TITLE: A generalization of Cayley-Bacharach Theorem
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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 $\mathcal{O}_C(E+p_1)\simeq\mathcal{O}_C(E+p_2)$.