Does finite projective dimension localize? Let $R$ be a commutative (but not necessarily Noetherian) ring with unity.  Let $M$ be an $R$-module.  Suppose that, for all $\mathfrak p \in\text {Spec}(R),$ $\text{pd}_{R_{\mathfrak p}}M_{\mathfrak p}&lt; \infty $.  Is it the case that $\text{pd}_RM &lt; \infty$?
Let $R$ be a ring with $\operatorname{gldim}(R)=\infty$. Then there exist $R$-modules of arbitrary large projective dimension. By taking their direct sum we can find an $R$-module $M$ of infinite projective dimension. Now, if the global dimension of all localizations of $R$ is finite, that is, $R_{\mathfrak p}$ is a regular local ring for all $\mathfrak p\in\operatorname{Spec}(R)$, it follows that the projective dimension of all localizations of $M$ is finite. 

Such an $R$ is the classical example of Nagata of a noetherian ring of infinite Krull dimension.
This is wrong, I'm leaving it up in case it helps anyone else avoid the mistakes I just made :) Thanks to QiL and YACP for the pointers!

Okay yeah you can carry out my first comment for affine schemes: namely, if you take an arbitrary product of rings $R_i$, $$Spec \; \prod R_i \simeq \coprod Spec \; R_i$$

Where $\coprod$ is disjoint union. With that, if we take a countable product of say $$R_i  = \mathbb{C}[x_1, \ldots, x_i]$$just so that $gldim \; R_i = i$, and take $M$ to be literally $(x_1, \ldots, x_i)$, inside $Spec \; R_i$ it has projective dimension $ i$ at the origin and is 0 elsewhere.