Lemma 42.29.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $X$ be integral and $n = \dim _\delta (X)$. Let $i : D \to X$ be an effective Cartier divisor. Let $\mathcal{N}$ be an invertible $\mathcal{O}_ X$-module and let $t$ be a nonzero meromorphic section of $\mathcal{N}$. Then $i^*\text{div}_\mathcal {N}(t) = c_1(\mathcal{N}) \cap [D]_{n - 1}$ in $\mathop{\mathrm{CH}}\nolimits _{n - 2}(D)$.

Proof. Write $\text{div}_\mathcal {N}(t) = \sum \text{ord}_{Z_ i, \mathcal{N}}(t)[Z_ i]$ for some integral closed subschemes $Z_ i \subset X$ of $\delta$-dimension $n - 1$. We may assume that the family $\{ Z_ i\}$ is locally finite, that $t \in \Gamma (U, \mathcal{N}|_ U)$ is a generator where $U = X \setminus \bigcup Z_ i$, and that every irreducible component of $D$ is one of the $Z_ i$, see Divisors, Lemmas 31.26.1, 31.26.4, and 31.27.2.

Set $\mathcal{L} = \mathcal{O}_ X(D)$. Denote $s \in \Gamma (X, \mathcal{O}_ X(D)) = \Gamma (X, \mathcal{L})$ the canonical section. We will apply the discussion of Section 42.26 to our current situation. For each $i$ let $\xi _ i \in Z_ i$ be its generic point. Let $B_ i = \mathcal{O}_{X, \xi _ i}$. For each $i$ we pick generators $s_ i \in \mathcal{L}_{\xi _ i}$ and $t_ i \in \mathcal{N}_{\xi _ i}$ over $B_ i$ but we insist that we pick $s_ i = s$ if $Z_ i \not\subset D$. Write $s = f_ i s_ i$ and $t = g_ i t_ i$ with $f_ i, g_ i \in B_ i$. Then $\text{ord}_{Z_ i, \mathcal{N}}(t) = \text{ord}_{B_ i}(g_ i)$. On the other hand, we have $f_ i \in B_ i$ and

$[D]_{n - 1} = \sum \text{ord}_{B_ i}(f_ i)[Z_ i]$

because of our choices of $s_ i$. We claim that

$i^*\text{div}_\mathcal {N}(t) = \sum \text{ord}_{B_ i}(g_ i) \text{div}_{\mathcal{L}|_{Z_ i}}(s_ i|_{Z_ i})$

as cycles. More precisely, the right hand side is a cycle representing the left hand side. Namely, this is clear by our formula for $\text{div}_\mathcal {N}(t)$ and the fact that $\text{div}_{\mathcal{L}|_{Z_ i}}(s_ i|_{Z_ i}) = [Z(s_ i|_{Z_ i})]_{n - 2} = [Z_ i \cap D]_{n - 2}$ when $Z_ i \not\subset D$ because in that case $s_ i|_{Z_ i} = s|_{Z_ i}$ is a regular section, see Lemma 42.23.2. Similarly,

$c_1(\mathcal{N}) \cap [D]_{n - 1} = \sum \text{ord}_{B_ i}(f_ i) \text{div}_{\mathcal{N}|_{Z_ i}}(t_ i|_{Z_ i})$

The key formula (Lemma 42.26.1) gives the equality

$\sum \left( \text{ord}_{B_ i}(f_ i) \text{div}_{\mathcal{N}|_{Z_ i}}(t_ i|_{Z_ i}) - \text{ord}_{B_ i}(g_ i) \text{div}_{\mathcal{L}|_{Z_ i}}(s_ i|_{Z_ i}) \right) = \sum \text{div}_{Z_ i}(\partial _{B_ i}(f_ i, g_ i))$

of cycles. If $Z_ i \not\subset D$, then $f_ i = 1$ and hence $\text{div}_{Z_ i}(\partial _{B_ i}(f_ i, g_ i)) = 0$. Thus we get a rational equivalence between our specific cycles representing $i^*\text{div}_\mathcal {N}(t)$ and $c_1(\mathcal{N}) \cap [D]_{n - 1}$ on $D$. This finishes the proof. $\square$

Comment #6643 by WhatJiaranEatsTonight on

I think the equation in the statement should be $i^*div_N(t)=c_1(N|_D)\cap[D]$. Otherwise, the element in the right is an element in $CH_{n-2}(X)$.(Though they are equal in the cycle level.)

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