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\DeclareMathOperator{\rank}{rank}

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\newcommand{\F}{\mathbb{F}}
\newcommand{\Fp}{\F_p}
\newcommand{\Gal}{\text{\rm Gal}}
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\newcommand{\N}{\mathbb{N}}
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\newcommand{\Z}{\mathbb{Z}}

\begin{document}

\title{Galois Module Structure of Galois Cohomology}

\author[Nicole Lemire]{Nicole Lemire$^\star$}
\address{Department of Mathematics, Middlesex College, \
University of Western Ontario, London, Ontario \ N6A 5B7 \ CANADA}
\thanks{$^\star$Research supported in part by NSERC grant R3276A01.}
\email{nlemire@uwo.ca}

\author[J\'{a}n Min\'{a}\v{c}]{J\'an Min\'a\v{c}$^{*\dagger}$}
\address{Department of Mathematics, Middlesex College, \
University of Western Ontario, London, Ontario \ N6A 5B7 \ CANADA}
\thanks{$^*$Research supported in part by NSERC grant R0370A01.}
\thanks{$^\dag$Supported by the Mathematical Sciences Research
Institute, Berkeley.} \email{minac@uwo.ca}

\author[John Swallow]{John Swallow$^\ddag$}
\address{Department of Mathematics, Davidson College, Box 7046,
Davidson, North Carolina \ 28035-7046 \ USA}
\thanks{$^\ddag$Research supported in part by National Security
Agency grant MDA904-02-1-0061.} \email{joswallow@davidson.edu}

\begin{abstract}
Let $F$ be a field containing a
primitive $p$th root of unity, and let $U$ be an open normal
subgroup of index $p$ of the absolute Galois group $G_F$ of $F$.
We determine the structure of the cohomology group $H^n(U,\Fp)$ as
an $\Fp[G_F/U]$-module for all $n\in\mathbb{N}$.  Previously this
structure was known only for $n=1$, and until recently the
structure even of $H^1(U,\Fp)$ was determined only for $F$ a local
field, a case settled by Borevi\v{c} and Faddeev in the 1960s.
\end{abstract}

\date{October 29, 2004}

\maketitle

\newtheorem{theorem}{Theorem}
\newtheorem{proposition}{Proposition}
\newtheorem{lemma}{Lemma}

\theoremstyle{definition}
\newtheorem*{remark*}{Remark}

\parskip=10pt plus 2pt minus 2pt

Let $F$ be a field containing a
primitive $p$th root of unity $\xi_p$. Let $G_F$ be the absolute
Galois group of $F$, $U$ an open normal subgroup of $G_F$ of index
$p$, and $G=G_F/U$. Let $E$ be the fixed field of $U$ in the
separable closure $F_\sep$ of $F$. Fix $a\in F$ such that
$E=F(\root{p}\of{a})$, and let $\sigma\in G$ satisfy
$\root{p}\of{a}^{\sigma-1}=\xi_p$.

In the 1960s Z.~I.~Borevi\v{c} and D.~K.~Faddeev classified the
possible $G$-module structures of the first cohomology groups
$H^1(U, \Fp)$ in the case $F$ a local field \cite{B}. Quite
recently this result was extended for all fields $F$ as above
\cite{MS}. For the study of Galois cohomology it is important to
extend these results to all cohomology groups $H^n(U, \Fp)$,
$n\in\N$, and a solution of this problem was out of reach until
now.

Recently, based on earlier work of A.~S.~Merkurjev, M.~Rost and
A.~A.~Suslin, V.~Voevodsky established the Bloch-Kato Conjecture
\cite{Vo1,Vo2}, and it turns out that some of the main theorems in
his proof are sufficient to determine the structure of all
$G$-Galois modules $H^n(U, \Fp)$, using only simple arithmetical
invariants attached to the field extension $E/F$. The theorems we
use (quoted as Theorems~\ref{th:bk} and \ref{th:es} in
section~\ref{se:bkktheory} below) had, in fact, been standard
conjectures on Galois cohomology. It is interesting to point out,
however, that the case $n=2$ could have already been settled some
20 years ago, thanks to the work of Merkurjev and Suslin
\cite{MeSu}.

The main ingredient for our determination of the $G$-module
structure of $H^n(U,\Fp)$ is Milnor $K$-theory. (See \cite{Mi} and
\cite[Chap.~IX]{FV}.)  For $i\ge 0$, let $K_iF$ denote the $i$th
Milnor $K$-group of the field $F$, with standard generators
denoted by $\{f_1,\dots,f_i\}$, $f_1, \dots, f_i\in F\setminus
\{0\}$. For $\alpha\in K_iF$, we denote by $\bar\alpha$ the class
of $\alpha$ modulo $p$, and we use the usual abbreviation $k_nF$
for $K_nF/pK_nF$. We write $N_{E/F}$ for the norm map $K_nE\to
K_nF$, and we use the same notation for the induced map modulo
$p$. We denote by $i_E$ the natural homomorphism in the reverse 
direction. We also apply the same notation $N_{E/F}$ and $i_E$ 
for the corresponding homomorphisms between cohomology groups.
The image of an element $\alpha\in K_i F$ in $H^i(G_F,\Fp)$ 
we also denote by $\alpha$.
Voevodsky's proof of the Bloch-Kato Conjecture
establishes a $G_F$-isomorphism $H^n(U,\Fp)\cong k_n E$.  We
formulate our results in terms of Galois cohomology for intended
applications, but we use Milnor $K$-theory in our proof.

We concentrate upon the case when $F$ is a perfect field, and at
the end of the paper we indicate how one may reduce the case of an
imperfect field $F$ to the case of characteristic $0$.

Our decomposition depends on four arithmetic invariants
$\Upsilon_1$, $\Upsilon_2$, $y$, $z$, which we define as follows.
First, for an element $\bar\alpha$ of $k_i F$, let
\begin{equation*}
    \ann_{k_{n-1}F}\bar\alpha = \ann
    \left(k_{n-1}F \xrightarrow{\bar\alpha \cdot -}
    k_{n-1+i}F\right)
\end{equation*}
denote the annihilator of the product with $\bar\alpha$. When the
domain of $\bar\alpha$ is clear, we omit the subscript on the map
and write simply $\ann\bar\alpha$.  Because we will often use the
elements $\overline{\{a\}}$, $\overline{\{\xi_p\}}$, $\overline{\{a,a\}}$, and
$\overline{\{a, \xi_p\}}$, we omit the bars for these elements.
We also omit the bar in the element $\overline{\{\root{p}\of{a}\}}\in k_n E$. 

Fix $n\in \N$ and $U$ an open normal subgroup of $G_F$ of index
$p$ with fixed field $E$.  Define invariants associated to $E/F$
and $n$ as follows:
\begin{align*}
    d &:= \dim_{\Fp}\ k_n F\ /\ N_{E/F}k_n E \\
    e &:= \dim_{\Fp}\ N_{E/F}k_n E \\
    \Upsilon_1 &:= \dim_{\Fp}\ \ann \{a, \xi_p\}\ /\ \ann \{a\} \\
    \Upsilon_2 &:= \dim_{\Fp}\ k_{n-1} F\ /\ \ann \{a,\xi_p\}
\end{align*}
\begin{equation*}
    y :=
    \begin{cases}
    \dim_{\Fp}\ (N_{E/F}k_n E)\ /\ \{a\}\cdot k_{n-1}F, & p>2  \\
    \dim_{\F_2}\ (N_{E/F}k_n E)\ /\ \{a\} \cdot
    \ann_{k_{n-1}F}\{a,-1\}, &p=2
    \end{cases}
\end{equation*}
\begin{equation*}
    z :=
    \begin{cases}
    \dim_{\Fp}\ (k_n F)\ /\ \left(\{\xi_p\}\cdot W + N_{E/F}k_n
    E\right), &p>2 \\
    \dim_{\F_2}\ (k_n F)\ /\ \left(\{a\} \cdot k_{n-1}F +
    N_{E/F}k_n E\right), &p=2,
    \end{cases}
\end{equation*}
where $W$ is a complement of $\ann \{a, \xi_p\}$ in $k_{n-1}F$.  (In
Lemma~\ref{le:windep} we show that $z$ is independent of the
choice of $W$.)

Our main results are then the following.

\begin{theorem}\label{th:pnot2}
If $p>2$, $F$ is perfect and $n\in\N$ then
\begin{equation*}
    H^n(U, \Fp) \cong X_1 \oplus X_2 \oplus Y \oplus Z
\end{equation*}
where
\begin{enumerate}
    \item $X_1$ is a trivial $\Fp[G]$-module of dimension
    $\Upsilon_1$
    \item $X_2$ is a direct sum of $\Upsilon_2$ cyclic
    $\Fp[G]$-modules of dimension 2
    \item $Y$ is a free $\Fp[G]$-module of rank $y$
    \item $Z$ is a trivial $\Fp[G]$-module of dimension $z$.
\end{enumerate}
Further we have
\begin{enumerate}
    \setcounter{enumi}{4}
    \item $Y^G = i_E N_{E/F}H^n(U,\Fp)$
    \item $N_{E/F}\colon X_1\oplus X_2 \to \{a\}\cdot
    H^{n-1}(G_F,\Fp)$ is surjective
    \item $\Upsilon_1+\Upsilon_2+y=e$
    \item $\Upsilon_2+z=d$
\end{enumerate}
\end{theorem}

\begin{theorem}\label{th:p2}
If $p=2$ and $n\in\N$ then
\begin{equation*}
    H^n(U, \F_2) \cong X_1 \oplus Y \oplus Z
\end{equation*}
where
\begin{enumerate}
    \item $X_1$ is a trivial $\F_2[G]$-module of dimension
    $\Upsilon_1$
    \item $Y$ is a free $\F_2[G]$-module of rank $y$.
    \item $Z$ is a trivial $\F_2[G]$-module of dimension $z$.
\end{enumerate}
Further we have
\begin{enumerate}
    \setcounter{enumi}{3}
    \item $Y^G = i_E N_{E/F}H^n(U,\F_2)$
    \item $N_{E/F}\colon X_1 \to \{a\}\cdot
    \ann \{a, -1\}$ is an isomorphism
    \item $\Upsilon_1+y=e$
    \item $\Upsilon_2+z=d$
\end{enumerate}
\end{theorem}

\begin{remark*}
As explained at the end of section~\ref{se:esext}, if $n\leq 2$ in
Theorem~\ref{th:pnot2}, we may remove the hypothesis that $F$ is
perfect. The case $n=1$ in the two theorems recovers the results
of \cite{MS}.
\end{remark*}

\section{Bloch-Kato and Milnor $K$-theory}\label{se:bkktheory}

Our proof relies on the following two results in Voevodsky's proof
of the Bloch-Kato Conjecture. Because we apply Voevodsky's results 
in the case when the base field contains a primitive $p$th root 
of unity we shall formulate Voevodsky's results restricted
to this case. The first is the Bloch-Kato Conjecture itself:

\begin{theorem}[{\cite[Def.~5.1]{Vo1} and \cite[Thm.~7.1]{Vo2}}]
\label{th:bk}\
    \begin{enumerate}
    \item Let $F$ be a field containing 
	a primitive $p$th root of unity and
        $m\in\N$.  Then the norm residue homomorphism
    \begin{equation*}
        k_m F\to H^m(G_F, \mu_p)
    \end{equation*}
    is an isomorphism.

    \item For any cyclic extension $E/F$ of degree $p$, the
    sequence
    \begin{equation*}
        K_mE \xrightarrow{\sigma-1} K_mE \xrightarrow{N_{E/F}}
        K_mF
    \end{equation*}
    is exact.
    \end{enumerate}
\end{theorem}

\noindent The second result establishes an exact sequence
connecting $k_mF$ and $k_mE$ for consecutive $m$.  (We translate
the statement of the original result to $K$-theory using the
previous theorem.)  In the following result $a$ is chosen to
satisfy $E=F(\root{p}\of{a})$.

\begin{theorem}[{\cite[Prop.~5.2]{Vo1}}]\label{th:es}
    Let $F$ be a field  
    containing a primitive $p$th root of unity with no
    extensions of degree prime to $p$.  Then for any cyclic
    extension $E/F$ of degree $p$ and $m\ge 1$, the sequence
    \begin{equation*}
        k_{m-1}E \xrightarrow{N_{E/F}} k_{m-1}F \xrightarrow{\{a\}
        \cdot -} k_m F \xrightarrow{i_E} k_m E
    \end{equation*}
    is exact.
\end{theorem}

Now if $F$ is a perfect field, we
observe that we may remove the hypothesis that the field $F$ has
no extensions of degree prime to $p$.  It is precisely to ensure
that the sequence above is exact that we require the hypothesis
that $F$ is perfect in Theorem~\ref{th:pnot2}.  We give a proof of
the following theorem in section~\ref{se:esext}.

\begin{theorem}[{Modification of
Theorem~\ref{th:es}}]\label{th:esext}
    Let $F$ be a field containing a primitive
    $p$th root of unity, and if $p>2$ assume that $F$ is perfect.
    Then for any cyclic extension $E/F$ of degree $p$ and $m\geq
    1$ the sequence
    \begin{equation*}
        k_{m-1}E \xrightarrow{N_{E/F}} k_{m-1}F \xrightarrow{\{a\}
        \cdot -} k_m F \xrightarrow{i_E} k_m E
    \end{equation*}
    is exact.
\end{theorem}

\section{Notation and Lemmas}

For the remainder of the paper, except for section~\ref{Red}, we assume that $F$ is perfect if
$p>2$.  We fix $n\in \N$ and the cyclic extension
$E=F(\root{p}\of{a})$, and we write $k_{n-1}F = \ann\{a\} \oplus V
\oplus W$, where $\ann\{a, \xi_p\} = \ann\{a\} \oplus V$. Observe
that $\Upsilon_1=\dim_{\Fp} V$ and $\Upsilon_2=\dim_{\Fp} W$. We
show that the invariant $z$ is independent of the choice of $W$:

\begin{lemma}\label{le:windep}
    The invariant $z$ is independent of the particular complement
    $W$ of $\ann \{a, \xi_p\}$ in $k_{n-1}F$.
\end{lemma}

\begin{proof}
    This is obvious for $p=2$, so assume that $p>2$. Let $W'$ be
    another $\Fp$-subspace of $k_{n-1}F$ with $k_{n-1}F = \ann\{a,
    \xi_p\}\oplus W'$.  Then we may choose bases $\{w_i\}_{i\in
    \Ic}$ and $\{w'_i\}_{i\in \Ic}$ of $W$ and $W'$, respectively,
    such that $w'_i=w_i+u_i$ for $u_i \in \ann \{a,\xi_p\}$. Then
    $\{\xi_p\}\cdot w'_i = \{\xi_p\}\cdot w_i + \{\xi_p\}\cdot
    u_i$, and $\{\xi_p\}\cdot u_i \in \ann\{a\}$. Because
    $\ann\{a\} = N_{E/F}k_n E$ by Theorem~\ref{th:esext}, we see
    that
    \begin{equation*}
        \{\xi_p\}\cdot W + N_{E/F}k_nE = \{\xi_p\}\cdot W' +
        N_{E/F}k_nE
    \end{equation*}
    and therefore $z$ does not depend on the choice of $W$.
\end{proof}

We denote by $i_E\colon K_nF\to K_nE$ the map induced by the
inclusion of $F$ in $E$. In what follows we will frequently refer
to the element $\root{p}\of{a}$, and so we abbreviate it by $A$.
We will also often use the observation that if $p=2$ then
$\{a,\xi_p\}=\{a,-1\}=\{a,a\}\in k_2F$, while if $p>2$ then
$\{a,a\} = 0\in k_2 F$.  Finally, we will use the projection
formula for taking the norms of standard generators of $K_i F$
(see \cite[p.~81]{FW}).

\begin{lemma}\label{le:VW}
    We have the vector space isomorphism
    \begin{equation*}
        V\oplus W \xrightarrow{\{a\}\cdot -} \{a\}\cdot
        k_{n-1}F
    \end{equation*}
    and, if $p>2$, the compositum of the maps $\{\xi_p\}\cdot-$
    and $i_E$
    \begin{equation*}
        W \xrightarrow{\{\xi_p\}\cdot - } \{\xi_p\} \cdot W
        \xrightarrow{i_E}
        i_E(\{\xi_p\}\cdot W)
    \end{equation*}
    is a vector space isomorphism as well.
\end{lemma}

\begin{proof}
    The first isomorphism follows from the fact that $V\oplus W$
    is a complement in $k_{n-1}F$ of the kernel of
    multiplication by $\{a\}$. For the second, assume $p>2$.
    Suppose that $\bar w\in W$ and $\bar\alpha = \{\xi_p\}\cdot
    \bar w\in \ker i_E$. Then by Theorem~\ref{th:esext}, $\bar\alpha
    = \{a\}\cdot \bar c$ for $c\in K_{n-1}F$.  Since
	$\{a,a\}=0$ we see that $\{a\}\cdot \bar\alpha = 0$.  But then
    $\bar w \in \ann \{a,\xi_p\}$ and so $\bar w=0$.
\end{proof}

For $\gamma\in K_nE$, let $l(\gamma)$ denote the dimension of the
cyclic $\Fp[G]$-submodule $\langle \bar \gamma \rangle$ of
$k_n E$ generated by $\bar\gamma$. Then we have
\begin{equation*}
    (\sigma-1)^{l(\gamma)-1}\langle \bar\gamma \rangle = \langle
    \bar\gamma\rangle^G \neq 0 \text{\ \ \ and \ \ }
    (\sigma-1)^{l(\gamma)}\langle \bar\gamma \rangle=0.
\end{equation*}

We denote by $N$ the map $(\sigma-1)^{p-1}$ on $k_n E$. Because
$(\sigma-1)^{p-1}=1+\sigma+\dots+\sigma^{p-1}$ in $\Fp[G]$, we may
use $i_EN_{E/F}$ and $N$ interchangeably on $k_n E$.

\begin{lemma}\label{le:extend}
    Suppose $p>2$ and $\gamma\in K_n E$.
    \begin{enumerate}
    \item If $3\le l(\gamma) \le p$, then there exists $\alpha \in
    K_nE$ such that
    \begin{equation*}
        \langle N\bar\alpha \rangle = \langle \bar
        \gamma \rangle^G.
    \end{equation*}
    \item If $l(\gamma)=2$ and
    \begin{equation*}
        \bar \gamma \not\in \{A\} \cdot i_E(k_{n-1}F) + (k_n E)^G
    \end{equation*}
    then there exist $\alpha\in K_nE$ and $b\in K_{n-1}F$ such
    that
    \begin{equation*}
        \langle N\bar\alpha \rangle = \langle \bar \gamma +
        \{A\}\cdot i_E(\bar b) \rangle^G.
    \end{equation*}
    \end{enumerate}
\end{lemma}

\begin{proof}
    Let $l=l(\gamma)$ and suppose $3\le l\le i\le p$.  We show by
    induction on $i$ that there exists $\alpha_i\in K_nE$ such
    that $\left<(\sigma-1)^{i-1} \bar\alpha_i\right> = \langle \bar\gamma
    \rangle^G$.  Then setting $\alpha:=\alpha_p$, the proof will
    be complete.  If $i=l$ then $\alpha_i=\gamma$ suffices.
    Assume now that $l\le i<p$ and that our statement is true for
    $i$.

    Set $c=N_{E/F}\alpha_i$.  Since $i_E\bar c = N\bar\alpha_i =
    (\sigma-1)^{p-1}\bar\alpha_i$ and $i<p$, $i_E\bar c = 0$.  By
    Theorem~\ref{th:esext}, $\bar c = \{a\} \cdot \bar b$ for $b\in
    K_{n-1}F$.  Equivalently, $c=\{a\}\cdot b + p f$ for $f\in
    K_nF$. Then
    \begin{equation*}
        N_{E/F}\big(\alpha_i-(\{A\}\cdot
        i_E(b) + i_E(f) )\big) = 0.
    \end{equation*}
    By Theorem~\ref{th:bk}, there exists $\omega\in K_nE$ such
    that
    \begin{equation*}
        (\sigma-1)\omega = \alpha_i - (\{A\}\cdot
        i_E(b) + i_E(f)).
    \end{equation*}
    Then $(\sigma-1)^2\omega = (\sigma-1)\alpha_i - \{\xi_p\}\cdot
    i_E(b)$. Since $i\ge 3$, $\left<(\sigma-1)^i\bar \omega\right> = \langle
    \bar\gamma \rangle^G$ and we can set $\alpha_{i+1}=\omega$.

    For the second part, suppose $l=2=i$.  Proceeding in the same
    way as above, we see that for $\alpha_2=\gamma$ we have
    $N_{E/F}\alpha_2 = \{a\}\cdot b + pf $ for $b\in K_{n-1}F$ and
    $f\in K_nF$.  As before, there exists $\omega\in K_nE$ such
    that $(\sigma-1)\omega = \alpha_2 - (\{A\}\cdot
    i_E(b) + i_E(f))$. Then
    \begin{equation*}
        (\sigma-1)^2\omega = (\sigma-1)(\alpha_2-\{A\}
        \cdot i_E(b)) = (\sigma-1)(\gamma-\{A\}
        \cdot i_E(b)).
    \end{equation*}

    Observe that $\bar\gamma-\{A\}\cdot i_E(\bar b)\not\in (k_n
    E)^G$ by hypothesis.  Therefore $l(\gamma-\{A\}\cdot
    i_E(b))=2$ and we can set $\alpha_3:=\omega$.  We may then
    continue by induction on $i$ as above, concluding that there
    exists an element $\alpha = \alpha_p\in K_nE$ such that
    $\left<N\bar\alpha_p\right> = \left<(\sigma-1)^{p-1} \bar\alpha_p\right> = \langle
    \bar\gamma - \{A\}\cdot i_E(\bar b)\rangle^G$, as required.
\end{proof}

In the following lemma we elongate the exact sequence of
Theorem~\ref{th:esext}.

\begin{lemma}\label{le:fixedes}
    The following sequence is exact:
    \begin{equation*}
	0\to\ann\{a\}\to
        k_{n-1}F \xrightarrow{\{a\}\cdot -} k_n F
        \xrightarrow{i_E} (k_n E)^G \xrightarrow{N_{E/F}}
        \{a\}\cdot \ann \{a,\xi_p\}\to 0.
    \end{equation*}
Here the map $\ann\{a\}\to k_{n-1}F$ is the natural inclusion.
\end{lemma}
\begin{proof}
    We show first that $N_{E/F}((k_n E)^G)\subset \{a\} \cdot \ann
    \{a,\xi_p\}$.  Let $\bar \alpha\in (k_n E)^G$ and $\beta =
    N_{E/F}\alpha$. Since $i_E(N_{E/F}\bar\alpha) =
    (\sigma-1)^{p-1} \bar\alpha= 0$ we
    have that $\bar\beta = N_{E/F}\bar \alpha = \{a\}\cdot \bar b$
    for some $b\in K_{n-1}F$ by Theorem~\ref{th:esext}.

    Suppose $p=2$. Since $\bar\beta$ is in the image of $N_{E/F}$,
    we have by Theorem~\ref{th:esext} that $\{a\} \cdot \bar \beta = \{a,a\}
    \cdot \bar b = 0$.  Since $\{a,a\} = \{a,-1\}$, we
    have $\bar b \in \ann \{a,-1\}$.

    Now suppose that $p>2$.  Write $\beta = \{a\} \cdot b + pf$
    for some $f\in K_nF$. Then by the projection formula
    \begin{equation*}
        N_{E/F}\big( \alpha - (\{A\} \cdot i_E(b)
        + i_E(f)) \big) = 0.
    \end{equation*}
    By Theorem~\ref{th:bk}, there exists $\omega\in K_nE$ such
    that
    \begin{equation*}
        (\sigma-1)\omega = \alpha - (\{A\} \cdot i_E(b) - i_E(f)).
    \end{equation*}
    Then $(\sigma-1)^2\bar\omega = \{\xi_p\} \cdot i_E(\bar{b})$.

    If $(\sigma-1)^2\bar\omega = 0$ then since by
    Theorem~\ref{th:esext}, $\ker i_E = \{a\} \cdot k_{n-1}F$,
    \begin{equation*}
        \{\xi_p\} \cdot \bar b = \{a\} \cdot \bar h
    \end{equation*}
    for some $h\in K_{n-1}F$.  Because $\{a,a\}=0$, the
    right-hand side of the preceding equation is annihilated by
    $\{a\}$.  Therefore $\bar b\in \ann \{a,\xi_p\}$.

    If $(\sigma-1)^2\bar \omega \neq 0$ then $l(\omega)=3$ and
    Lemma~\ref{le:extend} shows that
    \begin{equation*}
        i_E(\{\xi_p\}\cdot \bar b) = cN\bar\lambda =
        i_E(N_{E/F}(\overline{c\lambda}))
    \end{equation*}
    for some $\lambda\in K_nE$ and $c\in \Z$.  Since by
    Theorem~\ref{th:esext}, $\ker i_E = \{a\}\cdot k_{n-1}F$ we have
    \begin{equation*}
        \{\xi_p\} \cdot \bar b = N_{E/F}(\overline{c\lambda}) + \{a\}
        \cdot \bar h
    \end{equation*}
    for some $h\in K_{n-1}F$.  Now by Theorem~\ref{th:esext}
    and the fact that $\{a,a\}=0$, the right-hand side of
    the preceding equation is annihilated by $\{a\}$.  Then
    $\bar b\in \ann \{a,\xi_p\}$.   Hence in all cases
    $N_{E/F}\bar\alpha \in \{a\}\cdot \ann \{a, \xi_p\}$.

    Exactness at the first two terms is obvious, and exactness at the third term follows from
    Theorem~\ref{th:esext}.

    For exactness at the fourth term, suppose 
	$$\bar \gamma\in (k_n E)^G\mbox{ and }N_{E/F}\bar\gamma = 0.$$
    Then $N_{E/F}\gamma= pf$ for $f\in K_nF$.  Let $\beta=\gamma-i_E(f)$.  Then
    $N_{E/F}\beta=0$ and by Theorem~\ref{th:bk} there exists
    $\alpha\in K_nE$ such that $(\sigma-1)\alpha = \beta$.
	If $p=2$ then $\bar\beta=i_E(N_{E/F}\bar\alpha)\in i_E k_n F$
	and we are done. Thus assume $p>2$. 

    Now suppose $\bar\alpha \in \{A\}\cdot i_E(k_{n-1}F) + (k_n
    E)^G$. Then
    \begin{equation*}
        \bar\beta = (\sigma-1)\bar \alpha \in \{\xi_p\} \cdot
        i_E(k_{n-1} F) \subset i_E(k_n F),
    \end{equation*}
    and hence $\bar\gamma = \bar\beta + i_E(\bar f)\in i_E(k_n F)$
    as well. Otherwise $\bar\alpha\notin\{A\}\cdot i_E(k_{n-1}F)+(k_n E)^G$.
	Now if $(\sigma-1)\bar\alpha=\bar\beta=0$ we are done as then 
	$\bar\gamma=i_E(\bar f)$. Hence assume $(\sigma-1)\bar\alpha\neq 0$. Then 
	$l(\alpha)=2$ and by Lemma~\ref{le:extend} we see that 
	there exist $\delta\in K_n E,b\in K_{n-1}F$ and 
	$c\in\Z$ such that 
		\begin{equation*}
			c N \bar\delta = (\sigma-1)(\bar\alpha+\{A\}\cdot i_E(\bar{b}))=\bar\beta+\{\xi_p\}\cdot i_E(\bar{b}). 
		\end{equation*}
        Thus $\bar\beta = c N \bar\delta-\{\xi_p\}\cdot i_E(\bar{b})\in i_E(k_n F)$ and  		
	exactness at the fourth term is established. 

Finally we show the exactness at the fifth term. Since
	\begin{equation*} 
		\{a\}\cdot\ann\{a,\xi_p\}=\{a\}\cdot V
	\end{equation*}
\noindent it is enough to show that each element $\{a\}\cdot\bar{v}$ where
$\bar{v}\in V$ can be written as $N_{E/F}\bar\alpha$
for some $\bar\alpha\in(k_n E)^G$. Observe that 
$(\sigma-1)(\{A\}\cdot i_E\bar{v})=\{\xi_p\}\cdot i_E(\bar{v})$.	
Also we have
	\begin{equation*}
		N_{E/F}(\{A\}\cdot i_E (\bar v))=
		  \begin{cases}
			\{a\}\cdot\bar v &\mbox{ if }p>2 \\
			\{-a\}\cdot\bar v &\mbox{ if }p=2.
		  \end{cases}
	\end{equation*}
Therefore it is enough to show that there exists an element 
$\bar\gamma\in k_n E$ such that $(\sigma-1)\bar\gamma=\{\xi_p\}\cdot i_E(\bar{v})$
and 
	\begin{equation*}
		N_{E/F}\bar\gamma =
		  \begin{cases}
			0 &\mbox{ if }p>2 \\
			\{-1\}\cdot\bar v &\mbox{ if }p=2.
		  \end{cases}
	\end{equation*}
Indeed then we can set $\bar\alpha=\{A\}\cdot i_E(\bar{v})-\bar\gamma$. 

Because $\bar{v}\in\ann\{a,\xi_p\}$ we see that 
$\{\xi_p\}\cdot i_E(\bar{v})\in\ann\{a\}$. 
By Theorem~\ref{th:esext} there exists $\bar\beta\in k_n E$ such
that 
\begin{equation*}
	\{\xi_p\}\cdot\bar{v}=N_{E/F}\bar\beta\mbox{ and }
	i_E(N_{E/F}\bar\beta)=(\sigma-1)^{p-1}\bar\beta.
\end{equation*}
Then setting $\bar\gamma=(\sigma-1)^{p-2}\bar\beta$ 
we obtain our required element. The proof of our lemma has now been
completed.
\end{proof}

Finally, we need a general lemma about $\Fp[G]$-modules.

\begin{lemma}[Exclusion Lemma]\label{le:excl}
    Let $M_1$ and $M_2$ be $\Fp[G]$-modules contained in a common
    $\Fp[G]$-module.  Suppose that $M_1^G\cap M_2^G=\{0\}$.  Then
    $M_1+M_2=M_1\oplus M_2$.
\end{lemma}

\begin{proof}
    Let $M=M_1 \cap M_2$ and suppose that $m\in M\setminus \{0\}$.  Let
    \begin{equation*}
        \tilde m=(\sigma-1)^{l(m)-1}(m)\neq 0.
    \end{equation*}
    Then $\tilde m\in M_1^G\cap M_2^G$, a contradiction. Hence
    $M_1\cap M_2=\{0\}$ and $M_1+M_2= M_1\oplus M_2$.
\end{proof}

\section{Construction of Submodules}

\begin{proposition}\label{pr:x}
    $k_n E$ contains a submodule $X_1$ such that
    \begin{itemize}
        \item $X_1$ is a trivial $\Fp[G]$-module of dimension
        $\Upsilon_1$
        \item $X_1 \cap i_E k_n F = \{0\}$
        \item $N_{E/F}$ restricts to an isomorphism
        $X_1\to \{a\}\cdot V$.
    \end{itemize}

    Moreover, if $p>2$, then $k_n E$ contains a submodule
    $X_2$, independent of $X_1$, such that
    \begin{itemize}
        \item $X_2$ is a direct sum of $\Upsilon_2$ cyclic
        submodules of dimension $2$ and $\dim_{\Fp}X_2^G=\Upsilon_2$. 
        \item $(X_1+X_2) \cap i_E k_n F = (\sigma-1)X_2 = X_2^G
        = i_E(\{\xi_p\}\cdot W)$
        \item We have an exact sequence
        \begin{equation*}
            0\to \{\xi_p\} \cdot W \xrightarrow{i_E}
            X_1+X_2 \xrightarrow{N_{E/F}} \{a\}
            \cdot k_{n-1}F \to 0
        \end{equation*}
    \end{itemize}
\end{proposition}

\begin{proof}
    Let $\Ic$ be an $\Fp$-basis for $V$.  Let $\bar v$ be an
    arbitrary element of $\Ic$, and consider $\bar\alpha =
    \{A\} \cdot i_E\bar v$.  Now
    $(\sigma-1)\bar\alpha = i_E(\{\xi_p\}\cdot \bar v)$.

    Since $\bar v\in \ann \{a,\xi_p\}$ we see that
	$\{\xi_p\}\cdot\bar v\in\ann_{k_n F}\{a\}$. 
        By Theorem~\ref{th:esext}
    \begin{equation*}
        \{\xi_p\}\cdot \bar v = N_{E/F}\bar\beta\mbox{ and }i_E(N_{E/F}\bar\beta)=
        N\bar\beta=(\sigma-1)^{p-1}\bar\beta
    \end{equation*}
    for some $\beta\in K_n E$.  Set $\gamma =
    (\sigma-1)^{p-2}\beta$ and $\bar x_v= \bar\alpha - \bar\gamma
    \in k_n E$.

    If $p=2$ then 
    \begin{equation*}
        N_{E/F}\bar x_v =\{-a\} \cdot \bar v - N_{E/F}\bar\gamma =
        \{-a\} \cdot \bar v - \{-1\}\cdot \bar v =
        \{a\}\cdot \bar v.
    \end{equation*}

    If $p>2$, then observe that since $\gamma$ is in the image of
    $\sigma-1$ we have $\overline{N_{E/F}\gamma}=0$.  Then, by the
    projection formula
    \begin{equation*}
        N_{E/F}\bar x_v = \{a\}\cdot \bar v -
        \overline{N_{E/F}\gamma} = \{a\} \cdot \bar v.
    \end{equation*}

    Now in either case, since $(\sigma-1)^{p-1}\bar\beta =
    i_E(N_{E/F}\bar\beta)$,
    \begin{equation*}
        (\sigma-1)\bar x_v = i_E(\{\xi_p\}\cdot \bar v) -
        (\sigma-1)^{p-1} \bar\beta \\ = i_E(\{\xi_p\}\cdot \bar
        v) - i_E(\{\xi_p\}\cdot \bar v) = 0.
    \end{equation*}

    Set
    \begin{equation*}
        X_1 := \oplus_{\bar v\in \Ic} \langle \bar x_v\rangle.
    \end{equation*}
    We have shown that $X_1$ is a trivial $\Fp[G]$-module.
    Moreover, because $N_{E/F}\bar x_v=\{a\}\cdot \bar v$ and
    $\{a\}\cdot -$ is injective on $V$ by Lemma~\ref{le:VW},
    \begin{equation*}
        N_{E/F}\big\vert_{X_1}\colon X_1\to \{a\}\cdot V
    \end{equation*}
    takes a basis of $X_1$ to a basis of $\{a\}\cdot V$ and
    $\dim_{\Fp} X_1 = \dim_{\Fp} V = \Upsilon_1$.  Finally, since
    $N_{E/F}$ is trivial on $i_E k_n F$, we have $X_1\cap i_E k_n
    F = \{0\}$.

    Now suppose that $p>2$.  Set
    \begin{equation*}
        X_2 := \left(\{A\}\cdot i_EW\right) +
        i_E(\{\xi_p\} \cdot W).
    \end{equation*}
    Let $\bar w\in W$ and consider $\bar x_w=\{A\}
    \cdot i_E(\bar w)$.

    Since $(\sigma-1)\bar x_w = i_E(\{\xi_p\}\cdot \bar w)$ and
    $(\sigma-1)(\{\xi_p\} \cdot \bar w) = 0$, we obtain
    $(\sigma-1)X_2 = i_E(\{\xi_p\}\cdot W)$.  Hence on $\{A\}\cdot
    i_E W$, $\sigma-1$ acts as $i_E(\{\xi_p\}\cdot -)$, which by
    Lemma~\ref{le:VW} is an isomorphism of vector spaces.  Hence
    $\sigma-1$ is an isomorphism as well.  Moreover, if an
    arbitrary $\{A\}\cdot i_E (\bar w_1) + \{\xi_p\} \cdot i_E (\bar w_2)\in
    X_2$ lies in the kernel of $\sigma-1$, $\bar w_1=0$.  Hence
    $X_2^G = i_E(\{\xi_p\}\cdot W)$. Since we already observed that 
	$X_1\cap i_E k_n F=\{0\}$ we see that $X_2^G\cap X_1=\{0\}$
	and by Lemma~\ref{le:excl} we conclude that $X_1+X_2=X_1\oplus X_2$.

    By the projection formula $N_{E/F}\bar x_w = \{a\}\cdot \bar
    w$ and by the definition of $W$, $\{a\}\cdot \bar w = 0$
    implies $\bar w = 0$.  Since $N_{E/F}(\{\xi_p\}\cdot i_E (\bar w_2))
    = 0$ for all $\bar w_2\in W$, we deduce that restricted to
    $X_2$, $N_{E/F}$ surjects $X_2$ onto $\{a\}\cdot W$ with kernel
    $i_E(\{\xi_p\}\cdot W)$.  By Lemma~\ref{le:VW}, $\{a\}\cdot
    k_{n-1}F = \{a\}\cdot (V+W)$; hence on $X_1\oplus X_2$,
    $N_{E/F}$ is a surjection onto $\{a\}\cdot k_{n-1}F$ with
    kernel $i_E(\{\xi_p\} \cdot W)$.

    Finally observe that $N_{E/F}i_E k_n F = \{0\}$. Hence
    \begin{equation*}
        (X_1+X_2) \cap i_E k_nF \subset i_E(\{\xi_p\}\cdot W).
    \end{equation*}
    Since $i_E(\{\xi_p\}\cdot W) \subset i_E k_n F$, we have
    equality.

    Now we have shown that $\sigma-1$ is an isomorphism of vector
    spaces $\{A\}\cdot i_E W \to i_E(\{\xi_p\}\cdot
    W)$, and by Lemma~\ref{le:VW}, we have an isomorphism $W\to
    i_E(\{\xi_p\}\cdot W)$.  Therefore $X_2$ is a direct sum
    of cyclic submodules $\langle \bar x_w \rangle$ of dimension
    $2$, with $\bar x_w$ in one-to-one correspondence with basis
    elements of $W$.  Hence the direct sum contains $\Upsilon_2$
    cyclic summands.
\end{proof}

If $p=2$, let $X=X_1$ be a submodule of $k_n E$ satisfying the
conditions of the preceding proposition.  If $p>2$, let
$X=X_1+X_2$ for $X_1$, $X_2$ satisfying the conditions of the
same.

\begin{proposition}\label{pr:y}
    $k_n E$ contains a submodule $Y$ independent from $X$
    such that
    \begin{itemize}
        \item $Y$ is a free $\Fp[G]$-module of rank
        \begin{equation*}
            y =
            \begin{cases}
            \dim_{\Fp} (N_{E/F}k_n E)/\{a\}\cdot k_{n-1}F, &p>2 \\
            \dim_{\F_2} (N_{E/F}k_n E)/\{a\}\cdot \ann_{k_{n-1}F}
            \{a,-1\}, &p=2
            \end{cases}
        \end{equation*}
    \item $Y^G = i_E N_{E/F}k_nE$
    \item if $p>2$, $\Upsilon_1 + \Upsilon_2 + y = e$
    \item if $p=2$, $\Upsilon_1 + y = e$
    \end{itemize}
\end{proposition}

\begin{proof}
    Let $\Ic$ be a basis for the subspace $i_E(N_{E/F}k_n E)$.
    For each basis element $\bar y\in \Ic$, let $\alpha_y\in K_nE$
    satisfy $i_E(N_{E/F}\bar \alpha_y) = \bar y$.  Then $\langle
    \bar\alpha_y \rangle$ is a cyclic submodule of dimension $p$,
    hence isomorphic to $\Fp[G]$, with
    \begin{equation*}
        \langle \bar\alpha_y\rangle^G = (\sigma-1)^{p-1}\langle
        \bar\alpha_y\rangle = \langle N\bar\alpha_y\rangle
        = \langle \bar y\rangle.
    \end{equation*}
    Set
    \begin{equation*}
        Y = \sum_{\bar y\in \Ic} \langle \bar\alpha_y \rangle.
    \end{equation*}
    By Lemma~\ref{le:excl}, $Y=\oplus_{\bar y\in \Ic} \langle
    \bar \alpha_y \rangle$ and so $Y$ is a free $\Fp[G]$-module.
    Moreover,
    \begin{equation*}
        Y^G=(\sigma-1)^{p-1}Y=NY=\oplus_{\bar y\in \Ic}
        \langle \bar y\rangle = i_E(N_{E/F}k_n E).
    \end{equation*}

    Now the rank of $Y$ is equal to the dimension of
    $i_E(N_{E/F}k_n E)$, or
    \begin{equation*}
        \dim_{\Fp} (N_{E/F}k_n E) / \left((N_{E/F}k_n E) \cap
        \ker i_E\right).
    \end{equation*}
    Now by Theorem~\ref{th:esext}, $N_{E/F}(k_n E) = \ann\{a\}$,
    and by the same Theorem, $\ker i_E = \{a\} \cdot k_{n-1}F$.
    Hence
    \begin{equation*}
        N_{E/F}(k_n E) \cap \ker i_E =
        \ann_{k_n F}\{a\} \cap \{a\}\cdot k_{n-1}F.
    \end{equation*}

    Suppose that $p=2$.  Since $\{a,a\} = \{a,-1\}$ we
    deduce that
    \begin{equation*}
        N_{E/F}(k_n E) \cap \ker i_E = \{a\} \cdot \ann
        \{a,-1\}.
    \end{equation*}
    The dimension of this subspace is equal to $\dim_{\Fp} \ann
    \{a,-1\}/\ann \{a\}$, or $\Upsilon_1$.

    Now suppose that $p>2$.  Since $\{a,a\}=0$, 
    $\{a\}\cdot k_{n-1}F \subset \ann \{a\}$ and we deduce that
    \begin{equation*}
        N_{E/F}(k_n E) \cap \ker i_E = \{a\}\cdot k_{n-1}F,
    \end{equation*}
    which is of dimension $\dim_{\Fp} (k_{n-1}F)/\ann \{a\}$, or
    $\Upsilon_1+\Upsilon_2$.

    As $e=\dim_{\Fp} N_{E/F}k_n E$, we deduce that if $p=2$ then
    $\Upsilon_1 + \rank Y = e$ and if $p>2$,
    $\Upsilon_1+\Upsilon_2+\rank Y = e$.

    Now we claim that $Y$ is independent from $X$.  Suppose $\bar
    \beta\in X^G\cap Y^G$.  Now $Y^G = i_E(N_{E/F} k_n E) \subset
    i_E(k_n F)$, so $\bar\beta = i_E(\bar\alpha)$ where $\bar
    \alpha = N_{E/F}\bar\gamma$ for $\gamma\in K_n E$.  If $p=2$
    then by Proposition~\ref{pr:x}, $i_E(k_n F) \cap X = \{0\}$, and
    so $X\cap Y = \{0\}$ by Lemma~\ref{le:excl}.

    If $p>2$, Proposition~\ref{pr:x} tells us that
    \begin{equation*}
        i_E(k_n F) \cap X = X_2^G = \{\xi_p\} \cdot i_E W.
    \end{equation*}
    Hence $\bar \beta = \{\xi_p\} \cdot i_E(\bar w)$ for $\bar
    w\in W$.  Since $i_E(\{\xi_p\} \cdot \bar w)=i_E(\bar\alpha)$
    and by Theorem~\ref{th:esext}, $\ker i_E = \{a\} \cdot
    k_{n-1}F$,
    \begin{equation}\label{eq:y}
        \{\xi_p\} \cdot \bar w = \bar \alpha + (\{a\} \cdot
        \bar f)
    \end{equation}
    for $f\in K_{n-1}F$.  Now because $\bar\alpha\in N_{E/F} k_n
    E$, by Theorem~\ref{th:esext}, $\{a\}\cdot \bar\alpha=0$.
    Moreover, $\{a,a\} = 0$ since we have assumed that
    $p>2$.  Hence the right-hand side of \eqref{eq:y} is
    annihilated by multiplication by $\{a\}$.  Therefore $\bar
    w\in \ann \{a,\xi_p\}$, and by the definition of $W$, $\bar w=
    0$. By Lemma~\ref{le:excl}, $X+Y=X\oplus Y$.
    \end{proof}

Now let $X$ and $Y$ be submodules satisfying the conditions of the
preceding propositions.

\begin{proposition}\label{pr:z}
    $k_n E$ contains a submodule $Z$ independent from $X+Y$
    such that
    \begin{itemize}
    \item $Z$ is a trivial $\Fp[G]$-module of dimension
        \begin{equation*}
            z =
            \begin{cases}
            \dim_{\Fp}(k_n F)/(\{\xi_p\}\cdot W +
            N_{E/F}k_n E), &p>2 \\
            \dim_{\F_2} (k_n F)/(\{a\}\cdot k_{n-1}F + N_{E/F}k_n
            E), &p=2
            \end{cases}
        \end{equation*}
    \item $(k_n E)^G=X^G+Y^G+Z$
    \item $\Upsilon_2 + z = d$
    \end{itemize}
\end{proposition}

\begin{proof}
    Let $Z$ be a complement of $(X^G+Y^G)\cap i_E(k_n F)$
    in $i_E(k_n F)$.  By Lemma~\ref{le:excl}, $(X+Y)+Z =
    (X+Y)\oplus Z$.

    Clearly $X^G+Y^G+Z\subset (k_n E)^G$.  Now suppose $\bar
    \alpha\in (k_n E)^G$ and let $\beta = N_{E/F}\alpha$. By
    Lemma~\ref{le:fixedes}, $\bar\beta = \{a\} \cdot \bar b$ for
    some $\bar b\in \ann \{a,\xi_p\}$.

    Let $\bar v\in V$ be the component of $\bar b$ in the
    decomposition $\ann \{a\} \oplus V$ of $\ann \{a,\xi_p\}$.  By
    Proposition~\ref{pr:x}, there exists $\bar \gamma\in
    X_1\subset X^G$ such that
    \begin{equation*}
        N_{E/F}\bar \gamma = \{a\} \cdot \bar v = \{a\} \cdot
        \bar b = \bar \beta.
    \end{equation*}
    Then $N_{E/F}(\bar \alpha - \bar\gamma)=0$.  By
    Lemma~\ref{le:fixedes}, $\bar\alpha-\bar\gamma\in
    i_E(k_n F)$.  But $i_E(k_n F)\subset X^G+Y^G+Z$.  Hence
    $\bar \alpha\in X^G+Y^G+Z$ and we have shown that $(k_n E)^G =
    X^G+Y^G+Z$.

    For the dimension of $Z$, assume first that $p>2$.  By
    Theorem~\ref{th:esext}, $N_{E/F}k_n E = \ann_{k_n F} \{a\}$
    and $\ker i_E = \{a\} \cdot k_{n-1}F$.  Since $\{a,a\} = 0$ we
    see that $\ker i_E \subset N_{E/F}k_n E$.  Hence
    \begin{equation*}
        d = \dim_{\Fp} \frac{k_n F}{N_{E/F}k_n E} = \dim_{\Fp}
        \frac{i_E(k_n F)}{i_E(N_{E/F}k_n E)}= \dim_{\Fp}
        \frac{i_E(k_n F)}{Y^G},
    \end{equation*}
    where in the last equation we use Proposition~\ref{pr:y} to
    identify $Y^G$. By Propositions~\ref{pr:x} and \ref{pr:y},
    $(X^G+Y^G)\cap i_E(k_n F) = X_2^G\oplus Y^G$. Hence
    $d=\dim_{\Fp} (X_2^G \oplus Y^G \oplus Z)/Y^G$. By
    Proposition~\ref{pr:x}, $\dim_{\Fp} X_2^G = \Upsilon_2$.
    Hence $\Upsilon_2 + \dim_{\Fp} Z = d$ for $p>2$. Also we see that
    \begin{equation*}
        \dim_{\Fp}Z = \dim_{\Fp}\frac{i_E (k_n F)}{X_2^G\oplus
        Y^G} = \dim_{\Fp} \frac{k_n F}{\{\xi_p\}\cdot W+N_{E/F}k_n
        E} = z.
    \end{equation*}

    Now assume $p=2$.  By Propositions~\ref{pr:x} and \ref{pr:y},
    $i_E(k_n F)\cap (X^G+Y^G) = Y^G$.  Proceeding as in the last
    case,
    \begin{align*}
        \dim_{\F_2} Z &= \dim_{\F_2} \frac{i_E(k_n F)}{Y^G} =
        \dim_{\F_2} \frac{i_E(k_nF)}{i_E(N_{E/F}k_nE)} \\ &=
        \dim_{\F_2} \frac{k_nF}{N_{E/F}k_nE + \ker i_E} \\ &=
        \dim_{\F_2} \frac{k_n F}{N_{E/F}k_n E +
        \{a\}\cdot k_{n-1}F} = z,
    \end{align*}
    since $\ker i_E = \{a\}\cdot k_{n-1}F$, by
    Theorem~\ref{th:esext}.

    We then consider the filtration
    \begin{equation*}
        k_n F \supset \Big((\{a\}\cdot k_{n-1}F)
        + N_{E/F}k_n E\Big) \supset N_{E/F}k_n E.
    \end{equation*}
    The dimension of the quotient of the first and third modules
    is, by definition, $d$.  By Theorem~\ref{th:esext},
    $N_{E/F}k_n E = \ann_{k_n F} \{a\}$. Since $\{a,a\}=\{a,-1\}$
    we see that
    \begin{equation*}
        (\{a\}\cdot k_{n-1}F) \cap N_{E/F}k_n E = \{a\} \cdot V.
    \end{equation*}
    Hence
    \begin{equation*}
       \dim_{\F_2} \frac{(\{a\}\cdot k_{n-1}F) +
       N_{E/F}k_n E}{ N_{E/F}k_n E} = \dim_{\F_2} \frac{\{a\}
       \cdot k_{n-1}F}{\{a\} \cdot V} = \dim_{\F_2} \{a\}\cdot W.
    \end{equation*}
    By Lemma~\ref{le:VW}, $\dim_{\F_2} \{a\}\cdot W = \dim_{\F_2}
    W = \Upsilon_2$. Hence $\Upsilon_2 + \dim_{\F_2} Z = d$ for
    $p=2$ as well.
\end{proof}

\section{Proofs of Theorems \ref{th:pnot2} and \ref{th:p2}}

\begin{proof}[Proof of Theorem~\ref{th:pnot2}]
    By Propositions~\ref{pr:x}, \ref{pr:y}, and \ref{pr:z}, there
    exist independent submodules $X=X_1+X_2$, $Y$, and $Z$
    satisfying the conditions of the theorem.  All that remains
    is to show that $k_n E = X+Y+Z$.

    We proceed by induction on the length $l(\gamma)$ of the
    cyclic submodule $\langle \bar \gamma \rangle$ of $k_n E$
    generated by an arbitrary element $\bar\gamma\in k_n E$.  If
    $l(\gamma)=1$, then by Proposition~\ref{pr:z}, $\bar\gamma\in
    X^G+Y^G+Z$.  Assume then that $\bar \beta\in X+Y+Z$ if
    $l(\beta)\le i<p$ and that $l(\gamma)=i+1$.

    Suppose first that $l(\gamma)=2$ and
    \begin{equation*}
        \bar\gamma \in \{A\}\cdot i_E(k_{n-1}F) +
        (k_n E)^G.
    \end{equation*}
    Then $(\sigma-1)\bar \gamma = i_E(\{\xi_p\} \cdot \bar b)$ for
    some $b\in K_{n-1}F$.  In the decomposition $\ann \{a,\xi_p\}
    \oplus W$ of $k_{n-1}F$, write $\bar b = \bar f + \bar w$. By
    Proposition~\ref{pr:x} there exists $\bar\omega\in X_2$ such
    that $(\sigma-1)\bar\omega = i_E(\{\xi_p\}\cdot \bar w)$. We
    also have $\{\xi_p\}\cdot \bar f \in \ann\{a\}$ and therefore
    by Theorem~~\ref{th:esext} and Proposition~\ref{pr:y} there
    exists $\bar y\in Y$ such that $i_E(\{\xi_p\}\cdot
    \bar{f})=i_E(N_{E/F}(\bar y))$. Hence there exists
    $\overline{y'}\in Y$ such that $(\sigma-1) \bar\gamma =
    (\sigma-1)\bar\omega + (\sigma-1)\overline{y'}$. Hence
    $l(\gamma-\omega-y')\le 1$ and by the inductive hypothesis
    $\bar\gamma\in X+Y+Z$.

    Now since by the preceding arguments $\{A\}\cdot i_E(k_{n-1}F)\subset X+Y+Z$, in order to
    show that an arbitrary $\bar\gamma$ with $l(\gamma)=2$ lies in
    $X+Y+Z$ it is enough to show that $\bar\gamma + \{A\} \cdot
    i_E(\bar b)\in X+Y+Z$ for any $b\in K_{n-1}F$.

    Suppose then that $l(\gamma)=2$ and
    \begin{equation*}
        \bar\gamma \not\in \{A\}\cdot i_E(k_{n-1}F) + (k_n E)^G.
    \end{equation*}
    Then, by Lemma~\ref{le:extend}, there exist $b\in K_n F$ and $\alpha\in K_n E$ 
    such that $\bar\beta = \bar\gamma + \{A\}\cdot
    i_E(\bar b)$ satisfies $l(\bar\beta)\leq 2$ and $\langle \bar\beta \rangle^G = \langle
    N\bar\alpha \rangle$. Hence $(\sigma-1)^{l(\beta)-1}\bar\beta
    = cN\bar\alpha$ for some $c\in \Z$.  But $cN\bar\alpha =
    i_EN_{E/F}(\overline{c\alpha})\in Y^G$, by Proposition~\ref{pr:y}.
    Hence there exists $\bar\omega\in Y$ such that
    $(\sigma-1)^{p-1}\bar\omega = cN\bar\alpha$.  Now $\bar\lambda
    = (\sigma-1)^{p-l(\beta)}\bar\omega\in Y$ and
    $(\sigma-1)^{l(\beta)-1}(\bar\beta-\bar\lambda)=0$.  Hence
    $l(\beta-\lambda)<l(\gamma)$ and by the inductive hypothesis
    $\bar\beta$ and hence $\bar\gamma$ lie in $X+Y+Z$.

    If $l(\gamma)\ge 3$ then the same argument works again.  By
    Lemma~\ref{le:extend} $\langle \bar\gamma\rangle^G = \langle
    N\bar\alpha\rangle$ and so $(\sigma-1)^{l(\gamma)-1}\bar\gamma
    = cN\bar\alpha$ for some $c\in \Z$. But $cN\bar\alpha =
    i_EN_{E/F}(c\bar\alpha) \in Y^G$, by Proposition~\ref{pr:y}.
    Hence there exists $\bar\omega\in Y$ such that
    $(\sigma-1)^{p-1}\bar\omega = cN\bar\alpha$.  Now $\bar\lambda
    = (\sigma-1)^{p-l(\gamma)}\bar\omega \in Y$ and
    $(\sigma-1)^{l(\gamma)-1}(\bar\gamma-\bar\lambda) = 0$.  Hence
    $l(\gamma-\lambda)<l(\gamma)$ and by the inductive hypothesis
    $\bar\gamma\in X+Y+Z$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{th:p2}]
    By Propositions~\ref{pr:x}, \ref{pr:y}, and \ref{pr:z}, there
    exist independent submodules $X=X_1$, $Y$, and $Z$ satisfying
    the conditions of the theorem.  All that remains is to show
    that $k_n E = X+Y+Z$.

    Let $\bar\gamma\in k_n E$ be arbitrary.  If $l(\gamma)=1$,
    then by Proposition~\ref{pr:z}, $\bar\gamma\in X^G+Y^G+Z$.
    Otherwise $(\sigma-1)\bar\gamma = (\sigma+1)\bar\gamma =
    i_EN_{E/F}\bar\gamma\in Y^G$, by Proposition~\ref{pr:y}.
    Hence there exists $\bar\omega\in Y$ such that
    $(\sigma-1)\bar\omega = (\sigma-1)\bar\gamma$.  Therefore
    $l(\gamma-\omega)<2$ and by the inductive hypothesis
    $\bar\gamma\in X+Y+Z$.
\end{proof}

\section{Proof of Theorem~\ref{th:esext}}\label{se:esext}

For the case $p=2$ we have the long exact sequence of Galois
cohomology groups due to Arason \cite[Satz~4.5]{A}.  Suppose then
that $p>2$ and $F$ is perfect. Let $S$ be any $p$-Sylow subgroup
of $G_F=\Gal(F_{\sep}/F)$, and set $L$ to be the fixed field of
$S$. Because $F$ is perfect, the separable closure $F_{\sep}$ is
identical to the algebraic closure $\bar F$, and hence each finite
extension of $L$ has degree a power of $p$. In particular,
all of the hypotheses of Theorem~\ref{th:es} are valid for the
field $L$ in place of $F$. Furthermore, $([L:F],p)=1$. (Here we
use basic properties of supernatural numbers and Sylow
$p$-subgroups. See \cite[Chapter~1]{Ser}.) Therefore if
$E=F(\root{p}\of{a})$ is a cyclic extension of $F$ of degree $p$,
so is $EL=L(\root{p}\of{a})$ over $L$.  By Theorem~\ref{th:es} we
see that the sequence
\begin{equation*}
    k_{m-1}EL\xrightarrow{N_{EL/L}} k_{m-1}L
    \xrightarrow{\{a\}\cdot-} k_m L\xrightarrow{i_{EL}} k_m EL
\end{equation*}
is exact for each $m\in\N$.

We claim that $i_L\colon k_mF \to k_mL$ is injective. Indeed,
suppose that $i_L(\alpha)=0$ for some $\alpha\in k_m F$. Then
there exists a finite subextension $M/F$ of $L/F$ such that
$i_M(\alpha)=0$. Then
\begin{equation*}
    0=N_{M/F}(i_M(\alpha))=[M:F]\alpha,
\end{equation*}
(see \cite[p.~300]{FV}). Because $[M:F]$ is coprime with $p$, we
see that $\alpha=0$ and $i_L$ is injective as asserted. Similarly
we have that $i_{EL}\colon k_m E\to k_m EL$ is injective.

We then have the following commutative diagram:
\begin{equation*}
\xymatrix{ k_{m-1}EL \ar[r]^{N_{EL/L}} & k_{m-1}L
\ar[r]^{\{a\}\cdot -} & k_m L \ar[r]^{i_{LE}} & k_m EL \\ k_{m-1}E
\ar[r]^{N_{E/L}} \ar[u]_{i_{EL}} & k_{m-1}F \ar[r]^{\{a\}\cdot -}
\ar[u]_{i_{L}} & k_m F \ar[r]^{i_E} \ar[u]_{i_{L}} &
k_m E \ar[u]_{i_{EL}} \\
}
\end{equation*}
Because the vertical maps are injective, we see that the bottom
row of the diagram is a complex: the composition of any two
consecutive maps is the zero map. We now establish exactness at
the second and third terms of the complex.

Let $\alpha\in k_{m-1}F$ such that $\{a\}\cdot\alpha=0$. Then
$\{a\}\cdot i_L(\alpha)=0$ and therefore there exists an element
$\beta\in k_{m-1}EL$ such that $N_{EL/L}(\beta)=i_L(\alpha)$. Let
$M/F$ be a finite extension such that $\beta$ is defined over
$EM$. Then $N_{EM/M}(\beta)=i_M(\alpha)$, and we have
\begin{equation*}
    N_{EM/F}(\beta) = N_{M/F}(N_{EM/M}(\beta)) = N_{M/F}(i_M
    (\alpha)) = [M:F]\alpha
\end{equation*}
and $N_{EM/F}(\beta)=N_{E/F}(N_{EM/E}(\beta))$. Thus
\begin{equation*}
    N_{E/F}(N_{EM/E}(\beta))=[M:F]\alpha.
\end{equation*}
Because $([M:F],p)=1$ we see that $\alpha\in N_{E/F}(k_{m-1}E)$.
Therefore we have established the exactness of our complex at
$k_{m-1}F$.

Now assume that $\alpha\in k_m F$ such that $i_E(\alpha)=0\in k_m
E$. Then arguing as above, we see that there exist a finite
extension $M/F$ and $\beta\in k_{m-1}M$ such that
\begin{equation*}
    \{a\}\cdot \beta = i_M(\alpha)\in k_m M.
\end{equation*}
Applying $N_{M/F}$ and using the projection formula we see that
\begin{equation*}
    \{a\}\cdot N_{M/F}(\beta) = N_{M/F}(i_M(\alpha)) =
    [M:F]\alpha.
\end{equation*}
Because $[M:F]$ is coprime with $p$, $\alpha\in\{a\}\cdot
N_{M/F}(\gamma)$ for a suitable element $\gamma\in k_{m-1}M$. Hence
we see that our complex is also exact at $k_m F$ and the full
complex is exact. \qed

\begin{remark*}
Assuming as usual that $F$ contains a primitive root $\xi_p$, then
if $m\leq 2$, no further assumption on $F$ in
Theorem~\ref{th:esext} is necessary. For $m=1$ this claim follows
from basic Kummer theory, and for $m=2$ see \cite[Prop.~5]{Me} and
\cite[Chap.~5, Lemma~8.4]{Sr}.
\end{remark*}

\section{Reduction to the Case $\chr F = 0$}\label{Red}

Suppose now that $\chr F = q > 0$ and $q\neq p$. We also assume
that $F$ is infinite, because if $F$ is finite then $K_n F=0$ for
$n\geq 2$ and therefore this is a trivial case. (See
\cite[Prop.~IX.1.3]{FV}.)  Assume as before that $F$ contains a
primitive $p$th root of unity $\xi_p$ and $E/F$ is a cyclic
extension of degree $p$. We shall show that there exists an
explicitly defined cyclic extension $J/L$ of degree $p$ such that
$\chr L = 0$, so that $L$ is perfect, and $k_n J$ is naturally
isomorphic with $k_n E$ as a $G=\Gal(E/F) \cong \Gal(J/L)$ module.

Recall first that there exists a discrete valuation ring $A$ of
characteristic $0$ such that its maximal ideal $M$ is generated by
$q$ and $A/M\cong F$. (Such a ring is called a $q$-ring. See
\cite[p.~223]{Ma}.) By passing to a completion $\hat A$ of $A$
with respect to $M$-adic topology and observing that $\hat A/\hat
M\cong A/M\cong F$ and $\hat M=\hat A.q$ we see that we may and
will assume that $A$ is a complete local $q$-ring such that
$A/M\cong F$.

It is known that a complete $q$-ring is uniquely determined up to
its isomorphism by its residue field. (See \cite[Cor.,
p.~225]{Ma}.) Observe further that a complete discrete valued
field is henselian. (See \cite[Thm.~5]{R}.)

Now following \cite[\S IX.3]{FW} we have a natural construction
\begin{align*}
    R &=\varinjlim(A^{(1)}\subset A^{(2)}\subset
    A^{(3)}\subset\cdots), \mbox{ where } \\
    A^{(1)} &=A \mbox{ as above and } A^{(n+1)} :=
    A^{(n)}[t]/(t^p-\pi_n),
\end{align*}
where $\pi_n$ is a uniformizer of $A^{(n)}$ for $n\ge 1$. As was
noticed in \cite{FW}, this ring $R$ is a henselian valuation ring
of characteristic $0$, with value group the underlying group of
the ring $\Z[1/p]$.

For each $i\in\N$ let $L^{(i)}$ be the quotient field of
$A^{(i)}$, and let $L$ be the quotient field of $R$. Then from
Lemma~IX.3.5 in \cite{FW}, there is a natural isomorphism $k_n
F\cong k_nL$ for each $n\in\N$.

Let $T$ be the inertia subgroup of $G_{L^{(1)}}$. We have the
natural isomorphism $G_{L^{(1)}}/T\xrightarrow{\thicksim} G_F$.
Let $\varphi$ be the compositum of the natural surjections
\begin{equation*}
    G_{L^{(1)}}\to G_{L^{(1)}}/T \xrightarrow{\thicksim} G_F
    \to \Gal(E/F),
\end{equation*}
and let $J^{(1)}$ be the fixed field of $\ker\varphi$. Then
$\Gal(J^{(1)}/L^{(1)})$ is naturally isomorphic to $\Gal(E/F)$.
(This is a special case of a more general construction about
lifting certain Galois abelian extensions. See
\cite[Lemma~2.5]{K}.)

Thus we see that $J^{(1)}/L^{(1)}$ is a cyclic, purely inert
extension of degree $p$. Because the tower
\begin{equation*}
    L^{(1)}\subset L^{(2)}\subset\dots\subset L^{(n)}\subset
    \cdots
\end{equation*}
is a chain of totally ramified extensions $L^{(n+1)}/L^{(n)}$, we
see that $J^{(1)}\cap L=L^{(1)}$. Set $J:=J^{(1)} L$. Then $J/L$
is a cyclic, purely inert extension of degree $p$.

Now set $J^{(i)}:=J^{(1)}L^{(i)}$, and let $B^{(i)}$ be the unique
valuation ring in $J^{(i)}$ such that $B^{(i)}\cap
L^{(i)}=A^{(i)}$. ($B^{(i)}$ is unique because $A^{(i)}$ is
henselian.) Then again following the proofs of Lemma~IX.3.2 and
Lemma~IX.3.5 in \cite{FW}, we establish, in exactly the same way
we proved that $k_nF\cong k_nL$, that $k_n E\cong k_n J$ under a
$G$-equivariant isomorphism.  Our reduction is complete.

\section{Acknowledgements}

We are very grateful to M.~Rost and to J.-P. Serre for correspondence concerning the Bloch-Kato conjecture. 
We are also very grateful to J.-P.~Tignol for his kind and encouraging remarks concerning this paper. 

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\end{document}
