The Stacks project

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 42.27 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

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

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 42.24.2. Similarly,

The key formula (Lemma 42.27.1) gives the equality

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. $\square$

Proof. Let $\alpha \in Z_{k + 1}(X)$ and assume that $\alpha \sim _{rat} 0$. This means there exists a locally finite collection of integral closed subschemes $W_ j \subset X$ of $\delta $-dimension $k + 2$ and $f_ j \in R(W_ j)^*$ such that $\alpha = \sum i_{j, *}\text{div}_{W_ j}(f_ j)$. Set $X' = \coprod W_ i$ and consider the diagram

of Remark 42.29.7. Since $X' \to X$ is proper we see that $i^*p_* = q_*(i')^*$ by Lemma 42.29.8. As we know that $q_*$ factors through rational equivalence (Lemma 42.20.3), it suffices to prove the result for $\alpha ' = \sum \text{div}_{W_ j}(f_ j)$ on $X'$. Clearly this reduces us to the case where $X$ is integral and $\alpha = \text{div}(f)$ for some $f \in R(X)^*$.

Assume $X$ is integral and $\alpha = \text{div}(f)$ for some $f \in R(X)^*$. If $X = D$, then we see that $i^*\alpha $ is equal to $c_1(\mathcal{L}) \cap \alpha $. This is rationally equivalent to zero by Lemma 42.28.2. If $D \not= X$, then we see that $i^*\text{div}_ X(f)$ is equal to $c_1(\mathcal{O}_ D) \cap [D]_{n - 1}$ in $\mathop{\mathrm{CH}}\nolimits _{n - 2}(D)$ by Lemma 42.30.1. Of course capping with $c_1(\mathcal{O}_ D)$ is the zero map (Lemma 42.25.2). $\square$

Proof. This is immediate from the definition of $i_*$ on cycles and the definition of $i^*$ given in Definition 42.29.1. $\square$

Proof. With exactly the same proof as in Lemma 42.30.2 this follows from Lemmas 42.26.4, 42.28.3, and 42.30.1. $\square$

Proof. Denote $j : D \cap D' \to D$ and $j' : D \cap D' \to D'$ the closed immersions corresponding to $(\mathcal{L}|_{D'}, s|_{D'}$ and $(\mathcal{L}'_ D, s|_ D)$. We have to show that $(j')^*i^*\alpha = j^* (i')^*\alpha $ for all $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(X)$. Let $W \subset X$ be an integral closed subscheme of dimension $k$. Let us prove the equality in case $\alpha = [W]$. We will deduce it from the key formula.

We let $\sigma $ be a nonzero meromorphic section of $\mathcal{L}|_ W$ which we require to be equal to $s|_ W$ if $W \not\subset D$. We let $\sigma '$ be a nonzero meromorphic section of $\mathcal{L}'|_ W$ which we require to be equal to $s'|_ W$ if $W \not\subset D'$. Write

and similarly

as in the discussion in Section 42.27. Then we see that $Z_ i \subset D$ if $n_ i \not= 0$ and $Z'_ i \subset D'$ if $n'_ i \not= 0$. For each $i$, let $\xi _ i \in Z_ i$ be the generic point. As in Section 42.27 we choose for each $i$ an element $\sigma _ i \in \mathcal{L}_{\xi _ i}$, resp. $\sigma '_ i \in \mathcal{L}'_{\xi _ i}$ which generates over $B_ i = \mathcal{O}_{W, \xi _ i}$ and which is equal to the image of $s$, resp. $s'$ if $Z_ i \not\subset D$, resp. $Z_ i \not\subset D'$. Write $\sigma = f_ i \sigma _ i$ and $\sigma ' = f'_ i\sigma '_ i$ so that $n_ i = \text{ord}_{B_ i}(f_ i)$ and $n'_ i = \text{ord}_{B_ i}(f'_ i)$. From our definitions it follows that

as cycles and

The key formula (Lemma 42.27.1) now gives the equality

of cycles. Note that $\text{div}_{Z_ i}(\partial _{B_ i}(f_ i, f'_ i)) = 0$ if $Z_ i \not\subset D \cap D'$ because in this case either $f_ i = 1$ or $f'_ i = 1$. Thus we get a rational equivalence between our specific cycles representing $(j')^*i^*[W]$ and $j^*(i')^*[W]$ on $D \cap D' \cap W$. By Remark 42.19.6 the result follows for general $\alpha $. $\square$