most recent 30 from http://mathoverflow.net 2013-05-22T15:15:32Z http://mathoverflow.net/feeds/question/57946 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57946/freeness-of-modules-along-ring-homomorphisms Hailong Dao 2011-03-09T13:27:28Z 2011-03-23T17:50:51Z

<p>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. </p> <blockquote> <p>Question: If $M \otimes_R S$ is $S$-free, must $M$ be $R$-free? </p> </blockquote> <p>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. </p> <p>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? </p>

http://mathoverflow.net/questions/57946/freeness-of-modules-along-ring-homomorphisms/57970#57970 Andrew Parker 2011-03-09T17:08:32Z 2011-03-09T17:08:32Z

<p>So, I guess $Pic(R)$ doesn't need to be non-trivial, then? (Since all line bundles over $S^2$ are free...)</p> <p>EDIT: That was supposed to be in the comments section...</p>

http://mathoverflow.net/questions/57946/freeness-of-modules-along-ring-homomorphisms/59336#59336 Hailong Dao 2011-03-23T17:50:51Z 2011-03-23T17:50:51Z

<p>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. </p>