$(eR)_R$ is a semisimple right $R$-module $\implies$ $eRe$ is a semisimple ring

Question

Let $e$ be an idempotent in a semisimple ring $R$. I want to prove that if $(eR)_R$ is a semisimple right $R$-module, then $eRe$ is a semisimple ring.

The things I have done so far:

Since $(eR)_R$ is semisimple, $(eR)_R= \bigoplus_{j \in J} S_j$, where $S_j$ is the simple submodule and $J$ is the family of submodules. By Wedderburn-Artin theorem it suffices to show that $eRe \simeq \mathbb{M}_{n_{1}}(D_1)$ x ... x $\mathbb{M}_{n_{r}}(D_r)$ for suitable division rings $D_1,...,D_r$ and possible integers $n_1,...,n_r$.

Moreover, for any idempotent $e \in R$, there is a natural ring isomorphism $End_R(eR) \simeq eRe$.

Combining these instruments I need to show that $End_R(eR) \simeq \mathbb{M}_{n_{1}}(D_1)$ x ... x $\mathbb{M}_{n_{r}}(D_r)$.

How should I show desired assertion?

Answer 1

$\require{begingroup}\begingroup\DeclareMathOperator{\End}{End}\DeclareMathOperator{\Hom}{Hom}$Suppose you have $$ eR = \bigoplus_{i = 1}^r S_i $$ where each $S_i$ is a simple module. A map $eR \to eR$ is determined by a family of maps $S_i \to S_j$. That is, $$ \End_R(eR) \cong (\Hom_R(S_i, S_j) : (i, j) \in I^2). $$

Now, by Schur's lemma, either:

• $\Hom_R(S_i, S_j) = 0$ if $S_i \not\cong S_j$
• $\Hom_R(S_i, S_j) = \End_R(S_i)$ is a division ring, if $S_i \cong S_j$.

Now let us group by isomorphism classes:

$$ eR = \bigoplus_{i = 1}^r T_i^{n_i} $$

where $T_i \cong T_j$ iff $i = j$. So what you have is

$$ \End(eR) \cong \bigoplus_{i = 1}^r M_{n_i}(D_i)$$

where $D_i = \End_R(T_i)$. Note that $M_{n_i}(D_i)$ is simple. Thus $eRe \cong \End(eR)$ is semisimple. $\endgroup$

Answer 2

More generally, if $M_R$ is a finitely generated semisimple module, then its endomorphism ring is semisimple.

Write $M$ as a direct sum of simple modules and group them together by isomorphism classes; then we can assume $$ M=\bigoplus_{k=1}^n S_k^{n_k} $$ where each $S_k$ is simple and $S_k\not\cong S_{k'}$ for $k\ne k'$. Since $$ \operatorname{Hom}_R(S_k,S_{k'})=0 $$ for $k\ne k'$, we are reduced to prove the statement for $M=S^n$, where $S$ is simple. The endomorphism ring of $S$ is a division ring, so we are done.

Combine with $\operatorname{End}_R(eR)=eRe$ and you're finished.