The Stacks project

Proof. Part (1) follows from Divisors, Lemma 31.13.13 since the generic point of $W$ is not mapped into $D_0$ or $D_\infty $ under the projection $X \times _ S \mathbf{P}^1_ S \to \mathbf{P}^1_ S$ by assumption.

Since $X \times _ S D_0 \to X$ is a closed immersion, we see that $W_0$ is isomorphic to a closed subscheme of $X$. Similarly for $W_\infty $. The morphism $p : W \to X$ is proper as a composition of the closed immersion $W \to X \times _ S \mathbf{P}^1_ S$ and the proper morphism $X \times _ S \mathbf{P}^1_ S \to X$. By Lemma 42.18.2 we have $[W_0]_ k \sim _{rat} [W_\infty ]_ k$ as cycles on $W$. Hence part (2) follows from Lemma 42.20.3 as clearly $p_*[W_0]_ k = [W_0]_ k$ and similarly for $W_\infty $.

The only content of statement (3) is, given parts (1) and (2), that the collection $\{ (W_ i)_0, (W_ i)_\infty \} $ is a locally finite collection of closed subschemes of $X$. This is clear.

Suppose that $\alpha \sim _{rat} 0$. By definition this means there exist integral closed subschemes $V_ i \subset X$ of $\delta $-dimension $k + 1$ and rational functions $f_ i \in R(V_ i)^*$ such that the family $\{ V_ i\} _{i \in I}$ is locally finite in $X$ and such that $\alpha = \sum (V_ i \to X)_*\text{div}(f_ i)$. Let

be the closure of the graph of the rational map $f_ i$ as in Lemma 42.18.2. Then we have that $(V_ i \to X)_*\text{div}(f_ i)$ is equal to $[(W_ i)_0]_ k - [(W_ i)_\infty ]_ k$ by that same lemma. Hence the result is clear. $\square$

Proof. Let $\{ W_ i\} _{i \in I}$ be the collection of irreducible components of $Z$ which have $\delta $-dimension $k + 1$. Write

with $n_ i > 0$ as per definition. Note that $\{ W_ i\} $ is a locally finite collection of closed subsets of $X \times _ S \mathbf{P}^1_ S$ by Divisors, Lemma 31.26.1. We claim that

and similarly for $[Z_\infty ]_ k$. If we prove this then the lemma follows from Lemma 42.21.1.

Let $Z' \subset X$ be an integral closed subscheme of $\delta $-dimension $k$. To prove the equality above it suffices to show that the coefficient $n$ of $[Z']$ in $[Z_0]_ k$ is the same as the coefficient $m$ of $[Z']$ in $\sum n_ i[(W_ i)_0]_ k$. Let $\xi ' \in Z'$ be the generic point. Set $\xi = (\xi ', 0) \in X \times _ S \mathbf{P}^1_ S$. Consider the local ring $A = \mathcal{O}_{X \times _ S \mathbf{P}^1_ S, \xi }$. Let $I \subset A$ be the ideal cutting out $Z$, in other words so that $A/I = \mathcal{O}_{Z, \xi }$. Let $t \in A$ be the element cutting out $X \times _ S D_0$ (i.e., the coordinate of $\mathbf{P}^1$ at zero pulled back). By our choice of $\xi ' \in Z'$ we have $\delta (\xi ) = k$ and hence $\dim (A/I) = 1$. Since $\xi $ is not an embedded point by assumption (3) we see that $A/I$ is Cohen-Macaulay. Since $\dim _\delta (Z_0) = k$ we see that $\dim (A/(t, I)) = 0$ which implies that $t$ is a nonzerodivisor on $A/I$. Finally, the irreducible closed subschemes $W_ i$ passing through $\xi $ correspond to the minimal primes $I \subset \mathfrak q_ i$ over $I$. The multiplicities $n_ i$ correspond to the lengths $\text{length}_{A_{\mathfrak q_ i}}(A/I)_{\mathfrak q_ i}$. Hence we see that

and

Thus the result follows from Lemma 42.3.2. $\square$

Proof. Let $\{ W_ i\} _{i \in I}$ be the collection of irreducible components of $\text{Supp}(\mathcal{F})$ which have $\delta $-dimension $k + 1$. Write

with $n_ i > 0$ as per definition. Note that $\{ W_ i\} $ is a locally finite collection of closed subsets of $X \times _ S \mathbf{P}^1_ S$ by Lemma 42.10.1. We claim that

and similarly for $[\mathcal{F}_\infty ]_ k$. If we prove this then the lemma follows from Lemma 42.21.1.

Let $Z' \subset X$ be an integral closed subscheme of $\delta $-dimension $k$. To prove the equality above it suffices to show that the coefficient $n$ of $[Z']$ in $[\mathcal{F}_0]_ k$ is the same as the coefficient $m$ of $[Z']$ in $\sum n_ i[(W_ i)_0]_ k$. Let $\xi ' \in Z'$ be the generic point. Set $\xi = (\xi ', 0) \in X \times _ S \mathbf{P}^1_ S$. Consider the local ring $A = \mathcal{O}_{X \times _ S \mathbf{P}^1_ S, \xi }$. Let $M = \mathcal{F}_\xi $ as an $A$-module. Let $t \in A$ be the element cutting out $X \times _ S D_0$ (i.e., the coordinate of $\mathbf{P}^1$ at zero pulled back). By our choice of $\xi ' \in Z'$ we have $\delta (\xi ) = k$ and hence $\dim (\text{Supp}(M)) = 1$. Since $\xi $ is not an associated point of $\mathcal{F}$ by assumption (3) we see that $M$ is a Cohen-Macaulay module. Since $\dim _\delta (\text{Supp}(\mathcal{F}_0)) = k$ we see that $\dim (\text{Supp}(M/tM)) = 0$ which implies that $t$ is a nonzerodivisor on $M$. Finally, the irreducible closed subschemes $W_ i$ passing through $\xi $ correspond to the minimal primes $\mathfrak q_ i$ of $\text{Ass}(M)$. The multiplicities $n_ i$ correspond to the lengths $\text{length}_{A_{\mathfrak q_ i}}M_{\mathfrak q_ i}$. Hence we see that

and

Thus the result follows from Lemma 42.3.2. $\square$