Double Centralizer Property in Modules over Semisimple Rings Why a module $M$ over a semisimple ring $A$ has the double centralizer, that is, for every $\varphi\in\mathrm{Hom}_D(M,M)$, where $D=\mathrm{Hom}_A(M,M)$, there is $a\in A$ such that $\varphi(x)=ax$?
The book "Representation Theory of finite groups and associative algebras" by Curtis and Reiner suggests that I use the following lemma:
If $A$ is a ring with unity, and $N$ is a left $A$-module, then $M={}_AA\oplus N$ has double centralizer property.
If $A$ is simple, then, by Wedderburn structure theorem, $A$ is direct sum of minimal left ideals $A=L_1\oplus\dots\oplus L_n$ and $L_i\cong L_1$ as $A$-modules, so that every $A$-module is a direct sum of irreducible $A$-modules and they are all isomorphic to $L_1$, so $M^n=M\oplus\dots\oplus M\cong {}_AA\oplus N$ and $M^n$ has double centralizer property, so $M$ has double centralizer property.
However, if $A$ is semisimple, then every $A$-module $M$ is also a direct sum of irreducible $A$-modules, and they are isomorphic to minimal left ideals of $A$, but I cannot necessarily have a $k$ such that $M^k\cong{}_AA\oplus N$.
Another attempt was to use the Jacobson density theorem, that if $M$ is a semisimple $A$-module, then for $x_1,\dots,x_n\in M$ and $\varphi\in\mathrm{Hom}_D(M,M)$, then there exists $a\in A$ such that $\varphi(x_i)=ax_i$ for $i=1,\dots,n$. But I do not know how to prove that $M$ is a finitely generated $D$-module is $A$ is semisimple.
 A: Let $A$ be a semisimple ring. Then, by the structure theorem, we have $A=B_1\oplus\cdots\oplus B_r$, where $B_i$ is an ideal of $A$ and a simple ring with identity. Let $M$ be an $A$-module, then we have $M=\oplus_{k\in K}M_k$, where $M_k$ is an irreducible $A$-module. For $k\in K$, then $M_k$ is isomorphic to a minimal left ideal $L_k$ of $A$, so there is a unique $i_k\in\{1,\dots,r\}$ such that $L_k\subseteq B_{i_k}$, and for all $b\in B_i$ with $i\neq i_k$ we have $\forall x\in M_k:bx=0$. Thus, let $N_i=\oplus_{k\in K,i_k=i}M_k$, then for all $b\in B_j$ with $j\neq i$, we have $bN_i=0$, so, if $D_i=\mathrm{End}_A(N_i)$, then $D_i=\mathrm{End}_{B_i}(N_i)$. Moreover, $N_i$ is a $B_i$-module with double centralizer property. Now, if $D=\mathrm{End}_A(M)$, then for all $f\in\mathrm{End}_D(M)$ and for all $i$ we have $f\upharpoonright N_i\in\mathrm{End}_{D_i}(N_i)$, so there is a $b_i\in B_i$ such that $\forall x\in N_i:f(x)=b_ix$; moreover, for $j\neq i$, for $x\in N_i$ we have $b_jx=0$. So, let $a=b_1+\cdots+b_r$, then for $x\in M$, if we set $x=x_1+\cdots+x_r$ where $x_i\in N_i$, then:
$$ax=(b_1+\cdots+b_r)(x_1+\cdots+x_r)=b_1x_1+\cdots+b_rx_r=f(x_1)+\cdots+f(x_r)=f(x),$$
so $f(x)=ax$. Therefore $M$ is an $A$-module with double centralizer property.
