I'm studying intersection of curves with a fixed plane cubic, the first case I consider is of course lines, in particular lines intersecting the cubic at only one point. The problem is quite easy and I've solved it in terms of basic intersection theory on $\mathbb{P}^2$. Now I'm considering a new way of attacking the problem because it's easier to generalize to higher degrees but I'm a little bit unsure if I'm moving to the right direction.

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 $3dP \sim dH\vert_B$ (where $d \in \mathbb{N}$). With these notations I'm considering the exact sequence of sheaves

$0 \rightarrow \mathcal{O}_X(d-3) \rightarrow \mathcal{O}_X(d) \rightarrow \mathcal{O}_B(dH \vert_B) \rightarrow 0$

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 restriction again?
  2. Does it follow from this (case $d=1$ and $d=2$) that there is only one line (respectively a conic) meeting the cubic only at $P$? (Here I can choose $P$ such taht it has order 3 with respect to the addition in $B$ obviously)?

I'm thinking is it obvious regarding to the geometry but trying to writing down explicitly I'm getting a little bit of problems so I'm wondering if I'm misunderstanding something.

Thank you all

  • $\begingroup$ If $3P \sim dH_{|B}$ then $d = 1$, since the degree of the LHS is 3, while the degree of the RHS is 3d. The line touching $B$ with multiplicity 3 at a given flex point is unique. $\endgroup$ – Sasha Sep 25 '11 at 4:08
  • $\begingroup$ Sasha you're right, I've forgotten a d in $3dP$ (fixed). Then for the second sentence you're right again but I would like to have hints about an explicit derivation of that form the isomorphism of $H^0$... thanks anyway $\endgroup$ – Srks Sep 25 '11 at 9:46
  1. is ok. For 2: I don't think you can prove the existence of flexes this way, but assuming P is a flex, then unicity of the line does follow from the iso. BTW, the unique "conic" that intersects B in 6P (P a flex) is the tangent line, doubled.
  • $\begingroup$ BTW: I don't know offhand if there exist other points P (not flexes) where there is an irreducible conic Q with Q.B=6P. My guess is yes. $\endgroup$ – quim Sep 26 '11 at 9:42
  • $\begingroup$ One can find such conics passing through any point of order $6$ on $B$ (where we choose a flex as the zero for the group law). $\endgroup$ – ulrich Sep 26 '11 at 10:55
  • $\begingroup$ thank you quim for your answer. For my problem I can assume $P$ to be a point of order $3d$ for all $d$ so I have no problems about flexes. Anyway I can see the existence form the isomorphism of $H^0$ but how can you show that is unique? $\endgroup$ – Srks Sep 26 '11 at 12:04
  • 1
    $\begingroup$ The point is that you know $H|_B=3P$, and then the isomorphism tells you that $3P\in |3P|$, being a single divisor, has a single divisor in the preimage. In other words, the isomorphism in $H^0$ induces an isomorphism in the projectivizations $|H|={\mathbb P}(H^0({\mathcal O}_{X}(d)))\cong {\mathbb P}(H^0({\mathcal O}_{B}(3P)))=|3P|$. $\endgroup$ – quim Sep 26 '11 at 13:35
  • $\begingroup$ @quim Thank you! this is just what I was missing. $\endgroup$ – Srks Sep 27 '11 at 9:35

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