Lemma 82.21.2. In Situation 82.2.1 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
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 82.18.1 is zero. By Definition 82.15.1 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
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 82.19.4 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 82.21.1 we conclude that
However, by Lemma 82.18.2 we see that the right hand side is zero in \mathop{\mathrm{CH}}\nolimits _ k(X) as desired. \square
Comments (0)