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Let $CH_0(X)^0$ denote the group of zero cycles of degree zero modulo rational equivalence.

I am looking for a reference for the following fact: If $X$ and $Y$ are smooth and projective varieties over a finite field and we suppose they are birational, then the groups $CH_0(X)^0$ and $CH_0(Y)^0$ are isomorphic.

As a second question, let $k$ denote the base field and set $$ \pi_1^{ab}(X)^0 := \text{ker}[\pi_1^{ab}(X) \rightarrow \pi_1^{ab}(k)]. $$ Then, with the same assumptions on $X$ and $Y$, $\pi_1^{ab}(X)^0$ and $\pi_1^{ab}(Y)^0$ should be isomorphic.

For the first fact, I have found a reference for the case of dimension 2 in the answer to this question (which links to a 1979 paper of Colliot-Thélène and Coray). The cases of dimension 1 and 0 are obvious. However, I have failed to find a reference for the case of general dimension. Any help is greatly appreciated.

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    $\begingroup$ I do not know references. Both of these follow fairly quickly from well-known results: the Lang-Nishimura theorem for the first, and the "Purity Theorem" (SGA 2, Exp. X, Thm. 3.4, p. 118) for the second. $\endgroup$ – Jason Starr Jun 10 '16 at 13:13
  • $\begingroup$ Thanks @JasonStarr , I have indeed managed to solve it for $\pi_1$ using your comment! $\endgroup$ – Joachim Jun 12 '16 at 14:34
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A reference for birational equivalence of $CH_0$ is Fulton's Intersection Theory [1], Example 16.1.11. In the example, he makes the assumption that $k$ is algebraically closed, but he never uses it. Since the argument is fairly short, let me repeat it here.

Theorem. Let $k$ be a field, and let $X$ and $Y$ be smooth proper $k$-varieties. If $X$ and $Y$ are birational, then $CH_0(X) \cong CH_0(Y)$.

Proof. Let $f \colon X \dashrightarrow Y$ be a birational map. Let $\Gamma \subseteq X \times Y$ be the closure of the graph. Then $\Gamma$ defines maps \begin{align*} f_* \colon CH_0(X) &\to CH_0(Y) & f^* \colon CH_0(Y) &\to CH_0(X)\\ a &\mapsto \pi_{Y,*}(\Gamma \cdot \pi_X^* a), & b &\mapsto \pi_{X,*}(\Gamma \cdot \pi_Y^* b). \end{align*} The composition $f^* \circ f_*$ (resp. $f_* \circ f^*$) is given by the cycle $\Gamma^\top \circ \Gamma := \pi_{13, *} (\pi_{12}^* \Gamma \cdot \pi_{23}^* \Gamma^\top)$ on $X \times X$ (resp. by $\Gamma \circ \Gamma^\top$ on $Y \times Y$); see [loc. cit., Def. 16.1.1 and Prop. 16.1.2(a)] for details.

Let $V \subseteq Y$ be an open such that $f$ induces an isomorphism $f^{-1}(V) \stackrel\sim\to V$. Let $U = f^{-1}(V)$, $Z = X \setminus U$, and $W = Y \setminus V$. I claim that the cycle $\varepsilon := \Gamma^\top \circ \Gamma - \Delta_X$ is supported on $Z \times Z$. By the short exact sequence of [loc. cit., Prop. 1.8], it suffices to show that the restriction of $\varepsilon$ to $S := X \times X \setminus Z \times Z = X \times U \cup U \times X$ is zero. Let $$T := (X \times Y \times U) \cup (U \times Y \times X) = \pi_{13}^{-1} (S).$$ Then the restriction of $\varepsilon$ to $S$ is the pushforward along $\pi_{13}$ of $$\left.\left(\pi_{12}^* \Gamma \cdot \pi_{23}^* \Gamma^\top\right)\right|_T.\tag{1}$$ We can compute the latter as $$\left.\left(\pi_{12}^* \Gamma\right)\right|_T \cdot \left.\left(\pi_{23}^* \Gamma^\top\right)\right|_T.$$ This is a proper intersection, and the intersection is equal to the 'diagonal' $$\left\{(a,b,c) \in U \times V \times U\ \big|\ f(a)=b=f(c)\right\}.$$ Indeed, the intersection agrees with this set on both $X \times Y \times U$ and $U \times Y \times X$.

Then the pushforward of (1) is $\Delta_U$. Hence $\varepsilon$ vanishes on $S$, so it is supported on $Z \times Z$. In particular, the projections $\pi_{1,*} \varepsilon, \pi_{2,*} \varepsilon \in CH_*(X)$ are supported on $Z$.

The punchline is that $\varepsilon$ acts as the identity on $CH_0(X)$ by the moving lemma. Indeed, any $0$-cycle $a$ on a smooth variety can be moved away from $Z$, so the intersection $\pi_1^* a \cdot \varepsilon$ is zero.

This proves that $f^* \circ f_*$ is the identity, and by symmetry the same holds for $f_* \circ f^*$. $\square$

Remark. It is even true that $CH_0$ is a stable birational invariant: if $X$ and $Y$ are smooth and proper varieties with $X \times \mathbb P^n \stackrel\sim\dashrightarrow Y \times \mathbb P^m$, then $CH_0(X) \cong CH_0(Y)$. (The only thing left to prove is that $CH_0(X) \cong CH_0(X \times \mathbb P^1)$, since we already know the result for birational varieties.)


Remark. The reason Fulton only proves the theorem for algebraically closed fields is that Fulton only proves the moving lemma over those fields. For a proof of the moving lemma over arbitrary fields, see Roberts's appendix to the Oslo 1970 Algebraic Geometry proceedings [2] (finite fields are addressed separately).

References:

[1] Fulton, William. Intersection theory (second edition). Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. Springer-Verlag, Berlin, 1998. ISBN: 3-540-62046-X; 0-387-98549-2. MR1644323

[2] Roberts, Joel. Chow's moving lemma. Appendix 2 to: "Motives" by Steven L. Kleiman. Algebraic geometry, Oslo 1970 (Proc. Fifth Nordic Summer School in Math.), pp. 89--96. Wolters-Noordhoff, Groningen, 1972. MR0382269

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    $\begingroup$ Thanks a lot for your answer, this helps a lot since Fulton is really concise at this point! $\endgroup$ – Joachim Jun 12 '16 at 14:49
  • $\begingroup$ Can I ask you to clarify one point? We write $\epsilon$ for the difference between the correspondence $\Gamma^{\top} \circ \Gamma$ and the identity correspondence $\Delta_X$. Why are the projections $\pi_{1*}\epsilon, \pi_{2*}\epsilon$ supported on $Z$? $\endgroup$ – Joachim Jun 12 '16 at 15:04
  • $\begingroup$ The restrictions of both $\Gamma^\top \circ \Gamma$ and $\Delta_X$ to $U \times U$ are equal to $\Gamma_U$, hence their difference vanishes on $U \times U$. But the difference is supported on $U \times U \cup Z \times Z$, hence has to be supported on $Z \times Z$. $\endgroup$ – R. van Dobben de Bruyn Jun 12 '16 at 17:23
  • $\begingroup$ Great, thanks for the clarification. $\endgroup$ – Joachim Jun 12 '16 at 19:17
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    $\begingroup$ A subtle point came up. To say that the cycle $\Gamma^\top \circ \Gamma$ is supported on $U \times U \cup Z \times Z$ we are implicitly using that $\Gamma^\top$ and $\Gamma$ are meeting properly (otherwise it would only be defined as a cycle class, not as a cycle). Can you give an argument/hint on why this should hold? Thanks a lot! $\endgroup$ – Joachim Jul 31 '16 at 16:35
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Using Jason Starr's comment I was able (I think) to figure out the case of $\pi_1(X)^0$. For anyone who stumbles across this with the same question in mind I add a sketch of the proof in an answer. For the experts, if you feel like leaving a comment whether my reasoning is correct, that would be great.

Let $\phi: X \rightarrow Y$ be a rational map, i.e. defined on some open $U \subset X$. Because $X$ is normal, $Z := X \backslash U$ has codimension at least two. This means the dimenson of the ring $\mathcal{O}_{X,z}$ is at least two for each point $z \in Z$. But then this local ring is pure (SGA 2, Exp. X, Thm 3.4), which in turn implies the couple $(X,Z)$ is pure (SGA 2, Exp. X, prop. 3.3), so that the categories $\operatorname{FEt}(X)$ and $\operatorname{FEt}(U)$ are equivalent.

Since $X$ and $Y$ are birational, we can find an open $V \subset Y$ such that $U$ is isomorphic to $V$ and in the same way we show $V$ and $Y$ have the same \'etale coverings. This shows $\pi_1(X) = \pi_1(U) = \pi_1(V) = \pi_1(Y)$ as desired, which proves in particular what I wanted to show.

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  • $\begingroup$ This is incomplete. But it is in SGA 1 exp. X cor 3.4 anyway. $\endgroup$ – Joachim Jul 31 '16 at 16:31

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