ,g;N)a-'(gjN, h N ) , we obtain

as required.

15.6. Lemma. Let a E Z 2 ( G , F * ) ,let V be an FOG-module and, f o r any given cp E EndFaN(VN),g E G, define 'pg by

Then 'pg E E n d p N ( V N ) and the map

is an F-algebra automorphism. Proof. have

Given g E G, n E N , v E V and cp E E n d p N (V N ), we @ ( n v ) = gcpg-'(nv) = gcp[(g-'ng)(g-'v)]

= gg-'ngcp(g-'v) = f i c p g ( v ) ,

Induction from normal subgroups

250

proving that cpg E EndpN(VN). It is obvious that cp H cpg is F-linear and injective. Bearing in mind that -- 1 g-I = a(g,g-')g-' and 9-1 = a-1 ( g , g-'>ij, we obtain (cpg-l)g(w)

=

gcpg

-'--I

g

-

--

v-gg-'cpg-'

1

g

w

= g a ( g , g-l)g-lcpa-'(gl g-l)gij-lv

=

cp(4,

which shows that (&)g = cp. Thus cp H cpg is a bijection. Finally, if cp1,cpz E EndpN(VN), then (cpM)(v)

=

c p ~ g c p 2 s - 1= ~ scpls-19cp2s-1v

= s ( ' P ~ ' P & - ~=v ( c p 1 ~ 2 ) ~ ( v )

Hence

(cpI'p2)g

= cpfcp; and the result follows.

In what follows, we write Am for the F-algebra of all m x m matrices with entries in an F-algebra A.

15.7. Lemma. Let V be an F*G-module and let & and & be FPQ-modules, Q = G I N , ofequal dimension m with F-bases ( ~ 1 , .. . ,w m ) and { w:, . ., ,w;}) respectively. Put y = ( i n f p ) a and W;= inf (K), i = 1,2. (i) There is an injective F-linear map

j=1

then B(cp) = ( c p j ; ) . (ii) I f & = V2 and wi = w;)1 5 i 5 m, then

15. Inflated modules over twisted group algebras

251

is an injective homomorphism of F-algebras. (iii) Suppose that gw; =

m

m

j=1

j=1

C ajj(g)wj,gwI = C bj;(g)wi

If Q E H o ~ F ~ G ( W@ FI V,W2 @ F V ) and

ifO((p)

(ajj, bjj E F ) = (Fji), then

Proof. (i) We may write any element in W28 V uniquely in the form w(i 8 vj for some v j E V . Hence, given and v E V,

cp E H o r n ~ ( W 1 8V,W2 8 V ) there exists a unique

cpj;

E E n d F ( V ) for which m

P(Wj

@v) =

1w;8

(1 5 i

Cpjjv

5 rn)

j=1

It therefore suffices to verify that if Q is an FYN-homomorphism, then each Pjj E E n d F a N ( V N ) . To this end, assume that cp is an FYN-homomorphism and fix n in N . Then m

C

W;

8 v j j ( f i v > = v(wj 8 f i v ) = ~

( f i8~n ui ) = ~ ( i i i ( w8 j v))

j=1 m

= ncp(w; @I v) =

fi(Cw; 8 Qj;v) j=1

m

=

c w ;8 f i Q j j ( V ) j= 1

and thus 'pjj(fiv)= ncpj;(v),as required. (ii) Let Q , $ E E ~ ~ F - ~ N ( (@IWVIN))N and let O(Q) = ( $ j i ) . Then

(vji),O($) =

Induction from normal subgroups

252

which shows that O(cp$) = O(cp)O($). (iii) Fixing g E G, cp E H o ~ F ~ G@(FW V,~W, @ F V ) ,we have

m

m

j = 1 k=l

m

/ m

and

k=l

whence

m

m

aji(9)vkjgv j=1

=

bkj(g)&'jiv j=1

Substituting v for ij-lv, we therefore derive m bkj(g)gyji(g-'v)

15. Inflated modules over twisted group algebras

253

thus completing the proof. We have now accumulated all the information necessary to prove the following result essentailly due to Huppert and Willems (1975, Satz. 2.)

15.8. Theorem. Let N be a normal subgroup of a finite group G, let F be an arbitrary field and let V be an FaG-module. Then for Q = GIN and for FPQ-modules U, U1 and U2 the following properties hold: (i) If VN is totally indecomposable and U is indecomposable, then

i n f ( U )@ F V is an indecomposable FrG-module, where y = (inf pja. (ii) If V N is absolutel~irreducible and U is irreducible, then inf (Uj @ F V is an irreducible FYG-module. (iii) If V , is totally indecomposable, then inf ( & ) @ F V = inf ( U ~ ) @ F V implies U, U2.

Proof. We put W = inf(U) and W; = inf(V;),i = 1,2. (i) Suppose that cp is an idempotent of EndF7G(W @ V ) . It suffices to verify that cp = 0 or cp = 1. Choose an F-basis ( ~ 1 , .. . ,wn}of W and write m

Swi = C a j i ( g ) w j

(9 E G,aji(g) E F )

(6)

j=1

In the notation of Lemma 15.7(iii) with 1 5 i 5 rn, we have O(p) = ( c p j ; ) , where m

U1

=

U2 = U and w;

= wi,

m

C aji(g)Ykj = jC akj(9)v:; =1

(7)

j=l

Applying Lemma 15.7(iii)) ( c p j ; ) is an idempotent of (EndFaN(VN)),. Bearing in mind that VN is totally indecomposable, we have

EndpN(VN) = F - l v

+ J(EndpN(VN))

Induction from normal subgroups

254

Thus we may write

+

= X j i IV

vji

with

Xji

E

(1 5 i , j

xji

5 m)

E J ( E ~ ~ F ~ N ( VByN Lemma )). 15.6, it follows that

F , ~ j i

~ 3' 9

=. X j i * l v

+

for all 9 E G

~ ; i

where x;i f J ( E ~ ~ F ~ N ( VComparing N)). the coefficients of Iv in (7), we derive m m j=1

j=1

Applying (6), it now follows easily that the map cp' : W -+W given 1 5 i 5 n, is an F'"fflG-homomorphism. by cp'(w;) = & X j i w j , Note that, by Lemma 15.4, W is indecomposable. Observe also that the matrix ( X j i ) of cp' is obtained from ( c p j i ) by reduction modulo J(Endp,( V N ) ) .Since ( c p j i ) is an idempotent, so is 'p' and hence cp' = 0 or cp' = 1, by the indecomposability of W . Replacing cp by 1 - cpl if necessary, we may therefore assume that cp' = 0, in which case Qji

EJ(E~~F~N(VN for) )all

;,j

Since J(EndpN(V,)) is a nilpotent ideal, there is a positive integer s such that ( c p j i ) s = 0. Thus cp = pa = 0 and i n f ( V )8~V is indecomposable. (ii) Let M # 0 be an FYG-submodule of W @ F V and let u be a nonzero element of M . Then we may write k

=CS~@V (Vi ~ E V , SE~ W ) i= 1 s1 # 0 and 211,. . . ,vk linearly independent over F . u

with Since V , is absolutely irreducible, E n d p , ( V ~ ) = F . Hence, by the Jacobson's density lemma, given v E V ,we may find a = CxEN uzZ in F*N, ax E F such that avl = v and a V i = 0 , 2 5 i 5 k. Since the elements 5 , s E N , operate.trivial1y on W , we obtain a(si

@ Vi) =

c c

axZ(si

@ Vi)

ZEN

=

a,(& @ Z V J

SEN

= s; @ a v i

15. Inflated modules over twisted group algebras

255

and therefore k

We conclude that

s1

k

@

V 2 M and hence, for all g E G,

However, FaGsl is a nonzero submodule of W and W is irreducible, by Lemma 15.4(i). Thus M = W @F V and therefore W @ F V is irreducible. (iii) Assume that cp : Wl @ F V -+ W2 @ F V is an FYG-isomorphism. Then we have dirnFWl = dirn~W2and so we may apply Lemma 15.7(iii). If 6(cp) = ( c p j ; ) then in the notation of Lemma 15.7(iii), we have m m aji(g)Pkj

=

j=1

C

bkj(g)vTi

j=1

By repeating the argument of (i), we derive m

m

j=1

j=1

for some Xj; E F . The latter implies that the map cp' : W1 defined by

c

+

Wz

cp-l,

we

m

cp'(W1)

=

xjiw;

j=1

is an F'"fPG-homomorphism. Applying the same procedure to deduce that cp' has an inverse and therefore

Thus, by Lemma 15.4(ii), U 1 2 U2 as FPQ-modules. This completes the proof of the theorem.

In the notation of Theorem 15.8, assume that is absolutely irreducible. Then inf (U)@ F V is absolutely irreducible if and only if so is U. 15.9. Corollary.

VN

Induction from normal subgroups

256

Proof. Let E / F be a field extension. Applying Lemmas 15.4(iv) and 15.3, we have

( i n f ( U )@F

V)Eg

inf(U)E @ E V , g inf(UE)8, VE

(8)

Assume that i n f ( U ) @F V is absolutely irreducible. Then, by (8)) inf(UE) is irreducible. Hence UE is irreducible, by Lemma 15.4, and therefore U is absolutely irreducible. Conversely, assume that U is absolutely irreducible. Then UE is irreducible and since ( V E ) N = ( V N ) E , (V')N is absolutely irreducible. Applying (8) and Theorem 15.8(ii), we conclude that ( i n f ( U )@F V ) , is irreducible. Thus inf(U )@F V is absolutely irreducible, as we wished to show.

16. Induction of absolutely irreducible modules Throughout this section, N denotes a normal subgroup of a finite group G and F an arbitrary field. Given an absolutely irreducible FN-module V , our aim is to provide a detailed analysis of the decomposition of VG.

16.1. Lemma. Let A be a finite-dimensional algebra over F , let V be an absolutely irreducible A-module, and let W = U @ F V , where U is some finite-dimensional vector space over F . For each a E A, let cp, E End=(V) be defined b y cp,(v) = uv,v E V . (i) I f 6 E E n d ~ ( wis) such that

6(l@ 9,) = (1 @ ya)6

for all a E A

then 6 = II, @ 1 for some II, E EndF(U). (ii) If6 E GL(W) and T E GL(V) are such that 8-l(1 @ cpa)6 = 1 @ T - ' ~ , T

for all a E A

then 0 = $ 8 T for some II, E GL(U).

Proof. (i) Let {ul,. . . ,u m } be an F-basis for U . Then, for all v E V , we have m

o(ui 8 v) =

C uj 8 Oji(v) j=1

(1)

16. Induction of absolutely irreducible modules

257

for some 8ji E E n d ~ ( v )Given . a E A, we have

m

j=1

Because the {uj} are linearly independent, we infer that

or, equivalently, that 0,; E E n d ~ ( v ) . Since V is absolutely irreducible, it follows that Bj; = X j i - l v for some Xi; E F . Now define T,!J E EndF(U) by

Then II, E GL(U)and, by (l),8 = $ 8 1, proving (i). (ii) Setting 7 = 1 87,we have

and therefore, by (i), 87-' = II, 8 1 for some II, E G L ( U ) . Hence

as required.

We are now ready to prove the following result partially due to Clifford (1937). 16.2. Theorem. Let F be an arbitrayfield, let V be an absolutely irreducible G-invariant FN-module, and let w = wo(V) E Z2(&,F*),

Induction from normal subgroups

258

Q = GIN,be an obstruction cocycle of V . Let ezt(V) denote any extension of V to an FOG-module, where a = infw ( b y Theorem 15.2,such an extension always exists). '1 {Ul,Uz,.. . ,Un)is a f i l l set of nonisomorphic irreducible (rspectively, absolutely irreducible) F"-' &-modules, then {inf(Ui)@F ezt(V)I 1 5 i 5 n} is a full set of nonisomorphic irreducible (respectively, absolutely irreducible) FG-modules whose restriction to N is a direct sum of copies of v.

Proof. By Corollary 15.9, it suffices to treat the case where all

U;are irreducible. Let M denote an irreducible FG-module such that MN is a direct sum of copies of V and put S = ezt(V), Si = inf (Ui),1 5 i 5 n. We may clearly assume that M = W @ V , where W is an F-space such that for all n E N,w E W and v E V ,

the

If 7 : N

G + G L ( M ) and T : G + G L ( S ) are the representations afforded by V,M and S, respectively, then by (2) and by the equality SN = V, we obtain + GL(V),p :

p(n) = 1 @ 7 ( n ) and ~ ( n=r)(n) ) for all n E N

(3)

Also if ng = gnl for some n,nl E N and g E G, then T(n)T(g) = T(g)T(nl) because a(n,g)= a(g,n)= 1 for all n E N,g E G. Thus

and hence, by (3), we have

16. Induction of absolutely irreducible modules

259

Invoking Lemma 16.1(ii), for each g E G there exists $ ( g ) E GL(W) such that

P ( d = $(s>@ 4 7 ) Setting gw = $(g)w,w E W , it follows that W is an Fa-lG-module such that M g W ezt(V) Moreover, since M is irreducible, so is W . Finally, since $(n) = 1 for all n E N, we conclude that W i n f ( U ) for some irreducible F"-'Qmodule U . Thus

M

inf(Vi)@ ezt(V)

.

for some i E {I,. . ,n }

Conversely, assume that U is an irreducible F"-'Q-module and let {ul,. . . ,uk} be an F-basis of U . By assumption, V is absolutely irreducible. Because V = e z t ( V ) N , it follows from Theorem 15.8(ii) that

inf(U ) 8 ezt ( V ) is irreducible. Finally, write

i n f ( U )8 ezt(V)= u 1 @V

$ - a .

$uk @ V

where each ui @ V is an FN-module, by the definition of i n f ( U ) . Since the map v H u, @ o is an FN-isomorphism of V onto u; 8 V , we deduce that ( i n f ( U )@ ezt(V))N is a direct sum of copies of V . Since, by Theorem 15.8(iii),

the result is established.

16.3. Corollary. Let N be a normal subgroup ofG and let W be an irreducible FG-module such that some irreducible direct summand V of WN is absolutely irreducible. Let H be the inertia group ofV and let w = W H ( VE) Z 2 ( H / N ,F*)be an obstruction cocycle of V . Then (i) There exists an irreducible F"-'(H/N)-module U such that

w 2 ( i n f ( U )8 e z t ( V ) ) G

Induction from normal subgroups

260

(ii) I f U is absolutely irreducible, then so is W .

Proof. (i) This follows from Theorems 2.4.2(iii) and 16.2 and transitivity of the induction. (ii) By Theorem 16.2, i n f ( U ) @ e z t ( V ) is absolutely irreducible. Now apply Corollary 2.4.5(iv). We now come to the main result of this section. 16.4. Theorem. Let F be an algebraically closed field, let V be an irreducible FN-module, let H be the inertia group of V and let charF not divide ( H : N ) . Denote by w = O H ( V E ) Z 2 ( H / N ,F') an obstruction cocycle o f V and by e x t ( V ) an extension ofV to an F"Hmodule with Q = infw. Let U1,. . . ,U,, be all nonisomorphic irreducible F"-'(HIN)-modules and, for each U;,let inf (U;) be the Fa-' H-module

injlatedfim U;. Then the

are nonisomorphic irreducible FG-modules such that for m; = dimFU;,

Proof. By Theorem 2.1 and transitivity of the induction, it suffices to prove that the inf(U;) @ e z t ( V ) are nonisomorphic irreducible FHmodules such that

To this end, we first invoke Theorem 2.10.14 to deduce that V H is completely reducible. Thus there exist integers t l , t Z , . . . ,t , and nonisomorphic irreducble FH-modules W1,W2,.. . ,W, such that V H2 @&W; By Theorem 2.2.14, ti equals the multiplicity of V as an irreducible constituent of (W;)N.Invoking Theorem 16.2, we may thus assume that W; % i n f ( U j ) e z t ( V ) (1 5 i 5 s )

17. Applications

261

Note that the multiplicity of V as an irreducible constituent of ( i n f ( U i ) @ e s t ( V ) ) N is dimFU; = m;. Accordingly,

By Proposition 6.11, F"-'(H/N) is semisimple, because charF does not divide ( H : N ) . Taking into account that F is algebraically closed, we have n

Thus

dimFVH = ( g m : ) dimFV, proving that n = s and hence the result.

17.

Applications

In this section, we provide a number of applications of the preceding results. Throughout, F denotes a field and G a finite group.

17.1. Lemma. Let F be an algebraically closed field such that charF does not divide [GI, and let V be an irreducible module over a twisted group algebra F"G. Then dimFV divides the order of G.

Proof. By making a diagonal change of basis {glg E G} of FOG, we may replace cu by any cohomologous cocycle. Hence, we may assume that cy is of finite order, say m. Let E be a primitive m-th root of 1 in F and let G' =< e'glg E G, 1 5 i 5 m >. Then the map f : G' + G given by f ( E i g ) = g is a surjective homomorphism whose kernel is a central subgroup < E >. Moreover, if p : G --t G L ( V ) is an arepresentation of G afforded by V , then p* : G + G L ( V ) defined by p*(ciij) = Eip(g) is easily seen to be an irreducible representation of G'. Because charF does not divide ]GI and G*/ < E >%G, the result

Induction from normal subgroups

262

follows by applying the following standard fact (see Curtis and Reiner (1981)): If F is an algebraically closed field, charF does not divide IGl and 2 is a central subgroup of G , then the dimensions of irreducible FGmodules divide ( G : 2). Let p be a prime. A group G is said to be p-solvable if the composition factors of G are either pgroups or p’-groups. It is an immediate consequence of the definition that (i) Any extension of a psolvable group by a psolvable group is psolvable. (ii) Subgroups and homomorphic images of psolvable groups are psolvable. We are now ready to establish the following important result. 17.2. Theorem. Let N be a normal subgroup of G , let F be an algebraically closed field and let V be an irreducible FG-module. If charF = p > 0 divides ( G : N ) , assume that GIN is p-solvable. Then dimFV divides ( G : N ) d , where d is the dimension of an irreducible constituent of VN.

Proof. We first demonstrate that the result holds under either of the following hypotheses: (i) charF does not divide ( G : N ) . (ii) charF = p > 0 and G / N is a p-group. Denote by W an irreducible constituent of VN. Let H be the inertia group of W and let w E Z 2 ( H / N ,F*)be an obstruction cocycle of W . By Corollary 16.3(i), there exists an irreducible F”-’(H/N)-module S such that

v

( i n f ( S )C ~ ezt(W))G I

Because dimFS = dirnFinf ( S ) and dirnFW = dirnFezt(W), we have

dimFV = (dirnFS)(dirnFW)(G: H ) In case (i), dimFS divides ( H : N ) , by Lemma 17.1, so dirnFV divides

( G : N)dimFW. In case (ii), F”-l(H/N) is a local ring by Proposition 6.4. Hence

17. Applications

263

dimFS = 1 and so dimFV again divides ( G : N)dimFW. Turning to the general case, we use induction on /GI. Suppose that the result is true for groups of lower order than G. By the above, we may assume that charF = p > 0 divides ( G : N ) , in which case GIN is psolvable, by hypothesis. Because GIN is p-solvable, there exists a proper normal subgroup M of G containing N and such that G / M is either a p or p'-group. Denote by W an irreducible constituent of VM and let S be an irreducible constituent of WN. Then S is clearly an irreducible constituent of V N . By the foregoing, dimFV

divides (G : M)dimFW

and, by induction hypothesis,

dimFW

divides

( M : N)dimFS

Thus dimFV divides ( G : M ) ( M : N)dirnFS = (G : N ) d i m ~ S as , required. 17.3. Corollary. (Dude (1968), Swan (1965')). Let F be an algebraically closed field of characteristic p > 0 and let A be a normal abelian subgroup of a p-solvable group G. Then the dimensions of irreducible FG-modules divide ( G : A).

Proof. Note that all irreducible FA-modules are of dimension 1. The desired conclusion is therefore a consequence of Theorem 17.2.

As a prepararion for the proof of our next result, we now record the following properties. 17.4. Proposition. Let N be a normal subgroup ofG such that (G : N ) = pn, p prime, and let F be an algebraically closed field of characteristic p. If V is an irreducible FN-module, then

where U is an irreducible FG-module such that V is a constituent of UN.

Induction from normal subgroups

264

Proof. We know, from Theorem 2.12.4, that P ( V ) is an indecomposable FN-module. Hence, by Theorem 7.5, P(V)" is indecomposable. It follows from Proposition 2.12.5(i) that P(VG)E P(V)". By Corollary 2.2.5(i), there is an FG-epimorphism V G + U. Thus, by Lemma 2.12.2(ii), P ( U ) is isomorphic to a direct summand of P(V"). But, by the above, P(V") is indecomposable, hence the result. H Let N be a normal subgroup ofG and let V be an irreducible FG-module. If(G : N ) # 0 in F , then 17.5. Proposition.

Proof. By Theorem 2.10.14, J(FG) = F G J ( F N ) and so

where

with nonisomorphic irreducible FG-modules 2.12.3(iii)) F G = $idiP(x)

x. Then, by Theorem

and so (FG)N = $ i d ; P ( V i ) ~ (3) Bearing in mind that F G is a free (hence, projective) FN-module, we also have (FG)N = P((FG)N) = P ( (FG)N/ J ( FN)(FG)N) = ~ ( ( F G / J ( F G ) ) N ) (by (1)) = $;d;(P((K)N)) (by (2) and Lemma 2.12.l(iii)) Thus the desired conclusion follows by virtue of (3) and Proposition 2.12.5(ii). H Given a natural number n and a prime p , we denote by np the highest power of p dividing n and put np:= n/np.

17. Applications

265

Let G be a p-solvable group, let F be an algebraically closed field of characteristic p , and let V be an irreducible FG-module. Then (i) (Fong (1961)) dirnFP( V ) = /GIp(dirnFV),I. (ii) (Willems (1980)). If V = W Gfor some FH-module W , H is a subgrozlp of G, then P( V ) s P( W ) G 17.6. Theorem.

Proof. (i) We argue by induction on IGl. Let N be a maximal normal subgroup of G. Because G is psolvable, either [ G I N [= p" for some n 2 1 or G / N is a p'-group. Owing to Clifford's theorem, there is a decomposition

with nonisomorphic irreducible FN-modules K of equal dimension. Furthermore, by Theorem 17.2, es divides IG/NI. Suppose first that IG/NI = p". Because dirnFV = esdirnFV1, we have

(dirnFV),t = (dirnF&),t Note also that, by Proposition 17.4,

(5)

Thus

as required.

Next assume that GIN is a p'-group. Owing to Proposition 17.5, P ( V , ) S P(V)N and so dirnFP(V) = dirnFP(VN). On the other hand, by (4>1 ~(VN = )e ( P ( l 4 )CB @ P(K))

Induction from normal subgroups

266

and thus

where the last equality follows from the fact that [NIP = ]GIp and esl(G : N ) . Hence (i) is established. (ii) Owing to Proposition 2.12.5(i), P ( V ) is isomorphic to a direct summand of P ( W ) G .On the other hand, by (i), we have

Hence P ( V ) "= P(W)Gand the result follows. H 18.

The Loewy length of induced modules

Throughout this section, A denotes a finite-dimensional algebra over a field F . Unless explicitly stated otherwise, all A-modules are assumed to be left and finitely generated. The same assumption is made on FG-modules, where G is a finite group. Let V # 0 be an A-module. Then the descending chain

v 2 J(A)V 2 J(A)2V2 . . . of submodules of V is called the (lower) Loewy series of V . Since J ( A ) is nilpotent, there is an integer k, called the Loewy length of V such that J(A)"-'V # 0 but J(A)"V = 0 Observe that if J(A)'V = J(A)'+lV for i < k, then

J(A)"'V = J(A)k-i-'J(A)'V= J(A)"-'-'.J(A)'+'V = J(A)kV= 0,

18. The Loewy length of induced modules

267

which is impossible. Thus, if k is the Loewy length of V , then

V

J(A)V 3

18.1. Lemma. Let e

#0

-

a

-

3 J(A)k-lV 3 0

be an idempotent of a ring R. Then

J(eRe) = eJ(R)e Proof. Let I be a primitive ideal in R and let V be a faithful irreducible R/I-module. Then eV is an eRe-module. If eV = 0, then J ( e R e ) eRe C 1. Assume that eV # 0. Let 0 c W = e W eV where W is an eRe-module. Then V = RW since V is irreducible. Thus eV = e R e W W and so eV = W . It follows that eV is an irreducible eRe-module, hence J ( e R e ) V = J ( e R e ) e V = 0. Consequently, J ( e R e ) 5 I for any primitive ideal I in R. Thus J ( e R e ) J ( R ) and J(eRe) = eJ(eRe)e eJ(R)e. Assume that a E J ( R ) . Then eae E J ( R ) and so, by Proposition 1.4.10, ( 1 - b ) ( l - eae) = 1 for some b E R. Thus beae = eae b and multiplying by e on both sides yields eae + ebe = ebeae. Therefore ( e - ebe)(e- eae) = e so that e - eae is a left unit. Because e J ( R ) e is an ideal of eRe, it follows from Proposition 1.4.10 that e J ( R ) e C J ( e R e ) , as required.

c

c

+

If W # 0 is an A-module and E = EndA(W),then W will also be regarded as an E-module via 'pw = 'p(w) for all 'p E E , w E W . 18.2. Lemma. Let e # 0 be an idempotent of A, let W = A e and let E = EndA( W ) . Then

for all i 2 1

J(E)'W = Ae(eJ(A)e)'

In particular, the Loewy lentgh of W as an E-module is equal to the nilpotency index of J ( E ) . Proof. For each x E eAe, let By Lemma 1.5.9, the map

E E be defined by f3:(w) = wx.

f3:

eAe + E

Induction from normal subgroups

268

is an anti-isomorphism of rings. Moreover, by Lemma 18.1, J(eAe) =

eJ(A)e and so fz E J(E)' if and only if x E (eJ(A)e)i.Thus

J(E)'W = Ae(eJ(A)e)' Finally, J(E)'W = 0 if and only if (eJ(A)e)' = 0. Thus the Loewy length of the E-module W is equal to the nilpotency index of J(eAe) = eJ(A)e. Because eAe Z E", the result follows. We have now accumulated all the information necessary to prove the following result in which L ( V G )denotes the Loewy length of the induced FG-module V G . 18.3. Theorem.

(Clarke (1972)). Let F be a field of charucteristic p > 0, let N be a normal p'-subgroup of G , let H be the inertia group of an irreducible FN-module V and let E = EndFG(VG). Then (i) J(FG)"VG = FG J(FH)"VH = J(E)"VG for all n 2 1. (ii) L ( V c ) is equal to the nilpotency index o f J ( E n d F H ( V H ) ) . Proof. We first show that (ii) is a consequence of (i). Indeed, we may take V = F N e for some primitive idempotent e of F N , in which case V G = FGe. Hence, by Lemma 18.2, the Loewy length of V Gas an E-module is equal to the nilpotency index of J(E). Thus (ii) is a consequence of (i) and Lemma 12.1. To establish (i), we write

as a sum of primitive idempotents of F N with e = el. Then we have

J ( F G ) V G = J(FG)e = FG(J(FG)e) = FGe(J(FG)e)t - - - FGe,(J(FG)e)

+

(1)

as left FG-modules, where the sum is not necessarily direct. For each u E e;FGe, let f a E HOmFG(FGe;,FGe) be defined by fa(.) = xu for all x E FGei. It follows, from Lemma 1.5.8, that the map

1

e;FGe a

HomFG(FGei,FGe)

+ H

f

a

18. The Loewy length of induced modules

269

is an F-isomorphism. In particular, if FGe; E FGe, then there is an a E e;FGe such that fa : FGe; + FGe is an FG-isomorphism. For the sake of clarity, we divide the rest of the proof into three steps. Step I. Here we demonstrate that J(FG)e = FGeJ(FG)e. Because FGeJ(FG)e J ( F G ) e ,it fo