Lemma 114.23.1. Let $(S, \delta )$ be as in Chow Homology, Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $\{ i_ j : D_ j \to X \} _{j \in J}$ be a locally finite collection of effective Cartier divisors on $X$. Let $n_ j > 0$, $j\in J$. Set $D = \sum _{j \in J} n_ j D_ j$, and denote $i : D \to X$ the inclusion morphism. Let $\alpha \in Z_{k + 1}(X)$. Then

$p : \coprod \nolimits _{j \in J} D_ j \longrightarrow D$

is proper and

$i^*\alpha = p_*\left(\sum n_ j i_ j^*\alpha \right)$

in $\mathop{\mathrm{CH}}\nolimits _ k(D)$.

Proof. The proof of this lemma is made a bit longer than expected by a subtlety concerning infinite sums of rational equivalences. In the quasi-compact case the family $D_ j$ is finite and the result is altogether easy and a straightforward consequence of Chow Homology, Lemma 42.24.2 and Divisors, Lemma 31.27.5 and the definitions.

The morphism $p$ is proper since the family $\{ D_ j\} _{j \in J}$ is locally finite. Write $\alpha = \sum _{a \in A} m_ a [W_ a]$ with $W_ a \subset X$ an integral closed subscheme of $\delta$-dimension $k + 1$. Denote $i_ a : W_ a \to X$ the closed immersion. We assume that $m_ a \not= 0$ for all $a \in A$ such that $\{ W_ a\} _{a \in A}$ is locally finite on $X$.

Observe that by Chow Homology, Definition 42.29.1 the class $i^*\alpha$ is the class of a cycle $\sum m_ a\beta _ a$ for certain $\beta _ a \in Z_ k(W_ a \cap D)$. Namely, if $W_ a \not\subset D$ then $\beta _ a = [D \cap W_ a]_ k$ and if $W_ a \subset D$, then $\beta _ a$ is a cycle representing $c_1(\mathcal{O}_ X(D)) \cap [W_ a]$.

For each $a \in A$ write $J = J_{a, 1} \amalg J_{a, 2} \amalg J_{a, 3}$ where

1. $j \in J_{a, 1}$ if and only if $W_ a \cap D_ j = \emptyset$,

2. $j \in J_{a, 2}$ if and only if $W_ a \not= W_ a \cap D_1 \not= \emptyset$, and

3. $j \in J_{a, 3}$ if and only if $W_ a \subset D_ j$.

Since the family $\{ D_ j\}$ is locally finite we see that $J_{a, 3}$ is a finite set. For every $a \in A$ and $j \in J$ we choose a cycle $\beta _{a, j} \in Z_ k(W_ a \cap D_ j)$ as follows

1. if $j \in J_{a, 1}$ we set $\beta _{a, j} = 0$,

2. if $j \in J_{a, 2}$ we set $\beta _{a, j} = [D_ j \cap W_ a]_ k$, and

3. if $j \in J_{a, 3}$ we choose $\beta _{a, j} \in Z_ k(W_ a)$ representing $c_1(i_ a^*\mathcal{O}_ X(D_ j)) \cap [W_ j]$.

We claim that

$\beta _ a \sim _{rat} \sum \nolimits _{j \in J} n_ j \beta _{a, j}$

in $\mathop{\mathrm{CH}}\nolimits _ k(W_ a \cap D)$.

Case I: $W_ a \not\subset D$. In this case $J_{a, 3} = \emptyset$. Thus it suffices to show that $[D \cap W_ a]_ k = \sum n_ j [D_ j \cap W_ a]_ k$ as cycles. This is Lemma 114.22.10.

Case II: $W_ a \subset D$. In this case $\beta _ a$ is a cycle representing $c_1(i_ a^*\mathcal{O}_ X(D)) \cap [W_ a]$. Write $D = D_{a, 1} + D_{a, 2} + D_{a, 3}$ with $D_{a, s} = \sum _{j \in J_{a, s}} n_ jD_ j$. By Divisors, Lemma 31.27.5 we have

\begin{eqnarray*} c_1(i_ a^*\mathcal{O}_ X(D)) \cap [W_ a] & = & c_1(i_ a^*\mathcal{O}_ X(D_{a, 1})) \cap [W_ a] + c_1(i_ a^*\mathcal{O}_ X(D_{a, 2})) \cap [W_ a] \\ & & + c_1(i_ a^*\mathcal{O}_ X(D_{a, 3})) \cap [W_ a]. \end{eqnarray*}

It is clear that the first term of the sum is zero. Since $J_{a, 3}$ is finite we see that the last term agrees with $\sum \nolimits _{j \in J_{a, 3}} n_ jc_1(i_ a^*\mathcal{L}_ j) \cap [W_ a]$, see Divisors, Lemma 31.27.5. This is represented by $\sum _{j \in J_{a, 3}} n_ j \beta _{a, j}$. Finally, by Case I we see that the middle term is represented by the cycle $\sum \nolimits _{j \in J_{a, 2}} n_ j[D_ j \cap W_ a]_ k = \sum _{j \in J_{a, 2}} n_ j\beta _{a, j}$. Whence the claim in this case.

At this point we are ready to finish the proof of the lemma. Namely, we have $i^*D \sim _{rat} \sum m_ a\beta _ a$ by our choice of $\beta _ a$. For each $a$ we have $\beta _ a \sim _{rat} \sum _ j \beta _{a, j}$ with the rational equivalence taking place on $D \cap W_ a$. Since the collection of closed subschemes $D \cap W_ a$ is locally finite on $D$, we see that also $\sum m_ a \beta _ a \sim _{rat} \sum _{a, j} m_ a\beta _{a, j}$ on $D$! (See Chow Homology, Remark 42.19.6.) Ok, and now it is clear that $\sum _ a m_ a\beta _{a, j}$ (viewed as a cycle on $D_ j$) represents $i_ j^*\alpha$ and hence $\sum _{a, j} m_ a\beta _{a, j}$ represents $p_* \sum _ j i_ j^*\alpha$ and we win. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).