PERFECT RINGS FOR WHICH THE CONVERSE OF SCHUR ’ S LEMMA HOLDS

If M is a simple module over a ring R then, by the Schur’s lemma, the endomorphism ring of M is a division ring. However, the converse of this result does not hold in general, even when R is artinian. In this short note, we consider perfect rings for which the converse assertion is true, and we show that these rings are exactly the primary decomposable ones.


Introduction
Let M be a module over a ring R. If M is simple, then the Schur's lemma states that End R (M ) is a division ring (a skew field).The converse of this statement is false.For example, if R is an integral (commutative) domain which is not a field, then its quotient field Q, considered as an R-module, is not simple, although End R (Q) ∼ = Q is a division ring.
For an example in the artinian case, one can take: R = ( K K 0 K ), the ring of upper triangular 2 × 2 matrices over a field K. Then for the R-module M = Re, where e = ( 0 0 0 1 ), we have End R (M ) ∼ = K, but M is not simple.Definition 1.1.We shall say that a ring R has the CSL property (abreviation of: Converse of the Schur's Lemma), or that R is a CSL-ring, if every module is simple whenever its endomorphism ring is a division ring.
The CSL property, has been studied by some authors.In [4], Ware and Zelmanowitz, considered modules with simple endomorphism ring over a commutative ring.From their results, it can be shown that a commutative ring R is a CSL-ring iff every prime ideal of R is maximal.In [3] some classes of noncommutative von Neumann regular rings with the CSL property has been studied.
The full class of CSL-rings seems to be very hard to characterize, the present note deals with perfect CSL-rings.Our main result is: Theorem 1.2.For a perfect ring R, the following assertions are equivalent: (i) Every R-module with semiprime endomorphism ring is semisimple.
(ii) Every R-module with von Neumann regular endomorphism ring is semisimple.
(iv) R is isomorphic to a finite product of primary rings.

Preliminaries and notations
(For the terminology and notations used here we refer to [1], [2].)Throughout this paper, all rings are associative with identity, and all modules are left unitary modules.If M is a module over a ring R, the endomorphism ring of M is denoted by End R (M ).The socle of M , i.e. the sum of all simple submodules of M , is denoted by Soc(M ).
A ring R is said to be perfect if it is left and right perfect.Over a perfect ring, every nonzero module has a maximal and a simple submodule.
A ring R is said to be primary, if the factor ring R/J(R), where J(R) denotes the Jacobson radical of R, is simple artinian.Any primary left or right perfect ring is isomorphic to a full matrix ring over a local ring [2].
A right or left perfect ring R is said to be primary decomposable, if it is isomorphic to a (finite) product of primary rings.It can be shown that R is primary decomposable, if and only if, every idempotent which is central modulo the Jacobson radical is central.
A ring R is said to be von Neumann regular (abbreviated VNR), if for every x ∈ R there exists y ∈ R such that xyx = x.An important example of a VNR ring is the endomorphism ring of a semisimple module.

The proofs
(i) ⇒ (ii) is obvious since every VNR ring is semiprime.
(ii) ⇒ (iii).If End R (M ) is a division ring, then it is VNR.So M is semisimple by hypothesis.Since M is indecomposable, it is therefore simple.
(iv) ⇒ (i).It is easy to see that any direct product of a finite number of rings verifying (i) has this property.Hence to show that (iv) implies (i), it suffices to show that every perfect primary ring verifies (i).Let R be such a ring.If M is any nonzero R-module, then M has a maximal submodule N , and a simple submodule S. Since R is primary, R has a unique isomorphism class of simple modules, so there exists an R-module isomorphism σ : M/N → S. If π : M → M/N and ı : S → M denote respectively the canonical surjection and the canonical injection, then Now suppose that M is not semisimple, then M contains a proper essential submodule E which is contained in a maximal submodule N .By what has been proved previously, there exists a nonzero u ∈ End R (M ) such that u(N ) = 0 and u(M ) ⊂ Soc(M ).Since E is essential, we have Soc(M ) ⊂ E and then u( (iii) ⇒ (iv).To prove this implication, we need a preliminary result.Lemma 3.1.Let M be a finitely generated module over a perfect ring R. Suppose that Hom R (N, Soc(M )) = 0 for every nonsimple submodule N of M .Then End R (M ) is a division ring.
Proof: Suppose that End R (M ) is not a division ring, then there exists u ∈ End R (M ) such that u is nonzero and noninvertible.Since M is finitely generated over a perfect ring, u is not injective.Let N be a submodule of M such that Ker ⊂ N and N/ Ker u is simple.If v = u| N denotes the restriction of u to N , then Im v ∼ = N/ Ker v so Im v is simple.Thus Im v ⊂ Soc(M ).This proves that Hom(N, Soc(M )) = 0.
We are now going to prove the implication (iii) ⇒ (iv).Suppose on the contrary that R is a CSL-ring which is not primary decomposable.Then there exists an idempotent e ∈ R central modulo J = J(R) but not central.
Either R(1 − e)Re = 0 or ReR(1 − e) = 0. Without loss of generality, we can suppose that R(1 − e)Re = 0. Since R(1 − e)Re = J(1 − e)Re, we can pick an element x ∈ R(1 − e)Re\J(1 − e)Re, and consider the left ideal I maximal with respect to: J(1 − e)Re ⊂ I ⊂ Re and x / ∈ I.Then, the module M = Re/I is finitely generated with simple socle equal to S = Rx + I/I.Since J(1 − e)Re ⊂ I, we have J(1 − e)M = 0. Hence (1 − e)M ⊂ S. On the other hand, eR ⊂ Re + J, thus eR(1 − e)Re ⊂ J(1 − e)Re, implying eS = 0. Now let N be a submodule of M such that Hom R (N, S) = 0 and u : N → S a nonzero homomorphism.We have u(N ) = S and u((1 − e)N ) = (1 − e)S = 0. Since (1 − e)N ⊂ S, then u(S) = 0. Consequently Ker u = 0 and u is therefore an isomorphism.So N is necessarly simple.By Lemma 3.1, End R (M ) is a division ring.Since R is a CSL-ring, M is simple.So M = S and eM = eS = 0, a contradiction.