5.9 Semisimple Modules

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Let R be a ring and let M be an R-module. We say that M is semisimple

if given any submodule N _ M, there exists a submodule N0 _ M with

M = N _ N0.

Example 1. Let V be an F-vector space and let T 2 EndF(V ) be a semisimple

linear transformation. If V is given an F[x]-module structure in

the usual way, then it is clear that V is semisimple. (Recall that by

Exercise 6 of Section 5.5, a linear transformation T is semisimple if and

only if the minimal polynomial mT (x) is multiplicity-free in its prime

factorization. We’ll obtain the same result as a direct consequence of

Theorem 5.9.5, below.)

Example 2. Let A be a finite abelian group of exponent e. Assume that

the factorization of e into primes is multiplicity-free. Then A is a

semisimple Z-module. This can be seen in a number of ways, including

using Theorem 5.9.5, below.

Example 3. Let F be a field and let R be the ring

R =

__

a b

0 c

_

| a, b, c 2 F

_

.

R acts in the obvious way on the vector space M, where

M =

__

a

b

_

| a, b 2 F

_

.

If M0 _ M is the submodule defined by setting

M0 =

__

a

0

_

| a 2 F

_

,

then it is easy to verify that M 6= M0 _ M00, for any submodule

M00 _ M. Thus M is not a semisimple R-module.

Example 4. Obviously, any irreducible module is semisimple.

Lemma 5.9.1 Submodules and homomorphic images of semisimple modules

are semisimple.

5.9. SEMISIMPLE MODULES 143

Recall that an R-module M is called irreducible if and only if it contains

no nontrivial submodules.

Lemma 5.9.2 Any non-zero semisimple module contains a non-zero irreducible

submodule.

Lemma 5.9.3 (Schur’s Lemma) Let R be a ring and let M be an Rmodule.

If M is irreducible, then the ring E := HomR(M,M) is a division

ring. More generally, if _ : M ! N is an R-module homomorphism

between irreducible R-modules M,N, then _ is either the 0-map or is an

isomorphism.

Theorem 5.9.4 The following conditions are equivalent for the R-module

M.

(i) M is semisimple.

(ii) M =

P

i2I Mi, for some family {Mi| i 2 I} of irreducible submodules

of M.

(iii) M = _i2IMi, for some family {Mi| i 2 I} of irreducible submodules

of M.

The astute reader will realize that the following important theorem is as

much a theorem about rings, as about modules.

Theorem 5.9.5 The following are equivalent about the ring R.

(1) Every R-module is injective.

(2) Every R-module is projective.

(3) Every R-module is semisimple.

(4) The left R-module R is a direct sum of a finite number of irreducible

left ideals:

R =

Mn

i=1

Li.

Furthermore each Li = Rei, where e1, e2, . . . , en are orthogonal idempotents

(i.e., eiej = 0, whenever i 6= j) satisfying

Xn

i=1

ei = 1 2 R.

144 CHAPTER 5. MODULE THEORY

(5) R = _k

i=1Ai, where A1,A2, . . . ,Ak are the distinct minimal 2-sided

ideals in R, and where

Ai

_= Mni(_i), i = 1, 2 . . . , k,

for suitable division rings _i, i = 1, 2, . . . , k.

Corollary 5.9.5.1 Let R be a ring satisfying any of the equivalent conditions

above, and let M be an irreducible R-module. Then M _= I (as

R-modules), for some minimal left ideal I _ R.

Let us illustrate examples of the kinds of rings indicated above.

Example 1. If R = Z/(n), where the prime factorization of n is multiplicityfree,

then by the Chinese Remainder Theorem, R is isomorphic to the

direct sum of fields of the form Z/(p).

Example 2. Let F be a field and let f(x) 2 F[x], where f(x) admits

a multiplicity-free factorization. As above, the Chinese Remainder

Theorem shows that R = F[x]/(f(x)) is the direct sum of fields.

Example 3. Let F be a field, and let R = Mn(F).

Note that if R is any of the rings above, then any R-module is semisimple.

In particular, this gives a proof of Exercise 6 of Section 5.5,

Exercises

1. Let R be a ring and let e be an idempotent. Prove that the left Rmodule

Re is a projective R-module.

2. Let R be a ring satisfying any one of the conditions of Theorem 5.9.5,

and let M be an irreducible R-module. Prove that M _= L, for some

irreducible left ideal L of R. (Hint: Let 0 6= m 2 M, and define

_ : R ! M via _(r) = rm 2 M. Conclude that M _= R/(ker _). Now

what?)

3. Let _ be a division ring and let R = Mn(_). Prove that R is a simple

ring in that it has no proper 2-sided ideals.

4. Let R be as in Exercise 3. Prove that all irreducible R-modules are

isomorphic. (Indeed, any irreducible R-module is isomorphic with the

module _n of all n × 1 column vectors with entries in _.)

5.9. SEMISIMPLE MODULES 145

5. Let F be a field and let R be the ring

R =

__

a 0

0 b

_

| a, b 2 F

_

.

Define the R-modules

M1 =

__

a

0

_

| a 2 F

_

, M2 =

__

0

b

_

| b 2 F

_

.

Prove that M1 6_= M2. (Does this surprise you?) (Compare with

Exercise 8, Section 1.2.)

6. Let R be a ring and assume that R = _n

i=1Ii, where I1, . . . In are

minimal left ideals of R. If M is an irreducible left R-module, prove

that M _= Ij for some index j, 1 _ j _ n.

7. Let R be a simple ring and let C = Z(R), the center of R. Prove that

C is a field.

8. Prove the converse of Schur’s lemma in case the module M is completely

reducible. (Note that the unconditional converse of Schur’s

lemma fails; see Exercise 7 of Section 5.7.)

146 CHAPTER 5. MODULE THEORY

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