TITLE: When does a quasicoherent sheaf vanish?
QUESTION [11 upvotes]: 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.

REPLY [8 votes]: 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 $M$.

REPLY [5 votes]: What about $R=\mathbb Z$ and $M=\mathbb Q/\mathbb Z$ ? All fibers are zero.