The Stacks project

Lemma 81.21.2. In Situation 81.2.1 let $X/B$ be good. Let $\mathcal{L}$ be invertible on $X$. The operation $\alpha \mapsto c_1(\mathcal{L}) \cap \alpha $ factors through rational equivalence to give an operation

\[ c_1(\mathcal{L}) \cap - : \mathop{\mathrm{CH}}\nolimits _{k + 1}(X) \to \mathop{\mathrm{CH}}\nolimits _ k(X) \]

Proof. Let $\alpha \in Z_{k + 1}(X)$, and $\alpha \sim _{rat} 0$. We have to show that $c_1(\mathcal{L}) \cap \alpha $ as defined in Definition 81.18.1 is zero. By Definition 81.15.1 there exists a locally finite family $\{ W_ j\} $ of integral closed subspaces with $\dim _\delta (W_ j) = k + 2$ and rational functions $f_ j \in R(W_ j)^*$ such that

\[ \alpha = \sum (i_ j)_*\text{div}_{W_ j}(f_ j) \]

Note that $p : \coprod W_ j \to X$ is a proper morphism, and hence $\alpha = p_*\alpha '$ where $\alpha ' \in Z_{k + 1}(\coprod W_ j)$ is the sum of the principal divisors $\text{div}_{W_ j}(f_ j)$. By Lemma 81.19.4 we have $c_1(\mathcal{L}) \cap \alpha = p_*(c_1(p^*\mathcal{L}) \cap \alpha ')$. Hence it suffices to show that each $c_1(\mathcal{L}|_{W_ j}) \cap \text{div}_{W_ j}(f_ j)$ is zero. In other words we may assume that $X$ is integral and $\alpha = \text{div}_ X(f)$ for some $f \in R(X)^*$.

Assume $X$ is integral and $\alpha = \text{div}_ X(f)$ for some $f \in R(X)^*$. We can think of $f$ as a regular meromorphic section of the invertible sheaf $\mathcal{N} = \mathcal{O}_ X$. Choose a meromorphic section $s$ of $\mathcal{L}$ and denote $\beta = \text{div}_\mathcal {L}(s)$. By Lemma 81.21.1 we conclude that

\[ c_1(\mathcal{L}) \cap \alpha = c_1(\mathcal{O}_ X) \cap \beta . \]

However, by Lemma 81.18.2 we see that the right hand side is zero in $\mathop{\mathrm{CH}}\nolimits _ k(X)$ as desired. $\square$


Comments (0)


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Your email address will not be published.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