## Freeness of modules along ring homomorphisms

This question arises from my discussion with a Master student. It concerns with the following situation: let $\phi: R \to S$ be a homomorphism between Noetherian commutative rings. Suppose the $R$-module structure of $S$ has a presentation matrix with all entries in the Jacobson radical of $R$ (so $S$, as $R$-module, is the cokernel of such matrix). Let $M$ be a finitely generated, projective $R$-module.

Question: If $M \otimes_R S$ is $S$-free, must $M$ be $R$-free?

Remark: this is trivially true if $R$ is semi-local. It is easy and well-known if $\phi$ is surjective. Without some conditions on $S$, the assertion is false, for example if we take $S$ to be a residue field of a maximal ideal in $R$ and $M$ be some projective, non-free module.

This kind of statement sounds like it should be in Bourbaki or EGA or the stack project (if it is true!). Does anyone know a proof or counter example?

-
Your assumption on $S$ is satisfied in particular if $S$ is a free $R$-module. I wouldn't be surprised if you could find a counterexample in this case (e.g. with $R$ a ring of algebraic integers, and $M$ invertible?) – Laurent Moret-Bailly Mar 9 2011 at 13:47
Dear Laurent, that sounds promising! So we want $S$ to be $R$-free, $Pic(S)$ is trivial but $Pic(R)$ is not? – Hailong Dao Mar 9 2011 at 14:02
Actually, $R=\mathbb R[x,y]/(x^2+y^2-1)$, $S=R\otimes \mathbb C$ seems to work. – Hailong Dao Mar 9 2011 at 14:19
Nice and simple! – Laurent Moret-Bailly Mar 9 2011 at 14:41
Dear Laurent, thank you for the hint. In hindsight this one probably should be asked on math.stackexchange.com! – Hailong Dao Mar 9 2011 at 20:40

Let me answer and accept this in CW so that it will not be bumped periodically as not answered by the software. It was hoped that the case of $\phi$ surjective can be generalized, but as Laurent pointed out in the comments, one can not hope to get any reasonable statement.

-

So, I guess $Pic(R)$ doesn't need to be non-trivial, then? (Since all line bundles over $S^2$ are free...)

EDIT: That was supposed to be in the comments section...

-
 @Andrew, this is $S^1$. – Hailong Dao Mar 9 2011 at 17:29 Indeed. I misread, but you still have a counterexample for $R=S^2$ and $S= R \otimes \mathbb{C}$ with your module being the tangent bundle on $S^2$, no? – Andrew Parker Mar 9 2011 at 17:43 @Andrew: you are right. My comment about Pic(R) was a response to Laurent's suggestion. – Hailong Dao Mar 9 2011 at 18:20