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 $\mathcal{L}|_ Y$.

There are also:

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

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).