## 42.25 Intersecting with an invertible sheaf

In this section we study the following construction.

Definition 42.25.1. 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. We define, for every integer $k$, an operation

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

called intersection with the first Chern class of $\mathcal{L}$.

1. Given an integral closed subscheme $i : W \to X$ with $\dim _\delta (W) = k + 1$ we define

$c_1(\mathcal{L}) \cap [W] = i_*(c_1({i^*\mathcal{L}}) \cap [W])$

where the right hand side is defined in Definition 42.24.1.

2. For a general $(k + 1)$-cycle $\alpha = \sum n_ i [W_ i]$ we set

$c_1(\mathcal{L}) \cap \alpha = \sum n_ i c_1(\mathcal{L}) \cap [W_ i]$

Write each $c_1(\mathcal{L}) \cap W_ i = \sum _ j n_{i, j} [Z_{i, j}]$ with $\{ Z_{i, j}\} _ j$ a locally finite sum of integral closed subschemes of $W_ i$. Since $\{ W_ i\}$ is a locally finite collection of integral closed subschemes on $X$, it follows easily that $\{ Z_{i, j}\} _{i, j}$ is a locally finite collection of closed subschemes of $X$. Hence $c_1(\mathcal{L}) \cap \alpha = \sum n_ in_{i, j}[Z_{i, j}]$ is a cycle. Another, more convenient, way to think about this is to observe that the morphism $\coprod W_ i \to X$ is proper. Hence $c_1(\mathcal{L}) \cap \alpha$ can be viewed as the pushforward of a class in $\mathop{\mathrm{CH}}\nolimits _ k(\coprod W_ i) = \prod \mathop{\mathrm{CH}}\nolimits _ k(W_ i)$. This also explains why the result is well defined up to rational equivalence on $X$.

The main goal for the next few sections is to show that intersecting with $c_1(\mathcal{L})$ factors through rational equivalence. This is not a triviality.

Lemma 42.25.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{L}$, $\mathcal{N}$ be an invertible sheaves on $X$. Then

$c_1(\mathcal{L}) \cap \alpha + c_1(\mathcal{N}) \cap \alpha = c_1(\mathcal{L} \otimes _{\mathcal{O}_ X} \mathcal{N}) \cap \alpha$

in $\mathop{\mathrm{CH}}\nolimits _ k(X)$ for every $\alpha \in Z_{k + 1}(X)$. Moreover, $c_1(\mathcal{O}_ X) \cap \alpha = 0$ for all $\alpha$.

Proof. The additivity follows directly from Divisors, Lemma 31.27.5 and the definitions. To see that $c_1(\mathcal{O}_ X) \cap \alpha = 0$ consider the section $1 \in \Gamma (X, \mathcal{O}_ X)$. This restricts to an everywhere nonzero section on any integral closed subscheme $W \subset X$. Hence $c_1(\mathcal{O}_ X) \cap [W] = 0$ as desired. $\square$

Recall that $Z(s) \subset X$ denotes the zero scheme of a global section $s$ of an invertible sheaf on a scheme $X$, see Divisors, Definition 31.14.8.

Lemma 42.25.3. Let $(S, \delta )$ be as in Situation 42.7.1. Let $Y$ be locally of finite type over $S$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ Y$-module. Let $s \in \Gamma (Y, \mathcal{L})$. 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 $\mathcal{O}_{Y, \xi } \to \mathcal{L}_\xi$.

Write $[Y]_{k + 1} = \sum n_ i[Y_ i]$ where $Y_ i \subset Y$ are the irreducible components of $Y$ of $\delta$-dimension $k + 1$. Set $s_ i = s|_{Y_ i} \in \Gamma (Y_ i, \mathcal{L}|_{Y_ i})$. Then

42.25.3.1
$$\label{chow-equation-equal-as-cycles} [Z(s)]_ k = \sum n_ i[Z(s_ i)]_ k$$

as $k$-cycles on $Y$.

Proof. Let $Z \subset Y$ be an integral closed subscheme of $\delta$-dimension $k$. Let $\xi \in Z$ be its generic point. We want to compare the coefficient $n$ of $[Z]$ in the expression $\sum n_ i[Z(s_ i)]_ k$ with the coefficient $m$ of $[Z]$ in the expression $[Z(s)]_ k$. Choose a generator $s_\xi \in \mathcal{L}_\xi$. Write $A = \mathcal{O}_{Y, \xi }$, $L = \mathcal{L}_\xi$. Then $L = As_\xi$. Write $s = f s_\xi$ for some (unique) $f \in A$. Hypothesis (3) means that $f : A \to A$ is injective. Since $\dim _\delta (Y) \leq k + 1$ and $\dim _\delta (Z) = k$ we have $\dim (A) = 0$ or $1$. We have

$m = \text{length}_ A(A/(f))$

which is finite in either case.

If $\dim (A) = 0$, then $f : A \to A$ being injective implies that $f \in A^*$. Hence in this case $m$ is zero. Moreover, the condition $\dim (A) = 0$ means that $\xi$ does not lie on any irreducible component of $\delta$-dimension $k + 1$, i.e., $n = 0$ as well.

Now, let $\dim (A) = 1$. Since $A$ is a Noetherian local ring it has finitely many minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$. These correspond 1-1 with the $Y_ i$ passing through $\xi '$. Moreover $n_ i = \text{length}_{A_{\mathfrak q_ i}}(A_{\mathfrak q_ i})$. Also, the multiplicity of $[Z]$ in $[Z(s_ i)]_ k$ is $\text{length}_ A(A/(f, \mathfrak q_ i))$. Hence the equation to prove in this case is

$\text{length}_ A(A/(f)) = \sum \text{length}_{A_{\mathfrak q_ i}}(A_{\mathfrak q_ i}) \text{length}_ A(A/(f, \mathfrak q_ i))$

which follows from Lemma 42.3.2. $\square$

The following lemma is a useful result in order to compute the intersection product of the $c_1$ of an invertible sheaf and the cycle associated to a closed subscheme. Recall that $Z(s) \subset X$ denotes the zero scheme of a global section $s$ of an invertible sheaf on a scheme $X$, see Divisors, Definition 31.14.8.

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 $\mathcal{L}|_ Y$.

Comment #5374 by Jonathan Ng on

Hi, not sure if I am right but think you meant $Z_{k+1}(X)$ instead of $Z_{k-1}(X)$ in 02SP.

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