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 $g \in R(Y)^*$.
$\square$

## Comments (2)

Comment #5534 by Chunhui Liu on

Comment #5723 by Johan on