THE ALGEBRAIC STRUCTURE OF CROSSED PRODUCTS
NORTH-HOLIAND MATHEMATICS STUDIES Notas de Matematica (118)
Editor: Leop...
24 downloads
552 Views
9MB Size
Report
This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
Report copyright / DMCA form
THE ALGEBRAIC STRUCTURE OF CROSSED PRODUCTS
NORTH-HOLIAND MATHEMATICS STUDIES Notas de Matematica (118)
Editor: Leopoldo Nachbin Centro Brasileiro de Pesquisas Fisicas Rio de Janeiro and University of Rochester
NORTH-HOLLAND -AMSTERDAM
NEW YORK
OXFORD .TOKYO
142
THE ALGEBRAIC STRUCTURE OF CROSSED PRODUCTS
G regory KARPILOVSKY Department of Mathematics University of the Witwatersrand Johannesburg, South Africa
1987
NOTH-HOLLAND -AMSTERDAM
NEW YORK
OXFORD .TOKYO
@
Elsevier Science Publishers B.K, 1987
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 permission of the copyright owner.
ISBN: 0 444 70239 3
Publishers:
ELSEVIER SCIENCE PUBLISHERS B.V. P.O. BOX 1991 1000 BZ AMSTERDAM THE NETHERLANDS
Sole distributors of the U. S.A. and Canada:
ELSEVIER SCIENCE PUBLISHING COMPANY, INC 52 VAN DER BI LT AVE NUE NEW YORK, N.Y. 10017 U.S.A.
Lihrav of Congress Cncalo~ngin-PubliationLhts
Karpilovsky, Gregory, 1940The algebraic structure of crossed products. (North-Holland mathematics studies ; 142) (Notas de matematica ; 118) Bibliography: p. Includes index. 1. Von Neumann algebras--Crossed products. I. Title. 11. Series. 111. Series: Notas de matematica (Rio de Janeiro, Brazil) ; no. 118. QAl.N86 no. 118 [QA326] 510 s (512l.551 87-9033 ISBN 0-444-70239-3 (U.S.)
PRINTED IN THE NETHERLANDS
FOR HELEN, SUZANNE and E L L 1 0 E t
This Page Intentionally Left Blank
vii
Preface
In the past
15 years, the theory of crossed products has enjoyed a period of
vigorous development.
The foundations have been strengthened and reorganized
from new points of view, especially from the viewpoint of graded rings. The purpose of this monograph is to give, in a self-contained manner, an upto-date account of algebraic structure of crossed products.
Although no claim to
completeness is made here, one of our goals is to tie together various threads of the development in an effort to convey a comprehensive picture of the current state of the subject. The present monograph is written on the assumption that the reader has had the equivalent of a standard first-year graduate course.
Thus we assume a famil-
iarity with basic ring-theoretic and group-theoretic concepts and an understanding of elementary properties of modules, tensor products and fields.
For the con-
venience of the reader, a chapter on algebraic preliminaries is included. chapter provides a brief survey of topics required later in the book.
This
A syste-
matic description of the material is supplied by the introductions to individual chapters.
There is a fairly large bibliography of works which are either direc-
tly relevant to the text or offer supplementary material of interest. A
word about notation.
As is customary, Theorem 3 . 4 . 2 denotes the second
result of Section 4 of Chapter 3 ; however, for simplicity, a l l references to this result within Chapter 3 itself are designated as Theorem 4.2. 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.
I am also grateful to Nan Alexander, who smoothed over many technical
problems for me with her friendly advice.
My thanks to D S Passman for sending
viii
me some reprints of his fundamental work on crossed products.
Preface
Finally, I w i s h to
thank Lucy Rich for her excellent typing. Gregory Karpilovsky
ix
Contents
vii
PREFACE CHAP'FER 1.
PRELIMINARIES Notation and terminology 2. Projective, injective and flat modules 3. Artinian and Noetherian modules 4. Group actions Cohomology groups and group extensions 5. Some properties of cohomology groups 6. 7. Matrix rings and related results 1.
CHAPTER 2.
GROUP-GRADED ALGEBRAS AND CROSSED PRODUCTS: GENERAL THEORY 1. 2. 3. 4. 5. 6. 7. 8.
CHAPTER 3.
CHAPTER4.
1 11 19 25 28
41 46 59
59 71 78 92 103 119 125 147
THE CLASSICAL THEORY OF CROSSED PRODUCTS
151
1. Central simple algebras 2. The Brauer group Classical crossed products and the Brauer group 3.
161 169
CLIFFORD THEORY FOR GRADED ALGEBRAS
181
Graded modules Restriction to Al Graded homomorphism modules Extension from A , Induction from A ,
181
1. 2. 3. 4.
5.
CHAPTER 5.
Definitions and elementary properties Equivalent crossed products Some ring-theoretlc results The centre of crossed products over simple rings Projective crossed representations Graded and G-invariant ideals Induced modules Montgomery's theorem
1
151
186 191 196 205
PRIMITIVE AND PRIME IDEALS OF CROSSED PRODUCTS
223
Primitive, prime and semiprime ideals 2. Primitive ideals in crossed products 3. Prime coefficient rings 4. Incomparability and Going Down 5. A Going Up Theorem 6. Chains of prime and primitive ideals
223 225 230 246 259 267
1.
Contents
X
CHAPTER 6.
SEMIPRIME AND PRIME CROSSED PRODUCTS 1.
2. 3. 4.
5.
Coset calculus A-methods The main theorem and its applications Sufficient conditions for semiprimeness Twisted group algebras
2 17 285
294 305 3 11
BIBILIOGRAPHY
331
NOTATION
339
INDEX
345
1
1 Preliminaries
I n t h i s chapter we consider s e v e r a l p r i m a r i l y u n r e l a t e d b a s i c t o p i c s t h a t we s h a l l These i n c l u d e p r o j e c t i v e modules,
need i n v a r y i n g d e g r e e t h r o u g h o u t t h e book.
Later
group a c t i o n s , cohomology groups and group e x t e n s i o n s and m a t r i x r i n g s .
c h a p t e r s w i l l t r e a t v a r i o u s a s p e c t s of t h e s e t o p i c s i n g r e a t e r d e t a i l and d e p t h . Here w e s h a l l be i n t e r e s t e d o n l y i n d e r i v i n g e l e m e n t a r y p r o p e r t i e s , i n i n t r o d u c i n g v a r i o u s a u x i l i a r y c o n c e p t s , and i n developing some i m p o r t a n t n o t a t i o n a l conventions.
1. NOTATION AND TERMINOLOGY I t i s important t o e s t a b l i s h a t t h e o u t s e t v a r i o u s n o t a t i o n a l conventions t h a t
For convenience, we d i v i d e t h e r e l e v a n t i n f o r -
w i l l be used t h r o u g h o u t t h e book. mation i n t o s u b s e c t i o n s .
Maps and diagrams X
Let
Y
and
be a r b i t r a r y sets.
A
map
f :x-Y
x E X a unique element y
i s a f u n c t i o n t h a t a s s o c i a t e s w i t h each element
f
:
x
I-+
Let
y.
y,
This
Y
X+
f :
is denoted by
t h e image of
and
gof.
g
:
Y 4Z gaf
The map
(qof, Given
f
:
X
-f
Y
and
g
:
X
3:
-+
f,
be maps.
i s denoted by
The composite map
Y,
y = f(xc).
X+
Y--t Z
i s g i v e n by t h e r u l e (XI =
z, fzg
under
6
grf
(5) 1
t h e map :
x-
YXZ
for all
x
E
X
CHAPTER 1
2
i s d e f i n e d by
The
identity m p lX
:
x- x
i s d e f i n e d by
X (2)=
1
Y
If
f :
X
X'
of
X.
The
X,Y
a r e s e t s , we w r i t e
-+
f(X'
is any map, t h e n
i n c h s i o n map
i
:
X'+
for all
3:
denotes t h e
X
r e s t r i c t i o n of
i s d e f i n e d by
i(x)
f
= 3c
3: E
X
to a subset for a l l
x E X'. If
X is
elements of by
Y.
a proper s u b s e t of
1x1.
Y The
X If
n o t contained i n
5Y if X X 5Y , Y - X X.
i s a s u b s e t of
Y
and
X cY
if
d e n o t e s a s u s u a l t h e s e t of a l l
The c a r d i n a l i t y of t h e s e t
X
i s denoted
diagram
Z signifies that
X,Y,
and
Z a r e sets and t h a t f
:
X
+
Y, k
:
X - + Z,g
:
2
4Y
The diagram
commutes, o r i s s a i d t o be commutative, i n c a s e f = g o k .
An a r b i t r a r y diagram i s
commtative i f we g e t t h e same composite maps whenever w e
a r e maps.
f o l l o w d i r e c t e d arrows along d i f f e r e n t p a t h s from one s e t t o a n o t h e r s e t i n t h e diagram.
Groups Unless e x p l i c i t e l y s t a t e d o t h e r w i s e , a l l groups a r e assumed t o be m ultiplicat will denote the subgroup of G generated by X.
a subset of
over N
for both the identity element and identity subgroup of a
1
The multiplicative group of a field F
G.
3
understand the smallest normal subgroup of
G
X.
containing
As
X
G we
in
usual, Z ( G )
G.
denotes the centre of
G is finite, the least common multiple of the orders of the elements of
If
G is called the exponent of G. Given
Y are subsets of G , then
In particular,
X and
[X,Yl is defined by
G’ for
G.
As
is customary, we
[G,G].
G is of finite order, G is called a torsion group,
If every element of G
for the commutator ~ - ~ y - ~ qIf.
is the commutator subgroup of
IG,Gl
also frequently write
while
k,yI
z , y E G, we write
is torsion-free if all its elements, except for
1, are of infinite
order. Let H
be a subgroup of
from each left coset xH
G.
A
subset of G
containing just one element
is called a l e f t transversal for
H
in
G , and right
transversals are defined correspondingly. Let f
:
G-+
H
understand any map
be a surjective group homomorphism.
g
:
H
G
{Gi}iEI
be a family of groups and let
Then a typical element of
f
we
such that
f o g = lH and g(1) Let
By a s e c t i o n of
rIiEIGi is
(gi)
=
1
niEIGi
be their direct product set.
with
g
E
Gi
for all i E I and
CHAPTER 1
4
(9.)= ( g ; )
i f and o n l y i f
(gi)
two elements
and
gi
(g!)
=
g;
niEIGi @ G iEI i
d i s t i n c t from groups
1
of
Define t h e p r o d u c t of
(gig;)
=
d i r e c t product of t h e groups
becomes a group c a l l e d t h e
subgroup
I.
E
by t h e r u l e
( g i ) (g;) Then
i
f o r each
niE1Gi
c o n s i s t i n g of a l l
(g.1
Gi.
w i t h f i n i t e l y many
The
g
i
i s c a l l e d t h e d i r e c t sum ( o r r e s t r i c t e d d i r e c t product) of t h e
Gi
A sequence of groups and homomorphisms
i s s a i d t o be exact a t
Gi
if Kerfi
The above sequence i s c a l l e d
exact i f
Imfi-l
=
i t i s e x a c t a t e v e r y group.
L
1-G i s e x a c t i f and o n l y i f
In particular,
H
f i s i n j e c t i v e , while G - f t H - 1
i s e x a c t i f and o n l y i f
f
is surjective.
The e x a c t n e s s of
f i s an isomorphism.
i s equivalent t o the f a c t t h a t
I f w e a r e given a three-
term e x a c t sequence
l - G I L G also c a l l e d a
short exact sequence, w e see t h a t G1
Let
N
2
f(G1)
be a normal subgroup of
morphism.
A G - I
G
and
2
G and l e t f
G2/f(C1) : G+
G/N
G/N-+
1
Then t h e s h o r t e x a c t sequence
1-
N
A
G
i s a l s o c a l l e d t h e natural e m c t sequence.
b e t h e n a t u r a l homo-
NOTATION AND TERMINOLOGY
f
A homomorphism
:
G
-+
H
5
i s s a i d t o factor through a group
B
i n case
t h e r e i s a commutative diagram
G
f
\/" f
I n t h i s c a s e , one a l s o s a y s t h a t t h e homomorphism
G--t B .
morphism
G-+ G / N
homomorphism
f
Thus a homomorphism
G
H
-.+
f a c t o r s through t h e n a t u r a l
5 Kerf
N
i f and o n l y i f
:
f a c t o r s t h r o u g h t h e homo-
Kings and modules 1#
A l l r i n g s i n t h i s book a r e a s s o c i a t i v e w i t h
assumed t o have t h e same i d e n t i t y element a s assumed t o p r e s e r v e i d e n t i t y elements.
0 and s u b r i n g s o f a r i n g
R.
R
are
Each r i n g homomorphism w i l l be
R
w i l l b e denoted by
t-+
n - 1 i s a r i n g homo-
The c e n t r e of
Z(R).
R
Let
be a r i n g .
The map
-+
R
n
d e f i n e d by
morphism whose image i s c a l l e d t h e prime s u b r i n g of
d
m
f o r a unique
R
The r i n g
R
Let
be a r i n g .
n.
J
p o t e n t , while
of
e e 1
= 0, where
2
= e e 2
= 1
r
An element An i d e a l
J
6
R
Jn
i s t h e p r o d u c t of
e2
=
e.
J
Let
1c =
0
or
y = 0
for all
E
1
n
times.
e ,e
Two idempotents
r,y
n
with i t s e l f
A nonzero idempotent i s
0.
=
0
i s n i l i f e v e r y element of
R
in
rn
i s nilpotent i f
0 and
R.
f o r some
J
is nil-
such t h a t
e
An element
a r e orthogonal if
2
primitive i f it c a n n o t b e w r i t t e n as a
sum of two nonzero o r t h o g o n a l idempotents. implies
and denoted by c h a r R.
is nilpotent i f there is a positive integer
i s idempotent i f
R
R
c a l l e d t h e characteristic of
is s a i d t o b e simple i f t h e r e a r e no i d e a l s o t h e r t h a n
positive integer
J"
2 0,
i t s kernel i s an i d e a l
R;
We s a y t h a t
R
i s a domain i f
q = O
R.
{Rili 6 I} be a f a m i l y of r i n g s and l e t
R
be t h e d i r e c t p r o d u c t s e t
CHAPTER 1
6
ni&.
It is straightforward to verify that R direct product of the family R
of
R by the rules:
We can define addition and multiplication on
onto Ri
{Rili E I}.
is a ring homomorphism:
For all
i
as the
I, the projection pri
E
the injections
Xi
:
R preserve
Ri+
1 and s o are not ring homomorphisms.
addition and multiplication but not
Assume that I is a finite set, say I = {1,2, ...,n } . if and only if R
R
is a ring; we shall refer to
Then R
9
R1
x
... x R
contains pairwise orthogonal central idempotents I -
such that 1= e 1
+
...
ie
n
and
Rei
9
Ri qua ring (1 G i
G n)
~ l R-modules l in this book are left R-modules unless otherwise specified, and each
V
R-module
1.V
An R-module summand of
0 and If
...
is assumed to be unital, i.e.
V
=
v
for all v E V
is compZetely reducible if every submodule of
V.
A
V
is a direct
V is irreducible if its only submodules are
nonzero R-module
V. V
is an R-module, we shall write
End(V)
for the endomorphism ring of
R ! I . The
f,g
elements of
t End(V),
f+ g
End(V) are R-homomorphisms from R and fg are defined by
V to V ; given
R
(f+g, ( U ) = A
f(Vf
+
( f g f(V)
g(V)
{ V i l i E 11
direct sum of a family of modules
=
iv E V)
f(g(uf)
is written
@Vi
whenever the
index set I is not pertinent to the discussion. An R-module elements x. E V
'L
is said to have an R-basis such that each
a: E V 3: =
can be written as a finite sum C r .x
z i
with uniquely determined coefficients r . E R. R-modules.
Expressed otherwise, V
{xili E 11 if there exist
is free if and only if
a direct sum of copies of the left R-module
#.
V
Such modules V
are called free is isomorphic to
Here the additive group of
RR
7
NOTATION AND TERMINOLOGY
R,
coincides with t h a t of t h e product Let
r
and t h e p r o d u c t of
E
m
R with
E R
i s d e f i n e d as
rm i n t h e r i n g R.
V # 0
be an R-module.
A
composition s e r i e s f o r
V is
a series of sub-
modules of t h e form
v = v
3 v
0
such t h a t f o r each
i>
2
3...3vk=o vi-l/Vi
t h e f a c t o r module
1,
W
i r r e d u c i b l e R-module
3 v 1
is irreducible.
i s c a l l e d a composition factor of
V
w = vi-1/v i Let
R
be a r i n g .
V
be w r i t t e n a s a d i r e c t sum called
=
X@ Y
if
f o r some
V # 0
Then an R-module
is called
An
i
decomposable i f
of nonzero submodules;
i t can
P
otherwise,
is
indecomposable.
Given a p a i r of r i n g s
R,S,
V is
w e say t h a t
an
l e f t R-module and a r i g h t S-module, w i t h t h e a c t i o n s of
(R,S)-bimoduZe if V R
and
S
V
on
is a
commut-
ing
(r E R , v
E
V, s E
Tensors products Given a r i g h t R-module
V
and a l e f t R-module
W,
t h e a b e l i a n group
V@W R called the
tensor product of
f r e e &module w i t h
VxW
V
W,
and
as a basis;
i s defined a s follows.
t h e n e a c h element of
F
Let
F
be a
can be u n i q u e l y
w r i t t e n i n t h e form (Zij
CZij ( V i , W j ’
w i t h f i n i t e l y many
z
i.i
d i s t i n c t from
g e n e r a t e d by a l l e l e m e n t s of t h e form
0.
Let
T
E
z,vi
E V,W. E
be t h e subgroup of
s
F
W)
S)
CHAPTER 1
8
where
v,vi 6 V , w,w E W , i
group
F/T,
(v,w)
The image of V
R.
r
and
V 8W R
Then
under the natural homomorphism F
With this notation, the z-module V @ W R
8 w.
is defined to be the factor
-+
F/T
consists of all finite sums
zvi 8 wi and the elements V 8 U
is denoted by
v, w .
(Vi E
E
W)
satisfy the relations
(v+v)@w=v o w + v 8 w 1
2
V@(W+W) 1
v The map
f
:
v
w-+
X
I.'@ R
w
8
2
MJ =
=v@w
+v@w
vr 8 w
defined by
f(V,W) = V @ t:
satisfies the following
properties: (i)
f(vl+ V 2 , w )
=
~ ( u , 1w+ W 2 1
(ii)
(iii)
f(v
=
: I/ X
ism
f(v,wl)+ f(v,w2)
f(vr,w) = f ( v , r ~ )
Furthermore, if g g
,W) + f ( V 2 , W )
A
W4
is any baZanced map from
v
X
w
into A ,
i.e.
if
satisfies (i), (ii), and (iii), then there is a unique homomorph-
$ : V 8 W-+
A
which renders commutative the following diagram:
R
Assume that
V
is an (S,R)-bimodule.
Then
@ R
W can be regarded as a (left)
S-module by putting
In particular, if
V
and
W are
modules over a commutative ring
R,
then V 8 W R
NOTATION AND TERMINOLOGY
9
is an R-module. The following standard properties of tensor products are direct consequences of the definition. 1.1. PROPOSITION.
Let
be an (S,R)-bimodule.
The11
V @ R E V R 1.2.
PROPOSITION.
bimodule.
Let
1.3. PROPOSITION. (VA)
(vA PP w
I-r
If
v
V be an (S,R)-
V @ R
w1 0 ... a3 V
@
R
w,
(as S-modules)
are two free modules over a commutative ring R
W
and
(respectively, (w ) ) P
is an R-basis of
V
(respectively, W), then
S
and let
V @ W. R
Let R
1.4. PROPOSITION.
R-module with
1.5. PROPOSITION. module of
V
and
1
Then S @ W
as an R-basis.
R
2
{ l @ w w l , l @W
module with
be a subring of a ring
{ w ,w ,. . . ,w 1
and
... 0 W,’ =
is an R-basis of
)
be left R-modules and let
Then V @ (W1a3 R
and
WI,W*,...,W,
as S-modules
Let
V
2
,...,1 @ w 1
is a free left S-
as an S-basis.
be a right R-module, W
W’ a submodule of W.
W be a free left
If
V’
V’
a left R-module,
a sub-
is a direct summand of
V
W‘ a direct summand of W, then the canonical homomorphism V ’ Q W ’ - +V @
R
R
w
8’ 8 W ’ under this homomorphism is a direct
is injective and the image of
R
summand of the &nodule V @ W. R 1.6. PROPOSITION.
module, M rings.
(Associativity of the tensor product).
an (R,S)-bimodule, and
Then
L @ M
R
N
Let R
a left S-module, where
is a right S-module, M @ N
be a right R-
and
S are
a left R-module, and we have
S
(L@M) @ N r L @ (MQNN) R S R S If, furthermore, L
L
is an ( R ,R) bimodule and 1
above is an isomorphism of ( R ,S )-bimodules. 1
1
N
as &nodules
is an (S,S )-bimodule, then the 1
CHAPTER 1
10
Algebras Let
R
be a commutative r i n g .
By an R-algebra w e u n d e r s t a n d a r i n g
A
which i s
a t t h e same time an R-module such t h a t
rs = ( r * l ) x f o r
I t f o l l o w s d i r e c t l y from t h e d e f i n i t i o n t h a t
r
t h a t t h e map s e n d i n g of
A.
f : R-+
Conversely, i f
R
r-1 i s a homomorphism of
to
Z(A)
r E R, s E A
Z(A)
i n t o the centre
is a homomorphism, t h e n
A
and
can b e r e g a r d -
ed as an R-algebra by s e t t i n g
rx A
I n t h i s way any r i n g
2
morphism Let
v
Z(A), n
-+
=
f(r)z
for a l l
r E R , x f? A.
can be viewed a s a z - a l g e b r a w i t h r e s p e c t t o t h e homo-
!-+ n'l.
be an R-module.
R-+
Then t h e mapping
End(l.')
which sends
r
E R
R t o t h e endomorphism
ul-+
c e n t r e of
Hence t h e r i n g
End(V).
ru
i s a r i n g homomorphism whose image belongs t o t h e
End(V)
R
can b e r e g a r d e d a s an R-algebra by
R
setting
(rf)( u )
rf(u)
=
We s a y t h a t two R-algebras
A
r
for all and
B
E
R , f E End(V1, u E V R
a r e R-isomorphic
a l g e b r a s ) i f t h e r e e x i s t s a r i n g isomorphism
A
-+
E
( o r isomorphic a s R-
which i s a l s o an isomorphism
of R-modules. An R-algebra is s a i d t o be R-free i f it i s f r e e as an R-module.
If
R
is a
f i e l d , then c l e a r l y e v e r y R-algebra i s R - f r e e . Suppose t h a t
A
and
A
1
are R-algebras.
Then one c a n d e f i n e an R-algebra
2
s t r u c t u r e on t h e R-module
A
8 A 2 by t h e formula
1 R
( a 1 8 a 2 )(a' 8 a ' ) = a a' 8 a a' 1
The R-algebra
Al f A 2
A i , i = 1,2.
2
2
i s s a i d t o b e t h e t e n s o r p r o d u c t of a l g e b r a s
I t p o s s e s s e s an i d e n t i t y element e q u a l t o
element of
1 1
2
e
0 e 2 , where
Moreover, t h e mappings
ei
A
and
A
1
i s the identity
.
11
PROJECTIVE, INJECTIVE AND FLAT MODULES
lA1
x c - t xA 18 ?2e
and
a r e homomorphisms of R-algebras such t h a t
f,(a1)f2(a2) = f2(a2’f, (a1) fi are
The homomorphisms
a
for a l l
c a l l e d canonical.
‘A l ,
The t e n s o r p r o d u c t
a2E A2 A
8 A2
is
1 R
c h a r a c t e r i z e d up t o isomorphism by t h e f o l l o w i n g u n i v e r s a l p r o p e r t y : 1 . 7 . PROPOSITION.
i
homomorphism,
all
a
€
A
1
morphism
A
Let =
and
b e a n R-algebra and l e t @
1,2, such t h a t
a
E
A2.
(a ) 1
1
$
and
2
qi (a 1 2
:
Ai+
A
b e an R-algebra
are permutable i n
A
for
Then t h e r e e x i s t s one and o n l y one R-algebra homo-
1
f
:
A
8 A2+
1 R
A
which r e n d e r s commutative t h e diagram
2 . PROJECTIVE, INJECTIVE AND FLAT MODULES Let
R
be a r i n g and l e t
0
4
x-+U
be a s h o r t e x a c t sequence of R-modules.
a(X) i s a d i r e c t summand of ( i ) There e x i s t s a
Y.
y E HomR(Y,X)
Y
B --+
2-
0
We s a y t h a t ( * )
(*)
i s a s p l i t sequence i f
T h i s i s e q u i v a l e n t t o e i t h e r of t h e f o l l o w i n g : such t h a t
ya
=
lX, where
lX d e n o t e s t h e
1L
CHAPTER 1
identity map on
X.
(ii) There exists a The homomorphisms Let
V
y
6 E HomR(Z,Y) such that B6
=
1
z
6 are called splitting homomorphisms.
and
be an R-module.
Then
is called projective, if for any given
diagram with exact bottom row
V
there exists an R-homomorphism y
V
Proof.
=
ny.
is projective
(ii) Any exact sequence (iii) V
6
The following conditions are equivalent
2.1. PROPOSITION.
(i) An R-module
such that
a
0 -+ X - - t
R
Y --+
V
-+
0 of R-modules splits
is a direct summand of a free R-module. (i) =. (ii):
Since
B is surjective and 'I
the following commutative diagram:
is projective, we have
13
PROJECTIVE, INJECTIVE AND FLAT MODULES
y
Thus
i s a s p l i t t i n g homomorphism and (ii)f o l l o w s .
( i i ) * ( i i i ) : Take a s
Y
s p l i t t i n g homomorphism. ( i i i )* (i):
(y.1
E
:
F
V+
Y(V)
Then
Y
and l e t
F.
TI
:
t h e n a t u r a l i n j e c t i o n so t h a t
B
a r e R-modules and t h a t
v-+
:
Y and
s u r j e c t i v e homomorphism, r e s p e c t i v e l y .
F+
V
X-+
be a
F,
and l e t
be t h e n a t u r a l p r o j e c t i o n Now, suppose
TIE = ly.
ci :
Y
Y.
is a d i r e c t summand of
Then we can l e t
v4
:
be a d i r e c t summand of t h e f r e e R-module
be a b a s i s f o r
and
Y
V
Let
a s u i t a b l e f r e e R-module,
Y are a
X
and
homomorphism and a
W e t h e n have a n a t u r a l map
from
F
Y, and we can d e s c r i b e t h e s i t u a t i o n p i c t o r i a l l y a s f o l l o w s :
to
F
Now f o r each g e n e r a t o r
yi
x. E X
@ n ( y.)
exists
A(y.1
a6
=
=
with
6lT(y.) E Y and because
E F, =
a
a(x.). Hence, i f we d e f i n e
is surjective there
a
:
F
X
-+
by
xi, t h e n t h e o u t e r p o r t i o n of t h e diagram i s commutative, t h a t i s ,
an.
Finally,
E
:
V-+
F
so t h a t
6E :
V
-+
X
and
.
Thus t h e lower D o r t i o n of t h e diagram i s commutative, and by d e f i n i t i o n w e
V
conclude t h a t
2.2.
COROLLARY.
i s projective.
t i v e i f and o n l y i f Proof.
Let
V
Let
V
Vl,V2,
be a f i n i t e l y g e n e r a t e d R-module.
Then
V
is projec-
i s a d i r e c t summand of a f r e e R-module of a f i n i t e rank.
...,V n
be a g e n e r a t i n g s e t f o r
V
and l e t
F
be a f r e e
CHAPTER 1
14
module freely generated by
3 1
surjective homomorphism F
summand of
F-+
,x ,...,xn 2
V.
Then the map V
Thus if
by Proposition 2.1.
sition 8.1, the result follows.
.
X
.
e
Ui
determines a
is projective, then V
is a direct
The converse being true by virtue of Propo-
'
If we reverse the direction of all the arrows in the definition of a projective module, we then obtain the definition of an injective module. R-module
V
Thus a given
is i n j e c t i v e , if for any given diagram with exact bottom row
\ P
\ \y
\ a!
Y-0 there exists an R-homomorphism Y 2.3. PROPOSITION.
(i) An R-module
(i)
$ = YU.
The following conditions are equivalent: V
is injective
(ii) Any exact sequence 0 Proof.
such that
\ AX
-+ +P'
a
(ii): Since c1
B X-+ Y -+
0
of R-modules splits.
is injective, we have the following commutative
diagram:
V
15
PROJECTIVE, INJECTIVE AND FLAT MODULES
y
Hence
is a splitting homomorphism and (ii) follows.
(ii) =) (i):
Consider a commutative diagram with exact bottom row
\ P
\ \
0-Y
is the inclusion map, in which case Y
We may harmlessly assume that
X , by hypothesis.
direct summand of by
.
y(y+z) = B f y ) , y E V , z E 2.
required.
2.4. PROPOSITION. @
(i)
Vi
{vili
Let
X
Write
y
Then
X
=
Y @ Z
r
and define
is projective if and only if each
:
X+
V
8 = Ya, as
is an R-homomorphism and
6 11 be a family of R-modules.
is a
Then
Vi
is projective
V.
is injective.
iei
rlVi
(ii)
i€I Proof.
is injective if and only if each
(i) If
V = @ Vi,
if and only if so
ieI is each Vi.
and only if each
Vi
V =
(ii) Let
rlVi
iei
then
V
is a direct summand of a free R-module
CL
:
Vi and
and let TI : V -
exists an R-homomorphism B(Z) =
(B.(z)).
sition 2.3,
V
Then
B
8;
!?-homomorphism.
y
:
X-
:
X-
Vi
:
Vi-+
V
with
TIi
be the natural
Vi
Suppose that each
is injective and
Then, for each =
Bia.
Define
is an R-homomorphism with
=
i
8
E :
I, there
x--t
V
by
ly and so, by Propo-
is injective.
Conversely, assume that
Define
ui
X be an injective R-homomorphism.
V+
is projective if
is projective.
projections and injections respectively. let
V
Hence, by Proposition 2.1,
V
is injective and let a : Vi
Then there exists an R-homomorphism
Vi by y
=
~ ~ 8Then . ya =
TI.(@)
z i -- 1vi
= 7T.p
8
:
-+
X+
X be an injective V
with
ui
=
Ba.
CHAPTER 1
16
and thus, by Proposition 2.3,
@
:
-
Let
R be a ring, V,V'
V
V', $
:
W - + W'
.
Vi is injective.
two right R-modules, W,W'
two R-homomorphisms.
two left R-modules, and
Then the map
defined by
is a homomorphism of additive groups. modules and
@
Moreover, if
is an (S,R)-homomorphism,then
@ @ $
V and V '
are (S,R)-bi-
is an 5'-homomorphism of the
$
left S-modules V @ W and V ' W'. Let R be a ring and let V be a right R R-module. The module V is said to be f l a t if for every injective homomorphism
f
:
W'+
Fi of left R-modules, the homomorphism l v Q D f :
is injective.
{Vili
E
V@W'-+V@W R H
It is an easy consequence of the definition that, given a family
I} of right R-modules, the module
@ .
Vi is flat if and only if each V .
-LEI
is flat. The following elementary result will be useful for our subsequent investigations. 2.5. PROPOSITION.
Proof.
Every projective module is flat.
Since projective modules are isomorphic to direct summands of free
modules (Proposition 2 . 1 ) , flat.
So
modules.
assume that f
it suffices to show that the regular module :
W'-+
RR is hi is an injective homomorphism of left R-
The maps
are R-isomorphisms which render commutative the following diagram
PROJECTIVE, INJECTIVE AND FLAT MODULES
17
0
Hence
18f
is injective and RR
.
is flat.
We close this section by proving
2.6. PROPOSITION.(Dual basis lemma).
An R-module
v
of elements in Hom(V,R) such that every R
{Uili E I }
is projective if and only
{uili E I } of elements in V and a family {fili E I }
if there exists a family
with finitely many
V
0.
distinct from
f.(V)
E
V
can be written in the form
Moreover, if
V
is projective,
can be any generating set.
Proof.
{Uili E I }
Let
module freely generated by
be any generating set of
{ziliE 11, and let f
f(z.1
:
let F be a free R-
V,
F
+
V
be the surjective
Owing to Proposition 2.1, i for all i E I . V is projective if and only if f splits, that is, if and only if there exists a
homomorphism defined by
homomorphism g Assume that with
fg
=
lv
-
:
V
V If
+
= U
F such that fg
is projective.
u E V,
=
lv.
Then there exists a homomorphism g
then write
g(u) =
C iEI
with
ri
E R
depending on 0.
The family
r.z zi
{fili E 11,
where
:
V+
F
CHAPTER 1
18
satisfies
Conversely, let Then
{Ui
Ii E
11 and
{Ui\i E
I} generates V
and a homomorphism
g
V-+
:
{fili t I} be as described in the theorem.
and there is a surjective homomorphism f
:
F
-+
F, where
and
Since fg = l v , V
is projective.
.
We close this section by quoting the following standard facts for the proof
of which we refer to Bourbaki (1959). 2.7.
Let M
PROPOSITION.
(i) If f
:
be a flat right R-module.
Y is a homomorphism of left R-modules, then
X--t
Ker(1, @ f) = M @ (Kerf)
R
and Im(1, 8 f) = M @ ( I m f )
R
(ii) If N ’ , ” ’
are two submodules of a left R-module N ,
M 2 . 8 . PROPOSITION.
Let
Q (N’
n N ‘ ! ) = ( M BN ’ ) n ( M BN “ )
M be
a right R-module.
R
R
then
R
Then the following properties
are equivalent: (i) M
is flat
(ii) For any finitely generated submodule N ’
of a left R-module
ical homomorphism
1,ej
:
M@”-+M@N
R
R
(j being the inclusion map) is injective. (iii) For every exact sequence of left R-modules and homomorphisms
N
the canon-
ARTINIAN AND NOETHERIAN MODULES
19
t h e corresponding sequence
a M@
M 8 N' R
R
N -@?!+M@N" R
i s exact. For e v e r y f i n i t e l y g e n e r a t e d l e f t i d e a l
(iv)
A
of
R,
t h e c a n o n i c a l map
M@A--MA
m 8 a r m a i s an isomorphism.
3. ARTINIAN AND NOETHERIAN MODULES i s s a i d t o be a r t i n i a n ( r e s p e c t i v e l y , noetherian) i f e v e r y descen-
V
An R-module
d i n g ( r e s p e c t i v e l y , ascending) c h a i n of submodules of
is artinian (respectively, noetherian).
n Z RV
V =
i
f o r some
Vl,Vp
The module
,...,U n
as b e i n g f i n i t e l y cogenerated i f f o r e v e r y f a m i l y
V
n Vi
with
=
stops.
0, t h e r e e x i s t s a f i n i t e s u b s e t
in
i s s a i d t o be f i n i t e l y
V.
We s h a l l r e f e r t o
{V,li J
PROPOSITION.
V
Let
be an R-module.
RR
v
I]
E
I
of
V
of submodules of
n V . = 0. jEJ
'
such t h a t
iEI
3.1.
R
The r i n g
a r t h i a n ( r e s p e c t i v e l y , noetherian) i f t h e r e g u l a r R-module
i t s e l f is called
generated i f
V
Then t h e f o l l o w i n g c o n d i t i o n s a r e
equivalent (i) V
is artinian
( i i ) Every nonempty s e t of submodules of (iii) Every factor-module of
( i )* ( i i ) : Denote by
Proof. assume t h a t
x
W'
with
W
2
nonempty.
W'
f o r each
i n f i n i t e descending c h a i n
{Vili
i s f i n i t e l y cogenerated.
X
E
I}
V
a nonempty s u b s e t of submodules of Then, f o r e a c h
W E
and
X, t h e s e t
Hence, by t h e axiom of c h o i c e , t h e r e i s a f u n c t i o n
W E
X.
W 3 W' 2 W" 2
Fixing
...
(iii): It suffices t o verify that i f
(ii)
h a s a minimal e l e m e n t
does n o t have a minimal element.
{W' E XlW' c W} i s WC-L
V
V
i s a f a m i l y of submodules of
W E X,
w e t h e r e f o r e o b t a i n an
of submodules of
W is a
V with
submodule of
n V.
ieI
V.
'
=
W,
then
V
and
W =
n V j w j
20
CHAPTER 1
J
f o r some f i n i t e s u b s e t
of
I.
Put
Y = { n
v Kcr
E d '
Y
By assumption,
W =
Clearly,
is finite}
-
h a s a minimal element, s a y
n Vi
J
with
,jcJ
5I
J
and
finite.
n V ,j€J . i
(iii)* ( i ) : Suppose t h a t
V
h a s a descending c h a i n
v1 3 v 2
so t h e r e e x i s t s
n
n V
W =
of submodules and p u t
61 = V
with
.
n.
1 -
...
By assumption, Hence
V/W
Vn+i = Vn
i s f i n i t e l y cogenerated,
i
for
,..., ,
= 1,2
as
asserted. 3 . 2 . COROLLARY.
V
Let
be a nonzero a r t i n i a n module.
Then
V
has an irredu-
c i b l e submodule. Proof.
Apply P r o p o s i t i o n 3.1. f o r t h e s e t of a l l nonzero submodules of
3 . 3 . PROPOSITION.
V
Let
b e an R-module.
V.
=
Then t h e f o l l o w i n g c o n d i t i o n s a r e
equivalent : (i) V
i s noetherian
(ii) Every nonempty set of submodules of ( iii) Every submodule of
Proof.
V
V
h a s a maximal element
i s f i n i t e l y generated.
.
The proof of t h i s r e s u l t i s d u a l t o t h a t of P r o p o s i t i o n 3 . 1 and t h e r e -
f o r e w i l l be o m i t t e d .
3 . 4 . COROLLARY.
Let
i/
be a nonzero n o e t h e r i a n module.
V
Then
has a
maximal submodule. Proof.
Apply P r o p o s i t i o n 3 . 3 f o r t h e s e t of a l l p r o p e r submodules of
v. =
Note t h a t t h e p r e v i o u s r e s u l t i s a l s o a consequence of t h e f o l l o w i n g observation.
3.5. PROPOSITION.
Let
V
be a f i n i t e l y g e n e r a t e d R-module.
submodule of
V
is c o n t a i n e d i n a maximal submodule.
nonzero t h e n
V
h a s a maximal submodule.
Proof. submodule of
Let
V.
{x ,.. . , x p } be a g e n e r a t i n g s e t f o r Denote by
X
Then e v e r p r o p e r
In particular, i f
V
and l e t
t h e set of a l l p r o p e r submodules of
W
V
is
be a p r o p e r
V
which
21
ARTINIAN AND NOETHERIAN MODULES
W.
contain
wA
then
Vu
Let
x 1
This i s inductive:
i s a submodule.
Wx,
V =
If
,...,xp
W
=
lJ
V,
a contradiction.
W,
.
a s required.
X
Hence
WAi
Wu
Then
R
R
contains
R
1,
v
itself.
As
and t h e submodules a r e
Thus w e o b t a i n t h e
l e f t ideals.
3.6.
i s g e n e r a t e d by t h e s i n g l e element
xi.
f o r some
i s i n d u c t i v e a n d , by
I n p a r t i c u l a r , t h e r e s u l t above i s a p p l i c a b l e t o t h e r i n g l e f t R-module,
x,
i s a chain i n
T h i s i s a maximal submodule of
Z o r n ' s lemma, it h a s a maximal e l e m e n t . containing
E
r
1
and s o
xz.
t h e n each
,...,wA .
W.,,
be t h e l a r g e s t of t h e modules
(w,)
W and i f
it c o n t a i n s
R
Let
COROLLARY.
be an a r b i t r a r y r i n g .
Then any p r o p e r l e f t i d e a l of
I n particular,
R.
i s c o n t a i n e d i n a maxiinal l e f t i d e a l of
R
h a s a maximal
l e f t ideal.
3.7. PROPOSITION.
Assume that
v-
u-
0-
i s an e x a c t sequence of R-modules. i f both
U and
Proof.
U
W
be a r t i n i a n .
is a l s o a r t i n i a n .
V, W
module of
U i s isomorphic t o
Since
U
U and
5V
and
W be a r t i n i a n . W = V/U.
W
u
V,
i s isomorphic t o a f a c t o r -
v
3 -
3 2 -
To prove t h a t
V
i s a r t i n i a n , we
... -3 vn -3 ... V.
Since
V/U
is artinian, there exists
such t h a t
vm + u = vm + i + u Because
a submodule of
Let
be a descending c h a i n of submodules of m
i s a r t i n i a n ( n o e t h e r i a n ) i f and o n l y
S i n c e e v e r y factor-module of
v1 an i n t e g e r
V
Then
is artinian.
Conversely, l e t may assume t h a t
0
are artinian (noetherian).
V
Let
w-
i s a r t i n i a n , t h e r e i s an i n t e g e r
v
n 2 m
n u = vn+i n u
Taking i n t o account modul-arity and t h a t
V
M
3 V n+i'
(i = 1 , 2 ,
..., )
(i =
...,)
such t h a t 1,2,
we have f o r each i = l , Z ,
...,
CHWTER 1
22
Vn = Vn n
vn+i + (vn+i
=
V
Hence
3.8.
V
PROPOSITION.
is
(vn+< + u) = vn+i + (vn n u)
U) = vn+i
@
. .. @ Vn.
Then
is a r t i n i a n (noetherian)
V.
and o n l y i f each
3.9.
v=
Let
n
The proof of t h e n o e t h e r i a n case i s d u a l .
is artinian.
COROLLARY.
(vn+Li) = vn
Assume t h a t
v#
0
.
. v
i s a r t i n i a n (noetherian) i f
i s e i t h e r a r t i n i a n o r noetherian.
Then
a d i r e c t sum of f i n i t e l y many indecomposable submodules.
Proof.
V
For each nonzero module
t h a t does n o t have a f i n i t e indecompos-
a b l e decomposition choose a p r o p e r decomposition
V = where
V'
v'
8 X'
h a s no f i n i t e indecomposable decomposition.
n o t a f i n i t e d i r e c t sum of indecomposable modules.
V
proving t h a t
X'@ X"
C
...
is
Then
Therefore t h e r e e x i s t i n f i n i t e chains
i s a sequence of p r o p e r decompositions.
X'
V # 0
Suppose t h a t
V
and
3
V' 3 V" 3
...
So t h e p r o p o s i t i o n i s
is n e i t h e r a r t i n i a n nor n o e t h e r i a n .
verified.
Let
3.10. PROPOSITION.
,.$ = v1 @
... @ v
with
vi
# 0
i
for
=
1,2, ...,n ,
and w r i t e
l = e Then
1
{e
,. . . ,e 1
< i < n.
potents i n
+
...
+ e
(ei E V . )
i s a s e t p a i r w i s e o r t h o g o n a l idempotents i n
Conversely, i f
R,
1
{e
,. . ., e 1
R
and
Vi
=
Rei,
I s a s e t of p a i r w i s e o r t h o g o n a l idem-
then
n
n
R ( C e i ) = @ Rei i=1 i=l Proof.
Given
r E R,
r = r.1
w e have
=
1-r = r e
particular,
ei = e.e
? - 1
+
... + e.ez n
+
... + re .
Hence, i n
23
ARTINIAN AND NOETHERIAN MODULES
e.e = dijei for all i , j . Furthermore, R e . c V . and z j 2.z @ Ren which implies that V . = Rei for all i.
proving that
...
,,J?= R e l 8
{el,.. .,e 1
Conversely, suppose that
is a set of pairwise orthogonal idem-
n
c ei. Then e 2 = e and eei = eie = e i for all i. i=l Hence Re = c Re on the right by e i' If Eriei = 0 then multiplication n j i=l gives r.e = 0 for all j . Thus Re = 8 Rei, as required. e
potents and put
=
n
I j
i=l
Assume that R
3.11. COROLLARY.
set
1
=
is artinian or noetherian.
Then there exists a
{ e ,,...,e 1 of pairwise orthogonal primitive idempotents of R n 1 ei. Furthermore,
with
i=l
# where each Rei
=
Rel @
... 8 Ren
is an indecomposable module.
Proof. Direct consequence of Propositions 3.10. and 3.9.
.
We next provide a criterion for a completely reducible module to be artinian and noetherian.
3.12. LEMMA.
We need the following preliminary observations
V
Let
be completely reducible. V
(i) Every homomorphic image of V
submodule of
is isomorphic to a submodule of
is isomorphic to a homomorphic image of V
(ii) Every submodule and every homomorphic image of (iii) If
V # 0
then W
Proof. (i) If W'
of
V.
(ii) Let
W
Then
V
V. is completely reducible.
is a submodule of
V/W'
V/W
and
1
ki"
V,
then
U is completely reducible.
V = W 8 W'
for some submodule
as asserted.
So
= V/W.
By (i), it suffices to
assume that
U
V /W
= 1
module of
U.
Then
V = V @ V 1
U
=
U @ (V + W)/W
(iii) Let
=
W 8 W'
W' = W 1 8 W 2
I/
for some submodule 2
of
V
v2
of
is a sub-
1
V.
Hence
is completely reducible.
u # 0 be an element of
submodule W
V
and so
and every
contains an irreducible submodule.
W be a submodule of V and let U
verify that
V
V.
Owing to Zorn's lemma, there exists a
maximal with respect to the property that u
for some submodule W'
of
V.
If W'
f o r some nonzero submodules W I , W p
of
W.
Write
is not irreducible, then V.
Since
CHAPTER 1
24
(wewl)n v 9 W @ Wi
it f o l l o w s t h a t
W.
maximality of 3.13. LEMMA.
v
(i)
.
i
= 1 or
is irreducible.
For a nonzero module
w = 2,
contrary t o the
the following conditions a r e equivalent:
I/
i s completely reducible
(ii) V
i s a d i r e c t sum of i r r e d u c i b l e submodules
(iii) V
i s t h e sum of i r r e d u c i b l e submodules
Proof.
( i i ) : Consider t h e c o l l e c t i o n of sets of i r r e d u c i b l e submodules
(i)
whose sum i s d i r e c t .
'b
of
i
for either
W'
Hence
( w @ w2i =
By Lemma 3 . 1 2 ( i i i ) , it i s nonempty a n d , by Z o r n ' s
{Y.),
lemma, t h e r e i s a maximal e l e m e n t , s a y
V = W@
and l e t
W'.
V'.
i r r e d u c i b l e submodule
{Yi}
i m a l i t y of
.
W' # 0
If
Hence
t h e n by Lemma 3 . 1 2 ( i i ) ,
W
+
V' =
W' = 0 and
Thus
i n t h i s collection.
C" 8
V = W,
W
Let
(iiil,
61'
= @
V
i
c o n t a i n s an
Vi), c o n t r a r y t o t h e max-
(@
as r e q u i r e d .
(ii) * ( i i i ) : Obvious ( i i i ) * ( i ):
W' of
module
W
+ W'
=
W
Let
V
W 8 61'.
V .
3
Vi E
Vi
V
E V
V.
and
d u c i b l e , we i n f e r t h a t
with
v 4 FIB W'.
V . n ( W @ W') # Vi. 3
V . n (W @ W') 3
(W' 8 V . )
W :1
COROLLARY.
3
Let
V # 0
=
=
0.
+
V = V
V, 1 6 i Because
Assume the
< n.
V
j
...
(ii) V
i s irre-
61' @
vj
0, c o n t r a r y t o t h e maximality of
be c o m p l e t e l y r e d u c i b l e .
W'.
.
Then t h e f o l l o w i n g con-
is artinian is n o e t h e r i a n
(iii) V
Proof.
V
Hence
d i t i o n s a r e equivalent: (i) V
+
Thus
v. = W @ 3
Thus
V = W @ W'.
By h y p o t h e s i s ,
i s an i r r e d u c i b l e submodule of
W @ W' +
3.14.
W n W ' = 0.
I t t h e r e f o r e s u f f i c e s t o show t h a t
9 W @ W' f o r some j and so
and t h e r e f o r e
By Z o r n ' s lemma, t h e r e e x i s t s a sub-
maximal w i t h t h e p r o p e r t y t h a t
c o n t r a r y and choose where
Y.
be a submodule of
.
i s t h e d i r e c t sum of a f i n i t e number of i r r e d u c i b l e submodules D i r e c t consequence o f Lemma 3 . 1 3 and C o r o l l a r y 3 . 8 .
.
25
GROUP ACTIONS
3.15.
Proof.
i s completely r e d u c i b l e , t h e n so i s e v e r y R-module.
17 be an R-module and l e t
Let
which sends
RU
fiR
If
COROLLARY.
to
p
by Lemma 3 . 1 2 ( i i ) .
i s a homomorphism.
W
V
Since
C
=
Then t h e map from
Thus
RR
to
i s completely r e d u c i b l e
RV
t h e r e s u l t f o l l o w s by v i r t u e of P r o p o s i t i o n
RV
UEV
=
3.2.
V.
ti E
wetherian.
V
i s s a i d t o b e of f i n i t e length i f
V
An R-module
i s b o t h a r t i n i a n and
We c l o s e t h i s s e c t i o n by q u o t i n g t h e f o l l o w i n g s t a n d a r d f a c t s ( s e e
Bourbaki (1959) ) .
3.16.
R.
PROPOSITION.
V
Then
3.17.
be a f i n i t e l y g e n e r a t e d module o v e r an a r t i n i a n r i n g
i s of f i n i t e l e n g t h .
PROPOSITION.
length.
v
Let
Then
V
(Krull-Schmidt theorem).
can b e w r i t t e n as a d i r e c t sum
ipdecomposable subnicdules.
n = k
t h i s k i n d , then
v=
Moreover, i f
v
Let
be an R-module of f i n i t e
n I.' = 6 V i ,
where V . a r e i=l i s a n o t h e r decomposition of
k @ V'
j=1 j V!) 3
and ( a f t e r p o s s i b l y r e o r d e r i n g t h e
w e have .'L
%
2
V!
Z
i.
f o r each
4. GROUP ACTIONS
Let
G
X
be an a r b i t r a r y group and l e t
X,
symmetric group on By an
action
of
G
be a set.
X we
on
X, g
E
G.
X.
(G,X,f) ( o r simply t o X) f
a s a G-set.
'x
and write
It w i l l
f (g) ( x ) ,
i n s t e a d of
With t h i s c o n v e n t i o n , t h e f o l l o w i n g p r o p e r t i e s h o l d
1x abz Conversely, i f f o r each
=
x
for all
x E X
=
a(bx)
for a l l
x
x
E
X
and
g
E G,
which s a t i s f i e s (1) and (21, t h e n t h e map
f ( g ) !I) = g x , x
f o r the
u n d e r s t a n d a homomorphism
be c o n v e n i e n t t o s u p p r e s s t h e r e f e r e n c e t o E
Sym(X)
i . e . t h e group of a l l p e r m u t a t i o n s of
We s h a l l r e f e r t o t h e t r i p l e
x
We w r i t e
E
X, g E G i s a
E
(1)
X, a , b
E
G
t h e r e i s a unique element
f
:
homomorphism.
G+
Sym(X)
d e f i n e d by
(2)
'x
E
X
26
CHAPTER 1
X
Let
x
By t h e orbit of
be a G-sat.
X, we u n d e r s t a n d t h e set
defined
‘5
by =
‘3:
X
I t is clear that
G(r)
Igxlg
GI
E
i s a d i s j o i n t union of o r b i t s .
f o r t h e stabilizer of
z in
G
z 6 X, w e w r i t e
Given
d e f i n e d by
G(z1 = { g E
GIgx
=
xI
I t i s an immediate consequence of t h e d e f i n i t i o n t h a t
G(gx) = gG(z)g
-1
for a l l
X,g
Z . E
E C.
and t h a t
Now assume t h a t
X is a X.
morphism group of
X
G on
action of
group, module o r an a l g e b r a and l e t
Then c l e a r l y
:
We s h a l l r e f e r t o t h e t r i p l e
(G,X,f)
algebra according t o whether
X
a c t s on a f i e l d
t o be t h e kernel Of
X
i s a subgroup of
X
be t h e a u t o By an
Sym(X).
we u n d e r s t a n d a homomorphism
f
C
Aut
Aut
F,
G
+
AutX
( o r simply t o
X)
a s a G-module o r a G-
i s a module o r an algebra..
we s a y t h a t
the action.
F is a G-field.
Thus i f
G
In particular, i f
f
The k e r n e l of
X
a c t s on
and
is said
Go i s t h e k e r n e l
of the action, then = If
Go = 1,
t h e n we s a y t h a t
X,
f u l l y on
tg
then
Assume t h a t
A
G
E
G
~ l g x=
x
for a l l
a c t s faithfuZly on
z E XI
X.
i s i d e n t i f i a b l e w i t h a subgroup of
i s a G-algebra.
G acts faith-
Thus i f Aut
X.
Then t h e f i x e d s u b a l g e b r a
‘A
of
C is
d e f i n e d by
AG Thus
AG Let
=
{a E ~ l g a= a
i s t h e l a r g e s t s u b a l g e b r a of
F
be a G - f i e l d .
Then
FG
A
for all
g
on which
G
6
GI acts trivially.
i s o b v i o u s l y a s u b f i e l d of
F.
An
GROUP ACTIONS
27
G
i m p o r t a n t p a r t i c u l a r c a s e of G - f i e l d s i s t h e c a s e where t h e a c t i o n of is faithful.
G
I n t h i s case
i s i d e n t i f i a b l e w i t h a subgroup of
G on F i s given by
a c t i o n of
'1
=
g(A)
F.
s i d e r e d a s an e x t e n s i o n of v e c t o r space o v e r written
W e can view
We s a y t h a t
is f i n i t e or infinite.
F
i s a r o o t of a nonzero polynomial o v e r
b r a i c over
F
azgebraic o v e r
i s s a i d t o be A
f
polynomial
F
L
if
over a f i e l d
a l g e b r a i c over tension
F.
f).
f E F[XI
nomial Let
E/F
~1
F.
which h a s a r o o t i n
be a f i e l d e x t e n s i o n .
f i x e d f i e l d of t h e group A field
e v e r y element of Let
E/F
F
i s algebraic o v e r
E
The e l e m e n t s of F
F
alge-
in
E;
E
i f it is a
i s separable o v e r
E/F
F
F
i s separable.
i f it is An ex-
E i s separable over
i f e v e r y element of
E
s p l i t s over
E.
Then t h e group of a l l F-automorphisms of
E
i s a l g e b r a i c over
Gal(E/F).
F
and
F
E
We s a y
is the
Gal(E/F).
i s s a i d t o be p e r f e c t i f e i t h e r
F
and i s
i s s a i d t o b e normal i f any i r r e d u c i b l e poly-
i s a Galois extension i f
E/F
E
of
i s c a l l e d t h e Gulois group of t h e e x t e n s i o n and i s w r i t t e n that
F
a l l of which have o n l y simple r o o t s
An element of
E/F
An a l g e b r a i c e x t e n s i o n
over
= E.
F[Xl,
i s a separable extension
E/F
An element
and i f i t s minimal polynomial o v e r
F
E
i s c a l l e d separable o v e r
F
p r o d u c t of i r r e d u c i b l e polynomials i n ( i n a s p l i t t i n g f i e l d of
and hence a
c a l l e d t h e algebraic closure of
L,
form a s u b f i e l d
con-
i s a f i n i t e or an i n f i n i t e extension accord-
E/F
(E:F)
CY
and t h e
E
f o r the f i e l d
a s an F-algebra
ingly as if
E/F
i t s dimension i s c a l l e d t h e degree o f
F;
(E:F).
E
Aut F
F
g E G, A E F.
for a l l
E be a f i e l d and F a s u b f i e l d ; we w r i t e
Let
on
charF = 0
or
charF = p > 0
and
i s a p - t h power.
be a f i e l d e x t e n s i o n .
g e n e r a t e d by a s i n g l e element o v e r c a l l e d a p r i m i t i v e element o v e r
We s a y t h a t F;
E
i s simple i f
E
such a g e n e r a t i n g e l e m e n t f o r
is E
is
F.
We c l o s e t h i s s e c t i o n by q u o t i n g t h e f o l l o w i n g s t a n d a r d f a c t s ( s e e C o h n ( l 9 7 7 ) ) .
4.1. on
PROPOSITION.
F.
Then
IGl
Let =
F
G
be a G-field where
( F :F I .
G i s f i n i t e and a c t s f a i t h f u l l y
28
CHAPTER 1
The f o l l o w i n g p r o p e r t y i s a n immediate consequence of P r o p o s i t i o n 4 . 1 .
4.2.
F
Let
COROLLARY.
IG/G~I 4.3.
PROPOSITION.
E/F
i f and o n l y i f 4.4.
Let
E/F
Then
)
be a f i n i t e f i e l d e x t e n s i o n .
E/F
A f i n i t e f i e l d extension
4 . 5 . PROPOSITION.
4.6.
be f i n i t e .
Then
E/F
i s Galois
i s normal and s e p a r a b l e .
PROPOSITION.
F
C
(F:F
=
a s p l i t t i n g f i e l d o f a polynomial o v e r
of
G/Go
be a G - f i e l d and l e t
F
A field
B
i s normal i f and o n l y i f
is
F.
i s p e r f e c t i f and o n l y if e v e r y f i n i t e e x t e n s i o n
i s separable.
PROPOSITION.
element
A
E E
Let
E/F
E
such t h a t
L
number of f i e l d s
be a f i n i t e f i e l d extension. =
F(h) F
such t h a t
t h e r e e x i s t s such an element
Then t h e r e e x i s t s an
if and o n l y i f t h e r e e x i s t s o n l y a f i n i t e
5L
E.
If
E
i s s e p a r a b l e over
F,
then
A.
5 . COHOMOLOGY GROUPS AND GROUP EXTENSIONS
N
Throughout t h i s s e c t i o n ,
G
and
G
a r e f i x e d groups.
An extension o f
N
by
i s a s h o r t e x a c t sequence of groups C
E :
Y
B
l--tN--+X---+G---tl
(1)
Assume that
a'
c('
E'
1-
:
i s a n o t h e r such e x t e n s i o n .
a homomorphism
y
:
X+
Y
+G
N-Y
W e say t h a t
E
and
--+
E'
1
are congruent i f t h e r e e x i s t s
which r e n d e r s commutative t h e f o l l o w i n g diagram:
COHOMOLOGY GROUPS AND GROUP EXTENSIONS
29
The following lemma shows that y
is in fact an isomorphism, and therefore the
congruence relation is symmetric.
Because it is obviously reflexive and tran-
N by G; the
sitive, we may speak about the congruence classes of extensions of
E
congruence class of 5.1. LEMMA.
[El.
will be denoted by
Assume that the diagram
p,G31-
a
l-Gl
-G2
is commutative and that both the upper and lower rows are exact. are injective (surjective or bijective), then so is Proof.
and y 3
y
Assume that both
1
y2
is also injective.
Suppose that
y2-
are injective.
y (g)
=
We wish to show that g E GP.
1 for some
2
1 = B
B (g)
from which we get
=
[Y ( g ) I
2
since y 3
1
=
Then we have
Y [B ( g ) 1 3
2
and Y 3
y1
If
1
is injective.
By hypothesis, the upper
1
row is exact.
So we have
= ImCl
KerD 1
such that
g = Nl(gl).
1
.
Hence, there is an element g
GI
Accordingly,
1
=
y (gl
=
y [a (g ) I 2
1
1
which entails, by the commutativity of the diagram, that a by assumption, both
of 1
y1
and
N 2
are injective.
2
So we get
g , = 1 and g = a ( g ) = 1 1
This proves that
Y2
is injective.
Next, assume that both
Y
and 1
show that
h E y2 (C2).
1
Since
y3
y3
are surjective.
Let h E h'
2
.
We must
is surjective, there is an element g 3 of
G3
30
CHAPTER 1
R2(h)
such t h a t
g
that
= y3(g,).
By h y p o t h e s i s , t h e upper row i s e x a c t .
g
R1(g2) f o r some element
=
Bz (h)
=
G
of 2
.
2
h
Put
Because t h e lower row i s e x a c t , w e have
So
o 2 [Yz(9,)1
Y 3 [B1 ( g 2 )1 =
by t h e commutativity of t h e diagram.
=
Kerl?
Y2(g2)k2.
Ima
=
2
h
E
H
1
a (h
such t h a t 1
2
g
element
E GI
)
=
.
h
1
satisfying
y1
As
Yl(g)
=
.
B (h
Then we have
2
)
=1
2
Hence t h e r e i s a n element
i s a l s o s u r j e c t i v e , t h e r e is a n
.
h
We conclude
k2
Thus
=
a2[yl(gl)l= Y 2 [ a ( g l ) l ,
and w e have
.
h = ~ ~ I g ~ @ ~ (Eg Y~ ~l (IG ~ I T h i s proves t h a t
y2
is surjective.
Observe t h a t any e x t e n s i o n (1) i s c o n g r u e n t t o t h e one i n which i n c l u s i o n map.
is t h e
ci
For t h i s r e a s o n , from now on we s h a l l c o n c e n t r a t e on t h e exten-
sions
E :
i
where
l - N ~ X - f t G - - + i
d e n o t e s t h e i n c l u s i o n map.
t
u n d e r s t a n d any map
: G--t
X
I n what f o l l o w s by a section o f
we
such t h a t
t ( l )= 1 and
f't
= 1
G
O f c o u r s e , a p a r t i c u l a r c a s e of a s e c t i o n i s a s p l i t t i n g homomorphism
( i f it e x i s t s ) .
f
Note a l s o t h a t f o r a l l
G
----f
X
x,y E G,
f ( t (2)t ( y ) t ( x y ) - l ) =f ( t ( 2 ))f(t(Y))f(t(?g) 1-1 = x y ( q f ) - l = 1 which shows t h a t
Let
T
=
{T(g)lg
W e say t h a t
n
E N.
E
GI
be a f a m i l y of automorphisms of
( a , T ) i s a f a ct o r s e t of G o v e r
N
N
and l e t
i f for a l l
ci :
G xG-+
r,y,z E G
and
N.
31
COHOMOLOGY GROUPS AND GROUP EXTENSIONS
(The c o n d i t i o n s ( 5 ) are n o t e s s e n t i a l f o r t h e subsequent d i s c u s s i o n s , b u t t h e y
a r e s a i d t o be congruent i f t h e r e e x i s t s a map
(a',T')
(a,T) and
Two f a c t o r sets
h e l p t o s i m p l i f y some of t h e c o m p u t a t i o n s ) .
1 : G+
N
with
h ( l ) = 1 such t h a t
I t i s immediate t h a t t h e congruence of f a c t o r s e t s i s an e q u i v a l e n c e r e l a t i o n .
5.2.
LEMMA.
let
t
:
G-
Let E :
X
2' = { T ( g )ig E G}
A
1 --t N
X
f
be a s e c t i o n of
5 Autu
f , G-
and l e t
a
: GxC-
N
and
(x,y)
=
t (x)t ( y ) t (q)
(z,y
T ( g ) ( n ) = t(g)nt(g)-l
G)
(8)
(g E G,n E N )
(9)
(a,T) i s a f a c t o r s e t ( t o which w e r e f e r as b e i n g a s s o c i a t e d w i t h
t).
corresponding t o t h e s e c t i o n
Furthermore, i f a d i f f e r e n t c h o i c e of
proving ( 5 ) .
Since Given
a(x,l)
t(l) = 1, w e o b v i o u s l y have
n
E
N
x , g E G,
and
E,
or
f
is
(a,T).
made, then t h e corresponding f a c t o r s e t i s congruent t o Proof.
G,
by
be d e f i n e d by N
Then
N
1 b e an e x t e n s i o n of
=
a(l,x)
= 1
and
T(1)
=1.
w e a l s o have
a(x,y)T(xyy) ( n ) a ( x , y ) - l = t (32) t ( y ) t(xy)-lt(qf)nt (q)-lt(q t (y)-lt(x)-l )
proving
(3).
proving ( 4 ) .
Next, f o r a l l
Finally, l e t
Then t h e r e e x i s t s a map
=
t ( 2 )t ( y ) n t( y ) - l t (x)
=
[P(x)T(yIl( n ),
x,y,z E G,
t' 1
: :
G
G
-t
--+
N
w e have
X
b e a n o t h e r s e c t i o n of with
f.
t ' ( g ) = X ( g ) t ( g ) f o r all
g E G.
32
CHAPTER 1
proving ( 6 ) .
x,y E G ,
Finally, for a l l
proving ( 7 ) and hence t h e r e s u l t .
we have
9
I t i s a consequence o f Lemma 4 . 2 t h a t e a c h e x t e n s i o n
E : of
N
over
G
by
N.
N
1-
A
X
L G-
1
-rE
d e t e r m i n e s a unique congruence class o f
o f f a c t o r sets of
G
We a r e now r e a d y t o o b t a i n a s u r v e y of a l l congruence c l a s s e s of
e x t e n s i o n s of a p r e s c r i b e d p a i r of ( n o t n e c e s s a r i l y commutative) g r o u p s . 5.3.
THEOREM.
{El
The assignment
T
&+
E d e t e r m i n e s a b i j e c t i v e correspondence N
between t h e congruence c l a s s e s o f e x t e n s i o n s o f c l a s s e s of f a c t o r s e t s of (i) E
G over N .
(a,T)
a(r,y)
f o r which
= 1
Let
(Ci,T)
E :
define
-+
X
then
N
with
E
i s a s p l i t exten-
A ( 1 ) = 1 such t h a t
T(2)(A(y))-1A(r)-1A(2y)
be a f a c t o r s e t of
1-N
C over N . X f .
G
for all
r , y E G.
be t h e d i r e c t p r o d u c t s e t
N x G.
We c o n s t r u c t a n e x t e n s i o n
1
---L
(a,!!') i s a f a c t o r s e t a s s o c i a t e d w i t h
such t h a t Let
E,
E G.
For t h e sake of c l a r i t y , w e d i v i d e t h e proof i n t o t h r e e s t e p s .
Proof.
S t e p 1.
=
z,y
for a l l
1: G
s i o n i f and o n l y i f t h e r e e x i s t s a map
Cd(r,y)
and t h e congruence
i s t h e congruence c l a s s of a
TE
(a,T) i s a factor set associated with
(ii) I f
G
Furthermore,
i s a s p l i t e x t e n s i o n i f and o n l y i f
factor s e t
by
E.
Given
( n ,g 1
1
)
,f n2 , g p )
E X,
33
COHOMOLOGY GROUPS AND GROUP EXTENSIONS
( n ,g 1 ( n ,g I 1
W e w i l l show t h a t
X
2
1
2
=
(nlT(gl) (n2)a(g1,g2) ,g1g2)
i s a group under t h i s o p e r a t i o n .
(10)
Indeed
and
(by ( 4 ) ) The i d e n t i t y is t h e p a i r (1,l).
proving t h a t t h e given o p e r a t i o n i s a s s o c i a t i v e .
(n,g)
The i n v e r s e of
and t h u s
X
i s a group.
The map
s u r j e c t i v e homomorphism whose k e r n e l
BY
( l o ) , the
fore identify
t(g)
and
=
map
N
(n,l) and
W
K.
(n)-',g-')
(CX(g-',g)-'T(g-')
is
n
f
:
K
X---t G d e f i n e d by
i s an isomorphism of
f.
fl(n,g)l = g (n,l)
c o n s i s t s of a l l p a i r s
Furthermore, t h e map
(l,g) i s a s e c t i o n of
s i n c e by ( 8 )
Since
t
K :
N.
onto
G
-+
X
is a
:
W e may t h e r e -
d e f i n e d by
CHAPTER 1
34
(T,a) i s
we s e e t h a t
S t e p 2. if
H e r e w e show t h a t t h e e x t e n s i o n s
t
Let
corresponding t o
, we
G
+
(f,T)
and we d e n o t e by
and ( 9 )
:
X
E'
and
are congruent i f and o n l y
t
and
g E G.
a and a '
t ' , respectively.
t h e f a c t o r s e t s of
Y
Applying
(f,T,= (f',T'). Thus
see t h a t
and l e t
t'
:
G
Y
t' i s o b v i o u s l y a s e c t i o n
Then
(f',T')
and
6
be a s e c t i o n of
t ' ( g ) = y ( t ( g ) ) for a l l
be d e f i n e d by
O',
E
t.
Assume w e a r e g i v e n a commutative diagram ( 2 ) i n which
TE = T E l .
a r e i n c l u s i o n maps.
of
t h e f a c t o r s e t corresponding t o t h e s e c t i o n
G
over
N
t o t h e f o r m u l a s (8)
TE = TE,.
Conversely, assume t h a t
5X-
N
1-
E :
G
f,
1
and
E' a r e e x t e n s i o n s of of
f
and
l - N - - + iY - + G ~ l
:
N
by
f' and
G
such t h a t
(a,T) and
N
--+
can be uniquely w r i t t e n i n t h e form
:
A ( 1 ) = 1.
with
nt(g),n
Y
one immediately v e r i f i e s t h a t t h e map
y
t
If
= TEl.
and
t' a r e s e c t i o n s
(a',T') a r e t h e c o r r e s p o n d i n g f a c t o r sets, t h e n
A : G
( 6 ) and ( 7 ) hold f o r some
TE
nt (g)
E N, g E
X----t
:
Y
X
N o w each element of
G.
Applying ( 6 ) and ( 7 ) ,
g i v e n by
nX ( g ) - l t 1 ( g )
i s an isomorphism which makes t h e r e q u i r e d diagram commute.
Hence
E
and
E'
are congruent.
S t e p 3.
Completion of t h e p r o o f .
By S t e p s 1 and 2 ,
{ E l c-t
T~
i s a b i j e c t i v e correspondence between t h e
congruence c l a s s e s of e x t e n s i o n s of
G
f a c t o r sets of s i o n of
N
G,
by
corresponding t o
over
let
t.
If
T~
for a l l
G
by
N.
Assume t h a t
t
be a s e c t i o n of
E
a homomorphism i n which c a s e that
N
and t h e congruence c l a s s e s of
E : 1-
E
G.
If
*A
f and l e t
X+
f
G-+
1 i s an e x t e n -
( a , T ) be t h e f a c t o r s e t
is a s p l i t e x t e n s i o n , t h e n we may assume t h a t
a(x,y)
= 1
for a l l
i s t h e congruence c l a s s of a f a c t o r s e t
x,y
N
x , E~ G. (a',T')
t
is
Conversely, assume f o r which
a'(x,y)
= 1
mn
COHONOLOGY GROUPS
E'
N
i s t h e e x t e n s i o n of 2,
E
E'
and
N
1-
:
G
by
Y-
-+
f'
G
E'
Since
1
4
(a',T')
c o n s t r u c t e d from
a r e congruent.
35
GROUP EXTENSIONS
i n S t e p 1, t h e n by S t e p
E,
i s a s p l i t e x t e n s i o n , so i s
.
proving ( i )
By ( i ) , E factor s e t
(a',?")
.
a'(x,y)
such t h a t
=
by a p p l y i n g ( 7 ) and hence t h e r e s u l t . 5.4.
COROLLARY.
E'
*
p?,
G
:
x,y,g E G
N
-+
and
(a,T)
Let
x,y E G .
i s congruent t o
f
f'
'
respectively.
for all
This proves (ii)
l - - + N ~ Y - G - l
:
be s p l i t t i n g homomorphisms.
Proof.
x , y E G.
1 for all
l - n T ~ X - + G - - - t l
be t w o s p l i t t i n g e x t e n s i o n s of
for all
i s congruent t o a
Let
E :
e x i s t s a map
(Cr,T)
i s a s p l i t e x t e n s i o n i f and o n l y i f
N Then
with
G
by
E
and l e t
:
i s congruent t o
G -.+
E'
X
u'
and
G
:
Y
-+
i f and o n l y i f t h e r e
A ( 1 ) = 1 such t h a t
n E N. (a',T'1
and
Since
and
By Theorem 5 . 3 ,
(a?,T').
be t h e f a c t o r sets c o r r e s p o n d i n g t o
i.~'a r e homomorphisms w e have
E
i s congruent t o
E'
u
and
a(x,y)=a'(x,y)= 1
i f and o n l y i f
(a,T)
The d e s i r e d c o n c l u s i o n i s now a consequence o f (6)
and ( 7 ) . We n e x t examine t h e s p e c i a l c a s e where
5.5. LEMMA.
Let
G
(a T ) 2' x
l+ N
i s a c y c l i c group of f i n i t e o r d e r .
n
be a c y c l i c group of o r d e r
E : be an e x t e n s i o n of
G
by
G.
N"-
*
X
f
-+
Then, f o r any
G
g e n e r a t e d by
E
and l e t
1
-+
x
g
X
with
f(x)
= g,
the pair
g i v e n by
(0 G i , j
<M-
1)
36
CHAPTER 1
Tx(gi) ( y )
tx
The map
,
(0 G i
t x ( gi)
G--' X g i v e n by
f.
o b v i o u s l y a s e c t i o n of Then, by ( 8 )
:
ziy3=i
n - 1,y E N )
Q
E.
i s a f a c t o r set corresponding t o Proof.
=
La ,!?
Let
x x
i,
= 3:
0
< i Q n - 1,
is
be t h e corresponding f a c t o r set.
)
we have
i i i+j i + j -1 ax(g ,g ) = z t x ( g )
(0 4 i,j 4 n - 1)
. . If
i+j n
E N
i
(iil
Given
t o be
i,j
E
T(g).
{O,l,
E
N
(13)
(0G i , j G n-1)
(14)
(0G i G n-1)
(15)
are such t h a t (13) h o l d s , t h e n
(CL,T)
defined
G o v e r N.
( i f I n t h e n o t a t i o n of Lemma 5 . 5 , p u t
@
y
for all
1
$I
by (14) and (15) is a f a c t o r s e t of
and d e f i n e
(a,T)
and a f a c t o r s e t
a(gi,gj)=
Proof.
G, t h e n t h e r e
by
such t h a t
@ ( a )= a
(ii) I f
N
1 i s a n e x t e n s i o n of
(a,!?) = (a ,T
x x
)
and
a
=
xn,
Then ( 1 3 1 , (14) and (15) h o l d by v i r t u e of Lemma 5.5.
...,n-11
and
y E N,
w e have
37
COHOMOLOGY GROUPS AND GROUP EXTENSTONS
Property ( 5 ) being obvious, w e a r e l e f t t o v e r i f y t h a t f o r a l l
proving
(3).
be a c y c l i c group of o r d e r
a c t i n g on an
Then
i=0 In particular,
n
N 2 ( C , A ) = 1 i f and o n l y i f
A
n-1
= (
.
C gZ)A. i=0
Proof. induced by map
~1
a
:
Let
a E AG, G =
g.
Then
G x G-
A
is a c o c y c l e .
@(a) = a
and
and l e t = 1
0
be t h e automorphism of
A
and hence, by Theorem 5 . 6 ( i . i ) , t h e
g i v e n by
Hence w e o b t a i n a homomorphism
46
CHAPTER 1
which i s s u r j e c t i v e , by v i r t u e of Lemma 5.5.
a
that
a
i s a coboundary i f and o n l y i f L',
Let
:
X
3A
1
f -+
ding t o
aa.
Setting
j = 0, we o b t a i n
now p u t
z=
i=o G
-+
1 be a n e x t e n s i o n o f
U
Then t h e r e i s a s e c t i o n
!J ( g ) ,
i
!J(g
i
of
for a l l
i
) =
n-1 Pi(g)!J(g) = 1
s p l i t s i f and o n l y i f
. ..
( b ~= )bgb ~
n-i E
Hence
a E
s p l i t s i f and o n l y i f
(
1 g
E
f
by
G
correspon-
such t h a t
{ o , l , ...,a-
E
x
I f we
1).
n
f o r some
'
.
s p l i t s , t h e r e s u l t follows.
--
n- 1 bzn
)A.
Z=0
and only i f
-*
A
t h e n w e have
n- 1
E
x
: G
U(g)
=
a = aa(g,g The e x t e n s i o n
W e are t h e r e f o r e l e f t t o v e r i f y n- 1 E ( C gi)A.
B u t , by (2).
...
= b'b
Since
b E A.
i s a coboundary i f
0:
7 . MATRIX RINGS AND RELATED RESULTS
Let
R
be an a r b i t r a r y r i n g and, f o r any
R.
of a l l n r n - m a t r i c e s o v e r
Mn(R)
n 2 1, l e t R
We shall identify
w i t h i t s image i n
c o n s i s t i n g of a l l s c a l a r m a t r i c e s d i a g ( r , r , . - . , P ) , r E R.
i , j G n,
(i,j),1 elsewhere.
e
let
ij
e
The elements
denote t h e r i n g
For e v e r y o r d e r e d p a i r
b e t h e matrix with ( i , j ) - t h e n t r y
ij'
called the
mtriz units,
Mn(R)
1 and
0
s a t i s f y t h e following
properties (i) e . . e = 0 1.3 k s 1= e
(ill
11
+
if
...
j# k + e
and (iii) determine
7.1. PROPOSITION. {vij(l G i , j 4 n } set in
S.
e i j e k s = eis
j
if
=
k
nn teij)
( iii) The c e n t r a l i z e r of
Note a l s o t h a t
and
R
e
11
M (I?) Let
S
in
M (R)ell. n
M (R)
is
R.
Our f i r s t o b s e r v a t i o n shows t h a t ( i ),( i i ),
up t o isomorphism. be a r i n g t h a t c o n t a i n s a s e t o f elements
s a t i s f y i n g (i) and ( i i ) and l e t
Then t h e map
R
b e t h e c e n t r a l i z e r of t h i s
MATRIX RINGS AND RELATED RESULTS
47
Furthermore, R
is an isomorphism of rings and R-modules.
Sv
21 11
11
.
We first note that Mn(R) is a free 8-module freely generated by the
Proof. matrix units
ei j .
eij
Hence the map
$.
modules, which clearly coincides with
extends to a homomorphism of R-
Vij
t+
In view of (i), (ii) and (iii), $
is
easily seen to be a ring homomorphism.
c a . .U
Suppose that
t
=
i , j 23 i j
0 and fix k , s E 11 ,...,n l .
Then, for all
t {l, ...,n l ,
c
0 = Vtk(
s t - aksVtt
U..V..)V
i ,j
ZJ 1-3
and therefore
u t ks tt - a k s (tC u t t ) = a ks
Q=Ca
(a..)
= 0 and so ii, is injective. z3 Finally, fix s E S and for each i , j
Thus
Pij
Then, for all
Vrt,
put
= FkiSUjk
we have
vrtrij
=
yrtvkisvjk
=
vpisujt
and
r Hence
rij
u = L u s v v - u s v ij rt k i j k rt - p i j t and so r
commutes with all Vrt
proving that $
is surjective.
.
Given a (left) R-module direct power of
V.
V
E R.
Moreover, by the foregoing,
Taking into account that
$ ( e l l M n ( R ) e l l )= u the result follows.
ij
Su 11
and 11
e
11
M (R)ell n
and a positive integer n ,
2
R,
we write
v"
for n-th
The following result illustrates that matrix rings arise as
48
CHAPTER 1
endomorphism r i n g s of d i r e c t powers of modules.
7.2.
PROPOSITION. proof.
End(v")
Mn(End(V)) R i , j E {1,2 n}, d e f i n e
R
Given
2
,...,
E..(U
7-3
Then o b v i o u s l y
$ ( u l ,...,un) $
= $ 1
= 2
able w i t h
=
EijEks
1
,...,v
6jk~is
)
and
=
i' 0,...,0)
ZE..
7-7-
... = Qn.
End(?) R centralizes a l l E..
v"
in
t3
is identifi-
End(?)
R
v"
v"
a s an R-module.
But one a l s o
M (R)-module i n t h e f o l l o w i n g n a t u r a l way.
as an
A E M (R)
as column v e c t o r s and, f o r each
(i) I f
W
We v i s u a l i z e
x
and
E
v",
We a r e now r e a d y t o prove
as t h e m a t r i x m u l t i p l i c a t i o n .
v,
i s a submodule of
V
t h e n t h e map
wk-+ w"
o n t o t h e l a t t i c e of
M
i s an
(R)-sub-
v".
(ii) End([/?)
9
M (R)
End(V)
( i i i ) TEe map
R V-
i s m c l a s s e s of
R
W w e
Proof.
I€..}
'
isomorphism from t h e l a t t i c e of submodules of
g i v e n by
i f and o n l y i f
The d e s i r e d c o n c l u s i o n i s t h e r e f o r e a consequence of Propo-
1 . 3 . PROPOSITION.
modules of
and
E
$3
Hence t h e c e n t r a l i z e r of
End(V). R
t h e elements of
Ax
$
If
= 1.
I n t h e preceding d i s c u s s i o n w e regarded
define
by
1
s i t i o n 7.1.
can r e g a r d
End(?) R
E
ii
(0I . . . , u
(JI ( u l ) ,...,$ , ( u n ) ) , t h e n $
=
E
11
v" and
i n d u c e s a b i j e c t i v e correspondence between t h e isomorph-
M
(R)-modules.
The i n v e r s e of t h i s correspondence i s
W.
( i ) The correspondence
f
W
---f
w" i s
obviously order-preserving.
It
t h e r e f o r e s u f f i c e s t o show t h a t i t h a s an i n v e r s e which i s a l s o o r d e r - p r e s e r v i n g . Consider t h e p r o j e c t i o n on t h e f i r s t f a c t o r
:
'TT 1
module
X
of
v",
put
order preserving.
v", V
w e see t h a t of
V.
g(X)
=
71
(X).
It is clear that
X
Thus
proving t h e a s s e r t i o n .
V
and f o r any sub-
Then t h e correspondence
(gf)( W ) = W.
can be w r i t t e n i n t h e form
1
v"+
X
X r t g(X)
From t h e a c t i o n of =
v"
is
M (R)
on
f o r a s u i t a b l e submodule
49
MATRIX RINGS AND RELATED RESULTS
S
(ii) Put
=
M (R)
f
and f i x
E End(?).
S
Then, by t h e n a t u r e of a c t i o n of
S
J", f
on
h a s t h e same p r o j e c t i o n s , s a y
$ E End(!/)
d e t e r m i n e s an element of
End(p) S
R $.
a r e equal t o
e
(iii) W e need o n l y show t h a t
R
and
M
f+-+Xf
Thus t h e map
11
?
whose p r o j e c t i o n s on a l l f a c t o r s
(ellWIn
and
W,
V
where
e
By t h e d e f i n i t i o n of
F = V X...O X O~ P
11
Conversely, any
p r o v i d e s t h e d e s i r e d isomorphism.
V
(R)-modules, r e s p e c t i v e l y .
e
on a l l f a c t o r s .
f'
11
and
W
are
'
V
Observe a l s o t h a t
W = e Because
eklw
=
e
0 i f and o n l y i f w
e 11
1
+
11
. . _i
We ...
11
w
enl F/
@J
0, 1 Q k
=
f
e w t-ni n
LJ
(2 11
1
Q
a s R-modules
n , w E W,
,...,e
i s w e l l d e f i n e d and i s a t l e a s t an R-isomorphism of
W
11
t h e map
w 1 n
onto
(ellki)".
I t there-
fore suffices t o verify that
f (eijeklw)
=
e . . f ( e k w1 ZJ
(w E W, i , j , k E {1,2
1
.
The l a t t e r b e i n g a consequence of t h e a c t i o n of follows.
I
Given an i d e a l
7.4.
PROPOSITION.
of
R.
I+
Mn(I) i s a b i j e c t i o n between t h e s e t s of
R i s simple i f and o n l y i f so i s
In particular,
n
(ellW)n, t h e r e s u l t
on
we write
( i ) The map
R and M (R).
i d e a l s of
M,(R)
,...,n } )
M, (R). (ii)
Mn(R)/Mn(I) p M n ( R / I )
(ii) I f
R = I
@J
... @I,
i s a two-sided decomposition of
Mn(R) = M n ( I l ) 0 i s a two-sided decomposition of p o s a b l e i f and o n l y i f so i s Proof.
(il
Mn(R).
R,
then
... 63 Mn(Is)
Furthermore, t h e i d e a l
Ii
i s indecom-
Mn(Ii).
It is clear t h a t
M (I1
i s an i d e a l of
Mn(R)
and t h a t t h e
CHAPTER 1
50
given map is injective.
J be an ideal of M ( R ) and let I C R consist
Let
J.
of all entries of elements in
R
(ii) The natural map
-+
I is an ideal of R such that J = M (I).
Then
induces a surjective homomorphism M ( R )
R/I
Mn(R/T)
--f
whose kernel is M n ( I )
.
Z ( M ( R ) ) = Z(R)
(iii) This is a direct consequence of the fact that
r e R , Mn(R)r
f o r any
=
Mn(Rr).
and that,
n
To present our next result, we require the following preliminary observations. V
V =
Vi
7.5. LEMMA.
Let
ible and let
W be an irreducible submodule of V . i for which W
ranges over those Proof. V .
be an R-module, let
Choose nonzero
W
A
Rw
=
Rvi.
V. =
to
V
into rvi
Since
W,Vi V
completely reducible R-module
Then
V
V = 8 V
Let
+I i'
such that
V j , j E S,
I.
For each
Vi,i
E
58 'Lj
is irreduc-
j
where
for r E R
vi
E
Vi.
If
is a nonzero homomorphism
are irreducible, W
2
Vi
as required..
is said to be homogeneous if it can be The sum of all irreducible V
which are isomorphic to a given irreducible submodule of
called a homogeneous component of 7.6. LEMMA.
W
with
= cVi
written as a sum of isomorphic irreducible modules. submodules of
V.
where each
i'
and write w
w E W
# 0 then the map sending rw
from
2
@
is
V.
where each
Vi
is irreducible and let J
5I
be
are the representatives of all isomorphism classes of
j
J,
let
Xj
denote the sum of all
Vi
with
V.
z
2
V
j-
Then (1)
The X j E J, are all homogeneous components of
V
j'
(ii) V = G3
x
i€J j Proof. (i) Apply Lemma 7.5. (ii) Direct consequence of the definition of Let
V
be any R-module.
invariant in
V
7.7. LEMMA.
Let
V
V.
j*
Then a submodule W
V
of
if it admits all R-endomorphisms of
is said to be fuZZy
V.
V # 0 be a completely reducible R-module.
is fully invariant if and only if
ents of
X
W is
A submodule
W of
a sum of certain homogeneous compon-
MATRIX RINGS AND RELATED RESULTS
Proof.
By Lemma 3.13,
v
51
can be written as in Lemma 7.6.
If
f E End(V)
R
V.
and
V
then either f(V,l
j' Hence
z
ducible.
f(xj) 5 X.,
=
V. z
0 or f(V.)
X
proving that each
j'
is fully invariant.
F/
P+Q
so
Let f
:
P, then Q
Q
=
5 W.
Q
If
P @ Q and therefore V
=
It suffices to show that If
P, this is clear; otherwise P n Q
Q
= P @
U for some submodule U of
@
p--+ Q be the given isomorphism and define an endomorphism $ l J J ( z + y+ a )
COROLLARY.
7.8.
is irre-
W and Q is an irreducible submodule of V
is an irreducible submodule of
such that
j
Vi
is fully invariant.
Conversely, assume that P
since
and hence the sum of
J.'
3
X
certain
V
Let
v
=
+
f ( z ) + f-l(y,
V by
Q,%
E
# 0 be a completely reducible R-module and express V
the sum of its homogeneous components
6X
V =
:
0,
V.
of
(z E P,y E
z
=
U)
as
Then
j € J j' End(V) R Proof. V
of
(f.1
Any family
3
2
of endomorphisms of
and it is clear that the correspondence
morphism
rl End(X.1 3
jEJ
-
X
(f.1 3
j I-+
defines an endomorphism f
f is an injective homo-
End(V) R
is fully invariant (Lemma 7 . 7 ) , any endomorphism f of V maps j into itself. Setting fj = flXj,j E J , it follows that (fi) W f
Since each each
n End(X.) R
j€J
X
j
X
.
as required. 7.9. LEMMA.
Let
V be an R-module and let e be an idempotent of R. Hom(Re,V) R
Similarly, if
V
eV
Then
as additive groups
is a right R-module, then Hom(eR,V)
Ve
R Proof. map V E
If f E Hom(Re,Vl, then ef(e) = f(e2) = f(e) E eV.
R
fF+ f(el is a homomorphism from Hom(Re,V) to eV. R eV
then the mapping g,
:
xe-
33u
Hence the
Conversely, if
is an R-homomorphism from
Re
to
V.
52
CHAPTER 1
Because t h e map
V L-+ gv
f kf ( e ) ,
i s an i n v e r s e of
t h e f i r s t isomorphism
The second isomorphism can b e e s t a b l i s h e d by a similar argument.
follows.
Given a r i n g
R
groups of
R, R"
and
Ro
we denote by
i.e.
t h e opposite r i n g ,
Re
c o i n c i d e and m u l t i p l i c a t i o n i n
the additive
aob
i s g i v e n by
ba
=
a,b E R. 7.10.
LEMMA.
Let
e
be an idempotent of a r i n g
R.
Then
End(eR) EZ eRe
End(Re) 1 ( e R e ) " and
R
R
In particular, End( R )
R R
= R"
End(R ) h' R
V = Re,
Invoking Lemma 7 . 9 f o r
Proof.
and
an isomorphism of t h e a d d i t i v e group of
R
2
f W f(e)
w e see t h a t t h e map
EndtRe)
is
eRe.
o n t o t h e a d d i t i v e group of
R
f,g
Given
E EndfRe),
R
f(e)
write
=
er e
g ( e ) = er e
and
for some
,r E R.
P 1
2
Then
($9)(el = f ( e r e ) = er e r e 2
=
1
(er e ) (er e )
proving t h a t t h e given map r e v e r s e s t h e m u l t i p l i c a t i o n . t i t y element of t h e r i n g
eRe,
= g(e)f(e)
Since
e
i s t h e iden-
t h e above map p r e s e r v e s i d e n t i t y e l e m e n t s .
This
e s t a b l i s h e s t h e f i r s t isomorphism and t h e second f o l l o w s by a s i m i l a r argument.
7.11.
LEMMA.
( S c h u r ' s Lemma).
is a division ring Proof.
z e r o submodule of and
Kerf
7.12.
V
=
LEMMA.
R/X
V
b e a nonzero R-homomorphism.
and K e r f
# V
i s a p r o p e r submodule o f
0, p r o v i n g t h a t f i s an isomorphism. Let
V
be an R-module.
Clearly
R/X
Conversely, assume t h a t
RV
V
f o r some maximal l e f t i d e a l
Proof.
Then
V
Then
X
Then
f(V)
V.
End(V)
R
V
R-homomorphism.
Hence
R/X
V = RV. 2
=
v
a
i s i r r e d u c i b l e i f and o n l y i f
R.
of
and s o
f(V)
So t h e lemma i s t r u e .
is i r r e d u c i b l e and choose a nonzero
V
i s a non-
Thus
X
i s i r r e d u c i b l e f o r any maximal l e f t i d e a l
i s a nonzero submodule of
surjective
be an i r r e d u c i b l e module.
.
$ : V+
Let
V
Let
V
The map
R--t
f o r some l e f t i d e a l
v
of
V.
in
R. Then
V,r
++
rV i s a
X
of
R.
53
MATRIX RINGS AND RELATED RESULTS
Since
is irreducible, X
V
is maximal, as required
We have now accumulated all the information necessary to prove the following result. 7.13. PROPOSITION.
fl
Assume that
artinian and there exist primitive idempotents e ,e 1
,...,n,
integers n
Then R
is completely reducible. 2
,...,e
of
is
R and positive
such that are all nonisomorphic irreducible R-modules
(i) Rel,Re2,...,Rer
nMni P
(ii) R
(eiRei) and each
eiRei
is a division ring
i=l
Proof.
#,
1 lies in the sum of finitely many irreducible submodules of
Since
Hence R
the same is true for RR.
is artinian and so by Corollary 3.11,
there exist primitive orthogonal idempotents el,e2, ...,en ,@ = Rel @
... @ Ren
and each
Rei
of
R such that
is indecomposable (and hence irreducible).
.
We may assume that Re ,Re2,.. ,Rer
are all nonisomorphic among the Rei,1 Q i G n .
Then, by Lemmas 7.12 and 7.6, Re ,Re ,...,Rer 1
are all nonisomorphic irreducible
2
R-modules. (ii) Let
Xi
be the homogeneous component of
Then, by Lemma 7.6, RR = X @ integer n
i' 1
G
i
4
r.
... @ X,
fl
and X.
corresponding to Re
i'
(Re.)
"i
lGiQr.
for some positive
Invoking Lemma 7.10, Corollary 7.8 and Proposition 7.2,
we derive
and therefore
The fact that e.Rei is a division ring being a consequence of Lemmas 7.10 and 7.11, the result follows. The Jacobson rudical maximal left ideals of Let by
V
J(R)
R.
be an R-module.
of
R is defined to be the intersection of a11
The ring
R is said to be semisimpZe if J(R)
The annihilator of
V,
written
= 0.
ann(V), is defined
54
CHAPTER 1
{r E R ( r V
ann(V) = I t is clear that
module.
I
An i d e a l
of
i r r e d u c i b l e module.
is faithfuZ i f
V
W e say t h a t
V may be viewed a s an R/ann(V)-
i s an i d e a l and t h a t
ann(V)
01
=
ann(V) = 0.
i s s a i d t o be primitive i f t h e r i n g
R
has a f a i t h f u l
I is
i s p r i m i t i v e i f and o n l y i f
I
Thus
R/I
the annihilator
of an i r r e d u c i b l e R-module. 7.14.
Every maximal l e f t i d e a l of
LEMMA.
R
c o n t a i n s a p r i m i t i v e i d e a l and e v e r y
p r i m i t i v e i d e a l i s t h e i n t e r s e c t i o n of t h e maximal l e f t i d e a l s c o n t a i n i n g it. Proof.
X
Let
d e n o t e a maximal l e f t i d e a l of
an i r r e d u c i b l e R-module.
whose a n n i h i l a t o r i s
Mv
Clearly
Mv
M
0 # v E V,
J(R)
5I
J(R)
R/X
is a primitive ideal
R/X
is
R.
R
in
V
b e an i r r e d u c i b l e R-module
V,
put
MV
is i r r e d u c i b l e ,
.
I
Since
=
{r
E
Rlrv
V = RV
=
i s t h e i n t e r s e c t i o n of a l l
be a r i n g .
i s an i d e a l of
I
In
R.
i s t h e a n n i h i l a t o r of a c o m p l e t e l y r e d u c i b l e R-module. if
01.
R/Mv
i s t h e i n t e r s e c t i o n of t h e a n n i h i l a t o r s of i r r e d u c i b l e R-modules.
particular, J ( R ) (ii)
Since
the r e s u l t follows. Let
v
Given a nonzero
R.
V
and l e t
i s a maximal l e f t i d e a l i n
PROPOSITION.
(i) J(R1
I.
is a l e f t ideal i n
V
with
7.15.
R
I be a p r i m i t i v e i d e a l i n
Let
By Lemma 7.12,
X.
containing
and so
Hence t h e a n n i h i l a t o r of
R.
Furthermore,
i s t h e a n n i h i l a t o r of a completely r e d u c i b l e R-module.
( i ) D i r e c t consequence of Lemmas 7.12 and 7.14.
Proof.
Vi,
= n ann(Vi’i).
S e t t i n g {Viji E I) t o b e t h e i€I s e t of r e p r e s e n t a t i v e s of t h e isomorphism classes of i r r e d u c i b l e R-modules, i t (ii) I f
V = 69
then
ann(V1
iEI
follows t h a t
J ( R ) = ann(V).
The second s t a t e m e n t i s a consequence of t h e f a c t
.
t h a t any completelv r e d u c i b l e R-module i s e i t h e r of i r r e d u c i b l e submodules.
7.16.
PROPOSITION.
Then
f(JUi)l
5J ( S )
Let
f
:
R
4
S
with e q u a l i t y i f
0
o r a d i r e c t sum of a f a m i l y
be a s u r j e c t i v e homomorphism of r i n g s . Kerf
J(R).
55
MATRIX RINGS AND RELATED RESULTS
Proof. taining
{Mil
X Thus
hi = 1
+
Xwh-l,
,
xpx
f # 0 since =
0 f o r all fi-'(b)
be such t h a t
uX-
hu and s a t i s f i e s
v
commutes w i t h
Xu
=
X $2 F
A
1, g F
x
# 0
E
but
f ( c ) = 0, and
uX-Xu
f o r some
q
=
pn.
But t h e n
E
F.
Z(D1 = F .
and
D.
- Au)v-l = u v - 5 , - xuv -1 = w x - XW
wq E F
Xp
such t h a t
we d e r i v e
= (UX
but
Then
= 3:-hP -
c = Hence, p u t t i n g
purely inseparable, i . e .
We may t h e r e f o r e choose
= p(x).
f be t h e mapping
D is
of
3:
Choose
b
E
fi(b)
=
i.e.
c h = Xc.
0.
D
Then
THE BRALIER GROUP
q,
1 = 0,
and t h e r e f o r e
=
1
+ (hwX-l)q
=
1
+
E/F
if
But
CD(E) # E ,
then
D E
T h i s would imply t h a t Thus
2.9.
COROLLARY.
Let
2.10.
D
contains a proper
D of maximal d e g r e e .
F
D.
in
By Theorem l . l 5 ( i ) ,
i s separable over
i.e.
D
E
F,
D
such t h a t
i s simple Therefore,
by t h e f i r s t p a r t .
D.
i s a maximal s u b f i e l d of
in
cD (E)
is a
c o n t r a r y t o t h e maximality of
be a c e n t r a l d i v i s i o n F-algebra.
E/F
E
Then
CD(E).
E',
has a proper separable extension
separable f i e l d extension Proof.
E F)
U'
i s a d i v i s i o n a l g e b r a , hence so i s
E'
C (E) = E, D
E.
Awqh-l
Consequently,
be a s e p a r a b l e e x t e n s i o n i n
E.
+
F.
maximal s e p a r a b l e e x t e n s i o n of with c e n t r e
(since
W'
a contradiction.
s e p a r a b l e f i e l d e x t e n s i o n of Let
1
=
167
E
splits
Then t h e r e i s a
D.
Apply P r o p o s i t i o n 2 . 8 and C o r o l l a r y 1.19.
f i n i t e G a l o i s e x t e n s i o n of Proof.
F
Every Brauer c l a s s of
COROLLARY.
h a s a s p l i t t i n g f i e l d which i s a
F.
D be a d i v i s i o n a l g e b r a i n a g i v e n Brauer c l a s s of
Let
D
P r o p o s i t i o n 2.8, r a b l e extension.
c o n t a i n s a maximal s u b f i e l d
E
Then
E
such t h a t
E/F
F.
By
i s a sepa-
i s c o n t a i n e d i n a f i n i t e G a l o i s e x t e n s i o n of
F, and
D.
this w i l l also s p l i t
W e c l o s e t h i s s e c t i o n by p r o v i d i n g c i r c u m s t a n c e s under which t h e t e n s o r
This w i l l
p r o d u c t of two c e n t r a l d i v i s i o n a l g e b r a s i s a g a i n a d i v i s i o n a l g e b r a be achieved w i t h t h e a i d o f t h e f o l l o w i n g g e n e r a l r e s u l t .
2.11. THEOREM.
D
Let
and 1
t h e i r centres such t h a t
Do
F,E
II
be d i v i s i o n a l g e b r a s of d e g r e e s
E
r e s p e c t i v e l y , where
i s embedded i n
1
Do, we s e e t h a t such
q
P,S
over
2
M (D ) 4
>_
F.
Denote by
q d r2).
exists, in fact
Then
1
M ( D )= D o @ D 4
the l e a s t integer
( t a k i n g a r e g u l a r m a t r i x r e p r e s e n t a t i o n of
2
f o r some c e n t r a l d i v i s i o n E-algebra
q
' F
2 D
of d e g r e e 3
rt
=
sq
t,
and
CHAPTER 3
168
D1
Moreover, Proof.
:
t over E.
D 2 h a s index
Since
Do C M ( D ) 1 - 4 2
it f o l l o w s from P r o p o s i t i o n 1 . 1 2 t h a t
M ( D )g D " 8 D 4 2 ' F 3 where
U
Do
i s t h e c e n t r a l i z e r of
3
E,
with c e n t r e
4
8D
h a s index
t.
D3
Write
E
t d e n o t e t h e d e g r e e of
q 2 S 2 = t 2 r 2 , proving t h a t
We a r e t h e r e f o r e l e f t t o v e r i f y t h a t U
D 3 i s simple
The a l g e b r a
2
Let
i n (3) we find
Comparing F-dimensions
M (D 1.
in
by P r o p o s i t i o n 1.10.
(3)
.
D
rt
= Sq.
i s a d i v i s i o n a l g e b r a and t h a t
D3
Mk(D),
D
where
i s a s k e w f i e l d component of
' F D 3
.
Then
M (D )
Do 8 D ' F
4 2
>
R
(4)
' F
' F
Dr 8 D
Denote t h e s k e w f i e l d component of some
Do 8 M k ( D ) p M ( D o @ D)
.s
by
Db.
Then
D
Df
ML(Dlt)
for
F
1
M (D )
and s o
9 2
Mka(D
P
4 in
S u b s t i t u t i n g t h i s value f o r
D
By u n i q u e n e s s ,
).
(41, w e f i n d
Do 8 D
2 2
D
4
4 = kR.
and
Ma(D-).
I t follows
1 F
Do
that
c Mk(Dz)
1 -
ensures t h a t
k
Finally,
D
=
and s o , by t h e m i n i m a l i t y of 1 and t h e r e f o r e
z M (D ) d 5
€3 D
' F
M
D
E D
q, we have
d 2 1 and some s k e w f i e l d
f o r some
Mr2(F) 8
This
i s a skewfield.
( D ) p D 8 M ( D ) E D 8 5 I F 4 2 1 F
dq
q 4 R.
D5-
Hence
D; 8 D3 F
o3
F 2 M,2
dq
Thus
=
r 2 and
D
5
D
(D3)
h a s degree
t.
It follows t h a t
3
D1
:
D 2 h a s index
t, a s r e q u i r e d . From t h e proof above, w e deduce
2.12.
where
COROLLARY.
Let
D 1 , D 2 , F , E , q , r be a s
D i s a d i v i s i o n a l g e b r a and dq
d i v i s i o n a l g e b r a i f and o n l y i f
q
=
r2.
=
r2.
COROLLARY.
Then
Dl
Let
U ,D 1
2
Then
I n particular,
'
A s a f u r t h e r consequence of Theorem 2 . 1 1 ,
2.13.
i n Theorem 2.11.
w e prove
be c e n t r a l d i v i s i o n F - a l g e b r a s of coprime d e g r e e s .
D 2 i s a g a i n a c e n t r a l d i v i s i o n F-algebra.
169
CLASSICAL CROSSED PRODUCTS AND THE BRAUER GROUP
Proof.
2
rt
t, where q
=
-
1.
r e s p e c t i v e l y and choose
=
rst
rst
=
qs2
Then
=
r 2 p ; but
m
and so
D1
;
D2
minimal
D2
has index
( r , s ) = 1,
t
r2s2m, i . e .
=
tlrs.
D 2 , so by t h e d e f i n i t i o n of index,
D1
p,q
D1
Owing t o Theorem 2 . 1 1 ,
q
Thus
r 2 m f o r some rn
rs
M (D2).
rp.
= sq, s t =
t h e i n d e x of = 1,t =
C
1 -
1
p
r,s
be of d e g r e e s
D o c M (D ) , D o
subject t o
i.e.
D,,D2
Let
so
rsm.
Now
r2/q,
t is
I t follows t h a t
has index equal t o i t s degree.
D1
Hence
;
D 2 is
a d i v i s i o n F - a l g e b r a which i s a l s o c e n t r a l , by P r o p o s i t i o n 1.10.
3. CLASSICAL CROSSED PRODUCTS AND THE BRAUER GROUP
E/F
Throughout t h i s s e c t i o n ,
Br(E/F)
W e d e n o t e by
is a f i n i t e G a l o i s e x t e n s i o n and
G = Gal(E/F).
t h e k e r n e l of t h e homomorphism
Br(F) 1/11
-
Br(B)
ct 1.4 El El
F Hence
BrIE/F)
s p l i t by of
G
E.
on
E*.
Given G
GI
c1 E
over
Z2(G,E*), E.
c o n s i s t i n g of t h o s e
ZZ(G,E*)
Consider t h e group
crossed product of E
Br(F)
i s t h e subgroup of
EaG
which a r e
with respect t o the n a t u r a l action
w e d e n o t e by
Thus
LA1
EaG
t h e corresponding
i s a f r e e l e f t E-module w i t h b a s i s
such t h a t
-
-1 =
and
=
Ct(x,y)xy
where
gh By Theorem 2.2.3,
n a t u r a l a c t i o n of
any c r o s s e d p r o d u c t of
G
on
I n what f o l l o w s we w r i t e
F)
-
CL
=
g(X)
G over
is equivalent t o
E%
E
for a suitable
f o r t h e cohomology c l a s s of
CI.
F o r any
c1 E
Z2(G,E*),
a
E
Z2(G,E*).
O u r main r e s u l t of
i s an isomorphism. LEMMA.
E E,g E G
(with r e s p e c t t o t h e given
t h i s s e c t i o n a s s e r t s t h a t t h e map
3.1.
X
for a l l
t h e f o l l o w i n g p r o p e r t i e s hold:
CHAPTER 3
170
(E)=
C
i s a c e n t r a l simple F-algebra and
( i ) E"C
E
E"G
E
(ii)
dimfG = \ G I 2 F ( i ) T h i s i s a p a r t i c u l a r c a s e o f C o r o l l a r y 2.4.3.
EaG
i s a s p l i t t i n g f i e l d f o r t h e F-algebra
Proof.
and
The f i r s t a s s e r t i o n f o l l o w s from ( i )and C o r o l l a r y 1 . 1 6 ( i ) .
(ii)
d i d = F
second, n o t e t h a t by P r o p o s i t i o n 1 . 4 . 1 ,
IGI
.
To prove t h e
Hence
dimEaG = (GIdimE = / G I 2 F F a s required. 3.2.
0
For a l l
LEMMA.
a
E G
(i)
6a G
Proof.
So
(1)
as F - a l g e b r a s
8
and
(iii)
the following conditions a r e equivalent:
E'G
i s equivalent t o
( i i ) EaG
Since
Z2(G,E*),
cl,a E
a r e cohomologous.
The e q u i v a l e n c e of
( i l and ( i i r l f o l l o w s from C o r o l l a r y 2.2.4.
o b v i o u s l y i m p l i e s ( i l l , w e a r e l e f t t o v e r i f y t h a t (ii) i m p l i e s ( i ) .
$ : ERG
assume t h a t
-t
ERG i s an isomorphism of F - a l g e b r a s .
i?'
Lemma 3 . 1 and Theorem 1 . 1 3 , t h e isomorphism
E B G.
i n n e r automorphism of
$(i= )
T-
kx
f o r all
y E E DG,
for a l l 3.3.
1
(ii)
E"G
(iiil
N
E.
Hence, if
8
E
for a l l
rr. E Z 2 ( G , E * ) ,
M (PI, where
( i ) G'B
4 (E)
Thus t h e r e e x i s t s a u n i t
( 8 4 ) (1)= k
then
For any
LEMMA.
f E
+
a Aut(E G)
E.
.Q. E
Owing t o
c a n be extended t o an
B
E G
of
2
such t h a t
i s d e f i n e d by
N o w a p p l y Lemma 2.2.6.
t h e following conditions a r e equivalent:
n = /GI
i s a skew group r i n g of
G
E
over
i s a coboundary
Proof.
By C o r o l l a r y 2.2.5,
( i i ) and (iii)a r e e q u i v a l e n t .
Invoking Lemma
3 . 2 , we a r e l e f t t o v e r i f y t h a t ( i i ) i m p l i e s ( i ) . Assume t h a t
E"G
i s a skew group r i n g of
$(x) E End(E) be d e f i n e d by F
@(x)( y )
=
$ : G by
b(g)
=
g.
xy -+
G
over
for a l l
E.
y E E.
Given
z E E,
let
Define
Aut(E) F
Then c o n d i t i o n s (1) and ( 2 ) of P r o p o s i t i o n 2.3.1
are s a t i s f i e d
171
CLASSICAL CROSSED PRODUCTS AND THE BRUAER GROUP
and so the map
is a homomorphism of F-algebras.
End(E1
EiiG
Since
is simple and both
E0"G
and
are of the same P-dimension, we conclude that
F EaC
End(E)
P
M,(F)
F as required.
ExarnpZe.
Then
G
9
Let
=
E
{l,g}
=
Q ( < ), i 2=-1,
and let g E Aut(E) be defined by
is a group of automorphisms of
G
Thus E / a ) is a Galois extension and
F whose fixed field is 0.
Gal(E/'Q)
=
Let
a
:
G x G
--f
E*
be
defined by
a ( l , l ) = Ci(g,l) = Ci(l,g) = 1, Cx(g,g) = -1 Then obviously where
1
c1 E
Z2(G,E*1 and EaG
is a free E-module with basis
E"C
is the identity element of
{i,;},
and
-1
jxz By identifying
o@(& with
E
and
=
g(x:) ,
j'
j
with
i,
=
-1
(x E E)
we see that
EaG
is isomorphic
to the quaternion algebra
over Q.
On the other hand, if
is a coboundary, then
by virtue of Lemma 3.3.
3.4. LEMMA.
Let A
be a (finite-dimensional) central simple F-algebra such that
E is a self-centralizing subfield of A. Proof.
For each g E G,
subalgebras of A.
-
the map
g
-
Then A :
E
F
2
ERG
for some
c1
E Z2(G,E*).
is an F-isomorphism of simple
Hence, by the Skolem-Noether theorem (Theorem 1 . 1 3 ) , there
exists g E U(A) such that
CHAPTER 3
172
-1 ;A;
=
d i d = (did)
By C o r o l l a r y 1 . 1 7 , we have
*
g(X)
A E E,g E G
for a l l
(1)
I n o r d e r t o prove t h a t
= ]GI2.
F
F
BE;
A =
(2)
SfG it s u f f i c e s t o show t h a t t h e r i g h t hand e x p r e s s i o n i s a d i r e c t sum, f o r t h e n b o t h
’.
s i d e s w i l l have F-dimension
]GI
I f t h e sum i s n o t d i r e c t , l e t
-
s = A g
i
1 1
k.
be a r e l a t i o n w i t h minimal Choosing
E*
A
such t h a t
... f h k g k
I t is clear t h a t
g (A) # g ( A ) ,
(Ai E E*)
= 0
k > 1 s i n c e each
-
g E U(A.1.
one e a s i l y v e r i f i e s t h a t
g1(Xls-Sh = 0
-
g i v e s a s h o r t e r n o n t r i v i a l r e l a t i o n connecting
-
This i s a contra-
g2, ...,gk.
d i c t i o n and t h u s ( 2 ) i s e s t a b l i s h e d .
-1
x,y
Suppose now t h a t
E
G.
h E E.
commutes with each
Since
E
A
Let
LEMMA.
A
D and some n 2 1.
v e c t o r space over
LJ
on which
an i n n e r automorphism of
where
I,
A.
A,
is the i d e n t i t y
A
A
P
M (Dl n
c1 : c1
G
X
G
we o b t a i n
----f
E Z2(G,t’*).
E*
and
in
Br(F).
[A1
=
[El
is a
B
c e n t r a l simple F-algebra,
[ A ] = [eAel
Then
B y Wedderburn’s theorem,
s i o n algebra
f o r some
be a ( f i n i t e - d i m e n s i o n a l )
e be a nonzero idempotent of Proof.
implies t h a t
EG ‘
A
cocycle and t h u s we have shown t h a t
A,
i s i t s own c e n t r a l i z e r i n
The a s s o c i a t i v i t y of m u l t i p l i c a t i o n i n
3.5.
x y q
Then t h e e q u a t i o n (1) shows t h a t
and l e t
f o r some d i v i -
By changing t h e b a s i s of an n-dimensional a c t s i r r e d u c i b l y and f a i t h f u l l y , i . e .
w e may assume t h a t
rxr-matrix.
Then
applying
CLASSICAL CROSSED PRODUCTS AND THE BRAUER GROUP
eAe
and so
M,(D).
[eAel = [ D l = [ A ]
Thus
a s required.
173
.
The n e x t p r o p e r t y i s c r u c i a l f o r t h e proof of t h e main r e s u l t . t h e r e i s a s i m i l a r i t y of F - a l g e b r a s .
- Ea’G
E ~ G E’G F Proof.
Let
C
=
A €9 B ,
where
F
A = E% Then
‘L
=
B
and
8 E; SfG
=
E’G
=
i s a c e n t r a l simple F-algebra by Lemma 3 . 1 and P r o p o s i t i o n 1.10.
e
5
E E @ E
eCe
C such t h a t F y i e l d t h e d e s i r e d r e s u l t by a p p e a l i n g t o Lemma 3 . 5 . aim i s t o f i n d an idempotent
W e f i r s t note t h a t t h e s u b f i e l d s
mentwise.
Since
E
E 63 1 and
i s s e p a r a b l e over
F,
E = F(A) ( P r o p o s i t i o n s 1 . 4 . 3 and 1.4.6). nomial of
8 E :
SfG
A.
degf(X) = n ,
Then
n
E @ E
F E E
there exists Let
where
of
1 8 E
F aaG.
Our
This w i l l
commute e l e such t h a t
f C X ) E F [ X ] be t h e minimal poly-
= dimE.
Now d e f i n e
F
g E G - {l}.
where t h e s e p r o d u c t s a r e t a k e n o v e r a l l
# g(x)
a t e r i s not zero since d i s t i n c t from z e r o i n
E
@
E,
Observe t h a t t h e denomin-
g E G - {l}.
f o r each
s i n c e t h e elements
The numerator i s a l s o
{ A z @ 110 4
i
n - 11
are
F l i n e a r l y independent o v e r
1 @ E.
Hence
e # 0 in
E @ E.
F Next we observe t h a t
where
g
ranges over
which shows t h a t
G - {l}.
Consequently, w e must have
(1 8 Ale = (1@ l ) e
in
E @ E.
Thus, by i n d u c t i o n ,
F (1’ Q 1 ) e = (1
Az)e
for all
n-1 Because
E = @ FAi
i=o
and m u l t i p l i c a t i o n i n
E@ E F
i s commutative, we d e r i v e
i >0
174
CHAPTER 3
Thus
proving t h a t
e
E 8 E.
i s an idempotent i n
We are l e f t t o v e r i f y t h a t
eCe
BG.
eCe =
c
To t h i s end, n o t e t h a t
e ( E Q E) (Lt. 8 i ) e
x,$G =
c
e(E Q l ) e - e ( l Q ~ ~ e - eQ( y’)e x
XI YEC
But
e ( l @ E)e = e ( E @ l ) e by ( 3 1 , and
t o E.
W e now compute
e ( E Q l)e = E ‘
i s a f i e l d F-isomorphic
e ( ; 8 g ) e , by u s i n g t h e formulas
Then
where
z
ranges over a l l e l e m e n t s of
On t h e o t h e r hand, when
x
=
y
A s i m i l a r argument proves t h a t
The foregoing shows t h a t
where
G - {I}.
we o b t a i n
From ( 3 ) , we o b t a i n
175
CLASSICAL CROSSED PRODUCTS AND THE BRAUER GROUP
E'
=
e ( E @ 1 ) e and
3
=
(g E GI
e ( i @ $)e
and a l s o shows t h a t
Clearly
G and, f o r
Gal(E'/F)
3
Hence, c o n j u g a t i o n by
8's
Thus t h e
all
g
a c t s as
!J E
E,g
on
E'.
E G,
Furthermore, f o r a l l
multiply according t o t h e cocycle
eCe
9
E
x,y
E G
T h i s shows t h a t
C$.
aBG
and t h e r e s u l t f o l l o w s . We have now come t o t h e d e m o n s t r a t i o n f o r which t h i s s e c t i o n h a s been developed. Let
E/F
be a f i n i t e G a l o i s e x t e n s i o n , l e t
Z2(G,E*),
let
E'G
3.7.
THEOREM.
any
a
E
t o t h e n a t u r a l a c t i o n of
be t h e c r o s s e d p r o d u c t of
G on
E.
G = Gal(E/F) and f o r
G over E with r e s p e c t
Then t h e map
i s an isomorphism. Proof.
Consider t h e map
Then, by Lemma 3 . 6 ,
f
i s a homomorphism. Kerf
and t h u s
?(a)
f
= [E'"G].
=
Furthermore, by Lemma 3 . 3 ,
B' ( G , E * )
i n d u c e s an i n j e c t i v e homomorphism Now assume t h a t
[A] E Br(E/F).
7
:
H2(G,E*)
--+
BF(E/F) g i v e n by
Then, by P r o p o s i t i o n 2 . 7 ,
there
CHAPTER 3
176
exists
B
E
[A]
B
such t h a t
E
contains
Invoking Lemma 3.4, w e deduce t h a t
B
E"G
as a self-centralizing subfield. f o r some
c(
€
Z2(G,E*).
Thus
and t h e r e s u l t f o l l o w s . 3.8. COROLLARY.
A
Let
be a ( f i n i t e - d i m e n s i o n a l ) c e n t r a l simple F-algebra.
Then t h e r e e x i s t s a f i n i t e G a l o i s e x t e n s i o n
G
= Gal(E/F),
Proof.
such t h a t
and
a E Z2( G , E * ) ,
i s s i m i l a r t o t h e crossed product
By C o r o l l a r y 2.10,
[A] E B r ( E / F ) .
that
A
E/F
where
b"G.
t h e r e e x i s t s a f i n i t e G a l o i s extension
E/F
such
Now a p p l y Theorem 3.7.
The above r e s u l t t e l l s u s t h a t e v e r y Brauer c l a s s c o n t a i n s a c r o s s e d p r o d u c t . I n p a r t i c u l a r , a c e n t r a l d i v i s i o n a l g e b r a i s always similar t o a c r o s s e d p r o d u c t . A n a t u r a l q u e s t i o n i s t o ask whether a c e n t r a l d i v i s i o n a l g e b r a i s always a
crossed product.
T h i s q u e s t i o n was i n v e s t i g a t e d i n d e t a i l i n t h e 1930s
(by
Hasse, A l b e r t and o t h e r s ) and i t was d i s c o v e r e d t h a t e v e r y c e n t r a l d i v i s i o n algebra i s a crossed product.
e-
However, i t was o n l y i n 1 9 7 2 t h a t Amitsur gave
examples of c e n t r a l d i v i s i o n a l g e b r a s t h a t a r e n o t c r o s s e d p r o d u c t s . To examine t h e Brauer group i n more d e t a i l , w e need t h e f o l l o w i n g s i m p l e observation 3.9.
LEMMA.
[A] E R r ( P ) Proof. !e may w r i t e
The Brauer group h a s index
P,
Br(F)
then
[Alp
i s a t o r s i o n group. = 1.
Keeping t h e n o t a t i o n of C o r o l l a r y 3.8,
A = I
+)
... @ In
More p r e c i s e l y i f
where each
w e may assume t h a t
A
=
PG.
li i s a minimal l e f t i d e a l of
A.
Then, by C o r o l l a r y 1 . 1 7 and Lemma 3 . 1 ( i ) ,
r
= indA = (degA)/n = ( d i d ) / E = (ndiml
E Invoking P r o p o s i t i o n 2.5.6, Theorem 3.7,
[A]'
[A].
as a s s e r t e d .
[A] E Br ( F )
The o r d e r of simple a l g e b r a
= 1
A,
.
w e deduce t h a t
)/n
= dim1
E l
E l
a'
i s a coboundary.
i s c a l l e d t h e exponent of
[A].
Hence, by
For any c e n t r a l
i t s exponent i s d e f i n e d t o be t h e exponent of i t s Brauer c l a s s
Thus Lemma 3.9 may be e x p r e s s e d by s a y i n g t h a t f o r any Brauer c l a s s , t h e
CLASSICAL CROSSED PRODUCTS AND THE BRAUER GROUP
177
Although i t i s n o t t r u e i n g e n e r a l t h a t t h e exponent
exponent d i v i d e s t h e index.
i s e q u a l t o t h e i n d e x , w e do have t h e f o l l o w i n g g e n e r a l r e s u l t . 3.10.
[ A ] E B r ( P ) , t h e i n d e x and exponent of
For any
THEOREM.
[A1
have t h e
same prime f a c t o r s . Proof.
LA]
Let
r
have index
p, p / r
I t t h e r e f o r e s u f f i c e s t o v e r i f y t h a t , f o r any prime
t h e n o t a t i o n o f C o r o l l a r y 3.8,
G and
p-subgroup of
let
Then
3.9.
In
S b e a Sylow
Let
S.
b e t h e f i x e d f i e l d of
p(n.
implies
A = E'G.
w e may assume t h a t
Eo
n ( r , by Lemma
n so t h a t
and exponent
dimE
=
(G:S)= W
F o i s prime t o
A @ E,. F
p,
while
pk
IS] =
k
f o r some
1.
d e n o t e t h e index of
1J
Let
Then, by P r o p o s i t i o n 2.4,
lilrlliv Furthermore, s i n c e
E is a splitting field f o r A,
we have
(A@Eo) @ E = A @ E - E
F d i d = p
Taking i n t o account t h a t
Thus
If
11
= 1,
then
PIV,
p.
is a p o s i t i v e power of
p
,
F
w e deduce t h a t
plp
k
.
Hence
I-r
is a
EO
p.
power of
F k
'A C3 Zo F and t h e r e f o r e t h e exponent of
A
but
PIP
and
p k
W,
a contradiction.
Now by Lemma 3.9,
(A @ Eo)p F
i s d i v i s i b l e by
-1 p,
a s we wished t o show.
We now d e r i v e t w o consequences of i n t e r e s t . 3.11.
where t h e
where
Di
pi
D
be a c e n t r a l d i v i s i o n F-algebra of d e g r e e
a r e d i s t i n c t primes.
Then
i s a c e n t r a l d i v i s i o n F-algebra of d e g r e e
Proof.
where
Let
COROLLARY.
0 < mi
By Theorem 3.10 and Lemma 3 . 9 ,
Q
n
i'
1
< i < s.
the class
p
i '
[D]
h a s exponent
By t h e b a s i s theorem for a b e l i a n g r o u p s ,
[DI
9
CHAPTER 3
178
[Dl
can be written as a product of classes which are powers of Yower exponent.
mi exponent pi
,
Di be a division algebra similar to a power of
Let
1
with prime
Q
s.
with
Then
D
- D 1 ‘8 D2 63 ... 8 Ds F
Owing to Theorem 3.10, the
[Dl
F
U . have coprime degrees.
Hence, by Corollary 2.13,
we have a division algebra on the right, and therefore the two sides are isomorphic. A central simple F-algebra is called primary if it is # F
and contains no
proper central simple F-subalgebra. 3.12. COROLLARY.
(i) Every central simple F-algebra is a tensor product of
primary ones (ii) A primary P-algebra is either a division algebra of prime power degree or of the form M (F), where
P
Proof.
p
is a prime.
(i) Direct consequence of Proposition 1.12.
(ii) Apply Corollary 3.11.
= G over E
An important special case of crossed products of E/F
is a cyclic extension, i.e. the group
G
=
Gal(E/F)
is that where
is cyclic.
This is
due to the fact that the formulae become simpler, and secondly this case is important in applications.
cyclic o f order
For the rest of this section we assume that
generated by
n
g.
We say that an F-algebra A Fix
E
F*
and define
ax
G is
is cyczic if A Z2(G,E*)
E
EuG
for some
N E.
Z2(G,E*).
by
-
if
i +j9n
if
i + j > n
a
Z2(G,E*) is cohomologous to
1
(04 i,j G n - 1)
X We know, from Proposition 1.6.5, that each for a suitable
A
E
F.
E
In what follows, we write
h
E G
for E
G.
T h u s , from
what we have said above, it follows that each cyclic algebra is isomorphic to
E h i; for some h E E’. E-module with basis
Moreover, by the definition of {;‘I0
i
n
-
11
such that
BA G,EX G is a free left
CLASSICAL CROSSED PRODUCTS AND THE BRAUER GROUP
C o n v e r s e l y , t h e above d e s c r i p t i o n d e t e r m i n e s a u n i q u e asso-
gll = g ( u ) .
where
179
c i a t i v e F-algebra.
u
F o r any
NEIF(p)
It is clear that
3.13.
THEOREM.
h
( i i ) E’G
E G
(iii)
(iv)
-
F
E
and
p
of
i s d e f i n e d by
NEIF(E*) i s a subgroup o f
and t h a t
F*.
F*.
belong t o
1-I
F
over
h h E NEIF(E*)
E~~J.G
M (F)
%
NE,F(p)
i f and o n l y i f
Q EPG
h F
h
Let
E’G
(i) E C
t h e norm
E E,
i s a norm
i f and o n l y i f
The map
i s an isomorphism. Proof.
h
F G
( i ) By Lemma 3 . 2 ,
By P r o p o s i t i o n 1 . 6 . 5 ,
homologous.
E’G
i f and o n l y i f
the l a t t e r is equivalent t o
( i i ) T h i s i s a d i r e c t consequence of t h e f a c t t h a t
c1
X1-I
a
and
CLA
= clrclu
1-I
a r e co-
h/v and Lemma 3.6.
( i i i ) Apply ( i ) and Lemma 3.3.
(iv)
Apply P r o p o s i t i o n 1.6.5 and Theorem 3.7.
3.14.
THEOREM.
Let
( i ) The exponent o f
A
=
x
E G
f o r some
[A] E B r ( F )
[A]
(ii) I f t h e exponent o f
h E P.
t
i s the least positive integer
n
is equal t o
= dim??,
A
then
such t h a t
is a division
F algebra. Proof.
[AIt =
1
( i ) W e have
kt
i f and o n l y i f
( i i ) Since
n
= dirnE,
in
Br(E).
Thus, by Theorem 3.13,
i s a norm.
w e have dimA = n 2 .
A
If
M,(D),
where
D
in
Br(F),
is a
F
F d i v i s i o n a l g e b r a of i n d e x v i r t u e of Lemma 3.9.
[A]* = [EAt G]
m, t h e n
n
=
mr.
Hence t h e exponent of
But
[A]
[Aim
= 1
divides
m.
by
Hence, i f t h e
exponent of
A
is equal to
algebra, as asserted.
n,
then
m = n
and
r = 1.
Thus
A
is a division
181
4
Clifford theory for graded algebras
A
Let
b e a G-graded a l g e b r a .
An i m p o r t a n t method f o r c o n s t r u c t i n g i r r e d u c i b l e
A-modules c o n s i s t s i n t h e a p p l i c a t i o n ( p e r h a p s r e p e a t e d ) of t h r e e b a s i s o p e r a t i o n s : 1.
Restriction t o
A
2.
E x t e n s i o n from
A
3.
I n d u c t i o n from
A
1
1 1
T h i s i s t h e c o n t e n t of t h e s o - c a l l e d C l i f f o r d t h e o r y f o r g r a d e d a l g e b r a s o r i g i n a l l y developed by C l i f f o r d f o r t h e c a s e
A
= FN,
A = FG
where
FG
i s regarded
1
as a G//N-graded a l g e b r a o v e r a f i e l d
F
whose gN-component i s g i v e n by
(FG)gN =
Q
Fs
sfsm
The a i m of t h i s c h a p t e r i s t o p r e s e n t t h e g e n e r a l t h e o r y developed by Dade.
A
number of a p p l i c a t i o n s i s a l s o provided.
1. GRADED MODULES
Throughout t h i s s e c t i o n ,
A
d e n o t e s a G-graded a l g e b r a .
While w e s h a l l have
o c c a s i o n t o use r i g h t modules, any u n s p e c i f i e d module w i l l always b e understood t o be l e f t and unitary. W e s a y t h a t an A-module
of A -submodules of
M
M
i s graded p r o v i d e d t h e r e e x i s t s a f a m i l y
indexed by
G such t h a t t h e f o l l o w i n g two c o n d i t i o n s hold.
1
M = @M SfG AzMy c_ M
w
( d i r e c t sum of A -modules)
(1)
1
for all
x,y
The above d e f i n i t i o n c e r t a i n l y i m p l i e s t h a t t h e r e g u l a r module
E
AA
G
(2)
i s graded
182
CHAPTER 4
(with M = A for all g E GI. We refer to M g g g the g-component of m E M we understand unique m
as the g-component of E M
m E N
M
of a graded module
and all g E G, m
g
E N.
m =
By
C m
S f C g' is said t o be a graded submodule if for all g
A submodule N
defined by
M.
g
Expressed otherwise, N
is graded if
C B ( N ~ M )
N =
Sfc Thus if
N
is a graded submodule of N
M,
$7
=
then N
~
g
N
is a graded module with
M
A
A graded left A-submodule I of
for all g E G.
is called a graded l e f t
ideal of A. We call (11 the G-grading of
M
and refer to
as the y-component of
M
g
M.
When ( 2 ) can be replaced by the stronger condition
A M = M Y ZY M
we say that Let M
for all r , y E G
(3)
is a strongly graded A-module. N
and
be two graded A-modules.
An A-homomorphism
f:M---tN is said to be graded if
f ( M ) C_N g
In case
f
A-moduZes
.
is an isomorphism, we say that
1.1. LEMMA.
Let
M and N .
Then
(i) Kerf (ii) f ( M )
f
:
M
d N
M
and
(4)
are isomorphic a s graded
N
be a graded homomorphism of two graded A-modules
is a graded submodule of
is a graded submodule of
(iii) M/Kerf
for all g E G
g
M N
f(M) as graded A-modules, where the g-component of
M/Kerf
is
defined by (g E
(M/Kerf) = ( M +Kerf)/Kerf g g Proof.
Given m E M, we have
f(m)
=
f( C m ) sfGg
=
C f(m
&G
)
G)
where each f(m
g
)
GRADED MODULES
N
is in
T h i s shows t h a t
g'
f(m) g
rn
If
183
E Kerf,
f(m1
then
=
f(m
=
9
0 and hence, by ( 5 ) , f ( m
-
9
) =
p r o v i n g ( i ) . P r o p e r t y ( i i ) i s a d i r e c t consequence of To prove ( i i i ) ,l e t
f*
: M/Kerf
f(M)
g E G,m E M
for all
)
0.
Thus
m g
(5)
E Kerf,
(5).
f.
be t h e isomorphism induced by
Then
f*(m + Kerf) 9 which shows t h a t If
V
f*
=
f(m
) =
g
i s a graded isomorphism.
i s an A-module,
.
f(m)
t h e n we s h a l l d e n o t e by
by t h e r e s t r i c t i o n of a l g e b r a ;
V
thus
=
V
E f(M) g
g
m
for all
M
E
t h e A -module o b t a i n e d
VA
as a d d i t i v e g r o u p s b u t o n l y a c t i o n
Al
of
A
i s d e f i n e d on
T h i s p r o c e s s w i l l b e c a l l e d r e s t r i c t i o n and i t
VA-.
1
Y
p e r m i t s u s t o g o from any A-module
t o a u n i q u e l y determined A -module
A'
As i n t h e c a s e of modules o v e r group a l g e b r a s , t h e r e i s a d u a l p r o c e s s of induction.
T h i s p r o c e s s p e r m i t s us t o go from any A -module
determined graded A-module
@
d e f i n e d i n Chapter 2.
V
t o a uniquely
For convenience, l e t u s
r e c a l l t h e f o l l o w i n g p i e c e of i n f o r m a t i o n . Let
V
be any A -module.
Then t h e t e n s o r p r o d u c t
1
$ = A @ V A 1
i s n a t u r a l l y an A-module w i t h
y(x@v) W e s h a l l r e f e r to
@
=
yx '8 v
as t h e induced module.
f o r any
g E G.
and
v
E
V
In view of
and Lemma 2 . 1 . 1 ( i i ) , w e may i d e n t i f y t h e A -module
@,
f o r a l l x,y E A
A
9
@ V
w i t h i t s image i n
A,
With t h i s i d e n t i f i c a t i o n , w e have
8=
@ (A
'8 V)
( d i r e c t sum of A -modules)
Setting
(fl)g = A g
Q V A,
for a l l
g
E G
(6)
CHAPTER 4
184
i t i s immediate from (6) and t h e containment
@
that
i s a g r a d e d A-module.
fl
we s h a l l p u r s u e o u r s t u d y o f
i n o u r sub-
Here w e w i l l o n l y b e c o n t e n t t o p r o v i d e a u s e f u l
sequent i n v e s t i g a t i o n s .
c h a r a c t e r i z a t i o n of s t r o n g l y g r a d e d a l g e b r a s which i n v o l v e s
@
(see Theorem 1 . 2
below). Let
M
be a g r a d e d A-module.
g E G,
For e a c h
put
4 = M and d e f i n e
(@Ix Then
=
Mxg
for all
I& i s o b v i o u s l y a g r a d e d A-module which d i f f e r s from
fl
We s h a l l r e f e r t o
M.
a s a conjugate of
g E G,
we s a y t h a t
M
G
(7)
o n l y by q r a d i n g .
In c a s e
M=@ for a l l
M
zE
( a s g r a d e d A-modules)
i s G-invariant.
We a r e now r e a d y t o prove t h e f o l l o w i n g r e s u l t .
1.2.
THEOREM. (Dade(19801).
The f o l l o w i n g p r o p e r t i e s of a G-graded a l g e b r a
A
are e q u i v a l e n t t o each o t h e r : (i) A
i s s t r o n g l y graded
( i i ) Every g r a d e d A-module i s s t r o n g l y g r a d e d
M,
( iii) For any g r a d e d A-module
t h e n a t u r a l map
i s a graded isomorphism
M,
( i v ) For any graded A-module
t h e r e e x i s t s an A -module 1
Me (i)
proof. graded,
A A
zY
= A
( i i ) : Let
zY
for a l l
M
fl
be a g r a d e d A-module.
z,y E G,
V
such t h a t
( a s g r a d e d A-modules) Since
s o by Lemma Z . l . l ( i ) ,
A
is strongly
w e have
185
GRADED MODULES
Mq = AIMxy = AxAX-1M zy C - AzMy C Mq Hence A M
"Y
where we have used (2) twice. (ii) =. (iii):
so
di
Given g E G,
=
The kernel N
(Lemma l.l(i)) whose g-component N
9
A l @ MI A,
0 and thus $
=
A N
91 onto MI, so N g
Put 'I = M
(iii) =) (iv):
is a graded A-module
is given by
is the kernel of the natural isomorphism of =
$
of
is strongly graded so that for a11 g E G, N
By hypothesis, N
N
w'
i
as required.
we have
is a graded epimorphism.
Therefore
M
.
N1
Now =
0.
is a graded isomorphism.
and apply (iii)
(iv) * (i): Owing to Lemma 2.1.1(i) ,tii), we have
A A =A Hence, for any A -module 1
for all g
9
91
E
G
V,
Applying the hypothesis, we therefore conclude that
A M = M 91
for any graded A-module M. g-component A some
x
E
9
A
The regular A-module
for all g E G.
Letting
M
is a graded A-module with
to be the conjugate module
A"
for
G, we obtain
A A 9" by (7).
9
Hence
1.3. COROLLARY.
A
=
A (A")l
=
g
(AX) 9
=
A
92
is strongly G-graded, as asserted. Let A
be a strongly C-graded algebra, let M
N be a graded homomorphism.
graded A-modules and let f : M--+
and
N be two
Then f
monomorphism, epimorphism, or isomorphism if and only if its restriction
f
1
: M - N 1
1
is, respectively, a monomorphism, epimorphism, or isomorphism of
A -modules.
is a
CHAPTER 4
186
M
By Theorem 1 . 2 ,
Proof.
N
and
a r e s t r o n g l y graded so t h a t
A M = M g l g
A N
and
91
= N
gE G
for a l l
g
Hence
A
Because
f (M
91
)
=
1
A N
91
if and o n l y i f
1
epimorphism i f and o n l y i f s o i s
f
If
fl m
A
f(m ) 9
Then
Y'
since
)
= All,
m cM
g-l g -
=
f
w e infer that
i s an
1
fl.
i s a monomorphism, t h e n c l e a r l y so i s
i s a monomorphism and l e t
E M
f (M
f(m)
=
0
f,. m
f o r some
C o n v e r s e l y , assume t h a t
E
M.
Write
f ( A -lmg) = 0
0 f o r a l l g E G, hence
m
=
C m SfG
and t h u s
9
But t h e n
1'
m
which i m p l i e s t h a t
g
=
.
0, as r e q u i r e d .
2 . RESTRICTION TO A
Throughout t h i s s e c t i o n ,
A
d e n o t e s a G-graded a l g e b r a .
i n v e s t i g a t e t h e behaviour of t h e module module.
'A '
where
V
Our a i m i s t o
i s an i r r e d u c i b l e A-
The main r e s u l t i s a g e n e r a l i z a t i o n o f a c l a s s i c a l C l i f f o r d theorem
which d e a l s w i t h t h e c a s e
A = FG,A
=
FN
where
FG
i s r e g a r d e d as a G/N-
1
F
graded a l g e b r a o v e r a f i e l d
For any A -module
V
and any
whose gN-component i s g i v e n by
g E G,
put
1
Then, by Lemma 2 . 1 . 1 ( i i ) ,
conjugate of
V.
2.1.
Let
LEMMA.
module and l e t
A
'V
i s an A -mod:le.
W e shall refer to
be a s t r o n g l y G-graded a l g e b r a , l e t
U be any A-module.
M
-Y'
as a
be any g r a d e d A-
with
RESTRICTION TO A
187
G on the set of isomorphism
(i) The formula (1) provides an action of the group
classes of A -modules, i.e. for all x,y E G
'V
"=
V
and any A -module
and
x(yV)
M
gM
g
2
V,
qV
(2)
as A -modules
(3)
1
1
I& A B g M 1
(as graded A-modules)
(4)
A1 (iii) For any irreducible A -submodule W module
A W
of
and all
9
v
(i) Let
be any A -module and let x,Y
E
G.
By definition,
and z(yV)
€3 V )
= x(A
=
A
X
A,
is an isomorphism of A -modules. (ii) The isomorphism morphism ( 3 ) . 2
@.
$
Because
€3
(A
(@)
Thus x(YV)
€3
V)
4,
A1
Next by Theorem 1.2(iii), applied to M =
A €3 M
G, the A 1 -
(as A -modules)
'WEAW
'v v
E
is also irreducible and
g
Proof.
g
the map
xyV,
proving (2).
of Theorem l.2(iii) is graded and hence induces an iso= Mg,
(@I1
it follows from Theorem 1.2(iii) that
The isomorphism
A1
induced by ( 2 ) is obviously graded, proving ( 4 ) (iii] Assume that submodule of
X is a nonzero submodule of A W.
W and hence A
- 1X =
W.
?i
by
Ag,
we derive
X
=
A W. g
Then A -lX is a nonzero ?i n Y Multiplying both sides of this equality
Thus
A W is irreducible.
is obviously a surjective homomorphism. A
€3 W Al
is irreducible.
The natural map
g
So assume that
Therefore it suffices to show that
X is a nonzero submodule of A
g
€3 Al
W.
CHAPTER 4
188
get
5A
A -lX
Then
9
X
8
= A
t,
W,
so
X
A
-1
=
A
8 W.
A
Again, m u l t i p l y i n g by
we
9’
Al
as r e q u i r e d . g.
Al
A
Let
8 W
A,
V
be a s t r o n g l y G-graded a l g e b r a and l e t
be an A -module.
Consider
1
the s e t
H H
Then, by Lemma 2 . 1 ( i ) ,
i n e r t i a group
V;
of
E GIV
= {g
i s a subgroup of
H
i n case
=
G
2
’V}
G.
W e shall refer t o
we s h a l l s a y t h a t
V
H
as the
i s G-invariant.
We
have now come t o t h e d e m o n s t r a t i o n f o r which t h i s s e c t i o n h a s been developed. 2.2.
V
and l e t
VA
(i)
G
Let
THEOREM.
A
be a f i n i t e q r o u p , l e t
be an i r r e d u c i b l e A-module.
be a s t r o n g l y G-graded a l g e b r a
Then
U
c o n t a i n s a n i r r e d u c i b l e submodule, s a y 1
(iil
H
If
U
i s t h e i n e r t i a g r o u p of
and
W
t h e sum of a l l submodules of
VA 1
isomorphic t o
VA
(a)
U,
then there e x i s t s a p o s i t i v e integer
e(g1Ue
e
such t h a t
... o gn u)
1
where
{l
=
gl,g2,...,gn}
H
i s a l e f t transversal for
a r e p a i r w i s e nonisomorphic i r r e d u c i b l e A -modules.
G
in
and
In p a r t i c u l a r ,
9 1
gn
U,..., U
VA
is 1
c o m p l e t e l y r e d u c i b l e of f i n i t e l e n g t h . (b)
hi
i s an i r r e d u c i b l e A(H1-module such t h a t WA
2
eU
and
8
V
1
cU
(Here, of c o u r s e , Proof.
d e n o t e s a d i r e c t sum o f
(i) Since
V
sfGg V
in
g e n e r a t e d A -module.
V.
Because 1
We now s e t
M
of
u).
C A V
V.
G
(5)
SfG’
By P r o p o s i t i o n 2 . 1 . 1 0 ,
f i n i t e l y g e n e r a t e d A -module. p r o p e r A -submodule
copies of
is irreducible,
V = A v = ( C A ) v = f o r a l l nonzero
e
each
A
9
is a finitely
i s f i n i t e , i t f o l l o w s from ( 5 ) t h a t
Hence, by P r o p o s i t i o n 1 . 3 . 5 ,
V
is a
t h e r e i s a maximal
RESTRTCTION TO
Since A M so
C
Mo
for all g E G,
M
A
189
is an A-submodule of
90= 0 and thus there is an injective homomorphism
M
V
V.
8 [V/A M
--f
Mo
But
5M #
Y,
of A -modules
SfG We are therefore left to verify that each A -module V / A M is irreducible or, 9
equivalently, that each A M
is a maximal A -submodule of
9
V.
1
5 M' c I.' for some A -submodule M' of 9 plication by A -1 on the left yields g M 5 A -lM' C A -1V 9 9 so M = A M' and hence M' = A M, as required. g 9-1 c A U is a nonzero 4-submodule of V , we have (ii] Because assume that
So
A M
sfc
Thus, by Lemma 1.3.13,
9,U , ...,gn U
VA
Then multi-
v= c
A U.
SfG
is an irreducible A -module isomorphic to 'U.
A U
Owing to Lemma Z.l(iii), each
V.
9
is completely reducible.
Moreover, by Lemma Z.l(i),
1
U.
are all nonisomorphic conjugates of VA
all nonisomorphic irreducible submodules of submodules of
VA
isomorphic to A
i
.
U,. ..,A
U are gn be the sum of all
Hence A
g1
Wi
Let
U, 1
Q
n.
Then, by Lemma 1.7.6,
W1 Q
... Q Wn
(Wl = W)
gi
1
VA = 1
Thus, to prove (a) it suffices to verify that each of irreducible direct summands.
Now
Wi contains the
for each g E G, A V = V
same number
and the modules
g
ASWi,
A
9
W
i'
i # j, have no composition factor in common. W.=A 7,
Hence
W 9
1 and use
0 for i # j ,
=
=
0.
1
Applying the auxiliary assertion, there exists an idempotent e r
-
e' = e 1
and 1
ee'
=
1
gonal idempotents.
e r e = 0.
0 for
is an idempotent such that
e e = e e
-
=
i# j
In other words, we may assume that e e
e = e
e.e 1-j
such that
there is nothing to prove, so we may assume that
induction on n
i,j > 1.
=
--
are idempotents such that
It follows that
1
e',e 1
2
,...,e
such that
are pairwise ortho-
It is now an easy matter to prove the required assertion.
Indeed, by (.i)and the above, there exist pairwise orthogonal idempotents
e ,e 1
2
,...,e
-
such that e .= 1-
E
if
1G
i
G n.
Thus for
u
=
e
1
+ e + 2
... + en
we
212
CHAPTER 4
-
-
u = 1,
have
e E R
To prove t h e f i n a l a s s e r t i o n , assume t h a t
e # 0,
Since
R,
x,y
e
0.
- -
e = e
e
+
1
2
e
0 or
=
e
.
xf y
0.
i s a p r i m i t i v e idempotent.
f o r some o r t h o g o n a l idempotents
e 1 , e 2E R
le + e ) R ,
2
1
y
0 or
=
1 = xte2
=
which i s p o s s i b l e o n l y
2
cc
Hence e i t h e r
-
with
0 and t h e r e f o r e
=
is primitive.
+
= u
where
V,
u = 0
is primitive.
5.6.
LEMMA.
(i) R
e
is primitive.
-
or
2, =
0.
Thus e i t h e r
R
Let
I
0 and so
-
e =
then
0 or
U =
U
+
2, =
T h i s completes t h e proof of t h e lemma.
e # 0.
If
- - - -V,uV=
V U
=
0
0, p r o v i n g t h a t
8
be an a r t i n i a n r i n g .
0 and
i s l o c a l i f and o n l y i f
(ii) I f
e#
Then
u,V a r e o r t h o g o n a l idempotents,
c
and so e i t h e r
e
=
By ( i i ) , eR
=
Conversely, assume t h a t
e
e
If
t h e n t h e r e e x i s t o r t h o g o n a l idempotents
i n t h e case
-
e#
w e have
and hence w i t h
u = 1.
1 - u i s a n i l p o t e n t idempotent, and hence
so
i s a n i l p o t e n t i d e a l of
1 a r e t h e o n l y idempotents of
R,
i s l o c a l i f and o n l y i f s o i s
R
then
R
R/I. Proof.
R
(i) If
p o t e n t s of
R.
R.
Since
J(R)
of
H/J(R1,
i s l o c a l , then obviously
i s a n i l p o t e n t i d e a l of
by v i r t u e of Lemma 5 . 5 ( i ) .
matrix r i n g s over d i v i s i o n r i n g s . therefore
R
LEMMA.
G
Let
C
a l g e b r a of
Hence
and
and
1 a r e t h e o n l y idem-
1 a r e t h e o n l y idempotents of
R,
0 and
1
But
R/J(Rl
i s a d i r e c t p r o d u c t of f u l l
R/J(R)
a r e t h e o n l y idempotents
must be a d i v i s i o n r i n g and
is local
( i i ) D i r e c t consequence of
5.7.
0
Conversely, assume t h a t
0
Proof.
E:
Let
homomorphism A
be a p-group,
F.
over
( i l and Lemma 5 . 5 1 i ) .
A
Then
8
l e t charF = p
and l e t
A
b e a t w i s t e d group
i s a local ring.
be t h e a l g e b r a i c c l o s u r e of
F.
Then t h e r e i s an i n j e c t i v e
E @ A i s a t w i s t e d group a l g e b r a of S o v e r E. F i s l o c a l i f and o n l y i f 0 and 1 are t h e o n l y idempotents of
--i
E @ A and
F
A
By Lemma 5.6,
A.
Hence we may assume t h a t
1.6(iv), an F - b a s i s
A
p
FC.
{g
Let
- 11 1 #
I(G)
g E
GI
F
i s a l g e b r a i c a l l y closed.
be t h e augmentation i d e a l of and each
g
-1
Then, by P r o p o s i t i o n
FG.
i s nilpotent since
Then
I(G)
has
LNDUCTION FROM A
Hence
.
I(G) is a nilpotent ideal of FG.
is a local ring.
Let F
5.8. LEMMA. and let A
G over F.
be a twisted group algebra of
Proof.
FG/I(G) e F and therefore FG
G
=
charF = p
1 or
Then A
0 and G
Assume that f G
G has an element g of order n
local ring and assume that
H =,
H'F
we have
e
=
is a local ring
is a p-group.
The "if" part is a consequence of Lemma 5.7.
Then, for
G be a finite group
be an algebraically closed field, let
if and only if either
F.
But
213
3
1 where
is a
# 0 in
y1
FH (Proposition 1.6.5) and n-1 i (Vnl C g i=0
is a nontrivial idempotent of contrary to Lemma 5.6(i).
FH.
Thus
Hence G'F
charF = p
3
.
has a nontrivial idempotent,
0 and G
is a p-group.
The following result generalizes a classical theorem of Green(1959). 5.9. THEOREM.
be an absolutely indecomposable A -module and let H
V
Let
be
1
V.
the inertia group of
If
charF = p > 0 and
H is a p-group, then VG is
indecomposable. Proof. V
By Theorem 5.2(i), we may harmlessly assume that H = G, i.e. that
is G-invariant.
Put E = En 0 and
is a local ring, by virtue of Lemma 5.7.
)
But
E-J(E )
nilpotent ideal of Thus
is a pis a
1
1
(ii)).
G
fi
E
(Theorem 5.4(ii)), hence
.
E is a local ring (Lemma 5.6
is indecomposable, by virtue of Lemma 4.3.
The result above can be strengthened under the assumption that F
is algebra-
ically closed. 5.10. THEOREM.
Let F
be an algebraically closed field, let
posable A -module and let H 1
conditions are equivalent: (i)
fi
is indecomposable
be the inertia group of
V.
V
be an indecom-
Then the following
CHAPTER 4
214
?
(ii)
is indecomposable
H
(iii) Either
E
=
H is a p-group
and
That (i) is equivalent to (ii) is a consequence of Theorem 5.2(i).
Proof. Let
charF = p > 0
1 or
=
so that by Lemma 4 . 3 ,
End(#)
(ii) is equivalent to
E being local.
By Lemma 5.6(ii), the latter is equivalent to E/B*J(B1) being local.
is H-invariant, so by Theorem 5.4(iii,
H over F.
But
E/E*J(E1) is a twisted group algebra of
The desired conclusion is therefore a consequence of Lemma 5.8.
fl
To examine the decomposition of
in the general case, we shall need the
following ring-theoretic results. 5.11. LEMMA.
N be a nil ideal of a ring R and let the ring R'
Let
=
R/N
admit a decomposition
ii = L
@ L @...@Ln
1
2
Then there exists a decompo-
into finitely many indecomposable right ideals.
R,
sition of
R
=
I @ I2@ 1
... @ I n
into indecomposable right ideals such that (i) Li
is the image of
(ii) L .
z
=L
Proof.
-
with
1=
Ti
under the natural map
as right ??-modules if and only if
j
1
E
1
+
+ e +
E
9
+
...
t. E
and
E.E. z
L. =
z
2
... + e .
1
I. I z
j
R. as right R-modules.
z
1
into indecomposable right ideals of R.
z
2
E
i
E
R'
Owing to Lemma 5.5(iii), there 2
Setting I.= e . R ,
R = I @I
(i) holds.
c
-t.
By hypothesis, there exist orthogonal primitive idempotents
exist orthogonal primitive idempotents e , e 1 = e
R
,...,e
€
R
with
-
ei =
E.
z
and
we have the decomposition
0 . . . @In Since
72.
=
e. 2
= E.R =
L
i'
property
Property (ii) being a consequence of Lemma 5.5(ii), the result
follows. We next prove a general fact which provides the link between a module and its ring of endomorphisms.
Let R
any subset S of
,
EndfJ)
R
be a ring and let
let SV
V
be a left R-module,
denote the set of all finite sums
For
INDUCTION FROM A
215
ISi ( V i )
Then, clearly, SV
cipal right ideal of
End(V1 R
Let R
5.12. LEMMA.
V.
is an R-submodule of
V
I is
Note also that if
4,
generated by
be a ring, let
S,Vi E V )
(Si E
a prin-
@(W.
then I V =
be a left R-module, and let End(v)
R admit a decomposition
... @ I n
End(v) = I @
R
into finitely many indecomposable right ideals. (i)
=
... 0 I,V
I V @ I V 0 1
2
= 1.V
(ii) 1 . V a modules.
Then
is a decomposition into indecomposable submodules.
I.E I
as left R-modules if and only if
3
z
(i) Put E = End(v1
Proof.
as right End(Vi)-
R
j
and write
R 1
4;
Then
=
# j,Ii
= 0,;
$i,$.Q.
=
Q1 +
Q2
= QcE,
... Qn
+
(Qi E 1;)
Ii is indecomposable, each of
and since
Z J
the
idempotents Qi
If v
is primitive.
V,
E
then
and so
v= Suppose that
0 1
,v
2
,...,v
v+
I
I
2
... + InV
v+
are such that
1
0.
Thus
Q
Now put
=
Q .,I
=
.
Y
+ X..
tents such that for each
v
E
Then
X
=
X 0X
1
v
also have
=
or
1
2
Then
z
@(v) for each
must be
IT
for some R-modules X
V and we denote by
and
1
IV,v = L, E
n
(u)
TI
IV.
+
n
(v).
TI
0.
But
2
X. z
=
IT
i
E
2
IV
=
@ ( V ) , we must
2
Hence
= T
I#I
TI.(V), so
z
X
=
1
+ n
and therefore either 2
0 or X
=
0, proving that
2
IV is indecomposable. (ii) Assume that
I. 2 I
z
i
E
are orthogonal idempoBecause
1 1
= 0.
2
are direct summands of
I
T
2
1
the canonical projection
2
2
I., and assume that I V
X
X
... Q n ( V n )
1 + Q (v 1 +
V = 69 I.V.
and
and
1
@i, we can deduce that each Q . ( v . ) z z
Then, taking the images of both sides under is
(v
Q
as right E-modules.
We may regard
V
a s a left E-
2 16
CHAPTER 4
module via
,@
@V = $ ( V )
V.
E E,V
E '8 V becomes a left E-module via E
Then
$ ( $ '8 0) = $+ '8
v
E,v
($,+ E
E
V)
Now the canonical map
Since Ii
is an isomorphism of E-modules.
is a direct summand of
identify I.'8 V with its image in E @ V. " E E is I.V, we therefore derive
I. @ " E which in turn implies that 1.v z
8
Now suppose that E Ii, we have
I$
1.V
Im$C IiV
(8$)( V ) = @ . ( v
that
:
3
1
Lz
Ii 8 V
Because the image of
in
V
E
I.V
I.V. 3
I.V is an R-isomorphism and fix
-f
3
so Im(e@)
for some
)
v
E, we may
V
5I j V .
E V.
Because
I.V 3
= @
V E
v.
.(V),
3
Given
we deduce
Thus
@ .(84J)(v) = @ . ( V ) = (8@,( V ) 3 3 1 and so
8@ = @.(6$)
c2
3
If we now define A : Ti + I
j
morphism of right E-modules.
by
A(@)
=
I j
8$, then
is obviously a homo-
A similar argument shows that the mapping
)l:I-Ii j defined by A,
p($) = 8
-1
q)
the result follows.
5.13. LEMMA.
Let R
.
is also an E-homomorphism.
be a local ring.
Since 1.1
is the inverse of
Then all finitely generated projective
R-modules are free. Proof.
v
1
,...,v
Let
V
be a finitely generated projective R-module, and let
be a minimal generating set for
V.
Suppose that we are given a
relation
r 1v 1 + rz vz + . . . +
r n vn = O
(Pi E R )
INDUCTION FROM A l
We claim that all such r
i
E
217
Indeed, if pi 4 J ( R ) , then it is a unit in
3(R).
R, and we can solve for the corresponding
vi
as an R-linear sum of the
This, however, is impossible since otherwise the generator
V
j
can be deleted.
vi
r.E J(R).
Thus all
W be a free R-module freely generated by
Let
u 's.
by sending w i
vi.
to
contained in J ( R ) E / .
W/Vr
..., n
and map
W
By the foregoing the kernel of this map, say
By hypothesis, V
W Since S
W1,W2,
=
S 8 V'
S, is
is projective so that we have V'
with
it must be finitely generated.
V
Furthermore,
and hence the inclusion S 5 J ( R ) V forces S = J(RIS. lemma (Lemma 2.3.7) we conclude that
W onto
Invoking Nakayama's
S = 0 and hence that V
W.
So the
=
End(fi).
lemma is true.
Then
be an indecomposable G-invariant A -module and let
V
Let
E
g
End(V) is a local ring (Lemma 4 . 3 ) ,
F
so any finitely generated projec-
1
tive E -module is free (Lemma 5.13).
In particular, if a right ideal
I of E
1
is a direct summand of
then I is a projective E -module since
E,
free E -module (Proposition 5.1).
Thus
1
tifying
I is in fact a free E -module.
Iden-
1
V
and
5.14. THEOREM.
ideals.
7 63 V , Let
V
we now prove be an indecomposable G-invariant A -module, let
and let E = @ Ii be a decomposition into indecomposable right
E = End(@)
(i)
B is a
1
Then the following properties hold:
9 = @ 1.V
(ii) I.e! I z
i
is a decomposition into indecomposable A-modules as
E-modules if and only if
1.V 2
2
I,V 3
as A-modules
(iii) dimI,V = ( E -rankla)(dimV) F Z
(iv) If
V
1
Z
is irreducible and
F
charF
[ \GI,
A- module. Proof.
(i) Owing to Lemma 5.12(i),
vc
=
8 IiVG
then each
I .V
is an irreducible
CHAPTER 4
218
so we need only verify that
ly have IiV C_ Ii#.
IiVG for all i.
=
Iifl =
Because
-
spaces of the form g €3 V,
9
71 E
V,
V
5 VG,
we obvious-
be defined by
and since
fl
for
is a sum of sub-
we need only show that
CPc; Let x E E
Since
To prove the reverse inclusion, observe that I.= @/?
I$ E E .
some idempotent
IV.
8 V)
x ( v ) = $,v 9
E
5 IiV
.
V
g
for all
E
G
and for all
we have
G
and thus I.V = I.V
,
proving (i)
.
(ii) Direct consequence of Lemma 5.12(ii). (iii) We first claim that the map
is an F-isomorphism. IGi,
Indeed, by Proposition 5.1,
6' is a free E -module of rank
so
1
$ be a unit in E Then 0 (i8 V ) = g -1' 9 g the above map is surjective. This substantiates our claim.
For each g E G,
let
8
v
and so
li is a direct summand of E,I. €3 V is identifiable with its
Because
Z
image in E €3 V
E 1
and so the induced map
El
G
I . 8 v - V El
is injective.
Taking into account that dim(I 8 V) = ( E -rankI.) (dimV) F iEl Z F
and that
T.V
is the image of
I. 8 V ,
(iv) Direct consequence of (i) and Corollary 5.3. Assume that
E
=
End(l/;).
V
is an absolutely indecomposable G-invariant A -module and put
Then, by Theorem 5.4,
where the group
.
the required assertion follows.
1
E/E'J(E
)
e F"G
for some
a E Z2(G,F*),
Z2(G,F*) is defined with respect to the trivial action of
G
on
INDUCTION FROM A
F.
E.J(El)
Since, by Theorem 5.4,
219 1
is a nilpotent ideal of
E, it follows from
Lemma 5.11 that any decomposition F'G
= L
=
Q L
2
F"G
into indecomposable right ideals of
E
1
Q
into indecomposable right ideals of
L
~
can be lifted to a decomposition
...
I 0I Q 1
... Q
2
E.
In
We are at last in a position to prove
the following generalization of a theorem of Conlon (1964). V
Let
5.15. THEOREM.
be an absolutely indecomposable G-invariant A -module.
Then, in the notation above, the following properties hold: (i)
fi
=
0 IiV is a decomposition into indecomposable A-modules
L
(ii) L .
z
as FaG-modules if and only if
j
1.V
z
I.V as A-modules 3
(iii) dimI .V = (dimL.) (dimV) F " F" F Proof. Properties (i) and (ii) follow from Theorem 5.14(i), (ii) and L e m a 5.11(iil.
Thanks to Theorem 5.14(iii), to prove (iii) we need only show that for
1 = 1
i
E -rankI
=
dimL
1
Because
F"G
is identifiable with
F i
E/E*J(EI), the latter will follow provided we
prove that for n = E -rankI 1
dim I F Choose an E -basis
{dl,d
summands of
,dn}
{dl,d2,..
,...,dn,dn+l,...,dm 1
r+
of
E
m
and so
= 1
of
JCE
) 1
@
F and since
(6)
~1
I and extend it to an E -basis
(this is always possible since all direct
E are free E -modules). n
Since F
E - S ( E ~ ) ) / E * J ( E ~ )=I
Then
rn
CHAPTER 4
220
(I + E*J(E1)l/E*J(E1)is a unique F-linear
it follows that each element in
combination of the images of d l ,d2,.
..,d,
in
(I + E*J (El 1 1 / E * J (El> .
This
proves ( 6 ) and hence the result. The following simple observation will enable us to take f u l l advantage of some of our previous results. Let V
5.16. LEMMA.
an indecomposable direct summand of of
W, respectively.
V and
w
be an indecomposable G-invariantA -module and let
fi.
Then n
multiplicity of the isomorphism type of
n
Denote by
and m/n
divides m V
and m
be
the F-dimensions
is equal to the into
in the decomposition of
wA 1 In particular, V
indecomposable A -modules. 1
an indecomposable direct summand L Proof.
By hypothesis,
fl
=
of
W@X
4
is extendible to A
if there is
such that d i d = dimv.
F
for some A-module
F X and therefore
Note also that
where all the summands are isomorphic to
V
since
V
is G-invariant.
It
follows, from the Krull-Schmidt theorem, that all indecomposable components of hl A1
are isomorphic to
V.
t denotes the number of such components, then rn = n t
If
and the required assertion follows. 5.17. THEOREM.
Let
I: be an algebraically closed field and let
invariant indecomposable A -module. 1
charF
)
lG/,
Proof.
summand of
Then
'
V
Suppose that
is extendible to
A.
Owing to Lemma 5.16, it suffices to exhibit an indecomposable direct VG having the same F-dimension as
ideal of F-dimension 1.
V.
Now the assumption that
FG, while the assumption that charF
Since F
be a G-
G is cyclic and that
By Theorem 5.15(iii), the
latter holds provided the twisted group algebra F'G
FO"C
v
./'
IGI
has an indecomposable right G
is cyclic guarantees that
ensures that FG
is semisimple
is algebraically closed, all irreducible FG-modules (equivalently, all
INDUCTION FROM A
indecomposable right ideals of able right ideals of
F'G
FGI
221 1
are one-dimensional.
Thus all indecompos-
are one-dimensional and the result follows.
This Page Intentionally Left Blank
223
5 Primitive and prime ideals of crossed products
P r i m e and p r i m i t i v e i d e a l s a r e i m p o r t a n t b u i l d i n g b l o c k s i n noncommutative r i n g
theory.
I t i s t h e r e f o r e a p p r o p r i a t e t o i n v e s t i g a t e t h e r e l a t i o n s between prime
R
and p r i m i t i v e i d e a l s of
G
R.
over a r i n g
R*G,
and
R*G
where
i s a c r o s s e d p r o d u c t of a group
f i n i t e , s i n c e v i r t u a l l y n o t h i n g i s known i n t h e g e n e r a l case.
P
0
over a f i e l d .
P
of
R*G
R = 0 and G-prtme i d e a l s of a c e r t a i n f i n i t e - d i m e n s i o n a l a l g e b r a We t h e n a p p l y t h i s correspondence t o examine t h e r e l a t i o n s h i p
R*G
between t h e prime i d e a l s of
and t h e G-prime i d e a l s of
R.
The i n f o r m a t i o n
o b t a i n e d a l l o w s us t o prove t h e I n c o m p a r a b i l i t y and Going Down Theorems. u s i n g a d i f f e r e n t approach, w e a l s o e s t a b l i s h a Going Up Theorem. s e c t i o n is devoted t o some a p p l i c a t i o n s . t h a t t h e prime (or p r i m i t i v e ) r a n k of
1GI-I E R,
is
One of o u r a i m s
i s t o e s t a b l i s h a b i j e c t i v e correspondence between t h e prime i d e a l s satsifying
G
Throughout we d e a l e x c l u s i v e l y w i t h t h e case where
By
The f i n a l
These i n c l u d e t h e proof of t h e f a c t
R*G
i s e q u a l t o t h a t of
t h e n t h e prime ( o r p r i m i t i v e ) r a n k of
R
R
and t h a t i f
i s e q u a l t o t h a t of
RG,
1. PRIMITIVE, PRIME AND SEMIPRIME IDEALS Throughout t h i s s e c t i o n ,
R
denotes a ring.
Our aim is t o p r o v i d e r i n g - t h e o r e -
t i c i n f o r m a t i o n r e q u i r e d f o r subsequent i n v e s t i g a t i o n s . W e s t a r t w i t h a few well-known
i f f o r a l l nonzero i d e a l s c a l l e d a prime
idea2
if
A , B of R/P
definitions.
R
w e have
is a prime r i n g .
The r i n g
AB#O.
R
An i d e a l
i s s a i d t o be prime P
of
R is
Prime r i n g s and prime i d e a l s are
i m p o r t a n t b u i l d i n g b l o c k s i n noncommutative r i n g t h e o r y . W e say t h a t
R
is
semiprime i f i t i s a s u b d i r e c t p r o d u c t of prime r i n g s o r
e q u i v a l e n t l y i f t h e i n t e r s e c t i o n of a l l prime i d e a l s of
R
i s zero.
An i d e a l
I
224
CHAPTER 5
i s s a i d t o be s e m i p r i m e i d e a l i f
R
of
i s a semiprime r i n g .
i s s a i d t o be a p r i m i t i v e ring i f it h a s a f a i t h f u l i r r e d u c i b l e
R
The r i n g
R/I
Such a r i n g should r e a l l y be c a l l e d left
module.
a r e l e f t modules.
p r i m i t i v e f o r a l l modules u s e d
r i g h t p r i m i t i v e rings.
We could s i m i l a r l y d e f i n e
In f a c t
t h e s e c o n c e p t s are d i s t i n c t , a s examples show, b u t w e s h a l l o n l y b e d e a l i n g w i t h l e f t p r i m i t i v e r i n g s and i d e a l s and t h e r e f o r e omit t h e q u a l i f y i n g a d j e c t i v e .
I
ideal
i s s a i d t o be p r i m i t i v e i f
R
of
R/I
i s t h e i n t e r s e c t i o n of a l l p r i m i t i v e i d e a l s of
R
Assume t h a t
A
R
c o n t a i n s no nonzero
A
of
c o n t a i n s a nonzero i d e a l
1s a n a r b i t r a r y prime i d e a l In Thus
R.
0.
i d e a l of square Proof.
Thus J ( R )
is a primitive ring.
R i s semiprime i f and o n l y i f
The r i n g
1.1. LEMMA.
An
R,
A'A = 0 c -P
then
0.
square
P
If
A C_ P.
and t h e r e f o r e
i s c o n t a i n e d i n t h e i n t e r s e c t i o n of a l l such prime i d e a l s and so
R
is
n o t semiprime. Conversely, assume t h a t nonzero
R
R,
ci E
X
=
Bn
E
R
X n {@I= @.
so
CY = c1
A
and
B
properly contain If
1.2.
# 0,
ananan # 0, and we p u t
n n n.
Owing t o Z o r n ' s lemma, t h e r e e x i s t s an i d e a l
rn
be i d e a l s of
P,
We now c l a i m t h a t
ci
m+l
= max(i,j),
LEMMA.
Proof.
P,
Since
R
A
with
then c l e a r l y
=
we have
Let
R
so by t h e maximality of
'm+l Since
ax
P
i s prime.
P
of
Because
9 P, t h e r e s u l t w i l l f o l l o w .
Let
i,j.
of nonzero e l e m e n t s of
Second, g i v e n
we have
P n X = 0.
maximal w i t h r e s p e c t t o
..I
2
Given a
= a B a
n+1
0 @ X,
,a , ...,an,.
= c1.
R,
CY NGW
1
F i r s t put
Thus, f o r some
0.
c o n t a i n s no nonzero i d e a l of s q u a r e
d e f i n e a sequence
i n d u c t i v e l y a s follows.
(RanR) # 0.
R
CYm'marn AB
E
c1
m
P.
E P + A , CY
rn
5P
P which p r o v e s t h a t
is primitive,
R
Then
R
E P+B
+
P+A
Then
P, a , € P + A ,
(P+A)(P+B)
be a p r i m i t i v e r i n g .
R
P,B
CY
j
E P+B
P+B
f o r some
and t h e r e f o r e
AB
P
and
i s prime.
.
i s prime and semisimple.
h a s a f a i t h f u l i r r e d u c i b l e l e f t module
V.
PRIMITIVE, PRIME AND SEMIPRLME IDEALS
J ( R ) V = 0,
Since
A
suppose t h a t
we conclude t h a t
B
and
V
a c t s f a i t h f u l l y on
. V,
module of
i s prime. 1.3.
R.
proof.
RyR
B # 0
V = BV.
But t h e n
i s semisimple.
AB
and
BV
i s an i d e a l , w e s e e t h a t
x
E
Z(R) and l e t
i s a nonzero i d e a l of
x = 0,
a s required.
R
Let
LEMMA.
AB = 0,
with
R
that
0.
=
Since
R
i s a nonzero R-sub-
so t h a t
A V = ABV = 0
Now,
R
A = 0
and
R
i s not a zero
I n p a r t i c u l a r , t h e c e n t r e of a prime r i n g i s an i n t e g r a l domain.
Let
therefore 1.4.
B
i.e.
A nonzero element i n t h e c e n t r e of a prime r i n g
LEMMA.
divisor in
and
R
a r e i d e a l s of and
and s o
J ( R ) = 0,
225
. R.
=
0
0# y
f o r some
E
R.
Then
R, RZ = 0 and
Hence, b y t h e primeness of
A,B
be a semiprime r i n g and l e t
BA = 0.
then
q
R.
b e i d e a l s of
If
I n p a r t i c u l a r , l e f t and r i g h t a n n i h i l a t o r s of i d e a l s a r e
equal. proof.
If
AB = 0,
then
(BAl2
B(AB)B = 0
=
and s o
BA = 0.
2. PRIMITIVE IDEALS I N CROSSED PRODUCTS
Throughout t h i s s e c t i o n , Given a c r o s s e d p r o d u c t for a l l
g
E
-1
gr=Ljrg Let
X
G d e n o t e s a f i n i t e group and R R*G
G, and p u t
of
G o v e r R,
= 1.
an arbitrary ring.
r
A s usual, given
E
R
lG1-l
R
E
or
X
i s i n h e r i t e d from
G is solvable.
Let
H
be a subgroup of
i r r e d u c i b l e R*G-module, t h e n
VH
G
X is
Then
X
not i nher i t ed
The main r e s u l t of t h i s s e c t i o n R*G
to
R
and t h a t t h e converse
As a by-product of t h e p r o o f , w e
o b t a i n some i n f o r m a t i o n on t h e p r i m i t i v i t y of
Proof.
g E G,
(R*G) g’ we put
d e n o t e t h e c l a s s of r i n g s whose p r i m i t i v e i d e a l s a r e maximal.
a s s e r t s t h a t t h e property
LEMMA.
in
.
by s u b r i n g s and o v e r r i n g s as e a s y examples show.
2.1.
g
and
i s o b v i o u s l y s t a b l e under homomorphisms a l t h o u g h t h e p r o p e r t y
is true i f
-
w e choose a u n i t
R
and
such t h a t
R*G.
1Hl-l E R.
If
V
i s an
i s completely r e d u c i b l e o f f i n i t e l e n g t h .
Owing t o Theorem 4.2.2,
VR
i s completely r e d u c i b l e of f i n i t e l e n g t h .
226
CHAPTER 5
(VR)H
Therefore
s i t i o n 4.2.4.
i s completely r e d u c i b l e of f i n i t e l e n g t h , by v i r t u e of PropoNow t h e map
d e f i n e d by
i s a s u r j e c t i v e R*H-homomorphism.
Hence
VH
i s c o m p l e t e l y r e d u c i b l e of f i n i t e
length. 2.2.
THEOREM (Lorenz ( 1 9 7 8 ) ) .
R
t h e p r i m i t i v e i d e a l s of
1GI-l E R
or
Proof.
G
is solvable.
P
R
i s a p r i m i t i v e i d e a l of
-1
Note t h a t
P
hence
M.
-P C 3
a r e maximal, t h e n
The converse i s t r u e i f e i t h e r
a r e maximal.
Suppose t h a t
maximal i d e a l
R*G
If a l l p r i m i t i v e i d e a l s of
g E G,
Given
put
'P
=
;Pi , F
s i n c e e q u a l i t y would imply t h a t
would b e maximal.
P
=
3
properly contained i n a
&
and
E
=
-1 § M g f o r some
@ g E G
. and
T h e r e f o r e , by P r o p o s i t i o n 2 . 6 . 5 ( i f ) ,
(R*G)P c ( R m Z . Let
hi
Then
V
P
be an i r r e d u c i b l e R-module such t h a t
= ann
(v), and
let
W =
8.
i s o b v i o u s l y of f i n i t e l e n g t h and hence we may f i x a f i n i t e composition
series
Set
Q
i
=
ann(Wi-l/Wi),
1
< i Q n.
Then, by P r o p o s i t i o n 2 . 7 . 1 6 ( i i ) ,
ann(w) =
(R*G)F
and so
...
Q ~ Q ~Q c ann(w) nSince
=
(R*G)P c (R*G)E
(R*G)E i s a p r o p e r two-sided i d e a l of
N of R*G c o n t a i n i n g
i t follows t h a t
Wi-l/Wi
(R*G)&
and o b t a i n
R*G,
&z. C -N
c o n t a i n s a copy of
we can choose a maximal i d e a l
f o r some
@ V
i.
f o r some
Because
g E G
and hence
PRIMITIVE IDEALS I N CROSSED PRODUCTS
-
g 8 'J
b e i n g G - i n v a r i a n t c o n t a i n s a copy of
and hence
Q. C
N.
g
for a l l
IG/ = 1 i s t r i v i a l , w e may assmue t h a t f o r groups of s m a l l e r o r d e r .
If
N
P
To t h i s end, l e t
V
W.
and
3Q
and
L.
Let
H
t r e a t t h e cases
H # G
and
H
Assume t h a t
H # G.
V
module
1
G
R*H
=
Q
R*G.
= ann(W)
W e have t o show t h a t
.
P
Set
= ann(V ) . 1
VR
c o n t a i n s an i r r e -
L.
I n what f o l l o w s w e
separately.
I(
V
WH
The r e s t r i c t i o n
f o r some i r r e d u c i b l e of
W
E*H-
i s completely
1
I n case
1GI-l E R
t h i s follows from Lemma 2 . 1 and
i s s o l v a b l e it f o l l o w s from t h e f a c t t h a t i n t h i s c a s e
R.
P=Q.
f a r some i r r e d u c i b l e R*G-modules
b e t h e i n e r t i a group of = G
G, t h e n
G i s simple.
By Theorem 4 . 2 . 2 ( i i ) ,
r e d u c i b l e of f i n i t e l e n g t h . when
lC\ 3 1 and t h a t t h e a s s e r t i o n is t r u e
are p r i m i t i v e i d e a l s of
= annCV1
A.
Since the case
/GI.
By Theorem 4 . 2 . 2 ( i ) , t h e r e s t r i c t e d module
d u c i b l e submodule
It follows t h a t
i s a p r o p e r normal subgroup of
Hence, we may h a r m l e s s l y assume t h a t
P
G.
E
Qi i s a nonmaximal p r i m i t i v e i d e a l of
Therefore
To prove t h e second a s s e r t i o n , we use i n d u c t i o n on
Assume t h a t
227
H = 1 and
Write
f o r some i r r e d u c i b l e R*H-modules
W.
and p u t
Q. =
ann(W.).
Then
n
Therefore
Qi
=
P
.
Qi C_ P 1 ,
s a y , and by i n d u c t i o n
(IHI < /GI), w e
conclude t h a t
Consequently, i t may b e deduced t h a t
G
a n n ( 8 ) = ann(V ) = ann(V) = The embedding of
Wi i n t o
W
P
e x t e n d s t o an R*G-homomorphism
WE
-+
W
which i s
CHAPTER 5
228
W
surjective since
is irreductble.
P
Hence
ann(8) c ann(W) = Q
=
z--
proving the case H # G . Now assume that
VR is isomorphic to a finite direct sum of
Then
L (Theorem 4.2.2(ii)).
copies of
R.
H = G.
Moreover, by Theorem 4.2.2
P n R = ann(L)
Hence
is a maximal ideal of
R
and by assumption on Q
~
R
R,
we obtain
= ngM
SfG for some maximal ideal M (R*G) ( P
The left ideal
flR )
It follows immediately that Q
R.
of
is actually a graded ideal of
5
=
-
R.
automorphism of
E
Let
E
Go
Since G
1.
consist of all those
5.
and hence
Z(E)
E
of
Otherwise, by Corollary
Then Go
g E G
9
such that
is a field since
E
is an inner
for which conjugation
zg =
is simple.
G = G
.
Hence, for each
-
rgi centralizes R. Since fi*G =
ZCE)
c Z(E*G) and we may view E*G c1 : G X
G+
by
we g'
as an Z(E)-algebra.
~
Define
The
EZ
SfG have
G
is a normal subgroup of
is simple, we deduce that
G, there exists a unit r E 9
centre
n R).
G,g.# 1, such that conjugation by
is an inner automorphism of
distinct from
g
P = Q = (R*G) ( P
is simple, we have
2.3.23(i), there exists g
9'
P n R.
R / P n R.
If E*G
by
=
E*G
R*G/(R*G)(P n R ) where
R*C
nR
- - - -1 Cl(~,y) = " Y x ~ .
Then, for all
r,y E
G,
we
have
follows that Y
algebra of
E*G
and obviously
simple subalgebra R*G
K
8- A ,
Z (R) ideal of A .
E,
E*G
-
=
E
the algebra E*G
A.
Because
A
A
is a Z(6-sub-
centralizes the central
h a s the structure of a tensor product,
E
'8 I, where I is an Z(R) (See Proposition 3.1.10 and Theorem 3 . 1 . 7 ) . Obviously I has to and the ideals of E*G
are of the form
229
PRIMITIVE IDEALS LN CROSSED PRODUCTS
R'
@ I
i s prime. (R) deduce t h a t a l l prime i d e a l s i n be prime i f
A
Since
E*G
is a finite-dimensional
a r e maximal,
algebra, we
In particular,
=
Q
thus
8
has
completing t h e p r o o f .
2.3.
THEOREM (Loren2 ( 1 9 7 8 ) ) .
E*G
(i) I f
R
i s prime and
R
(ii) I f
i s prime and
Proof.
(i) L e t
'L
The f o l l o w i n g p r o p e r t i e s h o l d :
is p r i m i t i v e , t h e n
R*G
i s p r i m i t i v e , then
R*G
R
is primitive
is primitive.
b e a f a i t h f u l i r r e d u c i b l e R-module.
Then
w=
f i n i t e l e n g t h and, by P r o p o s i t i o n 2 . 7 . 1 6 ( i i i ) ann(u = A (
a n n ( v ) ) =0
g-=
Fi
F i x a composition s e r i e s
= W
3
W
0
Q. = Then
Q1Q2
... Qn 5 ann(W) =
i, p r o v i n g t h a t
f o r some
v
(ii) L e t
0.
R*G
...
3
W
=
and p u t
Q.
i s prime it f o l l o w s t h a t
R*G
=
0
is primitive.
be a f a i t h f u l i r r e d u c i b l e R*G-module.
n >1
W
0 of
ann(FJi-l/Wi).
Since
VR = f o r some
3
1
w1
@
w2
@
Owing t o Theorem 4.2.2,
... @ wn
and some i r r e d u c i b l e R-modules
wi.
If
Qi d e n o t e s t h e a n n i -
Wi, t h e n
h i l a t o r of
n
Q1Q2
Hence
therefore
Remarks. (e.9.
... Qn 5
R
n Qi i=1
0.
Since
R
i s prime,
R
does n o t i n g e n e r a l imply
c o n s i d e r t h e group a l g e b r a of a f i n i t e group
R = F x F , where
R
'(?I,)J)
f o r some
i and
R*G
F is a
= (p,?I),
G
over a f i e l d
does n o t imply t h e primeness of
field, let
1 , E~ F.
R*G
t o b e prime
F
with
G)
dividing t h e order of
( i i ) The p r i m i t i v i t y of
and p u t
Q. = 0
is primitive.
( i ) The p r i m i t i v i t y of
charp = p > 0
over
=
G =
R.
Indeed, l e t
Then t h e c o r r e s p o n d i n g skew group r i n g o f
i s s i m p l e , hence p r i m i t i v e , b u t
R
2
b e t h e c y c l i c group of o r d e r
i s n o t prime.
(iii) I t w i l l be shown ( s e e Theorem 6.4), t h a t t h e second s t a t e m e n t of Theorem
G
CHAPTER 5
230
G.
i s t r u e f o r any f i n i t e group
2.2
However, t h e proof o f t h i s r e q u i r e s a form-
i d a b l e t e c h n i q u e t o be developed i n subsequent i n v e s t i g a t i o n s .
3. PRIME COEFFICIENT RINGS
G
Throughout t h i s s e c t i o n ,
R
i s a prime r i n g .
G-prime i f B
or
C 2 -
ideals
B1-B
A.
P
C 2 -
R
Let
R.
i n v a r i a n t i d e a l of
A
d e n o t e s a f i n i t e group and, u n l e s s s t a t e d o t h e r w i s e
A c - ii*G
be an a r b i t r a r y r i n g and l e t
be a G -
A
Following Lorenz and Passman ( 1 9 7 9 ) , we s a y t h a t for
Bi
G-invariant i d e a l s
of
R
implies t h a t
B1
is
5A
Our a i m i s t o e s t a b l i s h a b i j e c t i v e correspondence between t h e prime of
R*G
?
satisfying
f i n i t e dimensional a l g e b r a
E
n R
0
=
and t h e G-prime i d e a l s of a c e r t a i n The r e s u l t s o b t a i n e d w i l l be
over a f i e l d .
a p p l i e d i n t h e n e x t s e c t i o n i n which w e examine t h e r e l a t i o n s h i p between t h e prime
R*G
i d e a l s of
R.
and t h e G-prime i d e a l s of
We s t a r t by r e c o r d i n g two g e n e r a l o b s e r v a t i o n s , namely Lemmas 3 . 1 and 3.2 below, which a r e v a l i d f o r an a r b i t r a r y r i n g 3.1. of
R.
ideal
p
R*G
of
A
I
P
=
A =
A
R,
R
A*G
and
i s a G-prime i d e a l
t h e n t h e r e e x i s t s a prime
R*G,
i s a two-sided
R*G.
B
I
then since
i s a G-invariant i d e a l of
i s a prime i d e a l of
Suppose
Pn
then
A.
i s an i d e a l of
A. (R*G) = (R*G) - A
l e a s t G-invariant.
c ? n R.
p n R
Conversely, i f
Assume t h a t
R*G,
i s a G-prime i d e a l of
such t h a t
Note t h a t i f
I n R.
2.6.3(ii),
AB
i s a prime i d e a l of
Conversely, i f
Proof.
so i s
p
If
LEMMA.
R.
i deal of
R,
i s G-invariant t h e n by Theorem
R*G
Then, by t h e above,
a r e G - i n v a r i a n t i d e a l s of
with
P R
n R
is a t
with
Then (A*G) (B*G) = (R*G)AB(R*G)
c
(R*G) (P n R ) (R*G)
C P so t h e primeness of
B 5 P n R,
P
proving t h a t
implies t h a t
P n R
A*G
5P
i s G-prime.
or
B*G
5 P.
Thus
A EP n R
or
PRIME COEFFICIENT RINGS
A
Conversely, l e t
R
i d e a l of
A x G
and s o
R.
be a G-prime i d e a l of i s an i d e a l of
R*G
P nR
to
A.
=
J n R
and
I,J be i d e a l s of
Let
R
a r e G - i n v a r i a n t i d e a l s of
(I n R ) ( J
G-prime t h i s y i e l d s
R*G
n R)
A
A
Then
i s a G-invariant
n R
(A*G)
with
P
Z o r n ' s l e m m a , we may t h e r e f o r e choose an i d e a l
231
of
fi*G
A.
=
maximal w i t h r e s p e c t
P.
properly containing
A.
properly containing
P.
IJ
and hence
Applying
Then I
A
Since
P is
Hence
R
is
prime
as r e q u i r e d .
T
If
LEMMA.
I
Let
*,...,gn}
T = {gl,g
i
E {1,2,
G
be a s u b s e t of
g
with
R*G
wlth
=
and l e t
Assume t h a t
= 1. 1
...,n } ,
In
and
for a l l
T'
C
T
E I with
r
=
r .}
(R*T') = 0
Bi by
define
n B.
R*G
be a nonzero l e f t and r i g h t R-submodule of
I n (R*T) # 0 For each
f o r t h e s e t of e l e m e n t s of
T.
support i n 3.2.
R*T
G, w e w r i t e
i s a subset of
{r E R l t h e r e e x i s t s
fi
a nonzero i d e a l o f
R
=
C r
.;
j=1 3 3
z
Then
Bi i s
( i ) Each
f 7,.
( i i ) There e x i s t s a n a t u r a l b i j e c t i o n
(a) f i b a t ) = r f . ( a ) g i ,
for all
:
B1-
Bi
r,t E R
such t h a t
a
and
E
B 1
i s t h e i d e n t i t y map
(b) f 1
In
( i i i ) The elements of
(R*Tl
a r e p r e c i s e l y t h e e l e m e n t s of t h e form n
Proof.
Because
B
t h a t each each
i
Bi i s
I
i s a l e f t and r i g h t R-submodule
R.
i s a n i d e a l of
nonzero.
Note a l s o t h a t f o r each
and
(3'
s m a l l e r s u p p o r t and so
-
fi
- 0'
=
By t h e f o r e g o i n g , f o r each
fi
b
= a. 1
7,
i'
ensures t h a t
t h e m i n i m a l i t y of
T
=
0.
a
E
B
-
n 1 b.;. E I i=l 2 Bi by s e t t i n g f . (a)= b ..
t h e r e e x i s t s a unique 1
with
b. E B
it is c l e a r
T h i s i s so C r j g j E I w i t h ri = b.. 2 j=1 a r e two such e l e m e n t s , t h e n 6 - B ' E I i s an element of
i m p l i e s t h a t t h e r e e i x s t s a unique
fi
T
Moreover, t h e m i n i m a l i t y of
n
s i n c e if
R*G,
of
W e may t h e r e f o r e d e f i n e
f
:
B1
a
=
232
CHaPTER 5
fi,f,
By the definition of
is the identity map and
n
Conversley, it is obvious that each element of
I n (R*T) is of this form.
Moreover, each fi is clearly an additive bijection. Finally, let a €5 B
and
be as above and let r,t
CY
E
R.
Then
-
and since g, = 1, this implies that
So the lemma is true.
.
For the r e s t o f t h i s section,
R
w i l l be a prime r i n g .
In order to make
further progress, we briefly discuss a certain ring of quotients
S =
which
QoL(R)
is defined in Martindale (1969).
X of all left R-module homomorphisms
Consider the set
where
f
I?
:
A
A
ranges over all nonzero ideals of
-3
with
R
g
and
C EA n B
:
B--+
of
and with
R, A n B
We call two such homomorphisms
R equivalent if there exists a nonzero ideal C of f(c1 = g ( c )
obviously reflexive and symmetric. ideals A , B
R.
c
for all
E
C.
It is a l s o transitive, since for all nonzero
is a nonzero ideal of
an equivalence relation on the set X.
R.
For each f E
We have therefore defined
X, let [ f , A j
no confusion can arise) denote the class of the function f define an arithmetic on the set Qo(R)
I f 4 where
fg
This relation is
IgrBI
above.
(or
=
called the Martindale ring of q u o t i e n t s of
We can
[fg,BA] In this way we obtain a ring,
R.
The following five lemmas provide some basic properties of the ring For any f : A
--+
if
of these classes by
is the composition first f then g.
3.3. LEMMA.
7
R in X,
7=
0 if and only if f
=
0.
Qo(R).
In
PRIME COEFFICIENT RINGS
particular if g
B-
:
R
233
x,
is another element of
-
-
f=
then
g
if and only if
on A n B .
f = g
Then there exists a nonzero ideal B
F=O.
7=
It is clear that f = 0 implies
Proof.
that f ( B ) = 0.
5B
Now BA
R
of
F=
.
For every a E R ,
let a
E HomW,R)
‘ 3.4. LEMMA.
contained in A
The second
if and only if
a
be defined by
(2)=
za.
R
4
defined by
a
is an element of
X, then
aTf
Qo(R1
R
(i) The map
and such
and hence that f = 0.
assertion follows from the first and the fact that
f-g=o.
Conversely, assume that
and so
is prime, it follows that f ( A ) = 0
Since R
0.
C--+
ar
is an injective ring
homomorphism (ii) If f
:
A
R
-+
(if If a,b E R ,
proof.
then
7r
=
1, a b
Hence the given map is a ring homomorphism. 3.3.
Hence
a
x
every
m,
(ab)
and
.r
ar = 0 ,
If
then
a
R
E A,
C A, a-
(a,f) proving that
hence
f
is defined on R
U
“,f
(5) =
cf -
=
=
f(a),.
.
f(xa)
=
xf(a)
In view of Lemma 3.4(i), we may identify R
=
a? 3.5. LEMMA.
Let
sE
S
Asi
2
=
S = Qo(R)
As
and
s ,s 1
each
(a+b),
0, by Lemma
.
Therefore for
f(e),(x)
with its image in
this identification, Lemma 3.4(1i) tells us that for f : A
(ii) If
=
U
=
R,
E
(i) If
ar+br
0 and (i) follows.
=
(ii) For every
=
r r
=
,...
,S
=
f(a1
for all
Qo(R).
R
in
a E A,
7~
-f
With
X, Q~(R)
(1)
be as above
0 for some nonzero ideal A
of
R,
E 5, then there exists a nonzero ideal
then s = 0 A
of
R
with
in R.
(lii) S is a prime ring (iv). If ism of
S I
S.
i s
an automorphism of R ,
then a
extends uniquely to an automorph-
CHAPTER 5
234
Z(S) = C S ( R )
(v)
ti)
Proof.
f(a)
=
0
s
fore
(ii) L e t a l l the
s
Let
a
for a l l =
Z(S)
and
E
7 = 0. sl,s ,...,s
fi
is a field.
A.
E S
a
-
afi
-
f,
=
t h e n (1) shows t h a t
Then we may h a r m l e s s l y assume t h a t
1 b e a nonzero i d e a l of
S.
i
A.
Then, by (11,
fi(a)
=
E
R
A.
S meets R Let
si.
=
=
0 # As E I n R
(iv)
si
with
a r e d e f i n e d on t h e common domain
€
(iii) L e t
s
If
f v a n i s h e s on an i d e a l i n i t s domain and t h e r e -
Hence
as for a l l
A s = 0.
with
E S
nontrivially.
R
Since
'f
0#
of
R.
I,
S E
and l e t
t h e n by (ii) and (i),
Thus e v e r y nonzero i d e a l of
S.
i s prime, so i s
R
be an automorphism of
0
A
f o r some nonzero i d e a l
If
f
A-
:
R.
Then t h e map
: A'-R
d e f i n e d by
f'(a9
=
i s s u r e l y a l e f t R-module homomorphism.
From t h i s it f o l l o w s e a s i l y t h a t t h e
-
map
f'
?I--+
f(aP
g i v e s r i s e t o a n automorphism of
S
extending
8.
To prove uniqueness of e x t e n s i o n , i t s u f f i c e s t o show t h a t i f S
morphism of and l e t
A
fixing
R
elementwise, t h e n
be a nonzero i d e a l of
R
i s an a u t o -
s
To t h i s end, l e t
T = 1.
A s C_R.
with
T
Then, f o r a l l
€
S
a E A,
we
R,
it i s
have T as = ( a s ) =
Hence (v)
A(s-s')
=
s E CS(R)
Assume t h a t
clear that
T
=
T = 0 by ( i ) . Then t h e map
f
l e f t R-modules.
0 and ( i ) i m p l i e s t h a t and t h a t
a Ts T = asT .sT = s .
s # 0.
Since
s
centralizes
0} i s a two-sided i d e a l of
R
and so w e must have
By (ii), w e may choose a nonzero i d e a l
A
of
{t
:
€
AIf
Rlts
R
=
d e f i n e d by
B = f(4).,
then
f(a)
=
as
B = As = sA
R
with
As
5 R.
i s a n i n j e c t i v e homomorphism of since
s
E CS(RI.
Now t h e r e
235
PRIME COEFFICIENT RINGS
exists an inverse map g 8,
R,
E
S
and
:
R so that fg
B-
is an inverse of -1
s
we must have
E
C3(R)
7=
s.
Because
lA.
B is an ideal of s centralizes
Furthermore, since
C (R) is at least a division ring.
and therefore
S
s induces an automorphism of S which is trivial on R.
Finally, conjugation by
S.
Hence by (iv) the automorphism must also be trivial on
Consequently s
Z(S)
E
and the result follows. Qo(R)
By the foregoing, the structure of next result will indicate that Q ( R )
is very close to t.hat of
R.
Our
i s large enough to contain certain needed
additional units. 3.6. LEMMA.
Let
elements.
u be an automorphism of R and let a , b E R be fixed nonzero
If for all
r
E R,
arb then there exists a unit
=
s E S = Qo(R)
CTu br a such that
-1
s rs=T Put A = RaR, B = RbR
Proof.
TO prove that
f
b
=
as and such that
U
for all
and define the maps
f
:
is well-defined, it suffices to show that
A
-f
r
E
R
B and
Cxsayi
=
0 implies
i that Zxiby:
i
=
0.
To this end, assume that
the formula arb = braaa
and so
Tx.ayi
=
0.
Then for all
r
E
R
27'
Cxiby:
then for all
=
yields
0 since a
U
#
0 and R
is prime.
Similarly, if
Czibyi i
=
r E R we have
since they are clearly homomorphisms of left R-modules, we have
7=
sE
S
and
0,
CHAPTER 5
236
9 E S.
gp
Moreover, fg = lA and
Note that arf
is defined on R (a f)(;c) = 21
.$
Thus
=
gr,
,
lB
=
-
g
so
-1
and
= S
S
and for all
z f R we have
f(m) = xb
br(x)
=
is a unit in
S.
or equivalently as = b .
Finally, let c
E
R.
Then gc f
I3
is defined on
and for all xby tE B
we
and let A , B
be
have
-
c 7 = 'e
Thus
3.7. LEMMA.
u xbye
=
and so S
Let
-1
s cs = ,'c
Qo(R),
=
let
as required.
. R
be an automorphism of
U
Assume that f
nonzero ideals of R.
a cp(xbyl
=
:
B
A+
is an additive bijection which
satisfies
for all r , t E R
r
E
and f ( a l
R
Proof.
and a E A .
for all a
as
=
are elements of
-1
S.
q
Hence gq,f
7
E R.
=
qz
we have
Let u
and
b
f
A
is also an
B,
g are left R-module homomorphisms so that
Furthermore, since g
is an inverse of
Then g q f is defined on
(gq,S) (b)
=
ra for all
S =
f
and
f, we have
.
Let
s
=
A.
Moreover, for all P,t E R
Note also that f and
-
S, s-lrs
is a unit in
Since f is a bilection, its inverse g : B-+
additive bijection.
g = s
e
7
s =
Then
0
=
f(g(b)ql
and this yields
f(a)
= as
for all
be an automorphism of
=
-1
s qs a
R.
B and for all b E B ,
(gf) (bq') =
=
bq'
=
we have
qz(b)
qa for all q
R.
Finally, because
A. Following Kharchenko (19751, we say that
is X-inner if it is induced by conjugation by a unit of
words, these automorphisms arise from those units s
E S
S = Q,(R).
with
s-lRs = R.
In other If
231
PRTME COEFFICIENT RINGS
s
s
and
a r e t w o s u c h u n i t s , t h e n c l e a r l y so i s
2
s1s2.
Invoking L e m m a 3.5
R
( i v l , we s e e immediately t h a t t h e s e t of a l l X-inner automorphisms of Aut ( R )
normal subgroup of
R*G
Now l e t Then
G*
R
G*/U(R)
Bearing i n mind t h a t
{ g E Glr
=
G* be t h e g r o u p o f g r a d e d u n i t s o f
i s a normal subgroup of
t--t gi-
G
w e deduce t h a t
G.
r
k-+
'r
{gig
be t h e f r e e S-module f r e e l y g e n e r a t e d by
E GI
R
of
which w e d e n o t e by t h e same symbol.
S
R1
i s an X-inner automorphism o f
Owing t o Lemma 3.5 ( i v ) , t h e automorphism automorphism of
P
R*G.
U ( R ) s u r e l y g e t a s X-
by c o n j u g a t i o n and t h e e l e m e n t s of
i n n e r automorphisms.
Ginn
.
b e g i v e n and l e t
a c t s on
is a
e x t e n d s t o a unique
We now d e f i n e
S*G
to
and w i t h m u l t i p l i c a t i o n g i v e n
by
(a5( b y ) = a"bEa(33,ylrcy for
a,b
E
S
x,y
and
H e r e o f course
E G.
c1 : G
--
a(r,y) = "y
G
3.8.
S
over Let
LEMMA.
E
=
Cs*,(S),
t w i s t e d group a l g e b r a of
H
(iii) I f
S*G
S*G.
L
If
(S*G)Ln
s
Proof.
F
5 S*Ginn Ginn
,
S*G. =
,
S*G i n n = S €3 F E
and
some
F.
E,
and
E
then
f3
t (S*H) = F H
L ( S * G ) = (S*G)L
( S * G ) L c o n s i d e r e d as a l e f t S-submodule of Furthermore,
t
E = F Ginn,
then
(E n (S*H))
i s a G - i n v a r i a n t i d e a l of
Moreover,
summand o f
= S €3
i s i n f a c t a unique c r o s s e d
S*G e x t e n d i n g R*G
over the f i e l d
'inn
i s a subgroup of
S*H (iv)
E
then
i s d e f i n e d by
F = Z(S).
and l e t
(i) There e x i s t s a unique c r o s s e d p r o d u c t (ii) I f
U(R)
R*G.
extending
S = Q (R)
-+
q
The f i r s t p a r t of t h e lemma below shows t h a t p r o d u c t of
x G
--1
(S*G)L n (S*Ginn) = SL
and i f
S*G
i s a n i d e a l of is a direct
L # E,
then
0. ( i ) B y Lemma 3 . 5 ( i v ) , t h e automorphisms of
R
extend uniquely t o
2 38
CIIAPTER 5
S.
automorphisms of R*G.
extension of
R
S*G is the only possible
It therefore suffices to verify the associativity of the
Let G’
multiplication. acts on
Hence, the above definition for
Note that G*
R*G.
be the group of graded units of
and therefore the uniqueness of extension implies that we obtain a
G* on S .
group action of
This fact will be used implicitely in the computa-
Given a , b , c
tions below.
E
S and x,y,z E G, we have [
(&I ( b y )I (cz)
=
-
s 1 xyz
and
s l , s z E S.
f o r some
An
Ccyz
[ ( b y )( G ) I= s 2
c&)
easy computation shows that
and s
where
r
and r
and
case when
a
Hence
s
=
b
=
c
and 2
(ii) Let Y E E (i),
(ii),
=
a,b
and c.
StG
Since
a(x,y)xyca ( x , y )-1 But in the special
the above products belong to R*G
1
and so r
= 1
r
2
.
is associative.
-
and let x E Suppy, say y = sx
there exists a E R
commutes with
independent of
c terms in the two expressions are equal.
= S 1
=
axb x P c ) r 2
R
are elements of
z(yc?l the a , b ,
=
R*G
ccy E
with
+
... .
and with
3:
Owing to Lemma 3.5
E Suppay.
Since y
r a , we have aray
=
ayra
for all
P E
R
We conclude therefore that
-
arasx Since b
=
as
=
-
asxra
is a nonzero element of
R,
=
as xrxax
we see that the identity
arb = bxyxa holds f o r all r E R.
Invoking Lemma 3.6, we see that
r
x~
is an X-inner
239
PRIME COEFFICIENT RINGS
R and thus
automorphism of
g
For each on R
E
and put
3:
E
This proves that E
Ginn.
Ginn choose a unit s =
for all g
s-ls.
Then the elements g The elements g, g E Ginn
basis for S*Ginn.
2
Applying Lemma 3.5(iv), we see that E
E
5 S*Ginn.
is clear that each S n
%E
E
=
E.
Yg?
centralizes S.
of
$
Since
2
E
.
E
with Y E S and it g g is a unit of E, we have
CY
form an F-basis for
E.
More-
E is an associative F-algebra with
Moreover, for z , y E Ginn
, %;€
E
and
2;
=
s
6
for
t is isomorphic to F Ginn, some twisted group
It follows that E
S.
E
=
s, i.e.
C_ Z ( E ) we must have S*Ginn = S @F E .
{ZIg E Ginn}.
s
some
Y
Then we can write
To prove the last assertion, note that basis
To this end,notethat each
must centralize all of
F and we deduce that the elements
since F
over,
Ginn form an S-
S*G which acts by conjugation on S centralizing all of R .
is a unit in
Assume that y
E
are obviously F-lineary indepen-
dent and we claim that they form an F-basis for
z
'r
S inducing the automorphism r
E
g
5 S*Ginn.
over the field F .
Ginn
(iii) Direct consequence of the way the algebra FtGinn is constructed. (iv) Let LS = S L
L
be a G-invariant ideal of
and the G-invariance yields
follows that
L
=
5 ,;
Then, by definition of for all g
SL
€
in E.
Since
S*Ginn = S @ E
=
G.
E, we have
It therefore
S*G.
is an ideal of
Y be a transversal for Ginn in G and let L'
Let for
L ( S * G ) = (S*G)L
E.
SL 0 S L ' ,
be the F-complement
we have
F S*G =
c
c
SLY @
YEY
SL'y
YEY
Furthermore, since
L
which implies that
Z (S*Ginn)LY = C SLY fiy yf y (S*G)L is an S-direct summand of S*G.
is G-invariant, (S*GIL =
Z SLY is
Moreover,
L@y
clearly a direct sum and thus
If
L # E,
But then
then we can choose L '
1 SL'y fi7
contains S
above to contain the identity element
and therefore
(S*G)L n S
=
1.
0, as required,
'
240
CHWTER 5
From now, w e p u t S
E = CS*,LS)
= Q0(R),
F = Z(S)
and
O u r aim i s t o a p p l y t h e f o r e g o i n g r e s u l t s i n o r d e r t o e s t a b l i s h a b i j e c t i v e c o r r e -
P
spondence between t h e prime i d e a l s
E.
prime i d e a l s of
R*G
of
satisfying
P n R = 0 and t h e G-
G a c t s on E by t h e r u l e
R e c a l l t h a t t h e group
-1
gx=gxg
, r E E , g € G
A
and t h a t , by d e f i n i t i o n , a G - i n v a r i a n t i d e a l f o r G-invariant
L
If
ideals
Ai
E
of
of
E
A
C 1 -
implies t h a t
i s a G - i n v a r i a n t i d e a l of
A A
i s G-prime i f
A
or
A
C
2 -
1
2
A
C -
A.
then w e set
E,
L~ = L(S*G) n R*G L'
so t h a t
R*G
i s an i d e a l of
by Lemma 3 . 8 ( i v ) .
I
For any i d e a l
R*G,
of
we s e t
P P
Observe t h a t
A y
with
C 1 1 -
zero ideal
=
2
of
5I
f o r some nonzero i d e a l
E.
is a G - i n v a r i a n t i d e a l of
I, A y
B
{y E E l A y
C
I
and l e t
t E E.
A
of
Rj.
Indeed, assume t h a t
€Id YlIYL
By Lemma 3 . 5 ( i i I , t h e r e e x i s t s a non-
2 -
R with
B t &R*G.
and
Because
t centralize R,
YlrY2
we have
and
B A (ty 1 1
y1
Hence
+ yl,ylt,tylE
#
so
(Bt)( A y
=
1
9
)
1 1
i s an i d e a l of
5I E
which i s c l e a r l y G-invar-
iant.
Let that of
h*G
I R , Ax
I be an i d e a l of
R+G.
Following Lorenz and Passman ( 1 9 7 9 a ) , w e s a y
i s R-cancelable i f f o r any
EI
satisfying
implies t h a t
P nR = 0
x E I.
x
E
R*G
and any nonzero G - i n v a r i a n t i d e a l A
I t is obvious t h a t any prime i d e a l
i s R-cancelable.
P
of
241
PRIME COEFFICIENT RINGS
3.9. LEMMA.
Let
L,L
and L 1
(i) L'
E.
be G-invariant ideals of
Then
2
is R-cancelable
(ii) L'*L" 1
2
c (L~L~)'
-
(i) Assume that AX c - L"
Proof.
variant ideal A
of
R.
(S*G)L
Lemma 3.8(iv),
for some x E R*G
Lu,
By definition of
and some nonzero G-in-
we have Ax
is a left S-module direct summand of
for some S-submodule K
of
S*G.
5 (S*G)L. S*G and so
x = x 1 + x 2 with x
Writing
Owing to
E
(S*G)L,x2E Ic
we have
proving (i). (ii)
(S*G)Li = L i ( S * G )
By Lemma 3.8(iv), we have
as required.
and hence
.
The next lemma is crucial. 3.10. LEMMA.
With the notation above, we have
E, L
(i) For any G-invariant ideal L
of
(ii) If I is an ideal of
then
Proof.
(i) Given
some nonzero ideal R therefore L
A of R.
5L . Ud
Hence
y of
R*G,
E L,
R.
=
I s
L
lid
flu.
we have by Lemma 3.5(ii) that By Then B y C L(S*G) n R*G = L",
Conversely, if
y E LZAd,
then Ay C_ L'
so
5 R*G y E LUd
for and
for a nonzero ideal
242
CHAPTER 5
by Lemma 3 . 8 ( i v ) .
y
=
L'
Denote by
y, + y 2 w i t h
y1 t L
Y
and
E L'.
A(y-Y
(ii) Fix
3:
E
i n d u c t i o n on
I. The c a s e Suppl.
YE I
elements
if
y E Y',
Now 2'y
n SL'
E SL
ISuppz = 0
Choose
I n R*T # 0.
= {g
=
1
l,g2,...,g A
There e x i s t nonzero i d e a l s :
Bi
A-
=
R
,...,Rn
,B2
1
.
1 E Ti1
Siii
E
R
f.(a)
and
= as i
-
=
x;
S i n c e it s u f f i c e s t o
T
and
R
of
Ti'.
by
and a d d i t i v e b i j e c t i o n s
R.
t r i v i a l l y on
0 E B.
Lemma 3 . 2 ( i i i )
R.
s
z.ii
that for a l l
a
such t h a t
- --1 = girgi
=
.s'
=
a
Id.
E
asi
for a l l
-
S
Sigi
a
E A,
E
A.
E
=
ax
f,
gi
r
is
x
-
aOr
and l e t
a
acting
F:
= CS*G(S)
i t f o l l o w s from
cf;(a)giE I
be t h e i d e n t i t y c o e f f i c i e n t of
Y
S-lrs=
r E R, we see t h a t
for a l l
E A
z i
A,r,t E R
1.
=
fi(a)
Moreover, s i n c e
E
Furthermore, s i n c e
Invoking Lemma 3 . 5 ( i v ) , w e conclude t h a t
T h e r e f o r e , by d e f i n i t i o n , w e have
r
r
E A.
S
of
t h e n y i e l d s an automorphism of
aa = Cas
Let
7
=
S
Since
But
a
a
for a l l
s1 ,s2,...,s
for a l l
g;
1sI,. gz. E S*G.
centralizes
and t h u s
Indeed,
-1
r f i ( a ) "t
=
there exist units
t h e i n d e n t i t y f u n c t i o n , we have
B
T.
1 E
satisfying
Hence, by Lemma 3 . 7 ,
Let
by
minimal w i t h
g.
r
@
t h e n Lemma 3 . 2 a p p l i e s and we employ i t s n o t a t i o n .
f(rat)
for a l l
E
then
p,
fi
7 ' C _ Suppx
W e may assume t h a t
a l s o h a s t h i s minimal p r o p e r t y and -1 show t h a t xy E w e can r e p l a c e x by
T
x
# 0 and t h a t t h e r e s u l t i s t r u e f o r all
3:
-1
If
0
=
b e i n g t r i v i a l , w e show t h a t
of s m a l l e r s u p p o r t s i z e .
respect t o the property t h a t
and w r i t e
L.
E
Assume t h a t
B
in
Then
Ay
=
)
y = y
and Lemma 3 . 5 ( ) i m p l i e s t h a t
L
a n F-complement f o r
Put
PRIME COEFFICIENT RINGS
Because
Suppa
y
Furthermore,
6E
I?
if
D =
=
I since
E
aRr
and
T C- Suppz, 1
n gA,
E
R*G
uar
D
s
and
w e see t h a t
= 1,
ax
and t h u s
6
i s a nonzero G - i n v a r i a n t
SfG
R
i d e a l of
.
r e s u l t now f o l l o w s by i n d u c t i o n .
y
p.
E
for a l l
E
i s R-cancelable and hence
But, by Lemma 3 . 9 ( i ) ,
)SuppyI*)Supp).
I, so by i n d u c t i o n we have
E
Ul?
so
then
!7'
243
x
E
u E A.
Cx
with
Idu.
Note t h a t Hence,
5 Id".
The d e s i r e d
We have now come t o t h e d e m o n s t r a t i o n f o r which t h i s s e c t i o n h a s been d e v e l oped. 3.11.
THEOREM (Lorenz and Passman ( 1 9 7 9 a ) ) .
G
t h e f i n i t e group i z e r of
S
=
R
o v e r t h e prime r i n g
Qo(R) i n
S*G
where
R*G
Let
and l e t
F = Z(S).
Let
P
G-prime i d e a l of
E.
Proof.
(i)
Pa i s
( i i )'L
P
E
Pa'.
E*G
with
E
=
E ii*G
and
R*G
D =
n
8".By Lemma and
x
Lu n R 3.1O(iilI
(S*G)Fd
E
L
* L'.
=
0, and l e t
and
L
=
L be a
=
0
P
we have
=
I: 0.6 i
6.D
t h e r e e x i s t s a nonzero i d e a l
5P B
for all of
R
ud
5 Pdu.
D
i.
Assume t h a t
Di of
R
with
S*G and 6
i
E
Moreover, by Lemma 3 . 5 ( i i ) ,
B
BB. C R*G z-
for a l l
i s G-invariant.
Pa
0.6 . = & . D S P. a z zz-
is a nonzero G - i n v a r i a n t i d e a l of
such t h a t
t h e f o r e g o i n g argument we may a l s o assume t h a t
ai E
with
i=l
it follows t h a t
L
so
Pd, t h e r e e x i s t nonzero i d e a l s n n gDi,
&G i=1 t h e prime r i n g R w i t h
sion is that
P n R = 0
with
P = Pdu
with
x
Setting
yields a
P nR
n
By d e f i n i t i o n of
P +-+ Pa
W e must prove t h a t
P
x
R*G
be a prime i d e a l o f
i s a prime i d e a l o f
Then
of
be t h e c e n t r a l -
The i n v e r s e of t h i s map i s g i v e n by
a G-prime i d e a l of
W e f i r s t show t h a t
x
E.
t
E = F Ginn
Then t h e map
b i j e c t i v e correspondence between t h e prime i d e a l s and t h e G-prime i d e a l s of
be a c r o s s e d p r o d u c t of
i
and by
The conclu-
244
CHRPTER 5
P
But
x
Thus
P
i s prime and E P,
Pdu
so
T o prove t h a t
pd
M, ,M2
E.
ideals
of
does n o t c o n t a i n e i t h e r
5P
i s G-prime,
Mi
E
Since
Pd i s
Hence
C
# L
( M1 2M ) ~ c _ ~ Z ~ = P cP
M?
L'
L
maximality of
C
R*G.
I
f o r some i d e a l
I
nR
Lu
P
=
E
F,
.
0,
r e s u l t follows.
L
we see t h a t
w e conclude t h a t
ud
Then
I
. 8 - Lud
2&=
Lu.
L
=
3
and
Hence t h e
G-
I n R # O.
and s o
c 11,L' c I
L'
i s prime, assume t h a t
Then, by t h e above,
=
L = L
R*G.
of
II n R # 0
R # 0,
I2
and
f o r some i d e a l s whence
5 1112n R
0 # (Il n 81 (I2n R ) L'
a n d , by Lemma 3 . l O ( i ) ,
G-prime and ( i ) i s e s t a b l i s h e d .
and by Lemma 3 . 1 0 ( i ) , we have
implies t h a t
To prove t h a t
Because
i
f o r some
2 -
s i n c e o t h e r w i s e Lemma 3 . 1 0 ( i i ) would y i e l d
11,12 of
f o r some G - i n v a r i a n t
1 2 -
Furthermore, it i s a n immediate consequence of Lemma 3.8
nR = 0
'L
Assume t h a t
d I
c Pd
M M
i s a finite-dimensional algebra over t h e f i e l d
i s i n f a c t G-maximal. (iv) that
P n R = 0.
since
Then, by Lemma 3 . 9 ( i i ) ,
P i s prime, t h i s y i e l d s
Cd - Pd.
=
D*G
p.
=
assume t h a t
MU#C 1 2 -
Because
P
and t h e r e f o r e
or
B*G
I I 9 Lu.
Thus
1 2 -
Lu
i s prime and t h e
We c l o s e t h i s s e c t i o n by p r o v i d i n g two consequences of Theorem 3.11. 3.12.
Let
COROLLARY.
prime r i n g an i d e a l of Proof.
R.
If
R*G
R*G
P
i s a prime i d e a l of
properly containing
By Theorem 3 . 1 2 ,
P
=
L'
P,
PROPOSITION.
t h e prime r i n g
Let
R*G
R*G
with
P nR = 0
if
I
and i f
is
I n 8 # 0.
then
d
L = P
f o r t h e G-prime i d e a l
t h e argument i n t h e proof of Theorem 3.12, 3.13.
G over t h e
be a c r o s s e d p r o d u c t of t h e f i n i t e group
I
3
P
= 'L
then
I
.
f?
Hence, by
R # 0.
be a c r o s s e d p r o d u c t of t h e f i n i t e g r o u p
G
over
R.
( i ) A prime i d e a l
P o€ R*G
i s minimal i f and o n l y i f
( i i ) There a r e f i n i t e l y many s u c h minimal p r i m e s , s a y
P
f-
Pl,P2,
R = 0.
...,P
and i n f a c t
245
PRIME COEFFICIENT RINGS
(iii). S = P n P 1
2
n
... n Pn
is the unique largest nilpotent ideal of R*G.
In
fact
J where
=
E.
J(E) is the Jacobson radical of Since E
Proof.
I 'inn I = o ,
and J
J(E)U
is a finite-dimensional algebra over a field, there are
only finitely many, say
...,Ln
,
Ll,L2, n
of G-prime ideals of
dimgtGinn = IGinn/ =
E.
Then clearly
/GI
and J(E) =
,
If Pi = L:
R*G
P
then by Theorem 3.11,
n nL
i=1 i
,...,Pn
are the unique prime ideals of
Jd 5 L':
=
which are disjoint from R - (01. Put J
=
P.. 2
i=1
Then
J
5 Lui
so
,
n
C
Invoking Lemma 3.lO(ii), we see that J
Jdu 5 J ( E ) ' .
pi
Since
J ( E ) U and
we conclude, from Lemma 3.9(ii), that both other hand, each
Li by Theorem 3.11, and thus
J
J(E)
is nilpotent
are nilpotent.
surely contains all nilpotent ideals of R*G.
On the
Thus
n J = np. i=l 2 and therefore J = J(E)' IG i n n Furthermore, since S(E)
contains J(E)' of
R*G.
1 'inn 1 J
=
is clearly the largest nilpotent ideal
I
0
=
(see Theorem 2.3.16)'
we have
0.
Finally, let P be any prime ideal of
Since J
R*G.
is nilpotent,
n
and hence
P 3
Pi for some i.
ma1 members of the set
{Pl,
Hence the minimal primes of R*G
...,Pn}.
However, P . 3 P 2 -
L . = Pci'3 2
and hence, since L PI,
...,P
i
2 -
pciJ
=
implies that
Lj
is G-maximal, we must have
are precisely the minimal primes of
j
are the mini-
i
R*G
=
j.
This proves that
and the result follows.
.
CHAPTER 5
246
4 . INCOMPARABILITY AND GOING DOWN
R*G
Let
R
We s a y t h a t
Thus
R
R.
over an a r b i t r a r y r i n g
A,B
i s G-prime i f f o r a l l nonzero G - i n v a r i a n t i d e a l s
AB # 0.
have
G
be a c r o s s e d p r o d u c t o f a f i n i t e group
0
i s G-prime i f and o n l y i f
is a
R,
of
we
R.
G-prime i d e a l of
The problems t h a t m o t i v a t e t h i s s e c t i o n are t h e f o l l o w i n g IncornparahiZity.
If
P cP 1
a r e prime i d e a l s of
2
P
Down.
Going
P
n R
=
A
? nR = A ?
R
of
d o e s t h e r e e x i s t a prime
2'
does it f o l l o w t h a t
n R c p 2 n R ?
Al c A2
Given G-primes
R*G,
PI
and of
a prime
R*G
P2
of
satisfying
R*G
? C 1
P
with 2
o u r aim i s t o p r o v i d e a p o s i t i v e answer t o b o t h problems.
and Even
more, it w i l l be shown t h a t f o r I n c o m p a r a b i l i t y t h e primeness of t h e l a r g e r i d e a l
P2
i s unnecessary.
P
and i f
PcI
Indeed, it w i l l be proved t h a t i f
a r e i d e a l s of
R*G
i s prime, t h e n
The f o l l o w i n g two simple o b s e r v a t i o n s w i l l j u s t i f y our r e s t r i c t i o n t o G-prime coe f f i c i e n t rings. 4.1.
The f o l l o w i n g c o n d i t i o n s a r e e q u i v a l e n t .
LEMMA.
P C I a r e i d e a l s of
(i) I f (ii) I f
p C
I are ideals
R*G R*G
of
P prime, t h e n
with
n R# InR
R a G p r i m e and
P prime,
with
?
n R
= 0,
I n R # 0.
then
p flR = 0 i m p l i e s t h a t
(Note t h a t , by Lemma 3.1, Proof. 2.6.3(ii),
Note t h a t
p
f'
R
i s &prime).
R i s a G i n v a r i a n t i d e a l of
( p n R ) * G i s a graded i d e a l of
R.
Hence, by Theorem
R*G so t h a t
(P n R ) * G
I t f o l l o w s t h a t i n d e a l i n g w i t h ( i )w e may f a c t o r o u t t h e i d e a l
R*G
and t h u s reduce t o t h e c a s e
4 . 2 . LEMMA.
If
P
R
PnR
=
0
of
which i s t h e c o n t e n t of ( i i ) .
The Going Down problem i s e q u i v a l e n t t o t h e f o l l o w i n g s t a t e m e n t :
i s a G-prime r i n g and
P
a prime i d e a l of
R*G,
then
P
c o n t a i n s a prime
INCOMPARABILITY AND GOING DOWN
Q
ideal
of
R*G
Proof. that
Q nR
with
=
241
0. At
D i r e c t consequence of t h e assumption
i s G - p r i m e and t h e f a c t
(R/R1)*G.
(R*G)/(A1*G)
F o r t h e r e s t of t h i s s e c t i o n , w e s h a l l assume t h a t R
i s a G-prime r i n g
Note t h a t t h e above c o n d i t i o n i s s a t i s f i e d i f t h e r e e x i s t s a prime i d e a l
n gQ
such t h a t
0.
=
A A 1
=
2
A.,
For i f
A ,A
A A
0, then
1
we i n f e r t h a t
2
A.
cQ
so
C
fl gQ
' --&G
2
A.C Q
LEMMA. ( i ) R
i.
f o r some
-L-
and hence t h a t
Ai
c o n t a i n s a prime i d e a l
Q
Using t h e G-invariance of
0.
=
R
i n g o b s e r v a t i o n shows t h a t i n any G-prime r i n g 4.3.
R with
a r e G - i n v a r i a n t i d e a l s of
1
6 G
R
of
Q
P a r t ( i l of t h e follow-
one can f i n d such a prime
n gQ
with
=
0.
Q.
In particular,
&G
R is
semiprime.
(111
Any prime i d e a l of
R
contains a conpgate
a r e p r e c i s e l y t h e minimal primes of (iii) L e t
H
be the s t a b i l i z e r of
h i l a t o r of
Q
in
R
(by Lemma 1 . 4 ,
gQ
of
Q
{gQlg
and s o
E G)
8. Q
in
G
and l e t
N = ann(Q) R
be t h e anni-
l e f t and r i g h t a n n i h i l a t o r s of
are equal).
Q
Then
and
f o r all (iv)
g
E G-H.
A
If
i s any nonzero i d e a l of
R
with
A C_N,
then
annA = Q.
R
(il
Proof.
Q of
e x i s t s an i d e a l Assume t h a t
G
Because
A A
C Q
B
and
B
f o r some i d e a l s
A ,A 1
of
R
B.
and p u t
2
a r e G - i n v a r i a n t i d e a l s of
=
a
R
and s i n c e
B B
R
i s G-prime,
so
Bi
=
=
0.
qEG
C
1 2 -
But
n gQ
R maximal w i t h r e s p e c t t o t h e p r o p e r t y t h a t
1 2 -
Then
is f i n i t e , it f o l l o w s from Z o r n ' s lemma t h a t t h e r e
n gA.,i 6 G
A"A 1
2
,
= 1,2.
we have
0 f o r some i E {1,2} and t h e n t h e maximality of
Q
CHAPTER 5
248
A.
implies t h a t
=
Q.
Q
Hence
R
(ii) Any prime i d e a l of
R.
0 = n 'Q and hence, s i n c e G SfG Moreover, t h e r e are no i n c l u s i o n r e l a t i o n s between
certainly contains
I
'Q.
i s f i n i t e , c o n t a i n s some
1 s a prime i d e a l of
n E GI.
t h e primes
n
taking
Q
For, i f
C
'Q,
then
NQ = 0 ,
(iii) Since
t h e n by (ii),
w e have
N C_ gQ
NQ C - 'Q
for a l l
N
C_ n ' Q , . &H
and t h e r e f o r e
for all
>
n
1,
so by
Q c Q.
/GI w e o b t a i n t h e c o n t r a d i c t i o n
=
Q
Q C
g E
G.
g E G-H,
Hence, i f
Conversely, because
w e have
N = annQ
2
n
'Q ,
&H N =
proving t h a t
n 'Q.
Q
I n any c a s e , by ( ii)we have
N
g
9 H,
then
N
C
'Q
and so
N
P 0.
n $'Q = R .
N =
&H
Furthermore,
~ n g n= g g = o SEG
-1
and i f
H = G, t h e n by d e f i n i t i o n
Note t h a t i f
@=
so
'N
5 Q.
Hence f o r
g 9 H
we have
and hence
(iv)
If
A
5N
If
A
# 0
implies t h a t
3
5 Q.
A R = 0.
AQ = 0, so 4 5 a n d .
then c l e a r l y
A
then
Thus
Q,
Q
since
= annA
Conversely, assume t h a t
N n Q = 0 by ( i i i ) . Hence
AB = 0
and t h e r e s u l t f o l l o w s .
For t h e rest of t h i s s e c t i o n , t h e f o l l o w i n g n o t a t i o n w i l l be used
Q
i s a minimal prime of t h e G-prime r i n g
N
= annlQ)
H
i s t h e s t a b i l i z e r of
R
R M
=
4.4.
C 'N
(so t h a t
LEMMA.
( i ) Let
V
M
Q
in
G
i s a nonzero G - i n v a r i a n t i d e a l of
R)
With t h e n o t a t i o n above, t h e f o l l o w i n g p r o p e r t i e s hold: be a nonzero r i g h t R-submodule
of
NZ
and l e t
T = tg, ,g2,.
..,gnl
5Q
249
INCOMPARABILITY AND GOING DOWN
be a subset of
G with g
=
Assume that
1.
1
for all 2"
c T.
(ii) Let I
T
Then
V
n R*T
# 0
V
but
n R*T' = 0
5 H.
be an ideal of R*G.
Then there exists a nonzero G-invariant ideal
E of R with EI c- %J(I n R + H ) E Proof.
!I
C_ M?
o#
(i) By hypothesis, there exists
r
we have
i
E N
for all
T ensures that all r . # 0.
i
E
Given
{1,2,.
n 3: =
..,nl
z r z. g- i
E V
i=l
fi
R*T.
Because
and the minimality condition on Er; Q, we have xq = ri qgi E V and
z
q
i=lg. rlq
E
NQ = 0.
Thus the minimality condition on -1
and we have r g i Q = 0, 1
i
< i S n.
Thus
gir.Q
T implies-lrizq =
0, so 9i r i
0 for all i
=
E N = ann(Q)
and
therefore, since r . E N, we have
i is in H, as required. Since I is G-invariant, we have
By Lemma 4.3(iii), we conclude that each (ii) First assume that NI = 0.
MI and so we may take Observe that right ideal of
V
R*G.
E
=
M
= (
V
Let
N.
V
o = ZV(I n R*H)Z We may therefore assume that V = N I # 0. is a
X be the set of all subsets i" of G such that
is a right ideal of
empty collection of subsets of
Since
~M)I =
satisfies the hypothesis of (i) and, furthermore, V
V n R*T # 0, V Since
c
gfG in this case.
g
R * T 1 = 0 for a11 !?'C T R*G,
G.
is an R-subbimodule of
it is easy to see that
X is a finite non-
Moreover, by (i), ! ? & H
NF, AT
1E T
and
for all
is a nonzero ideal of
R
T E X.
contained in
Put
D= n A EX Because X is finite and A C N for all T TD # O.
E
X, we have by Lemma 4.3(iii) that
CHAPTER 5
250
I t w i l l now be shown t h a t
gm+lz c - E N ( I n R*H)E
m
The c a s e
Y
E
V
E V
0 being t r i v i a l , we u s e i n d u c t i o n on m.
=
ISupplcI = m
given with
x
for
m
and
5 Suppx
Choose
T
1E
Tg-',
Of c o u r s e ,
r i g h t i d e a l of
R*G.
If
g
then
Tg-l
E
T,
Suppxi
-1
= (Suppx)g
a l s o has t h i s minimal p r o p e r t y s i n c e
is
3 5''g-l
V =
and
NI i s a
Because w e need o n l y show t h a t
$r+l --l
xg
we can r e p l a c e
E I/
minimal w i t h r e s p e c t t o t h e
-1 R*T # 0.
V
x
So assume
(1)
0 and suppose t h e a s s e r t i o n h o l d s f o r a l l e l e m e n t s
of s m a l l e r s u p p o r t s i z e .
property t h a t
= /SuppxI
--1
x by xg
T
and
c - ZN(r n R*H)E
by
Tg
-1
.
Thus w e c a n assume t h a t 1 E Suppx
T E X.
and
c
Let
trx
=
d e f i n i t i o n of
and, s i n c e
AT,
SuppU
x and l e t d
be t h e i d e n t i t y c o e f f i c i e n t of
6E V n
w e may choose
5 suppx
and
R*T
with
Y
=
&-Be E V
try
=
0
trR
I SuppY/
we have
=
E
d.
n-th direct power of
SUPP
support of
A
the g-component of
9
R*G
V
v
2
A
the crossed product of
G
over R
R
342
Notation
the unit group of A =
19 E G ~ P + g r is an X-inner automorphism of R ) u
the degree of
graded units of A the inner automorphism induced by
u
8 Ax x%N crossed system
=
=
@A
5supporting subgroup of
x
the natural projection projective crossed representation irreducible characters of
N
over F
X
the stabilizer of coinduced module restriction of =
c
'1
&G
V
R*H
to
the smallest G-invariant ideal of R contajlning I
trace map equivalence class of A
degree of
Brauer group of index of
A
F
A
subgroup of are split by
consisting of those
BY(?')
F ~ G
twisted group algebra of
FG
group algebra of
Hom(M,N) A g Grnom (M,N )
A Qo ( R )
g
1.41
which
E'
G
G over F
over F
R-submodule of Hom(M,N) consisting of all graded homomorphisms fromAM to fl Martindale ring of quotients of
R
343
Notation
Cohomology Theory
t h e group of a l l A-valued 2-cocycles of
G
a coboundary
t h e subgroup of =
Z2(G,A)
Z2 ( G , A ) / B 2 ( G , A )
t h e cohomology class of =
c o n s i s t i n g of a l l coboundaries
f E Z2(G,A)
2’ ( G , A ) /B’ (G,A)
t h e c o r e s t r i c t i o n map t h e r e s t r i c t i o n map o b s t r u c t i o n c o c y c l e corresponding t o
F i e l d Theory
f i e l d extension
(F
:
K)
F
d e g r e e of
K
over
G a l o i s group of
F
over
t h e smallest s u b f i e l d c o n t a i n i n g
S
and
t h e smallest s u b f i e l d c o n t a i n i n g
K
and
t h e a l g e b r a i c c l o s u r e of
N~/~(?”)
K
norm of
1-1
over
P
F
K 1
,...,a
This Page Intentionally Left Blank
345
Index
Action, 25 faithful, 26 Algebra, 10 central, 1 5 1 definable over a subfield, 3 2 0 enveloping, 1 5 4 isomorphism of, l o R-free, 10 similar, 162 skewfield component of, 162 splitting field of, 163 tensor product of, 10 Algebraic closure, 27 algebraic element, 27 almost centralizer, 286 Arnitsur, 8 9 annihilator, 53 of induced modules, 137 automorphism system, 63 Bergman, 3 1 0 bimodule, 7 Bourbaki, 18, 25 Brauer, 159 Brauer classes, 1 6 2 Brauer group, 162 canonical homomorphism, 11 canonical injection, 1 2 7 centralizer, 3 characterization of, artinian modules, 19 flat modules, 1 8 induced modules, 1 2 7 injective modules, 1 4 Jacobson radical, 5 4 noetherian modules, 2 0 projective modules, 12, 13 semisimple artinian rings, 57 simple artinian rings, 57 strongly graded algebras, 184 Clifford theorem for strongly graded algebras, 188
coboundary, 3 8 Cohn, 27 cohomology class, 3 8 cohomologous cocycles, 3 8 commutator, 3 commutator subgroup, 3 conjugate of a module, 1 8 6 core, 88 corestriction map, 4 4 criteria for algebra to be crossed product, 7 1 algebra to be skew group ring, 77 A / J I A ) to be separable, 3 2 0 complete reducibility of induced modules, 1 3 9 crossed product to be artinian, 7 9 crossed product to be noetherian, 7 9 crossed product to be simple, 9 1 direct sum to be artinian, 22 direct sum to be noetherian, 22 element to be in J ( R ) , 5 5 equality of prime ranks of B and
RG,
275
equivalence of crossed products, 7 2 existence of faithful projective representation, 3 2 6 existence of maximal submodules, 2 0 existence of irreducible submodules, 2 0 extendibility of modules, 2 0 0 extendibility of regular module, 2 0 4 extension to be Galois, 28 extension to be normal, 2 8 extension to be simple, 28 extension to be split, 45 field to be perfect, 28 finite decomposition, 2 2 finite irreducible decomposition, 24 @G to be central simple, 321, 323 to be semisimple, 314, 315 F"G to be separable, 3 1 9 @G to be simple, 323 P P to be commutative, 315 induced module to be indecomposable, 213
346
Index
[criteria for] isomorphism of extensions, 202 G-grading, 60 isomorphism of induced and G-module, 26 coinduced modules, 133 Cset, 25 module to be completely reducible, 24 g-component, 60 module to be flat, 16 general semilinear group, 95 module to be irreducible, 52 Going down, 246 module to be projective, 17 Going up, 259 module to be of finite length, 25 Going up theorem, 266 prime ideal to be maximal, 226, 276 graded homomorphism, 119 prime ideal to be minimal, 244, 256 graded ideal, 119 ring to he matrix ring, 46 graded unit, 61 ring to he regular, 313 Group, 2 R*E to he primitive, 229 cyclic, 3 R*G to have unique prime ideal, 258 direct product of, 4 RG to be semisimple, 310 direct sum of, 4 submodule to be fully invariant, 50 exponent of, 3 ZCFGl = F, 321 extension, 28 Galnis 27 Dade, 121, 184, 193, 195, 200, 202 locally finite, 85 degree of extension, 27 multiplicative, 2 degree of unit, 61 n-divisible, 41 DeMeyer, 323 p-component of, 41 diagram, 2 of finite exponent, 42 commutative, 2 of symmetric type, 327 domain, 5 restricted direct product of, 4 dual basis lemma, 17 splitting of, 3 torsion, 3, 42 endomorphism ring, 6 torsion-free, 3 equivalent crossed products, 72 uniquely divisible by n, 41 essential submodule, 305 essential version of Maschke's Higman, 143 theorem, 307 homomorphism, 5 extension, 27 factoring, 5 algebraic, 27 Howlett, 323 congruence class of, 29 Huppert, 208 congruent, 28 factor set of, 30 Ideal, 54 finite, 27 G-annihilator-free, 283 finite normalizing, 259 G-nilpotent-free, 283 Galois, 27 nil of bounded degree, 265 infinite, 27 prime, 223 normal, 27 primitive, 54, 224 normalizing, 259 R-cancelable, 240 primitive element o f , 27 semiprime, 224 separable, 27 idempotent, 5 simple, 27 orthogonal, 5 primitive, 5 Field, 27 incomparability, 246 perfect, 27 inertia group, 188 first cohomology group, 39 inflation map, 113 fixed-point space, 96 intertwinning number, 129 fixed subalgebra, 26 irreducible constituent, 129 Frobenius reciprocity, 128 Isaacs, 117, 310, 323 fully invariant submodule, 50 Iwahori, 323 C-algebra, 26 G-field, 26 G-graded algebra, 59 G-graded ring, 60
Jacobson radical, 53 Janusz, 323 Karpilovsky, 100
347
Index
kernel of the action, 26 Kharchenko, 236 Kngrr, 1 3 8 Krull-Schmidt theorem, 25 left transversal, 3 left unit, 55 length, 60 linear representation, 1 0 5 L-injective module, 80 Lorenz, 2 2 6 , 229, 2 3 0 , 2 4 0 , 243, 2 5 3 , 255, 256, 2 5 7 , 258, 259, 260, 2 6 3 , 265, 266, 272, 215, 276, 304 L-projective module, 80
Map, 1 balanced, 8 composite, 1 identity, 1 inclusion, 1 restriction, 1 Martindale, 232 Martindale ring of quotients, 232 Maschke's theorpm, R 3 for strongly graded rings, 3 0 1 matrix of semilinear transformation,lo9 matrix a-representation, 110 matrix unit, 46 module, 6 absolutely indecomposable, 208 absolutely irreducible, 322 artinian, 19 basis of, 6 coinduced, 1 2 8 completely reducible, 6 composition factor of, I composition series of, 7 decomposable, 7 direct sum of, 6 faithful, 54 finitely cogenerated, 1 9 finitely generated, 1 9 flat, 16 free, 6 homogeneous, 50 homogeneous component of, 5 0 indecomposable, I induced, 1 2 6 injective, 1 4 irreducible, 6 noetherian, 1 9 of finite length, 25 projective, 1 2 regular, 6 separable, 319 strongly indecomposable, 1 9 8 weakly G-invariant, 1 9 5 unital, 6 Monomial representation, 9 5 monomial space, 9 5
Montgomery, 5 9 , 1 4 7 , 150, 258, 317 Montgomery's theorem, 1 5 0 multiplicity, 1 2 9 Nakayama's lemma, 8 3 Nakayama reciprocity, 1 2 9 Nastacescu, 3 0 1 nilpotent element, 5 nil ideal, 5 nilpotent ideal, 5 norm, 1 7 9 normal closure, 3 normalizer, 3 normalizing basis, 8 4 normal subgroup, 3 obstruction cocycle, 116 orbit, 26 outer automorphism, 78 Par;, 259 Passman, 8 6 , 1 4 7 , 2 3 0 , 240, 243, 253, 255, 265, 296, 307,
256, 266, 299, 309,
257, 2 5 8 , 259, 260, 263, 272, 2 7 5 , 277, 286, 292, 3 0 1 , 3 0 2 , 3 0 3 , 304, 305, 316, 317 prime rank, 267 primitive rank, 267 principal crossed homomorphism, 39 projective crossed representation, 1 0 3 projective representation, 105
regular class, 100 regular element, 96 regular orbit, 96 restriction map, 4 3 representation, 105 completely reducible, 1 0 5 degree of, 1 0 5 indecomposable, 1 0 5 irreducible, 1 0 5 linear, 105 linearly equivalent, 1 0 6 projectively equivalent, 106 right transversal, 3 right unit, 55 ring, 5 artinian, 19 associative, 5 centre of, 5 characteristic of, 5 direct product of, 6 G-prime, 2 3 0 G-semiprime, 283 homomorphism, 5 left primitive, 224 matrix 46 noetherian, 1 9 opposite, 52 primitive, 224
348
[ring 51 regular, 313 right primitive, 224 semiprime, 223 semisimple, 53 Schelter, 259 Schur's lemma, 5 2 sequence, 4 exact, 4 natural exact, 4 short exact, 4 split, 11 second cohomology group, 38 section, 3 semilinear transformation, 94 group of, 94 nonsingular, 94 separable element, 27 separable polynomial, 27 set 2, cardinality of, 2 size of the form, 286 skew group ring, 63 strongly G-graded algebra, 6 0 strongly G-graded ring, 60 strong permutation of ideals, 277 splitting field, 323 splitting homomorphism, 12 support, 60 supporting subgroup, 86
Index
Tensor product, 7 associativity of, 9 characterization o f , 11 of modules, 7 Transitivity of induction, 126 Trivial intersection ideal, 301 Twisted group algebra, 63 Twisted group ring, 63 Unique product group, 296 Unit, 55 Villamayor, 87 Villamayor's theorem, 87 Weak divisibility, 195 Weak isomorphism, 195 Weiss, 43, 44 X-inner automorphism, 236 Yamazaki, 327