The converse of Schur's Lemma in group rings

Publ. Mat. 50 (2006), 203–209 THE CONVERSE OF SCHUR’S LEMMA IN GROUP RINGS M. Alaoui and A. Haily Abstract In this paper, we study the structure of group rings by means of endomorphism rings of their modules. The main tools used here, are the subrings fixed by automorphisms and the converse of Schur’s lemma. Some results are obtained on fixed subrings and on primary decomposition of group rings. 1. Introduction and Notations Let Mbe a module over a ring R. If Mis simple, then Schur’s lemma states that EndR(M) is a division ring (a skew field). However, the converse of this result does not hold in general, even when Ris artinian. Definition 1. We shall say that a ring Rhas the CSL property (abbreviation of: Converse of Schur’s Lemma), or that Ris a CSL-ring, if every module is simple whenever its endomorphism ring is a division ring. In [6], we have shown that a perfect ring Ris a CSL-ring if and only if Ris primary decomposable in the sense of [4]. In this note, we shall give some characterizations of perfect and primary decomposable group rings. To this aim, we shall use the results of [6] and [1] concerned by the converse of Schur’s lemma in perfect rings. We start this work with the following remark: Let Abe a ring, A[G] the group ring of a finite group Gover A. If M is an A[G]-module, then Mis an A-module and for every g∈G, the mapping Lg:M→Mdefined by Lg(x) = gx, is an automorphism of the A-module M. Let L:G→AutA(EndA(M)) such that g7→ Lgwhere Lgis the inner automorphism of the group AutA(EndA(M)) attached 2000 Mathematics Subject Classification. 16S50, 16S34. Key words. Modules, group ring, Schur’s lemma. This work has been partially supported by El Ministerio de Ciencia y Tecnolog´ıa, proyecto BFM 2001–2335. Spain. 204 M. Alaoui, A. Haily to Lg:Lg(u) = LguLg−1for all u∈EndA(M). For each u∈EndA(M) we have: u∈EndA[G](M)⇔u∈EndA(M) and u(gx) = gu(x),∀g∈G, ∀x∈M ⇔u∈EndA(M) and g−1ug =u, ∀g∈G ⇔u∈EndA(M) and LguLg−1=u, ∀g∈G. Consequently, EndA[G](M) = EndA(M)G′where G′={Lg|g∈G}, hence the results on fixed subrings can be used to study the properties of EndA[G](M) and it will provide some information on the structure of A[G]. The work is divided in two parts. In the first, we consider some ringtheoretical properties which remain true when passing from the fixed subring RGto the whole ring R. In this context, we study when regular elements of RGremain regular in R(Corollary 5). The results are obtained by imposing some condition on Gand on R. In the second part, we apply these results to study the EndA[G](M), the endomorphism ring of M. This enables us to derive condition on G that imply the primary decomposability of A[G]. We therefore pursue the study we made in [1]. (For the terminology and notations used here we refer to [2], [4].) All rings considered in this work are associative with identity, and all the modules are left unitary modules. If Mis a module over a ring R, the endomorphism ring of Mis denoted by EndR(M). •A ring Ris said to be perfect if it is left and right perfect. •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 [4]. •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, for a perfect and CSL-ring R, each factor ring of Ris primary decomposable. This is true because a CSL-property is conserved by passing to factors. •If Mis an abelian group, Ga group of automorphisms of M, we write MG={x∈M|σ(x) = x, ∀σ∈G}for the set of elements of M fixed by G. This is clearly a subgroup of M. If Lis a subgroup of M The Converse of Schur’s Lemma in Group Rings 205 stable by G(G-stable), we note also LG={x∈L|σ(x) = x, ∀σ∈G}. Notice that if G=HP, where Hand Pare two subgroups of Gsuch that His normal in G, then MG= (MH)P. •For every n∈N∗, we note Tn(M) = {x∈M|nx = 0}the n-torsion subgroup of M. This is a fully invariant subgroup of M, and if Tn(M) = 0, we say that Mis n-torsion-free.Mis said to be torsion free, if Mis n-torsion-free for every n∈N∗. 2. CSL-property in group rings Lemma 2. Let Mbe an abelian group, pa prime integer and Ga finite p-group of automorphisms of M. If MG= 0, then Mis p-torsion free. Proof: Assume that L=Tp(M)6= 0. Then Lis a G-stable p-torsion group, and it can be viewed as a Fp-vector space, where Fpis the finite field of pelements and Ga group of automorphisms of the Fp-space L. We are going to show that LG6= (0). •Suppose first that G=hσiis cyclic of order pk, then (σ−idM)pk=σpk−idM= 0 so σ−idMis nilpotent and hence ker(σ−idM)6= 0. This implies that LG6= 0. •We now argue by induction on |G|the order of G. If |G|=p, then Gis a cyclic p-group; we apply the last situation. Let us assume that the lemma holds for p-groups of order < pk where k > 1. If Gis a p-group of order pk, then by elementary group theory, Gcontains a normal subgroup Hof order pk−1. Hence G/H is cyclic. Put G/N =grhσi. We have G=N∪Nσ ∪Nσ2∪ · · ·∪ Nσp−1so G=N.H where H=grhσi, the subgroup generated by σ. Now using the induction hypothesis, one obtains LN6= 0. Thus, since His cyclic, the last case show that (LN)H6= (0) so LG= (LN)H6= (0). We will need the following version of Bergman-Isaacs’s theorem: Theorem 3 ([3] or [14, Corollary 2.5.53]).Let Rbe a ring not necessarily unitary and Ga finite group of automorphisms of R. If Ris |G|-torsion free and Ris not nilpotent, then RG6= 0. Recall that for a finite group Gand a prime integer pwe say that g∈Gis a p′-element, if the order of gis prime to p. Moreover, if the 206 M. Alaoui, A. Haily set Hof p′-elements is a subgroup of G, then His normal in Gand G=HP, where Pis a p-Sylow subgroup of G. Theorem 4. Let Rbe a semiprime ring, Ga finite group of automorphisms of R. Assume that for every prime integer pfor which Tp(R)6= 0, the set of p′-elements of Gis a subgroup of G. Then for every G-stable nonzero left or right ideal Iwe have IG6= 0. Proof: •Let Ibe a nonzero G-stable left (resp. right) ideal. If Iis torsion free, since Ris semi prime then the Bergman-Isaacs theorem (see [3] or Theorem 3 or [14, Theorem 2.5.52, p. 198]), says IG6= 0. •If Iis not torsion free, then T(I)6= 0 and so A=Tp(I)6= 0 for some prime number pdividing |G|. Thus, Ais a nonzero p-torsion left (resp. right ideal) of R. Now, since Tp(I)⊆Tp(R), by hypothesis, we have G=HP where His a normal p′-subgroup of Gand Pis a Sylow p-subgroup. Thus, AG= (IH)P. Since Ris semiprime, then Ais not nilpotent and the Bergman-Isaacs theorem implies L=AH6= 0. Now Lemma 2 implies AG=LP6= 0 as required. The same argument is valid if we change the word “left” by “right”. Recall that a ring Ris said to be a quotient ring, if every regular element of Ris invertible. Quotient rings are called classical rings in [9]. Corollary 5. Let Rbe a semiprime ring, Ga finite group of automorphisms of R. Assume that for every prime integer psuch that Rhas p-torsion elements, the set of p′-element of Gis a subgroup of G. Then: (i) Every left (resp. right) regular element in RGis left (resp. right regular) in R. (ii) If Ris a quotient ring then so is RG. Proof: (i) Let a∈RGleft regular in RGsuppose that ais not left regular in Rthen, I= Annd(a) = {b∈R:ab = 0}is right ideal, G-invariant since ab = 0 ⇒aσ(b) = σ(ab) = 0,∀σ∈G. Since Ris semiprime, by Theorem 3, IG6= 0. Hence there exists a nonzero b∈RGsuch that ab = 0, contradiction. (ii) If a∈RGis left and right regular in RG, then by (i), ais left and right regular in R. Since Ris a quotient ring, ais invertible in R. Thus, there is c∈Rsuch that ac =ca = 1. For every σ∈G,aσ(c) = σ(c)a=1. Thus σ(c) = cfor all σ∈Gand hence ais invertible in RG. We can now prove the main theorem: The Converse of Schur’s Lemma in Group Rings 207 Theorem 6. Let Aa commutative and perfect ring, Ga finite group. The following assertions are equivalents: (i) A[G]is a primary decomposable ring. (ii) A[G]is CSL-ring. (iii) For each prime number psuch that Tp(A)6= 0, there exists a p′-subgroup Hof G, and a p-Sylow subgroup Pof Gsuch that: G=HP . Proof: •Since Gis finite group and Ais perfect ring, then the group ring A[G] is perfect (cf. [13] or [15]). The equivalence of (i) and (ii) is a consequence of [6, Theorem 3]. •(i) ⇒(iii) Let pa prime number such that Tp(A)6= 0. We have pA is proper ideal of A. By Zorn’s lemma, pA is contained in a maximal ideal I. Hence the factor ring A/I =Kis a field (commutative) of characteristic p. Since A[G] is a primary decomposable ring, so is K[G] (see the Introduction). Now, we can apply the results of [1]: there exists a p′-subgroup Hof G, and a p-Sylow subgroup Pof Gsuch that: G=HP. •(iii) ⇒(ii) Let MaA[G]-module such that D= EndA[G](M) is a division ring. We are going to show that Mis a simple A[G]-module. First, put I= AnnA(M), the annihilator of Min Aand show, that Iis a maximal ideal of A. Since Ais commutative, then for each a∈A, the map ρa:M→M,m7→ ρa(m) = am is an endomorphism of the A[G]-module Mand so the map ρA →EndA[G](M), a7→ ρ(a) = ρais a homomorphism of rings. It is clear that Iis the kernel of ρ. It follow that the factor ring A/I is isomorph to a subring of a division ring D. Hence A/I is a domain so the ideal Iis prime. But Ais perfect by hypothesis; so Iis maximal and K=A/I is a commutative field. Now, let Nbe a nonzero submodule of the A[G]-module M, and consider the set I={u∈EndK(M)|u(N) = 0}. Then I 6= 0. However, for each u∈ I,g∈Gand x∈Nwe have: LguLg−1(x) = gu(g−1x) = 0 since Nis a A[G]-submodule of M. This show that Iis a nonzero left ideal G′-invariant of EndA(M) where G′={Lg|g∈G}. Let R=EndK(M)=EndA(M) (because K=A/I and I= AnnA(M)). Then Ris a von Neumann regular ring and hence semiprime. But G′= L(G) is a group of automorphisms of R, homomorphic image of G; by Theorem 3 we have IG′6= 0. Thus, it exists a nonzero ulie in Rsuch that uis G′-invariant: u∈EndK[G](M). Since EndA[G](M) is a division ring, so it is for EndK[G](M). Finally, N= 0 because u(N) = 0. 208 M. Alaoui, A. Haily As application, we can give a new characterization of nilpotent finite groups using a primary decomposition of the group ring (Z/nZ)[G], n∈N∗: Corollary 7. Let Gbe a finite group of order n. Then, (Z/nZ)[G]is primary decomposable, if and only if, Gis nilpotent group. Acknowlegement. The authors wish to thank the referee for her/his helpful suggestions. References [1] M. Alaoui and A. Haily, The converse of Schur’s Lemma in Noetherian rings and group algebras, Comm. Algebra 33(7) (2005), 2109–2114. [2] F. 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D´epartement de Math´ematiques et Informatique Facult´e des Sciences BP 20 El Jadida Morocco E-mail address:[email protected] E-mail address:[email protected] Primera versi´o rebuda el 30 de maig de 2005, darrera versi´o rebuda el 27 de setembre de 2005.