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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}< \infty $. Is it the case that $\text{pd}_RM < \infty$?

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hmm...if you replace ring with scheme and module with Qcoh, this is false: take a countable disjoint union of schemes and stick on the $n^{th}$ one a module $M_n$ with projective dimension $n$ hunch is this is false too – uncookedfalcon Jan 10 '13 at 0:33
stupid comment: if all the $pd_{R_p} M_p < N$, then $pd_R M < N$ – uncookedfalcon Jan 10 '13 at 0:36
up vote 9 down vote accepted

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.

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YACP: nice work! And thanks for the reference; this is a nice example to have in one's arsenal. – mbrown Jan 10 '13 at 17:05

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.

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pd$R_i=i$ where $R_i$ is a module over...? As far as I know gldim$R_i=i$. Moreover, I have some doubts that the prime specrum of an infinite direct product is the direct sum of the spectra. – user26857 Jan 10 '13 at 10:06
uh oh/thanks guys... hmm this is interesting... Rings is opposite to affine schemes, so products of rings correspond to coproducts of affine schemes. so we're saying coproduct for schemes isn't just disjoint union (with obv gluing of structure sheaves)? – uncookedfalcon Jan 10 '13 at 10:47
@YACP yeah I goofed - I meant global dimension. thanks for the tip! – uncookedfalcon Jan 10 '13 at 10:49
yeah of course: this couldn't be quasi-compact! my bad guys – uncookedfalcon Jan 10 '13 at 10:54
w.r.t. my earlier comment - the inclusion of a full subcategory (AffSch -> Sch) a priori has no reason to play nicely with products and coproducts! – uncookedfalcon Jan 10 '13 at 11:08

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