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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?

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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 '11 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 '11 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 '11 at 14:19
Nice and simple! – Laurent Moret-Bailly Mar 9 '11 at 14:41
So, I guess $Pic(R)$ doesn't need to be non-trivial, then? (Since all line bundles over $S^2$ are free...) – Andrew Parker Mar 9 '11 at 17:08
up vote 1 down vote accepted

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.

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