The Stacks project

Proof. Omitted. $\square$

Proof. Immediate from the explicit description of $\alpha \cap \beta$ given above. $\square$

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

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

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

Thus we we have

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

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 $S' = Y'$. $\square$

Proof. Observe that

It follows that this is a closed subset of $\delta$-dimension $\leq r + e$. Whence the expression $[\mathcal{F} \otimes _{\mathcal{O}_ X} f^*\mathcal{G}]_{r + e}$ makes sense.

We will use the notation $\beta = \sum n_ i[Z_ i]$, $y_ i \in Z_ i$, $\alpha _{y_ i} = \sum m_{ij} [V_{ij}]$, and $\overline{V}_{ij}$ introduced in the construction of $\alpha \cap \beta$. Since $\beta = [\mathcal{G}]_ e$ we see that the $Z_ i$ are the irreducible components of $\text{Supp}(\mathcal{G})$ which have $\delta$-dimension $e$. Similarly, the $V_{ij}$ are the irreducible components of $\text{Supp}(\mathcal{F}_{y_ i})$ having dimension $r$. It follows from this and the equation in the first paragraph that $\overline{V}_{ij}$ are the irreducible components of $\text{Supp}(\mathcal{F} \otimes _{\mathcal{O}_ X} f^*\mathcal{G})$ having $\delta$-dimension $r + e$. Thus to prove the lemma it now suffices to show that

By the first paragraph of the proof the left hand side is equal to the length of the $B = \mathcal{O}_{X, \xi _{ij}}$-module

Here $M = \mathcal{G}_{y_ i}$ is a finite length $A = \mathcal{O}_{Y, y_ i}$-module and $N = \mathcal{F}_{\xi _{ij}}$ is a finite $B$-module such that $N/\mathfrak m_ AN$ has finite length. Since $\mathcal{F}$ is flat over $Y$ the module $N$ is $A$-flat. The right hand side of the formula is equal to

Thus the right and left hand side of the formula are additive in $M$ (use flatness of $N$ over $A$). Thus it suffices to prove the formula with $M = \kappa _ A$ is the residue field in which case it is immediate. $\square$

Proof. This is a special case of Lemma 62.11.4 if we take $\mathcal{F} = \mathcal{O}_ Z$ and $\mathcal{F} = \mathcal{O}_ W$. $\square$

Proof. 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. Thus $Y'$ is quasi-compact. 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 $e$. Set $Z = g(Z')$. This is an integral closed subscheme of $Y$ of $\delta$-dimension $\leq e$. For simplicity we are going to assume $Z$ has $\delta$-dimension equal to $e$ and leave the other case (which is easier) to the reader. Let $y \in Z$ and $y' \in Z'$ be the generic points. Write $\alpha _ y = \sum m_ j[V_ j]$ with $V_ j \subset X_ y$ integral closed subschemes of dimension $r$.

Assume first $g$ is a closed immersion. Then $g_*\beta ' = [Z]$ and $(g^*\alpha )_{y'} = \sum n_ j[V_ j]$; this makes sense because $V_ j$ is contained in the closed subscheme $X'_{y'}$ of $X_ y$. Thus in this case the equality is obvious: in both cases we obtain $\sum m_ j[\overline{V}_ j]$ where $\overline{V}_ j$ is the closure of $V_ j$ in the closed subscheme $X' \subset X$.

Back to the general case with $\beta ' = [Z']$ as above. Set $W = Z \times _ X Y$ and $W' = Z' \times _{X'} Y'$. Consider the cartesian squares

Since we know the result for the first two squares with by the previous paragraph, a formal argument shows that it suffices to prove the result for the last square and the element $\beta ' = [Z'] \in Z_ e(Z')$. This reduces us to the case discussed in the next paragraph.

Assume $Y' \to Y$ is a generically finite morphism of integral schemes of $\delta$-dimension $e$ and $\beta ' = [Y']$. In this case both $f_*(g^*\alpha \cap \beta ')$ and $\alpha \cap g_*\beta '$ are cycles which can be written as a sum of prime cycles dominant over $Y$. Thus we may replace $Y$ by a nonempty open subscheme in order to check the equality. After such a replacement we may assume $g$ is finite and flat, say of degree $d \geq 1$. Of course, this means that $g_*\beta ' = g_*[Y'] = d[Y]$. Also $\beta ' = [Y'] = g^*[Y]$. Hence

as desired. The second equality is Lemma 62.11.3 and the third equality is Chow Homology, Lemma 42.15.2. $\square$