Lemma 42.67.3. In Situation 42.67.1 let $X \to S$ locally of finite type and let $X' \to S$ be the base change by $S' \to S$.

1. Let $Z \subset X$ be a closed subscheme with $\dim _\delta (Z) \leq k$ and base change $Z' \subset X'$. Then we have $\dim _{\delta '}(Z')) \leq k + c$ and $[Z']_{k + c} = g^*[Z]_ k$ in $Z_{k + c}(X')$.

2. Let $\mathcal{F}$ be a coherent sheaf on $X$ with $\dim _\delta (\text{Supp}(\mathcal{F})) \leq k$ and base change $\mathcal{F}'$ on $X'$. Then we have $\dim _\delta (\text{Supp}(\mathcal{F}')) \leq k + c$ and $g^*[\mathcal{F}]_ k = [\mathcal{F}']_{k + c}$ in $Z_{k + c}(X')$.

Proof. The proof is exactly the same is the proof of Lemma 42.14.4 and we suggest the reader skip it.

The statements on dimensions follow from Lemma 42.67.2. Part (1) follows from part (2) by Lemma 42.10.3 and the fact that the base change of the coherent module $\mathcal{O}_ Z$ is $\mathcal{O}_{Z'}$.

Proof of (2). As $X$, $X'$ are locally Noetherian we may apply Cohomology of Schemes, Lemma 30.9.1 to see that $\mathcal{F}$ is of finite type, hence $\mathcal{F}'$ is of finite type (Modules, Lemma 17.9.2), hence $\mathcal{F}'$ is coherent (Cohomology of Schemes, Lemma 30.9.1 again). Thus the lemma makes sense. Let $W \subset X$ be an integral closed subscheme of $\delta$-dimension $k$, and let $W' \subset X'$ be an integral closed subscheme of $\delta '$-dimension $k + c$ mapping into $W$ under $X' \to X$. We have to show that the coefficient $n$ of $[W']$ in $g^*[\mathcal{F}]_ k$ agrees with the coefficient $m$ of $[W']$ in $[\mathcal{F}']_{k + c}$. Let $\xi \in W$ and $\xi ' \in W'$ be the generic points. Let $A = \mathcal{O}_{X, \xi }$, $B = \mathcal{O}_{X', \xi '}$ and set $M = \mathcal{F}_\xi$ as an $A$-module. (Note that $M$ has finite length by our dimension assumptions, but we actually do not need to verify this. See Lemma 42.10.1.) We have $\mathcal{F}'_{\xi '} = B \otimes _ A M$. Thus we see that

$n = \text{length}_ B(B \otimes _ A M) \quad \text{and} \quad m = \text{length}_ A(M) \text{length}_ B(B/\mathfrak m_ AB)$

Thus the equality follows from Algebra, Lemma 10.52.13. $\square$

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).