## 41.26 Intersecting with an invertible sheaf and rational equivalence

Applying the key lemma we obtain the fundamental properties of intersecting with invertible sheaves. In particular, we will see that $c_1(\mathcal{L}) \cap -$ factors through rational equivalence and that these operations for different invertible sheaves commute.

Lemma 41.26.1. Let $(S, \delta )$ be as in Situation 41.7.1. Let $X$ be locally of finite type over $S$. Assume $X$ integral and $\dim _\delta (X) = n$. Let $\mathcal{L}$, $\mathcal{N}$ be invertible on $X$. Choose a nonzero meromorphic section $s$ of $\mathcal{L}$ and a nonzero meromorphic section $t$ of $\mathcal{N}$. Set $\alpha = \text{div}_\mathcal {L}(s)$ and $\beta = \text{div}_\mathcal {N}(t)$. Then

\[ c_1(\mathcal{N}) \cap \alpha = c_1(\mathcal{L}) \cap \beta \]

in $A_{n - 2}(X)$.

**Proof.**
Immediate from the key Lemma 41.25.1 and the discussion preceding it.
$\square$

Lemma 41.26.2. Let $(S, \delta )$ be as in Situation 41.7.1. Let $X$ be locally of finite type over $S$. 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

\[ c_1(\mathcal{L}) \cap - : A_{k + 1}(X) \to A_ k(X) \]

**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 41.23.1 is zero. By Definition 41.19.1 there exists a locally finite family $\{ W_ j\} $ of integral closed subschemes 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 41.24.4 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 41.26.1 we conclude that

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

However, by Lemma 41.23.2 we see that the right hand side is zero in $A_ k(X)$ as desired.
$\square$

Let $(S, \delta )$ be as in Situation 41.7.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{L}$ be invertible on $X$. We will denote

\[ c_1(\mathcal{L})^ s \cap - : A_{k + s}(X) \to A_ k(X) \]

the operation $c_1(\mathcal{L}) \cap - $. This makes sense by Lemma 41.26.2. We will denote $c_1(\mathcal{L}^ s \cap -$ the $s$-fold iterate of this operation for all $s \geq 0$.

Lemma 41.26.3. Let $(S, \delta )$ be as in Situation 41.7.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{L}$, $\mathcal{N}$ be invertible on $X$. For any $\alpha \in A_{k + 2}(X)$ we have

\[ c_1(\mathcal{L}) \cap c_1(\mathcal{N}) \cap \alpha = c_1(\mathcal{N}) \cap c_1(\mathcal{L}) \cap \alpha \]

as elements of $A_ k(X)$.

**Proof.**
Write $\alpha = \sum m_ j[Z_ j]$ for some locally finite collection of integral closed subschemes $Z_ j \subset X$ with $\dim _\delta (Z_ j) = k + 2$. Consider the proper morphism $p : \coprod Z_ j \to X$. Set $\alpha ' = \sum m_ j[Z_ j]$ as a $(k + 2)$-cycle on $\coprod Z_ j$. By several applications of Lemma 41.24.4 we see that $c_1(\mathcal{L}) \cap c_1(\mathcal{N}) \cap \alpha = p_*(c_1(p^*\mathcal{L}) \cap c_1(p^*\mathcal{N}) \cap \alpha ')$ and $c_1(\mathcal{N}) \cap c_1(\mathcal{L}) \cap \alpha = p_*(c_1(p^*\mathcal{N}) \cap c_1(p^*\mathcal{L}) \cap \alpha ')$. Hence it suffices to prove the formula in case $X$ is integral and $\alpha = [X]$. In this case the result follows from Lemma 41.26.1 and the definitions.
$\square$

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