Perfect rings for which the converse of Schur's lemma holds

Publ. Mat. 45 (2001), 219–222 PERFECT RINGS FOR WHICH THE CONVERSE OF SCHUR’S LEMMA HOLDS A. Haily and M. Alaoui Abstract If Mis a simple module over a ring Rthen, by the Schur’s lemma, the endomorphism ring of Mis a division ring. However, the converse of this result does not hold in general, even when Ris 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. 1. Introduction Let Mbe a module over a ring R.IfMis simple, then the Schur’s lemma states that EndR(M) is a division ring (a skew field). The converse of this statement is false. For example, if Ris an integral (commutative) domain which is not a field, then its quotient field Q, considered as an R-module, is not simple, although EndR(Q)∼ =Qis a division ring. For an example in the artinian case, one can take: R=( KK 0K), the ring of upper triangular 2 ×2 matrices over a field K. Then for the R-module M=Re, where e=(00 01 ), we have EndR(M)∼ =K, but Mis not simple. Definition 1.1. We shall say that a ring Rhas the CSL property (abreviation of: Converse of the Schur’s Lemma), or that Ris 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 Ris a CSL-ring iff every prime ideal of Ris maximal. In [3] some classes of noncommutative von Neumann regular rings with the CSL property has been studied. 2000 Mathematics Subject Classification. 16D60, 16K40. Key words. Schur’s lemma, perfect rings, simple module, uniform module. 220 A. Haily, M. Alaoui 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. (iii) Ris a CSL-ring. (iv) Ris isomorphic to a finite product of primary rings. 2. 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 Mis a module over a ring R, the endomorphism ring of Mis denoted by EndR(M). The socle of M, i.e. the sum of all simple submodules of M, is denoted by Soc(M). A ring Ris 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 Ris 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 Ris said to be primary decomposable, if it is isomorphic to a (finite) product of primary rings. It can be shown that Ris primary decomposable, if and only if, every idempotent which is central modulo the Jacobson radical is central. A ring Ris said to be von Neumann regular (abbreviated VNR), if for every x∈Rthere exists y∈Rsuch that xyx =x. An important example of a VNR ring is the endomorphism ring of a semisimple module. 3. The proofs (i) ⇒(ii) is obvious since every VNR ring is semiprime. (ii) ⇒(iii). If EndR(M) is a division ring, then it is VNR. So M is semisimple by hypothesis. Since Mis 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 Mis any nonzero R-module, then Mhas a maximal submodule N, and a simple submodule S. Since Ris primary, Rhas a On the Converse of Schur’s Lemma 221 unique isomorphism class of simple modules, so there exists an R-module isomorphism σ:M/N →S.Ifπ:M→M/N and ı:S→Mdenote respectively the canonical surjection and the canonical injection, then u=ı◦σ◦πis a nonzero endomorphism of Msuch that u(N) = 0 and u(M)⊂S. Now suppose that Mis not semisimple, then Mcontains a proper essential submodule Ewhich is contained in a maximal submodule N.By what has been proved previously, there exists a nonzero u∈EndR(M) such that u(N) = 0 and u(M)⊂Soc(M). Since Eis essential, we have Soc(M)⊂Eand then u(Soc(M)) ⊂u(N) = 0. Now for every v∈EndR(M), (u◦v◦u)(M)⊂(u◦v)(Soc(M)) ⊂u(Soc(M)) = 0. This proves that u◦v◦u= 0 for every v∈EndR(M); so that EndR(M)is not semiprime. (iii) ⇒(iv). To prove this implication, we need a preliminary result. Lemma 3.1. Let Mbe a finitely generated module over a perfect ring R. Suppose that HomR(N,Soc(M)) = 0 for every nonsimple submodule N of M. Then EndR(M)is a division ring. Proof: Suppose that EndR(M) is not a division ring, then there exists u∈EndR(M) such that uis nonzero and noninvertible. Since Mis finitely generated over a perfect ring, uis not injective. Let Nbe a submodule of Msuch that Ker ⊂Nand N/Ker uis simple. If v=u|N denotes the restriction of uto N, then Im v∼ =N/Ker vso Im vis 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 Ris a CSL-ring which is not primary decomposable. Then there exists an idempotent e∈Rcentral modulo J=J(R) but not central. Either R(1 −e)Re =0orReR(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 Imaximal 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,wehaveJ(1 −e)M= 0. Hence (1 −e)M⊂S. On the other hand, eR ⊂Re +J,thuseR(1 −e)Re ⊂ J(1 −e)Re, implying eS =0. Now let Nbe a submodule of Msuch that HomR(N,S)= 0 and u:N→Sa nonzero homomorphism. We have u(N)=Sand u((1 − e)N)=(1−e)S= 0. Since (1 −e)N⊂S, then u(S)= 0. Consequently Ker u= 0 and uis therefore an isomorphism. So Nis necessarly simple. 222 A. Haily, M. Alaoui By Lemma 3.1, EndR(M) is a division ring. Since Ris a CSL-ring, M is simple. So M=Sand eM =eS = 0, a contradiction. References [1] F. W. Anderson and K. R. Fuller,“Rings and categories of modules”, Graduate Texts in Mathematics 13, Springer-Verlag, New York, 1974. [2] C. Faith,“Algebra. II. Ring theory”, Grundlehren der Mathematischen Wissenschaften 191, Springer-Verlag, Berlin, 1976. [3] Y. Hirano and J. K. Park, Rings for which the converse of Schur’s lemma holds, Math. J. Okayama Univ. 33 (1991), 121–131. [4] R. Ware and J. Zelmanowitz, Simple endomorphism rings, Amer. Math. Monthly 77 (1970), 987–989. D´epartement de Math´ematiques Facult´e des Sciences B.P. 20, El Jadida Morocco E-mail address:[email protected] E-mail address:alaoui−[email protected] Primera versi´o rebuda el 6 de juny de 2000, darrera versi´o rebuda el 31 d’octubre de 2000.