# Why should the fibers of a surjective morphism of curves be finite?

Let $\phi : C_1 \rightarrow C_2$ be a nonconstant (and therefore surjective) morphism of smooth curves, and let $\phi^* : K(C_2)\rightarrow K(C_1)$ be the pullback homomorphism it induces on function fields.

Let $P\in C_1$; the ramification index of $\phi$ at $P$ is defined as $e_\phi (P) = \text{ord}_P (\phi^* (t_{\phi(P)})$, where

• $t_{\phi(P)}\in K(C_2)$ is a uniformiser at $\phi(P)$
• $\text{ord}_P (t) = \sup \left\{n\geq 0: t\in \mathfrak{m}_P^n\right\}$
• $\mathfrak{m}_P$ is the unique maximal ideal of the local ring $K[C_1]_P$

Silverman's Arithmetic of Elliptic Curves states in Prop.2.6 (proof is in Hartshorne) that for every point $Q\in C_2$, the sum over every $P\in\phi^{-1}(\left\{Q\right\})$ of all ramification indices $e_\phi (P)$ is equal to the degree of $\phi$. However it's not immediately clear to me why the fiber $\phi^{-1}(\left\{Q\right\})$ should only contain finitely many points (although I suspect it must be related to the fact that $K(C_1):\phi_* K(C_2)$ is an algebraic extension). Can anyone explain why?

## 2 Answers

If the (closed!) fiber $\phi^{-1}(q)$ of some point $q\in C_2$ were not finite it would have an infinite closed irreducible component.
But the only infinite closed irreducible subset of $C_1$ is $C_1$, so that we would have $\phi (C_1)=q$, which is contrary to the non-constancy hypothesis.

Hint: show that this set is an algebraic variety of dim 0