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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!

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2  
I am no expert here, but Hartshorne Exercise III 6.8 and Auslander-Buchsbaum formula seem to do the trick (for more general situation). –  Hailong Dao Aug 3 '11 at 15:00
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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. –  Francesco Polizzi Aug 3 '11 at 15:25
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Deleted my answer, and added comment instead. Actually this question, given its rather elementary nature, should have been posted at math.stackexchange.com –  Francesco Polizzi Aug 3 '11 at 15:32
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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. –  Donu Arapura Aug 3 '11 at 15:44
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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. –  Torsten Ekedahl Aug 3 '11 at 16:29
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