Lemma 42.62.1. The product defined above is associative. More precisely, let $k$ be a field, let $X$ be smooth over $k$, let $Y, Z, W$ be schemes locally of finite type over $X$, let $\alpha \in \mathop{\mathrm{CH}}\nolimits _*(Y)$, $\beta \in \mathop{\mathrm{CH}}\nolimits _*(Z)$, $\gamma \in \mathop{\mathrm{CH}}\nolimits _*(W)$. Then $(\alpha \cdot \beta ) \cdot \gamma = \alpha \cdot (\beta \cdot \gamma )$ in $\mathop{\mathrm{CH}}\nolimits _*(Y \times _ X Z \times _ X W)$.
42.62 Intersection products
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.
Let $X$ be a smooth scheme over $k$. The bivariant class $\Delta ^!$ of Section 42.60 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
\[ Y \times _ X Z = (Y \times _ k Z) \times _{X \times _ k X, \Delta } X \]
Hence we can consider the following sequence of maps
\[ \mathop{\mathrm{CH}}\nolimits _ n(Y) \otimes _\mathbf {Z} \mathop{\mathrm{CH}}\nolimits _ m(Z) \xrightarrow {\times } \mathop{\mathrm{CH}}\nolimits _{n + m}(Y \times _ k Z) \xrightarrow {\Delta ^!} \mathop{\mathrm{CH}}\nolimits _{n + m - *}(Y \times _ X Z) \]
Here the first arrow is the exterior product constructed in Section 42.61 and the second arrow is the gysin map for the diagonal studied in Section 42.60. If $X$ is equidimensional of dimension $d$, then we end up in $\mathop{\mathrm{CH}}\nolimits _{n + m - d}(Y \times _ X Z)$ and in general we can decompose into the parts lying over the open and closed subschemes of $X$ where $X$ has a given dimension. Given $\alpha \in \mathop{\mathrm{CH}}\nolimits _*(Y)$ and $\beta \in \mathop{\mathrm{CH}}\nolimits _*(Z)$ we will denote
\[ \alpha \cdot \beta = \Delta ^!(\alpha \times \beta ) \in \mathop{\mathrm{CH}}\nolimits _*(Y \times _ X Z) \]
In the special case where $X = Y = Z$ we obtain a multiplication
\[ \mathop{\mathrm{CH}}\nolimits _*(X) \times \mathop{\mathrm{CH}}\nolimits _*(X) \to \mathop{\mathrm{CH}}\nolimits _*(X),\quad (\alpha , \beta ) \mapsto \alpha \cdot \beta \]
which is called the intersection product. We observe that this product is clearly symmetric. Associativity follows from the next lemma.
Proof. By Lemma 42.61.5 we have $(\alpha \times \beta ) \times \gamma = \alpha \times (\beta \times \gamma )$ in $\mathop{\mathrm{CH}}\nolimits _*(Y \times _ k Z \times _ k W)$. Consider the closed immersions
\[ \Delta _{12} : X \times _ k X \longrightarrow X \times _ k X \times _ k X, \quad (x, x') \mapsto (x, x, x') \]
and
\[ \Delta _{23} : X \times _ k X \longrightarrow X \times _ k X \times _ k X, \quad (x, x') \mapsto (x, x', x') \]
Denote $\Delta _{12}^!$ and $\Delta _{23}^!$ the corresponding bivariant classes; observe that $\Delta _{12}^!$ is the restriction (Remark 42.33.5) of $\Delta ^!$ to $X \times _ k X \times _ k X$ by the map $\text{pr}_{12}$ and that $\Delta _{23}^!$ is the restriction of $\Delta ^!$ to $X \times _ k X \times _ k X$ 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 42.54.8 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 42.61.4. Similarly for the second equation. $\square$
Lemma 42.62.2. Let $k$ be a field. Let $X$ be a smooth scheme over $k$, equidimensional of dimension $d$. The map is an isomorphism. Via this isomorphism composition of bivariant classes turns into the intersection product defined above.
\[ A^ p(X) \longrightarrow \mathop{\mathrm{CH}}\nolimits _{d - p}(X),\quad c \longmapsto c \cap [X]_ d \]
Proof. Denote $g : X \to \mathop{\mathrm{Spec}}(k)$ the structure morphism. The map is the composition of the isomorphisms
\[ A^ p(X) \to A^{p - d}(X \to \mathop{\mathrm{Spec}}(k)) \to \mathop{\mathrm{CH}}\nolimits _{d - p}(X) \]
The first is the isomorphism $c \mapsto c \circ g^*$ of Proposition 42.60.2 and the second is the isomorphism $c \mapsto c \cap [\mathop{\mathrm{Spec}}(k)]$ of Lemma 42.61.2. From the proof of Lemma 42.61.2 we see that the inverse to the second arrow sends $\alpha \in \mathop{\mathrm{CH}}\nolimits _{d - p}(X)$ to the bivariant class $c_\alpha $ which sends $\beta \in \mathop{\mathrm{CH}}\nolimits _*(Y)$ for $Y$ locally of finite type over $k$ to $\alpha \times \beta $ in $\mathop{\mathrm{CH}}\nolimits _*(X \times _ k Y)$. From the proof of Proposition 42.60.2 we see the inverse to the first arrow in turn sends $c_\alpha $ to the bivariant class which sends $\beta \in \mathop{\mathrm{CH}}\nolimits _*(Y)$ 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. $\square$
Lemma 42.62.3. Let $k$ be a field. Let $f : X \to Y$ be a morphism of schemes smooth over $k$. Then the gysin map exists for $f$ and $f^!(\alpha \cdot \beta ) = f^!\alpha \cdot f^!\beta $.
Proof. Observe that $X \to X \times _ k Y$ is an immersion of $X$ into a scheme smooth over $Y$. Hence the gysin map exists for $f$ (Definition 42.59.4). To prove the formula we may decompose $X$ and $Y$ into their connected components, hence we may assume $X$ is smooth over $k$ and equidimensional of dimension $d$ and $Y$ is smooth over $k$ and equidimensional of dimension $e$. Observe that $f^![Y]_ e = [X]_ d$ (see for example Lemma 42.59.8). Write $\alpha = c \cap [Y]_ e$ and $\beta = c' \cap [Y]_ e$ and hence $\alpha \cdot \beta = c \cap c' \cap [Y]_ e$, see Lemma 42.62.2. By Lemma 42.59.7 we know that $f^!$ commutes with both $c$ and $c'$. Hence
\begin{align*} f^!(\alpha \cdot \beta ) & = f^!(c \cap c' \cap [Y]_ e) \\ & = c \cap c' \cap f^![Y]_ e \\ & = c \cap c' \cap [X]_ d \\ & = (c \cap [X]_ d) \cdot (c' \cap [X]_ d) \\ & = (c \cap f^![Y]_ e) \cdot (c' \cap f^![Y]_ e) \\ & = f^!(\alpha ) \cdot f^!(\beta ) \end{align*}
as desired where we have used Lemma 42.62.2 for $X$ as well.
An alternative proof can be given by proving that $(f \times f)^!(\alpha \times \beta ) = f^!\alpha \times f^!\beta $ and using Lemma 42.59.6. $\square$
Lemma 42.62.4. Let $k$ be a field. Let $f : X \to Y$ be a proper morphism of schemes smooth over $k$. Then the gysin map exists for $f$ and $f_*(\alpha \cdot f^!\beta ) = f_*\alpha \cdot \beta $.
Proof. Observe that $X \to X \times _ k Y$ is an immersion of $X$ into a scheme smooth over $Y$. Hence the gysin map exists for $f$ (Definition 42.59.4). To prove the formula we may decompose $X$ and $Y$ into their connected components, hence we may assume $X$ is smooth over $k$ and equidimensional of dimension $d$ and $Y$ is smooth over $k$ and equidimensional of dimension $e$. Observe that $f^![Y]_ e = [X]_ d$ (see for example Lemma 42.59.8). Write $\alpha = c \cap [X]_ d$ and $\beta = c' \cap [Y]_ e$, see Lemma 42.62.2. We have
\begin{align*} f_*(\alpha \cdot f^!\beta ) & = f_*(c \cap f^!(c' \cap [Y]_ e)) \\ & = f_*(c \cap c' \cap f^![Y]_ e) \\ & = f_*(c \cap c' \cap [X]_ d) \\ & = f_*(c' \cap c \cap [X]_ d) \\ & = c' \cap f_*(c \cap [X]_ d) \\ & = \beta \cdot f_*(\alpha ) \end{align*}
The first equality by the result of Lemma 42.62.2 for $X$. By Lemma 42.59.7 we know that $f^!$ commutes with $c'$. The commutativity of the intersection product justifies switching the order of capping $[X]_ d$ with $c$ and $c'$ (via the lemma). Commuting $c'$ with $f_*$ is allowed as $c'$ is a bivariant class. The final equality is again the lemma. $\square$
Lemma 42.62.5. Let $k$ be a field. Let $X$ be an integral scheme smooth over $k$. Let $Y, Z \subset X$ be integral closed subschemes. Set $d = \dim (Y) + \dim (Z) - \dim (X)$. Assume
$\dim (Y \cap Z) \leq d$, and
$\mathcal{O}_{Y, \xi }$ and $\mathcal{O}_{Z, \xi }$ are Cohen-Macaulay for every $\xi \in Y \cap Z$ with $\delta (\xi ) = d$.
Then $[Y] \cdot [Z] = [Y \cap Z]_ d$ in $\mathop{\mathrm{CH}}\nolimits _ d(X)$.
Proof. Recall that $[Y] \cdot [Z] = \Delta ^!([Y \times Z])$ where $\Delta ^! = c(\Delta : X \to X \times X, \mathcal{T}_{X/k})$ is a higher codimension gysin map (Section 42.54) with $\mathcal{T}_{X/k} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\Omega _{X/k}, \mathcal{O}_ X)$ locally free of rank $\dim (X)$. We have the equality of schemes
\[ Y \cap Z = X \times _{\Delta , (X \times X)} (Y \times Z) \]
and $\dim (Y \times Z) = \dim (Y) + \dim (Z)$ and hence conditions (1), (2), and (3) of Lemma 42.54.6 hold. Finally, if $\xi \in Y \cap Z$, then we have a flat local homomorphism
\[ \mathcal{O}_{Y, \xi } \longrightarrow \mathcal{O}_{Y \times Z, \xi } \]
whose “fibre” is $\mathcal{O}_{Z, \xi }$. It follows that if both $\mathcal{O}_{Y, \xi }$ and $\mathcal{O}_{Z, \xi }$ are Cohen-Macaulay, then so is $\mathcal{O}_{Y \times Z, \xi }$, see Algebra, Lemma 10.163.3. In this way we see that all the hypotheses of Lemma 42.54.6 are satisfied and we conclude. $\square$
Lemma 42.62.6. Let $k$ be a field. Let $X$ be a scheme smooth over $k$. Let $i : Y \to X$ be a regular closed immersion. Let $\alpha \in \mathop{\mathrm{CH}}\nolimits _*(X)$. If $Y$ is equidimensional of dimension $e$, then $\alpha \cdot [Y]_ e = i_*(i^!(\alpha ))$ in $\mathop{\mathrm{CH}}\nolimits _*(X)$.
Proof. After decomposing $X$ into connected components we may and do assume $X$ is equidimensional of dimension $d$. Write $\alpha = c \cap [X]_ n$ with $x \in A^*(X)$, see Lemma 42.62.2. Then
\[ i_*(i^!(\alpha )) = i_*(i^!(c \cap [X]_ n)) = i_*(c \cap i^![X]_ n) = i_*(c \cap [Y]_ e) = c \cap i_*[Y]_ e = \alpha \cdot [Y]_ e \]
The first equality by choice of $c$. Then second equality by Lemma 42.59.7. The third because $i^![X]_ d = [Y]_ e$ in $\mathop{\mathrm{CH}}\nolimits _*(Y)$ (Lemma 42.59.8). The fourth because bivariant classes commute with proper pushforward. The last equality by Lemma 42.62.2. $\square$
Lemma 42.62.7. Let $k$ be a field. Let $X$ be a smooth scheme over $k$ which is quasi-compact and has affine diagonal. Then the intersection product on $\mathop{\mathrm{CH}}\nolimits ^*(X)$ constructed in this section agrees after tensoring with $\mathbf{Q}$ with the intersection product constructed in Section 42.58.
Proof. Let $\alpha \in \mathop{\mathrm{CH}}\nolimits ^ i(X)$ and $\beta \in \mathop{\mathrm{CH}}\nolimits ^ j(X)$. Write $\alpha = ch(\alpha ') \cap [X]$ and $\beta = ch(\beta ') \cap [X]$ $\alpha ', \beta ' \in K_0(\textit{Vect}(X)) \otimes \mathbf{Q}$ as in Section 42.58. Set $c = ch(\alpha ')$ and $c' = ch(\beta ')$. Then the intersection product in Section 42.58 produces $c \cap c' \cap [X]$. This is the same as $\alpha \cdot \beta $ by Lemma 42.62.2 (or rather the generalization that $A^ i(X) \to \mathop{\mathrm{CH}}\nolimits ^ i(X)$, $c \mapsto c \cap [X]$ is an isomorphism for any smooth scheme $X$ over $k$). $\square$
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