%% Copyright (C) 1998, 1999 Markus Rost.
%% Copyright (C) 1996, 1998 Jean-Pierre Tignol (for the appendix).
%%
%% French parts translated from the English
%% and revised by Jean-Pierre Serre.
%%
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%% "The chain lemma for Kummer elements of degree 3"
%%
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%% Comptes rendus de l'Academie des sciences - serie I - Mathematique
%% Alg\`ebre/Algebra
%% Note pres\'ent\'ee par Jean-Pierre Serre.
%% (Re\c cu et accept\'e le 30 novembre 1998)
%% C. R. Acad. Sci. Paris S\'er. I Math. 328 (1999), no. 3, 185-190.
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%% Last Changes:
%% Nov 28 1998 (ready for submission)
%% Dec 17 1998 (change in comments)
%% Jan 05 1999 (revision following proofs)
%% Jan 28 1999 (2x stupid fields `F' changed to `k')
%% Feb 19 1999 (it appeared!)

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\begin{document}

\title{The chain lemma for Kummer elements of degree~$3$}
\author{Markus Rost}
\address{NWF I - Mathematik,
  Universit\"at Regensburg,
  D-93040 Regensburg, Germany}
\email{markus.rost@mathematik.uni-regensburg.de}
\urladdr{http://www.physik.uni-regensburg.de/\~{}rom03516}
\date{February 19, 1999}
\thanks{The text appeared as: \frenchspacing C. R. Acad. Sci. Paris
S\'er. I Math. 328 (1999), no. 3, 185--190.}
\maketitle

\begin{Abstracts}{Abstract}
  Let $A$ be a skew field of degree~$3$ over a field containing the
  $3^{\rm rd}$ roots of unity.  We prove a sort of chain equivalence
  for Kummer elements in~$A$.  As a consequence one obtains a common
  slot lemma for presentations of~$A$ as a cyclic algebra.
\end{Abstracts}

\frenchtitle{Cha{\^\i}nes d'\'el\'ements de Kummer en degr\'e~$3$}

\begin{Abstracts}{R\'esum\'e}
%%  Laissez $A$ \^etre une zone oblique du degr\'e~$3$ au-dessus d'une
%%  zone contenant les troisi\`eme racines de l'unit\'e.  Nous prouvons
%%  un tri de l'\'equivalence \`a cha\^\i nes pour des \'el\'ements de
%%  Kummer dans~$A$.  Par cons\'equent on obtient un lemme commun de
%%  fente pour des pr\'esentations de~$A$ comme alg\`ebre
%%  cyclique.\footnote{Traduction de l'anglais par M. Computer.}
%%
%%  Dear Markus,
%%
%%    I finally downloaded and printed your note on the chain lemma. So far,
%%  I have only looked at the hilarious french translation of the summary.
%%  Are you going to keep that ?
%%
%%  All the best,
%%
%%  JP
%%
  Soit $k$ un corps contenant les racines cubiques de l'unit\'e, et
  soit $A$ un corps gauche de centre~$k$, avec $[A\mathop:k]=9$.  Nous
  montrons que deux \'el\'ements de Kummer de~$A$ peuvent \^etre
  joints par une cha{\^\i}ne de longueur~$4$.
\end{Abstracts}

\begin{VersFrancAbr}
  Soit $k$ un corps contenant une racine primitive $n$-\`eme de
  l'unit\'e~$\zeta$, et soit $A$ une $k$-alg\`ebre centrale simple de
  degr\'e~$n$.  Un \emph{\'el\'ement de Kummer} de~$A$ est un
  \'el\'ement dont le polyn\^ome caract\'eristique est de la forme
  $t^n-a$, avec $a\in k^*$.  Par une \emph{$\zeta$-paire} on entend un
  couple~$(X,Y)$ d'\'el\'ements de Kummer de~$A$ tels que $YX=\zeta
  XY$.  Une telle paire donne une pr\'esentation de~$A$ comme produit
  crois\'e cyclique\,:
  \begin{equation*}
    A=\amida{X,Y}{X^n=a,\,Y^n=b,\, YX=\zeta XY}, \quad\text{avec}\quad
    a,b\in k^*.
  \end{equation*}

  Soient $X$,~$Y$ deux \'el\'ements de Kummer de~$A$, et soit $m$ un
  entier~$\geq1$.  Une \emph{cha{\^\i}ne de longueur~$m$ joignant $X$
  \`a~$Y$} est une suite de $m+1$ \'el\'ements de Kummer\,:
  \begin{equation*}
    X=Z_0,Z_1,\dots, Z_m=Y
  \end{equation*}
  tels que $(Z_{i-1}, Z_i)$ soit une $\zeta$-paire pour $i=1$,
  \dots,~$m$.

  Supposons que $A$ soit un corps gauche.  Si $n=2$ (i.e.\ si $A$ est
  un corps de quaternions), il est facile de voir que tout couple
  d'\'el\'ements de Kummer peut \^etre joint par une cha{\^\i}ne de
  longueur~$2$.  Si $n=3$, J.-P. Tignol a donn\'e des exemples de
  couples~$(X,Y)$ d'\'el\'ements de Kummer tels qu'il n'existe aucune
  cha{\^\i}ne de longueur~$2$ joignant $X$ \`a~$Y$ (ni m\^eme \`a un
  conjugu\'e de~$Y$, cf.~Appendice)\,; dans ce qui suit, nous montrons
  qu'un tel couple peut \^etre joint par une cha{\^\i}ne de
  longueur~$4$.  La d\'emonstration s'inspire de celle donn\'ee par
  Petersson-Racine~\cite{Petersson-Racine:96} pour un r\'esultat
  analogue dans les alg\`ebres de Jordan exceptionnelles.  Comme
  cons\'equence, on obtient un ``common slot lemma'' pour les
  alg\`ebres de degr\'e~$3$.
\end{VersFrancAbr}

\section*{Introduction}

The well known common slot lemma for quaternion algebras asserts that
if $(a,b)$ is split over~$k(\sqrt c)$, then $(a,b)\simeq (a,e)\simeq
(c,e)$ for some~$e$.

Till a few years ago not much has been known about similar statements
for algebras of degree~$>2$.  Tignol has given an example (cf.\
Appendix) which shows that a common slot lemma with just one
additional ``slot'' does not hold in general for algebras of
degree~$3$.  The first positive result was obtained by Petersson and
Racine~\cite{Petersson-Racine:96} who proved, taking up a suggestion
of J-P.~Serre, a common slot lemma for exceptional Jordan algebras
over quadratically closed fields.

The major purpose of this Note is to present the Petersson-Racine
arguments in the much simpler case of central simple algebras of
degree~$3$.  They yield a sort of chain equivalence for Kummer
elements.  As a consequence one obtains a common slot lemma for such
algebras.

I am indebted to Jean-Pierre Tignol for leaving his text on the
counterexample as an appendix to this Note.

\section{Kummer elements}

Let $n\geq 2$ and let $k$ be a field containing a primitive $n^{\rm
th}$ root of unity~$\zeta$.  For $a$, $b\in k^*$ we denote by $(a,b)$
the $k$-algebra defined by the presentation
\renewcommand{\theequation}{$*$}%
\begin{equation}
  \label{eq:1}
  \amida{X,Y}{X^n=a,\,Y^n=b,\, YX=\zeta XY}.
\end{equation}

Let $A$ be a central simple algebra of degree~$n$ over~$k$.  A
\emph{Kummer element} in~$A$ is an element $X\in A$ whose
characteristic polynomial~$\Pc_X$ is of the form $\Pc_X(t)=t^n-a$ for
some $a\in k^*$.

\begin{lemma}
  Let $X\in A$ be a Kummer element and let
  \begin{equation*}
    E(X,\zeta)=\bmidb{Z\in A}{ZX=\zeta XZ}.
  \end{equation*}
  \begin{enumerate}
    \renewcommand{\labelenumi}{(\theenumi)}
    \renewcommand{\theenumi}{\roman{enumi}}
    \item
    \label{item:a1}
    $L=k[X]$ is the centralizer of\/~$X$ in~$A$.
    \item
    \label{item:a2}
    There exists $Y\in A^*$ such that $YXY\inv=\zeta X$.
    \item
    \label{item:a3}
    For\/~$Y$ as in~\eqref{item:a2} one has $E(X,\zeta)=YL=LY$.
  \end{enumerate}
\end{lemma}

\begin{proof}
  \eqref{item:a1} follows from $\dim_kL=\deg A$, \eqref{item:a2} from
  the Skolem-Noether theorem, and \eqref{item:a3} from \eqref{item:a1}
  and~\eqref{item:a2}.
\end{proof}

By a \emph{$\zeta$-pair} we understand a pair $(X,Y)$ of invertible
elements $X$, $Y\in A$ such that $YX=\zeta XY$.

\begin{lemma}
  Let $(X,Y)$ be a $\zeta$-pair.
  \begin{enumerate}
    \renewcommand{\labelenumi}{(\theenumi)}
    \renewcommand{\theenumi}{\roman{enumi}}
    \item
    \label{item:1}
    $X$ and\/~$Y$ are Kummer elements.
    \item
    \label{item:2}
    If $A=M_n(k)$ and $X^n=Y^n=1$, then the pair~$(X,Y)$ is conjugate
    to the pair~$(X_0,Y_0)$, where $X_0$ is the diagonal matrix
    $\diag(1,\zeta,\zeta^2,\dots,\zeta^{n-1})$ and where $Y_0$ is the
    permutation matrix $e_i\mapsto e_{i-1}$ with $i$ taken $\amod n$.
    \item
    \label{item:3}
    The algebra~$A$ has the presentation~\eqref{eq:1}.
  \end{enumerate}
\end{lemma}

\begin{proof}
  Since $YXY\inv=\zeta X$, the $n$ different powers of~$\zeta$ are
  roots of~$P_X$, whence $P_X(t)=t^n-a$ for some $a\in k$.  Further,
  $X$ is invertible and therefore $a\neq0$.  Similarly one sees
  $P_Y(t)=t^n-b$ for some $b\in k^*$.  This proves~\eqref{item:1}.
  For~\eqref{item:2} note that any matrix~$X$ with $\Pc_X(t)=t^n-1$ is
  conjugate to~$X_0$ and we may therefore assume $X=X_0$.  Then
  necessarily $Y=UY_0$ where $U$ is in the centralizer $L=k[X]$
  of~$X$.  One has $N_{L/k}(U)=Y^n=1$.  Therefore there exist $V\in
  L^*$ such that $U=VY_0V\inv Y_0\inv$.  It follows that $V\inv
  YV=Y_0$, which proves the claim.  For~\eqref{item:3} one may assume
  that $k$ is algebraically closed and that $A=M_n(k)$.  The claim
  follows from~\eqref{item:2} after replacing $X$ by $X/\sqrt[n]a$ and
  $Y$ by~$Y/\sqrt[n]b$.
\end{proof}

\section{Chains}

Let $X$, $Y\in A$ be Kummer elements.  By a \emph{chain} from $X$
to~$Y$ of length~$m$ we understand a sequence $X=Z_0$, $Z_1$, \dots,
$Z_m=Y$ of Kummer elements in~$A$ such that $(Z_{i-1},Z_i)$ is a
$\zeta$-pair for $i=1$, \dots,~$m$.

Let $Z_0$, \dots,~$Z_m$ be a chain of Kummer elements in~$A$ and let
$a_i=Z_i^n$.  Then
\begin{equation*}
  A\simeq (a_{i-1},a_i)
\end{equation*}
for $i=1$, \dots,~$m$.  This shows that a chain of Kummer elements
gives rise to a sequence of presentations~\eqref{eq:1} with ``common
slots''.

If there exists a chain from $X$ to~$Y$ of length~$m$, then there
exists also a chain from $X$ to~$Y$ of length~$m'$ for any $m'\geq m$
(if $X$,~$Y$ is a chain of length~$1$, then $X$, $YX$,~$Y$ is a chain
of length~$2$).

Given Kummer elements $X$ and~$Y$, does there exist a chain from $X$
to~$Y$?

Let us consider the case $n=2$.  Then $A$ is a quaternion algebra and
$X\in A$ is a Kummer element if and only if $X$ is invertible and
$\trace(X)=0$.  Given Kummer elements $X$ and~$Y$, let $Z=XY-YX$.  If
$Z$ is invertible, then $X$, $Z$,~$Y$ is chain from~$X$ to~$Y$.  If
$Z=0$, then $X$ and~$Y$ are scalar multiples of each other and any
Kummer element~$Z'$ anti-commuting with~$X$ gives rise to a chain $X$,
$Z'$,~$Y$.  It follows that for quaternion skew fields there exist
always chains from~$X$ to~$Y$ of length~$2$.  In the case $A=M_2(k)$
is not difficult to see that there exist always chains of length~$3$
and to give examples of Kummer elements $X$,~$Y$ for which there does
not exist a chain of length~$2$.

We now assume $n=3$.

\begin{proposition}
  \label{prop:2.1}
  Let $A$ a skew field of degree~$3$ over a field containing a
  primitive $3^{\it rd}$ root of unity\/~$\zeta$.  Then for any two
  Kummer elements $X$, $Y\in A$ there exists a chain of length~$4$
  from $X$ to~$Y$.
\end{proposition}

As an immediate consequence of the proposition one obtains:

\begin{corollary}
  Suppose that $(a,b)$ is split over $k(\sqrt[3]c)$.  Then there exist
  $e$, $f$, $g\in k^*$ such that
  \begin{equation*}
    (a,b)\simeq (a,e)\simeq (f,e)\simeq (f,g)\simeq (c,g).
  \end{equation*}
\end{corollary}

\begin{proof}
  Let $A=(a,b)$.  If $A$ is split, one takes $e=f=g=1$.  Assume that
  $A$ is a skew field and choose Kummer elements $X$, $Y\in A$ with
  $X^3=a$ and $Y^3=c$.  By Proposition~\ref{prop:2.1} there exists a
  chain $X$, $Z_1$, $Z_2$, $Z_3$,~$Y$.  It suffices to take $e=Z_1^3$,
  $f=Z_2^{-3}$, and $g=Z_3^{-3}$.
\end{proof}

Tignol's example in the appendix shows that there exist an algebra~$A$
of degree~$3$ and Kummer elements $X$, $Y\in A$ for which there is no
chain of length~$2$ from~$X$ to any conjugate of~$Y$.  The question
for chains of length~$3$ is more delicate:  it turns out that for
generic $X$,~$Y$ there exist exactly $2$ chains of length~$3$ which
however might be defined only over a quadratic extension of the ground
field.  We hope to provide details for this at another occasion.

\section{Proof of Proposition~\ref{prop:2.1}}

Let $k$ be a field with $\car k\neq3$ containing a primitive $3^{\rm
rd}$ root of unity~$\zeta$.  Moreover let $A$ be a skew field of
degree~$3$ and let $X$, $Y\in A$ be Kummer elements.  Let
$L=k[X]\subset A$ be the subfield generated by~$X$.  Then
\renewcommand{\theequation}{$**$}%
\begin{equation}
  \label{eq:2}
  A=L\oplus E(X,\zeta)\oplus E(X,\zeta^2).
\end{equation}

We show that there exist invertible elements $Z_1$, $Z_2$, $Z_3\in A$
such that:
\begin{enumerate}
  \renewcommand{\labelenumi}{(\theenumi)}
  \item
  \label{item:c1}
  $Z_1\in E(X,\zeta)$,
  \item
  \label{item:c2}
  $Z_2Z_1=\zeta Z_1Z_2$,
  \item
  \label{item:c3}
  $Z_3Z_2=\zeta Z_2Z_3$,
  \item
  \label{item:c4}
  $Z_3\in E(Y,\zeta^2)$,
  \item
  \label{item:c5}
  $Z_2\in X^2k\oplus E(X,\zeta^2)$,
  \item
  \label{item:c6}
  $Z_3\in E(X,\zeta)\oplus E(X,\zeta^2)$.
\end{enumerate}

Conditions \eqref{item:c1}--\eqref{item:c4} mean that $X$, $Z_1$,
$Z_2$, $Z_3$,~$Y$ is a chain.  The additional conditions
\eqref{item:c5} and \eqref{item:c6} are taken
from~\cite{Petersson-Racine:96}.  Their significance lies in the fact
that for generic $X$,~$Y$ the system of equations
\eqref{item:c1}--\eqref{item:c6} has a solution $(Z_1,Z_2,Z_3)$,
$Z_i\neq 0$ which is \emph{unique} up to scalar factors of the~$Z_i$.
It would be interesting to understand more about the geometry of the
system \eqref{item:c1}--\eqref{item:c6}.  In the following we merely
present a solution.

\begin{lemma}
  \label{lem:3.1}
  There exist~$Z_3\neq0$ satisfying \eqref{item:c4}
  and~\eqref{item:c6}.
\end{lemma}

\begin{proof}
  One has $\dim_kE(Y,\zeta^2)=3$ and $\dim_k\bigl(E(X,\zeta)\oplus
  E(X,\zeta^2)\bigr)=6$.  Both vector spaces lie in the
  $8$-dimensional vector subspace of~$A$ of trace zero elements.
  Hence they have a nontrivial intersection.
\end{proof}

We choose $Z_3$ as in Lemma~\ref{lem:3.1}.  It remains to find $Z_1$,
$Z_2\in A^*$ satisfying \eqref{item:c1}, \eqref{item:c2},
\eqref{item:c3}, and~\eqref{item:c5}.

Let $Z\in E(X,\zeta)$, $Z\neq 0$.  Then $E(X,\zeta)=ZL$ and
$E(X,\zeta^2)=LZ\inv$.  Write
\begin{equation*}
  Z_3=Z\mu'+\mu''Z\inv
\end{equation*}
with $\mu'$, $\mu''\in L$.

If $\mu'=0$, then $Z_3\in E(X,\zeta^2)$ and $Z_1=Z_3\inv$, $Z_2=X^2$
do the job.

If $\mu''=0$, then $Z_3\in E(X,\zeta)$ and $Z_1=Z_3X$, $Z_2=Z_3^2X$ do
the job.

Assume that $\mu'\neq 0$ and $\mu''\neq 0$.  After replacing $Z$
by~$Z\mu''$ we have $Z_3=Z\mu+Z\inv$ for some nonzero $\mu\in L$.

\begin{lemma}
  \label{lem:3.2}
  Let $(X,Z)$ be a $\zeta$-pair, let $\mu=m_0+m_1X+m_2X^2$, $m_i\in
  k$, and let $T=Z\mu+Z\inv$.  Let further $c_2$ be the second
  coefficient of the characteristic polynomial of\/~$T$.  Then
  $c_2=-3m_0$.
\end{lemma}

\begin{proof}
  One has $\trace(T)=0$ and $\trace(T^2)=2\trace(\mu)=6m_0$.  Since
  $2c_2=\trace(T)^2-\trace(T^2)$, it follows that $2c_2=-6m_0$.  This
  proves the claim for $\car k\neq2$.  For $\car k=2$, consider
  $c_2=-3m_0$ as a polynomial identity in the variables~$m_i$.  It
  suffices to verify this identity for a standard $\zeta$-pair $(X,Z)$
  in~$M_3(\mathbb Z[\zeta])$.  This follows from the
  characteristic~$0$ case.
\end{proof}

For the Kummer element $T=Z_3$ one has $c_2=0$ and Lemma~\ref{lem:3.2}
shows that
\begin{equation*}
  \mu=m_1X+m_2X^2,\quad Z_3=Z(m_1X+m_2X^2)+Z\inv
\end{equation*}
for some $m_1$, $m_2\in k$.

If $m_1=0$, then $Z_1=Z$ and $Z_2=(ZX)\inv$ do the job.

Otherwise let
\begin{gather*}
  b=Z^{-3},\quad c= \zeta m_1b/N_{L/k}(\mu),\quad
  \lambda=c\mu X,\\
  Z_1=Z\lambda,\quad Z_2=X^2\bigl(1+(Z\lambda)\inv\bigr).
\end{gather*}
With these settings, \eqref{item:c1}, \eqref{item:c2},
and~\eqref{item:c5} are obvious.  It remains to
verify~\eqref{item:c3}:
\begin{equation*}
  (Z\mu+Z\inv)X^2\bigl(1+(Z\lambda)\inv\bigr)= \zeta
  X^2\bigl(1+(Z\lambda)\inv\bigr)(Z\mu+Z\inv).
\end{equation*}
To check this, one considers the components with respect to the
decomposition~\eqref{eq:2}.  For the first component one gets $Z\mu
X^2\lambda\inv Z\inv=\zeta X^2\lambda\inv Z\inv Z\mu$, which follows
from $\mu X^2\lambda\inv=Xc\inv$ and $ZX=\zeta XZ$.  For the third
component one gets $Z\inv X^2=\zeta X^2Z\inv$, which is immediate from
$ZX=\zeta XZ$.  For the second component one gets
\begin{align*}
  Z\mu X^2 +Z\inv X^2 \lambda\inv Z\inv&= \zeta X^2Z\mu + \zeta X^2
  \lambda\inv Z\inv Z\inv
  .\\
  \intertext{This is equivalent to both of the following equations:}
  X^2\mu +Z^{-2} X^2 \lambda\inv Z\inv&= \zeta^2 X^2\mu + \zeta^2 X^2
  Z\inv \lambda\inv Z^{-2}
  ,\\
  (1-\zeta^2) X^2\mu &= \zeta^2 X^2 Z\inv \lambda\inv Z^{-2} -Z^{-2}
  X^2 \lambda\inv Z\inv .
\end{align*}
For the right hand side of the last equation one computes
\begin{align*}
  {\rm r.~h.~s.}&=\zeta^2 X^2 Z\inv (c\mu X)\inv Z^{-2} -\zeta^2X^2
  Z^{-2} (c\mu X)\inv Z\inv
  \\
  &=c\inv X Z\inv \mu \inv Z^{-2}
  -c\inv\zeta X Z^{-2} \mu\inv Z\inv\\
  &= bc\inv X (Z\inv \mu Z)\inv -bc\inv\zeta X (Z^{-2} \mu Z^2)\inv .
\end{align*}
We multiply both sides with the conjugates $Z\inv \mu Z$ and $Z^{-2}
\mu Z^2$ of~$\mu$.  Then our equation reads as
\begin{align*}
  (1-\zeta^2) X^2N_{L/k}(\mu) &= bc\inv X (Z^{-2} \mu Z^2)
  -bc\inv\zeta X (Z\inv \mu Z)
  \\
  &= bc\inv X (m_1\zeta X+m_2\zeta^2X^2) -bc\inv\zeta X (m_1\zeta^2
  X+m_2\zeta X^2)
  \\
  &= bc\inv m_1\zeta X^2 (1-\zeta^2) .
\end{align*}
The equality is now clear.

%% J-P's file, slightly edited.

% From Tignol@agel.ucl.ac.be Wed Apr 15 16:22 MET 1998
% Date: Wed, 15 Apr 1998 16:21:46 +0200 (MET DST)
% From: Jean-Pierre Tignol <tignol@agel.ucl.ac.be>
% To: Markus Rost t4287 <markus.rost@mathematik.uni-regensburg.de>
% Subject: Re: Chain-lemma fails of length 2 fails

% Dear Markus,

%   Here are the notes I promised you:

% -----------------------------cut-----------cut---------------------------
% \documentclass{article}
% \usepackage{amsfonts}
% \renewcommand{\theequation}{\fnsymbol{equation}}

% \begin{document}

\section*{Appendix}

With the kind permission of Jean-Pierre Tignol we reproduce here his
text on

\bigskip

\begin{center}
  A ``common slot'' counterexample in degree~$3$
\end{center}

\begin{notation}
  For $a$,~$b$ nonzero elements in a field~$F$ containing a primitive
  cube root of unity~$\omega$, the symbol $(a,b)$ denotes the element
  of the Brauer group of~$F$ represented by the $F$-algebra generated
  by elements $\alpha$,~$\beta$ subject to
  \begin{equation*}
    \alpha^{3}=a,\qquad\beta^{3}=b,\qquad
    \beta\alpha=\omega\alpha\beta.
  \end{equation*}
\end{notation}

Let $a_{1}$, $b_{1}$, $a_{2}\in F^\times$.  If there exist $x$, $y\in
F^\times$ such that
\renewcommand{\theequation}{$*$}%
\begin{equation}
  \label{eq:jp}
  (a_{1},b_{1})=(a_{1},x)+(a_{1},y),\quad (a_1,x)=-(a_{2},x),\quad
  (a_{1},y)=(a_{2},y),
\end{equation}
then the additivity of symbols yields $(a_{1},b_{1})=(a_{2},x^{-1}y)$.
However, the next example shows that when $(a_{1},b_{1})$ is split by
$F(\sqrt[3]{a_{2}})$, there need not exist elements $x$,~$y$
satisfying~\eqref{eq:jp}.

\begin{example}
  A global field~$F$ containing a primitive cube root of unity and
  elements $a_{1}$, $b_{1}$, $a_{2}$,~$b_{2}$ such that
  $(a_{1},b_{1})=(a_{2},b_{2})$, but no couple of elements $x$,~$y$
  satisfying~\eqref{eq:jp}.  In particular \textup(taking
  $x=1$\textup), the field~$F$ does not contain any element~$y$ such
  that
  \begin{equation*}
    (a_{1},b_{1})=(a_{1},y)=(a_{2},y)=(a_{2},b_{2}).
  \end{equation*}
\end{example}

Let $F=\mathbb{F}_{7}(t)$, where $t$ is an indeterminate, $a_{1}=t$
and $a_{2}=t(1-t)$.  Note that $(a_{1},a_{2})=0$.  Therefore, for all
places~$v$ of~$F$, the local invariant $(a_{1},a_{2})_{v}$ is trivial.
It follows that in the completion $F_{v}$ of~$F$ at~$v$ we have either
$a_{1}\in F_{v}^{\times 3}$ or $a_{1}\equiv a_{2}\bmod F_{v}^{\times
3}$ or $a_{1}\equiv a_{2}^{2}\bmod F_{v}^{\times 3}$ or $a_{2}\in
F_{v}^{\times 3}$, since the (generalized) Hilbert symbol
$\emptysymbol_v\colon(F_v^\times/F_v^{\times 3})\times
(F_{v}^\times/F_{v}^{\times3})\to\frac{1}{3}\mathbb{Z}/\mathbb{Z}$ is
a nondegenerate alternating pairing.

Consider in particular $v_{1}$ the $t$-adic place and $v_{2}$ the
$(t+3)$-adic place.  Since $a_{1}$,~$a_{2}$ are uniformizing
parameters at~$v_{1}$, we have $a_{1}$, $a_{2}\notin
F_{v_{1}}^{\times3}$; but $a_{1}\equiv a_{2}\bmod F_{v_{1}}^{\times
3}$.  On the other hand, $a_{1}$ and~$a_{2}$ have non-cube residues
at~$v_{2}$, hence $a_{1}$, $a_{2}\notin F_{v_{2}}^{\times 3}$ but
$a_{1}\equiv a_{2}^{-1}\bmod F_{v_{2}}^{\times 3}$.

Let now $A$ be the central simple $F$-algebra with local invariants
$1/3$ at~$v_{1}$, $2/3$ at~$v_{2}$ and $0$ everywhere else.  If $v$ is
a place of~$F$ where $a_{1}\in F_{v}^{\times 3}$, then $v\neq
v_{1}$,~$v_{2}$ hence $[A]_{v}=0$.  It follows that $A$ is split by
$F(\sqrt[3]{a_{1}})$, hence we may find $b_{1}\in F^\times$ such that
$[A]=(a_{1},b_{1})$ in the Brauer group of~$F$.  Similarly, $A$ is
split by $F(\sqrt[3]{a_{2}})$ hence we may find $b_{2}\in F^\times$
such that $[A]=(a_{2},b_{2})$; thus,
\begin{equation*}
  (a_{1},b_{1})=(a_{2},b_{2}).
\end{equation*}
Suppose now $x$, $y\in F^\times$ satisfy~\eqref{eq:jp}.  Since
$a_{1}\equiv a_{2}\bmod F_{v_{1}}^{\times 3}$, the relation
$(a_{1},x)_{v_{1}}=-(a_{2},x)_{v_{1}}$ implies $(a_{1},x)_{v_{1}}=0$.
On the other hand, since $a_{1}\equiv a_{2}^{-1}\bmod
F_{v_{2}}^{\times 3}$, it follows from
$(a_{1},y)_{v_{2}}=(a_{2},y)_{v_{2}}$ that $(a_{1},y)_{v_{2}}=0$,
hence $(a_{1},x)_{v_{2}}=(a_{1},b_{1})_{v_{2}}=2/3$.

For $v\neq v_{1}$,~$v_{2}$, we consider four cases, according to the
relation between $a_{1}$ and~$a_2$ in the group of cube classes:
\begin{itemize}
  \item
  if $a_{1}\in F_{v}^{\times3}$, then clearly $(a_{1},x)_{v}=0$.
  \item
  if $a_{1}\equiv a_{2}\bmod F_{v}^{\times 3}$, then $(a_{1},x)_{v}=0$
  as for $v=v_{1}$ above.
  \item
  if $a_{1}\equiv a_{2}^{-1}\bmod F_{v}^{\times 3}$, then
  $(a_{1},x)_{v}=(a_{1},b_{1})_{v}$ as for $v=v_{2}$ above, hence
  $(a_{1},x)_{v}=0$.
  \item
  if $a_{2}\in F_{v}^{\times 3}$, then $(a_{1},x)_{v}=0$ follows from
  $(a_{1},x)=(a_{2},x^{-1})$.
\end{itemize}

Thus, the invariants of $(a_{1},x)$ are:
\begin{equation*}
  (a_{1},x)_{v_{2}}=2/3, \quad\mbox{and}\quad (a_{1},x)_{v}=0
  \quad\mbox{for $v\neq v_{2}$,}
\end{equation*}
a contradiction to the reciprocity law.

\bigskip

\begin{flushright}
  Jean-Pierre Tignol, June 1996.
\end{flushright}

%% end of J-P's file

%% Letter to J-P

% From markus.rost@rphs1.physik.uni-regensburg.de Sun Nov  8 19:57 MET 1998
% Date: Sun, 8 Nov 1998 19:57:41 +0100
% From: Markus Rost <markus.rost@mathematik.uni-regensburg.de>
% To: tignol@agel.ucl.ac.be
% Subject: counterexample
% Reply-To: markus.rost@mathematik.uni-regensburg.de
% Content-Type: text
% Content-Length: 227

% Dear Jean-Pierre,

% concerning your counterexample: One can give b_1 and b_2 the explicit
% values b_1=t+3 and b_2\inv=(t-3)(t+3), since one has

%  {t,t+3}={(t-3)(t+3),t(1-t)} in K_2\LF_7(t)/3.

% Proof: Check residues.

% Markus

%% End of letter

%\bibliography{general}
%\bibliographystyle{amsplainR}
%\end{document}

\begin{thebibliography}{1}

  \bibitem{Petersson-Racine:96} H.~P. Petersson and M.~L. Racine,
  \emph{An elementary approach to the Serre-Rost invariant of Albert
  algebras}, Indag. Mathem., N.S.  \textbf{7} (1996), no.~3, 343--365.

%\bibitem{3} {\sc J.-P. Serre}, {\it R\'esum\'e des cours de l'ann\'ee}
%  1990--1991, Coll\`ege de France.

\end{thebibliography}
\end{document}
