Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
@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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.