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%\newcommand{\af}{\mathbb{A}}   % affine line
\newcommand{\pl}{\mathbb{P}}   % projective line
%\newcommand{\gm}{{\mathbb{G}_m}} % multiplicative group scheme
\newcommand{\zz}{\mathbb{Z}}   % integers
%\newcommand{\hh}{\mathbb{H}}   % hyperbolic plane
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%\newcommand{\F}{\mathrm{F}} %
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\DeclareMathOperator{\Spec}{\mathrm{Spec}}  % Spectrum
\DeclareMathOperator{\Spin}{\mathrm{Spin}}  % Spin
\DeclareMathOperator{\Pic}{\mathrm{Pic}}    % Picard group
\DeclareMathOperator{\Br}{\mathrm{Br}}     % Brauer group
\DeclareMathOperator{\Gal}{\mathrm{Gal}}   % Galois group
\DeclareMathOperator{\CH}{\mathrm{CH}}      % Chow group
\DeclareMathOperator{\CHO}{\widetilde{\mathrm{CH}}_0}
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\DeclareMathOperator{\End}{\mathrm{End}}    % Endomorphisms
\DeclareMathOperator{\Ker}{\mathrm{Ker}}    % Kernel
\DeclareMathOperator{\im}{\mathrm{Im}}      % image
\DeclareMathOperator{\SB}{\mathrm{SB}}      % Severi-Brauer variety
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\DeclareMathOperator{\Gr}{\mathrm{Gr}} % Grassmann
\DeclareMathOperator{\Fr}{\mathrm{Fr}} % Fraction field
\DeclareMathOperator{\rdim}{\mathrm{rdim}} % reduced dimension
\DeclareMathOperator{\rk}{\mathrm{rk}} % rank
\DeclareMathOperator{\tr}{\mathrm{tr}} % trace
\DeclareMathOperator{\ind}{\mathrm{ind}} % index
\DeclareMathOperator{\sep}{\mathrm{sep}} % separable closure
\DeclareMathOperator{\ad}{\mathrm{ad}} % separable closure
\DeclareMathOperator{\Stab}{\mathrm{Stab}}  % Stabilizer
\DeclareMathOperator{\car}{\mathrm{char}}  % characteristic
\DeclareMathOperator{\diag}{\mathrm{diag}} % diagonal matrix

\title{Zero cycles on twisted Cayley plane}

\author{V.~Petrov, N.~Semenov, K.~Zainoulline}


%%%%%%%%%%%%%%%%%%%% DOCUMENT %%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{document}

\maketitle

\begin{abstract}
In the present paper we compute 
the group of zero-cycles modulo rational equivalence 
of a twisted form $X$ of a Cayley plane. 
More precisely, we show that the degree map 
$\deg: \CH_0(X) \to \zz$ is injective.   
\end{abstract}

\section{Introduction}
Let $G$ be a split exceptional group of type $\E_6$ and 
$G/P$ a projective $G$-homogeneous variety with a stabilizer
given by a maximal parabolic subgroup $P$ corresponding to
the first (or last) root of the respective Dynkin diagram.
This variety is called {\it Cayley plane} and is denoted by 
$\OP$.
Its geometry is a subject of
intensive investigations (see \cite{IM05}).

Let $X$ be a twisted form of the Cayley plane, i.e.,
a smooth projective variety over a field $k$ which becomes
isomorphic to $\OP$ over the separable closure of $k$.
In this case the variety $X$ can have no rational points
and its geometry turns to be very complicated.

In the present paper we compute the group of zero-cycles 
$\CH_0(X)$ which is an important geometric invariant of a variety.
Namely, we prove:
\begin{mainthm}
Let $k$ be a field of characteristic different from 2 and 3.
Let $X$ be a  twisted form of a Cayley plane
corresponding to a group of inner type $\E_6$ over $k$ with trivial
Tits algebras. Then the degree map $\CH_0(X) \to \zz$ is injective.
\end{mainthm}
  

The history of the question starts with the work of I.~Panin \cite{Pa84}
where he proved the injectivity of the degree map for Severi-Brauer varieties.
For quadrics this was proved by R.~Swan in \cite{Sw89}. 
The case of involutive varieties was considered by A.~Merkurjev in \cite{Me95}. 
For varieties of type $\F$ it was announced by M.~Rost.

Our work was mostly motivated by the paper of D.~Krashen \cite{Kr05} where
he reformulated the question above
in terms of $R$-triviality 
of certain symmetric powers and proved the injectivity
for a wide class of generalized Severi-Brauer varieties 
and some involutive varieties,
hence, generalizing the previously known results by Panin and Merkurjev.
Another motivating point was the result of V.~Popov \cite{Po05} 
where he gave a full classification
of generically $n$-transitive actions 
of a split linear algebraic group $G$ on
a projective homogeneous variety $G/P$. 
The case of $\OP=\E_6/P_6$ which is
the main point of our investigations 
provides an example of such an action for $n=3$.
As a consequence, one can identify the open orbit $X^o$ 
of this action with a
homogeneous variety $\E_6/T\cdot \Spin_8$, where 
$T$ is the torus which is complementary to $\Spin_8$.
Then the result of Krashen reduces 
the question of injectivity to the question of  $R$-triviality of the twisted form of $X^o/S_3$.

Apart from the main result concerning exceptional varieties 
we give shortened proofs for injectivity
of the degree map for quadrics and Severi-Brauer varieties as well.

As it was mentioned to us by V.~Chernousov and M.~Rost,
 it is possible to prove the same results
using Rost invariant and Chain Lemma. Our proof doesn't use these tools but
only the geometry and several classical results 
concerning exceptional groups.

\section{Zero cycles and symmetric powers}
In this section we collect some facts from \cite{Kr05} 
on the relations 
between zero cycles on a projective variety $X$ and 
$R$-equivalence classes on symmetric powers of $X$.

\begin{ntt}\label{descent}
We systematically use the Galois descent language, i.e., identify a (quasi-projective) variety $X$ over $k$ with the variety $X_{k_{\sep}}$ over the separable closure $k_{\sep}$ equipped with the action (by semiautomorphisms of $X_{k_{\sep}}$) of the absolute Galois group $\Gamma=\Gal(k_{\sep}/k)$. It means that for any $\sigma\in\Gamma$ we are given by an automorphism $\varphi_\sigma$ of $X_{k_{\sep}}$ over $k$ such that
the diagram
$$
\xymatrix{
X_{k_{\sep}}\ar[d]\ar[r]^{\varphi_\sigma}&X_{k_{\sep}}\ar[d]\\
\Spec k_{\sep}\ar[r]^{\sigma^\sharp}&\Spec k_{\sep}
}
$$
commutes, and  $\varphi_{\sigma\tau}=\varphi_\sigma\varphi_\tau$. The set of $k$-rational points of $X$ is precisely the set of $k_{\sep}$-rational points of $X_{k_{\sep}}$ stable under the action of $\Gamma$.
\end{ntt}

\begin{ntt}
Let $X$ be a variety over $k$. Two rational points $p,q\in X(k)$ are called \emph{elementary linked} if there exists a rational morphism $\varphi\colon\pl^1_k\dasharrow X$ such that $p,q\in\im(\varphi(k))$. The $R$-equivalence is the equivalence relation generated by this relation. A variety $X$ is called \emph{$R$-trivial} if there exists exactly one class of $R$-equivalence on $X$, and \emph{algebraically $R$-trivial} if $X_K=X\times_{\Spec k}\Spec K$ is $R$-trivial for any finite field extension $K/k$.

The $n$-th symmetric power of $X$ is by definition the variety $S^nX=X^n/S_n$ where $S_n$ is the symmetric group acting on $X^n$ via permutations.

Let $p$ be a prime number. A field $k$ is called \emph{prime to $p$ closed} if there is no finite field extension $K/k$ of degree prime to $p$. For any field $k$ we will denote by $k_p$ its \emph{prime to $p$ closure} that is the algebraic extension of $k$ which is prime to $p$ closed.

Let $X$ be a projective variety over $k$. By $\CHO(X)$ we denote
the kernel of the degree map:
$$\CHO(X)=\Ker(\deg\colon\CH_0(X)\to\zz).$$
\end{ntt}

The crucial role in the sequel will play the following results:

\begin{prop}\label{propPrime}(\cite[Lemma~1.3]{Kr05}) If $\CHO(X_{k_p})=0$ for each prime $p$ then $\CHO(X)=0$.
\end{prop}

\begin{prop}\label{propKr}(\cite[Theorem~1.4]{Kr05}) Suppose that $k$ is prime to $p$ closed and the following conditions are satisfied:
\begin{enumerate}
\item $S^{p^n}X$ is algebraically $R$-trivial for some $n\ge 0$,
\item For any field $K/k$ such that $X(K)\ne\emptyset$ $X_K$ is $R$-trivial.
\end{enumerate}
Then $\CHO(X)=0$.
\end{prop}

\begin{ntt}
As an easy application we give a sketch of a proof of the well-known result (first appeared in \cite[Theorem~2.3.7]{Pa84}) that $\CHO(\SB(A))=0$ where $A$ is a central simple algebra (a more general case of flag varieties is settled in \cite{Kr05}). For simplicity we assume that $\deg A=p$ is prime.

We may assume that $k$ is prime to $p$ closed by Proposition~\ref{propPrime} (for primes $q$ different from $p$ $A$ splits over $k_q$). By Proposition~\ref{propKr} it is enough to show that $S^p\SB(A_K)$ is $R$-trivial for every finite extension $K/k$ (the second hypothesis holds for any twisted flag variety). Making base change we may assume $K=k$. If $A$ splits, the assertion is trivial; so we assume that $A$ is not split.

According to our conventions (\ref{descent}) the variety $\SB(A)$ is the variety of all parabolic subgroups $P$ of type $P_1$ in $\PGL_1(A\otimes_kk_{\sep})$ with the action of $\Gamma$ coming from the action on $k_{\sep}$. Therefore $S^pX$ is the variety of all unordered $p$-tuples $[P^{(1)},\ldots,P^{(p)}]$ of parabolic subgroups of type $P_1$ in $\PGL_1(A\otimes_kk_{\sep})$. Let $U$ be the open subvariety of $S^pX$ defined by the condition that the intersection $P^{(1)}\cap\ldots\cap P^{(p)}$ is a maximal torus in $\PGL_1(A\otimes_kk_{\sep})$. Every maximal torus $T$ in $\PGL_1(A\otimes_kk_{\sep})$ is contained in precisely $p$ parabolic subgroups of type $P_1$, whose intersection is $T$. Therefore $U$ is isomorphic to the variety of all maximal tori in $\PGL_1(A)$. This variety is known to be rational
(and therefore $R$-trivial since it is homogeneous). Moreover, one can check that if $A$ is not split then the embedding $U\to S^pX$ is surjective on $k$-points. So $S^pX$ is $R$-trivial, and we are done.
\end{ntt}

\begin{ntt}
The same method can be applied to prove that $\CHO(Q)=0$ for a nonsingular projective quadric $Q$ over a field of characteristic not $2$ (the result of Swan \cite{Sw89}, see \cite{Me95} and \cite{Kr05} for generalizations).

As above, we may assume that $p=2$, $Q$ is anisotropic, and we have to prove that $S^2Q$ is $R$-trivial. Let $q$ be a corresponding quadratic form on a vector space $V$. $Q$ is the variety of all lines $\langle v\rangle$ where $v\in V\otimes_kk_{\sep}$ satisfies $q(v)=0$, with the obvious action of $\Gamma$. So $S^2Q$ can be identified with the variety of pairs $[\langle v_1\rangle,\langle v_2\rangle]$ of lines of this kind, with induced action of $\Gamma$. Consider the open subvariety $U$ defined by the condition $b_q(v_1,v_2)\ne 0$ ($b_q$ stands for the polarization of $q$). Clearly the embedding $U\to S^2Q$ is surjective on $k$-points (otherwise the subspace $\langle v_1,v_2\rangle$ defines a totally isotropic subspace of $V$ defined over $k$). So it is enough to check that $U$ is $R$-trivial.

Consider the open subvariety $W$ of $\Gr(2,V)$ consisting of planes
$H\le V\otimes_kk_{\sep}$ such that $q|_H$ is nonsingular. For every such a plane there exists up to scalar factors exactly one hyperbolic base $\{v_1,v_2\}$ over $k_{\sep}$. Therefore the map from $U$ to $W$ sending $[\langle v_1\rangle,\langle v_2\rangle]$ to $\langle v_1,v_2\rangle$ is an isomorphism. But any open subvariety of $\Gr(2,V)$ is $R$-trivial, and we are done.
\end{ntt}

We will use in the sequel the following observation.
\begin{lem}\label{lemSur}
Let $H\le K\le G$ be algebraic groups over $k$. Suppose that the map $\HH(k,H)\to\HH(k,K)$ is surjective. Then the morphism $G/H\to G/K$ is surjective on $k$-points.
\end{lem}
\begin{proof}
An element $x$ of $G/K(k)$ is presented by an element $g\in G(k_{\sep})$ satisfying the
condition $\gamma(\sigma)=g^{-1}\cdot{}^\sigma g$ lies in $K(k_{\sep})$ for all $\sigma\in\Gamma$. But $\gamma$ is clearly a $1$-cocycle with coefficients in $K$. Therefore by the assumption there exists some $h\in K$ such that $h^{-1}\gamma(\sigma)\cdot{}^\sigma h=(gh)^{-1}\cdot{}^\sigma(gh)$ is a $1$-cocycle with coefficients in $H$. But then $gh$ presents an element of $G/H(k)$ which goes to $x$ under the morphism $G/H\to G/K$, and the lemma is proved.
\end{proof}

\section{The twisted Cayley Plane}

From now on we assume that $\car k\ne 2,3$. In the present section
we prove the theorem of the introduction.

\begin{ntt}
Let $J$ be a simple exceptional $27$-dimensional Jordan algebra over $k$, $N_J$ be its norm (which is a cubic form on $J$). An invertible linear map $f\colon J\to J$ is called a \emph{similitude} if there exists some $\alpha\in k^*$ (called the \emph{multiplier} of $f$) such that $N_J(f(v))=\alpha N_J(v)$ for all $v\in J$. The group
$G=\Sim(J)$ of all similitudes is a reductive group of inner type $\E_6$, and every group of inner type $\E_6$ with trivial Tits algebras can be obtained in this way up to isogeny (see \cite[Theorem~1.4]{Ga01}).

The \emph{(twisted) Cayley plane} $\OP(J)$ is the variety obtained by the Galois descent from the variety of all parabolic subgroups of type $P_1$ (enumeration of roots follows Bourbaki) in $\Sim(J\otimes_kk_{\sep})$
(the action of $\Gamma=\Gal(k_{\sep}/k)$ comes from the action on $k_{\sep}$). This variety can be identified with the variety of all lines $\langle e\rangle$ spanned by elements
$e\in J\otimes k_{\sep}$ satisfying the condition $e\times e=0$ (see \cite[Theorem~7.2]{Ga01}).
\end{ntt}

\begin{mainthm}
$\CHO(\OP(J))=0$.
\end{mainthm}
\begin{proof}

\begin{ntt}
Note that if $\OP(J)(K)\ne\emptyset$ for some extension $K/k$ then $\OP(J\otimes_kK)$ is rational and, since it is homogeneous, is $R$-trivial.

Let $p$ be a prime number. If $p\ne 3$ then $\OP(J)(k_p)\ne\emptyset$. Indeed, choose any cubic \'etale subalgebra $L$ of $J$ (\cite[Proposition~39.20]{Inv}). It splits over $k_p$, and therefore $L\otimes_kk_p$ contains a primitive idempotent $e$. As an element of $J\otimes_kk_p$ it satisfies the condition $e\times e=0$ (see \cite[Lemma~5.2.1(i)]{SV}).

So we assume $p=3$. We have to prove that $S^3\OP(J\otimes_kK)$ is $R$-trivial for all finite field extensions $K/k$. Making the base change, it suffices to prove it for $K=k$.
Moreover, we assume that $J$ is not reduced (otherwise $\OP(J)$ is rational).
\end{ntt}

\begin{ntt}
$S^3(\OP(J))$ is the variety of all unordered triples
$[\langle e_1\rangle,\langle e_2\rangle,\langle e_3\rangle]$ where $e_i$ are elements of $J\otimes_kk_{\sep}$ satisfying the conditions $e_i\times e_i=0$, with the natural action of $\Gamma$. Denote by $U$ the open subvariety of $\OP(J)$ defined by the condition
$$
N_{J\otimes_kk_{\sep}}(e_1,e_2,e_3)\ne 0
$$
(we denote by $N$ the polarization of the norm as well).

The embedding $U\to S^3(\OP(J))$ is surjective on $k$-points. Indeed, if
$[\langle e_1\rangle,\langle e_2\rangle,\langle e_3\rangle]$ is stable under the action of $\Gamma$ and
$N_{J\otimes_kk_{\sep}}(e_1,e_2,e_3)=0$, then $\langle e_1,e_2,e_3\rangle$ gives by descent a $k$-defined subspace $V$ of $J$ such that $N|_V=0$. But then $J$ is reduced
(\cite[Theorem~5.5.2]{SV}), which leads to a contradiction. So it is enough to show that $U$ is $R$-trivial.
\end{ntt}

\begin{ntt}
Choose a cubic \'etale subalgebra $L$ in $J$. Over the separable closure we have
$$
L\otimes_kk_{\sep}=k_{\sep}e_1\oplus k_{\sep}e_2\oplus k_{\sep}e_3
$$
where $e_1,e_2,e_3$ are primitive idempotents. Then $e_i\times e_i=0$ in $J\otimes_kk_{\sep}$,
$$
N_{J\otimes_kk_{\sep}}(e_1,e_2,e_3)=N_{L\otimes_kk_{\sep}}(e_1,e_2,e_3)=1,
$$
and the triple $[e_1,e_2,e_3]$ is stable under $\Gamma$ since $L$ is so. Therefore $[\langle e_1\rangle,\langle e_2\rangle,\langle e_3\rangle]$ is a $k$-rational point of $U$.

By \cite[Proposition~3.12]{SV68} $G$ acts transitively on $U$ (in geometric sense, that is over a separable closure). Therefore
$$
U\simeq G/\Stab_G([\langle e_1\rangle,\langle e_2\rangle,\langle e_3\rangle]).
$$
This stabilizer is defined over $k$ (since it is stable under $\Gamma$) and coincides with $\Stab_G(L)$ (one inclusion is obvious, and the other one follows from the fact that $e_1,e_2,e_3$ are the only elements $e$ of $L\otimes_kk_{\sep}$ satisfying the condition $e\times e=0$, up to scalar factors (see \cite[Theorem~5.5.1]{SV}).
\end{ntt}

\begin{ntt}
Consider the Springer decomposition of $J$ respect to $L$: $J=L\oplus V$. Then the pair $(L,V)$ has a natural structure of a twisted composition, and there is a monomorphism
$\Aut(L,V)\to\Aut(J)$ sending a pair $(\varphi,t)$ (where $\varphi\colon L\to L$,
$t\colon V\to V$) to $\varphi\oplus t\colon J\to J$ (see \cite[\S~38.A]{Inv}). Note that
$\Aut(L,V)$ coincides with the stabilizer of $L$ in $\Aut(J)$.

Show that there exists an exact sequence of algebraic groups:
\begin{align*}
1\longrightarrow\Aut(L,V)\longrightarrow\Stab_G(L)&\longrightarrow
R_{L/k}(\mathbb G_m)\longrightarrow 1,\\
f&\mapsto f(1)
\end{align*}
where $R_{L/k}$ stands for the Weil restriction. Exactness at the middle term follows from the considerations above and the fact that the stabilizer of $1$ in $G$ coincides with $\Aut(J)$ (\cite[Proposition~5.9.4]{SV}).

Prove the exactness at the last term. A $k_{\sep}$-point of $R_{L/k}(\mathbb G_m)$ is a triple of scalars $(\alpha_0,\alpha_1,\alpha_2)\in k_{\sep}^*\times k_{\sep}^*\times k_{\sep}^*$. Multiplying $f$ by the scalar transformation with the coefficient
$(\alpha_0\alpha_1\alpha_2)^\frac{1}{3}$ (which is an element of $\Stab_G(L)(k_{\sep})$) we may assume that $\alpha_0\alpha_1\alpha_2=1$. Choose a \emph{related triple}
$(t_0,t_1,t_2)$ of elements of $\GO^+(\mathbb O_d,N_{\mathbb O_d})$ ($\mathbb O_d$ is the split Cayley algebra) such that $\mu(t_i)=\alpha_i$, $i=0,1,2$ (see \cite[Corollary~35.5]{Inv}). Now the transformation $f$ of $J$ defined by
$$
\begin{pmatrix}
\varepsilon_0&c_2&\cdot\\
\cdot&\varepsilon_1&c_0\\
c_1&\cdot&\varepsilon_2
\end{pmatrix}
\mapsto
\begin{pmatrix}
\alpha_0\varepsilon_0&t_2(c_2)&\cdot\\
\cdot&\alpha_1\varepsilon_1&t_0(c_0)\\
t_1(c_1)&\cdot&\alpha_2\varepsilon_2
\end{pmatrix}
$$
lies in $\Sim(J)$ by \cite[(7.3)]{Ga01}, stabilizes $L_{k_{\sep}}=\diag(k_{\sep},k_{\sep},k_{\sep})$, and sends $1\in J_{k_{\sep}}$ to $\diag(\alpha_0,\alpha_1,\alpha_2)$.

Since $\HH(k,L^*)=1$ by Hilbert '90, the map $\HH(k,\Aut(L,V))\to\HH(k,\Stab_G(L))$ is surjective. By Lemma~\ref{lemSur} the morphism
$$
G/\Aut(L,V)\to G/\Stab_G(L)\simeq U
$$
is surjective on $k$-points. Therefore it suffices to show that $G/\Aut(L,V)$ is $R$-trivial.
\end{ntt}

\begin{ntt}
Consider the morphism
$$
\psi\colon G/\Aut(L,V)\to G/\Aut(J).
$$
By \cite[Corollary~3.14]{Kr05} it suffices to show that
\begin{enumerate}
\item $\psi$ is surjective on $k$-points;
\item $G/\Aut(J)$ is $R$-trivial;
\item The fibers of $\psi$ (which are isomorphic to $\Aut(J)/\Aut(L,V)$) are unirational and $R$-trivial.
\end{enumerate}
\end{ntt}

\begin{ntt}
In order to prove surjectivity of $\psi$ on $k$-points it is enough by Lemma~\ref{lemSur} to prove surjectivity of the map $\HH(k,\Aut(L,V))\to\HH(k,\Aut(J))$. Now $\HH(k,\Aut(L,V))$ classifies all twisted compositions $(L',V')$ which become isomorphic to $(L,V)$ over $k_{\sep}$, and $\HH(k,\Aut(J))$ classifies all (exceptional $27$-dimensional) Jordan algebras $J'$. It is easy to verify that the morphism sends $(L',V')$ to the Jordan algebra $L'\oplus V'$, and hence the surjectivity follows from the fact that any Jordan algebra admits a Springer decomposition (cf. \cite[Proposition~38.7]{Inv}).
\end{ntt}

\begin{ntt}
Let $W$ be the open subvariety of $J$ consisting of elements $v$ with $N_J(v)\ne 0$. Then $G$ acts transitively (in geometric sense) on $W$ (see \cite[Proposition~5.9.3]{SV}), and the stabilizer of the point $1$ coincides with $\Aut(J)$.
So $G/\Aut(J)\simeq W$ is clearly $R$-trivial.
\end{ntt}

\begin{ntt}
Consider the variety $Y$ of all \'etale cubic subalgebras of $J$. By \cite[Proposition~39.20(1)]{Inv} there is a map from an open subvariety $J_0$ of regular elements in $J$ to $Y$ (sending $a$ to $k[a]$), surjective on $k$-points. Therefore $Y$ is unirational and $R$-trivial
$24$-dimensional irreducible variety.

$\Aut(J)$ acts on $Y$ naturally. Let $L'$ be any $k$-point of $Y$. The stabilizer of $L'$ in $\Aut(J)$ obviously equals $\Aut(L',V')$ ($J=L'\oplus V'$ is the Springer decomposition). So the orbit of $L'$ is isomorphic to $\Aut(J)/\Aut(L',V')$ and, in particular, has dimension $24$. Therefore it is open, and since $L'$ is arbitrary, the action is transitive. So we have $\Aut(J)/\Aut(L,V)\simeq Y$ is unirational and $R$-trivial, and the proof is completed.
\end{ntt}
\end{proof}

\begin{rem}
The main result of the paper
also implies the injectivity of the degree map for
the twisted form $Y$ of $\mathrm{F}_4/P_4$. To see this one observes
that (if the splitting field has degree $p=3$) 
both varieties $\OP(J)$ and $Y$ split completely over
the function fields of each other. This together with \cite[Cor.~3.6]{Me}
implies that the groups of zero-cycles are isomorphic. 
\end{rem}



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\end{document}
