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I am reading Hartshorne's book chapter 5 (on surface) and I have a question:

on page 371, proposition 2.3, it says:

Let $X$ be surface, $C$ curve, $\pi:X\to C$ ruled surface. $f$ be a fiber, $\sigma$ be a section of $\pi$, $C_0=\sigma(C)$. Let $D$ be a divisor on $X$, $D.f$=n, and set $D'=D-nC_0$. Then he claims:

$L(D')\cong \pi^*\pi_* L(D')$

I don't know how this comes from? Can someone helps me? Thanks in advance.

User X
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1 Answers1

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By Lemma 2.1 of the same section, $\pi_*(L(D'))$ is a locally free sheaf of rank $D'.f +1= D.f-nC_0.f +1= n-n+1=1$. By the counit of the adjunction $f^*\leftrightarrows f_*$, we have a map $\pi^*\pi_*L(D') \to L(D')$ - note that both of these object are line bundles on $X$. By the adjunction formula, we have that $\operatorname{Hom}_{\mathcal{O}_X}(\pi^*\pi_*L(D'),L(D'))\cong \operatorname{Hom}_{\mathcal{O}_C}(\pi_*L(D'),\pi_*L(D'))$. Since $\pi_*L(D')$ is a line bundle on $C$, our map has to be a global section of $\mathcal{O}_C$. But $C$ is a projective curve, so therefore the map is either $0$ or an isomorphism. It is clear the map is not $0$, so it is therefore an isomorphism.

KReiser
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    Isn't $L(D')$ a sheaf on $X$? – Andrew Jul 17 '18 at 05:34
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    @Andrew fixed! Thank you for your attention to detail. – KReiser Jul 17 '18 at 07:19
  • Since $\pi_* L(D')$ is a line bundle on $C$, our map has to be a global section of $O_C$ - I don't understand this part. How do global sections of $O_C$ correspond to morphisms $\pi^* \pi_* L(D') \rightarrow L(D')$ since those invertible sheaves are on $X$? – David Lui Sep 20 '22 at 23:59
  • @DavidLui The automorphisms of a line bundle are exactly the global sections of the structure sheaf, because $\operatorname{Hom}(A\otimes L,B)\cong\operatorname{Hom}(A,B\otimes L^{-1})$ for any line bundle $L$ and arbitrary $\mathcal{O}$-modules $A,B$ by clever use of the tensor hom adjunction. – KReiser Sep 21 '22 at 02:22