Equivalence of points on a smooth cubic in $\mathbb{P}^2$ Let $C \subset \mathbb{P}^2 (\mathbb{C})$ be a smooth cubic. Show that $(p)$~$(q)$ if and only if $p=q$.
$p$ and $q$ are points, and two divisors $D$ and $D'$ are ~ if $D-D' = (f)$ for some meromorphic function $f$. I have to show there is no $f$ with one pole and one zero.
I heavily suspect Riemann-Roch is involved in this exercise, but I can't find a way to conclude.

If there were such a meromorphic function $f$, we could extend it to a holomorphic map $F: C \rightarrow \mathbb{P}^{1}$ (the pole mapped to the infinity point). This map clearly has degree $1$, and it is surjective by the irreducibility of the projective line. Therefore, it is an isomorphism. This is a contradiction, because the projective line has genus $0$ and the smooth plane cubic $C$ has genus $1$.
