## 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.

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 $\mathop{\mathrm{CH}}\nolimits _*(Y \times _ X Z \times _ X W)$.

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

$A^ p(X) \longrightarrow \mathop{\mathrm{CH}}\nolimits _{d - p}(X),\quad c \longmapsto c \cap [X]_ d$

is an isomorphism. Via this isomorphism composition of bivariant classes turns into the intersection product defined above.

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

1. $\dim (Y \cap Z) \leq d$, and

2. $\mathcal{O}_{Y, \xi }$ 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 $k$). $\square$

Comment #7552 by Hao Peng on

[Typo]in the definition of intersection products, $CH_n(Y)\otimes CH_m(Y)$ should be changed to $CH_n(Y)\otimes CH_m(Z)$

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