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

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?

• The Artin-Wedderburn Theorem is a statement about Artinian semisimple rings. You need the Artinian assumption to ensure a finite sum of matrix rings. I made that mistake myself in my answer and just corrected it now. – Trevor Gunn Jun 18 '17 at 16:28

$\require{begingroup}\begingroup\DeclareMathOperator{\End}{End}\DeclareMathOperator{\Hom}{Hom}$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. $\endgroup$

• It may be easy but I don't see the isomorphism $End(eR) \cong \bigoplus_{i \in I}M_{n_{i}}(D_i)$. Can you explain one more time? – bozcan Jun 18 '17 at 20:11
• @bozcan Note that $\operatorname{Hom}(T_i^{n_i}, T_j^{n_j}) \cong \operatorname{Hom}(T_i,T_j)^{n_i \times n_j}$ where we view these as matrices whose entries are homomorphisms. Now apply Schur's lemma. – Trevor Gunn Jun 18 '17 at 20:21
• @bozcan I've thought about it and I believe it is necessary to assume $eR$ is Artinian (i.e. a finite sum) because otherwise you would have a product: $\operatorname{End}(eR) \cong \prod_{i \in I} M_{n_i}(D_i)$ from which you cannot conclude that $eRe$ is semisimple. – Trevor Gunn Jun 18 '17 at 20:50

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.