I was studying Schur-Weyl duality and the double centralizer theorem and following the proof of the double c.t. by Etingof. Now in the third part he decomposes a module over a semisimple algebra in some way. I can't undertant why. So let's start with full generality:
I have some problems understanding the decomposition of a semisimple module over a ring $R$ (semisimple).
Recall that an $R$-module $M$ is semisimple, or completely reducible, if it can be expressed as a direct sum of simple $R$-submodules.
Now, if $M$ is semisimple, it can also be expressed as a sum of its isotypic components. Let $U_1,\ldots,U_n$ the simple submodules of $M$, and $I_j$ the isotypic component associated to $U_j$ then
$M=\bigoplus I_j=\bigoplus U_j^{n_j},$ where $n_j$ is the multiplicity of $U_j$.
Let $S$ be the centralizer of $R$, i.e. $S=\text{End}_R(M)$.
My questions are:
(1) Is true that each $I_j=U_j \otimes \text{Hom}_R(U_j,M)$?
(2) $\text{Hom}_R(U_j,M)$ are $S$ simple module?
(1): Yes. Indeed, $$\hom_R(U_j,M)\simeq \bigoplus_i\hom_R(U_j,U_i)^{n_i}=R^{n_i},$$ by Schur orthogonality. The more invariant way to state the isomorphism is: $$\bigoplus_i U_i\otimes\hom_R(U_i,M)\simeq M:(u_i\otimes f_i)_i\mapsto \sum_{i}f_i(u_i).$$
(2): Note that by Schur orthogonality, $S\simeq\bigoplus \hom_R(U_j^{n_j},U_j^{n_j})\simeq\bigoplus M_{n_j}(R)$. Now $\hom_R(U_j,M)\simeq R^{n_i}$ has the obvious $S$-action via the projection $\bigoplus M_{n_j}(R)\to M_{n_j}(R)$. So, your question is equivalent to asking if the $M_n(R)$-module $R^n$ is irreducible. This is a standard fact.