The Stacks project

Lemma 42.62.7. Let $k$ be a field. Let $X$ be a smooth scheme over $k$ which is quasi-compact and has affine diagonal. Then the intersection product on $\mathop{\mathrm{CH}}\nolimits ^*(X)$ constructed in this section agrees after tensoring with $\mathbf{Q}$ with the intersection product constructed in Section 42.58.

Proof. Let $\alpha \in \mathop{\mathrm{CH}}\nolimits ^ i(X)$ and $\beta \in \mathop{\mathrm{CH}}\nolimits ^ j(X)$. Write $\alpha = ch(\alpha ') \cap [X]$ and $\beta = ch(\beta ') \cap [X]$ $\alpha ', \beta ' \in K_0(\textit{Vect}(X)) \otimes \mathbf{Q}$ as in Section 42.58. Set $c = ch(\alpha ')$ and $c' = ch(\beta ')$. Then the intersection product in Section 42.58 produces $c \cap c' \cap [X]$. This is the same as $\alpha \cdot \beta $ by Lemma 42.62.2 (or rather the generalization that $A^ i(X) \to \mathop{\mathrm{CH}}\nolimits ^ i(X)$, $c \mapsto c \cap [X]$ is an isomorphism for any smooth scheme $X$ over $k$). $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 42.62: Intersection products

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0FFF. Beware of the difference between the letter 'O' and the digit '0'.