# Answer to Question #17107 in Algebra for john.george.milnor

Question #17107

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

Expert's answer

First assume

*R*is semisimple,and let*S*1*, . . . , Sr*be a complete set of simple left*R*-modules.Then*M*=*M*1*⊕**· · ·**⊕**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, End*RM**∼*(product on i) End*RMi**∼*(product on i) M*ni*(End*RSi*). By Schur’s Lemma, all End*RSi*are division rings, so End*RM*is semisimple. If, in fact,*R*issimple artinian, we have*r*= 1 in the calculation above. Therefore, End*RM**∼*= M*n*1(End*RS*1) is alsoa simple artinian ring.Need a fast expert's response?

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