The Stacks project

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. $\square$

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