## 42.63 Exterior product over Dedekind domains

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 42.7.1; see Example 42.7.3. Observe that $S$ is normal (Algebra, Lemma 10.120.17) and hence a disjoint union of normal integral schemes (Properties, Lemma 28.7.7). Thus all of the arguments below reduce to the case where $S$ is irreducible. On the other hand, we allow $S$ to be nonseparated (so $S$ could be the affine line with $0$ doubled for example).

Consider a cartesian square

$\xymatrix{ X \times _ S Y \ar[r] \ar[d] & Y \ar[d] \\ X \ar[r] & S }$

of schemes locally of finite type over $S$. We claim there is a canonical map

$\times : \mathop{\mathrm{CH}}\nolimits _ n(X) \otimes _{\mathbf{Z}} \mathop{\mathrm{CH}}\nolimits _ m(Y) \longrightarrow \mathop{\mathrm{CH}}\nolimits _{n + m - 1}(X \times _ S Y)$

which is uniquely determined by the following rule: given integral closed subschemes $X' \subset X$ and $Y' \subset Y$ of $\delta$-dimensions $n$ and $m$ we set

1. $[X'] \times [Y'] = [X' \times _ S Y']_{n + m - 1}$ if $X'$ or $Y'$ dominates an irreducible component of $S$,

2. $[X'] \times [Y'] = 0$ if neither $X'$ nor $Y'$ dominates an irreducible component of $S$.

Lemma 42.63.1. The map $\times : \mathop{\mathrm{CH}}\nolimits _ n(X) \otimes _{\mathbf{Z}} \mathop{\mathrm{CH}}\nolimits _ m(Y) \to \mathop{\mathrm{CH}}\nolimits _{n + m - 1}(X \times _ S Y)$ is well defined.

Proof. Consider $n$ and $m$ cycles $\alpha = \sum _{i \in I} n_ i[X_ i]$ and $\beta = \sum _{j \in J} m_ j[Y_ j]$ with $X_ i \subset X$ and $Y_ j \subset Y$ locally finite families of integral closed subschemes of $\delta$-dimensions $n$ and $m$. Let $K \subset I \times J$ be the set of pairs $(i, j) \in I \times J$ such that $X_ i$ or $Y_ j$ dominates an irreducible component of $S$. Then $\{ X_ i \times _ S Y_ j\} _{(i, j) \in K}$ is a locally finite collection of closed subschemes of $X \times _ S Y$ of $\delta$-dimension $n + m - 1$. This means we can indeed consider

$\alpha \times \beta = \sum \nolimits _{(i, j) \in K} n_ i m_ j [X_ i \times _ S Y_ j]_{n + m - 1}$

as a $(n + m - 1)$-cycle on $X \times _ S Y$. In this way we obtain an additive map $\times : Z_ n(X) \otimes _{\mathbf{Z}} Z_ m(Y) \to Z_{n + m}(X \times _ S Y)$. The problem is to show that this procedure is compatible with rational equivalence.

Let $i : X' \to X$ be the inclusion morphism of an integral closed subscheme of $\delta$-dimension $n$ which dominates an irreducible component of $S$. Then $p' : X' \to S$ is flat of relative dimension $n - 1$, see More on Algebra, Lemma 15.22.11. Hence flat pullback along $p'$ is an element $(p')^* \in A^{-n + 1}(X' \to S)$ by Lemma 42.33.2 and hence $c' = i_* \circ (p')^* \in A^{-n + 1}(X \to S)$ by Lemma 42.33.4. This produces maps

$c' \cap - : \mathop{\mathrm{CH}}\nolimits _ m(Y) \longrightarrow \mathop{\mathrm{CH}}\nolimits _{m + n - 1}(X \times _ S Y)$

which sends $[Y']$ to $[X' \times _ S Y']_{n + m - 1}$ for any integral closed subscheme $Y' \subset Y$ of $\delta$-dimension $m$.

Let $i : X' \to X$ be the inclusion morphism of an integral closed subscheme of $\delta$-dimension $n$ such that the composition $X' \to X \to S$ factors through a closed point $s \in S$. Since $s$ is a closed point of the spectrum of a Dedekind domain, we see that $s$ is an effective Cartier divisor on $S$ whose normal bundle is trivial. Denote $c \in A^1(s \to S)$ the gysin homomorphism, see Lemma 42.33.3. The morphism $p' : X' \to s$ is flat of relative dimension $n$. Hence flat pullback along $p'$ is an element $(p')^* \in A^{-n}(X' \to S)$ by Lemma 42.33.2. Thus

$c' = i_* \circ (p')^* \circ c \in A^{-n}(X \to S)$

by Lemma 42.33.4. This produces maps

$c' \cap - : \mathop{\mathrm{CH}}\nolimits _ m(Y) \longrightarrow \mathop{\mathrm{CH}}\nolimits _{m + n - 1}(X \times _ S Y)$

which for any integral closed subscheme $Y' \subset Y$ of $\delta$-dimension $m$ sends $[Y']$ to either $[X' \times _ S Y']_{n + m - 1}$ if $Y'$ dominates an irreducible component of $S$ or to $0$ if not.

From the previous two paragraphs we conclude the construction $([X'], [Y']) \mapsto [X' \times _ S Y']_{n + m - 1}$ factors through rational equivalence in the second variable, i.e., gives a well defined map $Z_ n(X) \otimes _{\mathbf{Z}} \mathop{\mathrm{CH}}\nolimits _ m(Y) \to \mathop{\mathrm{CH}}\nolimits _{n + m - 1}(X \times _ S Y)$. By symmetry the same is true for the other variable and we conclude. $\square$

Lemma 42.63.2. Let $(S, \delta )$ be as above. Let $X$ be a scheme locally of finite type over $S$. Then we have a canonical identification

$A^ p(X \to S) = \mathop{\mathrm{CH}}\nolimits _{1 - p}(X)$

for all $p \in \mathbf{Z}$.

Proof. Consider the element $[S]_1 \in \mathop{\mathrm{CH}}\nolimits _1(S)$. We get a map $A^ p(X \to S) \to \mathop{\mathrm{CH}}\nolimits _{1 - p}(X)$ by sending $c$ to $c \cap [S]_1$.

Conversely, suppose we have $\alpha \in \mathop{\mathrm{CH}}\nolimits _{1 - p}(X)$. Then we can define $c_\alpha \in A^ p(X \to S)$ as follows: given $X' \to S$ and $\alpha ' \in \mathop{\mathrm{CH}}\nolimits _ n(X')$ we let

$c_\alpha \cap \alpha ' = \alpha \times \alpha '$

in $\mathop{\mathrm{CH}}\nolimits _{n - p}(X \times _ S X')$. To show that this is a bivariant class we write $\alpha = \sum _{i \in I} n_ i[X_ i]$ as in Definition 42.8.1. In particular the morphism

$g : \coprod \nolimits _{i \in I} X_ i \longrightarrow X$

is proper. Pick $i \in I$. If $X_ i$ dominates an irreducible component of $S$, then the structure morphism $p_ i : X_ i \to S$ is flat and we have $\xi _ i = p_ i^* \in A^ p(X_ i \to S)$. On the other hand, if $p_ i$ factors as $p'_ i : X_ i \to s_ i$ followed by the inclusion $s_ i \to S$ of a closed point, then we have $\xi _ i = (p'_ i)^* \circ c_ i \in A^ p(X_ i \to S)$ where $c_ i \in A^1(s_ i \to S)$ is the gysin homomorphism and $(p'_ i)^*$ is flat pullback. Observe that

$A^ p(\coprod \nolimits _{i \in I} X_ i \to S) = \prod \nolimits _{i \in I} A^ p(X_ i \to S)$

Thus we have

$\xi = \sum n_ i \xi _ i \in A^ p(\coprod \nolimits _{i \in I} X_ i \to S)$

Finally, since $g$ is proper we have a bivariant class

$g_* \circ \xi \in A^ p(X \to S)$

by Lemma 42.33.4. The reader easily verifies that $c_\alpha$ is equal to this class (please compare with the proof of Lemma 42.63.1) and hence is itself a bivariant class.

To finish the proof we have to show that the two constructions are mutually inverse. Since $c_\alpha \cap [S]_1 = \alpha$ this is clear for one of the two directions. For the other, let $c \in A^ p(X \to S)$ and set $\alpha = c \cap [S]_1$. It suffices to prove that

$c \cap [X'] = c_\alpha \cap [X']$

when $X'$ is an integral scheme locally of finite type over $S$, see Lemma 42.35.3. However, either $p' : X' \to S$ is flat of relative dimension $\dim _\delta (X') - 1$ and hence $[X'] = (p')^*[S]_1$ or $X' \to S$ factors as $X' \to s \to S$ and hence $[X'] = (p')^*(s \to S)^*[S]_1$. Thus the fact that the bivariant classes $c$ and $c_\alpha$ agree on $[S]_1$ implies they agree when capped against $[X']$ (since bivariant classes commute with flat pullback and gysin maps) and the proof is complete. $\square$

Lemma 42.63.3. Let $(S, \delta )$ be as above. Let $X$ be a scheme locally of finite type over $S$. Let $c \in A^ p(X \to S)$. Let $Y \to Z$ be a morphism of schemes locally of finite type over $S$. Let $c' \in A^ q(Y \to Z)$. Then $c \circ c' = c' \circ c$ in $A^{p + q}(X \times _ S Y \to X \times _ S Z)$.

Proof. In the proof of Lemma 42.63.2 we have seen that $c$ is given by a combination of proper pushforward, multiplying by integers over connected components, flat pullback, and gysin maps. Since $c'$ commutes with each of these operations by definition of bivariant classes, we conclude. Some details omitted. $\square$

Remark 42.63.4. The upshot of Lemmas 42.63.2 and 42.63.3 is the following. Let $(S, \delta )$ be as above. Let $X$ be a scheme locally of finite type over $S$. Let $\alpha \in \mathop{\mathrm{CH}}\nolimits _*(X)$. 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 $\mathop{\mathrm{CH}}\nolimits _*(X \times _ S Y)$ for any $\beta \in \mathop{\mathrm{CH}}\nolimits _*(Z)$. Namely, this follows by taking $c = c_\alpha \in A^*(X \to S)$ the bivariant class corresponding to $\alpha$, see proof of Lemma 42.63.2.

Lemma 42.63.5. 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 $\alpha \in \mathop{\mathrm{CH}}\nolimits _*(X)$, $\beta \in \mathop{\mathrm{CH}}\nolimits _*(Y)$, $\gamma \in \mathop{\mathrm{CH}}\nolimits _*(Z)$. Then $(\alpha \times \beta ) \times \gamma = \alpha \times (\beta \times \gamma )$ in $\mathop{\mathrm{CH}}\nolimits _*(X \times _ S Y \times _ S Z)$.

Proof. Omitted. Hint: associativity of fibre product of schemes. $\square$

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