14
$\begingroup$

Lazard proved that every flat module over a ring is a directed colimit of finite free modules (Lam, Lectures on Modules and Rings, Theorem 4.34). I wonder if there is a similar theorem about flat modules on schemes.

Question: Let $X$ be a scheme which satisfies some finiteness conditions (at least quasicompact and quasiseparated, perhaps even noetherian and separated). Is then every flat quasi-coherent module $M$ on $X$ a directed colimit of locally free modules of finite rank?

This would be a nice characterization of flat modules since there is no allquantor. It would be very useful in the context of Tannaka Duality. Reading the proof of Lazard's theorem "backwards", it can be shown that the question is equivalent to:

Does every homomorphism from a quasi-coherent module of finite presentation to a flat quasi-coherent module factor through a locally free module of finite rank?

EDIT: The dissertation of Philipp Gross deals with the question when every quasi-coherent sheaf is a quotient of a locally free one. In Remark (3.5.7) it is remarked this would be true if every flat module is the directed colimit of locally free modules. But it's already hard to prove the resolution property for nice schemes ...

Improve this question
$\endgroup$
6
  • $\begingroup$ The link to Tannaka duality doesn't work. $\endgroup$
    – Dmitri Pavlov
    Commented Feb 14, 2011 at 22:18
  • $\begingroup$ @Dmitri: I've corrected the link. $\endgroup$
    – Martin Brandenburg
    Commented Feb 14, 2011 at 23:01
  • 1
    $\begingroup$ I don't know the answer to your question but would like to point out that Deligne introduced the notion of local inductive limit (SGA IV:Exp V, App) that allows one to essentially pretend that it is always true (in any topos, those were the days...). $\endgroup$
    – Torsten Ekedahl
    Commented Feb 15, 2011 at 8:23
  • $\begingroup$ @Torsten: Interesting! This seems to be the whole motivation of Deligne's Appendix. In Theorem 8.2.12 Lazard's theorem is generalized in a rather trivial to ringed toposes. But in my application I need global inductive limits. $\endgroup$
    – Martin Brandenburg
    Commented Feb 15, 2011 at 11:27
  • $\begingroup$ Let me compare this situation with a similar one: It is easy to prove that every module over some ring is a directed colimit of finitely presented modules. Then we can conclude that every module over some ringed topos is a local directed colimit of locally finitely presented modules. But it is a more interesting fact that every quasi-coherent module on a qs qc scheme is a directed colimit of locally finitely presented modules. The proof in (EGA I, 6.9.12) is tricky. $\endgroup$
    – Martin Brandenburg
    Commented Feb 15, 2011 at 11:33

2 Answers 2

Reset to default
3
$\begingroup$

In the paper "A Lazard-like theorem for quasi-coherent sheaves" by Sergio Estrada, Pedro A. Guil Asensio and Sinem Odabasi (arXiv), the following Theorem is proven:

Let $X$ be a quasi-compact and semi-separated scheme having enough locally countably generated vector bundles (for instance if $X$ is noetherian, separated, integral and locally factorial). Let $F$ be a flat quasi-coherent sheaf on $X$. Then $F = \lim\limits_{\longrightarrow} F_i$, where $F_i$ is locally countably generated and flat with $\mathcal{V} \dim F_i \leq 1$ (where $\mathcal{V}$ is the class of all vector bundles on $X$).

This is already a strong result, going into the direction of my question.

Improve this answer
$\endgroup$
-1
$\begingroup$

This is a local property. [if $M$ comes from an algebra over $A$]. It reduces to consider the case that $X=\text{Spec}(A)$ is an affine scheme.

If the ring $A$ is a PID, I think the statement is true (although I did not give a proof indeed).

For an arbitary ring $A$, the statement may be false. Please check the remark, Page 22, in: Commutative Algebra, 2ed, by Matsumura, 1980.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ -1. Reason: a) It is not a local property. b) For affine schemes it follows from Lazard's Theorem. $\endgroup$
    – Martin Brandenburg
    Commented Mar 25, 2012 at 8:55
  • $\begingroup$ Dear Martin, Sorry. Because here $M$ is not necessarily an algebra over $A$. This is for a). $\endgroup$
    – Feng-Wen An
    Commented Mar 26, 2012 at 15:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .