primitive idempotents in semisimple rings

Question

In a graduate course text I have found the following statement:

Given a primitive idempotent e of a semisimple ring $A$ there is exactly one simple $A$-module $S$ such that $eS\neq 0$ and $eT=0$ for any other simple $A$-module not isomorphic to $S$.

My intuition is to use Artin-Wedderburn's result.

If $A$ is a finite dimensional $k$-algebra the result follows indeed from Artin-Wedderburn's theorem since $A$ can be written as a direct sum of matrix rings over division rings. Then $e$ belongs only to one of these matrix rings. After eventually changing bases in that matrix ring one may suppose that $e=e_{11}$ a matrix that has entries 1 on (1,1)-position and zero otherwise. Then e acts as identity on the simple module corresponding to this block and as zero on the other blocks.

My questions are the following:

1. What happens if $A$ is not a finite dimensional $k$-algebra? or not an artin ring? I have seen on wikipedia that Artin- Wedderburn's result works for semisimple artin rings as well.

2. What is the easiest way to explain that given an idempotent in a such matrix ring then one can always change bases such that $e=e_{11}$?

Answer 1

For 1), I assume that $R$ is commutative and contains a unit.

primitive idempotent means that $eR$ is indecomposable, i.e $eR$ is not the sum of submodule of $R$. But $eR$ is an ideal, $x\in eR$ implies $x=ea$ and for every $r\in R, r(ea)=e(ar)$. Since $R$ is semi-simple, $eR=\sum_{i\in I}R_i$ where $R_i$ is a simple ideal. This implies that $I$ has only one element $R_0$ since $eR$ is idecomposable and an ideal $R_i$ is a submodule. We have $e=e1\subset eR=R_0$.

Answer 2

I think I got the answer by myself at least in the case $A$ is artin ring. Since $A$ is artin ring one may decompose $1-e$ as a (finite) sum of primitive orthogonal idempotents $$1-e=f_1+f_2+\cdots f_r.$$ Since $(1-e)f_i=f_i(1-e)=f_i$ it follows as well that $ef_i=0=f_ie$. As explained in the comment $Af_i$ and $Ae$ are both indecomposable and semisimple modules, thus they are in fact simple modules. One may take $S=Ae$ and indeed $eS\neq 0$. On the other hand, some of the simple modules $Af_i$ might be isomorphic to $S$. A classical result (easy to prove) states that

$$\mathrm{Hom}_A(Ae, Af_i)\simeq eAf_i$$ as abelian groups.

Thus if $Af_i\cong S=Ae$ then $eAf_i\neq 0$. But if $Af_i\ncong S$ then $eAf_i=0$ which shows that $eT=0$ if $T\cong Af_i$.