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Hi, maybe this is a stupid question, however none of my mathematicians colleagues could answer it properly. It's know that a finitely generated projective $A$-module $M$ is free if $A$ is a local ring. Now, consider an arbitrary finitely generated $A$-module $M$. If you pick all the possibles localizations of $M$, you will get some free $A_{\mathfrak{p}}$-modules $M_{\mathfrak{p}}$. In what conditions one can glue (using the Zariski topology) the localizations to get a free module $A$-module $N$ such that $N_{\mathfrak{p}} = M_{\mathfrak{p}}$ (if $M = N$, this will be better)? Of course, all the localizations must have the same rank, however I don't know if this is sufficient to produce a free module. I friend of mine said that his advisor researched this question for Von Neumann regular rings (which implies that the space will be Boolean and, then, there is a injection from the global sections into the product of the fibers) and, in some conditions (he doesn't remember), the result will hold.

Thanks in advance.

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The question is not clear (to me). You write "to get a free module $A$-module $M$". Do you want to recover $M$? Do you ask if $M$ is free? Of course this is not the case, this is what $K_0$ measures (or simply the Picard group for rank $1$). Or do you want to know if there is some free $A$-module $N$ with isomorphisms $M_{\mathfrak{p}} \cong N_{\mathfrak{p}}$? –  Martin Brandenburg Apr 13 '13 at 1:19
    
Do you want to get the same module back after gluing (which you want to be free) or do you want to get another free module after glueing? I think you mean the first one, because the second can be done trivially. But if you mean the first one, then you are really asking when is a locally free module, free, right? –  Mahdi Majidi-Zolbanin Apr 13 '13 at 1:22
    
@Martin: You beat me on that! But there is always a free module $N$ with $M_{\mathfrak{p}}\cong N_{\mathfrak{p}}$. –  Mahdi Majidi-Zolbanin Apr 13 '13 at 1:23
    
Contrary to what is claimed, given a family of modules over the local rings, there is a free module with localization at each prime isomorphic to the given local module iff all of the localizations have the same rank. If this is not what the question means...then I don't know what it means. –  Pete L. Clark Apr 13 '13 at 1:27
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Sorry for the ambiguity, I tried to edit it properly. Just to clarify, I have considered a projective module $M$ since I don't really know the relationship between locally free and projective (which one is stronger ?). Sorry if the question is trivial, but I'm not an specialist in commutative algebra and sheaf theory (and I don't know any specialist in these fields). –  user17868 Apr 17 '13 at 20:48
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I am assuming your ring is commutative. First of all, there is currently an error in your question, because you neither assume $M$ to be locally free, not projective. Your questions reads: "Now, consider an arbitrary finitely generated $A$-module $M$." My guess is you inadvertedly deleted this assumption when you edited your question, but I suppose $M$ is projective, because that's what you say in your comment. Now, as mentioned by several people in the comments, including myself, if all localizations of $M$ have equal rank $k$, then you can take $N$ to be the free module $A^k$. There is no reason for $M$ itself to be free. In special cases, there are cohomological criteria that if satisfied, will imply $M$ is free.

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I committed a mistake. I thought locally free and projective would coincide in the finitely generated case (that´s not true). This is why I have deleted the "projective" part. So I´m only assuming the locally freeness. –  user17868 Apr 18 '13 at 17:24
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Even if you assume locally freeness my answer will stand valid. –  Mahdi Majidi-Zolbanin Apr 18 '13 at 19:34
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