Lemma 42.67.4. In Situation 42.67.1 let $X \to S$ be locally of finite type and let $X' \to S'$ be the base change by $S' \to S$. The map $g^* : Z_ k(X) \to Z_{k + c}(X')$ above factors through rational equivalence to give a map

of chow groups.

Lemma 42.67.4. In Situation 42.67.1 let $X \to S$ be locally of finite type and let $X' \to S'$ be the base change by $S' \to S$. The map $g^* : Z_ k(X) \to Z_{k + c}(X')$ above factors through rational equivalence to give a map

\[ g^* : \mathop{\mathrm{CH}}\nolimits _ k(X) \longrightarrow \mathop{\mathrm{CH}}\nolimits _{k + c}(X') \]

of chow groups.

**Proof.**
Suppose that $\alpha \in Z_ k(X)$ is a $k$-cycle which is rationally equivalent to zero. By Lemma 42.21.1 there exists a locally finite family of integral closed subschemes $W_ i \subset X \times \mathbf{P}^1$ of $\delta $-dimension $k$ not contained in the divisors $(X \times \mathbf{P}^1)_0$ or $(X \times \mathbf{P}^1)_\infty $ of $X \times \mathbf{P}^1$ such that $\alpha = \sum ([(W_ i)_0]_ k - [(W_ i)_\infty ]_ k)$. Thus it suffices to prove for $W \subset X \times \mathbf{P}^1$ integral closed of $\delta $-dimension $k$ not contained in the divisors $(X \times \mathbf{P}^1)_0$ or $(X \times \mathbf{P}^1)_\infty $ of $X \times \mathbf{P}^1$ we have

the base change $W' \subset X' \times \mathbf{P}^1$ satisfies the assumptions of Lemma 42.21.2 with $k$ replaced by $k + c$, and

$g^*[W_0]_ k = [(W')_0]_{k + c}$ and $g^*[W_\infty ]_ k = [(W')_\infty ]_{k + c}$.

Part (2) follows immediately from Lemma 42.67.3 and the fact that $(W')_0$ is the base change of $W_0$ (by associativity of fibre products). For part (1), first the statement on dimensions follows from Lemma 42.67.2. Then let $w' \in (W')_0$ with image $w \in W_0$ and $z \in \mathbf{P}^1_ S$. Denote $t \in \mathcal{O}_{\mathbf{P}^1_ S, z}$ the usual equation for $0 : S \to \mathbf{P}^1_ S$. Since $\mathcal{O}_{W, w} \to \mathcal{O}_{W', w'}$ is flat and since $t$ is a nonzerodivisor on $\mathcal{O}_{W, w}$ (as $W$ is integral and $W \not= W_0$) we see that also $t$ is a nonzerodivisor in $\mathcal{O}_{W', w'}$. Hence $W'$ has no associated points lying on $W'_0$. $\square$

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