when is a ring a free module over a subring?

Question

Let $S \subset R$ be rings, $S$ not necessarily an ideal of $R$, and $S \neq R$.

Is there anything that can be said about when $R$ is free as an $S$-module?

Answer 1

I doubt there are many spectacular answers, but here are some obvious ones.

When S is a division ring, this is the case.

When R is a flat extension of S and S is a local perfect ring.

When R is a projective extension of S and S is a local ring.

When S is a commutative principal ideal domain and R is finitely generated and torsionfree over S.

Answer 2

Here's one situation in which $R$ can be free over $S$ (I only deal with Noetherian local rings here):

If $S$ is a regular local ring, and $R$ is a finite $S$-module, then $R$ is a free $S$-module iff $R$ is a Cohen-Macaulay ring. This follows from the Auslander-Buchsbaum formula. If $R$ is complete and contains a field, then there does exist a regular subring $S \subseteq R$ such that $R$ is a finite $S$-module.