## 42.58 Rational intersection products on regular schemes

We will show that $\mathop{\mathrm{CH}}\nolimits _*(X) \otimes \mathbf{Q}$ has an intersection product if $X$ is Noetherian, regular, finite dimensional, with affine diagonal. The basis for the construction is the following result (which is a corollary of the proposition in the previous section).

Lemma 42.58.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be a quasi-compact regular scheme of finite type over $S$ with affine diagonal and $\delta _{X/S} : X \to \mathbf{Z}$ bounded. Then the composition

\[ K_0(\textit{Vect}(X)) \otimes \mathbf{Q} \longrightarrow A^*(X) \otimes \mathbf{Q} \longrightarrow \mathop{\mathrm{CH}}\nolimits _*(X) \otimes \mathbf{Q} \]

of the map $ch$ from Remark 42.56.5 and the map $c \mapsto c \cap [X]$ is an isomorphism.

**Proof.**
We have $K'_0(X) = K_0(X) = K_0(\textit{Vect}(X))$ by Derived Categories of Schemes, Lemmas 36.38.4 and 36.38.5. By Remark 42.56.12 the composition given agrees with the map of Proposition 42.57.1 for $X = Y$. Thus the result follows from the proposition.
$\square$

Let $X, S, \delta $ be as in Lemma 42.58.1. For simplicity let us work with cycles of a given codimension, see Section 42.42. Let $[X]$ be the fundamental cycle of $X$, see Remark 42.42.2. Pick $\alpha \in CH^ i(X)$ and $\beta \in CH^ j(X)$. By the lemma we can find a unique $\alpha ' \in K_0(\textit{Vect}(X)) \otimes \mathbf{Q}$ with $ch(\alpha ') \cap [X] = \alpha $. Of course this means that $ch_{i'}(\alpha ') \cap [X] = 0$ if $i' \not= i$ and $ch_ i(\alpha ') \cap [X] = \alpha $. By Lemma 42.56.6 we see that $\alpha '' = 2^{-i}\psi ^2(\alpha ')$ is another solution. By uniqueness we get $\alpha '' = \alpha '$ and we conclude that $ch_{i'}(\alpha ) = 0$ in $A^{i'}(X) \otimes \mathbf{Q}$ for $i' \not= i$. Then we can define

\[ \alpha \cdot \beta = ch(\alpha ') \cap \beta = ch_ i(\alpha ') \cap \beta \]

in $\mathop{\mathrm{CH}}\nolimits ^{i + j}(X) \otimes \mathbf{Q}$ by the property of $\alpha '$ we observed above. This is a symmetric pairing: namely, if we pick $\beta ' \in K_0(\textit{Vect}(X)) \otimes \mathbf{Q}$ lifting $\beta $, then we get

\[ \alpha \cdot \beta = ch(\alpha ') \cap \beta = ch(\alpha ') \cap ch(\beta ') \cap [X] \]

and we know that Chern classes commute. The intersection product is associative for the same reason

\[ (\alpha \cdot \beta ) \cdot \gamma = ch(\alpha ') \cap ch(\beta ') \cap ch(\gamma ') \cap [X] \]

because we know composition of bivariant classes is associative. Perhaps a better way to formulate this is as follows: there is a unique commutative, associative intersection product on $\mathop{\mathrm{CH}}\nolimits ^*(X) \otimes \mathbf{Q}$ compatible with grading such that the isomorphism $K_0(\textit{Vect}(X)) \otimes \mathbf{Q} \to \mathop{\mathrm{CH}}\nolimits ^*(X) \otimes \mathbf{Q}$ is an isomorphism of rings.

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