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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.

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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).

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Thanks a lot, that helps! – Evariste Mar 11 '12 at 0:46

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