Lemma 42.25.3. Let $(S, \delta )$ be as in Situation 42.7.1. Let $Y$ be locally of finite type over $S$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ Y$-module. Let $s \in \Gamma (Y, \mathcal{L})$. Assume

1. $\dim _\delta (Y) \leq k + 1$,

2. $\dim _\delta (Z(s)) \leq k$, and

3. for every generic point $\xi$ of an irreducible component of $Z(s)$ of $\delta$-dimension $k$ the multiplication by $s$ induces an injection $\mathcal{O}_{Y, \xi } \to \mathcal{L}_\xi$.

Write $[Y]_{k + 1} = \sum n_ i[Y_ i]$ where $Y_ i \subset Y$ are the irreducible components of $Y$ of $\delta$-dimension $k + 1$. Set $s_ i = s|_{Y_ i} \in \Gamma (Y_ i, \mathcal{L}|_{Y_ i})$. Then

42.25.3.1
$$\label{chow-equation-equal-as-cycles} [Z(s)]_ k = \sum n_ i[Z(s_ i)]_ k$$

as $k$-cycles on $Y$.

Proof. Let $Z \subset Y$ be an integral closed subscheme of $\delta$-dimension $k$. Let $\xi \in Z$ be its generic point. We want to compare the coefficient $n$ of $[Z]$ in the expression $\sum n_ i[Z(s_ i)]_ k$ with the coefficient $m$ of $[Z]$ in the expression $[Z(s)]_ k$. Choose a generator $s_\xi \in \mathcal{L}_\xi$. Write $A = \mathcal{O}_{Y, \xi }$, $L = \mathcal{L}_\xi$. Then $L = As_\xi$. Write $s = f s_\xi$ for some (unique) $f \in A$. Hypothesis (3) means that $f : A \to A$ is injective. Since $\dim _\delta (Y) \leq k + 1$ and $\dim _\delta (Z) = k$ we have $\dim (A) = 0$ or $1$. We have

$m = \text{length}_ A(A/(f))$

which is finite in either case.

If $\dim (A) = 0$, then $f : A \to A$ being injective implies that $f \in A^*$. Hence in this case $m$ is zero. Moreover, the condition $\dim (A) = 0$ means that $\xi$ does not lie on any irreducible component of $\delta$-dimension $k + 1$, i.e., $n = 0$ as well.

Now, let $\dim (A) = 1$. Since $A$ is a Noetherian local ring it has finitely many minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$. These correspond 1-1 with the $Y_ i$ passing through $\xi '$. Moreover $n_ i = \text{length}_{A_{\mathfrak q_ i}}(A_{\mathfrak q_ i})$. Also, the multiplicity of $[Z]$ in $[Z(s_ i)]_ k$ is $\text{length}_ A(A/(f, \mathfrak q_ i))$. Hence the equation to prove in this case is

$\text{length}_ A(A/(f)) = \sum \text{length}_{A_{\mathfrak q_ i}}(A_{\mathfrak q_ i}) \text{length}_ A(A/(f, \mathfrak q_ i))$

which follows from Lemma 42.3.2. $\square$

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