Lemma 114.23.2. Let $(S, \delta )$ be as in Chow Homology, Situation 42.7.1. Let $X$ be locally of finite type over $S$. Assume $X$ integral and $\dim _\delta (X) = n$. Let $D$, $D'$ be effective Cartier divisors on $X$. Assume $\dim _\delta (D \cap D') = n - 2$. Let $i : D \to X$, resp. $i' : D' \to X$ be the corresponding closed immersions. Then

1. there exists a cycle $\alpha \in Z_{n - 2}(D \cap D')$ whose pushforward to $D$ represents $i^*[D']_{n - 1} \in \mathop{\mathrm{CH}}\nolimits _{n - 2}(D)$ and whose pushforward to $D'$ represents $(i')^*[D]_{n - 1} \in \mathop{\mathrm{CH}}\nolimits _{n - 2}(D')$, and

2. we have

$D \cdot [D']_{n - 1} = D' \cdot [D]_{n - 1}$

in $\mathop{\mathrm{CH}}\nolimits _{n - 2}(X)$.

Proof. Part (2) is a trivial consequence of part (1). Let us write $[D]_{n - 1} = \sum n_ a[Z_ a]$ and $[D']_{n - 1} = \sum m_ b[Z_ b]$ with $Z_ a$ the irreducible components of $D$ and $[Z_ b]$ the irreducible components of $D'$. According to Chow Homology, Definition 42.29.1, we have $i^*D' = \sum m_ b i^*[Z_ b]$ and $(i')^*D = \sum n_ a(i')^*[Z_ a]$. By assumption, none of the irreducible components $Z_ b$ is contained in $D$, and hence $i^*[Z_ b] = [Z_ b\cap D]_{n - 2}$ by definition. Similarly $(i')^*[Z_ a] = [Z_ a \cap D']_{n - 2}$. Hence we are trying to prove the equality of cycles

$\sum n_ a[Z_ a \cap D']_{n - 2} = \sum m_ b[Z_ b \cap D]_{n - 2}$

which are indeed supported on $D \cap D'$. Let $W \subset X$ be an integral closed subscheme with $\dim _\delta (W) = n - 2$. Let $\xi \in W$ be its generic point. Set $R = \mathcal{O}_{X, \xi }$. It is a Noetherian local domain. Note that $\dim (R) = 2$. Let $f \in R$, resp. $f' \in R$ be an element defining the ideal of $D$, resp. $D'$. By assumption $\dim (R/(f, f')) = 0$. Let $\mathfrak q'_1, \ldots , \mathfrak q'_ t \subset R$ be the minimal primes over $(f')$, let $\mathfrak q_1, \ldots , \mathfrak q_ s \subset R$ be the minimal primes over $(f)$. The equality above comes down to the equality

$\sum _{i = 1, \ldots , s} \text{length}_{R_{\mathfrak q_ i}}(R_{\mathfrak q_ i}/(f)) \text{ord}_{R/\mathfrak q_ i}(f') = \sum _{j = 1, \ldots , t} \text{length}_{R_{\mathfrak q'_ j}}(R_{\mathfrak q'_ j}/(f')) \text{ord}_{R/\mathfrak q'_ j}(f).$

By Chow Homology, Lemma 42.3.1 applied with $M = R/(f)$ the left hand side of this equation is equal to

$\text{length}_ R(R/(f, f')) - \text{length}_ R(\mathop{\mathrm{Ker}}(f' : R/(f) \to R/(f)))$

OK, and now we note that $\mathop{\mathrm{Ker}}(f' : R/(f) \to R/(f))$ is canonically isomorphic to $((f) \cap (f'))/(ff')$ via the map $x \bmod (f) \mapsto f'x \bmod (ff')$. Hence the left hand side is

$\text{length}_ R(R/(f, f')) - \text{length}_ R((f) \cap (f')/(ff'))$

Since this is symmetric in $f$ and $f'$ 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).