13
$\begingroup$

Let $X$ be a scheme (or more generally a ringed space, if it works). Does $Qcoh(X)$, the category of quasi-coherent sheaves on $X$, admit a generating set? This would be useful, because then every cocontinuous functor on $Qcoh(X)$ has a right adjoint (SAFT).

If $X$ is affine, then $\mathcal{O}_X$ is a generator. I doubt that this is true in general. If $X$ is quasi-separated, perhaps the direct images of the $\mathcal{O}_U$, $U$ affine, do the job, but the naive proof does not work. If $Qcoh(X)$ does not have a generating set in general, what conditions on $X$ guarantee this?

EDIT: It is true when $X$ is concentrated, i.e. quasi-compact and quasi-separated, in particular when $X$ is noetherian (see Philipp's comment). This is already satisfying. Anyway, are there other (counter)examples?

PS: Note that this question is somehow unnatural with the background of this question; $\underline{Qcoh}(X)$, considered as a stack of abelian categories, always has a "stack-generator", namely $U \mapsto \mathcal{O}_U$. Nevertheless, I think the question above is interesting.

$\endgroup$
15
  • 1
    $\begingroup$ For a ringed space X the category QCoh(X) need not be abelian. The usual hypothesis on a scheme X which ensures that QCoh(X) is a Grothendieck category is that X be quasi-compact and quasi-separated (this is often called "concentrated"). See [Lipman-Notes on derived functors and Grothendieck duality, 4.1.3.1] (available on his website) for a proof in this case. Also Daniel Murfet's notes are useful and contain a proof, see [therisingsea.org/notes/ModulesOverAScheme.pdf, Proposition 66]. I haven't thought about counterexamples in other cases. $\endgroup$ Sep 25, 2010 at 11:43
  • $\begingroup$ Thank you Philipp ;). The notes of Daniel Murfet are very good (and remind me of the stacks project). $\endgroup$ Sep 25, 2010 at 12:37
  • 10
    $\begingroup$ There are no counterexamples. For an arbitrary scheme $X$ there exists an infinite cardinal $\kappa$ so that every quasi-coherent sheaf is generated by its quasi-coherent subsheaves of type $\kappa$, where the latter means that sections over some open affine cover (and then necessarily over any open affine) are generated by $\le \kappa$ elements as a module. This was explained to me long ago by Gabber, so ask him for the details. There is obviously a set of isomorphism class representatives for the quasi-coherent sheaves of type $\kappa$, so that settles it affirmatively in general. $\endgroup$
    – BCnrd
    Sep 25, 2010 at 14:33
  • 3
    $\begingroup$ Gabber has sent me a scan of a 11 year old letter, which is addressed to BCnrd ;). I will try to write it up. $\endgroup$ Sep 28, 2010 at 14:36
  • 3
    $\begingroup$ Gabber's argument also appears in print in Enochs and Estrada, "Relative homological algebra in the category of quasi-coherent sheaves," Adv. in Math. 194 (2005) 284--295. $\endgroup$ Sep 29, 2010 at 17:39

3 Answers 3

9
$\begingroup$

Gabber's argument also appears in print in Enochs and Estrada, "Relative homological algebra in the category of quasi-coherent sheaves," Adv. in Math. 194 (2005) 284--295.

$\endgroup$
2
  • 2
    $\begingroup$ This is a great article. It shows more generally that for every quiver $Q$ and every flat ring representation $R$ of $Q$, the category $Qcoh(R)$ is a Grothendieck category. $\endgroup$ Oct 3, 2010 at 7:53
  • 2
    $\begingroup$ Let me just mention that in the paper you have to replace quivers by small categories and introduce compatiblity conditions in order to get the correct notion of (quasi-coherent) modules. I've already emailed with Enochs about that. The arguments should carry over. $\endgroup$ Dec 3, 2010 at 14:09
12
$\begingroup$

Yes: if $X$ is quasi-compact and quasi-separated, then the category of quasi-coherent sheaves on $X$ is canonically equivalent to the category of ind-objects on the (essentially small) category of coherent sheaves of finite presentation; see the Appendix of Deligne in Hartshorne's Residues and duality.

$\endgroup$
1
  • $\begingroup$ Perhaps, you can see the work of Neeman, Henning Krause and others on "WELL GENERATED CATEGORY". $\endgroup$
    – kaddar
    Sep 27, 2010 at 6:35
12
$\begingroup$

As BCnrd already told us in the comments, $Qcoh(X)$ always has a generating set. Klick here for my write-up of Gabber's proof.

$\endgroup$
1
  • 2
    $\begingroup$ Link seems to be broken $\endgroup$ Oct 20, 2015 at 0:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.