Skip to content

# stacks/stacks-project

Exterior products in chow

• Loading branch information...
aisejohan committed May 20, 2019
1 parent ad13ed8 commit 5b5d1ef07770716c0023a3f1fbdd7779e5842def
Showing with 144 additions and 0 deletions.
1. +144 −0 chow.tex
144 chow.tex
 @@ -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}

#### 0 comments on commit 5b5d1ef

Please sign in to comment.
You can’t perform that action at this time.