# When does a quasicoherent sheaf vanish?

Let $F$ be a quasi-coherent sheaf on a scheme $X$. To check that $F$ vanishes it suffices to check that all the stalks of $F$ vanish. I would like to know whether it suffices to check that all the fibers of $F$ vanish.

(I think I am using standard terms: the tensor product over $O_X$ with the local ring at $x$ is the stalk, and the tensor product over $O_X$ with the residue field at $x$ is the fiber.)

It suffices to answer the question on an affine scheme. Let $R$ be a commutative ring. For each prime ideal $p$ of $R$ let $k(p)$ be the residue field of the local ring $R_p$. Let $M$ be an $R$-module, and suppose that $\mathrm{Tor}_i(k(p),M) = 0$ for all $i$ and all $p$. (Tor taken in the category of $R$-modules). Does it follow that $M = 0$?

If $M$ is finitely generated, the answer is yes. In that case $M$ vanishes even when $\mathrm{Tor}_0(k(p),M) = 0$ for all maximal $p$, by Nakayama's lemma. My question is whether there is a good replacement for Nakayama's lemma when $M$ is not finitely generated.

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I believe something similar will hold if M is simple. – Harry Gindi Jan 26 '10 at 0:44

## 2 Answers

If the scheme is locally noetherian, this is true and can be proved by noetherian induction. In fact, you can even replace $M$ with an object of bounded derived quasi-coherent category, if you are interested in such things.

The proof is relatively straightforward: For a complex of modules $M$ over the ring $R$, we may assume that any non-zero $f\in R$ acts by a quasi-isomorphism $f:M\to M$ (by the induction hypothesis), and then the cohomology of $M$ are defined over the field of fractions of $M$.

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Wonderful. Is this a famous argument, or off the top of your head? – David Treumann Jan 26 '10 at 1:19
I suspect that this is well known. If so, I would like references; the only one I know is Lemma 10 in Arinkin, Orthogonality of natural sheaves on moduli stacks of SL(2)-bundles with connections on P^1 minus 4 points. – t3suji Jan 26 '10 at 2:24

What about $R=\mathbb Z$ and $M=\mathbb Q/\mathbb Z$ ? All fibers are zero.

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You'll notice he has derived fibers. Those are non-zero in your example. – t3suji Jan 26 '10 at 1:02
Yes, you are right. I only answered the first part of the question. – Qing Liu Jan 26 '10 at 1:06