$S$ is a simple Artinian ring with unity. $S \subset R$ is a ring extension. I know an additional condition that $R$ is finitely generated as left or right $S$-module. Is it true that $R$ is free as an $S$-module?
I know that S has only one simple module V. And $R \cong V^{(n)}$ as S-modules. R must be a finite direct sum of V by decomposing its unit 1. Then I don't know how to prove R is free over S. I heard that it is about a lemma of Artin and Whaples. But I can't find this lemma.
I find Artin-Whaples' lemma. It's recorded in "Structure of rings" by Jacobson. Here's the script of the proof. There is a general propostion for idempotents : $e_1 R \cong e_2 R$ if and only if $\exists e_{12}, e_{21}\in R$,
$e_1 e_{12} e_2 = e_{12}, e_2 e_{21} e_1= e_{21}$
$e_{12}e_{21} = e_1, e_{21}e_{12} = e_2$.
In our case, $S = \oplus I_i$ as a direct sum of minimal right ideals. Decompose the unit as $1 = e_1 + ... + e_k, e_i\in I_i$. And construct right ideals $e_i R$ in R. Applying this proposition, we can find $e_{i,j},e_{j,i}\in S$ in our case. And thus, $e_iR$ are isomorphic as R-modules. Of course they are isomorphic S-modules.
Since $e_i$ are orthogonal idempotents, $R= \oplus e_iR$. We don't have to assume R is finitely generated as S-module at the beginning. The real field over the rational is a good example of infinitely generated module.