2
$\begingroup$

I have been struggling with this Representation Theory question for the past week:

Let $R$ be a ring and $S_1, \ldots, S_r$ simple $R$-modules with $S_i$ not isomorphic to $S_j$ whenever $i \neq j$ and fix positive integers $n_1, \ldots, n_r$. Show that we have ring isomorphisms

\begin{align*} End_R{({S_1}^{n_1} \oplus \ldots \oplus {S_r}^{n_r} )} & \cong End_R{({S_1}^{n_1})} \times \ldots \times End_R{({S_r}^{n_r})} \\ & \cong {M_{n_1}}{(End_R{(S_1)})} \times \ldots \times {M_{n_r}}{(End_R{(S_r)})} \end{align*}

If anyone can help, I would highly appreciate it. I think that to prove the second isomorphism, it might be useful to first show that $ End_R{({S_i}^{n_i})} \cong {M_n}{(End_R{(S_i)})}$, which I am not sure how to show or if it even helps. For the first part, there is a lemma we have proved in lectures that looks useful: If $M$ is an $R$-module and $V_1 , V_2$ are simple submodules with $V_1 \ncong V_2$, then $$ End_R{(V_1 \oplus V_2)} \cong End_R{(V_1)} \times End_R{(V_2)} .$$ I am not sure how to use this though, as ${S_i}^{n_i}$'s are not simple. Anyone could help me show both isomorphisms of my initial question? We have also covered Artin Wedderburn Decomposition and 3 versions of Schur's Lemma in lectures. Thanks for your help!

$\endgroup$
1
$\begingroup$

The first step is to adapt the proof that $$End_R(V_1\oplus V_2)\cong End_R(V_1)\oplus End_R(V_2)$$ to the case of arbitrary $V_1$ and $V_2$ satisfying $Hom_R(V_1,V_2)=0$.

Then, prove the following:

For $\phi\in\mathrm{End}_R(S^n)$, define $$\phi_{ij}=\pi_j\circ\phi\circ \iota_i$$ where $\pi_j:S^n\to S$ is the projection onto the $j$th factor and $\iota_i:S\to S^n$ is the natural embedding into the $i$th factor. Then, the map $$\mathrm{End}_R(S^n)\to M_n(\mathrm{End}_R(S))$$ given by $\phi\mapsto(\phi_{ij})_{i,j=1}^n$ is an isomorphism.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.