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
\[ \xymatrix{ X \times _ k Y \ar[r] \ar[d] & Y \ar[d] \\ X \ar[r] & \mathop{\mathrm{Spec}}(k) = S } \]
of schemes locally of finite type over $k$. Then there is a canonical map
\[ \times : \mathop{\mathrm{CH}}\nolimits _ n(X) \otimes _{\mathbf{Z}} \mathop{\mathrm{CH}}\nolimits _ m(Y) \longrightarrow \mathop{\mathrm{CH}}\nolimits _{n + m}(X \times _ k Y) \]
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
\[ [X'] \times [Y'] = [X' \times _ k Y']_{n + m} \]
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
\[ \alpha \times \beta = \sum n_ i m_ j [X_ i \times _ k Y_ j]_{n + m} \]
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
\[ c' \cap - : \mathop{\mathrm{CH}}\nolimits _ m(Y) \longrightarrow \mathop{\mathrm{CH}}\nolimits _{m + n}(X \times _ k Y) \]
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 $p \in \mathbf{Z}$.
\[ A^ p(X \to \mathop{\mathrm{Spec}}(k)) = \mathop{\mathrm{CH}}\nolimits _{-p}(X) \]
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
\[ c_\alpha \cap \alpha ' = \alpha \times \alpha ' \]
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
\[ \coprod X_ i \xrightarrow {g} X \to \mathop{\mathrm{Spec}}(k) \]
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
\[ g_* \circ \nu \circ f^* \]
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
\[ c \cap [X'] = c_\alpha \cap [X'] \]
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.
\[ \alpha \times (c' \cap \beta ) = c' \cap (\alpha \times \beta ) \]
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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)