Lemma 81.19.1. In Situation 81.2.1 let $X, Y/B$ be good. Let $f : X \to Y$ be a flat morphism of relative dimension $r$. Let $\mathcal{L}$ be an invertible sheaf on $Y$. Assume $Y$ is integral and $n = \dim _\delta (Y)$. Let $s$ be a nonzero meromorphic section of $\mathcal{L}$. Then we have

\[ f^*\text{div}_\mathcal {L}(s) = \sum n_ i\text{div}_{f^*\mathcal{L}|_{X_ i}}(s_ i) \]

in $Z_{n + r - 1}(X)$. Here the sum is over the irreducible components $X_ i \subset X$ of $\delta $-dimension $n + r$, the section $s_ i = f|_{X_ i}^*(s)$ is the pullback of $s$, and $n_ i = m_{X_ i, X}$ is the multiplicity of $X_ i$ in $X$.

**Proof.**
Using sleight of hand we will deduce this from Lemma 81.16.1. (An alternative is to redo the proof of that lemma in the setting of meromorphic sections of invertible modules.) Namely, 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 $X$. Then it suffices to prove

\[ (q')^*f^*\text{div}_\mathcal {L}(s) = \sum n_ i (q')^*\text{div}_{f^*\mathcal{L}|_{X_ i}}(s_ i) \]

Observe that $T'_ i = q^{-1}(X_ i)$ are the irreducible components of $T'$ and that $n_ i$ is the multiplicity of $T'_ i$ in $T'$. The left hand side is equal to

\[ h^*q^*\text{div}_\mathcal {L}(s) = h^*\text{div}_ T(q^*(s)) \]

by Lemma 81.17.4 (and Lemma 81.10.4). On the other hand, denoting $q'_ i : T'_ i \to X_ i$ the restriction of $q'$ we find that Lemma 81.17.4 also tells us the right hand side is equal to

\[ \sum n_ i \text{div}_{T_ i}((q'_ i)^*(s_ i)) \]

In these two formulas the expressions $q^*(s)$ and $(q'_ i)^*(s_ i)$ represent the rational functions corresponding to the pulled back meromorphic sections of $q^*\mathcal{L}$ and $(q'_ i)^*f^*\mathcal{L}|_{X_ i}$ via the isomorphism $\alpha : q^*\mathcal{L} \to \mathcal{O}_ T$ and its pullbacks to spaces over $T$. With this convention it is clear that $(q'_ i)^*(s_ i)$ is the composition of the rational function $q^*(s)$ on $T$ and the morphism $h|_{T'_ i} : T'_ i \to T$. Thus Lemma 81.16.1 exactly says that

\[ h^*\text{div}_ T(q^*(s)) = \sum n_ i \text{div}_{T_ i}((q'_ i)^*(s_ i)) \]

as desired.
$\square$

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