Theorem. (Quillen) Let $P$ be a finitely presented $A[X]$-module. If for each maximal ideal $m \subset A$ the module $P_m$ is extended from $A_m$, then $P$ is extended from $A$.
Proof: (The proof is from paper titled Quillen's solution of Serre's Problem)
Consider the Quillen's sets $I= \{a \in A : P_a$ is extended from $A_a \}$. Now the author wants to show that $I$ is an ideal and $I$ is not contained in any maximal ideal and hence $1 \in I$ and we are done. Can someone explain me the idea of the proof to show $I$ is an ideal which is not contained in any maximal ideal? The proof is given in the paper but i don't really follow it. Here is the link of the paper.
While easy as to how to use finite presentation, it is a bit long so I write it as an answer, though it is really not.
If $P$ is finitely presented as an $R=A[X]$ module, then so is $M[X]=M\otimes_A R$ where $M=P/XP$. The assumption says for a maximal ideal $\mathfrak{m}\subset A$, you have $P_{\mathfrak{m}}\cong M[X]_{\mathfrak{m}}$. Using finite generation, clearing denominators, you get a map $P\to M[X]$ which is an isomorphism when you localize at $\mathfrak{m}$. The cokernel is zero when you localize at $\mathfrak{m}$ and the cokernel is finitely generated, so you can make it zero by inverting just one $a\not\in\mathfrak{m}$. The kernel is finitely generated since $P$ is finitely presented and again, you can make it zero by inverting a $b\not\in\mathfrak{m}$. Now, take $ab\not\in\mathfrak{m}$ and when you invert, both kernel and cokernel are zero. That is, $P_{ab}\cong M[X]_{ab}$.
The general fact used is for a finitely presented module $M$ over a ring $S$ and any module $N$ and a prime ideal $\mathfrak{p}\subset S$, the natural map $(\mathrm{Hom}_S(M,N))_{\mathfrak{p}}\to \mathrm{Hom}_{S_{\mathfrak{p}}}(M_{\mathfrak{p}}, N_{\mathfrak{p}})$ is an isomorphism.