The Stacks project

Lemma 42.67.9. Let $(S, \delta )$, $(S', \delta ')$, $(S'', \delta '')$ be as in Situation 42.7.1. Let $g : S' \to S$ and $g' : S'' \to S'$ be flat morphisms of schemes and let $c, c' \in \mathbf{Z}$ be integers such that $S, \delta , S', \delta ', g, c$ and $S', \delta ', S'', g', c'$ are as in Situation 42.67.1. Let $X \to S$ be locally of finite type and denote $X' \to S'$ and $X'' \to S''$ the base changes by $S' \to S$ and $S'' \to S$. Then

  1. $S, \delta , S'', \delta '', g \circ g', c + c'$ is as in Situation 42.67.1,

  2. the maps $g^* : Z_ k(X) \to Z_{k + c}(X')$ and $(g')^* : Z_{k + c}(X') \to Z_{k + c + c'}(X'')$ of compose to give the map $(g \circ g')^* : Z_ k(X) \to Z_{k + c + c'}(X'')$, and

  3. the maps $g^* : \mathop{\mathrm{CH}}\nolimits _ k(X) \to \mathop{\mathrm{CH}}\nolimits _{k + c}(X')$ and $(g')^* : \mathop{\mathrm{CH}}\nolimits _{k + c}(X') \to \mathop{\mathrm{CH}}\nolimits _{k + c + c'}(X'')$ of Lemma 42.67.4 compose to give the map $(g \circ g')^* : \mathop{\mathrm{CH}}\nolimits _ k(X) \to \mathop{\mathrm{CH}}\nolimits _{k + c + c'}(X'')$ of Lemma 42.67.4.

Proof. Let $s \in S$ and let $s'' \in S''$ be a generic point of an irreducible component of $(g \circ g')^{-1}(\{ s\} )$. Set $s' = g'(s'')$. Clearly, $s''$ is a generic point of an irreducible component of $(g')^{-1}(\{ s'\} )$. Moreover, since $g'$ is flat and hence generalizations lift along $g'$ (Morphisms, Lemma 29.25.8) we see that also $s'$ is a generic point of an irreducible component of $g^{-1}(\{ s\} )$. Thus by assumption $\delta '(s') = \delta (s) + c$ and $\delta ''(s'') = \delta '(s') + c'$. We conclude $\delta ''(s'') = \delta (s) + c + c'$ and the first part of the statement is true.

For the second part, let $Z \subset X$ be an integral closed subscheme of $\delta $-dimension $k$. Denote $Z' \subset X'$ and $Z'' \subset X''$ the base changes. By definition we have $g^*[Z] = [Z']_{k + c}$. By Lemma 42.67.3 we have $(g')^*[Z']_{k + c} = [Z'']_{k + c + c'}$. This proves the final statement. $\square$

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