Modules, Sheaves and Vector bundles. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T21:36:03Z http://mathoverflow.net/feeds/question/5008 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/5008/modules-sheaves-and-vector-bundles Modules, Sheaves and Vector bundles. Csar Lozano Huerta 2009-11-11T09:01:32Z 2009-11-11T15:06:26Z <p>Given a graded ring S and a graded S-module <img src="http://latex.mathoverflow.net/png?M" alt="M" title="" /> we can carry out a construction in order to get <img src="http://latex.mathoverflow.net/png?%5Ctilde%7BM%7D" alt="\tilde{M}" title="" />, which is a sheaf over the scheme Proj S. With this in view, I have an equivalence of categories between the category of (quasi-finite generate) modules and the category of q-coherent sheaves over Proj S. </p> <p>On the other hand, given a locally free sheaf <img src="http://latex.mathoverflow.net/png?%5Cmathcal%7BF%7D" alt="\mathcal{F}" title="" /> of rank <img src="http://latex.mathoverflow.net/png?n" alt="n" title="" /> over Proj S we can get a vector bundle out of it and further we have a 1-1 correspondence between isomorphism classes of free sheaves of rank n and isomorphism classes of vector bundles of rank n.</p> <p>With the last two facts in view, my question is the following. if I start with a finite generated module <img src="http://latex.mathoverflow.net/png?M" alt="M" title="" /> then the sheaf <img src="http://latex.mathoverflow.net/png?%5Ctilde%7BM%7D" alt="\tilde{M}" title="" /> is locally free? If so, I can get from <img src="http://latex.mathoverflow.net/png?M" alt="M" title="" /> a locally free sheaf <img src="http://latex.mathoverflow.net/png?%5Ctilde%7BM%7D" alt="\tilde{M}" title="" /> and from such a sheaf a vector bundle (and perhaps backwards as well). Therefore, Is it the same having a vector bundle over Proj S than a S-Module? or what are the limits of such a relation described here among Vector Bundles &amp; S-Modules?. By "Is it the same" I mean, We have the same amount of information in such objects.</p> http://mathoverflow.net/questions/5008/modules-sheaves-and-vector-bundles/5027#5027 Answer by anton for Modules, Sheaves and Vector bundles. anton 2009-11-11T10:32:21Z 2009-11-11T10:32:21Z <p>The following is a counterexample: Consider the Ring R=Z[x] with its usual gradation and the module R/nR, where n is a natural number.</p> <p>The idea works for projective modules, ie a projective module will give a locally free sheaf. This is why algebraic K-theory is based on the category of projective modules.</p> http://mathoverflow.net/questions/5008/modules-sheaves-and-vector-bundles/5056#5056 Answer by David Speyer for Modules, Sheaves and Vector bundles. David Speyer 2009-11-11T15:06:26Z 2009-11-11T15:06:26Z <p>There are (at least) two details you are missing.</p> <p>(1) This is not an equivalence of categories between finitely generated graded modules and coherent sheaves. If your module is $0$ in all sufficiently large degrees, then the corresponding sheaf will be zero. For example, let $S=k[x,y]$ and $M=S/\langle x,y \rangle$. The sheaf on $\mathbb{P}^1$ corresponding to $M$ is the zero sheaf.</p> <p>The category you want to work with is the one whose objects are finitely generated graded modules, and where we formally invert any map which is an isomorphism in sufficiently large degree. (Alternative formulation: we formally invert a map $f:M \to N$ if, for any $s$ in the irrelevant ideal, $s^{-1} f: s^{-1} M \to s^{-1} N$ is an isomorphism.)</p> <p>(2) Not every coherent sheaf is a vector bundle. Correspondingly, not every finitely generated graded module will correspond to a vector bundle. If we were dealing with an affine variety, vector bundles would correspond to locally free modules. (Also called projective modules.) For projective varieties, things are a little trickier: the criterion is to be locally free away from the irrelevant ideal. </p> <p>Sadly, I believe that there exist examples of modules which are locally free away from the irrelevant ideal, but are not isomorphic (in the above category) to any module which is locally free at the irrelevant ideal. This should be related to my question <a href="http://mathoverflow.net/questions/4590/when-are-dual-modules-free" rel="nofollow">here</a>. But you can go a long while without paying attention to this detail.</p>