## 115.24 Commutativity of intersecting divisors

The results of this section were originally used to provide an alternative proof of the lemmas of Chow Homology, Section 42.28 and a weak version of Chow Homology, Lemma 42.30.5.

Lemma 115.24.1. 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 115.23.10.

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. $\square$

Lemma 115.24.2. 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. $\square$

Lemma 115.24.3. 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 $\{ D_ j\} _{j \in J}$ be a locally finite collection of effective Cartier divisors on $X$. Let $n_ j, m_ j \geq 0$ be collections of nonnegative integers. Set $D = \sum n_ j D_ j$ and $D' = \sum m_ j D_ j$. Assume that $\dim _\delta (D_ j \cap D_{j'}) = n - 2$ for every $j \not= j'$. Then $D \cdot [D']_{n - 1} = D' \cdot [D]_{n - 1}$ in $\mathop{\mathrm{CH}}\nolimits _{n - 2}(X)$.

Proof. This lemma is a trivial consequence of Lemmas 115.23.10 and 115.24.2 in case the sums are finite, e.g., if $X$ is quasi-compact. Hence we suggest the reader skip the proof.

Here is the proof in the general case. Let $i_ j : D_ j \to X$ be the closed immersions Let $p : \coprod D_ j \to X$ denote coproduct of the morphisms $i_ j$. Let $\{ Z_ a\} _{a \in A}$ be the collection of irreducible components of $\bigcup D_ j$. For each $j$ we write

$[D_ j]_{n - 1} = \sum d_{j, a}[Z_ a].$

By Lemma 115.23.10 we have

$[D]_{n - 1} = \sum n_ j d_{j, a} [Z_ a], \quad [D']_{n - 1} = \sum m_ j d_{j, a} [Z_ a].$

By Lemma 115.24.1 we have

$D \cdot [D']_{n - 1} = p_*\left(\sum n_ j i_ j^*[D']_{n - 1} \right), \quad D' \cdot [D]_{n - 1} = p_*\left(\sum m_{j'} i_{j'}^*[D]_{n - 1} \right).$

As in the definition of the Gysin homomorphisms (see Chow Homology, Definition 42.29.1) we choose cycles $\beta _{a, j}$ on $D_ j \cap Z_ a$ representing $i_ j^*[Z_ a]$. (Note that in fact $\beta _{a, j} = [D_ j \cap Z_ a]_{n - 2}$ if $Z_ a$ is not contained in $D_ j$, i.e., there is no choice in that case.) Now since $p$ is a closed immersion when restricted to each of the $D_ j$ we can (and we will) view $\beta _{a, j}$ as a cycle on $X$. Plugging in the formulas for $[D]_{n - 1}$ and $[D']_{n - 1}$ obtained above we see that

$D \cdot [D']_{n - 1} = \sum \nolimits _{j, j', a} n_ j m_{j'} d_{j', a} \beta _{a, j}, \quad D' \cdot [D]_{n - 1} = \sum \nolimits _{j, j', a} m_{j'} n_ j d_{j, a} \beta _{a, j'}.$

Moreover, with the same conventions we also have

$D_ j \cdot [D_{j'}]_{n - 1} = \sum d_{j', a} \beta _{a, j}.$

In these terms Lemma 115.24.2 (see also its proof) says that for $j \not= j'$ the cycles $\sum d_{j', a} \beta _{a, j}$ and $\sum d_{j, a} \beta _{a, j'}$ are equal as cycles! Hence we see that

\begin{eqnarray*} D \cdot [D']_{n - 1} & = & \sum \nolimits _{j, j', a} n_ j m_{j'} d_{j', a} \beta _{a, j} \\ & = & \sum \nolimits _{j \not= j'} n_ j m_{j'} \left(\sum \nolimits _ a d_{j', a} \beta _{a, j}\right) + \sum \nolimits _{j, a} n_ j m_ j d_{j, a} \beta _{a, j} \\ & = & \sum \nolimits _{j \not= j'} n_ j m_{j'} \left(\sum \nolimits _ a d_{j, a} \beta _{a, j'}\right) + \sum \nolimits _{j, a} n_ j m_ j d_{j, a} \beta _{a, j} \\ & = & \sum \nolimits _{j, j', a} m_{j'} n_ j d_{j, a} \beta _{a, j'} \\ & = & D' \cdot [D]_{n - 1} \end{eqnarray*}

and we win. $\square$

Lemma 115.24.4. 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 $X$. Then

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

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

First proof of Lemma 115.24.4. First, let us prove this in case $X$ is quasi-compact. In this case, apply Lemma 115.23.11 to $X$ and the two element set $\{ D, D'\}$ of effective Cartier divisors. Thus we get a proper morphism $b : X' \to X$, a finite collection of effective Cartier divisors $D'_ j \subset X'$ intersecting pairwise in codimension $\geq 2$, with $b^{-1}(D) = \sum n_ j D'_ j$, and $b^{-1}(D') = \sum m_ j D'_ j$. Note that $b_*[b^{-1}(D)]_{n - 1} = [D]_{n - 1}$ in $Z_{n - 1}(X)$ and similarly for $D'$, see Lemma 115.23.5. Hence, by Chow Homology, Lemma 42.26.4 we have

$D \cdot [D']_{n - 1} = b_*\left(b^{-1}(D) \cdot [b^{-1}(D')]_{n - 1}\right)$

in $\mathop{\mathrm{CH}}\nolimits _{n - 2}(X)$ and similarly for the other term. Hence the lemma follows from the equality $b^{-1}(D) \cdot [b^{-1}(D')]_{n - 1} = b^{-1}(D') \cdot [b^{-1}(D)]_{n - 1}$ in $\mathop{\mathrm{CH}}\nolimits _{n - 2}(X')$ of Lemma 115.24.3.

Note that in the proof above, each referenced lemma works also in the general case (when $X$ is not assumed quasi-compact). The only minor change in the general case is that the morphism $b : U' \to U$ we get from applying Lemma 115.23.11 has as its target an open $U \subset X$ whose complement has codimension $\geq 3$. Hence by Chow Homology, Lemma 42.19.3 we see that $\mathop{\mathrm{CH}}\nolimits _{n - 2}(U) = \mathop{\mathrm{CH}}\nolimits _{n - 2}(X)$ and after replacing $X$ by $U$ the rest of the proof goes through unchanged. $\square$

Second proof of Lemma 115.24.4. Let $\mathcal{I} = \mathcal{O}_ X(-D)$ and $\mathcal{I}' = \mathcal{O}_ X(-D')$ be the invertible ideal sheaves of $D$ and $D'$. We denote $\mathcal{I}_{D'} = \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{O}_{D'}$ and $\mathcal{I}'_ D = \mathcal{I}' \otimes _{\mathcal{O}_ X} \mathcal{O}_ D$. We can restrict the inclusion map $\mathcal{I} \to \mathcal{O}_ X$ to $D'$ to get a map

$\varphi : \mathcal{I}_{D'} \longrightarrow \mathcal{O}_{D'}$

and similarly

$\psi : \mathcal{I}'_ D \longrightarrow \mathcal{O}_ D$

It is clear that

$\mathop{\mathrm{Coker}}(\varphi ) \cong \mathcal{O}_{D \cap D'} \cong \mathop{\mathrm{Coker}}(\psi )$

and

$\mathop{\mathrm{Ker}}(\varphi ) \cong \frac{\mathcal{I} \cap \mathcal{I}'}{\mathcal{I}\mathcal{I}'} \cong \mathop{\mathrm{Ker}}(\psi ).$

Hence we see that

$\gamma = [\mathcal{I}_{D'}] - [\mathcal{O}_{D'}] = [\mathcal{I}'_ D] - [\mathcal{O}_ D]$

in $K_0(\textit{Coh}_{\leq n - 1}(X))$. On the other hand it is clear that

$[\mathcal{I}'_ D]_{n - 1} = [D]_{n - 1}, \quad [\mathcal{I}_{D'}]_{n - 1} = [D']_{n - 1}.$

and that

$\mathcal{O}_ X(D') \otimes \mathcal{I}'_ D = \mathcal{O}_ D, \quad \mathcal{O}_ X(D) \otimes \mathcal{I}_{D'} = \mathcal{O}_{D'}.$

By Chow Homology, Lemma 42.69.7 (applied two times) this means that the element $\gamma$ is an element of $B_{n - 2}(X)$, and maps to both $c_1(\mathcal{O}_ X(D')) \cap [D]_{n - 1}$ and to $c_1(\mathcal{O}_ X(D)) \cap [D']_{n - 1}$ and we win (since the map $B_{n - 2}(X) \to \mathop{\mathrm{CH}}\nolimits _{n - 2}(X)$ is well defined – which is the key to this proof). $\square$

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