Lemma 42.20.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$, $Y$ be schemes locally of finite type over $S$. Assume $Y$ integral with $\dim _\delta (Y) = k$. Let $f : X \to Y$ be a flat morphism of relative dimension $r$. Then for $g \in R(Y)^*$ we have

$f^*\text{div}_ Y(g) = \sum n_ j i_{j, *}\text{div}_{X_ j}(g \circ f|_{X_ j})$

as $(k + r - 1)$-cycles on $X$ where the sum is over the irreducible components $X_ j$ of $X$ and $n_ j$ is the multiplicity of $X_ j$ in $X$.

Proof. Let $Z \subset X$ be an integral closed subscheme of $\delta$-dimension $k + r - 1$. We have to show that the coefficient $n$ of $[Z]$ in $f^*\text{div}(g)$ is equal to the coefficient $m$ of $[Z]$ in $\sum i_{j, *} \text{div}(g \circ f|_{X_ j})$. Let $Z'$ be the closure of $f(Z)$ which is an integral closed subscheme of $Y$. By Lemma 42.13.1 we have $\dim _\delta (Z') \geq k - 1$. Thus either $Z' = Y$ or $Z'$ is a prime divisor on $Y$. If $Z' = Y$, then the coefficients $n$ and $m$ are both zero: this is clear for $n$ by definition of $f^*$ and follows for $m$ because $g \circ f|_{X_ j}$ is a unit in any point of $X_ j$ mapping to the generic point of $Y$. Hence we may assume that $Z' \subset Y$ is a prime divisor.

We are going to translate the equality of $n$ and $m$ into algebra. Namely, let $\xi ' \in Z'$ and $\xi \in Z$ be the generic points. Set $A = \mathcal{O}_{Y, \xi '}$ and $B = \mathcal{O}_{X, \xi }$. Note that $A$, $B$ are Noetherian, $A \to B$ is flat, local, $A$ is a domain, and $\mathfrak m_ AB$ is an ideal of definition of the local ring $B$. The rational function $g$ is an element of the fraction field $Q(A)$ of $A$. By construction, the closed subschemes $X_ j$ which meet $\xi$ correspond $1$-to-$1$ with minimal primes

$\mathfrak q_1, \ldots , \mathfrak q_ s \subset B$

The integers $n_ j$ are the corresponding lengths

$n_ i = \text{length}_{B_{\mathfrak q_ i}}(B_{\mathfrak q_ i})$

The rational functions $g \circ f|_{X_ j}$ correspond to the image $g_ i \in \kappa (\mathfrak q_ i)^*$ of $g \in Q(A)$. Putting everything together we see that

$n = \text{ord}_ A(g) \text{length}_ B(B/\mathfrak m_ AB)$

and that

$m = \sum \text{ord}_{B/\mathfrak q_ i}(g_ i) \text{length}_{B_{\mathfrak q_ i}}(B_{\mathfrak q_ i})$

Writing $g = x/y$ for some nonzero $x, y \in A$ we see that it suffices to prove

$\text{length}_ A(A/(x)) \text{length}_ B(B/\mathfrak m_ AB) = \text{length}_ B(B/xB)$

(equality uses Algebra, Lemma 10.52.13) equals

$\sum \nolimits _{i = 1, \ldots , s} \text{length}_{B/\mathfrak q_ i}(B/(x, \mathfrak q_ i)) \text{length}_{B_{\mathfrak q_ i}}(B_{\mathfrak q_ i})$

and similarly for $y$. As $A \to B$ is flat it follows that $x$ is a nonzerodivisor in $B$. Hence the desired equality follows from Lemma 42.3.2. $\square$

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