Let $A$ a commutative ring with 1 and $M$ a finitely generated $A$-module. Assume the localization $M_p$ is a free $A_p$-module for some prime ideal $p$. Is it true that there is some set of generators of $M$ whose image in $M_p$ freely generate $M_p$? Feel free to assume $A$ is Noetherian.
Fancy way to ask the same question: is there a surjective map $A^{\oplus n} \to M$ which induces an isomorphism $A_p^{\oplus n} \to M_p$?
The reason I'm asking this is: I was able to reduce some Exercise in Hartshorne (II.5.7) to this statement, and all solutions I've found online treat this step as it's obvious but I just don't see it.
Best Answer
No. For instance, if $p$ and $q$ are two different maximal ideals of $A$ and $M=A/q$, then $M_p=0$ so we would have to have $n=0$ but no map $A^0\to M$ can be surjective.
This result is not needed to prove Exercise II.5.7, though, since you only need such a map $A^n\to M$ locally (i.e., after localizing at some element $f\in A\setminus p$.). As a sketch of a proof, you can take a basis of $M_p$ and lift each element of this basis to $M_f$ for some $f\in A\setminus p$. This gives a homomorphism $A_f^n\to M_f$ which is an isomorphism after localizing at $p$. To make this homomorphism surjective, note that its cokernel is a finitely generated module whose localization at $p$ is $0$, so you can find some $g\in A\setminus p$ which annihilates the cokernel, and then the map $A_{fg}^n\to M_{fg}$ obtained by also localizing at $g$ will be surjective. (Assuming $A$ is Noetherian, you can also do the same thing for the kernel and find a localization at an element of $A\setminus p$ which makes the map an isomorphism.)