Let M be a finitely generated left R-module and E = End(RM). Show that if R is simple artinian, then E is simple artinian
First assume R is semisimple,and let S1, . . . , Sr be a complete set of simple left R-modules.Then M = M1 ⊕· · ·⊕Mr, where Mi is the sum of allsubmodules of M which are isomorphic to Si. Since M isfinitely generated, Mi ∼niSi for suitable integers ni.Therefore, EndRM ∼ (product on i) EndRMi ∼(product on i) Mni(EndRSi). By Schur’s Lemma, all EndRSiare division rings, so EndRM is semisimple. If, in fact, R issimple artinian, we have r = 1 in the calculation above. Therefore, EndRM∼= Mn1(EndRS1) is alsoa simple artinian ring.