Lemma 82.26.8. In Situation 82.2.1 let $X/B$ be good. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Then $c_1(\mathcal{L}) \in A^1(X)$ commutes with every element $c \in A^ p(X)$.
Proof. Let $p : L \to X$ be as in Lemma 82.25.2 and let $o : X \to L$ be the zero section. Observe that $p^*\mathcal{L}^{\otimes -1}$ has a canonical section whose vanishing locus is exactly the effective Cartier divisor $o(X)$. Let $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(X)$. Then we see that
\[ p^*(c_1(\mathcal{L}^{\otimes -1}) \cap \alpha ) = c_1(p^*\mathcal{L}^{\otimes -1}) \cap p^*\alpha = o_* o^* p^*\alpha \]
by Lemmas 82.19.2 and 82.24.1. Since $c$ is a bivariant class we have
\begin{align*} p^*(c \cap c_1(\mathcal{L}^{\otimes -1}) \cap \alpha ) & = c \cap p^*(c_1(\mathcal{L}^{\otimes -1}) \cap \alpha ) \\ & = c \cap o_* o^* p^*\alpha \\ & = o_* o^* p^*(c \cap \alpha ) \\ & = p^*(c_1(\mathcal{L}^{\otimes -1}) \cap c \cap \alpha ) \end{align*}
(last equality by the above applied to $c \cap \alpha $). Since $p^*$ is injective by a lemma cited above we get that $c_1(\mathcal{L}^{\otimes -1})$ is in the center of $A^*(X)$. This proves the lemma. $\square$
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