Lemma 42.61.6. Let $k$ be a field. Let $X$ be a scheme smooth over $k$. Let $i : Y \to X$ be a regular closed immersion. Let $\alpha \in \mathop{\mathrm{CH}}\nolimits _*(X)$. If $Y$ is equidimensional of dimension $e$, then $\alpha \cdot [Y]_ e = i_*(i^!(\alpha ))$ in $\mathop{\mathrm{CH}}\nolimits _*(X)$.

Proof. After decomposing $X$ into connected components we may and do assume $X$ is equidimensional of dimension $d$. Write $\alpha = c \cap [X]_ n$ with $x \in A^*(X)$, see Lemma 42.61.2. Then

$i_*(i^!(\alpha )) = i_*(i^!(c \cap [X]_ n)) = i_*(c \cap i^![X]_ n) = i_*(c \cap [Y]_ e) = c \cap i_*[Y]_ e = \alpha \cdot [Y]_ e$

The first equality by choice of $c$. Then second equality by Lemma 42.58.7. The third because $i^![X]_ d = [Y]_ e$ in $\mathop{\mathrm{CH}}\nolimits _*(Y)$ (Lemma 42.58.8). The fourth because bivariant classes commute with proper pushforward. The last equality by Lemma 42.61.2. $\square$

## 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Your email address will not be published.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 0FFE. Beware of the difference between the letter 'O' and the digit '0'.