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Let $X$ be a smooth variety, $x$ a closed point, and $\mathcal{O}_x$ the skyscraper sheaf of the residue field $k(x)$ at $x$. Recall that the homological dimension of a complex of coherent sheaves $\mathcal{F}^\bullet$ is the smallest $n$ such that there exists a quasi-isomorphism between $\mathcal{F}^\bullet$ and a complex of locally free sheaves of length $n+1$. (A quasi-isomorphism of complexes is a morphism such that the induced homology maps are isomorphisms.)

How can one show that the homological dimension of $\mathcal{O}_x$ (considered as a complex of length one) is equal to the dimension $n$ of $X$?

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Can you do this locally? There it becomes a statement about rings and modules... –  Mariano Suárez-Alvarez Nov 18 '12 at 19:53
    
Well, according to (Hartshorne, Ex. 6.5, (c)), the homological dimension of a sheaf is equal to the supremum of the projective dimensions of the stalks; in this case, it is equal to the projective dimension of $k(x)$ viewed as an $\mathcal{O}_{X,x}$-module. I don't know how to calculate this, though. –  Adeel Nov 18 '12 at 20:30
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Forget about schemes and what not and do the affinecase: rings and modules. –  Mariano Suárez-Alvarez Nov 18 '12 at 20:51

1 Answer 1

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As in the comments above we need to compute the projective dimension of $k(x)$, which is easily seen to be equal to the global dimension of the stalk $\mathcal{O}_{X,x}$. Since $X$ is smooth, this is a regular ring and by a theorem of Serre [S, Théorème 1] its global dimension is the same as its Krull dimension, which is of course equal to the dimension of $X$.

[S] Serre, Jean-Pierre. Sur la dimension homologique des anneaux et des modules noethériens. Proceedings of the international symposium on algebraic number theory, Tokyo & Nikko, 1955, pp. 175–189. Science Council of Japan, Tokyo, 1956.

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