$R$ semisimple ring and $M$ finitely generated $R$-mod implies that $End_R(M)$ is semisimple.

Question

Let $R$ be a semisimple ring and $M$ be a finitely generated $R$-mod. Show that $End_R(M)$ is semisimple.

I was thinking of somehow showing that I can get an isomorphism between $R$ and $End_R(M)$ since I know there is an injection to the endomorphisms of $M$ from $R$, and maybe using the property of $M$ being finitely generated, but I'm not sure if this is the direction to take.

Any suggestions or hints would be appreciated.

Answer 1

The endomorphism ring of any finitely generated semisimple $R$-module is semisimple. Every module over a semisimple ring is a semisimple module.

The module $M$ can be written as $N_1^{a_1}\oplus\cdots\oplus N_k^{a_k}$ where the $N_i$ are pairwise non-isomorphic simple modules. The endormophism ring of $M$ will be the direct product of those of the $N_1^{a_i}$ which are matrix rings over skew fields.