For (quasi-compact and quasi-separated) schemes there is a categorical way to characterise quasi-coherent sheaves of finite type using purely the abelian category $\operatorname{QCoh}(X)$. In an abelian category one can define a categorically finitely generated object as being an object M such that for any directed family of subobjects $M_\alpha \subset M$, such that the sum $\sum_\alpha M_\alpha$ is equal to the original $M$, then there exists an index $\alpha_0$ such that $M_{\alpha_0} = M$.

Now, in the category of $R$-modules this is easy to see, and the argument essentially relies on the fact that any module is the colimit of its finitely generated submodules. For quasi-compact and quasi-separated schemes the only reference I could find is Daniel Murfet's excellent notes. This is Corollary 64 from here.

I would like to know whether this result still holds for algebraic spaces or more generally for algebraic stacks. (I imagine that knowing this is essentially equivalent to proving Lemma 61 in Murfet's notes for a scheme but using the étale toplogy)

EDIT: Perhaps one cook up a variant of Proposition 15.4 of Laumon-Moret-Bailly. Unfortunately the argument there seems to use at the very end that a submodule of a finite type module is of finite type, which does not hold in the non noetherian case.

EDIT2: By the way, for qcqs algebraic spaces this was already known to Raynaud-Gruson [RG Proposition 5.7.8].

  • $\begingroup$ I would like to understand why this isn't automatically so simply because the theorem "every module is the colimit of its finitely generated submodules" is intuitionistically valid, hence true in every topos. Of course, one has to interpret "finite" correctly. $\endgroup$ Jul 11, 2013 at 21:41
  • $\begingroup$ It smells dangerously close to "every module is the colimit of finitely presented modules", which would then imply that the category of modules is always locally finitely presentable... but this is surely not true, because Grothendieck toposes are not always locally finitely presentable. $\endgroup$
    – Zhen Lin
    Jul 11, 2013 at 21:49
  • $\begingroup$ @AndrejBauer: I'm certainly very far from understanding any topos theory, but perhaps another caveat to add might be that I'm interested in quasi-coherent sheaves, not arbitrary modules in the topos of étale sheaves, which is what I think you had in mind. (alhough come to think about it, if a sheaf is of finite type then it's automatically quasi-coherent) $\endgroup$ Jul 11, 2013 at 22:26
  • $\begingroup$ @ZhenLin: it is crucial that "finite" be understood internally, and so you would have to take the internal notion of "finitely prsented". I think that makes a difference. $\endgroup$ Jul 12, 2013 at 16:30
  • 1
    $\begingroup$ There is a classical reference for the scheme case, EGA I (new), 6.9.12. For stacks I have asked almost the same question here mathoverflow.net/questions/83041 where I have also gathered some special cases. $\endgroup$ Jul 14, 2013 at 8:10

1 Answer 1


Note that on any qcqs algebraic space or Artin stack, a quasi-coherent sheaf is finite type if its pullback to any fppf scheme cover is finite type in the usual sense on schemes. We can similarly define the notion of "finite type" for a quasi-coherent sheaf of modules over any quasi-coherent sheaf of algebras on such an algebraic space or stack in the same way, and this is the notion that will be used below; it satisfies the categorical condition you want, so that is good enough.

In this paper, absolute noetherian approximation is proved for qcqs algebraic spaces, building on the version for schemes proved by Thomason and Trobaugh: according to Theorem 1.2.2, every qcqs algebraic space $X$ is an inverse limit (with affine transition maps) of finitely presented algebraic spaces over $\mathbf{Z}$. In particular, $X$ is affine over an algebraic space $X_0$ of finite presentation over $\mathbf{Z}$.

Hence, we have an affine morphism $f:X \rightarrow X_0$ to a noetherian algebraic space, so quasi-coherent $O_X$-modules are the "same" as quasi-coherent sheaves of modules over $X_0$ over the quasi-coherent $O_{X_0}$-algebra $A := f_{\ast}(O_X)$. Consequently, it suffices to show that for any noetherian algebraic space $X_0$ and quasi-coherent $O_{X_0}$-module $A$, every quasi-coherent $A$-module $F$ is the direct limit of its directed system quasi-coherent $A$-submodules of finite type. The underlying $O_{X_0}$-module $F'$ of $F$ is the direct limit of its directed system of coherent $O_{X_0}$-submodules $F'_i$. Hence, the image $F_i$ of the natural $A$-linear map $A \otimes_{O_{X_0}} F'_i \rightarrow F$ is a quasi-coherent $A$-submodule of $F$ of finite type (over $A$), and it contains $F'_i$, so clearly the inclusion $\varinjlim F_i \rightarrow F$ is an isomorphism.

That settles the case of qcqs algebraic spaces, and to prove the same for qcqs Artin stacks you just need absolute approximation for qcqs Artin stacks (which would give that any such stack is affine over a noetherian one, so you can bootstrap from the known noetherian case exactly as above). In the case that the Artin stack is qcqs with quasi-finite diagonal (e.g., any Deligne--Mumford stack), this is proved in this paper (which includes as a special case qcqs algebraic spaces, so it is an alternative to the initial reference at the top).

  • $\begingroup$ welcome to MathOverflow? $\endgroup$
    – user1504
    Jul 11, 2013 at 21:12
  • $\begingroup$ @user36938: thanks! that's excellent. Do you reckon for general Artin stacks it fails? Or is it just that a different strategy must be employed? (even for "simple" stacks such as $BG_m$) $\endgroup$ Jul 11, 2013 at 22:31
  • $\begingroup$ @John: I don't know if it is true for general qcqs Artin stacks, but if you believe that absolute noetherian approximation should hold for them then you should believe it to be true. $\endgroup$
    – user36938
    Jul 12, 2013 at 0:56
  • $\begingroup$ @user36938: I guess my question should have been: "do you believe in noetherian approximation? (or the existence of an alternative proof)" $\endgroup$ Jul 14, 2013 at 18:06
  • 2
    $\begingroup$ David Rydh eventually removed the assumptions on the diagonal, see arxiv.org/abs/1408.6698. $\endgroup$ Sep 23, 2014 at 12:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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