Lemma 81.18.3. In Situation 81.2.1 let $Y/B$ be good. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ Y$-module. Let $s \in \Gamma (Y, \mathcal{L})$ be a regular section and assume $\dim _\delta (Y) \leq k + 1$. 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

81.18.3.1
\begin{equation} \label{spaces-chow-equation-equal-as-cycles} [Z(s)]_ k = \sum n_ i[Z(s_ i)]_ k \end{equation}

as $k$-cycles on $Y$.

Proof. Let $\varphi : V \to Y$ be a surjective étale morphism where $V$ is a scheme. It suffices to prove the equality after pulling back by $\varphi$, see Lemma 81.10.3. That same lemma tells us that $\varphi ^*[Y_ i] = [\varphi ^{-1}(Y_ i)] = \sum [V_{i, j}]$ where $V_{i, j}$ are the irreducible components of $V$ lying over $Y_ i$. Hence if we first apply the case of schemes (Chow Homology, Lemma 42.25.3) to $\varphi ^*s_ i$ on $Y_ i \times _ Y V$ we find that $\varphi ^*[Z(s_ i)]_ k = [Z(\varphi ^*s_ i)] = \sum [Z(s_{i, j})]_ k$ where $s_{i, j}$ is the pullback of $s$ to $V_{i, j}$. Applying the case of schemes to $\varphi ^*s$ we get

$\varphi ^*[Z(s)]_ k = [Z(\varphi ^*s)]_ k = \sum n_ i[Z(s_{i, j})]_ k$

by our remark on multiplicities above. Combining all of the above the proof is complete. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).