most recent 30 from http://mathoverflow.net 2013-06-19T19:00:30Z http://mathoverflow.net/feeds/user/17732 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75150/sheaf-with-free-stalks maxim 2011-09-11T16:52:26Z 2011-09-12T15:19:34Z

<p>Say we are given a complex manifold $X$ and an $\mathcal{O}_X$-module $\mathcal{F}$. Assume that for any point $P\in X$ the stalk $\mathcal{F}_P$ is a free $(\mathcal{O}_X)_P$-module of finite rank. Does it imply that $\mathcal{F}$ is locally free? If not, what do you need to know additionally about $\mathcal{F}$ to make it true?</p> <p>Note that if we were looking at the case of schemes then it would be wrong in general. Mathoverflow answer to a related question is <a href="http://mathoverflow.net/questions/44839/wikipedias-definition-of-locally-free-sheaf" rel="nofollow">here</a></p> <p><strong>Remark:</strong> As it was pointed out by Francesco Polizzi, this is true if $\mathcal{F}$ is coherent. What if we do not know it apriori?</p>

http://mathoverflow.net/questions/75150/sheaf-with-free-stalks/75161#75161 maxim 2011-09-13T06:53:27Z 2011-09-13T06:53:27Z

Thank you! Nice example!

http://mathoverflow.net/questions/75150/sheaf-with-free-stalks maxim 2011-09-11T21:01:39Z 2011-09-11T21:01:39Z

Hartshorne's exercise is about coherent sheaves. As the remark above says, we do not assume this apriori.

http://mathoverflow.net/questions/75150/sheaf-with-free-stalks/75152#75152 maxim 2011-09-11T17:57:13Z 2011-09-11T17:57:13Z

Thank you for the answer! Right, this is because the support of a coherent sheaf is closed. What if the sheaf is not coherent?