# Proj construction and ample dualizing sheaf

my question concerns a smooth projective variety $X$ with dualizing sheaf $\omega_X$: if I have that this dualizing sheaf is ample, then I have read you can conclude that

$X\simeq Proj(\oplus_{k} H^{o}(X,\omega_X^{k}))$

as projective varieties.

Can someone explain why this is the case and perhaps give me some reference apart from Hartshorne where these projective things are treated?

Thanks

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The dualizing sheaf tends not to be ample (for a smooth projective scheme, it's the canonical sheaf, and by the adjunction formula that has a big non-ample $\mathcal{O}(-n-1)$ in it). Do you have a reference for this claim? – Akhil Mathew Aug 17 '11 at 3:58
In general it isn't, but a number of theorems by Bondal and Orlov, for example in mi.ras.ru/~orlov/papers/Compositio2001.pdf work with this assumption. – Descartes Aug 17 '11 at 7:15

I do believe that this question is not treated in Hartshorne. The correct statement is the following one:

Proposition: Over a complex smooth projective variety $X$, then for any sufficiently ample line bundle $L$ over $X$, we have $X \simeq \mathrm{Proj}(\oplus H^0(X,kL))$.

Proof.

To see this, you may suppose that the global sections of $L$ induce an embedding $i:X \rightarrow \mathbb P(H^0(X,L)^*)=\mathbb P^N$. The image of $X$ under this embedding is given by an ideal sheaf $I$ (which remains to be determined), and thus $X \simeq \mathrm{Proj} (\mathbb C[x_0, \ldots, x_N]/I)$. Notice also that $L \simeq i^* \mathcal O_{\mathbb P^n(1)}$.

Now there is a natural morphism of graded rings $\mathbb C[x_0, \ldots, x_N] \to \oplus H^0(X,kL)$ sending $x_i$ to $s_i$, for $(s_0, \ldots, s_N)$ a basis of $H^0(X,L)$. Its kernel is by definition the graded ideal $I$, and it just remains to prove the surjectivity, which comes from the exact sequence of sheaves

$0 \to I \otimes \mathcal O_{\mathbb P^N}(m) \to \mathcal O_{\mathbb P^N}(m) \to \mathcal O_{i(X)}(m)\to 0$

We obtain then a surjective map $H^0(\mathbb P^N, \mathcal O_{\mathbb P^N}(m)) \to H^0(X, mL)$ for any $m$ sufficiently large, thanks to Serre's vanishing theorem. Now you may change $L$ with one of its multiple, and you're done.

Remark: In this, we have hidden the (non-trivial) fact that for $L$ ample, its graded ring of section is finitely generated. This is proved in Lazarsfeld's "Positivity in Algebraic Geometry, I".

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I don't really understand that answer: first: what is a "suffifciently ample" line bundle? Next: why is the dimension of $H^{o}(X,L)$ the same as N? Furthermore: I don't see why $I$ is the kernel, because by Hartshorne, II,5.16 the ideal I is $\Gamma_{*}(J)$ with $J$ the ideal of $X$. – Descartes Aug 18 '11 at 15:28
And how do you get $H^{o}(X,O_{i(X)}(m))=H^{o}(X,mL)$? – Descartes Aug 18 '11 at 15:30
Sufficiently ample means a sufficiently high multiple of the line bundle. Moreover, $N$ is defined to be the dimension of $H^0(X,L)$. If $i(X)$ is defined by (homogeneous) polynomial equations $P_1=0, \ldots, P_r=0$, then the kernel of the aforesaid map consists in all the algebraic relations between the $s_i$'s, thus it is precisely given by $P_1(s_0, \ldots, s_N)=0, \ldots, P_r(s_0, \ldots, s_N)=0$, because the $s_i$'s are coordinates on $i(X)$. Finally, as $i$ is finite, $H^0(X,mL)=H^0(X,i^* O_{\mathbb P^N}(m)) =H^0(i(X), \mathcal O_{\mathbb P^N}(m))$ – Henri Aug 18 '11 at 16:28
thanks a lot, Henri – Descartes Aug 18 '11 at 18:05