Lemma 62.11.4. Let $(S, \delta )$ and $f : X \to Y$ be as above. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module with $\dim (\text{Supp}(\mathcal{F}_ y)) \leq r$ for all $y \in Y$. Let $\mathcal{G}$ be a coherent $\mathcal{O}_ Y$-module with $\dim _\delta (\text{Supp}(\mathcal{G})) \leq e$. Set $\alpha = [\mathcal{F}/X/Y]_ r$ (Example 62.5.2) and $\beta = [\mathcal{G}]_ e$ (Chow Homology, Definition 42.10.2). If $\mathcal{F}$ is flat over $Y$, then $\alpha \cap \beta = [\mathcal{F} \otimes _{\mathcal{O}_ X} f^*\mathcal{G}]_{r + e}$.

Proof. Observe that

$\text{Supp}(\mathcal{F} \otimes _{\mathcal{O}_ X} f^*\mathcal{G}) = \text{Supp}(\mathcal{F}) \cap f^{-1}\text{Supp}(\mathcal{G}) = \bigcup \nolimits _{y \in \text{Supp}(\mathcal{G})} \text{Supp}(\mathcal{F}_ y)$

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

$\text{length}_{\mathcal{O}_{X, \xi _{ij}}}( (\mathcal{F} \otimes _{\mathcal{O}_ X} f^*\mathcal{G})_{\xi _{ij}}) = \text{length}_{\mathcal{O}_{X_{y_ i}, \xi _{ij}}}((\mathcal{F}_{y_ i})_{\xi _{ij}}) \cdot \text{length}_{\mathcal{O}_{Y, y_ i}}(\mathcal{G}_{y_ i})$

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

$\mathcal{G}_{y_ i} \otimes _{\mathcal{O}_{Y, y_ i}} \mathcal{F}_{\xi _{ij}} = M \otimes _ A N$

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

$\text{length}_ B(N/\mathfrak m_ A N) \cdot \text{length}_ A(M)$

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. $\square$

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