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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.

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    $\begingroup$ 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. $\endgroup$
    – Philipp Hartwig
    Commented 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). $\endgroup$
    – Martin Brandenburg
    Commented Sep 25, 2010 at 12:37
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    $\begingroup$ 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
    Commented Sep 25, 2010 at 14:33
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    $\begingroup$ Gabber has sent me a scan of a 11 year old letter, which is addressed to BCnrd ;). I will try to write it up. $\endgroup$
    – Martin Brandenburg
    Commented Sep 28, 2010 at 14:36
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    $\begingroup$ 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. $\endgroup$
    – Thomas Nevins
    Commented Sep 29, 2010 at 17:39

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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.

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    $\begingroup$ 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. $\endgroup$
    – Martin Brandenburg
    Commented Oct 3, 2010 at 7:53
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    $\begingroup$ 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. $\endgroup$
    – Martin Brandenburg
    Commented Dec 3, 2010 at 14:09
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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.

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

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    $\begingroup$ Link seems to be broken $\endgroup$
    – Rachmaninoff
    Commented Oct 20, 2015 at 0:45

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