The Stacks project

42.22 Chow groups and envelopes

Here is the definition.

reference

Definition 42.22.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be a scheme locally of finite type over $S$. An envelope is a proper morphism $f : Y \to X$ which is completely decomposed (More on Morphisms, Definition 37.75.1).

The exact sequence of Lemma 42.22.4 is the main motivation for the definition.

Lemma 42.22.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be a scheme locally of finite type over $S$. If $f : Y \to X$ and $g : Z \to Y$ are envelopes, then $f \circ g$ is an envelope.

Proof. Follows from Morphisms, Lemma 29.41.4 and More on Morphisms, Lemma 37.75.2. $\square$

Lemma 42.22.3. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X' \to X$ be a morphism of schemes locally of finite type over $S$. If $f : Y \to X$ is an envelope, then the base change $f' : Y' \to X'$ of $f$ is an envelope too.

Proof. Follows from Morphisms, Lemma 29.41.5 and More on Morphisms, Lemma 37.75.3. $\square$

Lemma 42.22.4. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be a scheme locally of finite type over $S$. Let $f : Y \to X$ be an envelope. Then we have an exact sequence

\[ \mathop{\mathrm{CH}}\nolimits _ k(Y \times _ X Y) \xrightarrow {p_* - q_*} \mathop{\mathrm{CH}}\nolimits _ k(Y) \xrightarrow {f_*} \mathop{\mathrm{CH}}\nolimits _ k(X) \to 0 \]

for all $k \in \mathbf{Z}$. Here $p, q : Y \times _ X Y \to Y$ are the projections.

Proof. Since $f$ is an envelope, $f$ is proper and hence pushforward on cycles and cycle classes is defined, see Sections 42.12 and 42.15. Similarly, the morphisms $p$ and $q$ are proper as base changes of $f$. The composition of the arrows is zero as $f_* \circ p_* = (p \circ f)_* = (q \circ f)_* = f_* \circ q_*$, see Lemma 42.12.2.

Let us show that $f_* : Z_ k(Y) \to Z_ k(X)$ is surjective. Namely, suppose that we have $\alpha = \sum n_ i[Z_ i] \in Z_ k(X)$ where $Z_ i \subset X$ is a locally finite family of integral closed subschemes. Let $x_ i \in Z_ i$ be the generic point. Since $f$ is an envelope and hence completely decomposed, there exists a point $y_ i \in Y$ with $f(y_ i) = x_ i$ and with $\kappa (y_ i)/\kappa (x_ i)$ trivial. Let $W_ i \subset Y$ be the integral closed subscheme with generic point $y_ i$. Since $f$ is closed, we see that $f(W_ i) = Z_ i$. It follows that the family of closed subschemes $W_ i$ is locally finite on $Y$. Since $\kappa (y_ i)/\kappa (x_ i)$ is trivial we see that $\dim _\delta (W_ i) = \dim _\delta (Z_ i) = k$. Hence $\beta = \sum n_ i[W_ i]$ is in $Z_ k(Y)$. Finally, since $\kappa (y_ i)/\kappa (x_ i)$ is trivial, the degree of the dominant morphism $f|_{W_ i} : W_ i \to Z_ i$ is $1$ and we conclude that $f_*\beta = \alpha $.

Since $f_* : Z_ k(Y) \to Z_ k(X)$ is surjective, a fortiori the map $f_* : \mathop{\mathrm{CH}}\nolimits _ k(Y) \to \mathop{\mathrm{CH}}\nolimits _ k(X)$ is surjective.

Let $\beta \in Z_ k(Y)$ be an element such that $f_*\beta $ is zero in $\mathop{\mathrm{CH}}\nolimits _ k(X)$. This means we can find a locally finite family of integral closed subschemes $Z_ j \subset X$ with $\dim _\delta (Z_ j) = k + 1$ and $f_ j \in R(Z_ j)^*$ such that

\[ f_*\beta = \sum (Z_ j \to X)_*\text{div}(f_ j) \]

as cycles where $i_ j : Z_ j \to X$ is the given closed immersion. Arguing exactly as above, we can find a locally finite family of integral closed subschemes $W_ j \subset Y$ with $f(W_ j) = Z_ j$ and such that $W_ j \to Z_ j$ is birational, i.e., induces an isomorphism $R(Z_ j) = R(W_ j)$. Denote $g_ j \in R(W_ j)^*$ the element corresponding to $f_ j$. Observe that $W_ j \to Z_ j$ is proper and that $(W_ j \to Z_ j)_*\text{div}(g_ j) = \text{div}(f_ j)$ as cycles on $Z_ j$. It follows from this that if we replace $\beta $ by the rationally equivalent cycle

\[ \beta ' = \beta - \sum (W_ j \to Y)_*\text{div}(g_ j) \]

then we find that $f_*\beta ' = 0$. (This uses Lemma 42.12.2.) Thus to finish the proof of the lemma it suffices to show the claim in the following paragraph.

Claim: if $\beta \in Z_ k(Y)$ and $f_*\beta = 0$, then $\beta = \delta + p_*\gamma - q_*\gamma $ in $Z_ k(Y)$ for some $\gamma \in Z_ k(Y \times _ X Y)$. Namely, write $\beta = \sum _{j \in J} n_ j[W_ j]$ with $\{ W_ j\} _{j \in J}$ a locally finite family of integral closed subschemes of $Y$ with $\dim _\delta (W_ j) = k$. Fix an integral closed subscheme $Z \subset X$. Consider the subset $J_ Z = \{ j \in J : f(W_ j) = Z\} $. This is a finite set. There are three cases:

  1. $J_ Z = \emptyset $. In this case we set $\gamma _ Z = 0$.

  2. $J_ Z \not= \emptyset $ and $\dim _\delta (Z) = k$. The condition $f_*\beta = 0$ implies by looking at the coefficient of $Z$ that $\sum _{j \in J_ Z} n_ j\deg (W_ j/Z) = 0$. In this case we choose an integral closed subscheme $W \subset Y$ which maps birationally onto $Z$ (see above). Looking at generic points, we see that $W_ j \times _ Z W$ has a unique irreducible component $W'_ j \subset W_ j \times _ Z W \subset Y \times _ X Y$ mapping birationally to $W_ j$. Then $W'_ j \to W$ is dominant and $\deg (W'_ j/W) = \deg (W_ j/W)$. Thus if we set $\gamma _ Z = \sum _{j \in J_ Z} n_ j[W'_ j]$ then we see that $p_*\gamma _ Z = \sum _{j \in J_ Z} n_ j[W_ j]$ and $q_*\gamma _ Z = \sum _{j \in J_ Z} n_ j\deg (W'_ j/W)[W] = 0$.

  3. $J_ Z \not= \emptyset $ and $\dim _\delta (Z) < k$. In this case we choose an integral closed subscheme $W \subset Y$ which maps birationally onto $Z$ (see above). Looking at generic points, we see that $W_ j \times _ Z W$ has a unique irreducible component $W'_ j \subset W_ j \times _ Z W \subset Y \times _ X Y$ mapping birationally to $W_ j$. Then $W'_ j \to W$ is dominant and $k = \dim _\delta (W'_ j) > \dim _\delta (W) = \dim _\delta (Z)$. Thus if we set $\gamma _ Z = \sum _{j \in J_ Z} n_ j[W'_ j]$ then we see that $p_*\gamma _ Z = \sum _{j \in J_ Z} n_ j[W_ j]$ and $q_*\gamma _ Z = 0$.

Since the family of integral closed subschemes $\{ f(W_ j)\} $ is locally finite on $X$ (Lemma 42.11.2) we see that the $k$-cycle

\[ \gamma = \sum \nolimits _{Z \subset X\text{ integral closed}} \gamma _ Z \]

on $Y \times _ X Y$ is well defined. By our computations above it follows that $p_*\gamma _ Z = \beta $ and $q_*\gamma _ Z = 0$ which implies what we wanted to prove. $\square$


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