The Stacks project

Lemma 42.28.4. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $(\mathcal{L}, s, i : D \to X)$ be as in Definition 42.28.1. Let $\alpha $ be a $(k + 1)$-cycle on $X$. Then $i_*i^*\alpha = c_1(\mathcal{L}) \cap \alpha $ in $\mathop{\mathrm{CH}}\nolimits _ k(X)$. In particular, if $D$ is an effective Cartier divisor, then $D \cdot \alpha = c_1(\mathcal{O}_ X(D)) \cap \alpha $.

Proof. Write $\alpha = \sum n_ j[W_ j]$ where $i_ j : W_ j \to X$ are integral closed subschemes with $\dim _\delta (W_ j) = k$. Since $D$ is the zero scheme of $s$ we see that $D \cap W_ j$ is the zero scheme of the restriction $s|_{W_ j}$. Hence for each $j$ such that $W_ j \not\subset D$ we have $c_1(\mathcal{L}) \cap [W_ j] = [D \cap W_ j]_ k$ by Lemma 42.24.4. So we have

\[ c_1(\mathcal{L}) \cap \alpha = \sum \nolimits _{W_ j \not\subset D} n_ j[D \cap W_ j]_ k + \sum \nolimits _{W_ j \subset D} n_ j i_{j, *}(c_1(\mathcal{L})|_{W_ j}) \cap [W_ j]) \]

in $\mathop{\mathrm{CH}}\nolimits _ k(X)$ by Definition 42.24.1. The right hand side matches (termwise) the pushforward of the class $i^*\alpha $ on $D$ from Definition 42.28.1. Hence we win. $\square$


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Your email address will not be published.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 02T9. Beware of the difference between the letter 'O' and the digit '0'.