Linearly equivariant prime divisors on a curve

Question

Let $C$ be an algebraic curve, and $P_1,P_2$ be prime divisors on $C$. I would like to prove : $$P_1\sim P_2\Rightarrow C \text{ is rational.}$$ I tried to use Riemann-Roch theorem to prove that the genus is $0$, without success.

Answer 1

If $P_1 \sim P_2$ there exists a rational function on $C$ with a simple zero at $P_1$, a simple pole at $P_2$, and no other zeroes or poles. This function is a degree one map from $C$ onto ${\mathbb P}^1$, so it is bijective and thus is an isomorphism.