Lemma 42.25.4. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $Y \subset X$ be a closed subscheme. Let $s \in \Gamma (Y, \mathcal{L}|_ Y)$. Assume

1. $\dim _\delta (Y) \leq k + 1$,

2. $\dim _\delta (Z(s)) \leq k$, and

3. for every generic point $\xi$ of an irreducible component of $Z(s)$ of $\delta$-dimension $k$ the multiplication by $s$ induces an injection $\mathcal{O}_{Y, \xi } \to (\mathcal{L}|_ Y)_\xi$1.

Then

$c_1(\mathcal{L}) \cap [Y]_{k + 1} = [Z(s)]_ k$

in $\mathop{\mathrm{CH}}\nolimits _ k(X)$.

Proof. Write

$[Y]_{k + 1} = \sum n_ i[Y_ i]$

where $Y_ i \subset Y$ are the irreducible components of $Y$ of $\delta$-dimension $k + 1$ and $n_ i > 0$. By assumption the restriction $s_ i = s|_{Y_ i} \in \Gamma (Y_ i, \mathcal{L}|_{Y_ i})$ is not zero, and hence is a regular section. By Lemma 42.24.2 we see that $[Z(s_ i)]_ k$ represents $c_1(\mathcal{L}|_{Y_ i})$. Hence by definition

$c_1(\mathcal{L}) \cap [Y]_{k + 1} = \sum n_ i[Z(s_ i)]_ k$

Thus the result follows from Lemma 42.25.3. $\square$

[1] For example, this holds if $s$ is a regular section of $\mathcal{L}|_ Y$.

## Comments (0)

There are also:

• 2 comment(s) on Section 42.25: Intersecting with an invertible sheaf

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Your email address will not be published.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