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](https://mathoverflow.net/questions/81510/are-chow-groups-a-birational-invariant) 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.



Reference request on birational invariance of Chow group of zero cycles of degree zero
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](http://www.ams.org/mathscinet-getitem?mr=1644323) 


[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](http://www.ams.org/mathscinet-getitem?mr=0382269)


