Lemma 81.19.3. In Situation 81.2.1 let $X, Y/B$ be good. Let $f : X \to Y$ be a proper morphism. Let $\mathcal{L}$ be an invertible sheaf on $Y$. Assume $X$, $Y$ integral, $f$ dominant, and $\dim _\delta (X) = \dim _\delta (Y)$. Let $s$ be a nonzero meromorphic section $s$ of $\mathcal{L}$ on $Y$. 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

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

Proof. The last equation follows from the first since $f_*[X] = [R(X) : R(Y)][Y]$ by definition. Proof of the first equaltion. Let $q : T \to Y$ be the morphism of Lemma 81.17.3 constructed using $\mathcal{L}$ on $Y$. We will use all the properties of $T$ stated in this lemma. Consider the fibre product diagram

$\xymatrix{ T' \ar[r]_{q'} \ar[d]_ h & X \ar[d]^ f \\ T \ar[r]^ q & Y }$

Then $q' : T' \to X$ is the morphism constructed using $f^*\mathcal{L}$ on $X$. It suffices to prove the equality after pulling back to $T'$. The left hand side pulls back to

\begin{align*} q^*f_*\left(\text{div}_{f^*\mathcal{L}}(f^*s)\right) & = h_*(q')^*\text{div}_{f^*\mathcal{L}}(f^*s) \\ & = h_*\text{div}_{(q')^*f^*\mathcal{L}}((q')^*f^*s) \\ & = h_*\text{div}_{h^*q^*\mathcal{L}}(h^*q^*s) \end{align*}

The first equality by Lemma 81.11.1. The second by Lemma 81.19.1 using that $T'$ is integral. The third because the displayed diagram commutes. The right hand side pulls back to

$[R(X) : R(Y)]q^*\text{div}_\mathcal {L}(s) = [R(T') : R(T)]\text{div}_{q^*\mathcal{L}}(q^*s)$

This follows from Lemma 81.19.1, the fact that $T$ is integral, and the equality $[R(T') : R(T)] = [R(X) : R(Y)]$ whose proof we omit (it follows from Lemma 81.11.1 but that would be a silly way to prove the equality). Thus it suffices to prove the lemma for $h : T' \to T$, the invertible module $q^\mathcal {L}$ and the section $q^*s$. Since $q^*\mathcal{L} = \mathcal{O}_ T$ we reduce to the case where $\mathcal{L} \cong \mathcal{O}$ discussed in the next paragraph.

Assume that $\mathcal{L} = \mathcal{O}_ Y$. In this case $s$ corresponds to a rational function $g \in R(Y)$. Using the embedding $R(Y) \subset R(X)$ we may think of $g$ as a rational on $X$ 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 Lemma 81.14.1 we see this true since $\text{Nm}_{R(X)/R(Y)}(g) = g^{[R(X) : R(Y)]}$ as $g \in R(Y)^*$. $\square$

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