Lemma 42.25.3. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$, $Y$ be locally of finite type over $S$. Let $f : X \to Y$ be a proper morphism. Let $\mathcal{L}$ be an invertible sheaf on $Y$. Let $s$ be a nonzero meromorphic section $s$ of $\mathcal{L}$ on $Y$. Assume $X$, $Y$ integral, $f$ dominant, and $\dim _\delta (X) = \dim _\delta (Y)$. Then

$f_*\left(\text{div}_{f^*\mathcal{L}}(f^*s)\right) = [R(X) : R(Y)]\text{div}_\mathcal {L}(s).$

as cycles on $Y$. In particular

$f_*(c_1(f^*\mathcal{L}) \cap [X]) = c_1(\mathcal{L}) \cap [Y].$

Proof. The last equation follows from the first since $f_*[X] = [R(X) : R(Y)][Y]$ by definition. It turns out that we can re-use Lemma 42.18.1 to prove this. Namely, since we are trying to prove an equality of cycles, we may work locally on $Y$. Hence we may assume that $\mathcal{L} = \mathcal{O}_ Y$. In this case $s$ corresponds to a rational function $g \in R(Y)$, and we are simply trying to prove

$f_*\left(\text{div}_ X(g)\right) = [R(X) : R(Y)]\text{div}_ Y(g).$

Comparing with the result of the aforementioned Lemma 42.18.1 we see this true since $\text{Nm}_{R(X)/R(Y)}(g) = g^{[R(X) : R(Y)]}$ as $g \in R(Y)^*$. $\square$

Comment #5534 by on

In the statement of the Lemma 02ST, I believe the last line should be $f_*(c_1(f^*L)∩[X])=c_1(L)∩[Y]$.

Comment #6641 by WhatJiaranEatsTonight on

In the statement of the Lemma 02ST, The last line is $f_*(c_1(f^*L)\cap[X])=[R(X):R(Y)]c_1(L)\cap[Y]$. I think the equation missed the coefficient.

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