5
$\begingroup$

Hi everybody,

how does one prove the following statement:

given a coherent sheaf $\mathscr{F}$ on a smooth projective variety $X$, then $\mathscr{F}$ has a resolution by locally free sheaves of length $n=\dim(X)$.

Or does anybody know a reference for this fact?

Greetings!

Improve this question
$\endgroup$
7
  • 2
    $\begingroup$ I am no expert here, but Hartshorne Exercise III 6.8 and Auslander-Buchsbaum formula seem to do the trick (for more general situation). $\endgroup$
    – Hailong Dao
    Aug 3, 2011 at 15:00
  • 2
    $\begingroup$ This is the celebrated "Hilbert syzygy theorem". You can find a discussion in [Griffiths-Harris, Principles of Algebraic Geometry, Chapter 5 section 4]. The starting point of the proof is that the global sections $H^0(X, \mathscr{F}(k))$ generate each $\mathcal{O}_X$-module $\mathscr{F}(k)_x$ $(x \in X)$ for $k$ big enough. $\endgroup$
    – Francesco Polizzi
    Aug 3, 2011 at 15:25
  • 6
    $\begingroup$ Deleted my answer, and added comment instead. Actually this question, given its rather elementary nature, should have been posted at math.stackexchange.com $\endgroup$
    – Francesco Polizzi
    Aug 3, 2011 at 15:32
  • 4
    $\begingroup$ Francesco, while I agree that this is pretty standard, I'm not quite sure that I would call it Hibert's syzygy theorem exactly -- you do need Auslander-Buschsbaum-Serre to stop the process. $\endgroup$
    – Donu Arapura
    Aug 3, 2011 at 15:44
  • 3
    $\begingroup$ I on the other hand feel that I must disagree with Donu. Applying Hilbert's syszygy theorem to a (finitely generated) graded module whose $\widetilde{(-)}$ is the given $\mathbb F$ gives exactly what the OP asked (by applying $\widetilde{(-)}$ to the resolution). Auslander-Buchsbaum-Serre is generalising Hilbert's theorem to more general situations. $\endgroup$
    – Torsten Ekedahl
    Aug 3, 2011 at 16:29

0

Reset to default

Your Answer

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

Browse other questions tagged or ask your own question.