Lemma 42.18.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Assume $X$ is integral and $n = \dim _\delta (X)$. Let $f \in R(X)^*$. Let $U \subset X$ be a nonempty open such that $f$ corresponds to a section $f \in \Gamma (U, \mathcal{O}_ X^*)$. Let $Y \subset X \times _ S \mathbf{P}^1_ S$ be the closure of the graph of $f : U \to \mathbf{P}^1_ S$. Then

1. the projection morphism $p : Y \to X$ is proper,

2. $p|_{p^{-1}(U)} : p^{-1}(U) \to U$ is an isomorphism,

3. the pullbacks $Y_0 = q^{-1}D_0$ and $Y_\infty = q^{-1}D_\infty$ via the morphism $q : Y \to \mathbf{P}^1_ S$ are defined (Divisors, Definition 31.13.12),

4. we have

$\text{div}_ Y(f) = [Y_0]_{n - 1} - [Y_\infty ]_{n - 1}$
5. we have

$\text{div}_ X(f) = p_*\text{div}_ Y(f)$
6. if we view $Y_0$ and $Y_\infty$ as closed subschemes of $X$ via the morphism $p$ then we have

$\text{div}_ X(f) = [Y_0]_{n - 1} - [Y_\infty ]_{n - 1}$

Proof. Since $X$ is integral, we see that $U$ is integral. Hence $Y$ is integral, and $(1, f)(U) \subset Y$ is an open dense subscheme. Also, note that the closed subscheme $Y \subset X \times _ S \mathbf{P}^1_ S$ does not depend on the choice of the open $U$, since after all it is the closure of the one point set $\{ \eta '\} = \{ (1, f)(\eta )\}$ where $\eta \in X$ is the generic point. Having said this let us prove the assertions of the lemma.

For (1) note that $p$ is the composition of the closed immersion $Y \to X \times _ S \mathbf{P}^1_ S = \mathbf{P}^1_ X$ with the proper morphism $\mathbf{P}^1_ X \to X$. As a composition of proper morphisms is proper (Morphisms, Lemma 29.41.4) we conclude.

It is clear that $Y \cap U \times _ S \mathbf{P}^1_ S = (1, f)(U)$. Thus (2) follows. It also follows that $\dim _\delta (Y) = n$.

Note that $q(\eta ') = f(\eta )$ is not contained in $D_0$ or $D_\infty$ since $f \in R(X)^*$. Hence (3) by Divisors, Lemma 31.13.13. We obtain $\dim _\delta (Y_0) = n - 1$ and $\dim _\delta (Y_\infty ) = n - 1$ from Lemma 42.16.1.

Consider the effective Cartier divisor $Y_0$. At every point $\xi \in Y_0$ we have $f \in \mathcal{O}_{Y, \xi }$ and the local equation for $Y_0$ is given by $f$. In particular, if $\delta (\xi ) = n - 1$ so $\xi$ is the generic point of a integral closed subscheme $Z$ of $\delta$-dimension $n - 1$, then we see that the coefficient of $[Z]$ in $\text{div}_ Y(f)$ is

$\text{ord}_ Z(f) = \text{length}_{\mathcal{O}_{Y, \xi }} (\mathcal{O}_{Y, \xi }/f\mathcal{O}_{Y, \xi }) = \text{length}_{\mathcal{O}_{Y, \xi }} (\mathcal{O}_{Y_0, \xi })$

which is the coefficient of $[Z]$ in $[Y_0]_{n - 1}$. A similar argument using the rational function $1/f$ shows that $-[Y_\infty ]$ agrees with the terms with negative coefficients in the expression for $\text{div}_ Y(f)$. Hence (4) follows.

Note that $D_0 \to S$ is an isomorphism. Hence we see that $X \times _ S D_0 \to X$ is an isomorphism as well. Clearly we have $Y_0 = Y \cap X \times _ S D_0$ (scheme theoretic intersection) inside $X \times _ S \mathbf{P}^1_ S$. Hence it is really the case that $Y_0 \to X$ is a closed immersion. It follows that

$p_*\mathcal{O}_{Y_0} = \mathcal{O}_{Y'_0}$

where $Y'_0 \subset X$ is the image of $Y_0 \to X$. By Lemma 42.12.4 we have $p_*[Y_0]_{n - 1} = [Y'_0]_{n - 1}$. The same is true for $D_\infty$ and $Y_\infty$. Hence (6) is a consequence of (5). Finally, (5) follows immediately from Lemma 42.18.1. $\square$

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