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\newtheorem*{remark}{Remark}

\title{Quartic Exercises}
\author{Max-Albert Knus\and Jean-Pierre Tignol}
\address{Departement Mathematik\\
ETH Zentrum\\
CH-8092 Z\"urich\\
Switzerland}
\email{knus@math.ethz.ch}
\address{Institut de Math\'ematique Pure et Appliqu\'ee\\
Universit\'e catholique de Louvain\\
B-1348 Louvain-la-Neuve\\
Belgium}
\email{tignol@math.ucl.ac.be}
\thanks{The authors gratefully acknowledge support from the
RT Network ``$K$-theory, Linear Algebraic Groups and Related
Structures'' (contract HPRN-CT-2002-00287). The first author is
partially supported by the Swiss National Foundation under grant 21-064712.01.
  The second author is partially supported by the National Fund
  for Scientific Research
(Belgium).}

\begin{document}

\begin{abstract}
     A correspondence between quartic \'etale algebras over a field and
     quadratic \'etale extensions of cubic \'etale algebras is set up
     and investigated. The basic constructions are laid out in general
     for sets with a profinite group action and for torsors, and
     translated in terms of \'etale algebras and Galois algebras. In
     the final section, a parametrization of cyclic quartic algebras
     is given.
\end{abstract}

\maketitle

It is known since the XVIth century that the solution of quartic
equations can be obtained by means of auxiliary equations of
degree~$3$, called cubic resolvents. The situation is easily
understood in terms of Galois theory. For any integer $n\geq1$, let
$\Sym_{n}$ denote the symmetric group on $\{1, \ldots, n\}$. The
symmetric group $\Sym_{4}$ contains a normal subgroup of order~$4$,
Klein's Vierergruppe
\[
\Vier=\{I, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)\},
\]
which is the kernel of the action of $\Sym_{4}$ on its three Sylow
$2$-subgroups. Numbering
from~$1$ to $3$ these Sylow subgroups, we get an exact sequence of
groups
\begin{equation}
      \label{1.eq}
      1\to\Vier\to \Sym_{4}
      \xrightarrow{\rho}
      \Sym_{3}\to1.
\end{equation}
Let $F$ be an arbitrary field and $P\in F[X]$ be a separable
polynomial of degree~$4$. Let also $\Fs$ be a separable closure of $F$
and $Q\subset\Fs$ be the subfield generated by the roots of $P$. The
Galois group $\Gal(Q/F)$ can be viewed as a subgroup of $\Sym_{4}$
through its action on the roots of $P$. The subfield $L$ of $Q$ fixed
under $\Gal(Q/F)\cap\Vier$ is generated by the roots of a cubic
resolvent, as was shown by Lagrange. For a given quartic polynomial
$P$, there are actually many polynomials of degree~$3$ which
qualify as cubic resolvents; only the extension $L/F$ is an invariant
of $P$ (or of $Q$).

Galois cohomology provides another viewpoint on this construction.
Since $\Sym_{n}$ is the automorphism group of the \'etale $F$-algebra
$F^n=F\times\cdots\times F$, it is well-known that the Galois
cohomology set $H^{1}(F,\Sym_{n})$ is in canonical one-to-one
correspondence with the isomorphism classes of \'etale $F$-algebras
of degree~$n$, see \cite[(29.9)]{BoI}. The map $\rho$ in \eqref{1.eq}
induces a map
\[
\rho^{1}\colon\,H^{1}(F,\Sym_{4})\to H^{1}(F,\Sym_{3})
\]
which associates to every quartic \'etale $F$-algebra $Q$ a cubic
\'etale $F$-algebra $\resolv(Q)$ uniquely determined up to
isomorphism. If $P\in F[X]$ is a separable polynomial of degree~$4$
with cubic resolvent $R$, and if $Q$ is the factor algebra $Q=F[X]/P$,
then $\resolv(Q)\simeq F[X]/R$, see \S\ref{quartal.subsec}.

Our first aim is to make explicit the construction of $\resolv(Q)$
from $Q$. But this construction can be further extended. Each of the
three Sylow $2$-subgroups of $\Sym_{4}$ contains two transpositions,
and each transposition is in one and only one Sylow subgroup, hence
the set of transpositions can be viewed as a double covering of the
set of Sylow $2$-subgroups. Therefore, the conjugation action of
$\Sym_{4}$ on its six transpositions defines a map
\[
\lambda\colon\,\Sym_{4}\to\Sym_{2}\wr\Sym_{3},
\]
where the wreath product $\Sym_{2}\wr\Sym_{3}$ is viewed as the group
of automorphisms of a double covering of a set of three elements (see
\S\ref{perm.subsec}).
The map $\lambda$ extends to an isomorphism of groups
\[
\hat\lambda\colon\,\Sym_{2}\times\Sym_{4}\stackrel{\sim}{\to}
\Sym_{2}\wr\Sym_{3},
\]
see \S\ref{cohom4.subsec}. The set $H^{1}(F,\Sym_{2}\wr\Sym_{3})$ classifies
the quadratic \'etale extensions of cubic \'etale $F$-algebras
(see \S\ref{cohsets.subsec}), and the induced bijection
\[
\hat\lambda^{1}\colon\,
H^{1}(F,\Sym_{2})\times H^{1}(F,\Sym_{4})
\stackrel{\sim}{\to}
H^{1}(F,\Sym_{2}\wr\Sym_{3})
\]
associates to every pair consisting of a quartic \'etale $F$-algebra
$Q$ and a quadratic \'etale $F$-algebra a quadratic \'etale extension
of the cubic resolvent $\resolv(Q)$. In \S\ref{quartal.subsec}, we give
an explicit construction of this quadratic extension, and we describe
in \S\ref{quadextet.subsec} the
inverse of $\hat\lambda^{1}$, attaching a quartic \'etale $F$-algebra
and a quadratic \'etale $F$-algebra to any quadratic \'etale
extension of a cubic \'etale $F$-algebra.

In the final sections, we classify quartic \'etale algebras and their
associated quadratic extensions of cubic algebras according to their
decomposition into direct products of fields (see
\S\ref{classif.subsec}) and we parametrize cyclic quartic extensions.

\tableofcontents

\begin{quote}
      La recherche des extensions d'un corps $k$ dont le groupe de
      Galois sur $k$ est $\Sym_{4}$ ou $\Alt_{4}$ n'est pas autre
      chose, du point de vue des alg\'ebristes du XIX$^{\text{e}}$
      si\`ecle, que la th\'eorie de l'\'equation du $4^{\text{e}}$
      degr\'e. C'est un probl\`eme pour lequel ces alg\'ebristes
      n'avaient que du m\'epris. (A.~Weil)
\end{quote}

\section{$\Gamma$-sets and coverings}
\label{sets.sec}

\subsection{Basic constructions on $\Gamma$-sets}
\label{basicset.subsec}

Let $\Gamma$ be a profinite group, which will be fixed throughout this
section. Finite sets with a continuous action of $\Gamma$ are called
$\Gamma$-sets. We let $\vert X\vert$ denote the number of elements in
a $\Gamma$-set $X$. If $X$ is a $\Gamma$-set with $n$ elements, and
$k$ is a positive integer, $k\leq n$, we let $\Sigma_{k}(X)$ denote
the set of $k$-tuples of pairwise distinct elements of $X$ and
$\Lambda_{k}(X)$ the set of $k$-element subsets of $X$; thus
\begin{gather*}
\Sigma_{k}(X)=\bigl\{(\xi_{1},\ldots,\xi_{k})\in X^k\mid
\xi_{i}\neq\xi_{j}\text{ for $i\neq j$}\bigr\},\\
\Lambda_{k}(X)=\bigl\{\{\xi_{1},\ldots,\xi_{k}\}\subset X\mid
\xi_{i}\neq\xi_{j}\text{ for $i\neq j$}\bigr\}.
\end{gather*}
The action of $\Gamma$ on $X$ induces actions on $\Sigma_{k}(X)$ and
$\Lambda_{k}(X)$, hence $\Sigma_{k}(X)$ and $\Lambda_{k}(X)$ are
$\Gamma$-sets, and we have
\[
\vert\Sigma_{k}(X)\vert=k!\binom{n}{k},\qquad
\vert\Lambda_{k}(X)\vert=\binom{n}{k}.
\]
The symmetric group $\Sym_{k}$ acts on
$\Sigma_{k}(X)$ by permutation of the entries,
  and we may consider $\Lambda_{k}(X)$ as the set of
orbits of $\Sigma_{k}(X)$ under this action, i.e. as the quotient
$\Gamma$-set
\[
\Lambda_{k}(X)=\Sigma_{k}(X)/\Sym_{k}.
\]

For $k=n$,
we may also consider the action of the alternating group
$\Alt_{n}$ on $\Sigma_{n}(X)$. The quotient is called the
\emph{discriminant} of $X$ and denoted by $\Delta(X)$,
\[
\Delta(X)=\Sigma_{n}(X)/\Alt_{n},
\]
see \cite[p.~291]{BoI}. This is a $\Gamma$-set with $\vert\Delta(X)\vert=2$
if $n\geq2$.

If $n$ is even, $n=2m$, let
\[
\gamma_{X}\colon\,\Lambda_{m}(X)\to\Lambda_{m}(X)
\]
be the map which associates to every $m$-element subset of $X$ its
complementary subset. Since $\gamma_{X}^{2}=\Id$, this map defines an
action of $\Sym_{2}$ on $\Lambda_{m}(X)$. The map $\gamma_{X}$ is
$\Gamma$-equivariant (i.e.\ compatible with the action of $\Gamma$),
hence the quotient
\[
\resolv(X)=\Lambda_{m}(X)/\Sym_{2}
\]
is a $\Gamma$-set. It is the set of partitions of $X$ into
$m$-element subsets.

\begin{example}
      \label{set4.ex}
      If $X=\{1,2,3,4\}$, then
      \[
      \Lambda_{2}(X)=\bigl\{\{1,2\},\{3,4\},\{1,3\},
      \{2,4\},\{1,4\},\{2,3\}\bigr\}
      \]
      and
      \[
      \resolv(X)=
      \Bigl\{\bigl\{\{1,2\},\{3,4\}\bigr\},
      \bigl\{\{1,3\},\{2,4\}\bigr\},
      \bigl\{\{1,4\},\{2,3\}\bigr\}\Bigr\}.
      \]
\end{example}

If $\vert X\vert=2$, the map
\[
\gamma_{X}\colon\,X=\Lambda_{1}(X)\to\Lambda_{1}(X)=X
\]
interchanges the two elements of $X$. For $X$, $X'$ two $\Gamma$-sets
with $\vert X\vert=\vert X'\vert=2$, the map
\[
\gamma_{X}\times\gamma_{X'}\colon\,X\times X'\to X\times X'
\]
defines an action of $\Sym_{2}$ compatible with the $\Gamma$-action.
Let
\[
X*X'=(X\times X')/\Sym_{2},
\]
a $\Gamma$-set with $\vert X*X'\vert=2$. Thus, if $X=\{x_{1},x_{2}\}$
and $X'=\{x'_{1},x'_{2}\}$, then
\[
X*X'=\bigl\{
\{(x_{1},x'_{1}),(x_{2},x'_{2})\},
\{(x_{1},x'_{2}),(x_{2},x'_{1})\}
\bigr\}.
\]
The following observations are clear:

\begin{prop}
     \label{groupstar.prop}
      Let $X$, $X'$, be $\Gamma$-sets of two elements.
      \begin{enumerate}
         \item[(a)]
         The $\Gamma$-action on $X*X$ is trivial.
         \item[(b)]
         If the $\Gamma$-action on $X'$ is trivial, then $X*X'\simeq X$.
         (Note that the isomorphism is \emph{not} canonical.)
         \item[(c)] The operation $*$ defines a group structure on the
         set of isomorphism classes of $\Gamma$-sets of two elements.
      \end{enumerate}
\end{prop}

See \S\ref{cohsets.subsec}
for a cohomological  interpretation of the group structure
induced by $*$.

\subsection{Coverings}
\label{cov.subsec}

A morphism $Y\xleftarrow{\pi}Z$ of $\Gamma$-sets (i.e.\ a
$\Gamma$-equivariant map) is called a
\emph{covering} if the number of
elements in each fiber $\pi^{-1}(\eta)\subset Z$ does not depend on
$\eta\in Y$. This number is called the \emph{degree} of the covering.
Coverings of degree~$2$ are called \emph{double} coverings.
%%
A \emph{morphism of coverings}
\[
(Y_1\xleftarrow{\pi_1}Z_1) \to
(Y_2\xleftarrow{\pi_2}Z_2)
\]
is a pair $(\sigma\colon\, Y_1 \to Y_2,\tau\colon\, Z_1 \to Z_2)$
of morphisms such that $ \sigma \circ \pi_1 =
\pi_2 \circ \tau$. Given two coverings $Y\xleftarrow{\pi_1}Z_1$
and $Y\xleftarrow{\pi_2}Z_2$ of the same $\Gamma$-set $Y$,
an \emph{isomorphism over $Y$} is an isomorphism $\tau\colon\,Z_1 \to Z_2$
such that $\pi_1= \pi_2 \circ \tau$.
%%

For any covering $Y\xleftarrow{\pi}Z$
let  $\Omega(Z/Y)$ be  the set of sections of $\pi$,
%
\[
\Omega(Z/Y)=\bigl\{
\{\zeta_{1},\ldots,\zeta_{n}\}\subset Z\mid
\{\pi(\zeta_{1}),\ldots,\pi(\zeta_{n})\}=Y\bigr\}
\subset\Lambda_{n}(Z).
\]
This is a $\Gamma$-set with $\vert\Omega(Z/Y)\vert=d^n$. If $d=2$,
the morphism
\[
\gamma_{Z}\colon\Lambda_{n}(Z)\to\Lambda_{n}(Z)
\]
preserves $\Omega(Z/Y)$ and induces a $\Sym_{2}$-action compatible
with the action of $\Gamma$. Define
\[
{\mathcal S}(Z/Y)=\Omega(Z/Y)/\Sym_{2}.
\]

Note that every double covering $Y\xleftarrow{\pi}Z$ has an
involutive automorphism $\gamma_{Z/Y}$ which is the identity on $Y$
and interchanges the two elements in each fiber of $\pi$. Thus, for
$\{\zeta_{1},\ldots,\zeta_{n}\}\in\Omega(Z/Y)$,
\[
\gamma_{Z}\bigl(\{\zeta_{1},\ldots,\zeta_{m}\}\bigr)=
\bigl\{\gamma_{Z/Y}(\zeta_{1}),\ldots,\gamma_{Z/Y}(\zeta_{n})\bigr\}.
\]

\begin{example}
      \label{cov6.ex}
      Let
      $Z=\{z_{1},z_{1}',z_{2},
      z_{2}',z_{3},z_{3}'\}$,
      $Y=\{1,2,3\}$, and let $Y\xleftarrow{\pi}Z$
      be the map which carries $z_{i}$ and $z_{i}'$ to
      $i$ for $i=1$, $2$, $3$. Then
      \begin{align*}
      \Omega(Z/Y)=
      \bigl\{&
      \{z_{1},z_{2},z_{3}\},
      \{z_{1},z_{2}',z_{3}'\},
      \{z_{1}',z_{2},z_{3}'\},
      \{z_{1}',z_{2}',z_{3}\},\\
      &\{z_{1}',z_{2}',z_{3}'\},
      \{z_{1}',z_{2},z_{3}\},
      \{z_{1},z_{2}',z_{3}\},
      \{z_{1},z_{2},z_{3}'\}
      \bigr\}
      \end{align*}
      and
      \begin{align*}
         {\mathcal S}(Z/Y)=
         \Bigl\{
         &\bigl\{\{z_{1},z_{2},z_{3}\},
         \{z_{1}',z_{2}',z_{3}'\}\bigr\},
         \bigl\{\{z_{1},z_{2}',z_{3}'\},
         \{z_{1}',z_{2},z_{3}\}\bigr\},\\
         &\bigl\{\{z_{1}',z_{2},z_{3}'\},
         \{z_{1},z_{2}',z_{3}\}\bigr\},
         \bigl\{\{z_{1}',z_{2}',z_{3}\},
         \{z_{1},z_{2},z_{3}'\}\bigr\}
         \Bigr\}.
      \end{align*}
\end{example}

Let $Y\xleftarrow{\pi}Z$
be an arbitrary double covering of a $\Gamma$-set $Y$ of $n$
elements, so $\vert Z\vert=2n$, and let
$\{\zeta_{1},\ldots ,\zeta_{n}\}\in\Omega(Z/Y)$. Even though the
$n$-tuple $(\zeta_{1},\ldots,\zeta_{n})$ is not uniquely determined
by the set $\{\zeta_{1},\ldots,\zeta_{n}\}$, it turns out that the
orbit
\[
\bigl(\zeta_{1},\ldots,\zeta_{n},\gamma_{Z/Y}(\zeta_{1}),\ldots,
\gamma_{Z/Y}(\zeta_{n})\bigr)^{\Alt_{2n}}
\]
is well-defined, since every permutation of $\zeta_{1},\ldots,\zeta_{n}$
  induces a corresponding permutation of
$\gamma_{Z/Y}(\zeta_{1}),\ldots,\gamma_{Z/Y}(\zeta_{n})$, and the resulting
permutation of the $2n$ elements $\zeta_{1}, \ldots,
\gamma_{Z/Y}(\zeta_{n})$ is
necessarily even. Therefore, we may define
\[
\delta\bigl(\{\zeta_{1},\ldots,\zeta_{n}\})=
\bigl(\zeta_{1},\ldots,\zeta_{n},
\gamma_{Z/Y}(\zeta_{1}),\ldots,
\gamma_{Z/Y}(\zeta_{n})\bigr)^{\Alt_{2n}}\in\Sigma_{2n}(Z)/\Alt_{2n}
=\Delta(Z)
\]
and thus obtain a morphism of $\Gamma$-sets
\begin{equation}\label{deltaZ.eq}
\delta= \delta_{Z}\colon\,\Omega(Z/Y)\to\Delta(Z).
\end{equation}
On the other hand, since ${\mathcal S}(Z/Y)$ is a quotient of
$\Omega(Z/Y)$, there is a canonical map
\[
\varepsilon\colon\,\Omega(Z/Y)\to{\mathcal S}(Z/Y).
\]

\begin{prop}
      \label{cov6.prop}
      For $Y\xleftarrow{\pi}Z$ a double covering with $\vert Y\vert=n$ odd,
      the map
      \[
      (\delta,\varepsilon)\colon\,\Omega(Z/Y)\to
      \Delta(Z)\times{\mathcal S}(Z/Y)
      \]
      is an isomorphism of $\Gamma$-sets.
\end{prop}

\begin{proof}
      Since the map $(\delta,\varepsilon)$ is clearly
      $\Gamma$-equivariant, and since the sets $\Omega(Z/Y)$ and
      $\Delta(Z)\times{\mathcal S}(Z/Y)$ both have $2^n$ elements, it
      suffices to show that $(\delta,\varepsilon)$ is injective.
      Suppose $\{\zeta_{1},\ldots,\zeta_{n}\}$,
      $\{\zeta_{1}',\ldots,\zeta_{n}'\}\in\Omega(Z/Y)$ are distinct
      elements such
      that $\varepsilon\bigl(\{\zeta_{1},\ldots,\zeta_{n}\}\bigr)=
      \varepsilon\bigl(\{\zeta_{1}',\ldots,\zeta_{n}'\}\bigr)$; then
      $\{\zeta_{1}',\ldots,\zeta_{n}'\}=
      \gamma_{Z}\bigl(\{\zeta_{1},\ldots,\zeta_{n}\}\bigr)$ and we
      may assume the elements are numbered in such a way that
      $\pi(\zeta_{i}')=\pi(\zeta_{i})$ (i.e.\
      $\zeta'_{i}=\gamma_{Z/Y}(\zeta_{i})$) for $i=1,\ldots,n$. Since the
      permutation which interchanges $\zeta_{i}$ and $\zeta_{i}'$ for
      $i=1,\ldots,n$ is odd, we have
      \[
      (\zeta_{1},\ldots,\zeta_{n},
      \zeta_{1}',\ldots,\zeta_{n}')^{\Alt_{2n}}\neq
      (\zeta_{1}',\dots,\zeta_{n}',
      \zeta_{1},\ldots,\zeta_{n})^{\Alt_{2n}},
      \]
      hence $\delta\bigl(\{\zeta_{1},\ldots,\zeta_{n}\}\bigr)\neq
      \delta\bigl(\{\zeta_{1}',\ldots,\zeta_{n}'\}\bigr)$.
\end{proof}

If $Y\xleftarrow{\pi}Z$ and $Y\xleftarrow{\pi'}Z'$ are two double
coverings of the same $\Gamma$-set $Y$ with $n$ elements,
consider the \emph{fiber product}
\[
Z\times_{Y}Z'=\bigl\{
(\zeta,\zeta')\in Z\times Z'\mid
\pi(\zeta)=\pi'(\zeta')\bigr\}
\subset Z\times Z'.
\]
The group $\Sym_{2}$ acts on $Z\times_{Y}Z'$ by mapping
$(\zeta,\zeta')$ to
$\bigl(\gamma_{Z/Y}(\zeta),\gamma_{Z'/Y}(\zeta')\bigr)$. Let
\[
Z*_{Y}Z'=(Z\times_{Y}Z')/\Sym_{2}.
\]
The canonical map $Y\xleftarrow{\pi\times\pi'}Z\times_{Y}Z'$ induces
a map
\[
Y\xleftarrow{\pi*\pi'}Z*_{Y}Z'
\]
which is a double covering. In particular, for any
$\Gamma$-set $X$ of two elements and any covering
$Y\xleftarrow{\pi}Z$ of degree~$2$, we may consider
\[
Y\xleftarrow{\pi_{2}*\pi}(X\times Y)*_{Y}Z
\]
where $Y\xleftarrow{\pi_{2}}X\times Y$ is the projection map. Abusing
notation, we write simply
\[
Y\xleftarrow{\pi}X*Z
\]
for this double covering.
  The proof of the following easy proposition is omitted:

\begin{prop}
      \label{compcov.prop}
      Let $X$ be a set of two elements with trivial $\Gamma$-action,
      and let $Y\xleftarrow{\pi}Z$ be a double covering.
      \begin{enumerate}
         \item[(a)]
         The covering $Y\xleftarrow{\pi*\pi}Z*_{Y}Z$ is isomorphic to
         $Y\xleftarrow{\pi_{2}}X\times Y$.
         \item[(b)]
         The covering $Y\xleftarrow{\pi}X*Z$ is (non-canonically) isomorphic
         to $Y\xleftarrow{\pi}Z$.
         \item[(c)]
  The operation $*_Y$ defines a group structure on the set of isomorphism
classes over $Y$ of double coverings of $Y$. The neutral element
is the isomorphism class of $Y\xleftarrow{\pi_{2}}X\times Y$.
      \end{enumerate}
\end{prop}

\begin{prop} \label{deltahom.prop}
Let $Y\xleftarrow{\pi}Z$ and $Y\xleftarrow{\pi}Z'$ be two double
coverings. There is a canonical isomorphism
$\Delta(Z *_Y Z') \simeq \Delta(Z) *\Delta(Z')$.
\end{prop}

\begin{proof}
Recall from \eqref{deltaZ.eq} the map
\[
\delta_Z\colon\,\Omega(Z/Y)\to\Delta(Z)
\]
defined for any double covering $Y\xleftarrow{\pi}Z$ of a
$\Gamma$-set $Y$ with $|Y|=n$. In the sequel, we write simply
$\delta_Z(z_1,\ldots,z_n)$ for $\delta_Z(\{z_1,\ldots,z_n\})$. For
$\omega$, $\omega'\in\Omega(Z/Y)$, we have
$\delta_Z(\omega)=\delta_Z(\omega')$ if and only if
$|\omega\cap\omega'|\equiv n\bmod2$. In particular, the map $\delta_Z$
is onto.
Let $Y\xleftarrow{\pi}Z$ and $Y\xleftarrow{\pi'}Z'$ be two double
coverings of $Y$.
  Denote simply by $\invo$ the canonical automorphisms
$\gamma_{Z/Y}$ and $\gamma_{Z'/Y}$ and also the canonical
automorphisms of $\Delta(Z)$ and $\Delta(Z')$. For
$\{z_1,\ldots,z_n\}\in\Omega(Z/Y)$ and
$\{z'_1,\ldots,z'_n\}\in\Omega(Z'/Y)$, we have
\[
\delta_Z(\overline{z_1},z_2,\ldots,z_n)=
\overline{\delta_Z(z_1,z_2,\ldots,z_n)}
\]
and, similarly,
\[
\delta_{Z'}(\overline{z'_1},z'_2,\ldots,z'_n)=
\overline{\delta_{Z'}(z'_1,z'_2,\ldots,z'_n)}.
\]
Therefore, the element
\[
\Bigl\{
\bigl(\delta_Z(z_1,\ldots,z_n),\delta_{Z'}(z'_1,\ldots,z'_n)\bigr),
\bigl(\overline{\delta_Z(z_1,\ldots,z_n)},
\overline{\delta_{Z'}(z'_1,\ldots,z'_n)}\bigr)
\Bigr\}\in\Delta(Z)*\Delta(Z')
\]
depends only on
\[
\omega=\Bigl\{
\bigl\{(z_1,z'_1),(\overline{z_1},\overline{z'_1})\bigr\},\ldots,
\bigl\{(z_n,z'_n),(\overline{z_n},\overline{z'_n})\bigr\}
\Bigr\}\in\Omega(Z*_YZ'/Y).
\]
We thus have a canonical map
\[
\psi\colon\,\Omega(Z*_YZ'/Y)\to\Delta(Z)*\Delta(Z').
\]
If $\omega'\in\Omega(Z*_YZ'/Y)$ is obtained from $\omega$ by
substituting
\[
\bigl\{(z_1,\overline{z'_1}),(\overline{z_1},z'_1)\bigr\}
\quad\text{for}\quad
\bigl\{(z_1,z'_1),(\overline{z_1},\overline{z'_1})\bigr\},
\]
then $\psi(\omega)\neq\psi(\omega')$, hence $\psi$ is onto. On the
other  hand, if $\omega'$ is obtained from $\omega$ by an even number
of changes as above, then  $\psi(\omega)=\psi(\omega')$. Therefore,
$\psi(\omega)=\psi(\omega')$ if $|\omega\cap\omega'|\equiv n\bmod2$,
and it follows that $\psi$ factors through the map
\[
\delta_{Z*_YZ'}\colon\,\Omega(Z*_YZ'/Y)\to\Delta(Z*_YZ').
\]
This completes the proof.
\end{proof}

For later use, we record another case where the map $\delta$ of
\eqref{deltaZ.eq} can be used in the computation of a discriminant.
Let $X$ be a $\Gamma$-set of two elements. For any $\Gamma$-set $Y$,
we may consider the double covering $Y\xleftarrow{\pi_{2}}X\times Y$
given by the projection.

\begin{prop}
     \label{discprod.prop}
     If $\vert X\vert=2$ and $\vert Y\vert$ is odd, the composition
     of the $\Gamma$-equivariant map
     \[
     \iota\colon\,X\to\Omega(X\times Y/Y),
     \qquad
     \iota(x)=\bigl\{(x,y)\mid y\in Y\bigr\}
     \]
     and
     \[
     \delta_{X\times Y}\colon\,\Omega(X\times
     Y/Y)\to\Delta(X\times Y)
     \]
     defines an isomorphism
     \[
     \delta\circ\iota\colon\,X\iso\Delta(X\times Y).
     \]
\end{prop}

\begin{proof}
     Since $\vert Y\vert$ is odd, we have
     \[
     \delta\circ\iota\bigl(\gamma_{X}(x)\bigr)=
     \gamma_{\Delta(X\times Y)}\bigl(\delta\circ\iota(x)\bigr)
     \qquad\text{for $x\in X$}.
     \]
     Therefore, $\delta\circ\iota$ is surjective, hence bijective.
\end{proof}

\begin{prop}
    \label{coverinvariance.prop}
      Let $X$ be a $\Gamma$-set of two elements, and let
      $Y\xleftarrow{\pi}Z$ be a double covering. There is a
      canonical isomorphism
      \[
      {\mathcal S}(Z/Y)\simeq{\mathcal S}(X*Z/Y).
      \]
\end{prop}

\begin{proof}
      For simplicity of notation, let $X=\{+,-\}$ and denote by
      $\invo$ the canonical automorphism $\gamma_{Z/Y}$. We may then
      identify $X*Z$ with the set of formal polynomials
      $\zeta-\overline{\zeta}$, for $\zeta\in Z$. Note however that
      the $\Gamma$-action on these polynomials is not linear, since
      $\Gamma$ may act non trivially on $\{+,-\}$. The structure map
      $Y\leftarrow X*Z$ carries $\zeta-\overline{\zeta}$ to
      $\pi(\zeta)=\pi(\overline{\zeta})$. Therefore,
      \[
      \Omega(X*Z/Y)=\bigl\{\{\zeta_{1}-\overline{\zeta_{1}},\ldots,
      \zeta_{n}-\overline{\zeta_{n}}\}\mid
      \{\pi(\zeta_{1}),\ldots,\pi(\zeta_{n})\}=Y\bigr\}
      \]
      and
      \begin{multline*}
      {\mathcal S}(X*Z/Y)=\\
      \bigl\{\bigl\{
      \{\zeta_{1}-\overline{\zeta_{1}},\ldots,
      \zeta_{n}-\overline{\zeta_{n}}\},
      \{\overline{\zeta_{1}}-\zeta_{1},\ldots,
      \overline{\zeta_{n}}-\zeta_{n}\}\bigr\}\mid
      \{\pi(\zeta_{1}),\ldots,\pi(\zeta_{n})\}=Y\bigr\}.
      \end{multline*}
      On the other hand,
      \[
      {\mathcal S}(Z/Y)=
      \bigl\{\bigl\{
      \{\zeta_{1},\ldots,\zeta_{n}\},
      \{\overline{\zeta_{1}},\ldots,\overline{\zeta_{n}}\}\bigr\}
      \mid
      \{\pi(\zeta_{1}),\ldots,\pi(\zeta_{n})\}=Y\bigr\}.
      \]
      The map
      \[
      \bigl\{\{\zeta_{1},\ldots,\zeta_{n}\},
      \{\overline{\zeta_{1}},\ldots,\overline{\zeta_{n}}\}\bigr\}
      \mapsto
      \bigl\{
      \{\zeta_{1}-\overline{\zeta_{1}},\ldots,
      \zeta_{n}-\overline{\zeta_{n}}\},
      \{
      \overline{\zeta_{1}}-\zeta_{1},\ldots,
      \overline{\zeta_{n}}-\zeta_{n}\}
      \bigr\}
      \]
      is a $\Gamma$-isomorphism ${\mathcal S}(Z/Y)\iso {\mathcal S}(X*Z/Y)$,
      whatever the action of $\Gamma$ on $\{+,-\}$~is.
\end{proof}

\section{\'Etale algebras and extensions}
\label{etal.sec}

In this section, $F$ is an arbitrary field. We denote by $F_{s}$ a
separable closure of $F$ and let $\Gamma=\Gal(F_{s}/F)$. A
finite-dimensional $F$-algebra $E$ is called \emph{\'etale} if
$E\otimes_{F}F_{s}$ is isomorphic to a split $F_{s}$-algebra
$F_{s}\times\cdots\times F_{s}$. For any \'etale $F$-algebra $E$ of
dimension $n$, the set of $F$-algebra homomorphisms
\[
\funcX(E)=\Hom_{\text{$F$-alg}}(E,F_{s})
\]
is a $\Gamma$-set of $n$ elements since $\Gamma$ acts on $F_{s}$.

Conversely, starting from a $\Gamma$-set $X$ with $\vert X\vert=n$, we
may let $\Gamma$ act by semi-linear automorphisms on the
$F_{s}$-algebra of maps $\Map(X,F_{s})$. The fixed $F$-algebra
\[
\funcM(X)=\Map(X,F_{s})^\Gamma=\{f\colon\,X\to F_{s}\mid
\gamma f(\xi)=f({}^\gamma \xi) \text{ for $\gamma\in\Gamma$, $\xi\in
X$}\}
\]
is \'etale of dimension $n$. Moreover, there are canonical
isomorphisms
\[
\funcM\bigl(\funcX(E)\bigr)\simeq E,\qquad
\funcX\bigl(\funcM(X)\bigr)\simeq X
\]
(see \cite[(18.19)]{BoI}), so that the functors $\funcM$ and $\funcX$
define an
anti-equivalence between the category $\Et_{F}$ of \'etale $F$-algebras
(with
$F$-algebra homomorphisms) and the category $\Set_{\Gamma}$ of
$\Gamma$-sets. Under this anti-equivalence, the direct product (resp.\
tensor product) of $F$-algebras corresponds to the disjoint union
(resp.\ direct product) of $\Gamma$-sets: for $E_{1}$, $E_{2}$
\'etale $F$-algebras, there are obvious identifications
\[
\funcX(E_1 \tens E_2) = \funcX(E_1) \times \funcX(E_2) \quad \text{and}\quad
\funcX(E_1 \times E_2) =\funcX(E_1) \coprodd \funcX(E_2).
\]
Moreover, if $G$ is a group acting on an \'etale $F$-algebra $E$ by
$F$-automorphisms, then for the fixed subalgebra $E^{G}$ we have
\[
\funcX(E^{G})=\funcX(E)/G
\]
since $E^{G}$ is the equalizer of the automorphisms
$\sigma\colon\,E\to E$ for $\sigma\in G$, and $\funcX(E)/G$ is the
co-equalizer of the corresponding automorphisms of $\funcX(E)$.
Through the anti-equivalence $\Et_{F}\equiv\Set_{\Gamma}$, the
constructions on $\Gamma$-sets defined in \S\ref{basicset.subsec}
therefore have counterparts in the category of \'etale $F$-algebras.
The aim of this section is to make them explicit.

\subsection{Basic constructions on \'etale algebras}
\label{Basicet.subsec}

Let $E$ be an \'etale $F$-algebra of dimension~$n$. Under the
canonical isomorphism $E\simeq\funcM\bigl(\funcX(E)\bigr)$,
the idempotents of $E$ correspond to the characteristic functions
of $\Gamma$-subsets of $\funcX(E)$. If $e\in E$ is the characteristic
function of a subset $Y\subset \funcX(E)$, then multiplication by $e$
defines an isomorphism $E/(1-e)E\iso eE$. Moreover, $\funcX(eE)=Y$ and
under the anti-equivalence $\Et_{F}\equiv\Set_{\Gamma}$, the map
$E\to eE$ corresponds to the inclusion $\funcX(E)\leftarrow Y$.

\begin{example}
     \label{splitidemp.ex}
     If $E$ is the split \'etale $F$-algebra $E=F^n$, then $\funcX(E)$ is in
     duality with the set $e_{1}$, \ldots, $e_{n}$ of minimal idempotents
     of $E$, namely $\funcX(E)=\{\xi_{1},\ldots,\xi_{n}\}$ where
     \[
     \xi_{i}(e_{j})=
     \begin{cases}
         0&\text{if $i\neq j$},\\
         1&\text{if $i=j$}.
     \end{cases}
     \]
     The idempotent corresponding to a subset $Y=\{\xi_{i}\mid i\in
     I\}\subset \funcX(E)$ is $\sum_{i\in I}e_{i}$.
\end{example}

Let $E$ be an arbitrary \'etale $F$-algebra of dimension~$n$.
For any integer $k$ with $1\leq k\leq n$, we let $s_{k}\in E^{\otimes
k}$ be the idempotent corresponding to the characteristic function of
the subset
\[
\Sigma_{k}\bigl(\funcX(E)\bigr)=
\bigl\{(\xi_{1},\ldots,\xi_{k})\mid\text{$\xi_{i}\neq\xi_{j}$ for
$i\neq j$}\bigr\}\subset \funcX(E)^k=\funcX(E^{\otimes k}).
\]
Therefore, letting $\Sigma_{k}(E)=s_{k}E^{\otimes k}$, we have
\[
\funcX\bigl(\Sigma_{k}(E)\bigr)=\Sigma_{k}\bigl(\funcX(E)\bigr).
\]

In particular, for $k=2$ the idempotent $1-s_{2}$ is the
characteristic function of the diagonal of $\funcX(E)\times
\funcX(E)=\funcX(E\otimes E)$. It is the \emph{separability idempotent}
of $E$,
see \cite[p.~285]{BoI}. For $k\geq3$, the idempotent $s_{k}$ can also be
defined in terms of the separability idempotent of $E$, see
\cite[p.~42]{SaltLect}, \cite[p.~320]{BoI}.

The symmetric group $\Sym_{k}$ acts on
$E^{\otimes k}$ by permutation
of the factors, and the idempotent $s_{k}$ is fixed under this action,
so $\Sym_{k}$ also acts on $\Sigma_{k}(E)$. We consider the fixed
subalgebra
\[
\Lambda_{k}(E)=\Sigma_{k}(E)^{\Sym_{k}}=s_{k}(E^{\otimes k})^{\Sym_{k}}.
\]
We have
\[
\funcX\bigl(\Lambda_{k}(E)\bigr)=
\Lambda_{k}\bigl(\funcX(E)\bigr)
\]
since under the anti-equivalence $\Et_{F}\equiv\Set_{\Gamma}$ the
fixed algebra under $\Sym_{k}$ corresponds to the factor set under
the $\Sym_{k}$-action.

The \emph{discriminant} of $E$ is defined by
\[
\Delta(E)=\Sigma_{n}(E)^{\Alt_{n}},
\]
and we have
\[
\funcX\bigl(\Delta(E)\bigr)=\Delta\bigl(\funcX(E)\bigr).
\]
%
\begin{example}
      \label{splitS.ex}
      If $E=F^n$ is split with minimal idempotents $e_{1}$, \ldots,
      $e_{n}$, then
      \[
      s_{k}=\sum_{(i_{1},\ldots,i_{k})}e_{i_{1}}\otimes\cdots\otimes
      e_{i_{k}}
      \]
      where $(i_{1},\ldots,i_{k})$ ranges over the $k$-tuples of
      pairwise distinct integers from $1$ to $n$ (i.e.\
      $(i_{1},\ldots,i_{k})\in\Sigma_{k}\bigl(\{1,\ldots,n\}\bigr)$).
      The algebra $\Sigma_{k}(E)$ is then split, with minimal idempotents
      \[
      e_{i_{1}}\otimes\cdots\otimes e_{i_{k}}
      \]
      for $(i_{1},\ldots,i_{k})\in\Sigma_{k}\bigl(\{1,\ldots,n\}\bigr)$.
      Similarly, the algebra $\Lambda_{k}(E)$ is split, with minimal
      idempotents
      \[
      e_{\{i_{1},\ldots,i_{k}\}}=\sum_{\sigma\in\Sym_{k}}
      e_{i_{\sigma(1)}}\otimes\cdots\otimes e_{i_{\sigma(k)}}
      \]
      for $\{i_{1},\ldots,i_{k}\}$ a subset of $k$ elements of
      $\{1,\ldots,n\}$, i.e.\
      $\{i_{1},\ldots,i_{k}\}\in\Lambda_{k}\bigl(\{1,\ldots,n\}\big)$.

      We also have $\Delta(E)\simeq F\times F$ if $n\geq2$, with minimal
      idempotents
      \[
      \sum_{\sigma\in\Alt_{n}}e_{\sigma(1)}\otimes\cdots\otimes
      e_{\sigma(n)}
      \qquad\text{and}\qquad
      \sum_{\sigma\notin\Alt_{n}}e_{\sigma(1)}\otimes
      \cdots\otimes e_{\sigma(n)}.
      \]
\end{example}

For an arbitrary \'etale $F$-algebra $E$ of dimension~$n$, the algebra
$\Lambda_{k}(E)$ can also be viewed as an algebra of
linear transformations of the exterior power $\bigwedge^k E$ (where
$E$ is just regarded as a vector space), as we now show.

Multiplication on the left defines an $F$-algebra homomorphism (the
regular representation)
\[
E^{\otimes k}\to\End_{F}(E^{\otimes k}).
\]
As pointed out by Saltman \cite[Lemma~1.1]{S}, the image of
$(E^{\otimes k})^{\Sym_{k}}$ in $\End_{F}(E^{\otimes k})$ preserves
the kernel of the canonical map $E^{\otimes k}\to\bigwedge^k E$.
Therefore, there is an induced $F$-algebra homomorphism
\begin{equation}
\label{tensym.eq}
(E^{\otimes k})^{\Sym_{k}}\to\End_{F}({\textstyle\bigwedge^k E}).
\end{equation}

\begin{lem}
      \label{tensym.lem}
      The homomorphism \eqref{tensym.eq} maps $s_{k}$ to the identity
      map on $\bigwedge^k
      E$.
\end{lem}

\begin{proof}
      It suffices to check the assertion over an extension of $F$. We
      may thus assume that $E$ is split, $E=F^n$. Let $e_{1}$, \ldots,
      $e_{n}$ be the minimal idempotents of $E$; then $s_{k}$ is as in
      Example~\ref{splitS.ex} and its image in $\End_{F}(\bigwedge^k
      E)$  maps $e_{j_{1}}\wedge\ldots\wedge e_{j_{k}}$ to
      \[
      \sum_{(i_{1},\ldots,i_{k})}e_{i_{1}}e_{j_{1}}\wedge\ldots
      \wedge e_{i_{k}}e_{j_{k}}=e_{j_{1}}\wedge\ldots\wedge e_{j_{k}}.
      \]
\end{proof}

In view of the lemma, the homomorphism \eqref{tensym.eq} induces an
$F$-algebra homomorphism
\[
\varphi_{k}\colon\,
(E^{\otimes k})^{\Sym_{k}}/(1-s_{k})(E^{\otimes k})^{\Sym_{k}}=
\Lambda_{k}(E)\to\End_{F}({\textstyle\bigwedge^k E}).
\]
Saltman \cite[Lemma~1.3]{S} has shown that the image of this map has
dimension $\binom{n}{k}=\dim\Lambda_{k}(E)$, hence $\varphi_{k}$ is
injective.

For instance, for $a$, $x_{1}$, \ldots, $x_{k}\in E$,
\[
\varphi_{k}\bigl(s_{k}(a\otimes\cdots\otimes a)\bigr)(x_{1}\wedge\ldots\wedge
x_{k})=
ax_{1}\wedge\ldots\wedge ax_{k},
\]
and
\begin{multline*}
      \varphi_{k}\bigl(s_{k}(a\otimes1\otimes\cdots\otimes1+
      1\otimes a\otimes\cdots\otimes1+\cdots+
      1\otimes\cdots\otimes1\otimes a)\bigr)(x_{1}\wedge\ldots\wedge
      x_{k})=\\
      (ax_{1}\wedge x_{2}\wedge\ldots\wedge x_{k})+
      (x_{1}\wedge ax_{2}\wedge\ldots\wedge x_{k})+ \cdots +
      (x_{1}\wedge x_{2}\wedge\ldots\wedge ax_{k}).
\end{multline*}

Now, consider the case where $n$ is even, $n=2m$. Since
$\dim\bigwedge^n E=1$, and the exterior product $\bigwedge^m
E\times\bigwedge^m E\to\bigwedge^n E$ is a nonsingular bilinear
pairing, there is an adjoint involution $\gamma$ on
$\End_{F}(\bigwedge^m E)$, defined by the equation
\[
f(x)\wedge y=x\wedge\gamma(f)(y)
\qquad\text{for $x$, $y\in{\textstyle\bigwedge^mE}$,
$f\in\End_{F}({\textstyle\bigwedge^mE})$}.
\]

\begin{prop}
      The involution $\gamma$ preserves the image of $\varphi_{k}$.
      Therefore, there is an induced involutive automorphism
      $\gamma_{E}$ on $\Lambda_{m}(E)$ defined by
      $\varphi_{m}\circ\gamma_{E}=\gamma\circ\varphi_{m}$.
\end{prop}

\begin{proof}
      Extending scalars to a separable closure, we may assume $E$ is
      split. It is then spanned by its minimal idempotents $e_{1}$,
      \ldots, $e_{n}$, and $\Lambda_{m}(E)$ is spanned by the minimal
      idempotents $e_{\{i_{1},\ldots,i_{m}\}}$ defined in
      Example~\ref{splitS.ex} for
      $\{i_{1},\ldots,i_{m}\}\in\Lambda_{m}\bigl(\{1,\ldots,n\}\bigr)$.

      Computation shows that for
      $\{i_{1},\ldots,i_{m}\}$,
      $\{j_{1},\ldots,j_{m}\}\in\Lambda_{m}\bigl(\{1,\ldots,n\}\bigr)$,
      \[
      \varphi_{m}(e_{\{i_{1},\ldots,i_{m}\}})(e_{j_{1}}\wedge\ldots\wedge
      e_{j_{m}})=
      \begin{cases}
         e_{j_{1}}\wedge\ldots\wedge e_{j_{m}}&\text{if
         $\{i_{1},\ldots,i_{m}\}=\{j_{1},\ldots,j_{m}\}$},\\
         0&\text{if
         $\{i_{1},\ldots,i_{m}\}\neq\{j_{1},\ldots,j_{m}\}$}.
      \end{cases}
      \]
      It is then easily verified that
      \begin{equation}
         \label{gamma.eq}
         \gamma\circ\varphi_{m}(e_{\{i_{1},\ldots,i_{m}\}})=
          e_{\{k_{1},\ldots,k_{m}\}},
      \end{equation}
      where $\{k_{1},\ldots,k_{m}\}$ is the
      complementary subset of $\{i_{1},\ldots,i_{m}\}$ in
      $\{1,\ldots,n\}$.
\end{proof}

We may now define
\[
\resolv(E)=\Lambda_{m}(E)^{\Sym_{2}}=
\{x\in\Lambda_{m}(E)\mid\gamma_{E}(x)=x\}.
\]

\begin{thm}
      $\funcX\bigl(\resolv(E)\bigr)=\resolv\bigl(\funcX(E)\bigr)$.
\end{thm}

\begin{proof}
      It suffices to see that under the anti-equivalence
      $\Et_{F}\equiv\Set_{\Gamma}$, the automorphism $\gamma_{E}$ of
      $\Lambda_{m}(E)$ corresponds to the permutation $\gamma_{\funcX(E)}$
      of $\funcX\bigl(\Lambda_{m}(E)\bigr)=\Lambda_{m}\bigl(\funcX(E)\bigr)$.
      Again, we may extend scalars to a separable closure of $F$ and
      assume $E$ is split. Using the same notation as in the preceding
      proof, we may identify $\funcX(E)$ with the dual basis of $e_{1}$,
      \ldots, $e_{n}$. Equation~\eqref{gamma.eq} shows that
      \[
      \gamma_{E}(e_{\{i_{1},\ldots,i_{m}\}})=
          e_{\{k_{1},\ldots,k_{m}\}},
      \]
      where $\{k_{1},\ldots,k_{m}\}$ is the
      complementary subset of $\{i_{1},\ldots,i_{m}\}$ in
      $\{1,\ldots,n\}$, and the proof is complete.
\end{proof}

When $\dim E=2$, the algebra $E$ is called a \emph{quadratic} \'etale
$F$-algebra. In the notation above, we then have $m=1$,
so $\Lambda_{m}(E)=E$,
hence $E$ carries a canonical involutive automorphism $\gamma_{E}$.
Let $E$, $E'$ be two quadratic \'etale $F$-algebras, with
canonical involutive automorphisms $\gamma_{E}$, $\gamma_{E'}$. The
tensor product $\gamma_{E}\otimes\gamma_{E'}$ defines a
$\Sym_{2}$-action on $E\otimes_{F}E'$, and we let
\[
E*E'=(E\otimes_{F}E')^{\Sym_{2}}.
\]

\begin{prop}
     $\funcX(E*E')=\funcX(E)*\funcX(E')$.
\end{prop}

\begin{proof}
     \[
     \funcX\bigl((E\otimes_{F}E')^{\Sym_{2}}\bigr)=
     \funcX(E\otimes_{F}E')/\Sym_{2}=
     \bigl(\funcX(E)\times\funcX(E')\bigr)/\Sym_{2}.
     \]
\end{proof}

Let $\Quadr(F)$ be the set of isomorphism classes of quadratic
\'etale $F$-algebras, which is in bijection under $\funcX$ with the set
of isomorphism classes of $\Gamma$-sets of two elements.
The following analogue of Proposition~\ref{groupstar.prop} is easily
proved, either directly or by reduction to
Proposition~\ref{groupstar.prop} under the anti-equivalence
$\Et_{F}\equiv\Set_{\Gamma}$:

\begin{prop}
     \label{staretale.prop}
     Let $E$, $E'$ be two quadratic \'etale $F$-algebras.
     \begin{enumerate}
	\item[(a)]
	The $F$-algebra $E*E$ is split: $E*E\simeq F\times F$.
	\item[(b)]
	If the algebra $E'$ is split, then $E*E'\simeq E$ (not canonically).
	\item[(c)]
	The operation $*$ defines a group structure on the set $\Quadr(F)$.
     \end{enumerate}
\end{prop}

\subsection{Extensions of \'etale algebras}
\label{etex.subsec}

An \'etale $F$-algebra $B$ containing an $F$-algebra $A$
(necessarily \'etale) is called an \emph{extension of degree~$d$} of
$A$ if it is a free $A$-module of rank~$d$. Equivalently, this
condition means that the inclusion $A\xrightarrow{i}B$ corresponds
under the anti-equivalence $\Et_{F}\equiv\Set_{\Gamma}$ to a map
$\funcX(A)\xleftarrow{i^{*}}\funcX(B)$ which is a covering of degree~$d$.
Extensions of degree~$2$ are called \emph{quadratic} extensions.

Suppose $B/A$ is an extension of degree~$d$. Let $\dim_{F}A=n$
(hence $\dim_{F}B=nd$), and let $s_{n}^A\in (A^{\otimes
n})^{\Sym_{n}}$ be the idempotent defining $\Sigma_{n}(A)$, see
\S\ref{Basicet.subsec}. Then $i^{\otimes n}(s_{n}^A)$ is an idempotent
of $(B^{\otimes n})^{\Sym_{n}}$. Define
\[
\Omega(B/A)=i^{\otimes n}(s_{n}^A)\cdot(B^{\otimes n})^{\Sym_{n}}.
\]

\begin{prop}\label{xomega.prop}
      There is a canonical surjective map
      $\Lambda_{n}(B)\to\Omega(B/A)$, and
      \[
      \funcX\bigl(\Omega(B/A)\bigr)=\Omega\bigl(\funcX(B)/\funcX(A)\bigr).
      \]
      Therefore, $\dim_{F}\Omega(B/A)=d^n$.
\end{prop}

\begin{proof}
      The set $\funcX\bigl(\Omega(B/A)\bigr)$ is the set of $F$-algebra
      homomorphisms $(B^{\otimes n})^{\Sym_{n}}\to F_{s}$ which map
      $i^{\otimes n}(s_{n}^A)$ to $1$. Since
      \[
      \funcX\bigl((B^{\otimes n})^{\Sym_{n}}\bigr)=\funcX(B)^n/\Sym_{n},
      \]
      every such homomorphism is the orbit of an $n$-tuple
      $(\xi_{1},\ldots,\xi_{n})$ of elements of $\funcX(B)$; the condition
      that the homomorphism maps $i^{\otimes n}(s_{n}^A)$ to $1$ is
      equivalent to the following: the homomorphism $(A^{\otimes
      n})^{\Sym_{n}}\to F_{s}$ associated to the $n$-tuple
      $\bigl(i^{*}(\xi_{1}),\ldots,i^{*}(\xi_{n})\bigr)$ maps $s_{n}^A$
      to $1$. In view of the definition of $s_{n}^A$, this means that
      $i^{*}(\xi_{1})$, \ldots, $i^{*}(\xi_{n})$ are pairwise distinct,
      hence
      \[
      \{i^{*}(\xi_{1}),\ldots,i^*(\xi_{n})\}=\funcX(A)
      \]
      since $\vert \funcX(A)\vert=n$. Of course, this condition implies
      that $\xi_{1}$, \ldots, $\xi_{n}$ are pairwise distinct, hence
      $\{\xi_{1},\ldots,\xi_{n}\}\in\Lambda_{n}\bigl(\funcX(B)\bigr)$. Thus,
      \begin{multline*}
      \funcX\bigl(\Omega(B/A)\bigr)=
      \bigl\{\{\xi_{1},\ldots,\xi_{n}\}\subset \funcX(B)\mid
      \{i^{*}(\xi_{1}),\ldots,i^{*}(\xi_{n})\}=\funcX(A)\bigr\}\\
      =\Omega\bigl(\funcX(B)/\funcX(A)\bigr).
      \end{multline*}
      The inclusion
      $\Omega\bigl(\funcX(B)/\funcX(A)\bigr)\subset
      \Lambda_{n}\bigl(\funcX(B)\bigr)$
      yields the canonical surjective map
      $\Lambda_{n}(B)\to\Omega(B/A)$ under the anti-equivalence
      $\Et_{F}\equiv\Set_{\Gamma}$.
\end{proof}

Now, suppose $d=2$, so that $\dim_{F}B=2n$. The canonical involutive
automorphism $\gamma_{\funcX(B)/\funcX(A)}$ on $\funcX(B)$ corresponds
to a canonical involutive automorphism $\gamma_{B/A}$ of $B$ such that
\[
A=\{x\in B\mid\gamma_{B/A}(x)=x\}.
\]
On the other hand, there is also a ``complementary
subset'' map
\[
\gamma_{\funcX(B)}\colon\,\Lambda_{n}\bigl(\funcX(B)\bigr)\to
\Lambda_{n}\bigl(\funcX(B)\bigr).
\]
Since this map preserves
$\Omega\bigl(\funcX(B)/\funcX(A)\bigr)$, the corresponding map
$\gamma_{B}\colon\,\Lambda_{n}(B)\to\Lambda_{n}(B)$ induces an
involutive automorphism on $\Omega(B/A)$, which we also denote by
$\gamma_{B}$, and
we may consider the subalgebra of fixed points
\[
{\mathcal S}(B/A)=\Omega(B/A)^{\Sym_{2}}=
\{x\in\Omega(B/A)\mid\gamma_{B}(x)=x\}.
\]
By definition, it is clear that
\[
\funcX\bigl({\mathcal S}(B/A)\bigr)=
{\mathcal S}\bigl(\funcX(B)/\funcX(A)\bigr),
\]
hence $\dim_{F}{\mathcal S}(B/A)=2^{n-1}$.

\begin{example}
      \label{ext6.ex}
      Suppose $A$ and $B$ are split of dimensions~$3$ and $6$
      respectively, with minimal idempotents $e_{1}$, $e_{2}$, $e_{3}$
      and $f_{1}$, $f'_{1}$, $f_{2}$, $f'_{2}$, $f_{3}$, $f'_{3}$ such
      that
      \[
      e_{i}=f_{i}+f'_{i}\qquad\text{for $i=1$, $2$, $3$}.
      \]
      As observed in Example~\ref{splitS.ex},
      \[
      s_3^{A}=\sum_{\sigma\in\Sym_{3}} e_{\sigma(1)}\otimes e_{\sigma(2)}
      \otimes e_{\sigma(3)},
      \]
and $\{ e_{\sigma(1)}\tens e_{\sigma(2)}\tens e_{\sigma(3)}\mid
\sigma \in \Sym_{3}\}$
is the set of minimal idempotents of $\Sigma_{3}(A)$.
      Denoting in general by $\sum u\otimes v\otimes w$ the sum of the
      six products obtained by permuting the factors $u$, $v$, $w$ (so
      \[
      \sum u\otimes v\otimes w=
      \sum_{\sigma\in\Sym_{3}}u_{\sigma(1)}\otimes u_{\sigma(2)}\otimes
      u_{\sigma(3)}
      \]
      where $u_{1}=u$, $u_{2}=v$, $u_{3}=w$), the minimal
      idempotents of $\Omega(B/A)$ are
      \begin{align*}
         g_{0}& =\sum f_{1}\otimes f_{2}\otimes f_{3},&
         g'_{0}& =\sum f'_{1}\otimes f'_{2}\otimes f'_{3},\\
         g_{1}& =\sum f_{1}\otimes f'_{2}\otimes f'_{3},&
         g'_{1}& =\sum f'_{1}\otimes f_{2}\otimes f_{3},\\
         g_{2}& =\sum f'_{1}\otimes f_{2}\otimes f'_{3},&
         g'_{2}& =\sum f_{1}\otimes f'_{2}\otimes f_{3},\\
         g_{3}&  =\sum f'_{1}\otimes f'_{2}\otimes f_{3},&
         g'_{3}& =\sum f_{1}\otimes f_{2}\otimes f'_{3}.
      \end{align*}
      The involution $\gamma_{B}$ interchanges $g_{i}$ and
      $g_{i}'$ for $i=0$, \ldots, $3$, hence the minimal idempotents
      of ${\mathcal S}(B/A)$ are
      \[
      g_{0}+g'_{0},\qquad g_{1}+g'_{1},\qquad
      g_{2}+g'_{2},\qquad g_{3}+g'_{3}.
      \]
\end{example}

Let $B$, $B'$ be quadratic extensions of an \'etale $F$-algebra
$A$. The canonical map $B\otimes_{F}B'\to B\otimes_{A}B'$ induces an
injective map
\[
\funcX(B)\times\funcX(B')=\funcX(B\otimes_{F}B')
\leftarrow\funcX(B\otimes_{A}B')
\]
which identifies $\funcX(B\otimes_{A}B')$ to the fiber product
$\funcX(B)\times_{\funcX(A)}\funcX(B')$. The tensor product
$\gamma_{B/A}\otimes\gamma_{B'/A}$ defines an action of $\Sym_{2}$
on $B\otimes_{A}B'$ by $A$-automorphisms, and we let
\[
B*_{A}B'=(B\otimes_{A}B')^{\Sym_{2}}.
\]
The following result is clear:

\begin{prop}
     $\funcX(B*_{A}B')=\funcX(B)*_{\funcX(A)}\funcX(B')$.
\end{prop}

If $E$ is an \'etale $F$-algebra of dimension~$2$, then
$E\otimes_{F}A$ is a quadratic extension of $A$. For any quadratic
extension $B/A$, we have
$E\otimes_{F}B=(E\otimes_{F}A)\otimes_{A}B$ and we write simply
\[
E*B\qquad\text{for}\qquad (E\otimes_{F}A)*_{A}B.
\]

Let $\Quadr(A)$ be the set of isomorphism classes \emph{over $A$} of
quadratic extensions of $A$.
The following proposition is the analogue of
Proposition~\ref{compcov.prop}:

\begin{prop}
     Let $E$ be a split \'etale $F$-algebra of dimension~$2$ (i.e.\
     $E\simeq F\times F$), and let $B/A$ be a quadratic extension
     of \'etale $F$-algebras.
     \begin{enumerate}
	\item[(a)]
	The extension $(B*_{A}B)/A$ is isomorphic to $(E\otimes_{F}A)/A$
	(hence also to $(A\times A)/A$).
	\item[(b)]
	The extension $(E*B)/A$ is (non-canonically) isomorphic to $B/A$.
	\item[(c)]
	The operation $*_{A}$ defines a group structure on $\Quadr(A)$.
	The neutral element is the isomorphism class of $(A\times A)/A$.
     \end{enumerate}
\end{prop}

It is clear that Propositions~\ref{cov6.prop}, \ref{deltahom.prop}
and \ref{coverinvariance.prop} have analogues for \'etale algebras. We
record them below.

\begin{prop}
     Let $E$ be an \'etale $F$-algebra of dimension~$2$, and let $B/A$
     and $B'/A$ be quadratic extensions of an \'etale $F$-algebra $A$.
     There are canonical isomorphisms:
     \begin{enumerate}
     \item[(a)]
     $\Delta(B)\otimes\rescov(B/A)\simeq\Omega(B/A)$
     (if $\dim A$ is odd);
     \item[(b)]
     $\Delta(B*_{A}B')\simeq\Delta(B)*\Delta(B')$;
     \item[(c)]
     $\rescov(E*B/A)\simeq\rescov(B/A)$.
     \end{enumerate}
\end{prop}

\section{Cohomology of permutation groups}
\label{coh.sec}

\subsection{Permutations}
\label{perm.subsec}

For any finite set $X$,
let $\Sym_{X}$ be the symmetric group of $X$,
i.e.\ the group of all permutations of $X$.
Thus, $\Sym_{X}=\Sym_{n}$ for
$X=\{1,\ldots,n\}$.
Every permutation of a set $X$ of $n$ elements induces
a permutation of
the sets $\Sigma_{k}(X)$, $\Lambda_{k}(X)$ (for $k\leq n$), $\Delta(X)$,
and of $\resolv(X)$ if $n$ is
even. There are therefore canonical group homomorphisms
\[
\Sym_{X}\to\Sym_{\Sigma_{k}(X)},\quad
\Sym_{X}\to\Sym_{\Lambda_{k}(X)},\quad
\Sym_{X}\to\Sym_{\Delta(X)},\quad
\Sym_{X}\to\Sym_{\resolv(X)}\text{ (if $n$ is even)}.
\]
(If $n\geq2$, the map
$\Sym_{X}\xrightarrow{\sgn}\Sym_{\Delta(X)}=\Sym_{2}$ is the
\emph{signature} map.)
\medbreak
\par
If $Y\xleftarrow{\pi}Z$ is a covering of degree~$d$ of a set of $n$
elements, let
\[
\Sym_{Z/Y}=\bigl\{(\sigma,\tau)\in\Sym_{Y}\times\Sym_{Z}\mid
\pi\circ\tau=\sigma\circ\pi\bigr\}
\]
be the group of automorphisms of the covering. The map
$(\sigma,\tau)\mapsto\tau$ identifies $\Sym_{Z/Y}$ to a subgroup of
$\Sym_{Z}$. On the other hand,
the map $(\sigma,\tau)\mapsto\sigma$ defines a
surjective homomorphism
\[
\beta_{Z/Y}\colon\,\Sym_{Z/Y}\to\Sym_{Y}
\]
whose kernel is isomorphic to $\Sym_{d}^{n}$ upon
identifying each fiber of $\pi$ with $\{1,\ldots,d\}$. Therefore,
the group $\Sym_{Z/Y}$ has order $(d!)^nn!$ and can be identified to
a wreath product
\[
\Sym_{Z/Y}\simeq\Sym_{d}\wr\Sym_{n}.
\]
Automorphisms of the covering $Y\leftarrow Z$ induce permutations of
$\Omega(Z/Y)$, and of $\rescov(Z/Y)$ if $d=2$, hence there are
canonical group homomorphisms
\[
\omega_{Z/Y}\colon\,\Sym_{Z/Y}\to\Sym_{\Omega(Z/Y)},
\qquad
s_{Z/Y}\colon\,\Sym_{Z/Y}\to\Sym_{\rescov(Z/Y)}
\text{ if $d=2$}.
\]
For later use, note that the kernel of $s_{Z/Y}$ is the
``diagonal'' subgroup $\Sym_{2}$ of $\Sym_{Z/Y}$, whose nontrivial
element is $\gamma_{Z/Y}$. This diagonal subgroup is central in
$\Sym_{Z/Y}$.
\medbreak
\par
On the other hand, every permutation of a set $X$ with $n=2m$ elements
induces an automorphism of the covering
$\resolv(X)\xleftarrow{\varepsilon}\Lambda_{m}(X)$, hence there is a
canonical group homomorphism
\[
\lambda_{X}\colon\,\Sym_{X}\to\Sym_{\Lambda_{m}(X)/\resolv(X)}
\subset\Sym_{\Lambda_{m}(X)}.
\]

\begin{prop}
     \label{perm.prop}
     If $m\geq2$, the image of $\lambda_{X}$ is in the kernel of the
     signature map
     \[
     \sgn\colon\,\Sym_{\Lambda_{m}(X)}\to\Sym_{\Delta(\Lambda_{m}(X))}.
     \]
     Moreover, the composition of $\lambda_{X}$ and the canonical
     homomorphism $s_{\Lambda_{m}(X)/\resolv(X)}$ is an injective
     map
     \[
     \Sym_{X}\hookrightarrow\Sym_{\rescov(\Lambda_{m}(X)/\resolv(X))}.
     \]
\end{prop}

The proof is left to the reader.

\subsection{Cohomology and $\Gamma$-sets}
\label{cohsets.subsec}

As in \S\ref{sets.sec}, we denote by $\Gamma$ a profinite group,
which will be fixed throughout this subsection. The action of
$\Gamma$ on a set $X$ with $\vert X\vert =n$ can be
viewed as a group homomorphism
\[
\Gamma\to\Sym_{X}\simeq\Sym_{n}.
\]
Since the isomorphism $\Sym_{X}\simeq\Sym_{n}$ depends on the
indexing of the elements in $X$, the homomorphism $\Gamma\to\Sym_{n}$
is defined by $X$ up to conjugation by an element in $\Sym_{n}$.
Therefore, there is a canonical one-to-one correspondence between
isomorphism classes of $\Gamma$-sets of $n$ elements and the
cohomology set $H^{1}(\Gamma,\Sym_{n})$ (with the trivial action of
$\Gamma$ on $\Sym_{n}$), by definition of this cohomology set.
Under this correspondence,
the distinguished element of $H^{1}(\Gamma,\Sym_{n})$ is mapped to the
set with trivial $\Gamma$-action.

Since the symmetric group $\Sym_{2}$ is abelian, there is an abelian
group structure on $H^{1}(\Gamma,\Sym_{2})$. We leave it to the
reader to verify that the product of the isomorphism classes of the
$\Gamma$-sets $X$, $X'$ with $\vert X\vert=\vert X'\vert=2$ is the
isomorphism class of $X*X'$.

Similarly, every covering $Y\xleftarrow{\pi}Z$ of degree~$d$ of a
$\Gamma$-set $Y$ with $\vert Y\vert=n$ yields a group homomorphism
\[
\Gamma\to\Sym_{Z/Y}\simeq\Sym_{d}\wr\Sym_{n},
\]
and there is a canonical one-to-one correspondence between
isomorphism classes of coverings of degree~$d$ of $\Gamma$-sets of
$n$ elements and the cohomology set
$H^{1}(\Gamma,\Sym_{d}\wr\Sym_{n})$, which maps the distinguished
element of the cohomology set to the covering with trivial
$\Gamma$-action.

The basic constructions in subsections~\ref{basicset.subsec} and
\ref{cov.subsec} yield
canonical maps of cohomology sets through the induced homomorphisms of
permutation groups (see \S\ref{perm.subsec}). For instance, if $X$ is
a $\Gamma$-set of $n$ elements and $k\leq n$, the canonical
homomorphism $\sigma_{k}\colon\,\Sym_{X}\to\Sym_{\Sigma_{k}(X)}$
induces a morphism of pointed sets
\[
\sigma_{k}^{1}\colon\,H^{1}(\Gamma,\Sym_{X})\to
H^{1}(\Gamma,\Sym_{\Sigma_{k}(X)}).
\]
Since the $\Gamma$-action on $\Sigma_{k}(X)$ is
induced by the $\Gamma$-action on $X$ through $\sigma_{k}$, the
morphism $\sigma_{k}^{1}$
maps the isomorphism class of $X$ to the isomorphism class of
$\Sigma_{k}(X)$. A similar statement obviously holds for the morphisms
\begin{align*}
H^{1}(\Gamma,\Sym_{X})& \to H^{1}(\Gamma,\Sym_{\Lambda_{k}(X)}),
\\
H^{1}(\Gamma,\Sym_{X})& \to H^{1}(\Gamma,\Sym_{\Delta(X)}),
\\
H^{1}(\Gamma,\Sym_{X})& \to H^{1}(\Gamma,\Sym_{\resolv(X)})
\text{ if $n$ is even}.
\end{align*}

Similarly, if $Y\xleftarrow{\pi}Z$ is a covering of degree~$d$
of $\Gamma$-sets, the canonical homomorphisms $\omega_{Z/Y}$ and
$s_{Z/Y}$ of \S\ref{perm.subsec} induce morphisms of pointed sets
\begin{align*}
\omega_{Z/Y}^{1}\colon\,H^{1}(\Gamma,\Sym_{Z/Y})& \to
H^{1}(\Gamma,\Sym_{\Omega(Z/Y)}),
\\
s_{Z/Y}^{1}\colon\,H^{1}(\Gamma,\Sym_{Z/Y})& \to
H^{1}(\Gamma,\Sym_{\rescov(Z/Y)})
\text{ if $d=2$}.
\end{align*}
Since the $\Gamma$-action on $\Omega(Z/Y)$ and $\rescov(Z/Y)$ (if
$d=2$) are induced by the $\Gamma$-action on $Z/Y$ through
$\omega_{Z/Y}$ and $\sigma_{Z/Y}$ respectively, the morphisms
$\omega_{Z/Y}^{1}$ and $s_{Z/Y}^{1}$
map the isomorphism class of the covering $Z/Y$ to the isomorphism
class of the $\Gamma$-sets $\Omega(Z/Y)$ and $\rescov(Z/Y)$,
respectively.

Recall also from \S\ref{perm.subsec} the canonical homomorphism
$\beta_{Z/Y}\colon\,\Sym_{Z/Y}\to\Sym_{Y}$ which maps every
permutation of a covering to the induced permutation of the base.
Let $\mathfrak{T}_{Z/Y}=\ker\beta_{Z/Y}$. This is the group of
automorphisms over $Y$ of the covering $Z/Y$, hence
$H^{1}(\Gamma,\mathfrak{T}_{Z/Y})$ is in one-to-one correspondence
with the set of isomorphism classes over $Y$ of coverings of
degree~$d$ of $Y$, where the $\Gamma$-action on $Y$ is trivial.

The case of non-trivial $\Gamma$-action can be taken into account by
twisting, see \cite[\S28.C]{BoI}. If $Z/Y$ is a covering of
degree~$d$, we define a non-trivial action of $\Gamma$ on $\Sym_{Z/Y}$
by conjugation: the action of
$\Gamma$ on $Z/Y$ is a group homomorphism
\[
\alpha\colon\,\Gamma\to\Sym_{Z/Y},
\]
and we define, for $\gamma\in\Gamma$ and $f\in\Sym_{Z/Y}$,
\[
\gamma*f=
\alpha(\gamma)\circ f\circ\alpha(\gamma)^{-1}.
\]
Let $\Sym_{Z/Y}'$ be the group $\Sym_{Z/Y}$ with this action of
$\Gamma$, and define $\Sym_{Y}'$ similarly. By \cite[(28.8)]{BoI},
there are canonical bijections
\[
     H^{1}(\Gamma,\Sym_{Z/Y}')\iso H^{1}(\Gamma,\Sym_{Z/Y})
     \qquad\text{and}\qquad
     H^{1}(\Gamma,\Sym_{Y}')\iso H^{1}(\Gamma,\Sym_{Y})
\]
which map the distinguished element of $H^{1}(\Gamma,\Sym_{Z/Y}')$,
$H^{1}(\Gamma,\Sym_{Y}')$ to the isomorphism class of the covering
$Z/Y$ and to the isomorphism class of $Y$, respectively. The map
$\beta_{Z/Y}$ is also a $\Gamma$-group homomorphism
$\beta_{Z/Y}\colon\,\Sym_{Z/Y}'\to\Sym_{Y}'$. Let
$\mathfrak{T}'_{Z/Y}\subset\Sym_{Z/Y}'$ be the kernel of
$\beta_{Z/Y}$. Then $H^{1}(\Gamma,\mathfrak{T}_{Z/Y}')$ is in natural
one-to-one correspondence with the set $C^d(Y)$ of isomorphism
classes \emph{over $Y$} of coverings of degree~$d$ of $Y$. (The
distinguished element of $H^{1}(\Gamma,\mathfrak{T}_{Z/Y}')$
corresponds to the isomorphism class of $Z/Y$.)

The exact sequence of
$\Gamma$-groups
\[
1\to\mathfrak{T}'_{Z/Y}\to\Sym_{Z/Y}'\xrightarrow{\beta_{Z/Y}}
\Sym_{Y}'\to1
\]
yields an exact sequence in cohomology
\[
H^{0}(\Gamma,\Sym_{Y}')\to H^{1}(\Gamma,\mathfrak{T}'_{Z/Y})\to
H^{1}(\Gamma,\Sym_{Z/Y}')\xrightarrow{\beta^{1}_{Z/Y}}
H^{1}(\Gamma,\Sym_{Y}').
\]
The kernel of $\beta^{1}_{Z/Y}$ is the set of isomorphism classes of
coverings of degree $d$ of $Y$. By \cite[(28.4)]{BoI}, this kernel is
in canonical bijection with the orbit space of
$H^{1}(\Gamma,\mathfrak{T}_{Z/Y}')$ under the fixed-point group
$H^{0}(\Gamma,\Sym_{Y}')$. Note that
$H^{0}(\Gamma,\Sym_{Y}')$ is the group of permutations of $Y$ which
commute with the action of $\Gamma$; in other words, it is the group
of automorphisms of the $\Gamma$-set $Y$,
\[
H^{0}(\Gamma,\Sym_{Y}')=\Aut_{\Gamma}(Y).
\]
This group acts naturally on $C^d(Y)$, and
\[
\ker\beta^{1}_{Z/Y}\simeq C^d(Y)/\Aut_{\Gamma}(Y).
\]

When the $\Gamma$-action on $Y$ is
transitive,
let $\Gamma_{0}\subset\Gamma$ be the stabilizer of an arbitrary (but
fixed) element of $Y$, so that $Y\simeq\Gamma/\Gamma_{0}$.
Then we may identify $\mathfrak{T}_{Z/Y}'$ with
$\Map(\Gamma/\Gamma_{0},\Sym_{d})$ and get a canonical bijection in
the spirit of Shapiro's lemma
\[
H^{1}(\Gamma,\mathfrak{T}_{Z/Y}')\simeq H^{1}(\Gamma_{0},\Sym_{d}),
\]
see \cite[(28.20)]{BoI}.

Whatever the action of $\Gamma$ on $Y$, when $d=2$ the group
$\mathfrak{T}_{Z/Y}'$ ($\simeq\Sym_{d}^n$ where $n=\vert Y\vert$)
is abelian, hence the set
$H^{1}(\Gamma,\mathfrak{T}_{Z/Y}')$ is an abelian group. When $Z/Y$ is
the projection covering $\bigl(\{1,2\}\times Y\bigr)/Y$, the
bijection $H^{1}(\Gamma,\mathfrak{T}_{Z/Y}')\simeq C^{2}(Y)$ is a
group isomorphism for the group structure induced on $C^{2}(Y)$ by
the operation $*_{Y}$ of \S\ref{cov.subsec}. Note that this operation
is generally \emph{not} defined on the orbit set
$\ker\beta^{1}_{Z/Y}\simeq C^{2}(Y)/\Aut_{\Gamma}(Y)$.

\subsection{Torsors}
\label{torsors.subsec}

As in the preceding subsection, we fix a profinite group $\Gamma$.
Besides the correspondence between $H^{1}(\Gamma,\Sym_{n})$ and the
isomorphism classes of $\Gamma$-sets of $n$ elements explained in the
preceding subsection, there is also a one-to-one correspondence between
$H^{1}(\Gamma,\Sym_{n})$ and isomorphism classes of
$\Sym_{n}$-torsors, i.e.\ of $\Gamma$-sets of $n!= \vert \Sym_{n}\vert$
elements with a free action of $\Sym_{n}$ (on the right) compatible with
the $\Gamma$-action (on the left), see \cite[(28.14)]{BoI}.
Combining the correspondences, we obtain a bijection between
isomorphism classes of $\Gamma$-sets of $n$ elements and
$\Sym_{n}$-torsors.
To the isomorphism class of the
$\Gamma$-set $X$ with $\vert X \vert =n$ corresponds
the class of $\Sigma_{n}(X)$, which clearly is an $\Sym_{n}$-torsor.
Conversely, we associate to an $\Sym_{n}$-torsor $\Sigma$ the class of
$\Sigma/\Sym_{n-1}$.
  The $n$ projections
  \[
  \pi_i \colon\,\Sigma_{n}(X)
\to X,\qquad
(\xi_{1},\ldots,\xi_{n}) \mapsto \xi_i
\text{ for $i=1$, \ldots, $2$}
\]
are $\Gamma$-equivariant maps and satisfy
\[
\pi_i\bigl((\xi_1,\ldots,\xi_n)^\sigma\bigr) =
\pi_{\sigma(i)}(\xi_1,\ldots,\xi_n)
\qquad\text{for $\sigma\in\Sym_{n}$}.
\]

\begin{defn}
     \label{Galcloset.def}
     An \emph{$\Sym_{n}$-Galois closure} of a $\Gamma$-set $X$ of $n$
     elements is a pair $(\Sigma,\pi)$ where $\Sigma$ is an
     $\Sym_{n}$-torsor and $X\xleftarrow{\pi}\Sigma$ is a covering
     (necessarily of degree $(n-1)!$) such that $\pi(x^\sigma)=\pi(x)$
     for $x\in\Sigma$ and $\sigma\in\Sym_{n-1}$.
\end{defn}

Every $\Sym_{n}$-Galois closure of $X$ is isomorphic to
$(\Sigma_{n}(X),\pi_{n})$.
\medbreak
\par
A similar construction can be given for coverings, since
the set $H^{1}(\Gamma,\Sym_{d}\wr\Sym_{n})$ classifies
$\Sym_{d}\wr\Sym_{n}$-torsors as well as coverings
of degree~$d$ of $\Gamma$-sets of $n$ elements.
%
If $Y\xleftarrow{\pi}Z$ is a covering of degree~$d$ of a
$\Gamma$-set $Y$ of $n$ elements (so $\vert Z\vert=nd$), let
%%
$\Sigma(Z/Y)$ be the set of arrays
\[({\zeta_{ij}})_{\substack{1\leq i\leq d\\ 1\leq j\leq
n}}
\]
of pairwise distinct elements of $Z$ such that $\pi(\zeta_{ij})$
depends only on $j$ for $i=1,\, \ldots,\,n$. The set $\Sigma(Z/Y)$ is
a $\Sym_{d}\wr\Sym_{n}$-torsor. Its isomorphism class, viewed
as an element of $H^1(\Gamma, \Sym_{d}\wr\Sym_{n})$, corresponds
to the isomorphism class of the covering  $Y\xleftarrow{\pi}Z$.
The $nd$ projections
\[
\pi_{k\ell}\colon\, \Sigma(Z/Y) \to Z,\quad
({\zeta_{ij}})_{\substack{1\leq i\leq d\\ 1\leq j\leq
n}} \mapsto \zeta_{k\ell}
\]
are $\Gamma$-equivariant maps. The projection $\pi\colon\,Z \to Y$ induces a
$ \Sym_{n}$-equivariant projection
\[
\Sigma(\pi)\colon\, \Sigma(Z/Y) \to \Sigma_{n}(Y),\quad  (\zeta_{ij}) \mapsto
\bigl(\pi(\zeta_{i1}),\, \ldots,\,  \pi(\zeta_{in} )\bigr)
\]
and $ \Sigma_{n}(Y) \xleftarrow{\Sigma(\pi)}\Sigma(Z/Y)$ is a 
covering of degree
  $(d\,!)^n$. Moreover the diagram
\[
\begin{CD}
\Sigma_{n}(Y) @<\Sigma(\pi)<<\Sigma(Z/Y)\\
@V \pi_{j} VV  @VV \pi_{ij}
V \\
Y @<\pi<< Z
\end{CD}
\]
is commutative. We say that
$\Sigma_{n}(Y) \xleftarrow{\Sigma(\pi)}\Sigma(Z/Y)$
is an \emph{$\Sym_{d}\wr\Sym_{n}$-Galois closure of $Y\xleftarrow{\pi}Z$}.
(We leave it to the reader to formalize the definition of a
$\Sym_{d}\wr\Sym_{n}$-Galois closure of $Y\xleftarrow{\pi}Z$.)

Note that the set $\Omega(Z/Y)$ of sections of $Z/Y$ (see \S\ref{cov.subsec})
can be identified with the set $\Sigma(Z/Y)/( \Sym_{d-1}\wr\Sym_{n})$.

\subsection{Cohomology and \'etale algebras}
\label{cohet.subsec}

In this section, $F$ is an arbitrary field, $F_{s}$ is a separable
closure of $F$ and $\Gamma=\Gal(F_{s}/F)$ is the absolute Galois
group of $F$.
The anti-equivalence $\Et_{F}\equiv\Set_{\Gamma}$ induces a canonical
bijection between the set of isomorphism classes of \'etale
$F$-algebras of dimension~$n$ and the set of isomorphism classes of
$\Gamma$-sets of $n$ elements. Since the latter set is in one-to-one
correspondence with the cohomology set $H^{1}(\Gamma,\Sym_{n})$ (see
\S\ref{cohsets.subsec}), there is also a canonical bijection between
$H^{1}(\Gamma,\Sym_{n})$ and isomorphism classes of \'etale $F$-algebras
of dimension~$n$. This bijection can  be set up directly,
by identifying $\Sym_{n}$ with the group of automorphisms of the
split algebra $\Fs^n$.
More precisely,
given an \'etale algebra $A$ and an isomorphism
$\alpha\colon\,F^n \tens \Fs \isom  A \tens \Fs$,
the corresponding cocycle is $(f_{\gamma})_{\gamma\in\Gamma}$, where
\[
f_\gamma = \alpha^{-1} \circ
(1 \tens \gamma) \circ
\alpha \circ (1\tens\gamma^{-1}) \in  \Aut_{\Fs}(F^n\tens\Fs) = \Sym_{n}.
\]
Conversely, given a
cocycle $(f_\gamma)_{\gamma\in\Gamma}$ in $\Sym_{n}$,
the corresponding \'etale algebra
is
\[
  A_\gamma = \{ x \in\Fs^n \ | \ \gamma f_\gamma(x) = x\}
\]
where $\Gamma$ acts on $\Fs^n$ entrywise.

As in \S\ref{cohsets.subsec}, the basic constructions on \'etale
algebras of \S\ref{Basicet.subsec} can be interpreted in terms of
morphisms of cohomology sets. Details are left to the
reader, as well as the analogues for extensions of \'etale algebras
and the cohomology of wreath products. We simply note for later use
the canonical isomorphism
\[
\Quadr(F)\simeq H^{1}(F,\Sym_{2})
\]
where $\Quadr(F)$ is the group of isomorphism classes of quadratic
\'etale $F$-algebras (see Proposition~\ref{staretale.prop}). For any
\'etale $F$-algebra $A$, we also have canonical isomorphisms
\[
\Quadr(A)\simeq C^{2}\bigl(\funcX(A)\bigr)\simeq
H^{1}(\Gamma,\mathfrak{T}_{\funcX(A\times A)/\funcX(A)}'),
\]
see \S\ref{cohsets.subsec}. The set of isomorphism classes
\emph{over $F$} of quadratic extensions of $A$ is
$\Quadr(A)/\Aut_{F}(A)$. The operation $*_{A}$ is generally \emph{not}
defined on this set.

\subsection{Galois algebras}
\label{galoisalg.subsec}

As in the preceding subsection, $F$ is an arbitrary field and
$\Gamma$ is the absolute Galois group of $F$. Let $G$ be
a finite group. A $G$-Galois $F$-algebra is an \'etale $F$-algebra of
dimension $\vert G\vert$
with an action of $G$ by $F$-algebra automorphisms such that the
algebra of fixed points is $F$ (see \cite[(18.15)]{BoI}).
Equivalently, an \'etale $F$-algebra $E$ of dimension $\vert G\vert$
with an action of $G$ is $G$-Galois if and only if the $\Gamma$-set
$\funcX(E)$ is a $G$-torsor for the induced action of $G$. Therefore,
the discussion of torsors in \S\ref{torsors.subsec} has an analogue
in terms of Galois algebras, and the set $H^{1}(\Gamma,\Sym_{n})$ is
also in one-to-one correspondence with the set of isomorphism classes
of $\Sym_{n}$-Galois $F$-algebras.

If $E$ is an \'etale $F$-algebra of dimension~$n$, the algebra
$\Sigma_{n}(E)$ has a natural action of $\Sym_{n}$, for which it is
an $\Sym_{n}$-Galois algebra. There are $n$ embeddings
$\varepsilon_{i}\colon\,E\to\Sigma_{n}(E)$ corresponding to the
projections $\pi_{i}\colon\,\Sigma_{n}\bigl(\funcX(E)\bigr)\to
\funcX(E)$. They are defined explicitly as follows: for $x\in E$,
\[
\varepsilon_{i}(x)=s_{n}\cdot 1\otimes\cdots\otimes x\otimes\cdots
\otimes 1\qquad\text{($x$ in $i$-th position)}
\]
where $s_{n}\in E^{\otimes n}$ is the idempotent such that
$\Sigma_{n}(E)=s_{n}E^{\otimes n}$. Clearly, for $\sigma\in\Sym_{n}$
and $x\in E$,
\[
\varepsilon_{\sigma(i)}(x)=\sigma\bigl(\varepsilon_{i}(x)\bigr).
\]

\begin{defn}
     An \emph{$\Sym_{n}$-Galois closure} of an \'etale $F$-algebra $E$
     of dimension~$n$ is a pair $(\sigma,\varepsilon)$ where $\Sigma$
     is an $\Sym_{n}$-Galois $F$-algebra and
     $\varepsilon\colon\,E\to\Sigma$ is an embedding such that
     $\sigma\bigl(\varepsilon(x)\bigr)=\varepsilon(x)$ for
     $\sigma\in\Sym_{n}$ and $x\in E$.
\end{defn}

Every $\Sym_{n}$-Galois closure of $E$ is isomorphic to
$(\Sigma_{n}(E),\varepsilon_{n})$. This construction was suggested by
Saltman, see \cite[p.~42]{SaltLect}.

\begin{example}
     \label{cubicetale.ex}
     Let $A$ be a \emph{cubic \'etale $F$-algebra}, i.e., $\dim A=3$.
     The choice of any of the three canonical embeddings
     $\varepsilon_i\colon\, A \to \Sigma_{3}(A)$ induces an isomorphism
     \[
      A \tens \Delta(A) \simeq \Sigma_{3}(A).
     \]
     This follows from the fact
     that the corresponding map $\Sigma_{3}(X) \to X \times \Delta(X)$
     is bijective if $\vert X \vert =3$, see \cite[(18.27)]{BoI}.
\end{example}
\medbreak
\par
We next sketch an analogue of the Galois closure for extensions of
\'etale algebras, on the model of the corresponding construction for
coverings in \S\ref{torsors.subsec}.

Let $B/A$ be an extension of degree~$d$ of an \'etale $F$-algebra $A$
of degree~$n$.
%
Viewing $B$ as an \'etale $A$-algebra of degree $d$, we have
an $\Sym_d$-Galois closure $\Sigma_d^A(B)$ of $B$ which is \'etale
of degree
$d\,!$ over $A$,
\[
\Sigma_d^A(B) = s_d^{B/A}\cdot B^{\tens_A d}
\]
where $ s_d^{B/A}$ is the idempotent corresponding to the
characteristic function
of the subset
$\{(\xi_1,\ldots, \xi_d) \mid
\text{$\xi_i \neq \xi_j$ for $i\neq j$, and
  $\pi(\xi_i)=\pi(\xi_j)$ for $i=1$, \ldots, $d$}\} $
in
\[
\funcX(B) \times_{\funcX(A)} \cdots \times_{\funcX(A)}\funcX(B)
= \funcX(B  \tens _A \cdots \tens_A B).
\]
There are $d$ canonical embeddings
$\varepsilon_i^{A}\colon\, B \to \Sigma_d^A(B) $ and
a canonical embedding $j \colon\, A \to \Sigma_d^A(B)$ corresponding to
the $A$-algebra structure on $\Sigma_d^A(B)$.
Define
\[
\Sigma(B/A)= j^{\otimes n}(s_{n}^A)\cdot\Sigma_d^A(B)^{\otimes n}.
\]
As for Proposition~\ref{xomega.prop} we have
%
\begin{prop}
     \label{clotcouronn.prop}
     $\funcX\bigl(\Sigma(B/A)\bigr) = \Sigma\bigl(\funcX(B)/\funcX(A)\bigr)$.
\end{prop}

The algebra $\Sigma(B/A)$ is an extension of $\Sigma_{n}(A)$ of degree
$(d\,!)^n$ and there exist
$nd$ canonical embeddings
\begin{equation}\label{ijembeddings.eq}
     \varepsilon_{ij} \colon B \to \Sigma(B/A), \ 1\leq i\leq d,\  1\leq j\leq n
\end{equation}
such that the diagram
\[
\begin{CD}
  \Sigma_{n}(A) @>>> \Sigma(B/A) \\
@A {\varepsilon_i}AA @AA {\varepsilon_{ij}}A\\
A @>>> B
\end{CD}
\]
is commutative for all $i$ and $j$.
We say that the extension $\Sigma(B/A)/ \Sigma_{n}(A)$ is an
\emph{$\Sym_d \wr \Sym_n$-Galois closure}
of the extension $B/A$. Since
$\Sigma_d^A(B)^{\Sym_{d-1}} =B$, we have
\[
\Sigma(B/A)^{\Sym_{d-1}\wr \Sym_n} = \Omega(B/A).
\]
%

If $d=2$, each of the canonical embeddings $\varepsilon_{1}^A$,
$\varepsilon_{2}^A\colon\,B\to\Sigma_{2}^A(B)$ is an isomorphism
(and $\varepsilon_{2}^A=\varepsilon_{1}^A\circ\gamma_{B/A}$), and
$j\colon\,A\to\Sigma_{2}^A(B)=B$ is the inclusion. The algebra
\[
\Sigma(B/A)=j^{\otimes n}(s_{n}^A)\cdot B^{\otimes n}
\]
is an extension of degree~$2^n$ of $\Sigma_{n}(A)$, and
\[
\Omega(B/A)\simeq\Sigma(B/A)^{\Sym_{n}}.
\]

\begin{example}
     Suppose, as in Example~\ref{ext6.ex}, that $A$ and $B$ are split
     of dimensions~$3$ and $6$ respectively, with minimal idempotents
     $e_{1}$, $e_{2}$, $e_{3}$
     and $f_{1}$, $f'_{1}$, $f_{2}$, $f'_{2}$, $f_{3}$, $f'_{3}$ such
     that
     \[
     e_{i}=f_{i}+f'_{i}\qquad\text{for $i=1$, $2$, $3$}.
     \]
     The algebra $\Sigma(B/A)$ is split. Its $48$ minimal idempotents
     are
     \[
     \begin{array}{lll}
     f_{\sigma(1)}\tens f_{\sigma(2)}\tens f_{\sigma(3)},\quad &
     f'_{\sigma(1)}\tens f'_{\sigma(2)}\tens f'_{\sigma(3)}, \\
     %
     f_{\sigma(1)}\tens f'_{\sigma(2)}\tens f'_{\sigma(3)},\quad &
     f'_{\sigma(1)}\tens f_{\sigma(2)}\tens f_{\sigma(3)}, \\
     %
     f'_{\sigma(1)}\tens f_{\sigma(2)}\tens f'_{\sigma(3)},\quad &
     f_{\sigma(1)}\tens f'_{\sigma(2)}\tens f_{\sigma(3)}, \\
     %
     f'_{\sigma(1)}\tens f'_{\sigma(2)}\tens f_{\sigma(3)},\quad &
     f_{\sigma(1)}\tens f_{\sigma(2)}\tens f'_{\sigma(3)},\end{array}
     \]
     where $\sigma$ varies in $\Sym_{3}$. The action of $\Sym_{3}$ on
     these idempotents is clear, and the fixed subalgebra is
     $\Omega(B/A)$ as described in Example~\ref{ext6.ex}.
\end{example}

\begin{prop}\label{generatorgalclos.prop}
Let $B/A$ be an extension of \'etale algebras of degree $d$ and
  let $n =\dim_FA$. If $\,b \in B$ is a generator of $B$ as $F$-algebra,
  then the
$nd$ elements $\varepsilon_{ij}(b)$ generate $\Sigma(B/A)$ over
$\Sigma_{n}(A)$.
\end{prop}
%
\begin{proof}
  Let $\Sigma'$ be the subalgebra  of $\Sigma(B/A)$ generated
over $\Sigma(A)$ by the $\varepsilon_{ij}(b)$. We show that
$\Sigma'= \Sigma(B/A)$.
We may assume that $A$ and $B$ are split and we assume for
simplicity that $n=3$ and $d=2$. We use the notations of Example
\ref{ext6.ex}. Let
  \[
      b=\beta_{1}f_{1}+\beta'_{1}f'_{1}+\beta_{2}f_{2}+\beta'_{2}f'_{2}+
      \beta_{3}f_{3}+\beta'_{3}f'_{3}
      \]
      with $\beta_{1}$, \ldots, $\beta'_{3}\in F$, hence
      \[
      \overline{b}=
      \beta'_{1}f_{1}+\beta_{1}f'_{1}+\beta'_{2}f_{2}+\beta_{2}f'_{2}+
      \beta'_{3}f_{3}+\beta_{3}f'_{3}.
      \]
Since $b$ generates $B$,  the  6 elements $\beta_i,\,\beta'_j$ are pairwise
  different.  We have
\[
\varepsilon_{11}(b)  = s_3^A \cdot b\tens 1 \tens 1 \ \text{and} \
\varepsilon_{21}(b) = s_3^A \cdot \overline{b}\tens 1 \tens 1.
\]
Thus
\[
\begin{array}{lcl}
\varepsilon_{11}(b) (e_1 \tens e_2 \tens e_3)& =  &
  (\beta_{1}f_{1}+\beta'_{1}f'_{1}) \tens e_2 \tens e_3\quad \text{and} \\
\varepsilon_{21}(b) (e_1 \tens e_2 \tens e_3) &= &
  (\beta'_{1}f_{1}+\beta_{1}f'_{1}) \tens e_2 \tens e_3
\end{array}
\]
are elements of $\Sigma'$. It follows that
$ f_1 \tens e_2 \tens e_3$ and $ f'_1 \tens e_2 \tens e_3$
are in $\Sigma'$. Hence all the minimal idempotents of
$\Sigma(B/A)$ are in $\Sigma'$ and $\Sigma'=\Sigma(B/A) $.
\end{proof}

\section{The symmetric group on four elements}
\label{four.sec}

In the rest of this paper, we focus on various aspects of \'etale
algebras of dimension~$4$ (called \emph{quartic} \'etale algebras)
which, as explained in the preceding sections, can be viewed from the
perspective of $\Gamma$-sets of $4$ elements, or of the cohomology
of $\Sym_{4}$, or of $\Sym_{4}$-torsors, or of $\Sym_{4}$-Galois
algebras. It turns out that there is a group isomorphism
\[
\Sym_{2}\times\Sym_{4}\simeq\Sym_{2}\wr\Sym_{3}
\]
which relates the various ``quartic'' notions listed above to those
associated with the cohomology of $\Sym_{2}\wr\Sym_{3}$: quadratic
extensions of cubic \'etale algebras, double coverings of sets of $3$
elements, $\Sym_{2}\wr\Sym_{3}$-torsors and
$\Sym_{2}\wr\Sym_{3}$-Galois algebras. We explain this relation in
the simplest case, namely $\Gamma$-sets and coverings, and then give
the cohomological viewpoint in the next subsection. In the last two
subsections, we give explicit constructions of $\resolv(Q)$ for a
quartic algebra $Q$, making clear that this algebra is related to the
resolvent cubic of quartic equations, and of $\Omega(B/A)$ and
$\rescov(B/A)$ for a quadratic extension of a cubic algebra $A$.

\subsection{Sets of four elements and double coverings}
\label{fouranddouble.subsec}

In this subsection, $\Gamma$ is an arbitrary profinite group. Suppose
$X$ is a $\Gamma$-set with $\vert X\vert=4$, as in
Example~\ref{set4.ex}, where the constructions of $\Lambda_{2}(X)$
and $\resolv(X)$ are made explicit. Our first observation concerns
the discriminants of $\Lambda_{2}(X)$ and $\resolv(X)$:

\begin{prop}
      \label{discres.prop}
      The map which carries
      $(\xi_{1},\xi_{2},\xi_{3},\xi_{4})\in\Sigma_{4}(X)$ to
      \[
      \bigl(\bigl\{\{\xi_{1},\xi_{2}\},\{\xi_{3},\xi_{4}\}\bigr\},
      \bigl\{\{\xi_{1},\xi_{3}\},\{\xi_{2},\xi_{4}\}\bigr\},
      \bigl\{\{\xi_{1},\xi_{4}\},\{\xi_{2},\xi_{3}\}\bigr\}\bigr)
      \in\Sigma_{3}\bigl(\resolv(X)\bigr)
      \]
      induces a canonical isomorphism of $\Gamma$-sets
      \[
      \Delta(X)\iso\Delta\bigl(\resolv(X)\bigr).
      \]
      Moreover, the $\Gamma$-action on
      $\Delta\bigl(\Lambda_{2}(X)\bigr)$ is trivial.
\end{prop}

The proof is a straightforward verification. To see that the
$\Gamma$-action on $\Delta\bigl(\Lambda_{2}(X)\bigr)$ is trivial, it
suffices to observe that every transposition on $X$ ---hence every
permutation of $X$--- induces an even permutation of $\Lambda_{2}(X)$.
For another approach, see Proposition~\ref{discrescoh.prop}.
\medbreak
\par
To get a better grasp of the various constructions associated with $X$,
it is useful to think of $X$ as the set of diagonals of a
cube.\footnote{We are indebted to F.~Buekenhout for his suggestion to
use geometric language in this context.}
Each pair of diagonals determines a diagonal plane (passing through
an edge and its opposite), hence $\Lambda_{2}(X)$ is identified with
the set of diagonal planes of the cube. The map $\gamma_{X}$ carries
each diagonal plane to the plane through parallel edges, and
$\resolv(X)$ can therefore be identified with the set of directions of
the edges. The canonical map
$\resolv(X)\xleftarrow{\varepsilon}\Lambda_{2}(X)$ maps each diagonal
plane to the direction of the edges it contains. The set
$\Omega\bigl(\Lambda_{2}(X)/\resolv(X)\bigr)$ consists of
(unordered) triples of diagonal planes with different edge directions.
For each such triple $\tau$, either the intersection of the planes is
a diagonal, or the intersection is just the center of the cube.
However, if the intersection is a diagonal, then the intersection of
the complementary triple
$\overline{\tau}=\gamma_{\Lambda_{2}(X)/\resolv(X)}(\tau)$
is the center. Therefore, we may associate to the pair
$\{\tau,\overline{\tau}\}
\in {\mathcal S}\bigl(\Lambda_{2}(X)/\resolv(X)\bigr)$ a unique diagonal in
$X$, and obtain a map
\[
\Phi\colon\,{\mathcal S}\bigl(\Lambda_{2}(X)/\resolv(X)\bigr)
\to X.
\]

\begin{prop}
      \label{set4.prop}
      For $\vert X\vert=4$, the map
      \[
      \Phi\colon\,{\mathcal S}\bigl(\Lambda_{2}(X)/\resolv(X)\bigr)
      \to X
      \]
      is a canonical isomorphism of $\Gamma$-sets.
\end{prop}

\begin{proof}
      \relax From the definition, it is clear that $\Phi$ is
      $\Gamma$-equivariant. Bijectivity of $\Phi$ is checked by direct
      inspection.
\end{proof}

To put this result into perspective, consider the full subcategory
$\Set_{\Gamma}^{4}$ of $\Set_{\Gamma}$ whose objects are the
$\Gamma$-sets of four elements, and the category
$\Cov_{\Gamma}^{2\wr3}$ of double coverings of $\Gamma$-sets of three
elements, with morphisms of coverings. There are functors
\[
\funcL\colon\,\Set_{\Gamma}^{4}\to\Cov_{\Gamma}^{2\wr3}
\qquad\text{and}\qquad
\funcS\colon\,\Cov_{\Gamma}^{2\wr3}\to\Set_{\Gamma}^{4}
\]
defined by
\[
\funcL(X)=\Lambda_{2}(X)/\resolv(X)
\qquad
\text{and}\qquad
\funcS(Y\xleftarrow{\pi}Z)=\rescov(Z/Y).
\]
Proposition~\ref{set4.prop} yields a natural equivalence between
$\funcS\circ\funcL$ and the identity on $\Set_{\Gamma}^{4}$.
\medbreak
\par
To investigate the composition $\funcL\circ\funcS$, suppose
$Y\xleftarrow{\pi}Z$ is a double covering of a
$\Gamma$-set $Y$ with $\vert Y\vert=3$. (See Example~\ref{cov6.ex} for
an explicit description of $\Omega(Z/Y)$ and $\rescov(Z/Y)$.)
We may consider $Z$ as the
set of faces of a cube, $Y$ as the set of directions of edges, and
$\pi$ as the map which carries each face to the orthogonal direction.
Then $\Omega(Z/Y)$ is the set of (unordered) triples of faces which
are not pairwise parallel. Since the faces in each such triple meet
at one vertex, we may view $\Omega(Z/Y)$ as the set of vertices of
the cube. The map $\gamma_{Z/Y}$ carries each vertex to its opposite,
hence ${\mathcal S}(Z/Y)$ is the set of diagonals of the cube. As in the
discussion before Proposition~\ref{set4.prop}, we may then identify
$\Lambda_{2}\bigl({\mathcal S}(Z/Y)\bigr)$ with the set of diagonal planes
and $\resolv\bigl({\mathcal S}(Z/Y)\bigr)$ with the set of edge directions.
It is then clear that $\resolv\bigl({\mathcal S}(Z/Y)\bigr)$ is canonically
identified with $Y$, but there is no canonical identification of
$\Lambda_{2}\bigl({\mathcal S}(Z/Y)\bigr)$ with $Z$.

As we now show, we may however define a canonical bijection
\[
\Delta(Z)*\Lambda_{2}\bigl({\mathcal S}(Z/Y)\bigr)\iso Z,
\]
  hence an
isomorphism of coverings between $Y\xleftarrow{\pi}Z$ and
the covering
$\resolv\bigl({\mathcal S}(Z/Y)\bigr)\xleftarrow{\varepsilon}
\Delta(Z)*\Lambda_{2}\bigl({\mathcal S}(Z/Y)\bigr)$
induced by the canonical covering $\resolv\bigl({\mathcal S}(Z/Y)\bigr)
\xleftarrow{\varepsilon}\Lambda_{2}\bigl({\mathcal S}(Z/Y)\bigr)$.
(We denote both coverings by $\varepsilon$.)

Our first goal is to give a geometrical interpretation of the set
$\Delta(Z)$. Recall the map
\[
\delta_{Z}\colon\,\Omega(Z/Y)\to\Delta(Z)
\]
of \eqref{deltaZ.eq}. By Proposition~\ref{cov6.prop}, this map is onto.
It may therefore be used to consider $\Delta(Z)$ as a quotient of
$\Omega(Z/Y)$, the set of vertices of the cube. It is easily checked
that the four vertices which have the same image under $\delta_{Z}$
are the vertices of a regular tetrahedron whose edges are the
diagonals of the faces of the cube. Therefore, we may identify
$\Delta(Z)$ with the set $\{T_{1},T_{2}\}$ of such tetrahedra. Given
a diagonal plane $\lambda\in\Lambda_{2}\bigl(\rescov(Z/Y)\bigr)$ and
a tetrahedron $T\in\Delta(Z)$, there is a unique face $z\in Z$ whose
intersection with $\lambda$ is an edge of $T$. The same face $z$
intersects the ``complementary'' plane $\overline{\lambda}$ following
an edge of the ``complementary'' tetrahedron $\overline{T}$.
Therefore, the map $(T,\lambda)\mapsto z$ induces a well-defined map
\[
\Psi\colon\,\Delta(Z)*\Lambda_{2}\bigl({\mathcal S}(Z/Y)\bigr)\to Z.
\]

\begin{prop}
      \label{cov6bis.prop}
      The map $\Psi$ defines an isomorphism of coverings between
      \[
      \resolv\bigl({\mathcal S}(Z/Y)\bigr)\xleftarrow{\varepsilon}
      \Delta(Z)*\Lambda_{2}\bigl({\mathcal S}(Z/Y)\bigr)
      \qquad\text{and}\qquad
      Y\xleftarrow{\pi}Z.
      \]
\end{prop}

\begin{proof}
      The map $\Psi$ is clearly equivariant. The other properties are
      checked by direct inspection.
\end{proof}

This proposition shows that $\funcL\circ\funcS$ is \emph{not}
equivalent to the identity. However, when $\Delta(Z)$ is a trivial
$\Gamma$-set the proposition yields an isomorphism between $Z/Y$
and $\funcL\circ\funcS(Z/Y)$:

\begin{cor}
     \label{cov6bis.cor}
     If the $\Gamma$-action on $\Delta(Z)$ is trivial, then there is an
     isomorphism of coverings between
     \[
      \resolv\bigl({\mathcal S}(Z/Y)\bigr)\xleftarrow{\varepsilon}
      \Lambda_{2}\bigl({\mathcal S}(Z/Y)\bigr)
      \qquad\text{and}\qquad
      Y\xleftarrow{\pi}Z.
      \]
\end{cor}

\begin{proof}
     This readily follows from Proposition~\ref{cov6bis.prop} and
     Proposition~\ref{compcov.prop}(b).
\end{proof}

Corollary~\ref{cov6bis.cor} applies in particular to double coverings
of the form $\Lambda_{2}(X)/\resolv(X)$, for $X$ a $\Gamma$-set
with $\vert X\vert=4$, by Proposition~\ref{discres.prop}. Therefore,
$\funcL\circ\funcS(X)\simeq X$.

\begin{thm}
     \label{corresp1.thm}
     The functors $\funcL$ and $\funcS$ define a canonical one-to-one
     correspondence between the set of isomorphism classes of
     $\Gamma$-sets of $4$ elements and the set of isomorphism classes
     of double coverings $Z/Y$ of $\Gamma$-sets $Y$ of $3$ elements
     with trivial action on $\Delta(Z)$.
\end{thm}

\begin{remark}\footnote{The authors are indebted to F.~Borceux for
     enlightening comments
     about this remark.}
     For $X$, $X'\in\Set_{\Gamma}^{4}$, every morphism of coverings
     $f\colon\,\Lambda_{2}(X)/\resolv(X)\to\Lambda_{2}(X')/\resolv(X')$
     induces a morphism
     $\rescov\bigl(\Lambda_{2}(X)/\resolv(X)\bigr)\to
     \rescov\bigl(\Lambda_{2}(X')/\resolv(X')\bigr)$, hence, by
     Proposition~\ref{set4.prop}, a morphism $\tilde f\colon\,X\to
     X'$. The functor $\funcL$ carries $\tilde f$ to $f$, hence it is
     full. Since $\funcS\circ\funcL$ is equivalent to the identity, the
     functor $\funcL$ is also faithful. Moreover,
     Corollary~\ref{cov6bis.cor} shows that every covering
     $Z/Y\in\Cov_{\Gamma}^{2\wr3}$ such that the $\Gamma$-action on
     $\Delta(Z)$ is trivial is isomorphic to a covering of the form
     $\funcL(X)$. Therefore, it follows from \cite[Theorem~1,
     p.~93]{McL} that $\funcL$ is an equivalence of categories from
     $\Set_{\Gamma}^{4}$ to the full subcategory of
     $\Cov_{\Gamma}^{2\wr3}$ whose objects are the coverings $Z/Y$
     with trivial $\Gamma$-action on $\Delta(Z)$.
\end{remark}

In order to take into account the double coverings of $\Gamma$-sets
of three elements which have non-trivial action on the discriminant, we
consider the product category $\Set_{\Gamma}^{2}\times\Set_{\Gamma}^{4}$
whose objects are pairs $(U,X)$ of $\Gamma$-sets with $\vert
U\vert=2$ and $\vert X\vert=4$, and extend $\funcL$ and $\funcS$ to
functors
\[
\funchatL\colon\,\Set_{\Gamma}^{2}\times\Set_{\Gamma}^{4}\to
\Cov_{\Gamma}^{2\wr3}
\qquad\text{and}\qquad
\funchatS\colon\,\Cov_{\Gamma}^{2\wr3}\to
\Set_{\Gamma}^{2}\times\Set_{\Gamma}^{4}
\]
defined by
\[
\funchatL(U,X)=\bigl(U*\Lambda_{2}(X)\bigr)/\resolv(X)
\qquad\text{and}\qquad
\funchatS(Y\xleftarrow{\pi}Z)=
\bigl(\Delta(Z),\rescov(Z/Y)\bigr).
\]
Proposition~\ref{cov6bis.prop} yields a natural equivalence between
$\funchatL\circ\funchatS$ and the identity on $\Cov_{\Gamma}^{2\wr3}$.
\par
On the other hand, for $U$, $X$ with $\vert U\vert=2$ and $\vert
X\vert=4$, we have canonical isomorphisms
\[
\rescov\bigl(U*\Lambda_{2}(X)/\resolv(X)\bigr)
\simeq
\rescov\bigl(\Lambda_{2}(X)/\resolv(X)\bigr)\simeq X
\]
by Propositions~\ref{coverinvariance.prop} and \ref{set4.prop}, and
\begin{equation}
     \label{deltaU.eq}
     \Delta\bigl(U*\Lambda_{2}(X)\bigr)
     \simeq\Delta\bigl(U\times\resolv(X)\bigr)*
     \Delta\bigl(\Lambda_{2}(X)\bigr)
     \simeq U*\Delta\bigl(\Lambda_{2}(X)\bigr),
\end{equation}
by Propositions~\ref{deltahom.prop} and \ref{discprod.prop}. The
$\Gamma$-action on $\Delta\bigl(\Lambda_{2}(X)\bigr)$ is trivial by
Proposition~\ref{discres.prop}, hence the right-most $\Gamma$-set in
\eqref{deltaU.eq} is isomorphic to $U$ by
Proposition~\ref{groupstar.prop}(b). Note that the latter isomorphism
is \emph{not} canonical, hence $\funchatS\circ\funchatL$ is not
naturally equivalent to the identity on
$\Set_{\Gamma}^{2}\times\Set_{\Gamma}^{4}$. However, since
$\funchatS\circ\funchatL(U,X)\simeq(U,X)$, we have an isomorphism
between sets of isomorphism classes.

\begin{thm}
     \label{corresp2.thm}
     The functors $\funchatL$ and $\funchatS$ define a canonical
     one-to-one correspondence between the set of isomorphism classes
     of pairs of $\Gamma$-sets $(U,X)$ with $\vert U\vert=2$ and $\vert
     X\vert=4$ and the set of isomorphism classes of double coverings
     of $\Gamma$-sets with three elements.
\end{thm}

An alternative proof in cohomology can be derived from
Diagram~\eqref{H1.diag}. Theorems~\ref{corresp1.thm} and
\ref{corresp2.thm} have
analogues in terms of quartic \'etale algebras and double coverings
of cubic algebras, whose statements are left to the reader.

\begin{remark}
     The functor $\funchatS$ is faithful since
     $\funchatL\circ\funchatS$ is equivalent to the identity. Moreover,
     every $(U,X)\in\Set_{\Gamma}^{2}\times\Set_{\Gamma}^{4}$ is
     isomorphic to an object of the form $\funchatS(Z/Y)$ (namely,
     $Z/Y=\funchatL(U,X)$). Furthermore, every morphism
     $f\colon\,\funchatS(Z/Y)\to\funchatS(Z'/Y')$ induces a morphism
     $\funchatL(f)\colon\,\funchatL\circ\funchatS(Z/Y)\to
     \funchatL\circ\funchatS(Z'/Y')$, hence by
     Proposition~\ref{cov6bis.prop} a morphism $\tilde f\colon\,Z/Y\to
     Z'/Y'$. We may check that $f=\funchatS(\tilde f)$, hence the
     functor $\funchatS$ is full. By \cite[Theorem~1, p.~93]{McL}, it
     defines an equivalence of categories
     $\Cov_{\Gamma}^{2\wr3}\equiv\Set_{\Gamma}^{2}\times\Set_{\Gamma}^{4}$.
\end{remark}

\subsection{Cohomology}
\label{cohom4.subsec}

This subsection presents the cohomological perspective on
Theorem~\ref{corresp2.thm}. We use the same notation as in the
preceding subsection.

Let $U=\{1,2\}$ and $X=\{1,2,3,4\}$ with trivial $\Gamma$-action. The
group of automorphisms of $(U,X)$ in the category
$\Set_{\Gamma}^{2}\times\Set_{\Gamma}^{4}$ is
$\Sym_{2}\times\Sym_{4}$, and since
$\funchatS\circ\funchatL(U,X)\simeq(U,X)$, the functor $\funchatL$
yields an isomorphism
\begin{equation}
     \label{hatlambda.eq}
     \Hat\lambda\colon\,\Sym_{2}\times\Sym_{4}\iso
     \Aut\bigl(U*\Lambda_{2}(X)/\resolv(X)\bigr)\simeq
     \Sym_{2}\wr\Sym_{3}.
\end{equation}

For definiteness, consider $\Sym_{2}\wr\Sym_{3}$ as the group of
automorphisms of the covering
\[
Y=\{1,2,3\}\xleftarrow{\pi_{2}}\{1,2\}\times\{1,2,3\}=Z,
\]
where $\Gamma$ acts trivially on $Y$ and $Z$. The right isomorphism
in \eqref{hatlambda.eq} depends on the choice of an isomorphism
$U*\Lambda_{2}(X)/\resolv(X)\simeq Z/Y$.

Similarly, the functor $\funchatS$ yields an isomorphism
\begin{equation}
     \label{hats.eq}
     \Hat s\colon\,\Sym_{2}\wr\Sym_{3}\iso
     \Aut\bigl(\Delta(Z),\rescov(Z/Y)\bigr)\simeq\Sym_{2}\times\Sym_{4}
\end{equation}
where again the latter isomorphism is given by identifications
$\Delta(Z)\simeq\{1,2\}$ and $\rescov(Z/Y)\simeq\{1,2,3,4\}$. The
isomorphisms $\Hat\lambda$ and $\Hat s$ induce bijections
\[
H^{1}(\Gamma,\Sym_{2}\times\Sym_{4})\simeq
H^{1}(\Gamma,\Sym_{2}\wr\Sym_{3}).
\]
Since these cohomology sets are in one-to-one correspondence with the
sets of isomorphism classes in
$\Set_{\Gamma}^{2}\times\Set_{\Gamma}^{4}$ and $\Cov_{\Gamma}^{2\wr3}$
respectively (see \S\ref{cohsets.subsec}), we thus recover
Theorem~\ref{corresp2.thm}.
\medbreak
\par
The isomorphisms $\Hat\lambda$ and $\Hat s$ can also be described in
purely group-theoretical terms. The subgroup
$\Sym_{2}=\Sym_{2}\times\{1\}\subset\Sym_{2}\times\Sym_{4}$ is mapped
to the ``diagonal'' subgroup $\Sym_{2}\subset\Sym_{2}\wr\Sym_{3}$,
which is the center of $\Sym_{2}\wr\Sym_{3}$. On the other hand, the
restriction of $\Hat\lambda$ to $\Sym_{4}=\{1\}\times\Sym_{4}$ is a
homomorphism
\[
\lambda\colon\,\Sym_{4}\to\Sym_{2}\wr\Sym_{3}
\]
which may be described as the action of $\Sym_{4}$ by conjugation on
its transpositions. Indeed, $\Sym_{4}$ contains six transpositions,
which sit by pairs in the three Sylow $2$-subgroups of $\Sym_{4}$.
The map which carries each transposition to the unique Sylow
$2$-subgroup which contains it is a double covering of a set of three
elements. Note that the composition of $\lambda$ and the canonical
homomorphism $\beta\colon\,\Sym_{2}\wr\Sym_{3}\to\Sym_{3}$ is the
surjective homomorphism
\[
\rho\colon\,\Sym_{4}\to\Sym_{3}
\]
which is the action of $\Sym_{4}$ on its three Sylow $2$-subgroups.
(Alternately, the map $\rho$ may be identified with the canonical
homomorphism $\Sym_{X}\to\Sym_{\resolv(X)}$ for $X=\{1,2,3,4\}$,
since there is a canonical one-to-one correspondence between
$\resolv(X)$ and the Sylow $2$-subgroups of $\Sym_{X}$.) The kernel
of $\rho$ is the Vierergruppe $\Vier$.

By definition, it is clear that the first component of $\Hat s$ is the
signature map
\[
\sgn\colon\,\Sym_{2}\wr\Sym_{3}\subset\Sym_{6}\to\Sym_{2},
\]
since the map $\Sym_{Z/Y}\to\Sym_{\Delta(Z)}$ is the signature. The
second component is a homomorphism
\[
s\colon\,\Sym_{2}\wr\Sym_{3}\to\Sym_{4}
\]
which is the action of $\Sym_{2}\wr\Sym_{3}$ on its four Sylow
$3$-subgroups. (There is a natural one-to-one correspondence between
the Sylow $3$-subgroups of $\Sym_{Z/Y}$ and $\rescov(Z/Y)$.)
The image of $\lambda$ is the kernel of $\sgn$, by
Proposition~\ref{discres.prop} or by Proposition~\ref{perm.prop}, and
the map $s$ splits $\lambda$ (if the Sylow $3$-subgroups of
$\Sym_{2}\wr\Sym_{3}$ are suitably indexed). The maps $\rho$,
$\lambda$ and $\beta$, and the inclusions $\iota$, $\eta$, are part of
the following commutative diagram
with exact rows and columns:
\begin{equation}
      \label{3.diag}
      \begin{CD}
         {}@. 1@. 1\\
         @. @VVV @VVV\\
         1@>>> \Vier @>\iota>> \Sym_{4} @>{\rho}>> \Sym_{3}
         @>>> 1\\
         @. @VVV @VV{\lambda}V @\vert\\
         1 @>>> \Sym_{2}^{3}
         @>\eta>> \Sym_{2}\wr\Sym_{3}
         @>{\beta}>>
         \Sym_{3} @>>> 1\\
         @. @V{\sigma}VV @VV{\sgn}V \\
         {} @. \Sym_{2} @= \Sym_{2}\\
         @. @VVV @VVV\\
         {} @. 1 @. 1
      \end{CD}
\end{equation}
where $\sigma$ is the sum. Since the exact sequences in this diagram
are split, there is a corresponding commutative diagram of exact
sequences in cohomology:
\begin{equation}
      \label{H1.diag}
      \begin{CD}
         {}@. 1@. 1\\
         @. @VVV @VVV\\
         1@>>> H^{1}(\Gamma,\Vier)@> \iota^1>>
         H^{1}(\Gamma,\Sym_{4}) @>{\rho^{1}}>> H^{1}(\Gamma,\Sym_{3})
         @>>> 1\\
         @. @VVV @VV{\lambda^{1}}V @\vert\\
         1 @>>> H^{1}(\Gamma,\Sym_{2}^{3})
         @>\eta^1>> H^{1}(\Gamma,\Sym_{2}\wr\Sym_{3})
         @>{\beta^{1}}>>
         H^{1}(\Gamma,\Sym_{3}) @>>> 1\\
         @. @V{\sigma^{1}}VV @VV{\sgn^{1}}V \\
         {} @. H^{1}(\Gamma,\Sym_{2}) @= H^{1}(\Gamma,\Sym_{2})\\
         @. @VVV @VVV\\
         {} @. 1 @. 1.
      \end{CD}
\end{equation}

Cohomology yields an alternative proof of
Proposition~\ref{discres.prop}:

\begin{prop}
      \label{discrescoh.prop}
      For any $\Gamma$-set $X$ with $\vert X\vert=4$,
      \[
      \Delta(X)\simeq\Delta\bigl(\resolv(X)\bigr).
      \]
      Moreover, the $\Gamma$-action on
      $\Delta\bigl(\Lambda_{2}(X)\bigr)$ is trivial.
\end{prop}

\begin{proof}
      The commutative diagram
      \[
      \begin{CD}
         \Sym_{4} @>{\rho}>> \Sym_{3}\\
         @V{\sgn}VV @VV{\sgn}V\\
         \Sym_{2} @= \Sym_{2}
      \end{CD}
      \]
      induces a commutative diagram in cohomology
      \[
      \begin{CD}
         H^{1}(\Gamma,\Sym_{4}) @>{\rho^{1}}>> H^{1}(\Gamma,\Sym_{3})\\
         @V{\sgn^{1}}VV @VV{\sgn^{1}}V\\
         H^{1}(\Gamma,\Sym_{2}) @= H^{1}(\Gamma,\Sym_{2}).
       \end{CD}
       \]
      The first part of the proposition follows since $\rho^{1}$ maps the
      isomorphism
      class of $X$ to the isomorphism class of $\resolv(X)$, and
      $\sgn^{1}$ maps the isomorphism class of any $\Gamma$-set to the
      isomorphism class of its discriminant. The second part follows
      from the fact that
      \[
      H^{1}(\Gamma,\Sym_{4})\xrightarrow{\lambda^{1}}
      H^{1}(\Gamma,\Sym_{2}\wr\Sym_{3})\xrightarrow{\sgn^{1}}
      H^{1}(\Gamma,\Sym_{2})
      \]
      is a zero-sequence.
\end{proof}

As another application of cohomology, we describe the quartic \'etale
algebras which have a given resolvent cubic.

Let $R$ be a $\Gamma$-set of three elements, and let
$X_{0}=R\coprodd\{0\}$ be the $\Gamma$-set of four elements obtained
by adjoining to $R$ a fixed point $0$. To each partition of $X_{0}$
into $2$-element subsets, we may associate the unique element $r\in
R$ such that $\{0,r\}$ is in the partition, and thus identify
\[
\resolv(X_{0})=R.
\]
As in \S\ref{cohsets.subsec}, we let $\Gamma$ act by conjugation on the
groups $\Sym_{X_{0}}$, $\Sym_{R}$, and denote by $\Sym_{X_{0}}'$,
$\Sym_{R}'$ the $\Gamma$-groups thus defined. The inclusion
$R\hookrightarrow X_{0}$ yields a $\Gamma$-equivariant embedding
$\Sym_{R}'\hookrightarrow\Sym_{X_{0}}'$ which splits the map
$\rho\colon\,\Sym_{X_{0}}'\to\Sym_{R}'$. The split exact sequence
\[
1\to\Vier_{X_{0}}'\xrightarrow{\iota}\Sym_{X_{0}}'
\xrightarrow{\rho}\Sym_{R}'\to1
\]
yields an exact sequence in cohomology
\begin{equation}
     \label{splx.eq}
1\to H^{1}(\Gamma,\Vier_{X_{0}}')
\xrightarrow{\iota^{1}}H^{1}(\Gamma,\Sym_{X_{0}}')
\xrightarrow{\rho^{1}}H^{1}(\Gamma,\Sym_{R}')\to1,
\end{equation}
and the isomorphism classes of $X\in\Set_{\Gamma}^{4}$ such that
$\resolv(X)\simeq R$ are in one-to-one correspondence with
$\ker\rho^{1}=\image \iota^{1}$. They form a pointed set with the
isomorphism class of $X_{0}$ as distinguished element. Note that
exactness of the sequence \eqref{splx.eq} does \emph{not} mean that
$\iota^{1}$ is injective. In fact, the group
$\Aut_{\Gamma}(R)=H^{0}(\Gamma,\Sym_{R}')$ acts on
$H^{1}(\Gamma,\Vier_{X_{0}}')$, and $\image\iota^{1}$ is in canonical
bijection with the orbit set
$H^{1}(\Gamma,\Vier_{X_{0}}')/\Aut_{\Gamma}(R)$, by \cite[(28.4)]{BoI}.

To give a more explicit description, we use a variant of
Diagram~\eqref{H1.diag}.
First, observe that we may identify $\Lambda_{2}(X_{0})$ to
$\{1,2\}\times R$, as follows: we map a $2$-element subset $U\subset
X_{0}$ to $(1,r)$ if $0\notin U$ and $r\notin U$, to $(2,r)$ if
$U=\{0,r\}$. We may then identify the double covering
$\Lambda_{2}(X_{0})/\resolv(X_{0})$ to
\[
R\xleftarrow{\pi_{2}}\{1,2\}\times R.
\]
Let $Z_{0}=\{1,2\}\times R$. As above, we let $\Gamma$ act by
conjugation on $\Sym_{Z_{0}/R}$, and denote by $\Sym_{Z_{0}/R}'$ the
corresponding $\Gamma$-group. As in \S\ref{cohsets.subsec}, let
$\mathfrak{T}_{Z_{0}/R}'$ be the kernel of the canonical map
$\beta_{Z_{0}/R}\colon\,\Sym_{Z_{0}/R}'\to\Sym_{R}'$. The exact
sequence
\[
1\to\mathfrak{T}_{Z_{0}/R}'\xrightarrow{\eta}\Sym_{Z_{0}/R}'
\xrightarrow{\beta_{Z_{0}/R}}\Sym_{R}'\to1
\]
is split, and induces an exact sequence in cohomology
\[
1\to H^{1}(\Gamma,\mathfrak{T}_{Z_{0}/R}')\xrightarrow{\eta^{1}}
H^1(\Gamma,\Sym_{Z_{0}/R}')\xrightarrow{\beta^{1}}H^{1}(\Gamma,\Sym_{R}')
\to1.
\]
As above, there is a canonical bijection
\[
\image\eta^{1}=\ker\beta^{1}\simeq
H^{1}(\Gamma,\mathfrak{T}_{Z_{0}/R}')/\Aut_{\Gamma}(R).
\]
The orbit set on the right side may therefore be identified with
the set of isomorphism
classes of double coverings of $R$, see \S\ref{cohsets.subsec}. Consider
the commutative diagram analogous to \eqref{H1.diag},
\begin{equation}
     \label{H1'.diag}
      \begin{CD}
         1@. 1\\
         @VVV @VVV\\
         H^{1}(\Gamma,\Vier_{X_{0}}')@> \iota^1>>
         H^{1}(\Gamma,\Sym_{X_{0}}') @>{\rho^{1}}>>
         H^{1}(\Gamma,\Sym_{R}')
         \\
         @VVV @VV{\lambda^{1}}V @\vert\\
         H^{1}(\Gamma,\mathfrak{T}_{Z_{0}/R}')
         @>\eta^1>> H^{1}(\Gamma,\Sym_{Z_{0}/R}')
         @>{\beta^{1}}>>
         H^{1}(\Gamma,\Sym_{R}')\\
         @V{\sigma^{1}}VV @VV{\sgn^{1}}V \\
         H^{1}(\Gamma,\Sym_{2}) @= H^{1}(\Gamma,\Sym_{2})\\
         @VVV @VVV\\
         1 @. 1.
      \end{CD}
\end{equation}
The left vertical sequence is an exact sequence of groups. It shows
that $H^{1}(\Gamma,\Vier_{X_{0}}')$ can be identified with the kernel
of $\sigma^{1}$. Recall from \S\ref{cohsets.subsec} that
$H^{1}(\Gamma,\mathfrak{T}_{Z_{0}/R}')$ is in canonical bijection with
the set $C^{2}(R)$ of isomorphism classes over $R$ of double coverings
of $R$, and that $H^{1}(\Gamma,\Sym_{2})$ classifies $\Gamma$-sets of
two elements up to isomorphism. By commutativity of the lower square
in \eqref{H1'.diag}, the map $\sigma^{1}$ carries every double
covering to the isomorphism class of its discriminant. Therefore, we
may identify $H^{1}(\Gamma,\Vier_{X_{0}}')$ with the group
$C^{2}_{0}(R)$ of isomorphism classes over $R$ of double coverings of $R$
with trivial discriminant,
\[
H^{1}(\Gamma,\Vier_{X_{0}}')=C^{2}_{0}(R).
\]

We have thus shown:

\begin{prop}
     \label{givencubic1.prop}
     The set of isomorphism classes of sets $X$ of four elements such
     that $\resolv(X)\simeq R$ is in canonical bijection with the set
     $C_{0}^{2}(R)/\Aut_{\Gamma}(R)$ of isomorphism classes of double
     coverings of $R$ with trivial discriminant.
\end{prop}

Suppose now $\Gamma$ is the absolute Galois group of a field $F$ with
separable closure $F_{s}$, and let $A$ be a cubic \'etale $F$-algebra.
Using the anti-equivalence $\Set_{\Gamma}\equiv\Et_{F}$, we may
translate Proposition~\ref{givencubic1.prop} into the following
statement, where we denote by $\Quadr_{0}(A)$ the set of
isomorphism classes over $A$ of quadratic extensions of $A$ whose
discriminant (as $F$-algebra) is trivial:

\begin{prop}
     \label{givencubic2.prop}
     The set of isomorphism classes of quartic \'etale $F$-algebras $Q$
     with $\resolv(Q)\simeq A$ is in canonical
     bijection with the set $\Quadr_{0}(A)/\Aut_{F}(A)$ of
     $F$-isomorphism classes of quadratic extensions of $A$ with
     trivial discriminant.
\end{prop}

The next proposition gives an explicit description of the group
$\Quadr_{0}(A)$.

\begin{prop}
     \label{quadrA.prop}
     Let $N^{1}(A)$ be the (multiplicative) group of elements of $A$ of
     norm~$1$ and let $T^{0}(A)$ be the (additive) group of elements
     of $A$ of trace~$0$.
     \begin{enumerate}
	\item[(a)]
	If $\carac F\neq2$, $\Quadr_{0}(A)\simeq N^{1}(A)/N^{1}(A)^{2}$.
	\item[(b)]
	If $\carac F=2$, $\Quadr_{0}(A)\simeq
	T^{0}(A)/\wp\bigl(T^{0}(A)\bigr)$, where $\wp$ is the Artin-Schreier
	map $\wp(x)=x^{2}-x$.
     \end{enumerate}
\end{prop}

\begin{proof}
     If $A$ is a field, the action of $\Gamma$ on $\funcX(A)$ is
     transitive. Letting $\Gamma_{0}\subset\Gamma$ be the absolute
     Galois group of a copy of $A$ in $F_{s}$, we have $\Quadr(A)\simeq
     H^{1}(\Gamma_{0},\Sym_{2})$ as observed in \S\ref{cohsets.subsec},
     and the map $\sigma^{1}$ can be interpreted as the corestriction
     \[
     H^{1}(\Gamma_{0},\Sym_{2})\to H^{1}(\Gamma,\Sym_{2}).
     \]

     If $\carac F\neq2$, we identify $\Sym_{2}$ with $\{1,-1\}\subset
     F_{s}^\times$. The exact sequence
     \[
     1\to\Sym_{2}\to F_{s}^\times\xrightarrow{2}F_{s}^\times\to1
     \]
     yields isomorphisms
     \[
     H^{1}(\Gamma,\Sym_{2})\simeq F^\times/F^{\times2}
     \qquad\text{and}\qquad
     H^{1}(\Gamma_{0},\Sym_{2})\simeq A^\times/A^{\times2}
     \]
     under which the corestriction corresponds to a map induced by the
     norm. Its kernel is
     $N^{1}(A)/N^{1}(A)^{2}$ since if $y\in A^\times$ is such that
     $N_{A/F}(y)=z^{2}\in F^{\times2}$, then $N_{A/F}(y^{3}z^{-2})=1$.

     If $\carac F=2$, we identify $\Sym_{2}$ with $\{0,1\}\subset
     F_{s}$. The exact sequence
     \[
     0\to\Sym_{2}\to F_{s}\xrightarrow{\wp}F_{s}\to 0
     \]
     yields isomorphisms
     \[
     H^{1}(\Gamma,\Sym_{2})\simeq F/\wp(F)
     \qquad
     \text{and}\qquad
     H^{1}(\Gamma_{0},\Sym_{2})\simeq A/\wp(A)
     \]
     under which the corestriction corresponds to a map induced by the
     trace. Its kernel is $T^{0}(A)/\wp\bigl(T^{0}(A)\bigr)$ since if
     $T_{A/F}(y)=z^{2}-z$, then $T_{A/F}(y-z^{2}+z)=0$.

     If $A$ is not a field, it decomposes into a direct product of
     fields,
     \[
     A\simeq F\times K
     \qquad\text{or}\qquad
     A\simeq F\times F\times F.
     \]
     In the first case,
     \[
     \Quadr(A)\simeq\Quadr(F)\times\Quadr(K)\simeq
     \begin{cases}
	(F^\times/F^{\times2})\times(K^\times/K^{\times2})
	&\text{if $\carac F\neq2$},\\
	\bigl(F/\wp(F)\bigr)\times
	\bigl(K/\wp(K)\bigr)
	&\text{if $\carac F=2$},
     \end{cases}
     \]
     and the map $\sigma^{1}$ can be again interpreted as induced by
     the norm or the trace. We may then use the same arguments as above.
     The case where $A\simeq F\times F\times F$ is left to the reader.
\end{proof}

Following \S\ref{galoisalg.subsec}, the set $H^{1}(\Gamma,\Sym_{R}')$
for $R=\funcX(A)$ is also in one-to-one correspondence with the set of
isomorphism classes of $\Sym_{3}$-Galois $F$-algebras, where the
distinguished element corresponds to the isomorphism class of the
$\Sym_{3}$-Galois closure $\Sigma_{3}(A)$. Likewise, the set
$H^{1}(\Gamma,\Sym_{X_{0}}')$ classifies $\Sym_{4}$-Galois
$F$-algebras up to isomorphism, with the class of $\Sigma_{4}(F\times
A)$ as distinguished element. The upper exact sequence of
Diagram~\eqref{H1'.diag} shows that the $\Sym_{4}$-Galois
$F$-algebras $M$ which are the $\Sym_{4}$-Galois closure of an \'etale
quartic $F$-algebra $Q$ with $\resolv(Q)\simeq A$ are in one-to-one
correspondence with $\Quadr_{0}(A)/\Aut_{F}(A)$. Using
Proposition~\ref{quadrA.prop}, we may make this correspondence
explicit as follows:

\begin{prop}
     Let $A$ be a cubic \'etale $F$-algebra, identified with a
     subalgebra of its $\Sym_{3}$-Galois closure $\Sigma_{3}(A)$, and
     let $\rho\in\Sym_{3}$ be an element of order~$3$.
     \begin{enumerate}
	\item[(a)]
	If $\carac F\neq2$, let $a\in A^\times$ be such that $N_{A/F}(a)=1$,
	and set
	\[
	M=\Sigma_{3}(A)[\sqrt{a},\sqrt{\rho(a)},
	\sqrt{\rho^{2}(a)}].
	\]
	\item[(b)]
	If $\carac F=2$, let $a\in A$ be such that $T_{A/F}(a)=0$, and set
	\[
	M=\Sigma_{3}(A)\bigl[\wp^{-1}(a),\wp^{-1}\bigl(\rho(a)\bigr),
	\wp^{-1}\bigl(\rho^{2}(a)\bigr)\bigr].
	\]
     \end{enumerate}
     In each case, there is a $\Sym_{4}$-action on $M$
     which endows it with the structure of an $\Sym_{4}$-Galois algebra.
     The quartic subalgebra $Q=M^{\Sym_{3}}$ satisfies $\resolv(Q)\simeq
     A$. Moreover, every $\Sym_{4}$-Galois $F$-algebra which is the
     $\Sym_{4}$-Galois closure of a quartic \'etale $F$-algebra $Q$ with
     $\resolv(Q)\simeq A$ is of this form.
\end{prop}

\begin{remark}
Similar constructions are described by Serre (see
\cite{Serre85}\footnote{We are indebted to J-P.~Serre for calling our
attention to this reference.})
and by Weil (for the construction
of dyadic field extensions with Galois group $\Sym_4$, see
\cite[Section 31]{Weil74e}).
\end{remark}

\subsection{Quartic \'etale algebras}
\label{quartal.subsec}

In this subsection, our goal is to make explicit
the relation between resolvent cubics of quartic polynomials and the
construction of $\resolv(Q)$ for $Q$ a quartic \'etale $F$-algebra.
Our first observation is a direct consequence of
Proposition~\ref{discres.prop}
(see also Proposition~\ref{discrescoh.prop}).

\begin{prop}
      \label{discreset.prop}
      Let $Q$ be a quartic \'etale $F$-algebra. There is a canonical
      isomorphism
      \[
      \Delta\bigl(\resolv(Q)\bigr)\iso\Delta(Q).
      \]
      Moreover, $\Delta\bigl(\Lambda_{2}(Q)\bigr)\simeq F\times F$.
\end{prop}

\begin{proof}
      The proposition readily follows from
      Proposition~\ref{discres.prop} under the anti-equi\-valence
      $\Et_{F}\equiv\Set_{\Gamma}$, since the $\Delta$, $\resolv$ and
      $\Lambda_{2}$ construction commute with the functor~$\funcX$.
\end{proof}

Recall from \cite{VdWalg} that ``the'' resolvent cubic of a quartic
polynomial
\begin{equation}
      \label{grad4.eq}
      f(u) = u^4 -\alpha_1u^3 + \alpha_2u^2 -\alpha_3u +\alpha_4
\end{equation}
with roots $u_1$, $u_2$, $u_3$, $u_4$ in an algebraic closure,
is the polynomial $g(v)$ with roots
\[
v_1 = (u_1 +u_2)(u_3 +u_4), \quad v_2 =  (u_1 +u_3)(u_2 +u_4),
\quad v_3 = (u_1 +u_4)(u_2 +u_3).
\]
This polynomial has the form
\begin{equation}
      \label{grad3.eq}
      g(v) = v^3 -\beta_1v^2 +\beta_2v -\beta_3
\end{equation}
where
\begin{equation}
      \label{coefficients.eq}
      \begin{split}
         \beta_1 & = 2\alpha_2,\\
         \beta_2 & = \alpha_1\alpha_3 + \alpha_2^2 -4\alpha_4,\\
         \beta_3 & = \alpha_1\alpha_2\alpha_3 - \alpha_1^2\alpha_4 -\alpha_3^2.
       \end{split}
\end{equation}

An alternative resolvent cubic suggested by Lagrange \cite[(32),
p.~266]{L} in
characteristic different from~$2$ has roots
\[
w_{1}=(u_{1}+u_{2}-u_{3}-u_{4})^{2},\qquad
w_{2}=(u_{1}-u_{2}+u_{3}-u_{4})^{2},\qquad
w_{3}=(u_{1}-u_{2}-u_{3}+u_{4})^{2}.
\]
Since $w_{i}=\alpha_{1}^{2}-4v_{i}$ for $i=1$, $2$, $3$,
this polynomial has the form
\begin{equation}
      \label{gradd3.eq}
      h(w)=-4^{3}g\Bigl(\frac{\alpha_{1}^{2}-w}{4}\Bigr)=
      w^{3}-\varkappa_{1}w^{2}+\varkappa_{2}w-\varkappa_{3},
\end{equation}
where
\begin{equation}
      \label{coefficientss.eq}
      \begin{split}
         \varkappa_{1} & =3\alpha_{1}^{2}-8\alpha_{2},\\
         \varkappa_{2} & =3\alpha_{1}^{4}-16\alpha_{1}^{2}\alpha_{2}
         +16\alpha_{1}\alpha_{3}+16\alpha_{2}^{2}-64\alpha_{4},\\
         \varkappa_{3} &
         =(\alpha_{1}^{3}-4\alpha_{1}\alpha_{2}+8\alpha_{3})^{2}.
      \end{split}
\end{equation}

Now, let $Q$ be a quartic \'etale algebra over a field $F$ of
arbitrary characteristic. For $x\in Q$, let
\begin{equation}
      \label{phia.eq}
      \lambda_{x}=s_{2}\cdot(x\otimes1+1\otimes x)\in\Lambda_{2}(Q).
\end{equation}

\begin{prop}
      \label{propgenericet.prop}
      Suppose $x\in Q$ is a generating element with minimal polynomial
      $f(u)$ as in \eqref{grad4.eq}, so that the coefficient
$\alpha_1$ is the \emph{trace} $T_{Q/F}(x)$ of $x$. Then
\begin{enumerate}
  \item[(a)] $\lambda_{x} +  \gamma_{Q}(\lambda_{x}) = T_{Q/F}(x)$.
\item[(b)]
      $\gamma_{Q}(\lambda_{x})\lambda_{x}\in\resolv(Q)$ is a generating
      element with minimal polynomial $g(v)$ as in \eqref{grad3.eq}.
      Moreover, if the characteristic of $F$ is different from~$2$,
      then $\bigl(\lambda_{x}-\gamma_{Q}(\lambda_x)\bigr)^{2}\in\resolv(Q)$
      is a generating element
      with minimal polynomial $h(w)$ as in \eqref{gradd3.eq}.
\end{enumerate}
      In arbitrary characteristic, if the element $\varkappa_{3}$ of
      \eqref{coefficientss.eq} is not zero,\footnote{In
      characteristic~$2$, the condition is thus $T_{Q/F}(x)\neq 0$.}
      then $\lambda_{x}\in\Lambda_{2}(Q)$ is a generating element over
      $\resolv(Q)$, with minimal polynomial
      \[
      t^{2}-T_{Q/F}(x)t+\gamma_{Q}(\lambda_x)\lambda_x\in \resolv(Q)[t].
      \]
\end{prop}

\begin{proof}
      Extending scalars, we may assume that $Q$ is split, with a
      basis $(e_{1},e_{2},e_{3},e_{4})$ consisting of minimal
      (orthogonal) idempotents. Then $\resolv (Q)$ is split and
$e_1 \otimes e_2 +
          e_2 \otimes e_1 + e_3 \otimes e_4 + e_4 \otimes e_3$,
$e_1 \otimes e_3 +
         e_3 \otimes e_1 + e_2 \otimes e_4 + e_4 \otimes e_2$ and
$e_1 \otimes e_4 +
         e_4 \otimes e_1 + e_2 \otimes e_3 + e_3 \otimes e_2$
is a basis of $\resolv (Q)$  consisting of minimal idempotents.
          Let
      \[
      x=x_{1}e_{1}+x_{2}e_{2}+x_{3}e_{3}+x_{4}e_{4}
      \]
      with $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}\in F$. Since $x$ generates
      $Q$, the coefficients $x_{i}$ are pairwise distinct. Computation
      shows that
      \[
      \lambda_{x}=
      \sum _{1 \leq i < j\leq 4} (x_i +x_j)(e _i \otimes e_j + e_j \otimes e_i),
      \]
      hence
\[
\gamma_{Q}(\lambda_x) = \sum _{1 \leq i < j\leq 4} (x_i +x_j)
(e _{i'} \otimes e_{j'} + e_{j'} \otimes e_{i'})
\]
where $\{i,\,j,\,i',\,j'\}= \{1,\,2,\,3,\,4\}$. It follows that
\[
\lambda_{x} + \gamma_{Q}(\lambda_{x})= (x_1  +x_2 + x_3 +x_4)
\sum_{1 \leq i < j\leq 4}(e_i \tens e_j +e_j \tens e_i)= T_{Q/F}(x).
\]
Similarly
      \[
      \begin{split}
         \gamma_{Q}(\lambda_{x})\lambda_{x} & =
         (x_1 +x_2)(x_3 +x_4)(e_1 \otimes e_2 +
          e_2 \otimes e_1 + e_3 \otimes e_4 + e_4 \otimes e_3)\\
         &\quad  + (x_1 +x_3)(x_2 +x_4)(e_1 \otimes e_3 +
         e_3 \otimes e_1 + e_2 \otimes e_4 + e_4 \otimes e_2)\\
         &\quad  + (x_1 +x_4)(x_2 +x_3)(e_1 \otimes e_4 +
         e_4 \otimes e_1 + e_2 \otimes e_3 + e_3 \otimes e_2).
      \end{split}
      \]
      This shows that $\gamma_{Q}(\lambda_{x})\lambda_{x}$ is a root of
      a polynomial $g$ whose roots in $F$ are
      \[
      y_{1}=(x_{1}+x_{2})(x_{3}+x_{4}),
      \quad
      y_{2}=(x_{1}+x_{3})(x_{2}+x_{4}),
      \quad
      y_{3}=(x_{1}+x_{4})(x_{2}+x_{3}).
      \]
      These roots are distinct since an
      easy computation yields
      \[
      (y_{1}-y_{2})(y_{1}-y_{3})(y_{2}-y_{3})=
      -
      \prod_{i<j}(x_{i}-x_{j}).
      \]
      Therefore, $\gamma_{Q}(\lambda_{x})\lambda_{x}$
      is a generator
      of $\resolv(Q)$ and $g$ is its minimal polynomial.

      Similarly,
      \[
      \begin{split}
         \bigl(\lambda_{x}-\gamma_{Q}(\lambda_{x})\bigr)^{2} & =
         (x_{1}+x_{2}-x_{3}-x_{4})^{2}(e_1 \otimes e_2 +
          e_2 \otimes e_1 + e_3 \otimes e_4 + e_4 \otimes e_3)\\
         &\quad  + (x_1 - x_2+x_3 -x_4)^{2}(e_1 \otimes e_3 +
         e_3 \otimes e_1 + e_2 \otimes e_4 + e_4 \otimes e_2)\\
         &\quad  + (x_1 -x_2-x_3 +x_4)^{2}(e_1 \otimes e_4 +
         e_4 \otimes e_1 + e_2 \otimes e_3 + e_3 \otimes e_2),
      \end{split}
      \]
      and the same arguments show that
      $\bigl(\lambda_{x}-\gamma_{Q}(\lambda_{x})\bigr)^{2}$ is a
      generating element of $\resolv(Q)$ with minimal polynomial $h$ as
      in \eqref{gradd3.eq} if the characteristic of $F$ is different
      from~$2$.

      Since
      \[
      \varkappa_{3}=
      (x_{1}+x_{2}-x_{3}-x_{4})^{2}(x_{1}-x_{2}+x_{3}-x_{4})^{2}
      (x_{1}-x_{2}-x_{3}+x_{4})^{2},
      \]
      the condition $\varkappa_{3}\neq0$ implies that the elements
      $x_{i}+x_{j}$ for $1\leq i<j\leq4$ are pairwise distinct, hence
      $\lambda_{x}$ generates $\Lambda_{2}(Q)$. Since
      $\lambda_{x}+\gamma_{Q}(\lambda_{x})=T_{E/F}(x)$, the minimal
      polynomial of $\lambda_{x}$ over $\resolv(Q)$ is as stated in the
      proposition.
\end{proof}

\begin{remark}
      Allison gives in \cite[\S6]{Allison91} another description of the
      algebra $\resolv(Q)$, for $Q$ a quartic \'etale $F$-algebra. For
      $x\in Q$, he considers the image
      \[
      f_{x}=\varphi_{2}(\lambda_{x})
      \in\End_{F}({\textstyle\bigwedge^{2}Q})
      \]
      of $\lambda_{x}\in\Lambda_{2}(Q)$ under the map $\varphi_{2}$
      induced by the homomorphism in \eqref{tensym.eq} (see
      Lemma~\ref{tensym.lem}); thus
      \[
      f_{x}(a\wedge b) = xa\wedge b + a\wedge xb
      \qquad\text{for $a$, $b\in Q$}.
      \]
      Assuming that the characteristic of $F$ is different from~$2$,
      Allison defines $\resolv(Q)$ as the span of the products
      $f_{x}\circ f_{y}$, for $x$, $y\in Q$ of trace zero. This definition
      coincides with the definition in \S\ref{Basicet.subsec}
      under an isomorphism induced by $\varphi_{2}$.
\end{remark}

\subsection{Quadratic extensions of cubic \'etale algebras}
\label{quadextet.subsec}

Let $A$ be an \'etale $F$-algebra of dimension~$3$, and let $B$ be an
extension of degree~$2$ of $A$. In the same spirit as the preceding
subsection, we proceed to give explicit equations for generating
elements of ${\mathcal S}(B/A)$.

Our first observation is the analogue of Proposition~\ref{cov6.prop}
through the anti-equi\-valence between coverings of $\Gamma$-sets and
extensions of \'etale $F$-algebras.

\begin{prop}
      \label{DeltainLambda.prop}
      There is a canonical embedding
      $\Delta(B)\hookrightarrow\Omega(B/A)$ such that
      \[
      \Omega(B/A)\simeq\Delta(B)\otimes_{F}{\mathcal S}(B/A).
      \]
\end{prop}

In the case where $B$ (and therefore $A$) is split, the image of
$\Delta(B)$ in $\Omega(B/A)$ is spanned by the idempotents
\[
d=g_{0}+g_{1}+g_{2}+g_{3}
\qquad\text{and}\qquad
d'=g'_{0}+g'_{1}+g'_{2}+g'_{3},
\]
in the notation of Example~\ref{ext6.ex}.
\medbreak
\par
In the general case, for $b\in B$ we set for brevity
$\overline{b}=\gamma_{B/A}(b),$
and
\begin{equation}\label{bundeux.eq}
%
\begin{array}{lll}
b_1 = s_3^A\cdot (b \tens 1\tens 1) = \varepsilon_{11}(b), \quad &
b'_1 = s_3^A\cdot (\overline{b} \tens 1\tens 1) = \varepsilon_{21}(b) \\
%
b_2 = s_3^A\cdot (1 \tens b\tens 1) = \varepsilon_{12}(b), \quad &
b'_2 = s_3^A\cdot (1 \tens \overline{b} \tens 1) = \varepsilon_{22}(b) \\
b_3 = s_3^A\cdot ( 1\tens 1 \tens b) = \varepsilon_{13}(b), \quad &
b'_3 = s_3^A\cdot (1\tens 1\tens \overline{b} ) = \varepsilon_{23}(b)
\end{array}
\end{equation}
where $\varepsilon_{ij}\colon\, B \to \Sigma(B/A)$ are the embeddings
of \eqref{ijembeddings.eq}.
Hence we have $\Sigma(B/A) = \Sigma_3(A)[b_1,b'_1, \ldots, b_3,b'_3]$,
by Proposition \ref{generatorgalclos.prop}, and $\Sym_3$ acts
on $\Sigma(B/A)$ through the action on $\Sigma_3(A)$ and
by permuting the $b_i$ and the $b'_j$. The  algebra $\Omega(B/A)$ is
  generated over $F$ by all the
polynomials in  the  $b_i$ and the $b'_j$ which are symmetric under
$\Sym_3$.
In particular
\[
\begin{array}{lll}
\delta_{b}  & = & b_1b_2 b_3 +  b'_1b'_2 b_3 +
b'_1b_2b'_3 + b_1b'_{2}b'_3 \\
\omega_b & = & b_1 +b_2 +b_2\\
\end{array}
\]
are elements of $\Omega(B/A)$.
%
\begin{prop}
      \label{deltainLambda.prop}
      The element $\delta_{b}$ lies in the image of $\Delta(B)$ in
      $\Omega(B/A)$, and
      \[
      \gamma_{B}(\delta_{b})=\delta_{\overline{b}}.
      \]
      Moreover, the following conditions are equivalent:
      \begin{enumerate}
         \item[(a)]
         $b$ generates $B$ over $A$;
         \item[(b)]
         $\delta_{b}$ generates  $\Delta(B)$;
                     \end{enumerate}
      Similarly, the following conditions are equivalent:
      \begin{enumerate}
         \item[(a')]
         $\omega_{b}$ generates $\Omega(B/A)$ over $\Delta(B)$;
         \item[(b')]
         $(b-\overline{b})^{2}$ generates $A$.
      \end{enumerate}
\end{prop}

\begin{proof}
      It suffices to prove the assertions after scalar extension. We may
      therefore assume $B$ is split, and use the same notation as in
      Example~\ref{ext6.ex}. Let
      \[
      b=\beta_{1}f_{1}+\beta'_{1}f'_{1}+\beta_{2}f_{2}+\beta'_{2}f'_{2}+
      \beta_{3}f_{3}+\beta'_{3}f'_{3}
      \]
      with $\beta_{1}$, \ldots, $\beta'_{3}\in F$, hence
      \[
      \overline{b}=
      \beta'_{1}f_{1}+\beta_{1}f'_{1}+\beta'_{2}f_{2}+\beta_{2}f'_{2}+
      \beta'_{3}f_{3}+\beta_{3}f'_{3}.
      \]
      Computation yields
      \begin{equation}
         \label{delta.eq}
         \begin{split}
         \delta_{b}& =(\beta_{1}\beta_{2}\beta_{3}+
	\beta_{1}\beta'_{2}\beta'_{3}+
         \beta'_{1}\beta_{2}\beta'_{3}+
	\beta'_{1}\beta'_{2}\beta_{3})(g_{0}+g_{1}+g_{2}+g_{3})\\
         &\quad
         +(\beta'_{1}\beta'_{2}\beta'_{3}+\beta'_{1}\beta_{2}\beta_{3}+
	\beta_{1}\beta'_{2}\beta_{3}
         +\beta_{1}\beta_{2}\beta'_{3})(g'_{0}+g'_{1}+g'_{2}+g'_{3}),
         \end{split}
      \end{equation}
      proving $\delta_{b}\in\Delta_{B}$. Since $\overline{\gamma}_{B}$
      interchanges $g_{0}+\cdots+g_{3}$ and $g'_{0}+\cdots+g'_{3}$, it
      is clear that
      \[
      \overline{\gamma}_{B}(\delta_{b})=\delta_{\overline{b}}.
      \]
      We have $\delta_{b}\in F$ if and only if the coefficients of
      $g_{0}+\cdots+g_{3}$ and $g'_{0}+\cdots+g'_{3}$ in
      \eqref{delta.eq} above are equal, and this condition is equivalent
      to $\delta_{b}=\delta_{\overline{b}}$. On the other hand, $b$
      generates $B$ over $A$ if and only if $\beta_{i}\neq \beta'_{i}$ for
      $i=1$, $2$, $3$. Since
      \[
      \delta_{b}-\delta_{\overline{b}}=
      (\beta_{1}-\beta'_{1})(\beta_{2}-\beta'_{2})(\beta_{3}-\beta'_{3})
      (g_{0}+g_{1}+g_{2}+g_{3}-g'_{0}-g'_{1}-g'_{2}-g'_{3}),
      \]
      this condition holds if and only if
      $\delta_{b}\neq\delta_{\overline{b}}$. The equivalence of (a)
      and (b) is thus proved.

      To complete the proof, let
      \[
      \omega_{b}=u_{0}g_{0}+u_{1}g_{1}+u_{2}g_{2}+u_{3}g_{3}
      +u'_{0}g'_{0}+u'_{1}g'_{1}+u'_{2}g'_{2}+u'_{3}g'_{3}.
      \]
      This element generates $\Omega(B/A)$ over $\Delta(B)$ if and only
      if $u_{0}$, \ldots, $u_{3}$ are pairwise distinct and $u'_{0}$,
      \ldots, $u'_{3}$ are pairwise distinct. Computation yields
      \begin{align*}
         u_{0} =\beta_{1}+\beta_{2}+\beta_{3},\
         u_{1} =\beta_{1}+\beta'_{2}+\beta'_{3}, \
         u_{2} = \beta'_{1}+\beta_{2}+\beta'_{3},\
         u_{3} =\beta'_{1}+\beta'_{2}+\beta_{3},\\
         u'_{0} =\beta'_{1}+\beta'_{2}+\beta'_{3},\
         u'_{1} =\beta'_{1}+\beta_{2}+\beta_{3},\
         u'_{2} =\beta_{1}+\beta'_{2}+\beta_{3},\
         u'_{3} =\beta_{1}+\beta_{2}+\beta'_{3},
      \end{align*}
      and
      \begin{multline*}
         (u_{0}-u_{1})(u_{0}-u_{2})(u_{0}-u_{3})
         (u_{1}-u_{2})(u_{1}-u_{3})(u_{2}-u_{3})= \\
         \bigl((\beta_{2}-\beta'_{2})^{2}-(\beta_{3}-\beta'_{3})^{2}\bigr)
         \bigl((\beta_{1}-\beta'_{1})^{2}-(\beta_{2}-\beta'_{2})^{2}\bigr)
         \bigl((\beta_{1}-\beta'_{1})^{2}-(\beta_{3}-\beta'_{3})^{2}\bigr)=\\
         (u'_{0}-u'_{1})(u'_{0}-u'_{2})(u'_{0}-u'_{3})
         (u'_{1}-u'_{2})(u'_{1}-u'_{3})(u'_{2}-u'_{3}).
      \end{multline*}
      Therefore, $\omega_{b}$ generates $\Omega(B/A)$ over $\Delta(B)$
      if and only if $(\beta_{1}-\beta'_{1})^{2}$, 
$(\beta_{2}-\beta'_{2})^{2}$ and
      $(\beta_{3}-\beta'_{3})^{2}$ are pairwise distinct. Since
      \[
      (b-\overline{b})^{2}=
      (\beta_{1}-\beta'_{1})^{2}e_{1}+(\beta_{2}-\beta'_{2})^{2}e_{2}
      +(\beta_{3}-\beta'_{3})^{2}e_{3},
      \]
      this proves the equivalence of (a') and (b').
\end{proof}

Recall from \cite[p.~xviii]{BoI} the forms $T=T_{A/F}$, $S=S_{A/F}$ and
$N=N_{A/F}$ of degrees~$1$, $2$ and $3$ respectively on $A$, such
that the generic polynomial of every element $a\in  A$ has the form
\[
X^{3}-T(a)X^{2}+S(a)X-N(a)\in F[X].
\]
(The form $T$ is the \emph{trace}, and $N$ is the \emph{norm}.)
For $a \in A$, let $a_i = \varepsilon_i(A) \in \Sigma _3(A)$.
One has $T(a) = a_1+a_2+a_3$, $S(a)= a_1a_2 +a_1a_3 +a_2a_3$ and
$N(a)=a_1a_2a_3$.

Fix $b\in B$, and let
\[
\alpha_{1}=b+\overline{b}\in A,\qquad
\alpha_{2}=b\overline{b}\in A.
\]
Computation yields
\begin{equation}
      \label{normdelta.eq}
      \begin{split}
\delta_{b}+\delta_{\overline{b}}& =N(\alpha_{1}),\\
\delta_{b}\delta_{\overline{b}}& =
S(\alpha_{1}^{2}-2\alpha_{2})T(\alpha_{2})
-T(\alpha_{1}^{2}-2\alpha_{2})T\bigl((\alpha_{1}^{2}-2\alpha_{2})\alpha_{2}\bigr)\\
&\quad
+T\bigl((\alpha_{1}^{2}-2\alpha_{2})^{2}\alpha_{2}\bigr)+4N(\alpha_{2}).
\end{split}
\end{equation}

\begin{prop}
      \label{minpolambda.prop}
      If $\omega_{b}$ generates $\Omega(B/A)$ over $\Delta(B)$, its
      minimal polynomial is
      \begin{multline*}
         X^{4}-2T(\alpha_{1})X^{3}+
         \bigl(T(\alpha_{1}^{2})+2T(\alpha_{2})+3S(\alpha_{1})\bigr)X^{2}\\
         -\bigl(4\alpha_{b}+2\alpha_{\overline{b}}
+2T(\alpha_{1})T(\alpha_{2})+S(\alpha_{1})T(\alpha_{1})-3N(\alpha_{1})\bigr)X\\
         +(2\delta_{b}+\delta_{\overline{b}})T(\alpha_{1})
         +T(a_{1})^{2}T(\alpha_{2})-S(\alpha_{1})T(\alpha_{2})\\
-T(\alpha_{1}^{2}\alpha_{2})+T(\alpha_{2}^{2})-T(\alpha_{1})N(\alpha_{1})-2S(\alpha_{2}).
      \end{multline*}
\end{prop}
%
\begin{proof}
Use that in the split case, the four roots of the minimal polynomial
of~$\omega_b$ over $\Delta(A)$ are (with the notations of the proof
of Proposition~\ref{deltainLambda.prop}) the elements $u_i$, $i=0,\ldots,3$.
\end{proof}
%
If $\carac F\neq2$, we may simplify the results above by a specific
choice of generating element $b$.
  Let $b \in B$ be such that $\overline{b}= -b$ and assume that
$a=b^{2}\in A$ generates~$A$.
\begin{prop}
      \label{charneq2.prop}  ${\mathrm (\carac F \neq 2)}$
      With the notation above,
      \[
      \Delta(B)=F[\delta_{b}]
      \qquad\text{and}\qquad
      \Omega(B/A)=F[\delta_{b}, \omega_b]= \Delta(B)[\omega_{b}].
      \]
      Moreover, the minimal polynomial of $\delta_{b}$ over $F$ is
      \[
      X^{2}-16N(a),
      \]
      and the minimal polynomial of $\omega_{b}$ over $\Delta(B)$ is
      \[
      X^{4}-2T(a)X^{2}-2\delta_{b}X+T(a^{2})-2S(a).
      \]
\end{prop}

\begin{proof}
     Proposition~\ref{deltainLambda.prop} shows that $\delta_{b}$
     generates $\Delta(B)$ and that $\omega_{b}$ generates
     $\Omega(B/A)$ over $\Delta(B)$. This last fact can also be seen
     directly:
the algebra $\Omega(B/A)$ is generated  over $F$ by the elementary
symmetric functions in the~$b_i$; since
\[
2(b_1b_2 + b_1b_3 +b_2b_3) = \omega_b^2 -(b_1^2 +b_2^2 +b_3^2)=
\omega_b^2 - T(a)
\]
we have $\Omega(B/A)=F[\delta_{b}, \omega_b]= \Delta(B)[\omega_{b}]$.
The formula for the minimal polynomial of $\delta_{b}$ (resp.\
$\omega_{b}$) follows
from~\eqref{normdelta.eq} (resp.\ Proposition~\ref{minpolambda.prop}).
One can also repeat the proof
of Proposition~\ref{minpolambda.prop} with the special choice of $b$.
\end{proof}
%
\begin{cor}
      \label{charneq2.cor}  ${\mathrm (\carac F \neq 2)}$
      The discriminant $\Delta(B)$ is split if and only if $N(a)\in
      F^{\times2}$. If $N(a)=\nu^{2}$ for some $\nu\in F^\times$,
      then ${\mathcal S}(B/A)$ is generated over $F$ by an element whose
      minimal polynomial is
      \[
      X^{4}-2T(a)X^{2}-8\nu X+T(a^{2})-2S(a).
      \]
\end{cor}

\begin{proof}
      The first part readily follows from Proposition~\ref{charneq2.prop}.
      If $N(a)=\nu^{2}$, then $\Delta(B)\simeq F\times F$. Let $d$,
      $d'$ be the minimal idempotents of $\Delta(B)$. By
      Proposition~\ref{DeltainLambda.prop}, we have
      \[
      \Omega(B/A)\simeq{\mathcal S}(B/A)\times{\mathcal S}(B/A),
      \]
      and we may identify $d\Omega(B/A)$ and $d'\Omega(B/A)$
      with ${\mathcal S}(B/A)$. Since
      \[
      d\delta_{b}=\pm4\nu d
      \qquad\text{and}\qquad
      d'\delta_{b}=\mp4\nu d',
      \]
      the minimal polynomials of $d\omega_{b}$ and
      $d'\lambda_{b}$ are
      \[
      X^{4}-2T(a)X^{2}\pm8\nu X+T(a^{2})-2S(a).
      \]
\end{proof}


Assume now $\carac F=2$. Let $b$ be a generating element for $B$
over $A$ with $\overline{b}=b+1$, hence $b^{2}+b\in A$. Let
$a=b^{2}+b$, i.e., using the notation $\wp$ for the map $x\mapsto
x^{2}+x$,
\[
a=\wp(b)\in A.
\]
Assume moreover $a$ generates $A$, hence
\[
A=F[a]
\qquad\text{and}\qquad
B=F[\wp^{-1}(a)].
\]
In contrast with Proposition~\ref{charneq2.prop}, $\omega_{b}$ does
not generate $\Omega(B/A)$ since $(b-\overline{b})^{2}=1$ does not
generate $A$ (see Proposition~\ref{deltainLambda.prop}). One could
take for example  $\omega_{ab}$ as a generator of $\Omega(B/A)$,
since $(ab -\overline{ab})^2= a^2$ (assuming that $a^2$ also generates $A$,
which for a cubic \'etale algebra is in general the case).
However a simpler minimal
polynomial is obtained for the element
\[
\mu_{b}=b_1b_2 +b_1b_3 +b_2b_3
\]
(with the notations of Equation~\ref{bundeux.eq}).
Moreover we have \[
\delta_b = b_1 +b_2 +b_3.
\]
if $\carac F =2$ and $\overline{b} +b =1$.
%
\begin{prop}
      \label{char2.prop} ${\mathrm (\carac F = 2)}$
      With the notation above,
      \[
      \Delta(B)=F[\delta_{b}]
      \qquad\text{and}\qquad
      \Omega(B/A)=F[\delta_b,\mu_b]=\Delta(B)[\mu_{b}].
      \]
      Moreover, the minimal polynomial of $\delta_{b}$ over $F$ is
      \[
      X^{2}+X+T(a),
      \]
      and the minimal polynomial of $\mu_{b}$ over $\Delta(B)$ is
      \[
      X^{4}+X^{3}+(\delta_{b}^{2}+1)X^{2}+(\delta_{b}^{2}+S(a)+1)X
      +(\delta_{b}+S(a)+1)S(a)+N(a).
      \]
\end{prop}

\begin{proof}
      Since $b$ generates $B$ over $A$,
      Proposition~\ref{deltainLambda.prop} shows that
      $\Delta(B)=F[\delta_{b}]$, and Equations~\eqref{normdelta.eq} yield
      the minimal polynomial of $\delta_{b}$.

      To prove the rest, we extend scalars and assume $B$ is split.
      Using the same notation as in Example~\ref{ext6.ex}, we have
      \[
      b=\beta_{1}f_{1}+(\beta_{1}+1)f'_{1}
      +\beta_{2}f_{2}+(\beta_{2}+1)f'_{2}+\beta_{3}f_{3}+(\beta_{3}+1)f'_{3}
      \]
      for some $\beta_{1}$, $\beta_{2}$, $\beta_{3}\in F$. Then, letting
      \[
      d=g_{0}+g_{1}+g_{2}+g_{3}
      \qquad\text{and}\qquad
      d'=g'_{0}+g'_{1}+g'_{2}+g'_{3}
      \]
      be the minimal idempotents of $\Delta(B)\subset\Omega(B/A)$, we
      have
      \[
\delta_{b}=(\beta_{1}+\beta_{2}+\beta_{3})d+(\beta_{1}+\beta_{2}+\beta_{3}+1)d',
      \]
      and
      \[
      \mu_{b}=v_{0}g_{0}+v'_{0}g'_{0}+v_{1}g_{1}+v'_{1}g'_{1}
      +v_{2}g_{2}+v'_{2}g'_{2}+v_{3}g_{3}+v'_{3}g'_{3}
      \]
      where
    \begin{align*}
         v_{0}& =\beta_{1}\beta_{2}+\beta_{1}\beta_{3}+\beta_{2}\beta_{3},&
         v'_{0}& =v_{0}+1,\\
         v_{1}& 
=\beta_{1}\beta_{2}+\beta_{1}\beta_{3}+\beta_{2}\beta_{3}+\beta_{2}+\beta_{3}+1,&
         v'_{1}& =v_{1}+1,\\
         v_{2}& 
=\beta_{1}\beta_{2}+\beta_{1}\beta_{3}+\beta_{2}\beta_{3}+\beta_{1}+\beta_{3}+1,&
         v'_{2}& =v_{2}+1,\\
         v_{3}& 
=\beta_{1}\beta_{2}+\beta_{1}\beta_{3}+\beta_{2}\beta_{3}+\beta_{1}+\beta_{2}+1,&
         v'_{3}& =v_{3}+1.
      \end{align*}
      Then
      \[
      \prod_{0\leq i<j\leq3}(v_{i}-v_{i})=
      \prod_{1\leq i<j\leq 3}(\beta_{i}-\beta_{j})^{2}.
      \]
      Since $b$ generates $B$ over $A$, the elements $\beta_{1}$, $\beta_{2}$,
      $\beta_{3}$ are pairwise distinct, hence this equality shows that
      $v_{0}$, \ldots, $v_{3}$ are pairwise distinct. Similarly,
      $v'_{0}$, \ldots, $v'_{3}$ are pairwise distinct, hence $\mu_{b}$
      generates $\Omega(B/A)$ over $\Delta(B)$. We have
      \begin{align*}
         T(a)& 
=(\beta_{1}^{2}+\beta_{1})+(\beta_{2}^{2}+\beta_{2})+(\beta_{3}^{2}+\beta_{3}),\\
         S(a)& =(\beta_{1}^{2}+\beta_{1})(\beta_{2}^{2}+\beta_{2})
         +(\beta_{1}^{2}+\beta_{1})(\beta_{3}^{2}+\beta_{3})
         +(\beta_{2}^{2}+\beta_{2})(\beta_{3}^{2}+\beta_{3}),\\
         N(a) 
&=(\beta_{1}^{2}+\beta_{1})(\beta_{2}^{2}+\beta_{2})(\beta_{3}^{2}+\beta_{3}),
      \end{align*}
      and brute force computation shows that $v_{0}$, $v_{1}$, $v_{2}$
      and $v_{3}$ are roots of
      \begin{multline*}
      X^{4}+X^{3}+(\beta_{1}^{2}+\beta_{2}^{2}+\beta_{3}^{2}+1)X^{2}
      +(\beta_{1}^{2}+\beta_{2}^{2}+\beta_{3}^{2}+S(a)+1)X\\
      +(\beta_{1}+\beta_{2}+\beta_{3}+S(a)+1)S(a)+N(a).
      \end{multline*}
      Similarly, $v'_{0}$, $v'_{1}$, $v'_{2}$ and $v'_{3}$ are roots of
      \begin{multline*}
      X^{4}+X^{3}+(\beta_{1}^{2}+\beta_{2}^{2}+\beta_{3}^{2})X^{2}
      +\bigl(\beta_{1}^{2}+\beta_{2}^{2}+\beta_{3}^{2}+S(a)\bigr)X\\
      +\bigl(\beta_{1}+\beta_{2}+\beta_{3}+S(a)\bigr)S(a)+N(a),
      \end{multline*}
      hence the proof is complete.
\end{proof}

\begin{cor}
      \label{char2.cor} ${\mathrm (\carac F = 2)}$
      With the same notation as in Proposition~\ref{char2.prop}, the
      discriminant $\Delta(B)$ is split if and only if $T(a)\in\wp(F)$.
      If $T(a)=\wp(\nu)$ for some $\nu\in F$, then ${\mathcal S}(B/A)$ is
      generated over $F$ by an element whose minimal polynomial is
      \[
      X^{4}+X^{3}+\nu^{2}X^{2}+\bigl(\nu^{2}+S(a)\bigr)X+
      \bigl(\nu+S(a)\bigr)S(a)+N(a).
      \]
\end{cor}

\begin{proof}
      The first part readily follows from Proposition~\ref{char2.prop}.
      If $T(a)=\wp(\nu)$, then $\Delta(B)=F\times F$. Let $d$, $d'$ be
      the minimal idempotents of $\Delta(B)\subset\Omega(B/A)$. We may
      assume $d=\delta_{b}+\nu$ and $d'=\delta_{b}+\nu+1$, hence
      \[
      d\delta_{b}=(\nu+1)d.
      \]
      As in the proof of Corollary~\ref{charneq2.cor}, we may identify
      $d\Omega(B/A)$ and $d'\Omega(B/A)$ with ${\mathcal S}(B/A)$, and it
      follows from Proposition~\ref{char2.prop} that the minimal
      polynomial of $d\mu_{b}$ is as stated.
\end{proof}

Combining the results of subsections \ref{quartal.subsec} and
\ref{quadextet.subsec} we get:
%
\begin{prop}\label{represet4.prop}
Let $Q$ be a quartic \'etale algebra.
\begin{enumerate}
\item[(a)] If $\carac F\neq 2$, let $x \in Q$ be a generator
such that $T_{Q/F}(x)=0$. There exists an isomorphism
$\phi_1 \colon\, Q \isom {\mathcal S}\bigl(\Lambda_2(Q)/\resolv(Q)\bigr)$ such
that
  $\phi_1(x) = \omega_{\lambda_x}$.
\item[(b)]
If $\carac F= 2$, let $x \in Q$ be a generator
such that $T_{Q/F}(x)=1$. There exists an isomorphism
$\phi_2 \colon\, Q \isom {\mathcal S}\bigl(\Lambda_2(Q)/\resolv(Q)\bigr)$ such
that
  $\phi_2(x) =  \mu_{\lambda_x}$.
\end{enumerate}
\end{prop}

\section{Special actions on four elements}
\label{special.sec}

As in the preceding sections, $\Gamma$ denotes a profinite group. The
constructions on $\Gamma$-sets given in \S\ref{basicset.subsec} take a
special form when the $\Gamma$-action has some particular properties.
For instance, if the action on a set $X$ is not transitive, then the
orbits $X_{1}$, \ldots, $X_{r}$ under $\Gamma$ yield a $\Gamma$-set
decomposition
\[
X=X_{1}\coprodd\ldots\coprodd X_{r}.
\]
Even if the $\Gamma$-action on $X$ is transitive, the induced action
on $\Sigma_{n}(X)$ may not be transitive.

\begin{prop}
     \label{special.prop}
     Let $X$ be a $\Gamma$-set with $\vert X\vert =n$, and let
     $\alpha\colon\,\Gamma\to\Sym_{X}$ be the action of $\Gamma$. If
     $\bigl(\Sym_{X}:\alpha(\Gamma)\bigr)=r$, there is a $\Gamma$-set
     decomposition
     \begin{equation}
	\label{Sigmadec.eq}
	\Sigma_{n}(X)=\Omega_{1}\coprodd\ldots\coprodd\Omega_{r}.
     \end{equation}
     Each $\Gamma$-set $\Omega_{i}$ is a $G_{i}$-torsor for some
     subgroup $G_{i}\subset\Sym_{n}$ isomorphic to $\alpha(\Gamma)$,
     and the subgroups $G_{i}$ are conjugate in $\Sym_{n}$. Moreover,
     the $\Gamma$-sets $\Omega_{1}$, \ldots, $\Omega_{r}$ are
     isomorphic.
\end{prop}

\begin{proof}
     The $\Gamma$-orbits of $\Sigma_{n}(X)$ yield the decomposition
     \eqref{Sigmadec.eq}. To see that each $\Omega_{i}$ is a
     $G_{i}$-torsor, recall that for
     $(x_{1},\ldots,x_{n})\in\Sigma_{n}(X)$, $\gamma\in\Gamma$ and
     $\sigma\in\Sym_{n}$, we have by definition
     \[
     \gamma(x_{1},\ldots,x_{n})=\bigl(\alpha(\gamma)(x_{1}),\ldots,
     \alpha(\gamma)(x_{n})\bigr)
     \text{ and }
     (x_{1},\ldots,x_{n})^\sigma=(x_{\sigma(1)},\ldots,x_{\sigma(n)}).
     \]
     For each $\gamma\in\Gamma$, there is a unique $\sigma\in\Sym_{n}$
     such that
     \[
     \gamma(x_{1},\ldots,x_{n})=(x_{1},\ldots,x_{n})^\sigma,
     \]
     and the map $\gamma\mapsto\sigma$ defines a homomorphism
     $\Gamma\to\Sym_{n}$ which depends on the choice of
     $(x_{1},\ldots,x_{n})$ and factors through $\alpha$ to yield an
     injection $\alpha(\Gamma)\hookrightarrow\Sym_{n}$. If
     $(x_{1},\ldots,x_{n})\in\Omega_{i}$, let $G_{i}\subset\Sym_{n}$ be
     the image of this map. Then $\Omega_{i}$ is a $G_{i}$-torsor.
     Moreover, if $(x_{1},\ldots,x_{n})\in\Omega_{i}$ and
     $(y_{1},\ldots,y_{n})\in\Omega_{j}$, there exists
     $\sigma\in\Sym_{n}$ such that
     \[
     (y_{1},\ldots,y_{n})=(x_{1},\ldots,x_{n})^\sigma.
     \]
     Conjugation by $\sigma$ maps $G_{j}$ to $G_{i}$ and the action
     of $\sigma$ defines an isomorphism of $\Gamma$-sets
     $\Omega_{i}\isom\Omega_{j}$.
\end{proof}

\begin{remark}
     The $\Sym_{n}$-torsor $\Sigma_{n}(X)$ can be obtained as the
     induced torsor $\Ind_{G_{1}}^{\Sym_{n}}\Omega_{1}$, by mimicking
     the construction in \cite[(18.17)]{BoI}.
\end{remark}

Note that if the $\Gamma$-action on $X$ is transitive, then the map
$X\xleftarrow{\pi_{n}}\Sigma_{n}(X)$ restricts to each $\Omega_{i}$ to
define a covering $X\leftarrow\Omega_{i}$. Extending
Definition~\ref{Galcloset.def}, this covering may be regarded as a
$G_{i}$-Galois closure of $X$.
\medbreak
\par
Taking for $\Gamma$ the absolute Galois group of a field $F$, and
using the anti-equivalence $\Et_{F}\equiv\Set_{\Gamma}$ of
Section~\ref{etal.sec}, we may adapt the construction above to
\'etale algebras. Disjoint unions of $\Gamma$-sets correspond to
direct product decompositions of algebras, hence an \'etale
$F$-algebra is a field if and only if the $\Gamma$-action on
$\funcX(E)$ is transitive. If $\dim E=n$,
Proposition~\ref{special.prop} thus yields a direct product
decomposition of the $\Sym_{n}$-Galois closure $\Sigma_{n}(E)$ into
isomorphic fields
\[
\Sigma_{n}(E)\simeq L_{1}\times\cdots\times L_{r}.
\]
Each $L_{i}$ is a Galois extension of $F$ with Galois group
$G_{i}\subset\Sym_{n}$ isomorphic to the image of the action
$\Gamma\to\Sym_{\funcX(E)}$. If $E$ is a field, each $L_{i}$ can be
regarded as a Galois closure of $E/F$, see \cite[(18.22)]{BoI}.
\medbreak
\par
In the rest of this section, we consider the particular case where
$n=4$. To determine the various possibilities for the image of the
action $\Gamma\to\Sym_{4}$, we list the subgroups of $\Sym_{4}$.

\begin{prop}
     In the symmetric group $\Sym_{4}$,
     \begin{itemize}
	\item
	the alternating group $\Alt_{4}$ is the unique subgroup of order~$12$;
	\item
	there are four subgroups of order~$6$; they are conjugate to
	$\Sym_{3}$;
	\item
	there are three subgroups of order~$8$; they are pairwise conjugate
	and isomorphic to the dihedral group $D_{4}$.
     \end{itemize}
     Moreover, every proper subgroup of $\Sym_{4}$ is contained in at
     least one of the subgroups listed above.
\end{prop}

\begin{proof}
     Any subgroup of index~$2$ in $\Sym_{4}$ must contain all the
     Sylow $3$-subgroups of $\Sym_{4}$. Since these Sylow subgroups
     are generated by the cycles of length~$3$, the first claim is
     clear.

     A subgroup of order~$6$ in $\Sym_{4}$ cannot be transitive on
     $\{1,2,3,4\}$. On the other hand, it has an orbit of $3$ elements
     since it contains a Sylow $3$-subgroup, hence it must be the
     isotropy group of one of $1$, $2$, $3$, or $4$.

     The dihedral group $D_{4}$ acts on the four vertices of a square,
     hence it may be considered as a subgroup of $\Sym_{4}$. It is then
     identified with a $2$-Sylow subgroup of $\Sym_{4}$, and all the
     $2$-Sylow subgroups are conjugate.

     Finally, let $G\subset\Sym_{4}$ be a subgroup. If its order is
     divisible by~$3$, then it is $3$, $6$ or $12$, hence $G$ is
     contained in $\Alt_{4}$ or in a conjugate of $\Sym_{3}$. If its
     order is a power of $2$, then $G$ is contained in a $2$-Sylow
     subgroup.
\end{proof}

In the following subsections, we examine the additional information
on a $\Gamma$-set $X$ with $\vert X\vert=4$ (or on a quartic \'etale
$F$-algebra $Q$) when the $\Gamma$-action factors through a subgroup
$\Sym_{3}$ or $D_{4}$. We then collect the information to obtain a
classification of quartic \'etale $F$-algebras in
\S\ref{classif.subsec}.

\subsection{Action through $\Sym_{3}$}
\label{Sym3.subsec}

Suppose first the action of $\Gamma$ leaves an element $x \in X$ invariant.
It then preserves a disjoint union decomposition
\[
X= \{x\}\coprodd R,
\]
where $\vert R\vert  =3$. The $2$-element subsets of $X$ containing $x$
are in one-to-one correspondence with $R$, hence $\Lambda_2(X)$ decomposes as
\[
\Lambda_2(X) = R \coprodd\Lambda_2(R).
\]
Moreover, the ``complementary subset'' involution $\gamma_X$ on $\Lambda_2(X)$
interchanges $R$ and $\Lambda_2(R)$ and defines an isomorphism
$R \simeq \Lambda_2(R)$. Therefore, we have canonical isomorphisms
\[
\Lambda_2(X) \simeq R \coprodd R \quad \text{and} \quad \resolv{(X)} \simeq
R.
\]
(See also \S\ref{cohom4.subsec}.)
%

Assuming $\Gamma$ is the absolute Galois group of a field $F$, we may
translate the results above in the framework of \'etale $F$-algebras.

\begin{prop}
     \label{etsym3.prop}
     Let $Q$ be a quartic \'etale $F$-algebra. If the $\Gamma$-action
     on $\funcX(Q)$ factors through a subgroup
     $\Sym_{3}\subset\Sym_{\funcX(Q)}$, then there is a cubic field
     extension $L/F$ such that
     \[
     Q\simeq F\times L,\qquad
     \Lambda_{2}(Q)\simeq L\times L,
     \quad\text{and}\quad
     \resolv(Q)\simeq L.
     \]
     Moreover, the following conditions are equivalent:
     \begin{enumerate}
	\item[(a)]
	the $\Gamma$-action factors through a cyclic subgroup $C_{3}$;
	\item[(b)]
	the extension $L/F$ is Galois (hence cyclic);
	\item[(c)]
	$\Delta(Q)\simeq F\times F$.
     \end{enumerate}
\end{prop}

\subsection{Action through $D_{4}$}
\label{Dfour.subsec}

Suppose now that the action of $\Gamma$ factors through a Sylow
$2$-subgroup of $\Sym_{X}$, i.e.\ through a dihedral subgroup
$D_{4}$. Since the Sylow $2$-subgroups of $\Sym_{X}$ are the isotropy
groups of partitions of $X$ into $2$-element subsets, there is such a
partition which is invariant under $\Gamma$. This observation
characterizes the case where $\Gamma$ acts through $D_{4}$:

\begin{prop}
     \label{charDfour.prop}
     For a set $X\in\Set_{\Gamma}^{4}$, the following conditions are
     equivalent:
     \begin{enumerate}
	\item[(a)]
	the $\Gamma$-action factors through a Sylow $2$-subgroup of
	$\Sym_{X}$;
	\item[(b)]
	the $\Gamma$-action leaves a point of $\resolv(X)$ fixed;
	\item[(c)]
	$\resolv(X)\simeq\{*\}\coprodd\Delta(X)$;
	\item[(d)]
	$X$ is a double covering of a set of two elements, i.e.\ there exists
	a map $(D\leftarrow X)\in\Cov_{\Gamma}^{2\wr2}$.
     \end{enumerate}
\end{prop}

\begin{proof}
     The points of $\resolv(X)$ are the partitions of $X$ into
     $2$-element subsets, hence (a)$\iff$(b). The implication
     (c)$\Rightarrow$(b) is clear, and (b)$\Rightarrow$(c) follows
     from Proposition~\ref{discres.prop}. If
     $D=\bigl\{\{x_{1},x_{2}\},\{x_{3},x_{4}\}\bigr\}\in\resolv(X)$ is
     fixed under $\Gamma$, then the canonical map $D\leftarrow X$ which
     carries $x_{1}$, $x_{2}$ to $\{x_{1},x_{2}\}$ and $x_{3}$, $x_{4}$
     to $\{x_{3},x_{4}\}$ is a double covering, hence
     (b)$\Rightarrow$(d). Finally, (d)$\Rightarrow$(a) follows from
     $\Sym_{2}\wr\Sym_{2}\simeq D_{4}$.
\end{proof}

The following proposition establishes the existence of a ``dual''
$\Gamma$-set $\check X$:

\begin{prop}
     \label{dual.prop}
     If the equivalent conditions of Proposition~\ref{charDfour.prop}
     hold, then there exists a $\Gamma$-set $\check
     X\in\Set_{\Gamma}^{4}$, with $\Gamma$-action through a Sylow
     $2$-subgroup of $\Sym_{\check X}$, such that
     \begin{align*}
	\Lambda_{2}(X) & \simeq \Delta(\check X)\coprodd\check X,&
	\Lambda_{2}(\check X)& \simeq \Delta(X)\coprodd X,\\
	\resolv(X)& \simeq\{\Delta(\check X)\}\coprodd\Delta(X),&
	\resolv(\check X) & \simeq\{\Delta(X)\}\coprodd
	\Delta(\check X).
     \end{align*}
     Moreover, $X$ is a double covering of $\Delta(\check X)$, and
     $\check X$ is a double covering of $\Delta(X)$.

     If the $\Gamma$-action on $\Delta(X)$ is not trivial, the
     $\Gamma$-set $\check X$ is canonically determined. If the
     $\Gamma$-actions on $\Delta(X)$ and $\Delta(\check X)$ are not
     trivial, then there is a canonical isomorphism
     \[
     \Check{\Check X}= X.
     \]
\end{prop}

\begin{proof}
     Let $D\in\resolv(X)$ be a fixed point of $\Gamma$. Define $\check
     X$ as the complementary subset in $\Lambda_{2}(X)$ of the fiber
     $\varepsilon^{-1}(D)$ under the canonical map
     $\resolv(X)\xleftarrow{\varepsilon}\Lambda_{2}(X)$. The set
     $\check X$ is thus canonically determined if $\Gamma$ has a unique
     fixed point $D\in\resolv(X)$, or, equivalently by
     Proposition~\ref{charDfour.prop}, if the $\Gamma$-action on
     $\Delta(X)$ is not trivial.

     We proceed to prove that $\check X$ satisfies the stated
     properties. To clarify the discussion, we use geometric language.
     If $D=\bigl\{\{x_{1},x_{2}\},\{x_{3},x_{4}\}\bigr\}$, we identify
     $X$ with the set of vertices of a square, letting $\{x_{1},x_{2}\}$ and
     $\{x_{3},x_{4}\}$ be the pairs of opposite vertices. We
     may thus identify $D$ to the set of diagonals of the square, and
     we have a decomposition
     \begin{equation}
	\label{decompX.eq}
     \Lambda_{2}(X)=D\coprodd\check X
     \end{equation}
     where $\check X$ is the set of pairs of adjacent vertices, which
     may be identified with the set of edges of the square. (Note
     that $\check X$ may also be viewed as the \emph{dual square} of
     $X$ in the sense of polytope theory.) There is a ``dual''
     decomposition
     \begin{equation}
	\label{decompXcheck.eq}
	\Lambda_{2}(\check X)=M\coprodd X,
     \end{equation}
     where $X$ is identified with the set of pairs of adjacent edges
     (by mapping every such pair to their common vertex) and $M$ is
     the set of pairs of parallel edges, which may be identified with
     the medians of the square. The ``complementary subset''
     involutions $\gamma_{X}$ and $\gamma_{\check X}$ preserve the
     decompositions \eqref{decompX.eq} and \eqref{decompXcheck.eq},
     and the set of orbits of $\check X$ (resp.\ $X$) under
     $\gamma_{X}$ (resp.\ $\gamma_{\check X}$) can be identified with
     $M$ (resp.\ $D$), hence
     \begin{equation}
	\label{resolvX.eq}
	\resolv(X)=\{D\}\coprodd M
	\qquad\text{and}\qquad
	\resolv(\check X)=\{M\}\coprodd D,
     \end{equation}
     and there are natural maps $D\leftarrow X$ and $M\leftarrow\check
     X$ which show $X$ and $\check X$ are double coverings of $D$ and $M$
     respectively. By Proposition~\ref{discres.prop},
     Equations~\eqref{resolvX.eq} yield canonical isomorphisms
     \[
     M=\Delta(X)\qquad\text{and}\qquad
     D=\Delta(\check X).
     \]

     If the $\Gamma$-action on $\Delta(\check X)$ is not trivial, then
     $M$ is the unique fixed point of $\resolv(\check X)$, and
     $X\subset\Lambda_{2}(\check X)$ is the complementary subset of the
     fiber of $M$ under the canonical map $\resolv(\check
     X)\leftarrow\Lambda_{2}(\check X)$, hence $\Check{\Check X}=X$.
     This completes the proof.
\end{proof}

We may use the $\Gamma$-set $\check X$ to obtain information on the
$\Gamma$-action on $X$, as follows:

\begin{prop}
     \label{Cfour.prop}
     Let $X\in\Set_{\Gamma}^{4}$ be a $\Gamma$-set satisfying the
     equivalent properties of Proposition~\ref{charDfour.prop}, and
     suppose the $\Gamma$-action on $\Delta(X)$ is not trivial, so the
     dual set $\check X$ is uniquely determined. The following
     properties are equivalent:
     \begin{enumerate}
	\item[(a)]
	the $\Gamma$-action on $X$ factors through a cyclic subgroup $C_{4}$;
	\item[(b)]
	$\check X\simeq X$;
	\item[(c)]
	$\Delta(\check X)\simeq\Delta(X)$.
     \end{enumerate}
\end{prop}

\begin{proof}
     If the action of $\Gamma$ factors through $C_{4}$, we may regard
     $X$ as the set of vertices of an \emph{oriented} square, and use
     the orientation to define a canonical isomorphism $X\isom\check
     X$, proving (a)$\Rightarrow$(b). Since the implication
     (b)$\Rightarrow$(c) is clear, it only remains to prove
     (c)$\Rightarrow$(a). If the image of $\Gamma$ under the action
     contains the Vierergruppe $\Vier_{X}$, then there is an element
     in $\Gamma$ which acts trivially on $M$ and non-trivially on $D$,
     hence $\Delta(X)\not\simeq\Delta(\check X)$. Similarly, if some
     element of $\Gamma$ acts by a single transposition on $X$, then
     it acts trivially on $D$ and non-trivially on $M$, hence
     $\Delta(X)\not\simeq\Delta(\check X)$. Therefore, (c) implies that
     the image of the action of $\Gamma$ contains at most cycles of
     length~$4$ and one element of $\Vier_{X}$.
\end{proof}

\begin{remark}
     The $\Gamma$-set $D*M=\Delta(X)*\Delta(\check X)$ can be identified with
     the set of orientations of the square.
\end{remark}

For the following proposition, recall that the dihedral group $D_{4}$
contains two non-conjugate elementary abelian subgroups $C_{2}\times
C_{2}$. Viewing $D_{4}$ as a subgroup of $\Sym_{4}$, one of these
subgroups is $\Vier$ ($=D_{4}\cap\Alt_{4}$). The other one is
generated by two disjoint transpositions; it is not transitive on
$\{1,2,3,4\}$.

\begin{prop}
     \label{elemab.prop}
     Let $X\in\Set_{\Gamma}^{4}$ be a $\Gamma$-set satisfying the
     equivalent properties of Proposition~\ref{charDfour.prop}, and
     suppose the $\Gamma$-action on $\Delta(X)$ is not trivial, so the
     dual set $\check X$ is uniquely determined. The following
     properties are equivalent:
     \begin{enumerate}
	\item[(a)]
	the $\Gamma$-action on $X$ factors through an elementary abelian
	subgroup $C_{2}\times C_{2}\neq\Vier_{X}$;
	\item[(b)]
	the $\Gamma$-action on $\check X$ factors through $\Vier_{\check X}$;
	\item[(c)]
	the $\Gamma$-action on $\Delta(\check X)$ is trivial.
     \end{enumerate}
\end{prop}

The proof is left to the reader.
\medbreak
\par
Finally, we consider the case where the $\Gamma$-action on $X$
factors through $\Vier_{X}$.

\begin{prop}
     \label{Vier.prop}
     For a $\Gamma$-set $X\in\Set_{\Gamma}^{4}$, the following
     conditions are equivalent:
     \begin{enumerate}
	\item[(a)]
	the $\Gamma$-action on $X$ factors through the Vierergruppe
	$\Vier_{X}$;
	\item[(b)]
	the $\Gamma$-action on $\resolv(X)$ is trivial;
	\item[(c)]
	the $\Gamma$-set $\Lambda_{2}(X)$ has a decomposition into
	$2$-element subsets stable under the canonical involution of
	$\Lambda_{2}(X)/\resolv(X)$,
	\[
	\Lambda_{2}(X)=D_{1}\coprodd D_{2}\coprodd D_{3};
	\]
	\item[(d)]
	$X$ satisfies the equivalent conditions of
	Proposition~\ref{charDfour.prop} and $\Gamma$ acts trivially on
	$\Delta(X)$.
     \end{enumerate}
     Moreover, if these conditions hold, then the $\Gamma$-action on
     $D_{1}*D_{2}*D_{3}$ is trivial.
\end{prop}

\begin{proof}
     The Vierergruppe can be defined as the subgroup of $\Sym_{X}$
     which leaves invariant all the partitions of $X$ into $2$-element
     subsets, hence (a)$\iff$(b). The equivalence of (b) and (c) is
     clear: take for $D_{1}$, $D_{2}$ and $D_{3}$ the fibers of the
     canonical map $\resolv(X)\leftarrow\Lambda_{2}(X)$. The
     equivalence (b)$\iff$(d) readily follows from
     Proposition~\ref{charDfour.prop}.

     If the equivalent conditions of the proposition hold, then the
     set $\check X$ of Proposition~\ref{dual.prop} can be arbitrarily
     chosen as $D_{1}\coprodd D_{2}$, $D_{1}\coprodd D_{3}$ or
     $D_{2}\coprodd D_{3}$. If we choose $\check X=D_{1}\coprodd
     D_{2}$, Proposition~\ref{dual.prop} yields $\Delta(\check
     X)=D_{3}$. On the other hand, it is easily checked that
     \[
     \Delta(D_{1}\coprodd D_{2})\simeq D_{1}*D_{2},
     \]
     hence $D_{1}*D_{2}\simeq D_{3}$ and therefore the $\Gamma$-action
     on $D_{1}*D_{2}*D_{3}$ is trivial.
\end{proof}

Taking for $\Gamma$ the absolute Galois group of a field $F$, we may
translate in terms of \'etale $F$-algebras the results of this
subsection, by using the anti-equivalence $\Et_{F}\equiv\Set_{\Gamma}$
of \S\ref{etal.sec}. By a \emph{quartic $2$-algebra} we mean an
\'etale algebra which is a quadratic extension of a quadratic \'etale
algebra. These algebras can be characterized through
Proposition~\ref{charDfour.prop}:

\begin{prop}
     \label{quartic2.prop}
     For a quartic $F$-algebra $Q$, the following conditions are
     equivalent:
     \begin{enumerate}
	\item[(a)]
	the $\Gamma$-action on $\funcX(Q)$ factors through a Sylow
	$2$-subgroup of $\Sym_{\funcX(Q)}$;
	\item[(b)]
	$\resolv(Q)$ is not a field;
	\item[(c)]
	$\resolv(Q)\simeq F\times\Delta(Q)$;
	\item[(d)]
	$Q$ is a quartic $2$-algebra.
     \end{enumerate}
\end{prop}

Proposition~\ref{dual.prop} proves for every quartic $2$-algebra $Q$
the existence of a ``dual'' quartic $2$-algebra $\check Q$, which is
canonically determined if $\Delta(Q)$ is not split. This algebra is a
quadratic extension of $\Delta(Q)$, and $Q$ is a quadratic extension
of $\Delta(\check Q)$. Moreover, $Q$ and $\check Q$ satisfy the
following relations:
\begin{align*}
     \Lambda_{2}(Q) & \simeq \Delta(\check Q)\times\check Q,&
     \Lambda_{2}(\check Q)& \simeq \Delta(Q)\times Q,\\
     \resolv(Q)& \simeq F\times\Delta(Q),&
     \resolv(\check Q) & \simeq F\times
     \Delta(\check Q).
\end{align*}

We record a few special cases:

\begin{prop}
     \label{special2alg.prop}
     Let $Q$ be a quartic $2$-algebra over $F$.
     \begin{enumerate}
	\item[(1)]
	If $Q$ is a cyclic field extension of $F$, then $\check Q\simeq Q$,
	hence $Q$ is a quadratic extension of $\Delta(Q)$, and
	\[
	\Lambda_{2}(Q)\simeq\Delta(Q)\times Q,
	\qquad
	\resolv(Q)\simeq F\times\Delta(Q).
	\]
	\item[(2)]
	If $Q=K_{1}\times K_{2}$, where $K_{1}$ and $K_{2}$ are
	non-isomorphic quadratic $F$-algebras, then $\check Q\simeq
	K_{1}\otimes_{F}K_{2}$, $\Delta(Q)\simeq K_{1}*K_{2}$, and
	\[
	\Lambda_{2}(Q)\simeq F\times F\times(K_{1}\otimes_{F}K_{2}),
	\qquad
	\resolv(Q)\simeq F\times(K_{1}*K_{2}).
	\]
	\item[(3)]
	If $Q=K_{1}\otimes_{F}K_{2}$, where $K_{1}$ and $K_{2}$ are
	quadratic field extensions of $F$, then one may take
	$\check Q=K_{1}\times
	K_{2}$, or $K_{1}\times (K_{1}*K_{2})$, or $K_{2}\times(K_{1}*K_{2})$.
	Moreover, $\Delta(Q)$ is split, and
	\[
	\Lambda_{2}(Q)\simeq K_{1}\times K_{2}\times(K_{1}*K_{2}),
	\qquad
	\resolv(Q)\simeq F\times F\times F.
	\]
     \end{enumerate}
\end{prop}

These results are easily derived from Propositions~\ref{Cfour.prop},
\ref{elemab.prop} and \ref{Vier.prop}. Note that split quadratic
algebras are allowed in (2), and that the case where $Q=K\times K$
for some quadratic field extension $K$ of $F$ is covered by (3) since
$K\times K\simeq K\otimes_{F}K$.

\medbreak
\par
Since $\Lambda_{2}(Q)\simeq\Delta(\check Q)\times\check Q$, we may use
the computations of \S\ref{quartal.subsec} to give an explicit
description of $\check Q$.

\begin{prop}
     \label{quarticexpl.prop}
     Let $Q$ be a quartic $2$-algebra over $F$, and let $K\subset Q$
     be a quadratic \'etale $F$-algebra. Denote by $\invo$ the
     canonical involution of $K$ over $F$.
     \begin{enumerate}
	\item[(1)]
	Suppose $\carac F\neq 2$ and $Q=K(\sqrt{y})$ where $y\in K$
	generates $K$. Then
	\[
	\Delta(Q)\simeq F\bigl(\sqrt{N_{K/F}(y)}\bigr)
	\quad\text{and}\quad
	\check Q\simeq F(\sqrt{y}+\sqrt{\overline{y}})
	\quad(\text{and }\Delta(\check Q)\simeq K).
	\]
	\item[(2)]
	Suppose $\carac F=2$ and $Q=K\bigl(\wp^{-1}(y)\bigr)$ where $y\in K$
	generates $K$. Then
	\[
	\Delta(Q)\simeq F\bigl(\wp^{-1}(T_{K/F}(y))\bigr)
	\quad\text{and}\quad
	\check Q\simeq F\bigl(\wp^{-1}(y)\wp^{-1}(\overline{y})\bigr)
	\quad(\text{and }\Delta(\check Q)\simeq K).
	\]
     \end{enumerate}
\end{prop}

In the proof below, we write simply $T(y)$, $N(y)$ for $T_{K/F}(y)$
and $N_{K/F}(y)$.

\begin{proof}
     (1) By hypothesis, the element $x=\sqrt{y}$ generates $Q$ over
     $F$. Its minimal polynomial is
     \begin{equation}
	\label{minpolx.eq}
     u^{4}-T(y)u^{2}+N(y)\in F[u].
     \end{equation}
     Let $\lambda_{x}\in\Lambda_{2}(Q)$ be defined as
     in~\eqref{phia.eq}. By Proposition~\ref{propgenericet.prop},
     $\gamma_{Q}(\lambda_{x})\lambda_{x}$ generates $\resolv(Q)$ and
     its minimal polynomial is
     \[
     t\bigl((t+T(y))^{2}-4N(y)\bigr).
     \]
     Therefore,
     \begin{equation}
	\label{RQ.eq}
	\resolv(Q)\simeq F\times F\bigl(\sqrt{N(y)}\bigr),
     \end{equation}
     determining $\Delta(Q)$. Moreover,
     Proposition~\ref{propgenericet.prop} also shows that
     $\gamma_{Q}(\lambda_{x})=-\lambda_{x}$, hence regarding
     \eqref{RQ.eq} as an identification,
     \[
     \lambda_{x}^{2}=\bigl(0,T(y)-2\sqrt{N(y)}\bigr).
     \]
     Therefore, the projection $\check\lambda_{x}$ of $\lambda_{x}$
     to $\check Q$ under the isomorphism
     $\Lambda_{2}(Q)\simeq\Delta(\check Q)\times\check Q$ satisfies
     \begin{equation}
	\label{minpolambda.eq}
     \check\lambda_{x}^{2}=T(y)-2\sqrt{N(y)}=
     y+\overline{y}-2\sqrt{y\overline{y}}.
     \end{equation}
     If the minimal polynomial~\eqref{minpolx.eq} of $x$ has no root
     in $F$, computation shows that $T(y)-2\sqrt{N(y)}$ is
     not a square in $\Delta(Q)$, hence
     \[
     \check Q=F(\check\lambda_{x}).
     \]
     The proof is complete, since \eqref{minpolambda.eq} shows that we
     may identify $\check\lambda_{x}$ with
     $\sqrt{y}+\sqrt{\overline{y}}$, determining the square roots in
     such a way that $\sqrt{y}\sqrt{\overline{y}}=-\sqrt{N(y)}$.

     If the minimal polynomial \eqref{minpolx.eq} has a root in $F$,
     then $Q$ has a factor $F$ and we are in the situation of
     Proposition~\ref{special2alg.prop}(2) with $K_{1}$ or $K_{2}$ split.
     This case is left to the reader.
     \medbreak
     \par
     (2) Suppose now $\carac F=2$. The element $x=y\wp^{-1}(y)$
     generates $Q$ with minimal polynomial
     \[
     u^{4}+T(y)u^{3}+
     \bigl(T(y)^{3}+T(y)N(y)+N(y)\bigr)u^{2}
     +T(y)^{2}N(y)u+N(y)^{3}\in F[u].
     \]
     Consider again the element $\lambda_{x}\in\Lambda_{2}(Q)$ defined
     in \eqref{phia.eq}. By Proposition~\ref{propgenericet.prop},
     $\gamma_{Q}(\lambda_{x})\lambda_{x}$ generates $\resolv(Q)$ and
     has minimal polynomial
     \[
     \bigl(t-N(y)\bigr)
     \bigl(t^{2}-N(y)t+
     T(y)^{6}+T(y)^{2}N(y)+
     T(y)N(y)^{2}\bigr).
     \]
     If $w$ is a root of the quadratic factor, then
     \[
     \wp\bigl(N(y)^{-1}(w-T(y)^{3}-T(y)N(y))\bigr)=T(y),
     \]
     hence
     \[
     \resolv(Q)\simeq F\times F\bigl(\wp^{-1}(T(y))\bigr).
     \]
     Proposition~\ref{propgenericet.prop} also shows that the
     projection $\check\lambda_{x}$ of $\lambda_{x}$ onto $\check Q$
     satisfies
     \begin{equation}
	\label{minpolambda2.eq}
	\check\lambda_{x}^{2}+T(y)\check\lambda_{x}+w=0.
     \end{equation}
     If $\wp^{-1}(y)$ and $\wp^{-1}(\overline{y})$ are determined in
     such a way that
     $\wp^{-1}(y)+\wp^{-1}(\overline{y})=\wp^{-1}(T(y))$,
     computation shows that
     $\wp^{-1}(T(y))\bigl(T(y)+
     \wp^{-1}(y)\wp^{-1}(\overline{y})\bigr)$ also satisfies
     \eqref{minpolambda2.eq}, hence we may identify $\check Q$ with
     $F\bigl(\wp^{-1}(y)\wp^{-1}(\overline{y})\bigr)$.
\end{proof}

We refer to \cite{LLT} for a description of quartic $2$-extensions of
fields in characteristic different from~$2$.

\subsection{Classification of quartic algebras}
\label{classif.subsec}

Combining the results of \S\S\ref{Sym3.subsec} and
\ref{Dfour.subsec}, we obtain a classification of quartic \'etale
$F$-algebras $Q$ based on the action of the absolute Galois group
$\Gamma$ of $F$ on $\funcX(Q)$. We summarize the various
possibilities for $Q$, $\Delta(Q)$, $\resolv(Q)$ and $\Lambda_{2}(Q)$
in the table below. In this table,
$\alpha(\Gamma)\subset\Sym_{\funcX(Q)}\simeq\Sym_{4}$ is the image of
the $\Gamma$-action. The letters $N$ and $L$ are used for sextic and
cubic separable field extensions of $F$, and $K$, $\check K$,
$K_{1}$, $K_{2}$ for quadratic separable field extensions of $F$. A
quartic separable field extension is called an $\Sym_{4}$-quartic
(resp.\ $\Alt_{4}$-quartic) if its Galois closure has Galois group
isomorphic to $\Sym_{4}$ (resp.\ $\Alt_{4}$).

\[
\begin{array}{|l|l|l|l|l|}
\hline
  \alpha(\Gamma) &Q & \Delta(Q) & \resolv(Q) & \Lambda_2(Q)\\
\hline
& & \\[-2ex]
\{1\} & F^{4} & F^2 & F^3  & F^6 \\
C_2 \not \subset \Vier & F^2 \times K &  K & F\times K & F^2 \times K^2 \\
C_2 \subset \Vier & K \times K & F^2 & F^3 & F^2 \times K^2\\
C_3 & F \times L,  L\,\text{cyclic}& F^{2} & L & L \times L \\
C_2 \times C_2 \not  \subset \Vier & K_1 \times K_2 &
  K_1 *K_2 & F \times K_1*K_2&
F^2 \times (K_1\otimes K_2)\\
\Vier & K_1 \tens K_2  & F^2 & F^3  & K_1\times K_1 \times K_1*K_2\\
C_4 & \text{cyclic} & K\subset Q & F \times K  & K \times Q\\
S_3 & F \times L &  K & L & L \times L\\
D_4 & Q \supset \check K & K & F\times K  & \check K \times \check Q,\,
\check Q \supset K\\
\Alt_4 & \Alt_4 \text{-quartic} & F ^2 & L\,\text{cyclic}  & N \supset L\\
\Sym_4 & \Sym_4\text{-quartic} &K&L\, S_3 \text{-cubic}  & N \supset L\\
\hline
\end{array}
\]

\section{Cyclic quartic algebras}
\label{cycl.sec}

Let $F$ be an arbitrary field with absolute Galois group
$\Gamma=\Gal(F_{s}/F)$.
Quartic \'etale $F$-algebras $Q$ such that the $\Gamma$-action on
$\funcX(Q)$ factors through a cyclic group $C_{4}$ can be endowed
with the structure of a $C_{4}$-Galois algebra. (In the table of
\S\ref{classif.subsec}, they can be found in the lines
$\alpha(\Gamma)=\{1\}$, $\alpha(\Gamma)=C_{2}\subset\Vier$ and
$\alpha(\Gamma)=C_{4}$.) Fixing a generator of $C_{4}$ (or,
equivalently, choosing an isomorphism
$C_{4}\simeq\mathbb{Z}/4\mathbb{Z}$), we may consider a
$C_{4}$-Galois $F$-algebra as a pair $(Q,\nu)$ where $Q$ is a
quartic \'etale $F$-algebra and $\nu$ is an $F$-algebra automorphism
of $Q$ such that
\[
\{x\in Q\mid\nu(x)=x\}=F.
\]
The automorphism $\nu$ then satisfies $\nu^{4}=\Id$, and it yields
the action on $Q$ of the generator of $C_{4}$. An isomorphism of
$C_{4}$-Galois $F$-algebras $\beta\colon\,(Q,\nu)\to(Q',\nu')$ is an
isomorphism $\beta\colon\,Q\isom Q'$ such that
$\nu'\circ\beta=\beta\circ\nu$. Let $\Cycl_{4}(F)$ be the set of
isomorphism classes of $C_{4}$-Galois $F$-algebras. As observed in
\S\ref{galoisalg.subsec}, there is a canonical bijection
\[
\Cycl_{4}(F)\simeq H^{1}(\Gamma,C_{4}).
\]
If $C_{4}$ is embedded in $\Sym_{4}$, the corresponding map in
cohomology $H^{1}(\Gamma,C_{4})\to H^{1}(\Gamma,\Sym_{4})$ maps the
isomorphism class of $(Q,\nu)$ to the isomorphism class of $Q$.

Since $C_{4}$ is an abelian group, the set $H^{1}(\Gamma,C_{4})$ is
an abelian group. The group structure on $\Cycl_{4}(F)$ is induced by
the following composition law (see \cite{FPS}):
\[
(Q,\nu)\star(Q',\nu')=\bigl((Q\otimes Q')^{\nu^{-1}\otimes \nu'},
\nu\otimes\Id\bigr).
\]
The class of the split algebra $F^{4}$ with the cyclic permutation of
factors is the neutral element. The squaring map
$\rho\colon\,C_{4}\to\Sym_{2}$ fits into an exact sequence
\[
1\to\Sym_{2}\xrightarrow{\iota}C_{4}\xrightarrow{\rho}\Sym_{2}\to1.
\]
Since $H^{1}(\Gamma,\Sym_{2})\simeq\Quadr(F)$ (see
\S\ref{cohet.subsec}), the induced exact sequence in cohomology takes
the form
\begin{equation}
     \label{QC.eq}
1\to\Quadr(F)\xrightarrow{\iota^{1}}\Cycl_{4}(F)
\xrightarrow{\rho^{1}}\Quadr(F).
\end{equation}
The map $\iota^{1}$ is induced by $K\mapsto(K\times K,\nu)$ where
$\nu(x,y)=\bigl(y,\gamma_{K}(x)\bigr)$, and the map $\rho^{1}$
carries every $C_{4}$-Galois algebra $(Q,\nu)$ to its discriminant
$\Delta(Q)$ (which is isomorphic to the quadratic subalgebra
$Q^{\nu^{2}}$, see Proposition~\ref{special2alg.prop}).

\begin{remark}
     The algebra $K\times K$ contains $K$ and $F\times F$ as quadratic
     subalgebras. However, Galois theory shows that if $(Q,\nu)$ is a
     $C_{4}$-Galois algebra and $Q$ is a field, then $Q$ contains a
     unique quadratic extension of $F$.
\end{remark}
%
\medbreak
\par
In the rest of this section, we give an explicit description of
$H^{1}(\Gamma,C_{4})$ and use it to parametrize $C_{4}$-Galois algebras
up to isomorphism.
The description depends in an essential way on whether the
characteristic is~$2$ or not.

\subsection{Characteristic not $2$}

If $\carac F\neq2$, the group $\mu_{4}(F_{s})\subset F_{s}$ of fourth
roots of unity is cyclic of order~$4$. Let $S=F[X]/(X^{2}+1)$.
Twisting the $\Gamma$-action on $\mu_{4}(F_{s})$ by a cocycle whose
image in $H^{1}(\Gamma,\Sym_{2})$ defines $S$, we obtain a
$\Gamma$-module $\mu_{4[S]}$ with trivial $\Gamma$-action. Thus,
$\mu_{4[S]}\simeq C_{4}$, and $C_{4}$-Galois $F$-algebras are also
classified by $H^{1}(\Gamma,\mu_{4[S]})$. To give an explicit
description of this group, consider the homomorphism
\[
\Upsilon\colon\,F^\times\times S^\times\to F^\times\times S^\times
\]
defined by
\[
\Upsilon(\ell,z)=(N_{S/F}(z),\ell z^{2}).
\]

\begin{prop}
     \label{Hmu4.prop}
     There is a canonical isomorphism
     \[
     H^{1}(\Gamma,\mu_{4[S]})\simeq(F^\times\times
     S^\times)/\Upsilon(F^\times\times S^\times).
     \]
\end{prop}

\begin{proof}
     Let $i=\sqrt{-1}\in S$ be the image of $X$. We may identify
     $\mu_{4[S]}$ with
     \[
     \bigl\{
     (1,1),(1,-1),(-1,1\otimes i),(-1,-1\otimes i)\bigr\}
     \subset F_{s}^\times\times(F_{s}\otimes S)^\times,
     \]
     which is the kernel of the map $\Upsilon$ extended to $F_{s}$. The
     proposition follows from the cohomology exact sequence associated
     with
     \[
     1\to\mu_{4[S]}\to F_{s}^\times\times(F_{s}\otimes S)^\times
     \xrightarrow{\Upsilon}F_{s}^\times\times(F_{s}\otimes S)^\times
     \to 1,
     \]
     since Hilbert's Theorem~90 and Shapiro's lemma yield
     $H^{1}(\Gamma,F_{s}^\times\times(F_{s}\otimes S)^\times)=1$.
\end{proof}

\begin{remark}
     Another description of $H^{1}(\Gamma,\mu_{4[S]})$ is given in
     \cite[(30.13)]{BoI}.
\end{remark}

It follows from Proposition~\ref{Hmu4.prop} that $C_{4}$-Galois algebras
are classified by the group
\[
(F^\times\times S^{\times})/\Upsilon(F^\times\times S^\times).
\]
(See \cite{FPS} for a proof without cohomology and, more generally
for a class of commutative rings in which $2$ is invertible.)
We give an explicit description of this correspondence.

Let $i=\sqrt{-1}\in S$. For $\lambda\in F^\times$ and
$s=s_{1}+is_{2}\in S^\times$, let $d=N_{S/F}(s)=s_{1}^{2}+s_{2}^{2}$
and let
\[
Q_{\lambda,s}=F[W,X,Y]/I_{\lambda,s}
\]
where $I_{\lambda,s}$ is the ideal generated by
\[
W^{2}-d,\quad
X^{2}-\frac{\lambda}{2}(d+s_{1}W),\quad
Y^{2}-\frac{\lambda}{2}(d-s_{1}W),\quad
XY-\frac{\lambda}{2}s_{2}W.
\]
The automorphism $\nu$ of $F[W,X,Y]$ defined by
\[
\nu(W)=-W,\quad\nu(X)=Y,\quad \nu(Y)=-X
\]
preserves $I_{\lambda,s}$ and induces an automorphism of
$Q_{\lambda,s}$ which we
denote by $\nu_{\lambda,s}$.

\begin{prop}
     \label{classC4.prop}
     For any $\lambda\in F^\times$ and $s\in S^\times$, the pair
     $(Q_{\lambda,s},\nu_{\lambda,s})$ is a $C_{4}$-Galois $F$-algebra.
     The map $(\lambda,s)\mapsto(Q_{\lambda,s},\nu_{\lambda,s})$
     induces a group isomorphism
     \[
     \Phi\colon\,
     F^\times\times S^\times/\bigl\{(N_{S/F}(z),\ell z^{2})\mid
     \ell\in F^\times, z\in S^\times\bigr\}
     \isom\Cycl_{4}(F).
     \]
\end{prop}

\begin{proof}
     We first show that $Q_{\lambda,s}$ is a quartic \'etale
     $F$-algebra. Let $\omega$, $\xi$, $\eta\in Q_{\lambda,s}$ be the
     images of $W$, $X$, $Y$ respectively. The algebra $F[\omega]$ is
     quadratic, $F[\omega]\simeq F[\sqrt{d}]$.

     If $s_{2}\neq0$, then $d\neq s_{1}^{2}$, hence $d+s_{1}\omega$ is
     invertible. Computation shows
     $(\xi^{-1}\frac{\lambda}{2}s_{2}\omega)^{2}=
     \frac{\lambda}{2}(d-s_{1}\omega)$, so
     \[
     Q_{\lambda,s}=F[\omega,\xi]\simeq
     F[\sqrt{d}]\biggl[\sqrt{\frac{\lambda}{2}(d+s_{1}\sqrt{d})}\biggr].
     \]

     If $s_{2}=0$, then $d=s_{1}^{2}$, hence $F[\omega]\simeq F\times
     F$ and we may identify $\xi$ and $\eta$ to
     $s_{1}(\sqrt{\lambda},0)$ and $s_{1}(0,\sqrt{\lambda})$ in
     \[
     Q_{\lambda,s}\simeq F[\sqrt{\lambda}]\times F[\sqrt{\lambda}].
     \]
     Therefore, in each case $Q_{\lambda,s}$ is a quartic \'etale
     $F$-algebra, and the fact that the subalgebra fixed under
     $\nu_{\lambda,s}$
     is $F$ is easily verified.

     To prove that $\Phi$ is a group homomorphism, consider
     $(\lambda,s)$, $(\lambda',s')\in F^\times\times S^\times$ with
     $s=s_{1}+is_{2}$, $s'=s_{1}'+is_{2}'$. Let $d=N_{S/F}(s)$,
     $d'=N_{S/F}(s')$ and let $\omega$, $\xi$, $\eta\in Q_{\lambda,s}$
     and $\omega'$, $\xi'$, $\eta'\in Q_{\lambda',s'}$ be defined as
     above. The elements
     \[
     \omega_{\star}=\omega\otimes\omega',\quad
     \xi_{\star}=\xi\otimes\xi'-\eta\otimes\eta',\quad
     \eta_{\star}=\xi\otimes\eta'+\eta\otimes\xi'
     \]
     are in $(Q_{\lambda,s}\otimes
     Q_{\lambda',s'})^{\nu_{\lambda,s}^{-1}\otimes\nu_{\lambda',s'}}$
     and satisfy
     \begin{align*}
	\omega_{\star}^{2} & =dd',\\
	\xi_{\star}^{2} & =\frac{\lambda\lambda'}{2}\bigl[
	dd'+(s_{1}s'_{1}-s_{2}s'_{2})\omega_{\star}\bigr],\\
	\xi_{\star}\eta_{\star} & =\frac{\lambda\lambda'}{2}
	(s_{1}s_{2}'+s_{2}s_{1}')\omega_{\star},\\
	\eta_{\star}^{2} & = \frac{\lambda\lambda'}{2}\bigl[
	dd'-(s_{1}s'_{1}-s_{2}s_{2}')\omega_{\star}\bigr].
     \end{align*}
     Moreover,
     \[
     (\nu_{\lambda,s}\otimes\Id)(\omega_{\star})=-\omega_{\star},\quad
     (\nu_{\lambda,s}\otimes\Id)(\xi_{\star})=\eta_{\star},\quad
     (\nu_{\lambda,s}\otimes\Id)(\eta_{\star})=-\xi_{\star}.
     \]
     Therefore,
     \[
     (Q_{\lambda,s},\nu_{\lambda,s})\star
     (Q_{\lambda',s'},\nu_{\lambda',s'})\simeq(Q_{\lambda\lambda',ss'},
     \nu_{\lambda\lambda',ss'}).
     \]

     If $(\lambda,s)=(N_{S/F}(z),\ell z^{2})$ for some $\ell\in
     F^\times$, $z=z_{1}+iz_{2}\in S^\times$, then the homomorphism
     $F[W,X,Y]\to F^{4}$ defined by
     \begin{align*}
     &W\mapsto \ell N_{S/F}(z)(1,-1,1,-1),\\
     &X\mapsto \ell N_{S/F}(z)(z_{1},z_{2},-z_{1},-z_{2}),\\
     &Y\mapsto \ell N_{S/F}(z)(z_{2},-z_{1},-z_{2},z_{1})
     \end{align*}
     induces an isomorphism $(Q_{\lambda,s},\nu_{\lambda,s})\isom
     (F^{4},\sigma)$ where $\sigma$ is the cyclic permutation.

     Conversely, suppose that for some $(\lambda,s)\in F^\times\times
     S^\times$ there is an isomorphism $Q_{\lambda,s}\simeq F^{4}$, and
     let $(\omega_{i})_{1\leq i\leq4}$, $(\xi_{i})_{1\leq i\leq 4}$,
     $(\eta_{i})_{1\leq i\leq4}$ be the images of $\omega$, $\xi$ and
     $\eta$ respectively in $F^{4}$. Then from the relations between
     $\omega$, $\xi$ and $\eta$ it follows that
     $z=\omega_{1}^{-1}(\xi_{1}+i\eta_{1})\in S$ and
     $\ell=(\xi_{1}^{2}+\eta_{1}^{2})^{-1}\omega_{1}^{3}$ satisfy
     \[
     \lambda=N_{S/F}(z)\qquad\text{and}\qquad
     s=\ell z^{2}.
     \]
     Therefore, the homomorphism $\Phi$ is injective, and it only
     remains to prove its surjectivity.

     Let $(Q,\nu)$ be a $C_{4}$-Galois $F$-algebra. If $Q\simeq
     F[\sqrt{\mu}]\times F[\sqrt{\mu}]$ for some $\mu\in F^\times$,
     then $Q\simeq Q_{\mu,1}$, as was observed at the beginning of the
     proof. Therefore, for the rest of the proof we may assume $Q$ is a
     field. Let $K=Q^{\nu^{2}}\subset Q$ be the subfield fixed under
     $\nu^{2}$, let $K=F(\omega)$ with $\omega^{2}=d$ for some $d\in
     F^\times$, and let $Q=K(\xi)$ with $\xi^{2}=y$ for some $y\in
     K^\times$. We have $y\notin F$ since $Q/F$ is cyclic. Let
     $y=a+b\omega$ with $a$, $b\in F$, $b\neq0$. Substituting for $y$
     an element of the form $u^{2}y$ with $u\in K$, we may assume
     $a\neq0$. Letting $\lambda=2ad^{-1}$ and $s_{1}=a^{-1}bd$, we may
     then write $y$ in the form
     \[
     y=\frac{\lambda}{2}(d+s_{1}\omega).
     \]
     Let $\eta=\nu(\xi)$. Since $\xi^{2}\in K$ and $\xi\notin K$, we
     have $\nu^{2}(\xi^{2})=\xi^{2}$ and $\nu^{2}(\xi)\neq\xi$,
     hence $\nu^{2}(\xi)=-\xi$. Therefore, $\nu(\xi\eta)=-\xi\eta$,
     and it follows that $\xi\eta\in\omega F^\times$. Let
     \begin{equation}
	\label{xieta.eq}
	\xi\eta=\frac{\lambda}{2}s_{2}\omega
	\qquad\text{for some $s_{2}\in F^\times$}.
     \end{equation}
     \relax From the equation
     $\xi^{2}=\frac{\lambda}{2}(d+s_{1}\omega)$, we obtain
     \[
     \eta^{2}=\frac{\lambda}{2}(d-s_{1}\omega).
     \]
     Therefore, \eqref{xieta.eq} yields
     \[
     \frac{\lambda^{2}}{4}s_{2}^{2}d=\frac{\lambda^{2}}{4}
     (d+s_{1}\omega)(d-s_{1}\omega),
     \]
     hence $d=s_{1}^{2}+s_{2}^{2}$. It is then clear that
     $(Q,\nu)\simeq(Q_{\lambda,s},\nu_{\lambda,s})$ with
     $s=s_{1}+is_{2}$.
\end{proof}

\begin{cor}
     A quadratic \'etale $F$-algebra $F[\sqrt{d}]$ can be embedded in
     a $C_{4}$-Galois algebra $(Q,\nu)$ as $F[\sqrt{d}]\simeq
     Q^{\nu^{2}}$ if and only if $d$ is a sum of two squares in $F$.
\end{cor}

\begin{proof}
     The ``only if'' part was shown in the last lines of the proof of
     Proposition~\ref{classC4.prop}. (Alternately, it follows from
     Propositions~\ref{special2alg.prop} and \ref{quarticexpl.prop}.)
     The ``if'' part follows from the observation that
     $F[\sqrt{d}]\simeq Q_{\lambda,s}^{\nu^{2}_{\lambda,s}}$ whenever
     $N_{S/F}(s)=d$.
\end{proof}

In other words, this corollary shows that the exact sequence
\eqref{QC.eq} can be extended to
\[
1\to\Quadr(F)\xrightarrow{\iota^{1}}\Cycl_{4}(F)
\xrightarrow{\rho^{1}}\Quadr(F)\xrightarrow{\delta}\Br(F),
\]
where $\Br(F)$ is the Brauer group of $F$ and $\delta$ maps
$K=F[\sqrt{d}]$ to
the Brauer class of the quaternion algebra $(-1,d)_{F}$.
Of course, this result is well-known and has an easy cohomological
proof.

\subsection{Characteristic $2$}

Cyclic Galois $C_{p^n}$-algebras over fields over characteristic $p$ were
constructed by Witt \cite[Satz 13]{Witt37}, using Witt vectors.  The group
$H^1(F,C_{p^n})$ over a field of characteristic $p$ was computed by Serre
in \cite[Chap. X, \S 3]{corpslocaux}, also in terms of  Witt vectors.
%
We recall explicitly the  results of Serre and Witt
for the group $C_4$ over a field $F$ of
characteristic $2$.

Let $W_2(F)$ be the additive group of Witt vectors of length~$2$.
By definition we have
$W_2(F) = \{ (t,s) \ | \ t,s \in F\}$
with the addition
\[
  (t_1,s_1)   \ldotplus  (t_2,s_2) = (t_1 +t_2, s_1 +s_2 + t_1t_2).
\]
Alternately,
\[
W_{2}(F)=\left\{\left.
\begin{pmatrix}
     1&t&s\\
     0&1&t\\
     0&0&1
\end{pmatrix}
\right\vert
s,t\in F\right\}
\subset\GL_{2}(F).
\]
The neutral element is $(0,0)$ and $\ldotminus\!\! (t,s)= (t,s+t^2)$.
  The map
\[
\wp_2\colon\, W_2(F) \to W_2(F), \ (t,s) \mapsto
  (t^2,s^2) \ldotminus (t,s)=(t^2 +t, s^2 +s +t^2 +t^3)
\]
is a group homomorphism and there is an exact sequence of
$\Gamma$-modules:
\begin{equation}
     \label{exseqC4.eq}
0 \to C_4 \to W_2(\Fs) \xrightarrow{\wp_{2}}  W_2(\Fs) \to 0,
\end{equation}
where $C_{4}$ is identified with the subgroup of $W_{2}(\Fs)$
generated by $(1,0)$.
%
\begin{prop}
     $ H^1\bigl(\Gamma, W_2(F_{s})\bigr) = 0$ and
     $H^1(\Gamma, C_4) \simeq  W_2(F)/ \wp_2\bigl( W_2(F)\bigr)$.
\end{prop}
%
\begin{proof}
The first claim follows from the exact sequence of (additive)
$\Gamma$-modules:
\[
0 \to F_{s} \stackrel{\iota}{\to} W_2(F_{s}) \stackrel{\pi}{\to} F_{s} \to 0
\]
where $\iota(s) = (0,s)$ and $\pi(t,s) =t$, and the fact that
$H^1(\Gamma,F_{s})=0$
(by the additive version of Hilbert's Theorem~90). The last claim follows
from the exactness of \eqref{exseqC4.eq} and from the first claim.
\end{proof}
%
For $(t,s)\in W_{2}(F)$, let $I_{(t,s)}\subset F[W,X]$ be the ideal
generated by the polynomials $f_{1}(W)$, $f_{2}(W,X)$ such that
\[
\bigl(f_{1}(W),f_{2}(W,X)\bigr)=\wp_{2}(W,X)\ldotminus(t,s),
\]
and let
\[
Q_{(t,s)}=F[W,X]/I_{(t,s)}.
\]
Letting $\omega$, $\xi$ be the images of $W$, $X$ in $Q_{(t,s)}$, the
relations defining $Q_{(t,s)}$ can be rewritten as
\[
\omega^{2}+\omega=t,\qquad
\xi^{2}+\xi=t\omega+s.
\]
It is therefore clear that $Q_{(t,s)}$ is a quartic $2$-algebra over
$F$.
The automorphism
of  $ F[W,X]$ given by
   \[
(W,X) \mapsto
   (W,X)  \ldotplus (1,0)
\]
induces an automorphism $\nu_{(t,s)}$ of $Q_{(t,s)}$;
we have by definition $\nu_{(t,s)}(\omega)=\omega+1$ and
$\nu_{(t,s)}(\xi) =\xi + \omega$, hence the fixed subalgebra of
$Q_{(t,s)}$ is $F$. Thus, $(Q_{(t,s)},\nu_{(t,s)})$ is a
$C_{4}$-Galois $F$-algebra.

\begin{prop}
     \label{classC42.prop}
     The map $(t,s)\mapsto(Q_{(t,s)},\nu_{(t,s)})$ induces a group
     isomorphism
     \[
     \Psi\colon\,W_{2}(F)/\wp_{2}\bigl(W_{2}(F)\bigr)\isom
     \Cycl_{4}(F).
     \]
\end{prop}

\begin{proof}
     Let $(t,s)$, $(t',s')\in W_{2}(F)$, and let $\omega$, $\xi\in
     Q_{(s,t)}$ and $\omega'$, $\xi'\in Q_{(s',t')}$ be the elements
     defined as above. In $Q_{(t,s)}\otimes Q_{(t',s')}$, consider the
     elements
     \[
     \omega_{\star}=\omega\otimes1+1\otimes\omega',
     \qquad
     \xi_{\star}=\xi\otimes1+1\otimes\xi'+\omega\otimes\omega'.
     \]
     Computation shows that $\omega_{\star}$ and $\xi_{\star}$ are
     invariant under $\nu_{(t,s)}^{-1}\otimes\nu_{(t',s')}$ and satisfy
     \[
     \omega_{\star}^{2}+\omega_{\star}=t+t',\qquad
     \xi_{\star}^{2}+\xi_{\star}=(t+t')\omega_{\star}+s+s'+tt'.
     \]
     Moreover,
     $(\nu_{(t,s)}\otimes\Id)(\omega_{\star})=\omega_{\star}+1$ and
     $(\nu_{(t,s)}\otimes\Id)(\xi_{\star})=
     \xi_{\star}+\omega_{\star}$. Therefore,
     \[
     (Q_{(t,s)},\nu_{(t,s)})\star(Q_{(t',s')},\nu_{(t',s')})\simeq
     (Q_{(t,s)\ldotplus(t',s')},\nu_{(t,s)\ldotplus(t',s')}).
     \]

     If $(t,s)=\wp_{2}(x,y)=(x^{2}+x,y^{2}+y+x^{3}+x^{2})$ for some
     $x$, $y\in F$, then the map $F[W,X]\to F$ defined by
     \[
     W\mapsto(x,x+1,x,x+1),\qquad
     X\mapsto(y,x+y,y+1,x+y+1)
     \]
     induces an isomorphism $(Q_{(t,s)},\nu_{(t,s)})\isom
     (F^{4},\sigma)$ where $\sigma$ is the cyclic permutation of
     factors.

     Conversely, if $Q_{(t,s)}\simeq F^{4}$  for some $t$, $s\in F$, then
     letting $(\omega_{i})_{1\leq i\leq4}$ and $(\xi_{i})_{1\leq
     i\leq4}$ denote the images of $\omega$ and $\xi$ in $F^{4}$, it
     is readily verified that
     \[
     t=\omega_{1}^{2}+\omega_{1}\qquad\text{and}\qquad
     s=\xi_{1}^{2}+\xi_{1}+\omega_{1}^{3}+\omega_{1}^{2},
     \]
     hence $(t,s)=\wp_{2}(\omega_{1},\xi_{1})$. This shows that the
     map $\Psi$ is an injective homomorphism of groups, and it only
     remains to prove its surjectivity.

     Let $(Q,\nu)$ be a $C_{4}$-Galois algebra. If $Q\simeq
     F[\wp^{-1}(u)]\times F[\wp^{-1}(u)]$ for some $u\in F$, then
     $(Q,\nu)\simeq(Q_{(0,u)},\nu_{(0,u)})$. For the rest of the
     proof, we may thus assume $Q$ is a field. Let
     $K=Q^{\nu^{2}}\subset Q$ be the subfield fixed under $\nu^{2}$
     and let $Q=K(\xi)$ with $\xi^{2}+\xi\in K$. Then
     $\nu^{2}(\xi)=\xi+1$, and it follows that the element
     $\omega=\xi+\nu(\xi)$ satisfies
     \[
     \nu^{2}(\omega),\qquad\nu(\omega)=\omega+1,
     \]
     hence $K=F(\omega)$. We then have $\xi^{2}+\xi=t\omega+s$ for
     some $t$, $s\in F$, and
     \[
     \omega^{2}+\omega=(\xi^{2}+\xi)+\nu(\xi^{2}+\xi)=t.
     \]
     Therefore, $(Q,\nu)\simeq(Q_{(t,s)},\nu_{(t,s)})$.
\end{proof}

\begin{cor}
     Every quadratic \'etale $F$-algebra can be embedded in a
     $C_{4}$-Galois $F$-algebra $(Q,\nu)$ as
     $Q^{\nu^{2}}\simeq\Delta(Q)$.
\end{cor}

\begin{proof}
     For any $t\in F$, we have $F[\wp^{-1}(t)]\simeq
     Q_{(t,s)}^{\nu^{2}_{(t,s)}}$ for all $s\in F$.
\end{proof}

The corollary shows that the last map in the exact sequence
\eqref{QC.eq} is onto when $\carac F=2$. This is clear from a
cohomological viewpoint, since the cohomological $2$-dimension of
$\Gamma$ is at most $1$, see \cite[p.~86]{CohGal}.

\begin{thebibliography}{xx}

\bibitem{Allison91}
B.N. Allison.
\newblock Construction of $3 \times3$-matrix algebras and some
{L}ie algebras of type ${D}_4$.
\newblock {\em J. Algebra}. 143:63--92, 1991.


\bibitem{FPS}
M.~Ferrero, A~Paques, and A.~Solecki.
\newblock On cyclic quartic extensions with normal basis.
\newblock {\em Bull. Sci. Math.}, 116:487--500, 1992.


\bibitem{BoI}
M.-A. Knus, A.S. Merkurjev, M. Rost, J.-P. Tignol, \textit{The Book of
Involutions}, Amer. Math. Soc. Coll. Pub. 44, AMS, Providence, RI, 1998.

\bibitem{L}
J.-L. Lagrange, \textit{\OE uvres, tome~3}. (Publi\'ees par
J.-A.~Serret), Gauthier-Villars, Paris, 1869.


\bibitem{LLT}
T.~Y. Lam, D.~B. Leep, and J.-P. Tignol.
\newblock Biquaternion algebras and quartic extensions
\newblock {\em  Inst. Hautes \'Etudes Sci. Publ. Math.}, 77:63--102, 1993.


\bibitem{McL}
S. Mac Lane, \textit{Categories for the Working Mathematician}, 2d
edition, Graduate Texts Math.~5, Springer-Verlag, New York Berlin
Heidelberg, 1998.

\bibitem{S}
D.J. Saltman.
\newblock  Exterior powers of fields and subfields.
\newblock{\em Nagoya Math. J.}, 89:119--127, 1983.

\bibitem{SaltLect}
D.J. Saltman.
\newblock \textit{Lectures on Division Algebras.} CBMS Regional
Conference Series in Math. \textbf{94}, Amer. Math. Soc., Providence,
RI, 1999.

\bibitem{corpslocaux}
J.-P. Serre.
\newblock {\em Corps Locaux}, Deuxi\`eme \'edition,
\newblock {\em Publications de l'Universit\'e de Nancago}, No VIII.
\newblock Hermann, Paris, 1968.

\bibitem{Serre85}
J.-P. Serre.
\newblock Corps quartiques associ\'es \`a un corps cubique donn\'e
(Compl\'ements au cours du 29/10/1984, Janvier 1985).

\bibitem{CohGal}
J.-P. Serre.
\newblock \textit{Cohomologie galoisienne}, Cinqui\`eme \'edition,
r\'evis\'ee et compl\'et\'ee, Lecture Notes in Math. \textbf{5},
Springer-Verlag Berline-Heidelberg-New York, 1994.

\bibitem{VdWalg}
B.L. van~der Waerden, \textit{Algebra {I}}, Grundlehren der Math.
Wiss. 33, Springer-Verlag, Berlin, 1930.


\bibitem{Weil74e}
A.~Weil.
\newblock Exercices dyadiques.
\newblock {\em Invent. Math.}, 27:1--22, 1974.


\bibitem{Witt37}
E.~Witt.
\newblock Zyklische K\"orper und Algebren der Charakteristik $p$ vom
  Grad $p^n$.
\newblock{\em J. reine angew. Math.}, 176: 126--140, 1937.

\end{thebibliography}


\end{document}