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\begin{document}

\title[Galois Module Structure of Milnor $K$-theory]
{Galois Module Structure of Milnor $K$-theory in Characteristic $p$}

\author[Bhandari]{Ganesh Bhandari}
\address{Department of Mathematics, Middlesex College, \
University of Western Ontario, London, Ontario \ N6A 5B7 \ CANADA}
\email{gbhandar@uwo.ca}

\author[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[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[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 NSA grant MDA904-02-1-0061.}
\email{joswallow@davidson.edu} \subjclass[2000]{19D45, 12F10}
\keywords{Milnor $K$-groups modulo $p$, cyclic extensions, Galois
modules}

\begin{abstract}
Let $E$ be a cyclic extension of degree $p^n$ of a field $F$ of
characteristic $p$.  We determine $k_mE$, the Milnor $K$-groups
modulo $p$, as $\Fp[\Gal(E/F)]$-modules for all $m\in \N$.
\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

Bloch, Gabber, and Kato established an isomorphism between, on one
hand, Milnor $K$-theory modulo $p$ in characteristic $p$ and, on
the other, the kernel of the Artin-Schreier operator defined on
the exterior algebra on K\"ahler differentials \cite{BK, K}. In
order to better understand Milnor $K$-theory, it is a natural
problem to determine the Galois module structure of Milnor
$K$-theory when the field is a Galois extension. Two significant
papers by Izhboldin consider Milnor $K$-theory in characteristic
$p$ and, using the previous work of Bloch, Gabber, and Kato,
ascertain its important Galois module properties \cite{I1, I2}.

In this paper we establish the remarkable fact that two results of
Izhboldin are enough to determine precisely the Galois module
structure of the Milnor groups $K_mE$ mod $p$ when $E$ is cyclic
extension of $p$th-power degree of a field $F$ of characteristic
$p$.  The result is simple and useful: these modules are direct
sums of trivial modules and modules free over some quotient of the
Galois group. Equivalently, the dimensions over $\Fp$ of the
indecomposable $\Fp[\Gal(E/F)]$-modules occurring as direct
summands of $K_m E/pK_m E$ are all powers of $p$.  Such a
description represents a culmination of results of Bloch-Kato,
\linebreak Gabber, and Izhboldin, available since 1987, concerning
the Galois module structure of Milnor $K$-groups mod $p$.

The precise description of the $\Fp[\Gal(E/F)]$-module
$K_mE/pK_mE$ depends on a natural filtration of $K_mF/pK_mF$
obtained from images of norm maps defined on Milnor $K$-groups mod
$p$ of intermediate fields of $E/F$.  This description, together
with the arguments in the proofs, reveals a flavor of abstract
class field theory.  In place of the axioms of class formations,
we use Izhboldin's theorems below, together with some technical
but relatively short consequences of Hilbert 90, which one might
call ``Hilbert 90 algebra.''

We assume in what follows that all fields have characteristic $p$
and that $m$ is a fixed natural number. For a field $F$, let
$K_mF$ denote the $m$th Milnor $K$-group of $F$ and $k_mF=K_mF/
pK_mF$. (See, for instance, \cite{Mi} and \cite[IX.1]{FV}.) If
$E/F$ is a Galois extension of fields, let $G=\Gal(E/F)$ denote
the associated Galois group. We write $i_E\colon K_mF\to K_mE$ and
$N_{E/F}\colon K_mE\to K_mF$ for the natural inclusion and norm
maps, and we use the same notation for the induced maps modulo $p$
between $k_mF$ and $k_mE$. In order to avoid possible confusion,
in a few instances we write $i_{F,E}$ instead of $i_E$.

Izhboldin's results are as follows.

\begin{theorem}[{\cite[Lemma 2.3]{I2}}]\label{th:isop}
Suppose $E/F$ is cyclic of degree $p$.  Then $i_E\colon k_mF\to
(k_mE)^G$ is an isomorphism.
\end{theorem}

It is useful to observe that Theorem~\ref{th:isop} generalizes to
cyclic extensions of degree $p^n$. For $\alpha\in K_mE$ we write
$\bar{\alpha}$ for the class of $\alpha$ in $k_mE$.

\begin{theorem}\label{th:isopn} Suppose that $E/F$ is cyclic of
degree $p^n$, $n\in\N$. Then $i_E\colon k_mF\to (k_mE)^G$ is an
isomorphism.
\end{theorem}

\begin{proof}
We use induction on $n$. By Theorem~\ref{th:isop} our statement is
true for $n=1$. Consider the tower $F\subset E_{n-1}\subset E$
where $E/E_{n-1}$ is a cyclic extension of degree $p$. Set
$H_{n-1} := \Gal(E/E_{n-1})$ and $G_{n-1} := \Gal(E_{n-1}/F)$.
Assume by induction that $i_{E_{n-1}}\colon k_mF \to
(k_mE_{n-1})^{G_{n-1}}$ is an isomorphism. From Theorem~
\ref{th:isop} we see that $i_{E_{n-1},E}\colon k_mE_{n-1}\to
(k_mE)^{H_{n-1}}$ is an isomorphism. Hence $i_E\colon k_mF\to
k_mE$ is injective.

Because $i_{F,E}(k_mF)\subset(k_mE)^G$ it is sufficient to show
that each $\bar\alpha\in(k_mE)^G$ can be written as $i_{F,E}
(\bar{f})$ for some $f\in K_mF$.

Since $\bar\alpha\in(k_mE)^G\subset(k_mE)^{H_{n-1}}$, by
Theorem~\ref{th:isop} we see that there exists $\bar\gamma \in
k_mE_{n-1}$ such that $i_{E_{n-1},E}\bar\gamma=\bar \alpha$.
Because $\bar\alpha\in(k_mE)^G$ we see also that $\bar\gamma
\in(k_mE_{n-1})^{G_{n-1}}$ and therefore there exists $f\in K_mF$
such that $i_{F,E_{n-1}}(\bar{f})= \bar\gamma$. Then
$i_{F,E}(\bar{f}) = \bar \alpha$ as required.
\end{proof}

\begin{theorem}[Hilbert 90 for Milnor $K$-theory: {\cite[Corollary of
Proposition~5]{I1}}, {\cite[Theorem~D]{I2}}]\label{th:exseq}
    Suppose $E/F$ is cyclic of $p$th-power degree. Then the
    sequence
    \begin{equation*}
        K_mE \xrightarrow{1-\sigma} K_mE \xrightarrow{N_{E/F}}
        K_mF
    \end{equation*}
    is exact.
\end{theorem}

Now suppose $E/F$ is cyclic of degree $p^n$.  For $i=0, \dots, n$,
let $E_i/F$ be the subextension of degree $p^i$ of $E/F$,
$H_i=\Gal(E/E_i)$, and $G_i=\Gal(E_i/F)$. Our main result is the
following
\begin{theorem}\label{th:main}
    There exists an isomorphism of $\Fp[G]$-modules
    $k_mE \cong \oplus_{i=0}^n Y_i$, where
    \begin{itemize}
    \item $Y_n$ is a free $\Fp[G]$-module of rank $\dim_{\Fp}
    N_{E/F}k_mE$,
    \item $Y_i$, $0<i<n$, is a free $\Fp[G_i]$-module of rank
    \begin{equation*}
        \dim_{\Fp} N_{E_i/F} k_mE_i/N_{E_{i+1}/F}k_mE_{i+1},
    \mbox{ and }
    \end{equation*}
    \item $Y_0$ is a trivial $\Fp[G]$-module of rank $\dim_{\Fp}
    k_mF/N_{E_1/F} k_mE_1$.
    \end{itemize}
\end{theorem}

Our proof of Theorem~\ref{th:main} proceeds by induction on $n$.
At each stage we use the norm map images $N_{E_i/F}k_mE_i$,
$i=1,\dots,n$, to construct the modules $Y_i$.  The most
interesting portion is the construction of $Y_{n-1}$. It is easy
to show that the modules $Y_i$ are all independent, and we then
use Theorems~\ref{th:isopn} and \ref{th:exseq} to show that the
sum of the $Y_i$ exhausts $k_mE$.

The case $n=1$ is section~\ref{se:nequal1}, and in
section~\ref{se:induct} we present the inductive argument.  This
argument relies on an extension lemma, Lemma~\ref{le:fullext},
which shows that certain elements in $k_mE$ are expressible as
norms.  That the various norm groups $N_{E_i/F}k_mE_i$ contain
enough elements is, in fact, the crucial step in our induction.
Lemma~\ref{le:fullext}, together with two more basic extension
lemmas used in the proof of that lemma, are presented in
section~\ref{se:nl}, along with technical results on
$\Fp[G]$-modules.

This work uses some ideas previously developed in investigations
of the structure of $p$th-power classes of field extensions of
$p$th-power degree \cite{MSS}.

\section{Notation and Lemmas}\label{se:nl}

\subsection{$\Fp[G]$-modules} For the reader's convenience,
after introducing some notation, we recall in this section some
basic elementary facts about $\Fp[G]$-modules.

Let $G$ be a cyclic group of order $p^n$ with generator $\sigma$.
For an $\Fp[G]$-module $U$, let $U^G$ denote the submodule of $U$
fixed by $G$.  For an arbitrary element $u\in U$, let $l(u)$
denote the dimension over $\Fp$ of the $\Fp[G]$-submodule $\langle
u\rangle$ of $U$ generated by $u$. Then we have
\begin{equation*}
    (\sigma-1)^{l(u)-1}\langle u \rangle =
    \langle u \rangle^G \neq \{0\} \text{\quad and\quad}
    (\sigma-1)^{l(u)}\langle u \rangle = \{0\}.
\end{equation*}
As usual, if $U$ is a free $\Fp[G]$-module with $U =
\oplus_{i\in\Ic} \Fp[G]$, we say that $U$ is a module of rank
$\vert\Ic\vert$. Denote by $N$ the operator $(\sigma-1)^{p^n-1}$
acting on module $U$.

\begin{lemma}[Inclusion Lemma]\label{le:incl}
    Let $U$ and $V$ be $\Fp[G]$-modules contained in an
    $\Fp[G]$-module $W$.  Suppose that $(U+V)^G\subset U$ and for each
    $w \in (U+V)\setminus (U+V)^G$ there exists $u\in U$ such that
    \begin{equation*}
        (\sigma-1)^{l(w)-1}(w)=N(u).
    \end{equation*}
    Then $V\subset U$.
\end{lemma}

\begin{proof}
    Let $\{T_i\}_{i=1}^s$ be the socle series of $U+V$:
    $T_1=(U+V)^G$ and $T_{i+1}/T_i=((U+V)/T_i)^G$, and let $s$ be
    the least natural number such that $T_s=U+V$. Observe that
    since $(\sigma-1)^{p^n}=0$, we have $s\le p^n$.  We prove the
    lemma by induction on the socle series.

    By hypothesis, $T_1\subset U$.  Assume now that $T_i\subset U$
    for some $i<s$. Then for each $w\in T_{i+1}\setminus T_i$ we
    have $l(w)= i+1$ and $(\sigma-1)^{l(w)-1}(w) = N(u) =
    (\sigma-1)^{p^n-1}(u)$ for some $u\in U$.  Therefore
    \begin{equation*}
        (\sigma-1)^{l(w)-1} \big(w-(\sigma-1)^{p^n-l(w)}(u)\big) =
        0.
    \end{equation*}
    Therefore $w-(\sigma-1)^{p^n-l(w)}(u)\in T_i\subset U$.  Hence
    $w\in U$ and $T_{i+1}\subset U$ as required.
\end{proof}

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

\begin{proof}
    Let $Z=U\cap V$ and suppose that $y\in Z\setminus \{0\}$.  Let
    \begin{equation*}
        z=(\sigma-1)^{l(y)-1}(y)\neq 0.
    \end{equation*}
    Then $z\in U^G\cap V^G$, a contradiction. Hence $U\cap
    V=\{0\}$ and $U+V=U\oplus V$.
\end{proof}

The following lemma follows from the fact that each free
$\Fp[G]$-module is injective.  (See \cite[Theorem~11.2]{Ca}.) We
shall, however, provide a direct proof.

\begin{lemma}[Free Complement Lemma]\label{le:freecomp}
    Let $V\subset U$ be free $\Fp[G]$-modules.  Then there exists
    a free $\Fp[G]$-submodule $\tilde V$ of $U$ such that $V\oplus
    \tilde V=U$.
\end{lemma}

\begin{proof}
    Let $Z$ be a complement of $V^G$ in $U^G$ as $\Fp$-vector
    spaces, and let $\Zc$ be an $\Fp$-base of $Z$.  For each $z\in
    \Zc$, there exists $u(z)$ such that $z=N(u(z))$.  Let $M(z)$
    be the $\Fp[G]$-submodule of $W$ generated by $u(z)$. Then
    $M(z)$ is a free $\Fp[G]$-submodule.  Moreover, its fixed
    submodule $M(z)^G$ is the $\Fp$-vector subspace generated by
    $z$. By the Exclusion Lemma, the set of modules $M(z)$, $z\in
    \Zc$, is independent.

    Set $\tilde V := \oplus_{z\in \Zc} M(z)$.  Then $\tilde V$ is
    a free $\Fp[G]$-submodule of $U$ and $\tilde V^G=Z$.  By the
    Exclusion Lemma $V+\tilde V = V\oplus \tilde V$ and $(V\oplus
    \tilde V)^G=V^G\oplus \tilde V^G = U^G$.

    Now let $u\in U$ be arbitrary and let $M$ be the cyclic
    $\Fp[G]$-submodule of $U$ generated by $u$.  Then $(M+V+\tilde
    V)^G \subset U^G \subset V+\tilde V$.  Moreover, for any $m\in
    M+V+\tilde V \setminus (M+V+\tilde V)^G$,
    $(\sigma-1)^{l(m)-1}(m)\in (M+V+\tilde V)^G \subset U^G =
    (V+\tilde V)^G = N(V + \tilde V)$ by the freeness of $V$
    and $\tilde V$. By the Inclusion Lemma, then, $M\subset
    V + \tilde V$. Hence $U=V\oplus \tilde V$.
\end{proof}

\subsection{Milnor $k$-groups}\

Recall that if $\alpha\in K_mE$ we write $\bar\alpha$ for the
class of $\alpha$ in $k_mE$.  For $\gamma\in K_mE$, let
$l(\gamma)$ denote the dimension over $\Fp$ of the
$\Fp[G]$-submodule $\langle \bar\gamma \rangle$ of $k_mE$
generated by $\bar\gamma$. Because $(\sigma-1)^{p^n-1} =
1+\sigma+\dots+\sigma^{p^n-1}$ in $\Fp[G]$, we may use
$i_EN_{E/F}$ and $N$ interchangeably on $k_mE$.

Our proof of Theorem~\ref{th:main} depends on
Lemma~\ref{le:fullext}.  This lemma, in turn, requires for its
induction two basic extension lemmas, the first for the case
$p>2$, $n=1$, and the second for the case $p=2$, $n=2$. (The case
$p=2$, $n=1$ is clear.)  It is in only these two basic lemmas that
Theorem~\ref{th:exseq} is used in this paper.

\begin{lemma}\label{le:key}
    Suppose $E/F$ is a cyclic extension of degree $p>2$ and
    $\gamma\in K_mE$ with $2\le l(\gamma)\le p$. Then there
    exists $\alpha\in K_mE$ such that
    \begin{equation*}
        \langle N\bar\alpha \rangle = \langle \bar\gamma\rangle^G.
    \end{equation*}
\end{lemma}

\begin{proof}
    Let $l=l(\gamma)$ and suppose $2\le l\le i\le p$. We show by
    induction on $i$ that there exists $\alpha_i\in K_mE$ such
    that
    \begin{equation*}
        \langle (\sigma-1)^{i-1}\bar\alpha_i\rangle = \langle
        \bar\gamma \rangle^G.
    \end{equation*}
    Then setting $\alpha := \alpha_{p}$, the proof will be complete.
    If $i=l$ then $\alpha_i=\gamma$ suffices.  Assume now that
    $l\leq i<p$ and 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 $i<p$, we conclude that
    $i_E\bar c = 0$.  By the injectivity of $i_E$ from
    Theorem~\ref{th:isop}, $\bar c = 0$.  Therefore there exists
    $f\in K_mF$ such that $c=pf$ in $K_mF$.

    We calculate
    \begin{equation*}
        N_{E/F}(\alpha_i-i_E(f)) = c-pf = 0.
    \end{equation*}
    By Theorem~\ref{th:exseq}, there exists $\omega\in K_mE$ such
    that
    \begin{equation*}
        (\sigma-1)\omega = \alpha_i - i_E(f).
    \end{equation*}
    Hence $(\sigma-1)^2\omega = (\sigma-1)\alpha_i$.  Since $i\ge
    2$,
    \begin{equation*}
        \langle (\sigma-1)^i \bar\omega \rangle = \langle
        (\sigma-1)^{i-1}\bar\alpha_i\rangle = \langle \bar\gamma
        \rangle^G
    \end{equation*}
    and we can set $\alpha_{i+1}=\omega$.
\end{proof}

\begin{lemma}\label{le:key2}
    Suppose $p=2$ and $E/F$ is a cyclic extension of degree $4$.
    Suppose $\gamma\in K_mE$ with $3\le l(\gamma)\le 4$. Then
    there exists $\alpha\in K_mE$ such that
    \begin{equation*}
        \langle N\bar\alpha \rangle = \langle \bar\gamma\rangle^G.
    \end{equation*}
\end{lemma}

\begin{proof}
    The case $l(\gamma)=4$ is clear.  Assume then that
    $l(\gamma)=3$, and set $\beta := (\sigma-1)\gamma$. Then
    \begin{equation*}
        (\sigma^2-1)\bar\beta = (\sigma-1)^3\bar\gamma = 0.
    \end{equation*}
    By Theorem~\ref{th:isop} applied to $E/E_1$, $\bar\beta \in
    i_E(k_mE_1)$. Let $b\in K_mE_1$ such that $i_E\bar b =
    \bar\beta$.

    Set $c := N_{E_1/F} b$. Then
    \begin{equation*}
        i_E\bar c = i_E(\sigma+1)\bar b = (\sigma^2-1)\bar\gamma =
        i_EN_{E/E_1}\bar\gamma,
    \end{equation*}
    and since $l(\gamma)=3$, $\langle \bar\gamma\rangle^G =
    \langle \bar c\rangle$.  Now $N_{E/E_1}\gamma = c+2e$ for some
    $e\in K_mE$, and $e_1 := 2e\in K_mE_1 \cap 2K_mE$.  Then
    $i_E(\bar e_1)=0\in k_mE$, and by the injectivity of $i_E$
    from Theorem~\ref{th:isop}, $e_1=2g$ for some $g\in K_mE_1$.
    Hence
    \begin{equation*}
        N_{E/E_1}\gamma = c + 2g, \quad g\in K_mE_1.
    \end{equation*}

    Now $N_{E/F}\gamma = N_{E_1/F}(c+2g) = 2c + 2N_{E_1/F}g$. Set
    $\delta := b + g$. We calculate
    \begin{equation*}
        N_{E/F}\delta = 2c+2N_{E_1/F}g,
    \end{equation*}
    so that $N_{E/F}(\gamma-\delta) = 0$. By
    Theorem~\ref{th:exseq} there exists an $\alpha\in K_mE$ with
    $(\sigma-1)\alpha = \gamma-\delta$. Then
    \begin{align*}
        i_EN_{E/F}\bar\alpha &= (\sigma-1)^3\bar\alpha =
        (\sigma-1)^2\overline{(\gamma-\delta)} = \\
        &i_E\overline{N_{E/E_1}(\gamma-\delta)} =
        i_E\overline{(c+2g-2b-2g)} = \\
        &i_E\bar{c}=(\sigma-1)^2\bar\gamma.
    \end{align*}
    Because $\langle (\sigma-1)^2\bar\gamma\rangle=
    \langle\bar\gamma\rangle^G$ our statement follows.
\end{proof}

Our full extension lemma is then
\begin{lemma}\label{le:fullext}
    Suppose that $E/F$ is a cyclic extension of degree $p^n$,
    $n\in\N$, and let $\bar \gamma\in k_mE\setminus k_mE_{n-1}$.
    Then there exists $\chi\in K_mE$ such that
    \begin{equation*}
        (\sigma-1)^{l(\gamma)-1}\bar\gamma =
        (\sigma-1)^{p^n-1}\bar \chi = i_E N_{E/F}\bar\chi.
    \end{equation*}
\end{lemma}

\begin{proof}
    We shall first prove our statement in the case $p>2$. If $n=1$
    then our statement follows from Lemma~\ref{le:key}.

    Assume therefore that $n>1$.  Let $H := H_{n-1}$, and write
    $l_H(\gamma)$ for the dimension over $\Fp$ of the
    $\Fp[H]$-submodule of $k_mE$ generated by $\bar\gamma$. From
    Theorem~\ref{th:isop} we see that $l_H(\gamma)\geq 2$.
    Therefore Lemma~\ref{le:key} tells us that there exists an
    element $\alpha\in K_mE$ such that
    \begin{equation*}
        (\sigma-1)^{p^n-p^{n-1}} \bar\alpha =
        \left(\sigma^{p^{n-1}}-1\right)^{p-1} \bar\alpha =
        \left(\sigma^{p^{n-1}}-1 \right)^{l_H(\gamma)-1}
        \bar\gamma \neq 0.
    \end{equation*}
    Set $s := l(\gamma)-p^{n-1}(l_H(\gamma)-1)$. Then
    \begin{equation*}
        (\sigma-1)^{p^n-p^{n-1}+s-1} \bar\alpha =
        (\sigma-1)^{p^{n-1} \left(l_H(\gamma) -1\right)+s-1}
        \bar\gamma \neq 0.
    \end{equation*}
    Furthermore, this element belongs to $(k_mE)^G$. Set $\lambda
    := (\sigma-1)^s\alpha$.  Then
    \begin{equation*}
        (\sigma-1)^{p^n-p^{n-1}-1} \bar\lambda =
        (\sigma-1)^{p^n-p^{n-1}+s-1} \bar\alpha,
    \end{equation*}
    whence $l(\lambda)=p^n-p^{n-1}$.

    Now we consider $l_H(\lambda)$.  On one hand,
    \begin{equation*}
        \left(\sigma^{p^{n-1}}-1\right)^{p-1} \bar\lambda = 0,
    \end{equation*}
    and on the other,
    \begin{equation*}
        (\sigma-1)^{p^{n-1}(p-2)} \bar\lambda =
        (\sigma-1)^{p^n-2p^{n-1}} \bar\lambda \neq 0.
    \end{equation*}
    We deduce that $l_H(\lambda) = p-1\geq 2$. Applying
    Lemma~\ref{le:key} again, there exists $\chi\in K_mE$ with
    \begin{equation*}
        (\sigma-1)^{p^n-p^{n-1}} \bar\chi =
        (\sigma-1)^{p^n-2p^{n-1}} \bar\lambda.
    \end{equation*}
    In particular, $l(\chi)=l(\lambda)+p^{n-1}=p^n$.

    Summarizing, we have obtained
    \begin{equation*}
        i_EN_{E/F}\bar\chi = (\sigma-1)^{p^n-1} \bar\chi =
        (\sigma-1)^{p^n-p^{n-1}-1} \bar\lambda =
        (\sigma-1)^{p^n-p^{n-1}+s-1} \bar\alpha.
    \end{equation*}
    Because
    \begin{equation*}
        (\sigma-1)^{p^n-p^{n-1}+s-1} \bar\alpha =
        (\sigma-1)^{p^{n-1}(l_H(\gamma)-1)+s-1} \bar\gamma =
        (\sigma-1)^{l(\gamma)-1} \bar\gamma,
    \end{equation*}
    we have established our equality in the case $p>2$.

    Now we shall consider the case $p=2$. The case $n=1$ is
    trivial, and the case $n=2$ is handled by Lemma~\ref{le:key2}.
    Assume therefore that $n\geq 3$. Let $H := \Gal(E/E_{n-2})$,
    and write $l_H(\gamma)$ for the dimension over $\F_2$ of the
    cyclic $\F_2[H]$-submodule of $k_mE$ generated by $\bar\gamma$.

    Because $\bar\gamma\notin k_mE_{n-1}$, we have $l_H(\gamma)
    \geq 3$. From Lemma~\ref{le:key2} we see that there exists
    $\alpha\in K_mE$ such that
    \begin{equation*}
        \left(\sigma^{2^{n-2}}-1\right)^3 \bar\alpha =
        \left(\sigma^{2^{n-2}}-1\right)^{l_H(\gamma)-1}
        \bar\gamma.
    \end{equation*}
    Hence
    \begin{equation*}
        \left(\sigma-1\right)^{2^n-2^{n-2}} \bar\alpha =
        \left(\sigma-1\right)^{2^{n-2}(l_H(\gamma)-1)} \bar\gamma
        \neq 0.
    \end{equation*}
    Let $s := l(\gamma)-2^{n-2}\left(l_H(\gamma)-1\right)$. Then
    we have
    \begin{equation*}
        \left(\sigma-1\right)^{2^n-2^{n-2}+s-1} \bar\alpha =
        \left(\sigma-1\right) ^{2^{n-2}
        \left(l_H(\gamma)-1\right)+s-1} \bar\gamma \neq 0.
    \end{equation*}
    Furthermore, this element belongs to $(k_mE)^G$.  Set $\lambda
    := (\sigma-1)^s\alpha$. Then
    \begin{equation*}
        (\sigma-1)^{2^n-2^{n-2}-1} \bar\lambda =
        (\sigma-1)^{2^n-2^{n-2}+s-1} \bar\alpha,
    \end{equation*}
    whence $l(\lambda)=2^n-2^{n-2}$.

    Now we consider $l_H(\lambda)$.  On one hand,
    \begin{equation*}
        \left(\sigma^{2^{n-2}}-1\right)^3 \bar\lambda =
        (\sigma-1)^{2^n-2^{n-2}} \bar\lambda = 0,
    \end{equation*}
    and on the other,
    \begin{equation*}
        \left(\sigma^{2^{n-2}}-1\right)^2\bar\lambda \neq 0.
    \end{equation*}
    We deduce that $l_H(\lambda)=3$. By Lemma~\ref{le:key2} there
    exists $\chi\in K_mE$ with
    \begin{equation*}
        \left(\sigma^{2^{n-2}}-1\right)^3 \bar\chi =
        \left(\sigma^{2^{n-2}}-1\right)^2 \bar\lambda.
    \end{equation*}
    Equivalently, $(\sigma-1)^{2^n-2^{n-2}} \bar\chi =
    (\sigma-1)^{2^{n-1}} \bar\lambda$, and therefore $l(\chi) =
    l(\lambda)+2^{n-2}=2^n$.

    Summarizing, we have obtained
    \begin{align*}
        i_EN_{E/F}\bar\chi &= (\sigma-1)^{2^n-1} \bar\chi
        =(\sigma-1)^{2^n-2^{n-2}-1} \bar\lambda \\
        &=(\sigma-1)^{2^n-2^{n-2}+s-1} \bar\alpha
        =(\sigma-1)^{l(\gamma)-1} \bar\gamma
    \end{align*}
    as required.
\end{proof}

\begin{remark*}
    Observe that since $\langle\bar\gamma\rangle^G$ is a
    one-dimensional $\Fp$-vector space and $N$ is additive, the
    equality $\langle N\bar\alpha\rangle = \langle \bar\gamma
    \rangle^G$ holds if and only if any $\bar\delta \in
    \langle\bar\gamma\rangle^G \setminus \{0\}$ is expressible as
    $N\bar\beta$ for some $\bar\beta\in k_mE$.
\end{remark*}

\begin{lemma}[Submodule-Subfield Lemma]\label{le:subsub}
    Suppose $E/F$ is a cyclic extension of degree $p^n$, and let
    $U$ be a free $\Fp[G]$-sub\-module of $k_mE$. Then for each
    $i$ in $\{0,1,\dots,n-1\}$ we have:
    \begin{equation*}
        U^{H_i} = (\sigma-1)^{p^n-p^i}U = U\cap
        i_E(N_{E/E_i}k_mE) = U\cap i_E(k_mE_i).
    \end{equation*}
\end{lemma}

\begin{proof}
    Suppose $\bar\alpha\in U^{H_i}$.  Then ${(\sigma^{p^i}-1)}
    \bar\alpha = (\sigma-1)^{p^i}\bar\alpha =0$, so $l(\alpha)\le
    p^i$. Since $U$ is free, $\bar\alpha =
    (\sigma-1)^{p^n-l(\alpha)} \bar\beta$ for some $\bar\beta \in
    U$. In particular,
    \begin{equation*}
        \bar\alpha=(\sigma-1)^{p^n-p^i}(\sigma-1)^{p^i-l(\alpha)}
        \bar\beta.
    \end{equation*}
    Hence $U^{H_i}\subset (\sigma-1)^{p^n-p^i}U$.  Now suppose
    $\bar\alpha=(\sigma-1)^{p^n-p^i}\bar\gamma$.  Then since
    \begin{equation*}
        (\sigma-1)^{p^n-p^i}\bar\gamma = i_EN_{E/E_i}
        \bar\gamma,
    \end{equation*}
    we have $(\sigma-1)^{p^n-p^i}U\subset U\cap i_E(N_{E/E_i}
    k_mE)\subset U\cap i_E(k_mE_i)$.

    Now suppose that $\bar\alpha \in U\cap i_E(k_mE_i)$.  Then
    $\bar\alpha \in U^{H_i}$ and hence all of the inclusions are
    equalities.
\end{proof}

\begin{remark*}
    Alternatively one can derive the first equality in the lemma
    above, as follows. If $U$ is a free $\Fp[G]$-module, then
    $U$ is also a free $\Fp[H_i]$-module.  But then $H^2(H_i,U)
    =\{0\}$. Hence $U^{H_i}=N_i(U):=$ the image of the norm
    operator $N_i$. Thus $U^{H_i}=U^{(\sigma-1)^{p^n-p^i}}$
    as required.
\end{remark*}

\section{$E/F$ Cyclic of Degree $p$}\label{se:nequal1}

\begin{proposition*}
    Theorem~\ref{th:main} holds for $n=1$.
\end{proposition*}

\begin{proof}
    We first construct submodules $Y_0$ and $Y_1$ of $k_mE$.  Let
    $\Ic$ be an $\Fp$-basis for the subspace $i_E(N_{E/F}k_mE)$.
    For each basis element $\bar y\in \Ic$, let $\alpha_y\in K_mE$
    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 $Y_1 := \sum_{\bar y \in \Ic} \langle \bar\alpha_y
    \rangle$. By the Exclusion Lemma, $Y_1 = \oplus_{\bar y\in
    \Ic} \langle \bar\alpha_y \rangle$ and so $Y_1$ is a free
    $\Fp[G]$-module. Moreover,
    \begin{equation*}
        Y_1^G = (\sigma-1)^{p-1}Y_1 = NY_1 = \oplus_{\bar y\in\Ic}
        \langle \bar y \rangle = i_E(N_{E/F}k_mE).
    \end{equation*}
    The rank of $Y_1$ is equal to the dimension of
    $i_E(N_{E/F}k_mE)$, and by the injectivity of $i_E$ from
    Theorem~\ref{th:isop}, we conclude
    \begin{equation*}
        \rank Y_1 = \dim_{\Fp} N_{E/F}k_mE.
    \end{equation*}

    Now let $Z$ be any $\Fp$-module complement of $N_{E/F}k_mE$ in
    $k_mF$, and set $Y_0 := i_EZ$.  By the injectivity of $i_E$
    from Theorem~\ref{th:isop}, we obtain
    \begin{equation*}
        \rank Y_0 = \dim_{\Fp} k_mF/N_{E/F}k_mE.
    \end{equation*}

    By the Exclusion Lemma, $Y_0+Y_1=Y_0\oplus Y_1$.  Moreover,
    $Y_0+Y_1$ contains $(k_mE)^G$:
    \begin{align*}
        (k_mE)^G &= i_E(k_mF) = i_E((N_{E/F}k_mE) \oplus Z) \\ &=
        i_E(N_{E/F}k_mE) \oplus Y_0 = Y_1^G \oplus Y_0 \subset
        Y_0\oplus Y_1.
    \end{align*}

    If $p=2$, we already have enough to prove the proposition, as
    follows. Let $\bar\gamma\in k_mE$.  If $(\sigma-1) \bar\gamma
    = 0$ then $\bar\gamma\in (k_mE)^G\subset Y_0\oplus Y_1$ as
    above. Otherwise, observe that since $p=2$, the operator
    $\sigma-1$ is equivalent to $i_EN_{E/F}$ on $k_mE$.  Let
    $\beta = i_EN_{E/F}\gamma$.  Then $(\sigma-1) \bar\gamma =
    \bar\beta \in Y_1^G$. Since $Y_1$ is free there exists
    $\bar\alpha\in Y_1$ such that $(\sigma-1) \bar\alpha =
    \bar\beta$.  Then $(\sigma-1) (\bar\gamma - \bar\alpha) = 0$,
    so $l(\gamma-\alpha)\le 1$ and $\bar\gamma-\bar\alpha\in
    (k_mE)^G\subset Y_1\oplus Y_0$. Since $\bar\alpha\in Y_1$, we
    conclude that $\bar\gamma\in Y_1\oplus Y_0$ as well.  Hence
    $k_mE = Y_1\oplus Y_0$.

    If $p>2$, we instead prove by induction on $l(\gamma)$ that
    each $\bar\gamma\in k_mE$ lies in $Y_1\oplus Y_0$, whence we will
    obtain the proposition.  The case $l(\gamma)=1$ follows from
    $(k_mE)^G\subset Y_1\oplus Y_0$. Suppose now that for each
    $\bar\beta \in k_mE$ with $l(\beta)\le i<p$ we have $\bar\beta
    \in Y_1\oplus Y_0$, and assume that $l(\gamma)=i+1\ge 2$.  By
    Lemma~\ref{le:key} there exists an $\alpha\in K_mE$ with
    \begin{equation*}
        \langle N\bar\alpha\rangle = \langle \bar\gamma \rangle^G.
    \end{equation*}
    Hence $(\sigma-1)^i\bar\gamma = cN\bar\alpha$ for some $c\in
    \Z$. Now $cN\bar\alpha = i_EN_{E/F}(c\bar\alpha)\in Y_1^G$.
    Since $Y_1$ is free, there exists $\bar\omega\in Y_1$ such
    that $(\sigma-1)^{p-1} \bar\omega = cN\bar\alpha$. Set
    $\bar\lambda = (\sigma-1)^{p-l(\gamma)} \bar\omega \in Y_1$,
    so that $(\sigma-1)^{l(\gamma)-1} (\bar\gamma - \bar\lambda) =
    0$. Hence $l(\gamma-\lambda)<l(\gamma)$ and by induction
    $\bar\gamma = (\bar\gamma - \bar\lambda) + \bar\lambda \in
    Y_1\oplus Y_0$. Thus again we conclude that $k_m E=Y_1\oplus
    Y_0$.
\end{proof}

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

We proceed by induction.  The case $n=1$ is the preceding
proposition.  Assume then that Theorem~\ref{th:main} holds
when degree of $E/F$ is $\leq p^{n-1}$, and suppose
that $E/F$ is a cyclic extension of degree $p^n$ with
$n\geq 2$.

\subsection{Constructing the $Y_i$}\

Let $\Ic$ be an $\Fp$-basis for $i_E(N_{E/F}k_mE)$. For each basis
element $\bar y\in \Ic$ construct a free $\Fp[G]$-module $\langle
\bar \alpha_y \rangle$, such that $i_EN_{E/F}\bar\alpha_{y} =
\bar{y}$. We see by the Exclusion Lemma that the modules
$\langle\bar\alpha_{y}\rangle$, $\bar{y}\in \Ic$, are independent.
Set $Y_n := \oplus_ {y\in \Ic} \langle \bar\alpha_y\rangle$. Hence
$Y_n$ is a direct sum of cyclic $\Fp[G]$-modules of dimension
$p^{n}$, and $Y_n^G = i_E(N_{E/F}k_mE)$.  By the injectivity of
$i_E$ from Theorem~\ref{th:isopn},
\begin{equation*}
    \rank_{\Fp[G]} Y_n = \dim_{\Fp} Y_n^G = \dim_{\Fp}N_{E/F}k_mE.
\end{equation*}

By induction, we have a $\Fp[G_{n-1}]$-module decomposition
\begin{equation*}
    k_mE_{n-1} = \tilde Y_{n-1}\oplus \tilde Y_{n-2} \oplus \dots
    \oplus \tilde Y_0
\end{equation*}
into direct sums $\tilde Y_i$ of cyclic $\Fp[G_{n-1}]$-modules of
dimension $p^i$, $i=0, 1, \dots, n-1$. Let $\bar\sigma$ denote the
image of $\sigma$ under the natural projection $G\to G_{n-1}$.
Because $i_{E_{n-1}} N_{E_{n-1}/F}$ acts on $k_mE_{n-1}$ as
$(\bar\sigma-1)^{p^{n-1}-1}$ and $i_{E_{n-1}}$ is injective, we
see that $N_{E_{n-1}/F}$ annihilates the sum $\tilde Y_{n-2}\oplus
\dots \oplus \tilde Y_0$.  Therefore
\begin{equation*}
    i_{E_{n-1}}(N_{E_{n-1}/F}k_mE_{n-1}) = \tilde Y_{n-1}^{G_{n-1}}.
\end{equation*}

Now consider $k_mE$ as an $\Fp[H_{n-1}]$-module. By the
injectivity of $i_E$ from Theorem~\ref{th:isop}, the images of
$\tilde Y_i$, $i=0, \dots, n-1$, under $i_E$ are direct sums of
cyclic modules of dimension $p^i$ and are independent. Because the
modules $\tilde Y_i$ are direct sums of cyclic
$\Fp[G_{n-1}]$-modules, the images $i_E\tilde Y_i$ are direct sums
of cyclic $\Fp[G]$-modules. Set $Y_i := i_E\tilde Y_i$ for
$i<n-1$. Observe that $\rank_{\Fp[G_i]} Y_i=\rank_{\Fp[G_i]}
\tilde Y_i$.

Set $W := Y_n^{H_{n-1}}$. By the Submodule-Subfield Lemma,
\begin{equation*}
    W = (\sigma-1)^{p^n-p^{n-1}} Y_n = Y_n \cap i_Ek_mE_{n-1}.
\end{equation*}
Since $Y_n$ is a direct sum of cyclic $\Fp[G]$-modules of
dimension $p^n$, $W$ is a direct sum of cyclic modules of
dimension $p^{n-1}$ and hence is free as an $\Fp[G_{n-1}]$-module.
Because $W\subset i_Ek_mE_{n-1}$, we may consider the image $P$ of
the projection map $\pr \colon W \to i_E\tilde Y_{n-1}$ from $W$
to the summand $i_E\tilde Y_{n-1}$ in the decomposition
\begin{equation*}
    i_Ek_mE_{n-1} = i_E\tilde Y_{n-1} \oplus Y_{n-2} \oplus
    \cdots \oplus Y_0.
\end{equation*}

Observe that $W\cong P$ as $\Fp[G_{n-1}]$-modules.  Indeed, since
$W$ is a free $\Fp[G_{n-1}]$-module, each $w\in W\setminus \{0\}$
may be written as $(\bar\sigma-1)^s\tilde w$ for some $0\le s\le
p^{n-1}-1$ and $\tilde w\in W$ with $l(\tilde w)=p^{n-1}$.  We
have
\begin{equation*}
    (\bar\sigma-1)^{p^{n-1}-1}\pr(\tilde w) =
    (\bar\sigma-1)^{p^{n-1}-1}\tilde w \neq 0,
\end{equation*}
since all other components of $\tilde w$ are killed by
$(\bar\sigma-1)^{p^{n-1}-1}$.  (Since $n\ge 2$, $p^{n-1}-1 \ge
p^{n-2}$.)  Therefore $(\bar\sigma-1)^{s} \pr(\tilde w) = \pr(w)
\neq 0$.  We conclude that the kernel of the projection map is
$(0)$, as required.

Since $M^{G_{n-1}}=(\sigma-1)^{p^{n-1}-1}M$ for free
$\Fp[G_{n-1}]$-modules, we have further obtained that $W^{G_{n-1}}
= P^{G_{n-1}}$. Observe that
\begin{equation*}
    W^{G_{n-1}} = (\sigma-1)^{p^{n-1}-1}W \subset
    i_E(N_{E_{n-1}/F}k_mE_{n-1}) = (i_E\tilde{Y}_{n-1})^G.
\end{equation*}

By the Free Complement Lemma, there exists a free
$\Fp[G_{n-1}]$-module complement $Y_{n-1}$ in $i_E\tilde Y_{n-1}$
of $P$.  Since $W=Y_n\cap i_Ek_m E_{n-1}$, we obtain $Y_n^G =
W^{G_{n-1}} = P^{G_{n-1}}$.  By the Exclusion Lemma,
$P^{G_{n-1}}\cap Y_{n-1}^{G_{n-1}} = \{0\}$ implies that $Y_{n-1}
+ Y_n = Y_{n-1} \oplus Y_n$.  Then, since $P^G+Y_{n-1}^G =
(i_E\tilde Y_{n-1})^G$, we obtain $(Y_{n-1} + Y_n)^{G}=(i_E\tilde
Y_{n-1})^G$. By induction and the injectivity of $i_E$,
\begin{equation*}
    \rank_{\Fp[G_{n-1}]} i_E\tilde Y_{n-1} = \dim_{\Fp} (i_E\tilde
    Y_{n-1})^{G} = \dim_{\Fp} N_{E_{n-1}/F}k_mE_{n-1}.
\end{equation*}
Since $Y_{n-1}^G\oplus Y_n^G = (i_E\tilde Y_{n-1})^G$,
\begin{equation*}
    \rank_{\Fp[G_{n-1}]} Y_{n-1} = \dim_{\Fp} Y_{n-1}^G =
    \dim_{\Fp} N_{E_{n-1}/F}k_mE_{n-1}/N_{E/F}k_mE.
\end{equation*}
Finally, by the Exclusion Lemma, $Y_{n-1}+ Y_n$ is independent
from $Y_{n-2} + \dots + Y_0$. Hence we have a submodule
\begin{equation*}
     Y= Y_n\oplus Y_{n-1} \oplus \dots \oplus  Y_0 \subset k_mE
\end{equation*}
with ranks satisfying the claims of Theorem~\ref{th:main}.

\subsection{Showing $k_mE=\oplus Y_i$}\

We prove first that $Y^{H_{n-1}} = i_E k_mE_{n-1}$.
Theorem~\ref{th:isop} tells us that $Y^{H_{n-1}}\subset i_Ek_m
E_{n-1}$, and we have the decomposition $i_Ek_m E_{n-1} = i_E
\tilde Y_{n-1} \oplus Y_{n-2} \oplus \cdots \oplus Y_0$. Therefore
it is sufficient to show that
\begin{equation*}
    i_E \tilde Y_{n-1} \subset Y^{H_{n-1}} = Y_n^{H_{n-1}} + Y_{n-1} +
    \cdots + Y_0.
\end{equation*}
Because $i_E\tilde Y_{n-1} = Y_{n-1}+P$ it is enough to show that
\begin{equation*}
    P \subset Y_n^{H_{n-1}} + Y_{n-1} + \cdots + Y_0 = W + Y_{n-1}
    + \cdots + Y_0.
\end{equation*}
But by the definition of the projection, $P\subset W + Y_{n-2} +
\dots + Y_0$.  Hence $P\subset W+Y_{n-1}+\dots + Y_0$, and we
conclude that $Y_n^{H_{n-1}} = i_Ek_mE_{n-1}$.

We adapt the proof of the Inclusion Lemma to show that
$k_mE\subset Y$, by induction on the socle series $V_i$ of $k_mE$.
Our base case is $V_{p^{n-1}}$.  Observe that $V_{p^{n-1}}$ is the
kernel of $(\sigma-1)^{p^{n-1}} = \sigma^{p^{n-1}}-1$, which is
$(k_mE)^{H_{n-1}}$. Hence $V_{p^{n-1}}\subset Y$.

For the inductive step, assume that $V_i\subset Y$ for all $i<t$
for some $p^{n-1} < t \le p^n$, and let $\bar\gamma\in
V_{t}\setminus V_{t-1}$.  Hence $l(\gamma)=t$. Therefore
$\bar\gamma \not\in i_Ek_mE_{n-1}$, and by Lemma~\ref{le:fullext}
there exists $\chi\in K_mE$ such that
\begin{equation*}
    (\sigma-1)^{t-1}\bar\gamma = i_EN_{E/F}\bar\chi \in Y_n^G.
\end{equation*}
Since $Y_n$ is a free $\Fp[G]$-module, there exists $\bar{\nu}\in
Y_n$ such that
\begin{equation*}
    i_EN_{E/F}\bar{\nu} = (\sigma-1)^{p^n-1}\bar{\nu} =
    (\sigma-1)^{p^n-1}\bar\chi=(\sigma-1)^{t-1}\bar\gamma.
\end{equation*}
Set $\bar\delta := (\sigma-1)^{p^n-t}\bar{\nu} \in Y_n \subset Y$.
Then $(\sigma-1)^{t-1}(\bar\gamma-\bar\delta)=0$ and hence
$l(\gamma-\delta)<t$. By induction $\bar\gamma-\bar\delta\in Y$,
and since $\bar\delta\in Y$, we see that $\bar\gamma\in Y$ as
well. \qed

\section*{Acknowledgment}

We are very grateful to Andy Schultz, as some of his ideas,
developed in \cite{MSS}, proved quite useful to us during the
investigations leading to this paper.

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