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 ERNST-AUGUST 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: 76-154366 AMS(M0S) 1970 Subject Classifications: 16-02, 06-A50, 20-M25, 17-A99 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 R-Linear mappings of an R-module 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 R-module 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 self-injective 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 VI-XI 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 Arens-Kaplansky 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 rings-a 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 self-injective 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 a-l be a solution of x 0 a = e, . It follows that a-1 0 a 0 a-1 Therefore if y is a solution of x 0 a-l = e, ,
e, o a o a-1
=y o
a-1 o a o a-1
=y o
a-1
= a-l.
= ec
This implies
a 0 a-l
=
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 , a-l is unique: for suppose a‘ is also an inverse of a, i.e., 0 a = e. Then
a-1
= ea-l(a’a)
a-1
= a’(aa-1) = 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~a-l=b~a-l
where a-1 is as above. It is clear that a semigroup with identity e is a group if each element a has an inverse a-l. 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 real-valued 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 one-to-one 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$) * ( b 4 )
for all a, b E GY
(10)
4 and we say that a is mapped to ad. In (lo), the dot on the left-hand side denotes the operation in GY and on the right-hand side the operation in GY'. If the only element mapped to a$ 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:<-2<
forall
<EZ
The mapping 4 does not preserve multiplication. For example, 2 . (2 . 3) # (2 * 2) . (2 . 3).
Example p. Let n be a fixed natural number. Consider the residue classes mod n of the integers H, where two integers i1and C2 belong to the same residue class if and only if l1- C2 is divisible by n. Let [ denote the residue and . on class to which 5 belongs. We may define two binary operations, the set of residue classes as follows:
+
51
+ 4,
+
= 51
and
52
__ 5152
=
5112
These operations are well defined (i.e. the sums and products are independent of the choice of representatives 5 from the residue classes) and the system consisting of the set of residue classes together with these operations is a ring denoted by Z/n. The mapping
is an epimorphism of the ring Z onto the ring Zln which is not a monomorphism because = ({ n)4.
<+
+
Example y. Not every ring possesses an identity, i.e. an element e such that
ae = ea
=a
for all U E R
For example, 2H has no identity. The identity endomorphism Id: m - t m
for all
EM
is the identity for the ring Hom(M, M ) . Let R be an arbitrary ring. W e may “adjoin” an identity to R as follows: Define two binary operations on the set
R* by
(C1
a,>
=
((5, a ) I 5 E Z ,
+ (5,
a2>= (5,
a1NL-2 9
az)
+ C2 al + a2)
and
(5,
9
R)
= (5152
7
51%
+
52a1
+a
d
8
I.
BASIC CONCEPTS
+ +
where [ a = 0, the zero element of R if 5 = 0, [ a = a ... a to [ terms if 5 2 1 and [ a = I 5 I (-a) if [ < -1. Under these operations we may easily verify that R* is a ring with (1,O) as identity. The mapping
#: a + (0, a)
for all a E R
is a monomorphism of R into R* and an isomorphism of R onto the subring of elements of R* of the form (0,a). We say that R is isomorphically embedded in R*. rf is an isomorphism of R onto R’, we may define a unique structurepreserving inverse mapping from R’ to R by
Here
4-l
4-l: a‘ + a ,
(1 1)
= a’
satisfies
#-l(a’ a’
where #a
+ b’) = #-la’ + #-lb’,
#-l(a‘b‘)= (#-lu’)(+-lb’)
+
Proof of (12). The image of +-l(a‘ b’) under # is $#-l(a’ b‘. On the other hand, since # is an isomorphism
+
(12)
+ b’) =
+ #-lb’) #+-la’ + +#-lb’ = a’ + b‘ Hence #-l(a’ + b’) and +-la’ + +-lb’ have the same image under 4. Since # +(#-la’
=
is a monomorphism, the first equality in (12) is proved. The second follows similarly. If, from an algebraic point of view, we do not distinguish between isomorphic rings, the remarks preceding Definition 6 may be more precisely stated in the following:
Theorem 4. 1. Any ring R is isomorphic to a subring of the ring of endomorphisms Hom(M, M ) of a suitable abelian group M . 2. If R possesses an identity 1, then R may be embedded in the ring of endomorphisms of the group (R,+).The embedding mapping # is defined by
#: a
---f
C,
where C, is the endomorphism of (R,+) given by C,: x + a x
for all
XE
R
C, is called the left multiplication of R by a.
3. If R has no identity we may first embed R in R* (cf. Example y ) and then apply Statement 2 to R*.
1.
EMBEDDING
9
Proof. a. The left distributivity in R gives
which proves Cu E Hom((R, +),( R , +)) p. That 4 is a homomorphism follows from
+
y. If R has an identity, is a monomorphism. For if f, = C, ax = bx for all x E R.In particular, a . 1 = b . 1, i.e. a = b. 4
, then
The image of R under 4 is a subring of Hom(M, M ) . In general, however, the image is not the whole of Hom(M, M ) . In other words, is not in general an epimorphism. In fact, let R be any noncommutative ring with 1 and let f, g be elements with f g # gJ Because of the right distributivity in R, the mapping 3tg:x+xg for ~ E ( R+) ,
+
is an endomorphism of (R, +).Suppose a is an element in R with +a i.e. C, = Yig . Then a=C;l
and hence Cs = C,
= Sg,
=%.,.I = g
. But then
= Sg
i.e. g f =f g , a contradiction. Thus 3ig is not the image of any element in R under 4. The existence of noncommutative rings was established after the end of Theorem 3. Definition 7. a. A division ring is a ring in which the set of nonzero elements form a group under multiplication. b. A field is a division ring with commutative multiplication. The rationals Q, the real numbers Iw and the complex numbers CC are examples of fields. In Section VI.3 we shall give an example of a division ring which is not a field, namely the quaternions.
10
I.
BASIC CONCEPTS
2. R-linear mappings of an R-module into itself According to Theorem 1.1.4, an arbitrary ring is isomorphic to a subring of the ring of endomorphisms Hom(M, M ) of a suitable abelian group M. We must therefore find means of describing such subrings. Particularly suited to this purpose are the R-linear mappings of a module. Both the concept of a module and that of an R-linear map have arisen as direct generalizations of a vector space and its usual linear transformations.
+
Definition 1. Let M be an abelian group with operation and let R be a ring. a. M is an R-module, or, more precisely, a left R-module, if to each ordered pair (a, m), a E R and m E M , there corresponds a uniquely determined element am in M such that
+
+
1. a(ml mz) = am, am, for all m, , m2 E M 2. (a b ) m = am bm for all m E M 3. (a . b ) m = a(bm) for all m E M
+
+
If R has an identity 1, and if lm = m for all m E M , we call M unital. p. We define a right R-module similarly. In this case, the elements a, b of R act on M from the right and hence we must have
+ +
+
for all m, , m2 E M 1'. (m, mz) a = m,a m2a 2'. m(a b) = ma mb for all m E M 3'. m(a . b) = (ma)b for all m E M
+
y. If R is a division ring, we call an R-module M a left (resp. right) vector space over R. 6. The R-module M is faithful if no nonzero element of R annihilates the whole of M , i.e. am = 0 for all m E M implies a = 0. Because of Statement 1 (resp. l'), each element of R acts as an operator on M . More precisely, each element of R gives rise to an endomorphism of the abelian group ( M , +) of the R-module M . An abelian group M is a Hom(M, M)-module and also a E-module where Z is the ring of integers. We make (R, +) into a left or right R-module, defining the action of R on (R, by ax=a.x and xa=x.a
+>
respectively, where . is the multiplication in R. It is convenient to denote these left and right R-modules 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 K-module, K a ring, then it is natural to consider
2. R-LINEARMAPPINGS OF
AN
R-MODULE INTO
ITSELF
11
those endomorphisms of the group ( M , +) which preserve the K-module structure of M in the sense of the following: Definition 2.
Let K be a ring and M a right K-module. A mapping
a E Hom(M, M ) is said to be K-linear 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 K-module, it is convenient to write the mappings of Hom(M, M ) on the right of M . Theorem 1. Let K be a ring. The K-linear mappings of the right K-module M into itself form a subring Horn&$, M ) of Hom(M, M ) . We call Hom,(M, M ) the centralizer of the K-module M . A similar theorem may be proved for left modules. Proof. We have only to show that the difference and product of two K-linear maps are also K-linear. 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 K-module 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 K-linear mappings of a suitable right K-module M . This result is a corollary to the following:
12
I.
BASIC CONCEPTS
Theorem 2. Let M be a left R-module and K its centralizer. Then the elements of R induce K-linear 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 R-module M forms a ring K. Definition 2, applied to the left R-module 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 R-module and Eq. (3), read from right
to left, completes the proof of the K-linearity of a.
I
Corollary An arbitrary ring R is isomorphic to a subring of the ring Hom,(M, M ) of K-linear mappings of a suitable right K-module 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 R-module ,R is the ring of right multiplications 31, of R and 2. R is isomorphic to the full ring of K-linear 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 K-linear 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 well-known theorems about finite-dimensional 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 K-basis {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= -cAl-l 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 K-basis, 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 K-linearly 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 set-theoretic union of the sets occurring in this chain. V is also a K-linearly 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 set-theoretic 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 one-one 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,
<
; Cu)fll G (3”; (3,”
*.*
in this partial order H , possesses an upper bound (33;C), in H , where 33 =
u (33, I
(3, ; C,) fzC},
c=
u
{C” I (33” ; e”)E
c>
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', ..., bh-l, b, . Thus {b,', ..., bh-l, b,,,}is a basis for (b, , ..., bm). Now construct a linear mapping i,h from (b,', ..., bA-l) to (b,, ..., bm-,) in the same way 4 was constructed above. Let i,h(bi')= b;, i = 1, ..., n - 1. Then as above, {b;, ..., bi-l} is a linearly independent set and ( b ; , ..., b,"-,) = (b, , ..., bm-l), 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 K-linear mappings by means of matrices with entries from K. By the K-linearity 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 K-basis {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 K-linear 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 K-linear 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 K-module over the division ring K. Let I I I be the cardinality of a K-basis 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 K-module, K a ring, it is natural to ask whether the K-linear mappings Hom,(M, M ) of A4 into itself may be given the structure of a right K-module 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 K-linear. 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 K-module, this condition implies
k'k = kk'
for all k'
E
K
4. ALGEBRAS
19
Conversely, if k satisfies (2), Ak is K-linear 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 K-module and let 2 be the center of the ring K. Then the ring Hom,(M, M ) may be given the structure of a right 2-module 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 K-module M . A similar result holds for algebras. To prove this we introduce the concept of an algebra-homomorphism (-isomorphism, -epimorphism, etc.): Definition 3. A homomorphism of the R-module M to the R-module 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 R-linear mappings of M into itself are therefore R-homomorphisms of M into itself. We denote the abelian group of R-homomorphisms 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 ring-homomorphism which is also a @-module homomorphism satisfying
+
(acx) 4 = (a4)cx
for all a E R,
01
E
@
(6)
Algebra-monomorphisms, 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 K-module 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 Z-module 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 finite-dimensional) 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 one-one 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 K-homomorphisms of a suitable R-module 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 R-module. 1. An R-submodule 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 R-module 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 R-submodule of the left R-module 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 R-module 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 R-submodule 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 R-submodule 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 R-module 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 R-submodule 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 R-submodule 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 k-I of k exists. It belongs to K, for
((am)k-l) k
= am
and
(a(mk-l)) k
= a(mk-lk) = am
proving (am) k-l = a(mk-l) since k is one-one. Clearly kk-’ and K is a division ring.
=
1
= k-lk
According to Schur’s lemma, an irreducible left R-module M is a vector space over K = HomR(M,M). M therefore possesses a K-basis. If dim,M is finite and {u, , ..., u,} is a K-basis of M , then R will induce the whole ring Hom,(M, M ) of K-linear 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 R-module, 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 K-subspace 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 R-module, 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 R-module 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 R-submodule 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 well-defined 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 R-linear:
+
(llUn-1
I2~n-d
4 = (I1
+
12)
un-14 = I1un
+
ZzUn
and
(4llun-J) 4 = ((all) un-1) 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(un-l$
-
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,
= .'. = bun-2 =
0,
b(un-14
-
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 R-module 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 R-module.
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 K-linear mappings of a right K-module 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 R-module, 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 K-linear 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 K-linear 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 K-vector 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 R-module 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 K-vector 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
R-module 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 R-module of finite dimension over its centralizer K . Then the mappings
+:B-*(O:B),
#:U-t(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 R-module. 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 R-submodule 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.
l@
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 R-module RR and the K-subspaces of the right vector space M . We may further show without much difficulty that the submodules of an R-module 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 R-submodules of an R-module 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 R-submodules 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 R-module 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 one-one 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 one-one mapping of the lattice V of left ideals B of R onto the lattice V’ of K-subspaces of a faithful, irreducible R-module 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 R-module 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 K-subspaces 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 R-module
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 finite-dimensional 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 (p-classes). 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 p-classes 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) . $(a) = 0 = +(a . x) for all x E R. This shows that ker (b is an ideal in R . Conversely, every ideal 02 in R defines a congruence relation cy. by actb
if a
-
bE02
and gives rise to the quotient ring RILE The null relation (only equal elements are congruent) and the all relation (any two elements are congruent) are called the trivial relations on R . Corresponding to these we have the zero ideal (0) and R , respectively. We have proved the following:
4.
HOMOMORPHISMS OF SEMIGROUPS, RINGS, A N D MODULES ; RELATIONS
39
Theorem 1. Let R2 # (0), R a ring. Then the following are equivalent: 1. R is simple. 2. The null and all relations are the only congruence relations on R. 3. A homomorphism of R is either a monomorphism or the zero homomorphism, i.e. the mapping which sends every element of R to zero. I Similar results may be proved for R-modules. To this end, let us recall Definition 1.4.3: Let A4 and M’ be two R-modules. An R-homomorphism of M into M’ is an additive map 4 from ( M , +) to ( M ‘ ,+) satisfying
Analogously we define R-module monomorphisms, epimorphisms, etc. Such a mapping 4 not only preserves addition but also the structure of M as an R-module. The kernel of 4 is the set ker C$
= {m 1
m E M,
= 0)
Since $(urn) = a$m, ker #J is an R-submodule of M . Conversely, every R-submodule U of M is the kernel of an R-homomorphism, 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 set-theoretic 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 R-module, there is a one-one correspondence between the congruence relations p on M and the R-submodules of M ; for let p be a congruence relation on M and let the kernel ker p of this relation be the pre-image 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 two-sided 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 two-sided 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 R-module 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 R-module. 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 K-linear 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 so-called row-finite 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 K-vector space M onto the K-subspace U of M is a K-linear 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 K-space
(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+(m-Em)
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 R-module 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 K-linear transformations on the R-module 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 K-space 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 R-module M onto the finite-dimensional 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 K-basis 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 R-module 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 R-module, 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 one-dimensional 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 R-module. We now ask if there is any relationship between L and the other faithful, irreducible R-modules. 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 R-module M is isomorphic to the left R-module L. Proof. An isomorphism of an R-module M onto an R-module 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 R-submodule 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 R-module 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 R-modules are isomorphic. The same result also holds for the centralizers of two faithful, irreducible R-modules. Theorem 2. Let R be a primitive ring with a minimal left ideal L. Let M and M' be faithful, irreducible R-modules. Then the centralizer K of M is isomorphic to the centralizer K' of M ' . Proof. By Theorem 1, M' and M are isomorphic as R-modules 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((a-lm') k )
for all m'
E
M
(1)
Clearly k" is additive. Since k is in the centralizer of M and a is an R-module isomorphism, (am')k" = a((a-'(am')) k) = ~((~(u-lm')) k) = a(a(o-lm') k ) = ao((a-lm')k ) = a(m'k') Therefore ka E HomR(M', M ' ) = K'. By using the R-module homomorphism T = a-l 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 one-one mapping of K onto K'. The additivity of 4 follows from the additivity of the R-module isomorphism u. This may be seen by substituting k, k, for k in (1). 4 is also multiplicative since
+
(m'k")A"
[~((a-lm')k ) ] P = o((u-~[u((u~~wz') k ) ] )A) k ) A) = u((a-lm'(k . A)) = m'(k . A)" =
= u(((u-1m')
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 R-module isomorphism u of M onto M' is not K-linear but only K-semilinear 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 one-one semilinear map u of the K-space M onto the K-space M' which is related to S in the following way:
as
=
uau-l
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 R-module by defining a 0 m'
= asm'
for a E R, m'
E
M'
(5)
By Theorem 1 there exists an R-module isomorphism u of the R-module A4 onto the R-module M'. Therefore
u(am) = a 0 urn and (4) is proved. If we set um
aau-lm'
=
=
aSum
for all m E M
m', we get
= asm'
for all m'
E
M'
6.
ISOMORPHISM THEOREMS
49
showing cracr-l =aS. CI is a semilinear map of the K-space M onto the K-space M' since the centralizer of the R-module M' is the same as the centralizer K' of M' as an R'-module by (5). 1 Conversely, if a is a one-one semilinear map of the K-space 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 -+ua0-l
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 R-module 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 one-one 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 ae-ae
*
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 R-module. 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 R-module 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 set-theoretic 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 R-module since there is a one-one correspondence between
u
7.
MODULAR MAXIMAL LEFT IDEALS
51
the submodules of RR containing L and the submodules of .R/L. If all irreducible R-modules were isomorphic, R R / Lwould be a representative of the isomorphism class. If, however, R does not contain minimal left ideals, the irreducible R-modules need not be isomorphic to one another. Nevertheless, we shall show that each irreducible R-module 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 R-module 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 R-module 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 R-module 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 R-module 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 0-map 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 (two-sided) 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 R-module 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 R-module. I Theorem 5. Every division ring K is a primitive ring. Proof.
K itself is a faithful, irreducible R-module.
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 (two-sided) 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 R-module 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 algebra-module 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 algebra-module 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 ring-module M over R is also an algebra-module and every endomorphism of M considered as a ring-module 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 algebra-module over the algebra R.Let R M = M . Then every endomorphism of A4 considered as a ring-module is also an endomorphism of M considered as an algebra-module, 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
l
Let C be an endomorphism of M considered as a ring-module. By (1)
1 ((a,m,)a:)C= C ((aia)mi)C = 1 (aia)(miC)= 1 (ai(miC))a: = 1 ((aimi)C)a:= (mC)a I
(ma:)C =
In order to extend the theorems of Section 7 to cover the case that R is an algebra, we make the following:
Definition 3. A left ideal L in the @-algebra R is an algebra-submodule of R R as defined in Definition 1. Analogously we define right and two-sided ideals in an algebra R. Theorem 2. Let R be an algebra over @. 1. A left ideal L in the algebra R is modular with e as right identity modulo L if and only if L, considered as a left ideal in the ring R, is modular. 2. A modular left ideal L of the algebra R is maximal if and only if L is maximal as a left ideal in the ring R . ProoJ 1. The condition { x - x e } C L is independent of the @-structure of R 2. Let L be a maximal left ideal of the ring R and let { x - xe I x E R} C L. Assume La: L for some a: E 0.Then
L$La:=R Hence e
=
1,
+ 12a
for li E L
Therefore by Definition I.4.2(5) of an algebra it follows that e2 = el,
+ elz.
=
el,
+ ( e a ) l2
But e - e2 E L and e2 E L imply e E L and hence L = R , a contradiction. Therefore LO C L and L is a left ideal of R considered as an algebra. A modular maximal left ideal of the algebra R is a modular left ideal of the ring R . By Zorn’s lemma L is therefore contained in a modular left ideal L’ of the ring R. As we have already shown, L’O _C L’ and therefore L = L’. I
56
11.
PRIMITIVE RINGS
Based on this result we may prove the following theorem which says essentially that there is no distinction between irreducible ring- and algebramodules. Theorem 3. I . Let R be an algebra over @ and M an irreducible ringmodule over R. We may define multiplication of elements of M by elements of @ in such a way that M becomes an irreducible algebra-module. 2. An R-module M is irreducible as a ring-module if and only if it is irreducible as an algebra-module.
Proox 1. Since Ru expressible as
=M
for some element u E M , every m E M is
m =au,
UER
We may then define ma
for a E R, a E @
= (au) a = (aa)u
(3)
If the mapping m + m a is well defined, we have (au)(a
. P)
=
[a431
=
( b l 4P
First, however, we must show that au E R. It is sufficient to show
= a’u
= ( ( a 4 a>P
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 ring-module 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 ring-module, clearly it is irreducible as an algebra-module. Conversely, let M be an irreducible algebra-module and let 0 # u E M . Since R@ = R, the ring-module Ru = R@u = Ru@ is also an algebra-module. Therefore Ru = 0 or Ru = M . Suppose Ru = 0. Consider the set M , = {m I Rm = 0} This is clearly an algebra-submodule 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 ring-module.
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 R-module M. By Theorem 3, M is irreducible as a ringmodule if and only if it is irreducible as an algebra-module and hence the theorem is proved. I
Since the ring Z of integers is a commutative ring with 1, every ring R and every R-module may be considered as an algebra and an algebra-module (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 R-module: J
=
n (0:M ) ,
M an irreducible R-module
If R has no irreducible R-modules, 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 R-module M is an irreducible RIJ-module since J C (0: M ) . Since n(0:M ) = J, = 1 a E R / J , a M = O} = (0). The family 3; of irreducible R-modules 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 R-module.
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 R-modules 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 R-module. 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 R-module 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 R-module 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 R-moales 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 R-modules, so chosen that any irreducible R-module 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 so-called 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 Z-modules. 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 so-called 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 pre-assigned 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
But (a, - b,)(i,) = 0. Therefore dl(il) = ail and dl(ij) = 0, j = 2, ..., n. Similarly for each 2 < m < n, there exists d, E LJ! such that dm(im)= aim and d,(ij) = 0 if j # m. Hence (d,
+ d2 + ... + dn)(ij)= a i , ,
j
=
1, ..., n
(1 3)
and GY is dense. 1 Later on in Theorem 111.3.6 we shall show that condition (10) may be simplified. Let {ai1 i E Z} be a family of pairwise nonisomorphic simple rings and let LJ! = Cs 6Ti . We show that (10) is satisfied in this case: Since @/Ai w GYi , the Ai are distinct maximal ideals in 02.Furthermore, LY2 $ Ai since GY: # (0). To show (10) is satisfied it is sufficient to prove
n
Aik
Ail
2
since the Ai's are maximal. (14) will follow if we can show A i , ..* AS, $ Ail
2.
SUBDIRECT SUMS OF PRIMITIVE RINGS
67
n2GRG1E
since AiZ ... Ain C Ad.. When n = 2, Ai2 # Ail and Aiz ,Ail maximal imply Aiz g Ail . Assume inductively that Aiz * * * Ai n-1 $ A i l . Now 012 Ail and therefore 6Y = GP Ad,
+
+ [Ai, ..*
=
(Ail
==
AizAis
Ai,,
Ain-ll)(Ail
+ Ail
+ Ai,) + Ail
proving that Ai2 .*.Ain $ Ail . By induction (15) follows and we have
Theorem 4. A subdirect sum of pairwise nonisomorphic simple rings is dense. I
By the same method we may prove: Theorem 5. If a semisimple ring R is a subdirect sum of simple ring R E , i E Z, then there exists a subset A of Z such that R is a dense subdirect sum of simple rings R, , h E A. Proof.
By hypothesis
(0)
Ai ,
=
Ai
=
(0 : Mi), and
R / A , simple
(16)
iel
If an ideal Ai appears more than once we may discard all repetitions and retain only one index h for which A A = A i . Let A denote the set of all such indices. Under these circumstances, (0) = A , and (15) is still valid with A replacing I. Theorem 5 therefore follows in the same way as Theorem 4. I If in addition the simple rings GYi are primitive, as is the case, for example, when GYi contains a minimal left ideal (Theorem 11.5.2), we may prove (15) using the concept of a prime ideal. Definition 4.
The ideal P in R is prime if BC
CP
implies either B C P
or C C P
where B and C are arbitrary ideals in R .
Theorem 6. A primitive ideal P is prime. Proof. Let P
= (0: M ) ,
M an irreducible R-module. If C $ P , there
68
111.
RINGS WITH A FAITHFUL FAMILY OF IRREDUCIBLE MODULES
exists 0 # u E M such that Cu # 0. Therefore Cu = M. If B P, there exists v E M and that Bv # 0. Hence B * Cu # 0 and BC $ P. I Now if mil is a primitive ring, Ail is a primitive ideal by Definition 11.7.2 and therefore if Ai,-..Atn C Ail it follows that at least one Aib , k 3 2 must be contained in Ail . By maximality, Aik = Ail , a contradiction, proving (15). By Theorem 11.7.6 a commutative ring is primitive if and only if it is a field. Therefore Theorem 5 implies the following:
Theorem 7. A semisimple commutative ring is a dense subdirect sum of fields. I 3. Semisimple artinian rings If a ring R is a dense subdirect sum of finitely many rings R , , ..., R , , then
since, given n elements ai E Ri , i = 1, ..., n, there exists a E R such that a(i) = ai , i = 1, ..., n. Suppose now that R is semisimple artinian. Semisimplicity implies there exists a family of primitive ideals {Ai I i E I} such that (0) = Ai . Since R is artinian on left ideals, it is, in particular, artinian on ideals. Therefore, we may choose a finite subset of the Ai’s, say A , , ..., A , , such that
nis,
(0) = A , n A, n ... n A,
(1)
Furthermore, we may assume that the set { A , , ..., A,} is irredundant, i.e. no member of the set may be omitted without affecting the intersection. We shall call (1) a representation of (0) as the intersection of an irredundant set of primitive ideals. RIA, is artinian since there is a one-one correspondence between the left ideals of RIAi and the left ideals of R containing Ai . By Theorems 11.5.1 and 11.3.5, R / A i is a simple ring isomorphic to the ring of ni x ni matrices over a division ring Ki. By Theorem 111.2.4 and the irredundancy of (1) R is a dense subdirect sum and hence R
=
c, RIA1
=
RIA, @
*.*
@ RIA,
Each RIA, is simple artinian and therefore isomorphic to K : ( i ) x n iThis . proves the following structure theorem first proved by Wedderburn for
3. SEMISIMPLEARTINIAN
RINGS
69
finite-dimensional algebras over their ground fields and later generalized by Artin to artinian rings (The proof of the uniqueness of the direct summands will be given in Theorem 3): Theorem 1 (Wedderburn-Artin). A semisimple artinian ring R is a direct sum of finitely many simple artinian rings Ri each isomorphic to a ring of n, x ni matrices over a division ring Ki.The ni and Ki are uniquely determined up to order and isomorphism (cf. Theorem 3).
a
Of course if the artinian ring R is an algebra over the commutative ring dj this theorem is also valid. This fundamental structure theorem of classical ring theory has been proved in the much more general context of dense subdirect sums of primitive rings. If we are only interested in arriving at the result, there is a much shorter proof of Theorem 1. By Theorem 11.3.5 we have only to show that an artinian ring with no nonzero nilpotent left ideals is the complete direct sum of finitely many primitive artinian rings. This may be done as follows: a. A minimal left ideal L in R is an irreducible R-module since RL = (0) implies L2 = (0), a contradiction. Furthermore, L is a faithful, irreducible module over the ring R/(O : L), where
(0 : L )
= {U
I u E R, UL = 0}
(2)
R/(O : L ) is therefore a primitive ring. p. LR is an ideal different from zero since L2# (0). Let
D
= LR
n (0 : L)
Then
D2L (0 : L ) . LR and therefore D Hence
= (0)
since R contains no nonzero nilpotent left ideals. LR n (0 : L)
y . LR
= (0)
=
(0)
(3)
+ (0 : L ) is an ideal in R. Suppose LR 4-(0 : L) C R
+
Then (LR (0 :L))/(O:L) C R/(O :L), since the left ideals of R/(O : L) are in one-one correspondence with the left ideals of R containing (0 : L). R/(O :L )
70
111.
RINGS WITH A FAITHFUL FAMILY OF IRREDUCIBLE MODULES
is primitive as was shown in a. Therefore by Theorem 11.5.1 R/(O : L ) is a simple artinian ring showing that (LR
+ (0 : L))/(O: L )
is the zero submodule of R/(O : L). Thus LR
+ (0 : L ) C (0 : L )
(4)
and LR C (0 : L). Therefore LR = LR n (0 : L ) = (0),contradicting LR # (0). Hence LR (0 : L ) = R (5 )
+
6 . By (3) and (5) every element in R is uniquely expressible as the sum of an element in LR and an element in (0 : L). Since LR and (0 : L ) are ideals, LR . (0 : L) C LR n (0 : L) = (0). It follows that R is isomorphic to the (complete) direct sum
icai
i=l
of the two rings Ol1 = LR and G12 = (0 : L). The inverse images of the G& under this isomorphism give rise to a decomposition of R: R
=
LR @ (0 : L )
(6)
[Without appealing to the concept of a complete direct sum, we say that a ring R is the direct sum of the ideals LR and (0 : L ) if (3) and (5) hold]. E. By (6), the ring R, = LR is isomorphic to the primitive artinian ring R/(O : L). The ring (0 : L ) is again artinian without nonzero nilpotent left ideals since by (6) a left ideal in (0 :L) is also a left ideal in LR 0(0 :L) = R . Apply the same reasoning as above to the semisimple artinian ring R"?-) = (0 : L), etc. This gives rise to a descending chain R
=
R, @ RI2)3 R'2'
=
R, @ R'3' 3 '.
By the descending chain condition, we obtain R(q+l)= (0) for some q. Therefore
R
= R,
0 ... @ Rq ,
Ri primitive artinian
(7)
and Theorem 1 is proved once again except for uniqueness of summands. This proof just given will be better understood once we have introduced the concept of a homogeneous component of R R in Section IV.2 and proved Theorem IV.2.3. It will be seen that LR is in fact a homogeneous component of RR.
3.
SEMISIMPLE ARTINIAN RINGS
71
Theorem 2. A semisimple artinian ring has an identity. Proof. Each ring Knx, occurring in the sum has an identity and it is clear that the sum of these identities is an identity for R . I
In the next chapter we shall show that R may be expressed as the direct sum of minimal left ideals in many ways. This is in sharp contrast with the uniqueness of the simple direct summands occurring in (7). We now proceed with the proof of the uniqueness of the representation (7). The proof depends on Theorem 2 and the following:
Theorem 3. A ring R 3 1 may be decomposed in at most one way as the direct sum of simple rings R, , i.e. if R
and
=
R1 @ ... @ R , ,
R R,' simple, then q (1, .**, 4). Proof.
=s
R, simple
(8)
R1' @ ... @ R,'
=
= Rh(,, ,
and Ri
i
=
(9)
1, ..., q,
T
a permutation of
Since Ri2 = R, and RiRj = (0) if i # j , Ri
=
RiR
=
Ri(R1' @ ... @ R8')
But RiRk' n RiRlt C Rk' n R1' = (0) if k # 1. Therefore at most one RiRk', say R,R&,, is different from zero. Since R, is simple, R,RA,,, = R, . Hence Ri = Rh(i,. Thus to each i E (1, ..., q } there exists a unique ~ ( iE)(1, ..., s} such that R, = Rh(,). Conversely, to each j E (1, ..., s} there exists a unique ~ ( jE ){ 1, ..., q } such that Rj' = R o ( j )and the theorem is proved. I
Theorem 4. Let R be semisimple artinian. By Theorem 1 R
=
R1 @ ... @ R , ,
R, simple,
i
=
1, ..., q
(8)
Let 6Y # (0) be an ideal in R . Then there exists Ril , R i 2 ,..., Rik such that GY
=
Ri, @ ... @ Ri,
Proof. Since 1 E R , GY = 6 Y . R . Now Gl . R, is either (0) or Ri since GTRi C Ol n Ri is an ideal in the simple ring R, . 6Y * Ri = R, implies Ri C Gl and the result follows. I
The information obtained in Theorem 4 on the ideals of a semisimple
72
111. RINGS WITH A FAITHFUL FAMILY OF IRREDUCIBLE MODULES
artinian ring R allows us to prove the following important theorem on the lattice of ideals of R [cf. Section 11.3, (lo1)]:
Theorem 5. The lattice Vof ideals in a semisimple artinian ring is distributive, i.e. A n(B C ) = ( A nB) ( A n C ) (9)
+
+
A, B, and C ideals in R . ProoJ Let @ be an ideal in R . By Theorem 4, there is a subset @ Ri, . Consider now the { i l , ..., i,} C (1, ..., q } such that A = Ril @
mapping
A
-+
{il , ..., i,}
(10)
of the lattice of ideals of R onto the lattice of subsets of (1, ..., q } with set theoretic inclusion as partial ordering. Clearly the mapping is a lattice isomorphism and since the lattice of subsets of (1, ..., q } is distributive, so is V . If V is a distributive lattice, the dual of (9) is valid: A
+ (B n C ) = ( A + B) n( A + C )
We prove this as follows: (A
+ B) n( A + C )
=
[(A
=A
(1 1)
+ B ) n A ] + [ ( A + B ) n Cl
+ [ ( A + B ) n Cl
=A+(AnC)+(BnC) =A+(BnC)
Dually, (9) follows from (1 1).
I
In this context, let us examine once again the fundamental criterion for a subdirect sum to be dense (cf. Theorem 111.2.3). If the lattice V of ideals in Ol is distributive we may replace condition (10) of Theorem 111.2.3 by 2sZkqn
+
(Ai, AiJ
=@
for every finite subset {il , ..., in} C I. This in turn is equivalent to requiring that Ol=Ai+A,, i f k , i,kEI We have therefore proved the following:
Theorem 6. Hypothesis: Let the lattice V of ideals of the subdirect sum @ =
3.
SEMISIMPLE ARTINIAN RINGS
73
Cs6Zi be distributive. Conclusion: Ct is dense in Cc Cti if and only if the kernels
A i = {a I a E Ct,
a(i) = 0}
are comaximal, i.e. if and only if Ai
+ A k = CPI
for every pair i, k
E
I,
i #
k. I
Let us apply this criterion to the ring of integers b. The lattice of ideals in Z is distributive and Z is a subdirect sum of fields Z / ( p ) ,p a prime. The kernels are the ideals ( p ) , p a prime and clearly if p # q, ( p ) ( q ) = Z. Therefore Z is a dense subdirect sum of the fields Z / ( p ) (cf. Theorem IV.2.7).
+
Completely Reducible Modules CHAPTER IV
The complete direct sum of a family M i , i E Z of R-modules is defined (as in the case of rings) to be the set of functions m from Z to M iwith m(i) E M i. On the other hand, the direct sum of the modules M i, i E I is that submodule of the complete direct sum consisting of those functions m for which m(i) = 0 for all but finitely many i E I. A module M is said to be completely reducible if it is the direct sum of irreducible modules. The concept of complete reducibility is of particular importance for semisimple rings because of the following reason: In Theorem 11.1.5 we characterized primitive rings as dense rings of linear transformations and in Theorem 111.3.1 we showed that a semisimple artinian ring is a direct sum of finitely many simple rings each isomorphic to a full matrix ring over a division ring. In each case, the information provides us with a clear picture of the structure of these two classes of rings. However, the structure theorem for (Jacobson) semisimple rings (cf. Theorem 111.2.1) does not provide us with as sharp a picture; in fact, as we have seen, a semisimple ring is only a subdirect sum of primitive rings and, if these primitive rings are not simple, (cf. Theorem 111.2.5), the subdirect sum is not even dense in the complete direct sum. In the Jacobson density theorem for a primitive ring R, we saw that R is a dense subring of the ring R o ( M ) of K-linear mappings of the faithful, irreducible R-module M considered as a right K-module over its centralizer K = Hom,(M, M ) . Following Bourbaki, we call R"(M) the double centralizer of the R-module M (see Definition IV.3.1 below). Now a semisimple ring R , as a subdirect sum of primitive rings R, , is a subring of the complete direct sum of these rings. Each Ri has a faithful, irreducible
uip,
I4
1.
DIRECT SUMS OF MODULES
75
R,-module M i .Form the direct sum M of the h-modules M iand turn this into an R-module by defining
am
= Cairni
for
a=(ai.
iEZ)
and
m
=
xmi
Again, as before, we first form the centralizer K of the completely reducible R-module M and then the double centralizer Ro(M) of K-linear mappings of M to itself. We shall see (Theorem IV.3.3) that we can describe the structure of Ro(M) quite explicitly. Moreover, we shall be able to show that a semisimple ring with identity is again dense in its double centralizer relative to the finite topology. 1. Direct sums of modules
Definition 1. Let R be a ring and { M i 1 i E I } a family of R-modules. The complete direct sum
c, Mi
(1)
consists of the set of functions m defined on I with values m(i)E M i. Addition and the operation by elements of R are defined by
(m
+ m’)(i) = m(i) + m’(i)
(2)
and
(am)
=
a(m(i))
for all a E R and m
E
2,
Mi
(3)
respectively. Clearly CcM iis an R-module.
Definition 2. Let R be a ring and ( M i 1 i E Z} a family of R-modules. The (discrete) direct sum C@Mi,
i E I
(4)
is the submodule of the complete direct sum consisting of those functions m which are nonzero for at most finitely many i E Z. If Z is a finite set, the complete and discrete direct sums coincide. It is for this reason that we used the notation of Definition 2 in Section 111.3. If all the R-modules M i , i E Z are submodules of a fixed R-module M , we shall denote the smallest R-module contained in M and containing each M iby
1 Mi
ie1
(5)
76
IV.
COMPLETELY REDUCIBLE MODULES
It consists of sums mi,
+ mi,+ + mi,
where mi,E Mik . If each element m E M is uniquely expressible as in (6), C M iis isomorphic to the direct sum of the M i, i E Z and conversely. In this case we shall write C 0M ifor C M i . To familiarize ourselves with these concepts, let us take as an example a semisimple artinian ring. A semisimple artinian ring is the direct sum
R = R,
0 ... @ R,
(7)
of finitely many simple artinian rings R, . By Theorem 11.3.3 the lattice V of left ideals L in a simple artinian ring is antiisomorphic to the lattice V’ of K-subspaces U of a faithful irreducible R-module M over its centralizer K = HomR(M,M ) . Since dim,(M) = n < co, the 0-subspace may be represented as the intersection of n “hyperplanes,” i.e. (n - 1)-dimensional K-subspaces Ui, i = 1, ..., n,
n ujgui,
O = U,n...nU,,
i#a
i=
1, ..., n
(8)
By the antiisomorphism, the U i s correspond to the minimal left ideals
L,
=
(0 : U,),
i = 1, ..., n
and we get
R=L,+**-+L,,
L i $ c Lj,
i=l,
...,n
i#i
and therefore Lin Cjzi Lj = 0. This implies RR
=
L 1 0 . * . @ L,
and the left R-module RR is expressible as the direct sum of minimal left ideals. Conversely, a representation of RR as the direct sum of minimal left ideals gives rise to a representation of the 0-subspace of the K-space as the intersection of n’ hyperplanes U , = (0 : Li),i = 1, ..., n’. Therefore n’ = n. Hence by (7) and Theorem 11.3.5 we have the following:
Theorem 1. I . A semisimple artinian ring R is a direct sum of minimal left ideals: R = L, @ ... @ L , (9) 2. If R is simple artinian, the number of summands in (9) is equal to the
1. DIRECT SUMS OF MODULES
77
dimension n of a faithful irreducible R-module over its centralizer K. Hence
Knxn
R
where Knxnis the ring of n x n matrices over K.
I
In the following discussion we no longer require the condition R M # (0) included in the definition of an irreducible R-module M . All we need is for A4 to have no submodule other than (0) and itself. We shall call these modules minimal R-modules. Theorem 2.
Let M be an R-module. The following are equivalent:
1. M = CielM i , M iminimal for all i E I 2. M = CAEA @ M A, MAminimal for all h E A 3. Every R-submodule U of M is a direct summand, i.e. there exists an R-module U’ of M such that M = U @ U’. (We say the lattice of submodules of M is complemented,) I
In the proof that Part 1 implies Part 2 we may choose
ACI We first prove a more general result:
Theorem 3. Let M = Xiel M , , M i minimal R-modules and let U be a submodule of M . Then there exists A C I such that
Proof. Let S be the family of subsets A of Z such that the sum
is direct. If M iC U for all i E Z, U = M and there is nothing to prove. Otherwise there exists some M iwith M i$Z U . By the minimality of M i , U in M i = (0) and the sum U @ M iis direct, showing that S # a . Let
.*.A i_c A , c ...
(*)
be a chain of subsets in S. The union A, of these sets is in S. For if m E U iC A E A , MA and if
m
= u1 4- mi,
+ ..-+ m i l ,
u1 E U ,
mi&E Mik
78
IV.
COMPLETELY REDUCIBLE MODULES
and
m
=
ul'
+ mi; + + misf,
ul'
E
mik,E Mi;
U,
are two representations of m as sums of elements from U and CAE*, M A, there exists A , in the chain (*) such that {il , ..., i, , il', ..., isr} C A , . Therefore u1 = ul', r = s and mi = mi; , k = I , ..., r. By Zorn's lemma, there exists a maximal set A, in S. Lit i, E Z be an arbitrary index. If Mi, U @ CAen, 0M A ,
by minimality of Mi, . Therefore
is a direct sum. But by associativity
(UO
1
O M A ) + M ~ , = U O ( X
AEA,
MAOMio)
AEA,
Contradicting the maximality of A , . Therefore, for all i E Z,
M,CU@
1 OM,
AEA,
and hence M=U@
1 @MA.
1
AEAo
Proof of Theorem 2 . Part 1 implies Part 2 by Theorem 3 with U = (0). Part 2 implies Part 3 again by Theorem 3. To prove that Part 3 implies Part 1 we must first show that every submodule 0 # U C M contains a minimal submodule of M . Let 0 # u E U . Then W = Zu Ru is a submodule of U different from (0). The union of a chain of submodules of W not containing u is also a submodule of W not containing u. Hence by Zorn's lemma, there exists a submodule W' maximal in the family of submodules of W not containing u. Every submodule of W properly containing W' must contain u and hence W . Therefore W' is maximal in W. By Part 3, there exists W" C M such that M = W' @ W". Let N = W n W". Then 01. W ' n N = W ' n W n W" = (0) and /3. W ' + N = W ' + ( W " n W ) = ( W ' + W " ) n W = M n W = W where the second equality in /3 is obtained by an application of the modular law (cf. Theorem 11.3). Therefore
+
W = W'ON
1. DIRECT SUMS OF MODULES
79
showing that N is a minimal submodule since WlW’ is minimal by the maximality of W‘. Now let S be the sum of all minimal R-submodules of M and let S’ be a complement of S in M . Then
If S’ # (0) there exists a minimal submodule of M contained in S ’ , contradicting (1 1). I Definition 3. An R-module M is completely reducible if any one of the three equivalent conditions of Theorem 2 holds. A semisimple artinian ring R considered as a left R-module is completely reducible by Theorem 1. Now let M be a unital module over the semisimple artinian ring R . Since
M =
RM mtM
and R
= L,
0 ... @ L , by Theorem
1,
where m ranges over M and L over the minimal left ideals of R. Since 1 E R, every minimal left ideal in R is irreducible. Therefore, for a fixed m and L, the R-homomorphism
I-Irn,
IEL
(13)
of L onto Lm is either an isomorphism (in which case Lm is irreducible) or the zero homomorphism. In the latter case we may omit Lm from (12b) and obtain a representation of A4 as a sum of irreducible modules. This proves the “only if” part of the following: Theorem 4. An artinian ring with identity is semisimple if and only if each unital R-module is completely reducible. For the sufficiency it is enough to assume that RR is completely reducible. Furthermore, the irreducible R-modules are isomorphic to the minimal left ideals of R. We still have to show that the condition is sufficient. Since R is artinian, this entails showing that there exist no nonzero nilpotent left ideals. By hypothesis, given a left ideal L # (0), there exists L‘ such that R = L 0L‘. In particular
1
=
e
4-e’
+
with 0 # e E L, 0 #
e’E
It follows that e = e2 ee‘ and hence e - e2 E L’ n L e = e2. Since 0 # e = e2 E L , L is not nilpotent. I
L’ = (0).
Therefore
80
IV.
COMPLETELY REDUCIBLE MODULES
2. Homogeneous components of an R-module Definition 1. Let R be a ring with identity. (Every minimal unital R-module is therefore irreducible.) An R-module U is homogeneous of type I if U is the direct sum of a family {MAI h E A } of irreducible R-submodules M A each isomorphic to the irreducible module I. Theorem 1. Let R be a ring with identity and A4 a completely reducible 0M A be a decomposition of M as the sum of module. Let M = CAEA irreducible submodules. Partition the set {MAI h E A} into classes L, , p E P, of mutually isomorphic submodules. Let H , = C M A E L pMA. Then M=C@H, (rEA
The H , are called the homogeneous components of M. Pro05 M = CpEP C M A e L p M A = CaEA Ha . The Ha therefore generate M. The directness of the sum follows from
Theorem 2. Let R be a ring with identity. Let U be an irreducible R-submodule of the R-module M
=
Then
10M i ,
M iirreducible
u c Mi,+
.**
(1)
+ Mi,
where
U w Mi,,
k
=
1, ..., r
Conversely, let
u c Mi,0 0Mi, *.*
and let some element u E U be expressible as u = uil
+ + ui,, ..*
ui,E Mi,,
k
=
with ui,# 0. Then U w Mil . ProoJ
Let 0 # u E U.Then by the definition of sum,
1, ..., r
2.
HOMOGENEOUS COMPONENTS OF AN
R-MODULE 81
with uniquely determined ui,, ..., ui,. Now U
=
RUC Ruil
+ + Ruir
Ruik = Mi,and therefore By the irreducibility of the M i ,
+ + Mi,
U C Mi,
* * a
The mapping of U to Mi, defined by au --+ uik
is an isomorphism. [That it is well defined follows from the directness of (l).] Hence U w Mi,. The converse is proved similarly. I The following theorem gives an indication of the significance of the homogeneous components in the theory of completely reducible modules. Theorem 3. Let R be a semisimple artinian ring and let R
=
R1 @ .*.@ R,
(3)
be a representation of R as a direct sum of simple artinian rings Ri . 1. The Ri are the homogeneous components of RR of type U i ,say. In other words, Ri is the sum of all minimal left ideals of R of type Uiand every minimal left ideal of R is contained in some Ri and is therefore of one of the types Ui. 2. Every irreducible R-module is faithful for exactly one R, and is annihilated by the other components. Proof.
Let M be an irreducible R-module. Then R M M
=
RIM
+ ... + R,M
=M.
Therefore
# (0)
Hence R,M # (0) for some j . Since R,M is an R-submodule, R,M = M . Suppose if possible that R,M = M , i # j. Then RiR,M = R,M = M . But RiR, C Ri n R, = (0). Therefore M = (0), a contradiction. Hence R,M = M
and
RiM
=
(0),
i # j
(4)
Moreover, M is a faithful R,-module. For suppose for some 0 # a E R, , aM = (0). Then the set S = {b I b E R, , bM = (0)) # (0). But S is an ideal and by simplicity of R, , S = R, . Therefore R,M = (0), a contradiction. The representation RR = L1 @ *.- @ L , as a direct sum of irreducible left ideals was obtained in Theorem IV. 1.1 by the further decomposition
82
IV.
COMPLETELY REDUCIBLE MODULES
of each R6 in (3) as the direct sum of minimal left ideals. If L is a minimal left ideal in R, there is some R j in (3) such that Rj 1RjL
=
L
by (4)
Moreover, it is clear by (4) that any minimal left ideal R-isomorphic to L is contained in Rj and that Rj is the sum of isomorphic left ideals. I Let M be a completely reducible R-module and let a E K = Hom,(M, M ) . Each homogeneous component H of M is mapped into itself by a since an irreducible R-submodule U contained in H is either mapped to (0) or to a submodule isomorphic to U . By Theorem 2, H a C H. In general, however, a submodule T # (0) of HI is not mapped into itself by all a E K . For T contains an irreducible submodule M I (Theorem IV. 1.3) which may be mapped isomorphically onto any irreducible submodule M , in HI by some isomorphism 4. This R-isomorphism 4 may be extended to an R-homomorphism +' of M into itself in the following way: As in Theorem IV.1.2, let M
=
c HI,
HI, =
and
kEK
Let W
=
c M D , M Dirreducible
DEPk
M , . Then the mapping +' defined by Hh+'
=
MD+'
m+'
=
(0)
k# 1
(O)
p f
m+
' 9
for m
PEP]
is an endomorphism of M . This discussion proves the following characterization of the homogeneous components of a completely reducible R-module.
Theorem 4. Let the unital R-module M be completely reducible and K = Hom,(M, M ) its centralizer. Then the homogeneous components of M are the smallest R-submodules H of M which are also right K-modules. The next four theorems are aimed at proving that if M is a completely reducible R-module, then M is also a completely reducible K-module where K = Hom,(M, M ) .
Theorem 5. If R is a ring with identity and M a unital, completely reducible R-module with A4 = CveN@ N , = EWE& @ M u , N , and Mu irreducible, then card A = card N , Proof If both sets A?'and JV are finite, the theorem is an easy consequence of the Jordan-Holder theorem (Theorem V.3.2). We therefore consider only the case where one of the sets, say N , is infinite. Let 0 # Xu E M , p E A.
2.
HOMOGENEOUS COMPONENTS OF AN
R-MODULE
83
zvGA
Then X , E N, for some finite subset A of M . Since M , is irreducible, this implies Mu C Even N , . By the directness of the sum, there exists a unique minimal finite subset A ( p ) of M such that M, C CvE4(,) N, . This determines a mapping p A(p). We claim that M = (JuGA A ( p ) . Suppose v,, @ A ( p ) for any p E 4,We have N,,, C Mui for suitable p iE A‘. Hence --f
@ N, . Therefore card A! is contrary to the directness of the sum CVEM greater than or equal to the cardinality of a set of finite subsets of Jlr whose union is .N. The latter has the same cardinality as N. Therefore card A’ 2 card Jlr. Reversing the argument we obtain card M 2 card A! and the theorem is proved. I
Theorem 6. Let M be a completely reducible R-module, R a ring with identity. Then any isomorphism of an irreducible submodule M , into M may be extended to an automorphism, i.e. an isomorphism of M onto itself. Proof. Let M = CpeP 0H, be the decomposition 3f M into its homogeneous components and suppose M , C Hl . Let u1 be an isomorphism of M , onto M,’, say. M,’ is also contained in HI . Now by Theorem IV. 1.3
and
By Theorem 5 card A = card A’. There is therefore a one-one correspondence between A and A’, say A t - t A’. Let a,,: M A-+ MA‘ be an isomorphism of M A onto MA’ and let u denote the resulting isomorphism of H , onto H , . Clearly u extends u1 . Now define T by = mlu, mir = m i ,
mlT
T
m, E Hl miEHi,
i# 1
extends u (and hence q) and is an automorphism of M .
I
Theorem 7. Let 1 E R and let M be a unital, completely reducible R-module, K its centralizer. If m # 0 belongs to an irreducible submodule U of M , mK is an irreducible K-module.
84
Iv.
COMPLETELY REDUCIBLE MODULES
Proof. Let 0 # m‘ E mK. Then there exists a E K such that m’ = m a The mapping u + u a of U onto U a is an isomorphism. By Theorem 6 this may be extended to an automorphism /3 of M . Therefore m’ = m/3 and hence m‘P-l = m showing that m’K
= mK
I
Theorem 8. Let 1 E R and let M be a unital, completely reducible R-module. Then M is also a completely reducible K-module, K = Hom,(M, M ) . Proof. M = C mK where m ranges over all nonzero m belonging to irreducible R-submodules. I
3. The double centralizer of a completely reducible module Not all rings have a faithful completely reducible R-module M. When this is the case, the last theorem of the preceding section suggests that we investigate M as a right K(= Hom,(M, M))-module and again form Hom,(M, M ) with a view to determining its structure and the extent to which R “fills” the double centralizer Hom,(M, M ) (cf. the introduction to Chapter IV). Definition 1. Let R be a ring and M a left R-module. Let K = Hom,(M, M ) . The centralizer of the right K-module M is called the double centralizer of M and is denoted by Ro(M) (or just RO when there is no risk of confusion). The elements of Ro(M) are written on the left of M . Of course R C RO
for every faithful R-module M
The following theorem provides us with a sufficient condition for an R-module isomorphism between two submodules of M to be an Ro-module isomorphism. Theorem 1. Let the R-submodule U of M be a direct summand of M then 1. U is an Ro-submodule of M and 2. If T is an R-submodule of M , every R-homomorphism $ of U into T is also an Ro-homomorphism of U into T.
3.
THE DOUBLE CENTRALIZER
85
Proof. 1. Let M = U 0 U' and let n - € K = HomR(M,M ) be the homomorphism defined by n-;m=u+u'-+u
for U E U , u ' E U '
(1)
i.e. n- is the projection of the R-module M onto the R-submodule U. Let a E Ro = Hom,(M, M ) . Then
2. Let
+
aU E
= a(Mn-) = ( a M ) n-
CU
HornR(U, T ) and form n-+. Clearly 774 E
c
HOmR(M, T ) HOmR(kf7M )
Hence for an arbitrary a E Ro and each u
= mn- in
U,
a(u+) = a(mn-4) = (am) n-# = (amn-)# = (uu)+
I
If the R-module M is a direct sum of submodules each isomorphic to a module U , then using Theorem 1 we may show that Ro(M) depends only on U and not on the number of summands occurring in the decomposition of M :
Theorem 2. Let the R-module M be a direct sum of isomorphic submodules Ui, i E I. Let U, be one of the direct summands. By Theorem 1 U, is an Ro-submoduleof M . Let a E Ro(M)and let a' denote the restriction of u to U, , i.e. for u1 E U ,
(2)
for a E Ro(M)
(3)
a'u, = au,
Then the mapping
a -+ a'
is an isomorphism of Ro(M) onto the double centralizer Ro(Ul) of Ui . Proof. 1. We assert that a' is in the double centralizer of U , . Let k, E HomR(U,, U,). By Theorem 1 with U = U1 = T, k , is also an ROhomomorphism of Ul and therefore for a E Ro(M) (aul) kl
= a(ulkl),
Hence a' E Ro(Ul). 2. The mapping a
-+
ul
u1
7
kl E HomR(u1
7
ul)
(4)
a' is a monomorphism. Let
u, ui
+i:
-+
be an R-isomorphism of U, onto Uiand let uibe an arbitrary element in Ui.
86
IV.
COMPLETELY REDUCIBLE MODULES
Let ui = u1+2 for some element u1 E U, . By Theorem I applied to U = U , , T = Ui,
au,
= a(ul4J = (aul) +i
(5)
Let a' be the zero of Rn(Ul) for some a E Ro(M). Then aUl = a'U, = 0. By (9,aUi = 0 for all i E I. Since A4 is the sum of the Uithis implies aM = (0) and hence a = 0. The mapping a a' is therefore a monomorphism of ---f
Rn(M)
into Ro(Ul)
3. To show that the mapping in (3) is onto we shall construct in (6) from a given element b E Ro(Ul) an element bi E Ro(Ui) for each i using the isomorphism diof U , onto Ui, 4, = identity. In (7) an element a E Ro(M) will be constructed from these bi in such a way that under the mapping a a', a' = b. Now for the calculations: Let b E Rn(Ul). We assert that the mapping ---f
bi: ui
di
+ (b[ui+;l])
for uiE Ui
(6)
is an element of the double centralizer Ro(Ui). Proof. kl + &lkl$i is an isomorphism of HomR(Ul , U,) onto HOmR(Ui , Ui). Let kiE HOmR(Ui, ui)and k t ki = $i1k14i for some k , E HomR(Ul , 17,Then ). by (6)
For the given b E Ro(U,) define the mapping a by
a
UA AEl
We must show a E Ro(M).Since M
=
c
(7)
bAuA
AEI
= C @ M Athe
mapping
is a projection of the R-module A4 onto Ui. To a given u E Uithere exists = u. Let k E HomR(M,M ) . Then by definition (6) u1 E U , such that
3.
THE DOUBLE CENTRALIZER
87
On the other hand
and again by (6)
a(uk) =
(b(ul+ikzA+il))
(9)
+A
A
Now +iknA+;l E HomR(Ul , U,) and therefore
a(uk) =
c (bu,) +ikTrA+;l+A
=
(bu,) +ik
=
(au)k
A
The last equality follows from (8). 4. It is clear that the mapping in (3) is a homomorphism.
I
If in particular R is a ring with identity and M a homogeneous R-module of type U , this theorem shows that the double centralizer of M is isomorphic to the double centralizer Ro(U) of the module U. Since U is irreducible, the double centralizer of M is isomorphic to a ring of linear transformations of the vector space U over the division ring Hom,(U, U ) . This proves the second assertion of the following: Theorem 3. Hypotheses: Let R be a ring with identity. Let M be a completely reducible R-module and let M = CDEP @ H, be a decomposition of M into its homogeneous components H , . Let H, be of type U, and let Ro(M) and Ro(H,), p E P, be the double centralizer of the R-modules M and H , , respectively. Conclusion: 1. Ro(M)is isomorphic to the complete direct sum CDEP Ro(H,) of the rings Ro(H,). 2. Ro(H,) is (up to isomorphism) the ring Hom ( U , , U,) of K,-linear K P transformations of the vector space U p over the division ring KO
=
HomR(U, U,> 3
Pro05 Only Part I remains to be proved. The complete direct sum of rings was defined in Section 111.2. By Theorem 1 the H , are RO(M)-submodules of M . Each element a E Ro(M) therefore maps H , into itself. Denote the restriction of a to H , by a , . Then the mapping
a -+ (a, 1 p E P)
for a E Ro(M)
(10)
is a ring isomorphism of R o ( M )into the complete direct sum of the double centralizers Ro(H,), p E P. To show that the mapping in (10) is an
88
IV.
COMPLETELY REDUCIBLE MODULES
epimorphism, let (b, 1 p E P, b, E Ro(H,))be an arbitrary element of CcRo(H,). Define b by
b
(cm,) c b,m, =
DEP
P
9
m, E H ,
(1 1)
b is clearly an endomorphism of M . (Note that by the definition of a direct sum only finitely many m, in (11) are different from zero.) We must show b E Ro(M). By Theorem IV.2.4 the H, are K-modules, K = Hom,(M, M ) . Hence for k E K it follows from Definition (1 1) that
i.e. b E Ro(M).The image of b under the mapping in (10) is the given element (b, I P E PI. I The precise description of the double centralizer Ro(M) of a completely reducible R-module M enables us to determine the structure of its centralizer K = Hom,(M, M ) completely. To this end we first form the centralizer Hom,,(,,(M, M ) of the Ro(M)-module M and prove that HomRo(M, M ) = Hom,(M, M ) . Theorem 4. Let M be an R-module and K = Hom,(M, M ) its centralizer. Then the double centralizer K o ( M ) of the right K-module M is K . Proof. Clearly K C K o ( M ) .By definition, the centralizer of the K-module M is the double centralizer Ro(M) of the R-module M . Hence for each 7 E K o ( M )and all a E R o ( M )
am)^
Therefore, in particular, 7 M). I
= a(m7)
the
for r n ~ M
equality is valid for a E R. Hence
By Theorem IV.2.8, a completely reducible R-module M is also completely reducible as a K-module, K = Honi,(M, M ) . Therefore the last two theorems imply the following: Theorem 5. Let R be a ring with identity and M a unital, completely reducible R-module. Then the centralizer of M is the complete direct sum of full rings of linear transformations of vector spaces U , over division rings K,,p~p. I
3.
THE DOUBLE CENTRALIZER
89
In general the ring R is not equal to the double centralizer of an R-module. However, we may prove that under certain conditions R is dense in Ro(M) (relative to the finite topology). Theorem 6 (Bourbaki). Let R be a ring with identity and let M be a unital, completely reducible R-module. Then R is dense in Ro(M), i.e. to each finite set of elements x1 , ..., x , E A4 and to a given a E Ro(M) there exists b E R such that bxi
=
axi,
i
=
1, ..., n
Proof. By Theorem 3 the double centralizer Ro(M) is the complete direct sum of the double centralizers of the homogeneous components of M . Therefore it is sufficient to prove Theorem 6 in the case that A4 is homogeneous. Let T = M @ ... @ M , the direct sum of n copies of M . To a given a E Ro(M) we define a mapping A of T into itself by A(m, , ..., m,)
=
( a m l , ..., am,)
(12)
A E Ro(M) since T is a homogeneous module and by Theorem 2 its double centralizer is isomorphic to Ro(M) under the mapping constructed in (6) and (7). Now let X denote the R-submodule of T generated by the element ( x l , ..., xn). T is clearly a completely reducible R-module and therefore, by Theorem IV.1.2, X is a direct summand of T. By Theorem 1, X is also an RO(T)-submodule of T. Therefore A(xl , ..., x,) E X . But X
=
{ ( b x , , ..., bx,) I b E R}
Therefore there exists b E R such that
(bxl , ..., bx,) and Theorem 6 is proved.
=
A(x1 , ..., x,)
= (axl, ..., ax,)
I
If R contains an identity and M is a faithful, irreducible R-module Theorem 4 reduces to the Jacobson density theorem on primitive rings. In this case the double centralizer is the full ring of linear transformations of the vector space M over the division ring K = Hom,(M, M ) .
Theorem 7. If the ring R contains an identity, then RR is a completely reducible R-module if and only if R is artinian and semisimple.
90
IV.
COMPLETELY REDUCIBLE MODULES
Proof. The sufficiency of the condition has already been shown in Theorem IV. 1.1. On the other hand, if RR is completely reducible, then RR =
1 0Li ,
Li minimal left ideals
iSl
In particular 1
=
eil
Therefore R
=
Rei,
+ .*- + ein ,
ei,
E Lik
+ + Rein = Lil 0
0Lin
By Theorem IV.1.3 applied to M = RR, every descending chain of left ideals in R stops after finitely many steps. Therefore R is artinian. On the other hand, by Theorem 6, R is the direct sum of finitely many full rings R, of linear transformations of the vector spaces Lik over the division rings Kk = HomR(Lik, Lik).This implies the simplicity of the artinian rings Rk . I 4. Modules with Boolean lattice of submodules
Let R be a ring with identity, M a faithful, unital, completely reducible R-module. Let M
=
1 @H,,
H,, of type U
PEP
be the decomposition of M into its homogeneous components. Let HD
0
=
' 0 ,
PEM
be the decomposition of H,, into the direct sum of isomorphic irreducible submodules U,, . By Theorem IV.3.3 the direct sum M*
=
1 @ U,
(3)
DEP
of mutually nonisomorphic irreducible R-modules is a faithful completely reducible R-module. The homogeneous components of M * are the U, . By Theorem IV.1.3 the lattice V(M*) of R-submodules of M * is complemented, i.e. to each U E V ( M * ) there exists V E V(M*) such that U nV
=
(0)
and
U+ V
=
M*
Theorem 1. Let R be a ring with identity and let M be a faithful, unital, completely reducible R-module whose decomposition is given in (1). Then
4.
MODULES WITH BOOLEAN LATTICE OF SUBMODULES
91
the R-module M* defined in ( 3 ) is also faithful and completely reducible. Moreover, the submodules of M* form a Boolean lattice, i.e. a complemented lattice satisfying the distributive law: A n(B
+ C)
=
(A nB)
+ ( A nC),
A , B, C E V(M*)
(4)
V(M*) is isomorphic to the lattice S ( P ) of subsets of P with inclusion as partial ordering. To complete the proof of Theorem 1 we first prove: Theorem 2. Every submodule V and every factor module of a completely reducible module M is completely reducible. Proof. Let the decomposition of M be as in (1). By Theorem IV.1.3, given a submodule V of M there exists a complementary R-submodule V’. Again by Theorem IV. 1.3
M=V’@
1 u,
AEA
But V N M/V’ rn Clan @ U, . Therefore V is completely reducible. By interchanging the roles of V and V’ we get the complete reducibility of MIV.
I
Because of Theorem 2 we now know that every R-submodule V of M* is completely reducible. Let V=C@V, AEA
be a representation of V as a direct sum of irreducible R-submodules V, . V, C M * = CDEP @ U,and by Theorem IV.2.2 it follows that V, is one of the U, since the homogeneous components of M* are the U , . Therefore
The mapping
#:
V+Py={pIp€P,
UDCV}
is a one-one mapping of the lattice V(M*)into the lattice T ( P ) of subsets of P. Given S C P,the submodule
92
Iv.
COMPLETELY REDUCIBLE MODULES
maps into S under since Pv+vt
+. Hence + is onto. Moreover =
Pv v Pvp
and
i,h
is a lattice isomorphism
Pvnvr = Pv n Pv*
The subsets of a set P under set theoretic intersection and union form a Boolean lattice S ( P ) and the theorem is proved. I Definition 1. A ring is distributively representable if it has a module whose lattice of submodules is distributive. R is then said to be distributively represented by M . In terms of Definition 1 we have: Theorem 3. If R is a ring (with identity) having a faithful completely reducible R-module M , then R is distributively representable by a module M*. Moreover M* may be chosen so that V(M*)is a Boolean lattice. I
Tensor Products, Fields, and Matrix Representations CHAPTER V
In the first four chapters we dealt with the theory of semisimple rings, with particular emphasis on semisimple artinian rings. In Chapters VI-IX we shall investigate rings with radical different from zero and it is the purpose of this chapter to forge the tools necessary for the development of the theory. The reader will therefore find the following miscellaneous topics: polynomials in an arbitrary number of indeterminates, topics in field theory, and a treatment of the more elementary properties of the tensor product. In addition we shall discuss the representation of an algebra by means of matrices over the ground field while the deeper theory of tensor products will be developed in the next chapter.
1. Polynomials
The definition of the ring of polynomials in an arbitrary number of indeterminates over a commutative ring with identity may be based on the following: Definition 1. Let S be a semigroup and @ a commutative ring with identity. Then the semigroup ring @IS] consists of those functions
94
v. TENSOR
PRODUCTS, FIELDS, AND MATRIX REPRESENTATIONS
defined on S with values in @ which are zero for all but a finite number of s E S. The sum and product of elements in @[S] are defined by
a+b:s+ocs+ps ab:s+
for S E S
C
(2,)
for S E S
nr=s
(22)
@[S] is a @-algebra in which S may be embedded by the monomorphism which sends soE S to the function s+OIs=
1 if s = so 0 ifsfs,
(3)
We denote this function by soalso. With this identification each element of @[S]is uniquely expressible as a sum a =
(4)
01,s SES
For the product we have
Now let X = { X i I i E I } be a set and S, the free commutative semigroup with identity generated by X.It consists of all expressions of the form
where the X i j E X are finite in number and the ui are nonnegative integers. The product of two such expressions is obtained by adding the exponents of the same symbols. The identity of S, is the element all of whose exponents are zero. The degree of the monomial in (6) is the sum u1 ... uN of its exponents.
+ +
Definition 2. Let @ be a commutative ring with identity. Let 36 = {Xi [ i E I} be a set. The ring @[XI of polynomials in the indeterminates Xi , i E I is the semigroup ring @[S,], where S, is the free commutative semigroup with identity generated by the Xi , i E I. By (4) each element f~ @[XIis uniquely expressible in the form
1.
POLYNOMIALS
95
with 01 E @. The degree o f f (deg f ) is the maximum of the degrees of the monomials occurring in (7). The degree of the zero polynomial is set equal to -a; f is normalized if the monomial of the highest degree in f has its coefficient equal to 1 E 4. Definition 3. Let R be a ring containing in its center the ring @ with identity. Let u be an element in R . Then the value of the polynomial
at X
=u
is the element
f ( u ) = a.
+ alu + ... + anunE R
Theorem 1. Let @ be a ring with identity contained in the center of R. Let a E R . The mapping
.Ia:f--f(a)
for
f E @[XI
is a homomorphism of the ring @[XI to the ring R . The proof is an immediate consequence of (2,) and (22).
(9)
I
Note: For polynomials f in more than one indeterminate we define the value off at (a) = (ai j i E I ) analogously. However, if R is not commutative the mapping y a is in general not multiplicative. Theorem 2. If @ contains no zero divisors, neither does @[XI. Proox If @ contains no zero divisors, the product of the monomials in f and g of highest degree is the monomial of highest degree infg. I Theorem 3. If the commutative ring @ with identity has no divisors of zero then neither has any polynomial ring @ [ X I in an arbitrary number of indeterminates. Proof. Let f and g be in @ [ X I . Only a finite number of indeterminates occur in f and only a finite number in g. Suppose { X I ,..., X,} is the set of indeterminates occurring either in f or g. Then the assertion of the theorem follows by induction and Theorem 2 since
If @ is a field we construct the quotient field of @[XI in the same way the
96
V.
TENSOR PRODUCTS, FIELDS, AND MATRIX REPRESENTATIONS
field of rationals is constructed from the ring of integers. More generally we have:
Theorem 4. Let R be a commutative ring with identity having no zero divisors. (Such rings are called domains.) Define on the set of pairs (a, b), a, b E R, b # 0, an equivalence relation by
(a, b)
-
(al ,b,)
if and only if ab,
=
ba,
(10)
Let a / b denote the equivalence class of (a,b). Define on the set of these equivalence classes two operations as follows: a.c
b
ac
d = a
and
a c a d + bc -+-=b d bd
(1 1)
The set of equivalence classes together with these two operations form a field Q. The mapping a + a l l , a E R,is an embedding of R in Q. Q is called the quotient field of R . I The proof depends on the fact that if b and d are different from zero so is bd. It is easily shown that the definitions of the operations in (11) are independent of the representatives of the equivalence classes. If a # 0, alb . b/a
=
1
Theorem 4 will be generalized in a suitable way to noncommutative rings in Theorem XI. 1. Definition 4. If we set R = @[XI in Theorem 4, where @ is a field, we obtain the field of polynomial quotients in the indeterminates X = {Xi1 i E Z} with coefficients in @. We denote this by @(X).
Theorem 5. The ring R = @[XI of polynomials in one indeterminate X over a field @ is a principal ideal domain, i t . each ideal in @[XI can be generated by one element. Proof. Let @be an ideal in R and a # 0 an element of lowest degree in @. Let b E U.Then there are polynomials q and r in R = @[XI such that b=aq+r
(12)
where either r is the zero polynomial or the degree of r is less than the degree of a. (This is the so-called division algorithm.) Since a, b E @[XI,
1.
POLYNOMIALS
97
r = b - ag E a. It follows that r = 0 since otherwise deg r < deg a, contradicting the choice of a. Hence consists of polynomials which are multiples of a. I
@[a.
Theorem 6 . Let @ be a field and R the principal ideal domain A polynomial f E @[XI is said to be @-reducible if there are polynomials g and h in @[XI with deg g < degf, deg h < degf, such that f = gh. Otherwise we say f is @-irreducible. The following are equivalent: 1. ( f ) is a maximal ideal in R 2. R / ( f ) is a field 3. J’is @-irreducible.
Proof. The equivalence of Parts 1 and 2 is a consequence of Theorems 11.7.4-11.7.7. Calculating modulo ( f ) , f = gh implies gh = 8. Therefore Part 2 implies Part 3. If (f)is not a maximal ideal then there exists g E @[XI with ( f ) C (g) C (1) = R.Hence f = gh, where h is a polynomial of positive degree, showing that f is @-reducible. Therefore Part 3 implies Part 1. I The division algorithm introduced in (12) allows us to compute the greatest common divisor o f f and g as follows: If g is not a divisor o f f in K [ X ] , K a field, there exist polynomials gi and ri in K [ X ] such that deg r2 < deg g and deg ri k1 < deg ri , and satisfying
rN-3
= rN-2qN-2
rN-2
= rN-lqN-l
rN-l
= rNqN
+
+
rN-l rN
This process (the so-called Euclidean algorithm) ends after a finite number of steps since the degree of the remainders ri decreases at each step. Now Y N is a divisor of rN--l in K [ X ] ,hence also of rN--2and climbing up step by step, T N is a divisor of both f and g. On the other hand any divisor of both f and g in K [ X ] is also a divisor of r2 and hence of r 3 . Again, this time climbing down step by step, we have that any common divisor of both f and g is a divisor of r N . This common divisor rN is unique up to unit factors and by (13) is expressible in the form rN
= Uf+
Vg
with
U, 21 E
K[X]
(14)
98
v. TENSOR
PRODUCTS, FIELDS, AND MATRIX REPRESENTATIONS
2. Fields
Let K be a division ring with identity e. Then the mapping u:z-+ze,
ZEZ
is a homomorphism of the ring E of integers into the ring K. Since K has no zero divisors, the kernel of u is either (0) or a principal ideal ( p ) generated by a prime p . The following definition therefore makes sense:
Definition 1. The characteristic char K of a division ring K is 0 if K contains a subfield @ isomorphic to the rationals Q. Char K is p if K contains a subfield @ isomorphic to the prime field C , = Z / ( p ) . In either case @ is the smallest subfield of K . It is the intersection of all subfields of K. In the following let K be a field. Definition 2. Let K be a subfield of the field L and let a E L. The element a is transcendental over K (K-transcendental), if the set {1, a, a2, ...} is K-linearly independent. If (1, a, a2,...} is linearly dependent over K then a is algebraic over K. In this case, let
f(X)= X"
+ a,X"-l + ... + a,
be the polynomial of least degree with leading coefficient 1 for whichf(a) = 0. Then f is called the minimal polynomial of a over K . If a is transcendental over K, set the minimal polynomial of 01 over K equal to the zero polynomial. The minimal polynomial f of a over K is uniquely determined. In fact, if fi and fi are minimal polynomials of a over K, then by the division algorithm there exist qi , ri such that
fl = f q 1
+ rl
f z =f142
+
and
r2
But,f,(a) = f 2 ( a ) = rl(a) = r2(a)= 0. Hencef, = f 2 q 1 andf, =f& . Therefore iffl # 0, q1q2= 1 showing that fi and& differ by a unit. In the definition above we took the minimal polynomial to be monic, i.e. with leading coefficient 1. Hence q1 = q2 = 1. Thereforef, = f 2 .
Theorem 1. Let K be a subfield of a field L and let a E L be algebraic over K withfas minimal polynomial. Then the smallest subfield K(a) of L containing K and a is isomorphic to K [ X ] / ( f )and f is irreducible.
2. FIELDS 99
Proof. By Theorem V.l.l, YE:
g
-
g(4,
g E K[XI
(1)
is an epimorphism of K[X] onto K[a]. The kernel is the principal ideal (f) in K[X] generated byf. Since L has no zero divisors, f is K-irreducible and hence by Theorem V.1.6, K[X]/(f) is a field. This shows K(a) = K[a]. I Conversely, let f be a given K-irreducible polynomial. Then K[X]/(f) is a field L and the mapping j : k + kXo
+ (f),
kEK
is an embedding of K in L. Identify the image of K in L with K. For g E K[X], let g be the residue class of g modulo (f).Then the polynomial f E K[XI evaluated at X is f ( ? ) = f(x> = 0 Hence f has a root
X
in L and L is generated by X over K. This proves:
Theorem 2. Let f be a polynomial in X irreducible over K. Then, up to isomorphism, there is a uniquely determined extension L of K containing a root a o f f and generated by a over K, i.e. L = K(a). Moreover, L K[Xl/(f>. I
-
If the polynomial f E L[X] has a root a in L then X - a is a factor off in L[X] since by the division algorithm, f = ( X - a ) g r , deg r < 0 and hence r(a) = 0 showing that r = 0. Now let fl be an L-irreducible factor of g in L[X]. f l exists since either g itself is irreducible or g can be factored into two factors of lower degree than degg in L[X]. Applying Theorem 2 to fl we obtain an extension field L(a,) w LIX]/(fl) of L. L(a,) is of course an extension of K. Proceeding in this fashion we eventually obtain an extension field E of K over which f factors completely into linear factors:
+
f
=
(X- al)(X - a,) ... ( X - a,)
in E [ X ]
The smallest field contained in E and containing K and cil , ..., a , is called the splittingfield off over K. In the following we prove that the splitting field of a polynomial is unique up to isomorphism.
+
be an isomorphism of the field K onto the field K*. Theorem 3. Let Then the mapping
4':a,
+
UlX
+ ... + a
x
-
(+a,)
+
(+l)
X
+ ... + ( 4 4 Xi
(2)
100
v. TENSOR
PRODUCTS, FIELDS, AND MATRIX REPRESENTATIONS
is an isomorphism of K[X] onto K*[X]. Let f ( x ) E K[X] and f * = $y and let E be a splitting field off over K and E* a splitting field off * over K*. Then the isomorphism $ of K onto K* can be extended to an isomorphism t,4 of E onto E*, i.e.
t+hk = $k
for all k E K
Proof. Clearly, $’ is an isomorphism. If E
= K,
E*
= K*
since
1 ($aJ($a)i = $ 1aiai = $0 = 0 i
whenever f ( a ) = 0. Iff =fl ...fs is a decomposition off into K-irreducible factors, then f * =fl* ...fs*, where h* = $3,i = 1, ..., s and the A* are also K*-irreducible. Suppose deg fl 2 2, fl(al) = 0 and .fl*(al*) = 0, where a l e E and a1E E*. By Theorem 2 the fields K(a,) C E and K*(al*) C E* are isomorphic to K[X]/(f,) and K*[X]/(f,*), respectively. By (2), KIX]/(fl) m K * [ X ] / ( f l * )under the isomorphism *1: d . 1 )
-+
g*(a,*)
(3)
where g* = $’g. t+hl is therefore an extension of $. Next suppose there are k > 0 roots ai off in E but not in K . We make the inductive hypothesis that for every polynomial with coefficients in a field @ which has fewer than k roots outside @ in a splitting field Y containing @, we can extend every isomorphism of @ to an isomorphism of the splitting field Y. Now E and E* are clearly splitting fields for f and 4’fover K(a,) and K*(a,*), respectively. But f has fewer than k roots in E outside K(a,). By induction, there exists an isomorphism T, of E onto E* which is an extension of$. I If we set K = K* in the above we have that the splitting field of a polynomial is unique up to isomorphism. To each polynomial f E K[X] there exists an extension field L of K such that f factors completely into linear factors over L. For example, the splitting field off will do for L. The same result is true if instead ofjust one polynomial we consider a finite family 3 of polynomials in K[X]. In this case we merely form the product of the polynomials in F and construct the splitting field of this product. For infinite families of polynomials we prove the following: Theorem 4 (Steinitz). Every field can be embedded in an algebraically closed field L , i.e. a field over which every polynomial in K[X] of positive degree splits completely into linear factors over L. L is also algebraically closed
2. FIELDS
101
over L, i.e. every polynomial of positive degree in L[X] splits completely into linear factors in L[X]. Proof. Let 3; = {f} be an arbitrary family of polynomials in K [ a . Denote the degree off by 1 f 1. Let
R
= K [ Y f ,11
di
< If
I,f€FI
(4)
R is the ring of polynomials in the indeterminates Yfi, where we have I f I of them for each f E 9 . In particular R contains the elementary symmetric polynomials
c
Sfi=
l<jl<...<ji
Y f jl.** Y f j i,
1
< i < If1
(5)
For a fixed f these polynomials may be defined by
1 (-lpsfixlfl-i=
o
where we set S,,
=
n
lGk
( X - Yfk)
(6)
afi E K
(7)
1. Now let
f(x)
=
1If 1 (-l)'afiXlfl-l,
o
Suppose that the f have been normalized so that af,= 1. Let GY be the ideal generated by the polynomials S f i - afi, 0 < i < I f 1, f E 9. It consists of polynomials of the form
where the sum extends over finitely many i andf. This ideal is not the whole of R . In fact the only element in common with both GY and K is 0. Otherwise, suppose 0 # k E GY n K . Then
In the splitting field L, of the polynomial f l ( X ) f 2 ( X )...fN( X ) over K, let L N , by the very definition (6) of the elementary symmetric functions we have a f j l, ..., af,lf,lbe the roots ofh(X), j = 1, ..., N . Then in
afjij= Sfjij(aj31 >
...?
afj1tjl)
Substituting a f j i for Yfjt in (9) we have k = 0, a contradiction. Since R contains an identity there is a maximal ideal GYu in R containing a. The factor ring L = R/GYw is a field since it is a commutative simple ring with
102
v. TENSOR
PRODUCTS, FIELDS, AND MATRIX REPRESENTATIONS
+
identity. Set Ffi = Y f i am. Since sfi - afi E GYu and using (6) we have the following equality in L [ X ] : f(X)
=
c
OGiflfj
(-1yafixIfl-i
n.
=
1
If I
(X -
&>
for allfE 3. The field K can be embedded in L since OZu n K = 0 (otherwise au= R). The fact that L is algebraically closed over L is a consequence of the following: Let x be a root of the polynomial A,
+ A,X + '.. + AnX"
E
L[X]
and set A = K(h, , ..., An). Then dim, A ( a ) < n dim, A < 00 since A, , ..., A, are algebraic over K. Hence a is algebraic over K and is therefore already in L. I Theorem 3 enables us to prove an important theorem about finite fields, i.e. fields containing only a finite number of elements. Clearly a finite field K cannot contain a copy of the rationals. Hence by Theorem 1 it must contain a copy of C , = Z / ( p ) , p a prime. K is a finite-dimensional vector space over C , and therefore the number of elements in K, or the order of K, o(K), is a power of p : o(K) = p n . The multiplicative subgroup KX of nonzero elements of K contains p n - 1 elements. Now the order of every subgroup U of a group G is a divisor of the order of the group o(G) as can be seen by decomposing the group G into its left cosets aU. In particular the order of an element a of a group (i.e. the order of the subgroup generated by a) is a divisor of pn - 1. Hence each element of K Xsatisfies the equation XP-1
=
1
Therefore every element of K is a root of the polynomial
X""- X
E
C,[X]
(10)
On the other hand, a polynomial f(X)of degree N has at most N roots 01, , ..., aN in a field as may be seen by an application of the division algorithm. Therefore the finite commutative field Kconsists of the roots of the polynomial in (10) and is therefore the splitting field of this polynomial over C , . By Theorem 3, K is uniquely determined up to isomorphism. We have therefore proved: Theorem 5. Every finite field K is of p power order, p a prime and is the splitting field of the polynomial XP" - X , where o(K) = p n . Moreover, for a given p n , p a prime, there is essentially only one field of order p". Finite fields are called Galoisjelds. I
2.
FIELDS
103
We may easily show that given pn, p a prime, there is a finite field of that order. This will not be needed in the sequel. However, later on we shall need to know that every finite field contains a primitive element, i.e. an element 6 such that K = c,(e) (1 1) where by C,(O) we mean the smallest subfield of K containing C , and 6. We prove this result by proving:
Theorem 6. The multiplicative group K X of a finite field K is cyclic. Any generator 8 of K Xis called a primitive element of K over C , . I We first prove the following:
Lemma. Let G be an abelian group and let u and b be elements of G of finite order, o(u) = a and o(b) = p. Then there is an element c in G whose order is the least common multiple (Icm) of a and p: o(c)
=
Icm{a, p}
(12)
Proof. Let C be written multiplicatively and let e be the identity. 1. Let ( a , A) denote the greatest common divisor (gcd) of 01 and A. Then a = ( a , A) a‘
and
A
Therefore uA = e - a a
(Here by “ a j
p”
= ( a , A)
1 At+
we mean “ a divides
a’
p.”)
O(d) =
A‘
with (a’, A’)
=
I
I A’coa’ 1 E Therefore
a/(&!,A)
2. Suppose ( a , p) = 1. Since G is commutative, caB = e, where c Hence y = o(c) 1 a . p. On the other hand by Part 1 y / ( y , a ) = o(c“) = ~ ( ~ “ b = o )o(b“) = p/(p, a ) = /3
(13) = ab.
(14)
and hence p 1 y. Similarly a j y. Since ( a , p) = 1, ap j y . 3. Let a = o(a) = n i p ? and p = o(b) = n i p > 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 finite-dimensional 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 K-irreducible 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 K-irreducible 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 finite-dimensional 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"n-l __ 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 =.f@- 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 K-irreducible 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 K-irreducible. 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 finite-dimensional 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 R-module 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 R-module. 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
together with componentwise addition is the free abelian group 9 with generators {(m',m) I m' E M ' , m E M } . Let U be the subgroup of 9 generated by all elements of the form hl'
+
(m', m1
h', m) - (m2', m ) (m', m1) - (m!, m2) (m'a, m) - (m', am)
m2', m) -
+ mz>
(1)
-
+
where mi' E M ' , m iE M and a E @. Denote the coset (m', m) U E FjU by m' @ m. The factor group S j U is called the tensorproduct of M' and M over @ consists of finite sums and is denoted by M' @ &4. M' @
108
V.
TENSOR PRODUCTS, FIELDS, AND MATRIX REPRESENTATIONS
Because of (l), we have
+
+ +
(m,’ m,’) @ m = m,’ @ m m2’ @ m m‘ @ (m, m2) = m’ @ m , m’ @ m2 (rn‘a) 0m = m’ @ (am)
+
(2)
In the context of relations the tensor product is the factor group of 9 by that relation T on 9 which is the finest congruence relation coarser than the relation defined by the pairs, Kml
+
m2 >
((4m1
+
+ 4) + (m, m2))
m), (m, m> 2
m2), (m, m d
(m2 >
((ma,4,(m, am))
(3)
This “minimality” of the relation T leads to a characterization of the tensor product which will be frequently used later on.
Definition 3. Let @ be a ring with identity and let ,M be a left and M,’ a right unital @-module. Let N be an additively written abelian group. Let q be a mapping from the Cartesian product M ’ x M to N . The mapping 7 is said to be balanced with respect to @ if it satisfies the following:
+
+ +
(4)
0m
(5)
dm1’ m2‘,m) = 7(m,’, m) 7(m2’,m> q h ’ , ml m2) = q h ’ , ml) q h ’ , m,) q(m’a, m) = q(m’, am), a E @
+
The canonical map rr:
(m’, m) + m‘
from the Cartesian product M‘ x M to the tensor product M ’ @ ,M is an example of such a map. The following theorem states that every balanced map from M ’ x M to an abelian group can be “factored through” M’ @ 4 4 .
Theorem 1. Let 0 be a ring with identity and let M,’ and @ Mbe right and left @-modules, respectively. Let N be an additively written abelian group and q a balanced map from M ’ x M to N. Then there is a unique additive mapping a: M‘ @ ,A4 + N making the following diagram commutative:
N
3.
TENSOR PRODUCTS
109
Here n- is the mapping defined in (5). (Note: by the diagram being commutative we mean crr = -q.) Proof. Since we want crr to be the same mapping as 7, and since the elements m’ @ m generate M’ @ ,J4, the only possible additive candidate for cr is the mapping defined by a:
1mi‘@ mi C -q(mi’,mi) -+
i
i
We must therefore prove that u is well-defined, i.e. we must show that
C Xi’
@ xi =
1yi‘ @ y3
implies
2 r l ( X i ’ , X i ) = 1-q(Yj’, Y3) The set of pairs of elements of 3;
is a relation T* on 3;. Let T‘ be the finest congruence relation on 3; which is coarser than T*. Since -q is balanced, T‘ 1 T, where T is the congruence relation induced by the pairs in (3). But T is the defining relation of the tensor product M‘ 0 = PIT and in the language of relations (7) becomes C (xZ’, xi)T C (yi’,y j ) . Hence the equality in (7) implies the equality in (8) and cr is well defined. I Let M and M I be right @-modules and N and Nl left @-modules. If A : M --+ Ml and B B: N Nl are @-homomorphisms then the mapping -+
-+
(m, n)
-+
mA @ nB
(10)
is a balanced map from M x N to the tensor product M I @ .Nl. By Theorem 1 there exists a homomorphism denoted by A @ B from M @ N to M , @ Nl . If in addition A and B are epimorphisms we have the following: Theorem 2. Let M , M I , N and Nl be as in the previous paragraph. Let A: M Ml and B: N - t Ni be epimorphisms with kernels U and W , respectively. Then A @ B is an epimorphism of the additive group M @ N onto Ml @ Nl such that ---f
A @ B: m @ n + m A @ n B
(1 1)
110
v. TENSOR
PRODUCTS, FIELDS, AND MATRIX REPRESENTATIONS
Moreover, ker A @ B is the subgroup S of M @ N generated by the elements u On, u E U , n E N and m @ w,m E M , w E W.
Proof. The mapping q : ( m , n ) + m A @ n B is balanced. Hence by Theorem 1 there exists exactly one homomorphism A @ B from M @ N onto q(M @ N ) = M , @ N , such that ( A 0B ) T = 7) where T :(m, n ) -+ m @ n. Clearly ker A @ B 2 S. Consider now the mapping ( A @ B)* induced by A @ B on M @ N / S . In particular we have ( A @ B)*: m @ n
+ S+mA
@nB
We show ( A @ B)* is an isomorphism of the additive group M @ N / S onto M , @ N , . The mapping 5 : (mA, nB) + m @ n S is well defined: For if m,A = m2A and n,A = n2B, then m, - m2 E ker A and n, - n2 E ker B. But
+
m,
0n,
- m2 0n2 = (m, - m2) 0n,
+
+
+ m2 0(n,
-
n2)
Therefore m, @ n, S = m2 @ n2 S and 5 is well defined. Clearly 5 is balanced. Hence by Theorem 1 there exists a homomorphism 7 : M , 0N , M @ N / S such that --f
7:
mA @ nB+ m @ n
+S
Now T ( A 0B)* is the identity on M @ N / S and ( A 0B)* 7 is the identity on MI @ N , . Therefore S = ker A 0B. I Theorem 2 may also be stated as follows: The mapping (m, n) mA 0nB induces a group homomorphism A @ B from M @ N to M , 0Nl . If both A and B are epimorphisms so is A @ B. The analogous result does not hold in general if A and B are monomorphisms as the following example shows: Let M = Z/2Z be the prime field C, of characteristic 2, considered as a Z-module and let A be the identity on M . Let N = Z and B the embedding of ,Z into @, Q the rationals. Then M @ ,N = C2 @ ,Z w C2 under the . On the other hand C2 @ Q , = 0 since mapping Z, @ z2 iF1z2= ---f
--f
-
z, @ z,
Hence A
@jB
=
__
22, @ +z2 = 0
is not a monomorphism.
From this discussion it follows that in general if N , C N , M @ @N1is not isomorphically contained in M 0 and therefore in @ n1 E M 0N , cannot be identijed with m @ n, E A4 @ N . However, if N , is a direct summand of N we may prove the following:
3.
TENSOR PRODUCTS
111
Theorem 3. Let M be a right @-module and let the left @-module N be the direct sum of its submodules N , and N,: N = N , @ N , . Then M @ wM @ 0A4 0@ N , under an isomorphism u, where
Proof. Let 7 be the mapping defined by
Clearly 7 is a balanced map from M @ N to the direct sum M @ N , 0M @ N , . Therefore by Theorem 1 there exists a homomorphism u of M @ N onto M @ Nl @ A4 @ N , satisfying U ~ T= 7. Under this homomorphism m @ (n, n,) ( m @ n1 , m @ n2). Now let ri, i = 1 , 2 be the canonical balanced map from M x Ni to M 0 Ni and let vi:M x Ni + M @ N be the mapping defined by
+
-
7i:(m,ni)-+ m @ ni E M @ Ni ,
i
=
1,2
Clearly rli is balanced and therefore by Theorem 1 there exists a homomorp h i s m 7 i : M @ N i + M O N s u c h t h a t r i 7 T i = y i , i = 1 , 2 . S e t ~= T , +T,. Then r is a homomorphism from M @ N , 0 A4 @ N , to M @ N such that 7: (m 0nl , m @ n2) + m @ n1 m @ n2 = m @ (n, n2). Therefore TU is the identity on M @ N and u is one-one. I
+
+
Suppose now that M is an (R, @)-bimodule and N a left @-module. We may give the free abelian group 9 with generators (m, n) E M x N the structure of a left R-module by defining
The subgroup U o f 9 generated by the elements in (1) is an R-submodule of 9. To prove this it is clearly sufficient to show that if x is a generator of U , ax E U. We show this when x is of the form (m, m, ,n) - (m, ,n) - (m, , n) and leave the other cases to the reader:
+
This enables us to make the following:
112
v. TENSOR
PRODUCTS, FIELDS, AND MATRIX REPRESENTATIONS
Definition 4. Let M be an ( R , @)-bimodule and N a left @-module. Then M @ @Nmay be given the structure of a left R-module by defining a C (mi@ n,) =
2 (ami) 0ni
(13)
If @ is a commutative ring and M a right @-module we make M into a left @-module by defining
It is easily checked that with this definition M is actually a (@, @)-bimodule. Therefore the tensor product of two @-modules over the commutative ring @ is a @-module. An algebra GZ over the commutative ring @ may be considered as a (@, @)bimodule. Therefore the tensor product of two @-algebras is a @-module. Moreover, if we define multiplication as below, the tensor product is again a @algebra. Definition 5. Let GZ‘ and @ be two algebras over the commutative ring @ with 1. Then the tensor product a’ @ &Y of the two algebras is an algebra if we define
To prove that this definition is independent of the representatives
xi(ai’,a,) and C j (bj’,bj) we need only point out that because of the distributive laws in a ring, we have, for example, (a’(&’
for all a’, b,’
E
+ b2’)a) - (a’bl’, a) - (a’b2’,a)
E
U
a’ and a E GZ.
Theorem 4. Let GZ be an algebra over the commutative ring @. Let A be a commutative ring containing @. Then the tensor product GZ oA is an algebra over A if we define i
i
for a, E a, A E A . We denote the algebra GZ @ @Aby GZA . Prooj GZ @ ,Jis an algebra over @. The operation of A on GZ 0$ A defined in (16) is independent of the choice of representative for a, @ A, .
x
4. REPRESENTATIONS BY MATRICES
113
This may be seen in exactly the same fashion as indicated above. Furthermore [(a1 0U a 2 0X,>l
= (0, = [(a1
since each of the three expressions is XJh, = h,X,h. I
0Xl", 0A), A1 04) hl(a2 0A,) equal to a1a2 0A1X2A
because
We are now in position to solve the embedding problem posed at the beginning of this section: Theorem 5. Let @ be a subfield of the ring A with identity. Let M be an (R, @)-bimodule. Then M is isomorphic as an (R, @)-bimodule to M 0 under the mapping u:m+m@l
It then follows that M is isomorphically embedded in the (R, A)-bimodule M 0@A. Proof. 01 is a subspace of the @-vector space A . Hence A = @ 0Y, a direct sum as a @-vector space. By Theorem 3 the mapping u is a monomorphism. 1
In a similar fashion we may prove: Theorem 6. Let R be an algebra over the field @. Let A be a commutative ring containing 0.Then u:a+a@l
for a e R
is a monomorphism of the algebra R over @ into the algebra R over A. 4
0@A
From Definition 1 of the tensor product M' 0@A4,it follows without difficulty that if N' is a right @-module isomorphic to M' and N a left @-module isomorphic to M then M' 0@A4m N' @ @ N . 4. Representations by matrices
Let R be an algebra over the field @ and let M be a left (algebra) module over R, that is to say, according to Definition 11.8.1, M is an ( R , @)-bimodule and in addition (aa) m = ama for all a E R,01 E 0,and m E M . The elements a of R induce @-linear mappings A of the @-vector space M into itself: am
=
Am,
a € R and
mE
M
114
V.
TENSOR PRODUCTS, FIELDS, AND MATRIX REPRESENTATIONS
These @-linear maps form an algebra over @ since @ is commutative (Theorem 1.4.1). The mapping for a E R
a+A
is therefore an algebra homomorphism.
Definition 1. Let R be an algebra over @. A @-representation of R is a homomorphism of the algebra R into the algebra of @-linear mappings of a @-vector space M into itself. M is called a @-representation module for this @-representation of R. Definition 2. Let A be an extension field of the field @ and R an algebra over @. A A-representation of the algebra R over @ is a A-representation of the algebra R @ ,A (cf. Theorem V.3.4). The theory of representations of @-algebras R in extension fields A of @ is therefore the same as the theory of R @ .A-modules about which something will be said in Section VI.1. In this chapter we shall discuss the matrix representations of algebras. These are obtained by choosing a @-basis in the representation module M . For the remainder of this section let us fix our notation: Let R denote an algebra over the field @ and let M be a representation module for the algebra R, i.e. M is an (R,0)-bimodule so that (aa) m = amci for all a E R, m E M , and 01 E a. Let dim, M = n < co and let { u l , ..., u,} be a @-basis of M . The action of a on M is determined by its action on the uk : aulz =
l
uiolik,
k
1, ..., n,
=
ciik
E
@
(1)
and the mapping a
---f
A
=
(aik)E Q n x n ,
aER
(2)
is a matrix representation of R in 0,i.e. an algebra homomorphism of R into Q n X n . Conversely let (2) be a given matrix representation of the @-algebra R and let M be a vector space over of dimension n. Let {ul, ..., u,} be a @-basis of M . Then M may be turned into an R-algebra module by defining
a
UkPk
k
=
uiciikpk i.k
9
p k
@
(32)
Since (3,) agrees with (l), relative to the @-basis { u l , ..., u,} of the algebra module M , we get back the original @-representation of R .
4.
REPRESENTATIONS BY MATRICES
115
Suppose {ul‘,..., un‘} is another @-basis of M where
uk’
uirik,
=
rile E @,
1, ..., n
k
=
1,
..., n
i
Let
auk’
=
1 ui’aik,
k
=
i
Then
Ci ui’
=
1
ujrjiolik=
auk’
1auirik = 1
=
i
i,j
ujolji~ik
i.j
Hence, relative to the new basis a corresponds to the matrix A’, where A’
=
P-lAP,
P
(4)
= (rile)
Two matrix representations a+A
and
a+A’
of the algebra R are said to be equivalent if there exists a matrix P E QnXn such that (4) holds for all a E R. The module M immediately determines a @-representation of R in Hom,(M, M ) . However, to obtain a matrix representation of R we must first choose a basis in M . Let R be a @-algebra and let M and M‘ be isomorphic R-modules under the isomorphism $: m + m’. [Thus $(am) = a$(m)
and
$(ma) = #(m) a.]
Then M and M’ give rise to the same equivalence classes of @-representations of the algebra R. In fact, if {ul , ..., u,} is a @-basis of M then {$ul , ..., $un} is a @-basis of M’ and
a$uk
=
$(auk) =
1 uiaik = 1 ($ui) i
i
aik
,
k = 1, ..., n
In other words: Theorem 1. Let R be a @-algebra. The equivalence classes of matrix representations of R are in one-one correspondence with the isomorphism classes of R-modules which are finite dimensional over @. These (R, @)modules are called the representation modules of R. I
This correspondence between matrix representations of a @-algebra R and representation modules enables us to apply the theory of the preceding chapters to matrix representations. We also take over the terminology
116
V.
TENSOR PRODUCTS, FIELDS, AND MATRIX REPRESENTATIONS
introduced. Thus if R is an algebra over the field @ we call a matrix representation of R: 1. irreducible if its corresponding representation module M is an irreducible R-module, 2. completely reducible if M is completely reducible, 3 . reducible if M contains a submodule different from 0 and M .
Since the radical J of a ring R annihilates every irreducibIe module M , we may always turn an irreducible R-module M into an RIJ-module. If RIJ is a finite-dimensional algebra over @, R / J is artinian and therefore RIJ
=
d l + .’. + d,
where the d iare ideals. By Theorem IV.2.3, M is a faithful module for exactly one d,and is annihilated by the other di . Therefore an irreducible matrix representation of the @-algebra R induces a representation of RIJ which is faithful on exactly one of the simple direct summands of RIJ and which annihilates all the others. This does not mean that the matrices of this representation “fill out” the whole matrix ring ,@ ,, since equality may not hold in @ C Hom,(M, M ) Representations in which this happens will be discussed later in the context of absolutely irreducible representations. If the representation module M is the direct sum of two submodules each different from 0, say M=U@V then we may complete a @-basis {ul, ..., u,> of U to a @-basis of M by adjoining a @-basis {ol , ..., v,} of V. The resulting representation matrices relative to the basis {ul , ..., u, , vl , ..., v,} have the form A
=
(ou oV ) ,
where
UE
V EQ s X s
We may express this as follows: The representation module M of the representation C#J: a + A is the direct sum of two submodules each different from 0 if and only if there exists a matrix P E Q n X n such that P-IAP
r
> 0, s > 0, r
=
(oU
VO) ’
+ s = n for all matrices +(a), a E R.
4.
REPRESENTATIONS BY MATRICES
117
In particular, this happens when M is completely reducible. In this case the matrices of the representation are of the form
where a + U j ,j = 1, ..., t is an irreducible @-representation of the @-algebra R and the square r j x r j matrices Ui lying along the diagonal of the n x n matrix P-lAP are the only submatrices of P-lAP having entries different from 0. Naturally, n = r1 ... rt . By Theorem IV.1.4 every R-module of a semisimple algebra R is completely reducible and therefore so is every matrix representation of R. If M is not completely reducible let
+ +
u, = (0)c u1c ... c ut = M
(7)
be a maximal chain of R-submodules Ui so that each U j is maximal in U,,, . Choose a @-basis of M as follows: Choose a basis of U , and extend successively to a basis of U, , U, , ..., Ut = M . Relative to this basis, the matrices A of the representation which M gives rise to are of the form
where Uji = Ujj(a),j = 1, ..., t are the matrices corresponding to the representation module Uj/Uj-,and the submatrices Ui, ,j < k lie in Onjxnk where ni = dim,(Ui/Ui-l). All entries under the Uij are zero. The submatrices U,, , ..., U,, occurring in (8) are uniquely determined up to order and equivalence by the representation module M . This implies that, up to equivalence and order, they are independent of the way the maximal chain in (7) was chosen. This is a consequence of the purely module-theoretic Theorem 2 below. Definition 3. Let M be an R-module and let U, C U , C ... C Ut be a chain of submodules. The length of this chain is defined to be the number of proper inclusions, i.e. t. Theorem 2 (Jordan-Holder). Let M be an R-module and suppose all chains of submodules are finite in length. Let 0
=
U, C U , C ... C U ,
=
M
and
0
=
V, C ... C V , = M
(9)
118
V.
TENSOR PRODUCTS, FIELDS, AND MATRIX REPRESENTATIONS
be two chains of maximal length. Then r of the numbers 1, 2, ..., r such that
=s
and there is a permutation
7r
This theorem is a corollary to a much more general theorem the proof of which depends on a number of intermediate results.
Theorem 3 (isomorphism theorem for R-modules and rings). V be R-submodules of the R-module M . The mappings #:u+(UnV)-.u+V
forall
1. Let U and
UEU
+
is an isomorphism of UjU n V onto ( U V)/K 2. The submodules X of A4 lying between U n V and U are in one-one correspondence with the submodules Y lying between V and U V under the mapping q!I’:X-+X+ V
+
#’ is a lattice isomorphism with inverse
3. Similar results hold if U is a subring and V an ideal of a ring R.
+
Proof of the isomorphism theorem. u u V is a module homomorphism of U onto U VjV. The kernel is clearly U n V and hence # is an isomorphism. In the same way we easily see that #’ is a lattice isomorphism of the “interval”
+
{XI U n V C X C U }
---f
onto
{Yl V C Y C U + V }
The assertion about the inverse of $’ follows from the modular law (Theorem 11.3.2) for the R-submodules of a module A4
(X+V)nU=X+(VnU)=X for all X such that U n V Z X Z U . Part 3 is proved in exactly the same way, I The proof of the following theorem due to Zassenhaus depends on the isomorphism theorem just proved.
4.
REPRESENTATIONS BY MATRICES
1 19
Theorem 4. Let A’ _C A and B‘ C B be submodules of M . Then the following three factor modules are isomorphic:
+ ( A n B’)) (A’ + ( A n B))/(A’ + ( A n B’)) (B’ + ( A n B))I(B’ + (A’ n B ) ) ( A n B)I((A’ n B )
(10) (11)
(12)
Proof. By symmetry it is sufficient to prove (10) and (11) isomorphic. Since the submodules of M form a modular lattice. (A’ n B )
+ ( A n B‘) = ( A n B’) + (A’ n ( A n B ) ) =
( ( A n B’)
Furthermore, A‘
Put U
=A
+ A’) n ( A n B )
+ ( A n B ) = (A’ + ( A n B’)) + ( A n B )
n B and V
= A’
+ ( A n B’) in Theorem 3 and notice that
U+ V=A’n(AnB)
1
Definition 4. 1 . We say that the chain of submodules U = U,, C U1 C ... C U, = W is a rejnement of the chain U = VoC ... C V,,, = W if the terms of the first chain include all of the terms that occur in the second chain. 2. The two chains in Part 1 are said to be equivalent if it is possible to set up a one-one correspondence between the factors UilUi-, and Vj/Vj-l such that the paired factors are isomorphic. Theorem 5 (Schreier).
Let
and be two chains of submodules of the module M . Then the chains have equivalent refinements. Proof. Let A,, B,,
A, = B, =
+ (A,,, n B,),
+ (B,,,
n A,),
i
= 0,1,
i
=
..., n
0, ..., M
Then since (1 1) and (1 2) of Theorem 4 are isomorphic, AZ,,+lIAZ,
B2+1,3/B23
[each factor module in (17) could be the 0-module].
1
120
v. TENSOR
PRODUCTS, FIELDS, AND MATRIX REPRESENTATIONS
In particular if the chains in (13) and (14) are maximal (we call maximal chains composition series) with U = 0 and W = M , neither can be refined. Hence Theorem 2, due to Jordan and Holder, follows. If the R-module M has a composition series of finite length, then all chains are of finite length. Indeed, assume M has a maximal chain of length r and suppose if possible that there is a chain of infinite length from 0 to M . Then there is a chain of length r 1. But the first chain cannot be refined and therefore the two chains do not have equivalent refinements, contradicting Theorem 5.
+
Definition 5. The trace of the matrix A diagonal elements of A : Tr A
=
mll
= (aiR)E
+ aZ2+ ... + unn
OnXnis the sum of the
E
@
It is easily checked that if A and P are matrices in GnXn, Tr(AP)
=
Tr(PA)
In particular, therefore, if P-l exists, Tr(P-lAP)
=
Tr A
(19)
We may therefore speak of the trace of an element a of R in a representation of R by means of @-linear mappings of the corresponding representation module M to itself without having to refer to any specific @-basis. We also speak of the trace of a @-representation a A of R. By this we mean the function Tr: R + @ defined by -j
Tr(a)
= Tr(A)
Let a A be a completely reducible representation and suppose that a basis has been so chosen that the matrices A are of the form ---f
-
where a Uj(a),j = 1, ..., t is an irreducible @-representation of R [cf. ( 6 ) ] . Let Trj(a) denote Tr Uj(a).Then Tr(a)
Tr,(a),
= l<j
aER
5.
THE TRACE OF A SEPARABLE FIELD EXTENSION
121
Now let R be artinian and let
be the decomposition of the semisimple artinian ring RIJ into its simple components d i. The irreducible representation induced by a + Uj(a) on RIJ is a faithful representation of &,(j) , say, and annihilates the other 0,. Moreover, the identity element e,(j) of is mapped onto the n j x nj identity matrix Uj(ek(j)).Hence Trj(t?k(j))= nj
and
Trj(ei)
=0
if i # k ( j )
(21)
By Theorem 2 a completely reducible @-representation a+ A of the semisimple artinian ring 8 = RIJ is determined up to equivalence once we know the number of irreducible components which are isomorphic to any given irreducible component occurring in the direct sum decomposition of the corresponding representation R-module M into irreducible submodules. By Theorem IV. 1.4 there is a one-one correspondence between the isomorphism classes of 8-modules and the isomorphism classes of minimal left ideals. But each of the simple components Biis the sum of all mutually isomorphic minimal left ideals. There are therefore q isomorphism classes of left ideals, represented by L, , ..., L, , say. Let M
=
u, 0 ... 0 u,
(22)
and let qi denote the number of U , isomorphic to Li.From (21) it follows that the trace of the identity ei of d irelative to the representation a+ A satisfies Tr(ei) = qini We may solve this equation for q i provided the characteristic of the field @ is either 0 or relatively prime to ni . Theorem 4. Let R be a semisimple artinian ring and @ a field of characteristic 0. A matrix @-representation of R is determined, up to equivalence, by the traces of the matrices of the @-representation.
5. The trace of a separable field extension Let L be a field containing a subfield K. Then L may be considered as an algebra over K. If dim, L is finite, L. is a K-representation module of L (Definition V.4.1) and we may therefore speak of the trace of this represen-
v. TENSOR
122
PRODUCTS, FIELDS, AND MATRIX REPRESENTATIONS
tation. We refer to it more briefly as the truce of L over K and denote it TrL,, . In this section we prove a theorem on the trace of a separable extension L over K. This theorem will be used in Chapter VIII to show that a semisimple @-algebra remains semisimple when we extend the ground field to L provided L is a separable extension of @. Let L be a field containing the subfield K and let dim, L = n < co. Let be a basis of L over K. Let h E L and let w, , w1 , ...,
Then A
-
h
- wj = C ~
wiuij(h)
where uij(h)E K
(1)
O
(uij(h))is the K-representation referred to above and TrLIK(h)
aidX)
= i
By V.4, (19), TrL,, is independent of the choice of basis.
Theorem 1. Let L be a separable extension of K. Then there is an element h E L such that Tr(h) # 0.
Proof.
1. By Theorem V.2.7, L has a primitive element 8 over K. Let
f(X)E K[X] be the minimal polynomial of 8 over K. By hypothesis f ( X ) is separable. Let 8 = 8,, O1 , ..., be the roots off(X). Adjoin O1 , ..., to L
= K(8)
to obtain a splitting field M off over K. The mapping
where the ci are in K, is an isomorphism of the field L = K(0) = K(8,) onto K(B,), q = 0, ..., n - 1, since by Theorem V.1.2, each field K(8,) is isomorphic to K[X]/(f) under the mapping X (f).+ 8, , q = 0, ..., n - 1. Since Tg8 = 8, and 8 = 8, , e l , ..., are pairwise distinct elements, the T~ are also n distinct isomorphisms. 2. In (1) choose as K-basis of L = K(8) the set {I, 8, ..., P-l}and apply T , , T~ , ..., T , - ~ to (1). Since uij(h)E K we get
+
( ~ , h ). (Ta8j)
C
=
0GiGn-l
But Tg8 = O P . Therefore
(T,8i)
uij(h),
j,q
=
0,I , ..., n
-1
(4)
5.
123
THE TRACE OF A SEPARABLE FIELD EXTENSION
The matrix
lies in Mnxnand the determinant of T is
n
(0, - 0),
o
by a theorem of Van der Monde. Hence det T # 0 and T has an inverse T-l E M n x n. The n2 equations in (4) may be written as a single matrix equation:
Thus multiplying on the right by T-’ we have
Taking traces of both sides we get
T ~+ A ...
+ ~ , - ~ =h Tr(T(aij(A)) T-l) = Tr(ai,(X)) = Tr(X)
(7)
3. Suppose now if possible that Tr(h) = 0 for all h E L . Then the distinct isomorphisms T ~ ...,, T , - ~ are linearly dependent over A4 since
But this is impossible as we show in the following proof due to Dedekind: Each singleton { T , ] is linearly independent; for ~ T ~ ( = A )0 for some fixed u , E M and for all h E L implies p = 0 since T , ( 1 ) = 1 . Now assume that { T ~ ..., , T n - l > is linearly dependent and let plTql
+ +
pt7gt
0,
1
pi
E
M , pi f 0
be a nontrivial dependence of minimal length t >, 2 between the
(9) T.
Since
v. TENSOR
124 #T we get T~~
~
PRODUCTS, FIELDS, AND MATRIX REPRESENTATIONS
there ~ , exists A.
pl(Tq>O)(Tq&
E
L such that ~ ~ , # h ,T,),
+ + ***
Multiplication of (9) by T
~ yields ~ A
+ +
p1(TYlh2)(Tq14
.
p t ( ~ ~ $ (T&) ~ )
. Applying (9) to Aoh
=
0
for all X
L
(10)
=
0
for all A E L
(1 1)
E
~
pdTgl~o)(Tq$)
Finally, subtracting (1 1) from (10) we obtain a linear dependence of length less than t . Moreover, the dependence is nontrivial since P ~ [ T ~ , A, T,,ho] # 0. This contradicts the minimality of the length of the dependence in (9). Thus (8) cannot hold and there is some h E L such that Tr(h) # 0. I
Separable Algebras CHAPTER VI
1. Central, simple algebras In Section V.4 we dealt with the @-representations of a @-algebra R. We shall now investigate what happens to an R-module M (and hence to the associated matrix representation) when we extend 0 to an overfield d. The following example shows that a simple @-algebra R does not necessarily remain simple when we extend the ground field @ to A . In fact we shall see that RA is not even semisimple. In Theorem V111.3.4 however, we shall prove that RA is a “Frobenius algebra.” Example. Let f(X)be a polynomial in X with coefficients in a field @. Suppose ,f is irreducible over Q but not separable, i.e. f has multiple roots in some splitting field; for example, we could take @ to be the field discussed in V. 1.2( 17). Let d be a splitting field off over @. Sincefis irreducible over @, the factor ring R = @ [ X ] ( f ( X ) )is a field. Let n be the degree off. The elements of R are of the form
whereX
=X
+ (f(X))and a E @ is identified with a + (f(X)). The mapping
induces an isomorphism of R
0&lonto A [ X ] / ( f ( X ) ) . 125
126
VI.
SEPARABLE ALGEBRAS
Proof. # is clearly an epimorphism. Now let over @ and suppose
#:
2 (
l
c
UJk)
O
{a,, ..., 8,)
be a basis of
@ai = CXkCaikai = 0 k
i
Since { X o ,...,X"-'} is linearly independent over A , it follows that
1 aikai = 0,
k
=
0, ..., n - 1
l
But
{a,, ..., a,}
is a @-basis of d and hence
aik
=0
for all i and k . Now if
f(X)= ( X - A J t 1 ... ( X - A k ) t k in A [ X ] , let g ( X ) = ( X - A,) ... ( X - Ak). Since some ti 2 2, g ( X ) #f(X)and hencef(X) does not divide g ( X ) . There-
fore g ( X ) # 0. On the other hand ( g ( X ) ) mis divisible byf(X) for some m. Thus A [ X ] / ( f ( X ) ) contains a nonzero, nilpotent element and since A [ X ] / ( f ( X ) rn ) R @ .A, so does R @ @ A . By Theorem 111.1.5, R @ &l is not semisimple. In the above example the @-algebra R is a field and consequently, when considered as a left R-module RR, it is irreducible. On the other hand, such is not the case with R @ since the radical of this ring is neither zero nor the whole of R 0.A and so is a proper R @ ,A-submodule of R 0@ A . This hints at the importance of the following concept:
Definition 1. Let R be a @-algebra. A left R-module M is absolutely irreducible if the R 0.A-module M 0.A is irreducible whenever A is an algebraic extension of @, i.e. whenever A is an overfield of @ consisting entirely of elements which are algebraic over @. A matrix @-representation of R is said to be absolutely irreducible if the associated representation module is absolutely irreducible. Theorem 1. Hypotheses: Let the artinian ring R be a @-algebra and let M be an irreducible R-module. Conclusion: 1. If Hom,(M, M ) rn @
(1)
M is absolutely irreducible. 2. If A4 is absolutely irreducible and dim,R < 03, then (1) follows. Property (1) is of particular importance with regard to the Jacobson density theorem (Theorem 11.1.4, also see Theorem 2). Proof of 1. 1. Let RIJ = d, @ ' " @ d, be the decomposition of the semisimple artinian ring R / J as a direct sum of the simple rings d i. Then M
1.
CENTRAL, SIMPLE ALGEBRAS
127
is a faithful di-module for some i. Suppose, without loss of generality, M is a faithful @,-module. Hence M is isomorphic as an R-module to a minimal left ideal El of @, . 2. From (1) and the density theorem applied to the simple artinian ring d,, it follows that
@, w
L, w
and hence
QnXn
c
Cd,Q
l
where Cik is the n x n matrix with 1 in the (i, k ) position and 0 elsewhere (cf. Section 11.5, Eq. (1)). Let d be an overfield of 0. The mapping
#: CilcO Si,<
+
i, k
SilcCik,
=
1, ..., n
induces an isomorphism # of the algebra Onxn@ @ A onto A,,, easily checked. Under this isomorphism
#
c Cil 0A i
as can be
CdId
= 2
xi
and Gild is a minimal left ideal in A,,, . Hence it is an irreducible R @ .A-module which is isomorphic to M 0 since
be an 3. Conversely assume that M is absolutely irreducible and let is isomorphic algebraically closed overfield of 0.The simple @-algebra to Knxn, where K is the centralizer of M :
K
=
Hom,(M, M )
By Schur’s lemma, K is a division ring. It follows that
O OQ w Knxn O @Q w (K O ,Qn)nxn
(2)
We assert that K Oq,Qis a division ring. For suppose not. Then the @-algebra K f2 contains a proper left ideal A . Then C Cd,A is a proper left ideal contained in the left ideal C14iGn C,,K@& of the ring Ci,kCiI,K 0Q . w d,
[email protected] M 0@Qw CIGiGn CilK @ ,&a s d , 0oQmodules. Therefore M @ oQ is not irreducible contradicting the absolute irreducibility of the R-module M . It now follows from (2) that d,0Q . is a simple algebra over Q. From (2) we obtain dim&,
=
dim,(@,
0,Q)
=
n2 dim,(K @ @Q)
where the first equality follows from the definition of the tensor product
128
VI.
SEPARABLE ALGEBRAS
and the second from the embedding of the field SZ in K@,SZ. Since dim,d, < dim,R and the latter is finite by hypothesis 2, it follows that dim,K @ ,SZ < co. Each element p E K @ ,SZ is therefore the root of a polynomial with coefficients in SZ and hence lies already in SZ. This proves
K@,SZ~SZ and hence dimOK= 1, i.e. K
= @.
(3)
I
Applying Theorem 1 and the density theorem (Theorem 11.3.1) to a simple artinian ring we have Theorem 2. Let R be a simple @-algebra, dim, R < co. Then R has a faithful absolutely irreducible module M if and only if R is isomorphic to Q n X n for a suitable n. I
The structure of the algebra Onxnover the field @ is, of course, wellknown. This explains our interest in absolutely irreducible algebra modules. The following discussion leads naturally to an important example. Definition 2. Let R be a @-algebra, @ a field. The elements a of R induce @-linear maps of R to itself by left and right multiplication by a:
Ca:x+a.x,
%,:x+x.a
for X E R
The set of elements C , and 3, with a E R does not form a subring of Hom,(R, R) but it generates a subring, the so-called multiplication algebra T(R). The ring R is a T(R)-module and the T(R)-submodules are precisely the ideals of R. The centralizer
r = Hom,c,)(R,
R)
is called the centroid of R. Theorem 3. If R2 = R, the centroid T is a commutative algebra over @.
1.
129
CENTRAL, SIMPLE ALGEBRAS
In connection with the Jacobson density theorem, the structure of simple @-algebras with centroid equal to the ground field is easily determined.
Definition 3. A @-algebra R is said to be central if the centroid of R is @. From Theorems 1 and 2 it follows that a central simple algebra R of finite dimension over the ground field @ is a faithful absolutely irreducible T(R)-module where T(R) is the multiplication algebra of R . Moreover, if Q is an algebraically closed overfield of @, the tensor product Qnxn T(R) O OQ for a suitable n. Every ring is an algebra over its center 2. If R is a simple ring with identity, Z is a field. For if 0 # z E 2, Rz is an ideal of R different from (0). Therefore Rz = R and there exists z’ E R such that z’z = zz’ = 1. But clearly z’ E Z and Z is a field. The relationship between the center and the centroid of a simple artinian ring is described in the following:
Theorem 4. The centroid T of a simple algebra R {Ss I z E Z } , where Z is the center of Knxn.
= Knxn
Proof. Clearly {%, I z E Z } C r. Conversely let y Then for all a E R ,
r and
ay
Hence y
=
=
( a . 1)y = (S,l)y = f,(ly)
=
E
over @ is
set c
=
ly.
a .c
flt, . Moreover ay = (1 ’ a)y = (Xlal)y = S L , ( l y )= f l t d = ca
for all a E 3.Therefore c E 2.
I
The following theorem is of basic importance in the theory of central algebras.
Theorem 5. Let GZ be a central, simple algebra over the field @. Let 33 be an arbitrary @-algebra with identity 1. Then if b is an ideal in 33,
over @. The mapping #: b GZ @ ob is an ideal in the algebra GZ @ is a lattice isomorphism of the lattice V(33) of ideals in 33 onto the lattice V ( a @ *33) of ideals in GZ @ @33. ---f
130
VI.
SEPARABLE ALGEBRAS
Proof. From the very definition of multiplication in the tensor product @ Qb, it is clear that a @@b is an ideal of 6t'@ @33 if b is an ideal of 3. Moreover, the implication b,
C b, * a @b, C 02 @ b,
(4)
is trivial. To show # is one-one, let b, $ b, and let b, E b, , b, $ b, . Then a @ b, 4 @ b, if a # 0. Otherwise, let {ui1 i E Z} be a @-basis of b, (cf. Theorem 1.3.1). Then {b, , zii I i E Z} is a linearly independent family over @. a 0b, E 0b, implies the existence of elements a, , ...,aN in 6t' such that a@b,~j@u,~=O
C
1<3
The linear independence of {b, , ui,, ..., ui,} and the definition of the tensor product imply a = a, = ... = aN = 0, a contradiction. Hence, if b, $ b, , #b, g #b, . This, together with (4) proves that # is one-one and preserves the partial ordering C of the lattices. Finally we must show # is onto. Let W' = {b 1 b E 33, a 0b E W"for all a E a},where W is an ideal of a @ The vector space W' over @ is contained in 33. Each V E W is expressible in the form V=
C
1
a,@b,,
aifO,
i=l,
..., N
(5)
where, without loss of generality, we may assume {a, , ...,aN} is linearly independent over @ and b i E 33. We now show b i E W ' . 02 is a faithful irreducible T(GPI)-module and by hypothesis
a) = @
Hom,,a,(6t',
Hence given a E .G?there exists T E T(02)such that Ta,=a,
Ta,=O,
j = 2
,..., N
Since T @LZ1 is an element of T(02@@33),the multiplication algebra of since W is an ideal of a @33, it follows from (5) that and b, E W ' . ( T 02'l)V = a 0b, E W. Thus a @ b, E W for all U E Similarly we show bi E W' for all i. Hence # W' = W . To complete the proof we merely observe that since
a@.%,and
a @ ub = (a 0u)(l @ 6 ) E W
for all a E a, o E W' and b E 33
a @ bu = (1 @ b)(a @ u ) E W
for all a E a, u E W and b E 33
and W' is an ideal of 33.
I
We now give two applications of this theorem.
1.
CENTRAL, SIMPLE ALGEBRAS
131
Theorem 6. Hypotheses: Let GZ and 33 be subalgebras of @-algebra @. Let Ol be central and simple and let 33 contain an identity. Suppose a . b = b . a for all a E GZ and b E 33, and assume that if b E 33 and GZ . b = 0, then b = 0. Conclusion: GZ . 33 is isomorphic to GZ @ ,33 under the isomorphism
Proof. Clearly $ is an epimorphism. Let ker $ be the kernel. If ker $ # 0, then by the preceding theorem ker$=@@b where b # 0 is some ideal of 33. Thus for each b # 0 in b and all a E a, = 0. By hypothesis b = 0 for all b E 33, a contradiction. #
ab
Theorem 7. Let GZ be a central, simple algebra over @ and A an arbitrary overfield of @. Then GZ @ ,A is a simple A-algebra. Proof. The lattice of ideals of G Z @ A is isomorphic to the lattice of ideals of A by Theorem 5 . The latter consists only of (0) and A . I The last theorem enables us to complete the investigation of algebras with absolutely irreducible modules begun in Theorems 1 and 2.
Definition 4. Let R be a @-algebra. An overfield A of 0 is a splittingfield of the algebra R if each irreducible module of the A-algebra R 0,A is absolutely irreducible. Theorem 8. Let @ be the center of the simple algebra R , let dim, R < co and let A be an overfield of 0.Then A is a splitting field of the @-algebra R if and only if R 0,A A,,,
-
for a suitable r. If in addition dim, d is finite, dim, R
= r2.
Proof. By Theorems 4 and 7, R @ @A is a simple A-algebra. Hence R
04
R5 K r x r
where K = HomRod(M, M ) , M an irreducible R @ A-module by Theorem 11.6.2. By Theorem 1, M is an absolutely irreducible module for the
132
VI.
SEPARABLE ALGEBRAS
A-algebra R @ A if and only if K as follows: dim, R . dim, A
= dim,(R
= A.
The @-dimension of R is obtained
@ A) = dim, A,,, = r2 dim,
A 1
As an immediate consequence we have: Theorem 9. Let R be a central simple @-algebra with dim, R < co. Then each irreducible R-module is absolutely irreducible if and only if R is isomorphic to Q n X n for a suitable n. 1 From Theorems 2 and 8 it follows that an algebraically closed field Q containing @ is a splitting field of the simple artinian ring R with center @ if R has faithful absolutely irreducible R-module. We have come across the concept of a splitting field of a field @ on two occasions: once in Section V.l as the splitting field or rootfield of a polynomialf(X) in @[XI and above, in Definition 4, as the splitting field of an algebra R over @. The reason for the importance of a splitting field A of a @-algebra R (if d exists) is due in part to the fact that R may be embedded in R @ ,A m A,,, , the structure of which is well known. The embedding map is a+a@l
Theorem 10. 1. A simple artinian ring R = K,,, is an algebra over its center @. If E is a maximal subfield of the division ring K , then E is a splitting field of the @-algebra R. Moreover, every overfield of E is a splitting field of R over @. 2 . Let n = 1 in Part 1. Then R = K , a division ring. We know K @ ,E = Erx,for some r. Then dim, K = r . Furthermore, if dim, K < co, then dim K = r 2 and dim E = r. Proof.
1. The set theoretic union of a chain of subfields of K containing @ @ = A,C.,,CAiCAkC...
is a subfield of K. Hence by Zorn’s lemma there exists at least one maximal subfield of K . 2. A field E lying between @ and K is maximal in the family of fields containing @ and contained in K if and only if E contains all the elements of K which commute with E, i.e. if and only if
E
=
{k I k
E
K, ek
= ke
for all e E E )
(6)
1.
CENTRAL, SIMPLE ALGEBRAS
133
Proof. Denote the set on the right of (6) by 6. Every overfield L3 of E lies in G. Hence if (6) holds, E is maximal. On the other hand if (6) does not hold, then there is an element z E G, z 6 E. But the division ring generated by E and z is a field strictly containing E. Thus E is not maximal. 3. Let M be a faithful irreducible R-module with centralizer HOmR(M,M )
=K
and let E be a maximal subfield of K containing 0.Since by hypothesis R is central, simple, R 0@Eis a simple E-algebra by Theorem 7. We turn M into an R @ .E-module by defining
(1a, 0 ei) m =
aimei
for m E M , a, R, ei c E
(7)
(7) is independent of the choice of representative since [(a, e [(a
+ e’) - (a, e)
+ a’, e)
-
-
(a, e’)]m = am(e
(a, e) - (a’, e)]m = (a
+
+ e’)
ame - ame‘ = 0 a’) me - ame - a‘me = 0 -
and [(aa:,e) - (a, a:e)]m = (ama:)e - (am)(a:e)= 0
for all
a: E @
Furthermore, (a @ e)[(a’ @ e’)m] = (a @ e)(a’me’) = aa‘me‘e = (ad
0ee‘)m
since E is commutative. By hypothesis M is an irreducible R-module and since R = R @ 1 C R @ E, it is also an irreducible R @ E-module. Hence, since R @ E is simple, M is faithful as an R @ E-module. By the density theorem, R @ E is isomorphic to D,,, , where D = HOmR@E(M,M ) and r
= dim,
(8)
M . By the first assertion in Theorem 1 , it is sufficient to prove E e D
(9)
Proof of (9). By definition 7, 1 @ E w E is contained in D, since for all e E E
134
VI.
SEPARABLE ALGEBRAS
On the other hand, D C K since M is an R-module and K = HOmR(M, M ) . Let 8 E D. From m8e = (1 @ e)(m8) = [(I @ e)m]8 = me8 it follows that m8e = me8 for all e E E and all m E M . Thus 8e = e8 for all e E E. By the characterization of maximal subfields of K proved in 2, it follows that 6 E E and hence D C E. 4. That each overfield A’ of a splitting field d is a splitting field follows from
@ J’
M
dixr
5. The assertions in Part 2 concerning the various dimensions follow from the fact that if R = K is a division ring, the module A4 = RR is a faithful irreducible R-module. Thus by (8) and (9), r = dim, M = dim, R = dim, R. If in addition dim, R < co, then dim, E = r by Theorem 8 and since r2 = dim, R = dim, R . dim, E = r * dim, E. 1 To generalize the above investigation of simple artinian rings to the semisimple case we introduce the following: Definition 5. A @-algebra R is separable over @ if for each overfield d of @, the d-algebra R @ .A is semisimple. We note that, in particular, a separable algebra is semisimple (d = @).
Theorem 11. Let the artinian ring R be a @-algebra. R is separable over @ if there is an overfield E of such that
Moreover the condition is also necessary if dimGR
< 00.
Proof. 1 . If R is separable, dim, R < co and 8 is an algebraically closed field containing @, the 8-algebra R 0 is separable and hence isomorphic to a direct sum
K&.l
0
@ K$krU
of simple 8-algebras. Since 8 is algebraically closed, then by (3) in the proof of the second assertion of Theorem 1, each K‘j)= 8. 2. Conversely, suppose condition (10) holds. We must show that for any overfield d of @ (and not only of E ! ) , R @ is semisimple. We first construct a field A containing as subfields E and d : The mappings e + e @ 1 and 8 + 1 @ 8 are embeddings of the fields E and A , respectively in E 0,A. By Zorn’s lemma there exists a maximal ideal A in the commutative ring
1. CENTRAL, SIMPLE ALGEBRAS
135
E @ ,& since l E 0 contains an identity 1 @ 1. The quotient ring E 0,d/A is a field A in which E and A are embedded by means of the embeddings
6+1@6+A
and
e+e@l+A By Theorem V.3.6 and (10) we have
Hence R
0&lis semisimple. On the other hand
If R @ ,A is not semisimple, its radical contains a nilpotent ideal W # 0. But then W @ ,A is a nilpotent ideal in ( R .A) @ ,A w R @ $A, contradicting (1 1). Therefore R @ od is semisimple. I
Theorem 12. Let R be a semisimple artinian ring R which is an algebra over @. Let
R
= Otl @
... @ Ota
be the decomposition of R as a direct sum of simple algebras Otj , j and suppose the center of Olj for each j is 0.Then R is separable.
j
=
1, ..., q
Proof. Let E be a field containing the splitting fields E ( j ) of the Otj, 1, ..., q. By Theorem 10, E is also a splitting field of each Olj and hence
=
By Theorem 11, R is separable over 0.
I
It may be shown that if d is a separable extension of @ in the field-theoretic sense, (this is the case, for example, when char 0 = 0) then R @ od is semisimple if R is. This result will be proved in the context of Frobenius algebras in Chapter VIlI (cf. Theorem VIII.6.2). A proof which is independent of the theory of Frobenius algebras may be found in van der Waerden [1959], Section 153.
136
VI.
SEPARABLE ALGEBRAS
2. The commutativity of rings satisfying x"(')
=
x
It was known to Wedderburn that a finite division ring K is a field. (We shall prove this result in Theorem 3.) The multiplicative group K X= K\O is finite and therefore each element k E K X has finite order, i.e. there exists a natural number m such that k" = 1. If we include the zero of K we may say quite generally that an(a)
=a
(1)
for all a E K and suitable n(a) E N. Wedderburn's theorem was generalized by Jacobson to the following: Theorem 1 (Jacobson). Let R be a ring such that for each a E R there exists a natural number n(a) > 1 dependent on a for which an(a) =
Then R is commutative.
a
I
The proof of this theorem depends on a number of intermediate results. First, we may write an(u)= a as la = a. It follows that an(a)-lak= ak, k = 1, 2, ... . Taking k = n(a) - 1 we have
Since e E (a) and a E (e), (e) = (a). If R is not semisimple, the radical J contains an idempotent e different from 0. But e must then be quasiregular and there exists b E R such that e b - eb = 0. Multiplying both sides by e we get e = 0, a contradiction. Hence R is semisimple and as such is a subdirect sum of primitive rings:
+
where each RIP has the same property as R. We now show that a primitive ring R with the property expressed in (1) is a division ring: Let M be a faithful, irreducible 8-module and let K be its centralizer. If dim, M > 1, there exist two K-linearly independent elements u1 , u2 E A4 and an element U E R such that au, = u2 and au2 = 0. This implies gnu1 = 0 # t12 for all n 2 2, a contradiction since an(a)= a. - Thus dim, A4 = 1 and R m K. The commutatitivity of the subdirect sum R follows from the commutativity of the direct summands RIP. Therefore to complete the proof of Theorem 1 it is sufficient to consider the special case that R is a division ring K. We start our investigation of this special case with the following:
2.
THE COMMUTATIVITY OF RINGS SATISFYING X n ( ‘ ) = x
137
Theorem 2. Let K be a division ring and let @ be the center of K. Clearly K is a @-algebra. Let S1 and S, be @-subalgebras of K which are division rings. Suppose dim, S , = dim, S, < co and let
4: s1+ 4sl
for s, ES,
be an isomorphism of S1onto S, . Then there exists an element k , E K such that +s1
We say that
=
k;’s,k,
for all s,
E S,
4 may be extended to an inner automorphism of K.
Proof. Let K* denote the division ring consisting of the same elements as K with the same additive structure as K but with multiplication in K* defined by
a*b=b.a
(3)
Both K and K* are central over @. By Theorem VI.1.5 applied to (72 = K* and 33 = S1, the tensor product S1@ ,K* is a simple ring. It contains a minimal left ideal since each left ideal of S , @ ,&* is a K*-module and by the very definition of the tensor product dim,*(& @ ,K*)
=
dim, S ,
< 00
(4)
It follows from Theorem 11.6.1 that all irreducible S1@ ,K*-modules are isomorphic to each other. Now K itself can be turned into an Si@ K*-module for i = 1 , 2 by defining (si @ ki)k = sikki ,
i
=
1, 2
(5)
and extending the definition to Si @ K* by linearity. Since distributivity is built into the definition we need only check
We note that we used (3) in our calculations. On the other hand, using the isomorphism 4 in the hypothesis of the theorem, K may be turned into an S1@ ,K*-module by defining (sl @ k,)k
= +(s,)
. kk,
for s, @ k, E S, @ ,K*
and
k
E
K (6)
as may be easily checked. Since K is a division ring it is an irreducible S1@ K*-module relative to the definitions in (5) and (6). Hence, as we
138
VI.
SEPARABLE ALGEBRAS
remarked above, there exists an S , @ K*-isomorphism i,h of the module K as defined in (5) onto the module K as defined in (6): Since # is an S10K*homomorphism,
for all s1 E S , , k
Since s1 E S , we obtain
E
K , k , E K*. Set k ,
=k =
1 in (7) and we get
C K , s1 is an element of K*. Setting (s, , k , k,) #(s1)
=
+(I) . W )
*
$1 =
= (1,
#(I) . s1
1 , s,) in (7) (9)
Comparing (8) and (9) it follows that
where k;'
=
#(1).
I
The proof we now give of Wedderburn's theorem is based on the previous result. Theorem 3. A finite division ring is a field.
Proof. Let 2 be the center of the finite division ring K . Since Q $ K, the characteristic of K is a prime p . K therefore contains the field C , as the smallest field in K . Assume by way of contradiction that Z # K . Since C , consists of multiples of the identity element l K of K, C,C 2. By Theorem VT.1.10 there is a subfield E of K containing Z which is maximal in the family of subfields of K . Furthermore, applying Theorem VI.1.10 with @ = 2 we have r 2 = dim, K
and
dim, E
=
r
It follows that all maximal subfields E of K have the same dimension over C , , namely n = dimcp D
=
dim, E dimcp Z
=
r . dimcp 2
The number of elements in E is therefore p n . Now by Theorem V.1.5, finite fields of the same order are isomorphic. Since 2 # K, there exist elements 8, and 0, in K such that OIOz # B , O , . Hence there exist at least two distinct (but isomorphic) maximal subfields El and E, of K. By Theorem 2 applied
2.
THE COMMUTATIVITY OF RINGS SATISFYING X n ( ' ) = X
139
to Si= Ei , i = 1,2, each isomorphism q5 of El onto E, can be extended to an inner automorphism of K:
+el
=
k;le,k,
for all e,
E
E
(10)
where k, is a suitable element of K. Every maximal subfield E j of K is the image under an inner automorphism of El :
Ej
=
kjElki1,
j
1, ..., N , N 2 2
=
(11)
The set theoretic union of the Ej is the whole of K since each element k E K is contained in the subfield Z ( k ) which in turn is contained in a maximal subfield of K. If we restrict our attention to the multiplicative group K X of nonzero elements of K, we see that (1 1) expresses the fact that the subgroups EjX are conjugate to one another, i.e.
EjX = k j E I X k i l ,
,j
=
1, ..., N , N 3 2
(12)
The set theoretic union of the EjXif KX. We show that this leads to a contradiction: Let k' and k" be elements in the same right coset of EiX in KX. Then there exists el E El such that k" = k'e, . Hence
k"E,Xk"-l
=
k'e E Xe;lk'-f 1
1
=
k'EIXk'-l
showing that conjugation of EIX by k' and k" gives rise to the same subgroup of K X . The number N of distinct maximal subfields E j , j = 1, ..., N of K is therefore at most j K X : EIX 1, the number of right cosets of Elx in K X . But the union of the E j x is the whole of K X and hence the E j x are pairwise disjoint, a contradiction since they each contain the identity of K X . I We now return to the proof of Jacobson's theorem. We recall that, from our discussion preceding Theorem 2 , it is sufficient to prove the theorem in the case the ring is a division ring K. Assume therefore that K is a division ring with the property that for each a E K an(a)
~
a,
n(a) 2 2
It follows that the smallest subfield contained in the center Z of K cannot be the field of rationals Q since, for example, 28 # 2 for any s > 1; hence it must be C, = C where p is prime (Theorem V.2.1). Suppose now Z # K and let u E K\Z. Letf(X) = a, a,X ... 3. X N be the monic polynomial of least degree with coefficients in 2 having u as a root; f(X)exists since u is a root of X n ( u )- X . Let F be the field generated by C and a,, , ..., aNPl. F is a C-space of finite dimension t over C. Since the order of C is p , the
+
+
140
VI.
SEPARABLE ALGEBRAS
C-space F contains exactly q = p t elements. The multiplicative group FXof F therefore contains q - 1 = p t - 1 elements. Each element f EFX satisfies f"-' = 1 and hence each element of F is a root of the polynomial X g - X . Since K is a division ring containing F, u cannot be a root of Xg - X since a polynomial o f degree q in the field F(u) can have at most q roots. Therefore 214
i .u
(1 3)
On the other hand u is a root of the polynomialf(X) with coefficients in F = C(a,, ..., aNP1)C Z sincef(X) is irreducible over 2.
-wa/(f(x>) * F(u) by Theorem V.2.2. Since q in K and hence By (14), the mapping
4:g ( 4
= pt, where
p is the characteristic of K, p . 1 = 0
W )= (f(.Nq
-
g(u9
(14)
=0
for g ( X ) E W
I
(15 )
is an isomorphism of Z(u) onto itself. Since dim, Z(u) = d e g f = N < co, we now have the situation described in Theorem 2 with S, = S, = Z(u), @ = 2, 4as in (15). Hence there exists an inner automorphism of K induced by some k, E K which extends 4. In particular therefore and hence This last equality implies that the division ring C(u, k,) generated over C by u and k , is finite. Proof: Because of (16) each element of the ring C[u, k,] generated by u and k, over C is of the form h(u, k,,), where h(X, Y ) is a polynomial in X and Y with coefficients in C. Since un = u and k,t = k, for suitable n and t E N, it follows that the degrees of X and Y in h ( X , Y ) are at most n - 1 and t - 1, respectively. Hence since C is finite, the ring generated by u and k, over C is finite. Now if h(u, k,) E C[u,k,], there exists s E N such that ( N u , ko)Is = h(u, ko) Therefore h(u, k,)"-l = I and C[u,k,] = C(u, k,). By Theorem 3 (Wedderburn), C(u, k,) is commutative. It follows from (16) that
kou = u . k, and hence
II =
uQ,contradicting (13).
=
koUQ
3.
THE QUARTERNIONS
141
Jacobson’s Theorem 1 has been generalized (cf. appendix) and has found application in areas outside ring theory, for example, in the foundations of geometry. The proof of the theorem is interesting in that use is made of the subdirect sum of rings. In general the fact that a ring S is a subdirect sum of rings Ri, i E Z gives us little information about the ring S. In this case, however, the relevant property, namely commutativity, is inhereited by S.
3. The quarternions Let R be the field of real numbers and C the field of complex numbers. Then the only finite-dimensional field extension of R is C. For it is well known that C is the smallest algebraically closed overfield of R and the adjunction of a transcendental element, i.e. an indeterminate X , to R gives rise to an infinite-dimensional extension. Now let K be a noncommutative division ring containing R as center and suppose dim, K is finite. By Theorem V.1.10 if E is a maximal subfield of K, dim, K
dim, E
= r2,
= dim,
K
=
r
(1)
From the remarks made above, E is either isomorphic to R or to C,i.e. = 1 or r = 2. If r = 1, dim, K = 1 and K = R showing that K is commutative. If r = 2, dim, K = 4 (2)
r
Now let E w C be a fixed maximal subfield of K. Let 4: e + i? be the automorphism of E which maps a complex number e onto its complex conjugate 2. By Theorem VI.2.2 with @ = R, S1= S, = E m C and 4: e + 2, there exists an element k, E K which induces an inner automorphism of K which coincides with 4 on E:
k
+ k,kki1
(3)
Let i denote the square root of -1 in E. Thus E = R(i), i2 and since +e = i?
k,ik;l
=
-i
and hence
k,i # ik,
=
-1. By ( 3 )
(4)
Now since E is a maximal subfield of K, each element of k commuting with E is already in E (cf. Eq. VI.l(l)). Thus by (4), R C R n E f E
This implies
R(k,) n E
=
R
142
VI.
SEPARABLE ALGEBRAS
since dim, E that e
= 2. On =
the other hand it follows from (3) and # ? e= Z
k,"ekG2
and hence
=e
ekO2= kO2e, V e E E
Therefore by VI.l(l), k,2 E E. But k,2 E R(k,) and hence k,2 E E n R(k,) = R. If kO2> 0, k , E R C E, a contradiction. Thus there exists w E R such that k,2 = -w2. Let
il
=
i
1 i2 = - k ,
and
W
Then iZ2= w-2k,2 = -1 and by (4),
i2ii-1 2 showing that i2il j,2
=
w-lk 0 jk-lw 0
=
-iliz. Finally, set i,
=
jlizjlj2 = -i
12 j22
=
= -i
= iliz. Then
. (-1)
-(-I)
=
-1
The following relationships hold between il , i 2 , and i3 :
ilz iliz = i3 = -&i1 ,
= j r 2 = j32 =
-
i2iB= il = -i,i2
1
,
i3il
=
i2
=
-i1i3
(6)
It follows that the set of elements of the form.
k
= a,
4-a,i,
+ a,i, + a,i3 ,
a? E R,
j
= 0,
1, 2, 3
(7)
constitute a subring K' of K. Moreover, K' is even a division ring; for if -
k
= a, - a,i, - a,i, - a,i3
then
kk
=
kk
=
ao2
+ a12+ a22+ a?
(8) (9)
The equalities in (6) show that R is the center of K'. Hence, as in (11, dim, K' = 4. From this and the fact that R C K' C K i t follows that K' == K . In particular, (1, i, , iz , i,} generates K' over R. Definition 1. The ring of real quarternions K is an algebra over the reals R with basis (1, il , i2 , i3} and identity 1. Multiplication of elements in K is defined by (6). Our discussion above proves: Theorem 1 (Frobenius). The ring of real quaternions is the only noncommutative division ring of finite dimension over the reals. I
4.
ALGEBRAS WITH SEGREGATED RADICAL
143
4. Algebras with segregated radical As a final application of the results on central, simple and on separable algebras obtained in Section VI.1, we shall prove in this section a theorem of Wedderburn’s which deals with the following question: If R is an algebra over the field @ and J its radical, does there exist a subalgebra S such that R is a direct sum of S and J ? If by “direct sum” we mean a direct sum as rings then we require S to be an ideal and not only a subring of R. In this case such an S very rarely exists. If however by “direct sum” we mean a direct sum as @-vector spaces, we can prove the existence of S if we impose the restriction that the semisimple algebra RIJ be separable in the sense of Definition VI. 1.5. Theorem 1 (Wedderburn). Let R be a finite-dimensional algebra over the field @. Let J be the radical of R and let RIf be separable. Then there exists a subalgebra isomorphic to R / J such that R , considered as a @-vector space, is the direct sum of S and J . Proof: 1. Define the exponent of the radical J (exp J ) as the smallest exponent e for which I“ = 0. If exp J = 1, J = 0 and there is nothing to prove. We now show that the proof in the case exp J > 2 can be reduced to the case exp J = 2. This we do by induction on d = dim, J. If d = 0, J = 0. Let 67 denote R / J and let
$:a+atf, and
t,h:a+a
aER
+J2
be the natural homomorphisms of R onto Ol Let 4’:a
+ J2+a + J
=R/J
for a
and RIJ2, respectively.
+J2ER/J2
be the natural homomorphism of R I J 2 onto RIJ = a. The radical of R / J 2 is J / J 2and thus dim ( J / J 2 )< d since otherwise J 2 = 0 and hence exp J = 2. By the induction hypothesis on d, there exists a subalgebra S of R I J 2 with w Ol = R / f and such that R / J 2 = J / J 2 , where the sum is a direct sum of @-vector spaces. Let S = $-l(S).Then S is a h u b a l g e b r a of R and
s
s+
R / J 2 = S / J 2@ J / J 2
Since S I P rn GZ it follows that J 2 is the radical of S. Moreover, dim J 2 < d, otherwise J = J 2 = ... = J - 0. Again by the induction hypothesis, there
144
VI.
SEPARABLE ALGEBRAS
exists a @-subalgebra T of S (and hence a subalgebra of R) such that T M 6T and S = T 0J2. It follows from R/Jz = ( T
+ J2)/J2@ J/Jz
t h a t R = T + J . But T n J C S n J C J 2 a n d h e n c e T n J = T n J 2 = 0 , showing that R = T @ J. 2. Now let exp J = 2 and let {b,, ..., b N }be a set of elements in R which map to a @-basis {h1, ..., bN}of R / J = GZ under the natural homomorphism ofRtom i = 1, ..., N 6, = +bi, The mapping
is a one-one @-linear mapping of the @-vector space QZ onto the @-subspace L of R generated by (6, , ..., b N } with r$ as inverse. Each element a E R is uniquely expressible in the form f(a)
+w
where S E Ol and
since r$a = r$f(a)(mod J ) for all a E R. Since exp J
(f(a;) + w d ( f ( g 2 ) + W2)
=f
( 4
w E
(3)
J
= 2,
.f(az) + W,f(.,)
+f(a;> . w2
(4)
Iff(.;) . f ( u 2 ) =f(alu2),f is an algebra-isomorphism of QZ onto L. In this case, by (3), R = L @ J and L M GZ proving the theorem. If, however, f(.;) .f(z2)# f ( i i , . a,), we must alter thef(Z) by the addition of elements p ( u ) from J in such a way as to make the mappingf p multiplicative. Thus we must find p E Hom-(QZ, J ) such that
+
(f(Z1)
+ P(.l)>
.(f(4 + P(a2))
= f(S1 *
Because of (4), this is equivalent to finding p f(al) p(a2) - p(a, . a,)
+ p(a,)f(a,) = h(ul
E
+ P(.,
.%)
Horn,(@ J) such that
. a,)
for all a, , a; E GZ
(5)
where 451 ’ ‘zz) = f ( ~ I ) f ( Q
-f@@2)
(6)
The existence of such a p is guaranteed by the fact that R / J is separable. 3. By Theorem VI.l.ll there exists an overfield E of @ such that
RIJ @ O E
=
ETlxrl0
0E T , p v r
(7)
4 . ALGEBRAS WITH SEGREGATED RADICAL
145
In this third part of the proof we show that if the assertion of the theorem holds for R 0@Eit also holds for R . J 0@Eis a nilpotent ideal in R @ @E since J is nilpotent. Hence J 0@Eis contained in the radical of R @ @E. On the other hand,
4‘‘:
aj @ ej +
1 ($aj) @ ej
for aj E R
and
ej E E
is a homomorphism of the algebra R @ @Eonto the algebra ( R / J )@ @E. The kernel of 4’‘ is J @ @E.By hypothesis R / J is separable over @ and therefore R/J @ @Eis semisimple. This implies that the radical of R @ @E is contained in J
[email protected] the radical of R 0@Eis J @ @E.The @-spaces GT @ @Eand L @ @Emay be turned into E-spaces by defining
[c
aj @ ei] e =
ej , e
for aj E GT and
aj @ (eje)
E
E
(8)
CY @ @Eis moreover an algebra over E since E is commutative. The mapping E Hom,(GZ @ E, L @ E ) defined by
f E Homo(@, L ) defined in (2) induces f ’
e, E E
f ’ : 1 6 , @ e i + l (f6i) @ e , ,
The set{& 0 1, ..., 6, 0 I} is an E-basis of GZ @ E and f is the identity on L
[email protected] shows that the mappings r#~“ and f‘ play the same roles with respect to the E-algebra R @ @Eand the corresponding semisimple algebra GY 0E as do $ and f with respect to the @-algebra R . I n the fourth part of this proof we shall show that if E is chosen so that (7) holds, there exists p’ E Hom,(CPI @ E, J @ E) satisfying the analog of ( 5 ) , namely ’$’I
f ’(A,)p’(A,) for all A1 , A,
E
- p’(A1Az)
+ p’(A1)f ’(&I
=
h’(A
Y
22)
(5’)
GZ @ E. Of course, h’ is the analog of h defined in (6):
h’(Al , A,) Since by definition f ’(a@ 1)
h’(a; @ 1, a2 @ 1)
=
f’(Al) .f ’(Az)-f ’(Al . A2)
=f (a)
=
(6’)
01
h’(q , a,) @ 1
for all iil, a, E GZ
We now show how, given ap‘ satisfying ( 5 7 , we may construct p E Homo(@,J ) satisfying (5): Relative to the E-basis (6, @ 1, ..., 6, @ l} the E-linear mapping p’ gives rise t o a system of linear forms with coefficients in E. On substitution of these linear forms in (5’) we obtain a system of defining linear equations over E for the coefficients of the linear forms. Restricting our attention to 0 1 which we identify with GT, we see that f ’ restricts tof, h’ to h, (5’) becomes (5) and the system of linear equations satisfied by the
146
VI.
SEPARABLE ALGEBRAS
coefficients of the linear forms reduces to a system of linear equations over CP. Since we are assuming that p‘ exists, these equations can be solved in E. But since these solutions are expressible in terms of determinants with entries from @, they can be solved in @. Hence there exists p E Homo(@, J ) satisfying (5). 4. This shows that we can restrict our attention to the case that the separable @-algebra RIJ = OL?is isomorphic to OTIXTI0 .*. 0@ T q X V q
(9)
Let Cik) denote the matrix from the Ith summand with 1 in the (i, k ) position and 0 elsewhere. (9) can now be written in the form
c
1 c;p
1<2
l
Let f E Homo(6Y, L ) be such that fq5 is the identity on L , where q5 is the natural homomorphism of R onto 02. We now construct p E Homo(02, J ) satisfying (5). Of fundamental importance in this construction is the fact that exp J = 2. Indeed, because of this, we may turn the radical J of R into an (02,02)-bimodule by defining and
aw = f ( a ) w Set p(5)
1
=
for
wa = wf(a)
w E
J , if€ 02
c,’;’(f(c;f’)* f ( a )- f(c;:’ a)), *
(10)
a E 02
1<2<4 1<2<1.1
To show p satisfies (5) it is sufficient to verify that (5) holds in the case
a1 = CW, ht
ii2
=
aq5
since p is linear,
f(CE’) p(a> -);&; = fee;:))
*
+ p(C;;’)f(a> Cf(c,’:’>rf(c::’> .m .a)
-
f(C::’a
2.1
=
Cf(cg’) [f(c;:’) *f(cg)) - f(c;:’c;’)] .f(a) *
i.1
where the expressions i n the square brackets are in the radical J. Since Definition (10) turns J into a left @-module and since the mutually orthogonal
4.
ALGEBRAS WITH SEGREGATED RADICAL
147
a, the right-hand side of the above equation is equal to f(cg’>[f(C::)) *f(4 - f(c:,.’ .all
C::’ are contained in
Here the first and third lines cancel. In the second line the expression in square brackets is the element -h(Cg), a) contained in the left a-module J . The second line is therefore equal to
In particular, since, by Theorem VI.1.12, the @-algebra R / J is separable if @ is the center of the simple algebras in R/J = @ ... @ d,, Theorem 1 is valid in this case. I
Malcev has shown that the semisimple subring S in the direct sum = S @ J is uniquely determined up to an inner automorphism of R (cf. Curtis and Reiner [1962], Theorem 72.19).
R
Rings with Identity CHAPTER VII
The theory of semisimple artinian rings was brought to a satisfactory conclusion in the preceding chapters. In this and the next two chapters we shall present some aspects of the theory of rings with nonzero radical. The theory developed will not be as complete as in the semisimple case and to avoid complications we shall assume that rings have an identity. If R has the descending chain condition on left ideals, R R is a direct sum of finitely many indecomposable left ideals, i.e. left ideals which cannot be expressed as a nontrivial direct sum of two left ideals. In the first section we shall investigate this decomposition fully. We shall see that, here too, idempotents will play the most important part. The main results obtained will be for the class of semiperfect rings with identity, i.e. the class of rings R such that R/J(R) is artinian and each idempotent of R / J ( R ) can be ‘‘lifted.’’ Every artinian ring is semiperfect but not conversely. For example, the ring of power series in one indeterminate over a division ring is semiperfect but not artinian. If in addition R / J ( R )is a simple ring, R can be represented as a matrix ring over a local ring T , i.e. a ring T such that T / J ( T )is a division ring. Theorem VII. 1.10 deals with the structure of a general semiperfect ring and later i n Theorem VI1. I . 19 we give a proof of the essential uniqueness of the decomposition obtained in Theorem 10. The indecomposable left ideals are the prototypes of the so-called projective modules. These modules will be discussed in Section 2. As an application we shall develop the theory of Asano orders in a quotient ring. This is a generalization to noncommutative rings of the classical theory of orders of the algebraic integers in an algebraic number field. 148
1.
SEMIPERFECT RINGS
149
The lattice of left ideals of a semisimple artinian ring is complemented. This gives rise to the concept of an injective module, i.e. a module which is a direct summand of any containing module. These modules will be dealt with in Section 3 and it will be shown that each module is embeddable in a smallest injective. The theory developed will be applied to self-injective rings R, i.e. rings R such that RR is an injective module. If in addition R has the descending chain condition on both left and right ideals, R is a quasiFrobenius ring.
1. Semiperfect rings
Although a nonsemisimple artinian ring R with 1 is not completely reducible (Theorem IV.1.4), RR is still the direct sum of a finite number of indecomposable left ideals. Definition 1. 1. The direct summands of the R-module M are called components of M . 2. An R-module M If (0) is indecomposable if the only components of M are (0) and M . 3. A decomposition of M into components Mi is a representation of M as the direct sum of the M i . The existence of a decomposition RR = L1 @ ... @ L, is a consequence of the minimal condition on left ideals in R: Let 5 be the family of left ideals of R which are not expressible as a direct sum of a finite number of indecomposable left ideals. Assume by way of contradiction that 9 # 0. Then by the minimal condition on left ideals there exists Lo minimal in 9. Lo is decomposable, say Lo = Lo' L". Since Lo' and L" are properly contained in L o , L,' and L" are not members of 5. Hence each is a finite direct sum of indecomposable modules showing that Lo E 5, a contradiction. Therefore 9 = @ . If the ring has 1 we can establish a connection between the components of RRand the representations of 1 as the sum of a finite number of orthogonal idempotents. Let the ring R, considered as a left R-module, be expressed as a direct sum of left ideals Li, i = 1, ..., r ,
+
The identity 1 of R is then expressible in the form, 1 = el
+ ... + e,
where ei E Li
(2)
150
VII.
RINGS WITH IDENTITY
The e, are idempotent and mutually orthogonal, i.e. e.e. 2 3 = Sije, ,
where
S i j is the Kronecker delta
(3)
The e, generate the L, , L,
ProoJ
=
Re,,
i
=
1, ..., r
(4)
Since e, E Li , Re, C L, . Conversely, if 1, E L,
By the directness of the sum in ( l ) ,
Hence by ( 5 ) , li E Re, and (4) is proved. Substituting e, for li in (6) we get the orthogonality, eiefc= 0 if i # k The idempotence of the ei is proved by observing that e,
+ ... + e . = 1
=
e12 + ...
+ er2
and again using the fact that the sum in ( 1 ) is direct. Conversely, if 1 = e, ... e, , where the e, are mutually orthogonal idempotents, then R R = Re, ... Re,
+ +
+ +
Moreover, this sum is direct since, if a = xlel
+ ... + x,e,
(7)
then aeA = x k e k . Hence the Re, are components of RR. Finally, if L is any component of RR, i.e. R R = L 0L’ for some left ideal L’, then by what we have shown above,
L =Re
and
L’
=
R(l - e)
for some idempotent e. We collect these results in the following:
Theorem 1. Let R be a ring with identity. 1. Each decomposition of R R into a direct sum of finitely many left
1.
151
SEMIPERFECT RINGS
ideals gives rise to a representation of the identity 1 of R as a sum of orthogonal idempotents and conversely: 1
=
e,
+ ... + e , o RR = Re, @ ... @ Re,
2. The left ideal Re is an indecomposable component of RR if and only if e cannot be written as the sum of two nonzero orthogonal idempotents. Such idempotents are called primitive. 3. If R is artinian, R R may be decomposed as a direct sum of indecomposable left ideals. I The primitivity of an idempotent of a ring R may be equivalently expressed as follows:
Definition 2. An idempotent e of a semigroup is primitive if for any idempotent f, the equalities f = ef = f e imply e =f: Theorem 2. An idempotent E # 0 in a ring R is primitive in the ringtheoretic sense if and only if it is primitive in the multiplicative semigroup S of R . ProoJ Let e = g + h, with g # 0, h # 0 orthogonal idempotents. Then ge = g 2 gh = g and eg = g 2 hg = g . Therefore ge = eg = g but e # g showing that e is not primitive in the multiplicative semigroup of R . Conversely, if there exists g E R such that 0 g2 = g # e and g = ge = eg, then g and e - g are nonzero idempotents whose sum is e . Therefore e is not primitive in the ring theoretic sense. I
+
+
+
Let J = J(R) be the radical of the ring R with 1 . Suppose that RIJ is artinian and hence completely reducible. The indecomposable left ideals ill RIJ are the minimal left ideals. Denote the coset u J by ii. Then a = R / J = Re, ... &$ where the 2% are primitive orthogonal idempotents whose sum is 1.If we can ‘‘lift’’ this set of idempotents { & , ..., 2), to a set { e l , ..., e,} of orthogonal idempotents in R with Zi = e, + J and such that e, ... e, = 1, then we can deduce that the ei are also primitive; for if e, = g h, g and h nonzero orthogonal idempotents, then 5 = S A. But neither g nor h can be zero since the radical can contain no idempotent. In fact, if 0 # g = g 2 E J and g y - y g = 0, then g2 y g - y g 2 = 0 and hence g = 0, a contradiction. In the following we shall show that if for each idempotent ii E RIJ we can find an idempotent e E R such that 2 = ii, then we can lift any countable family of mutually orthogonal idempotents
+
+ +
+ +
+
+
+
+
152
VII.
RINGS WITH IDENTITY
of R to a family of orthogonal idempotents of R . This discussion motivates the following definition: Definition 3. A ring R with identity 1 and radical J is semiperfect if 1 . RIJ is artinian and 2. each idempotent U E can be lifted to an idempotent e E R. Theorem 3. Let M C J = J ( R ) be an ideal of R and suppose idempotents can be lifted from RIM. Let g = g2E R and suppose u2 = u(mod M ) . If 0 = ug = gu(mod M ) , there exists an idempotent e E R such that e = u(mod M ) and eg = ge = 0. Proof. By hypothesis there exists f E R such that f =f = u(mod M ) . Therefore fg = 0 mod A4. Since 1 E R and fg E J, 1 -,fg is a unit. Let f ’ = (1 -fg>-’f(l -fg). f ’ is also idempotent and since
f ( l -fg)g
=fg -f2g2
=
0,
f‘g
= 0.
Furthermore, f ’ = (1 - f g ) f ’ = f ( l -fg) = f (mod M ) . Let e = (1 - g)f ’. Then ge = eg = 0 and e2 = e(l - g ) f ’ = ef ’ = (1 - g)f’ = e. I The proof that every countable family {U1 , U 2 ,...} of mutually orthogonal nonzero idempotents can be lifted follows easily: Theorem 4. Let M C J, R a ring with 1, A4 an ideal. Assume idempotents modulo A4 can be lifted. Then, given any countable family (ill ,U, , ...} of orthogonal nonzero idempotents in RIM, there exists an orthogonal family (el , e 2 ,...} of idempotents in R such that ei = uimod M , i = 1,2, ... . Proof. By induction. Assume that the idempotents U1 , ..., U,-l have been lifted to the mutually orthogonal idempotents e l , ..., ePp1.Let g = el
Then T-1
u,g
= i=l
urei = f
+ ... + erp1 r-1
u,ui
= 0 (mod M )
i=l
and similarly gu, = O(mod M ) . By Theorem 3, there exists e, er2= e, = u,(mod M ) and erg = ge, = 0. But e,ei = ergei
=0 =
eieT
E
R such that
I
In particular Theorem 4 implies that in a semiperfect ring R any finite
1.
SEMIPERFECT RINGS
153
family of orthogonal primitive idempotents {El , ..., U,} in R/J such that Ul + ... + U, = T can be lifted to orthogonal idempotents el , ..., e, in R . We can show under these circumstances that e, e, = 1. To this end set e
= el
+ +
+ ... + e, . Then
Hence 1 - e is an idempotent in J showing that 1 fore proved the following:
-
e
= 0.
We have there-
Theorem 5. A semiperfect ring R contains a family of orthogonal primitive idempotents whose sum is 1. R R is then expressible as the direct sum of finitely many indecomposable left ideals generated by these idempotents. I Theorem 6. Every artinian ring with identity is semiperfect. This is a corollary to the following: Theorem 7. Let M be a nil ideal in the ring R with 1. Let X idempotent:
x2 = x mod M
=x
+ A4 be (8)
Then there exists an idempotent e E R which is expressible as a linear combination of the elements x, x2, x3,..., with coefficients in Z and such that e=xmodM. Proof (due to Herstein). Since x2 - x E M and M is nil, there exists a natural number n such that (x2
- x)" = 0
(9)
The various powers of x commute with each other and we may therefore apply the binomial theorem to get
where
Let e
=
xngn
154
VII.
RINGS WITH IDENTITY
From (10) we obtain the following equalities: e2
= X2ng2n = Xn-1Xn+lgg2n-l - X2n-1
g
= Xn-l
=
2n-1
xng2n-1 .'. = xngn = e
The idempotent e is congruent to x mod M as may be seen from the following:
> 0. But
since xs = x mod M for all s
0
=
(1 - 1)" = 3=0
=
1
(-1)j
-
(-1)j
n (.) .I
n (J
j=l
Therefore
5
(?) = 1 .I
(-1)j-1
j=1
Hence xg
3
x mod M and xrLgn= xn mod M from which it follows that e = xmodM
since x2 = x m o d M
I
Theorem 8. Let R be a ring with identity. Let e be an idempotent in R . Then J(eRe) = eRe n J(R) = eJ(R)e
Proof. Let a E J(eRe). Then a = eae and exa E J(eRe) for all x E R. Let y be a quasi-inverse of exu in eRe so that (e - y)(e - exa) = e. It follows that yxa - y = exa. Therefore (1 - y)(l - xa)
=
1 -y
-
xa
+ yxa = 1 - (1 - e) xa
+
Since (1 [l - el xa)(l - [l - el xu) = 1 and a(l - e) = 0, 1 - xa has [I - el xa)(l - y ) as a left inverse. Hence for all x E R , 1 xa has a left inverse proving that a E J(R). Therefore u = eae E eJ(R)e. Conversely, let a E eJ(R)e and let x E eRe. Since eJ(R)e C J(R), 1 - xu has a left inverse 1 - y E R . Hence
(1
+
~
(e
-
eye)(e - xa)
=
e(l
-
y) e(e - xu)
=
e.I
=e
1.
SEMIPERFECT RINGS
155
Theorem 9. Let e # 0 be an idempotent in the semiperfect ring R. Then eRe is also semiperfect. Proof. By the isomorphism theorem and the previous result eRe eJe
eRe f J J
- B
+
Let LleJe C L’leJe be left ideals in eReleJe. Consider the left ideals (RL J ) / J and (RL’ J ) / J in RIJ. Suppose if possible that RL J = RL’ J. Then e(RL J)e = e(RL‘ J)e. Hence L eJe = L‘ eJe since eRLe = L and eRL’e = L’. Therefore L = L’, a contradiction proving that ( R L J ) / J C (RL’ J)/J. Hence eRe/eJe is artinian. Now let u E Re and u2 - u E eJe C J . Since R is semiperfect and u(l - e) = (1 - e)u = 0, there exists by Theorem 3 an idempotent f E R with f - u E eJe and f (1 - e) = (1 - e)f = 0. Thus f =f e = ef =f and it follows that f E eRe. I
+ +
+
+ +
+
+ +
+
Since the factor ring RIJ of a semiperfect ring R is artinian semisimple
R, 0 ... @ R, where the R, are simple rings Let 1 = gl + ... + g, , gzE R, . The set (8, , gz , ...,g,} of orthogonal idempotents of RIJ may be lifted to a set of orthogonal idempotents { g , , ..., g,} in R . Furthermore, g = g, + ... + g, is the identity of R since 1 - g is an RIJ
=
idempotent contained in J. By Theorem 9, the rings g,Rg, are semiperfect w 8% where R, is simple. We therefore obtain a decomposiand g7Rg2/gZJg2 tion of R as follows:
R
=
1R1
=
1
lGZC4
g,&
+
g,Rg, zf 3
The first sum on the right-hand side is a direct sum of rings g,Rgz since g,g, = &,g, . The second sum is a subgroup M of the additive group of the radical since g,Rg, = g,g3R zz 0 mod J. In general M i s not an ideal because, for example, g,Rg, . g,Rg, C g,Rgl . We have proved the following important theorem: Theorem 10. Let R be a semiperfect ring. Then considered as an additive group, R = S @ M , where M is a subgroup of the additive group of the radical of R and S is a subring of R with the following properties: S , considered as a ring, is a finite direct sum of semiperfect rings R1 , ..., R, whose factor rings R,/J,, i = 1, ..., q are simple. Furthermore, Ri = g,Rg, , J , = giJ(R) g, , M = Cizi giRgj and { g , , ..., g,} is a family of orthogonal idempotents with sum 1, lifted from the set of identities of the rings R, in the decomposition of R / J as a direct sum of the simple rings Ri . I
156
VII.
RINGS WITH IDENTITY
The structure of the rings Ri above will be further investigated in Theorem 21. We shall also show that the subring S and the subgroup of {J, +} are uniquely determined up to an inner automorphism of R. To this end we first prove that if e is an idempotent in R, the ring eRe is antiisomorphic to the endomorphism ring of the R-module Re. More generally we have:
Theorem 11. Let e be an idempotent of R and let a E eRS, the mapping +a:
xe + xea
R. Then for each
for all x E R
is an element of Horn,(&, Rf) and for all a ~ e R f
q: a+
is an isomorphism of the additive group of eRf onto the additive group of Hom,(Re, Rf). If e =f,q is a ring antiisomorphism. Proof. Let 4 E Horn,(&, Rf).4 is uniquely determined by the image of e under 4. Since $(e) E Rf, 4(ez) = e$(e) = euf for some a E R. q clearly preserves addition and therefore eRf and Horn,(&, Rfj are isomorphic as groups. The second assertion follows from [q(ab)]xe
=
y5ab(xe) = xeub
= +bSba(Xe) =
= (xea)b
M b ) d 4 l xe
I
In a semiperfect ring the primitivity of an idempotent e is equivalent to eRe being a local ring: Definition 4. A ring with identity is local if RIJ is a division ring. Theorem 12. Let e be a nonzero idempotent in the ring R with identity. Then the following two assertions are equivalent: 1. e is primitive 2. Re is an indecomposable left ideal Furthermore, Condition 3 implies Assertion 1 : 3. eRe is a local ring If R is semiperfect, Assertions 1 , 2, and 3 are equivalent. Proof: The equivalence of 1 and 2 was proved in Theorem 1. Now assume Re = L1 0L, and let ri be the projection of L, @ L, onto Li. Then r1 and r2 are nonzero orthogonal idempotents in Hom,(Re, Re)
1.
SEMIPERFECT RINGS
157
whose sum is the identity of the ring. By Theorem 11, eRe contains two nonzero idempotents el and e2 such that e, e2 = e. Since the radical eJe of eRe contains no idempotent, e, eJe and e2 eJe are nonzero orthogonal idempotents in eReleJe. Hence this ring is not a division ring. Conversely, if R is semiperfect and if eReleJe is not a division ring, it contains two orthogonal idempotents with sum C since eRe is semiperfect by Theorem 9. These idempotents may be lifted to orthogonal idempotents el and e2 with sum e. I
+
+
+
Theorem 13. Let e be an idempotent in the semiperfect ring R with radical J. Let v: x X = x J be the natural homomorphism of R to RIJ. Then Re is an indecomposable left ideal in R if and only if Rt? is a minimal left ideal in = RIJ. -+
+
Proof. C&? = ZZ = (eRe + J ) / J m eRe/eJe. By Theorem 12, Re (resp. Re) is indecomposable if and only if eRe (resp. ZRZ) is local. Since
ERC FX eRe/eJe and the latter ring is semisimple artinian, so is I%?. Hence eRe is local if and only if 2 8 2 is. The indecomposable left ideals in are minimal left ideals since the lattice Vj&R) is complemented, by Theorem IV. 4.1. Therefore Re is indecomposable if and only if RC is minimal. 1
w
Theorem 14. Let e and f be idempotents in the ring R. The R-modules Re and R f are isomorphic if and only if there exist u, v in R such that uv = e and vu = f .
Proof. Suppose Re FX R f and let xe -+ xeulf, x E R, be the isomorphism (cf. Theorem 11). Let y f -+ yfv’e, y E R be its inverse. Then e . eu’ffv‘e = e and f fv‘eeu’j’ = f . Let u = eu‘f and v =fv’e. Conversely, suppose uv = e and vu = J Then the homomorphism xe + xeu = xuv . u = x u . vu = xuf E R f for xe E Re has the mapping y f yfv for y f E R f as inverse. I -+
If Re w R f for idempotents e and f in R it follows that RZ m Ef in RIJ. The converse is a consequence of the following: Theorem 15. Let e and f be idempotents in the ring R with 1. Let J be the radical of R. Suppose u’u’ = e mod J and v’u’=f mod J. Then there exist elements u and u in R such that uv = e and vu = f .
Proof. fv’ G u‘u’v’ = v’e and eu‘ = u‘v’u’ = u’f mod J imply eu’jiv’ E u‘jiv‘ = u’v’e = e2 = c mod J. Therefore x = e - eu‘fv’ E J and since x = ex, eu’jiv’ = e(l - x). Now x E J and hence there exists y E J such that (1 - x)(l - y ) = 1. Let v =fv’(1 - y ) and u = e d f . Then
158
VII.
RINGS WITH IDENTITY
uu = eu’fv’(1 - y ) = e(l - x )( 1 - y ) = e and hence ( U U ) ~= vuuu = veu = vu. This implies ( f - V U ) ~= f - fuu - uuf uu = f - uu. But since uu =fv’(1 - y ) eu’f =fu‘eu‘f =fu’u’f =f 3 =f (mod J), f - uu E J. Hence f = uu. 1
+
Theorem 16. Let e a n d f b e idempotents in the ring R with identity and let J be the radical. The left ideals Re and Rf are isomorphic R-modules if and only if RZ and Rf are isomorphic R/J-modules. Proof. Theorems 14 and 15.
1
By Theorem 13 the indecomposable left ideals in a semiperfect ring R are precisely those whose homomorphic images in RIJ under the natural homomorphism are minimal left ideals of the artinian ring R/J. By Theorem IV.3.3, these are isomorphic to the irreducible R-modules and therefore as a simple consequence of Theorem 16 we have:
Theorem 17. Let R be semiperfect. Then there is a one-one correspondence between the isomorphism classes of irreducible R-modules and the isomorphism classes of indecomposable left ideals. I In Theorems 11, 14, and 15 we did not assume that R is semiperfect and in Theorem 12 we only needed the assumption to prove the implication 1 3. We may therefore apply these results in the proof of the following theorem:
Theorem 18. Let {el , ...,em} and {fi, ...,f n } be two orthogonal families of local idempotents in the ring R. Let Cz, ei = Zy=lh= 1. Then m = n and there is an inner automorphism of R mapping the set {el, ..., em}onto ...,fin}. the set {fif
+
Proof. 1. Let x X = x J be the natural homomorphism of R onto R/J. We first show that RZi is a minimal left ideal in R = R / J (cf. Theorem 13). Set e = ei and assume m > 1. Since R = K(T - a) Re, it follows that R(T - 2) # R . Applying Zorn’s lemma we obtain a left ideal Z C R maximal subject to containing R(T - a). Clearly R = L Be. This sum is also direct: For suppose (Zn R2) # (0). By semisimplicity of R, (Ln # (0). Let x in R be such that Ti2 E L and & . X2 # 8. The element a22 # 0 has an inverse j i n the division ring ZRZ. Hence j .Z%Z = 2 E 1. Therefore R = L, a contradiction. As the complement of a maximal left ideal in R, Rt; is minimal. ---f
+
+
1.
SEMIPERFECT RINGS
159
2. By hypothesis and what'we have just shown in Part 1,
Re,
+ ... + Re,
=
Rfl
+ ... + Rfn
are two decompositions of as direct sums of minimal (and hence irreducible) R-modules. By the Jordan-Holder Theorem (Theorem V.4.2), m = n and there is a permutation 7~ of the numbers 1, 2, ..., n such that R Z i m Rfni, i = 1, 2, ..., n. Theorem 14 guarantees the existence of elements Ui, iii in R with UiVi = 2, and ViBi = f n i . By Theorem 15 we may assume that uivi = ei and u p i =fni , i = 1, ..., n. It now follows that vieiui= viuiviei =fni and uiLivi = ei . Set
u
=
C eiuifmi
and
v
fniviei
=
i
i
xi
Then uv = ei = 1 =fmi = vu and vej =fnjvjej=.fnjv. The element v is therefore a unit in the ring R and it induces the inner automorphism x + vxv-I which maps ej onto f n j ,j = 1, 2, ..., r. 1 As a first application of this theorem we prove:
Theorem 19. Let R be a semiperfect ring and let R = S @ M where S and M are as in Theorem 10. Then this representation of R is unique up to an inner automorphism of R. Proof. Each idempotent gi in Theorem 10 is the sum of an orthogonal family of primitive (and hence local) idempotents gij . The g i j are mutually orthogonal for all i, j and gij = C i g i = 1. Let { g , , ..., g,} and {h, , ..., h,} be two sets of idempotents which give rise to a decomposition of R as in Theorem 10. By Theorem 18 there exists an inner automorphism x -+ vxu-l mapping { g i j }onto {hij). This mapping takes the summands g,, and gi, of gi onto the summands h,, and hrt of h, since g,gij = S,,gij . This shows the existence of a permutation 7~ of the numbers 1, 2, ..., n such that vg,v-l = h m i . I
xi,$
A further application of Theorem 8 is given in the following characterization of semiperfect rings. Theorem 20 (B*Muller). A ring with 1 is semiperfect if and only if 1 is the sum of a finite number of orthogonal, local idempotents. Proof. I . We have already shown the necessity in Theorem 5. 2. Conversely, let 1 = e, + ... + en , ei local orthogonal idempotents.
160
VII.
RINGS WITH IDENTITY
As was shown in the first part of the proof of Theorem 18, RE$, i = 1, ..., n is a minimal left ideal in R . R therefore has a finite maximal chain of left ideals RZ, @ ... @ RZi, i = 1, ..., II. By Theorem V.4.2 (Jordan-Holder) all such chains have the same length, or, in other words, R is artinian. 3. L e t f = f 2 m o d J a n d f = f l + . . . + f , , T - f = f , + l + * . . + f n bea decomposition of T in R as a sum of orthogonal primitive idempotents in the artinian ring R = R/J. By Theorem 18 there is an inner automorphism X iiXV with UV = Vii = T which maps the set of Zi onto the set of . By , i = I , 2 , ..., n and therefore renumbering we may assume that iieiV ii(Zl ... Zr)V =f. Set e = el ... e, . If u and u are representatives of the cosets U and V respectively, uv = 1 - o for some o E J. Therefore (1 - w’) uv = 1 where (1 - w’)(l - w) = I . Let u1 = (I - w’)u and v1 = v. Then u1 = u mod J and ulul = I . It follows that ululvlul= vlul _= vu = I mod J and hence I - ulul = (1 - ul@ E J. This implies vlul = 1. The mapping x -+ulxvl is therefore an inner automorphism of R and in particular ulevl = ul(el ... e,) u1 E f mod J. Hence idempotents may be lifted modulo J. I ---f
=Ti + +
+ +
+ +
The structure of the direct summands Ri of S in Theorem 10 may be described in terms of the following concept: Definition 5. An R-module M is free if it is a direct sum of modules each isomorphic to RR. Theorem 21. Let R be a semiperfect ring, J its radical and R/J a simple ring. Then R is isomorphic to the full endomorphism ring Hom,(M, M ) of a finitely generated free right T-module M over the local ring T = eRe, where e is a suitable primitive idempotent of R. Proof. Let I be the image of the element x of R under the natural homomorphism of R onto R = RIJ. Theorem 5 applied to R, guarantees the existence of an orthogonal family of primitive idempotents {el, ..., eTZ}such that el ... en = 1. Let e = el . By Theorem 13, ZR is a minimal right ideal of R. By hypothesis, is a simple artinian ring and therefore the ZiR, i = 1, ..., n are isomorphic to each other (Theorem IV.3.3). By Theorems 14 and 15 there exist elements ui , vi in R such that uivi = ei , u p i = e. Since 1 E R, R is isomorphic (as a ring) to HomR(RR, RR), the ring of left multiplications of R by elements of R (cf. Theorem 1.2.3). We first show
+ +
H o m R ( R R RR) , m HomPRe(Re, Re)
(12)
where eRe is the centralizer of the R-module Re and Re is considered as a right eRe-module. (If J = 0, (12) has already been proved in Theorem IV.3.2.)
1.
SEMIPERFECT RINGS
161
Proof of (12). Let 4 E HomR(RR, RR) and set d'(xe) = +(xe) = $(xe2) = $(xe)e E Re for x E R. 4' is clearly an element of Hom(Re, Re). Furthermore, $'(xe . eye) = +(xe) eye = +'(xe) eye for all y E R and hence 4' E HomeRe(Re,Re). Let 7 be the mapping defined by 7: --sr 4' for 4 E HOmR(RR , RR). We show 7 iS an isomorphism onto HOm,R,(Re, Re). Clearly 7 is additive and preserves multiplication. Given 4' as the image of 4, we may recover q5 as follows: uiui = e implies uieui = uiui ' uiui = ei2 = ei for *i = I , 2, ..., n. Let a E R. Then a = e,a ... e,a = uievia. It follows that C$
+ +
+a =
4'
C y5(uievia) = C (4(uie))via = C +'(uie) via
= 7 4 = 0 we therefore have
4 =0
and 7 is one-one. Now let C $(uie) uieia for a E R is e,R by the orthogonality of the e, . Since ax = C eiax and eiax E eiR, #o is clearly an endomorphism of RR . To show that ~ q =5 $,~ let a = ae = C eiae. Since $ is an endomorphism of the right eRe-module Re and $(uie) = $(uie2) = #(u,e)e, the product $(uie) uieiae = $(uievieiae) = $(eiae). Therefore (740)(ae)= q50(ae) = $(ae) and 7 is an epimorphism. Hence 7 is an isomorphism. We therefore have If
$ E HomeR,(Re, Re). The mapping well-defined since RR = e,R + ...
R
%
HOmR(RR, RR)
%
c$~:a
+
= eia -+
HOm,R,
(01eiRe, @ 1eiRe) i
(13)
i
We must now show that the eiRe are isomorphic as right eRe-modules to T, , where T = eRe. The mappings
eixe + uieixe,
x E R,
i = 1, 2,
..., n
(14)
are the required isomorphisms since uieixe
=
uiuiuixe = evixe E eRe
In other words, we have proved that R is isomorphic to the full ring of eRe-linear mappings of a direct sum of n copies of the ring eRe, considered as a right eRe-module. A free right T-module M with generators m, , ..., m, is the direct sum of the right T-modules miT. A T-linear mapping A of M into itself is (as in the case of vector spaces) uniquely determined by the images Am,, k = 1 , ..., n. Let Am, = mitik . Then Hom,(M, M ) is isomorphic to the ring of n x n matrices, T n x n ,over T. I
xi
We now prove:
Theorem 22. Let P be a ring with 1 and let J be its radical. Let P,,, be the
162
VII.
RINGS WITH IDENTITY
ring of n x n matrices over P and let Cik denote the matrix with I in the ith row and kth column and zeros elsewhere. Then the radical J(Pnxn)
=
C J(P) C i k
i.k
+ +
ProoJ Let E = C,, ... C,, be the identity of R = P,,, . Let Z k = zikCik where zilCE J , i = 1 , ..., n. In particular there exists (1 - Z k k ) - l E P. Set z k ' = - z k ( 1 - Z k k ) - l . Then
xi
+C
( E - Zk')(E - 2,) = E
~ i k ( l- ? k k ) - l C i k
-
=
2i
zik(l
C
~ikCik
i
i - Zkk)-lZkkCik
since c i k Cjk= S k j C i k .Therefore ( E - Z,')(E - 2,) = E and the left ideals JCik are quasiregular for each k = 1, ..., n. Hence x i , k JCik is contained in the radical J(R). Conversely, let 2 = Z&k E J(R). For each a E P, J(R) also contains aE . Z = azikCik.Then
xi
9
2 h
There exists Z' E
=zikc(jc E =
'hi
c
i,k
xi,k
xi,k =
azikcikcLh
azLhE
R such that
( E - Z')(E - az&)
=
( E - Z')(I - azjz).
Therefore 1 = ( 1 - zlJ1 - a z j z ) for all a E R and azj is quasiregular. JCil, . I Hence J(R) C
xi,k
Now let Tnxndenote the ring of n x n matrices over the local ring T and let J ( T ) be the radical of T. By Theorem 22, the radical of Tnxn consists of those matrices all of whose entries lie in J ( T ) . It follows that the factor ring Tnx,/J(Tnxn) is isomorphic to the ring of n x n matrices over the division ring T/J(T)and is therefore a simple ring. Moreover, the identity C1, ... C,, of T,,, is the sum of the orthogonal family of local idempotents Cii , i = 1, ..., n. By Theorem 20, Tnxnis therefore a semiperfect ring with simple factor ring Tnxn/J(Tnx,). In conjunction with Theorem 21 we have the following characterization of these rings:
+ +
Theorem 23. R is a semiperfect ring with R/J simple if and only if it is isomorphic to the ring of n x n matrices over a local ring. I As an illustration of this theorem we prove:
Theorem 24. Let @ be a field and @ *
= @[[w]]
the ring of formal power
1. SEMIPERFECT RINGS
163
series in the indeterminate w . Then the ring @$xn of n x n matrices is semiperfect and factored by its radical w@zxn is the simple ring G n X n . However, , @ ,; is not artinian.
@zxn
Proof. The radical of @* is w @ * and @ * / w @ * = @. However, @* is not artinian since urn@*, m = 0, 1, ..., is an infinite, strictly descending chain of ideals and this gives rise to the descending chain in . I
@zxn
In Theorem 18 we proved that if we are given 2 families of orthogonal, local idempotents whose sum is 1, we can find an inner automorphism which maps one family onto the other. We now use this fact in the proof of the following theorem due to Krull and Schmidt which we shall need in the next section: An R-module M is said to be indecomposable if M has no nontrivial direct summands. Now let M = Ml @ ... @ M , and M = Nl @ ..* @ N , be two representations of M as direct sums of indecomposable submodules. It will be shown that m = n and that for a suitable permutation ?T of the numbers 1, 2, ..., n, M im Ngi if M is both artinian and noetherian. The proof depends on the fact that such a module is local if and only if M is indecomposable. Theorem 25. Let 4 E Hom,(M, M ) . 1. If M is artinian, 4 is an epimorphism if is a monomorphism. 2. If M is noetherian, 4 is a monomorphism if 4 is an epimorphism.
+
Proofi 1. There is a natural number n such that M 3 Mq5 3 ... 3 M$n= M+"+l. Let x E M . Then xqP = y+"fl for some y E M . But 4%is a monomorphism since 4 is and therefore x = y+. 2. Similarly O C 04-1 C ... C 04-. = O+-n-l for some n. Let x+ = 0 for some EM. is an epimorphism since 4 is. Hence there exists y e M such that y4" = x . Then y+"+l = X#J = 0 and y E O+-"--l = O$-.. This implies x = y+n = 0. I
Every linear transformation T of a finite-dimensional vector space gives rise to a decomposition of the space as a direct sum of the kernel of the transformation and a subspace isomorphic to the image of T. A similar situation obtains when M is artinian and noetherian Theorem 26 (Fitting). Let A4 be an artinian and noetherian R-module and let 4 E Hom,(M, M ) . Then for some natural number n,
164
VII.
RINGS WITH IDENTITY
Proof. Since M is artinian, there exists n with M 3 A443 ... 3 M4" = M P + l = ... - M # P . Thus c#Pinduces an epimorphism of the noetherian module M4". By Theorem 25 this restriction of 4" to A44n is a monomorphism. Hence the sum MqP @ OC#-. is direct. Moreover, if X E M , x$" = yq52n for some Y E M . Therefore x - yqP E O C # - ~ and x = y+" (x - y4") proving that M~#P0O+-" = M . I
+
Theorem 27. Let M be an artinian and noetherian R-module. Its endomorphism ring Hom,(M, M ) is local if and only if M is indecomposable. In this case each endomorphism is either an automorphism or nilpotent.
Proof. Let 4 € H o m R ( M ,M ) and let M be indecomposable. By Theorem 26 either M 4 n = 0 or M+" = M . In the latter case 4" is onto and therefore 4 is a monomorphism by Theorem 25. In the former case, c $ ~ is nilpotent. If C$ is not an automorphism, neither is a 4 for any a E Hom,(M,M). Therefore every element in the left ideal of Hom,(M, M ) generated by 4 is nilpotent. This implies 4 is in the radical of the endomorphism ring. On the other hand no automorphism of M is in the radical of Hom,(M, M ) since the automorphisms are units in Hom,(M, M ) . Therefore the radical consists precisely of the nilpotent elements of the endomorphism ring. It follows that each nonzero element of Hom,(M, M)/J(Hom,(M, M ) ) is a unit and therefore Hom,(M, M ) is local. Conversely, if M is decomposable, say M = M I 0M , , then the projections x1 and x 2 of M onto M , and M, , respectively, are two orthogonal idempotents in Hom,(M, M ) , showing that this ring is not local. I Theorem 28 (Krull-Schmidt). Let M be an artinian and noetherian R-module. Let M = M , @ ... @ M , and A4 = Nl 0 ... @ N,s be two decompositions of M as the direct sum of indecomposable submodules. Then r = s and for some permutation x of the numbers I , ..., r, M i m M T i , i = 1 , 2,..., r .
Proof. Set E = Hom,(M, M ) . Let x , , i = 1, ..., r be the projections of M onto M i , qj , , j = 1, ..., s the projections of A4 onto N j . Then { x l , ..., 7 ~ and {vl, ..., v,) are two orthogonal families of idempotents of E, each with sum 1, where 1 denotes the identity automorphism of M . By Theorem 27 the endomorphism rings of the M iand N , are local. To apply Theorem 18 we must first show that these rings are isomorphic to the x i E x i and q j E q j , respectively. Since x i M i = M i , for each 4 E E, r i + x i induces an endomorphism of M i .On the other hand let 4, E Hom,(M, , M i ) . Set ('& m k )$i= tn& . Then $iE Hom,(M, M ) and x i $ i x i = q& . Therefore the mapping +%+ $i is an isomorphism of Hom,(M, M ) onto the subring
~ )
1.
SEMIPERFECT RINGS
165
riErTTi of E and hence niEri is local. Similarly yjEyj is local. By Theorem 18, r = s a n d there is a unit 5 E E (i.e. an automorphism of M ) and a permutation r of 1, 2, ..., r such that c-'vic = qni for i = 1,2, ..., r. Since Nni = Mqni = M[-%ri< = Mri5 = Mi[, M iis mapped onto Nniunder this automorphism f o r i = 1 , 2 ,..., r.
a
Theorem 28 was proved under the assumption that the R-module is artinian and noetherian. In applications the following is often useful:
Theorem 29. Let R be artinian with identity and M a unital R-module. Then M is artinian if and only if it is noetherian. Proof. Consider the following chain of submodules:
M 3 J M 3 ...3 JNM
=
0,
where J is the radical of R. The factor modules JiM/Ji+lM are RIJ-modules and hence completely reducible. If M is artinian (noetherian) so is JiM/Ji+lM. A completely reducible module is artinian (noetherian) if and only if it is a direct sum of finitely many irreducible modules. Therefore if M is artinian (noetherian), M has a finite composition series and by Theorem V.4.2 (Jordan-Holder), M is noetherian (artinian).
a
Theorem 30. Let R be a ring with 1 and U a submodule of M . Then 1. M is artinian (noetherian) if and only if both U and M / U are. 2. Let M be a finitely generated R-module. Then M is artinian (noetherian) if and only if R is.
Proof. l(a) Clearly if M is artinian (noetherian), so are U and MlU. l(b) Let U and MIU be artinian and
M
=
Mo1M,2M,2...
be a descending chain of submodules of M . Then U = M , n U 2 M I n U 2 M2 n U 2 . * . and MIU = ( M , U ) / U2 (M1 U ) / U 2 *.. are descending chains of submodules of U and MIU, respectively. By hypothesis there exists N such that MN n U = M , n U and ( M N U ) / U = ( M , U ) / U for all n 2 N . Since M , 2 M , and since the lattice of R-submodules is modular we have M , = M , (U n M,) = M , ( U n MN)= ( M , U ) n MN = (M, U ) n M , = M , for all n 3 N . Similarly we may show that if U and MIU are noetherian so is M . 2. Let M be finitely generated, say M = Rm, Rm, ... Rm, and let R be artinian (noetherian). If n = 1, M is cyclic and therefore isomorphic
+
+
+
+
+
+ +
+
+
+ +
166
VII.
RINGS WITH IDENTITY
to aR/L, where L = {x E R 1 xm, = O}. Since RR is artinian (noetherian) so is every factor module. Assume inductively that the theorem holds for modules which can be generated by n - 1 or fewer elements. Then Rm, is artinian (noetherian) as we have just shown and MIRm, w (Rm,
+ ..' + RmJRm,
n (Rm, i...
+ Rm,)
is artinian (noetherian) by induction hypothesis. Hence M is artinian (noetherian) by the first assertion of the theorem. I
2. Projective modules and Asano orders
In the last section the indecomposable direct summands of RR and in Theorem 1.21 the free R-modules played the most important roles. All these modules have the property expressed in Definition l(1) below. Definition 1. 1. Let ... -+ A BG C %D ... be a sequence of modules and homomorphisms. The sequence is said to be exact at B if image f = kernelg. The sequence is said to be exact if it is exact at each module. 2. Let R be a ring with 1 and M an R-module. M is called a projective R-module if the left diagram can be embedded in the right diagram -+
M I4
A
2 B +o
n
exact
A-B-0
in such a way that T $ = +. Roughly speaking M is projective if homomorphisms of M can be lifted from factor modules B to modules A. Theorem 1. The R-module RR is projective. Proof. Let 16 = b E B and b = UT for some a E A. Then x+ = x b and ( x a ) = ~ xb for X E R. Set xy5 = xu. Then $ E HomR(R, A ) and X$T = ( x u )= ~ x b = ~ ( 1 4=) ~ 4 I . Theorem 2. Let M be the direct sum of the R-modules M i ,i E I . A4 is projective if and only if the Mi are. Proof 1. Assume each M iis projective and consider the diagram M
I+
A L M - 0
2.
PROJECTIVE MODULES A N D ASANO ORDERS
where the row is exact. Let 4 I M Mi = 4ibe the restriction of projectivity of M i there exists $i E HomR(Mi, A ) with # E HomR(M, A ) . Then #T = 4 and M is projective. $= 2. Conversely, if M is projective and we have the diagram
xi
with exact row, extend to 4 E Hom,(M, B ) by defining $T = 4 for some $ E HomR(M,A ) . Then the restriction #A
=
$ IMA
makes the diagram commutative, i.e. T$,,
9
=
4 to i
= ~
167
M i . By
di. Let
4=
Let
A)
4A. I
The following is an immediate consequence of Theorems 1 and 2:
Theorem 3. Every free R-module and every indecomposable left ideal in a ring with 1 is projective. I Theorem 4. Every R-module is the homomorphic image of a free (and hence projective) R-module P . Proof. Let {bi I i E I}be a set of generators of the R-module B. Let Ri , i E I be a family of R-modules each equal to RR and let P = @ C Ri . The mapping xi -+ C xibi is an epimorphism of the free R-module P onto B. I
xi
Theorem 5. An R-module M is projective if and only if for every exact sequence A + M 0 there exists a submodule U of A such that A = U 0M ’ , where M’ w M . ---f
Proof. 1. Sufficiency: Let P be the module in Theorem 4. Then M is isomorphic to a direct summand of the projective module P and therefore by Theorem 2, M is projective. 2. Necessity: Take = 1 , the identity of M in Definition 1. Then there exists a monomorphism $ from M to A such that #T = l,,,, . Moreover, T$T$ = T $ and the mapping T $ from A onto M$ is an idempotent endomorphism of A . Hence A = M T $ 0A(1, - T $ ) since a = UT$ a(1, - T # ) for all a E A I
+
+
168
VII.
RINGS WITH IDENTITY
Theorems 2, 4,and 5 imply the first assertion of the following: Theorem 6. 1 . An R-module is projective if and only if it is isomorphic to a direct summand of a free module. 2. Let R be a semiperfect ring and let M be a finitely generated R-module. M is projective if and only if M is isomorphic to a direct sum of finitely many R-modules each isomorphic to some indecomposable left ideal of R . Proof. All has been proved except for the necessity of the second assertion.
R = M / J M is a finitely generated R = R/J-module and is therefore a direct sum li? = C @ U, of irreducible R-modules U, since R is semisimple artinian. These are images of indecomposable left ideals Re,in Runder epimorphisms 4%.
+
The epimorphism whose restriction to Rei is +t therefore maps the direct sum P = 0Re, onto M . Since P is projective by the first assertion of the theorem, there exists $ E HomR(P, M ) with i,!m = 4, where T :M -+ Since both 4 and $ 7 ~ are epimorphisms, we have M = P# JM. By the following lemma, J M is small in M , i.e., M = JM N implies M = N. This proves P$ = M . Since M is projective, ker t,h is a direct summand of P by Theorem 5. Moreover, #I = $T and hence ker $ C ker #I. By the definition of 4, ker = C @ J e t = JP and is therefore small in P. Hence ker t,h is small in P and, as a direct summand of P, is the zero module. In other words, $ is an isomorphism of P onto M.
+
+
a.
+
Lemma. (Nakayama). Let R be a ring with identity and J its radical. Let M be a finitely generated R-module. Then J M is small in M , i.e., if A4 = J M N for some submodule N of M , then M = N .
+
+ + + + + +
+
... Rm, and M = JM N i t follows that Proof. From M = Rm, m, = wlml + ... w,m, v , where w i E J and v , E N . Therefore (1 - w,)m, = wlml ... ~ , - ~ m , - , us . Multiplying this equation on the left by the inverse (1 - w,’)of (1 - w,),we get
+
m,
+
=
wl’ml
+ ... + ~ ~ - ~ m +, -us’ , + N . Applying the same process to m,-l
Thus A4 = J M , ... 4-Jnz,-, as we applied to m, and continuing in this fashion, we eventually arrive at M=N. I A vector space M over a division ring R is a free R-module and, given a basis {b, I i E I } each m E M is expressible in the form M = C(rn$,)6, , where $, E Hom,(M, ,R) maps the vector m onto its ith coordinate relative to the basis {b, 1 i E 11. The following is a generalization of this situation: Theorem 7. Let R be a ring with 1 and M and R-module generated by
2.
PROJECTIVE MODULES AND ASANO ORDERS
169
{bi I i E Z } . Then M is projective if and only if there exists a set 1 $iE HomR(M,RR), i E I } such that
{$t
m where m$i
=0
=
1 (m$J bi
for all m E M
i
for all but a finite number of
(1)
&.
xi
Proof. Let P = 0 Rei be the free R-module with basis lei 1 i E I } and rr an epimorphism of P onto M taking ei to 6, (Theorem 4). M is projective if and only if it is isomorphic to a direct summand of P. As in Theorem 5, this is the case if and only if there exists a homomorphism $ of M to P such that $rr is the identity automorphism 1 on M . If $ exists, each image
m$ is uniquely expressible in the form
m*
=
1 (m*,>ei
(2)
where m$i is an element of R depending on m and i. The mapping Qi:m -+my$ is an R-module homomorphism of M to RR. Applying rr to both sides of (2) we get (1) since $7r = l Mand e p = bi . Conversely, if the condition is satisfied, then we may define an R-module homomorphism $ of M to P by m$ = (m$) ei for all m E M . Applying rr to both sides of (1) we get m$n = (m$J bi = m for all m E M since ein = bi . I
xi
xi
We may now base the theory of noncommutative Asano orders (due to Robson) on this theorem. We start with the following: Definition 2. An element a of a ring R is regular if it is neither a right nor a left zero divisor. A ring Q with 1 is a quotient ring if each regular element has an inverse. The subring R of Q is an order (more precisely, a left order) in the quotient ring Q if each element of Q is expressible in the form a-lb, where a, b E R and a is regular. An additive subgroup I of Q is a left R-ideal provided (i) RZC I, (ii) I contains a regular element, and (iii) there exists a regular element b E R such that Ib C R. We define right R-ideals and (twosided) R-ideals in Q analogously. An R-ideal is integral if it is contained in R. Definition 3. The order R with 1 in the quotient ring Q is an Asano order if the R-ideals in Q form a group under multiplication. In this case the group is abelian, the prime ideals in R are the maximal ideals of R and each ideal in R is uniquely expressible as a product of powers of prime ideals (cf. Theorem 11). The integers are an example of an order in the quotient ring Q of rationals. The Z-ideals in Q are of the form Za/b where a # b. The aim of the following discussion is to obtain a characterization of the Asano orders R in Q. It will be shown that they are precisely
170
VII.
RINGS WITH IDENTITY
the “maximal orders” in Q with the property that each integral ideal in R is projective as a left R-module. The expression “maximal order” is precisely defined as follows: Definition 4. An order S in the quotient ring Q is equivalent to the order R in Q if there exist regular elements c, d , e , f in Q such that cRd C S and eSf C R. An order R in Q is maximal if it is maximal (with respect to set theoretic inclusion) among all orders equivalent to R. Theorem 8. 1. Let R be an order in the quotient ring Q. Then for each pair (a, b) of elements in R with a regular, there exists a pair ( a , , b,) in R with a, regular, such that alb = b,a is a common left multiple of a and b. Note: In Chapter XI, we shall show that this condition (due to Ore) is a sufficient condition for the existence of a quotient ring Q containing R as an order. Proof. The product ba-l is in Q . Hence there exist a, ,6, E R such that ba-l = ay’b, . 1
Addition and multiplication in Q can be reduced to the corresponding operations in R once we have established the existence of common denominators. Theorem 9. Let a , , a 2 ,..., a, be regular elements in the order R of the quotient ring Q. Then there exist regular elements a, 6, , ..., b, in R such that a;’ = a-lb, ,i = 1, ..., n. Proof. By induction on n: When n = 1, take a = a12,b, = a, . Assume inductively that regular elements a,, b,’, ..., bhPl can be found so that a;’ = aklb,’, i = 1, ..., n - 1. By Ore’s condition (Theorem 8) there exist elements b and b, in R with b, regular such that bnan = ba, . Solving for b we get b = bnana;‘ and as a product of three units in Q, b is also regular. It follows that a = bnan = ba, is regular and a;’ = a-lb, a;’ = a-lb, Finally, set bi = bbi’, i = I , 2, ..., n - 1. Then a;’ = aG1bi’ = a-lbb,‘ = a-lbi . 1
.
Let I be an R-ideal of Q. It may happen that Z is also a left ideal with respect to an order R’ properly containing R and contained in Q. To understand this we need the following: Definition 5. Let R be an order in the quotient ring Q and Z an R-ideal in Q. Then the set Ol(Z) = {q E Q I qZ C Z} is the left order ofZ and the set Or(Z) = {q E Q I Zq C I } is the right order of I . Finally, we define I* = { q E Q 1 ZqZ L I } as the inverse of I.
2.
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171
Since R is contained in both Ol(Z) and Or(Z), each is an order in Q. Moreover, they are equivalent to R: In fact, let x be a regular element of I and b be a regular element in R with Zb S R. (x and b exist by definition). Then for each q E Ol(Z), I . q . xb C Zb C R and 1 . R . 1 C Ol(Z). Similarly we may show that Or(Z) is equivalent to R. Since
I*
= {q E
Q 114 C Ol(Z))
= (4 E
Q I qZC Or(Z))
(3)
and since &(I) and Or(Z) are subrings of Q, I* is a subgroup of the additive group of Q for which Or(Z) . I*
C I*
and
I* . Ol(l)
L I*
Let b be a regular element with Zb C R C Ol(Z). Then b is in I*. Let x be a regular element in I. Then Z*x C Or(Z) and xZ* C Ol(Z). Hence I* is a left Or(Z)-ideal and a right OI(Z)-ideal. It follows from (3) that Z*I C Or(Z) for arbitrary I. The following theorem characterizes those R-ideals Z for which Z*Z = Or(Z).
Theorem 10. Let Z be an R-ideal in the order R of the quotient ring Q. Then Z*Z = Or(Z) if and only if Z is a projective left O1(I)-module. In this case Z is a finitely generated left Ol(Z)-ideal. If, in addition, R is a maximal order in the quotient ring Q then Or(Z) = Ol(Z) = R. Proof. 1. Since 1 E Or(Z) and Z*Z = Or(Z), there exist finite sets {q, E I* 1 i E J } and {b, E I I i E J } such that C qibi = I . Therefore, for each a E I, C aq,bi = a. Set a+, = aqi for a E Z and i E J. By (3), the form a set of homomorphisms of the left O1(Z)-module Z to the left Ol(I)-module Ol(Z) such that C (a#,) bi = a for all a E I. It follows from the criterion established in Theorem 7 that I is projective as a left OI(Z)-module and is generated by finitely many 6, . 2. Conversely, let Z be a projective Ol(Z)-module. Again, by Theorem 7, let {bi I i E J } be a set of generators of Z considered as a Ol(Z)-module and let {+, j i E J } be a set of Ol(Z)-module homomorphisms of Z to Oi(Z) with the property that Xi(a+,) b, = a for all a E I. Since 1 E Q , the endomorphism ring of ,Q consists of the set of right multiplications by elements q of Q. The mapping x xq for all x E Q induces an O1(Z)-module homomorphism of I to Ol(Z) if and only if Zq C Ol(1); in other words, if and only if q E I*. Since Z contains a regular element, it follows that distinct elements of I* give rise to distinct elements of Homo,c,)(Z, Ol(Z)). Moreover, this monomorphism of the additive group of I* to the additive group of Homo,c,)(Z, Oi(Z)) is an epimorphism as can be seen from the following: Since Z contains a
+,
---f
172
VII.
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regular element of Q (and hence a unit of Q), QZ = Q. Furthermore, Ol(Z) is an order in Q and therefore each element of Q is of the form c-lx, where c, x E Ol(Z) and c is regular. Now Q = QZ = {CZ1c-lxiai I ai E I , ci , xi E Ol(Z)}. The existence of common denominators proved in Theorem 9 gives Q = {c-la 1 c E Ol(Z), c regular, a E I}. Now let 8 be an Ol(I)-module homomorphism of Z to Ol(l). Then
8*: c-la
-,c-l(a8)
for c E Ol(Z), a E I
is a well-defined mapping of Q into itself. Proof. Suppose c-la = cylal. Then a = ccyla, . Since Ol(Z) is a (left) order in Q, there exist elements d and dl in Ol(Z) such that cc;" = d-ld 1 . Then a = d-ld1a, and hence du = dlal . It follows that d(a8) = (da)8 = (dlal)O = dl(alO) and therefore a0 = d-ld1(al8) = ccT1(a,8) proving that c-l(a8) = c;l(a18). Hence 8* is well-defined. That 8* is additive again follows from the existence of common denominators. Moreover, 8* is a Q-module homomorphism. Proof: cy1al . c-la = c;' * d-lb . a, where b E Z,d E Ol(Z) with a1c-l = d-lb. Then (cyla, . c-la) 8* = (dc,)-l ((ba) 8) = cy1d-lb(a8) = cy1a1c-l(a8) = ~;'a,((c-~a)8*). The mapping 8 E Homol(,,(Z, Ol(Z)) is the restriction of 8* E Hom,(,Q, ,Q) to Z since (1-la) 8* = a8 for all a E I. It follows that to each 8 E Homol(,)(Z, Ol(Z)), there exists q E I* such that a8 = aq for all a E Z. Let q i E I*, i E J correspond to the +i, i E J occurring in the equation a = (a&) b, for all a E I
c i
For a given a E I , aq, = = 0 for all but a finite number of i. Since Z contains a regular element this implies that q i = 0 for almost all i and hence almost all are the zero map. It follows that since
+<
a
=
1(a+i) bi = 1aqibi = a c Bibi 2
for all a E I
z
I is generated as an O1(I)-module by the finitely many b, for which q i # 0. Moreover, if we choose a regular in Z, we get 1 = C Bib, E I*I and hence Z*Z C Or(Z) C Z*ZOr(Z) = Z*Z. I
Theorem 11. Let R be a maximal order in the quotient ring Q. Let R contain the identity of Q and let each integral R-ideal in Q be projective as an R-module. Then the lattice of integral R-ideals in Q is noetherian and we have the following: 1. To each integral R-ideal Z in Q there corresponds an R-ideal I-1 in Q such that Z-lZ = ZZ-1 = R.
2.
PROJECTIVE MODULES AND ASANO ORDERS
173
2. Each integral R-ideal in Q is a product of maximal ideals in R. In particular, each prime ideal is maximal. 3. The R-ideals in Q form an abelian group under multiplication and therefore R is an Asano order in Q.
Furthermore, each R-ideal I in Q is expressible as a product of powers of maximal ideals in R: I
=
JLFJL:z
A,:
k, E Z
This product is unique up to order.
Proof. 1. By Theorem 10, Ol(I) = O,(I) = R for each R-ideal I in Q since R is a maximal order in Q. The lattice of R-ideals in Q is noetherian since each R-ideal is finitely generated, by Theorem 10. Hence, a fortiori, the lattice of integral R-ideals is noetherian. In Theorem 10 we have already shown 1*1 = R. We now prove that if I is a maximal ideal in R, then 11* = R. Assume R = I*. Then R = 1*1 = R I = I, contradicting the maximality of I. Hence R # I*. Since 1 E I*, R C I* and hence I C 11* C R. Assume I = II*. Then R = 1*1 = RI* = I*, a contradiction. Now let I be an arbitrary integral R-ideal. There exists a maximal R-ideal A, containing I . If I = I&,,* C &&* = R then &* = R&* = I*I&* = 1*1 = R, a contradiction. Therefore I C I&*. If I&,* # R, I&,* is contained in a maximal R-ideal A,. Again, as before, we have I&,*&,* _C R and I C I&,* C IJL,*jlt,* C R. After a finite number of steps we get IGl*JL2* ..*A,*
=R
Set I-, = A,* ... A,*. Then 11-1 = I-lI = R and the first assertion is proved. Multiplying (4)on the right by A, ... A, we get I = A,, ... A, . This proves the second assertion since if P is prime, P = JL, ..* JL, implies that P must be one of the A+. 2. If A, and A, are maximal ideals, JL1JL2= A, n A,. Proof JCl n A, C A, if JL, A, . Therefore, as we have already shown in 1 , A, n A, = A1S for some integral R-ideal S . This implies JLilc,GC A, and therefore 'S C A, since A, JL, and A, is prime because it is maximal. Hence A, n A, = A19C &,A2C A, n A , . In other words: multiplication of two maximal ideals is commutative and therefore by Part 1 , multiplication of arbitrary integral ideals is commutative. 3. Each integral ideal 'S possesses an inverse R-ideal 'S-l in Q. Let I be an R-ideal in Q. Let J = { x E R 1 Ix C R}. Then J is a subgroup of the additive group of R for which R J = J = JR. By the definition of an R-ideal, R contains a regular element a such that l a C R. Clearly a lies in J and therefore J is an integral R-ideal as well as the product IJ. By what we proved
174
VII.
RINGS WITH IDENTITY
in Part I, Z = IR = IJJ = Z J . J-1and hence I ( J . (ZJ)-l) = (ZJ)(ZJ)-l = R. Therefore each R-ideal in Q has an inverse in the multiplicative semigroup of R-ideals in Q with R as identity. We have therefore proved the first part of the third assertion of the theorem. Finally, if Z J = A1JLG, ... A, and J= ... M r , then
I
=
IJ * J-'
=
A1 .M,,JL;tl
a * -
A;'
It follows that the representation of Z as a product of maximal ideals is uniquely determined up to order since the JLi are prime. I We have just proved that if R is a maximal order, fulfilling the other hypotheses in Theorem 11, in a quotient ring Q it is an Asano order. The converse is also true: Theorem 12. Let R be an Asano order in the quotient ring Q. Then R is a maximal order and each integral R-ideal in Q is a projective R-module.
Proof. 1. We first show that R is a maximal order in Q . Suppose not and let T be equivalent to R with R C T. By Definition 1, there exist regular elements a-lb and c-ld in Q with a-lbTc-ld C R where a, b, c, and d are regular elements of R. It follows that bTd
+
=
bTcc-ld C bTc-ld C aR C R
TdR is a subring of T containing R and therefore U is Hence U = R an order in Q . Moreover, bU = bR bTdR C R. If R # U , set S = U . If R = U , then Td C TdR C U. Set S = T in this case. In either case S is an order in Q which is equivalent to R. Furthermore, R C S and bS C R (or Sd C R) for a suitable element b (resp. d) in R. Suppose bS C R. Then RbS is an integral R-ideal. By hypothesis, there exists an R-ideal C in Q which is the inverse of RbS. Therefore C * RbS = R. It follows that S = R S = C . RbS . S = CRbS = R, a contradiction. 2. Now let I be an R-ideal in Q. As we have just shown in Part I, R is a maximal order in Q and therefore by (3), I*ZC Or(Z) = R . By hypothesis, Z has an inverse R-ideal I-1 in Q. Now 11-1 = R implies ZZ-lZ = I from which it follows that Z-l C I*. Therefore, R = I-ll C I*I C R and hence Z*Z = R = Or(Z). By Theorem 10 this proves that Z is projective as an R-module. I
+
3. Injective modules and self-injective rings The following definition is the "dual" of the definition of a projective module given in Section 2:
3.
INJECTIVE MODULES AND SELF-INJECTIVE RINGS
175
Definition 1. Let R be a ring with identity and let B be an arbitrary R-module. Then an R-module M is injective if homomorphisms of submodules of B into M can be lifted to homomorphisms of B into M. In terms of diagrams, M is injective if the diagram
where i is the embedding of A into B, can be embedded in the commutative diagram
O + A ~ - B
M
If M is injective, take L
=A,
L a left ideal of R,and take R
= B.
Then
xt,h = (x . l)$ = x( 13) for all x E R. Hence there exists m, E M (namely, 13)
such that 14 = Im, for all I E L. The converse of this remark is also true and hence we have the following characterization of injectivity: Theorem 1. The R-module M is injective if and only if for each left ideal L in R and each R-module homomorphism 4 of L into M , there exists m, E M such that I+ = Im, for all I E L.
Proof. Suppose the condition is satisfied. Let B be an R-module and let A C B be a submodule of B. Let 4 E Hom,(A, M ) . Consider the family 3; of pairs (A’, +’), where A’ is a submodule of B containing A and 4‘E Hom,(A’, M ) is an extension of 4 to A ’ . Define a relation C on 3; by (A’, 4’) C ( A ” , 4”) if A’ C A” and 4’’ is an extension of 4’. Clearly C partially orders 3;. Moreover, if { ( A , , 4A)}AEA is a chain in 3; and 4 is the homomorphism defined on UAEA A,, by
$:a+a+,
if a E A , ,
PEA
the pair ((JAG,, AA , 4) E 3; is an upper bound for the chain. By Zorn’s lemma, there exists a maximal element (B, , 3,) in 3;. We must now show that B, = B. Suppose not and let b, E B\B, . Then the set { x E R I xb, E B,} is a left ideal L of R and the mapping x + (xb,) 3, for x E L is an R-module homomorphism of L to M. By hypothesis there exists m, E A4 such that (xb,) 3, = xm, . Set 3,: b, ybl +b0$, ym, for b , B,~ and Y E R. We show is well
+
+
176
VII.
RINGS WITH IDENTITY
+
+ +
defined: Suppose b, yb, = b,’ y’b, . Then b, - b,’ = (y‘ - y ) b, and therefore y’ - y E L. It follows that (b, - b,’) #, = ((y’ - y ) b,) I), = (y’ - y ) m, and hence b,+, ym, = b0’#, y‘m, , showing that #, is well defined. Rb, , M ) extends #, and B, Rb, 3 B, , contraClearly I), E Hom,(B, dicting maximality of (B, , #,). Therefore B, = B. I
+
+
+
The following is analogous to Theorem VII.2.2: Theorem 2. The complete direct sum M of modules M i, i E I is injective if and only if each M iis injective.
Proof. Suppose each M , is injective and let 4E Hom,(A, Cc M,) where A is a submodule of a given module B. Let rz be the projection of M onto k f z . Now $rzE Hom,(A, M,) and by injectivity of M , there exists E Horn,(& M,) such that #%I A = 4r,. Then # = C @ I),E Horn,(& M ) is an extension of 4 to B and hence M is injective. Conversely, suppose M is injective and let $ z Hom,(A, ~ M,). Denote the embedding of M , in M by p, . Then = +%pZ E Hom,(A, M ) . By hypothesis there exists #€Hom,(B, M ) with # ( A = 4 = dZpt and I)wZ is an extension of = (bZPL,T, to B. I
4
4,
Injective P-modules (i.e. abelian groups considered as modules over the integers) are characterized by the following: Theorem 3. An Abelian group M is injective as a Z-module if and only if it is divisible, i.e. for each m E M and each natural number N , there exists m, E M such that Nnz, = m.
Proof. Let L = PI, # 0 and let 4 be an additive map of L to M . Let lo+ = m E M . If M is divisible there exists m, E M such that m = lorno. Clearly /I#I = lm, for all 1 E L and by Theorem 1, M is injective. Conversely, suppose M is Z-injective. The mapping (b from Zl, to M defined by 4:z/, + zm is additive for arbitrary m E M , I, E Z.If M is injective there exists I):Z M which is an extension of 4.Let 14 = m, . Then lm = 1,(1#) = (I,#) = /,$ = m. Hence M is divisible. I ---f
The rational numbers under addition form an abelian group Q containing
as a subgroup the additive group of integers H. The group Q/Z is divisible
and hence Z-injective. Using this fact we may construct more injective modules. We begin with the following:
Definition 2. Let M be a right R-module. Then the additive group
3.
INJECTIVE MODULES AND SELF-INJECTIVE RINGS
177
Hom,(M, O/Z) can be made into an R-module M * by defining for a E R and x E M*
m(ax) = (ma)x
for all m E M
(1)
The R-module M* is called the character module of M . The R-module property of M * under (1) follows from
m((ab)x )
=
(mab) x
=
((ma)b) x
= (ma)(bx) =
rn(a(bx)) for all m E M (2)
Theorem 4. If F is a free, right R-module, its character module F* is an injective left R-module. Proof. Let L be a left ideal in R and let 4 E Hom,(L, F*). By Theorem 1 , it is sufficient to show that 4 can be extended to ,R. Let { f i I i E Z} be an R-basis of the free right R-module F. Then F L = {hill .’. +&,I, I li E L} is a subgroup of the additive group of F. In the representation of an element of FL as a sum of fili , the li are uniquely determined since {fi I i E Z} is an R-basis of F. Define a mapping # from F L to Q/Z by
+
Clearly $ E Hom,(FL, OjZ). Since by Theorem 3, O/Z is injective as a Z-module, there exists x E Hom,(F, Q/Z) = F* which extends $. By the definition of F* the mapping
q:a+a#
for U E R
is an R-module homomorphism of R R to F* and hence an element of Hom,(R, F*). By (3) the mapping q is such that
f(b> = f ( l x >= (f0x = (fl># Therefore q extends 4 to the whole of RR. I
=
fU4>
In Theorem 7 we shall show that an R-module is injective if and only if it is a direct summand of the character module of some free right R-module. To this end we first prove:
Theorem 5 (Baer). Every right R-module M is isomorphic to a submodule of the character module of some free module. Hence every right R-module is a submodule of an injective module.
178
VII.
RINGS WITH IDENTITY
Proof. 1. We first prove if 0 # m, E M , there exists x E M* such that moX # 0. Suppose first that zmo # 0 for all 0 Z z E Z.Set (zmo) = (z/2)rr where 77 is the natural homomorphism of (a,+) onto Q/Z. Since c$ E Hom,(Zm, , Q/Z), 4 can be extended to x E Hom(M, Q/Z) = M * and mox = mo$ = &r # 0. Next suppose zm, = 0 for some 0 f z E H. Let zo be the natural number for which z,m, = 0. Define by
4: (zmo) (zlz,) +
77
Clearly 4 E Hom,(Zm,, Q/Z) and again it can be extended to x E M*. Moreover, m,X = (I/z,)rr f 0. 2. M is isomorphically embedded as a submodule of M**. Proof For each m E M define the mapping & by
&:X+mX
for X E M *
Clearly & E Hom(M*, Q/Z) and the mapping m + E% is additive. E% implies m x = 0 for all x E M * and hence by Part 1, m = 0. Moreover, A
max
=
=
6
(ma)x = rn(ax) = &(ax) = (&a)x
for all x E M* A Therefore ma = h a and m & is an R-module monomorphism of M into M**. 3. Let r#~ be an R-module homomorphism of the right module U to the right module Wand let +* be the mapping of W* to U* defined by ---f
X+*:U-(+U)X
for
U E
U
+* is an R-module homomorphism since for U E U and a E R we have u((ax) $*) = (&)(ax) = (+(ua)) x = u(a(x$*)). If 4 is an epimorphism, c$* is a monomorphism. Proof Let x E W* and suppose XC$* = 0. Then ( 4 U ) x = 0 and hence Wx = 0 proving that x = 0. 4. As we showed in Part 2, M is isomorphic to a submodule of M** = (M*)*. But M* is the homomorphic image of a free module RF. By Part 3 it follows that ( M * ) * is isomorphic to a submodule of F*. I Theorem 6. The R-module M is injective if and only if A4 is a direct summand of any containing module.
Proof. (Cf. Proof of Theorem 2.5.) 1 . Suppose A4 is injective and let M C A . By injectivity of M the identity map 1 on M extends to a homo- #) = A#(# - 1,) = 0 and hence # is morphism # of A to M . Now idempotent. Therefore A = A # @ A(1, - #)
,
3.
INJECTIVE MODULES AND SELF-INJECTIVE RINGS
is a representation of A as a direct sum of the submodules A# 41.4 -
$1.
=M
179 and
2. If M satisfies the condition of the theorem then in particular M is a direct summand of the module F* of the previous theorem. By Theorem 2, M itself is injective. I
As an immediate corollary to Theorem 6 we have: Theorem 7. The R-module M is injective if and only if it is isomorphic to a direct summand of the character module of a free right R-module. I This theorem is analogous to the first assertion of Theorem 2.6 which states that an R-module is projective if and only if it is isomorphic to a direct summand of a free module. If R is an algebra over the field @ and if M is a right algebra R-module we can substitute @ for Q/Z in Definition 2. The elements of Hom,(M, 0) are called linear functionals on M . They form an R-module, again denoted by M * , if we define for a E R and x E Hom,(M, @)
m(ax) = (ma) x
for all m E M
Except for minor alterations in the terminology, Theorem 4 remains unchanged and the same proof carries through since the field @ is an injective @-module. The same is true of Theorems 5 and 6. In the context of @-algebras, Theorem 7 becomes: Theorem 7’. Let R be an algebra with identity over the field @. The R-(algebra)-module M is injective if and only if it is isomorphic to a direct summand of the dual module F* of a free R-(algebra)-module F. I By Theorem 5 every R-module can be embedded in an injective R-module. Our next objective is to prove that an R-module can be embedded in a minimal injective and that any two minimal injectives are isomorphic. For this purpose we introduce the following concept: Definition 3. The R-module U is an essential extension of the submodule M if every nonzero submodule of U intersects M nontrivially. The following theorem shows that every injective extension I of M isomorphically contains every essential extension U of M . Theorem 8. Let Z be an injective extension of the R-module M and let U be an essential extension of M . Then there exists a monomorphism of U into Z whose restriction to A4 is the identity 1 .
180
VII.
RINGS WITH IDENTITY
Proof. From the exactness of the sequence 0 + M U and the injectivity of Z follows the existence of 4 E Hom,(U, I ) with 4 , 1 = 1,. This implies that ker 4 n M = (0). Since U is an essential extension of M it follows that ker 4 = (0). Hence 4 is a monomorphism. I ---f
Theorem 9. The R-module M is injective if and only if it has no proper essential extension, i.e. no essential extension properly containing M . Proof. 1. If M is injective and U is an extension of M , then M is a direct summand of U. Hence M = U if we assume U is essential over M . 2. Suppose M has no proper essential extensions and let I be an extension of M . Let M’ be a submodule of I , maximal subject to M n M‘ = (0.) (M’ exists by Zorn’s lemma.) Now I / M ’ C ( M M’)/M’ = M . Moreover, if M’ C U C Z and ( U / M ’ ) n ( M M’)/M’ = (0), then U n ( M M’) C M’ and hence U n M C M‘ n M = (0). By the maximality of M ’ this implies U = M‘ proving that I/M’ is an essential extension of ( M M’)/M’. By hypothesis this implies that I = M @ M‘. If we choose I to be an injective extension of M , M will be a direct summand of I and hence, by Theorem 2, M is injective. I
+
+
+
+
Now let M be a submodule of the module I . The set theoretic union of an ascending chain of essential extensions of M in I is itself an essential extension of M . By Zorn’s lemma there exists a submodule U of Z maximal subject to containing M essentially. If U‘ is an essential extension of U contained in I then any nonzero submodules S of U’ intersects U nontrivially. Since U is an essential extension of M and since S n U # (0),
S n U n M = S n M # (0)
Hence any essential extension of U in I is also an essential extension of M . Therefore U has no proper essential extensions in I. Moreover, if I is injective, we can show that U has no proper essential extensions in any containing module: Let U‘ be any essential extension of U. By Theorem 8 applied to the essential extension U‘ of U , there exists a monomorphism 4 of U ‘ into I such that U C Or‘+. Clearly U‘4 is an essential extension of U in I. But U is maximal essential in Z and therefore U = U‘4. Since U has no proper essential extension it is an injective R-module by Theorem 9. We have therefore shown that if I is an injective extension of M , each maximal essential extension U of M in Z is an injective R-module. On the other hand, by Theorem 9, no proper submodule 0of U containing M is injective since U is a proper essential extension of 0.Therefore U is a minimal injective extension of M and is uniquely determined (up to isomorphism) by M : In fact, if U‘ is any essential extension of M , then by Theorem 8 there exists
3.
INJECTIVE MODULES AND SELF-INJECTIVE RINGS
18 1
a monomorphism C$ of U‘ into U . If U ‘is also a maximal essential extension of M , so is U’C$.Hence U‘C$= U . Before collecting the results proved in the above discussion we make the following: Definition 4. The R-module Z is an injective hull of its submodule M if 1. Z is injective 2. M C L C Z,L injective implies L = I.
Theorem 10. 1. Every R-module M has an injective hull. 2. Every injective extension of M contains an injective hull of M . 3. An R-module U containing M is an injective hull of M if and only if it is a maximal essential extension of M . 4. If Z and I’ are injective hulls of M then there exists an isomorphism of Z onto I’ restricting to the identity on M . I Since the injective hulls of a given module M are isomorphic we speak of “the” injective hull H ( M ) of M . This last theorem can be used as the basis of an investigation of those rings R with are injective considered as R-modules over themselves. Let us assume that 0is an indecomposable component in such a ring and suppose 0 contains at least one minimal left ideal M of R. By hypothesis RR is injective and since 0 is a direct summand, 0 itself is also injective. By Theorem 10, 0contains an injective hull H ( M ) of M . By Theorem 6 , H ( M ) is a direct summand of 0and hence H ( M ) = 0since 0was assumed to be indecomposable. Again by Theorem 10, 0 is an essential extension of M . If 0 contained another minimal left ideal M‘, then M‘ n M = (0) contradicting the fact that 0 contains M essentially. Therefore 0contains exactly are two isomorphic indecomposable one minimal left ideal. Now if 0and 0’ components of RR each containing minimal left ideals M and M‘, respectively, then M and M‘ are uniquely determined by 0and Or,respectively, and since 0 rn 0, M m M‘. Conversely if M w M‘ it follows that 0 w 0’since the injective hulls = H ( M ) and 0’ = H ( M ’ ) are uniquely determined up to isomorphism by Theorem 10. Finally, let M be a minimal left ideal in R and let H ( M ) be an injective hull of M contained in RR. Then H ( M ) = S
