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.

Let $S_1$ and $S_2$ be smooth projective surfaces over $\mathbb{C}$ and let $f: S_1 \to S_2$ be a morphism which is generically finite of degree $d$. How does one prove that $f_* f^* D = dD$ for all divisors $D$ on $S_2$? This shouldn't be hard, but I see no obvious way to do this.

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

As divisors are determined by their multiplicities at codimension $1$ points, it is enough to compare $f_*f^*D$ and $dD$ at such points. But above such points $f$ is finite (equivalently, the locus $Z$ in $S_1$ where $f$ is not finite is mapped to a finite subset of $S_2$ because $1\ge \dim Z > \dim f(Z)$), and you are reduced to the case of finite morphism (above a non necessarily projective, but this doesn't matter).

So do you know the proof in the case of finite morphisms ? It is the formula which relates the degree $d$ of a finite extension to the sum of the (ramification index) $\times$ (degree of residue extension).

share|improve this answer
    
Thanks a lot, that helps! –  Evariste Mar 11 '12 at 0:46
add comment

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.