Lemma 42.17.2. Let (S, \delta ) be as in Situation 42.7.1. Let X, Y be locally of finite type over S. Assume X, Y are integral and n = \dim _\delta (Y). Let f : X \to Y be a flat morphism of relative dimension r. Let g \in R(Y)^*. Then
f^*(\text{div}_ Y(g)) = \text{div}_ X(g)
in Z_{n + r - 1}(X).
Proof.
Note that since f is flat it is dominant so that f induces an embedding R(Y) \subset R(X), and hence we may think of g as an element of R(X)^*. Let Z \subset X be an integral closed subscheme of \delta -dimension n + r - 1. Let \xi \in Z be its generic point. If \dim _\delta (f(Z)) > n - 1, then we see that the coefficient of [Z] in the left and right hand side of the equation is zero. Hence we may assume that Z' = \overline{f(Z)} is an integral closed subscheme of Y of \delta -dimension n - 1. Let \xi ' = f(\xi ). It is the generic point of Z'. Set A = \mathcal{O}_{Y, \xi '}, B = \mathcal{O}_{X, \xi }. The ring map A \to B is a flat local homomorphism of Noetherian local domains of dimension 1. We have g in the fraction field of A. What we have to show is that
\text{ord}_ A(g) \text{length}_ B(B/\mathfrak m_ AB) = \text{ord}_ B(g).
This follows from Algebra, Lemma 10.52.13 (details omitted).
\square
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