I am reading the proof of Castelnuovo's contractibility criterion ( Hartshorne's Algebraic geometry). In the proof, he claim that there must be a very ample divisor $H$ on a projective surface $X$ such that $H^1(X,\mathscr L(H))=0$.

By Serre's theorem

Let $X$ be a projective scheme over a noetherian ring $A$, $\mathcal O_X(1)$ a very ample invertible sheaf on $X$ over $Spec A$ , and $\mathscr F$ a coherent sheaf on $X$. Then there is an integer $n_0$ such that for each $ i>0 $ and each $n\geqslant n_0$, $H^i(X,\mathscr F(n))=0$

Here $X$ is a projective surface over an algebraically closed field $\mathrm k$, a divisor $H$ corresponds to an invertible sheaf $\mathscr L(H)$, so, does he use $\mathscr L(H)\otimes_{\mathcal O_X}\mathcal O_X(n)\cong \mathscr L(nH)$ to give that claim ? I am not sure. But why we have this isomorphism ? Thank you!

  • $\begingroup$ We can take $H$ to be the divisor corresponding to the invertible sheaf $\mathcal O_X(n)$ for a large enough $n$. $\endgroup$ – danneks Apr 23 '18 at 6:03
  • $\begingroup$ @danneks Let $\mathscr F=\mathcal O_X$,$H=\mathscr F(n)=\mathcal O_X\otimes_{\mathcal O_X} \mathcal O_X(n)\cong \mathcal O_X(n)$ for a sufficiently large $n$? $\endgroup$ – Jiabin Du Apr 23 '18 at 8:50
  • $\begingroup$ Yes, the theorem says that higher cohomology of this sheaf are zero. $\endgroup$ – danneks Apr 24 '18 at 6:14
  • $\begingroup$ @danneks thank you $\endgroup$ – Jiabin Du Apr 24 '18 at 6:41

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