I'm currently trying to understand Quillen's original proof of the Quillen-Suslin theorem on projective modules over polynomial rings and I'm having difficulty justifying a claim in the paper.
On the second page, he makes a claim that I believe amounts to the following:
Claim: Let $P$ be a finitely presented module over a commutative ring $R$ and $\mathfrak{m}$ a maximal ideal of $R$. Then if $Q$ is another $R$-module for which $P_{\mathfrak{m}} \cong Q_{\mathfrak{m}}$ as $R_{\mathfrak{m}}$-modules, there exists an $f \in R$ such that this isomorphism comes from an isomorphism $P_f \cong Q_f$ of $R_f$-modules.
I've been having trouble proving this, especially since the isomorphism $P_{\mathfrak{m}} \cong Q_{\mathfrak{m}}$ is not assumed to be induced from a map $P \rightarrow Q$, nor does there seem to be any restriction on $R$ (e.g. that it is noetherian). How would one go about proving this claim?
EDIT: The way I stated the claim is false. In the paper, it is assumed that we have an isomorphism
$$A_{\mathfrak{m}}[T] \otimes_{A_{\mathfrak{m}}} N_{\mathfrak{m}} \cong M_{\mathfrak{m}}$$
where $M$ is finitely presented over $A[T]$ and we want to find an $f \notin \mathfrak{m}$ with
$$A_f[T] \otimes_{A_f} N_f \cong M_f.$$
Best Answer
$\newcommand{\fm}{\mathfrak{m}}$This is not true without additional hypotheses. For instance, let $R=\mathbb{Z}$, $\fm=(2)$, $P=\mathbb{Z}/(2)$, and $Q=\bigoplus_p \mathbb{Z}/(p)$ where $p$ ranges over all primes. Then $P_\fm\cong Q_\fm\cong\mathbb{Z}/(2)$, but if we invert just a single odd $f\in\mathbb{Z}$ then $Q$ still has infinitely many summands.
It is true if you assume $Q$ is finitely generated and projective. Indeed, since $P$ is finitely presented we can lift an isomorphism $P_\fm\to Q_\fm$ to a morphism $F:P_f\to Q_f$ for some $f\in R\setminus\fm$ (first choose $f$ such that we can lift the images of all the generators of $P$, then such that all the relations are preserved). Since $Q$ is finitely generated, we can further choose $f$ such that this $F$ is surjective. But then since $Q_f$ is projective over $R_f$, $F$ will split. So $\ker F$ is a direct summand of $P_f$ and hence finitely generated since $P$ is. Thus we can modify $f$ to make sure each generator of $\ker F$ dies, and thus we obtain $F:P_f\to Q_f$ which is an isomorphism.
Note that neither assumption on $Q$ (finitely generated or projective) is enough on its own. For counterexamples to both, let $R$ be the ring of eventually constant sequences in some field $k$. There is a maximal ideal $\fm$ of sequences which are eventually $0$, and the localization $R_\fm$ is just $R/\fm\cong k$ (sending a sequence to its eventually constant value).
Now let $P=R$ and $Q=R/\fm$. Then $P_\fm\cong Q_\fm$ and $Q$ is finitely generated, but $P_f\not\cong Q_f$ for any $f\in R\setminus\fm$ (inverting a single $f\in R\setminus \fm$ will only eliminate the finitely many coordinates on which $f$ is $0$).
Alternatively, let $P=0$ and let $Q=\fm$. Note that $Q$ is projective, since it is the direct sum of the ideals which are supported on just a single coordinate, and each of those ideals is a direct summand of $R$. We have $P_\fm\cong Q_\fm\cong 0$, but $Q_f$ is nonzero for any $f\in R\setminus\fm$.