## 42.17 Principal divisors

The following definition is the analogue of Divisors, Definition 31.26.5 in our current setup.

Definition 42.17.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Assume $X$ is integral with $\dim _\delta (X) = n$. Let $f \in R(X)^*$. The *principal divisor associated to $f$* is the $(n - 1)$-cycle

\[ \text{div}(f) = \text{div}_ X(f) = \sum \text{ord}_ Z(f) [Z] \]

defined in Divisors, Definition 31.26.5. This makes sense because prime divisors have $\delta $-dimension $n - 1$ by Lemma 42.16.1.

In the situation of the definition for $f, g \in R(X)^*$ we have

\[ \text{div}_ X(fg) = \text{div}_ X(f) + \text{div}_ X(g) \]

in $Z_{n - 1}(X)$. See Divisors, Lemma 31.26.6. The following lemma will be superseded by the more general Lemma 42.20.2.

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).
$\square$

## Comments (0)