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.

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