This is a follow-up to Akhil Mathew’s blog post which explains that the category of quasi-coherent sheaves on a scheme X is a Grothendieck abelian category. The key is a result of Gabber: given a scheme X there exists a cardinal κ such that every quasi-coherent sheaf is the directed colimit of its κ-generated quasi-coherent subsheaves. It follows by a standard argument that the embedding QCoh(O_X) —> Mod(O_X) has a right adjoint, whence limits exist in QCoh(O_X) and QCoh(O_X) has enough injectives.

Earlier today I wrote this up for the stacks project (it is in the chapter on properties of schemes) and it occurred to me that the exact same results hold for algebraic stacks, with the exact same proof. (The proof is one of these “randomly pick elements and see what happens” arguments, kinda like this post.) I’ll check the details and write out the proof some time later this week; keep watching this feed to see it appear.

Anyway, I guess it is just one of those general facts… easy to prove but hard to use.

Edit 10/17/2011. Beware of the following facts on quasi-coherent modules:

- It isn’t true that a product of quasi-coherent modules is quasi-coherent.
- An injective object in QCoh(O_X) is not always injective O_X-module.
- Cohomology using resolutions in QCoh(O_X) does not agree with cohomology.
- There exists a ring A and an injective A-module I such that the quasi-coherent sheaf I~ associated to I isn’t flasque, I~ isn’t an injective O_X-module, and there exists an open U of Spec(A) such that I~|_U isn’t an injective object of QCoh(O_U).
- D^+_{QCoh}(O_X) isn’t equivalent to D^+(QCoh(O_X)) in general.
- The coherator Q : Mod(O_X) —> QCoh(O_X) isn’t exact in general.
- And so on.

I think you can argue more simply (as someone pointed out in the comments to my post; I just figured out the reason) that every presentable category is complete, by the adjoint functor theorem. Namely, according to the version of the adjoint functor theorem stated on the nLab, a functor between presentable (they say “locally presentable”) categories has a right adjoint if and only if it preserves small colimits. So if $\mathcal{C}$ is a presentable category and $I$ any category, this applies to the diagonal functor $\mathcal{C} \to \mathcal{C}^I$.

Yeah, OK. But the adjoint functor theorem isn’t in the stacks project and the argument you gave is quite natural to the situation for quasi-coherent modules as a subcategory of the category of all modules. The stacks project is missing a discussion of Brown representabilty, as well as a general discussion of Grothendieck categories… and much, much more. I like knowing the general Brown representability and adjoint functor results, but I’m happy also if there is a more direct argument for specific categories.

I’ve been interested in presentability of categories of quasicoherent sheaves for a while now. Is it obvious that $Qcoh(X)$ is presentable for an algebraic stack $X$? My own argument was that using a smooth hypercover of $X$ by schemes, one can write $Qcoh(X)$ as “qcoh sheaves on a scheme + descent data”, which translates to a 2-limit of a diagram of accessible functors between presentable categories of the form “quasicoherent sheaves on a scheme” (what’s usually called a pseudolimit). Such a limit is always presentable, but that’s not an easy result.

I’ve been wondering if there’s an easier way to do it.

A Grothendieck abelian category is presentable! See this post by Akhil Matthew. So: Yes, because the category of quasi-coherent sheaves on an algebraic stack is a Grothendieck abelian category, see Proposition Tag 0781.

Yep, Grothendieck catgories are presentable all right, but I don’t think the argument is substantially simpler (the proof that I know for the presentability of Grothendieck categories again uses non-trivial results on presentale categories, such as their characterization as locally generated categories).

And by the way, here’s a natural question (albeit a bit off-topic for this blog post, I guess): is it the case that an abelian category that happens to be presentable is automatically Grothendieck? I remember thinking about this at one point, when I realized that Grothendieck $\Rightarrow$ presentable, but I never figured it out..

Yeah, sorry! I didn’t mean that it was simpler. Note, though, that the way we prove this in the stacks project (following Akhil’s blog post) is by constructing a “coherator”, so that it follows that in QCoh we have all limits because in Mod we have all limits.