%%% % Archiv version: copied from main copy on 25/7/2002; 
%%% % 'further ideas' section deleted.
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%%% % Galois cohomology for fields without roots of unity, and applications.
%%% % Started 6-7/2001, cont and written 1-7/2002.
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\begin{document}
%\baselineskip = 18pt


\title%
[Galois Cohomology without unity roots] {Galois Cohomology of
Fields Without Roots of Unity}

\author{ Uzi Vishne }

\address{Department of Mathematics, Yale University, 10 Hillhouse Avenue, New-Haven CT 06520, USA}

\email{uv2@math.yale.edu}

\thanks{I thank J.-P.~Tignol and L.H.~Rowen for many stimulating
discussions. I gratefully acknowledge a post-doctoral grant from
the Fulbright program, United States Department of State.}

\subjclass[2000]{16K20, 12G05}
% 16K20: Division rings, Finite-dimensional
% 12G05: Galois cohomology
%\keywords{linear operator, invariant subspace, transitive algebra}

\date{\today}
%\date{Jul 15, 2002.}

%\dedicatory{}




\begin{abstract}
\iffurther {\large Notice: this version has the 'further ideas'
section.\\ } \fi %
 There is a standard correspondence between elements of
the cohomology group $\H1{F}{\mu_n}$ (with the trivial action of
$\G[F] = \Gal(\sep{F}/F)$ on $\mu_n$) and cyclic extensions of
dimension $n$ over $F$. We extend this to a correspondence between
the cohomology groups $\H1{F}{\mu_n}$ where the action of $\G[F]$
on $\mu_n$ varies, and the extensions of dimension $n$ of $K$
which are Galois over $F$, where $K = F[\mu_n]$ and
$\dimcol{K}{F}$ is prime to $n$. The cohomology groups are also
related to eigenspaces of $\H1{K}{\Zn}$ with respect to the
natural action of $\Gal(K/F)$.

As a result, we extend Albert's cyclicity criterion, stated in the
1930s for division algebras of prime degree, to algebras of
prime-power degree over $F$, under the assumption stated above. We
also extend the Rosset-Tate result on the corestriction of cyclic
algebras in the presence of roots of unity, to extensions in which
roots of unity live in an extension of dimension $\leq 3$ over the
base field. In particular if roots of unity live in a quadratic
extension of the base field, then corestriction of a cyclic
algebra along a quadratic extension is similar to a product of two
cyclic algebras.  Another application is that $F$-central algebras
which are split by a certain semidirect product extension of $F$,
are cyclic. In particular, if $\dimcol{K}{F}=2$ then algebras over
$F$ which are split by an odd order dihedral extension, are
cyclic.

We also construct generic examples of algebras which become cyclic
after extending scalars by roots of unity, and show the existence
of elements for which most powers have reduced trace zero.
\end{abstract}

\maketitle

%-----------------------------------------------------------------



\section{Introduction}

The purpose of this paper is to generalize some standard
constructions of Galois cohomology to fields without roots of
unity. Let $n$ be a prime power, $F$ a field of characteristic
prime to $n$, and $K = F[\rho]$, where $\rho$ is a primitive unity
root of order $n$. Let $\sep{F}$ denote the separable closure of
$F$. %
In \cite{Merk83} and \cite{Merk86} A.~Merkurjev use decomposition
into eigenspaces of the cohomology groups of $K$ to study the
relation between the Brauer group of $F$ to the Brauer group of
$K$, which is better understood. We apply his technique first to
the first cohomology group $\H1{K}{\Zn}$, and use the cup product
to study the Brauer groups. This somewhat different approach
provides numerous new results on central simple algebras over $F$.

We start with a decomposition of the cohomology group
$\H1{K}{\Zn}$ to eigenspaces under the natural action of
$\Gal(K/F)$. It turns out that the components are naturally
isomorphic to the cohomology groups $\H[\chi]1{F}{\mu_n}$, where
this notation stands for the cohomology group of $\G[F]$ with an
action on $\mu_n$ which is twisted by a character $\chi$ of
$\Gal(K/F)$ (so in particular the restricted action of $\G[K]$ is
trivial).

It is well known that the cocycles of order $n$ in
$\H1{F}{\mu_n}$, with the trivial action, correspond to cyclic
Galois extensions of dimension $n$ over $F$. In a similar manner
we interpret the other groups $\H[\chi]1{F}{\mu_n}$, in terms of
field extensions of dimension $n$ over $K$ which are Galois over
$F$.

The theorem of Merkurjev-Suslin is often quoted as saying that if
$K$ contains $n$ roots of unity, then $\Brn(K) \isom \Kn{n}{K}$
(where $\Brn(K)$ is the exponent $n$ part of the Brauer group and
$\KK(K)$ is the second Milnor $\KK[]$-group of $K$). In fact, the
theorem holds more generally, giving
 the isomorphism $\H2{F}{\mu_n} \isom \Kn{n}{F}$ whenever $F$ has
characteristic prime to $n$ (see \cite[Sec.~8]{Srinivas}). Since
$\Brn(F)=\H2{F}{\mu_n^{\tensor 2}}$, this is useful (in terms of
Brauer group theory)  mainly when the two actions coincide, \ie\
when $\mu_n \sub F$.

Our description of the various cohomology groups $\H1{F}{\mu_n}$
can be used, taking cup products, to provide information on the
groups $\H[\chi]2{F}{\mu_n}$, including $\H2{F}{\mu_n}$ and
$\H2{F}{\mu_n^{\tensor 2}}$ which are both isomorphic to
eigenspaces of $\H2{K}{\mu_n}$. This approach is similar to that
of \cite{HKRT}, where the authors study cohomology groups twisted
by quadratic extensions of $F$. They handle arbitrary quadratic
extensions, while here the focus is on the extension of $F$ by the
$n$-roots of unity; however, the main technique here,
decomposition into eigenspaces of characters (from the Galois
group to $U_n$), should work for arbitrary Abelian Galois
extensions with dimension prime to $n$. The price paid in
\cite{HKRT} for the generality is that the characteristic of $F$
is assumed to be different than $2$, while here we do not make
this assumption. More recently, decomposition into eigenspaces
with respect to the action of $\Gal(K/F)$ is used in \cite{Wad} to
study cohomology groups. The main focus of this paper is in the
Galois groups of maximal $p$-extensions, and it also develops
methods to avoid the assumption that $\dimcol{K}{F}$ is prime to
$n$. Our Theorem \ref{Albert} is Theorem 3.6 in \cite{Wad}. % And their Prop. 3.4 is the characterization of cyclic algebras.

 Using a projection formula for the corestriction of cyclic
algebras from $K$ to $F$, we show that if $D$ is a division
algebra of degree $n$ over $F$ which restricts to a cyclic algebra
$\symbalg[K]{a}{\beta}$ with $a\in K$ and $\beta \in F$, then $D$
is cyclic over $F$. One application is a generalization of
Albert's cyclicity criterion, that a central simple algebra of
degree $n$ with an $n$-central elements is cyclic, from prime
degree to any prime power degree, assuming $\dimcol{K}{F}$ is
prime to $n$.

An algorithm due to Rosset and Tate shows that the corestriction
of a cyclic algebra from $K_2$ to $K_1$, when $K_1$ has enough
roots of unity, is similar to at most $\dimcol{K_2}{K_1}$ cyclic
algebras over $K_1$. A similar result follows for separable
extensions $L/F$ when we assume $\dimcol{K}{F}\leq 3$, namely that
in this case the corestriction of a cyclic algebra over $L$ is
similar to at most $\dimcol{K}{F}\cdot (\dimcol{L}{F}-1)+1$ cyclic
algebras over $F$. If $\dimcol{K}{F} = 2$ and $L/F$ is a quadratic
extension, then the corestriction is a product of two (and not
three) cyclic algebras.

If roots of unity are present in the base field, then it is known
\cite{RSsemi} that algebras with certain semidirect product
splitting fields are cyclic. It is easy to show that $\Brn(F)$ is
generated by algebras which we call \defin{quasi-symbols}, which
are algebras of degree $n$ over $F$, which when restricted to $K$
are cyclic of some special form. We show that a cyclicity result
for algebras with semi-direct product splitting field is
equivalent to the cyclicity of quasi-symbols, which is not known
to hold. We find splitting fields of smaller dimension for certain
quasi-symbols, and in particular show that algebras with splitting
field containing $K$ whose Galois group is $G =
\sg{\s,\tau\subjectto \s^n = \tau^d = 1, \tau\s\tau^{-1} = \s^t}$
(where $\tau$ is a generator of $\Gal(K/F)$ such that $\tau(\rho)
= \rho^t$), are cyclic. In a related direction, we extend the
theorem of Rowen and Saltman \cite{RSdi} that dihedral algebras
are cyclic, where they assume the base field contains $n$-roots of
unity, to dihedral algebras over $F$ in which the roots of unity
live in a quadratic extension of the base field. Moreover,
$F$-algebras with dihedral splitting field which intersects $K$
non-trivially are quasi-symbols.

The paper is organized as follows. After setting the notation of
decomposition into eigenspaces in Section \ref{sec2}, we decompose
the isomorphic $\Gal(K/F)$-modules $\md{K}$ and $\H1{K}{\Zn}$ in
Sections \ref{sec3} and \ref{Kco}, respectively. We derive some
technical results on representatives of classes in $\md{K}$, which
are frequently used later on.

In Section \ref{Fco} we explicitly describe an isomorphism from an
eigenspace of $\md{K}$ to a group $\H[\chi]1{F}{\mu_n}$, for a
suitable character $\chi$. Also, the eigenspaces of $\H1{K}{\Zn}$
are shown to be isomorphic to those $\H[\chi]1{F}{\mu_n}$. The
isomorphisms are natural in the sense that they commute with the
restriction map. The correspondence between the first cohomology
groups of $F$ with coefficients in $\mu_n$ to Galois extensions of
$F$ which are cyclic over $K$ is presented in Section
\ref{fields}, where we also give explicit formulas for the inverse
of the isomorphisms presented in previous sections. We always
assume that $\dimcol{K}{F}$ is prime to $n$. An example in which
this is not the case is given in Section \ref{d=2}, where we
explain how the constructions fail without this assumption.

The cup products enables to lift the information we gather on the
first cohomology groups, to the groups $\H2{F}{\mu_n}$ (again for
various actions of $\G[F]$ on $\mu_n$), and in Section \ref{csa}
we compute the corestriction of cyclic algebras from $K$ to $F$,
and show that the various restriction maps take the groups
$\H2{F}{\mu_n}$ to distinct eigenspaces of $\H2{K}{\Zn}$, which is
the Brauer group of $K$. In Section \ref{secact} we extend the
action of $\Gal(K/F)$ to certain cyclic algebras over $K$.

The action of $\Gal(K/F)$ on cyclic algebras is applied in
Sections \ref{sec:alb}--\ref{sec:sd} to derive the results
described above.

Finally in Section \ref{secgeneric} we present a candidate for
being a non-cyclic algebra of prime degree over fields without
roots of unity. We also show that some generic algebras of this
type have elements $w$ for which the reduced trace $\Tr(w^\ell)$
is zero for most values of $\ell < n$.

\smallskip

We use the notation suggested by J.-P.~Tignol and S.A.~Amitsur
\cite{TA85}, that a central simple algebra $A$ is \defin{split by}
a group $G$ is $A$ is similar (in the Brauer sense) to a crossed
product with respect to a subgroup of $G$. In particular $A$ is
split by the cyclic group $C_n$ iff it is similar to a cyclic
algebra of degree dividing $n$.

Frequent use is made of the fact that if the index $\dimcol{G}{H}$
is prime to $n$ and $M$ is a $G$-module of exponent $n$, then the
restriction of cohomology groups (in particular Brauer groups)
from $G$ to $H$ is injective, and the corestriction (from $H$ to
$G$) is onto.




\section{Decomposition into Eigenspaces}\label{sec2}

Let $G$ be a finite Abelian group, and $M$ a faithful $G$-module
of finite torsion $n$ (\ie\ $n\cdot M = 0$). Since we are
interested in decomposing $M$, we assume that $n = p^m$ is a prime
power.

Let $U_n = (\Zn)^{\times}$ denote the Euler group of $n$. For a
multiplicative character  $\phi \colon G \ra U_n$, we define a
homomorphism of $G$-modules $\tr_\phi \co M \ra M$, by
\begin{equation}
\tr_\phi(a) = \sum_{\tau \in G} \phi(\tau)^{-1} \tau(a).
\end{equation}
Let
$$M^{(\phi)} = \set{a \in M\suchthat \tau(a) = \phi(\tau) \cdot
a}$$ be the eigenspace of $\phi$, and observe that $\Im(\tr_\phi) \sub M^{(\phi)}$. % Checked.
Note that for $a \in M^{(\phi)}$ we have $\tr_\phi(a) = \power{G}
\cdot a$, so in particular if $\power{G}$ is prime to $n$,
$\tr_{\phi} \co M \ra M^{(\phi)}$ is onto.

The following easy observation will be frequently used.
\begin{rem}\label{gmodp}
If $1\neq g \in U_n$ has order prime to $p$, then $g \not\equiv 1
\pmod{p}$.
\end{rem}
\begin{proof}
The kernel of the projection $U_n \ra U_p$ is of order $n/p$ which
is a $p$-power, and by assumption $g$ is not in this kernel.
%% Old, direct, proof:
%
% Write $g = 1+t p^i$ for maximal $i\geq 0$, and let $d$ denote the
% order of $g$ in $U_n$. We have $p^i<n$ since $g \neq 1$. If $i>0$,
% then modulo $p^{i+1}$ we have $1 \equiv g^d = (1+t p^i)^d \equiv
% 1+tdp^i$ so $p\divides t$, a contradiction to the maximality of
% $i$.
\end{proof}

\begin{cor}\label{Hsg}
If $H \leq U_n$ is a non-trivial subgroup and $d = \power{H}$ is
prime to $p$, then $s = \sum_{h\in H}{h} \equiv 0 \pmod{n}$.
\end{cor}
\begin{proof}
Choose some $1 \neq g \in H$, then $g s = \sum{gh} = s$, so
$(g-1)s \equiv 0 \pmod{n}$, but $g-1$ is prime to $p$ by the last
remark.
\end{proof}

\begin{prop}\label{decompose}
If $d = \power{G}$ is prime to $n$, then
$$M = \bigoplus_{\phi} M^{(\phi)}.$$
\end{prop}
\begin{proof}
% This is obvious using idempotents, but here is the precise formula:
Applying the above corollary to the subgroups
$\set{\phi(\tau)}_{\phi}$ (for every $\tau \in G$), we have  for
arbitrary $a \in M$ that
\begin{eqnarray*}
d^{-1} \sum_{\phi} {\tr_\phi(a)} & = &
    d^{-1} \sum_{\phi}\sum_{\tau} \phi(\tau) \tau(a) \\
    & = & d^{-1} \sum_{\tau} (\sum_{\phi}\phi(\tau)) \tau(a) \\
    & = & \sum_{\tau} (\delta_{\tau,1}) \tau(a)  = a,
\end{eqnarray*}
where $\tr_{\phi}(a) \in M^{(\phi)}$ as we noted above.
\end{proof}




\section{The module $\md{K}$}\label{sec3}

For the rest of this paper, let $F$ be a field of characteristic
prime to $n$, where $n = p^m$ is a prime power, and let $K =
F[\rho]$ where $\rho = \rho_n$ is a primitive root of unity of
order $n$.

The dimension $d = \dimcol{K}{F}$ obviously divides $\power{U_n} =
(p-1)p^{m-1}$. We assume throughout that $d$ is prime to $n$. In
particular, $p$ is odd, and $G = \Gl$ is cyclic. For every $n =
p^m$, setting $K = F[\rho_n]$, we have that
$\dimcol{K}{F[\rho_p]}$ is a power of $p$, so the following holds:
\begin{rem}\label{noproots} % 3.1
$\dimcol{K}{F}$ is prime to $n$ iff $K = F[\rho_n] = F[\rho_p]$.
\end{rem}
For a given field $F$, there is a maximal $\alpha \leq \infty$
such that the dimension of $K = F[\rho_{p^\alpha}]$ over $F$ is
prime to $p$.

The decomposition given in the last section applies to $M =
\md{K}$, which is an $n$-torsion module over $G = \Gal(K/F)$. For
a character $\Kpsi \co G \ra U_n$, the lift to $\mul{K}$ of the
eigenspace $(\md{K})^{(\Kpsi)}$ defined in Section \ref{sec2}, is
$$\K{\Kpsi} = \set{a \in \mul{K} \suchthat  \forall\tau,\quad \tau(a) \equiv
a^{\Kpsi(\tau)} \pmod{\mul{K}^n}},$$ and the twisted norm
$$\N{\Kpsi} \co \mul{K} \ra \K{\Kpsi}/\mul{K}^n$$ is defined by
\begin{equation}\label{normdef}
\N{\Kpsi}{a} = \prod_{\tau \in G} \tau(a)^{\Kpsi(\tau^{-1})}.
\end{equation}

For every $a \in \mul{K}$, Proposition~\ref{decompose} gives a
decomposition
\begin{equation}\label{decomposeK}
a = \Pi a_{\Kpsi}
\end{equation}
where $\Kpsi$ runs over the characters of $\Gl$, and each
$a_{\Kpsi} \in \K{\Kpsi}$ is unique modulo $\mul{K}^n$.

 It is
obvious that $\mul{F} \sub \K{1}$. On the other hand,
\begin{prop}\label{K0F}
We have that $\K{1} = \mul{F} \mul{K}^{n}$.
\end{prop}
\begin{proof}
The inclusion $\mul{F} \mul{K}^{n} \sub \K{1}$ is obvious. Fix a
generator $\tau$ of $\Gl$ and let $a \in \K{1}$. then by
assumption there is some $\mu \in K$ such that $\tau(a) = \mu^n
a$. Taking norms, we see that $\Norm[K/F](\mu)^n = 1$, but since
the norm is in $F$ which does not contain roots of unity even of
order $p$ (Remark \ref{noproots}), we have that $\Norm[K/F](\mu) =
1$. Thus, there is some $g \in K$ such that $\mu = \tau(g)g^{-1}$.
Let $\alpha = g^{-n}a$. Then $\tau(\alpha) = \tau(g)^{-n} \tau(a)
= \mu^{-n}g^{-n}\mu^na = \alpha$, so that $\alpha\in F$ and $a \in
\mul{F}\mul{K}^n$, as asserted.
\end{proof}

Of similar nature is the following observation:
\begin{rem}\label{FcapKn}
We have that $\mul{F}\cap \mul{K}^n = \mul{F}^n$.
\end{rem}
\begin{proof}
Let $a \in \mul{K}$ be an element such that $a^n \in F$. Let
$\tau$ be a generator of $\Gl$. Since $\tau(a^n) = a^n$, we have
that $\tau(a) = \rho^i a$ for some $i$. Write $\tau(\rho) =
\rho^t$. Since $\sg{t}$ is a subgroup of order $d$ of $U_n$,
$t-1$ is prime to $p$ (Remark \ref{gmodp}). Let $b =
\rho^{i/(1-t)}a$ (where the inverse is taken mod $n$), and compute
that $\tau(b) = \rho^{it/(1-t)+i}a = \rho^{i(t/(1-t)+1)}a = b$,
showing that $b \in F$, hence $a^n = b^n \in \mul{F}^n$.
\end{proof}


Since $K$ has $n$ roots of unity, the cyclic extensions of
dimension $n$ over $K$ are classified, by Kummer correspondence,
by elements of $\md{K}$. In a similar manner, the subspaces
$\K{\psi}$ classify cyclic extensions of dimension $n$ over $K$,
which are Galois over $F$. The following is in \cite[p.~211]{Alb0}
for $p$ prime. The condition that $\dimcol{K}{F}$ is prime to $n$
is not needed here.

\begin{prop}\label{whenGal} % 3.4
Let $a \in K$, $\alpha = \sqrt[n]{a}$, and assume $K_1 =
K[\alpha]$ is a field. Then $K_1$ is Galois over $F$ iff $a \in
K^{(\Kpsi)}$ for some character $\Kpsi$.
\end{prop}
\begin{proof}
Let $\sfs$ denote the generator of $\Gal(K_1/K)$, defined by
$\sfs(\alpha) = \rho \alpha$.
%
Assume $K_1/F$ is Galois; then every $\tau \in \Gl$ extends to
$K_1$. Fix $\tau$. Since $\tau(\alpha)^n = \tau(\alpha^n) =
\tau(a) \in K$, we have that $\sfs(\tau(\alpha))^n =
\tau(\alpha)^n$, and $\sfs(\tau(\alpha)) =
\rho^{\Kpsi(\tau)}\tau(\alpha)$ for some number $\Kpsi(\tau) \in
\Z$. %
Now $\sfs(\frac{\tau(\alpha)}{\alpha^{\Kpsi(\tau)}}) =
\frac{\sfs\tau(\alpha)}{\sfs(\alpha)^{\Kpsi(\tau)}} =
\frac{\rho^{\Kpsi(\tau)}\tau(\alpha)}{(\rho \alpha)^{\Kpsi(\tau)}}
= \frac{\tau(\alpha)}{\alpha^{\Kpsi(\tau)}}$, so that
$\tau(\alpha)\alpha^{-\Kpsi(\tau)} \in \mul{K}$ and $\tau(a) \in
\mul{K}^n a^{\Kpsi(\tau)}$. It is then immediate that $\Kpsi$ is
multiplicative mod $n$, so that $\Kpsi$ is a character and $a \in
\K{\Kpsi}$.

Now assume that $a \in \K{\Kpsi}$. Then for every $\tau \in \Gl$,
we have $\tau(a) = k^n a^{\Kpsi(\tau)}$ for some $k \in \mul{K}$.
For every $i \in \Zn$, $\alpha \mapsto \rho^i k
\alpha^{\Kpsi(\tau)}$ is a well defined extension of $\tau$ to
$K_1$ (since it preserves the defining relation $\alpha^n = a$),
so there are $n \power{\Gl} = \dimcol{K_1}{F}$ distinct
automorphisms, showing that $K_1/F$ is Galois.
\end{proof}

We can omit the assumption that $K_1$ is a field. If the order
$n'$ of $a$ in $\md{K}$ is a proper divisor of $n$, then $K_1 =
\joinroot{K}{a}$ is a direct product of $n/n'$ copies of the field
$\joinroot[n']{K}{a}$. Letting $\Kpsi_1 \co \Gl \ra U_{n'}$ denote
the composition of $\Kpsi$ with the projection $U_n\ra U_{n'}$, we
see that $K_1$ is a Galois ring over $F$ iff $a \in \K{\Kpsi_1}$
for some character $\Kpsi$.

\medskip

The following remark allows to choose $a \in \K{\Kpsi}$ with the
convenient property $\Norm[K/F](a) = 1$. In the case
$\dimcol{K}{F} = 2$ this is \cite[Prop.~25]{HKRT}.

\begin{rem}\label{normone} % 3.5
Let $\Kpsi \neq 1$. Every class in $\K{\Kpsi}/\mul{K}^n$ has a
representative $a$ such that $\Norm[K/F](a) = 1$.
\end{rem}
\begin{proof}
Let $\Norm = \Norm[K/F]$ denote the usual norm. Let $\tau$ be a
generator of $\Gal(K/F)$, and $r \equiv \Kpsi(\tau) \pmod{n}$. By
Remark \ref{gmodp} there are $i,j \in \Z$ such that $i(r-1)+jn=1$.
Let $a \in \K{\Kpsi}$, then by assumption $\tau(a) = \mu^n a^r$
for some $\mu \in \mul{K}$. The equivalent element $a_1 = (\mu^i
a^{-j})^n a$ satisfies $\Norm(a_1) = \Norm(\mu)^{ni}
\Norm(a)^{1-jn} = (\Norm(\mu)^{n} \Norm(a)^{r-1})^i =
(\Norm(\tau(a))/\Norm(a))^i = 1$.
\end{proof}

If $d = \dimcol{K}{F}$ is even and $K_0/F$ the quadratic
intermediate field, let $\mu$ denote the character of order $2$ of
$\Gl$. A very similar argument proves the following
\begin{rem}\label{normK0one}
Let $\Kpsi \neq 1, \mu$. Every class in $\K{\Kpsi}/\mul{K}^n$ has
a representative $a$ such that $\Norm[K/K_0](a) = 1$.
\end{rem}
\begin{proof}
As in the previous remark let $\tau$ be a generator of
$\Gal(K/F)$, $r \equiv \Kpsi(\tau) \pmod{n}$, and denote
$\Norm'(a) = \Norm[K/K_0](a)\cdot \tau\Norm[K/K_0](a)^{-1}$. Since
$r \not\equiv -1$ by assumption, there are $i',j' \in \Z$ such
that $i'(r+1)+j'n=1$. Given an element $a \in \K{\Kpsi}$, write
$\tau(a) = \mu^n a^r$, so that $\tau(\Norm'(a)) = \Norm'(\mu)^n
\Norm'(a)^r$. Now the element $a_1  = (\mu^{i'} a^{-j'})^n a$
satisfies $\Norm'(a_1) = \Norm'(\mu)^{ni'} \Norm'(a)^{1-j'n} =
(\Norm'(\mu)^{n} \Norm'(a)^{r+1})^{i'} =
(\tau\Norm'(a)\Norm'(a))^{i'} = 1$, so that $\Norm[K/K_0](a_1) \in
F$.

Since $\Kpsi \neq 1$ we can by the previous remark assume
$\Norm(a) = 1$ to begin with, then $\Norm(\mu)^n = \Norm(a)^{1-r}
= 1$ so that $\Norm(\mu) = 1$, and $\Norm[K/K_0](a_1)^2 =
\Norm[K_0/F]\Norm[K/K_0](a_1) = \Norm(a_1) = 1$. Finally if
$\Norm[K/K_0](a_1) = -1$ we can replace $a_1$ by $a_1^{n+1}$ which
will have norm $1$.
\end{proof}

% The argument of the last two remarks works for every $\Norm^{s} = \pi{\tau(a)^{s^i}}$,
% as $\Norm^{s}(\tau(a))^s = a^{s^d-1}\Norm^s(a)$. Taking $i,j$ such that
% $i(rs-1)+jn=1$ and $a_1 = (\mu^{ri}a^{-j})^n a$, we have that $\Norm^{s}(a_1) = a^{(s^d-1)i}$.
% The above remarks are the cases $s = 1, -1$.
% Notice that we can apply the argument again and again, showing that if $a\in \K{\Kpsi}$
% then it is of the form $\tau(a^s)/a$ for every $s \neq \Kpsi(\tau)$.


% There is a paper (from 2002 or 2003) by Minac and Swallow's in which they describe
%  the structure of $\mul{L}/\mul{L}^p$ as a module over $\Gal(L/K)$,
%  assuming that $\dimcol{L}{K} = p$ and $K$ contains roots of unity.
% I do not think it is relevant.





\section{Decomposition of $\H1{K}{\mu_n}$}\label{Kco}

Let $\sep{F}$ denote the separable closure of $F$, and let
$$
\G[F] = \Gal(\sep{F}/F)
$$
denote the absolute Galois group of
$F$. As usual, we denote the cohomology groups of $\G[F]$ by
$\H{i}{F}{M} = \H{i}{\G[F]}{M}$. % the elements these are classes of continues functions where $\G[F]$ is given the pro-finite topology.

The long exact sequence of cohomology groups applied to the Kummer
sequence
\begin{equation}\label{Kummer}
1 \lra \mu_n \lra \mul{\sep{F}} \lra \mul{\sep{F}} \lra 1,
\end{equation}
gives an isomorphism
\begin{equation}\label{isom1}
j_F \co \md{F} \lra \H1{F}{\mu_n}
\end{equation}
sending  $a\mul{F}^n$ to $(a)$, which is defined by $(a) \co
\s\mapsto \s(\alpha)\alpha^{-1}$ where $\alpha = \sqrt[n]{a}$ is a
fixed root.

As before, let $K = F[\rho]$ where $\rho$ is a primitive $n$-root
of unity. The action of $\G[K]$ on $\mu_n$ is always taken to be
the trivial action; this is often expressed by writing
$\H1{K}{\Zn}$ instead of $\H1{K}{\mu_n}$. Considering \eq{Kummer}
as a sequence of $\G[K]$-modules (rather than $\G[F]$-modules),
gives an isomorphism
$$j_K \co \md{K} \lra \H1{K}{\Zn}$$ defined similarly to
$j_F$.

Let $a \in \mul{K}$ and $\alpha = \sqrt[n]{a}$. Since $K$ has
enough roots of unity, $K[\alpha]$ is Galois over $K$, and the
value $(a)_{\sigma}$ is determined by the restriction of $\sigma$
to $K[\alpha]$. In other words, $(a)$ is an element of
$\H1{K}{\Zn}$ induced from $\H1{\Gal(K[\alpha]/K)}{\Zn}$.
%
% (A general observation in this spirit::)
% General nonsense: if $A\normal G$, $M$ is a $G$-module and $A$
% acts trivially, then an element of $\H1{G/H}{M}$ defines an
% element of $\H1{G}{M}$.

The Galois group $G = \Gl$ acts on $\H1{K}{\Zn}$ by
$$
\tau(a) \co \s \mapsto \tau((a)_{\tau^{-1} \s \tau}).
$$
Since the Kummer sequence \eq{Kummer} is a sequence of
$G$-modules, $j_K$ is an isomorphism of $G$-modules. Indeed, let
$a \in \mul{K}$ with $\alpha = \sqrt[n]{a}$, $\tau\in \Gl$ and $\s
\in \G[K]$, then by definition
\begin{eqnarray*}
\tau(a) \co {\s} & \mapsto & \tau((a)_{\tau^{-1}\s\tau}) \\
    & = & \tau(\tau^{-1}\s\tau(\alpha)\alpha^{-1}) \\
    & = & \s \tau(\alpha) \tau(\alpha)^{-1} \\
    & = & (\tau a)_{\s}.
\end{eqnarray*}

In particular, $j_K$ is compatible with the decomposition of
$\md{K}$ and $\H1{K}{\Zn}$ into eigenspaces which is given in
Proposition~\ref{decompose}, and restricts to an isomorphism
$$j_K \co \K{\Kpsi}/\mul{K}^n \lra \H1{K}{\Zn}^{(\Kpsi)}$$
for every character $\Kpsi \co \Gl \ra U_n$. Since $\N{\Kpsi}$ of
\Eq{normdef} is the twisted trace function defined on $\md{K}$,
the following diagram commutes:
%\begin{prop}
%The following diagram commutes.
\ifXY
\begin{equation}\label{KHK}
\xymatrix{ % @R=30pt@C=36pt, C = length of --- lines.
    {\K{\Kpsi}/\mul{K}^n} \ar[d]^{j_K}\ar@{^(->}[r]
    & {\md{K}}\ar[d]^{j_K}           \ar[r]^{\N{\Kpsi}}
    & {\K{\Kpsi}/\mul{K}^n}\ar[d]^{j_K}
\\
    {\H1{K}{\Zn}}^{(\Kpsi)}        \ar@{^(->}[r]
    & {\H1{K}{\Zn}}               \ar[r]^{\tr_{\Kpsi}}
    & {\H1{K}{\Zn}}^{(\Kpsi)}
}
\end{equation}
\else % For \ifXY
\bigskip
$j_K$ commutes with $\res$ and $\cores$.
\bigskip
\fi % For \ifXY
%\end{prop}




\section{First Cohomologies of $F$}\label{Fco}

Let $F$ be a field of characteristic prime to $n$, where $n =
p^m$. Let $\mu_n$ denote the group of roots of unity of order $n$
in $\sep{F}$. Fix a primitive root $\rho \in \mu_n$.

In this section we study the Galois groups $\H1{F}{\mu_n}$ for
various actions of $\G[F]$ on $\mu_n$, which restrict to the
trivial action of $\G[K]$. One case is well understood: if we let
$\G[F]$ act naturally on $\mu_n$ (that is, by the usual action of
automorphisms on $\mu_n \sub \mul{\sep{F}}$), then $\H1{F}{\mu_n}
\isom \md{F}$ by the Kummer sequence mentioned earlier.

% The restriction map::    $$\res\co \H1{F}{\mu_n} \lra \H1{K}{\Zn},$$
Let $K = F[\rho]$. A character   $\chi \co \Gl \ra U_n$ determines
an action of $\G[F]$ on $\mu_n$ by $\s_{*}(\rho) =
\rho^{\chi(\s)}$ (the star reminds us that this is not the usual
action); and of course every extension of the trivial action of
$\G[K]$ to $\G[F]$ is determined in this way. We use
$\H[\chi]1{F}{\mu_n}$ to denote the cohomology group with respect
to the action determined by $\chi$. Since the action of $\G[K]$ is
trivial, we use $\H1{K}{\Zn}$ for the cohomology group with
respect to $\G[K]$, and we have the restriction map $\res \co
\H[\chi]1{F}{\mu_n} \ra \H1{K}{\Zn}$.

Denote $G = \Gl$, and view $\mu_n$ as a $\G[F]$-module with the
action via $\chi$. As a subgroup, $\G[K] \leq \G[F]$ acts
trivially on $\mu_n$, and the short exact sequence
$$
1\lra \G[K] \lra \G[F] \lra G \lra 1,
$$
gives rise to the Serre-Hochshild spectral sequence
% $$\H1{G}{\mu_n} \lra \H[\chi]1{F}{\mu_n} \lra \H1{K}{\mu_n}^{G} \lra \H2{G}{\mu_n}.$$
$$\H1{G}{\mu_n} \lra \H[\chi]1{F}{\mu_n} \lra \H1{K}{\mu_n}^{G} \lra \H2{G}{\mu_n}.$$

Since $\power{\G[F]/\G[K]} = \dimcol{K}{F}$ is by assumption prime
to $n$, we have that
$$
\H1{G}{A} = \H2{G}{A} = 0,
$$
so we obtain an isomorphism
$$
\H[\chi]1{F}{\mu_n} \isom \H1{K}{\mu_n}^G,
$$
which we will explicitly describe below. In particular we will see
that for the action of $G$ on $\H1{K}{\mu_n}$ via $\chi$, the
invariant subgroup $\H1{K}{\mu_n}^G$ is the component
$\H1{K}{\Zn}^{(\nu\chi^{-1})}$ of the previous section.

{}From Proposition~\ref{K0F} and Remark~\ref{FcapKn} we see that
$$
\K{1}/\mul{K}^n = \mul{K}^n \mul{F} / \mul{K}^n \,\isom\,
\mul{F}/\mul{F}^n,
$$
so that $\K{1}/\mul{K}^n \isom \H1{F}{\mu_n}$ with the usual
action. This motivates an attempt to express other cohomology
groups in terms of the components $\K{\Kpsi}/\mul{K}^n$.

% OLD paragraphs:
%
% 3. An excuse for using eigen-elements $a$, and not the general element.
% The following is not so precise; the value is determined by the action on
% $F[\alpha]$, not by the restriction to that field (which need not
% be preserved):
%
% Note that the value of the cocycle $(a)$ at $\s\in \G[F]$ is determined by
% the restriction of $\s$ to $F[\alpha]$, or, in other words, that $(a)$
% is an element of $\H1{F}{\mu_n}$  induced from
% $\H1{\Gal(F[\alpha]/F)}{\mu_n}$.
%

Identify $U_n$ with $\Aut(\mu_n)$ is the usual way (where the
action is by exponentiation). Fix a character
$$
\Kpsi \co \Gl \ra U_n,
$$
and let $a \in \K{\Kpsi}$ be an element such that $K_1 =
K[\alpha]$ is a field, where $\alpha = \sqrt[n]{a}$ is a fixed
root (the assumption that $K_1$ is a field will eventually be
removed). By Proposition \ref{whenGal}, $K_1/F$ is a Galois
extension, and so every $\s \in \G[F]$ restricts to an
automorphism of $K_1/F$. The idea is to use this restriction to
define the value of the cocycle at $\s$, making the cocycle
induced from $\H1{\Gal(K_1/F)}{\mu_n}$ (with an action which is
yet to be determined), just as $(c)\in \H1{K}{\Zn}$ is induced
from $\H1{\Gal(\joinroot{K}{c}/F)}{\Zn}$ for $c \in \mul{K}$. The
analogy to the special case $\Kpsi = 1$ still holds: if $a \in
\K{1}$ then by Proposition~\ref{K0F} we can assume that $a \in
\mul{F}$, and then the value of $(a) \in \H1{F}{\mu_n}$ (with the
usual action) at $\s \in \G[F]$ is determined by the restriction
of $\s$ to $K_1 = K \tensor F[\alpha]$, where $\alpha =
\sqrt[n]{a}$ is a fixed root.


%Back to an arbitrary character,
Let $\sfs$ denote a generator of $\Gal(K_1/K)$ defined by
\begin{equation}\label{sfsdef}
\sfs(\alpha) = \rho\alpha.
\end{equation}
Let $\nu \co \Gl \ra U_n$ be the distinguished character defined
by
\begin{equation}\label{nudef}
\tau(\rho) = \rho^{\nu(\tau)}.
\end{equation}
Note that $\nu$ is uniquely determined by $F$, and that $\nu$
extends to a character $\nu \co \G[F]\ra U_n$ in the obvious way.
The natural action of $\G[F]$ on $\mu_n$ is the action via $\nu$.

Consider the short exact sequence
\begin{equation}\label{K1ses}
1 \lra \Gal(K_1/K) \lra \Gal(K_1/F) \lra \Gl \lra 1.
\end{equation}
Since $\Gl$ is Abelian, the action of
$$
H = \Gal(K_1/F)
$$
on $\Gal(K_1/K)$ by conjugation reduces to an action of $\Gl$. Let
$$
\chi \co \Gl \ra U_n
$$ be the character associated to this action,
namely
\begin{equation}\label{chidef}
\tau \sfs \tau^{-1} = \sfs^{\chi(\tau)}.
\end{equation}

%% An unnecessarily long version of the last paragraph:
%
% Since $\Gal(K_1/K)$ is normal in $G = \Gal(K_1/F)$, $G$ acts on
% $\Gal(K_1/K)$ by conjugation; but $\Gl$ is Abelian, so this action
% reduces to an action of $\Gl$. Let $\chi \co \Gl \ra U_n$ be the
% character defined by
% $$\tau \sfs \tau^{-1} = \sfs^{\chi(\tau)}.$$

The three characters $\Kpsi,\nu,\chi$ are related by the following
equation:
\begin{rem}\label{chinupsi} % 5.1
We have that $\chi\Kpsi = \nu$.
\end{rem}
\begin{proof}
Writing $\tau^{-1}(\alpha) = k \alpha^{\Kpsi(\tau^{-1})}$ for some
$k \in \mul{K}$, we have that
\begin{eqnarray*}
\rho^{\chi(\tau)} \alpha & = & \sfs^{\chi(\tau)} (\alpha) \\
    & = & \tau\sfs(\tau^{-1}(\alpha)) \\
    & = & \tau\sfs(k \alpha^{\Kpsi(\tau^{-1})}) \\
    & = & \tau(k \rho^{\Kpsi(\tau^{-1})} \alpha^{\Kpsi(\tau^{-1})}) \\
    & = & \tau(\rho)^{\Kpsi(\tau^{-1})} \tau(k \alpha^{\Kpsi(\tau^{-1})}) \\
    & = & \rho^{\nu(\tau) \Kpsi(\tau^{-1})}\alpha.
\end{eqnarray*}
\end{proof}

Let $N = \Gal(K_1/K) = \sg{\sfs}$. Repeating the spectral sequence
argument given above, we obtain an isomorphism
$$
\H[\chi]1{F}{\mu_n} \isom \H1{K}{\mu_n}^{H/N},
$$
which we will use to construct the co-cycle in
$\H[\chi]1{F}{\mu_n}$ associated to the element $a$.
%
%%
\forget %% A detailed version of the 'using the spectral sequence argument give above'
%%
 Consider the Serre-Hochshild spectral sequence
$$\H1{H/N}{A^N} \lra \H1{H}{A} \lra \H1{N}{A}^{H/N} \lra \H2{H/N}{A^N},$$
with
$$H = \Gal(K_1/F),$$
$$N = \Gal(K_1/K) = \sg{\sfs},$$
$$A = \mu_n$$
and the action of $H$ on $A$ is via $\chi$ (!). As a subgroup, $N$
acts trivially on $\m$. Since $\power{H/N} = \dimcol{K}{F}$ is by
assumption prime to $n$, we have that
$$
\H1{H/N}{A^N} = \H2{H/N}{A^N} = 0,
$$
%
% (On $\H1{}{}$ in general::)
% More generally, $\H[\chi]1{\Gl}{\mu_n}$ is computed in Theorem 8.3 of \cite{CFG}:
% $\H{1}{G/H}{A} = \Aut(G,A,G/A)/(A\cap \Inn(G))$, where by definition $\Aut(G,A,G/A)$
% is the group of automorphisms of $G$ which fix $A$ pointwise and act trivially on $G/A$.
%
so there is an isomorphism
$$
\H[\chi]1{H}{\mu_n} = \H1{N}{\mu_n}^{H/N},
$$
which we will explicitly describe below. %
\forgotten % End of detailed version. ---------------------------------
The group
$$
\H1{N}{\mu_n} = \Hom(N,\mu_n)
$$
is cyclic of order $n$, generated by an element
$c$ which is defined by $c(\sfs^s) = \rho^s$. Moreover, the action
of
$$
H/N = \Gal(K/F)
$$
on this group is trivial: for every $\tau \in \Gl$, we have $(\tau
c)(\sfs^s) = \tau_{*}(c(\tau^{-1} \sfs^s \tau)) =
\tau_{*}(c(\sfs^{s\chi(\tau)^{-1}})) =
\tau_{*}(\rho^{s\chi(\tau)^{-1}}) = \rho^{s} = c(\sfs^s)$, where
we used the definition $\tau_*(\rho) = \rho^{\chi(\tau)}$.

We will now define a map
$$
J^\Kpsi \co \K{\Kpsi}/\mul{K}^n \ra \H[\chi]1{F}{\mu_n}
$$ where
$\Kpsi$ and $\chi = \nu\Kpsi^{-1}$ are the characters fixed above.
\begin{defn}\label{defdef}
Let $F,n,K$ be as above, $G = \Gl$, $\Kpsi \co G \ra U_n$ a
character, and $a\in \K{\Kpsi}$. Let $\nu$ be the character
defined by \eq{nudef}, and $\chi = \nu \Kpsi^{-1}$.

First assume that $a$ is of order $n$ in $\md{K}$, and let $\alpha
= \sqrt[n]{a} \in \sep{F}$, $K_1 = K[\alpha]$ and $H =
\Gal(K_1/K)$. Choose a splitting of \eq{K1ses} which maps $\tau
\mapsto \tau' \in \Gal(K_1/F)$
(this is possible since $\H[\chi]2{G}{H} =0$).%; for this case this is Schur-Zassenhaus lemma

Let $\s \in \G$ be given, and let $\tau \in \Gal(K/F)$ be the
restriction of $\s$ to $K$. Then for a unique $s \in \Zn$, the
restriction of $\s$ to $K_1$ is $\s = \sfs^s \tau'$. Now, define
$c_a \in \H[\chi]1{F}{\mu_n}$ by
\begin{equation}\label{eqdef}
c_a(\s) = \rho^s.
\end{equation}

If the order $n'$ of $a$ in $\md{K}$ strictly divides $n$, then
repeating the argument for $b = \sqrt[n']{a} \in K$ and $n/n'$
(noting that $b \in \K{\Kpsi}$ when $\Kpsi$ is viewed as a
character modulo $n/n'$), we define
$$c_a(\sfs^s \tau') = \rho^{n's}.$$

Finally, we define $J^\Kpsi \co \K{\Kpsi}/\mul{K}^n \ra
\H[\nu\Kpsi^{-1}]1{F}{\mu_n}$ by
$$J^\Kpsi(a) = c_a.$$
\end{defn}

We need to show that the definition does not depend on the
splitting of \eq{K1ses}. Indeed, since $\H[\chi]1{G}{H} = 0$,
every other splitting $\tau \mapsto \tau''$ is of the form
$$
\tau'' = \tau_{*}(\sfs^i)\sfs^{-i}\tau' =
\sfs^{i(\chi(\tau)-1)}\tau'
$$ for some $i$. Let $c_a'$ be the
cocycle defined using this splitting. If $\s = \sfs^{s} \tau' =
\sfs^{s-i(\chi(\tau)-1)}\tau''$, then $c_a(\s) = \rho^{s}$ and
$c_a'(\s) = \rho^{s-i(\chi(\tau)-1)}$, and the quotient is
$$
g(\s) = \rho^{i(\chi(\tau)-1)} = \tau_*(\rho^i)\rho^{-i} =
\s_*(\rho^i)\rho^{-i},
$$ which is cohomologous to $1$.

%% Long proof:
%
%Indeed, let $\tau \mapsto \tau''\in
%\Gal(K_1/F)$ be another splitting. Then $\tau'' =
%\sfs^{k_\tau}\tau'$, and $k$ is a cocycle in $\H[\chi]1{G}{\Zn}$
%(as the action of $\Gl$ on $\sg{\sfs}$ is determined by
%\eq{chidef}). By the assumption \eq{Hassump1}, $k$ is cohomologous
%to $1$, so for some $i$ we have $\sfs^{k_\tau} =
%\tau^{*}(\sfs)\sfs^{-i} = \sfs^{i(\chi(\tau)-1)}$.
%
%This changes the cocycle $c_a$ from \eq{eqdef} to
%$c_a'(\sfs^s\tau') = \rho^{s-k_\tau}$.
%But $g(\sfs^s\tau') = (\sfs^s\tau')^{*}(\rho^{i})\rho^{-i}
%= \rho^{i(\chi(\sfs^s\tau')-1)} = \rho^{i(\chi(\tau)-1)} = \rho^{k_\tau}$
%is trivial in $\H[\chi]1{\Gal(K_1/F)}{\mu_n}$, so that $c \equiv c'$.
%
%----------------------



\begin{thm}\label{KHFthm}
The following diagram commutes:
%
\ifXY
\begin{equation}\label{KHF}
\xymatrix{ % @R=30pt@C=36pt, C = length of --- lines.
    {\K{\Kpsi}/\mul{K}^n} \ar[d]^{J^{\Kpsi}}     \ar@{^(->}[r]
    & {\md{K}} \ar[d]^{j_K} \ar[r]^{\N{\Kpsi}}
    & {\K{\Kpsi}/\mul{K}^n}\ar[d]^{J^{\Kpsi}}
\\
    {\H[\chi]1{F}{\mu_n}}        \ar[r]^{\res}
    & {\H1{K}{\Zn}}               \ar[r]^{\cores}
    & {\H[\chi]1{F}{\mu_n}}
}
\end{equation}
\else % For \ifXY
\bigskip
$\res$ and $\cores$ for arbitrary action.
\bigskip
\fi % For \ifXY
\end{thm}
\begin{proof}
Let $a \in \K{\Kpsi}$, and let $\alpha,K_1,\sfs,\chi$ be as above.
Then $j_K \co a \mapsto (a)$, where $(a)_\s =
\s(\alpha)\alpha^{-1}$ for every $\s \in \G[K]$. On the other
hand, if $\s = \sfs^s$ on $K_1$, then
$$
J^{\Kpsi}(a) = c_a \in \H[\chi]1{F}{\mu_n}
$$
has value $\rho^s$ at
$\s$ by definition, and $\s(\alpha)\alpha^{-1} = \rho^s =
c_a(\s)$.

Now consider the right hand square. Since $\mul{K}$ is generated
by the subgroups $\K{\Kpsi'}$, it is enough to check that $J^\Kpsi
\circ \N{\Kpsi} = \cores\circ j_K$ on an element $a \in
\K{\Kpsi'}$. The corresponding element in $\H1{K}{\Zn}$ is by
definition $\sigma \mapsto \frac{\sigma(\alpha)}{\alpha}$ where
$\alpha$ is a fixed $n$th root of $a$. Let $K_1 = K[\alpha]$, and
$\tau \mapsto \tau'$ a splitting of \eq{K1ses} for this $K_1$.
Define $\sfs$ by \eq{sfsdef}, and set
$$
\chi' = \nu\Kpsi'^{-1},
$$
so that $G = \Gal(K/F)$ acts on $\sg{\sfs}$ via $\chi'$.

We need to show that $J^\Kpsi(\N{\Kpsi}(a))$ and $\cores(a)$ agree
on every $\s \in \G[F]$. Let $\sfs^s \tau_0'$ be the restriction
of $\s$ to $K_1$. Since $\Gamma_F = \cup_{\tau \in G}{\Gamma_K
\tau}$, we have by definition of the corestriction that
\begin{eqnarray*}
\cores(a) \co \s & \mapsto &
\prod_{\tau}\tau_*^{-1}((a)_{\tau\s(\tau\tau_0)^{-1}}) \\
& = & \prod_{\tau}\tau_*^{-1}((a)_{\tau\sfs^s\tau^{-1}}) \\
& = & \prod_{\tau}\tau_*((a)_{\sfs^{s\chi'(\tau)^{-1}}})
\\
& = & \prod_{\tau}\tau_*(\rho^{s\chi'(\tau)^{-1}}) \\
& = & \prod_{\tau}\rho^{s\chi(\tau)\chi'(\tau)^{-1}},
\end{eqnarray*}
where $\tau_*(\rho) = \rho^{\chi(\tau)}$ since the computation is
in $\H[\chi]1{F}{\mu_n}$.

If $\chi \neq \chi'$, then
$$
\sum_{\tau} \chi(\tau)\chi'(\tau)^{-1} = 0
$$
by Remark \ref{Hsg}, so that $\cores(a) = 1$. This is what we need
since $\N{\Kpsi'}(a) \equiv 1$ modulo $\mul{K}^n$, again by the
same remark.

Now assume $\chi' = \chi$ (\ie\ $\Kpsi' = \Kpsi$), then $\cores(a)
\co \s \mapsto \rho^{sd}$ (where $d = \dimcol{K}{F}$), and
likewise $\N{\Kpsi}(a) \equiv a^d$ so that
$$
J^\Kpsi(\N{\Kpsi}(a)) = J^{\Kpsi}(a)^d
$$ while
$$
J^{\Kpsi}(a)= c_a \co \s \mapsto \rho^s,
$$
and the maps coincide.
\end{proof}

\begin{cor}\label{Jisisom}
For $\Kpsi \co \Gl\ra U_n$ and $\chi = \nu \Kpsi^{-1}$,
$$
J^\Kpsi \co \K{\Kpsi}/\mul{K}^n \lra \H[\chi]1{F}{\mu_n}
$$
is an isomorphism.
\end{cor}
\begin{proof}
Since
$$
\cores\circ \res \co \H[\chi]1{F}{\mu_n} \ra \H[\chi]1{F}{\mu_n}
$$
is multiplication by $d = \dimcol{K}{F}$ which is prime to $n$,
$\res$ is injective and $\cores$ surjective. Our claim then
follows from the commutativity of \eq{KHF}.
\end{proof}

\begin{cor}\label{diff}
If $\chi \neq \chi'$, the composition
$$\H[\chi]1{F}{\mu_n} \stackrel{\res}{\lra} \H1{K}{\Zn}
\stackrel{\cores}{\lra} \H[\chi']1{F}{\mu_n}$$ is the zero map.
\end{cor}
\begin{proof}
Use the left hand square of \eq{KHF} for $\Kpsi = \nu \chi^{-1}$
and $\chi$, and the right hand square for $\Kpsi' = \nu \chi^{-1}$
and $\chi'$, and the fact (following from Proposition
\ref{decompose}) that $\N{\Kpsi'}$ is zero on
$\K{\Kpsi}/\mul{K}^n$.
\end{proof}


Notice that Equation \eq{KHF} provides an expression for
$J^{\Kpsi}$ in terms of the restriction and corestriction, namely
$J^{\Kpsi} = d^{-1}\cdot \cores \circ j_K$, where the
corestriction here is $\cores \co {\H1{K}{\Zn}} \ra
{\H[\chi]1{F}{\mu_n}}$.


For $\Kpsi = 1$ we have $\chi = \nu$, so that $\H[\chi]1{F}{\mu_n}
= \H1{F}{\mu_n}$ is the group with usual action, and the
isomorphism $J^1$ is %$\K{1}/\mul{K}^n \lra \H1{F}{\mu_n}$,
the composition of $\K{1}/\mul{K}^n \isom \mul{F}/\mul{F}^n$ with
$j_F$, which was discussed at the beginning of this section. In
particular, if $K = F$, then $J^{1} = j_K$.

\begin{cor}\label{H1KH1F}
For every character $\Kpsi \co \Gl \ra U_n$,
$$\H1{K}{\Zn}^{(\Kpsi)} \isom \H[\chi]1{F}{\mu_n},$$
where $\chi = \nu \Kpsi^{-1}$. Moreover, the following diagram
commutes: %
\ifXY
\begin{equation}
\xymatrix{ % @R=30pt@C=36pt, C = length of --- lines.
    {\H1{K}{\Zn}}^{(\Kpsi)} \ar[d]     \ar@{^(->}[r]
    & {\H1{K}{\Zn}} \ar@{=}[d] \ar[r]^{\tr_{\Kpsi}}
    & {\H1{K}{\Zn}}^{(\Kpsi)}\ar[d]
\\
    {\H[\chi]1{F}{\mu_n}}        \ar[r]^{\res}
    & {\H1{K}{\Zn}}               \ar[r]^{\cores}
    & {\H[\chi]1{F}{\mu_n}}
}
\end{equation}
\else % For \ifXY
\bigskip
$j_K$ commutes with $\res$ and $\cores$.
\bigskip
\fi % For \ifXY
where the left and right
arrows are $J^{\Kpsi} \circ j_K^{-1}$.
\end{cor}
\begin{proof}
Apply the commutativity of \eq{KHK} and \eq{KHF} to the following
diagram: %
\ifXY
\begin{equation}\nonumber
\xymatrix@C=-9pt { % @R=30pt@C=36pt, C = length of --- lines.
    {\K{\Kpsi}/\mul{K}^n} \ar[dd]^{J^{\Kpsi}}     \ar@{^(->}[rr]
    \ar[rd]^{j_K}
    & {}
    & {\md{K}} \ar[dd]^(0.3){j_K} \ar[rr]^{\N{\Kpsi}} \ar[rd]^{j_K}
    & {}
    & {\K{\Kpsi}/\mul{K}^n}\ar[dd]^(0.3){J^{\Kpsi}} \ar[rd]^{j_K}
    & {}
\\
    {}
    & {\H1{K}{\Zn}}^{(\Kpsi)} \ar[ld]
    \ar@{^(->}'[r][rr]
    & {}
    & {\H1{K}{\Zn}} \ar@{=}[ld] \ar'[r]^(0.8){\tr_{\Kpsi}}[rr]
    & {}
    & {\H1{K}{\Zn}}^{(\Kpsi)}\ar[ld]
\\
    {\H[\chi]1{F}{\mu_n}}        \ar[rr]^{\res}
    & {}
    & {\H1{K}{\Zn}}               \ar[rr]^{\cores}
    & {}
    & {\H[\chi]1{F}{\mu_n}}
    & {}
}
\end{equation}
\else % For \ifXY
\bigskip
$j_K$ commutes with $\res$ and $\cores$.
\bigskip
\fi % For \ifXY
\end{proof}

In general, if $M$ is a $G$-module and $H\normal G$, then the
image of the restriction $\H1{G}{M} \ra \H1{H}{M}$ is invariant
under the action of $G/H$; one might thus expect that the image of
$$
\res \co \H[\chi]1{F}{\mu_n} \ra \H1{K}{\mu_n}
$$
would be contained in $\H1{K}{\mu_n}^{(1)}$, and not
$\H1{K}{\mu_n}^{(\Kpsi)}$ as the corollary indicates. In fact,
every character $\chi$ defines a different action of $\Gal(K/F)$
on $\H1{K}{\mu_n}$, compatible with $\mu_n$ being a $\G[F]$-module
with the action induced by $\chi$ (and indeed $\Im(\res)$ is
invariant under this action). However, notice that we decompose
$\H1{K}{\mu_n}$ to eigenspaces with respect to a fixed action of
$\G[F]$ on $\mu_n$, namely the natural one, and $\Im(\res)$ is
invariant under this action iff $\chi = \nu$.

\medskip

We end this section by showing that the maps $J^{\Kpsi}$ commute
with restriction when $F$ is being changed. For that, we need to
connect eigenspaces of $\md[n]{K}$ with respect to the action of
$\Gl$ and a subgroup.
\begin{lem}\label{decKL}
Let $F \sub L \sub K$ be an intermediate field, and $\Kpsi_0 :
\Gal(K/L) \ra U_n$ a character. Let $\K{\Kpsi_0}$ be the
$\Kpsi_0$-component of $\mul{K}$ with respect to the action of
$\Gal(K/L)$.

Then $\K{\Kpsi_0}$ is the product of the components $\K{\Kpsi'}$
for all the extensions of $\Kpsi_0$ to characters $\Kpsi' \co \Gl
\ra U_n$.
\end{lem}
\begin{proof}
Let $\tau$ be a generators of $\Gal(K/F)$, so that $L =
K^{\tau^e}$ and $\Gal(K/L) = \sg{\tau^e}$. When viewed as a
$\Gal(K/L)$-module, the eigenspace of $\mul{K}$ with respect to
$\Kpsi_0$ is
$$
\K{\Kpsi_0} = \set{a \in \mul{K}\suchthat \tau^e(a) \equiv
a^{\Kpsi_0(\tau^e)} \pmod{\mul{K}^n}}.
$$

Now let $\Kpsi' \co \Gl\ra U_n$ be an arbitrary character, and
compare $\K{\Kpsi'}$ and $\K{\Kpsi_0}$: if $a$ belongs to the
intersection, then $\tau^e(a)\equiv a^{\Kpsi_0(\tau^e)} \equiv
a^{\Kpsi'(\tau^e)}$, so if $\Kpsi'$ restricts to $\Kpsi_0$ we have
that $\K{\Kpsi'} \sub \K{\Kpsi_0}$, while otherwise the
intersection is $\mul{K}^n$.
\end{proof}

%
Let $L$ be a separable extension of $F$ (viewed as a subfield of
$\sep{F}$), and let $L_0 = L \cap K$. Let
$$
G_0 = \Gal(K/L_0)
$$
be the corresponding subgroup of $G = \Gal(K/F)$, and $\Kpsi_0$
the restriction of $\Kpsi \co G \ra U_n$ to $G_0$. By the lemma,
$\K{\Kpsi} \sub \K{\Kpsi_0}$. Since $T = L[\rho] = L
\tensor[L_0]K$, there is a natural identification
$$
\Gal(T/L) \stackrel{\isom}{\lra} \Gal(K/L_0),
$$
which makes $\Kpsi_0$ a character on
$\Gal(T/L)$. Likewise let $\chi_0$ be the restriction of $\chi =
\nu \Kpsi^{-1}$ to $\Gal(T/L)$.
\begin{prop}
Let $L/F$ be a separable extension, $T = L[\rho]$, and let
$\Kpsi_0,\chi_0 \co \Gal(T/L) \ra U_n$ be defined as above. Then
the
following diagram commutes. %
\ifXY
\begin{equation}\label{JresJ}
\xymatrix{ % @R=30pt@C=36pt, C = length of --- lines.
    {\K[T]{\Kpsi_0}/\mul{T}^n} \ar[r]^{J^{\Kpsi_0}}
    & {\H[\chi_0]1{L}{\mu_n}}
\\
    {\K{\Kpsi}/\mul{K}^n} \ar[r]^{J^{\Kpsi}} \ar@{^(->}[u]
    & {\H[\chi]1{F}{\mu_n}} \ar[u]^{\res}
}
\end{equation}
\else % For \ifXY
\bigskip
$J^{\Kpsi}$ and $J^{\Kpsi'}$.
\bigskip
\fi % For \ifXY
\end{prop}
\begin{proof}
It is enough to treat the extensions $L/L_0$ and $L_0/F$, that is,
to show that the following diagram commutes. %
\ifXY
\begin{equation}\nonumber
\xymatrix{ % @R=30pt@C=36pt, C = length of --- lines.
    {\K[T]{\Kpsi_0}/\mul{T}^n} \ar[r]^{J^{\Kpsi_0}}
    & {\H[\chi_0]1{L}{\mu_n}}
\\
    {\K{\Kpsi_0}/\mul{K}^n} \ar[r]^{J^{\Kpsi_0}} \ar@{^(->}[u]
    & {\H[\chi_0]1{L_0}{\mu_n}} \ar[u]^{\res}
\\
    {\K{\Kpsi}/\mul{K}^n} \ar[r]^{J^{\Kpsi}} \ar@{^(->}[u]
    & {\H[\chi]1{F}{\mu_n}} \ar[u]^{\res}
}
\end{equation}
\else % For \ifXY
\bigskip
$J^{\Kpsi}$ and $J^{\Kpsi'}$.
\bigskip
\fi % For \ifXY
For the lower square, let $a \in \K{\Kpsi}$, $K_1 =
\joinroot{K}{a}$, and assume $K_1$ is a field (the general case
will then easily follows). We need to prove that the restriction
of $J^{\Kpsi}(a)$ to $\G[L_0]$ coincides with $J^{\Kpsi_1}(a)$.
Recall that $c_a = J^{\Kpsi}(a)$ is not well defined as a function
on $\G[F]$, as it depends on the splitting of \eq{K1ses}. Choose a
splitting $G \ra \Gal(K_1/F)$ which maps $\tau \mapsto \tau'$. Let
$\s \in \G[L_0]$, and write the restriction of $\s$ to $K_1$ as
$\sfs^s \tau'$ where $\tau \in \Gal(K/L_0) \sub \Gal(K/F)$; then
$c_a(\s) = s$ either as a cocycle on $\G[F]$ or on $\G[L_0]$.

Next, for the upper square, let $a \in \K{\Kpsi_0}$. Again, it is
enough to assume that $T_1 = \joinroot{T}{a}$ is a field. Let
$\tau \mapsto \tau'$ be a splitting of \eq{K1ses} (for $K/L_0$
instead of $K/F$), and extend each $\tau'$ to $T = L\tensor[L_0]K$
as $1\tensor \tau'$. For $\s \in \G[L]$, write the restriction to
$T_1$ as $\sfs^s \tau'$ where $\sfs$ is defined as usual. Then
$\s$ restricts to $\sfs^s\tau'$ on $K_1$, so $J^{\Kpsi_0}(\s) =
\rho^s$ either over the extension $T/L$, or over $K/L_0$.
\end{proof}

Note that for the special case $L = K$, \eq{JresJ} is the left
hand side of \eq{KHF}.





\section{Subfields of $K_1$}\label{fields}

As before, let $F$ be a field of characteristic prime to $n =
p^m$, $K = F[\rho]$ where $\rho$ is a primitive root of unity of
order $n$. %, and $G = \Gl$.
We assume that $\dimcol{K}{F}$ is prime to $n$.

Let $\Kpsi \co \Gl \ra U_n$ be a character, $\chi = \nu
\Kpsi^{-1}$ (where $\nu$ is defined by \eq{nudef}) and $a \in
\K{\Kpsi}$ an element such that $K_1 = K[\alpha]$ is a field for
$\alpha = \sqrt[n]{a}$. We discuss subfields of $K_1$, and give an
explicit computation of the inverse of the map $J^{\Kpsi}$ of
Definition \ref{defdef}.

Let $\sfs$ be the generator of $\Gal(K_1/K)$ defined by
$\sfs(\alpha) = \rho\alpha$. By Proposition \ref{whenGal}, $K_1$
is Galois over $F$, with $\Gal(K_1/F)$ the semidirect product of
$$
\Gal(K/F) = \Zn[d]
$$
acting (via $\chi$, by Remark \ref{chinupsi}) on
$$\Gal(K_1/K)=\sg{\sfs} \,\isom\, \Zn.$$

Let $\tau \mapsto \tau'$ be a splitting of \eq{K1ses}, and $G'$
the copy of $G = \Gl$ in $\Gal(K_1/F)$ under this map. Let $F_1$
denote the invariant subfield of $K_1$ under $G'$, so that
$\dimcol{F_1}{F} = n$. Notice that $K_1 = F_1 \tensor[F] K$ as the
dimensions are co-prime. The following is essentially in
\cite[Sec.~2]{Saltman82}.

\begin{prop}\label{whicharecyclic}
If $\chi = 1$ then $F_1 = K_1^{G'}$ is cyclic over $F$. Every
cyclic extension of dimension $n$ over $F$ has this form.
\end{prop}
\begin{proof}
If $\chi = 1$ then
$$\
\Gal(K_1/F) = \Zn \times \Gal(K/F)
$$
by \Eq{chidef}, so that $G'$ is normal in $\Gal(K_1/F)$, with
cyclic quotient.

Now let $F_1/F$ be cyclic extension of dimension $n$. Let $K_1 =
F_1 \tensor[F] K$, then $K_1$ is Galois over $F$ (since $F_1 \cap
K = F$ as the dimensions are co-prime) and thus of the form $K_1 =
K[\alpha]$ where $\alpha^n = a$ and $a\in\K{\Kpsi}$ for some
character $\Kpsi \co G \ra U_n$. Let $\chi = \nu \Kpsi^{-1}$
(where $\nu$ is defined by \Eq{nudef}). Let $\sfs$ be the
automorphism defined by $\sfs(\alpha) = \rho\alpha$. Let $1 \neq
\tau \in \Gl$, and let $\tau'$ be the extension of $\tau$ to
$K_1$. By the normality of $G'$ in $\Gal(K_1/F)$,
$\sfs\tau'\sfs^{-1} =\sfs^{1-\chi(\tau)}\tau'$ is in $G'$, so that
$\chi(\tau)= 1$ (otherwise  $1-\chi(\tau)$ is prime to $n$ by
Remark \ref{gmodp}).
\end{proof}

\begin{exmpl}
When $\dimcol{K}{F} = 2$ we can, by Remark~\ref{normone}, choose a
representative of $[a] \in \K{\nu}/\mul{K}^n$ such that $\tau(a) =
a^{-1}$. Let $\alpha = \sqrt[n]{a}$, then $K_1 = K[\alpha]$ with
$\tau(\alpha) = \alpha^{-1}$ and $\sfs(\alpha) = \rho \alpha$.
Then
$$
F_1 = F[\alpha+\alpha^{-1}].
$$

Writing $a = a_0+a_1\rho$ for $a_0,a_1\in F$, the condition
$\tau(a) = a^{-1}$ is equivalent to
$$a_0^2 +
(\rho+\rho^{-1})a_0a_1+a_1^2 = 1.
$$

Of course the general solution to this quadratic equation is
obtained by taking $a = b\tau(b)^{-1}$ for arbitrary $b =
b_0+b_1\rho \in K$, namely
$$a_0 =
(b_0^2-b_1^2)/N\quad \mbox{and} \quad a_1 =
(2b_0b_1+(\rho+\rho^{-1})b_1^2)/N
$$
where $b_0,b_1 \in F$ and $N = b_0^2+(\rho+\rho^{-1})b_0b_1+b_1^2
= \Norm[K/F](b_0+b_1\rho)$.
\end{exmpl}

Since $\H[\chi]1{G}{\Zn} = 0$, the splitting of \eq{K1ses}, and
hence $G'$, is unique up to conjugation by elements of
$\sg{\sfs}$.
\begin{rem}
If $\chi \neq 1$, then there are $d = \dimcol{K}{F}$ distinct
conjugates of $G'$ in $\Gal(K/F)$.
\end{rem}
\begin{proof}
Let $\tau$ be a generator of $\Gal(K/F)$. Applying Remark
\ref{gmodp} to $\chi(\tau)$, we see from the equality $\sfs^i
\tau' \sfs^{-i} = \sfs^{i(\chi(\tau')-1)}\tau'$ that $\sg{\tau'}$
is its own normalizer in $\Gal(K_1/K)$, and the result follows.
\end{proof}

 The various copies $G'$ of $G$ correspond to $d$ isomorphic subfields of
$K_1$, all having dimension $n$ over $F$. This can be used to
invert the map
$$
 J^{\Kpsi} \co \K{\Kpsi}/\mul{K}^n \ra \H[\chi]1{F}{\mu_n}
$$
in the following way. Suppose we are given a cocycle $c \in
\H[\chi]1{F}{\mu_n}$. By assumption $c(\s_1\s_2) =
{\s_1}_*(c(\s_2))c(\s_1)$, so that
$$
H = \set{\s \in \G[F] \suchthat c(\s) = 1}
$$
is a subgroup of $\G[F]$ (though in general
not a normal subgroup). It also follows that $c$ is well defined
on right cosets of $H$, so that $\dimcol{\G[F]}{H}$ divides $n$.
Adjusting $n$ if necessary, we may assume $H$ is of index $n$.
Now, $F_1 = \sep{F}^{H}$ has dimension $n$ over $F$, and
$$
K_1 = F_1 \tensor K
$$
is a cyclic extension, generated by some
$\alpha \in K$ such that $a = \alpha^n \in K$. Choosing $\alpha$
so that $\sfs$ defined by \eq{sfsdef} will satisfy $c(\sfs) = 1$,
it is immediate that $a \in \K{\Kpsi}$ and $c = J^{\Kpsi}(a)$.

%% The action on $K_1$.

\medskip

We now describe explicitly how the action of $\Gl$ on $K$ extends
to $K_1$.
 Let $\tau$ be a generator of $\Gal(K/F)$, and fix $a\in
\K{\Kpsi}$. Let
$$
r \equiv \Kpsi(\tau) \pmod{n}
$$
be an integer. By the assumption $\tau(a) \equiv a^r
\pmod{\mul{K}^n}$, so for some $\mu \in \mul{K}$ we have $\tau(a)
= \mu^n a^r$. Let $\alpha = \sqrt[n]{a}$ be a generator of $K_1$
over $K$. For every $j$, $\tau_1(\alpha) = \rho^j \mu \alpha^r$ is
a well defined automorphism of $K_1$. The fact that \eq{K1ses}
splits ensures that $j$ can be chosen so that $\tau_1^d(\alpha) =
\alpha$. Changing $\mu$ accordingly, we may assume that
\begin{equation}
\tau'(\alpha) = \mu \alpha^r,
\end{equation}
where $\tau \mapsto \tau'$ splits \eq{K1ses}. In particular,
choosing $F_1 = K_1^{\tau'}$, we can decompose $K_1 = K\tensor[F]
F_1$. The action of $\Gal(K/F)$ on $K_1$ will be used in Section
\ref{secact} to define an action of $\Gal(K/F)$ on cyclic
$K$-algebras.

The action of $\Gl$ on $K_1$ which we just described can be
refined in some cases, namely when $\Kpsi$ does not generate the
full character group $\ch{\Gal(K/F)}$. This refinement will be
used in the construction of generic examples in Section
\ref{secgeneric}.

\begin{lem}\label{decomposeL}
Let $F \sub L \sub K$ be an intermediate subfield and $e =
\dimcol{L}{F}$, a divisor of $d = \dimcol{K}{F}$. Then
$$\mul{L} \mul{K}^n = \K{1}\K{\nu^{d/e}}\dots \K{\nu^{(e-1)d/e}}.$$
\end{lem}
\begin{proof}
Using \Pref{K0F}, this is a special case of \Lref{decKL} with
$\Kpsi_0 = 1$.
%%
% Let $\tau$ be a generators of $\Gal(K/F)$, so that $L =
% K^{\tau^e}$ and $\Gal(K/L) = \sg{\tau^e}$. In particular, when
% viewed as a $\Gal(K/L)$-module, the eigenspace of $\mul{K}$ with
% respect to the trivial character is $$\set{a \in \mul{K}\suchthat
% \tau^e(a) \equiv a \pmod{\mul{K}^n}},$$ which contains
% $\K{\nu^{i}}$ if $d/e$ divides $i$, and intersects with it
% trivially (mod $\mul{K}^n$) otherwise. We are thus done by
% applying Proposition \ref{K0F} to the extension $K/L$.
\end{proof}

Let $a \in \K{\Kpsi}$ as before, and let $e$ denote the order of
$\Kpsi$ in the character group $\ch{\Gal(K/F)}$; equivalently,
that $\Ker(\Kpsi) = \sg{\tau^e}$; or that $\Kpsi \in
\sg{\tau^{d/e}}$. {}From the lemma it then follows that $a$ can be
taken to be an element of $L = K^{\tau^e}$.

\begin{prop}\label{goodx}
%Suppose $n$ is a prime power, and $d$ prime to $n$.
Let $L = K^{\tau^e}$, and assume $a \in \K{\Kpsi}\cap \mul{L}$ for
a character $\Kpsi$ of order $e$, a proper divisor of $d =
\dimcol{K}{F}$. Let $\alpha = \sqrt[n]{a}$ and $K_1 = K[\alpha]$.
Then $\Gal(K/F)$ can be extended to $K_1$ in a way that satisfies
$\tau^{e}(\alpha) = \alpha$. Moreover, if $\tau(\alpha) = \mu
\alpha^{c}$ for $\mu \in K$ and $c \equiv \Kpsi(\tau) \pmod{n}$,
then $\mu \in L$.

If $e = 2$ we can choose $c = -1$, and then $\mu \in F$.
\end{prop}
\begin{proof}
Choose $c = \Kpsi(\tau)$, and let $\mu \in \mul{K}$ be an element
such that $\tau(a) = \mu^n a^c$. Extend $\tau$ to $K_1$ by setting
$\tau(\alpha) = \mu \alpha^c$. Since $\tau^e(a) = a$,
$(\tau^e(\alpha)\alpha^{-1})^n = 1$. Replacing $\alpha$ by
$\rho^j\alpha$, we still have $\alpha^n = a$, but
$\tau^{e}(\alpha)\alpha^{-1}$ is now multiplied by
$\tau^{e}(\rho)\rho^{-1} = \rho^{j(\Kpsi(\tau^e)-1)}$. By
Remark~\ref{gmodp}, $\Kpsi(\tau)^e-1$ is invertible mod $n$, so
there is a (unique) $j$ for which $\tau^{e}(\alpha) = \alpha$.

Now compute that
$$\tau^{e-1}(\mu) \tau^{e-2}(\mu)^{c} \dots \mu^{c^{e-1}} \alpha^{c^e-1} = \tau^{e}(\alpha)\alpha^{-1} = 1,$$
so applying $\tau$ and dividing by the $c$-power of this equality,
we find that $\tau^{e}(\mu) = \mu$.

Finally, assume $c = -1$, so that $\tau(\alpha) = \mu
\alpha^{-1}$. Then $\mu = \alpha\tau(\alpha)$ which is a norm from
$K^{\tau^{2}}$ to $F$.
\end{proof}
This proposition can be viewed (in the case $e = n$) as a direct proof that \Eq{K1ses} is
a split extension.





\section{An example}\label{d=2}

Many of the statements in this paper hold only if $d =
\dimcol{K}{F}$ is prime to $n$, and providing counterexamples in
this direction is an interesting and worthy task. In this section
we give one such example, and hint on some of the problem it
arises.

Let $F =  \joinroot[]{\Q}{-2}$, and $n = 4$. Adding the forth root
of unity $\rho = \sqrt{-1}$, we obtain the field $K = F[\rho]$,
where of course $d = \dimcol{K}{F} = 2$. Let $\tau$ denote the
non-trivial automorphism of $K$ over $F$. Notice that $K$ contains
an eighth root of unity, namely $\rho_8 =
\frac{\sqrt{-2}}{2}(1-\rho)$ (and $\tau(\rho_8) = -\rho_8^{-1}$).

Let $a = 1 + \sqrt{2}$, then $\Norm[K/F](a) = -1$ and $\tau(a) =
-a^{-1} = \rho^2a^{-1}$, so that $a \in \K{\nu}$ (where $\nu$ is
the non-trivial character from $\Gal(K/F) = \sg{\tau}$ to $U_4 =
\set{\pm 1}$). Let $\alpha$ be the forth root of $a$. Then by
Proposition \ref{whenGal}, $K_1 = K[\alpha]$ is Galois over $F$.
Indeed, $\tau(\alpha) = \rho_8\alpha^{-1}$ preserves the defining
relation $\alpha^4 = a$, and so extends from $K$ to an
automorphism of $K_1$ over $F$. In fact, $\tau^2(\alpha) =
\tau(\rho_8\alpha^{-1}) = -\rho_8^{-1}(\rho_8\alpha^{-1})^{-1} =
\rho \alpha$, $\tau^4(\alpha) = \tau^2(\rho\alpha)=-\alpha$, and
$\tau^8(\alpha) = \alpha$; thus, $\Gal(K_1/F) = \sg{\tau}$ is a
cyclic group of order $8$.

%% Subfields of $K_1$ (by dimension):
%
%       K_1
%
%   K_1^{\tau^4}=K[\alpha^2]=\Q[i,\sqrt{-2},\alpha^2], \Q[\alpha]
%
%   K_1^{\tau^2}=K=\Q[i,\sqrt{-2}], \Q[\alpha^2], \Q[i\alpha^2]
%
%   \Q[i],\Q[\sqrt{2}], F=\Q[\sqrt{-2}]=K_1^\tau
%
%     \Q

In particular, the short exact sequence \eq{K1ses} does not split
in this case, and there is no subfield $F_1$ of $K_1$ such that $K
= K\tensor[F] F_1$. Also, The character $\chi$, which comes from
the action of $\Gal(K/F)$ on $\Gal(K_1/K)$, is trivial.

Finally, suppose $c \in \H[\chi]1{F}{\mu_4} = \Hom(\G[F],\Zn[4])$
is defined with the basic properties of Section \ref{Fco}, namely
that $c(\s)$ only depends on $\s|_{K_1}$, and that $c(\s) = 1$ for
$\s \in \G[K]$. Since $\tau^2$ is the identity on $K$ we have that
$c(\tau^2) = 1$, so $c$ has order at most $2$, and cannot be of
order $n = 4$.

%% The Hochshild-Serre sequence become in our case:
% 2-->4-->4-->?; the first map is injective, so the second takes 4-->2,
% and this is why there is no element of order 4 in its image.

For the rest of the paper we again assume that $d = \dimcol{K}{F}$
is prime to $n$.







\section{Central Simple Algebras}\label{csa}

Denote by $\Brn(F)$ the exponent $n$ part of the Brauer group of
$F$. Applying the long exact sequence of cohomology groups to
the Kummer sequence \eq{Kummer}, % Actually we need the sequence for F <---!
and using the identification $\H2{F}{\mul{\sep{F}}} = \Br(F)$ of
the crossed products construction \cite{Jacobson}, we obtain the
isomorphism $\H[\nu]2{F}{\mu_n} = \Brn(F)$, where $\nu$ is the
natural action of $\G[F]$ on $\mu_n$, defined by \eq{nudef}. Since
in general the action of $G$ on tensor products $M\tensor N$ is
diagonal, the cup product $\cup \co \H{i}{G}{M} \tensor
\H{j}{G}{N} \ra \H{i+j}{G}{M\tensor
N}$ is in our notation a map %
$$
\H[\Kpsi]{1}{F}{\mu_n} \tensor \H[\Kpsi']{1}{F}{\mu_n}
\stackrel{\cup}{\lra} \H[\Kpsi\Kpsi']2{F}{\mu_n}.
$$
In particular, for every character
$\Kpsi$, letting $\chi = \nu\Kpsi^{-1}$, we have a map
$$
\H[\Kpsi]{1}{F}{\mu_n} \tensor \H[\chi]{1}{F}{\mu_n}
\stackrel{\cup}{\lra} \Brn(F),
$$
which we explicitly describe below. The case $\Kpsi =
1$ is well
known (\eg\ \cite[p.~555]{RT}% This is Section 1 of that paper.
): an element $\gamma \in \H[1]{1}{F}{\mu_n} = \Hom(\G[F],\Zn)$
has kernel $\G[]_1 = \Ker(\gamma)$, of index equal to the order of
$\gamma$. Assuming $\gamma$ has order $n$, let $F_1 =
\sep{F}^{\G[]_1}$ be the invariant subfield, then $F_1$ is a
cyclic extension of dimension $n$ over $F$. Now, given $(b) \in
\H[\nu]{1}{F}{\mu_n}$ (which corresponds to an element $b \in
\mul{F}$ by \eq{isom1}), $\gamma\cup (b)$ is the Brauer class of
the cyclic algebra
$$(F_1/F,\sfs,b) = F_1[z \subjectto z f z^{-1} = \sfs(f),\, z^n=b],$$
where $\sfs$ is the generator of $\Gal(F_1/F)$ specified by
$\gamma(\sfs) = 1$. This is always an $F$-central simple algebra
of degree $n$. When the construction is applied to $K = F[\rho_n]$
instead of $F$, we get what is called a symbol algebra,
$$(a) \cup (b) = \symbalg[K]{a}{b} = K[x,y \subjectto x^n=a,\,
y^n=b,\, yxy^{-1} = \rho x],$$ %
which is another notation for $(\joinroot{K}{a}/K,\sfs,b)$.

%
We call a pair of generators $x,y$ a \defin{standard pair}  if
$x^n = a$, $y^n = b$ and $yxy^{-1} = \rho x$. The symbol
$\symbalg[K]{a}{b}$ is multiplicative in both variables, that is
$$
\symbalg[K]{a}{b_1b_2} \sim \symbalg[K]{a}{b_1} \tensor
\symbalg[K]{a}{b_2},$$
$$
\symbalg[K]{a_1a_2}{b} \sim \symbalg[K]{a_1}{b} \tensor
\symbalg[K]{a_2}{b}.
$$
By Merkurjev-Suslin theorem the symbol algebras of degree $n$
generate $\Brn(K)$. Strictly speaking, not every algebra of the
form $\symbalg[K]{a}{b}$ is a cyclic algebra, since $K[x] =
\joinroot{K}{a}$ need not be a field. However it is always split
by the cyclic group $C_n$ (in the sense of \cite{TA85}, given at
the end of the introduction).

Let $\Kpsi,\Kpsi' \co\Gl\ra U_n$ be characters such that
$\Kpsi\Kpsi' = \nu$. To describe the cup product
$\H[\Kpsi]1{F}{\mu_n} \cup \H[\Kpsi']1{F}{\mu_n}$, recall that the
maps
$$
J^\Kpsi \co \K{\Kpsi}/\mul{K}^n \ra \H[\Kpsi']1{F}{\mu_n},
$$
$$
J^{\Kpsi'} \co \K{\Kpsi'}/\mul{K}^n \ra \H[\Kpsi]1{F}{\mu_n}
$$ are onto (Corollary \ref{Jisisom}).

\begin{prop}\label{cupcomp}
% Assume $n$ is a prime power, and $d = \dimcol{K}{F}$ prime to $n$.
Let $\Kpsi,\Kpsi'\co\Gl\ra U_n$ be characters such that
$\Kpsi\Kpsi' = \nu$. Then for every $a \in \K{\Kpsi}$ and $b \in
\K{\Kpsi'}$, we have that
$$J^{\Kpsi}(a)\cup J^{\Kpsi'}(b) \sim
\cores[K/F]\symbalg[K]{a}{b}^{\tensor[] d'},$$ %
where $d'$ is the inverse of $d$ modulo $n$.
\end{prop}
\begin{proof}
{}From Theorem~\ref{KHFthm} we have that $(a) = \res[K/F]
J^{\Kpsi}(a)$ and $(b) = \res[K/F] J^{\Kpsi'}(b)$, where $(a),(b)
\in \H1{K}{\Zn}$. Thus, the
projection formula gives %
%
%% The general projection formula is:  cor(a \cup \res(b)) = \cor(a) \cup b.
%
\begin{eqnarray*}
\cores \symbalg[K]{a}{b} & = & \cores ((a) \cup (b)) \\
    & = & \cores (\res J^{\Kpsi}(a) \cup \res J^{\Kpsi'}(b)) \\
    & = & J^{\Kpsi}(a) \cup \cores (\res J^{\Kpsi'}(b)) \\
    & = & d\cdot J^{\Kpsi}(a) \cup J^{\Kpsi'}(b).
\end{eqnarray*}
\end{proof}

Notice that if $D$ has degree $n$ over $K$, then $\cores[K/F] D$
has degree $n^{\updimcol{K}{F}}$ over $F$. From time to time we
will use degree argument to write that the corestriction is equal
(and not just similar) to a given algebra.

\begin{cor}\label{projnu}
Under the assumptions of Proposition \ref{cupcomp},
%if $a \in \K{\Kpsi}$, $b \in \K{\Kpsi'}$ and $\Kpsi\Kpsi' = \nu$, then
$$\res[K/F]\cores[K/F]\symbalg{a}{b} =\symbalg{a}{b}^{\tensor[] d}.$$
\end{cor}
\begin{proof}
By the commutativity of restriction with the cup product,
\begin{eqnarray*}
\res\ \cores \symbalg[K]{a}{b} & = & d\cdot \res (J^{\Kpsi}(a)
\cup
J^{\Kpsi'}(b)) \\
    & = & d \cdot (a)\cup (b) \\
    & = & [\symbalg[K]{a}{b}^{\tensor[] d}].
\end{eqnarray*}
\end{proof}

We now want to compute the algebra $\cores[K/F] \symbalg[K]{a}{b}$
for $a \in \K{\Kpsi}$ and $b \in \K{\Kpsi'}$, without the
assumption $\Kpsi\Kpsi' = \nu$. Just as above, the general
projection formula yields
$$\cores ((a) \cup (b)) = d\cdot J^{\Kpsi}(a) \cup J^{\Kpsi'}(b),$$
but the corestriction here is from $\Brn(K) = \H2{K}{\Zn}$ to
$\H[\Kpsi\Kpsi']2{F}{\mu_n}$, which is in general {\it not} the
corestriction of central simple algebras. To compute the algebra
$\cores[K/F]\symbalg[K]{a}{b}$, we need to know what are the
conjugate algebras $\tau \symbalg{a}{b}$ for $\tau \in \Gl$ (see
\cite[p.~220]{rowenbook}).

%The absolute Galois group of $F$ acts on the collection of central simple algebras over $K$ by its action on defining relations.
If $\s \in \G[K]$, then $\s \symbalg[n;K]{a}{b}  =
\symbalg[n;K]{\s a}{\s b}$. More generally, let $\s \in \G[F]$ be
an automorphism. Letting
$$D = K[x,y \subjectto x^n=\s a,\, y^n=\s
b,\, yxy^{-1}=\s \rho = \rho^t]
$$
where $t = \nu(\s)$ and $tt' \equiv 1 \pmod{n}$, we see that
$x^{t'},y$ generate $D$, with $yx^{t'}y^{-1} = \rho$, so by our
definition $\s \symbalg[K]{a}{b} = D$ is the symbol algebra
$\symbalg[K]{\s a^{t'}}{\s b}$.


\begin{prop}\label{corF} % This is 8.3
Suppose $a \in \K{\Kpsi}$ and $b \in \K{\Kpsi'}$, where $\Kpsi
\Kpsi' \neq \nu$.  Then the corestriction of algebras is
$$\cores[K/F]\symbalg[K]{a}{b} \sim F.$$% = \symbalg{a}{b}^{\tensor[]\Sigma_{\psi\psi'\Kpsi'^{-1}}}.$$
\end{prop}
\begin{proof}
Let $\phi = \Kpsi \Kpsi' \nu^{-1}$ ($\phi \neq 1$ by the
assumption). For $a\in\K{\Kpsi}, b \in \K{\Kpsi'}$ we have $\s
\symbalg[K]{a}{b} \sim \symbalg[K]{\s a}{\s b}^{\tensor[]
\nu(\s)^{-1}} \sim \symbalg[K]{a}{b}^{\tensor \phi(\s)}$.

It is known (\eg\ \cite{Srinivas}) that
$$
\res[K/F]\cores[K/F]\symbalg{a}{b} = \prod_{\tau \in \Gl}
\t\symbalg{a}{b},
$$
so we have
\begin{eqnarray*}
\res[K/F]\cores[K/F]\symbalg{a}{b}
    & = & \prod\symbalg{a}{b}^{\tensor[] \phi(\tau)} \\
    & = & \symbalg{a}{b}^{\tensor[] \sum \phi(\tau)}
\end{eqnarray*}
and the result follows from Corollary \ref{Hsg}, and the
injectivity of $\res[K/F]$.
\end{proof}

Let $\RR{K} \co \KK(K)/n \ra \Brn(K)$ denote the norm residue map
$$
\RR{K} \co \symb{a}{b} \mapsto \symbalg[K]{a}{b}.
$$
\begin{cor}
There is no map making the diagram \eq{predef} commutative, unless
$\KK(F)/n = 0$. \ifXY
\begin{equation}\label{predef}
\xymatrix{
    \KK(F)/n \ar@{.>}[d] \ar[r] ^{\res}
        & \KK(K)/n \ar[d]^{\RR{K}} % &
\\
    \Brn(F) \ar[r]^{\res}
        & \Brn(K) % \ar[r]^{\cores} & \Brn(F)
}
\end{equation}
\else\nonumber
\medskip\nonumber
Extended diagram.\nonumber
\medskip\nonumber
\fi % For ifXY
\end{cor}
\begin{proof}
Let $\RRh{F}$ denote a map making the diagram commute. Then for
$\alpha,\beta \in F$ we have that
$$
\res[K/F] \RRh{F}\symb[F]{\alpha}{\beta} =
\RR{K}\res[K/F]\symb[F]{\alpha}{\beta} =
\symbalg[K]{\alpha}{\beta}.
$$

Taking corestriction,  we get
$$\(\RRh{F} \symb[F]{\alpha}{\beta}\)^{\tensor d} =
 \cores[K/F]
\symbalg[K]{\alpha}{\beta} \sim F,$$  so that $\RRh{F} = 0$.
\end{proof}



By the above computation, the $\Gal(K/F)$-eigenspace of $\Brn(K)$
with respect to a character $\psi$ is the subgroup
$$
\sg{[\symbalg{a}{b}] \suchthat a\in \K{\Kpsi},\,b \in \K{\Kpsi'},
\Kpsi\Kpsi' = \psi}.$$ Combining Corollary~\ref{projnu} and the
proposition, we obtain the projection formula for $K/F$: given
$a,b\in \mul{K}$, decompose $a = \prod{a_\psi}$ and $b =
\prod{b_\psi}$ mod $\mul{K}^n$ with $a_\psi,b_\psi \in \K{\psi}$;
then
\begin{equation}\label{genproj}
\res[K/F]\cores[K/F]\symbalg{a}{b} \sim
 \tensor_{\Kpsi\Kpsi'=\nu}
\symbalg{a_\Kpsi}{b_{\Kpsi'}}^{\tensor[] d}.
\end{equation}

\begin{cor}\label{res=cor}
Let $\Kpsi,\Kpsi'$ be characters with $\Kpsi\Kpsi' = \nu$, and let
$a \in \K{\Kpsi}$, $b \in \K{\Kpsi'}$, and $D =
\symbalg[n,K]{a}{b}$.

Let $D_0$ be a central simple algebra of degree $n$ over $F$. The
following conditions are equivalent.

a. $\res_{K/F} D_0 = D$.

b. $D_0 \sim \cores[K/F] D^{\tensor[] d'}$ where $d'$ is an
inverse of $d$ modulo $n$.
\end{cor}
%\begin{proof}
%If $\res[K/F] D_0 = D$, then $\cores[K/F]D^{\tensor[] d'} =
%\cores[K/F]\res[K/F] D_0 ^{\tensor[]d'} \sim D_0^{\tensor dd'} \sim D_0$
%so b. holds. The other direction follows from \Eq{genproj}.
%\end{proof}

\begin{cor}\label{atob}
Let $D$ be a central simple algebra of degree $n$ over $F$. If $D
\tensor[F] K = \symbalg[K]{a}{b}$ and $a \in \K{\Kpsi}$, then we
can assume $b \in \K{\Kpsi'}$ where $\Kpsi\Kpsi' = \nu$.
\end{cor}
\begin{proof}
Let $b_{\Kpsi'}$ be the component of $b$ in $\K{\Kpsi'}$, then
$\cores[K/F]\symbalg[K]{a}{b} =
\cores[K/F]\symbalg[K]{a}{b_{\Kpsi'}}$ by Proposition \ref{corF}.
By the previous corollary $D \sim \cores[K/F]
\symbalg[K]{a}{b}^{\tensor[] d'} = \cores[K/F]
\symbalg[K]{a}{b_{\Kpsi'}}^{\tensor[] d'}$, and then also
$\res_{K/F}D = \symbalg[K]{a}{b_{\Kpsi'}}$.
\end{proof}







\section{Second cohomologies of $F$} \label{sec:H2}

In this section we consider eigenspaces of the second cohomology
group $\H2{K}{\mu_n}$ for $K = F[\rho]$. Applying the cup product
to the decomposition $\H1{K}{\mu_n} =
\oplus_{\Kpsi}\H1{K}{\mu_n}^{(\Kpsi)}$, we obtain the subspaces
$$\H1{K}{\mu_n}^{(\Kpsi)} \cup \H1{K}{\mu_n}^{(\Kpsi')} \sub
\H2{K}{\mu_n}^{(\Kpsi\Kpsi')}.$$
The 'onto' part of the
Merkurjev-Suslin Theorem \cite[Thm~8.5]{Srinivas}, applied to $K$,
shows that $\H1{K}{\mu_n} \cup \H1{K}{\mu_n} = \H2{K}{\mu_n}$, so
in fact
\begin{equation}\label{H2dec}
\H2{K}{\mu_n} = \oplus_{\psi}(\sum_{\Kpsi\Kpsi' = \psi} \H1{K}{\mu_n}^{(\Kpsi)} \cup \H1{K}{\mu_n}^{(\Kpsi')}).
\end{equation}
and in particular for every $\psi$,
\begin{equation}\label{H2dec+}
\H2{K}{\mu_n}^{(\psi)} = \sum_{\Kpsi\Kpsi' = \psi} \H1{K}{\mu_n}^{(\Kpsi)} \cup \H1{K}{\mu_n}^{(\Kpsi')}.
\end{equation}

Similarly to Corollary \ref{H1KH1F}, we have the following
isomorphism:
\begin{thm}\label{H2KH2F}
For every character $\psi \co \Gl \ra U_n$,
$$\H2{K}{\mu_n}^{(\psi)} \isom \H[\theta]2{F}{\mu_n},$$
where $\theta = \nu^2 \psi^{-1}$.
\end{thm}
\begin{proof}
We will show that the restriction $\res \co \H[\theta]2{F}{\mu_n}
\ra \H2{K}{\mu_n}$ (which is obviously injective) covers
$\H2{K}{\mu_n}^{(\psi)}$.

Recall the general projection formula for cohomology groups: let
$M,N$ be $\G[F]$-modules, $a\in \H1{F}{M}$ and $b \in \H1{K}{N}$.
Then
$$\cores(\res(a)\cup
b) = a \cup \cores(b),$$ where the restriction is $\H1{K}{M}\ra\H1{F}{M}$, the
first corestriction is $\H2{F}{M\tensor N}\ra\H2{K}{M \tensor N}$
and the second %corestriction
is $\H1{F}{N}\ra\H1{K}{N}$.

\forget
 \ifXY
\begin{equation*}% \label{}
\xymatrix{ % @R=30pt@C=36pt, C = length of --- lines.
    \H1{K}{\m}^{(\Kpsi)} \quad\cup\quad
    \H1{K}{\m}^{(\Kpsi')} \ar[r]
\ar@<10ex>[d]^{\cores[\theta\chi^{-1}]}
    & \H2{K}{\m}^{(\Kpsi\Kpsi')} \ar[d]^{\cores[\theta]}
\\
    \H[\chi]1{K}{\m} \quad\cup\quad
    \H[\chi]1{K}{\m} \ar@<10ex>[u]_{\res_{\chi}} \ar[r]
    & \H[\theta]2{K}{\m}
}
\end{equation*}
\else % For \ifXY
\bigskip
$\H1{}{}$, $\H1{}{}$ and $\H2{}{}$.
\bigskip
\fi % For \ifXY
\forgotten


Let $\Kpsi,\Kpsi'$ be arbitrary characters. We choose $M = N =
\mu_n$ with the actions of $\G[F]$ on $M,N$ via $\chi =
\nu\Kpsi^{-1}$ and $\theta \chi^{-1}$, respectively, so our
restriction and corestriction are the  maps
\begin{eqnarray*}
\res[\chi]      & : & \H[\chi]1{F}{\m}\ra\H1{K}{\m},\\
\cores_{\theta} & : & \H2{K}{\m} \ra\H[\theta]2{F}{\m},\\
\cores_{\theta\chi^{-1}} & : & \H1{K}{\m}\ra \H[\theta\chi^{-1}]1{F}{\m}.
\end{eqnarray*}

We apply the projection formula to an element
$$
u = c \cup c' \in \H1{K}{\mu_n}^{(\Kpsi)} \cup
\H1{K}{\mu_n}^{(\Kpsi')}
$$
in the decomposition \eq{H2dec}. % Note: we do not assume $\Kpsi\Kpsi' = \psi$.
{}From Corollary \ref{H1KH1F} we
have that
\begin{equation}\label{reson}
\res[\chi] \H[\chi]1{F}{\m} = \H1{K}{\m}^{(\Kpsi)}
\end{equation}
and similarly
\begin{equation}\label{reson'}
\res[\chi'] \H[\chi']1{F}{\m} = \H1{K}{\m}^{(\Kpsi')}
\end{equation}
for $\chi' = \nu\Kpsi'^{-1}$, so we can write $c = \res[\chi]a$
and $c' = \res[\chi']a'$ where $a \in \H[\chi]1{F}{\m}$ and $a'
\in \H[\chi']1{F}{\m}$. It follows that
$$
\cores_\theta(c\cup c') = \cores_{\theta}(\res[\chi](a) \cup
\res[\chi'](a')) =
 a \cup \cores[\theta\chi^{-1}]\res_{\chi'}(a').
$$

If $\chi\chi' = \theta$, then we find that $\cores_{\theta}(c \cup
c') = d \cdot (a\cup a')$ since $\cores_{\chi'}\res[\chi']$ is
multiplication by $d$. On the other hand if $\chi\chi' \neq
\theta$ then by Corollary \ref{diff} we have that
$\cores[\theta\chi^{-1}]\res_{\chi'}(a') = 0$, so that
$\cores_{\theta}(c\cup c') = 0$.

Applying $\cores_{\theta}$ to \eq{H2dec} and recalling that
$\cores_{\theta}$ is onto $\H[\theta]2{F}\m$, we find that
\begin{equation}\label{H2Fdec}
\H[\theta]2{F}\m  = \sum_{\chi\chi' = \theta} \H[\chi]1{F}{\m}\cup \H[\chi']1{F}{\m}.
\end{equation}

Finally applying $\res[\theta] = \res[\chi]\cup \res[\chi']$ (for
every $\chi\chi' = \theta$), and using Equations \eq{H2dec+},
\eqs[ and ]{reson}{reson'}, we obtain
$$\res[\theta] \H[\theta]2{F}\m  = \sum_{\Kpsi\Kpsi' = \psi} \H1{K}{\m}^{(\Kpsi)} \cup \H1{K}{\m}^{(\Kpsi')} = \H2{K}{\m}^{(\psi)}.$$
\end{proof}

The onto part of Merkurjev-Suslin for the field $F$ gives the
equality $\H[\nu^2]2{F}{\mu_n} = \H[\nu]1{F}{\mu_n} \cup
\H[\nu]1{F}{\mu_n}$, which is of course much better than
Equation~\eq{H2Fdec} for $\psi = 1$ (and $\theta = \nu^2$). It is
not known if $\H[\nu]2{F}{\mu_n} = \H[\nu]1{F}{\mu_n} \cup
\H[1]1{F}{\mu_n}$ in general (see Question \ref{Q:gen} below). A
related negative result is given in the remark after Lemma 29 in
\cite{HKRT}.



\forget
Since $\cores_{K/F}$ is onto $\Brn(F)$ and $d$ is prime to $n$, we
proved [due to Merkurjev]
\begin{cor}\label{genF}
If $n = p^m$ and $(d,n)=1$, then the $n$-part of $\Br(F)$ is
generated by algebras of the form $\cores[K/F]\symbalg[n,K]{a}{b}$
for $a \in \K{\Kpsi}$ and $b \in \K{\Kpsi'}$, $\Kpsi\Kpsi' = \nu$.
\end{cor}
\forgotten







\section{Action of $\Gal(K/F)$ on $\symbalg[K]{a}{b}$}\label{secact}

In Section \ref{csa} we have seen that if $a \in \K{\Kpsi}$ and $b
\in \K{\Kpsi'}$ where $\Kpsi\Kpsi' = \nu$,
%and $d = \dimcol{K}{F}$ is prime to $n$, % Does not seem to be necessary.
then $\tau \symbalg[K]{a}{b} =\symbalg[K]{a}{b}$ for every $\tau
\in \Gal(K/F)$.

\begin{prop}\label{GactD}
Let $\Kpsi,\Kpsi'$ be characters such that $\Kpsi \Kpsi' = \nu$,
$a \in \K{\Kpsi}$ and $b \in \K{\Kpsi'}$.

There is a natural extension of the action of $\Gl$ on $K$ to an
action on $D = \symbalg[K]{a}{b}$.
\end{prop}
\begin{proof}
As in Section \ref{fields}, fix a generator $\tau$ of $\Gal(K/F)$,
let $r \equiv \Kpsi(\tau)$, $r' \equiv \Kpsi'(\tau)$, and $t = rr'
\equiv \nu(\tau)$. By the assumption on $a,b$, there exist
$\mu,\eta \in \mul{K}$ such that
$$
\tau(a) = \mu^n a^r\qquad \mbox{and}\qquad \tau(b) = \eta^n
b^{r'}.
$$
Define $\tau$ on
$$
D = K[x,y\subjectto x^n = a,\, y^n = b,\, yxy^{-1} = \rho x]
$$
by
$$\tau(x) = \mu x^r,$$
$$\tau(y) = \eta y^{r'},$$
where $\mu,\eta$ are chosen (\ie\ multiplied by appropriate powers
of $\rho$) so that $\tau^d(x) = x$ and $\tau^d(y) = y$. Since
$\tau(yxy^{-1}) = \mu y^{r'}x^ry^{-r'} = \mu \rho^{rr'} x^r =
\tau(\rho x)$, this action respects the defining relations, and is
thus well defined on $D$. Finally, $\tau^d$ is the identity on
$K[x,y] = D$.
\end{proof}

\begin{lem}\label{res=cor=fixed}
Let $\Kpsi,\Kpsi'$ be characters with $\Kpsi\Kpsi' = \nu$, and let
$a \in \K{\Kpsi}$, $b \in \K{\Kpsi'}$. Let $D =
\symbalg[n,K]{a}{b}$ be a cyclic algebra over $K$.

Let $D_0$ be a central simple algebra of degree $n$ over $F$. The
following conditions are equivalent.

a. $\res_{K/F} D_0 = D$.

b. $D_0 \sim \cores[K/F] D^{\tensor[] d'}$ where $d'$ is an
inverse of $d$ modulo $n$.

c. $D_0$ is the invariant subalgebra $D^{\Gl}$ under the action
defined in Proposition \ref{GactD}.
\end{lem}
\begin{proof}
The equivalence of a. and b. is Corollary \ref{res=cor}.

Let $G =
\Gal(K/F)$. In general the invariant subalgebra $D^{G}$ is central
simple of degree $n$ over $F = K^{G}$ \cite[Cor.~7.2.15]
{rowenbook}. Since $\res_{K/F} D^G = D^G \tensor[F] K = D$ and the
restriction $\Brn(F) \ra \Brn(K)$ is injective, a. is equivalent to c.
\end{proof}

The following theorem is proved in \cite{Merk83} for $n$ prime.
\begin{thm}\label{cyclicF} % 10.3
Let $F$ be a field of characteristic prime to $n = p^m$, $K =
F[\rho_n]$, and assume that $d = \dimcol{K}{F}$ is prime to $n$.

Let $D_0$ be a central simple algebra of degree $n$ over $F$.
$D_0$ is cyclic if and only if there are $a \in \K{\nu}$  and
$\beta\in \mul{F}$ where $K_1 = \joinroot{K}{a}$ is a field, such
that $\res[K/F] D_0 = \symbalg[n,K]{a}{\beta}$.
\end{thm}
\begin{proof}
If $a,\beta$ are as assumed, then $G$ acts on $D =
\symbalg[K]{a}{\beta}$ (by Proposition \ref{GactD}), $K_1 =
\joinroot{K}{a}$ is a subfield of $D$, and $F_1 = K_1^G$ is a
cyclic subfield of $D^G$ (Proposition \ref{whicharecyclic}).

On the other hand, if $D_0$ is cyclic of degree $n$ over $F$, it
has a cyclic subfield $F_1$ of dimension $n$ over $F$, which again
by Proposition \ref{whicharecyclic} has the form $F_1 = K_1^G$
where $K_1 = \joinroot{K}{a}$ is a field, for some $a \in
\K{\nu}$. As $D_0$ is cyclic, we can write $D_0 =
(F_1/F,\sfs,\beta)$ for some $\beta \in F$, and then $D = D_0
\tensor K = (K_1/K,\sfs,\beta) = \symbalg[K]{a}{\beta}$.
\end{proof}

\begin{defn}\label{defqs}
An algebra $D_0$ of degree $n$ over $F$ is called a
\defin{quasi-symbol} (of type $(\Kpsi,\Kpsi')$) iff it satisfies
the conditions of Lemma \ref{res=cor=fixed}.
\end{defn}
We give a description of quasi-symbols in terms of their splitting
fields in \Pref{propqs} below.

Under the assumption that $\dimcol{K}{F}$ is
prime to $n$, the corestriction from $\Brn(K)$ to $\Brn(F)$ is onto, so by the
projection formula \eq{genproj}, $\Brn(F)$ is generated by quasi-symbols.
\begin{ques}\label{Q:cyc}
Are quasi-symbols cyclic over $F$?
\end{ques}
By the \Tref{cyclicF}, quasi-symbols of type $(1,\nu)$ (or
$(\nu,1)$) are cyclic, but nothing is known in general about the
cyclicity of other quasi-symbols.

Merkurjev proved in \cite{Merk83} that
$\Brn(F)$ is generated by cyclic
algebras (of degree dividing $n$) if $\dimcol{K}{F} \leq 3$.
This is not known if $\dimcol{K}{F} \geq 4$.
\begin{ques}\label{Q:gen}
Is $\Brn(F)$ generated by (the classes of) algebras which are split by $C_n$?
\end{ques}
Of course, a positive answer to the first question will answer the
second. One approach is to compute the corestriction of cyclic
algebras from $K$ to $F$; we return to this in Section
\ref{sec:cores}.


\smallskip

We conclude this section with a remark concerning involutions of
the second kind. We touch this subject again in Proposition
\ref{NE}. In Proposition \ref{corF} we saw that algebras of the
form $A = \symbalg[K]{a}{b}$ where $a \in \K{\Kpsi}$ and $b \in
\K{\Kpsi'}$ and $\Kpsi \Kpsi' \neq \nu$, have trivial
corestriction. If $\dimcol{K}{F} = 2$, this is known to be
equivalent to the existence of an involution of the second kind on
$A$ for which the invariant subfield  of the center is $F$. Since
we assume $\dimcol{K}{F} = 2$, there are only two cases to
consider: either $a,b\in \K{1} = \mul{F}\mul{K}^n$, or
$a,b\in \K{\nu}$. %In the following proposition,
%which is similar in nature to Proposition~\ref{secact}, we
%construct the involutions in these cases.
A similar configuration is studied in \cite[Thm. 3.5]{Nir}; it is shown there
that if $\dimcol{K}{F} = 2$ and a cyclic algebra $A = \symbalg[K]{\alpha}{b}$
has an involution of the second kind over $F$ where $\alpha \in F$, then
it can be assumed that $b \in F$. This theorem covers also
the case of even $n$ (where it is assumed that $\tau(\rho) = \rho^{-1}$).

Similarly to Proposition~\ref{GactD}, we now use the action of
$\Gl$ on $K[x]$ and $K[y]$ (where $x,y$ are a standard pair of
generators of $A$) to define an involution of the second kind on
$A$.

Assume that $\dimcol{K}{F} = 2$, and let $\tau \in \Gl$ be the
non-trivial automorphism. Note that $\tau(\rho) = \rho^{-1}$. Let
$\Kpsi (\tau) = \epsilon$ define a character, where $\epsilon =
\pm 1$ is fixed. Let $a,b\in \K{\Kpsi}$ and
$$
A = \symbalg[K]{a}{b} = K[x,y\subjectto x^n=a,\,y^n=b,\,yx=\rho
xy].
$$
By assumption there are $\mu,\eta \in \mul{K}$ such that $\tau(a)
= \mu^n a^{\epsilon}$ and $\tau(b) = \eta^n b^{\epsilon}$. Define
a linear map $(*)$  on $A$ by
$$
(\sum{k_{ij}x^i y^{j}})^* =
\sum{\tau(k_{ij})(y^*)^j (x^{*})^i}
$$
where $x^{*} = \mu x^{\epsilon}$ and $y^{*} = \eta y^{\epsilon}$.
The relations $x^n = a$ and $y^n = b$ are automatically preserved,
and for both possible values of $\epsilon$, $(yx)^{*} = (\rho x
y)^* = \rho^{-1} \mu y^\epsilon \eta x^\epsilon = \mu \eta
x^{\epsilon} y^{\epsilon} = x^* y^*$. Finally, $x^{**} = x$ and
$y^{**} = y$ by the choice of $\mu,\eta$, so that $u \mapsto u^*$
is an involution.







\section{Albert's cyclicity criterion} \label{sec:alb}

Every cyclic algebra $D = (F_1/F, \sfs, b)$ has by definition an
element $z \in D $ such that $z^n = b$ is central. In the "special
results" chapter (Chapter XI) of his seminal book \cite{Albert},
Albert proves the converse, which is known as Albert's cyclicity
criterion: a central simple algebra $D$ of prime degree $p$ over a
field $F$ of characteristic prime to $p$ is cyclic, if and only if
$u^p \in F$ for some non-central $u\in D$. Of course the assertion
follows from Kummer theory if $F$ has $p$-roots of unity. Albert
remarks \cite[p.~175]{Albert} that the criterion fails if the
degree is not a prime, and indeed in \cite{Albp} he provides an
example of a non-cyclic algebra of degree (and exponent) $4$, with
an element $u$ such that $u^4$ is central but $u^2$ is not. In
Theorem \ref{Albert} below we generalize Albert's criterion, and
show that it holds for algebras of degree $n = p^m$ over a field
$F$, assuming that $d = \dimcol{K}{F}$ is prime to $n$, where as
always $K = F[\rho_n]$.

The following is a slightly more general version of
Theorem~\ref{cyclicF}.
\begin{prop}\label{degcor} % 11.1
Let $n$ be a prime power, $K = F[\rho]$ where $\rho$ is a
primitive $n$-root, and assume $d=\dimcol{K}{F}$ is prime to $n$.

Let $a \in \mul{K}$ and $\beta \in \mul{F}$. Then $A =
\cores[K/F]\symbalg[n,K]{a}{\beta}$ is split by $C_n$.
\end{prop}
\begin{proof}
Since $\beta \in \mul{F} \sub \K{1}$, we can by \Cref{atob}
replace $a$ by its $\nu$-component, so we assume $a \in \K{\nu}$.
Let $D = \symbalg[K]{a}{\beta^d}$. The projection formula for
$K/F$ (Equation~\eq{genproj}) gives $\res[K/F] A \sim D$, which is
an algebra of degree $n$. Since $d$ is prime to $n$, the index of
$A$ divides $n$.

Let $D_0$ be a the algebra of degree $n$ over $F$ which is similar
to $A$. Then $\res_{K/F} D_0 = D$, so if $\joinroot{K}{a}$ is a
field, $D_0$ is cyclic by Theorem \ref{cyclicF}.
%
For the general case, let $n'$ denote the order of $a$ in
$\md{K}$; equivalently, $n'$ is the maximal divisor of $n$ such
that $a \in \mul{K}^{n/n'}$. Write $a = c^{n/n'}$ for some $c \in
\mul{K}$; then
$$D = \symbalg[n,K]{a}{\beta^d} =
\symbalg[n,K]{c^{n/n'}}{\beta^d} \sim
\symbalg[n',K]{c}{\beta^d},$$
%
and $\joinroot[n']{K}{c}$ is a field. In particular,
$$
A \sim \cores[K/F] D^{\tensor d'} \sim
\cores[K/F]\symbalg[n',K]{c}{\beta},
$$
where $d'$ is the inverse of $d$ modulo $n$. Let $D_1$ be the
algebra of degree $n'$ similar to $A$. By
Lemma~\ref{res=cor=fixed} we have that $\res_{K/F}D_1 =
\cores[K/F]\symbalg[n',K]{c}{\beta}$. In order to show that $D_1$
is cyclic, we need to know that $c$ is in the $\nu$-component of
the decomposition of $\md[n']{K}$. Indeed, Let $\tau$ be a
generator of $\Gl$, and $t = \nu(\tau)$. For some $\mu \in
\mul{K}$ we have $\tau(a) = \mu^n a^t$, so that $\tau(c) =
(\rho^{i}\mu)^{n'}c^t$ for some $i$. We are now done by Theorem
\ref{cyclicF}.
\end{proof}


\begin{exmpl}
Assume that the $p$-part of the group of units in $K$ is of order
$n$, and take $a = \rho^{n/n'}$ in the above corollary. Then $a$
is of order $n'$ in $\md{K}$, and
$$\cores[K/F] \symbalg[K]{\rho^{n/n'}}{\beta} \sim (F_1/F,\sfs,\beta^d),$$
a cyclic algebra of degree $n'$, with $F_1 \sub K_1 =
\joinroot[n']{F}{\rho}$.
\end{exmpl}


\begin{cor}\label{resFcyc}
With $F,n,K$ as in the proposition, let $D_0$ be a division
algebra of degree $n$ over $F$. If $\res[K/F] D_0 =
\symbalg{a}{\beta}$ for some $a\in \mul{K}$ and $\beta \in
\mul{F}$, then $D_0$ is cyclic.
\end{cor}
\begin{proof}
 Let $d'$ be an
inverse of $d$ modulo $n$. By Lemma \ref{res=cor}\,\, $D_0 \sim
\cores[K/F] \symbalg{a}{\beta^{d'}}$, so by the proposition $D_0$
is similar to a cyclic algebra of degree dividing $n$, and being a
division algebra it is cyclic.
\end{proof}

This leads directly to the general form of Albert's theorem.
\begin{thm}\label{Albert}
Let $F$ be a field as above. Let $D$ be a central division algebra
of degree $n$ over $F$, with an element $u \in D$ such that
$\dimcol{F[u]}{F} = n$ and $\beta = u^n \in \mul{F}$. Then $D$ is
cyclic.
\end{thm}
\begin{proof}
Since $F[u]$ splits $D$, we have that $K[u] = K\tensor[] F[u]$
splits $K \tensor[] D$. Thus, there exists an element $a \in
\mul{K}$ such that $\res[K/F]D = \symbalg[K]{a}{\beta}$, and we
are done by the corollary.
\end{proof}

A standard argument using idempotents gives the following more
general result.
\begin{cor}
Let $F$ be a field as above. Let $D$ be a central simple algebra
of degree $n$ over $F$, with an element $u \in D$ such that
$\dimcol{F[u]}{F} = n$ and $\beta = u^n \in \mul{F}$. Then $D$ is
split by $C_n$.
\end{cor}
\begin{proof}
If $F[u]$ is a field then $D$ is cyclic by the proof given above.
Otherwise, let $n'$ be the order of $u$ in $\md{F}$. Let $D_1 =
D\tensor[F]{K}$. The subring $K[u^{n'}]$ of $D_1$ is isomorphic to
a direct product of $n/n'$ copies of $K$. Let $C =
\Cent_{D_1}(K[u^{n'}])$. Taking a minimal idempotent $e_1$ of
$K[u^{n'}]$, $Ce_1$ is a central simple algebra of degree $n'$
over $Ke_1 \isom K$ and $Ke_1[ue_1]$ is a subfield of dimension
$n'$ of $Ce_1$, so $Ce_1$ is cyclic. We are done since $D_1 \sim C
e_1$.
\end{proof}

It would be interesting to have examples of non-cyclic algebras of
odd degree $n = p^2$ with an $n$-central element, when
$\dimcol{K}{F}$ is divisible by $p$. As mentioned above, Albert
gave such an example for $p = 2$.

We prove another generalization of Albert's theorem, due to
N.~Elkayam \cite[Thm. 3.9]{Nir}. We give only the version of prime
degree, as the generalization to prime-power degree along the
lines drawn above is easy.
\begin{prop}\label{NE}
Let $D$ be a central division algebra of degree $p$ over $F$, with an involution
of the second kind over $F/F_0$, and assume $F_0[\rho] = F[\rho]$.

If $D$ has a symmetric element $u \not \in F$ such that $u^p \in
F$, then $D$ contains a subfield $F_1$ Galois over $F_0$, such
that $\Gal(F_1/F_0)$ is dihedral.
\end{prop}
\begin{proof}
Let $d = \dimcol{K}{F}$, and let $\mu$ be the element of order $2$
in the character group of $\Gal(K/F_0) \isom \Zn[2d]$. Let $\nu_0
\co \Gal(K/F_0) \ra U_p$ be the character defined by \Eq{nudef}
for the extension $K/F_0$, and $\nu$ the restriction to $\Gl$;
there are two characters of $\Gal(K/F_0)$ whose restriction is
$\nu$, namely $\nu_0$ and $\nu_0\mu$. By \Lref{decKL} we have that
$\K{\nu} = \K{\nu_0}\K{\nu_0\mu}$, where $\K{\nu}$ is the
component with respect to the action of $\Gal(K/F)$ on $\mul{K}$,
and $\K{\nu_0}, \K{\nu_0\mu}$ are components with respect to the
action of $\Gal(K/F_0)$.

By the assumption, $\alpha = u^p \in F_0$. Moreover, $K[u] \sub
D\tensor[F] K$ is a cyclic subfield, and we can write $D_1 = D
\tensor[F] K = \symbalg[K]{\alpha}{b}$ for some $b \in \mul{K}$.
By Corollary \ref{atob}, we may assume $b\in \K{\nu}$. Let  $b =
b_{\nu_0} b_{\nu_0 \mu}$ be the decomposition to eigenvectors with
respect to the action of $\Gal(K/F_0)$. Since $\cores[F/F_0] D
\sim F_0$, we have that $F_0 \sim \cores[F/F_0] D^{\tensor[] d} =
\cores[F/F_0] \cores[K/F] D_1 = \cores[K/F_0]
\symbalg[K]{\alpha}{b} = \cores[K/F_0]
\symbalg[K]{\alpha}{b_{\nu_0}}$, so that in fact $D^{\tensor[] d}
\sim \cores[K/F]\res_{K/F} D \sim \cores[K/F]
\symbalg[K]{\alpha}{b_{\nu \mu}}$. Letting $D_2 =
\symbalg[K]{a}{b_{\nu \mu}}$ we have that $D = D_2^{\Gal(K/F)}$ by
Lemma \ref{res=cor=fixed}.

But now, $K_1 = \joinroot[p]{K}{b_{\nu\mu}}$ is Galois over $F_0$
with Galois group $\Zn[d]\semidirect\Zn[p]$, where the action is
by the character $\mu$. In particular, letting $\tau$ be a
generator of $\Gal(K/F_0)$, $\tau^2$ is central in $\Gal(K/F_0)$,
and $K_1^{\tau^2} \sub D_2^{\tau^2} = D$ is Galois over $F_0$,
with $\Gal(K_1^{\tau^2}/F_0) \isom D_p$.
\end{proof}





\section{Corestriction of cyclic algebras}\label{sec:cores}

Rosset and Tate proved in \cite{RTsymb} that if $K_2/K_1$ is an
extension of fields in which $K_1$ has $n$-roots of unity, then
the corestriction of a cyclic algebra of degree $n$ over $K_2$ to
$K_1$ is similar to a product of at most $\dimcol{K_2}{K_1}$
cyclic algebras of the same degree. No similar result is known if
$K_1$ does not have roots of unity (see \cite{rowenbook}, the
remark after Corollary 7.2.39). A general result in this direction
should not be easy to obtain, as it would imply a positive answer
to Question \ref{Q:gen}.

In \cite[Lemma~2]{Merk83}, Merkurjev proved that if
$\dimcol{K}{F}\leq 3$ (for an arbitrary fields extension $K/F$),
then $\KK(K)$ is generated by symbols of the form $\symb{K}{F}$
(\ie\ with one entry in $F$; this is how Question \ref{Q:gen} is
solved in this case). We give an explicit version of this useful
lemma.
\begin{lem}\label{KK=KF}
Let $K/F$ be any extension of fields, and let $a,b\in K$, $a,b
\not \in F$. Then the following identities hold in $\KK(K)$.

a. Assume $\dimcol{K}{F} = 2$, and write $b = \alpha + \beta a$,
$\alpha,\beta \in F$. Then
\begin{equation}\label{symb2}
\symb{a}{b} =
    \begin{cases}
        \symb{\frac{b}{\beta}}{-\frac{\beta}{\alpha}} + \symb{a}{\alpha} & \mbox{if $\alpha \neq 0$}
        \\
        \symb{a}{-\beta} & \mbox{if $\alpha = 0$}
    \end{cases}
\end{equation}

b. Assume $\dimcol{K}{F} = 3$, and write $b = g(a)$, $g \in
F[\lam]$ a polynomial of degree $\leq 2$. Since the minimal
polynomial of $a$ is of degree $3$, there exist $\alpha+\beta\lam,
\alpha'+\beta'\lam \in F[\lam]$ such that $(\alpha+\beta a)\cdot
g(a) = \alpha'+\beta' a$. Then we have (where in each case we only
indicate the non-zero coefficients):
\begin{equation*}%\label{symb3}
\symb{a}{b} =
    \begin{cases}
\symb{a}{-\frac{\beta'}{\alpha}} & \mbox{$\alpha'  = \beta = 0$}
\\
\symb{a}{-\frac{\alpha'}{\beta}} & \mbox{$\alpha = \beta' = 0$}
\\
\symb{a}{-\beta'} - \symb{b}{-\frac{\beta}{\alpha}} - \symb{\beta a}{\alpha} & \mbox{$\alpha' = 0$}
\\
\symb{a}{\alpha'} - \symb{b}{-\frac{\beta}{\alpha}} - \symb{\beta a}{\alpha} & \mbox{$\beta' = 0$}
\\
\symb{b}{-\frac{\beta'}{\alpha'}} + \symb{\beta' a}{\alpha'}-\symb{a}{-\beta} & \mbox{$\alpha = 0$}
\\
\symb{b}{-\frac{\beta'}{\alpha'}} + \symb{\beta' a}{\alpha'}-\symb{a}{\alpha} & \mbox{$\beta = 0$}
\\
       \symb{a}{\frac{\alpha'}{\alpha}} + \symb{\frac{\alpha'+\beta'a}{\alpha'}}{\frac{-\beta'}{\alpha'}} + \symb{\frac{\alpha+\beta a}{\alpha}}{\frac{-\alpha}{\beta}} & \mbox{} % CHECKED FEW TIMES...
\end{cases}
\end{equation*}
\end{lem}
\begin{proof}
For the $\dimcol{K}{F} = 2$ case, compute that
\begin{eqnarray*}
\symb{a}{b} - \symb{a}{\alpha} & = & \symb{a}{1+\frac{\beta a}{\alpha}} \\
    & = & \symb{a}{1+\frac{\beta a}{\alpha}} - \symb{-\frac{\beta a}{\alpha}}{1+\frac{\beta a}{\alpha}} \\
    & = & \symb{-\frac{\alpha}{\beta}}{\frac{b}{\alpha}} = \symb{\frac{b}{\alpha}}{-\frac{\beta}{\alpha}}
    = \symb{\frac{b}{\beta}}{-\frac{\beta}{\alpha}},
\end{eqnarray*}
where the last equality follows from
$\symb{\frac{\beta}{\alpha}}{-\frac{\beta}{\alpha}} = 0$. The case
$\dimcol{K}{F} = 3$ follows from $\symb{a}{b} =
\symb{a}{\alpha'+\beta' b} - \symb{a}{\alpha+\beta b}$, applying
the case of dimension $2$ for each of these symbols.
\end{proof}

We can now prove a version of the Rosset-Tate theorem for the
corestriction from $K$ to $F$, if $\dimcol{K}{F}\leq 3$ (under the
usual assumptions that $\dimcol{K}{F}$ is prime to $n$).
\begin{prop}\label{RTnoroots}% 12.2
Let $D = \symbalg[n,K]{a}{b}$ be a cyclic algebra over $K$. If
$\dimcol{K}{F} \leq 3$, then $\cores[K/F]D$ is similar to a
product of at most $\dimcol{K}{F}$ algebras which are split by
$C_n$.
\end{prop}
\begin{proof}
The corestriction of an algebra of the form $\symbalg{c}{\gamma}$,
$c\in K$ and $\gamma \in F$, is split by $C_n$ by Proposition
\ref{degcor}. Expressing $\symbalg[K]{a}{b}$ as a product of at
most $\dimcol{K}{F}$ symbols of this form (using the previous
lemma), we are done by taking corestriction.

We remark that the case $\dimcol{K}{F} = 2$ also follows from
decomposing $a = a_1 a_{\nu}$ and $b = b_1
b_{\nu}$ as in \Eq{genproj}, for then $\cores\symbalg[K]{a}{b}
\sim \cores \symbalg[K]{a_1}{b_\nu} \tensor
 \cores \symbalg[K]{a_\nu}{b_1}$.
\end{proof}

\begin{cor} % 12.3
Let $L/F$ be a separable extension of fields  of characteristic
prime to $n$, such that $d = \dimcol{K}{F} \leq 3$ where $K =
F[\rho]$. If $D$ is cyclic of degree $n$ over $L$, then
$\cores[L/F] D$ is similar to a product of at most
$$d\cdot (\dimcol{L}{F}-1) + 1$$
algebras which are split by $C_n$.
\end{cor}
\begin{proof}
Let $m = \dimcol{L}{F}$. If $K \sub L$ then in fact we only need
$\dimcol{L}{F}$ cyclic algebras, since $\cores[L/K] D$ is similar
to a product of $\dimcol{L}{K}$ cyclics of degree $n$ over $K$,
and the corestriction down to $F$ requires $\dimcol{K}{F}$ cyclics
to each of these by the proposition.

So suppose $L$ does not contain $K$, and let $T = L[\rho] = L
\tensor[F] K$. Since $D$ is cyclic over $L$, we can (by
\Pref{cyclicF}) write $\res_{T/L} D = \symbalg[n]{a}{\beta}$ for
$a \in T$ and $\beta \in L$. Let $D_2 =
\symbalg[n]{a^{d'}}{\beta}$, where $dd' \equiv 1 \pmod{n}$, and
note that
$$
\cores[L/F] D \sim \cores[L/F] \cores[T/L] D_2 = \cores[K/F] \cores[T/K] D_2.
$$

By the Rosset-Tate algorithm, $\cores[T/K] D_2$ is a product of
$\dimcol{T}{K} = m$ symbols, the first of which is
$\symbalg[K]{c}{\Norm[T/K](\beta)}$ for some $c\in K$ \cite[Cor.
7.2.38]{rowenbook}. Since $\Norm[T/K](\beta) = \Norm[L/F](\beta)
\in F$, the corestriction from $K$ to $F$ of this first symbol is
split by $C_n$ (Proposition \ref{degcor}). The corestriction of
each of the other $m-1$ symbols is similar to a product of $d$
cyclic algebras over $F$ by the previous proposition.
\end{proof}

The bound $d \cdot (\dimcol{L}{F}-1)+1$ of the corollary is higher
than the $\dimcol{L}{F}$ value suggested by the Rosset-Tate
result. In the proof we write $\res[T/L] D = \symbalg{a}{\beta}$
for $a\in T$ and $\beta \in L$, and ignores the fact that $a \in
\K[T]{\nu}$. This can be used to improve the bound if
$\dimcol{L}{F} = \dimcol{K}{F} = 2$.

\begin{thm}\label{betterbound} % 12.4
Suppose that $K = F[\rho]$ is quadratic over $F$, and let $L/F$ be
any separable quadratic extension. Let $D_1$ be a cyclic algebra
of degree $n$ (an odd prime power) over $L$.

Then $\cores[L/F] D_1$ is similar to a
product of at most two algebras over $F$ which are split by $C_n$.
\end{thm}
\begin{proof}
If $L = K$ then the result follows from Proposition
\ref{RTnoroots}, and we assume this is not the case. Let
$$
T = L[\rho] = L \tensor[F] K,
$$
a quadratic extension of $K$. By Theorem \ref{cyclicF},
$$
 D_2 = \res[T/L] D_1 = \symbalg[T]{a}{\beta}
$$
where $a \in \K[T]{\nu}$ and $\beta \in L$; in fact we can assume
$\Norm[T/L](a) = 1$ (see Remark \ref{normone}). We compute
$\cores[T/K] D_2$ using \Eq{symb2}. If $\beta \in F \sub K$ then
$\cores[T/K] D_2 = \symbalg[K]{\Norm[T/K](a)}{\beta}$ is cyclic,
and the corestriction from $K$ to $F$ is split by $C_n$ by
Proposition \ref{degcor}. We thus assume $\beta \not \in F$, so
that $K[\beta] = T$.

Choose $c,d \in K$ such that $a = d+ c\beta$. By the above
argument we may assume $a \not \in K$, so that $c \neq 0$.
Similarly if $d = 0$ then $\symbalg{a}{\beta} =
\symbalg{c\beta}{\beta} = \symbalg{-c}{\beta}$, and we are back in
the case $a\in K$.
We thus assume $d \neq 0$.

By \Eq{symb2} (after taking opposite of all the symbols involved),
$$
D_2 = \symbalg[T]{a}{\beta} \sim
\symbalg[T]{-\frac{c}{d}}{\frac{a}{c}} \tensor[T]
\symbalg[T]{d}{\beta},
$$
so that
\begin{equation}\label{corTKD2}
\cores[T/K] D_2 = \symbalg[K]{-\frac{c}{d}}{\Norm[T/K](\frac{a}{c})} \tensor[K] \symbalg[K]{d}{\Norm[L/F](\beta)}.
\end{equation}
The right-hand symbol has at least one entry, namely
$\Norm[L/F](\beta)$, in $F$, and so its corestriction from $K$ to
$F$ is split by $C_n$. Let $\s$ denote the non-trivial
automorphism of $L/F$ (which extends to $T/K$ as the identity on
$K$). We will show that $\Norm[T/K](a/c) \in F$. Consider the
polynomial
$$
g(\lam) = \Norm[K/F](\lam+\frac{d}{c}) - \Norm[K/F](\frac{1}{c}), % 'gilama' -- Rotem's favorite (her version of "pijamas", 6/2002).
$$
which is quadratic over $F$. We have that
$$g(\beta) = \Norm[T/L](a/c) - \Norm[K/F](1/c) = 0,
$$
so that $g$ is the minimal
polynomial of $\beta$, and $\s(\beta)$ is its other root. It
follows that $g(\lam) = \lam^2 - (\beta+\s(\beta))\lam +
\beta\cdot\s(\beta)$, so comparing coefficients we obtain
$$\beta+\s(\beta) = -\(\frac{\tau(d)}{\tau(c)}+\frac{d}{c}\),$$
$$\beta\cdot\s(\beta) = \frac{\tau(d)d-1}{\tau(c)c},$$
where $\tau$ generates $\Gal(K/F)$.
%
We can now compute that
\begin{eqnarray}
\Norm[T/K](a/c) & = & \Norm[T/K](d/c+\beta) \label{NTKac}\\
    & = & (d/c+\beta)(d/c+\s(\beta)) \nonumber \\
%    & = & \frac{d^2}{c^2} + \frac{d}{c}(\beta+\s(\beta)) + \beta\s(\beta) \nonumber \\
    & = & \frac{d^2}{c^2} - \frac{d}{c}\(\frac{\tau(d)}{\tau(c)}+\frac{d}{c}\) + \frac{\tau(d)d-1}{\tau(c)c} \nonumber \\
    & = &  -\frac{1}{\tau(c)c}. \nonumber
\end{eqnarray}

It follows that $\Norm[T/K](a/c) \in F$, and the left-hand symbol
in \Eq{corTKD2} also has one entry in $F$. The corestriction from
$K$ to $F$ of $\cores[T/K]D_2$ is thus similar to the product of
two algebras split by $C_n$.
\end{proof}


By induction, we obtain
\begin{cor}\label{better22}
Let $L/F$ be a separable extension and assume there is a chain of
quadratic extensions from $F$ to $L$.  Also assume that
$\dimcol{K}{F} = 2$ where $K = F[\rho]$.

For every cyclic algebra $D_1$ of degree $n$ over $L$,
$\cores[L/F] D_1$ is similar to a product of at most
$\dimcol{L}{F}$ algebras split by $C_n$.
\end{cor}

In the situation described in \Tref{betterbound}, it is obvious
that $D_2 = \res_{T/L}D_1$ is invariant under the action of
$\Gal(K/F)$, and that the corestriction $\cores[T/K]D_2$ is a
product of two symbol algebras over $K$. This can be improved:
\begin{rem}
The algebra $D_2$ of \Tref{betterbound} satisfies
\begin{equation}\label{corD2}
\cores[T/K]D_2 = \symbalg[K]{\tau(c)}{c} \tensor[K]
\symbalg[K]{d}{1-\tau(d)d},
\end{equation}
where both symbols in this expression are invariant under
$\Gal(K/F)$, and their corestriction to $F$ is split by $C_n$.
\end{rem}
\begin{proof}
We continue the computation from \Eq{corTKD2},  using the equality
$\Norm[T/K](a/c) = -1/\tau(c)c$\ \ proved in \Eq{NTKac} and the
value $\Norm[L/F](\beta) = (\tau(d)d-1)/\tau(c)c$.
\begin{eqnarray*}
\cores[T/K] D_2 & = &
\symbalg[K]{-\frac{c}{d}}{-\frac{1}{\tau(c)c}} \tensor[K]
\symbalg[K]{d}{\Norm[L/F](\beta)} \\
    & = &
\symbalg[K]{-c}{-\frac{1}{\tau(c)c}} \tensor[K]
\symbalg[K]{d}{-\tau(c)c\Norm[L/F](\beta)} \\
    & = &
\symbalg[K]{\tau(c)}{c} \tensor[K] \symbalg[K]{d}{1-\tau(d)d},
\end{eqnarray*}
where we use the fact that $\symbalg{c}{-c}\sim K$  and that $-1$
is an $n$-power in $K$ (since $n$ is odd). %
Writing $c = c_1 c_\nu$ for $c_1 \in \mul{F}$ and $c_1 \in
\K{\nu}$, one can check that $\symbalg{\tau(c)}{c} =
\symbalg{c_1^2}{c_\nu}$; likewise $1-\tau(d)d\in F$, so the
corestriction of both symbols is split by $C_n$ by \Pref{degcor}.

Recall that $\tau(\rho) = \rho^{-1}$ (since $\dimcol{K}{F} = 2$),
so in general $\tau\symbalg[K]{u}{v} =
\op{\symbalg{\tau(u)}{\tau(v)}}$. %
Now
$$
\tau \symbalg[K]{\tau(c)}{c} = \op{\symbalg[K]{c}{\tau(c)}} =
\symbalg[K]{\tau(c)}{c},
$$
and
\begin{eqnarray*}
\tau \symbalg[K]{d}{1-\tau(d)d} & = &
\op{\symbalg[K]{\tau(d)}{1-\tau(d)d}} \\
& \sim & \op{\symbalg[K]{\tau(d)d}{1-\tau(d)d}} \tensor
\symbalg[K]{d}{1-\tau(d)d} \\
& \sim & \symbalg[K]{d}{1-\tau(d)d}.
\end{eqnarray*}
\end{proof}

%% On the case $\dimcol{K}{F} = 3$ (in \Pref{betterbound}):
%
% The polynomial g(\lam) has three roots: $\beta,\s(\beta)$ and a certain $\gamma \in F$.
% The computation of $\Norm[T/K](a/c)$ is a little bit more complicated, and
% results in $\cores[T/K]D_2 = (-c/d,\frac{d'd''}{c'c''}+(\frac{d'}{c'}+\frac{d''}{c''})\gamma+\gamma^2) \tensor (the other guy of (23)).
% Does not seem to be of the form $(K,F)$ in general.

%% On the case $\dimcol{K}{F} = 4$:
%
% About the same proof yields $\cores[T/K]D_2 = (-c/d,-1/c\tau^2(c))\tensor (d,(\tau^2(d)d-1)/c\tau^2(c))$,
%  which has the form (K,F_1)\tensor (K,F), where $F\sub F_1 \sub K$.
% The corestriction is, thus, a product of two cyclic algebras and one quasi-symbol...
% The condition $c\tau^2(c) \in F$ (which will result in two cyclics) turns out to be equivalent to $c/d \in F$.




\section{Semidirect crossed products} \label{sec:sd} % Section 13

Rowen and Saltman \cite{RSsemi} proved that if $D$ is a division
algebra of prime degree $p$ over a field $L$ with $p$ roots of
unity, and $D$ is split by a field $T$ such that $\Gal(T/L) =
\Zn[e] \semidirect \Zn[p]$, where $e = 2,3,4$ or $6$ and divides
$p-1$, then $D$ is cyclic.
% Also compare to \cite[Prop.~7.2.41]{rowenbook}, where a similar
% situation is considered. %% That Prop. has a mistake: one must assume $n$ is a prime power.
Our version (Corollary \ref{SD} below) holds for any prime-power
degree $n = p^k$ and arbitrary $e$ (dividing $p-1$), and rather
than assuming the base field contains roots of unity, we assume
that the splitting field contains them. On the other hand, the
cyclicity result only holds for specific semidirect products. To
make the notation clear, if $\tau$ is a generator of $\Zn[d]$ and
$\Kpsi\co \Zn[d] \ra U_n$ is a character defined by $\Kpsi(\tau) =
t$, then
$$
\Zn[d]\semidirect_{\Kpsi} \Zn = \sg{\sfs,\tau \subjectto \sfs^n =
\tau^d = 1,\, \tau\sfs\tau^{-1} = \sfs^t}.
$$

The cyclicity of semidirect crossed products is strongly related
to  Question \ref{Q:cyc}, as the following observation on
quasi-symbols (see Definition \ref{defqs}) shows.
\begin{prop}\label{propqs}
A central simple algebra $D$ of degree $n$ over $F$ is a
quasi-symbol, iff it has a splitting field $S \supset K$ which is
Galois over $F$ with Galois group a semidirect product
$\Zn[d]\semidirect \Zn$.

The splitting field $S$ has $\Gal(S/F) = \Zn[d]\semidirect_{\Kpsi}
\Zn$ iff $D$ is of type $(\Kpsi,\nu\Kpsi^{-1})$.
\end{prop}
\begin{proof}
If $\res[K/F] D = \symbalg{a}{b}$ for $a \in \K{\Kpsi}$ and $b \in
\K{\Kpsi'}$, then $S = \joinroot{K}{a}$ is a splitting field
satisfying the assumptions (by Proposition \ref{whenGal}).

On the other hand, let $S$ be a splitting field satisfying the
assumptions. Then $S$ is Galois over $K$, and $\Gal(S/K) \isom
\Zn$ since this is the unique normal subgroup of order $n$ of
$\Zn[d]\semidirect\Zn$. We can thus write $S = K[\alpha]$ where
$\alpha^n = a \in K$, and again by Proposition \ref{whenGal}, $a
\in \K{\Kpsi}$ for some character $\Kpsi$. By Corollary \ref{atob}
there is an element $b \in \K{\Kpsi'}$ such that $D \tensor[F] K =
\symbalg[K]{a}{b}$, where $\Kpsi\Kpsi' = \nu$. Finally, in this
case $\Gal(S/F) =
\Zn[d]\semidirect_{\Kpsi'} \Zn$ by %Equation \eq{chidef} and
Remark \ref{chinupsi}.
\end{proof}

It follows that semidirect product splitting fields come in pairs:
\begin{cor} \label{tsuzamen}
Let $\Kpsi,\Kpsi'$ be characters with $\Kpsi \Kpsi' = \nu$. A
central simple algebra of degree $n$ over $F$ has a splitting
field containing $K$ with Galois group $\Zn[d]\semidirect_{\Kpsi}
\Zn$ iff it has a splitting field containing $K$ with Galois group
$\Zn[d]\semidirect_{\Kpsi'} \Zn$.
\end{cor}

Proposition \ref{propqs} can be extended to characterize
quasi-symbols with splitting fields of smaller dimension.
\begin{prop}
Let $F \sub L \sub K$ be an intermediate field of dimension $e =
\dimcol{L}{F}$, and let $D$ be a central simple algebra of degree
$n$ over $F$.

Then $D$ has a splitting field $S \supset L$ such that $\Gal(S/F)
= \Zn[e]\semidirect\Zn$, iff $D$ is a quasi-symbol of type
$(\Kpsi,\Kpsi')$ for $\Kpsi$ of order $e$ in $\ch{\Gl}$.
\end{prop}
\begin{proof}
Let $S \supset L$ be a splitting field which is Galois over $F$.
We want to check that $S_1 = S \tensor[L] K$ is also Galois over
$F$, and for that we need to count automorphisms of $S_1$ over
$F$.

Let $\sfs$ denote a generator for $\Gal(S/L)$, and let $\tau$ be a
generator of $\Gal(K/L)$. Since $S/F$ is Galois, $\tau$ extends to
an automorphism (of order $e$) of $S$. But $\tau$ is also defined
on $K$, and thus on $S_1$. Likewise $\sfs$ extends to $S_1$ as the
identity on $K$. Finally, we have that $\tau \sfs \tau^{-1} =
\sfs^r$ on $S$ for some $r$, and the same relation trivially holds
on $K$ (since $\sfs$ is the identity on $K$) and thus on $S_1$. It
follows that $\power{\sg{\sfs,\tau}} = nd$ and $S_1/F$ is Galois.
In particular $S = S_1^{\tau^{e}}$.

Now let $\Kpsi$ denote the character defining the action of
$\Gal(K/F)$ on $\Gal(S_1/K)$. Since $S/F$ is Galois, $\sg{\tau^e}$
is normal in $\Gal(S_1/F)$; as easily seen, this is equivalent to
$\tau^e$ being central; which is equivalent to $\Kpsi^e = 1$.

Writing $S_1 = K[\alpha]$ for $\alpha^n = a \in \mul{K}$, we have
by Proposition \ref{whenGal} that $a \in \K{\nu\Kpsi^{-1}}$, and
when we write $D \tensor[F] K = \symbalg[K]{a}{b}$ by Corollary
\ref{atob} we have $b \in \K{\Kpsi}$, so that $D$ is a
quasi-symbol of type $(\Kpsi,\nu\Kpsi^{-1})$.
\end{proof}

In particular, for the case $e = 1$, we have

\begin{cor}\label{SD} % 13.4
Let $D$ be a central simple algebra of degree $n$ over $F$. If $D$
is split by a Galois extension $S \supset K$ such that
$\Gal(S/F)=\Zn[d]\semidirect_{\nu}\Zn$, then $D$ is split by
$C_n$.
\end{cor}
\begin{proof}
By the above proposition $\res_{K/F} D = \symbalg{a}{b}$ for $a
\in \K{\nu}$ and $b \in \K{1}$, so we are done by Proposition
\ref{degcor}.
\end{proof}

Though essentially the same, we can formulate the result more
generally for $\dimcol{K}{F} = 2$.
\begin{prop}\label{dicyc}
Let $m$ be an arbitrary odd number, $F$ a field of characteristic
prime to $m$, $\rho \in \sep{F}$ a primitive $m$-root of unity and
$K = F[\rho]$, and assume that $\dimcol{K}{F}$ is prime to $m$.

Let $D$ be a division algebra of degree $m$ over $F$. If $D$ is
split by a field $S \supset K$ which is dihedral of dimension $2m$
over $F$, then $D$ is cyclic.
\end{prop}
\begin{proof}
The only normal subgroups of the dihedral group $$D_m =
\sg{\s,\tau \subjectto \s^m=\tau^2=1, \tau\s\tau^{-1}=\s^{-1}}$$
are contained in $\sg{\s}$, so from $S$ being dihedral over $F$ it
follows that $\dimcol{K}{F} = 2$, $K = S^{\s}$ and $S$ is cyclic
over $K$.

Let $m = n_1 \dots n_t$ be the factorization of $m$ into prime
power factors. Decompose $S = K_1 \tensor[K] \dots \tensor[K] K_t$
where $\dimcol{K_i}{K} = n_i$, and $D = D_1 \tensor[F] \dots
\tensor[F] D_t$ where $\deg D_i = n_i$. The fields $K_i$ are
Galois over $F$, and since $\dimcol{K}{F} = 2$, there are only two
possibilities for every $i = 1,\dots,t$: $\Gal(K_i/F)$ is either
the dihedral group of order $2n_i$, or the direct product $\Zn[2]
\times \Zn[n_i]$. Applying the proposition in the first case and
standard arguments in the second, we see that every $D_i$ is
cyclic, and (tensoring the cyclic splitting fields over $F$) $D$
is cyclic as well.
\end{proof}



Our final result on dihedral algebras is to remove the assumption
that the splitting field contains roots of unity, made in the last
proposition. For that we need to assume that $\dimcol{K}{F} = 2$.
This theorem is proved in \cite[Cor.~30]{HKRT} under the
assumption $\mychar F \neq 2$.

\begin{thm}\label{di:KF=2} % 13.6
Let $n$ be an odd prime power, $K = F[\rho]$ where $\rho$ is an
$n$-root of unity, and assume $\dimcol{K}{F}\leq 2$. Let $D$ be a
central simple algebra of degree $n$ over $F$.

If $D$ is split by a dihedral extension of dimension $2n$ over
$F$, then $D$ is cyclic.
\end{thm}
\begin{proof}
The result was proved under the assumption $K = F$ in \cite{RSdi},
so we assume $\dimcol{K}{F} = 2$. Let $S$ be the dihedral
splitting field. The case $K\sub S$ is covered by \Pref{dicyc}, so
we assume $S$ does not contain $K$.

Let $\sfs,\s$ be generators of the Galois group $\Gal(S/F)$, such
that $\sfs^n=\s^2=1$ and $\s\sfs\s^{-1}=\sfs^{-1}$. Let $L =
S^{\sfs}$ be the quadratic subfield, over which $S$ is cyclic. By
our assumption $K$ is linearly independent with $L$, and we let $T
= L \tensor[F] K =  L[\rho]$.

Since $D_1 = D \tensor[F] L$ is split by $S$ and thus cyclic,
Theorem \ref{cyclicF} shows that
$$
D_2 = \res[T/L] D_1 = \symbalg[T]{a}{\beta}
$$
for some $a \in T$ and $\beta \in L$, which is the situation
analyzed in \Tref{betterbound}. In fact (Section \ref{fields}),
letting $\alpha = \sqrt[n]{a}$ we have that $S_1 = T[\alpha]$
decomposes as $S_1 = S \tensor[L] T$. Moreover, letting $\tau$ be
the non-trivial automorphism of $T/L$ (so that $\tau(\rho) =
\rho^{-1}$), we may by Remark~\ref{normone} assume that $\tau(a) =
a^{-1}$, and then $\tau$ extends to $S_1$ by $\tau(\alpha) =
\alpha^{-1}$. We can also extend $\sfs$ to $S_1$ by $\sfs(\alpha)
= \rho \alpha$, and then $T = S_1^{\sfs}$.

The main feature of the proof is that to understand $D$, we first
take the restriction to $L$ and from there to $T$, and then the
corestriction to $K$ and down back to $F$. Indeed, we will apply
the  computations of \Tref{betterbound} to $\res[T/L]D_1$; {}from
this it will follow that $D^{\tensor 4} \sim
\cores[T/F]\res[T/L]D_1$ is split by $C_n$, which proves the
claim.

For that we need to know more about $a$. Since $S_1 = K \tensor[F]
S$, it is Galois over $F$ (with $\Gal(S_1/F) = D_n \times \Z/2$),
and $\s$ is extended to $S_1$ as the identity on $K$. Since the
relations $\s^2 = 1$ and $\sfs \s = \s \sfs^{-1}$ hold both in $K$
and $S$, they also hold in $S_1$. Now let $\alpha' = \s(\alpha)$
and note that $\s(\alpha') = \alpha$. We have that
$$
\sfs(\alpha') = \sfs\s(\alpha) = \s\sfs^{-1}(\alpha) =
\s(\rho^{-1}\alpha) = \rho^{-1}\alpha'.
$$

It follows that $g = \alpha\alpha' \in S_1^{\s,\sfs} = K$, and
then $a \s(a) = g^n$. Checking that $\tau(\alpha') =
\tau\s(\alpha) = \s\tau(\alpha) = \s(\alpha^{-1}) =
{\alpha'}^{-1}$, we see that $\tau(g) = g^{-1}$. Now, consider the
element $a_1 = (g^{(n-1)/2}a^{-1})^na$: since $\tau(a_1) =
a_1^{-1}$ and $\joinroot{K}{a_1} = \joinroot{K}{a}$ we can replace
$a$ by $a_1$; but $\s(a_1) = a_1^{-1}$, so from now on we may
assume $\s(a) = a^{-1}$ and $g = 1$. % Then we also have $\s(\alpha) = \alpha' = \alpha^{-1}$.


Recall that $D_2 = \res[T/F] D = \symbalg[T]{a}{\beta}$. If $\beta
\in F$ then
$$
\cores[T/K] D_2 \sim \symbalg[K]{\Norm[T/K]a}{\beta}
= \symbalg[K]{1}{\beta} \sim K,
$$
so $D$ is already split.  We thus assume $\beta \not \in F$, and
in particular $T = K[\beta]$. As in \Tref{betterbound}, let $c,d
\in K$ be such that $a = c \beta+d$. We proved in \Eq{NTKac} that
$\Norm[T/K](a/c) = -1/\Norm[K/F](c)$, and adding the fact that
$\Norm[T/K](a) = 1$, we obtain the equality $\tau(c) = -c$.
Plugging this in \Eq{corD2}, we immediately obtain
$$
\cores[T/K]D_2 \sim \symbalg[K]{d}{1-\tau(d)d},
$$
and since  $1-\tau(d)d \in F$, we have that $D^{\tensor 4} \sim
\cores[T/F]\res[T/F]D = \cores[T/F]D_2 = \cores[K/F]\cores[T/K]D_2
= \cores[K/F] \symbalg[K]{d}{1-\tau(d)d}$ is split by $C_n$.
\end{proof}



%\subsection{Cyclicity after restriction}

We end this section with a curious property of $\Brn(F)$ when
$\dimcol{K}{F} = 6$, which is related to the fact that in the
equation
$$
3+4 \equiv 1 \pmod{6},
$$
the common divisor with $6$ with each of the summands is greater
than $1$. Let $F \sub L_2,L_3 \sub K$ be the intermediate
subfields (with $\dimcol{L_t}{F} = t$). Since $\dimcol{K}{L_t}
\leq 3$ ($t = 2,3$), we have from \cite{Merk83} that $\Brn(L_t)$
is generated by (classes of) cyclic algebras of degree $n$. Since
$\cores[L_t/F]$ is onto, we immediately get that $\Brn(F)$ is
generated by algebras which become cyclic after extending scalars
to $L_2$, and also by algebras which become cyclic after extending
scalars to $L_3$. But in fact $\Brn(F)$ is generated by algebras
with a better set of splitting fields:
\begin{prop}
Assume that $\dimcol{K}{F} = 6$ and let $L_2,L_3$ be as above.
Then $\Brn(F)$ is generated by (the classes of) algebras which are
split by both $\Zn[2]\semidirect\Zn$ and $\Zn[3]\semidirect\Zn$.
\end{prop}
\begin{proof}
By \Lref{KK=KF}, $\Brn(K)$ is generated by cyclic algebras of the
form $\symbalg{a}{b}$ where $a \in \mul{K}$ and $b \in \mul{L_2}$.
We can decompose $a = a_0 a_1 \dots a_5$ and $b = b_0 b_3$ where
$a_i,b_i \in \K{\nu^{i}}$ (see \Lref{decomposeL}). By Proposition
\ref{corF},
$$
\cores[K/F]\symbalg{a}{b} \sim \cores[K/F]\symbalg{a_1}{b_0}
\tensor \cores[K/F]\symbalg{a_4}{b_3}.
$$
The algebras $\cores[K/F] \symbalg{a_1}{b_0}$ are split by $C_n$,
by \Pref{degcor}.

By Lemma \ref{decomposeL} we may assume that $a_4 \in L_3$ and
$b_3 \in L_2$, so the same argument works for
$\res_{L_t/F}\cores[K/F]\symbalg{a_4}{b_3}$, and the result
follows from \Cref{tsuzamen}.
\end{proof}

A similar result holds by the same proof whenever $d =
\dimcol{K}{F}$ is divisible by $6$: $\Brn(F)$ is generated by
algebras which are split by both $\Zn[(d/2)]\semidirect\Zn$ and
$\Zn[(d/3)]\semidirect\Zn$.




\section{Generic constructions}\label{secgeneric}

In Section \ref{csa} we have seen that $\Brn(F)$ is generated by
quasi-symbols, which are algebras $D_0$ such that $D_0 \tensor[F]
K = \symbalg[K]{a}{b}$ for $a \in \K{\Kpsi}$ and $b \in
\K{\Kpsi'}$, where $\Kpsi,\Kpsi'$ range over the pairs of
characters with product $\nu$. If either $\Kpsi = 1$ or $\Kpsi' =
1$, then $D_0$ is cyclic, and we will assume this is not the case.
We will describe a generic construction for quasi-symbols, and
present elements $w$ with the property that $\Tr(w^i) = 0$ for
most values of $i$.

According to Theorem~\ref{cyclicF}, $D_0$ is split by $C_n$ iff
there are $\alpha \in \mul{F}$ and $c \in \K{\nu}$ of order $n$,
such that $J^{\Kpsi}(a) \cup J^{\Kpsi'}(b) = (\alpha) \cup
J^{\nu}(c)$. It is generally believed that there exist non-cyclic
algebras of prime degree $p \geq 5$, though no one was yet able to
construct such an algebra. Based on the experience with other
crossed products, it seems even more plausible that non-cyclic
algebras exist over fields without roots of unity of order $p$,
and the equation given above suggests that a generic quasi-symbol
is a reasonable candidate to be non-cyclic.

Let $n = p^m$ be a prime power, and $k_0$ a field of
characteristic prime to $n$. Let $\rho$ be an $n$-root of unity in
the separable closure $\sep{(k_0)}$, and $k=k_0[\rho]$. We assume
that $d = \dimcol{k}{k_0}$ is prime to $n$. We then define
$$
K = k(a,b,\mu_0,\dots,\mu_{d-2},\eta_0,\dots,\eta_{d-2}),
$$
a transcendental extension of degree at most $2d$ (in some cases
there are equations on the $\mu_i,\eta_i$, see below). Let $\tau$
be a generator for $\Gal(k/k_0)$. The extension of $\tau$ to $K$
will depend on our choice of characters. Define $\nu \co
\Gal(k/k_0) \ra U_n$ by \Eq{nudef}, and let $\Kpsi,\Kpsi'$ be two
characters such that $\Kpsi\Kpsi' = \nu$. Let
$$
r = \Kpsi(\tau), \qquad r' = \Kpsi'(\tau).
$$
We extend $\tau$ to $K$ by setting
\begin{eqnarray*}
\tau(a) & = & \mu_0^n a^r \\
\tau(b) & = & \eta_0^n b^{r'} \\
\tau(\mu_i) & = & \mu_{i+1 \pmod{d}} \\
\tau(\eta_i) & = & \eta_{i+1 \pmod{d}},
\end{eqnarray*}
where $\mu_{d-1}$ and $\eta_{d-1}$ are defined in $K$ by the
following formulas: %
\begin{eqnarray*}
\mu_{d-1} & = & \mu_{d-2}^{-r} \mu_{d-3}^{-r^2} \dots
\mu_1^{-r^{d-2}} \mu_0^{-r^{d-1}} a^{-\frac{r^d-1}{n}}, \\
\eta_{d-1} & = & \eta_{d-2}^{-r'} \eta_{d-3}^{-{r'}^2} \dots
\eta_1^{-{r'}^{d-2}} \eta_0^{-{r'}^{d-1}} b^{-\frac{{r'}^d-1}{n}}.
\end{eqnarray*}
This definition makes $\tau$ an automorphism of order $d$ of $K$,
and $F = K^{\tau}$ satisfies $F[\rho] = K$.

If the order $e$ of $\Kpsi$ is strictly less than $n$, we can
apply \Lref{decomposeL} and Proposition \ref{goodx} to simplify
$K$ without losing the generality of the construction.
Specifically, if $L = K^{\Ker(\Kpsi)}$, we can have $a \in L$ and
$\mu_0 \in L$, so that $\mu_{e+i} = \mu_i$. In particular if
$\Kpsi^2 = 1$ we choose $r = -1$ and $\mu_i = 1$, and then
$\tau(a) = a^{-1}$. More generally we can alter the definition so
that $\Norm[K/F](a) = 1$ (Remark \ref{normone}), which can be used
to express $\mu_{d-2}$ in terms of $\mu_0,\dots,\mu_{d-3},a$, and
similarly $\eta_{d-2}$ in terms of $\eta_0,\dots,\eta_{d-3},b$. If
$d = \dimcol{K}{F}$ is even (and $F \subset K_0 \sub K$ the
intermediate quadratic extension), we can have $\Norm[K/K_0](a) =
a\tau^2(a)\dots\tau^{d-2}(a)=1$ by Remark \ref{normK0one}, and
this can be used to express $\mu_{d-3}$ in terms of
$\mu_0,\dots,\mu_{d-4},a$.

After $\tau$ is defined on $K$, we let
\begin{equation}\label{Ddef}
D = \symbalg[K]{a}{b} = K[x,y \subjectto x^n=a,\, y^n=b,\,
yxy^{-1} = \rho x],
\end{equation}
and extend $\tau$ to $D$ by
$$
\tau(x) = \mu_0 x^{r},\qquad \tau(y) = \eta_0 y^{r'}.
$$
The generic quasi-symbol $D_0 = D^{G}$ is, according to Lemma \ref{res=cor=fixed},
similar to $\cores[K/F](D)$.

\begin{exmpl}\label{d=4}
Consider the case $d = \dimcol{K}{F} = 4$ (so in particular $n
\equiv 1 \pmod{4}$). There are only two options for the pair
$\Kpsi,\Kpsi'$: either they are $(1,\nu)$ (in which case the
quasi-symbol is surely cyclic), or $\Kpsi = \nu^2$, $\Kpsi' =
\nu^{-1}$.

We construct a quasi-symbol of type $(\nu^2,\nu^{-1})$. Let
$k,k_0$ be as above, $\Gal(k/k_0)= \sg{\tau}$, and $t \equiv
\nu(\tau) \pmod{n}$. We then choose $r = -1$ and $r' = -t$; if
$\tau(b) = \eta_0^n b^{-t}$ and $\tau(\eta_0) = \eta_1$ then the
condition $b \tau^2(b) = 1$ gives $\eta_1 = \eta_0^{t}
b^{-(t^2+1)/n}$. We thus set
$$
K = k(a,b,\eta)
$$
and define $\tau$ on $K$ by
\begin{eqnarray*}
\tau(a) & = & a^{-1}, \\
\tau(b) & = & \eta^n b^{-t}, \\
\tau(\eta) & = & \eta^{t} b^{-(t^2+1)/n}.
\end{eqnarray*}

Now define $D = \symbalg[K]{a}{b}$ by \Eq{Ddef}, and extend $\tau$
to $D$ by $\tau(x) = x^{-1}$, $\tau(y) = \eta y^{-t}$. Notice that
$\tau^2(\eta) = \eta^{-1}$ and $\tau^2(y) = y^{-1}$.
%
%In fact $K[y]^{\tau^2}$ is cyclic over $L$, so that $D^{\tau^2}$ is cyclic.
\end{exmpl}

The problem we already stated (Question \ref{Q:cyc}) is whether or
not $D^{\tau}$ is cyclic. By Theorem \ref{Albert}, this is
equivalent to the existence of $u \in D^{\tau}$ such that
$\dimcol{F[u]}{F}=n$ and $u^n \in F$. Naturally, we view the
candidates $u$ as elements of $D \tensor[F] K =
\symbalg[K]{a}{b}$.

We now change perspective, and study some special elements in
quasi-symbols. Let $w$ be an element in a central simple algebra.
If the characteristic of the field is zero, the coefficients of
the minimal polynomial of $w$ are expressed by Newton formulas in
terms of the reduced traces $\Tr(w^i)$. In particular, assuming
$\dimcol{F[w]}{F} = n$, $w^n \in F$ iff $\Tr(w) = \dots =
\Tr(w^{n-1}) = 0$. One traditional approach to cyclicity problems
is to find elements for which the trace of powers is often zero.
The same technique works in our case: by \Tref{Albert}\ \
$D_0=D^G$ is cyclic over $F$ iff there is $u \in D$ such that
$\tau(u) = u$ and $u^n \in K$ (in which case $u^n \in F$). Rowen
has shown \cite{rowentr} that in an algebra of odd degree, there
always exist an element $u$ such that $\Tr(u) = \Tr(u^2) = 0$. A
similar result was obtained by Haile \cite{Haile}: in any division
algebra there exists an element $u$ with $\Tr(u) = \Tr(u^{-1}) =
0$. The technique of power traces was effectively used by Rowen
and Saltman \cite{RSdi} to prove their above mentioned result on
cyclicity of dihedral algebras. Our computations can be viewed as
a partial analog to fields without roots of unity.

{}From now on, assume $d = \dimcol{K}{F}$ is even, and let $\mu =
\nu^{d/2} $ denote the character of order $2$ in $\ch{\Gl}$.
Consider a quasi-symbol $D_0$ of type $(\mu,\nu\mu)$. There exist
$a \in \K{\mu}$ and $b\in \K{\nu\mu}$ such that $\res_{K/F}D_0 =
\symbalg{a}{b}$; letting $\tau$ be a generator of $\Gl$, we set $t
= \nu(\tau)$, so that $\tau(a) \equiv a^{-1}$ and $\tau(b) \equiv
b^{-t}$.

Let $W = \set{\tr_\tau(kxy) \suchthat k \in K}$. This is a
subspace of dimension $d$ of $D_0$ (its elements are all different
since the coefficient of $xy$ in $\tr_\tau(kxy)$ equals $k$). We
will show that powers of elements $w  \in W$, often have zero
reduced trace. To this end, note that the reduced trace in $D_0$
is the same as in $D$, which is easily computed: writing $D =
\sum_{i,j=0}^{n-1}{Kx^iy^j}$, the reduced trace of an element is
$n$ times the free coefficient.

\begin{prop}\label{lotsof0}
Let $W$ be as above. Then for every $w \in W$,
$$
\Tr(w) = \Tr(w^3) = \Tr(w^5) = \Tr(w^7) = \dots = \Tr(w^{n-2}) =
0, $$ and
$$
\Tr(w^2) = 0.
$$
If, moreover, $\mychar{F} = 2$, then $\Tr(w^\ell) = 0$ for every
$\ell = 0,\dots,n-1$.
\end{prop}
\begin{proof}
Write $D = K[x,y]$ for a standard pair of generators. Let $w =
\tr_\tau(k xy) \in \sum_{i = 0}^{d-1} K x^{(-1)^i} y^{s^i}$ where
$s = -\nu(\tau)$ and $k \in K$ is arbitrary. For odd $\ell<n$,
$w^\ell \in \sum_{i=1,3,\dots,n-2} (K[y] x^{i} + K[y] x^{-i})$, so
that $\Tr(w^\ell) = 0$.

To see that $\Tr(w^2) = 0$, notice that
$$w^2 \in K[y]x^{-2} + K[y] + K[y]x^{2},$$
and the component in $K[y]$ is in $\sum K y^{s^i+s^j}$ where the
sum is over $i = 0,2,\dots,d-2$ and $j = 1,3,\dots,d-1$. To have
non-zero reduced trace there must be a pair $i,j$ such that
$s^{i}+s^{j}\equiv 0\pmod{n}$, but then $(-s)^i \equiv (-s)^j$,
which is impossible since $(-s)$ is of order $d$ in $U_n$.

\sloppy{%
Finally, if $\mychar{F} = 2$, the claim follows by induction on
$\ell$ from Lemma \ref{char=2} below, whose easy proof was pointed
out to me by A.~Wadsworth.%
}
\end{proof}

\begin{lem}\label{char=2}
Let $D$ be a central simple algebra over a field of characteristic $2$.
The reduced trace satisfies $\Tr(u^2) = \Tr(u)^2$.
\end{lem}
\begin{proof}
Extending scalars to the separable closure of $F$, every element of $D$ is
conjugate to an upper triangular matrix, where the reduced trace is the sum of
diagonal entries. But if $u_{11},\dots,u_{nn}$ is the diagonal of $u$, the
diagonal of $u^2$ is $u_{11}^2,\dots,u_{nn}^2$ and $\Tr(u^2) = \sum {u_{ii}^2} = (\sum u_{ii})^2 = \Tr(u)^2$.
\end{proof}


%% An attempt to say something for cases other than $d=4$:
%
% For the case $d = 6$, $\Kpsi = \nu^2$, $\chi = \nu^5$, all we
% have for $w$ is that the trace of $w^i$ vanish for $i = 1,2,4,5$ (and that's it).

The result $\Tr(w^2) = 0$ of the proposition can be significantly
improved if $\dimcol{K}{F} = 4$. Note that if $\dimcol{K}{F} = 4$,
then $\Brn(F)$ is generated by algebras which are split by $C_n$,
and the quasi-symbols of type $(\mu,\nu\mu)$.
\begin{prop} \label{lotsof0.4}
Let $W$ be the space defined above, where we assume $d = 4$. Let
$\ell \geq 1$ be odd. If $n > \ell^2$, then for every $w \in W$,
$$
\Tr(w^{2\ell}) = 0.
$$
\end{prop}
\begin{proof}
Let $\tau$ be a generator of $\Gl$ and let $t$ be a number such
that $\tau(\rho) = \rho^t$. Since
$$
w = \tr_{\tau}(kxy) \in Kxy + Kx^{-1}y^{-t} + Kxy^{-1} + K x^{-1}
y^{t},
$$
we have that
$$
w^2 \in (Ky^{\pm 2t}+K)x^{-2} + Ky^{\pm 1 \pm t} + (Ky^{\pm 2}
+K)x^2.
$$
The trace $\Tr(w^{2\ell})$ can be non-zero only if there is a
product of $\ell$ monomials of $w^2$ which is in $K$. Let
$g_{i,j}$ denote the number of monomials coming from $Kx^iy^j$ in
such a product, so that $\sum {g_{ij}} = \ell$. Let $h_{i}$ denote
the sum of $g_{i,j}$ over all possible values of $j$. Since the
exponent of $x$ in the product is $2(h_{2}-h_{-2})$, we have that
$h_{-2} = h_{2}$, and in particular $h_0 = \ell - 2h_2$ is odd.

It then follows that the exponent of $y$ in the product is of the
form $d_1 t + d_2$, where the only condition on the $d_i$ is that
they are odd and $\power{d_i} \leq \ell$. We thus have that $d_1 t
\equiv -d_2 \pmod{n}$, and since $t^2 \equiv -1$, we obtain\ \ $n
\divides d_1^2+d_2^2$, which is even. But $n$ is odd, so $n
\divides (d_1^2+d_2^2)/2\leq \ell^2$.
\end{proof}

\begin{exmpl}\label{usetrig}
Let $D$ be the algebra of Example \ref{d=4}, and let
\begin{equation*}\label{w} w = \tr_{\tau}(kxy) = k xy+
\tau(k) \eta x^{-1} y^{-t} + \tau^2(k) x y^{-1} + \tau^{3}(k)
\eta^{-1} x^{-1} y^t.\end{equation*} We proved that
$$
\Tr(w) = \Tr(w^3) = \Tr(w^5) = \dots = \Tr(w^{n-2}) = 0
$$
and
$$
\Tr(w^2) = \Tr(w^6) = \Tr(w^{10}) = \dots = \Tr(w^{2\ell}) = 0$$
for $\ell \leq \sqrt{n}$ odd.

The only way to combine four monomials of $w$ so that the product
is in $K$, is if they are different. Summing all the possible
products, we obtain
$$ \Tr(w^4) = 4n ((\rho+\rho^{-1})(\rho^{t}+\rho^{-t})+2) \cdot
\Norm[K/F](k).$$ The coefficient is never zero by the final
proposition below, so that $\Tr(w^4) \neq 0$ whenever $w \neq 0$.
\end{exmpl}

\begin{rem}
A similar computation shows that if $\dimcol{K}{F} = 6$ and $n>7$,
and $D_0$ is a quasi-symbol of type $(\nu^3,\nu^4)$, then for
every $w \in W$,
$$\Tr(w^4) = 0.$$
\end{rem}

We end this paper with a bit of trigonometry (which is needed for
the Example \ref{usetrig}).
\begin{prop}
Let $n$ be an odd prime power and $t$ an integer such that $t^2
\equiv -1\pmod{n}$, and let $\rho$ be an $n$-root of unity. Then
$$
(\rho+\rho^{-1})(\rho^t+\rho^{-t}) + 2 \neq 0.
$$
\end{prop}
\begin{proof}
Let $\theta_1 = \frac{2(t-1)\pi}{n}$ and $\theta_2 =
\frac{2(t+1)\pi}{n}$. Since $(\rho+\rho^{-1})(\rho^t+\rho^{-t}) =
\rho^{t+1}+\rho^{t-1}+\rho^{1-t}+\rho^{-1-t} =
2\cos{\theta_1}+2\cos{\theta_2}$, we need to show that
$\cos{\theta_1}+\cos{\theta_2} \neq -1$. Otherwise, the average of
$\cos{\theta_1},\cos{\theta_2}$ is $-1/2$, so for some
$\set{i,j}=\set{1,2}$ we have $-1/2\leq\cos{\theta_i}\leq 0$ and
$\cos{\theta_j}\leq -1/2$; in other words
$$\theta_i \in [\pi/2,2\pi/3] \cup [4\pi/3,3\pi/2],$$
$$\theta_j \in [2\pi/3,4\pi/3].$$

First assume $j<i$. Then since $\theta_1<\theta_2$ we have
$4\pi/3\leq \theta_2$, and (by changing $t$ to $n-t$ if necessary)
$\theta_1 < \pi$.
 It follows that
$4\pi/n = \theta_2 - \theta_1 > 4\pi/3-\pi = \pi/3$, so $n < 12$
and $n \in \set{5,9}$.

Now assume $i <j$: again since $\theta_1 < \theta_2$, we have that
$\theta_1 \leq 2\pi/3 < \theta_2$, so substituting we obtain $n-3
< 3t \leq n+3$. Writing $n = 3p+u$ for $u = 0,1,2$, we get $t \in
\set{p,p+1}$. Recall that by assumption $n\divides t^2+1$. If $t =
p$, then $n \divides 9(t^2+1)-n(3p-u) = 9+u^2$; otherwise $t =
p+1$, and $n \divides 9(t^2+1)-(3p+6-u)n = 9+(3-u)^2$. In both
cases $n \in \set{5,9,13}$.

A direct computation now excludes the few possible values of
$(n,t)$. Note that for $n = 5$,
$(\rho+\rho^{-1})(\rho^2+\rho^{-2}) = -1$.
\end{proof}


 \iffurther


\fi % For: \iffurther







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\end{document}
