Lemma 42.61.1. The map $\times : \mathop{\mathrm{CH}}\nolimits _ n(X) \otimes _{\mathbf{Z}} \mathop{\mathrm{CH}}\nolimits _ m(Y) \to \mathop{\mathrm{CH}}\nolimits _{n + m}(X \times _ k Y)$ is well defined.

## 42.61 Exterior product

Let $k$ be a field. In this section we work over $S = \mathop{\mathrm{Spec}}(k)$ with $\delta : S \to \mathbf{Z}$ defined by sending the unique point to $0$, see Example 42.7.2.

Consider a cartesian square

of schemes locally of finite type over $k$. Then there is a canonical map

which is uniquely determined by the following rule: given integral closed subschemes $X' \subset X$ and $Y' \subset Y$ of dimensions $n$ and $m$ we have

in $\mathop{\mathrm{CH}}\nolimits _{n + m}(X \times _ k Y)$.

**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. $\square$

Lemma 42.61.2. Let $k$ be a field. Let $X$ be a scheme locally of finite type over $k$. Then we have a canonical identification

for all $p \in \mathbf{Z}$.

**Proof.**
Consider the element $[\mathop{\mathrm{Spec}}(k)] \in \mathop{\mathrm{CH}}\nolimits _0(\mathop{\mathrm{Spec}}(k))$. We get a map $A^ p(X \to \mathop{\mathrm{Spec}}(k)) \to \mathop{\mathrm{CH}}\nolimits _{-p}(X)$ by sending $c$ to $c \cap [\mathop{\mathrm{Spec}}(k)]$.

Conversely, suppose we have $\alpha \in \mathop{\mathrm{CH}}\nolimits _{-p}(X)$. Then we can define $c_\alpha \in A^ p(X \to \mathop{\mathrm{Spec}}(k))$ as follows: given $X' \to \mathop{\mathrm{Spec}}(k)$ and $\alpha ' \in \mathop{\mathrm{CH}}\nolimits _ n(X')$ we let

in $\mathop{\mathrm{CH}}\nolimits _{n - p}(X \times _ k X')$. To show that this is a bivariant class we write $\alpha = \sum n_ i[X_ i]$ as in Definition 42.8.1. Consider the composition

and denote $f : \coprod X_ i \to \mathop{\mathrm{Spec}}(k)$ the composition. Then $g$ is proper and $f$ is flat of relative dimension $-p$. Pullback along $f$ is a bivariant class $f^* \in A^ p(\coprod X_ i \to \mathop{\mathrm{Spec}}(k))$ by Lemma 42.33.2. Denote $\nu \in A^0(\coprod X_ i)$ the bivariant class which multiplies a cycle by $n_ i$ on the $i$th component. Thus $\nu \circ f^* \in A^ p(\coprod X_ i \to X)$. Finally, we have a bivariant class

by Lemma 42.33.4. The reader easily verifies that $c_\alpha $ is equal to this class and hence is itself a bivariant class.

To finish the proof we have to show that the two constructions are mutually inverse. Since $c_\alpha \cap [\mathop{\mathrm{Spec}}(k)] = \alpha $ this is clear for one of the two directions. For the other, let $c \in A^ p(X \to \mathop{\mathrm{Spec}}(k))$ and set $\alpha = c \cap [\mathop{\mathrm{Spec}}(k)]$. It suffices to prove that

when $X'$ is an integral scheme locally of finite type over $\mathop{\mathrm{Spec}}(k)$, see Lemma 42.35.3. However, then $p' : X' \to \mathop{\mathrm{Spec}}(k)$ is flat of relative dimension $\dim (X')$ and hence $[X'] = (p')^*[\mathop{\mathrm{Spec}}(k)]$. Thus the fact that the bivariant classes $c$ and $c_\alpha $ agree on $[\mathop{\mathrm{Spec}}(k)]$ implies they agree when capped against $[X']$ and the proof is complete. $\square$

Lemma 42.61.3. Let $k$ be a field. Let $X$ be a scheme locally of finite type over $k$. Let $c \in A^ p(X \to \mathop{\mathrm{Spec}}(k))$. Let $Y \to Z$ be a morphism of schemes locally of finite type over $k$. Let $c' \in A^ q(Y \to Z)$. Then $c \circ c' = c' \circ c$ in $A^{p + q}(X \times _ k Y \to Z)$.

**Proof.**
In the proof of Lemma 42.61.2 we have seen that $c$ is given by a combination of proper pushforward, multiplying by integers over connected components, and flat pullback. Since $c'$ commutes with each of these operations by definition of bivariant classes, we conclude. Some details omitted.
$\square$

Remark 42.61.4. The upshot of Lemmas 42.61.2 and 42.61.3 is the following. Let $k$ be a field. Let $X$ be a scheme locally of finite type over $k$. Let $\alpha \in \mathop{\mathrm{CH}}\nolimits _*(X)$. Let $Y \to Z$ be a morphism of schemes locally of finite type over $k$. Let $c' \in A^ q(Y \to Z)$. Then

in $\mathop{\mathrm{CH}}\nolimits _*(X \times _ k Y)$ for any $\beta \in \mathop{\mathrm{CH}}\nolimits _*(Z)$. Namely, this follows by taking $c = c_\alpha \in A^*(X \to \mathop{\mathrm{Spec}}(k))$ the bivariant class corresponding to $\alpha $, see proof of Lemma 42.61.2.

Lemma 42.61.5. Exterior product is associative. More precisely, let $k$ be a field, let $X, Y, Z$ be schemes locally of finite type over $k$, let $\alpha \in \mathop{\mathrm{CH}}\nolimits _*(X)$, $\beta \in \mathop{\mathrm{CH}}\nolimits _*(Y)$, $\gamma \in \mathop{\mathrm{CH}}\nolimits _*(Z)$. Then $(\alpha \times \beta ) \times \gamma = \alpha \times (\beta \times \gamma )$ in $\mathop{\mathrm{CH}}\nolimits _*(X \times _ k Y \times _ k Z)$.

**Proof.**
Omitted. Hint: associativity of fibre product of schemes.
$\square$

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