Lemma 81.21.2. In Situation 81.2.1 let $X/B$ be good. 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

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

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 81.18.1 is zero. By Definition 81.15.1 there exists a locally finite family $\{ W_ j\}$ of integral closed subspaces 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 81.19.4 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 81.21.1 we conclude that

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

However, by Lemma 81.18.2 we see that the right hand side is zero in $\mathop{\mathrm{CH}}\nolimits _ k(X)$ as desired. $\square$

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