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Let $X$ be a complex projective surface, and let $D \in \mathrm{Div}(X)$. Assume that $D=aE$ for some $a \in \mathbb N$, where $E \in \mathrm{Div}(X)$ is a divisor such that $|E|$ is a pencil (i.e. $h^0(E)=2$) with empty base locus.

Is it true or false that $\dim |D| = h^0(D) - 1 = a$?

Francesco Genovese
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  • The linear system $|E|$ gives a morphism $f:X \rightarrow \mathbf{P}^1$ such that $E=f^* O(1)$. Hence $D=f^* O(a)$, and the issue is to show that all sections of $D$ are pullbacks of sections of $O(a)$. It's easy to see by intersection numbers that any such section must be supported in a union of fibres. It's a bit more delicate, but still not too hard, I think, to check that actually such a section must be a union of fibres, and that the coefficients match up correctly, so that it really does come from a section of $O(a)$. –  Sep 09 '13 at 08:58
  • By the way, if the earlier comment isn't clear, please let me know! –  Sep 09 '13 at 14:29
  • @AsalBeagDubh Can you explain a little bit? I could not catch your idea well. Thanks –  Apr 13 '14 at 09:44
  • Sorry I am a beginner, how do we imply |E| is a pencil from $h^0(O_X(E))=2$? And free base locus? –  Apr 13 '14 at 09:51

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