Lemma 42.59.12. Let $(S, \delta )$ be as in Situation 42.7.1. Let $f : X \to Y$ be a morphism of schemes locally of finite type over $S$ such that both $X$ and $Y$ are quasi-compact, regular, have affine diagonal, and finite dimension. Then $f$ is a local complete intersection morphism. Assume moreover the gysin map exists for $f$. Then
\[ f^!(\alpha \cdot \beta ) = f^!\alpha \cdot f^!\beta \]
in $\mathop{\mathrm{CH}}\nolimits ^*(X) \otimes \mathbf{Q}$ where the intersection product is as in Section 42.58.
Proof.
The first statement follows from More on Morphisms, Lemma 37.62.11. Observe that $f^![Y] = [X]$, see Lemma 42.59.8. Write $\alpha = ch(\alpha ') \cap [Y]$ and $\beta = ch(\beta ') \cap [Y]$ where $\alpha ', \beta ' \in K_0(\textit{Vect}(X)) \otimes \mathbf{Q}$ as in Section 42.58. Setting $c = ch(\alpha ')$ and $c' = ch(\beta ')$ we find $\alpha \cdot \beta = c \cap c' \cap [Y]$ by construction. By Lemma 42.59.7 we know that $f^!$ commutes with both $c$ and $c'$. Hence
\begin{align*} f^!(\alpha \cdot \beta ) & = f^!(c \cap c' \cap [Y]) \\ & = c \cap c' \cap f^![Y] \\ & = c \cap c' \cap [X] \\ & = (c \cap [X]) \cdot (c' \cap [X]) \\ & = (c \cap f^![Y]) \cdot (c' \cap f^![Y]) \\ & = f^!(\alpha ) \cdot f^!(\beta ) \end{align*}
as desired.
$\square$
Comments (0)