# Gluing free modules to get a finitely generated free module

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.

• 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 $M_{\mathfrak{p}} \cong N_{\mathfrak{p}}$? Apr 13 '13 at 1:19
• @Martin: You beat me on that! But there is always a free module $N$ with $M_{\mathfrak{p}}\cong N_{\mathfrak{p}}$. Apr 13 '13 at 1:23
• 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). Apr 17 '13 at 20:48
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.