The Stacks project

Proof. A first remark is that if $\alpha = \sum n_ i[X_ i]$ and $\beta = \sum m_ j[Y_ j]$ with $X_ i \subset X$ and $Y_ j \subset Y$ locally finite families of integral closed subschemes of dimensions $n$ and $m$, then $X_ i \times _ k Y_ j$ is a locally finite collection of closed subschemes of $X \times _ k Y$ of dimensions $n + m$ and we can indeed consider

as a $(n + m)$-cycle on $X \times _ k Y$. In this way we obtain an additive map $\times : Z_ n(X) \otimes _{\mathbf{Z}} Z_ m(Y) \to Z_{n + m}(X \times _ k Y)$. The problem is to show that this procedure is compatible with rational equivalence.

Let $i : X' \to X$ be the inclusion morphism of an integral closed subscheme of dimension $n$. Then flat pullback along the morphism $p' : X' \to \mathop{\mathrm{Spec}}(k)$ is an element $(p')^* \in A^{-n}(X' \to \mathop{\mathrm{Spec}}(k))$ by Lemma 42.33.2 and hence $c' = i_* \circ (p')^* \in A^{-n}(X \to \mathop{\mathrm{Spec}}(k))$ by Lemma 42.33.4. This produces maps

which the reader easily sends $[Y']$ to $[X' \times _ k Y']_{n + m}$ for any integral closed subscheme $Y' \subset Y$ of dimension $m$. Hence the construction $([X'], [Y']) \mapsto [X' \times _ k Y']_{n + m}$ factors through rational equivalence in the second variable, i.e., gives a well defined map $Z_ n(X) \otimes _{\mathbf{Z}} \mathop{\mathrm{CH}}\nolimits _ m(Y) \to \mathop{\mathrm{CH}}\nolimits _{n + m}(X \times _ k Y)$. By symmetry the same is true for the other variable and we conclude. $\square$