7 Rings with Semisimple Generators.

Transcription

1 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 R M is simple in case it is a non-zero module with no non-trivial submodules 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( R M) is a division ring. A module R M 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 R M the following are equivalent: (a) M is semisimple; (b) M is a direct sum of simple submodules; (c) Every submodule of M is a direct summand. Let s begin with a particularly nice example. Indeed, let D be a division ring. Then the regular module D D is a simple module, and hence, a simple progenerator. In fact, every finitely generated module over D is just a finite dimensional D-vector space, and hence, free. That is, the progenerators for D are simply the non-zero finite 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 R T. But then T generates R R, 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 finite set of these simple left ideals that sum to R. Thus, there is a minimal such set, say, I 1,...,I n. By minimality, these left ideals must be independent, and so this sum must be direct. That is, R = I 1 I n

2 54 Section 7 with I k = T for each k =1,...,n. That, in turn, implies that R R has finite 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( R T )=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 iff D is equivalent to S. But as we saw above, this means that R is equivalent to S iff S = M n (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 Theorem. For a ring R the following are equivalent: (a) R is simple left Artinian; (b) R has a simple left generator; (c) R is left equivalent to a division ring; (d) R = M n (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, 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 first 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 D 1,...,D n be division rings (with possible repetitions) and let R = D 1 D n be their ring product. If ι j : D j R is the coordinate injection (j =1,...,,n), then each ι j (D j )isa simple left ideal (actually an ideal) of R, and so R = ι 1 (D 1 ) ι n (D n )

3 Non-Commutative Rings Section 7 55 so R R is semisimple. Then, of course, R is a semisimple left generator. Also, (see Corollary 6.14) R is equivalent to S iff S = S 1 S n with each S i equivalent to D i. But D i is a division ring, and so (see Theorem 7.3) it is equivalent to S i iff S i = Mni (D i ) for some n i N. So for this example we conclude that R is equivalent to S iff there exist n 1,...,n n N with S = M n1 (D 1 ) M nn (D n ). Let s now look at the general case. So suppose that R has a semisimple left generator, say R G. Then G generates R R and G is a sum of simple submodules, so R is a sum of simple submodules. This means that R R is semisimple (such a ring is left semisimple ), so by Proposition 7.2 R R is a direct sum of simple left ideals. But R R is finitely generated, so there is a finite set I 1,...,I m of simple left ideals with RR = I 1 I m. This tells us, among other things, that R R has finite composition length and hence is both left Artinian and left Noetherian, and that each of the simple left ideals I k is projective. Let T 1,...,T n be a complete set of representatives of the simple modules I 1,...,I k. (That is, each I j is isomorphic to one and only one of the T i.) Let T be the coproduct of T 1,...,T n. Then T is certainly finitely generated projective and it generates each I j, so it generates R R. That is, T is a progenerator for R. Nowifi j, then (see Lemma 7.1) Hom R (T i,t j )=0,so End( R T ) = End( R T 1 ) End( R T n ). But each T i is simple, so each End( R T i )=D i is a division ring. Thus, R is equivalent to S iff S is equivalent to D 1 D n, and we re back to our above simple example. This gives us the following characterizations of semisimple rings Theorem. For a ring R the following are equivalent: (a) R is left semisimple; (b) R has a semisimple left generator; (c) R is left equivalent to a finite product of division rings; (d) R = M n1 (D 1 ) M nn (D n ) for D 1,...,D n division rings and n 1,...,n n N; (e) R is isomorphic to a product of a finite number of simple Artinian rings; (f) R is right equivalent to a finite 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-

4 56 Section 7 ple rings. There is another slick bonus to this. The ring R is semisimiple iff it is equivalent to a finite 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 Corollary. For a ring R the following are equivalent: (a) R is semisimple; (b) Every module in RMod is semisimple; (c) Every short exact sequence 0 K M N 0 in RMod splits; 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 If R is a semisimple ring, then it is isomorphic to a finite product of simple Artinian rings. There is a version of this that holds for any semisimple module. Thus, let R M be a semisimple module over a ring R (not itself necessarily semisimple), and let S = End( R M) acting, as usual, on the right. Let (T α ) α Ω be a complete set of representatives of the simple submodules of R M. (That is, each simple submodule R K of M is isomorphic to exactly one T α.) For each α Ω let H α = {K R M : K = T α }. So H α is the trace of T α in M; it is called the T α -homogeneous component of M. Prove that (a) Each H α is an (R, S) sub bimodule of R M S. (b) There is a central idempotent e α S with H α = Me α for each α Ω. (c) End((H α ) eαse α ) = End( R T α ) is a division ring. (d) Dα (H α ) eαse α is faithfully balanced where D α = End( R T α ). (In particular, each e α Se α is the endomorphism ring of some vector space over the division ring D α.) (e) S is isomorphic to the product α Ω e αsa α.

5 Non-Commutative Rings Section The converse of Schur s Lemma 7.1 fails. Indeed, (a) Prove that for every n N there exists a module R M of length n with End R M a division ring. [Hint: Try upper triangular matrix rings.] (b) On the other hand suppose that R I is a left ideal of R with End( R I) a division ring. Prove that if x 2 = 0 implies that x = 0 for all x I, then R I is simple 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 iff I 2 0. [Hint: Recall that I is a direct summand iff 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 = Here we consider the dual of those rings with a simple generator. A ring R has a simple left cogenerator R C 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.