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Exterior products in chow

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aisejohan committed May 20, 2019
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@@ -9773,6 +9773,150 @@ \section{Higher codimension gysin homomorphisms}



\section{Exterior product}
\label{section-exterior-product}

\noindent
Let $k$ be a field and let $(S, \delta)$ be as in Example \ref{example-field}
which is a special case of our general Situation \ref{situation-setup}.
Consider a cartesian square
$$
\xymatrix{
X \times_k Y \ar[r] \ar[d] & Y \ar[d] \\
X \ar[r] & \Spec(k) = S
}
$$
of schemes locally of finite type over $k$. Then there is a canonical map
$$
\times :
A_n(X) \otimes_{\mathbf{Z}} A_m(Y)
\longrightarrow
A_{n + m}(X \times_k Y)
$$
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 $A_{n + m}(X \times_k Y)$.

\begin{lemma}
\label{lemma-exterior-product-well-defined}
The map
$\times : A_n(X) \otimes_{\mathbf{Z}} A_m(Y) \to A_{n + m}(X \times_k Y)$
is well defined.
\end{lemma}

\begin{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.

\medskip\noindent
Let $i : X' \to X$ be the inclusion morphism of
an integral closed subscheme of dimension $n$.
Then flat pullback along the morphism $p' : X' \to \Spec(k)$ is an element
$(p')^* \in A^{-n}(X' \to \Spec(k))$ by
Lemma \ref{lemma-flat-pullback-bivariant}
and hence $c' = i_* \circ (p')^* \in A^{-n}(X \to \Spec(k))$ by
Lemma \ref{lemma-push-proper-bivariant}.
This produces maps
$$
c' \cap - : A_m(Y) \longrightarrow A_{m + n}(X \times_k Y)
$$
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
$Z_n(X) \otimes_{\mathbf{Z}} A_m(Y) \to A_{n + m}(X \times_k Y)$.
By symmetry the same is true for the other variable and we conclude.
\end{proof}

\begin{lemma}
\label{lemma-chow-cohomology-towards-point}
Let $k$ be a field and let $(S, \delta)$ be as in Example \ref{example-field}.
Let $X \to \Spec(k) \to S$ be morphisms locally of finite type.
Then we have a canonical identification
$$
A^p(X \to \Spec(k)) = A_{-p}(X)
$$
for all $p \in \mathbf{Z}$.
\end{lemma}

\begin{proof}
Observe that $A_*(\Spec(k))$ is nonzero only in degree $d$
with generator $[\Spec(k)]$. Hence we get a map
$A^p(X \to \Spec(k)) \to A_{-p}(X)$ by sending $c$ to
$c \cap [\Spec(k)]$.

\medskip\noindent
Conversely, suppose we have $\alpha \in A_{-p}(X)$.
Then we can define $c_\alpha \in A^p(X \to \Spec(k))$ as
follows: given $X' \to \Spec(k)$ and $\alpha' \in A_n(X')$
we let
$$
c_\alpha \cap \alpha' = \alpha \times \alpha'
$$
in $A_{n - p}(X \times_k X')$. To show that this is a bivariant
class we write $\alpha = \sum n_i[X_i]$ as a locally finite sum
of cycles on $X$. Consider the composition
$$
\coprod X_i \xrightarrow{g} X \to \Spec(k)
$$
and denote $f : \coprod X_i \to \Spec(k)$ the composition.
Then $g$ is proper and $f$ is flat of relative dimension $-p$.
Pullback along $f$ is a bivariant class
$f^* \in A^p(\coprod X_i \to \Spec(k))$ by
Lemma \ref{lemma-flat-pullback-bivariant}.
Denote $c' \in A^0(\coprod X_i)$ the bivariant class
which multiplies a cycle by $n_i$ on the $i$th component.
Thus $c' \circ f^* \in A^p(\coprod X_i \to X)$.
Finally, we have a bivariant class
$$
g_* \circ c' \circ f^*
$$
by Lemma \ref{lemma-push-proper-bivariant}. The reader easily
verifies that $c_\alpha$ is equal to this class and hence
is itself a bivariant class.

\medskip\noindent
To finish the proof we have to show that the two constructions
are mutually inverse. Since $c_\alpha \cap [\Spec(k)] = \alpha$
this is clear for one of the two directions. For the other, let
$c \in A^p(X \to \Spec(k))$ and set $\alpha = c \cap [\Spec(k)]$.
It suffices to prove that
$$
c \cap [X'] = c_\alpha \cap [X']
$$
when $X'$ is an integral scheme locally of finite type over $\Spec(k)$,
see Lemma \ref{lemma-bivariant-zero}. However, then $p' : X' \to \Spec(k)$
is flat of relative dimension $\dim(X')$ and hence
$[X'] = (p')^*[\Spec(k)]$. Thus the fact that the bivariant classes
$c$ and $c_\alpha$ agree on $[\Spec(k)]$ implies they
agree when capped against $[X']$ and the proof is complete.
\end{proof}










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