# stacks/stacks-project

Intersection products for smooth / Dedekind

 @@ -10149,7 +10149,7 @@ \section{Exterior product} \section{Intersection products for smooth varieties} \section{Intersection products} \label{section-intersection-product} \noindent @@ -10291,7 +10291,7 @@ \section{Intersection products for smooth varieties} \section{Exterior product over a $1$-dimensional base} \section{Exterior product over Dedekind domains} \label{section-exterior-product-dim-1} \noindent @@ -10500,6 +10500,197 @@ \section{Exterior product over a $1$-dimensional base} Some details omitted. \end{proof} \begin{remark} \label{remark-commuting-exterior-dim-1} The upshot of Lemmas \ref{lemma-chow-cohomology-towards-base-dim-1} and \ref{lemma-chow-cohomology-towards-base-dim-1-commutes} is the following. Let $(S, \delta)$ be as above. Let $X$ be a scheme locally of finite type over $S$. Let $\alpha \in A_*(X)$. Let $Y \to Z$ be a morphism of schemes locally of finite type over $S$. Let $c' \in A^q(Y \to Z)$. Then $$\alpha \times (c' \cap \beta) = c' \cap (\alpha \times \beta)$$ in $A_*(X \times_S Y)$ for any $\beta \in A_*(Z)$. Namely, this follows by taking $c = c_\alpha \in A^*(X \to S)$ the bivariant class corresponding to $\alpha$, see proof of Lemma \ref{lemma-chow-cohomology-towards-base-dim-1}. \end{remark} \begin{lemma} \label{lemma-exterior-product-associative-dim-1} Exterior product is associative. More precisely, let $(S, \delta)$ be as above, let $X, Y, Z$ be schemes locally of finite type over $S$, let $\alpha \in A_*(X)$, $\beta \in A_*(Y)$, $\gamma \in A_*(Z)$. Then $(\alpha \times \beta) \times \gamma = \alpha \times (\beta \times \gamma)$ in $A_*(X \times_S Y \times_S Z)$. \end{lemma} \begin{proof} Omitted. Hint: associativity of fibre product of schemes. \end{proof} \section{Intersection products over Dedekind domains} \label{section-intersection-product-dim-1} \noindent Let $S$ be a locally Noetherian scheme which has an open covering by spectra of Dedekind domains. Set $\delta(s) = 0$ for $s \in S$ closed and $\delta(s) = 1$ otherwise. Then $(S, \delta)$ is a special case of our general Situation \ref{situation-setup}; see Example \ref{example-domain-dimension-1} and discussion in Section \ref{section-exterior-product-dim-1}. \medskip\noindent Let $X$ be a smooth scheme over $S$. The bivariant class $\Delta^!$ of Section \ref{section-gysin-for-diagonal} allows us to define a kind of intersection product on chow groups of schemes locally of finite type over $X$. Namely, suppose that $Y \to X$ and $Z \to X$ are morphisms of schemes which are locally of finite type. Then observe that $$Y \times_X Z = (Y \times_S Z) \times_{X \times_S X, \Delta} X$$ Hence we can consider the following sequence of maps $$A_n(Y) \otimes_\mathbf{Z} A_m(Y) \xrightarrow{\times} A_{n + m - 1}(Y \times_S Z) \xrightarrow{\Delta^!} A_{n + m - *}(Y \times_X Z)$$ Here the first arrow is the exterior product constructed in Section \ref{section-exterior-product-dim-1}. If $X$ is equidimensional of dimension $d$, then $X \to S$ is smooth of relative dimension $d - 1$ and hence we end up in $A_{n + m - d}(Y \times_X Z)$. In general we can decompose into the parts lying over the open and closed subschemes of $X$ where $X$ has a given dimension. Given $\alpha \in A_*(Y)$ and $\beta \in A_*(Z)$ we will denote $$\alpha \cdot \beta = \Delta^!(\alpha \times \beta) \in A_*(Y \times_X Z)$$ In the special case where $X = Y = Z$ we obtain a multiplication $$A_*(X) \times A_*(X) \to A_*(X),\quad (\alpha, \beta) \mapsto \alpha \cdot \beta$$ which is called the {\it intersection product}. We observe that this product is clearly symmetric. Associativity follows from the next lemma (as well as the one following). \begin{lemma} \label{lemma-associative-dim-1} The product defined above is associative. More precisely, with $(S, \delta)$ as above, let $X$ be smooth over $S$, let $Y, Z, W$ be schemes locally of finite type over $X$, let $\alpha \in A_*(Y)$, $\beta \in A_*(Z)$, $\gamma \in A_*(W)$. Then $(\alpha \cdot \beta) \cdot \gamma = \alpha \cdot (\beta \cdot \gamma)$ in $A_*(Y \times_X Z \times_X W)$. \end{lemma} \begin{proof} By Lemma \ref{lemma-exterior-product-associative-dim-1} we have $(\alpha \times \beta) \times \gamma = \alpha \times (\beta \times \gamma)$ in $A_*(Y \times_S Z \times_S W)$. Consider the closed immersions $$\Delta_{12} : X \times_S X \longrightarrow X \times_S X \times_S X, \quad (x, x') \mapsto (x, x, x')$$ and $$\Delta_{23} : X \times_S X \longrightarrow X \times_S X \times_S X, \quad (x, x') \mapsto (x, x', x')$$ Denote $\Delta_{12}^!$ and $\Delta_{23}^!$ the corresponding bivariant classes; observe that $\Delta_{12}^!$ is the restriction (Remark \ref{remark-restriction-bivariant}) of $\Delta^!$ to $X \times_S X \times_S X$ by the map $\text{pr}_{12}$ and that $\Delta_{23}^!$ is the restriction of $\Delta^!$ to $X \times_S X \times_S X$ by the map $\text{pr}_{23}$. Thus clearly the restriction of $\Delta_{12}^!$ by $\Delta_{23}$ is $\Delta^!$ and the restriction of $\Delta_{23}^!$ by $\Delta_{12}$ is $\Delta^!$ too. Thus by Lemma \ref{lemma-gysin-commutes} we have $$\Delta^! \circ \Delta_{12}^! = \Delta^! \circ \Delta_{23}^!$$ Now we can prove the lemma by the following sequence of equalities: \begin{align*} (\alpha \cdot \beta) \cdot \gamma & = \Delta^!(\Delta^!(\alpha \times \beta) \times \gamma) \\ & = \Delta^!(\Delta_{12}^!((\alpha \times \beta) \times \gamma)) \\ & = \Delta^!(\Delta_{23}^!((\alpha \times \beta) \times \gamma)) \\ & = \Delta^!(\Delta_{23}^!(\alpha \times (\beta \times \gamma)) \\ & = \Delta^!(\alpha \times \Delta^!(\beta \times \gamma)) \\ & = \alpha \cdot (\beta \cdot \gamma) \end{align*} All equalities are clear from the above except perhaps for the second and penultimate one. The equation $\Delta_{23}^!(\alpha \times (\beta \times \gamma)) = \alpha \times \Delta^!(\beta \times \gamma)$ holds by Remark \ref{remark-commuting-exterior}. Similarly for the second equation. \end{proof} \begin{lemma} \label{lemma-identify-chow-for-smooth} Let $(S, \delta)$ be as above. Let $X$ be a smooth scheme over $S$, equidimensional of dimension $d$. The map $$A^p(X) \longrightarrow A_{d - p}(X),\quad c \longmapsto c \cap [X]_d$$ is an isomorphism. Via this isomorphism composition of bivariant classes turns into the intersection product defined above. \end{lemma} \begin{proof} Denote $g : X \to S$ the structure morphism. The map is the composition of the isomorphisms $$A^p(X) \to A^{p - d + 1}(X \to S) \to A_{d - p}(X)$$ The first is the isomorphism $c \mapsto c \circ g^*$ of Proposition \ref{proposition-compute-bivariant} and the second is the isomorphism $c \mapsto c \cap [S]_1$ of Lemma \ref{lemma-chow-cohomology-towards-base-dim-1}. From the proof of Lemma \ref{lemma-chow-cohomology-towards-base-dim-1} we see that the inverse to the second arrow sends $\alpha \in A_{d - p}(X)$ to the bivariant class $c_\alpha$ which sends $\beta \in A_*(Y)$ for $Y$ locally of finite type over $k$ to $\alpha \times \beta$ in $A_*(X \times_k Y)$. From the proof of Proposition \ref{proposition-compute-bivariant} we see the inverse to the first arrow in turn sends $c_\alpha$ to the bivariant class which sends $\beta \in A_*(Y)$ for $Y \to X$ locally of finite type to $\Delta^!(\alpha \times \beta) = \alpha \cdot \beta$. From this the final result of the lemma follows. \end{proof}