%% Copyright (C) 1998 by Jean-Pierre Tignol.

%% 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---------------------------
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\begin{center}
A ``common slot'' counterexample in degree~$3$
\end{center}
\bigskip
\par
\emph{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
\[
\alpha^{3}=a,\qquad\beta^{3}=b,\qquad
\beta\alpha=\omega\alpha\beta.
\]
Let $a_{1}$, $b_{1}$, $a_{2}\in F^\times$. If there exist $x$, $y\in 
F^\times$ such that
\begin{equation}
(a_{1},b_{1})=(a_{1},x)+(a_{1},y),\quad
(a_1,x)=-(a_{2},x)\mbox{ and }
(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 $(*)$.
\medskip
\par
\noindent
\textbf{Example:}
\emph{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 $(*)$. In particular (taking $x=1$), the field $F$ does not 
contain any element $y$ such 
that
\[
(a_{1},b_{1})=(a_{1},y)=(a_{2},y)=(a_{2},b_{2}).
\]
}
\medskip
\par
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 
place $v$ 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 
$(\;,\;)_{v}\colon\,(F_{v}^\times/F_{v}^{\times 3})\times
(F_{v}^\times/F_{v}^{\times3})\to\frac{1}{3}\mathbb{Z}/\mathbb{Z}$ 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,
\[
(a_{1},b_{1})=(a_{2},b_{2}).
\]
Suppose now $x$, $y\in F^\times$ satisfy $(*)$. 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:
\[
(a_{1},x)_{v_{2}}=2/3, \qquad\mbox{and } (a_{1},x)_{v}=0
\mbox{ for $v\neq v_{2}$,}
\]
a contradiction to the reciprocity law.
\bigskip
\par
\begin{flushright}
Jean-Pierre Tignol, June 1996.
\end{flushright}

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