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If $X$, $Y$ are irreducible smooth projective curves over an algebraically closed field and $\alpha:X\rightarrow Y$ is a morphism, how do we prove that $\mathrm{deg}(\alpha)=\Sigma_{x:\alpha(x)=y}(e_x(\alpha))$ for any $y\in Y$ (where $e_x(\alpha)$ denotes ramification degree)?

This is a really important theorem for calculating the degrees of morphisms, but I haven't been able to find a proof anywhere! I've looked in Hulek, Reid, Hartshorne, Shafarevich, ed., 'Algebraic Geometry', and Shafarevich, 'Basic Algebraic Geometry', 2nd ed., but without success.

Many thanks for any help with this!

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@QiL: I have seen this theorem called the finiteness theorem! But I've now changed the title to make it clearer. – Harry Macpherson Jan 17 '13 at 15:31
When $K=\mathbb{C}$, we can do this by the theory of Riemann surfaces. But is there a more algebraic-geometry proof? – Harry Macpherson Jan 17 '13 at 15:32

Note that the pull-back $\alpha^*[y]$ of the divisor $[y]$ on $Y$ is the divisor $\sum_{x\in \alpha^{-1}(y)} e_{x}(\alpha)[x]$ on $X$.

So what you want is $\deg (\alpha^*[y])=\deg\alpha$ or, more generally, for any divisor $D$ (instead of $[y]$) on $Y$, $$ \deg(\alpha^*D)=(\deg \alpha)\deg D.$$

You can find this equaliy for instance in "Algebraic geometry and arithmetic curves", Prop. 7.3.8.

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