Lemma 62.11.3. Forming $\alpha \cap \beta$ is compatible with flat base change and flat pullback (see proof for elucidation).

Proof. Let $(S, \delta )$, $(S', \delta ')$, $g : S' \to S$, and $c \in \mathbf{Z}$ be as in Chow Homology, Situation 42.67.1. Let $X \to Y$ be a morphism of schemes locally of finite type over $S$. Denote $X' \to Y'$ the base change of $X \to Y$ by $g$. Let $\alpha$ be a family of $r$-cycles on the fibres of $X/Y$. Let $\beta \in Z_ e(Y)$. Denote $\alpha '$ the base change of $\alpha$ by $Y' \to Y$. Denote $\beta ' = g^*\beta \in Z_{e + c}(Y')$ the pullback of $\beta$ by $g$, see Chow Homology, Section 42.67. Compatibility with base change means $\alpha ' \cap \beta '$ is the base change of $\alpha \cap \beta$.

Proof of compatibility with base change. Since we are proving an equality of cycles on $X'$, we may work locally on $Y$, see Lemma 62.11.2. Thus we may assume $Y$ is affine. In particular $\beta$ is a finite linear combination of prime cycles. Since $- \cap -$ is linear in the second variable (Lemma 62.11.1), it suffices to prove the equality when $\beta = [Z]$ for some integral closed subscheme $Z \subset Y$ of $\delta$-dimension $e$.

Let $y \in Z$ be the generic point. Write $\alpha _ y = \sum m_ j [V_ j]$. Let $\overline{V}_ j$ be the closure of $V_ j$ in $X$. Then we have

$\alpha \cap \beta = \sum m_ j[\overline{V}_ j]$

The base change of $\beta$ is $\beta ' = \sum [Z \times _ S S']_{e + c}$ as a cycle on $Y' = Y \times _ S S'$. Let $Z'_ a \subset Z \times _ S S'$ be the irreducible components, denote $y'_ a \in Z'_ a$ their generic points, and denote $n_ a$ the multiplicity of $Z'_ a$ in $Z \times _ S S'$. We have

$\beta ' = \sum [Z \times _ S S']_{e + c} = \sum n_ a[Z'_ a]$

We have $\alpha '_{y'_ a} = \sum m_ j [V_{j, \kappa (y'_ a)}]_ r$ because $\alpha '$ is the base change of $\alpha$ by $Y' \to Y$. Let $V'_{jab} \subset V_{j, \kappa (y'_ a)}$ be the irreducible components and denote $m_{jab}$ the multiplicity of $V'_{jab}$ in $V_{j, \kappa (y'_ a)}$. We have

$\alpha '_{y'_ a} = \sum m_ j [V_{j, \kappa (y'_ a)}]_ r = \sum m_ j m_{jab} [V'_{jab}]$

Thus we we have

$\alpha ' \cap \beta ' = \sum n_ a m_ j m_{jab} [\overline{V}'_{jab}]$

where $\overline{V}'_{jab}$ is the closure of $V'_{jab}$ in $X'$. Thus to prove the desired equality it suffices to prove

1. the irreducible components of $\overline{V}_ j \times _ S S'$ are the schemes $\overline{V}'_{jab}$ and

2. the multiplicity of $\overline{V}'_{jab}$ in $\overline{V}_ j \times _ S S'$ is equal to $n_ a m_{jab}$.

Note that $V_ j \to \overline{V}_ j$ is a birational morphism of integral schemes. The morphisms $V_ j \times _ S S' \to V_ j$ and $\overline{V}_ j \times _ S S' \to \overline{V}_ j$ are flat and hence map generic points of irreducible components to the (unique) generic points of $V_ j$ and $\overline{V}_ j$. It follows that $V_ j \times _ S S' \to \overline{V}_ j \times _ S S'$ is a birational morphisms hence induces a bijection on irreducible components and identifies their multiplicities. This means that it suffices to prove that the irreducible components of $V_ j \times _ S S'$ are the schemes $V'_{jab}$ and the multiplicity of $V'_{jab}$ in $V_ j \times _ S S'$ is equal to $n_ a m_{jab}$. However, then we are just saying that the diagram

$\xymatrix{ Z_ r(V_ j) \ar[r] & Z_{r + c}(V_ j \times _ S S') \\ Z_0(\mathop{\mathrm{Spec}}(\kappa (y))) \ar[r] \ar[u] & Z_ c(\mathop{\mathrm{Spec}}(\kappa (y)) \times _ S S') \ar[u] }$

is commutative where the horizontal arrows are base change by $\mathop{\mathrm{Spec}}(\kappa (y)) \times _ S S' \to \mathop{\mathrm{Spec}}(\kappa (y))$ and the vertical arrows are flat pullback. This was shown in Chow Homology, Lemma 42.67.5.

The statement in the lemma on flat pullback means the following. Let $(S, \delta )$, $X \to Y$, $\alpha$, and $\beta$ be as in the construction of $\alpha \cap \beta$ above. Let $Y' \to Y$ be a flat morphism, locally of finite type, and of relative dimension $c$. Then we can let $\alpha '$ be the base change of $\alpha$ by $Y' \to Y$ and $\beta '$ the flat pullback of $\beta$. Compatibility with flat pullback means $\alpha ' \cap \beta '$ is the flat pullback of $\alpha \cap \beta$ by $X \times _ Y Y' \to Y$. This is actually a special case of the discussion above if we set $S = Y$ and $S' = Y'$. $\square$

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