Let $M \cong R^n$ be a free $R$-module of rank $n$, where $R$ is an integral domain. Let $S$ be an integral domain such that $(S,+) \cong \Bbb Z^n$ as abelian groups.
Assume that we have a ring morphism $S \to (End(M), +, \circ)$ commuting with the action of $R$ (that is, $r \cdot (s \cdot m) = s \cdot (r \cdot m)$), i.e. $M$ is then a $(R,S)$-bimodule.
Is it true that $M$ is free of rank $1$ over $R \otimes_{\Bbb Z} S$ ?
I'm not sure about the freeness, actually. If this is not true, what conditions ensure that this is valid (e.g. $R$ is a DVR, $S$ is Dedekind, $n \leq 2$, etc.)?
Thank you!
Best Answer
No, this is usually not true. For instance, suppose that $S$ is actually a subring of $R$, and the action of $S$ on $M$ is just given by restricting the action of $R$. Then $M$ will not be free over $R\otimes_\mathbb{Z} S$ (unless $S$ is just $\mathbb{Z}$), since for each $s\in S$, $s\otimes 1$ and $1\otimes s$ act on $M$ in the same way.
I don't know of any reasonable conditions that would make the answer yes. Certainly it's not enough to just put conditions on $R$ and $S$ themselves, since $R$ and $S$ could be arbitrarily nice in the example above (for instance, $R$ could a number field and $S$ could be its ring of integers). What you need is for the actions of $R$ and $S$ on $M$ to be "independent" in some strong way.