RING THEORY
This is Volume 44 in PURE AND APPLIED MATHEMATICS A Series of Monographs and Textbooks Editors: PAULA. SMITHAND SAMUEL EILENBERG A complete list of titles in this series appears at the end of this volume
R I N G THEORY ERNSTAUGUST BEHRENS Department of Mathematics McMaster University Hamilton, Ontario, Canada
TRANSLATED BY
CLIVE REIS Department of Mathematics University of Western Ontario London, Ontario, Canada
1972
ACADEMIC PRESS
New York and London
COPYRIGHT 0 1972, BY ACADEMIC PRESS, INC. ALL RIGHTS RESERVED NO PART O F THIS BOOK MAY BE REPRODUCED IN ANY FORM, BY PHOTOSTAT, MICROFILM, RETRIEVAL SYSTEM, OR ANY OTHER MEANS, WITHOUT WRITTEN PERMISSION FROM THE PUBLISHERS. THIS IS THE ONLY AUTHORIZED ENGLISH EDITION OF Ring Theory, ORIGINALLY PUBLISHED AND COPYRIGHTED IN THE GERMAN LANGUAGE BY BIBLIOGRAPHISCHES INSTITUT, MANNHEIM, 1972.
ACADEMIC PRESS, INC.
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United Kingdom Edition published by ACADEMIC PRESS INC. (LONDON) LTD.
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LIBRARY OF CONGRESS CATALOG CARDNUMBER: 76154366 AMS(M0S) 1970 Subject Classifications: 1602, 06A50, 20M25, 17A99 PRINTED IN THE UNITED STATES OF AMERICA
Contents
Preface
Chapter I 1. 2. 3. 4.
5.
6. 7. 8.
Basic Concepts
Embedding a ring R in the ring of endomorphisms of an abelian group RLinear mappings of an Rmodule into itself Vector spaces Algebras
Chapter I1 1. 2. 3. 4.
vii
Primitive Rings
Dense rings of linear transformations of a vector space into itself The finite topology The lattice of left ideals in a primitive artinian ring Homomorphisms of semigroups, rings, and modules ; relations Simple rings with minimal left ideals Isomorphism theorems Modular maximal left ideals Primitive algebras
Chapter I11
58 63 68
Completely Reducible Modules
1 . Direct sums of modules 2. Homogeneous components of an Rmodule 3. The double centralizer of a completely reducible module 4. Modules with boolean lattice of submodules
Chapter V
23 28 31 37 40 46 50 54
Rings with a Faithful Family of Irreducible Modules
1 . The radical of a ring 2. Semisimple rings as subdirect sums of primitive rings 3. Semisimple artinian rings
Chapter IV
1 10 13 18
75 80 84 90
Tensor Products, Fields, and Matrix Representations
1. Polynomials 2. Fields 3. Tensor products
93 98 106 V
vi 4. 5.
CONTENTS
Representations by matrices The trace of a separable field extension
Chapter VI
Separable Algebras
1. Central, simple algebras 2. The commutativity of rings satisfying xnlz)= x 3. The quarternions 4. The Wedderburn principal theorem
Chapter VII
The dual of an algebra module Characterization of Frobenius algebras Examples Injective modules of a Frobenius algebra R 5. Behavior of semisimple algebras under separable extensions of the ground field
Chapter XI Chapter XI1 1. 2. 3. 4.
186 189 199 203 208
Distributively Representable Rings
1. Modules with distributive lattice of submodules 2. Rings with distributive lattice of left ideals 3. Arithmetic rings
Chapter X
149 166 174
Frobenius Algebras
I. 2. 3. 4.
Chapter IX
125 136 141 143
Rings with Identity
1. Semiperfect rings 2. Projective modules and Asano orders 3. Injective modules and selfinjective rings
Chapter VIII
113 121
Noetherian Ideal Theory in Nonassociative Rings Orders in Semisimple Artinian Rings
21 1 222 229 249 260
Rings of Continuous Functions
Biregular right algebras The structure space of a ring The theorem of Arens and Kaplansky Boolean rings
Guide to the Literature Bibliography Index
282 291 300 306 309 3 14 317
Preface
The first five chapters of this book give an intrcrduction to the theory of rings (the Jacobson density theorem, semisimple artinian rings, completely reducible modules, and representation theory). Chapters VIXI are devoted to various aspects of the theory of associative rings with radical different from zero, and the last chapter (Chapter XII) deals with rings of continuous functions, culminating with the theorems of ArensKaplansky and of Stone. Except for some minor changes and corrections, the first part of the book is a direct translation of the author’s “Algebren,” Mannheim, 1965. The author is indebted to Dr. E. Hotzel for pointing out some errors in the German edition. In the second half of the book, semiperfect rings play the most important role, particularly in the chapter dealing with distributively representable rings. The section on arithmetic rings which forms part of this chapter contains a number of hitherto unpublished results. In order to give the broadest possible view of ring theory compatible with keeping the book to a reasonable size, I have had to omit a number of important topics, among them division ringsa treatment of which will be found in Jacobson’s “Structure of Rings”and the application of homological methods to ring theory. However, projective and injective modules are introduced in Chapter VII and the results obtained are applied to the theory of Asano orders and selfinjective rings. Once again, I would like to express my appreciation to the “Bibliographisches Institut” in Mannheim. I would also like to thank Academic Press for undertaking to publish the book and for their patience in the face of many delays. Finally, my sincere thanks go to Professor Clive Reis of the University of Western Ontario for translating the German manuscript.
vii
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RING THEORY
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Basic Concepts CHAPTER I
1. Embedding a ring R in the ring of endomorphisms of an abelian group
A mapping of a set M into itself is a correspondence which assigns to each element m in M a uniquely determined element m' also in M . The image of m under the mapping a is denoted by am. If a and b are mappings of M into itself, the product (or composite) mapping ab is obtained by the successive application of the mappings b and a in that order: rn
f
bm + a(bm)
for all m E M
(1)
The associative law,
a(bc) = (ab)c holds for composition of mappings, since by the definition of the product,
(a(bc))m
= a((bc) m ) = a(b(cm))
((ab)c) m
= (ab)(cm) = a(b(cm))for
and all m E M
Definition l(a). A binary operation defined on a set S is a correspondence which assigns to each ordered pair of elements a, b from M a uniquely determined element denoted ab in M. Definition l(b). A semigroup is a system consisting of a nonempty set S and an associative binary operation defined on S. 1
2
I.
BASIC CONCEPTS
The associative law (2) holds for composition of mappings of a set M into itself. Hence: Theorem 1. The mappings a, b,... o f a set M into itself form a semigroup with composition of mappings as binary operation. We may also define a sum of mappings which will have the usual properties of ordinary addition, provided the set M itself possesses a suitable algebraic structure. To this end, let us recall the following: Definition 2. A group G is a semigroup with the additional property that for any arbitrary pair of elements a, b E G, both the equations
aox
=b
yoa
and
=b
have solutions in G. Here o denotes the operation of the semigroup. Let e be an element of a semigroup S such that
eoa = a
aoe = a
and
forall a E S
Then e is called an identity of the semigroup. We may show that a group possesses a unique identity and that the solutions of a 0 x = b and y 0 a = b are unique. This may be seen as follows: Let c and a be arbitrary elements of G. Let e, be a solution of e, 0 c = c and x a solution of c o x = a. Then
e,oa=e,ocox=cox=a and hence
e, 0 a
=a
for all a E G
Let al be a solution of x 0 a = e, . It follows that a1 0 a 0 a1 Therefore if y is a solution of x 0 al = e, ,
e, o a o a1
=y o
a1 o a o a1
=y o
a1
= al.
= ec
This implies
a 0 al
=
e,
which in turn gives
a~e,=aoa~oa=e,oa=a
forall a E G
The uniqueness of the identity e = e, in G follows from the fact that if e and e' are identity elements in a semigroup S, then
e
= e o e ' = e'
(3)
3
1. EMBEDDING
a’
Given a E G , al is unique: for suppose a‘ is also an inverse of a, i.e., 0 a = e. Then
a1
= eal(a’a)
a1
= a’(aa1) = a’e
a‘
I
The uniqueness of the solutions of the equations a 0 x follows from
x = a  l ~ a ~ x = a  l ~ band
=b
and y o a = b
y=y~a~al=b~al
where a1 is as above. It is clear that a semigroup with identity e is a group if each element a has an inverse al. An abelian group is a group in which the operation 0 is commutative:
a0b
=b
0
a
for all a, b
+.
We often call the operation of an abelian group addition and denote it by The identity element is then generally denoted by 0. We may define the sum of two mappings of an additively written abelian group into itself as in the case of realvalued functions:
Definition 3. Let A4 be an additively written abelian group and let a and b be mappings of M into itself. Their sum a b is defined to be the mapping
+
a
+ b: m
f
am
+ bm
for all m E M
(4)
This addition satisfies the usual rules, i.e. we have
Theorem 2. The mappings a, b, ... of an abelian group M into itself, under the operation of addition defined above, form an abelian group 02.
Proof. The associativity of the addition in a follows from the associativity of the operation in M :
As a consequence of the commutativity in M we have (a
+ b ) m = am + b m
= bm
+am
= (b
+a)m
4
I.
BASIC CONCEPTS
Hence the addition in @ is commutative. If 0 denotes the identity of M , the zero mapping defined by
m is the identity for
f
for all m E A4
0
+ in @ and the mapping. for all m E M
a: m + (am)
is the additive inverse of a. We have therefore shown that @ under abelian group. 1
+ is an
Theorems 1 and 2 show that the mappings of an abelian group into itself almost form a ring in the sense of the following: Definition 4. A ring R is a system consisting of a set R and two binary operations in R called addition (+) and multiplication (.) such that 1. R together with addition is an abelian group 2. R together with multiplication is a semigroup 3. a . (b c) = a . b a . c and (b c) . a = b . a c . a (distributivity of . over +)
+
+
+
+
+
The two distributive laws in 3 link together the two operations and . We often omit the dot in a . b and write simply ab. We have already proved Properties 1 and 2 of Definition 4 for the system consisting of the set of mappings of an abelian group into itself under addition and multiplication. We may further show right distributivity (b
For, ((b
+
C)
u = ba
+ cu
+ c) a) in = (b + c)(am) = bum + cam
for all m E M
On the other hand, if we try to prove left distributivity a(b
+
C)
= ab
+ ac
we find that the proof breaks down since, in general, a(b
+ c) m = a(@ + c) m) = a(bm + cm)
is not the same as abm a(ml
+ acm because the equality
+ mz) = am, + am,
for all m, , m, E M
need not hold. This suggests that we should single out those mappings
a, b, ... of M into itself which are additive in the sense of the following:
1. EMBEDDING
5
Definition 5. A mapping of the abelian group M into itself is said to be additive if
a(m,
+ m,)
= am,
+ am,
for all rn, , m,
E
M
(8)
Such mappings are called endomorphisms of M . The left distributivity of multiplication over addition in the set of endomorphisms of M now follows from (7). In order to show that this subset of (2together with addition and multiplication is a ring, we still have to show closure of the two operations; i.e. we must show that if a and b are arbitrary endomorphisms, then both a  b and ab are also endomorphisms. In the proof that a  b is an endomorphism, the commutativity of the group M is essential. 1. (a  b)(ml
+ mz) = a(m, + m,)
 b(ml
k am, = (a  b) m, = am,
2.
ab(m,
+ m,)
+ mz)
 bm,  bm,
+ (a  b) m, = a(b(m, + m,)) = a(bm, + bm,) = abm, + abm,
We have therefore proved the following: Theorem 3. The endomorphisms of an abelian group M form a ring, denoted Hom(A4, M ) . The endomorphism
e: m + m
for all m E M = ae = a
is an identity in Hom(M, M ) , i.e. ea
(9)
for all a E Hom(M, M ) .
Multiplication in the ring of endomorphisms of an abelian group M is not in general commutative as the following example shows: Let M be the set of ordered pairs (m, , m,) of integers. Define addition by (m, , m,)
+ (m,’,m,‘)
The mappings a and b defined by
+ ml’, m2 + m,’)
= (m,

a: (m, mz) b: (4, mz) 9
are both endomorphisms of M . Now
(m, m,) (m19 0) 9
6
I.
BASIC CONCEPTS
but
(ba)(m, m,) 9
= b(m2 Y
ml)
= (mz 9
0)
Rings of endomorphisms of abelian groups are not the best known examples of rings. The integers E , for example, form a ring. However, we shall soon show that, from an algebraic point of view, an arbitrary abstract ring may be considered as a subring of the ring of endomorphisms of a suitable abelian group. A subring of a ring R is, of course, a subset of R which is itself a ring under the operations inherited from R. Two rings R and R' are indistinguishable "from an algebraic point of view" if they have the same algebraic structure, i.e. if there is a onetoone correspondence between their elements which preserves the operations. This concept, fundamental to the whole of algebra, is precisely formulated in the following:
Definition 6. A homomorphism from a semigroup GY to a semigroup 02' is a correspondence 4 which assigns to each element a E GY a uniquely determined element a+ in 02'such that (a * b) 4
=
(a$) * ( b 4 )
for all a, b E GY
(10)
4 and we say that a is mapped to ad. In (lo), the dot on the lefthand side denotes the operation in GY and on the righthand side the operation in GY'. If the only element mapped to a$ is a, i.e. if for all a, b E a, a4 is called the image of a under
a4
=
b$
implies
a
=b
then we call 4 a monomorphism. Under a homomorphism not every element of r3" need occur as the image of some element in a. If, however, this does occur, then we call 4 an epimorphism of GY onto GY'. A homomorphism which is both a monomorphism and an epimorphism is called an isomorphism of a onto GY'. Since a group is a semigroup, the above definitions apply to groups as well. As an example, consider the isomorphism between the additive group of real numbers and the multiplicative group of positive real numbers given by
4: a + r a
where
0 < r # 1 , r fixed
Every endomorphism of an abelian group M is a homomorphism of A4 to M . The same concepts as those introduced above are defined for rings R except that we require (10) to hold for both operations. Thus we require (a
+ b) 4 = a+ + b+
and
(a . b) 4
=
(a+) . (64)
for all a, b E R
1.
EMBEDDING
7
Example a. The additive groupZ of integers may be mapped isomorphically onto the additive group Z‘ of even integers by means of 4:
respectively, where . is the multiplication in R. It is convenient to denote these left and right Rmodules by ,R and R, , respectively. We have already made use of ,R in Theorem 1.1.4 without explicitly mentioning it. If M is a Kmodule, K a ring, then it is natural to consider
2. RLINEARMAPPINGS OF
AN
RMODULE INTO
ITSELF
11
those endomorphisms of the group ( M , +) which preserve the Kmodule structure of M in the sense of the following: Definition 2.
Let K be a ring and M a right Kmodule. A mapping
a E Hom(M, M ) is said to be Klinear if it satisfies the further condition
1. a(mk) = (am)k
for all m E M
and k
E
K
In a similar fashion, if M is a left Kmodule, it is convenient to write the mappings of Hom(M, M ) on the right of M . Theorem 1. Let K be a ring. The Klinear mappings of the right Kmodule M into itself form a subring Horn&$, M ) of Hom(M, M ) . We call Hom,(M, M ) the centralizer of the Kmodule M . A similar theorem may be proved for left modules. Proof. We have only to show that the difference and product of two Klinear maps are also Klinear. Let a, b E Hom,(M, M ) . Then
(a  b)(mk) = a(mk)  b(mk) = (em)k
 (bm)k = ((a  b) m)k
and
(a . b)(mk) = a(b(mk)) = a((bm)k ) for all m
E
M and k
E
K.
= (a(bm))k =
((a . b) m)k
I
The reason for calling Hom,(M, M ) the “centralizer” of the Kmodule M becomes apparent if we write the elements a E Hom,(M, M ) on the right of M . Then, instead of a(mk) = (am)k
(1)
we get mka
= mak
for all m E M
and k E K
The elements of Hom,(M, M ) thus centralize K, i.e. ka
= ak
for all aEHom,(M, M ) and k E K
We now show that an arbitrary ring R is isomorphic to a subring of the ring of all Klinear mappings of a suitable right Kmodule M . This result is a corollary to the following:
12
I.
BASIC CONCEPTS
Theorem 2. Let M be a left Rmodule and K its centralizer. Then the elements of R induce Klinear mappings on M , i.e. we have a(m1k1 for all
U E
+ m2kd
R, m i E M , and
=
(am,) k,
+ (am2) k2
(2)
k i E K.
Proof. By Theorem 1, the centralizer of the left Rmodule M forms a ring K. Definition 2, applied to the left Rmodule M , gives
(am)k = a(mk)
for all a € R, m E M , and k e K
(3)
a is clearly additive since M is an Rmodule and Eq. (3), read from right
to left, completes the proof of the Klinearity of a.
I
Corollary An arbitrary ring R is isomorphic to a subring of the ring Hom,(M, M ) of Klinear mappings of a suitable right Kmodule M. Proof. See Theorem I. 1.4.
I
A closely related theorem of interest is the following:
Theorem 3. Let R be a ring with 1. Then 1. The centralizer K = Hom,(,R, ,R) of the left Rmodule ,R is the ring of right multiplications 31, of R and 2. R is isomorphic to the full ring of Klinear mappings of RR into itself under the isomorphism a + C , . Proof. 1. Let k be an element of the centralizer K and let lk Then for all a E R ak = (a . 1) k = a . (lk) = as
=sE
Hence k = 31, . Conversely, for each s E R , the right multiplication is an element of Horn,(&, ,R) as may be seen from (ax) 31, = (ax) s = a(xs)
R.
as
for all a, x E R
The second equality follows from the associativity of R. 2. By Theorem 1.1.4 (2), the mapping a + C , is a monomorphism of R into Horn,(& RR).Let A be a Klinear map of ,R into itself and let A1 = a. Then A x = A(1 * x ) = A(l . = ( A 1) %$ = ax = f,x for all x E R. Therefore A
=C,
. I
In actual fact, Theorem 2 gives little information about the structure of R unless we know more about K and the subring of HomK(M,M ) which
3.
VECTOR SPACES
13
R induces. In Chapter I1 we shall characterize those rings R with the following
two properties: 1. R possesses a faithful module, the centralizer of which is a division ring; 2. the subring of Hom,(M, M ) which R induces is (in a sense to be precisely defined) “dense” in Hom,(M, M ) . To arrive at this characterization we must first extend the wellknown theorems about finitedimensional vector spaces over division rings to vector spaces of arbitrary dimension.
3. Vector spaces Definition 1. Let K be a division ring and M a right vector space over K. The set {mi I i E Z} of elements of M is linearly dependent over K if there is a finite subset I,, of the index set Z together with a corresponding set {ki I i E I,,} of elements of K, not all zero, such that
C miki = 0 itl,
Otherwise, the set {mi I i E I } is linearly independent. In the next theorem we shall prove that every vector space M possesses a maximal linearly independent set {mi I i E I } , i.e. a linearly independent set with the property that any set properly containing it is linearly dependent. Each element m of the vector space is then uniquely expressible as aJinite linear combination of elements of the Kbasis {mi 1 i E I } , i.e.
m=xmAkA, AEA
ProoJ
Acz,
1A1<W
(1)
Since m u {mi1 i E I} is linearly dependent, there exist elements
l A X, E (1C Z, 1 A I < co,not all zero, such that m5
+ C mAL
=
5,
0
AEA
{mi I i E Z} is linearly independent and therefore 5 # 0. Since K is a division ring, m = m A k A , where kA= cAll AEA
To show the uniqueness of (1) we may assume without loss of generality that
m
=
1 mAkA = C mAkA’ AEA
A€A
14
I.
BASIC CONCEPTS
where each summation is taken over the same index set. This implies
1 mA(kA 
kh’) =
0
A€A
and hence k ,
= k,‘
for all A E A .
1
Theorem 1. Every vector space M # (0) over a division ring K possesses a Kbasis, i.e. a maximal linearly independent set of vectors. The cardinality of this set is called the dimension of M over K and is denoted by dim, M .
Proof. We first show the existence of a basis. M contains at least one element m, # 0. Clearly {m,} is a linearly independent set. Let * * a
C A,c A,C ...
(2)
be a chain of Klinearly independent subsets A,of M . That is to say, for each pair A, and A6 of members of (2), either A,C A. or A, C A,. Let V denote the settheoretic union of the sets occurring in this chain. V is also a Klinearly independent set since for each finite set of elements u, , u2 , ..., u, E V, there is a member A i ( Tof) the chain (2) which contains each ut , t = 1, ..., r. The existence of a maximal linearly independent set {mi 1 i E Z> in M now follows. I This last step in the argument is a consequence of Zorn’s lemma. Zorn’s lemma, an equivalent form of the axiom of choice, is of great importance in modern algebra and we shall often make use of it in this book. We pave the way for a statement of the lemma with the following: Definition 2. 1. A relation defined on a set H i s a subset p of the set H x H of ordered pairs (h, , h,), h, , h, E H. If (h, ,h,) is in p we say hl is in the relation p to h, or h,ph, . 2. Apartial order ( H , C) , is a set H together with a relation C satisfying the following:
h C h for all h E H (reflexivity) /3. if h, C h, and h, C h, , then h, = h, (antisymmetry) y. if h, C h, and h, C h, , then h, C h, (transitivity). a.
3. Let H be a partial order and U a subset of H . An element s E H is an upper bound of U if u C s for all u E U. 4. An element m in a partial order H is maximal if m C h for some h E H implies m = h.
The set P(&) of subsets of a set A, together with set theoretic inclusion as the relation, is an example of a partial order. The set of natural numbers
3.
VECTOR SPACES
15
with hl C h, if h, is not greater than h, is another example. In this case the partial order has the additional property that any two elements are related. Accordingly, we have:
Definition 3. A partial order is a chain (also an order) if any two of its elements are comparable, i.e. either h, C h, or h, C h, for any h, , h, E H . Using the terminology introduced, we may now state: Zorn’s lemma If every nonempty chain of elements of a nonempty partial order H has an upper bound in H , then H contains at least one maximal element. 4 Example. In the proof of Theorem 1 , the elements of H are the linearly independent subsets of M partially ordered by settheoretic inclusion. An arbitrary chain (2) has an upper bound, namely V , which is in H. The equivalence of Zorn’s lemma and the axiom of choice will not be proved here. The reader is referred to B. H. Hermes (1955) for a proof. Many authors [e.g. N. Bourbaki (1947), Chapter 1111 call a partial order an order and a chain a total order. We shall now use Zorn’s lemma to complete the proof of Theorem 1 by showing that the dimension of M is independent of the choice of basis for M . If {b, 1 h E A } is a set of elements of M , (b, 1 h E A ) will denote the subspace generated by {b, I X E A } over K . This subspace consists of all finite linear combinations of the b, with coefficients from K. Two subsets {b, 1 h E A>, {c, I p E M } are said to be linearly equivalent if (b, I h E A ) = (c, I p E M ) . Let At and A’ be two bases for M and let 33 and C be subsets of At and A”, respectively. We shall write (33;C), if i. 33 and C are linearly equivalent and ii. there exists a fixed oneone mapping f from 33 onto e. We partially order the set ((33; C , I) 33 C At, C C A’} by the relation < defined as follows: (3;C),
if
d
(33’; C ’ h
i. 33 _C a’, C C C’ and ii. f restricted to 33 is precisely$ A chain C *.*
< (33,
I6
I.
BASIC CONCEPTS
and f is the mapping defined by
f ( b ) =f,(b)
if b e $ ,
Clearly f is well defined, since for any pair of mappings occurring in the chain C, one is the restriction of the other. By Zorn's lemma there exists a maximal element (a*; C*),* in H . We must now show that 33* = A and C* = N. Assume there is an element b, E JL  33*. Since (a*)= (C*) and A" is a basis for A, there are elements c l , ..., ci in N' such that b, E (C*, c, , ..., ci). Let i be minimal for b, . We see that i > 1, since otherwise (%* u b, ;C* u c,)~~E H, where fi restricted to 33* is f * and fl(b,) = c1 . This contradicts the maximality of (a*; C*),* in H . Now l e t j be minimal subject to
, ..., c i ) C (a*,bl , ..., bj),
( ~ 1
bi E A
I f j > I , let (b, , ..., b j ) C (C*, c1 , ..., ci , ci+, , ..., cg), cifl , ..., c k E N, again choosing k minimal. Continue in this fashion. The process either continues without stopping or it stops after finitely many steps. In the former case we obtain two infinite sequences b , , b , , ... and cl,c 2 , ... in A and N, respectively. Clearly (33*
U bl U
* * a ;
C* U c1 U ..*)fo
E
H
where fo restricted to 33* is f * and fo(bi) = ci and hence we arrive at a contradiction. In the latter case, we obtain the following equality after a finite number of steps
(a*, b l , ..., bm) = (C*,
c,, ..., c,)
for some pair of natural numbers m and n. To complete the proof we need only show that m = n. It then follows that
(a*u bl u *.. u 6,
; C* u c1 u ... u c,)~E H
where
is defined in the obvious way, contradicting the maximality of To show that m = n, we shall use induction on n. For each i = 1, ,.., n,
(a*, C*),*.
m
ci = bi*
+ C bjkij, j=1
bi* E (a*), k i j E K
for uniquely determined kij and bi*. Define a mapping to (b1, * . * > bm) by m
$(ci) = j=1
bjkij = bi',
C$
i = 1,..., n
from ( c l , ..., c,)
3.
VECTOR SPACES
17
and extend linearly. The linear mapping 4 is well defined and {bl', ..., bn'} is a linearly independent set since A and N are bases. Furthermore, 4 is onto ( b , , ..., bm), since each element of (b, , ..., b,) is expressible as a sum of an element from (C*> and an element from (c, , ..., c,). Hence {b,', ..., bn'} is a basis for (b, , ..., bm).Therefore b, is a linear combination of b,', ..., b,', where we may assume, without loss of generality, that the coefficient of b,' is not zero. We may then solve for b,' in terms of bl', ..., bhl, b, . Thus {b,', ..., bhl, b,,,}is a basis for (b, , ..., bm). Now construct a linear mapping i,h from (b,', ..., bAl) to (b,, ..., bm,) in the same way 4 was constructed above. Let i,h(bi')= b;, i = 1, ..., n  1. Then as above, {b;, ..., bil} is a linearly independent set and ( b ; , ..., b,",) = (b, , ..., bml), Assume inductively that if a vector space has a basis of n  1 vectors, then every basis has IZ  1 vectors. It follows from the induction hypothesis that n  1 = m  1 and hence m = n. As we remarked above, this leads to a contradiction and therefore the assumption that 33* is not the whole of A is false. Similarly, we may show that e* = ."and the theorem is proved. I The existence of a basis in an arbitrary vector space not only allows us to express an arbitrary element of the space as a finite linear combination of the basis elements, but also to describe Klinear mappings by means of matrices with entries from K. By the Klinearity of A E Hom,(M, M ) and on account of (l), the action of A is completely determined once we know the images Am, of a fixed Kbasis {m,,I X E I } . Since Am, E M
Am,
=
m,a,,, ,
a,,E K,
AEI
(3)
where, for a fixed A, only finitely many a,,, are different from zero. With respect to a fixed basis { m , [ L E I } , each Klinear mapping of M into itself gives rise to a matrix (a',), L , X E I, a',, E K under the correspondence A
f
(4)
(a~A)
(cY,,) is an I Z I x I I [ matrix ( 1 11 possibly infinite) with only finitely many
nonzero entries in each column. Conversely, it is easily verified that each such matrix represents (again, relative to a fixed basis) a Klinear mapping A of M into itself, where A is defined as in (3). The sum and the product in Hom,(M, M ) correspond, respectively, to the sum and the product of the corresponding matrices: A
+B

(%A
K
+
P
L
J
18
I.
BASIC CONCEPTS
zK
We note that the sum a,,/IK, makes sense since only finitely many terms are nonzero. Of course we define the sum of infinitely many zeros to be zero. We have therefore proved
Theorem 2. Let M be a right Kmodule over the division ring K. Let I I I be the cardinality of a Kbasis of M . Then the mapping +: A
f
(a,,),
A
E
Hom,(M, M )
(5)
is an isomorphism of the ring Hom,(M, M ) onto the ring of I I 1 x 1 I I matrices having only finitely many nonzero entries in each column. If dim, A4 = y1 is finite, Hom,(M, M ) is isomorphic to the full ring of n x n matrices over K. I 4. Algebras
In Theorem 1.1.3 we used the additive structure of an abelian group M to show that the endomorphisms of A4 form a ring Hom(M, M ) . If M is also a right Kmodule, K a ring, it is natural to ask whether the Klinear mappings Hom,(M, M ) of A4 into itself may be given the structure of a right Kmodule by defining the operation of k E K on A E Hom,(M, M ) by
Ak:m+Amk
for all mE M
(1)
Clearly Ak is an additive mapping of M into itself and hence an element of Hom(M, M ) . In general, however, it is not Klinear. In fact, if k' E K ,
(Ak)(mk') = Amk'k but
((Ak)m ) k'
=
Amkk'
In particular, if Id is the identity mapping of M , (Id . k)(mk') = ((Id * k ) m) k'
for all k'
E
K, m E M
if and only if
m(k'k  kk')
=0
for all k'
E
K and m E M
If M is a faithful right Kmodule, this condition implies
k'k = kk'
for all k'
E
K
4. ALGEBRAS
19
Conversely, if k satisfies (2), Ak is Klinear and hence an element of HOmK(M, M ) . As a consequence of our discussion, we are led to the following:
Definition 1. The center 2 of a ring R is the subring {z I z
E
R, za
=
Z is indeed a subring, for if z, , z, a(zlzz)= (az,) z2
=
az E
for all a E R}
Z and a E R,
(zla)z, = zl(azz) = zl(zza) = (zlzz)a
Using the concept of the center of K we are led by our discussion above to the following: Theorem 1. Let M be a right Kmodule and let 2 be the center of the ring K. Then the ring Hom,(M, M ) may be given the structure of a right 2module by defining A5: m + A m 5
for all A E H ~ ~ , ( MM,) , ~ E Z and , m e : M (3)
The following also holds: (AB)c
ProoJ
5EZ
for all A , B E Hom,(M, M ) and
= A(B5) = (A5)B
(4)
The module properties (A
+ B) 5 + 5,)
4 5 1
+ B5 A51 + A52
= A5 =
and
A(5152)
=
(&I)
52
follow immediately from Definition (3); (4) follows from ((AB) 5) m
and
= (AB) m5 =
A(Bm5)
=
A(B()m
((4 B )) m = (A5)(Bm) = ABmS I If the center
Z of K possesses an identity, we have shown that HOmK(M, M )
is an algebra over 2 in the sense of the following:
Definition 2. Let R be a ring and @ a commutative ring with 1. R is an algebra over @ if R is a unital right @module satisfying (ab) 01 = a(ba) = (aa)b
for all a, b E R and
01
E
@
(5)
In Chapter XI1 we shall consider functions defined on a topological space
20
I.
BASIC CONCEPTS
with values in a noncommutative ring R. We shall then generalize the concept of an algebra to that of a right algebra over R (cf. Definition XII. 1.1). The fact that Hom,(M, M ) may be considered as an algebra over the center of K has application in the theory of linear transformations of a vector space over a field. For example, the minimal polynomial of a linear transformation is best introduced in this context. According to the corollary to Theorem 1.2.2, every ring R may be isomorphically embedded in the ring Hom,(M, M ) of a suitable right Kmodule M . A similar result holds for algebras. To prove this we introduce the concept of an algebrahomomorphism (isomorphism, epimorphism, etc.): Definition 3. A homomorphism of the Rmodule M to the Rmodule M’ is a homomorphism of the additive group ( M , +) into ( M ’ , +) satisfying
(am)
=
a(m+)
for all a E R, m E M
Monomorphisms, epimorphisms, and isomorphisms are defined similarly. The Rlinear mappings of M into itself are therefore Rhomomorphisms of M into itself. We denote the abelian group of Rhomomorphisms of M into M’ by Hom,(M, M ’ ) Of course, the sum
m(+l
+1
+
+ +2)
+2
of two elements of Hom,(M, M ’ ) is defined by =
m+l
+ m+2
for all m E M
Definition 4. Let R and R‘ be algebras over @. An (algebra) homomorphism of R into R’ is a ringhomomorphism which is also a @module homomorphism satisfying
+
(acx) 4 = (a4)cx
for all a E R,
01
E
@
(6)
Algebramonomorphisms, etc. are defined similarly. We may now state and prove: Theorem 2. 1. A @algebra R with identity is isomorphic to the @algebra Hom,(M, M ) of a suitable right Kmodule M . 2. If R has no identity we may isomorphically embed R in a @algebra R(1)with identity.
ProoK If R possesses an identity, we may take the right @module RR for M and the centralizer of RK for K . Then by Theorem 1.2.3, R is ringisomorphic to the full ring Hom,(M, M ) under the mapping a +C, .
4. Therefore
ALGEBRAS
Hom,(M, M ) = {Ca 1 a E R }
If we define
21
(7)
for a € R, a € @
Caa = C,,
condition (5) may be verified thus showing that Hom,(M, M ) is an algebra over @: (CaC,) a = UGP) = (Caa) C b Furthermore, the mapping a + C , is an algebra isomorphism by (7). 2. If R has no identity we may embed it as a subalgebra of an algebra R ( l ) with identity. We proceed as in Example y of Section 1.1: Let R'l) = ((a, a) 1 a E @, a E Rj and define the operations by (011
9
ad
+ (a,
(a1 , a,)(% (011
9
9
4)= (011
,a,)
a,) a2
+
=
2
+ + + a1ad
a1 a21 4% 4%
a2
9
= (a1 9
3
0)
The identity of the algebra R(l)over @ is (1,O) and we embed R in means of the mapping a (0, a).
by
f
We have defined an algebra in such a way that every ring may be considered as an algebra over the ring of integers Z.The additive group of R is a right Zmodule and ( 5 ) is easily verified for all a E Z and a, b E R . The definition we have given of an algebra is more general than that found in the classical theory. There, an algebra or hypercomplex system means an algebra (usually finitedimensional) over a field @. With this restriction we have at our disposal, by Theorem 1.3.1, a @basis {b, 1 L E I}. (5) implies a(b,a,
+ b,ol,)
= (ab,) a,
+ (ab,) a,
for all a, b, E R, a E 0
showing that the multiplication in R is completely determined once we know the product of every pair of basis elements. With respect to a fixed @basis of R we have b,b,
b,y,,,
=
for
L, K
EI
(8)
A€ I
where yLKA E @ and for fixed ( L , K ) , at most finitely many yLKA are different from zero. The associativity of multiplication in R implies ,
(b',) b,
= b,(b,b,)
for all
L, K ,
uEI
22
I.
BASIC CONCEPTS
and this in turn gives
C
=
Y'KAYA~T
A
1
for all
YLI\TYKOI\
6, K , 0
(9)
,E ~
AEI
The sums occurring on the left and right of (9) make sense since for fixed ( L , (resp. ( K , u)) only finitely many y L K(resp. A yKOJare different from zero. Conversely, if we are given a field @, a set Z and a mapping (1, K ,
A)
+
YLKA
9 YLKI\
E @,
1, K ,
K)
AEI
such that the following conditions hold: h different from zero a. for each fixed pair ( 1 , K ) only finitely many y L Kare and p. (9) is satisfied, we may use (8) to define an algebra over CP with basis {b, 1 L E Z}, where {b, I L E Z} is a set of symbols in oneone correspondence with Z. In Theorem 1.4.2.1, we used the mapping a +C, to prove that a @algebra R is isomorphic to Hom,(,R, RR). Relative to the basis { b , I L E I } , f, may be described by
(10) and it follows from (9) that the mapping
LEI
'
is a homomorphism of the @algebra R into the algebra of I Z I x I Z I matrices over @. The image of R under this mapping is the set of matrices with all but finitely many entries zero in each column. For the definitions of sum and product of these matrices see Section 1.3. We call the mapping (1 I) theJirst regular representation of the algebra R by matrices over the ground field @ relative to the @basis {b, I L E I } . If R has an identity the representation is faithful, i.e. no nonzero element of R is mapped to zero. The general theory of representations of rings by matrices over a field A will be presented in Chapters V and VI.
Primitive Rings CHAPTER I1
1. Dense rings of linear transformations of a vector space into itself We showed in Theorem 1.2.2 and its corollary that an arbitrary ring R is isomorphic to a subring of the full ring of Khomomorphisms of a suitable Rmodule M over its centralizer K. This result, however, gives us little information about the structure of R if we know neither the extent to which the isomorphic image of R “fills” Hom,(M, M ) nor the structure of K. By assuming that R possesses an irreducible module M we shall be able to obtain more precise information about K. Definition 1. Let M be an Rmodule. 1. An Rsubmodule U of M is a subgroup of the additive group of A4 which is closed under multiplication by elements of R, i.e. for all a E R and u E U , au E U . 2. An Rmodule M is irreducible if a. A4 and the zero submodule (0) are the only submodules of M and p. R M = {Cy=,aLmbI a, E R, inLE M } f (0).
If we take M
=
RR we have the following:
Definition 2. A left ideal L of a ring R is an Rsubmodule of the left Rmodule R R , i.e. L is a subgroup of the additive group of R for which al E L for all a E R and I E L. Similarly, a right ideal is defined to be a submodule of R, . A left ideal which is also a right ideal is called an (twosided) ideal. 23
24
11.
PRIMITIVE RINGS
Theorem 1. Let R be a ring with identity. Then the left Rmodule RR is irreducible if and only if R is a division ring. Proof. 1. Let a # 0 be an element of the division ring R. Then Ra = {x . a I x E R} = R is the smallest Rsubmodule of RR containing a. Furthermore, RRR = R2 # (0) since 1 E R. 2. Let RR be irreducible and let 0 # a E R. Since 0 # a = 1 . a E Ra, the Rsubmodule Ra of RR is the whole of R. Hence the equation xa = 1 has a solution x in R ; x is clearly different from 0 and as above, the equation yx = 1 also has a solution y . Since xa = 1 , it follows that xax = x and therefore a . x = lax = y x a x = y x = 1
Hence x is an inverse of a.
I
Theorem 2. (Schur’s lemma). The centralizer K irreducible Rmodule M is a division ring.
= HomR(M, M
) of an
Proof. Let 0 and 1 be the zero and identity, respectively, of K. Let k # 0 be an element o f K . The subset (0:k )
= {m I
m E M, mk
= 0}
is an Rsubmodule since mlk = m,k = 0 imply (ml  m,) k = 0 and (aml) k = a(m,k) = 0 for all a E R and m, E (0: k). Hence (0:k ) is either the whole of M or (0). The first possibility is untenable since k # 0. Therefore (0: k ) = 0 and k is a monomorphism. The image M k of M under k is an Rsubmodule of M since R ( M k ) = (RM) k C Mk. Hence, either M k = (0) or Mk = M . Since k # 0, mk # (0) and therefore M k = M , proving that k is an isomorphism of M onto itself. Thus the inverse mapping kI of k exists. It belongs to K, for
((am)kl) k
= am
and
(a(mkl)) k
= a(mklk) = am
proving (am) kl = a(mkl) since k is oneone. Clearly kk’ and K is a division ring.
=
1
= klk
According to Schur’s lemma, an irreducible left Rmodule M is a vector space over K = HomR(M,M). M therefore possesses a Kbasis. If dim,M is finite and {u, , ..., u,} is a Kbasis of M , then R will induce the whole ring Hom,(M, M ) of Klinear transformations if, given arbitrary vectors u1 , ..., u, of u, we can find an element a E R such that
au,
= u,
, au,
= u,
, ..., au,
=
v,
1.
DENSE RINGS OF LINEAR TRANSFORMATIONS
25
If dim, M is not finite we would like to have a measure of how well a subring “fills” Hom,(M, M ) . To this end we introduce the concept of density. Definition 3. Let K be a division ring and A4 a right vector space over K. Let @ be a subring of Hom,(M, M ) . If for each finite set {u, , ..., u,} of linearly independent vectors and each arbitrary finite set {vl, ..., v,} of vectors there exists a E Hom,(M, M ) such that
au,
au,
=vl,
=
v 2 , ..., au,
=
v,
(1)
then we say 0 2is dense in Hom,(M, M ) . Clearly, if dim, M is finite, CPG is dense in Hom,(M, M ) if and only if CPG = Hom,(M, M ) . The connection between the concept of density as defined above and the topological concept of density will be established in the next section. The following Theorem 3, due to Jacobson, implies that when M is a faithful, irreducible Rmodule, R is dense in Hom,(M, M ) . We may prove a similar result for left ideals B in R . In this case, however, it is clear that B is not necessarily dense in Hom,(M, M ) , since an element 0 # u1 E A4 may be annihilated by the whole of B, i.e. u1 may be in the set ( 0 :B )
= {m
I bm
for all b E B }
=0
(2)
Clearly, since b(mk) = (bm)k for all k E K, (0 : B ) is a Ksubspace of M . We therefore replace the linearly independent set {ul, ..., u,} of Definition 3 by one which is linearly independent modulo (0 : B ) in the sense of the following: Definition 4. Let V be a subspace of the vector space M over the division ring K . A set {u, , ..., u,} of elements of M is linearly independent modulo V if
ulXl
+ + u,X, ***
implies 1 
/) 2

E
V,
hi E K
... = h , = O
Theorem 3. (Jacobson). Let M be an irreducible Rmodule, Kits centralizer and B a left ideal in R. Let {ul, ..., u,} be an arbitrary set which is linearly independent modulo (0 : B ) and let {vl , ..., v,} be an arbitrary set of vectors. Then there exists b E B such that bui
=
vi ,
i
=
1, ..., n
(3)
26
11.
PRIMITIVE RINGS
Proof. 1. If (0 : B ) = M there is no set of elements which is linearly independent modulo (0 : B ) and the theorem is trivially true. If (0 : B ) C M , it is sufficient to show that, for each given set {u, , ..., u,} which is linearly independent modulo (0 : B), there exist n elements b, , ...,b, in B such that
bLuK= 0
if
i
#
and
K
b,u, # 0
(4)
Suppose we have shown the existence of the b , . The irreducibility of the Rmodule M implies that the submodule Rb,u, is either (0) or M . Now (0 : R )
=
{m I am
for all a E R )
=0
is a submodule of M since m, and m 2E (0 : R) imply rn,  m2 E (0 : R ) and Rxm, C Rm, = (0) for arbitrary x E R. If Rb,u, = (0) then 0 # b,u, E (0 : R) and (0 : R) = M , contradicting the irreducibility of M . Hence Rb,u, = M and there exists an element a, E R such that
a,b,u,
=
a,b,uK = 0
and
v,
if
1
#
K
(5)
+ +
The element b = alb, ... anbn E B satisfies (3). 2. We shall prove the existence of the elements 6, , ..., b, satisfying (4) by induction on n. When n = 1 the existence of b, such that b,u, # 0 follows since u1 4 (0 : B ) . Assume inductively that for every modulo (0 : B ) linearly independent set {w, ,..., w,~} there exists b E B such that
bw,
=
..'
=
bwnPz= 0 ,
bw,, # 0
(6)
We must show the existence of an element bn E B such that
The subset L The subset
= {I
I I E B, Iu,
=
Lu,,
... = Iu,~ = {Iun,
= 0} is
I 1E L}
clearly a left ideal of R. (8)
of M is an Rsubmodule of M since RLu,, C Lu,, . {u, , ..., u+,} is linearly independent modulo (0 : B) and by the induction hypothesis there exists b E B such that bu1  ... = bun4 = 0, bun, # 0 i.e. there exists an element b E L with bun, # 0. Thus the submodule Lu,, is not zero and hence Lu,,
=
M
(9)
1.
DENSE RINGS OF LINEAR TRANSFORMATIONS
Assume by way of contradiction that there is no element b i.e. assume
IU,~
=
lu,
implies
0
=
E
B satisfying (7),
for all Z E L
0
27
(10)
Then
4:
+
for all I
lu,
E
(11)
L
is a welldefined mapping of M = LunPl into itself since ZlunPl = Z2u,l implies (Il  12) unP1 = 0, giving llun = 12u, by our assumption. The mapping 4 is Rlinear:
+
(llUn1
I2~nd
4 = (I1
+
12)
un14 = I1un
+
ZzUn
and
(4llunJ) 4 = ((all) un1) for all I1 , I2 the form,
E
L. Therefore
4EK
Zu,lq3
4
=
= Hom,(M,
=
lu,
(all) un
= 4llun)
M ) . We may write (11) in
for all 1 E L
or as Z(unl$

u,) = 0
for all
ZE L
(12)
On the other hand, any nontrivial linear combination of { u l , ..., u n T Z , unP14 u,} may be considered as a nontrivial linear combination of {ul , ..., u,~ , u,} since 4 E Hom,(M, M ) . Hence {ul , ..., unP2, u,.~+ u,} is linearly independent modulo (0 : B ) and we may apply the induction hypothesis to obtain an element b E B such that bu,
= .'. = bun2 =
0,
b(un14

un) # 0
Such an element b lies in L and fails to satisfy (12). Hence our assumption (10) is false and an element 6, E B satisfying (7) exists. I Put R = B in Theorem 3 and assume that M is a faithful module. Then R is a subring of Hom,(M, M ) and we have the following corollary to Theorem 11.1.3: Theorem 4 (Jacobson density theorem). Let K be the centralizer of the faithful, irreducible Rmodule M . Then R is dense in Hom,(M, M ) . I
This theorem, of fundamental importance in the theory of rings, suggests that we should give a special name to the class of rings possessing a faithful irreducible module.
28
11.
PRIMITIVE RINGS
Definition 5. A ring R is primitive if it possesses a faithful, irreducible Rmodule.
In Theorem 4 we have proven the harder half of Theorem 5. Theorem 5. A ring R is primitive if and only if it is isomorphic to a dense ring of linear transformations of a vector space M # (0) into itself.
To prove the implication the other way we need only observe that by the definition of equality of mappings, M is a faithful module and the density implies Ru = M for any element u # 0. 2. The finite topology
We shall now justify the use of the term “dense.” The toplogical concepts introduced in this section are not essential to the development of the theory in the next seven chapters. In Chapter XII, however, these concepts will play an essential role. Until then the reader is only required to know the following: 1. By a neighborhood U(a; u l , ..., u,) of a linear transformation a of a vector space A4 into itself we mean the set of all linear transformations c such that i = 1 , ...,n cui = a u i , 2. A subset R of the set R‘(M) of all linear transformations of M into itself is dense in Ro(M)if for each element x of Ro(M) and for each neighborhood U ( x ;u l , ..., u,) of x, U ( x ; u l , ..., u,) n R # o. Although this section may be omitted, the connection between Section 11.1 and the topological ideas introduced here will shed light on the concept of a subdirect sum in the next two chapters. We may endow a set T with a topological structure as follows (cf. van der Waerden [1959] Vol. 11, Chap. XII, or Franz [1960]): To each point p E T assign a family 9 ( p ) of subsets U ( p ) such that
U, . 3 ( p ) # o and p E U ( p ) for all U ( p ) E 9 ( p ) . U,. Given U ( p ) and V ( p ) in 9 ( p ) , there exists W ( p )E F ( p ) such that
W P ) c U(P)n VP).
U, . For each q E U ( p ) there is V(q)E S(q)such that V(q)C U ( p ) . A set T with a topological structure is called a topological space. Each family T ( p ) is called a neighborhood basis of p and U ( p ) E 3 ( p ) is called a basic neighborhood. A neighborhood of p is any set containing at least one
2. THE
FINITE TOPOLOGY
29
basic neighborhood of p. A subset 0 of T is said to be open if it is the empty set or if for each point p E 0 there exists a neighborhood of p contained in 0 . A subset S of T is closed if its complement eS is open. In the Euclidean plane we may take the set of open disks with center p as a neighborhood basis of p. Definition 1. A topological ring R is a ring such that a. the set R is a topological space p. the functions 4 and bz, from R x R to R defined by
are continuous, i.e. for each pair of elements a, b E R and each neighborhood U(a  b) (resp. U(ab)) there exist neighborhoods V(a) and W(b) such that x E V(a) and y E W(b) imply x  y E U(a  b) [resp. x y E U(ab)]. We may endow the ring R = Hom,(M, M ) of Klinear mappings of a right Kmodule M with a topological structure by taking for the family F(a) of basis neighborhoods all sets of the form U(a;u l , ..., u,) = {C I c E R, cul
=
aul , ..., C U ,
= au,}
(1)
where ( u l , ..., u,) is an arbitrary finite set of elements of M. The basic neighborhood (1) of a consists therefore of those elements c E Hom,(M, M ) whose restriction to {ul , ..., u,} coincides with that of a. Condition Ul holds trivially for these families of subsets. For U, , let U = U ( p ;u1 , ...,u,) and V = U ( p ;u l , ..., u,) be two basic neighborhoods of p . Then W = U ( p ; u1 , ..., u, , u l , ..., v,) _C U n V. In order to show U, let q E U ( q ;u1 , ..., u,) and take V = U(q; u1 , ..., u,) E 3;(q). If x E V(q), xui
=
qui =pui
for i
=
1, ..., n
and hence x E U(p). The continuity of subtraction relative to this topological structure follows immediately. For let U(a  b; u l , ..., u,) be a given neighborhood of a  b and let x E U(a; u1 , ..., u,) and y E U(b;u1 , ..., u,). Then ( X  y ) ui =
xui  yui
= aui  bui =
(a  b) ui ,
i = 1,
...,n
showing that x  y E U(a  b; u1 , ..., u,). To prove the continuity of multiplication we note first that the product ab of two elements in Hom,(M, M ) is defined by m ta(bm). Therefore,
30
11. PRIMITIVE
RINGS
if U(ab; u, , ..., u,) is given, let W(b) = U(b;u1 ,... , u,) U(a;bu, , ..., bu,). Then if x E V(a) and y E W(b) for i
x(yu,) = x(bu,) = abu,
=
and
V(a) =
1, ..., n
and hence xy E U(ab). We have proved the following:
Theorem 1. R = Hom,(M, M ) is a topological ring if, for each a E R we take the family of sets
U(a; u1 , ..., u,)
= {c
1
c E R,
cui
= au,
, i
=
1 , ..., n}
(2)
as a neighborhood basis of a. This topology is called the finite topology on Hom,(M, M ) . I
Corollary 2. Hom(M, M ) is a topological ring relative to the topology defined by (2). To prove this we merely observe that Hom(M, M ) = Hom,(M, M ) , Z the ring of integers. I Let Ol be a subset of the topological space T. An element t E T is adherent to Ol if each neighborhood of t contains at least one element of Ol (which may be t itself). The set d of all elements t E T adherent to Ol is called the closure of Ol. A subset 3 of T is dense in T if 3 = T. In the terminology introduced in the preceding paragraphs, the density theorem (Theorem 11.1.3) may be stated as follows: If M is a faithful, irreducible Rmodule, K its centralizer and a an arbitrary element of Hom,(M, M ) , then each neighborhood U(a;u, , ..., u,) of a intersects R nontrivially. In other words, R is dense in Hom,(M, M ) in the usual topological sense. This may be seen by observing that from each finite set of elements {ul , ..., u,} we may extract a maximal linearly independent subset { u i z ,..., ui,>.By the density theorem there exists b E R such that buil = m i l ,..., bui = Q U , . It follows that bu, = au, , i = 1, ..., n since each ui is a Klinear combination of u,, , ..., u i 8 . Theorem 11.1.4may be restated as follows:
Theorem 3. A primitive ring R is a dense ring of Klinear transformations of a vector space M relative t o the finite topology on Hom,(M, M ) . I Theorem 11.1.3 gives us some information about the left ideals of R .
3.
THE LATTICE OF LEFT IDEALS IN A PRIMITIVE ARTINIAN RING
31
This information may be used to characterize the closure in Hom,(M, M ) of an arbitrary left ideal. To this end we define [complementary to Definition 11.I , (2) of (0 : B)]
(0 : U) = ( A 1 A
E
Hom,(M, M ) , Au
=
0 for all u E U }
(3)
where U is a subspace of the right Kvector space M. Clearly (0 : U ) is a sctbgroup of the additive group of the ring Hom,(M, M ) and furthermore, since (CA) u = C(Au) = CO = 0 for a11 C E Hom,(M, M ) and A E (0 : U ) , it is a left ideal in Hom,(M, M ) . Using this notation, (0 : (0 : B)) consists of all those elements of Hom,(M, M ) annihilating all those elements of M annihilated by B. Hence
B C (0 : (0 : B))
(4)
Theorem 4. Let K be the centralizer of the faithful irreducible Rmodule M . Then the closure (relative to the finite topology) of a left ideal B is the left ideal (0 : (0 : B)) in Hom,(M, M ) .
Proof. 1. Let A E B. If (0 : B ) = 0, (0 : (0 : B ) ) = Hom,(M, M ) contains A . Hence assume 0 # u1 E (0 : B). Then the neighborhood U(A; ul) contains an element b E B. But bu, = Au, , by the definition of the finite topology and bu, = 0. Therefore Aul = 0. u, is arbitrary in (0 : B ) and hence A E (0 : (0 : B)). 2. Conversely, let A E (0 : (0 : B ) ) . We must show that every neighborhood U ( A ; u1 , ..., u,) of A intersects B nontrivially. Let {ui,, ..., ail) be a maximal modulo (0 :B ) linearly independent subset of {u, , ..., u,}. By Theorem 11.1.3, there exists b E B such that bui, = Auil ,j = 1, ..., s. But ut = uilk,t
+ ... + tri,k,t + ~
t
,
t = 1, ..., n
for suitable lcjt E K and vt E (0 : B ) . Therefore but and hence b E U ( A ; u l , ..., u,). I
= Aut
for t
=
1, ..., n
3. The lattice of left ideals in a primitive Artinian ring
We may simplify the theorems of the last two sections considerably if we assume that the Kvector space M is finite dimensional. In fact, if dim, M = n < co and { u l , ..., u,} is a basis of M , an element A E Horn,(M, M ) is in the neighborhood U(B; u l , ..., u,) if and only if A = B. In this case the finite topology is the discrete topology, i.e. each singleton { A } is a neighborhood. Therefore, if A4 is a faithful, irreducible
32
11.
PRIMITIVE RINGS
Rmodule of finite dimension over its centralizer, the density theorem (Theorem 11.1.4) implies that R = Hom,(M, M ) and 3 = B, B a left ideal. Hence (0 : (0 : B)) = B by Theorem 11.2.4. The mapping B+(O:B) of the set of left ideals into the set of subspaces of M is onto, i.e. every subspace U is of the form (0 : B ) for a suitable left ideal B. This follows from (0 : (0 : U ) ) = u (2) which is proved as follows: (0 : (0 : U ) ) consists of those elements m annihilated by (0 : U ) . Therefore
E
A4
u 2 (0 : (0 : U ) ) Assume that w E (0 : (0 : U ) )  U.Since dim, M < 03, let {ul , ..., u,} be a basis of U. Then {u, , ..., u, , w} is a linearly independent set since w $ U. By Theorem 11.1.4 there exists a E R = Hom,(M, M ) such that all, =
...  au,,
aw # 0
This implies a E (0 : U ) and w $ (0 : (0 : U ) ) , a contradiction. Let U , and U, be two subspaces of M . Then
u, c u,

(0 : U,) I ( 0 : U,)
(3)
* (0 : B,) 2 (0 : B,)
(4)
and similarly for left ideals
B, C B,
The set of subspaces of M and the set of left ideals of R form partial orders under set theoretic inclusion. The implications ( 3 ) and (4) together with (0 : (0 : U ) ) = U and (0 : (0 : B ) ) = B prove the following: Theorem 1. Let M be a faithful, irreducible Rmodule of finite dimension over its centralizer K . Then the mappings
+:B*(O:B),
#:Ut(O:U)
are inverses of each other and reverse inclusion, i.e.
B, C B, o (0 : B,) 2 (0 : B,).
I
(5)
3.
THE LATTICE OF LEFT IDEALS IN A PRIMITIVE ARTINIAN RING
33
The mapping 4 is called an antiisomorphism of the partial order of left ideals in R onto the partial order of subspaces of M .
Definition 1. Let M be an Rmodule. The meet Ul n U, of two submodules U, and U, is the set theoretic intersection
Uln U,
= (m
I m E Ui,
i = 1,2}
(6)
Their join U1 u U, is the smallest Rsubmodule of M containing Ul and U, , namely Ulu U, = {u, u2 1 uiE U i , i = 1 , 2) (7)
+
The join is also called the sum because of (7) and is denoted by U,
+ U, .
We may easily verify that U, u U, is the smallest submodule containing U, and U, and that dually, U, n U, is the largest submodule contained in both U , and U , . It follows that the mapping 4 of Theorem 1 has the following two properties: 01.
W
P.
[email protected] l
n B,) = Wl) u W,) u B,) = ml)n 4(BJ
We are of course applying Definition 1 to the submodules B of the left Rmodule RR and the Ksubspaces of the right vector space M . We may further show without much difficulty that the submodules of an Rmodule M under the operations of join and meet defined above form a lattice in the sense of the following:
Definition 2. A lattice C is a set of elements u1 , u2 , ... with two operations u and n satisfying 01
u
01 u 0, = u, u u1 , u1 n v2 = u2 n fil ,
01 = 01,
u1 n u1
=
u1 , U,
n (ul u 0,)
=
u1 ,
u1 u (u2 u u3) = (0, u v,) u v3 u1 n (u, n u3) = (ul n u,) n v3 (8) U, u (ol n 0,) = u1
In the lattice V ( M ) of Rsubmodules of an Rmodule M , we may define the partial ordering C,i.e. set theoretic inclusion, in terms of the operations n and u as follows:
u, C u, o U ,
=
Ul n U , o U ,
=
U1 u U ,
(9)
It has been shown that V ( M ) is a partial order (C, C) with the following properties: a. if u1 and 0, E f, there exists an element denoted v1 u u2 E C such that (i) u1 C o1 u u2 and u2 C u1 u v2 and (ii) v1 C u and v2 C u, implies u1 U u2 C u @. the dual of Property a, i.e. the statement obtained from a by rep!acing U by n and C by 1.I: is a lattice under n and u.
34
11.
PRIMITIVE RINGS
The set of subsets of an arbitrary set M under set theoretic union and intersection forms a lattice P(M). The corresponding partial ordering is set theoretic inclusion. In P ( M ) the distributive laws
and the corresponding dual
hold for arbitrary subsets Vl , Vz , V,
In the lattice of Rsubmodules these laws do not in general hold. However, (10,) is valid in the particular case that V , C V, .
Theorem 2. The lattice V ( M )of submodules of an Rmodule A4 is modular, i.e. U , C U,
for Ui E V ( M ) ,i
=
3
U,
+ ( U , n U,)
=
(u, + U,) n U,
(1 1)
1,2, 3.
+
+
Proof. Since U , C U, and U , n U, C U, C U, U, , U, (U, n U,) C (Ul U,) n U, . Conversely, let u E ( U , U,) n U, . Then u = u, U, , ui E Ui , i = 1,2. It follows that uz = u  u1 E U, since Ul C U, . Therefore u = u, u, with u2 E U , n U , and u E U1 (U, n U,). 1
+
+ +
+
+
Definition 3. Let 4 be a oneone mapping of the lattice V onto the lattice V’. 4 is an (lattice) isomorphism of V onto V’ if 4 preserves the operations u and n, i.e. if
(a u b) 4
=
a+ u b 4
and
(a n b) 4
=
(a n b) 4
= a+
a$ n b 4 for all a, b E V (12,)
4 is an antiisomorphism if (a u 6 ) 4
=
a 4 n bI$ and
u b 4 for all a, b E V (12,)
We showed in Theorem 1 that I$: B + (0 : B ) is a oneone mapping of the lattice V of left ideals B of R onto the lattice V’ of Ksubspaces of a faithful, irreducible Rmodule M of finite dimension over its centralizer K . Furthermore, 4 reverses inclusion. We state the following corollary to Theorem 1 in the new terminology:
Theorem 3. Hypotheses: Let M be a faithful, irreducible Rmodule of finite dimension over its centralizer K . Conclusion: The lattice of left ideals
3.
THE LATTICE OF LEFT IDEALS IN A PRIMITIVE ARTINIAN RING
35
of R is antiisomorphic to the lattice of Ksubspaces of M under the mapping
(6: B The inverse of
f
(0: B ) ,
B a left ideal in R
is the mapping
4: U +
(0: U ) ,
U a subspace of A4
I
Theorems 11.1.5 and 1.3.2 show that a primitive ring with a faithful and irreducible module M of finite dimension n over its centralizer K is isomorphic to the ring of n x n matrices over K. It is natural, therefore, to try to characterize those primitive rings with modules of finite dimension over their centralizers in terms of some internal property of the rings. Theorems 1 and 3 of this section give us a clue as to the property required. Theorem 1 tells us that in these rings, every descending chain of left ideals B, 3 B, I)...
(1 3)
stops after a finite number of steps, i.e. for some n, B, = B,,, = ..., since the corresponding ascending chain of subspaces U , C U , C ... stops after finitely many steps by the finite dimensionality of the vector space. A ring with the property that every descending chain of left ideals stops after finitely many steps is called a ring with descending chain condition (abbreviated, dcc) on left ideals. The dcc on left ideals is equivalent to the minimal condition on left ideals as defined here:
Definition 4. A ring R has the minimal condition on left ideals if every nonempty family 9 of left ideals of R possesses a minimal left ideal in 9, i.e. if there exists in 9 a left ideal Lo such that L E 9 and L C L, implies L = L, . Such a ring is called artinian. Proof of the equivalence. Let 9 be a nonempty family of left ideals having no minimal member. Then there is a descending chain of left ideals of infinite length since for each left ideal L, E 9,there exists at least one L,+l E 9 such that L,,, C L, . Conversely, a descending chain of infinite length forms a family with no minimal member. It should be noted that the minimal left ideal L, of Definition 4 must belong to 9. Otherwise (0) would always be a minimal left ideal for any family of left ideals in any ring [in fact, (0) is the minimum left ideal; i.e. it is contained in all left ideals]. Our discussion above shows that if dim,(M) < co, Hom,(M, M ) is artinian. The converse is also true.
Theorem 4. Let R be an artinian ring, A4 a faithful, irreducible Rmodule
36
11. PRIMITIVE
RINGS
+
and K the centralizer of M . Then dim,(M) = n is finite and n 1 is the number of left ideals in a maximal chain, i.e. a chain which cannot be refined by the interpolation of further left ideals. Proof. Let {ui1 i E I } be a basis of M . If I is infinite, {ui1 i E I } contains a countable subset {u, l j = 1,2, ...}. By Theorem 11.1.3 applied to the left ideal B = R , there exists a, E R such that anul =
=
an%,
= 0,
a,u, # 0
The left ideals L,
= {u
I a E R , au,
=
... = au,
= 0},
n
=
1,2, ...
form an infinite chain, contradicting the fact that R is artinian. The last assertion of the theorem follows from Theorem 3. I Theorem 11.1.4 enables us to state the following corollary to Theorem 4: Theorem 5. A primitive artinian ring is isomorphic to a full ring Hom,(M, M ) of linear transformations of a finitedimensional vector space M over a division ring K. Hence, by Theorem 1.3.2, a primitive artinian ring is also isomorphic to a full ring of matrices Knxn, n = dim,(M). I
Since we have already shown the converse, we have the following characterization: Theorem 6. A ring is primitive artinian o it is a ring of n x n matrices over some division ring K. I As we shall show in Section 5, we may also characterize the primitive rings in the class of artinian rings as those which are simple in the following sense:
Definition 5. A ring R is simple if a. R2 # (0) and b. R has no ideals other than (0) and R It is convenient to exclude from our investigations those rings in which the product of any two elements is 0. Such a ring R is of little interest to us since its additive structure plays the only significant role and the ideals are precisely the subgroups of (R, +). If ( R , +) has (0) and itself as the only two subgroups, then (R, +) must be cyclic of prime order, i.e. ( R , +) must be the additive group of Z / ( p ) for some prime p (cf. Example /3 in Section 1.1).
4.
37
HOMOMORPHISMS OF SEMIGROUPS, RINGS, AND MODULES; RELATIONS
In anticipation of our investigation of simple rings, we shall introduce some concepts of a general nature in Section 4. 4. Homomorphisms of semigroups, rings, and modules; relations
Let H be a set. An equivalence relation p on H is a relation (cf. Definition 1.3.2) which gives rise to a decomposition of H into mutually disjoint equivalence classes (pclasses). The class containing a is defined by
ii
= {h 1
h E H , hpa}
This means that p satisfies the following conditions apa
(reflexivity of p ) (symmetry of p )
apb tj bpa apb
(1 a)
and
bpc
* apc
(Ib) (transitivity of p )
(1c)
If H is a semigroup under 0, a congruence relation on H is an equivalence relation on H which is compatible with the algebraic structure of H. A congruence relation p on the semigroup H is an equivalence relation on H for which
alpaz 3 (a, 6 ) p(az 0 6 ) and (b a,) p(b 0 az) for all a, , a, , b E H 0
0
(2)
If p is a congruence relation on H , we may define a semigroup operation on the set Hlp of pclasses of H by
0
From (2) it follows that if a,pa and b,pb
i.e. definition (3) is independent of the choice of representative of the classes. The 0 operation in Hlp is clearly associative since the 0 operation in H is. The mapping $1
a +a,
aEH
(4)
is an epimorphismof the semigroup H onto the semigroup Hlp since $a 0 $b = 6 = a 0 b = $(a 0 b). We call it the natural epimorphism of H onto Hlp. 0
38
TI.
PRIMITIVE RINGS
Conversely, if (b is an epimorphism of the semigroup H onto the semigroup H’ we obtain a decomposition of H into disjoint subsets a, 6, ..., where
a
= {h 1
h E H , (bh = ( b ~ }
The relation defined on H by
ap,b
if
$a
=
(bb
(5)
is an equivalence relation. Since +(a 0 b) = (ba 0 (bb, it is also a congruence relation. We therefore have a correspondence between congruence relations and epimorphisms of H , namely, each epimorphism gives rise to a congruence relation and each congruence relation gives rise to an epimorphism. The inverse images of $a E H’ all belong to the same p,class and therefore the mapping a + (ba for all a E Hip, is an isomorphism of H/p, onto H’. Similar results hold for rings provided we take into account both operations and define a congruence relation on R to be compatible with the additive as well as the multiplicative structure. Therefore, p is a congruence relation on R if and
alp4 a1pa2
(a, . b) p(a2 b) ‘
* (a1
and
(b .a,) P(b . 4
+ b) p(a2 + b)
(64 (6b)
There is a very close connection between the ideals of a ring R, the congruence relations on R and the epimorphisms of R . Let (b be an epimorphism of R onto R’. We define the kernel of (b by ker (b
= {a
I a E R , (ba = 0} = {a I ap,O}
(7)
If a, b E ker (b, $(a  6 ) = (ba  (bb = 0 and (b(x . a ) = (b(x) . $(a) = 0 = +(a . x) for all x E R. This shows that ker (b is an ideal in R . Conversely, every ideal 02 in R defines a congruence relation cy. by actb
if a

bE02
and gives rise to the quotient ring RILE The null relation (only equal elements are congruent) and the all relation (any two elements are congruent) are called the trivial relations on R . Corresponding to these we have the zero ideal (0) and R , respectively. We have proved the following:
4.
HOMOMORPHISMS OF SEMIGROUPS, RINGS, A N D MODULES ; RELATIONS
39
Theorem 1. Let R2 # (0), R a ring. Then the following are equivalent: 1. R is simple. 2. The null and all relations are the only congruence relations on R. 3. A homomorphism of R is either a monomorphism or the zero homomorphism, i.e. the mapping which sends every element of R to zero. I Similar results may be proved for Rmodules. To this end, let us recall Definition 1.4.3: Let A4 and M’ be two Rmodules. An Rhomomorphism of M into M’ is an additive map 4 from ( M , +) to ( M ‘ ,+) satisfying
Analogously we define Rmodule monomorphisms, epimorphisms, etc. Such a mapping 4 not only preserves addition but also the structure of M as an Rmodule. The kernel of 4 is the set ker C$
= {m 1
m E M,
= 0)
Since $(urn) = a$m, ker #J is an Rsubmodule of M . Conversely, every Rsubmodule U of M is the kernel of an Rhomomorphism, namely the natural epimorphism defined by x:m+m+U={m+u1u~U)
for m E M
+
The coset % i =m U is an element of the factor module M/U. Addition and multiplication by elements of R are defined by 
m,

+ m2 = m, + m,
and

am
=

am
By definition a relation on a set H is a subset of the Cartesian product H x H = {(h, , h,) I hi E H , i = 1, 2). Since the subsets of any given set form a partial order under set theoretic inclusion C,this is also the case for the set of all relations on H. Call this set P. P contains, in particular, the null relation v = {(h, h) 1 h E H ) and the all relation 01 = H x H. We say the relation p, is finer than the relation p z on H if p1 C p, . In this case we also say that p, is coarser than p l . In terms of elements from H this means that hlp,h, => hlp,h,. The set of equivalence relations on H form a partial order contained in P. This set contains v and 01. Now given a set pi , i E I of equivalence relations on H, the set theoretic intersection
a=npi i€I
is also an equivalence relation on H . Therefore, given an arbitrary relation o
40
11.
PRIMITIVE RINGS
on H there exists a uniquely determined finest equivalence relation on H which is coarser than u, namely
n{ p I p is an equivalence relation on H
and
(T
C p}
This set is not empty since a belongs to it. If H is endowed with some algebraic structure (e.g. if H is a semigroup) then the set theoretic intersection of congruence relations pi , i E Z is also a congruence relation. The all relation is still a congruence relation on H . Therefore, as in the case of equivalence relations, given an arbitrary relation u on H there exists a uniquely determined congruence relation on H which is finer than any congruence relation coarser than u, namely
n { p I p is a congruence relation on H
and
(T
C p}
Since the set P of all relations on H consists of the set of all subsets of H x H , P forms a distributive lattice. In general, however, this is not the case for the set of equivalence relations. It is true, as we have seen above, that the intersection of two equivalence relations is an equivalence relation and that to given equivalence relations p1 and pz there exists an equivalence relation finer than any equivalence relation coarser than the settheoretic union p1 u pz , namely, p1 U p2 =
n{ p I p is an equivalence relation
and p1 u p2 c p}
Nevertheless, in general p1 u pz is different from p1 u pz . The set of equivalence relations form a lattice under n and U. However, this lattice is not necessarily a sublattice of P since in general u and u do not coincide. Similar remarks apply to congruence relations in the case H is endowed with an algebraic structure. If H = M , an Rmodule, there is a oneone correspondence between the congruence relations p on M and the Rsubmodules of M ; for let p be a congruence relation on M and let the kernel ker p of this relation be the preimage of the 0 element of M under the natural epimorphism of H to H l p associated with p. Then ker(pl n pz) = ker p1 n ker pz and ker(pl u pz) = ker p1 ker p2 . The analogous result holds for congruence relations on rings. In this case ker p is an ideal. It is to be noted, however, that not all congruence relations on a semigroup are uniquely determined by their kernels.
+
5. Simple rings with minimal left ideals We shall first characterize the simple rings in the class of artinian rings:
5.
SIMPLE RINGS WITH MINIMAL LEFT IDEALS
41
Theorem 1 (Artin). An artinian ring R is primitive if and only if R is simple. Proof. 1. By Theorem 11.3.5, we may take R = K,,, . The n x n matrices Cij with 1 in the (i,j)th entry and 0 elsewhere form a basis of K,,, over K. Let n
a For some pair ( j , ,k,), contains
=
C
Cjkajk
j.k=l
cij,k,
# 0,
oljk E
# 0 and any ideal
K
(1)
a C Knxncontaining a also
Ci,= CijQa CkQ&,iQ Clearly 02 contains every linear combination of such elements and hence LT 2 R. Therefore R is simple. 2. To prove the converse we weaken the hypotheses and prove a more general result. Assume therefore that R is simple and possesses at least one minimal left ideal L in the family of left ideals different from (0). We shall show that L, considered as a submodule of RR, is faithful and irreducible. Clearly the only submodules of L are (0) and L. The irreducibility of L is a consequence of (2) below since L2 # (0) implies RL # (0). To prove L is faithful, consider the set (0 : L )
= {a I
a E R, aL
= (0))
RaL = (0) if a E (0 : L ) and since RL C L, aRL = 0 showing that (0 : L ) is a twosided ideal. By the simplicity of R, either (0 :L ) = (0)or (0 :L ) = R. If the latter, then RL = (0) and L2 = (0) also. This is a contradiction since R simple
=. L2 #
(0),
L a nonzero left ideal
(2)
Proof of (2). L f LR is a twosided ideal since
R ' (L
+ LR) C L + LR
and
( L f LR) R
CL
+ LR
Assume L2= 0. Then R2 = ( L f LR)2 = L2 f LR f LRL f LRLR C L2f L2R = (0)
contradicting the simplicity of R. From (2) it follows that RL # 0 and therefore (0 : L ) = (0).Hence L is a faithful, irreducible Rmodule and R is primitive. Since an artinian ring has at least one minimal left ideal, Theorem 1 is proved. I However, we have proved something stronger, namely:
42
11.
PRIMITIVE RINGS
Theorem 2. A simple ring with at least one minimal left ideal L is primitive and L itself is a faithful irreducible Rmodule. I We may characterize such subrings of Hom,(M, M ) with the help of the following:
Definition 1. a E Hom,(M, M ) has finite rank if dim,(aM) is finite. Theorem 3. The set of elements of finite rank in Hom,(M, M ) form an ideal GZ in Hom,(M, M ) . Proof. A linearly independent set of elements of M may be mapped by a Klinear mapping to a linearly dependent set. Therefore we have the following inequalities:
+
dim,(a  b) M < dim, aM dim, bM dim,(ac) M = dim, a ( c M ) < dim, aM dim,(ca) M = dim, c(aM) < dim, aM for a, b E C!? and c E Hom,(M, M ) .
I
Corollary 4. Let R be a subring of Hom,(M, M). Let GZ be as above. Then R n C!? is an ideal in R. Proof. Let a, b E R n 02 and c E R . Then a  b E R n 02.Furthermore, since Ol is an ideal in Hom,(M, M ) , ca, ac E R n a. I
The elements in Hom,(M, M ) of finite rank may be represented (as in Section 1.3) by socalled rowfinite matrices, i.e. matrices with at most finitely many nonzero rows. This is a stronger condition than the one discussed in Section 1.3, where each column had only finitely many entries different from zero. In the characterization of simple rings with at least one minimal left ideal as primitive rings all of whose elements are of finite rank (cf. Theorems 6 and 7), we make use of “idempotent elements.” For a better understanding of the significance of these elements, we introduce the following notion. A projection of the Kvector space M onto the Ksubspace U of M is a Klinear transformation E of M onto U which leaves each element of U fixed. In symbols E E H ~ ~ , ( MM), ,
EM
=
U,
Eu
=
u,
for all
U E
U
(3)
5.
SIMPLE RINGS WITH MINIMAL LEFT IDEALS
43
It follows that
Em
= E2m
for all m E M
and hence E = E2. Conversely, if E E Hom,(M, M ) and E2 = E, then E is a projection of A4 onto EM since for each u = Em E EM,
Eu
= E2m = Em = u
Projections are closely related to the decomposition of vector spaces as sums of subspaces. Let E = E 2 and let
(0 : E ) = {m I Em
=
0)
+
Then M = (0 : E ) E M , (0) = (0 : E ) n EM, i.e. M is the sum of two subspaces whose intersection is (0). When this is the case we say M is the direct sum of the subspaces and write
M
=
(0 : E ) @ E M
as a Kspace
(4)
Since (Em) k = E(mk), EM and (0 :E ) are subspaces. Let m E EM n (0 : E). Then m is both fixed and annihilated by E and hence is zero. Since m=Em+(mEm)
for m E M
(5)
the sum of the two subspaces is M . We now make the following:
Definition 2. An element e of a ring R is idempotent if e2 = e. Theorem 5. Let M be a vector space over the division ring K. An element E E Hom,(M, M ) is a projection of A4 onto EM if and only if E is idempotent. In this case M is the direct sum of the two subspaces EM and (0 : E). Theorem 6 (Jacobson). Let R be a simple ring and L a minimal left ideal in R. Let K be the centralizer of the Rmodule L. Then R is a dense subring of Hom,(L, L) all of whose elements are of finite rank. Furthermore, there exists an idempotent e E R which generates L, i.e. for which L = Re. Proof. By the simplicity of R and (2), it follows that L2 # (0). Since L is minimal and the left ideal L2is contained in L , L2 = L. Hence there exists I, E L such that LIo # (0). L1, is also a left ideal contained in L. Therefore L1, = L. It follows that there exists e E L such that el, = Z, and hence e21, = el, . Thus e2  e E L n (0 : lo).But L n (0 : Z,) is a left ideal contained in L and is therefore either the whole of L or (0). If L n (0 : 1,) = L, then
44
11.
PRIMITIVE RINGS
L C (0: lo), a contradiction since LI, # (0). Therefore L n (0 : lo) = (0) and e2 = e. Now Re C L and Re # (0) since 0 # e = e2 E Re. Hence Re = L. The idempotent e has finite rank. In fact dim,(eL)
=
1
(6)
We prove (6) by contradiction. Assume there are two linearly independent elements, el', el" in eL. By Theorem 2, R induces a dense ring of Klinear transformations on the Rmodule L. Hence there exists a E R such that
ael'
=
The left ideal
ael" # 0
0,
(7)
L n (0 : 1')
contains the element ae # 0. Therefore L n (0 : 1') = L showing that LI' = 0. This contradicts our assumption that el' is a member of a linearly independent set and therefore different from 0. We have proved (6) and also Theorem 6 since, by Corollary 4,the elements of finite rank in R form an ideal and e # 0 is in this ideal. I Theorem 6 and the following Theorem 7 characterize the simple rings with at least one minimal left ideal. Theorem 7. A dense ring of linear transformations of finite rank of a Kspace M over a division ring K is simple and possesses a minimal left ideal. Proof. 1. Let 02 # (0) be an ideal in R and c # 0 an arbitrary element of R. The basic idea of the proof is to find an idempotent eE02 which projects the faithful, irreducible Rmodule M onto the finitedimensional subspace c M . It then follows that
ecm
=
for all m E M
cm
and hence c = ec E 02 showing that 02 = R. 2. Let x # 0 and a # 0 be two elements of M and 02, respectively, and let { y , , ..., y,} be a Kbasis of aM. Then there exists x1 E Msuch that ax, = y , and by the density of R, there exists b E R such that by, = X, ,
by,
=
= byn = 0
Therefore baM
=
and
x,K
bax,
= x1
Since x # 0 and x1 # 0, there exist c1 , c2 E R such that C1X =
x,,
czx, = x
5.
SIMPLE RINPJS WITH MINIMAL LEFT IDEALS
45
Form
e Then
c,bac,M
= c,bac, E @
C c,baM
= c,xlK = xK
and each element xk, k E K, is left fixed by e. Therefore e projects M onto xK and e E @. 3. At this point we may prove the existence of a minima1 left ideal in R. By Theorem 5 M = eM @ (0 : e ) (8) Since eM = xK, a subspace of minimal positive dimension, it is natural to conjecture that the annihilator of the maximal subspace (0 : e), i.e. L = (0 : (0 :e)), is a minimal left ideal. To prove this, let 1, and 1, E L, 1, # 0. Since the Rmodule M is faithful, it follows that llM # 0. 1, annihilates (0 :e ) and therefore by (8) 0 # 1,M = 1,eM = 1,xK The density of R implies the existence of a E R such that
allx
=
l,x
The element all  l2 annihilates xK = eM as well as (0 : e ) and therefore, by (8), annihilates M . Hence all = 1, proving that if 1, E L , ll # 0, then Rl, = L. L is therefore irreducible as a left Rmodule, or, in other words, L is a minimal left ideal. 4. Now let c E R. If dim, cM = 1, we have shown in Part 2 that there exists an idempotent e E Gi? such that eM = cM. Let U be a subspace of M and let dim, U = n. Assume inductively that, given an arbitrary (n  1)dimensional subspace U",there exists e" E @such that e" projects M onto U". U contains a onedimensional subspace U' and by Part 2, there exists an idempotent e' which projects M onto U ' : U' = e'M. Consider U" = U n (0 : el)
a subspace of U. Clearly U" n U'
= (0). Furthermore
u = U" @ U' since each eIement of U is expressible as u = e'u
and
+ (u  e'u)
e'(u  e'u) = 0
(9)
46
11.
PRIMITIVE RINGS
But dim, U" = dim, U  dim, U' = n  1 and therefore by induction there exists an idempotent errE @ which projects M onto U". Consider now the element e = e' + e"  ene' contained in Gl. We claim that e projects M onto U. Proof eM and ex ex
e'x = e'x =
C e'M
+ e"M + e"e'M = U' @ U"
+ e"x  e"e'x = e'x = x + e"x  e"e'x = e"x = x
for all x E U' for all x E U" C (0 : e')
We have shown that for each finite dimensional subspace U of M we can find a projection e E @ such that eM = U . By the remarks made in Part 1, the theorem is proved. I The above generalization of simple artinian rings is not only of interest in the field of algebra but also in functional analysis. 6. Isomorphism theorems
We have already remarked in Theorem 11.5.1 that every minimal left ideal L of a simple ring R is a faithful irreducible Rmodule. We now ask if there is any relationship between L and the other faithful, irreducible Rmodules. The following theorem gives us some information in this direction. Theorem 1. Let R be a primitive ring with a minimal left ideal L. Then each faithful, irreducible Rmodule M is isomorphic to the left Rmodule L. Proof. An isomorphism of an Rmodule M onto an Rmodule M' (cf. Definition 11.4.1) is an isomorphism C$ of the additive group ( M , +) onto ( M ' , +) which satisfies the additional condition
+(am) = a+(m)
for all m E M and a € R
Since L # (0) and M is faithful, there exists an element u E M such that Lu # (0). RL C L and therefore Lu is an Rsubmodule of A4 which, by the
irreducibility of M , is the whole of M . Consider the mapping +:E+lu
forall
IEL
6. Since (al) u The kernel
= a(lu),
4
ISOMORPHISM THEOREMS
47
is an Rmodule epimorphism of L onto M = Lu. ker
4 = { I 1 I E L,
Zu = 0}
is a left ideal contained in L. By the minimality of L and since Lu # (0), ker 4 = (0). This theorem shows that any two faithful, irreducible Rmodules are isomorphic. The same result also holds for the centralizers of two faithful, irreducible Rmodules. Theorem 2. Let R be a primitive ring with a minimal left ideal L. Let M and M' be faithful, irreducible Rmodules. Then the centralizer K of M is isomorphic to the centralizer K' of M ' . Proof. By Theorem 1, M' and M are isomorphic as Rmodules under some isomorphism a. Let k E K. We must assign to k an endomorphism k n E K'. We define kn by k": m' + u((alm') k )
for all m'
E
M
(1)
Clearly k" is additive. Since k is in the centralizer of M and a is an Rmodule isomorphism, (am')k" = a((a'(am')) k) = ~((~(ulm')) k) = a(a(olm') k ) = ao((alm')k ) = a(m'k') Therefore ka E HomR(M', M ' ) = K'. By using the Rmodule homomorphism T = al of M ' to M , we may show that each element of K' occurs as the image of exactly one element k E K:
k'
,k'r
+klro
 k'
for all k'
E
K'
Thus 4:k 4 k' is a oneone mapping of K onto K'. The additivity of 4 follows from the additivity of the Rmodule isomorphism u. This may be seen by substituting k, k, for k in (1). 4 is also multiplicative since
+
(m'k")A"
[~((alm')k ) ] P = o((u~[u((u~~wz') k ) ] )A) k ) A) = u((alm'(k . A)) = m'(k . A)" =
= u(((u1m')
This shows that the mapping k + ko for k E K is an isomorphism of the division ring K onto the division ring K'. 1
48
11.
PRIMITIVE RINGS
If we put m'
= urn
in (1) for a suitable m E M , we obtain
u(mk) = (um)ka
for m E M and k E K
(2)
This shows that in general the Rmodule isomorphism u of M onto M' is not Klinear but only Ksemilinear in the following sense:
Definition 1. Let M and M' be vector spaces over the division rings K and K , respectively. Let k + k* be an isomorphism of K onto K . A semilinear map $ of M into M' is a homomorphism of ( M , +) to ( M ' , +) together with an isomorphism k 3 k* of K onto K* such that *(mk)
=
*Wk*
The isomorphism k + k* is said to be associated with #. So far we have considered different modules over the same ring R. If R and R' are isomorphic rings under some isomorphism S, we may define a semilinear map from A4 onto M '. This map will naturally depend on S. Theorem 3. Let M and M' be faithful, irreducible modules over the rings R and R', respectively. Let the division rings K = Hom,(M, M ) and K' = Hom,,(M', M ' ) be their centralizers. Let each ring contain a minimal left ideal and let S:a+aS, aER be an isomorphism of R onto R'. Then there exists a oneone semilinear map u of the Kspace M onto the Kspace M' which is related to S in the following way:
as
=
uaul
for aE R
(3)
i.e.
o(am) = aSum
for aE R, m E M
(4)
Proof. M' may be made into a faithful, irreducible Rmodule by defining a 0 m'
= asm'
for a E R, m'
E
M'
(5)
By Theorem 1 there exists an Rmodule isomorphism u of the Rmodule A4 onto the Rmodule M'. Therefore
u(am) = a 0 urn and (4) is proved. If we set um
aaulm'
=
=
aSum
for all m E M
m', we get
= asm'
for all m'
E
M'
6.
ISOMORPHISM THEOREMS
49
showing cracrl =aS. CI is a semilinear map of the Kspace M onto the Kspace M' since the centralizer of the Rmodule M' is the same as the centralizer K' of M' as an R'module by (5). 1 Conversely, if a is a oneone semilinear map of the Kspace M onto the K'space M', we may define an isomorphism of the ring Hom,(M,M) onto the ring HOm,,(M', M ' ) by for a E Hom,(M, M )
S: a +ua0l
By Theorem 1, the faithful, irreducible modules over a primitive ring R containing a minimal left ideal L are all isomorphic to L. In this case, Theorem 2 shows that the centralizers of these modules are also uniquely determined up to isomorphism. These facts suggest that we should try to describe the centralizer in terms of the internal structure of the ring. Let e be an idempotent generating the minimal left ideal L. We showed in Theorem 11.5.6 that dim,(el) = 1
where K
=
HomR(L,L)
This equality gives us a clue as to how to proceed. From L that eL = eRe and hence dim,(eRe) = 1.
= Re
it follows
Theorem 4. Let e be an idempotent in a ring R. Then the centralizer K of the left Rmodule Re is isomorphic to the ring eRe under the mapping
4: k + ek,
k EK
(6)
Proof. Since (ae) k = a(ek) for all a E R, k is uniquely determined by the image of e; #J is therefore a oneone mapping of K into Re. Now ek = re for some r E R. But ek = (e . e) k = e(ek) = ere and therefore ek E eRe. To show that #J is onto, let ere E eRe. Then the mapping
t,k aeae
*
ere
(7)
for a E R
is an element of HomR(re,Re) since by (7) x(ae)
= (xu) e +
xae . ere
= x(ae
*
ere)
The image of e under 4 is ere showing that #J is onto. We have still to show that #J is a homomorphism. Let ki E K and eki = er,e. Then #J(kl
+ k,)
= e(kl
+ k,) +
=
ek,
+ ek, = erle + erze
= 4kl #Jkz #J(klk,) = (ek1) k, = @le> k2 = (er,e)(ek,) = (er1e>(erze)= (#Jk1)(4kz)
1
50
11.
PRIMITIVE RINGS
Theorem 5. A minimal left ideal L in a primitive ring R is generated by an idempotent e. R is a dense ring of linear transformations of the vector space Re over the division ring eRe. Proof. In the proof of Theorem 11.5.6 we showed that L is generated by an idempotent if L2 # (0). We therefore prove L2 # (0). 1. The product of two ideals A # 0, B # 0 is diferent from zero. Proof: AM = M and BM = M if M is a faithful, irreducible Rmodule. Therefore
(AB)M
=
A ( B M ) = M # (0)
2. R contains no nilpotent left ideals L, # (0), i.e. no left ideal L1 # (0) such that L," = (0) for some natural number n (cf. Definition 111.1.3). Proof LIn = (0) implies (L,R)n+l
=
L,(RL,)" F
C LY+lR = (0)
Therefore by Part 1 it follows that the ideal L,R C={alaER,
= (0).
But the set
aR =0}
is not zero since C 1 L, . Therefore C and R are two ideals with 0 product, contradicting Part 1. 3. Hence L2 # (0) and L = Re for some idempotent e. By Theorem 1 L is a faithful, irreducible Rmodule and by Theorem 4 its centralizer K is the ring eRe. I 7. Modular maximal left ideals
In Section 6 we showed that all faithful, irreducible modules over a primitive ring R containing a minimal left ideal L are isomorphic to L. If we do not know of the existence of minimal left ideals in R but assume instead that R contains an identity I , we may first prove the existence of maximal left ideals as follows: Let 9 be the family of left ideals not containing 1 and let ... C Li c Lk c ... be a chain of members of 3;. Then the settheoretic union (J Li does not contain 1 and hence (J Li E 3;. Clearly Liis an upper bound of the chain and therefore by Zorn's Lemma (cf. Section 1.3), there exists a maximal left ideal L # R in 9. Every ring with identity therefore contains a maximal left ideal. The maximality of L implies that the factor module ,R/L is an irreducible Rmodule since there is a oneone correspondence between
u
7.
MODULAR MAXIMAL LEFT IDEALS
51
the submodules of RR containing L and the submodules of .R/L. If all irreducible Rmodules were isomorphic, R R / Lwould be a representative of the isomorphism class. If, however, R does not contain minimal left ideals, the irreducible Rmodules need not be isomorphic to one another. Nevertheless, we shall show that each irreducible Rmodule is isomorphic to .R/L for some maximal left ideal L. Furthermore, we shall see that we may drop the assumption that R contain 1 and replace it by the weaker hypothesis that R contain an element e which is a right identity modulo L, i.e. an element such that x  xeE L
for all
X E
R
(1)
Thus, let M be an irreducible Rmodule and u # 0 an element of M such that Ru # 0. Then Ru = M and the mapping a
+ (0 : u )
f
for all a E R
au
(2)
is an Rmodule isomorphism of the factor module RR/(O: u ) onto M as may be verified without difficulty. From Ru = M it follows that there exists e E R such that
eu This implies xeu
= xu
x
(3)
=u
and hence 
xe E (0 : u )
for all x
E
R
(4)
The element e is therefore a right identity modulo (0 : u). Furthermore, (0 : u) is a maximal left ideal in R since any left ideal L' properly containing (0 : u ) is mapped under the isomorphism in (2) onto the whole of M and hence L' = R. The left ideal (0 : u ) is a maximal modular left ideal in the sense of Definition 1. A left ideal L of a ring R is modular if there exists an element e which is a right identity modulo L , i.e. an element e E R such that
x
= xe(mod L)
for all x
E
R
(5)
or, equivalently, such that {x

xe I X
E
R} C L
If R has an identity, every left ideal in R is modular. Corresponding to a given element e E R there is a modular left ideal L with.e as right identity modulo L, namely L = ( x  xe I x E R}.
52
11. PRIMITIVE RINGS
Theorem 1. Every modular left ideal L different from R is contained in a modular. maximal left ideal L'. Proof. Let e be a right identity modulo L and 0 the union of the members of the chain L C ... c Li c Lk C ...
(7)
of ideals containing L but not containing e. Then e $0.By Zorn's lemma there exists a left ideal L' in R containing L but not e and maximal subject to these conditions. L' is also maximal in R since any left ideal L" properly containing L' contains e and hence R. The connection between modular, maximal left ideals and irreducible modules is established in the following:
Theorem 2. The Rmodule M is irreducible if and only if there exists a modular maximal left ideal L such that R/L and M are isomorphic. Proof. The necessity has already been shown above. Let L be a modular, maximal left ideal in R. By the maximality of L the Rmodule RIL contains only itself and (0) as submodules. We must show that R(R/L) # 0, i.e. R2 $ L. Let e be a right identity modulo L. Then { x  xe I x E R} C L . If Re C L, then x = xe ( x  xe) E L for all x E R showing R C L, a contradiction. Therefore R2 $ L. I
+
Let L be a left ideal in R. The element c E R induces, by left multiplication, the 0map on R / L if and only if c belongs to the ideal (L: R) = { C I c E R, CR C L}
This set is a (twosided) ideal since caR C cR C L and acR C aL C L. R/L is then a faithful module over the factor ring R / ( L : R) as we show in the following more general theorem:
Theorem 3. 1 . Let M be an Rmodule and CPG an ideal contained in the ideal (0 : M ) . Then M is an R/C&module. 2. M is a faithful module over RICPG if and only if a = (0 : M ) . Proof. 1. Define multiplication by elements of RICPG by (x+CPG)m=xm,
XER, mEM
7.
MODULAR MAXIMAL LEFT IDEALS
53
Since (2 C (0 : M ) , the multiplication is well defined. The module properties are clear. 2. If O? = (0 : M ) and x $ (0 : M ) , there exists m, E M such that (x
+ ( 0 :M ) ) m , = xm, # 0 +
3. If O? C (0 : M ) , let b E (0 : M ) , b $ (2. Then b @ is not the zero element of RlGT but (b (2) M = 0. Therefore M is not a faithful R/O?module. It follows that ( L : R) is the smallest ideal Ipd in R such that RIL is a faithful R/O?module. I
+
Definition 2. An ideal P in a ring R is primitive if RIP is a primitive ring. In this terminology, our discussion above and Theorem 2 show that ( L : R) is a primitive ideal if and only if L is a modular maximal left ideal. This proves the following: Theorem 4. 1. The ideal P in a ring R is primitive if and only if there exists a modular maximal left ideal L in R such that P = ( L : R). 2. A ring R is primitive if and only if there exists a modular maximal left ideal L in R such that (0) = ( L :R). If such an L exists, RIL is a faithful, irreducible Rmodule. I Theorem 5. Every division ring K is a primitive ring. Proof.
K itself is a faithful, irreducible Rmodule.
I
Theorem 6. A commutative ring R is primitive if and only if it is a field. Proof. The sufficiency has already been established in Theorem 5. To prove the necessity we first prove
Theorem 7. If L is a modular left ideal, ( L : R ) is the largest ideal of R contained in L. Proof. Let e be a right identity modulo L. If a E ( L : R) = {c 1 CR _C L}, then ae E L .But a  ae E L. Therefore a = ae (a  ae) E L and ( L : R) C L. Let @ be an ideal of R contained in L. Since @R Z O? C L, GY C ( L : R). Now if R is a commutative primitive ring and L a modular maximal left ideal such that ( L : R) = (0), L = ( L : R) by Theorem 7 and hence L = (0). Therefore if e is a right identity modulo L, e is a (twosided) identity in R . The set W' = {ay I y E R} is an ideal in R and W' # (0) since ae = a.
+
54
11. PRIMITIVE RINGS
Therefore W’ in R. I
=R
showing that the equation ay
= b has
at least one solution
As a corollary to Theorem 7 we may state: Theorem 8. A simple ring with identity is primitive.
Proof. The identity 1 of the simple ring R is a right identity modulo every left ideal of R. By Zorn’s lemma there exists a left ideal L maximal in the family of left ideals not containing 1 . Clearly L is also maximal in R. The irreducible Rmodule R/L is also faithful. For, x . (R/L) = 0 implies xR C L and consequently RxR C L showing that x = 0. I
8. Primitive algebras Results analogous to those proved in the preceding sections of this chapter may also be proved for algebras over a commutative ring di. Definition 1. Let R be an algebra over 0.Let M be a module over R considered as a ring. Then M is an algebramodule over R if M is also a right @module such that
(am) a
=
a(ma) = (am) m,
ml
=
m
(1)
for all a E R, m E M, and a E di. Definition 2. An endomorphism C of an algebramodule M over an algebra R is an endomorphism of M , considered as a module over the ring R, such that (ma) C
=
(rnC)a
for all m E M and
a
E @
(2)
If 1 E R, l a E R for all a E @. Therefore every ringmodule M over R is also an algebramodule and every endomorphism of M considered as a ringmodule is also an endomorphism of M considered as an algebramodule. The last assertion is valid if we drop the hypothesis that 1 E R and merely assume that R M = M. Theorem 1. Let M be an algebramodule over the algebra R.Let R M = M . Then every endomorphism of A4 considered as a ringmodule is also an endomorphism of M considered as an algebramodule, and conversely.
8.
Proof. The converse is clear. Since M expressible in the form m
=
1
aimi,
PRIMITIVE ALGEBRAS
= RM,
aiE
55
each element m E M is
R , m i €M
lP
implies (aa)u
= (a’a)u
for
a, a’
au
Since M mapping
=
0
= Ru,
(aa) u = 0
implies
for a~ @,
U E
R
(4)
R/(O: u ) is isomorphic as a ringmodule to M under the
a
+ (0: u )
+ au
Therefore (0 : u) is a modular maximal left ideal of the ring R [cf. Eq. II.7(4)]. By the preceding theorem, (0 : u ) is also a left ideal in the algebra R, i.e. a E (0 : u ) implies aa E (0 : u). This proves (4). 2. If A4 is irreducible as a ringmodule, clearly it is irreducible as an algebramodule. Conversely, let M be an irreducible algebramodule and let 0 # u E M . Since
[email protected] = R, the ringmodule Ru =
[email protected] =
[email protected] is also an algebramodule. Therefore Ru = 0 or Ru = M . Suppose Ru = 0. Consider the set M , = {m I Rm = 0} This is clearly an algebrasubmodule of M and M o # (0) since u E A4, . Therefore M,, = M , implying R M = (0), a contradiction. Therefore Ru = M proving the irreducibility of M as a ringmodule.
8.
PRIMITIVE ALGEBRAS
57
Theorem 4. Let R be an algebra over @. The ideal of the algebra R is primitive if and only if it is primitive considered as an ideal of the ring R . Proof. By Section 11.7 a is primitive if and only if (2 = (0 : M ) for some irreducible Rmodule M. By Theorem 3, M is irreducible as a ringmodule if and only if it is irreducible as an algebramodule and hence the theorem is proved. I
Since the ring Z of integers is a commutative ring with 1, every ring R and every Rmodule may be considered as an algebra and an algebramodule (over Z),respectively.
Rings with a Faithful Family of Irreducible Modules CHAPTER I11
1. The Radical of a Ring Not all rings possess irreducible modules. For example, consider the ring R consisting of two elements 0 and a with addition and multiplication defined by a+a=O, a2=0 L = (0) is the only proper left ideal in R. There is, however, no right identity modulo L (cf. Theorem 11.7.4). Even if R does possess irreducible modules, there need not exist a faithful one and therefore R need not be primitive. In this chapter we shall investigate those rings possessing a family of irreducible modules which is faithful as a whole, i.e. a family of irreducible modules Ma such that no nonzero element of R is contained in every (0 : Ma). We first introduce the concept of the Jacobson radical of a ring.
Definition 1 (Jacobson). The (Jacobson) radical J of the ring R is the intersection of all ideals of the form (0: M ) , where A4 is an irreducible Rmodule: J
=
n (0:M ) ,
M an irreducible Rmodule
If R has no irreducible Rmodules, we set J = ~ @ = R
R is semisimple if J 58
= (0).
(1)
1. THE RADICAL OF A RING
59
Theorem 1. The radical of the factor ring R / J is zero.
Proof. Every irreducible Rmodule M is an irreducible RIJmodule since J C (0: M ) . Since n(0:M ) = J, = 1 a E R / J , a M = O} = (0). The family 3; of irreducible Rmodules is therefore a faithful family for the factor ring R / J provided 3; # a. 1
n{z
) :o(n
Just as in the case of primitive rings, we may define the radical of a ring internally, i.e. without involving the notion of an Rmodule.
Theorem 2. The radical of a ring is the intersection of its modular, maximal left ideals. Proof. By Theorem 11.7.2, there is a correspondence between the isomorphism classes of irreducible Rmodules and the modular maximal left ideals L of R defined by M + R/L
(2)
From our discussion preceding Definition 11.7.1, the modular, maximal left ideals are precisely those of the form L
=
(0 : u)
where u is a nonzero element of some irreducible Rmodule. Since
it follows that
We may also describe the radical of a ring as the sum of all left ideals which consist entirely of “quasiregular” elements. To introduce the concept of quasiregularity we proceed as follows: Let z $ J . Then there exists an irreducible Rmodule M o and an element 0 # u E Mo such that zu # 0. By the irreducibility of M,, it follows that Rzu = M o . Hence there exists a E R such that azu
=
u
(3)
First assume that R contains an identity 1. Then (3) may be written in the form (1  a z ) u = 0
60
111.
RINGS WITH A FAITHFUL FAMILY OF IRREDUCIBLE MODULES
This implies that 1  az is not a right unit in R, i.e. there is no element b E R such that b . ( l UZ) = 1 otherwise u = b(l  az) u = 0. Therefore if z $ J there is an element az in the left ideal Rz such that 1  az is not a right unit. In other words, the set { z I 1  az is a right unit for all az E Rz} is contained in J. In a ring without identity it is meaningless to talk about units. However, if we substitute the notion of quasiregularity we obtain, by essentially the same reasoning as in the preceding paragraph, the analogous conclusion, namely { z I rz is quasiregular for all r E R} C J To motivate the definition of quasiregularity let us for a moment assume that R has an identity. If 1  y , y E R , is a right unit, there exists b E R such that b(1  y) = 1. This element b may be expressed as 1  y’ for some y’ E R. Therefore there exists y’ E R such that (1  y’)(l y) = 1
Multiplying this out and canceling 1 from both sides we get y  y’
+ y’y
= 0.
Hence, for a giveny, 1  y is a right unit if and only if there exists y’ such that y y’  y‘y = 0.
+
Definition 2. The element z E R is quasiregular (qr) if there exists y‘ E R such that y y’  Y’Y = 0 (4)
+
A left ideal is quasiregular if all its elements are quasiregular. Now suppose R is an arbitrary ring (with or without identity). Let z E J. There exists an irreducible Rmodule M a and an element 0 # u E Ma such that azu = u for some a E R. Suppose now that the left ideal Rz is qr. Then, in particular, there exists z’ E R such that az
But since u
= azu = 0,
+ z‘  z’az
=
0
we have
0 = (u  azu)  z‘(u

azu)
= u  (az
+ z’  z’az) u = u
1. THE RADICAL OF A
RING
61
a contradiction. Therefore
{zIzER, Rzqr}CJ
(5)
To show the reverse inclusion we use Theorem 2. If z E R is not qr then the modular left ideal L = {x  xz 1 x E R># R. For otherwise we would have z‘ E R such that z‘  z‘z = z and z would be qr. The modular left ideal L is contained in a modular maximal left ideal L‘. If z E J , then z E L‘ by Theorem 2 and since L C L‘,then L‘ = R, a contradiction. Therefore z $ J. Now if the left ideal R z is not qr, there exists a E R such that az is not qr. Therefore az 4 J as we have just shown. Since J is an ideal this implies z $ J. We have proved
Theorem 3. The radical of a ring is a qr ideal containing all qr left ideals. Theorem 4.
The element y
E
I
R is qr if and only if the left ideal
L ={x

XY
I X E R }= R
Proof. If L = R, there exists x E R such that x  xy = y. Conversely, if y is qr, x  xy = y for some x E R and therefore y E L . But y E L implies Ry C L and hence R _C L. I Not every qr element z E R is in the radical J. In the ring of integers the only qr elements z are those for which
(1  z‘)(l  z ) = 1 for some z‘ E Z.Therefore z = 0 or z = 2. The ideal generated by 2 consists of all the even integers. However, 2 is the only one which is qr showing that 2 $ J. Therefore the radical of Z is zero. From y E J it follows that y y‘  y’y = 0. Since y’ = y y’y E J , y’ is also qr. Let y’ y”  y”y’ = 0. Then
+
y”
= Y” =y
+
+
+
+ (y‘ + y Y’Y)  Y”(Y’ + y  Y‘Y> + (y’ + y”  y”y’)  (y’ + y”  y”y‘)y 
=y
Hence y‘ y  yy’ = 0. Therefore each element of J is also “right quasiregular” and we obtain the same radical if we use right modules and substitute “right” for “left” throughout. It is important to note, however, that not all rings primitive on the left are also primitive on the right as Bergmann [1964] showed. It should be further noted that Jacobson uses right modules in his book [1956]. Sometimes it is difficult to determine whether an element is qr. Theorem 5
62
111. RINGS WITH A FAITHFUL FAMILY OF IRREDUCIBLE MODULES
provides us with a sufficient condition for an element to be qr in terms of the following notion:
Definition 3. An element z E R is nilpotent if there exists a natural number n such that zn = 0. The left ideal L is nil i f all its elements are nilpotent. The left ideal L is nilpotent if there exists a natural number n such that the product of any n elements of L is zero, i.e. L" = (0). Theorem 5. 1. Every nilpotent element z E R is qr. 2. Every nil left ideal is contained in the radical J of R.
+
z'  z'z = 0 for z' = z  z2  ..*  ~"1. Proof. If z n = 0, z Therefore every left ideal consisting of nilpotent elements is contained in J . I In an artinian ring the converse of Theorem 5.2 holds.
Theorem 6. Let R be artinian. Then the radical of R is nilpotent. Proof. Consider the descending chain
J 1J22 * . . Since R is artinian, there exists a natural number n such that J n = Jn+1 =
... =
J27Z
=
=
J271+1
...
(6)
Assume Jn # (0). Then by (6) J n is contained in the family 9 = { L j L is a left ideal in R, L
C J", J " L
# (0))
Hence 9 f a. Let L, be minimal in F. There exists 1 E L , such that J"I # 0. Since J" = J2", J n l is a member of 9. Furthermore J"1C L, and therefore JnI = L, . Hence there exists z E J n C J such that zl
Since z is qr, there exists z' by (7), that
0
=
1 lz '

E
=
I
R such that 0
z'(1  lz) = I  ( z
a contradiction since J"1 # (0). Therefore J"
(7)
=z
+ z'  z'z. This implies,
+ z'  z'z) I = I = (0).
I
In a commutative ring R every nilpotent element generates a nilpotent
2.
SUBDIRECT SUMS OF PRIMITIVE RINGS
63
ideal. However, it should be observed that the absence of nonzero nilpotent elements does not imply that J is zero. 2. Semisimple rings as subdirect sums of primitive rings
If the radical
J= M
n
( 0 : ~ )
irreducible
of a ring R is zero, the family of irreducible Rmoales is faithful as a whole. We may of course restrict ourselves to a family of pairwise nonisomorphic irreducible modules since
Thus let
M m M'
implies (0 : M )
=
(0 : M ' )
3i = {Mi IiEZ}
(1)
(2)
be a family of pairwise nonisomorphic irreducible Rmodules, so chosen that any irreducible Rmodule is isomorphic to some member of 9.Then J
=
n (0 : M ~ )
(3)
iCI
Let i be an arbitrary fixed element of Z. Let
a(i) = a The mapping
a+a(i),
+ (0 : Mi) R/(O : Mi) E
a € R,
ifixed
(4) (5)
is clearly an epimorphism of R onto the primitive ring Ri = R/(O: Mi). In general it is not possible to recover R from Ri in any easy way. However, we can embed R in the socalled direct sum of the R j (cf. Definition 1 below). Again, as in the case of primitive rings, the embedding need not be an epimorphism. Definition 1. Let {ai1 i E Z } be a family of rings. The complete direct sum, & ai, consists of the set of functions
a:i+a(i)eai,
iEZ
(6)
with Z as domain and their ranges in the set theoretical union of the ai, i E Z. The sum and the product are defined by a
+ b: i = a(i) + b(i),
ab(i) = a(i)b(i)
(7)
64
111.
RINGS WITH A FAITHFUL FAMILY OF IRREDUCIBLE MODULES
Clearly Cc a, is a ring. The operations in this ring depend on the operations in the aj. Given an arbitrary choice of at E a, , i E I, there exists exactly one element a E Ccai such that a(!)
=

ai
for
i E Z
Here (4) implies that the mapping which assigns to each element a E R the element i a(i) = a (0 : Mi) for i E I (8)
+
of the complete direct sum CcR, of primitive rings Ri, is a homomorphism. Furthermore, since J = (0), the mapping is a monomorphism of R into Z c Ri . This monomorphism is not in general an epimorphism. For example, consider the ring Z of integers. Let {piI i E I } denote the set of primes and (pi) the ideal generated by p i . Take {Z/(p,) I i E I } as a complete set of representatives of the isomorphism classes of irreducible Zmodules. J = (0) and the mapping a (i a (pi)) (9)
+
is a monomorphism of Z into the complete direct sum C c Z / ( p i )of fields. However, for given ai E Z it is not always possible to solve the congruences a = a, mod p i simultaneously. The image R' of R in CcR,is therefore in general only a subring of Cc Ri . We may, nevertheless say a little more: Each element a, of each subring R, occurs as the image of i under some a E R'. This implies that R is isomorphic to a subdirect sum of the Ri in the sense of the following:
zs
Definition 2. A subdirect sum mi of the rings mi, i E I, is a subring of the complete direct sum CCaisatisfying the condition that for each i,, E I and each ai, E Rio there exists a E CsGTi such that a(&) = aio
We have proved the following:
Theorem 1. A semisimple ring R is isomorphic to a subdirect sum of primitive rings Ri . I

Let GY be a subdirect sum of rings GTi , i E I. Then the mapping a a(i) for a E GY is an epimorphism of the ring GT onto the ring GYi with kernel the ideal Q, = {a E GT I a(i) = 0} in GY. Since is embedded in the complete
2.
SUBDIRECT SUMS OF PRIMITIVE RINGS
65
n
direct sum C, 0lZi of the LTi , it follows that {MiI i E I}= (0). Conversely, given a ring R and a family {ai I i E I}of ideals of R such that
then R is isomorphic to a subdirect sum of the factor rings R / a i under the mapping a + (a GZi I i E I ) for a E R .
+
Once we have shown that a ring is isomorphic to a subdirect sum of rings R i , we are faced with the problem of describing more exactly how this subdirect sum is contained in the complete direct sum. Subdirect sums which are dense in Ri relative to the topology defined below play a particularly important role.
zc
Definition 3. 1 . Let 0 2 be a ring. Define a topology on Gi! by choosing as a neighborhood basis for each a E lZ the family of sets consisting only of the subset (a}. The resulting topology is called the discrete topology. 2. Let 6Yi,i E I be a family of rings with the discrete topology. We topologize 0Tiby means of the socalled product topology of the discrete topologies on the individual (Xi’s.This product topology is defined by taking as neighborhood basis of a E CclZi the sets
xc
1 c,
U(u) = [ b b E
G& , b(ij) = u(ij), j = 1,
..., n
t
U(u) consists of those functions b on I which take on the same values as a on a preassigned finite subset {il , ..., in} of I. If I is finite, the product topology is none other than the discrete topology on 23, O l i . It is easily shown that the families U(a) defined above satisfy conditions U , , U,, and U, of Section 11.2 and that CcGZi is a topological ring under the product topology. The following theorem is an immediate consequence of the definition: Theorem 2. A subdirect sum Gi! of topological rings (Xiwith the discrete topology is dense in CcCZi if and only if for each finite subset {il , ..., in} of I and for each corresponding set of elements aik E Olik , k = 1, ..., n, there exists an element a E Gi! such that a&) =
ai, ,
Ol is then called a dense subdirect sum.
k
=
1, ..., n
If I is finite, Gi! is dense in Cc (Xiif and only if Gi! = Cc Oli
.
66
111. RINGS WITH
A FAITHFUL FAMILY OF IRREDUCIBLE MODULES
Theorem 3. Let 0' = CBGYi and let
Ai
= {a
1 a E a, a(i) = 0}
be the kernels of the epimorphisms a + a(i) of GY onto ai. Then GY is dense in CcRi if and only if for each finite subset { i , , ..., i,} of Z,
Proof. 1. Let GY be dense. For given a E GY and { i , , ..., in}C I, there exists b E GY such that b(i,) = a(il) and b(ik)= 0, k = 2, ..., n. Since (a  b)(i,) = 0,
a

bE
and
b E Ail
fi Aik
k=2
Therefore (10) is proved. 2. Suppose conversely that (10) is satisfied. Let ail be an arbitrary element in GYi, , j = 1, ..., n. Since GY is a subdirect sum, there exists at least one elemint a, E LJ! such that adid
BY (lo), a, = b,
+ d,
=
(1 1)
ai,
where b, E Ail and d, E
n
Aik
(12)
2 with p i , ui E 0 u N , p i prime and p i # p k . Then setting ai = apYpi, pi = p * p p , uai and bpi have orders p,”i and ppi, respectively. Thus one of these two elements, say ci has order p y ( p i * u r By ) . Part 2 the product of these elements ci has order the Icm of 01 and p. I Proof of Theorem 6 . Let 6 E K Xbe an element of maximal order and let k,
V.
104
TENSOR PRODUCTS, FIELDS, AND MATRIX REPRESENTATIONS
be any other element of KX. Then o(k,) I o(8) since otherwise by the lemma we could find an element of order greater than o(B). Hence every element of KX is a root of the polynomial XO(e) 1 E C,[X]. This implies that KX contains at most o(8) elements. Therefore o(KX) = o(8) and 8 generates KX. I Let L be a finitedimensional extension of K. We may prove the existence of a primitive element in L if L is a separable extension of K in the sense of the following: Definition 1. Let K be a field. 1. A Kirreducible polynomial f ( X ) in K[X] is separable ouer K if f(X)has no repeated linear factors in any splitting field off. 2. f( X )in K[X] is separable over Kif its Kirreducible factors are separable. 3. An element a in L 2 K is separable over K if a is algebraic over K and the minimal polynomial of a over K is separable over K. 4. An extension L of K with dim, L < cx) is separable over K if each element of L is separable over K.
If dim, L = n < co we say L is finite over K. In this case all elements of L are algebraic over K since if a E L, {I, a, a2,..., a"} is linearly dependent over K. Theorem 7. Let L be a separable extension of K. Then there exists a (separable) primitive element 8 in L, i.e. L = K(8) for some 8 E L.
Proof. If K contains a finite number of elements, a finitedimensional extension L of K also contains only a finite number of elements. In this case the existence of 19 is guaranteed by Theorem 6. Moreover, every a E L, is a root of (15)
X""  XE C,[X]
If X

a
is a repeated factor of (15), then
X"" x = ( X  a)Zh(X),
h ( X ) E L[X]
Differentiating formally with respect to X we have
1XO
= pnX"nl __ 1 = 2(X  a)h(X)
+ ( X  a)2h'(X)
showing that X  a is a factor of 1XO. This contradiction establishes that X""  X has no repeated linear factors in any extension field. But the minimal polynomial of a over C , divides Xu"  X and therefore it is separable. (Note: It is easily shown that the ordinary product rule of the
2.
FIELDS
105
calculus for the differentiation of the product of two functions holds also for the formal differentiation of the product of two polynomials.) 2. Suppose now that K contains infinitely many elements. Since dim, L < 00, L is obtained by the successive adjunction to K of a finite number of algebraic elements. Therefore the theorem will follow by induction if we can show the existence of a primitive element when L = K(a:, 6). Let f ( X ) and g ( X ) be the minimal polynomials of a: and p, respectively over K. Let M be the splitting field off(X) g ( X ) over L and let a: = a1 , ..., a:r and /3 = pl, ..., ps be the roots off and g in M , respectively. Since a: and p are separable both the mi and the pi are distinct. Since K contains infinitely many elements there exists c E K such that ai
+ cpj f a: + c/3
for all (i, j ) # (1, 1)
(16)
+
We show that 8 = 01 c p is the element we are looking for: Clearly K(8) C L. The polynomial
fXX) =
m  CX)
E
K(wa
has /3 as a root since f ( a ) = 0. fl and g have no further common roots in M since, if
g(Bb
=0
and
fi(BJ
[email protected] 4) =0
then 6  cpj = olio for some i o , contradicting (16). Hence the greatest common divisor of fl(X) and g ( X ) in M is X  p. But, using the fact that the greatest common divisor offi and g in K(e)[X]is expressible in the form ufl vg (cf. Section 1, Eq. (14)), it follows that X  p E K(B)[X].Therefore /3 E K(6). But cp and a: cp E K(8). Hence 01 c p  cp = a: E K(8) and therefore L = K(a, p) C K(6) C L. Since L is separable over K , 6 is a separable element.
+
+
+
Not all extensions are separable. As a counterexample, consider the following: Let K = C,( Y ) , the quotient field of C,[ Y ] . Let L be the splitting field of XY  Y E K [ X ] and let a: be a root of this polynomial in L. Thus X p

Y =X
p
 a:, = ( X  a:)”
in L [ X ]
(17)
where the last equality follows from the binomial theorem and the fact that all the binomial coefficients (3 except for (3and (@are divisible by p, the characteristic of L. Now a: does not lie in K for if OL E K , then a: = f ( Y ) [g( Y)]l, hence a:, = Y =f ( Y)” g( Y )  p and p degf  p deg g = p[degf  deg g] = 1, a contradiction. Now let h ( X ) be a Kirreducible factor of X P  Y with leading coefficient 1. Then X P  Y is a power of h ( X ) :
106
v. TENSOR PRODUCTS, FIELDS,
A N D MATRIX REPRESENTATIONS
For if j ( X ) is an irreducible factor of X p  Y different from h ( X ) , j ( X ) and h ( X ) are relatively prime. Hence lXo = a(X) h ( X ) b ( X ) j ( X ) by Sectionl, Eq. (14). This, however, leads to a contradiction since by (17) both la and j have a: as a root. Hence x*  Y = (h(X))"z
+
and therefore p = m deg h. It follows that since 01 $ K, m = 1 and X p  Y is Kirreducible. By (17), a: is an inseparable element over K and hence L is not a separable extension of K.
Theorem 8. Let K be a field of characteristic 0. Then every polynomial in K [ X ] is separable and therefore every finitedimensional extension of K is separable. ProoJ say
Suppose f ( X ) E K [ X ] has a repeated factor in its splitting field L,
f(X)= ( X  C X ) ~ ~ ( X )in t[X]
By formal differentiation we have
f'(X)
= 2(X  a ) g ( X )
+ (X

a:)z g ' ( X )
The polynomials f and f' in K[X] have the common factor ( X  a) and hencefand f' are not relatively prime. By Section 1, Eq. (14), the greatest common divisor offandf' already lies in K [ X ] .Sincefis irreducible over K, this implies that f is a factor off'. But degf' = degf  1. Therefore f ' is the zero polynomial. Since char K = 0, f ' = 0 implies f = cXo, c E K , a contradiction to the fact that deg f > 0. I
3. Tensor products Let A4 be a left Rmodule and @ = Horn,(M, M ) its centralizer. Then M is an (R, @)bimodule in the sense of the following:
Definition 1. Let R and @ be rings and let M be a right @module as well as a left Rmodule. We call A4 an (R,@)bimodule if in addition
(am)a = a(ma)
for all
C I ER,
m E M , and
a€@
Let M be an ( R , @)bimodule and let (1 be a ring containing @. We now ask if it is possible to construct an (R, A)bimodule containing an isomorphic copy of M . In the particular case that @ is a field we may proceed as follows: Let {bi 1 i E I } be a @basis. Form the set S of formal sums C bi 0h i ,
3.
TENSOR PRODUCTS
107
hi E A and hi = 0 for all but a finite number of i E I. Two such expressions are defined to be equal if the corresponding hi are equal. We define addition componentwise on S thus making it into an abelian group. For h E A define
(cbi @ X i ) h
=
1bi @ hih
and for a E R define
where
S together with the operations defined above is an (R,A)bimodule denoted by M 0A. The mapping
is an embedding of the (R,@)bimodule into the (R,A)bimodule M @ A. We shall prove this and the independence of the construction of the choice of basis in Theorem 2. Indeed, the approach we choose to take does not appeal to the existence of a basis and moreover we can do away with the assumption that @ is a field. Definition 2. Let M' be a right and M a left unital module over the ring @ with identity. Denote the elements of M ' by m' and those of M by m. The set of all expressions of the form
1
zi(mi',mi),
n E N, ziE Z
l