Let $R$ be a Noetherian ring. Is it true that a module $M$ is finitely generated if and only if $M_{\mathfrak p}$ is finitely generated over $R_{\mathfrak p}$ for all prime/maximal ideal $\mathfrak{p}$?
When the ring is not Noetherian, I can think of some pathological examples. For Noetherian case it seems true.
After a while, I could find this counter-example: $M:=\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/3\mathbb{Z}\oplus\mathbb{Z}/5\mathbb{Z}\oplus\cdots$ over $\mathbb{Z}$. As $\mathbb{Z}/p\mathbb{Z}\otimes\mathbb{Z}_{(q)}=\mathbb{Z}/p\mathbb{Z}$ if $p=q$ or $0$ otherwise. So $M_{\mathfrak{p}}$ is finitely generated for each prime $\mathfrak{p}$.
Maybe now, I strengthen the condition that $M_{\mathfrak p}$ is finite free over $R_{\mathfrak p}$ not just finitely generated. Is it possible to find a counter-example that $M$ is not finitely generated?
After a while, I could find this counter-example: $M:=\{a/b\mid b \text{ is square free.}\}$ over $\mathbb{Z}$. $M_{(p)}\cong\mathbb Z_{(p)}$ and $M_{(0)}\cong\mathbb Q$. So $M$ is stalkwise finite free, but $M$ is definitely not finitely generated or projective or free over $\mathbb Z$.
Ps, my intuition was to study the difference between projective module and locally free module and stalk-wise locally free module. Locally free means the information is on Zariski open set, and a lot of information can be glued up. But stalk-wise was on each prime ideal. So in general, it is weaker and a lot of information can not be glued up.