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%% Israel J. Math. 96 B (1996) 299--340
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\begin{document}

\begin{center}
  { \large \bf ALGEBRAS OF ODD DEGREE WITH INVOLUTION,
  \\[2mm]
  TRACE FORMS AND DIHEDRAL EXTENSIONS\footnote{The text appeared as:
  {\frenchspacing Israel J. Math. 96 B, 299-340 (1996) - Amitsur
  Volume}}}
  \\[4mm]
  {\sc Darrell Haile, Max-Albert Knus, Markus Rost and Jean-Pierre
  Tignol\footnote{Supported in part by the
  National Fund for Scientific Research (Belgium)}}\\[2mm]
\end{center}

\begin{center}
  Dedicated to the memory of \,S.A. Amitsur
\end{center}

\begin{abstract}
  A $3$-fold Pfister form is associated to every involution of the
  second kind on a central simple algebra of degree~$3$.  This
  quadratic form is associated to the restriction of the reduced trace
  quadratic form to the space of symmetric elements; it is shown to
  classify involutions up to conjugation.  Subfields with dihedral
  Galois group in central simple algebras of arbitrary odd degree with
  involution of the second kind are investigated.  A complete set of
  cohomological invariants for algebras of degree~$3$ with involution
  of the second kind is given.
\end{abstract}

{\bf 1. Introduction.} \ Let $B$ be a central simple algebra over a
field $K$ with an involution $\sigma$ of the second kind and let $F$
be the fixed subfield of $K$.  Let $\Trd$ be the reduced trace and
$\Nrd$ the reduced norm of $B$.  The restriction $Q_\sigma $ of the
trace form $Q\colon (x,y) \mapsto \Trd(xy) $ to the $F$--space $(B,
\sigma)_+$ of symmetric elements of $B$ is a quadratic form with
values in~$F$.  It is an invariant of $\sigma$ and the aim of this
paper is to study this invariant.  We first describe the general form
of $Q_\sigma$ for algebras of arbitrary odd degree and then restrict
our attention to central simple algebras of degree~$3$.

Consider a cubic \'etale $F$-subalgebra $L\subset(B,\sigma)_+$.  The
restriction of the trace form to $L$ is nonsingular, hence we have an
orthogonal decomposition:
$$
(B,\sigma)_+=L\perp V.
$$
We give an explicit description of the restriction of the form
$Q_\sigma$ to $V$, using a special case of a construction introduced
by T.~Springer \cite{Sp}, in connection with exceptional Jordan
algebras, and generalized by Petersson--Racine \cite{PR} to algebras
of degree~$3$.  Our next goal is to show that the trace form
$Q_\sigma$ determines the involution up to isomorphism.  As a
consequence we get a parametrization of all involutions of the second
kind on a central simple algebra of degree~$3$ which leave elementwise
invariant a given cubic separable subalgebra.  We also associate a
$3$-fold Pfister form $\pi(B,\sigma)$ to $Q_\sigma$, which determines
$\sigma$ up to isomorphism, and characterize the class of involutions
for which this 3-fold Pfister form is hyperbolic.  The existence of
such involutions, which we call distinguished, follows from Springer's
construction, but is also related to a crossed product construction
given by A.A.~Albert \cite{Alb}.  A distinguished involution is
characterized by the fact that the space $(B,\sigma)_+$ contains up to
isomorphism every cubic \'etale $F$-subalgebra of~$B$.

In the last section, we use Galois cohomology and symbols to analyze
\'etale subalgebras of dihedral central simple algebras. As
applications, we get on one hand different proofs of previous results
of the paper and on the other hand we show that a dihedral algebra of
degree $2n$, $n$ odd, is cyclic if a quadratic extension of~$F$
contains a primitive $n^{\rm th}$--root of~$1$ (see
Corollary~\ref{dihRcyc.cor} for the precise statement).  This is due
to L.~Rowen and D.~Saltman \cite{RS} if $F$ contains a primitive
$n^{\rm th}$--root of $1$.  As a last application, we give a complete
set of cohomological invariants for algebras of degree three with
involution of the second kind.

The second author is indebted to H.P.~Petersson for useful comments on
the subject.

\medskip

{\bf 2. Some General Results.} \ Throughout the paper, $B$ denotes a
central simple algebra over a field $K$ of characteristic different
from 2 and $\sigma$ denotes an involution of the second kind on $B$,
i.e.\ a map $\sigma\colon B\to B$ such that
$$
\sigma(x+y)=\sigma(x)+\sigma(y),\qquad
\sigma(xy)=\sigma(y)\sigma(x),\qquad \sigma^2(x)=x
$$
for all $x,y\in B$, and $\sigma|_K\neq I_K$.  We let $F$ denote the
subfield of $K$ elementwise invariant under $\sigma$ and denote by
$\invo\colon K\to K$ the restriction of $\sigma$ to $K$.

Under a scalar extension of $F$, the field $K$~---~hence also the
algebra $B$~---~may split into a direct product of two factors.
Therefore, we shall also allow $K$ to be a split quadratic \'etale
$F$-algebra:
$$
K=F\times F.
$$
In that case, $ B=A\times A' $ for some central simple $F$-algebras
$A$, $A'$ which are exchanged under the involution $\sigma$.
Therefore, there is an isomorphism of $K$-algebras with involution
(i.e.\ an isomorphism which commutes with the involutions):
$$
(B,\sigma)\simeq(A\times A^{\op},s)
$$
where $A^{\op}$ is the opposite algebra of $A$ and $s$ is the switch
involution:
$$
s(a_1,a_2^{\op})=(a_2,a_1^{\op}).
$$
Abusing the terminology, we shall also consider $(B,\sigma)$ as a
central simple algebra with involution in this case.  (It is indeed
simple as an algebra-with-involution: see Jacobson's definition in
\cite[p.~208]{Jacjord}).

We let $\alpha\in F^\times$ be such that $
K=F(\sqrt{\alpha})=F[X]/(X^2-\alpha).  $ In particular, we have
$\alpha\in {F^{\times}}^2$ if $K=F\times F$.

Let $(B,\sigma)_+$ denote the $F$-vector space of $\sigma$-symmetric
elements:
$$
(B,\sigma)_+=\{b\in B\colon\sigma(b)=b\}.
$$
We denote by $Q_\sigma$ the restriction to $(B,\sigma)_+$ of the
reduced trace quadratic form:
$$
Q_\sigma(x)=\Trd_B(x^2)\qquad\mbox{for }x\in(B,\sigma)_+.
$$
For any $u \in (B,\sigma)_+ \cap B^\times$, $\sigma' = \Int(u) \circ
\sigma$ is again an involution of the second kind of $B$ and
conversely, if $\sigma, \sigma'$ are involutions of the second kind of
$B$, there exists $u \in (B,\sigma)_+ \cap B^\times$ such that
$\sigma' = \Int (u) \circ \sigma$.  Let $\langle u \rangle_B$ be the
$B$-hermitian form on $B$ (as a right $B$--module) given by
$$
\langle u \rangle_B (x, y) = \sigma(x) u y,
$$
for $u\in (B,\sigma)_+\cap B^\times$ and $x,y \in B$.  A right
$B$-module automorphism $r_v\colon x \mapsto vx,\, v \in B^\times$, of
$B$ is an {\bf isometry} $\langle u_1 \rangle_B \isom \langle u_2
\rangle_B$ if $\sigma(v) u_2v=u_1$ and is a {\bf similarity\/} if
there is $\lambda \in F^\times$ such that $\lambda\sigma(v) u_2v=u_1$.

\begin{lem}
  \label{isom.lem}
  Let $u_1, u_2 \in (B,\sigma)_+ \cap B^\times$ and let $\sigma_i =
  \Int(u_i) \circ \sigma$.  Then
  \begin{enumerate}
    \item[(1)]
    An isomorphism $(B,\sigma_1) \isom (B,\sigma_2)$ induces an
    isometry $Q_{\sigma_1} \isom Q_{\sigma_2}$.
    \item[(2)]
    $(B, \sigma_1)$ and $(B, \sigma_2)$ are isomorphic (as
    $K$-algebras with involution) if and only if the hermitian spaces
    $\langle u_1 \rangle_B$ and $\langle u_2 \rangle_B$ are similar.
  \end{enumerate}
\end{lem}

\proof (1): If $\Int (v) \colon (B, \sigma_1) \isom (B, \sigma_2)$ is
an isomorphism, then $\Int (v) [ (B, \sigma_1)_+ ] = (B, \sigma_2)_ +$
and
$$
Q_{\sigma_2} (v x v^{-1}, vyv^{-1}) = \Trd _B (v x y v^{-1}) =
Q_{\sigma_1} (x,y)
$$
for $x, y \in (B, \sigma)_+$.\\
(2):\ The automorphism $\Int(v)$ of $B$ is an isomorphism $(B,
\sigma_1) \isom (B, \sigma_2)$, if and only if $\sigma_2 (v x v^{-1}
)= v \sigma_1 (x) v^{-1}$ for all $x \in B$, if and only if $u_2 =
\lambda v u_1 \sigma (v)$ for some $\lambda \in F^\times$, hence
$\langle u_1 \rangle_B$ and $ \langle u_2 \rangle_B$ are similar.
Conversely any such similitude induces an isomorphism $(B, \sigma_1)
\isom (B, \sigma_2)$.  \qed

Let $B = {\rm M}_n (K)$ be split and let $\tau (x_{ij}) =
(\overline{x}_{ij})^t$, where $t$ is transpose and $x \mapsto
\overline{x}$ is conjugation on $K$.  Any $u \in ({\rm M}_n (K),
\tau)_+$ is a hermitian matrix, hence there is $v \in~{\rm GL}_n (K)$
such that $ \sigma (v)uv = a = {\rm diag} (\alpha_1, \ldots,
\alpha_n), \, \alpha_i \in F^\times$.  Thus any involution of ${\rm
M}_n(K)$ of the second kind is isomorphic to an involution of the form
$\sigma = \Int (a) \circ \tau$ with $a = {\rm diag}(\alpha_1,\cdots,
\alpha_n),\alpha_i \in F^\times$.

Let $h_a= \langle \alpha_1,\ldots, \alpha_n \rangle_K$ be the
hermitian form on $K^n$ determined by ${\rm
diag}(\alpha_1,\cdots,\alpha_n)$, i.e.  $h_a(x,y)=\sum
\overline{x}_i\alpha_iy_i= \overline{x}^t a y,\, x,y \in K^n$.  Any
isometry $h_a \isom h_{a'}$ of the $K$--space $K^n$ can be viewed as
an isometry $\langle a\rangle_{{\rm M}_n(K)} \isom \langle
a'\rangle_{{\rm M}_n(K)}$ of the ${\rm M}_n(K)$--space ${\rm M}_n(K)$.

For $\alpha_1,\ldots,\alpha_n\in F^\times$, we denote by
$\langle\alpha_1,\ldots,\alpha_n\rangle$ the quadratic form on $F^n$
determined by ${\rm diag}(\alpha_1,\cdots,\alpha_n)$, and by
$\Pf\alpha_1,\ldots,\alpha_n\pF$ the $n$-fold Pfister form:
$$
\Pf\alpha_1,\ldots,\alpha_n\pF=\langle1,-\alpha_1\rangle
\cdot\,\cdots\,\cdot\langle1,-\alpha_n\rangle.
$$
The next proposition follows by straightforward computation.

\begin{prop}
  \label{split.prop}
  Let $K=F(\sqrt{\alpha})$.  For $a={\rm
  diag}(\alpha_1,\cdots,\alpha_n)\in {\rm M}_n(K)$ and $\sigma = \Int
  (a) \circ \tau$ we have
  $$
  Q_\sigma \simeq n \langle 1 \rangle \perp \langle 2 \rangle \cdot
  \langle\!\langle \alpha \rangle\!\rangle \cdot (\perp_{1 \leq i < j
  \leq n} \; \langle \alpha_i \alpha_j \rangle).
  $$
\end{prop}

In order to get a similar result for arbitrary central simple algebras
of odd degree, we first prove:

\begin{lem}
  \label{qadform.lem}
  Let $L/F$ be a field extension of odd degree and let $q$ be a
  quadratic form over~$F$.  Let also $q_L$ denote the quadratic form
  over $L$ derived from $q$ by extending scalars to $L$, and let
  $\alpha\in F^\times\smallsetminus {F^{\times}}^2$.  If
  $q_L\simeq\Pf\alpha\pF\cdot h$ for some quadratic form $h$ over $L$,
  of determinant~$1$, then there is a quadratic form $t$ of
  determinant~$1$ over $F$ such that
  $$
  q\simeq\Pf\alpha\pF\cdot t.
  $$
\end{lem}

\proof Let $K=F(\sqrt{\alpha})$ and $E=L\cdot K=L(\sqrt{\alpha})$.
Let also $q_{\rm an}$ denote an anisotropic form over $F$ which is
Witt-equivalent to $q$.  The form $(q_{\rm an})_E$ is Witt-equivalent
to the form $(\Pf\alpha\pF\cdot h)_E$, hence it is hyperbolic.  Since
the field extension $E/K$ has odd degree, Springer's theorem on the
behaviour of quadratic forms under field extensions of odd degree
\cite[Theorem~2.5.3]{SchQHF} shows that $(q_{\rm an})_K$ is
hyperbolic, hence, by \cite[Remark~2.5.11]{SchQHF},
$$
q_{\rm an}=\Pf\alpha\pF\cdot t_0
$$
for some quadratic form $t_0$ over $F$.  Let $\dim q=2d$, so that
$\dim h=d$, and let $w$ denote the Witt index of $q$, so that
\begin{equation}
  q\simeq w{\Bbb H}\perp\Pf\alpha\pF\cdot t_0,
  \label{qsimeq.eq}
\end{equation}
where $\Bbb H$ is the hyperbolic plane.  We then have $\dim t_0=d-w$,
hence
$$
\det q=(-1)^w\,(-\alpha)^{d-w}\cdot {F^{\times}}^2\in F^\times/
{F^{\times}}^2.
$$
On the other hand, the relation $q_L\simeq\Pf\alpha\pF\cdot h$ yields:
$$
\det q_L=(-\alpha)^d\cdot {L^{\times}}^2\in L^\times/{L^{\times}}^2.
$$
Therefore, $\alpha^w\in F^\times$ becomes a square in $L$; since the
degree of $L/F$ is odd, this implies that $\alpha^w\in
{F^{\times}}^2$, hence $w$ is even.  Letting $t_1=\frac{w}{2}{\Bbb
H}\perp t_0$, we then derive from (\ref{qsimeq.eq}):
$$
q\simeq\Pf\alpha\pF\cdot t_1.
$$
It remains to prove that we may modify $t_1$ so as to satisfy the
determinant condition.  Since $\dim t_1=d$, we have modulo the square
$I^2F$ of the fundamental ideal of the Witt ring of $F$:
$$
t_1\equiv\left\{
\begin{array}{cl}
  \Pf(-1)^{d(d-1)/2}\det t_1\pF&\mbox{if $d$
  is even}\\
  \langle(-1)^{d(d-1)/2}\det t_1\rangle&\mbox{if $d$ is odd.}
\end{array}
\right.
$$
We may use these relations to compute the Clifford algebra of
$q\simeq\Pf\alpha\pF\cdot t_1$ (up to Brauer-equivalence): in both
cases we get the same quaternion algebra:
$$
C(q)\sim (\alpha,(-1)^{d(d-1)/2}\det t_1)_{F}.
$$
On the other hand, since $\det h=1$ we derive from
$q_L\simeq\Pf\alpha\pF\cdot h$:
$$
C(q_L)\sim (\alpha,(-1)^{d(d-1)/2})_{L}.
$$
Therefore, the quaternion algebra $(\alpha,\det t_1)_F$ is split,
since it splits over the extension $L/F$ of odd degree.  Therefore, if
$\delta\in F^\times$ is a representative of $\det t_1\in
F^\times/F^{\times 2}$, we have
$$
\delta\in n_{K/F}(K^\times).
$$
Let $\beta\in F^\times$ be a represented value of $t_1$, so that
$$
t_1\simeq t_2\perp\langle\beta\rangle
$$
for some quadratic form $t_2$ over $F$, and let
$$
t=t_2\perp\langle \delta\beta\rangle.
$$
Then
$$
\det t=\delta\cdot\det t_1=1.
$$
On the other hand, since $\delta$ is a norm from the extension $K/F$
we have
$$
\Pf\alpha\pF\cdot\langle \delta\beta\rangle \simeq\Pf\alpha\pF\cdot
\langle\beta\rangle,
$$
hence
$$
\Pf\alpha\pF\cdot t\simeq \Pf\alpha\pF\cdot t_1\simeq q.
$$
\qed

\begin{prop}
  \label{gen.prop}
  Let $B$ be a central simple $K$-algebra of odd degree $n=2m-1$ with
  an involution $\sigma$ of the second kind.  There is a quadratic
  form $q_\sigma$ of dimension $n (n -1)/2$ and determinant $1$ over
  $F$ such that
  $$
  Q_\sigma\simeq n\,\langle1\rangle\perp\langle2\rangle
  \cdot\Pf\alpha\pF\cdot q_\sigma.
  $$
\end{prop}

\proof Suppose first $K=F\times F$.  We may then assume $
(B,\sigma)=(A\times A^{\op},s) $, where $s$ is the switch involution.
In that case
$$
(B,\sigma)_+=\{(a,a^{\op})\colon a\in A\}\simeq A,
$$
and $Q_\sigma$ is isometric to the reduced trace quadratic form on
$A$.  Since $\alpha\in {F^{\times}}^2$, we have to show that this
quadratic form is Witt-equivalent to $n\langle1\rangle$.  By
Springer's theorem, it suffices to prove this relation over an
odd-degree field extension.  Since the degree of $A$ is odd, we may
therefore assume $A$ is split: $A={\rm M}_n(F)$.  In that case, the
relation is easy to check.  (Observe that the upper-triangular
matrices with zero diagonal form a totally isotropic subspace).

For the rest of the proof, we may thus assume $K$ is a field.  Let $D$
be a division $K$-algebra Brauer-equivalent to $B$ and let $\tau\colon
D\to D$ be an involution of the second kind on $D$.  Let also $L$ be a
field contained in $(D,\tau)_+$ and maximal for this property.  The
field $E=L\cdot K$ is then a maximal subfield of $D$, otherwise the
centralizer $C_DE$ contains a symmetric element outside $E$,
contradicting the maximality of $L$.  We have $ [L:F]=[E:K]=\deg D, $
hence the degree of $L/F$ is odd, since $D$ is Brauer-equivalent to
the algebra $B$ of odd degree.  Moreover, the algebra
$$
B\otimes_{F}L=B\otimes_KE
$$
splits, since $E$ is a maximal subfield of $D$.  By Proposition 2 the
quadratic form $[Q_\sigma]_{L}$ obtained from $Q_\sigma$ by scalar
extension to $L$ has the form
\begin{equation}
  \label{Qsigma.eq}
  [Q_\sigma]_{L}\simeq n\,\langle1\rangle\perp
  \langle2\rangle\cdot\Pf\alpha\pF\cdot h
\end{equation}
where $h=\perp_{1\leq i< j\leq n}\langle\alpha_i\alpha_j\rangle$ for
some $\alpha_1,\ldots,\alpha_n\in L^\times$.  Therefore, the Witt
index of the form $[Q_\sigma]_{L}\perp n\,\langle-1\rangle$ is at
least $n$:
$$
w([Q_\sigma]_{L}\perp n\,\langle-1\rangle) \geq n.
$$
By Springer's theorem the Witt index of a form does not change under
an odd-degree scalar extension.  Therefore,
$$
w(Q_\sigma\perp n\,\langle-1\rangle)\geq n,
$$
and it follows that $Q_\sigma$ contains a subform isometric to
$n\,\langle1\rangle$.  Let
$$
Q_\sigma\simeq n\,\langle1\rangle\perp q
$$
for some quadratic form $q$ over $F$.  Relation~(\ref{Qsigma.eq})
shows that
$$
(q)_{L}\simeq\langle2\rangle\cdot \Pf\alpha\pF\cdot h.
$$
Since $\det h=1$, we may apply Lemma~\ref{qadform.lem} to the
quadratic form $\langle2\rangle\cdot q$ and get a quadratic form
$q_\sigma$ over $F$, of determinant~$1$, such that
$$
\langle2\rangle\cdot q\simeq\Pf\alpha\pF\cdot q_\sigma,
$$
hence
$$
Q_\sigma\simeq n\langle1\rangle\perp\langle2\rangle
\cdot\Pf\alpha\pF\cdot q_\sigma.
$$
\qed

We conclude with a result which will be used in the next section for
algebras of degree~$3$.

\begin{lem} \label{etsubalg.lem}
  Let $L \subset (B,\sigma)_+$ be \'etale of dimension $n$ over $F$
  and let $R=L\otimes_F L\otimes_F K$.  Then $B$ is a free $R$-module
  of rank one via left and right multiplication; the action is
  equivariant with respect to the involution $\sigma$ on $B$ and the
  action $\sigma\colon R \to R$ given by $\sigma(\lambda\otimes \mu
  \otimes x )= \mu \otimes \lambda \otimes \bar x$.  In particular we
  have an induced action of $R^\sigma$ on $(B,\sigma)_+$
\end{lem}

\proof We may assume that $F$ is separably closed, that
$$
B={\rm M}_n(F)\times {\rm M}_n(F)^{\op},
$$
$\sigma$ is the switch involution, and
$$
L=\{(d,d^{\op})\colon d\mbox{ is diagonal}\}.
$$
Let $L'$ be the set of diagonal matrices in ${\rm M}_n(F)$.  We have

$$
{\rm M}_n(F)=L'\oplus xL'\oplus\cdots\oplus x^{n-1}L',
$$
with $x$ a permutation matrix of order $n$.  In this case
$\xi=1+x\cdots+ x^{n-1}$ is a free generator of ${\rm M}_n(F)$ as a
$L'\otimes L'$-module.  Thus $(\xi, \xi^{\op})$ is a free generator of
$B$ as $L\otimes L$-module.  The last claim follows by a direct
verification.  \qed

\medskip

{\bf 3. A Construction of Springer.} \ In this section, we restrict
attention to central simple algebras of degree~3 over a field of
characteristic different from $2,3$.  We use the same notation as in
the preceding section; in particular, we denote by $B$ a central
simple $K$-algebra with involution $\sigma$ of the second kind.
Consider a cubic \'etale $F$-subalgebra $L\subset(B,\sigma)_+$ and
denote by $t_{L/F}\colon L\to F$ and $n_{L/F}\colon L\to F$ the trace,
resp. the norm of $L$.  The restriction of the trace form to $L$ is
nonsingular, hence we have an orthogonal decomposition:
$$
(B,\sigma)_+=L\perp V.
$$
For $v\in V$ let
$$
\textstyle N(v)=\textstyle{\frac{1}{2}}\Trd_B(v^2)-p_L(v^2)\in L,
$$
where $p_L\colon(B,\sigma)_+\to L$ denotes the orthogonal projection.
We define an $L$-action on $V$ such that $(V,N)$ is a nonsingular
quadratic space of rank $2$ over $L$.  We give two descriptions of the
action of $L$ on $V$.  The first was introduced by T.~Springer
\cite{Sp} in connection with exceptional Jordan algebras and
generalized by Petersson--Racine \cite{PR} to Jordan algebras of
degree~3.  The second uses Lemma~\ref{etsubalg.lem} and is specific to
central simple algebras of degree $3$.

\begin{lem}{\mbox{\sc (Springer's Construction)}}
  \label{N.lem}
  The space $V$ is a free $L$-module of rank~2 through the operation
  $$
  (\ell,v)\mapsto\ell\circ v=t_{L/F}(\ell)v-\ell v-v\ell\in V
  $$
  and $N\colon V\to L$ is a nonsingular quadratic form for this
  structure.  Moreover, for all $v\in V$,
  $$
  Q_\sigma(v)=2\,t_{L/F}(N(v)).
  $$
  If $v\in V$ is invertible in $B$, then
  $$
  \Trd_B(v^{-1}\ell)= -\Nrd_B(v)^{-1}t_{L/F}(N(v)\ell)
  $$
  for all $\ell\in L$.
\end{lem}

\proof Extending scalars from $F$ to an algebraic closure, we may
assume that $B$ is of the form $ {\rm M}_3(F)\times {\rm
M}_3(F)^{\op}, $ $\sigma$ is the switch involution, and $
L=\{(d,d^{\op})\colon d\mbox{ is diagonal}\}.  $ In this case the
lemma follows by explicit computation.  \qed

We now describe the second construction: Let $D=F(\sqrt\delta)$ be the
discriminant algebra of $L$, i.e. $\delta$ is the determinant of the
trace form $t_{L/F}(x^2)=Q_\sigma|_L(x)$.  There exists a
decomposition

\begin{equation} \label{cub.disc}
  L\otimes_F L=L\times L\otimes_F D,
\end{equation}
such that the twist $\sigma$ of $L\otimes_F L$ restricts on $D$ to
$a\mapsto \bar a$.  Note that there are three natural imbeddings $L\to
L\otimes_FD$.  Two of them are given by $\lambda\mapsto {\rm
pr}(\lambda\otimes1)$ and $\lambda\mapsto{\rm pr}(1\otimes\lambda)$,
respectively (${\rm pr}$ is the projection $L\otimes_F L \to
L\otimes_F D$) and $L\to L\otimes_FD$,
$\lambda\mapsto\lambda\otimes1$, is the third one, which is
$\sigma$-invariant.  We have an induced decomposition
\begin{equation}\label{4.equ}
  R=L\otimes_F L\otimes_F K =L\otimes_FK\times L\otimes_FD\otimes_FK.
\end{equation}

The $\sigma$-action on $R$ (see Lemma \ref{etsubalg.lem}) restricts on
$ L\otimes_FD\otimes_FK$ to $\lambda\otimes d\otimes a \mapsto
\lambda\otimes \bar d\otimes \bar a$, so the fixed subalgebra is
$$
R^\sigma=L\times L\otimes_FH,
$$
where $H=F(\sqrt{\alpha\delta})$.

\begin{lem} \label{N2.lem}
  The decomposition of the $R$-module $B$ induced by the decomposition
  (\ref{4.equ}) reduces to the decomposition
  $$
  B= L\otimes _F K \perp V\otimes _F K
  $$
  over $K$.  In particular $ R^\sigma=L\times L\otimes_FH $ acts on
  $(B,\sigma)_+=L\perp V$ componentwise and $V$ is a free $L\otimes
  H$-module of rank one.  Moreover the action given by the restriction
  $L\subset L\otimes H$ coincide with the Springer action, hence
  $N\colon V\to L$ is a nonsingular quadratic form on the free
  $L$-module $V$ of rank $2$.  We have $N(hv)=n_{H/F}(h)N(v)$ for $h
  \in H$, so that $N$ extends to a hermitian form on the $L\otimes_F
  H$-space $V$ of rank one.
\end{lem}

\proof As for Lemma~\ref{N.lem} it suffices to check the split case,
where the claims follow by explicit computations.  \qed

Let, as above, $\delta$ denote the discriminant of $L$.  We recall
that $\delta \in {F^{\times}}^2 $ if and only if $L$ is cyclic and
that
$$
Q_\sigma|_L=\langle1,2,2\delta\rangle;
$$
this is easy to check if $L$ is not a field, since then $L=F\times
F(\sqrt{\delta})$.  The general case follows by extending scalars from
$F$ to $L$ and applying Springer's theorem.  Combining this result
with Lemma~\ref{N.lem}, we get
\begin{equation}
  \label{qs.eq}
  Q_\sigma=\langle1,2,2\delta\rangle\perp\langle2 \rangle\cdot
  t_{L/F}(N),
\end{equation}
where $t_{L/F}(N)$ denotes the Scharlau transfer of $N$.

\begin{prop}
  \label{N.prop}
  Let $K=F(\sqrt{\alpha})$.  The following conditions are equivalent:
  \begin{enumerate}
    \item[(a)] $N$ is hyperbolic.
    \item[(b)] $\delta=\alpha$ in
    $F^\times/{F^{\times}}^2$.
    \item[(c)] $LK$ is a cyclic extension of $K$ and is Galois
    with Galois group ${\cal S}_3$ over $F$.
  \end{enumerate}
  Moreover, if these conditions hold, then $B$ is a crossed product:
  $$
  B=LK\oplus LKx\oplus LKx^2
  $$
  for some $x$ such that $x^3\in F$ and $\sigma(x)=x$.
\end{prop}

\proof {\em(a)~$\Leftrightarrow$~(b):\/} follows immediately from
Lemma~\ref{N2.lem}.  Alternatively, if we only want to use Springer's
Construction (Lemma~\ref{N.lem}), we get {\em(a)~$\Rightarrow$~(b)\/}
as follows: If $N$ is hyperbolic, equation~(\ref{qs.eq}) yields:
$$
\det Q_\sigma=-\delta.
$$
On the other hand, Proposition~\ref{gen.prop} shows that $\det
Q_\sigma=-\alpha$.  Therefore, {\em(b)\/} follows.

{\em(b)~$\Rightarrow$~(c):\/} If $\delta=\alpha$, then the
discriminant of $L$ becomes a square in $K$, hence $LK/K$ is cyclic.
Let $\rho$ denote a generator of the Galois group $\Gal(LK/K)$.  The
restriction of $\sigma$ to $LK$ is an automorphism of order~2 of $LK$
over $F$, hence $\sigma|_{LK}$ and $\rho$ generate a group of
automorphisms of order at least~6 of $LK/K$ which shows that $LK/F$ is
Galois.

If $\alpha\neq1$, then $L/F$ is not cyclic, hence the group generated
by $\rho$ and $\sigma|_{LK}$ is not cyclic.  It is therefore
isomorphic to ${\cal S}_3$.

If $\alpha=\delta=1$, then $B\simeq A\times A^{\op}$ and $\sigma$ is
isomorphic to the switch involution.  Moreover, $LK\simeq L_0\times
L_0^{\op}$ for some cyclic extension $L_0/F$.  If $\gamma$ is a
generator of the Galois group $\Gal(L_0/F)$, then we choose for $\rho$
the automorphism of $L$ given by:
$$
\rho(\ell_1,\ell_2^{\op})= (\gamma(\ell_1),\gamma^2(\ell_2)^{\op}).
$$
Thus, $\sigma$ and $\rho$ do not commute; they generate a group
isomorphic to ${\cal S}_3$.

Our next goal is to show that $B$ contains an invertible element $x$
such that $\sigma(x)=x$ and $x\ell=\rho(\ell)x$ for all $\ell\in LK$.
These relations imply that $x^3$ centralizes $LK$, is
$\sigma$-symmetric and commutes with $x$, hence $x^3\in F$.  Let
$$
S=\{x\in(B,\sigma)_+\colon x\ell=\rho(\ell)x \mbox{ for all }\ell\in
LK\};
$$
this is a vector space over $F$ in which the invertible elements form
a Zariski open set.  In order to prove that this set is non-empty, we
may extend scalars from $F$ to an algebraic closure, and assume
$$
B={\rm M}_3(F)\times {\rm M}_3(F)^{\op},
$$
$\sigma$ is the switch involution, and
$$
L=\{(d,d^{\op})\colon d\mbox{ is diagonal}\}.
$$
We may further assume that $\rho$ maps
$$
\left(
\left(
\begin{array}{ccc}
  d_1& &\\
  &d_2& \\
  & &d_3
\end{array}
\right),
\left(
\begin{array}{ccc}
  d'_1& &\\
  &d'_2& \\
  & &d'_3
\end{array}
\right)^{\op} \right)
$$
to
$$
\left( \left(
\begin{array}{ccc}
  d_3& &\\
  &d_1& \\
  & &d_2
\end{array}
\right), \left(
\begin{array}{ccc}
  d'_2& &\\
  &d'_3& \\
  & &d'_1
\end{array}
\right)^{\op} \right);
$$
we may then choose {\small
$$
x=\left( \left(
\begin{array}{ccc}
  0&0&1\\
  1&0&0\\
  0&1&0
\end{array}
\right), \left(
\begin{array}{ccc}
  0&0&1\\
  1&0&0\\
  0&1&0
\end{array}
\right)^{\op}\right) \in S\cap B^\times,
$$
}%
proving the claim.  It follows that $B$ is a crossed product, as
required.

{\em(c)~$\Rightarrow$~(a):\/} For every element $x$ as above, we have
$x,x^2\in V$ (as is easily seen by scalar extension to an algebraic
closure of $F$), hence $ N(x)=0, $ proving that $N$ is isotropic,
hence hyperbolic since its dimension over $L$ is $2$.  \qed

{\it Remark.\/} Proposition \ref{N.prop} can also be deduced from the
Corollary of \cite[Theorem~1]{PR1} in relation with
\cite[Proposition~2.2]{PR}.

In order to give an explicit description of the form $N$ for general
$L$, we need the following lemmas:

\begin{lem}
  \label{formN.lem}
  Let $p_V\colon(B,\sigma)_+\to V$ denote the orthogonal projection.
  For all $v\in V$,
  $$
  n_{L/F}(N(p_V(v^2)))=n_{L/F}(N(v))^2.
  $$
\end{lem}

\proof The lemma follows by explicit computation after extending
scalars from $F$ to an algebraic closure.  \qed

\begin{lem}
  \label{trhyp.lem}
  For any $\lambda\in L^\times$ such that $n_{L/F}(\lambda)\in
  {F^{\times}}^2$, the quadratic form
  $$
  \Pf\delta\pF\cdot[t_{L/F}(\langle\lambda\rangle)
  \perp\langle-1\rangle]
  $$
  is hyperbolic.
\end{lem}

\proof By Springer's theorem, it suffices to prove that the quadratic
form above is hyperbolic after extending scalars from $F$ to $L$.  We
may thus assume $L=F\times F'$ where $F'=F(\sqrt{\delta})$.  Let
$\lambda=(\lambda_0,\lambda_1)$ with $\lambda_0\in F$ and
$\lambda_1\in F'$; then
\begin{equation}
  \label{dect.eq}
  t_{L/F}(\langle\lambda\rangle)= \langle\lambda_0\rangle\perp
  t_{F'/F}(\langle \lambda_1\rangle).
\end{equation}
By \cite[p.~50]{SchQHF}, the image of the transfer map from the Witt
ring $WF'$ to $WF$ is killed by $\Pf\delta\pF$, hence
\begin{equation}
  \label{lone.eq}
  \Pf\delta\pF\cdot t_{F'/F}(\langle\lambda_1\rangle)=0 \qquad\mbox{in
  }WF.
\end{equation}
On the other hand,
$$
n_{L/F}(\lambda)=\lambda_0\,n_{F'/F}(\lambda_1) \in {F^{\times}}^2,
$$
hence $\lambda_0$ is a norm from $F'$ to $F$, and therefore
\begin{equation}
  \label{lzero.eq}
  \Pf\delta\pF\cdot\langle\lambda_0,-1\rangle=0 \qquad\mbox{in }WF.
\end{equation}
The lemma follows from (\ref{lone.eq}) and (\ref{lzero.eq}), in view
of the decomposition~(\ref{dect.eq}).  \qed

\begin{prop}
  \label{diagN.prop}
  The quadratic form $N$ has a diagonalization
  $$
  \Pf\alpha\delta\pF\cdot\langle\lambda\rangle
  $$
  for some $\lambda\in L^\times$ such that $n_{L/F}(\lambda)\in
  {F^{\times}}^2$.  Consequently,
  \begin{eqnarray*}
    Q_\sigma&=&\langle1,2,2\delta\rangle\perp
    \langle2\rangle\cdot\Pf\alpha\delta\pF\cdot
    t_{L/F}(\langle\lambda\rangle)\\
    &=&\langle1,1,1\rangle\perp\langle2\delta\rangle
    \cdot\Pf\alpha\pF\cdot t_{L/F}(\langle\lambda \rangle)
  \end{eqnarray*}
  for some $\lambda\in L^\times$ such that $n_{L/F}(\lambda)\in
  {F^{\times}}^2$.
\end{prop}

\proof Let $v\in V$ be such that $N(v)$ is invertible in $L$.
Lemma~\ref{formN.lem} shows that $N(p_V(v^2))$ also is invertible.
The element $p_V(v^2)$ is a basis of $V$ as a $L\otimes H$-module, so
that by Lemma~\ref{N2.lem}, letting $\lambda=N(p_V(v^2))$, we get
$$
N=\Pf\alpha\delta\pF\cdot\langle\lambda\rangle
$$
as wanted.  Alternatively, without using Lemma~\ref{N2.lem}, and
letting $\lambda=N(p_V(v^2))$ as above, we have
$$
N=\langle \lambda,\lambda'\rangle
$$
for some $\lambda'\in L^\times$, and
$n_{L/F}(\lambda)=n_{L/F}(N(v))^2\in {F^{\times}}^2$, by
Lemma~\ref{formN.lem}.  On the other hand, Proposition~\ref{N.prop}
shows that $N$ becomes hyperbolic over $L(\sqrt{\alpha\delta})$, so
$\det N=-\alpha\delta$, and therefore
$$
N=\Pf\alpha\delta\pF\cdot\langle\lambda\rangle.
$$
The first formula for $Q_\sigma$ then follows from (\ref{qs.eq}) by
Frobenius reciprocity.  Since
$$
\Pf\alpha\delta\pF=\Pf\alpha\pF\cdot\langle\delta \rangle+\Pf\delta\pF
\qquad\mbox{in }WF,
$$
the following relation in $WF$ follows:
$$
Q_\sigma=\langle1,2,2\delta\rangle +
\langle2\delta\rangle\cdot\Pf\alpha\pF\cdot
t_{L/F}(\langle\lambda\rangle)+ \langle2\rangle\cdot\Pf\delta\pF\cdot
t_{L/F}(\langle\lambda\rangle).
$$
Lemma~\ref{trhyp.lem} shows that the last term on the right-hand side
is equal to $\langle2\rangle\cdot\Pf\delta\pF$.  Since
$$
\langle2\delta\rangle+\langle2\rangle\cdot
\Pf\delta\pF=\langle2\rangle \qquad\mbox{and}\qquad
\langle1,2,2\rangle=\langle1,1,1\rangle,
$$
we get
$$
Q_\sigma=\langle1,1,1\rangle+\langle2\delta\rangle
\cdot\Pf\alpha\pF\cdot t_{L/F}(\langle\lambda\rangle) \qquad\mbox{in
}WF.
$$
Since both sides have the same dimension, these two quadratic forms
are isometric, proving the second formula for $Q_\sigma$.  \qed

\begin{cor}
  The form $q_\sigma$ defined in Proposition~\ref{gen.prop} may be
  chosen as
  $$
  q_\sigma =\langle\delta\rangle\cdot t_{L/F}(\langle\lambda \rangle)
  $$
  for some $\lambda\in L^\times$ such that $n_{L/F}(\lambda)\in
  {F^{\times}}^2$.
\end{cor}

\proof According to \cite[p.~51]{SchQHF}, we have
$$
\det(t_{L/F}(\langle\lambda\rangle)) = \delta\,n_{L/F}(\lambda).
$$
\qed

So far, the involution $\sigma$ has been fixed, as well as the \'etale
subalgebra $L\subset(B,\sigma)_+$.  In the last proposition, we
compare the quadratic forms $Q_\sigma$ and $Q_{\sigma'}$ associated to
two involutions of the second kind which leave $L$ elementwise
invariant.  We denote by
$$
q\colon L\to L
$$
the quadratic map such that $\ell q(\ell)=n_{L/F}(\ell)$ for all
$\ell\in L$.

\begin{prop}
  \label{chginv.prop}
  Let $\sigma,\sigma'$ be two involutions on $B$ such that
  $\sigma|_L=\sigma'|_L=I_L$:
  $$
  \sigma'=\Int(z)\circ\sigma
  $$
  for some $z\in L^\times$.  If $\lambda\in L^\times$ is such that
  $$
  Q_\sigma=\langle1,2,2\delta\rangle\perp
  \langle2\rangle\cdot\Pf\alpha\delta\pF\cdot
  t_{L/F}(\langle\lambda\rangle),
  $$
  then
  $$
  Q_{\sigma'}=\langle1,2,2\delta\rangle\perp
  \langle2\rangle\cdot\Pf\alpha\delta\pF\cdot t_{L/F}(\langle
  q(z)\lambda\rangle).
  $$
\end{prop}

\proof The map $b\mapsto zb$ is an isomorphism
$(B,\sigma)_+\to(B,\sigma')_+$ which maps $L$ to $L$; therefore, it
also maps the orthogonal $V$ of $L$ in $(B,\sigma)_+$ to the
orthogonal $V'$ of $L$ in $(B,\sigma')_+$.  Let $N'\colon V'\to L$
denote the quadratic form
$$
\textstyle N'(v')=\frac{1}{2}\Trd(v'^2)-p'_L(v'^2) \qquad\mbox{for
}v'\in V',
$$
where $p'_L\colon(B,\sigma')_+\to L$ denotes the orthogonal
projection.  An explicit computation, after extending scalars to an
algebraic closure of $F$, shows that
$$
N'(zv)=q(z)N(v) \qquad\mbox{for all }v\in V.
$$
Therefore, multiplication by $z$ defines a similarity $N\isom N'$ with
similarity factor $q(z)$.  \qed

Proposition \ref{chginv.prop} leads to the following converse of
Proposition \ref{diagN.prop}:
\begin{cor}
  \label{chginv.cor}
  Let $L$ be an arbitrary cubic \'etale $F$-subalgebra in $B$.  For
  every $\lambda\in L^\times$ such that $n_{L/F}(\lambda)\in
  {F^{\times}}^2$, there is an involution $\sigma$ on $B$ leaving $L$
  elementwise invariant such that
  \begin{eqnarray*}
    Q_\sigma&=&\langle1,2,2\delta\rangle\perp
    \langle2\rangle\cdot\Pf\alpha\delta\pF\cdot
    t_{L/F}(\langle\lambda\rangle)\\
    &=& \langle1,1,1\rangle\perp\langle2\delta\rangle \cdot\Pf\alpha\pF\cdot t_{L/F}(\langle\lambda\rangle).
  \end{eqnarray*}
\end{cor}

\proof By a theorem of Albert \cite[p.~157]{SofA}, there is an
involution $\tau$ leaving $L$ elementwise invariant.  Let
$$
Q_\tau=\langle1,2,2\delta\rangle\perp
\langle2\rangle\cdot\Pf\alpha\delta\pF\cdot
t_{L/F}(\langle\mu\rangle)
$$
for some $\mu\in L^\times$ such that $n_{L/F}(\mu)\in F^{\times 2}$.
If $\lambda\in L^\times$ is such that $n_{L/F}(\lambda)\in
{F^{\times}}^2$, let $\xi\in F^\times$ be such that
$$
n_{L/F}(\lambda\mu^{-1})=\xi^2.
$$
Since $n_{L/F}(\lambda\mu^{-1})=\lambda\mu^{-1} q(\lambda\mu^{-1})$,
the preceding equality yields:
$$
\lambda\mu^{-1}=\xi^2q(\lambda\mu^{-1})^{-1}= q(\xi\lambda^{-1}\mu).
$$
Therefore, the preceding proposition shows that
$\sigma=\Int(\xi\lambda^{-1}\mu)\circ\tau$ satisfies the required
properties.  \qed

\medskip

{\bf 4. A Classification of Involutions in Degree 3.}\ In this
section we continue to assume that $B$ has degree $3$ and that $F$
has characteristic different from $2,3$.  By
Proposition~\ref{gen.prop}, the trace form of any involution
$\sigma$ of the second kind has the form
$$
Q_\sigma = \langle 1, 1, 1\rangle \perp \langle 2\rangle \cdot
\langle\!\langle \alpha \rangle\!\rangle \cdot \langle - b, -c,
bc\rangle
$$
where $K = F (\sqrt{\alpha})$ and $b, c \in F$.  Our goal is to show
that the 3-fold Pfister form $\Pf\alpha,b,c\pF$ determines the
involution $\sigma$ up to isomorphism and to characterize the
involutions for which this 3-fold Pfister form is hyperbolic.

For any Pfister form $\Pf a_1,\ldots,a_n\pF$, we let $\Pf
a_1,\ldots,a_n\pF^\sharp=\langle1\rangle^\perp$, so that
$$
\Pf b,c\pF^\sharp=\langle -b,-c,bc\rangle.
$$

\begin{thm}
  \label{clasinv.thm}
  Let $\sigma, \sigma'$ be involutions of the second kind on a
  central simple $K$--algebra $B$ of degree 3.  Let
  $$
  Q_\sigma= \langle 1,1,1\rangle \perp \langle 2\rangle \cdot
  \langle\!\langle \alpha\rangle\!\rangle \cdot \langle\!\langle b,
  c\rangle\!\rangle^\sharp
  $$
  and
  $$
  Q_{\sigma'}= \langle 1,1,1\rangle \perp \langle 2\rangle\cdot
  \langle\!\langle \alpha\rangle\!\rangle \cdot \langle\!\langle b',
  c'\rangle\!\rangle^\sharp.
  $$
  The following conditions are equivalent:
  \begin{enumerate}
    \item[(a)]
    The involutions $ \sigma$ and $\sigma'$ are isomorphic.
    \item[(b)]
    The quadratic forms $Q_\sigma $ and $Q_{\sigma'}$ are isometric.
    \item[(c)]
    The quadratic forms $\langle\!\langle \alpha\rangle\!\rangle
    \cdot \langle\!\langle b, c\rangle\!\rangle^\sharp$ and $
    \langle\!  \langle \alpha\rangle\!\rangle \cdot \langle\!\langle
    b', c'\rangle\!\rangle^\sharp$ are isometric.
    \item[(d)]
    Either $K=F\times F$ or the $K$-hermitian forms $\langle - b, -
    c, b c\rangle_K$ and $\langle - b', -c', b' c' \rangle_K$ are
    isometric.
    \item[(e)]
    The quadratic forms $ \langle\!\langle \alpha, b,
    c\rangle\!\rangle$ and $ \langle\!\langle \alpha, b',
    c'\rangle\!\rangle $ are isometric.
  \end{enumerate}
\end{thm}

\proof {\em(a) $\Rightarrow$ (b)} is already in
Lemma~\ref{isom.lem}, and {\em (b)$\iff$(c)$\iff$(e)} follows by
Witt cancellation.  Moreover, {\em(c)~$\Rightarrow$~(d)} is a
theorem of Jacobson \cite{Jacherm} if $K$ is a field, and is clear
if $K=F\times F$.  We finally check that {\em(d)~$\Rightarrow$~(a)}.
If $K=F\times F$, all involutions on $B\simeq A\times A^{\op}$ are
isomorphic to the switch involution, so {\em(a)\/} holds trivially.
Thus we are reduced to the case $K$ a field.  Assume next that $B =
{\rm M}_3 (K)$ is split.  Up to automorphisms of $(B,\sigma)$,
resp. $(B,\sigma')$, we have $\sigma = \Int (a) \circ \tau$, resp.
$\sigma' = \Int (a') \circ \tau$ with $a = {\rm diag}( \alpha_1,
\alpha_2, \alpha_3)$ and $a'={\rm diag}( \alpha_1', \alpha_2',
\alpha_3')$.  We may assume that $\alpha_1 \alpha_2 \alpha_3 = 1 =
\alpha_1' \alpha_2' \alpha_3'$.  By Proposition~\ref{split.prop}, we
have
$$
Q_\sigma \simeq \langle 1, 1, 1\rangle \perp \langle 2\rangle \cdot
\langle\!\langle \alpha\rangle\!\rangle \cdot \langle \alpha_2
\alpha_3^{-1}, \alpha_3 \alpha_1^{-1}, \alpha_1 \alpha_2^{-1}
\rangle
$$
and
$$
Q_{\sigma'} \simeq \langle 1,1,1\rangle \perp \langle 2\rangle \cdot
\langle\!\langle \alpha\rangle\!\rangle \cdot \langle \alpha_2'
{\alpha_3}'^{-1}, \alpha'_3 {\alpha'_1}^{-1}, \alpha_1'
{\alpha'_2}^{-1}\rangle.
$$
Thus
$$
\langle \alpha_2 \alpha_3^{-1}, \alpha_3 {\alpha_1}^{-1}, \alpha_1
{\alpha_2}^{-1}\rangle_K \simeq \langle \alpha_2' {\alpha_3'}^{-1},
\alpha'_3 {\alpha'_1}^{-1}, \alpha_1' {\alpha'_2}^{-1}\rangle_K.
$$
Since
$$
\langle\alpha_1 \alpha_2 \alpha_3\rangle\cdot \langle \alpha_2
\alpha_3^{-1}, \alpha_3 \alpha_1^{-1}, \alpha_1 \alpha_2^{-1}
\rangle_K \simeq \langle \alpha_1, \alpha_2, \alpha_3\rangle_K,
$$
we get isometries $h_a\simeq h_{a'}$ or $\langle a\rangle_{{\rm
M}_3(K)} \simeq \langle a'\rangle_{{\rm M}_3(K)}$ and $({\rm M}_3
(K), \sigma)$ is isomorphic to $({\rm M}_3 (K), \sigma')$ by
Lemma~\ref{isom.lem}.

If $B$ is not split and if $\sigma' = \Int (u) \circ \sigma$, we
have to check by Lemma~\ref{isom.lem} that the hermitian spaces
$\langle u\rangle_B$ and $\langle 1\rangle_B$ are similar.
Replacing $u$ by $u \cdot \Nrd (u)$, we may assume that $\Nrd (u)
\in {F^{\times}}^2$.  Let $L = F(x)$ with $x \in (B, \sigma)_+, x
\not\in F$, so that $L$ is a field extension of $F$ of degree $3$.
The algebra $B \otimes_F L$ is split over $K \otimes _F L^\times$
and there is $v' \in {\rm G L}_3 (K \otimes_F L)$ such that
$$
u \otimes 1 = \lambda v' (\sigma \otimes 1) (v')
$$
Thus, denoting by $\invo\colon K\to K$ the non-trivial automorphism
of $K/F$,
$$
\Nrd (u\otimes 1)=\lambda^3 \Nrd (v') \overline{\Nrd (v')}=\mu^2
$$
with $\mu \in F$, since $\Nrd (u)\in {F^{\times}}^2$, and we can
write
$$
\lambda= (\mu\lambda^{-1} \Nrd (v')^{-1})(\overline{\mu\lambda^{-1}
\Nrd (v')^{-1}}) =\nu \overline{\nu}.
$$
It follows that $u \otimes 1 = v (\sigma \otimes 1) (v)$ with $v =
\nu v'$, so $\langle u\rangle_{B\otimes L} \simeq \langle
1\rangle_{B\otimes L}$.  By the Bayer--Lenstra \cite{BL}
generalization of Springer's theorem to hermitian spaces, we have
$\langle u \rangle_B \simeq \langle 1 \rangle_B$ and, by
Lemma~\ref{isom.lem}, $(B, \sigma) \simeq (B, \sigma')$.  \qed

In view of the equivalence {\em(a)$\iff$(e)} in
Theorem~\ref{clasinv.thm}, the Pfister form $\langle\!\langle
\alpha, b, c\rangle\!\rangle$ classifies involutions $\sigma$ on
$B$.  We denote it by $\pi(B, \sigma)$.  Note that $\pi(B,\sigma)$
is isometric to the norm of the octonion algebra ${\rm Oct} A$
associated in \cite{PRac2} to any simple Jordan algebra $A$ of
degree~3 and dimension~9, if $A= (B,\sigma)_+$.  Let
$$
(B, \sigma)_+^\circ = \{ x \in (B, \sigma)_+\colon \Trd (x) = 0\} =
1^\perp \subset (B, \sigma)_+.
$$
Since $Q_\sigma (1) = 3$, the restriction $Q_\sigma^\circ$ of
$Q_\sigma$ to $(B, \sigma)_+^\circ$ is given by
$$
Q_\sigma^\circ \simeq \langle 2, 6\rangle \perp \langle 2\rangle
\cdot \langle \!  \langle \alpha\rangle\!\rangle\cdot
\langle\!\langle b, c\rangle\!\rangle^\sharp\simeq \langle 2\rangle
\cdot (\langle 1, 3\rangle \perp\langle \!  \langle
\alpha\rangle\!\rangle\cdot \langle\!\langle b,
c\rangle\!\rangle^\sharp ).
$$
In the Witt ring $WF$, we have
$$
\Pf\alpha\pF\cdot\Pf b,c\pF^\sharp=\Pf\alpha,b,c\pF-\Pf\alpha\pF;
$$
therefore,
$$
Q_\sigma^\circ=\langle2\rangle\cdot(\langle3,\alpha\rangle
+\Pf\alpha,b,c\pF) \qquad\mbox{in }WF.
$$
Comparing dimensions on both sides, we see that the Witt indices of
both sides are related by:
\begin{equation}
  \label{WittQ.eq}
  w(Q_\sigma^\circ)=w(\langle3,\alpha\rangle\perp
  \Pf\alpha,b,c\pF)-1.
\end{equation}
Isotropic elements of $Q_\sigma^\circ$ are elements $u \in (B,
\sigma)_+$ such that $\Trd (u) = \Trd (u^2) = 0$.  Since the reduced
characteristic polynomial of any $a \in B$ has the form
$$
\textstyle X^3 - \Trd (a) X^2 + \frac{1}{2} \; [\Trd (a)^2 - {\rm
Trd} (a^2)] \; X - \Nrd (a)\cdot 1,
$$
it follows that $u$ is isotropic if and only if $u^3 = \Nrd (u) \in
F$.

\begin{thm}
  \label{distinv.thm}
  The following conditions are equivalent:
  \begin{enumerate}
    \item[(a)]
    $\pi(B,\sigma)$ is hyperbolic;
    \item[(b)]
    Either $K=F\times F$ or $\langle -b, -c, bc\rangle_K \simeq
    \langle 1, -1, -1\rangle_K$;
    \item[(c)]
    $w(Q_\sigma^\circ)\geq2$;
    \item[(c')]
    $(B,\sigma)_+$ contains a subspace $U$ of dimension~2 whose
    elements satisfy: $u^3=\Nrd(u)$;
    \item[(d)]
    $w(Q_\sigma^\circ)\geq3$;
    \item[(d')]
    $(B,\sigma)_+$ contains a subspace $U$ of dimension~3 whose
    elements satisfy: $u^3=\Nrd(u)$;
    \item[(e)]
    $(B,\sigma)_+$ contains an \'etale cubic $F$-algebra $L$ of
    discriminant $\alpha$.
    \item[(f)]
    $B$ is a crossed product:
    $$
    B=M\oplus M\,x\oplus M\,x^2,
    $$
    where $M$ ($\supset K$) is a Galois extension of $F$ with Galois
    group ${\cal S}_3$, the involution $\sigma$ preserves $M$ and
    leaves $x$ invariant.
  \end{enumerate}
\end{thm}

\proof {\em(a)~$ \Rightarrow $~(b)} is a straightforward consequence
of Theorem~\ref{clasinv.thm}, and {\em(c)$\iff$(c')},
{\em(d)$\iff$(d')} follow from the preceding observations on
isotropic elements in $Q_\sigma^\circ$.  Moreover,
{\em(a)~$\Rightarrow$~(d)} follows from~(\ref{WittQ.eq}),
{\em(e)$\iff$(f)\/} from Proposition~\ref{N.prop}, and
{\em(d)~$\Rightarrow$~(c)} is clear.  We now show
{\em(c)~$\Rightarrow$~(a):\/} if the Witt index of $Q_\sigma^\circ$
is at least~2, then (\ref{WittQ.eq}) shows that
$\langle3,\alpha\rangle\perp \Pf\alpha,b,c\pF$ contains isotropic
subspaces of dimension~3.  Therefore, $\Pf\alpha,b,c\pF$ is
isotropic, hence hyperbolic, proving {\em(a)}.  Assuming {\em(e)},
the formula for $Q_\sigma$ in Proposition~\ref{diagN.prop} shows
that $w(Q_\sigma^\circ)\geq3$, hence {\em(e)~$\Rightarrow$~(d)}.  To
complete the proof, we show {\em(c')~$\Rightarrow$~(e)}, using
results from \cite{HK}.  We first consider the easy special case
where $B$ is not a division algebra.  If $\alpha=1$ in
$F^\times/{F^{\times}}^2$, then we may assume
$$
B=A\times A^{\op}
$$
for some central simple $F$-algebra $A$ of degree~3, and $\sigma$ is
the switch involution.  A theorem of Wedderburn \cite{SofA} shows
that $A$ contains a cyclic extension $L$ of $F$.  We may then choose
$$
M=\{(\ell,\ell^{\op})\colon\ell\in L\}.
$$
If $B$ is split, then since {\em(c')~$\Rightarrow$~(b)\/}, it
follows that $\sigma$ is the adjoint involution with respect to an
isotropic hermitian form.  We may thus assume $B={\rm M}_3(K)$ and
$\sigma=\Int(u)\circ\tau$, where $u= \left(
\begin{array}{ccc}
  1&0&0\\
  0&0&1\\
  0&1&0
\end{array}
\right)$.  We may then choose
$$
M= \left\{ \left(\begin{array}{ccc}
f&0&0\\
0&k&0\\
0&0&\overline{k}
\end{array}
\right)\colon f\in F,\;k\in K\right\}.
$$
For the rest of the proof of {\em(c')~$\Rightarrow$~(e)}, we now
assume that $B$ is a division algebra.  Let $U\subset(B,\sigma)_+$ be
a subspace of dimension~2 such that $u^3=\Nrd(u)$ for all $u\in U$.
According to Lemma~2 of \cite{HK}, one can find a basis $(w_1,w_2)$ of
$U$ such that
$$
\Trd(w_1^{-1}w_2)=0.
$$
Following \cite{HK}, we let
$$
\theta_1=w_1^{-1}w_2,\qquad \theta_2=w_1^{-1}\theta_1w_1,\qquad
\theta_3=w_1^{-1}\theta_2w_1.
$$
Note $\theta_2=w_1^{-3}(w_1w_2w_1)$, so $\sigma(\theta_2)=\theta_2$.
We let also $E=K(\theta_2^{-1}\theta_3)$ if
$\theta_2^{-1}\theta_3\not\in K$ and $E=K(\theta_2)$ if
$\theta_2^{-1}\theta_3\in K$.  Theorem~3 of \cite{HK} shows that $E/K$
is cyclic and $E/F$ is Galois with Galois group ${\cal S}_3$.
Moreover, one of the order~2 automorphisms of $E/F$ is the restriction
of the involution
$$
\sigma'=\Int(\theta_2^{-1})\circ\sigma.
$$
Let $L'\subset E$ denote the subfield of $\sigma'$-invariant elements.
Since $L'K=E$ is cyclic over $K$ and Galois over $F$ with group ${\cal
S}_3$, Proposition~\ref{N.prop} shows that the discriminant of $L'/F$
is $\alpha$.

Observe now that
$$
\theta_2=w_1^{-3}w_2^{-3}(w_1w_2^2)(w_2^2w_1),
$$
hence $ \theta_2^{-1}=\lambda v\sigma(v) $ for $\lambda=w_1^3w_2^3$
and $v=w_1^{-1}w_2^{-2}$, so that
$$
\Int(v)\colon(B,\sigma)\isom(B,\sigma')
$$
is an isomorphism of algebras with involution.  Pulling
$L'\subset(B,\sigma')_+$ back gives the wanted extension
$L\subset(B,\sigma)_+$.  \qed

\medskip

{\it Remark.}\ Theorem \ref{distinv.thm} gives a positive answer to a
question about Tits constructions asked in \cite[(2.12)]{PRac2}.

\medskip An involution $\sigma$ satisfying the equivalent conditions
of Theorem~\ref{distinv.thm} is called {\bf distinguished}.  By
Theorem~\ref{clasinv.thm}, two distinguished involutions on $B$ are
isomorphic.  The existence of distinguished involutions is clear if
$B\simeq {\rm M}_3(K)$ is split: the adjoint involution with respect
to any isotropic hermitian form on $K^3$ is distinguished.  If
$K=F\times F$, the switch involution on $B\simeq A\times A^{\rm op}$
is distinguished: in fact we have $\alpha=1$, so that $\langle
\!\langle \alpha\rangle\!\rangle= \langle1,-1\rangle$ and
$Q^\circ_\sigma$ has Witt index at least $3$.  The existence in
general of distinguished involutions for algebras of degree $3$ is
shown next.

\begin{prop}
  \label{exdist.prop}
  For every cubic \'etale $F$-algebra $L\subset B$, there is a
  distinguished involution $\sigma$ such that $L\subset(B,\sigma)_+$.
\end{prop}

\proof Let $\lambda_0\in L^\times$ be such that
$t_{L/F}(\lambda_0)=0$, and let
$$
\lambda=\lambda_0/n_{L/F}(\lambda_0);
$$
then $n_{L/F}(\lambda)=n_{L/F}(\lambda_0)^{-2}\in {F^{\times}}^2$ and
$t_{L/F}(\lambda)=0$.  Corollary~\ref{chginv.cor} shows that there is
an involution $\sigma$ on $B$ such that $L\subset(B,\sigma)_+$ and
$$
Q_\sigma=\langle1,1,1\rangle\perp
\langle2\delta\rangle\cdot\Pf\alpha\pF\cdot t_{L/F}(\langle
\lambda\rangle),
$$
hence
$$
Q_\sigma^\circ=\langle2\rangle\cdot(\langle1,3\rangle \perp
\langle\delta\rangle\cdot\Pf\alpha\pF\cdot t_{L/F}(\langle
\lambda\rangle)).
$$
Since $t_{L/F}(\lambda)=0$, the form $t_{L/F}(\langle\lambda\rangle)$
is isotropic, hence the Witt index of $\Pf\alpha\pF\cdot
t_{L/F}(\langle\lambda\rangle)$ is at least 2.  Therefore,
condition~{\em(c)\/} of Theorem~\ref{distinv.thm} holds.  \qed

{\it Alternative proof.\/} Pick an involution $\sigma_0$ such that
$L\subset(B,\sigma_0)_+$ and denote by $V$ the orthogonal of $L$ with
respect to $Q_{\sigma_0}$, as in the preceding section.  By Zariski
density, we may find an invertible element $v\in V$ such that
$n_{L/F}(N(v))\neq0$, where $N\colon V\to L$ is the quadratic form
defined in section~3.  Lemma~\ref{N.lem} shows:
$$
\Trd(v^{-1}\ell)=-\Nrd(v)^{-1}t_{L/F}(N(v)\ell)
$$
for all $\ell\in L$.  Since $N(v)$ is invertible in $L$, we may, again
by Zariski density, find $\ell\in L^\times$ such that
$$
\Trd(v^{-1}\ell)=0.
$$
Following \cite[Proposition~1]{HK}, the vector space
$$
U=v^{-1}L\cap\ker\Trd
$$
is at least $2$-dimensional over $F$ and satisfies: $u^3=\Nrd(u)$ for
all $u\in U$.  We then have $v^{-1}\ell\in U$ for $\ell\in L$ as
above. Let
$$
\sigma=\Int(\ell^{-1})\circ\sigma_0.
$$
Since $L\subset(B,\sigma_0)_+$ and $\ell\in L^\times$, it follows that
$L\subset(B,\sigma)_+$.  We claim that $\sigma$ is distinguished.  Let
$y=\ell^{-1}v\in B^\times$, then
$$
yU=\ell^{-1}(vU)\subset L\subset(B,\sigma)_+
$$
and
$$
\sigma(y)=\ell^{-1}\sigma_0(y)\ell=\ell^{-1} v\ell^{-1}\ell=y,
$$
hence $(B,\sigma)_+$ contains $y^2Uy=y(yU)y$.  Since $v^{-1}\ell \in
U$, we have $y^{-1}\in U$, hence $y^3=\Nrd(y)\in F^\times$.  For $u\in
U$, we have $u^3=\Nrd(u)\in F$, hence
\begin{eqnarray*}
  (y^2uy)^3&=&y^2uy\cdot y^2uy\cdot y^2uy
  =y^2u^3y\,\Nrd(y)^2\\
  &=&\Nrd(u)\Nrd(y)^3 =\Nrd(y^2uy).
\end{eqnarray*}
Therefore, $y^2Uy\subset(B,\sigma)_+$ is a subspace of
dimension~$\geq2$ whose elements satisfy $x^3=\Nrd(x)$, hence $\sigma$
is distinguished.  \qed

A third proof of the existence of distinguished involutions is given
in Proposition~\ref{degthree.prop}.

\begin{cor}
  The space $(B,\sigma)_+$ contains an isomorphic copy of every cubic
  \'etale $F$-subalgebra $L$ of $B$ if and only if $\sigma$ is
  distinguished.
\end{cor}

\proof Since all the distinguished involutions on a central simple
algebra $B$ are isomorphic, the if direction follows from
Proposition~\ref{exdist.prop}.  Conversely, by
Theorem~\ref{distinv.thm},~{\em(e)\/}, the only involutions which
leave elementwise invariant \'etale cubic $F$-subalgebras of
discriminant $\alpha$ are the distinguished involutions.  \qed

The fact that the only involutions which leave elementwise invariant
\'etale cubic $F$-subalgebras of discriminant $\alpha$ are the
distinguished involutions also follows from the following general
result:

\begin{prop}
  Let $\sigma$ be an involution on $B$ and let $L$ be a cubic \'etale
  $F$-algebra such that $L\subset(B,\sigma)_+$.  Let $\delta\in
  F^\times$ represent the discriminant of $L/F$.  The Pfister form
  $\pi(B,\sigma)$ has a factorization:
  $$
  \pi(B,\sigma)=\Pf\alpha\pF\cdot\varphi
  $$
  where $\varphi$ is a 2-fold Pfister form whose pure subform
  satisfies:
  $$
  \varphi^\sharp=t_{L/F}(\langle\ell\rangle)
  $$
  for some $\ell\in L^\times$ such that
  $$
  n_{L/F}(\ell)\in\delta\cdot {F^{\times}}^2.
  $$
  In particular,
  $$
  \varphi\cdot\Pf\delta\pF=0.
  $$
\end{prop}

\proof Proposition~\ref{diagN.prop} shows that
$$
Q_\sigma=\langle1,1,1\rangle\perp
\langle2\rangle\cdot\Pf\alpha\pF\cdot
t_{L/F}(\langle\delta\lambda\rangle)
$$
where $\lambda\in L^\times$ is such that $n_{L/F}(\lambda)\in
{F^{\times}}^2$.  Letting $\ell=\delta\lambda$, we get
$n_{L/F}(\ell)\in\delta\cdot {F^{\times}}^2$ and
$t_{L/F}(\langle\ell\rangle)$ is a 3-dimensional form of
determinant~1.  Comparing with the form of $Q_\sigma$ in
Theorem~\ref{clasinv.thm}, we obtain:
$$
\pi(B,\sigma)=\Pf\alpha\pF\cdot(\langle1\rangle\perp
t_{L/F}(\langle\ell\rangle)).
$$
The relation $(\langle1\rangle\perp
t_{L/F}(\langle\delta\lambda\rangle))\cdot\Pf\delta\pF=0$ follows from
Lemma~\ref{trhyp.lem}, using the fact that
$\langle-\delta\rangle\cdot\Pf\delta\pF=\Pf\delta\pF$.  \qed

The condition $\varphi\cdot\Pf\delta\pF=0$ in the proposition above
yields some restriction on the discriminants of cubic $F$-algebras $L$
which lie in $(B,\sigma)_+$.  It does not yield any information on
cyclic extensions however, since in this case $\Pf\delta\pF=0$.
Nevertheless, there are examples where $(B,\sigma)_+$ does not contain
any cyclic cubic extension of $F$.  Consider for instance the ``symbol
algebra'' $ A_\omega(s,t) $ generated over the iterated powerseries
field $\CC((s))((t))$ by two indeterminates $i,j$ subject to the
relations:
$$
i^3=s,\qquad j^3=t,\qquad ji=\omega ij,
$$
where $\omega$ is a primitive cube root of unity.  This algebra
carries an involution $\sigma$ extending conjugation on $\CC$ such
that
$$
\sigma(i)=i,\qquad\sigma(j)=j.
$$
The subfield consisting of the $\sigma$-invariant elements in the
center $\CC((s))((t))$ is $\RR((s))((t))$, which does not have any
non-trivial cyclic extension of odd degree.  Therefore, the space of
symmetric elements $(A_\omega(s,t),\sigma)_+$ does not contain any
cyclic extension of $F$ of degree~3.  We are indebted to
H.P.~Petersson (see \cite{PR2}) for suggesting this example.

\medskip

Let $L$ be an arbitrary cubic \'etale $F$-subalgebra in $B$.  Let
$\Sigma(B,L)$ be the pointed set of isomorphism classes of involutions
$\sigma$ of $B$ such that $\sigma|_L=I_L$, the point being given by
the class of distinguished involutions.

\begin{cor}
  The map $\mu \in L^\times \mapsto \sigma_{\mu}$, where
  $\sigma_{\mu}$ is the involution determined up to isomorphism by the
  trace form
  $$
  Q_{\sigma_{\mu}} = \langle 1,1,1\rangle \perp \langle2 \delta
  \rangle \cdot \Pf\alpha\pF\cdot t_{L/F}(\langle q(\mu) \rangle_L),
  $$
  induces a surjective map of pointed sets
  $$
  L^\times/n_{LK/L}(LK^\times) \cdot F^\times \to \Sigma(B,L).
  $$
  If $L/F$ is cyclic, the map is well-defined on the set of orbits of
  the group $\Gal(L/F)$ in

  $L^\times/n_{LK/L}(LK^\times) \cdot F^\times$.
\end{cor}

\proof The proof of Corollary~\ref{chginv.cor} shows that the elements
$\lambda \in L$ such that $n_{L/F}(\lambda) \in {F^{\times}}^2$ are
exactly the elements of the form $q(\mu)$ for $\mu \in L^\times$.
Thus, by Corollary~\ref{chginv.cor}, $Q_{\sigma_{\mu}}$ is the trace
form of an involution $\sigma_{\mu}$.  Elements $\mu,\,\mu' \in
L^\times$ such that $ \mu'= \xi n_{LK/L}(\eta)\mu $ for $\eta \in
{LK}^\times$ and $\xi \in F^\times$ give isomorphic involutions in
view of Theorem~\ref{clasinv.thm}{, \it(d)}, since $q(\mu')=
\xi^2\,q(\eta)\,q(\mu) \,\overline{q(\eta)} $ implies that the
hermitian forms $\langle q(\mu)\rangle_{LK}$ and $\langle
q(\mu')\rangle_{LK}$ are isomorphic.  \qed

\medskip

{\bf 5. Cohomology and Symbols.} \ Let $F$ be a field of
characteristic different from~2; let $F_s$ denote a separable closure
of $F$ and $\Gamma=\Gal(F_s/F)$.  For any integer $n$ relatively prime
to the characteristic of $F$, let
$$
\mu_n=\{x\in F_s\colon x^n=1\}
$$
denote the $\Gamma$-module of $n$-th roots of $1$.  The Kummer exact
sequence
$$
1\to\mu_n\to F^{\times}_s\stackrel{n}{\to} F_s^\times\to1
$$
and Hilbert's Theorem~90 yield canonical isomorphisms:
$$
H^1(\Gamma,\mu_n)=F^\times/F^{\times n} \qquad\mbox{and}\qquad
H^2(\Gamma,\mu_n)=\nBr(F),
$$
where $\nBr(F)$ denotes the subgroup killed by $n$ in the Brauer group
$\Br(F)$.  Since the action of $\Gamma$ on $\mu_2$ is trivial, we also
have
$$
H^1(\Gamma,\mu_2)=\Hom(\Gamma,\{\pm1\}).
$$
For any $\alpha\in F^\times$, let $\phi_\alpha\colon\Gamma\to\{\pm1\}$
be the (continuous) homomorphism corresponding to $\alpha\cdot
{F^{\times}}^2$ under the identification:
$F^\times/{F^{\times}}^2=\Hom(\Gamma,\{\pm1\})$; it is explicitly
defined by:
$$
\phi_\alpha(\gamma)=\left\{
\begin{array}{ll}
  +1&\mbox{if $\gamma$ leaves invariant}\\
  -1&\mbox{if $\gamma$ permutes}
\end{array}
\right\} \mbox{the square roots of $\alpha$ in $F_s$.}
$$
We then define a $\Gamma$-module $\ZZ(\alpha)$ by twisting the trivial
action of $\Gamma$ on $\ZZ$: for $\gamma\in\Gamma$ and $z\in\ZZ$, we
let $\gamma\cdot z=\phi_\alpha(\gamma)z$.  Multiplication in $\ZZ$
yields a canonical isomorphism:
$$
\ZZ(\alpha)\otimes_{\Bbb Z}\ZZ(\beta)= \ZZ(\alpha\beta)
\qquad\mbox{for $\alpha,\beta\in F^\times$,}
$$
since $\phi_\alpha(\gamma)\phi_\beta(\gamma)=
\phi_{\alpha\beta}(\gamma)$ for all $\gamma\in\Gamma$.

For any $\Gamma$-module $M$, we define a twisted module $M(\alpha)$
by:
$$
M(\alpha)=M\otimes_{\Bbb Z}\ZZ(\alpha).
$$
The Galois cohomology groups $H^\ell(\Gamma,M)$ are also denoted
$H^\ell(F,M)$.

\begin{prop}
  \label{cohexseq.prop}
  Let $K=F(\sqrt{\alpha})$ be a quadratic field extension of~$F$.  For
  all $\ell\geq1$ and all $\Gamma$-modules $M$ on which multiplication
  by~2 is invertible, the following sequence is split exact:
  $$
  0\to H^\ell(F,M(\alpha)) \stackrel{\res}{\longrightarrow}
  H^\ell(K,M) \stackrel{\cores}{\longrightarrow} H^\ell(F,M)\to 0
  $$
  where $\res$ (resp.\ $\cores$) is the restriction (resp.\ the
  corestriction) map.

  The restriction map identifies $H^\ell(F,M(\alpha))$ with the
  subgroup of $H^\ell(K,M)$ consisting of the classes of cocycles
  which are cohomologous to the negative of their conjugate under the
  action of $\Gal(K/F)$:
  $$
  H^\ell(F,M(\alpha))=\{\varphi\in H^\ell(K,M)\colon
  \varphi+\overline{\varphi}=0\}.
  $$
\end{prop}

\proof From the definition of $M(\alpha)$, it is clear that
$M=M(\alpha)$ as modules over $\Gal(F_s/K)$.  Let
$G=\Gal(K/F)=\{1,\iota\}$ and let $IG=\ZZ\cdot(1-\iota)$ denote the
augmentation ideal in the group ring $\ZZ G$.  Denote by $\Sigma$ the
augmentation map:
$$
\Sigma\colon\ZZ G\to\ZZ
$$
and define $\Gamma$-module homomorphisms $s\colon\ZZ\to\ZZ G$ and
$t\colon\ZZ G\to IG$ by:
$$
s(z)=z(1+\iota),\qquad t(x)=(1-\iota)x\qquad \mbox{for
}z\in\ZZ,\;x\in\ZZ G.
$$
These maps fit into a commutative diagram:

$$
\begin{array}{ccccccccc}
  0&\leftarrow&IG&\stackrel{t}{\longleftarrow}& \ZZ
  G&\stackrel{s}{\longleftarrow}&\ZZ&
  \leftarrow&0\\
  & &\Big\uparrow\vcenter{\rlap{$\scriptstyle 2\cdot$}}& &\Big\|&
  &\Big\downarrow\vcenter{\rlap{
  $\scriptstyle2\cdot$}}& & \\
  0&\rightarrow&IG&\longrightarrow&\ZZ G&
  \stackrel{\Sigma}{\longrightarrow}&\ZZ& \rightarrow&0
\end{array}
$$
Observe that $IG\simeq\ZZ(\alpha)$; therefore, tensoring the diagram
above with $M$, we get the following commutative diagram:

$$
\begin{array}{ccccccccc}
  0&\leftarrow&M(\alpha)& \stackrel{t}{\longleftarrow}& M\otimes_{\ZZ}
  \ZZ G&\stackrel{s}{\longleftarrow}&M&
  & \\
  & &\Big\uparrow\vcenter{\rlap{$\scriptstyle 2\cdot$}}& &\Big\|&
  &\Big\downarrow\vcenter{\rlap{
  $\scriptstyle2\cdot$}}& & \\
  & &M(\alpha)&\stackrel{i}{\longrightarrow}& M\otimes_{\ZZ}\ZZ G&
  \stackrel{\Sigma}{\longrightarrow}&M& \rightarrow&0
\end{array}
$$
and since multiplication by 2 is invertible on $M$, it follows that
the sequence
$$
0\to M(\alpha)\stackrel{i}{\longrightarrow} M\otimes_{\ZZ} \ZZ G
\stackrel{\Sigma}{\longrightarrow} M\to 0
$$
is split exact.  It yields a split exact sequence in cohomology:
$$
0\to H^\ell(F,M(\alpha))\stackrel{i_*}{\longrightarrow}
H^\ell(F,M\otimes\ZZ G) \stackrel{\Sigma_*}{\longrightarrow}
H^\ell(F,M)\to 0.
$$
By the lemma of Eckmann--Faddeev--Shapiro, there is an isomorphism
$$
\psi\colon H^\ell(F,M\otimes\ZZ G)\isom H^\ell(K,M)
$$
such that $\Sigma_*=\cores\circ\,\psi$ and $\psi\circ i_*=\res$.  This
proves the first part.

Let $\res'\colon H^\ell(F,M)\to H^\ell(K,M)$ denote the restriction
map.  For $\varphi\in H^\ell(K,M)$, we have:
\begin{equation}
  \label{rescor.eq}
  \res'\circ\cores(\varphi)=\varphi+ \overline{\varphi},
\end{equation}
hence $\varphi+\overline{\varphi}=0$ if $\cores(\varphi)=0$.
Conversely, applying $\cores$ to both sides of (\ref{rescor.eq}), we
get:
$$
2\,\cores(\varphi)=\cores(\varphi+ \overline{\varphi}),
$$
since $\cores\circ\res'=2$.  Since multiplication by 2 is invertible
on $M$, it follows that $\cores(\varphi)=0$ if
$\varphi+\overline{\varphi}=0$.  Therefore, the image of
$H^\ell(F,M(\alpha))$ in $H^\ell(K,M)$ can be indifferently described
as the kernel of the corestriction map or as the kernel of the map
$\varphi\mapsto\varphi+\overline{\varphi}$.  \qed

{\it Remark.\/} If multiplication by 2 is not invertible on $M$, the
restriction and corestriction maps fit into a long exact sequence
described by Arason and Elman in \cite[Appendix]{AE}.

\medbreak

We shall apply this proposition to the special cases where
$M=\ZZ/n\ZZ$ ($n$ odd) with trivial action and $M=\mu_n=\{x\in
F_s\colon x^n=1\}$ (assuming $n$ odd and relatively prime to the
characteristic of $F$).  We first review how, for an arbitrary finite
group $\cal G$, the first cohomology set $H^1(F,{\cal G})$, where the
Galois action on $\cal G$ is trivial, is related to Galois extensions
of $F$ with Galois group $\cal G$.  We recall that, in this case,
$$
H^1(F,{\cal G})=\Hom(\Gamma,{\cal G})/\sim,
$$
where $\Hom(\Gamma,{\cal G})$ is the set of continous homomorphisms
$\Gamma \to {\cal G}$ and $\chi' \sim \chi$ if
$\chi'(\gamma)=g\chi(\gamma)g^{-1}$ for some $g \in {\cal G}$.  In
particular we have $H^1(F,{\cal G})=\Hom(\Gamma,{\cal G})$ if ${\cal
G}$ is abelian.

Let $L$ be a finite-dimensional commutative algebra over $F$ and let
$$
X(L)=\Alg_F(L,F_s)
$$
denote the set of $F$-algebra homomorphisms $L\to F_s$.  Let $\cal G$
be an arbitrary finite group acting on $L$ by $F$-algebra
automorphisms $\ell\mapsto g*\ell$.  We call $L$ a {\bf Galois $\cal
G$-algebra\/} if $L$ is \'etale of dimension $n=|{\cal G}|$ over $F$
and the action of $\cal G$ on $X(L)$ is simply transitive.

Now, assume $L$ is a Galois $\cal G$-algebra and fix some $\xi\in
X(L)$.  For every $\gamma\in\Gamma$, there is a unique
$\chi_\xi(\gamma)\in{\cal G}$ such that
$$
\gamma(\xi(\ell))=\xi(\chi_\xi(\gamma)*\ell) \qquad\mbox{for all
}\ell\in L.
$$
The map $\chi_\xi\colon\Gamma\to{\cal G}$ is a continuous
homomorphism:
$$
\chi_\xi\in\Hom(\Gamma,{\cal G}).
$$

If $L$, $L'$ are Galois $\cal G$-algebras over $F$ and $\xi\in X(L)$,
$\xi'\in X(L')$, an isomorphism $\psi\colon(L,\xi)\isom(L',\xi')$ is
an isomorphism of algebras over $F$ which commutes with the action of
$\cal G$ and such that $\xi'\circ\psi=\xi$.  An isomorphism of Galois
$\cal G$-algebras $\psi\colon L \isom L'$ is an isomorphism of
algebras over $F$ which commutes with the action of $\cal G$.

\begin{prop}
  \label{Galois.prop}
  The map $(L,\xi) \mapsto \chi_\xi$ defines a 1--1 correspondence
  between the set of isomorphism classes of couples $(L,\xi)$, where
  $L$ is a Galois $\cal G$-algebra and $\xi\in X(L)$, and the set
  $\,\Hom(\Gamma,{\cal G})$.  Moreover $L\mapsto\chi_\xi$ induces a
  1--1 correspondence between the set of isomorphism classes of Galois
  $\cal G$-algebras and the set $H^1(F,{\cal G})$.
\end{prop}

\proof This proposition is presumably well-known (see for example
\cite{Se}), but we include a sketch of proof for the reader's
convenience.  For $\chi\in\Hom(\Gamma,{\cal G})$, we define an action
of $\Gamma$ on the algebra ${\frak F}({\cal G},F_s)$ of maps ${\cal
G}\to F_s$ by:
$$
(\gamma\diamond f)(x)=\gamma(f(\chi(\gamma)^{-1} x))\qquad\mbox{for
}\gamma\in\Gamma,\;x\in{\cal G}.
$$
Let ${\frak F}_\chi$ denote the algebra of invariant elements under
this action:
$$
{\frak F}_\chi=\{f\colon{\cal G}\to F_s\colon
\gamma(f(x))=f(\chi(\gamma)x)\mbox{ for }\gamma \in\Gamma,\;x\in{\cal
G}\}.
$$
This algebra carries a natural $\cal G$-action defined by:
$$
(g*f)(x)=f(xg)\qquad\mbox{for }x,g\in{\cal G},\; f\in{\frak F}_\chi.
$$
Moreover, for $g\in{\cal G}$, there is an $F$-algebra homomorphism
$$
\varepsilon_g\colon{\frak F}_\chi\to F_s
$$
defined by:
$$
\varepsilon_g(f)=f(g)\qquad\mbox{for }f\in{\frak F}_\chi,
$$
and $X({\frak F}_\chi)=\{\varepsilon_g\colon g\in{\cal G}\}$.  It is
then straightforward to check that ${\frak F}_\chi$ is a Galois $\cal
G$-algebra, and that $\chi_{\varepsilon_1}=\chi$.

For any Galois $\cal G$-algebra $L$ and any $\xi\in X(L)$, there
exists a unique isomorphism $(L,\xi)\isom({\frak
F}_{\chi_\xi},\varepsilon_1)$: this isomorphism maps $\ell\in L$ to
$f_\ell\in{\frak F}_\chi$ defined by:
$$
f_\ell(g)=\xi(g*\ell) \qquad\mbox{for }g\in{\cal G}.
$$

Let now $\xi,\eta\in X(L)$.  By definition of Galois algebra, there is
a unique element $g\in{\cal G}$ such that $\eta(\ell)=\xi(g*\ell)$ for
all $\ell\in L$.  Then, for $\gamma\in\Gamma$ and $\ell\in L$,
$$
\gamma(\eta(\ell))= \gamma(\xi(g*\ell))=\xi(\chi_\xi(\gamma)g*\ell) =
\xi(g*(g^{-1}\chi_\xi(\gamma)g*\ell)),
$$
hence
$$
\chi_\eta(\gamma)=g^{-1}\chi_\xi(\gamma)g.
$$
This implies the last claim of Proposition \ref{Galois.prop}.  \qed

We now consider the particular case where $\cal G$ is dihedral: let
${\cal D}_n$ be the group generated by two elements $r,s$ subject to
the relations:
$$
r^n=1,\qquad s^2=1 \qquad\mbox{and}\qquad rsr=s,
$$
and let ${\cal Z}_n$ denote the cyclic (normal) subgroup of ${\cal
D}_n$ generated by $r$.  The group ${\cal D}_n$ is the semidirect
product ${\cal D}_n= {\cal Z}_n \rtimes \mu_2 $ and we have a split
exact sequence
\begin{equation}
  \label{dih.eq}
  1\to{\cal Z}_n\to{\cal D}_n \stackrel{\rho}{\longrightarrow}\mu_2\to
  1.
\end{equation}

\begin{prop}
  \label{dih.prop}
  Let $L$ be a Galois ${\cal D}_n$-algebra over $F$ and let
  $$
  L_0=\{\ell\in L\colon r*\ell=\ell\},\qquad L_1=\{\ell\in L\colon
  s*\ell=\ell\}.
  $$
  Then $L_0\simeq F(\sqrt{\alpha})$ for some $\alpha\in F^\times$.
  For all $\xi\in X(L)$, the homomorphism
  $\rho\circ\chi_\xi\in\Hom(\Gamma,\mu_2)$ corresponds to $\alpha\cdot
  {F^{\times}}^2$ under the identification
  $\Hom(\Gamma,\mu_2)=H^1(F,\mu_2)= F^\times/{F^{\times}}^2$.
  Moreover, if $n$ is odd, $\alpha\cdot {F^{\times}}^2$ is the
  discriminant of $L/F$, so that the map $H^1(F, {\cal D}_n) \to
  H^1(F, \mu_2)$, induced in cohomology by $\rho$, associates to $L$
  the discriminant of $L$.  The discriminant of $L_1/F$ is $1$ if
  $n\equiv1$~{\rm mod }$4$ and $\alpha\cdot F^{\times 2}$ if
  $n\equiv3$~{\rm mod }$4$.
\end{prop}

\proof The first statement is clear, since Galois theory shows $L_0$
is a quadratic \'etale algebra over $F$.  Since the action of ${\cal
Z}_n$ on $L_0$ is trivial, the action of ${\cal D}_n$ factors through
$\mu_2$: we may set
$$
\rho(g)*\ell=g*\ell\qquad\mbox{for }g\in{\cal D}_n\mbox{ and }\ell\in
L_0.
$$
For all $\xi\in X(L)$, we then have
$$
\gamma(\xi(\ell))=\xi(\rho\circ\chi_\xi(\gamma)* \ell) \qquad\mbox{for
$\gamma\in\Gamma$ and $\ell\in L_0$},
$$
proving that $\rho\circ\chi_\xi\in H^1(F,\mu_2)$ is the homomorphism
corresponding to the Galois {$\mu_2$-algebra} $L_0/F$.

For the rest of the proof, assume $n$ is odd.  The discriminant of
$L_1/F$ is represented by the determinant
$$
\det(\tr_{L_1/F}(e_ie_j))_{1\leq i,j\leq n},
$$
where $(e_i)_{1\leq i\leq n}$ is a basis of $L_1$ over $F$.  Since,
for $\ell\in L_1$,
$$
\tr_{L_1/F}(\ell)=\sum_{i=1}^n\xi(r^i*\ell),
$$
we have
$$
(\tr_{L_1/F}(e_ie_j))_{1\leq i,j\leq n}=m^t\cdot m
$$
where
$$
m=(\xi(r^i*e_j))_{1\leq i,j\leq n}.
$$
Therefore, if $\delta_1=\det m\in\xi(L)$, the discriminant of $L_1/F$
is represented in $F^\times/{F^{\times}}^2$ by~$\delta_1^2$.  For
$\gamma\in\Gamma$, we have
$$
\gamma(\delta_1)=\xi(\det(\chi_\xi(\gamma)r^i* e_j)_{1\leq i,j\leq
n})= \xi(\det(\chi_\xi(\gamma)r^is*e_j)_{1\leq i,j\leq n}).
$$
If $\chi_\xi(\gamma)\in{\cal Z}_n$, multiplication by
$\chi_\xi(\gamma)$ is an even permutation of ${\cal Z}_n$, hence
$$
\det(\chi_\xi(\gamma)r^i* e_j)_{1\leq i,j\leq n}=\det(r^i*e_j)_{1\leq
i,j\leq n},
$$
and therefore $ \gamma(\delta_1)=\delta_1.  $ If
$\chi_\xi(\gamma)\not\in{\cal Z}_n$, the map
$z\mapsto\chi_\xi(\gamma)zs$ is a permutation of ${\cal Z}_n$ of
signature $(-1)^{(n-1)/2}$, hence
$$
\gamma(\delta_1)=(-1)^{(n-1)/2}\delta_1.
$$
Therefore, if $n\equiv1$~mod~$4$ we have $\delta_1\in F$, hence the
discriminant of $L_1/F$ is trivial.  If $n\equiv3$~mod~$4$, we have
$$
\gamma(\delta_1)=\rho\circ\chi_\xi(\gamma)\, \delta_1,
$$
hence the discriminant of $L_1/F$ corresponds to $\rho\circ\chi_\xi\in
H^1(F,\mu_2)$, hence it is $\alpha\cdot {F^{\times}}^2$.

The discriminant of $L/F$ is calculated in a similar way:
multiplication by $g\in{\cal D}_n$ defines on ${\cal D}_n$ a
permutation of signature $\rho(g)$, hence the discriminant of $L/F$
corresponds to $\rho\circ\chi_\xi\in H^1(F,\mu_2)$.  \qed

Suppose now $K=F(\sqrt{\alpha})$ is a quadratic field extension of $F$
contained in $F_s$.  We denote by $\Gamma_K$ the Galois group
$\Gamma_K=\Gal(F_s/K)$ of $F_s$ over $K$.

\begin{prop}
  \label{clasdih.prop}
  Suppose $n$ is odd.  The group $H^1(F,{\cal Z}_n(\alpha))$
  classifies the Galois ${\cal Z}_n$-algebras $L$ over $K$ which can
  be endowed with a Galois ${\cal D}_n$-algebra structure over $F$
  extending the action of ${\cal Z}_n$.
\end{prop}

\proof Let $\sigma\in\Gamma\smallsetminus\Gamma_K$.
Proposition~\ref{cohexseq.prop} shows that
\begin{eqnarray*}
  H^1(F,{\cal Z}_n(\alpha))&=&\{\chi\in H^1(K,{\cal
  Z}_n)\colon \chi\overline{\chi}=1\}\\
  &=& \{\chi \in \Hom(\Gamma_K, {\cal Z}_n) \colon
  \chi\overline{\chi}=1\} ,
\end{eqnarray*}
where $\overline{\chi}$ is defined by:
$$
\overline{\chi}(\gamma)= \chi(\sigma\gamma\sigma^{-1}) \qquad\mbox{for
}\gamma\in\Gamma_K.
$$
Let $L$ be a Galois ${\cal Z}_n$-algebra over $K$ which also has a
Galois ${\cal D}_n$-algebra structure over $F$ extending the action of
${\cal Z}_n$.  Let $\xi\in X(L)$.  We may assume $\xi$ is a
$K$-algebra homomorphism; the restriction to $\Gamma_K$ of the
associated homomorphism $\chi_\xi\colon\Gamma\to{\cal D}_n$ is then
the element $\chi_\xi|_{\Gamma_K}\in H^1(K,{\cal Z}_n)$ associated to
$L$, viewed as a Galois ${\cal Z}_n$-algebra over $K$.

If $\chi_\xi(\sigma)\in{\cal Z}_n$, then $\rho$ maps the image of
$\chi_\xi$ to $1\in\mu_2$, hence Proposition~\ref{dih.prop} shows that
the subalgebra $L_0$ of invariant elements under $r$ splits as
$F\times F$.  This is a contradiction since $L_0=K$.  Therefore,
$\chi_\xi(\sigma)\not\in{\cal Z}_n$, hence, for $\gamma\in\Gamma_K$,
$$
\chi_\xi|_{\Gamma_K}(\sigma\gamma\sigma^{-1})= \chi_\xi(\sigma)
\chi_\xi(\gamma)\chi_\xi(\sigma)^{-1}=
\chi_\xi|_{\Gamma_K}(\gamma)^{-1}.
$$
This shows that $\chi_\xi|_{\Gamma_K}
\overline{\chi_\xi|_{\Gamma_K}}=1$, hence $\chi_\xi|_{\Gamma_K}\in
H^1(F,{\cal Z}_n(\alpha))$.

Conversely, to every $\chi\in H^1(F,{\cal Z}_n(\alpha))$ we associate
the Galois ${\cal Z}_n$-algebra over $K$:
$$
{\frak F}_\chi=\{f\in{\frak F}({\cal Z}_n,F_s) \colon
\gamma(f(x))=f(\chi(\gamma)x) \mbox{ for all }x\in{\cal Z}_n\}.
$$
Since $\chi\overline{\chi}=1$, we may extend the ${\cal Z}_n$-action
on ${\frak F}_\chi$ to an action of ${\cal D}_n$ by letting:
$$
(s*f)(x)=\sigma(f(x^{-1})) \qquad\mbox{for }f\in{\frak
F}_\chi,\;x\in{\cal Z}_n.
$$
\qed

{\em Remark.\/} Another proof of the proposition above can be obtained
by defining a twisted action of $\Gamma$ on ${\cal D}_n$ extending the
action on ${\cal Z}_n$.  There is an exact sequence corresponding to
(\ref{dih.eq}):
$$
1\to{\cal Z}_n(\alpha)\to{\cal D}_n(\alpha)\to \mu_2\to1,
$$
and the associated cohomology sequence:
$$
1\to H^1(F,{\cal Z}_n(\alpha))\to H^1(F,{\cal D}_n(\alpha))\to
H^1(F,\mu_2)
$$
yields an alternative (equivalent) description of $H^1(F,{\cal
Z}_n(\alpha))$.

As observed in the proof, all Galois ${\cal Z}_n$-algebras $L$ over
$K$ which can be endowed with a Galois ${\cal D}_n$-algebra structure
over $F$ extending the ${\cal Z}_n$-action have discriminant
$\alpha\cdot {F^{\times}}^2$.  It is also clear from the proof that
the ${\cal D}_n$-algebra structure on $L$ is not uniquely determined,
since it depends on the choice of $\sigma$.  However, the subalgebra
$L_1$ of invariant elements under the element $s$ of ${\cal D}_n$ does
not depend on the choice of the ${\cal D}_n$-algebra structure up to
$F$-isomorphism.  Therefore, for $\chi\in H^1(F,{\cal Z}_n(\alpha))$,
we may denote by $F_\chi$ an $F$-algebra which is isomorphic to the
subalgebra of $s$-invariant elements in the algebra $L$ corresponding
to $\chi$ under the correspondence of Proposition~\ref{clasdih.prop}.
According to Proposition~\ref{dih.prop}, the discriminant of $F_\chi$
is $1$ if $n\equiv1$~mod~$4$ and $\alpha\cdot {F^{\times}}^2$ if
$n\equiv3$~mod~$4$.

Any $\chi\in H^1(F,{\cal Z}_n)$ defines a Galois ${\cal Z}_n$-algebra
over $F$ which we will also denote by $F_\chi$.  The discriminant of
this extension is $1$.

We now turn to the Galois module $\mu_n$ of $n$-th roots of unity,
assuming that $n$ is relatively prime to the characteristic of $F$.
As above, we denote
$$
\mu_n(\alpha)=\mu_n\otimes_{\ZZ}\ZZ(\alpha).
$$
\begin{prop}
  \label{Gacomu.prop}
  Let $K=F(\sqrt{\alpha})$ be a quadratic field extension of $F$.  For
  any odd integer $n$ relatively prime to the characteristic of $F$,
  $$
  H^1(F,\mu_n(\alpha))=\{x\cdot K^{\times n}\in K^\times/K^{\times n}
  \colon x\overline{x}=1\}
  $$
  and $H^2(F,\mu_n(\alpha))$ classifies up to Brauer-equivalence
  central simple $K$-algebras of exponent $n$ which admit involutions
  of the second kind leaving $F$ elementwise invariant.
\end{prop}

\proof Applying Proposition~\ref{cohexseq.prop} with $M=\mu_n$ and
$\ell=1$ yields:
$$
H^1(F,\mu_n(\alpha))=\{x\cdot K^{\times n}\in K^\times/K^{\times
n}\colon x\overline{x}\in F^{\times n}\}.
$$
If $x\in K^\times$ is such that $x\overline{x}=\lambda^n$ with
$\lambda\in F^\times$, then
$$
x'=x(\lambda^{(n-1)/2}x^{-1})^n
$$
represents the same element of $K^\times/K^{\times n}$ and satisfies:
$x'\overline{x'}=1$.  This proves the first claim.

For $\ell=2$, Proposition~\ref{cohexseq.prop} yields a split exact
sequence:
$$
1\to H^2(F,\mu_n(\alpha))\to\nBr(K) \stackrel{\cores}{\longrightarrow}
\nBr(F)\to 1.
$$
The proposition follows, since by the theorem of
Albert--Riehm--Scharlau \cite{SchQHF} a central simple $K$-algebra
admits an involution of the second kind leaving $F$ elementwise
invariant if and only if its corestriction is trivial.  \qed

For $a \in K^\times$ with $a\overline{a}=1$, we denote the class of
$a$ in $H^1(F,\mu_n(\alpha))$ by $[a]$ and we define a commutative
\'etale $F$-algebra $F_a$ of dimension $n$ as follows: on the
$K$-algebra $K(y)$ where $y$ is an indeterminate subject to $y^n=a$,
define an $F$-automorphism $\theta$ extending $\invo$ on $K$ by
letting $\theta(y)=y^{-1}$.  Then $F_a$ is the $F$-subalgebra
$K(y)^\theta$ of invariant elements under $\theta$.  It is easily seen
that $F_a \simeq F_{a'}$ if $[a]=[a'] \in H^1(F,\mu_n(\alpha))$.  For
$[a]\in H^1(F,\mu_n)=F^\times/F^{\times n}$, $a\in F^{\times}$, we
also denote by $F_a$ the $F$-algebra $F(y)$ where $y^n=a$.

We now relate by cup-product the special cases ${\cal Z}_n$ and
$\mu_n$ considered above.  There is a canonical isomorphism: ${\cal
Z}_n\otimes_{\ZZ}\mu_n=\mu_n$ defined by:
$r^i\otimes\zeta\mapsto\zeta^i$.  Therefore, for $\delta_1,\delta_2\in
F^\times$,
$$
{\cal Z}_n(\delta_1)\otimes_{\ZZ}\mu_n(\delta_2) = {\cal
Z}_n\otimes\mu_n\otimes\ZZ(\delta_1)
\otimes\ZZ(\delta_2)=\mu_n(\delta_1\delta_2).
$$
Thus the cup-product defines a map:
$$
H^1(F,{\cal Z}_n(\delta_1))\times H^1(F,\mu_n(\delta_2))\to
H^2(F,\mu_n(\delta_1\delta_2)).
$$

\begin{prop}
  \label{constpc.prop}
  Let $K=F(\sqrt{\alpha})$ be a quadratic field extension of $F$ and
  let $\delta_1,\delta_2\in F^\times$ be such that
  $\delta_1\delta_2\equiv\alpha$~{\rm mod}~${F^{\times}}^2$.  Let
  $\chi\in H^1(F,{\cal Z}_n(\delta_1))$ and $a\in
  F(\sqrt{\delta_2})^\times$ such that
  $n_{F(\sqrt{\alpha_2})/F}(a)=1$, with cohomology class $[a]\in
  H^1(F,\mu_n(\delta_2))$.  There exists a central simple $K$-algebra
  $B(\chi,a)$ of degree~n such that
  \begin{enumerate}
    \item[(1)] $[B(\chi,a)] =
    \chi\cup [a]\in H^2(F,\mu_n(\alpha))$;
    \item[(2)] $B(\chi,a)$ admits an involution $\sigma$
    of the second kind such that $(B(\chi,a),\sigma)_+$ contains
    subalgebras isomorphic to $F_\chi$ and to $F_a$.
  \end{enumerate}
\end{prop}

\proof We first consider the special cases where $\delta_1\in
{F^{\times}}^2$ or $\delta_2\in {F^{\times}}^2$.  If $\delta_1\in
{F^{\times}}^2$, we have $\chi\in H^1(F,{\cal Z}_n)$ and $[a]\in
H^1(F,\mu_n(\alpha))$ with $a\in K^\times$ such that
$a\overline{a}=1$.  The algebra $F_\chi$ is a Galois ${\cal
Z}_n$-algebra over $F$.  The central simple algebra $B(\chi,a)$ is the
crossed product:
$$
B(\chi,a)=\bigoplus_{i=0}^{n-1}(F_\chi\otimes_FK) z^i,
$$
where the indeterminate $z$ is subject to the relations:
\begin{eqnarray*}
  z(\ell\otimes k)&=&[(r*\ell)\otimes k]z \quad\mbox{for }\ell\in
  F_\chi,\;k\in K, \quad\mbox{and }\ z^n=a.
\end{eqnarray*}
An involution of the second kind $\sigma$ on $B(\chi,a)$ is defined
by:
\begin{eqnarray*}
  \sigma(\ell\otimes k)&=&\ell\otimes\overline{k} \quad\mbox{for
  }\ell\in F_\chi,\;k\in K, \quad\mbox{and }\ \sigma(z)=z^{-1}.
\end{eqnarray*}
Clearly, $(B,\sigma)_+$ contains $F_\chi$ and the subalgebra
$F_a\subset K(z)$.

If $\delta_2\in {F^{\times}}^2$, then $\chi\in H^1(F,{\cal
Z}_n(\alpha))$ and $a\in F^\times$, $[a]\in H^1(F,\mu_n)$.  Let $L$ be
a Galois ${\cal Z}_n$-algebra over $K$ corresponding to $\chi$.
According to Proposition~\ref{clasdih.prop}, we may also choose a
Galois ${\cal D}_n$-algebra structure on $L$, viewed as an algebra
over $F$.  The algebra $B(\chi,a)$ is the crossed product:
$$
B(\chi,a)=\bigoplus_{i=0}^{n-1}L\,z^i
$$
where the indeterminate $z$ is subject to the relations:
\begin{eqnarray*}
  z\ell&=&(r*\ell)z\quad\mbox{for }\ell\in L, \quad\mbox{and }\ z^n=a.
\end{eqnarray*}
An involution of the second kind $\sigma$ on $B(\chi,a)$ is defined
by:
\begin{eqnarray*}
  \sigma(\ell)&=&s*\ell\quad\mbox{for }\ell\in L,\quad\mbox{and }\
  \sigma(z)=z.
\end{eqnarray*}
By definition, it is clear that $F_a=F(z)\subset(B,\sigma)_+$ and that
the subalgebra $L_1\subset L$ of invariant elements under $s$, which
is isomorphic to $F_\chi$, lies in $(B,\sigma)_+$.

Suppose now $\delta_1,\delta_2\not\in {F^{\times}}^2$.  Let
$F'=F(\sqrt{\delta_1})$ and $K'=K(\sqrt{\delta_1})$.  Let $L$ be a
Galois ${\cal Z}_n$-algebra over $F'$ which corresponds to $\chi\in
H^1(F,{\cal Z}_n(\delta_1));$ according to
Proposition~\ref{clasdih.prop}, we may endow it with a Galois ${\cal
D}_n$-algebra structure over $F$.  Since
$\delta_1\delta_2\equiv\alpha$~mod~${F^{\times}}^2$, we may identify
$F(\sqrt{\delta_2})$ with a subfield of $K'$.  Consider the following
crossed product algebra over $K'$:
$$
B'=\bigoplus_{i=0}^{n-1}(L\otimes_FK)\,z^i
$$
where $z$ is subject to:
\begin{eqnarray*}
  z(\ell\otimes k)&=&[(r*\ell)\otimes k]z\quad \mbox{for }\ell\in
  L,\;k\in K\ \ \mbox{and} \ \ z^n=a.
\end{eqnarray*}
This algebra represents the restriction $\res(\chi\cup [a])\in
H^1(F',\mu_n(\alpha))$.  We define an involution $\sigma'$ on $B'$ by:
\begin{eqnarray*}
  \sigma'(\ell\otimes k)&=&(s*\ell)\otimes\overline{k}\quad\mbox{for
  }\ell\in L,\;k\in K, \ \ \mbox{and} \ \ \sigma'(z)=z.
\end{eqnarray*}
On the other hand, we define a $K$-algebra automorphism $\theta$ on
$B'$ by:
\begin{eqnarray*}
  \theta(\ell\otimes k)&=&(s*\ell)\otimes k\quad\mbox{for }\ell\in
  L,\;k\in K,\ \ \mbox{and}\ \ \theta(z)=z^{-1}.
\end{eqnarray*}
The restriction of $\theta$ to the center $K'$ is the non-trivial
automorphism over $K$, and $\theta^2=I$; therefore, the subalgebra
$B={B'}^\theta$ of invariant elements under $\theta$ is a central
simple {$K$-algebra} such that
$$
B'=B\otimes_KK'.
$$
Since the restriction map $\res\colon H^2(F,\mu_n(\alpha))\to
H^2(F',\mu_n(\alpha))$ is injective (since $n$ is odd), it follows
that $B$ represents the cup product $\chi\cup [a]\in
H^2(F,\mu_n(\alpha))$ and we define $B(\chi,a)=B$.  Moreover, $\theta$
and $\sigma'$ commute, hence the restriction of $\sigma'$ to
$B(\chi,a)$ defines an involution $\sigma$ of the second kind.  The
subalgebra $L_1$ of $L$ elementwise invariant under $s$ and the
subalgebra $K'(z)^\theta$ invariant under $\theta$ both lie in
$(B(\chi,a),\sigma)_+$; the involution $\sigma$ thus satisfies the
required properties, since $L_1\simeq F_\chi$ and $K'(z)^\theta\simeq
F_a$.  \qed

\begin{prop}
  \label{Rpc.prop}
  Let $B$ be a central simple $K$-algebra of degree $n$ (odd) which
  admits involutions of the second kind leaving $F$ elementwise
  invariant, and let $\delta\in F^\times$.  If $B$ contains a subfield
  isomorphic to $F_\chi$ for some $\chi\in H^1(F,{\cal Z}_n(\delta))$,
  then $B\simeq B(\chi,a)$ for some $a\in
  F(\sqrt{\alpha\delta})^\times$ such that
  $n_{F(\sqrt{\alpha\delta})/F}(a)=1$.
\end{prop}

\proof Suppose first $\delta\in {F^{\times}}^2$.  Then $F_\chi\otimes
K$ is a maximal subfield of $B$ which is cyclic over $K$, hence $B$ is
a cyclic algebra:
\begin{equation}
  \label{Beqres.eq}
  [B]=\res_{K/F}(\chi)\cup[b]\in H^2(K,\mu_n)
\end{equation}
for some $b\in K^\times$ (so that $[b]\in H^1(K,\mu_n)$), where $[B]$
denotes the image of $B$ in $H^2(K,\mu_n)$.  Since $B$ admits
involutions of the second kind over $F$, we have $\cores_{K/F}(B)=0$;
therefore,
\begin{equation}
  \label{chicor.eq}
  \chi\cup\cores_{K/F}([b])=0 \quad\mbox{in }H^2(F,\mu_n).
\end{equation}
Since $n$ is odd, we may find $m\in\ZZ$ such that $2m\equiv1$~mod~$n$.
Then
$$
b\equiv b^{2m}= (b{\overline{b}}^{-1})^m n_{K/F}(b)^m \quad\mbox{mod
}{K^{\times}}^n;
$$
therefore, letting $a=(b{\overline{b}}^{-1})^m$, we have $[a]\in
H^1(F,\mu_n(\alpha))\,$ and
$$
[b]=[a]+m\,\res_{K/F}\circ\cores_{K/F}([b]) \quad\mbox{in
}H^1(K,\mu_n).
$$
Substituting in equation~(\ref{Beqres.eq}), we get:
$$
[B]=\res_{K/F}(\chi)\cup[a]+m\,\res_{K/F}(\chi \cup\cores_{K/F}([b])),
$$
hence $B\simeq B(\chi,a)$, in view of (\ref{chicor.eq}).  Suppose next
$\delta\not\in {F^{\times}}^2$ and let $F'=F(\sqrt{\delta})$.  Since
$\delta\in {{F'}^\times}^2$, the case already considered yields:
$$
[B\otimes F']=\res_{F'/F}(\chi)\cup [a']
$$
for some $[a']\in H^1(F',\mu_n(\alpha\delta))$.  Applying the
corestriction map to both sides of this equation, we get by the
projection formula:
$$
2[B]=\chi\cup\cores_{F'/F}([a']).
$$
Therefore, if $2m\equiv1$~mod~$n$,
$$
[B]=m\,\chi\cup\cores_{F'/F}([a']),
$$
hence $B\simeq B(\chi,a)$ for $[a]=m\,\cores_{F'/F}([a'])\in
H^1(F,\mu_n(\alpha\delta))$.  \qed

In a different direction, we have the following result on the Galois
cohomology of the twisted modules $\mu_n(\alpha)$:

\begin{prop}
  \label{Gaco.prop}
  Let $\delta_1,\delta_2\in F^\times$.  In
  $H^2(F,\mu_n^{\otimes2}(\delta_1\delta_2))$, we have:
  $$
  H^1(F,\mu_n(\delta_1))\cup H^1(F,\mu_n(\delta_2)) \subset
  H^1(F,\mu_n)\cup H^1(F,\mu_n(\delta_1\delta_2)),
  $$
  where the left-hand side is
  $$
  \{x\cup y\colon x\in H^1(F,\mu_n(\delta_1)) , \,y\in
  H^1(F,\mu_n(\delta_2))\}.
  $$
\end{prop}

\proof Assume first that $\delta_1,\delta_2,\delta_1\delta_2\not\in
F^{\times 2}$.  For $i=1,2$, let $F_i=F(\sqrt{\delta_i})\subset F_s$
and let $a_i\in F_i^\times$ be such that $n_{F_i/F}(a_i)=1$, so that
$[a_i]\in H^1(F,\mu_n(\delta_i))$.  If $a_i\in F^{\times n}$, then
$[a_i]=0$, hence $[a_1]\cup [a_2]=0\in H^1(F,\mu_n)\cup
H^1(F,\mu_n(\delta_1\delta_2))$.  For the rest of the proof, we may
thus assume $a_i\neq\pm1$ for $i=1,2$.  Let
$$
u=\frac{a_1-1}{a_1+1}\in F_1^\times \quad\mbox{and}\quad
v=\frac{a_2+1}{a_2-1}\in F_2^\times,
$$
so that
$$
a_1= \frac{1+u}{1-u} \quad\mbox{and}\quad a_2=\frac{v+1}{v-1}.
$$
Since $n_{F_i/F}(a_i)=1$, it follows that $u$ and $v$ have trace zero.
Therefore, $u^2,v^2\in F^\times$ and $uv^{-1}\in
F(\sqrt{\delta_1\delta_2})^\times$.  Let
$$
[b]=\frac{1-v^2}{1-u^2}\cdot F^{\times n}\in H^1(F,\mu_n)
$$
and
$$
[c]=\frac{1+uv^{-1}}{1-uv^{-1}}\cdot
F\left(\sqrt{\delta_1\delta_2}\right)^{\times n}\in
H^1(F,\mu_n(\delta_1\delta_2)).
$$
We claim that
$$
[a_1]\cup [a_2]=[b]\cup [c]\quad\mbox{in
}H^2(F,\mu_n^{\otimes2}(\delta_1\delta_2)).
$$
Let $M=F(\sqrt{\delta_1},\sqrt{\delta_2})\subset F_s$.  Since the
degree of the extension $M/F$ is relatively prime to $n$, the
restriction map
$$
\res_{M/F}\colon H^2(F,\mu_n^{\otimes2}(\delta_1\delta_2)) \to
H^2(M,\mu_n^{\otimes2}(\delta_1\delta_2))= H^2(M,\mu_n^{\otimes2})
$$
is injective; therefore, it suffices to prove the claim over $M$.  The
identity
$$
\left\{ \frac{1+u}{1-u}; \frac{v+1}{v-1} \right\} = \left\{
\frac{1-v^2}{1-u^2}; \frac{1+uv^{-1}}{1-uv^{-1}} \right\}.
$$
holds for symbols $\{a,b\}$ in Milnor's $K_2$, a proof is given in the
following Lemma \ref{beautiful.lem}.

Applying the norm residue homomorphism $K_2M\to
H^2(M,\mu_n^{\otimes2})$ to both sides, we get
$$
[a_1]\cup [a_2]=[b]\cup [c]\quad\mbox{in } H^2(M,\mu_n^{\otimes2})=
H^2(M,\mu_n^{\otimes2}(\delta_1\delta_2)),
$$
completing the proof in the case where
$\delta_1,\delta_2,\delta_1\delta_2\not\in {F^{\times}}^2$.  If
$\delta_1$ or $\delta_2\in {F^{\times}}^2$, the proposition is
obvious.  If $\delta_1\delta_2\in {F^{\times}}^2$, we may use the same
arguments as above, substituting $F$ for $F(\sqrt{\delta_1\delta_2})$,
except if $u=\pm v$.  In this case however, we have $a_1=-a_2^{\pm1}$,
hence $[a_1]=[a_2]^{\pm1}$ and therefore $[a_1]\cup [ a_2]=0$.  \qed

\begin{lem} \label{beautiful.lem}
  For $u,v \neq \pm 1 \in F^\times, u\neq \pm v$, the relation
  $$
  \left\{ \frac{1+u}{1-u}; \frac{v+1}{v-1} \right\} = \left\{
  \frac{1-v^2}{1-u^2}; \frac{1+uv^{-1}}{1-uv^{-1}} \right\}
  $$
  holds in $K_2F$.
\end{lem}

\proof We have $\{1-u;1+uv\inv\} + \{1-u;v\}= \{1-u;u+v\} $.  On the
other hand the basic relation $\{a;b\} = \{a;b-a\}+ \{a-b; b\}$ in
$K_2$ (see for example \cite[p.~75]{K}) implies that
\begin{eqnarray*}
  \{1-u;u+v\}&=& \{1-u;1+v\} -\{-u-v;1+v\} \\
  &=& \{1-u;1+v\} + \{1+v; -u-v\} .
\end{eqnarray*}
Subtracting $0=\{1+v; -v\}$ we get
$$
\{1-u; 1+uv\inv\}- \{1+v; 1+uv\inv\}= \{1-u;1+v\} - \{1-u;v\}.
$$

Replacing $u,v$ by $\pm u,\pm v$, we get four expressions which adds
to the wanted relation.  Another proof is to check the formula in the
function field $F(u,v)$ by using the exact sequence of Milnor for
$K_2F(t)$ (see \cite{Miln}) to show that
$$
\left\{{{1+u} \over{1-u}}; {{v+1} \over{v-1}} \right\} - \left\{
{{1-v^2} \over{1-u^2}}; {{1+uv^{-1}} \over{1-uv^{-1}}} \right\}
$$
lies in $K_2F(u)$, $K_2F$ and then to show that it is $0$ by
specialization at some places.  \qed

{\em Remark.\/} The Galois module $\mu_n$ plays a special r\^ole in
the preceding proposition.  Suppose that $\delta_1$, $\delta_2$,
$\varepsilon_1$, $\varepsilon_2\in F^\times$ are such that
$$
\mu_n(\delta_1)\otimes\mu_n(\delta_2)=\mu_n(\varepsilon_1)
\otimes\mu_n(\varepsilon_2),
$$
and that the sets $\{\mu_n(\delta_1),\mu_n(\delta_2)\}$ and
$\{\mu_n(\varepsilon_1),\mu_n(\varepsilon_2)\}$ are disjoint (up to
isomorphism).  If $\mu_n(\varepsilon_1)$ and $\mu_n(\varepsilon_2)$
are different from $\mu_n$, then there is an extension $\tilde F$ of
$F$ such that
$$
H^1(\tilde F,\mu_n(\delta_1))\cup H^1(\tilde F,\mu_n(\delta_2))
\not\subset H^1(\tilde F,\mu_n(\varepsilon_1))\cup H^1(\tilde
F,\mu_n(\varepsilon_2))
$$
Indeed, after a biquadratic extension of $F$ we may assume that
$\delta_1,\delta_2\in {F^{\times}}^2$.  Let then $\tilde F$ be the
iterated powerseries extension $F((s))((t))$.  Then
$$
s\cdot\tilde F^{\times n}\cup t\cdot \tilde F^{\times n}\in H^1(\tilde
F,\mu_n)\cup H^1(\tilde F,\mu_n)
$$
has non-trivial residue.  On the other hand,
$$
\partial(H^1(\tilde F,\mu_n(\varepsilon)))\subset H^0( F,{\cal
Z}_n(\varepsilon))=0
$$
for nontrivial $\varepsilon$, hence $ s\cdot\tilde F^{\times n}\cup
t\cdot\tilde F^{\times n}$ cannot be a sum of cup-products of elements
of $H^1(\tilde F,\mu_n(\varepsilon_1))$ and $H^1(\tilde
F,\mu_n(\varepsilon_2)).  $

\begin{cor}
  \label{dihRcyc.cor}
  Let $n$ be an odd integer and let $F$ be a field of characteristic
  relatively prime to $n$.  Assume that $\mu_n={\cal
  Z}_n(\varepsilon)$ for some $\varepsilon\in F^\times$; then every
  central simple $F$-algebra split by a Galois extension of degree
  $2n$ with dihedral Galois group is cyclic.
\end{cor}

\proof Let $A$ be a central simple $F$-algebra split by a Galois
extension $L/F$ with dihedral Galois group of order $2n$.  The index
of $A$ then divides $2n$.  Since $n$ is odd, we may decompose the
Brauer class of $A$:
$$
[A]=\alpha_1+\alpha_2
$$
where $\alpha_1$ has index $1$ or $2$ and $\alpha_2$ has index
dividing $n$.  The element $\alpha_1$ is split by an cyclic extension
$C_1/F$ of degree $1$ or $2$.  If we show that $2[A]$ is split by a
cyclic extension $C_2/F$ of degree dividing $n$, then $\alpha_2$ is
also split by $C_2$, hence $A$ is split by the cyclic extension
$C_1\otimes C_2/F$.  Therefore, it suffices to show that
$A^{\otimes2}$ is cyclic.  Let $\delta\in F^\times$ be a
representative of the discriminant of $L$.  Proposition~\ref{dih.prop}
shows that $L$ is a cyclic extension of $F(\sqrt{\delta})$.  Choosing
a Galois ${\cal Z}_n$-algebra structure of $L$ over
$F(\sqrt{\delta})$, we may associate to $L$ an element $\chi\in
H^1(F,{\cal Z}_n(\delta))$ by Proposition~\ref{clasdih.prop}.  Since
$A\otimes F(\sqrt{\delta})$ is split by $L$, we have
$$
[A\otimes_FF(\sqrt{\delta})]= \res_{F(\sqrt{\delta})/F}(\chi)\cup [a]
\quad\mbox{in }H^2(F(\sqrt{\delta}),\mu_n)
$$
for some $[a]\in H^1(F(\sqrt{\delta}),\mu_n)=
H^1(F(\sqrt{\delta}),\mu_n(\delta))$.  Taking the corestriction of
both sides, we get
\begin{equation}
  \label{twoA.eq}
  2[A]=\chi\cup [a']\quad\mbox{in }H^2(F,\mu_n),
\end{equation}
where $[a']=\cores_{F(\sqrt{\delta})/F}([a])\in H^1(F,\mu_n(\delta))$.
Since $\mu_n={\cal Z}_n(\varepsilon)$, we have ${\cal
Z}_n(\delta)=\mu_n(\varepsilon\delta)$, hence relation~(\ref{twoA.eq})
yields:
$$
2[A]\in H^1(F,\mu_n(\varepsilon\delta))\cup H^1(F,\mu_n(\delta)).
$$
From Proposition~\ref{Gaco.prop}, it follows that
$$
2[A]=[b]\cup [c]
$$
for some $[b]\in H^1(F,\mu_n)$ and some $[c]\in
H^1(F,\mu_n(\varepsilon))=H^1(F,{\cal Z}_n)$.  The element $[c]$ then
defines a cyclic extension of $F$ which splits $A^{\otimes2}$.  \qed

We conclude by discussing the special features of the case where $n=3$
(assuming that the characteristic of $F$ is not $2$ nor $3$).  Every
cubic field extension $L$ of $F$ is of the form $F_\chi$ and also of
the form $F_a$: more precisely, if the discriminant of $L/F$ is
represented by $\alpha\in F^\times$, then the field
$L\otimes_FF(\sqrt{\alpha})$ is cyclic over $F(\sqrt{\alpha})$, hence
$L=F_\chi$ for any $\chi\in H^1(F,{\cal Z}_3(\alpha))$ representing
$L\otimes_FF(\sqrt{\alpha})$.  Since $\mu_3={\cal Z}_3(-3)$, we may
also view $L=F_a$ for $[a]=\chi\in H^1(F,\mu_3(-3\alpha))$.

\begin{prop}
  \label{degthree.prop}
  Let $K=F(\sqrt{\alpha})$ be a quadratic field extension of $F$ and
  let $B$ be a central division $K$-algebra of degree~3 which admits
  involutions of the second kind leaving $F$ elementwise invariant.
  Let also $L$ be a cubic field extension of $F$ with discriminant
  $\delta$ and let $\chi\in H^1(F,{\cal Z}_3(\delta))$ be such that
  $L=F_\chi$.  The algebra $B$ contains an isomorphic image of $L$ if
  and only if $B\simeq B(\chi,a)$ (i.e.\ $[B]=\chi\cup [a]$ in
  $H^2(F,\mu_3(\alpha))$) for some $[a]\in
  H^1(F,\mu_3(\alpha\delta))$.  The algebra $B$ then contains a cubic
  field extension $L'=F_a$ of discriminant $-3\alpha\delta$.
  Moreover, there exists an involution $\sigma$ such that
  $L,L'\subset(B,\sigma)_+$.

  The algebra $B$ always contains cubic field extensions $L_*$, $L'_*$
  of discriminant $-3$, $\alpha$ respectively, and carries an
  involution $\sigma_*$ such that $L_*,L'_*\subset(B,\sigma_*)_+$.
\end{prop}

\proof The first part follows from Propositions~\ref{constpc.prop} and
\ref{Rpc.prop}.  In order to prove that $B$ always contains extensions
$L_*$, $L'_*$ as stated, we use Proposition~\ref{Gaco.prop}: let $L$
be any cubic field extension of $F$ contained in $B$; then, letting
$\delta\in F^\times$ denote a representative of the discriminant of
$L/F$, we have
$$
B\simeq B(\chi,a)
$$
for some $\chi\in H^1(F,{\cal Z}_3(\delta))$ and some $[a]\in
H^1(F,\mu_3(\alpha\delta))$.  Since ${\cal
Z}_3(\delta)=\mu_3(-3\delta)$, it follows that
$$
[B]\in H^1(F,\mu_3(-3\delta))\cup H^1(F,\mu_3(\alpha\delta)),
$$
hence Proposition~\ref{Gaco.prop} shows that $B \simeq B(\chi_*,a_*)$
for some $\chi_*\in H^1(F,\mu_3)=H^1(F,{\cal Z}_3(-3))$ and some
$[a_*]\in H^1(F,\mu_3(-3\alpha))=H^1(F,{\cal Z}_3(\alpha))$.  We then
let $L_*=F_{\chi_*}$ and $L'_*=F_{a_*}$.  \qed

Note that, according to condition~{\em(e)\/} of
Theorem~\ref{distinv.thm}, the involution $\sigma_*$ is distinguished;
the proposition above thus yields another proof of the existence of
distinguished involutions.

Note also that the pair $(-3,\alpha)$ is the only pair of
discriminants that occurs for every central division
$F(\sqrt{\alpha})$-algebra of degree~3 which admits involutions of the
second kind: given a pair $(\varepsilon_1,\varepsilon_2)$ with
$\varepsilon_1\varepsilon_2 \in (-3\alpha)\cdot {F^\times}^2$, the
remark following Proposition 25 together with Proposition 27 shows
that there is an algebra $B$ over some extension of $F$ which does not
contain any cubic separable extension of discriminant $\varepsilon_1$
or $\varepsilon_2$.  For example, there are algebras $B$ which do not
contain a cyclic cubic field extension, i.e.\ a cubic field extension
of discriminant~$1$, as already observed in the example at the end of
section~4.

Since $3$-fold Pfister forms are classified up to isometry by their
Arason invariant:
$$
{\rm Ar}(\Pf a,b,c\pF)= (a\cdot {F^{\times}}^2)\,\cup\, (b\cdot
{F^{\times}}^2)\,\cup\, (c\cdot {F^{\times}}^2)\,\in H^3(F,\mu_2),
$$
(see \cite{EL}), Theorem~\ref{clasinv.thm} (together with
Proposition~\ref{Gacomu.prop}) yields a classification of central
simple algebras of degree~3 with involution by cohomological
invariants:

\begin{cor}
  Triples $(K,B,\sigma)$ where $K=F(\sqrt{\alpha})$ is a quadratic
  \'etale $F$-algebra, $B$ is a central simple $K$-algebra of degree~3
  and $\sigma$ is an involution of the second kind on $B$ leaving $F$
  elementwise invariant, are classified over $F$ by the three
  cohomological invariants:
  $$
  \begin{array}{lcl}
    f_1(K,B,\sigma)&=&\alpha\cdot {F^{\times}}^2\in H^1(F,\mu_2),
    \\
    g_2(K,B,\sigma)&=&[B]\in H^2(F,\mu_3(\alpha)),
    \\
    f_3(K,B,\sigma)&=&{\rm Ar}(\pi(B,\sigma))\in H^3(F,\mu_2).
  \end{array}
  $$
\end{cor}

\proof

Indeed we have seen that $f_1$ determines $K$, $g_2$ determines $B$
and $f_3 $ determines the involution $\sigma$.  \qed

{\em Remark.\/} Let $K_0/F_0$ be a separable quadratic field extension
and let $B_0$ be a central division $K_0$-algebra of degree $3$ which
admits an involution of the second kind.  Let $F=F_0((t)),\,K=K
_0((t))$ and $B= B_0((t))$.  Let $\sigma$ be any involution on $B$,
not necessarily defined over $B_0$.  We claim that the second residue
$\varphi^t(Q_\sigma)$ is always trivial, i.e.\ $Q_\sigma$ is induced
from $F_0$.  A cubic \'etale $F$-subalgebra $L$ of $B$ is either
unramified or totally ramified, i.e.\ $L=F[X]/(X^3-ut)$ for $u \in
F_0^\times$.  Thus in both cases the discriminant of $L$ is a class
defined over $F_0$.  On the other hand, an element $\lambda\in L$ such
that $n_{L/F}(\lambda)$ is a square in $F^\times$ must be a unit in
$F_0[[t]]$.  This, together with Proposition~\ref{diagN.prop}, implies
that $\varphi^t(Q_\sigma)=0$.  It follows that the invariant
$f_3(B,\sigma)$ cannot in general be arbitrary.  In particular the
three invariants $f_1,\, g_2$ and $f_3$ are not independent.  A
corresponding question for invariants of exceptional Jordan algebras
is discussed in \cite{Se2}.

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\end{thebibliography}

\medskip
\begin{tabular}{l}
  Darrell E. Haile\\
  Department of Mathematics\\
  Indiana University\\
  Bloomington, IN 47405\\
  USA\\
  haile@ucs.indiana.edu
\end{tabular}

\medskip
\begin{tabular}{l}
  Max-Albert Knus\\
  Departement Mathematik\\
  ETH-Zentrum\\
  CH-8092 Z\"urich\\
  Switzerland\\
  knus@math.ethz.ch
\end{tabular}

\medskip
\begin{tabular}{l}
  Markus Rost\\
  NWFI-Mathematik\\
  Universit\"at Regensburg\\
  D-93053 Regensburg\\
  Germany\\
  markus.rost@mathematik.uni-regensburg.de
\end{tabular}

\medskip
\begin{tabular}{l}
  Jean-Pierre Tignol\\
  Institut de Math\'ematique Pure et Appliqu\'ee\\
  Universit\'e catholique de Louvain\\
  B-1348 Louvain-la-Neuve\\
  Belgium\\
  tignol@agel.ucl.ac.be
\end{tabular}
\end{document}
