Lemma 43.19.4. Let X be a nonsingular variety. Let \mathcal{F} and \mathcal{G} be coherent sheaves on X with \dim (\text{Supp}(\mathcal{F})) \leq r, \dim (\text{Supp}(\mathcal{G})) \leq s, and \dim (\text{Supp}(\mathcal{F}) \cap \text{Supp}(\mathcal{G}) ) \leq r + s - \dim X. In this case [\mathcal{F}]_ r and [\mathcal{G}]_ s intersect properly and
[Chapter V, Serre_algebre_locale]
Proof. The statement that [\mathcal{F}]_ r and [\mathcal{G}]_ s intersect properly is immediate. Since we are proving an equality of cycles we may work locally on X. (Observe that the formation of the intersection product of cycles, the formation of \text{Tor}-sheaves, and forming the cycle associated to a coherent sheaf, each commute with restriction to open subschemes.) Thus we may and do assume that X is affine.
Denote
Consider a short exact sequence
of coherent sheaves on X with \text{Supp}(\mathcal{F}_ i) \subset \text{Supp}(\mathcal{F}), then both LHS(\mathcal{F}_ i, \mathcal{G}) and RHS(\mathcal{F}_ i, \mathcal{G}) are defined for i = 1, 2, 3 and we have
and similarly for LHS. Namely, the support condition guarantees that everything is defined, the short exact sequence and additivity of lengths gives
(Chow Homology, Lemma 42.10.4) which implies additivity for RHS. The long exact sequence of \text{Tor}s
and additivity of lengths as before implies additivity for LHS.
By Algebra, Lemma 10.62.1 and the fact that X is affine, we can find a filtration of \mathcal{F} whose graded pieces are structure sheaves of closed subvarieties of \text{Supp}(\mathcal{F}). The additivity shown in the previous paragraph, implies that it suffices to prove LHS = RHS with \mathcal{F} replaced by \mathcal{O}_ V where V \subset \text{Supp}(\mathcal{F}). By symmetry we can do the same for \mathcal{G}. This reduces us to proving that
where W \subset \text{Supp}(\mathcal{G}) is a closed subvariety. If \dim (V) = r and \dim (W) = s, then this equality is the definition of V \cdot W. On the other hand, if \dim (V) < r or \dim (W) < s, i.e., [V]_ r = 0 or [W]_ s = 0, then we have to prove that RHS(\mathcal{O}_ V, \mathcal{O}_ W) = 0 1.
Let Z \subset V \cap W be an irreducible component of dimension r + s - \dim (X). This is the maximal dimension of a component and it suffices to show that the coefficient of Z in RHS is zero. Let \xi \in Z be the generic point. Write A = \mathcal{O}_{X, \xi }, B = \mathcal{O}_{X \times X, \Delta (\xi )}, and C = \mathcal{O}_{V \times W, \Delta (\xi )}. By Lemma 43.19.1 we have
Since \dim (V) < r or \dim (W) < s we have \dim (V \times W) < r + s which implies \dim (C) < \dim (X) (small detail omitted). Moreover, the kernel I of B \to A is generated by a regular sequence of length \dim (X) (Lemma 43.13.3). Hence vanishing by Lemma 43.16.2 because the Hilbert function of C with respect to I has degree \dim (C) < n by Algebra, Proposition 10.60.9. \square
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