7 Rings with Semisimple Generators

7 Rings with Semisimple Generators.

It is now quite easy to use Morita to obtain the classical Wedderburn and Artin-Wedderburn characterizations of simple Artinian and semisimple rings. We begin by reminding you of a few elementary facts about semisimple modules.

Recall that a module RM is simple in case it is a non-zero module with no non-trivial submodules.

7.1. Lemma. [Schur’s Lemma] If M is a simple module and N is a module, then

1. Every non-zero homomorphism M → N is a monomorphism;

2. Every non-zero homomorphism N → M is an epimorphism;

3. If N is simple, then every non-zero homomorphism M → N is an isomorphism;

In particular, End(RM) is a division ring.

A module RM is semisimple if it is the sum of its simple submodules. Then an easy, but important, characterization of semisimple modules is given in [1] (see Theorem 9.6):

7.2. Proposition. For left R-module RM the following are equivalent:

(a) M is semisimple;

(b) M is a direct sum of simple submodules;

Let’s begin with a particularly nice example. Indeed, let D be a division ring. Then the regular module DD is a simple module, and hence, a simple progenerator. In fact, every nitely generated module over D is just a nite dimensional D-vector space, and hence, free. That is, the progenerators for D are simply the non-zero nite dimensional D-vector spaces, and so the rings equivalent to D are nothing more than the n n matrix rings over D. That leads us to investigate what can be said about an arbitrary ring that has a simple generator.

Well, suppose that R is a ring and that it has a simple generator RT . But then T generates RR, and so R is a sum of (by Lemma 7.1) simple left ideals all isomorphic to T . Since 1 ∈ R, this means that there is a nite set of these simple left ideals that sum to R. Thus, there is a minimal such set, say, I1,...,In. By minimality, these left ideals must be independent, and so this sum must be direct. That is,

R = I1 In 54 Section 7

with Ik = T for each k =1,...,n. That, in turn, implies that RR has nite composition length, and so R is both left Artinian and left Northerian. It also implies that T is projective and hence actually a progenerator for R. But

End(RT )=D is a division ring (see Lemma 7.1). So (see Corollary 6.9), R is equivalent to D, and hence, R is equivalent to S i D is equivalent to S. But as we saw above, this means that R is equivalent to S i S = Mn(D) for some n ∈ N. Finally, we note that D is a simple ring, and so (see Corollary 6.14) R is a simple ring. This gives us most of the following characterization of simple Artinian rings.

7.3. Theorem. For a ring R the following are equivalent:

(a) R is simple left Artinian;

(b) R has a simple left generator;

(d) R = Mn(D) for some division ring D and some n ∈ N;

(e) R is right equivalent to a division ring;

(f) R has a simple right generator;

(g) R is simple right Artinian.

Proof. As we saw above (b) =⇒ (c) =⇒ (d) =⇒ (a). For (a) =⇒ (b), since R is left Artinian, it has a minimal left ideal I. By Lemma 7.1 for each a ∈ R, the left ideal Ia is either zero or isomorphic to the simple module I. But since R is a simple ring, X R = IR = {Ia : a ∈ R} is a sum of simple modules isomorphic to I.SoI is a simple generator. Finally, the equivalence of the last four statements follows from the rst four and the left-right symmetry of (d).

Next, let’s try to extend this and look at rings with a semisimple generator. We begin with a particularly nice example. Indeed, let D1,...,Dn be division rings (with possible repetitions) and let

R = D1 Dn be their ring product. If j : Dj → R is the coordinate injection (j =1,...,,n), then each j(Dj)isa simple left ideal (actually an ideal) of R, and so

R = 1(D1) n(Dn) Non-Commutative Rings Section 7 55

so RR is semisimple. Then, of course, R is a semisimple left generator. Also, (see Corollary 6.14) R is equivalent to S i S = S1 Sn with each Si equivalent to Di. But Di is a division ring, and M ∈ N so (see Theorem 7.3) it is equivalent to Si i Si = ni (Di) for some ni . So for this example we ∈ N M M conclude that R is equivalent to S i there exist n1,...,nn with S = n1 (D1) nn (Dn).

Let’s now look at the general case. So suppose that R has a semisimple left generator, say RG.

Then G generates RR and G is a sum of simple submodules, so R is a sum of simple submodules. This means that RR is semisimple (such a ring is left semisimple ), so by Proposition 7.2 RR is a direct sum of simple left ideals. But RR is nitely generated, so there is a nite set I1,...,Im of simple left ideals with

RR = I1 Im.

This tells us, among other things, that RR has nite composition length and hence is both left Artinian and left Noetherian, and that each of the simple left ideals Ik is projective. Let T1,...,Tn be a complete set of representatives of the simple modules I1,...,Ik. (That is, each Ij is isomorphic to one and only one of the Ti.) Let T be the coproduct of T1,...,Tn. Then T is certainly nitely generated projective and it generates each Ij, so it generates RR. That is, T is a progenerator for R.Nowifi =6 j, then (see

Lemma 7.1) HomR(Ti,Tj)=0,so End(RT ) = End(RT1) End(RTn).

But each Ti is simple, so each End(RTi)=Di is a division ring. Thus, R is equivalent to S i S is equivalent to D1 Dn, and we’re back to our above simple example. This gives us the following characterizations of semisimple rings.

7.4. Theorem. For a ring R the following are equivalent:

(a) R is left semisimple;

(b) R has a semisimple left generator;

M M ∈ N (d) R = n1 (D1) nn (Dn) for D1,...,Dn division rings and n1,...,nn ;

(e) R is isomorphic to a product of a nite number of simple Artinian rings;

(f) R is right equivalent to a nite product of division rings;

(g) R has a semisimple right generator;

(h) R is right semisimple.

One consequence of this is that we usually drop the left-right adjectives when talking about semisim- 56 Section 7 ple rings. There is another slick bonus to this. The ring R is semisimiple i it is equivalent to a nite product S of division rings, so the categories RMod and SMod are equivalent. But the category SMod is just a product of vector space categories. So we have yet another way to characterize semisimple rings.

7.5. Corollary. For a ring R the following are equivalent:

(a) R is semisimple;

(b) Every module in RMod is semisimple;

Proof. The implications (a) =⇒ (b), (c) follow from Theorem 7.4 since conditions (b) and (c) hole when R is a division ring. The implications (b) =⇒ (a) and (c) =⇒ (a) follow from Proposition 7.2.

Exercises 7.

7.1. If R is a semisimple ring, then it is isomorphic to a nite product of simple Artinian rings. There

is a version of this that holds for any semisimple module. Thus, let RM be a semisimple module

over a ring R (not itself necessarily semisimple), and let S = End(RM) acting, as usual, on the

right. Let (T)∈ be a complete set of representatives of the simple submodules of RM. (That

is, each simple submodule RK of M is isomorphic to exactly one T.) For each ∈ let X H = {K R M : K = T}.

So H is the trace of T in M; it is called the T-homogeneous component of M. Prove that

(a) Each H is an (R, S) sub bimodule of RMS.

(b) There is a central idempotent e ∈ S with H = Me for each ∈ . (c) End((H)eSe ) = End(RT) is a division ring.

(d) D (H)eSe is faithfully balanced where D = End(RT). (In particular, each eSe is the

endomorphism ring of some vector space over the division ring D.) Q (e) S is isomorphic to the product ∈ eSa. Non-Commutative Rings Section 7 57

7.2. The converse of Schur’s Lemma 7.1 fails. Indeed,

(a) Prove that for every n ∈ N there exists a module RM of length n with End RM a division ring. [Hint: Try upper triangular matrix rings.]

(b) On the other hand suppose that RI is a left ideal of R with End(RI) a division ring. Prove 2 that if x = 0 implies that x = 0 for all x ∈ I, then RI is simple.

7.3. Here is an important characterization of semisimple rings. It begins in (a) with a simple lemma.

(a) Prove that a minimal left ideal I of a ring R is a direct summand of R i I2 =6 0. [Hint: Recall that I is a direct summand i I = Re for some idempotent e ∈ R.]

(b) Let R be a left Artinian ring. Prove that the following are equivalent:

i. R is semisimple; ii. R has no non-zero nilpotent left ideals; iii. For all x ∈ R,ifxRx = 0, then x =0.

7.4. Here we consider the dual of those rings with a simple generator. A ring R has a simple left

cogenerator RC if C is a simple module that cogenerates every left R-module. (See Exercise 2.7.) Say that a ring R is co-simple in case it has a simple left cogenerator.

(a) Prove that every cosimple ring is simple.

(b) For a left Artinian ring R prove that the following are equivalent:

i. R is simple Artinian; ii. R is cosimple; iii. Every non-zero R module is a generator; iv. Every non-zero R module is a cogenerator.