Lemma 42.62.5. Let $k$ be a field. Let $X$ be an integral scheme smooth over $k$. Let $Y, Z \subset X$ be integral closed subschemes. Set $d = \dim (Y) + \dim (Z) - \dim (X)$. Assume
$\dim (Y \cap Z) \leq d$, and
$\mathcal{O}_{Y, \xi }$ and $\mathcal{O}_{Z, \xi }$ are Cohen-Macaulay for every $\xi \in Y \cap Z$ with $\delta (\xi ) = d$.
Then $[Y] \cdot [Z] = [Y \cap Z]_ d$ in $\mathop{\mathrm{CH}}\nolimits _ d(X)$.
Proof. Recall that $[Y] \cdot [Z] = \Delta ^!([Y \times Z])$ where $\Delta ^! = c(\Delta : X \to X \times X, \mathcal{T}_{X/k})$ is a higher codimension gysin map (Section 42.54) with $\mathcal{T}_{X/k} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\Omega _{X/k}, \mathcal{O}_ X)$ locally free of rank $\dim (X)$. We have the equality of schemes
and $\dim (Y \times Z) = \dim (Y) + \dim (Z)$ and hence conditions (1), (2), and (3) of Lemma 42.54.6 hold. Finally, if $\xi \in Y \cap Z$, then we have a flat local homomorphism
whose “fibre” is $\mathcal{O}_{Z, \xi }$. It follows that if both $\mathcal{O}_{Y, \xi }$ and $\mathcal{O}_{Z, \xi }$ are Cohen-Macaulay, then so is $\mathcal{O}_{Y \times Z, \xi }$, see Algebra, Lemma 10.163.3. In this way we see that all the hypotheses of Lemma 42.54.6 are satisfied and we conclude. $\square$
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