Structure of semisimple rings Let $R$ be a ring and $M$ a completely reducible artinian $R$ module.  How do we show that the ring $\operatorname{End}_{R}(M)$ semisimple artinian?
Being $M$ completely reducible, then $M$  is the direct sum of irreducible submodules $\{M_{\alpha}\}_{\alpha \in A}$ of $M$. Since $M$ is artinian, then there exist $\alpha_{1}, \alpha_{2},\dots,\alpha_{n}\in A$ such that $M$ is the direct sum of $\{M_{\alpha_{i}}\}$. Then how can we continue to relate this with the ring   $\operatorname{End}_{R}(M)$ or use the Wedderburn-Artin theorem?
 A: Your work is good; now, you can divide the simple submodules $M_{\alpha_1}=M_1,\dots,M_{\alpha_n}=M_n$ into blocks where submodules in the same block are isomorphic with each other and submodules in different blocks are not isomorphic with each other. Reordering them it is not restrictive to write this direct sum as
$$
M=
\underbrace{
  M_1\oplus\dots\oplus M_{k_1}
}_{\cong N_1}
\oplus
\underbrace{
  M_{k_1+1}\oplus\dots\oplus M_{k_1+k_2}
}_{\cong N_2}
\oplus\dots\oplus
\underbrace{
  M_{k_1+\dots+k_{r-1}+1}\oplus\dots\oplus M_{k_1+\dots+k_r}
}_{\cong N_r}
$$
(where $k_1+k_2+\dots+k_r=n$) or
$$
M=N_1^{k_1}\oplus N_2^{k_2}\oplus\dots N_r^{k_r}
$$
and the modules $N_i$ are pairwise not isomorphic.
This implies that $\operatorname{Hom}_R(N_i,N_j)$ is zero for $i\ne j$, so
$$
\operatorname{End}_R(M)\cong
\prod_{i=1}^r\operatorname{End}_R(N_i^{k_i})
$$
Note that, if $S$ is a simple $R$-module, then $\operatorname{End}_R(S^k)$ (for a positive integer $k$) is a matrix ring over a division ring, so semisimple.
