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This is exercise 19.4.B on Ravi Vakil's notes.

Let $C$ be a regular plane curve of degree $e>2$, and $D_1,D_2$ be two plane curves of same degree $d$ not containing $C$. By Bezout's theorem $D_i$ and $C$ meet at $de$ points. Suppose both $D_i$ meet $C$ at the same $de-1$ points. Show that the remaining point is the same as well.

In other words, let $E$ be a divisor on $C$ of degree $de-1$ such that $D_i\cap C=E+p_i$ for some closed points $p_1,p_2$ of degree 1. Show $p_1=p_2$.

I think I'm supposed to use 19.4.2 which says if $C$ is not isomorphic to $\mathbb{P}^1$, then $\mathscr{O}_C(p_1)\cong\mathscr{O}_C(p_2)$ iff $p_1=p_2$. We know $C$ is not $\mathbb{P}^1$ in this case since $e>2$. It would be great to show $\mathscr{O}_C(E+p_1)\cong\mathscr{O}_C(E+p_2)$ but I don't know how?

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Since $D_1$ and $D_2$ are linearly equivalent as divisors ($\mbox{Pic}(\mathbb{P}^2)\simeq\mathbb{Z}$), we have that their restrictions to $C$ are also linearly equivalent (restrict the rational function that gives the linear equivalence). This means that $E+p_1$ is linearly equivalent to $E+p_2$, and so $\mathcal{O}_C(E+p_1)\simeq\mathcal{O}_C(E+p_2)$.

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  • $\begingroup$ Thank you. Can you add a little about why the restriction/pullback of $\mathscr{O}_{\mathbb{P}^2}(D_i)$ to $C$ is $\mathscr{O}_C(D_i\cap C)$? It makes sense intuitively but can't show rigorously. $\endgroup$ – Gazerun Dec 25 '13 at 17:45
  • $\begingroup$ You can define the restriction of $\mathcal{O}_{\mathbb{P}^2}(D_i)$ to $C$ by $\mathcal{O}_C\otimes\mathcal{O}_{\mathbb{P}^2}(D_i)$, and if $D_i$ meets $C$ transversely, for instance, this sheaf has the same support as $\mathcal{O}_C(D_i\cap C)$ and so they must be equal. $\endgroup$ – rfauffar Dec 25 '13 at 18:04
  • $\begingroup$ Okay, I interpreted your answer as this: The sections of $\mathscr{O}_C\otimes\mathscr{O}_{\mathbb{P}^2}(D_i)$ over $U$ can be described as rational sections $f$ of $\mathscr{O}_{\mathbb{P}^2}$ such that $\mathrm{div}|_{V} f+ D_i|V\geq0$ for any open $V\subset \mathbb{P}^2$ containing $U$, modded by rational sections vanishing at $C$. Conversely, the sections of $\mathscr{O}_C(E+p_i)$ over $U$ are rational sections $f$ of $\mathscr{O}_{\mathbb{P}^2}$ modded out by rational sections vanishing at $C$, such that the image $\overline{f}$ has $\mathrm{div}|_{U} \overline{f}+ E+p_i\geq0$. $\endgroup$ – Gazerun Dec 26 '13 at 6:00
  • $\begingroup$ So I guess the question is whether taking div and the modding out is the same as modding out first and then taking div. i.e. that $\mathrm{div}|_V f+D_i|_V\geq0$ iff $\mathrm{div}|_U\overline{f}+E+p_i\geq0$. Is this correct? Or is there a better way to think about this? $\endgroup$ – Gazerun Dec 26 '13 at 6:11
  • $\begingroup$ Take the exact sequence $0\to\mathcal{O}(D_i-C)\to\mathcal{O}_{\mathbb{P}^2}(D_i)\to\mathcal{O}_C\otimes\mathcal{O}_{\mathbb{P}^2}(D_i)\to 0$. The map $\mathcal{O}_{\mathbb{P}^2}(D_i)(\mathbb{P}^2)\to(\mathcal{O}_C\otimes\mathcal{O}_{\mathbb{P}^2}(D_i))(\mathbb{P}^2)$ is surjective, and so a global section of $\mathcal{O}_C\otimes\mathcal{O}_{\mathbb{P}^2}$ is just a global section of $\mathcal{O}_{\mathbb{P}^2}(D_i)$ mod the ideal sheaf of $C$. One such global section would be the polynomial that defines $D_i$. $\endgroup$ – rfauffar Dec 26 '13 at 11:09

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