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I'm starting with $X=\mathbb{P}^2(\mathbb{C})$ and a cubic curve $B \subset X$ and a flex $P$ on $B$ such that for a hyperplane section $H$ I have $3P \sim dH\vert_B$ (where $d \in \mathbb{N}$). With these notations I'm considering the exact sequence of sheaves

$0 \rightarrow \mathcal{O}_X(-2) \rightarrow \mathcal{O}_X(1) \rightarrow \mathcal{O}_B(dH \vert_B) \rightarrow 0$

In the case where $d=1$ (and of course when $d=2$ too), passing to the long exact sequence associated to it I get the isomorphism $H^0(X, O_X(1)) \cong H^0(B,\mathcal{O}_B(dH \vert_B))$. I have then two questions:

  1. Is this correct? I mean, is the natural exact sequence obtained by the restriction map from $\mathcal{O}(1)$ to the rational functions on the cubic with at most a triple pole on $P$ and is the isomorphism I get from the long exact sequence precisely the restrction again?
  2. Does it follow from this (case $d=1$) that there is only one line meeting the cubic only at $P$? (Here I can choose $P$ such taht it has order 3 with respect to the addition in $P$ obviously).

Thank you all

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Dear Srks, What do you mean when you write $dH_{| B}$; in particular, what is the role of $d$ in this notation? Assuming that your notation means what I think it means, then the answers to your questions are "yes". Regards, – Matt E Sep 21 '11 at 13:09
Dear Matt, here d is a natural number. So $d H \vert_B$ means the divisor on $B$ equivalent to $3P$ where $P$ is a point of order $3d$ in the cubic $B$. For the isomorphism of the zero-cohomological groups and in the second question $d$ is set to be 1 (I've edit the question). I'm thinking that both the answer are yes but I can't undestand in detail why, specially for the second one. Thanks – Srks Sep 24 '11 at 14:43

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