Lemma 81.17.2. In Situation 81.2.1 let $X/B$ be good. Assume $X$ is integral and $n = \dim _\delta (X)$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $s \in \Gamma (X, \mathcal{L})$ be a nonzero global section. Then

\[ \text{div}_\mathcal {L}(s) = [Z(s)]_{n - 1} \]

in $Z_{n - 1}(X)$ and

\[ c_1(\mathcal{L}) \cap [X] = [Z(s)]_{n - 1} \]

in $\mathop{\mathrm{CH}}\nolimits _{n - 1}(X)$.

**Proof.**
Let $Z \subset X$ be an integral closed subspace of $\delta $-dimension $n - 1$. Let $\xi \in |Z|$ be its generic point. To prove the first equality we compare the coefficients of $Z$ on both sides. Choose an elementary étale neighbourhood $(U, u) \to (X, \xi )$, see Decent Spaces, Section 67.11 and recall that $\mathcal{O}_{X, \xi }^ h = \mathcal{O}_{U, u}^ h$ in this case. After replacing $U$ by an open neighbourhood of $u$ we may assume there is a trivializing section $s_ U$ of $\mathcal{L}|_ U$. Write $s|_ U = f s_ U$ for some $f \in \Gamma (U, \mathcal{O}_ U)$. Then $Z \times _ X U$ is equal to $V(f)$ as a closed subscheme of $U$, see Divisors on Spaces, Definition 70.7.6. As in Spaces over Fields, Section 71.7 denote $\mathcal{L}_\xi $ the pullback of $\mathcal{L}$ under the canonical morphism $c_\xi : \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi }^ h) \to X$. Denote $s_\xi $ the pullback of $s_ U$; it is a trivialization of $\mathcal{L}_\xi $. Then we see that $c_\xi ^*(s) = fs_\xi $. The coefficient of $Z$ in $[Z(s)]_{n - 1}$ is by definition

\[ \text{length}_{\mathcal{O}_{U, u}}(\mathcal{O}_{U, u}/f\mathcal{O}_{U, u}) \]

Since $\mathcal{O}_{U, u} \to \mathcal{O}_{X, \xi }^ h$ is flat and identifies residue fields this is equal to

\[ \text{length}_{\mathcal{O}_{X, \xi }^ h} (\mathcal{O}_{X, \xi }^ h/f\mathcal{O}_{X, \xi }^ h) \]

by Algebra, Lemma 10.52.13. This final quantity is equal to $\text{ord}_{Z, \mathcal{L}}(s)$ by Spaces over Fields, Definition 71.7.1, i.e., to the coefficient of $Z$ in $\text{div}_\mathcal {L}(s)$ as desired.
$\square$

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