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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?

edited Mar 9 2011 at 13:32

asked Mar 9 2011 at 13:27

Hailong Dao
15.4k●29●69

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

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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

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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

Hailong Dao · Accepted Answer · 2011-03-23 17:50:51Z UTC

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.

Andrew Parker · Answer 2 · 2011-03-09 17:08:32Z UTC

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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...

answered Mar 9 2011 at 17:08

Andrew Parker
205●1●7

	@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

Freeness of modules along ring homomorphisms

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