# Is the ring $R$ a free module over $S$?

$$S$$ 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.

• Are your rings commutative? Commented Feb 26, 2022 at 3:09
• Noncommutative rings with unit 1. Commented Feb 26, 2022 at 4:12
• $R$ need not be a finite direct sum of $V$. The infinite product of rings $S^{\mathbb{Z}}$ has infinite rank over $S$. Commented Feb 26, 2022 at 4:21
• In the finite-dimensional case, this follows from Artin-Wedderburn; see this question. Commented Feb 26, 2022 at 4:25
• Thanks, I add the condition that R is finitely generated as S-modules. But I can only apply Artin-Wedderburn theorem to S. I still can not see that R is free over S. Commented Feb 26, 2022 at 4:47

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.