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
\dim _\delta (Y) \leq k + 1,
\dim _\delta (Z(s)) \leq k, and
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
\begin{equation} \label{chow-equation-equal-as-cycles} [Z(s)]_ k = \sum n_ i[Z(s_ i)]_ k \end{equation}
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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