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
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
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