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
the projection morphism p : Y \to X is proper,
p|_{p^{-1}(U)} : p^{-1}(U) \to U is an isomorphism,
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),
we have
\text{div}_ Y(f) = [Y_0]_{n - 1} - [Y_\infty ]_{n - 1}
we have
\text{div}_ X(f) = p_*\text{div}_ Y(f)
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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