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Given a graded ring $S$ and a graded S-module $M$ we can carry out a construction in order to get $\tilde{M}$, which is a sheaf over the scheme $\mathrm{Proj}~ S$. 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 $\mathrm{Proj}~S$.

On the other hand, given a locally free sheaf $\mathcal{F}$ of rank $n$ over $\mathrm{Proj}~S$ 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 $n$.

With the last two facts in view, my question is the following. if I start with a finite generated module $M$ then the sheaf $\tilde{M}$ is locally free? If so, I can get from $M$ a locally free sheaf $\tilde{M}$ and from such a sheaf a vector bundle (and perhaps backwards as well). Therefore, Is it the same having a vector bundle over $\mathrm{Proj}~ S$ than a $S$-Module? or what are the limits of such a relation described here among Vector Bundles & $S$-Modules?. By "Is it the same" I mean, We have the same amount of information in such objects.

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up vote 6 down vote accepted

There are (at least) two details you are missing.

(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.

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.)

(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.

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 here. But you can go a long while without paying attention to this detail.

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thanks a lot! your answer was very clear. – Csar Lozano Huerta Nov 11 '09 at 16:37

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.

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.

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