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.

somefree $A$-module $N$ with isomorphisms $M_{\mathfrak{p}} \cong N_{\mathfrak{p}}$? – Martin Brandenburg Apr 13 '13 at 1:19anotherfree 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