%%[8] la 3eme edition revisité en 2002 est chez springer \documentclass[CRMATH,Unicode,biblatex,XML]{cedram} \TopicFR{Algèbre} \TopicEN{Algebra} \addbibresource{CRMATH_Gupta_20230884.bib} \usepackage{mathtools} \DeclareMathOperator{\rk}{rk} \DeclareMathOperator{\alt}{Alt} \DeclareMathOperator{\tr}{Tr} \DeclareMathOperator{\gal}{Gal} \DeclareMathOperator{\mo}{mod} \DeclareMathOperator{\ord}{ord} \DeclareMathOperator{\gc}{gcd} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \makeatletter \let\save@mathaccent\mathaccent \newcommand*\if@single[3]{% \setbox0\hbox{${\mathaccent"0362{#1}}^H$}% \setbox2\hbox{${\mathaccent"0362{\kern0pt#1}}^H$}% \ifdim\ht0=\ht2 #3\else #2\fi } %The bar will be moved to the right by a half of \macc@kerna, which is computed by amsmath: \newcommand*\rel@kern[1]{\kern#1\dimexpr\macc@kerna} %If there's a superscript following the bar, then no negative kern may follow the bar; %an additional {} makes sure that the superscript is high enough in this case: \newcommand*\widebar[1]{\@ifnextchar^{{\wide@bar{#1}{0}}}{\wide@bar{#1}{1}}} %Use a separate algorithm for single symbols: \newcommand*\wide@bar[2]{\if@single{#1}{\wide@bar@{#1}{#2}{1}}{\wide@bar@{#1}{#2}{2}}} \newcommand*\wide@bar@[3]{% \begingroup \def\mathaccent##1##2{% %Enable nesting of accents: \let\mathaccent\save@mathaccent %If there's more than a single symbol, use the first character instead (see below): \if#32 \let\macc@nucleus\first@char \fi %Determine the italic correction: \setbox\z@\hbox{$\macc@style{\macc@nucleus}_{}$}% \setbox\tw@\hbox{$\macc@style{\macc@nucleus}{}_{}$}% \dimen@\wd\tw@ \advance\dimen@-\wd\z@ %Now \dimen@ is the italic correction of the symbol. \divide\dimen@ 3 \@tempdima\wd\tw@ \advance\@tempdima-\scriptspace %Now \@tempdima is the width of the symbol. \divide\@tempdima 10 \advance\dimen@-\@tempdima %Now \dimen@ = (italic correction / 3) - (Breite / 10) \ifdim\dimen@>\z@ \dimen@0pt\fi %The bar will be shortened in the case \dimen@<0 ! 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It is known that the space $\mathrm{Alt}_K(L)$ of alternating $K$-bilinear forms (skew-forms) on $L$ decomposes into a direct sum of $K$-subspaces $A^{\sigma^i}$ indexed by the elements of $\mathrm{Gal}(L/K) = \langle \sigma \rangle$. It is also known that the components $A^{\sigma^i}$ can have nice constant-rank properties. We enhance and enrich these constant-rank results and show that the component $A^\sigma$ often decomposes directly into a sum of constant rank subspaces, that is, subspaces all of whose non-zero skew-forms have a fixed rank $r$. In particular, this is always true when $-1 \not \in L^2$. As a result we deduce a decomposition of $\mathrm{Alt}_K(L)$ into subspaces of constant rank in several interesting situations. We also establish that a subspace of dimension $\frac{n}{2}$ all of whose nonzero skew-forms are non-degenerate can always be found in $A^{\sigma^i}$ where $\sigma^i$ has order divisible by $2$. \end{abstract} \begin{altabstract} Soit $L/K$ une extension cyclique de degré $n = 2m$. On sait que l'espace $\mathrm{Alt}_K (L)$ des formes bilinéaires $K$-formes bilinéaires alternées (skew-forms) sur $L$ se décompose en une somme directe de $K$-sous-espaces $A^{\sigma^i}$ indexés par les éléments de $\mathrm{Gal}(L/K ) = \langle \sigma\rangle$. Il est également connu que les composants $A^{\sigma^i}$ peuvent avoir de belles propriétés de rang constant. Nous améliorons et enrichissons ces résultats de rang constant et montrons que la composante $A^{\sigma}$ se décompose souvent directement en une somme de sous-espaces de rang constant, c'est-à-dire des sous-espaces dont toutes les formes asymétriques non nulles ont un rang fixe $r$. En particulier, ceci est toujours vrai lorsque $-\not\in L^2$. En conséquence, nous déduisons une décomposition de $\mathrm{Alt}_K (L)$ en sous-espaces de rang constant dans plusieurs situations intéressantes. Nous établissons également qu'un sous-espace de dimension $\frac{n}{2}$ dont toutes les formes asymétriques non nulles sont non dégénérées peut toujours être trouvé dans $A^{\sigma^i}$ où $\sigma^i$ a un ordre divisible par $2$. \end{altabstract} \begin{document} \maketitle \section{Introduction}\label{Sect1} Let $K$ be a field of characteristic other than two and $\alt_K (V)$ denote the space of all alternating bilinear forms (skew-forms) on a $K$-space $V$ of dimension $n$. Suppose $K$ admits a Galois extension $L$ of degree $n$. Taking the $n$-dimensional $K$-space $L$ as a model for $V$ it was shown in~\cite{GQ09} that ideas from Galois Theory can be fruitfully applied for studying skew-forms on $V$. Notably, this approach sheds light on the subspaces of $\alt_K (V)$ whose nonzero skew-forms all have the same rank equal to $k$, say. Such ``$k$-subspaces'' besides being interesting in their own right play an important role in coding theory (see~\cite{PD75},\cite{PD78}). Of particular importance are the $n$-subspaces of $\alt_K(V)$, that is, subspaces all of whose nonzero skew-forms are non-degenerate. Replacing $V$ by the $K$-space $L$, we begin with some definitions and facts given in~\cite[Lemma~2]{GQ09}. For each $\sigma \in G : = \gal(L/K)$ and $b \in L$ we may define the skew-form \begin{equation} \label{defn-sk-frm} f_{b,\sigma}(x,y) = \tr^L_K(b(x\sigma(y)-\sigma(x)y)), \qquad \forall x, y \in L. \end{equation} where $\tr^L_K : L \to K$ is the Galois-theoretic trace map defined by \[ \tr^L_K(a) = \sum_{\sigma \in \gal(L/K)} \sigma(a), \quad \forall a \in L. \] With each $\sigma \in G$ we can thus associate a subspace $A^\sigma$ of $\alt_K(L)$ defined as $A^\sigma : = \{f_{b, \sigma} : b \in L \}$. Each $ A^{\sigma}$ has dimension $n$ unless $\sigma$ has order $2$ (see \cite[Theorem~1]{GQ09}). %The dimensions of the subspaces $A^\sigma$ were determined in~\cite{GQ09} and It was shown in~\cite{GQ09} that $\alt(L)$ decomposes as a direct sum of the spaces $A^\sigma$ with $\sigma$ ranging over the elements of the Galois group $G$ (see Theorems 1 and 2 below). Let $ \ord(\sigma) $ denote the order of $ \sigma \in G $. Interestingly, for $n$ odd, each $A^\sigma$ is an $ n - n/{\ord(\sigma)}$-subspace (Theorem 1). However when $n$ is even the situation is less clear as in this case we only know that the subspace $A^\sigma$ has a constant rank property only when $\sigma$ is either an involution or else it has odd order (see Section~\ref{section2}). When $\sigma$ has even order it is only known that a skew form $f_{b, \sigma} \in A^{\sigma}$ may have rank either $n$ or $n - 2n/\ord(\sigma) $ and that both of these values are attained as ranks of suitable skew forms in $A^{\sigma}$. We study this last case more closely here and show that there are constant-rank subspaces in $A^{\sigma}$. In fact, $A^{\sigma}$ always has an $n$-subspace of dimension $\frac{n}{2}$ and moreover decomposes as a direct sum of $k$-subspaces for suitable $k$ (see Theorems A-D). %aim to this situation here by establishing a decomposition (in Theorems ... ) as a direct sum of constant-rank subspaces for $A^\sigma$ %in the case when $n$ is even, $L/K$ is cyclic with $G = \langle \sigma \rangle$. %we can do it for all elements %In particular we can always find an $n$-subspace inside $A^\sigma$. %For the sake of simplicity we assume that $L/K$ is cyclic with $G = \langle \sigma \rangle$. Then we have the following. \begin{theorem}[\cite{GQ09}] \label{odd decomposition} Suppose that $ n = [L : K] $ is odd and the Galois group $G = \{1,\sigma_1,\ldots,\sigma_{m}, \sigma_{1}^{-1}\sigma_{m}^{-1}\}$ where $ m = (n - 1)/2 $. Then there is a direct decomposition \begin{equation}\label{odd direct decom} \alt_K(L)=A^{1}\oplus A^{2}\oplus \cdots\oplus A^{m}, \end{equation} where $ A^{ i} \coloneqq A^{\sigma_i}$ has dimension $n$ ($1 \le i \le m$). Moreover, if $\ord(\sigma_{i})$ = $2r_i +1$, the non zero skew-forms in $A^i$ all have rank $n - \frac{n}{2r_i +1} $. \end{theorem} \begin{theorem} \emph{(\cite{GQ09})} Suppose that $ n = [L : K] $ is even and the Galois group $$G = \{1, \tau_1,\ldots,\tau_{k},\sigma_1,\ldots,\sigma_{m}, \sigma_{1}^{-1},\ldots,\sigma_{m}^{-1}\},$$ where $ \{\tau_{1},\tau_{2} ,\ldots,\tau_{k} \}$ are the involutions of $ G $, then there is a direct decomposition \begin{equation}\label{ev decompose} \alt_K(L)= B^{1} \oplus B^{2} \oplus \cdots\oplus B^{k}\oplus A^{1} \oplus A^{2} \oplus \cdots \oplus A^{m}. \end{equation} where $B^{i}: =A^{\tau_i}$ is an $n$-subspace of dimension $n/2$ for all $1 \le i \le k$ and $A^{j}: = A^{\sigma_j}$ ($1 \le j \le m$) has dimension $n$. Moreover if $\ord(\sigma_i)$ is odd then $A^{\sigma}$ is an $n - n/{\ord(\sigma_i)}$-subspace of dimension $n$. \end{theorem} %\begin{notation} If $ L/K$ is cyclic Galois extension of degree $ n $ with $G= \gal(L/K) = \langle \sigma \rangle$ we define $A^i \coloneqq A^{\sigma^{i}}$. Thus $ A^{i}= \{f_{b,\sigma^i}: b \in L\}$. If $n$ is even then there is a unique involution $ \tau_1 = \sigma^{n/2}$ and in this case we denote $B^1 : =A^{\tau_1}=\{f_{b,\sigma^{n/2}}: b\in L\}$. Then the decomposition~\eqref{ev decompose} becomes \begin{equation}\label{ dcmpstn for cyclic} \alt_K(L)= B^{1} \oplus A^{1}\oplus A^{2}\oplus \cdots\oplus A^{m}, \end{equation} \begingroup \setcounter{cdrthm}{0} \renewcommand{\thecdrthm}{\Alph{cdrthm}} \begin{theorem} \label{odd dcmposn for any field} Let $K$ be a field and $n = 2k$, where $ k\geq 1$ is odd. Let $L$ be any cyclic extension of $K$ of degree $n$ with Galois group $G = \langle \sigma \rangle$. Then \begin{equation} \label{odd dcmposn for any field eq} A^1 = \mathcal{U}_{1} \oplus \mathcal{V}_{1}, \end{equation} where $ \mathcal{U}_{1}$ is an $n$-subspace of dimension $k $ and $\mathcal{V}_{1}$ is an $ (n-2)$-subspace of dimension $k$. % More generally if $\ord(\sigma^i)$ is even in $G$ %Then there is a direct-sum decomposition into constant rank subspaces % \[A^1= \mathcal{U}_{1} \oplus \mathcal{V}_{1},\] where the subspace $ \mathcal{U}_{1}$ is an $n$-subspace of dimension $k $ and the subspace $\mathcal{V}_{1}$ is an $ (n-2)$-subspace of dimension $k$. In general if the order of $\sigma^{i}$ is $ \mathfrak{o}(\sigma^{i}) $ then if $ \mathfrak{o}(\sigma^{i}) \equiv 1 (\mo 2)$ then $A^{i}$ is an $(n - n/ \mathfrak{o}(\sigma^{i}))$-subspace of dimension $ n $ and $ \mathfrak{o}(\sigma^{i}) \equiv 0~(\mo 2)$ %then there is a direct-decomposition into constant rank subspaces \begin{equation*} % A^i = \mathcal{U}_i \oplus \mathcal{V}_i, %\end{equation*} where the subspace $ \mathcal{U}_i$ is an $n$-subspace of dimension $k $ and the subspace $\mathcal{V}_i$ is an $ (n-2n/ \mathfrak{o}(\sigma^{i}))$-subspace of dimension $k$.\\ \item[(3)] Finally \label{odd dcmpsn fr any field} \begin{equation} % \alt(L) = B^1 \oplus \left(\bigoplus_{\substack{ \ord(\sigma^{i}) ~\equiv~ 0~ (\mo 2)\\\ord(\sigma^{i})\neq 2}}\left(\mathcal{U}_i \bigoplus \mathcal{V}_i \right)\right) \bigoplus \left( \bigoplus_{\ord(\sigma^{i}) ~\equiv~ 1~ (\mo 2)} A^i \right) %\end{equation} %\end{itemize} \end{theorem} In view of Theorem~\ref{odd dcmposn for any field} in the following theorems we focus on the case where $ n$ is divisible by $ 4 $. \begin{theorem}\label{theoB} Suppose $ n = 2^{\alpha} k $ where $ \alpha \geq 2$ and $k$ is odd. Let $ K $ be an algebraic number field such that $ -1$ is not a square in $ K$. Then there exists a cyclic extension $L$ of $ K $ of degree $ n $ with the Galois group $G = \langle \sigma \rangle$ such that \begin{equation}\label{decmpsn A1 algebraic} A^{1} = \mathcal{E}_1 \oplus \cdots \oplus \mathcal{E}_{\alpha - 1} \oplus \mathcal{V}_1 \oplus \mathcal{V}_2 , \end{equation} where \begin{enumerate}[label=(\roman*)] \item %[(i)] $\mathcal{E}_{i}$ is an $n$-subspace of dimension $n/2^i$ for $ 1 \leq i \leq \alpha - 1$, \item %[(ii)] $\mathcal{V}_j$ is an $( n-2)$-subspace of dimension $ k $ for $ 1 \leq j \leq 2$. \end{enumerate} \end{theorem} \begin{theorem}\label{th for finite field} Let $ K $ be a finite field with $ q$ elements such that $ -1$ is not a square in $ K$. Let $ q + 1 = 2^a l$ (l odd) where $a \geq 1$ and $ n = 2^{\alpha}k$ (k odd) where $ \alpha \geq 2$. Suppose $ L $ is a cyclic extension of $ K $ of degree $ n $ with $\gal(L/K)=\langle\sigma_f\rangle$ where $\sigma_f$ is the Frobenius map of $L$ defined by $\sigma_f: b \to b^q$. \begin{enumerate}[label=(\arabic*)] \item %[(1)] If $\alpha \leq a+1 $ then \begin{equation}\label{ A1 finite field} A^{1} = \mathcal{V}_1 \oplus \mathcal{V}_2 \oplus \mathcal{E}_1 \oplus \cdots \oplus \mathcal{E}_{\alpha - 1}, \end{equation} where \begin{enumerate}[label=(\roman*)] \item %[(i)] $\mathcal{E}_{i}$ is an $n$-subspace of dimension $n/2^i$ for $ 1 \leq i \leq \alpha - 1$, \item %[(ii)] $\mathcal{V}_j$ is an $ (n-2)$-subspace of dimension $ k $ for $ 1 \leq j \leq 2$. \end{enumerate} \item %[(2)] If $\alpha >a+1 $ and $ l=1$, that is, $ q = 2^a - 1$, then \begin{equation}\label{decmpsn finite field2} A^{1} = \mathcal{V}_1 \oplus \mathcal{V}_2 \oplus \mathcal{E}_1 \oplus \cdots \oplus \mathcal{E}_{\alpha - 1}, \end{equation} where \begin{enumerate}[label=(\roman*)] \item %[(i)] $\mathcal{E}_{i}$ is an $n$-subspace of dimension $n/2^i$ for $ 1 \leq i \leq a$ and an $ (n-2)$-subspace of dimension $n/2^i$ for $ a+1 \leq i \leq \alpha -1$, \item %[(ii)] $\mathcal{V}_j$ is an $( n-2)$-subspace of dimension $ k $ for $ 1 \leq j \leq 2$. \end{enumerate} \end{enumerate} \end{theorem} \begin{theorem}\label{theoD} Let $p$ be a prime and $K = \mathbb{Q}_p$ be the $p$-adic completion of $ \mathbb{Q}$ such that $ -1$ is not a square in $ K$. Let $ p + 1 = 2^a l$ (l odd) where $a \geq 1$ and $ n = 2^{\alpha}k$ (k odd) where $ 2 \leq \alpha \leq a+1$. Then there exists a cyclic extension $ L $ of $ K $ of degree $ n $ such that the decomposition~\eqref{ A1 finite field} holds. %there is direct decomposition into constant rank subspaces % \begin{equation*} %A^{1} = \mathcal{V}_1 \oplus \mathcal{V}_2 \oplus \mathcal{E}_1 \oplus \cdots \mathcal{E}_{\alpha - 1}, % \end{equation*} % where %\begin{itemize} % \item[(i)] $\mathcal{E}_{i}$ is an $n$-subspace of dimension $n/2^i$ for $ 1 \leq i \leq \alpha - 1$, % \item[(ii)] $\mathcal{V}_j$ is an $ n-2$-subspace of dimension $ k $ for $ 1 \leq j \leq 2$. % \end{itemize} \end{theorem} \endgroup \setcounter{cdrthm}{0} \section{Skew forms and Galois extensions}\label{section2} Retaining the notation of the previous section we now collect some basic results from~\cite{GQ09} concerning the application of Galois theory to the study of some crucial properties of bilinear forms over $K$. In the following $L/K$ is a (not necessarily cyclic) Galois extension and $1 \ne \sigma \in \gal{L/K}$ is arbitrary. \begin{lemma}[{\cite[Lemma~2]{GQ09}}] \label{GQ degeneracy condition} Let $ f = f_{ b,\sigma} $ be an alternating bilinear form as defined above with $ b \neq 0 $ and let $F$ be the fixed field of the automorphism $\sigma^{2}$. If $\sigma(b)b^{-1}$ is expressible in the form $ \sigma^{2}(c)c^{-1} $ for some $ c \in L^{\times}$ then $\rk(f_{ b,\sigma}) = n - n/[L:F]$. Otherwise $ \rk(f_{ b,\sigma}) = n $. \end{lemma} \begin{lemma}[{\cite[Lemma~4]{GQ09}}] \label{existance a non degenarate} Suppose that the automorphism $\sigma $ has even multiplicative order $ 2r $, say. Then there exist elements $ b \in L^{\times} $ such that the equation $ \sigma(b)b^{-1} = \sigma^{2}(c)c^{-1} $ has no solution for all $ c\in L^{\times} $. \end{lemma} %We can now introduce our main object of focus. %Let $A^{\sigma} \coloneqq \{f_{b,\sigma} : b\in L \}$. Clearly $A^{\sigma}$ is a %subspace of $\alt(L)$ and it can be checked that $A^{\sigma} = A^{\sigma^{-1}}$. %Also we have from~\cite{GQ09} %that if $\sigma $ is a non-identity element of the Galois group $G $ of $L$ over $K$ then, unless $\sigma $ has order %$2$, $ \dim A^{\sigma} = n $. In the exceptional case that $ \sigma $ has order $2 $, $ A^{\sigma} $ has dimension $ \frac{n}{2} $. \begin{rema}\label{isomorphism} If $\sigma$ is not an involution then the map $ b \to f_{b,\sigma}$ defines an isomorphism of $K$-spaces between $A^\sigma$ and $L$ \cite[Theorem~1]{GQ09}. \end{rema} \begin{lemma}[{\cite[Lemma~3]{GQ09}}] \label{rank of f b sigma for odd case } Suppose that the automorphism $ \sigma $ has odd multiplicative order $ 2r + 1 > 1 $, say. Then, if $ b \neq 0 $, the rank of the skew-form $ f = f_{ b,\sigma } $ is $ n - n/{2r+1} $. \end{lemma} \begin{lemma}[{\cite[Lemma~4]{GQ09}}] \label{rank of f b sigma for even case } Suppose that the automorphism $\sigma$ has even multiplicative order $2r \geq 2$, say. Then, if $ b \neq 0 $, the rank of the skew-form $f=f_{b,\sigma} $ is either $ n-\frac{n}{r}$ or $ n $. \end{lemma} \section{Preliminary results}\label{Sect3} Our aim in this section is to establish certain facts which will be found useful in the subsequent sections and are also interesting in their own right. Recall that if $F$ is an intermediate subfield and $a \in L$ then the $L/F$-norm $N_{L/F}(a)$ of $a $ is defined as $N_{L/F}(a) = \prod_{\theta \in \gal(L/F)} \theta(a)$. \begin{nota} Throughout this section $L/K$ denotes a cyclic extension with Galois group $\gal(L/K) = \langle \sigma \rangle$. For the sake of convenience in what follows we shall denote the subfield $L^{\langle \sigma^{i} \rangle}$ as $L_i$. \end{nota} We begin by noting the following restatement of the degeneracy criterion Lemma~\ref{GQ degeneracy condition}. \begin{prop} \label{GMdegen-crit} Let $b \in L$. Then the skew-form $f_{b,\sigma}$ is degenerate if and only if \begin{equation}\label{form1} N_{L/L_{2}} (\sigma(b)/b) = 1, \end{equation} that is, $f_{b,\sigma}$ is degenerate if and only if \begin{equation} \label{form2} N_{L/L_{2}} (b) = b\sigma^{2}(b)\cdots \sigma^{n-2}(b) \in K. \end{equation} \end{prop} \begin{proof} By Lemma~\ref{GQ degeneracy condition}, the skew form $f_{b, \sigma}$ is degenerate if and only if $\sigma(b)/b = \sigma^2(c)/c$ for some $c \in L$. The first assertion is now clear in view of the Hilbert Theorem 90. Moreover the condition $ N_{L/L_2 }(\sigma(b)/b) = 1 $ is easily seen to be equivalent to the product $b\sigma^{2}(b)\cdots \sigma^{n-2}(b) $ being $\sigma$-invariant. \end{proof} %The Proposition~\ref{GMdegen-crit} above provides the nondegeneracy condition for the skew-form $f_{b,\sigma} \in A^1 \subseteq \alt_K(L)$. Now, we will interpret the nondegeneracy condition Suppose that $\sigma^i$ is not an involution. By Lemma~\ref{GQ degeneracy condition} the skew-form $f_{b,\sigma^i} \in A^i \subseteq \alt_K(L) $ is degenerate if and only if $\sigma^i(b)/b= \sigma^{2i}(c)/c$. As $\sigma^{2i}$ is a generator for $\gal(L/L_{2i})$, in view of Hilbert Theorem 90, $f_{b,\sigma^i}$ is degenerate if and only if $N_{L/L_{2i} }(\sigma^i(b)/b) = 1$. A glance at Proposition~\ref{GMdegen-crit} above shows that this is precisely the condition for the skew-form $ \displaystyle f^{\sim}_{b,\sigma^{i}} \in \alt_{L_i}(L)$ defined by \[ f^{\sim}_{b,\sigma^{i}} = \tr^L_{L_i} (b(x\sigma(y)-\sigma(x)y)), \qquad \forall x, y \in L. \] to be degenerate (we write $f^{\sim}_{b,\sigma^{i}}$ instead of $f^{}_{b,\sigma^{i}}$ to emphasize the fact that we are now considering $L$ as $L_i$-space). Let us write $\displaystyle A^{\sim 1} \coloneqq \{ \displaystyle f^{\sim}_{b,\sigma^{i}} \mid b \in L\}$. In view of Remark~\ref{isomorphism} we then have a K-isomorphism $A^i \equiv L$ via $f_{b,\sigma^i} \mapsto b$ and an $L_i$-isomorphism $L \equiv A^{\sim 1}$ via $b \mapsto f^{\sim}_{b,\sigma^{i}}$. The composition of these maps clearly yields a $K$-isomorphism $A^i \cong A^{\sim 1}$. The following is then clear. \begin{rema}\label{crspn} With respect to the above isomorphism if an $L_i$-subspace $\mathcal W \le A^{\sim 1}$ has all its non-zero skew forms non-degenerate (or all its non-zero skew forms degenerate) then the same is true for the corresponding (K-) subspace in $A^i$. %Set $ H\coloneqq \gal(L/L_i)=\langle\sigma^i\rangle$ then suppose $ T: L \to L_i$ be the trace form defined by \[T(x) =\displaystyle\sum_{\rho \in H}(\rho(x))\] $\forall x \in L$. Let $ \tau_1,\ldots,\tau_k$ be a complete set of coset representatives of the subgroup $H$ in $ G$ and let $ T_2 : L_i \to K$ be the $K$-linear form defined by \[S(z)= \displaystyle \sum_{j=1}^{k} S(z)\] $\forall z \in L_i$. Then we have $ Tr(x) =S(T(x))$ $ \forall x \in L$. Thus \[f_{b,\sigma^{i}}(x,y)=S f^{\sim}_{b,\sigma^{i}}(x,y) \] $\forall x,y \in L$. Then in view of Proposition~\ref{GMdegen-crit} $f_{b,\sigma^{i}}$ is degenerate if and only if $f^{\sim}_{b,\sigma^{i}}$ is degenerate. \end{rema}\label{chng2Ai} \begin{lemma}\label{Egnspcdecomp} Let $n= 2^{\alpha}k$ where $ \alpha \geq 2$ and $k$ is odd. Suppose that $L$ is a cyclic extension of a field $K$ of degree $n$ with Galois group $\gal(L/K) = \langle \sigma \rangle$. Then the following hold. \begin{enumerate}[label=(\roman*)] \item %[(i)] For $1 \leq i \leq \alpha - 1 $ the subspace $ E_i\coloneqq \{b \in L : \sigma^{n/2^i}(b)= -b\} \le L$ has dimension $n/2^i$. \item %[(ii)] Let $V_1 \coloneqq \{b \in L : \sigma^{k}(b) = b\}$ and $V_2 \coloneqq \{b \in L : \sigma^{k}(b) = -b\}$. Then $\dim(V_1) = \dim (V_2)= k$. \end{enumerate} \end{lemma} \begin{proof} Let $1 \le i \le \alpha - 1$. As the order of the automorphism $ \sigma^{n/2^i}$ is $ 2^i$ so the fixed field $L_{n/2^i}$ of $ \sigma^{n/2^i}$ has dimension $ n / 2^i$ over $K$. We can view $\sigma^{n/2^i}$ as a $K$-linear map of $L$. By the Dedekind independence theorem the minimal polynomial of $\sigma^{n/2^i}$ is $x^{2^i} - 1$. Let $ j_i\in L$ be an eigenvector of $ \sigma^{n/2^i}$ corresponding to the eigenvalue $ -1$. It is easily checked that the corresponding eigenspace is $E_i \coloneqq j_i L_{n/2^i}$. It follows that $\dim(E_i)= n/2^i$. The proof of (ii) is similar. \end{proof} \begin{lemma}\label{ degeneracy for E_i} Let $n= 2^{\alpha}k$ where $ \alpha \geq 2$ and $k$ is odd. Suppose that $L$ is a cyclic extension of a field $K$ of degree $n$ with Galois group $\gal(L/K) = \langle \sigma \rangle$. Then $\forall b_i \in E_i \setminus \{0\} $ \begin{equation} N_{L/L_2}(b_i)= (-1)^{n/2^2} w_i^{2^i}, \end{equation} where $w_i \coloneqq b_i \sigma^{2}(b_i)\cdots \sigma^{n/2^{i} - 2} (b_i)$. Moreover, $f_{b_i,\sigma}$ is degenerate if and only if $\eta_i \coloneqq \sigma(w_i)/w_i$ is a $ 2^i$-th root of unity in $L$ such that $\sigma(\eta_i)= - {\eta_i}^{-1}$. In particular, $f_{b_1,\sigma}$ is non-degenerate for all $b_1 \in E_1 \setminus \{0\}$. \end{lemma} \begin{proof} In view of the chain of inclusions \[ L \supset L_{n/2} \supset \cdots \supset L_{n/2^{i-1}} \supset E_i,\] we have for $ b_i \in E_i \setminus \{0\}$ \begin{align*} N_{L/L_2}(b_i)&= b_i \sigma^{2}(b_i)\cdots \sigma^{n-2}(b_i)\\ &= \left( b_{i} \sigma^2(b_i) \cdots \sigma^{n/2 -2 }(b_i)\right) \left( \sigma^{n/2}(b_i) \sigma^{n/2 + 2}(b_i) \cdots \sigma^{n/2 + n/2 -2}(b_i)\right)\\ &=\left(b_i \sigma^{2}(b_i)\cdots \sigma^{n/2 - 2} (b_i)\right)^{2}\\ &=\left(b_i \sigma^{2}(b_i)\cdots \sigma^{n/4 - 2} (b_i)\right)^{2^2}\\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots \\ &=\left(b_i \sigma^{2}(b_i)\cdots \sigma^{n/2^{i-1} - 2} (b_i)\right)^{2^{i-1}}\\ &=\left[\left(b_i \sigma^{2}(b_i)\cdots \sigma^{n/2^{i} - 2} (b_i)\right)\left(\sigma^{n/2^{i}}(b_i) \sigma^{n/2^{i} + 2}(b_i)\cdots \sigma^{n/2^{i} +n/{2^{i}} -2}(b_i) \right)\right]^{2^{i-1}}\\ &= \left[\left(b_i \sigma^{2}(b_i)\cdots \sigma^{n/2^{i} - 2} (b_i)\right)\left((-b_i) (-\sigma^{2}(b_i))\cdots (-\sigma^{n/2^{i} - 2} (b_i))\right)\right]^{2^{i-1}}\\ &= \left[(-1)^{n/2^{i+1}}\left(b_i \sigma^{2}(b_i)\cdots \sigma^{n/2^{i} - 2} (b_i)\right)^2\right]^{2^{i-1}}\\ &= (-1)^{n/{2^2}}[b_i \sigma^{2}(b_i)\cdots \sigma^{n/2^{i} - 2} (b_i)]^{2^i}\\ &=(-1)^{n/2^2} w_i^{2^i}. \end{align*} Then \begin{align*} \frac {N_{L/L_2}(\sigma(b_i))}{N_{L/L_2}(b_i)}= \left( \frac{(\sigma((-1)^{n/{2^2}}w_i))}{(-1)^{n/{2^2}}w_i}\right)^{2^i} = \left(\frac{\sigma(w_i)}{w_i}\right)^{2^i}= \eta_i^{2^i}. \end{align*} Set $\eta_i \coloneqq \frac{\sigma(w_i)}{w_i}$. By Proposition~\ref{GMdegen-crit}, $f_{b_i , \sigma}$ is degenerate if and only if $ \eta_i $ is a $ 2^{i}$-th root of unity $ \eta_i $. Moreover, \[ -w_i =\sigma^2(w_i)=\sigma(\eta_i w_i)=\sigma(\eta_i) \eta_i w_i,\] whence $ \sigma(\eta_i) \eta_i =-1$, that is, $ \sigma(\eta_i) = -\eta_i^{-1}$. The last assertion in the theorem is now clear. \end{proof} \begin{lemma}\label{degeneracy for V1 ,V2} Let $n= 2^{\alpha}k$ where $ \alpha \geq 2$ and $k$ is odd. Suppose that $L$ is a cyclic extension of $K$ of degree $n$ with $\gal(L/K) = \langle \sigma \rangle$. Then $\forall b \in V_1 \cup V_2 $, $f_{b,\sigma}$ is degenerate. \end{lemma} \goodbreak \begin{proof}\ \relax \begin{proof}[Case I] Let us first assume that $ k>1$. Then the field $V_1 = L_k$ has dimension $ k $ over $ K$. Again by Dedekind's independence theorem it follows that the minimal polynomial of $\sigma^k$ is $ x^{2^{\alpha}} -1$. Let $ j_\alpha $ be an eigenvector of $ \sigma^k$ corresponding to the eigenvalue $-1$ and it is easily checked that the corresponding eigenspace is $V_2 = j_\alpha L_k$. Thus $\dim(V_1)=\dim(V_2)=k$. Note that $V_1$ and $V_2$ are $\sigma$-invariant. Again in view of the inclusions \[L \supset L_{n/2} \supset \cdots \supset L_{n/2^{\alpha-1}}=L_{2k} \supset L_k =V_1,\] we have $\forall b \in V_1 \setminus \{0\}$, \begin{align*} N_{L/L_2}(b) &= b \sigma^{2}(b) \cdots \sigma^{n-2}(b)\\ &=\left(b \sigma^{2}(b) \cdots \sigma^{n/2^{\alpha-1}-2}(b)\right)^{2^{\alpha -1}}\\ &=\left( b \sigma^2(b) \cdots \sigma^{2k-2}(b)\right)^{2^{\alpha -1 }}\\ &=\left[\left(b \sigma^2(b)\cdots \sigma^{k-1}(b)\right)\left( \sigma^{k+1}(b) \cdots \sigma^{2k-2}(b)\right)\right]^{2^{\alpha - 1}}\\ &=\left[\left(b \sigma^2(b) \cdots \sigma^{k-1}(b)\right)\left( \sigma(b) \cdots \sigma^{k-2}(b)\right)\right]^{2^{\alpha -1}}\\ &=[b \sigma(b) \sigma^2(b)\cdots \sigma^{k-1}(b)]^{2^{\alpha-1}}\\ &= N_{L/L_2}(\sigma(b)). \end{align*} On the other hand in view of the inclusions \[L \supset L_{n/2} \supset \cdots \supset L_{n/2^{\alpha-1}}=L_{2k} \supset j_{\alpha}L_k =V_2,\] we have $ \forall b \in V_2 \setminus \{0\}$, \begin{align*} N_{L/L_2}(b) &= b \sigma^{2}(b) \cdots \sigma^{n-2}(b)\\ &=\left(b \sigma^{2}(b) \cdots \sigma^{n/2^{\alpha-1}-2}(b)\right)^{2^{\alpha -1}}\\ &=\left( b \sigma^2(b) \cdots \sigma^{2k-2}(b)\right)^{2^{\alpha -1 }}\\ &=\left[\left(b \sigma^2(b)\cdots \sigma^{k-1}(b)\right)\left( \sigma^{k+1}(b) \cdots \sigma^{2k-2}(b)\right)\right]^{2^{\alpha - 1}}\\ &=\left[\left(b \sigma^2(b) \cdots \sigma^{k-1}(b)\right) \left( (-\sigma(b)) \cdots (-\sigma^{k-2}(b)\right)\right]^{2^{\alpha -1}}\\ &=[b \sigma(b) \sigma^2(b)\cdots \sigma^{k-1}(b)]^{2^{\alpha-1}}\\ &= N_{L/L_2}(\sigma(b)). \end{align*} Consequently $N_{L/L_{2}} (\sigma(b)/b) = 1$ and thus by Proposition~\ref{GMdegen-crit} $ \forall b \in V_1 \cup V_2$, $ f_{b,\sigma}$ is degenerate. \let\qed\relax \end{proof} \begin{proof}[Case II] We now assume that $ k = 1$ (thus $n = 2^\alpha$ and $ L_{2k} = L_2$). Then $ V_1 \coloneqq K$ and it is easily checked that $ V_2 \coloneqq j_{\alpha}K$, where $ j_{\alpha}$ is an eigenvector of $\sigma$ corresponding to the eigenvalue $-1$. %It is straight forward to check that the corresponding eigenspace is $j_\alpha K$. Thus $\dim(V_1)=\dim(V_2)=1$. Clearly if $ b \in L_2^{\times} $ then $ N_{L/L_2}(b) = b^{2^{\alpha-1}}$ and $N_{L/L_2}(\sigma(b)) = (\sigma(b))^{2^{\alpha -1}} $ as $L_2$ is $\sigma$-invariant. By definition if $b \in V_1 \cup V_2$ then $\sigma(b)= \pm b$ and in either case \[ N_{L/L_{2}} \biggl (\frac{\sigma(b)}{b} \Biggr ) = { \Biggl (\frac{\sigma(b)}{b} \Biggr )}^{2^{\alpha -1}} = 1. \] Thus by Proposition~\ref{GMdegen-crit} if $ b \in V_1 \cup V_2$, $ f_{b,\sigma}$ is degenerate. \end{proof} \let\qed\relax \end{proof} \section{Proofs of Theorems~\texorpdfstring{\ref{odd dcmposn for any field}}{A} and~\texorpdfstring{\ref{theoB}}{B} }\label{Sect4} \subsection{Proof of Theorem~\texorpdfstring{\ref{odd dcmposn for any field}}{A}} \begin{proof} Let $V \coloneqq L_{k}$ and $0 \ne v \in V$. Clearly \[ \sigma^2(v), \sigma^4(v),\ldots, \sigma^{2k-2}(v) \in V. \] It follows that \[ N_{L/L_2}(v) \in L_2 \cap V = L_2 \cap L_{k} = K. \] \label{ground field} By Proposition~\ref{GMdegen-crit} the skew-form $f_{v, \sigma}$ is degenerate and by Lemma~\ref{rank of f b sigma for even case } it has rank $n-2=2k-2$. By Lemmas~\ref{GQ degeneracy condition} and~\ref{existance a non degenarate} there exists a $j \in L$ such that $f_{j, \sigma}$ is non-degenerate. Then for $0 \ne v \in V$ \[ N_{L/L_2}(jv) = N_{L/L_2}(j)N_{L/L_2}(v) \not \in K.\] It thus follows by proposition~\ref{GMdegen-crit} that all the nonzero skew-forms $ f_{b,\sigma}$ where $b$ lies in the subspace $U = jV$ (of dimension $k$) are non-degenerate. Clearly $U \cap V = \{0\}$ so $L= U \oplus V$. By Remark~\ref{isomorphism} the subspace $U$ of $L $ corresponds to a subspace $\mathcal{U}$ of $\alt_K(L)$ with the same dimension defined by $\mathcal{U}\coloneqq \{f_{b,\sigma}: b \in U\}$. Similarly $V$ corresponds to $\mathcal{V} \le \alt_K(L)$ such that $\dim(V)=\dim(\mathcal{V})$. Then the decomposition~\eqref{odd dcmposn for any field eq} follows. \end{proof} \begin{coro}\label{dcompsn-Ai-for-odd} Let $K $ be a field and $n$ be even. Suppose $L$ is a cyclic Galois extension of a field $K$ of degree $n$ with Galois group $\gal(L/K) = \langle \sigma \rangle$. If $ \ord(\sigma^{i}) \equiv 2~(\mo 4)$ and $ \ord(\sigma^{i}) \ne 2 $ then \[A^i = \mathcal{U}_i \oplus \mathcal{V}_i,\] where $ \mathcal{U}_i$ is an $n$-subspace of dimension $n/2 $ and $\mathcal{V}_i$ is an $ (n-2n/ \ord(\sigma^{i}))$-subspace of dimension $n/2$. \end{coro} \begin{proof} This follows from Theorem~\ref{odd dcmposn for any field}, noting Remark~\ref{crspn} and the fact (Lemma~\ref{rank of f b sigma for even case }) that a skew form in $A^i$ is either non-degenerate or has rank equal to $n - 2n/\ord(\sigma^i)$. \end{proof} Consequently we obtain the following. \begin{coro} \label{odd dcmpsn fr any field} Let $K$ be a field and $n = 2k$, where $ k\geq 1$ is odd. Let $L$ be any cyclic Galois extension of $K$ of degree $n$ with Galois group $G = \langle \sigma \rangle$. Then \begin{equation} \alt_K(L) = B^1 \bigoplus \left(\bigoplus_{\substack{ \ord(\sigma^{i}) ~\equiv~ 0~ (\mo 2)\\\ord(\sigma^{i})\neq 2}}\left(\mathcal{U}_i \bigoplus \mathcal{V}_i \right)\right) \bigoplus \left( \bigoplus_{\ord(\sigma^{i}) ~\equiv~ 1~ (\mo 2)} A^i \right) \end{equation} \end{coro} \begin{proof} Clear in view of Corollary~\ref{dcompsn-Ai-for-odd}, Lemma~\ref{rank of f b sigma for odd case } as well as the decomposition~\eqref{ dcmpstn for cyclic}. \end{proof} \begin{rema} \label{n-subs-exists-in} Let $n= 2^{\alpha}k$ where $ \alpha \geq 1$ and $k$ is odd. Suppose that $L$ is a cyclic extension of a field $K$ of degree $n$ with Galois group $\gal(L/K) = \langle \sigma \rangle$. If $\ord(\sigma^i)$ is even then there always exists an $n$-subspace of dimension $n/2$ inside $A^i$. If $ \alpha = 1$ this follows from Corollary~\ref{odd dcmpsn fr any field}. Otherwise if $ \alpha > 1$ then it follows from Lemma~\ref{ degeneracy for E_i} that $\mathcal{E}_1: =\{f_{b,\sigma}: b\in E_1\} $ is the desired subspace for $A^1$. The corresponding assertion for $A^i$ now follows in the light of Remark~\ref{crspn}. \end{rema} \subsection{ Proof of Theorem~\texorpdfstring{\ref{theoB}}{B}} \begin{proof} Firstly we will construct a cyclic extension $L$ of $ K $ such that $ i \notin L$ where $ i $ is a primitive $ 2^2$-th root of unity. Let $p$ be a prime such that $ p \equiv 1 (\text{mod}~ n) $ and consider the cyclotomic extension $\mathbb{Q}(\eta_p)$ where $ \eta_p$ is a primitive $p$-th root of unity. As is known (e.g.,~\cite[Lemma~4]{GQ06}) it is possible to pick the prime $p$ as above such that $ \mathbb{Q}(\eta_p) \cap K(i) =\mathbb{Q} $. %Then also $ \mathbb{Q}(\eta_p) \cap K =\mathbb{Q} $. Let $L$ be the unique intermediate field $\mathbb{Q} \subseteq L \subseteq \mathbb{Q}(\eta_p)$ such that $[L:\mathbb{Q}] =n$. Clearly $ L \cap K(i) =\mathbb{Q}= L \cap K$. By a well known fact (e.g.,~\cite[Chapter~6, Theorem~1.12]{SL1993}) the extensions $ LK(i)/K(i)$ and $ LK/K$ are Galois and %\begin{equation}\label{x} \[ \gal(LK(i)/K(i)) \cong \gal(L/{L \cap K(i)})= \gal(L/\mathbb{Q})= \gal(L/{L \cap K})\cong \gal(LK/K). \] %\end{equation} If $i \in LK$ then by the last equation \[ [LK:K]=[LK:K(i)][K(i):K]=[LK(i):K(i)][K(i):K], \] whence $[K(i):K]=1$ thus contradicting the hypothesis on $K$. Redefining $L\coloneqq LK$ yields the desired cyclic extension $L/K$ with degree $n$.\newline Let $E_i \coloneqq \{b \in L : \sigma^{n/2^i}(b)= -b\} \ (1 \le i \le \alpha -1)$. By Lemma~\ref{Egnspcdecomp} we obtain $ L_{n/2^{i-1}}= L_{n/2^i} \oplus E_i$ and $L_{2k} = V_1 \oplus V_2$, where $V_1$ and $V_2$ denote the eigenspaces of $\sigma^k$ with respect to the eigenvalues $1$ and $-1$ respectively. Consequently, we obtain \begin{equation} \label{A1decom} L = L_{n/2} \oplus E_1 = L_{n/4} \oplus E_2 \oplus E_1 = L_{2k} \oplus E_{\alpha - 1} \oplus \cdots \oplus E_1 = V_1 \oplus V_2 \oplus E_{\alpha - 1} \oplus \cdots \oplus E_1. \end{equation} Let $\mathcal{E}_i$ be the subspace of $A^1$ corresponding to $E_i \coloneqq \{b \in L : \sigma^{n/2^i}(b)= -b\}$ under the isomorphism of Remark~\ref{isomorphism}, that is, $\mathcal{E}_i= \{ f_{b,\sigma}: b \in E_i\}$ ( $1 \leq i \leq \alpha -1$ ). By our construction, the only $2^i$-th roots in $L$ are $\pm 1$. As $\sigma$ fixes both these roots, it follows from Lemma~\ref{ degeneracy for E_i} that $\mathcal E_i$ is an $n$-subspace for all $i$ in the above range. Similarly, let $\mathcal V_j$ correspond to the subspace $V_j$ of $L$. By Lemma~\ref{degeneracy for V1 ,V2} the nonzero skew-forms in $\mathcal V_j$, where $j = 1,2$ are degenerate whence these are $(n - 2)$-spaces by Lemma~\ref{rank of f b sigma for even case }. The required decomposition~\eqref{decmpsn A1 algebraic} is now immediate from~\eqref{A1decom}. \end{proof} \begin{coro}\label{decmpsn Ai algebraic} In the situation of Theorem~\ref{theoB} if $ \ord(\sigma^{i}) \equiv 0~(\mo 4)$, say $\ord(\sigma^{i})= 2^{\beta}k' \ (\beta \ge 2)$ then \begin{equation} A^{i} = \mathcal{V}_1^i \oplus \mathcal{V}_2^i \oplus \mathcal{E}_1^i \oplus \cdots \mathcal{E}_{\beta - 1}^i, \end{equation} where \begin{enumerate}[label=(\roman*)] \item %[(i)] $\mathcal{E}_{k}^{i}$ is an $n$-subspace of dimension $n/2^i$ for $ 1 \leq k \leq \beta - 1$, \item %[(ii)] $\mathcal{V}_j^i$ is $(n -2n/\ord (\sigma^i))$-subspace % $\mathcal{V}_j^i$ is $(n-2)$-subspace of dimension $ k'n/{\ord(\sigma^{i})} $ for $ 1 \leq j \leq 2$. \end{enumerate} \end{coro} \begin{proof} This follows from proof of Theorem~\ref{theoB}, noting Remark~\ref{chng2Ai} and the fact (Lemma~\ref{rank of f b sigma for even case }) that a skew form in $A^i$ is either non-degenerate or has rank equal to $n - 2n/\ord(\sigma^i)$. \end{proof} \begin{coro}\label{algebraic dcmpsn Alt L} %Suppose $ n = 2^{\alpha} k $ where $ \alpha \geq 2$ and $k$ is odd. Let $ K $ be an algebraic number field such that $ -1$ is not a square in $ K$. Then there exists a cyclic Galois extension $L$ of $ K $ of degree $ n $ with the Galois group $G = \langle \sigma \rangle$ such that In the situation of Theorem~\ref{theoB} there is direct-decomposition \begin{equation} \begin{aligned}[b] \alt_K(L) = & B^1 \oplus \left(\bigoplus_{\substack{ \ord(\sigma^{i}) ~\equiv~ 2~ (\mo 4)\\\ord(\sigma^{i})\neq 2}} \left(\mathcal{U}_i \oplus \mathcal{V}_i \right)\right) \oplus \left( \bigoplus_{\ord(\sigma^{i}) ~\equiv~ 1~ (\mo 2)} A^i \right) \\ &\bigoplus_{ \ord(\sigma^{i}) ~\equiv~ 0~ (\mo 4)}\left(\mathcal{V}_1^i \oplus \mathcal{V}_2^i \oplus \mathcal{E}_{\beta - 1}^i \oplus \cdots \oplus\mathcal{E}_1^i\right) \end{aligned} \end{equation} \end{coro} \begin{proof} Using Corollaries~\ref{dcompsn-Ai-for-odd},~\ref{decmpsn Ai algebraic} and Lemma~\ref{rank of f b sigma for odd case } as well as the decomposition~\eqref{ dcmpstn for cyclic}, we can deduce the required decomposition. \end{proof} \begin{rema}\label{theorem B rema} As its proof shows, Theorem~\ref{theoB} as well as its corollaries remain valid for an arbitrary cyclic extension $L/K$ of degree $n = 2^{\alpha}k$ ($\alpha \ge 2$) such that $-1$ is not a square in $L$. Similarly, let $K$ be a field such that $f(X): = X^{4} + 1 $ is irreducible in $K[X]$ (it is not difficult to show that $K$ has this property if and only if none of $ -1 ,2$ and $-2$ is a square in $K$). Then Theorem $B$ holds true for any cyclic extension $L/K$ of degree $n = 2^\alpha k$. Indeed, if $\eta_i$ is a $2^i$-root of unity for $i \ge 1$ then the conditions $-1 \not \in K^2$ and $\sigma(\eta_i) = -\eta_i^{-1}$ mean that $\eta_i \not \in \{\pm 1, \pm i \}$, where $i$ denotes a primitive $4$-th root of unity in $L$. Thus $\eta$ must have order $2^s$ where $s \ge 3$. Since $\eta \in L_2$ this would mean that $L_2$ contains an element of order $8$ and thus a root of $f$ implying $f$ has a quadratic factor in $K[X]$. %If $ n = 2^{\alpha}k$ where $ \alpha \geq 2$ and $ k $ is odd then then for any cyclic Galois extension $ L$ over $ K $ the above decomposition~\eqref{decmpsn A1 algebraic} %of $ A^{1}$ remain valid. %If possible let there exists some %$ b_i \in E_i \setminus \{0\}$ %such that $f_{b_i,\sigma}$ be degenerate % then by Lemma~\ref{ degeneracy for E_i} $\eta_i \in L_2$ with $\sigma(\eta_i)=-\eta_i^{-1}$. % By the hypothesis in $K$, $\eta_i$ must be lies in the set $\{\pm 1, \pm i\}$. Then $\eta_i= \pm i$ as $\sigma$ fixes $ \pm 1$. Consequently \[ \sigma(\eta_i) = - \eta_i^{-1}= \eta^2 \eta_i^{-1} = \eta_i,\] contradicting the fact that $-1$ is not a square in $K$. %Since we have seen that in the proof that if $ b_i \in E_i \setminus \{0\}$ then $ f_{b,\sigma}$ is degenerate if and only if $ \eta_i$ is a $ 2^i$-th root of unity and $ \sigma(\eta_i)= -\eta_i^{-1}$. But we have already seen that then $i\geq 3$ and $\sigma(\eta_i) = - \eta_i^{-1}$ which implies $ \eta_i \in L_2$. But by the hypothesis $ x^{4} +1$ is irreducible over $ K $ so any primitive $8$-th root of unity, say $ \zeta \notin L_2$ which implies $ \eta_i \notin L_2$, a contradiction. \end{rema} \section{Proofs of Theorems~\texorpdfstring{\ref{th for finite field}}{C} and~~\texorpdfstring{\ref{theoD}}{D}}\label{Sect5} \subsection{Proof of Theorem~\texorpdfstring{\ref{th for finite field}}{C}} \begin{proof} % Let $\sigma: b \to b^q$ be the Frobenius map of $L$. Then $\gal(L/K) = \langle \sigma_f \rangle$. Let $ E_i$ ( $1 \leq i \leq \alpha -1$ ) and $ V_j$ ( $1 \leq j \leq 2$ ) be as in Lemma~\ref{Egnspcdecomp}. As in the proof of Theorem~\ref{theoB}, we have \begin{equation*} L = V_1 \oplus V_2 \oplus E_{\alpha - 1} \oplus \cdots \oplus E_1. \end{equation*} By the hypothesis $-1$ is not a square in $K$ from which it easily follows that $a \geq 2$. Let $ w_i$ and $ \eta_i$ be as in Lemma~\ref{ degeneracy for E_i}. Note that $ \sigma_f^2(w_i) = -w_i$ and thus $w_i^2\in L_2$ but $ w_i \notin L_2$. Consequently $ w_i^{2(q^2 -1)} = 1$ and $w_i^{(q^2 -1)}= -1$. Since $\sigma_f(w_i) = w_i^q$ hence $ \eta_i = w_i^{q-1}$. %\begin{align*} %\et%a = w^{q-1} %mplies & \eta^{q+1} \ \ \% =w^{q^2 -1}\\ %\implies &\eta^{q+1}\ \ \ = -1\\ % \implies &\eta^{2(q+1)} = 1. % \end{align*} It follows that $ \eta_i$ is a $2(q+1)$-th root of unity but not a $ (q+1)$-th root of unity. \begin{proof}[(1)] Suppose $ \alpha \leq a+1$. Since $1 \leq i\leq \alpha -1$ therefore $1\leq i \leq a$. Again by Lemma~\ref{ degeneracy for E_i}, $f_{b_i,\sigma_f}$ is degenerate if and only if $ \eta_i $ is a $2^{i}$-th root of unity. Since $ i \leq a$, this would mean that $\eta_i^{q+1}= \eta_i^{2^a l}=1$, a contradiction. Let $\mathcal{E}_i$ be the subspace of $ A^1$ corresponding to $ E_i$ under the isomorphism of Remark~\ref{crspn}. It follows that $\mathcal{E}_i$ is an $n$-subspace of dimension $n/2^i$. \let\qed\relax \end{proof} \begin{proof}[(2)] Suppose $\alpha > a+1$. Pick $ i \in [1,\alpha -1] $. If $ 1 \leq i\leq a$ it follows from part (1) above that $ E_i$ is an $n$-subspace for $1 \leq i \leq a$. So we assume that $i \geq a+1$. By the hypothesis $l = 1$, whence $ \eta_i^{2^{a+1}}= \eta_i^{2(q+1)}=1$. It follows that if $ a+1 \leq i \leq \alpha -1 $ then $ \eta_i^{2^i}=1$. Thus in view of Lemma~\ref{ degeneracy for E_i} all the skew-forms in $ \mathcal{E}_i $ are degenerate and in this case by Lemma~\ref{rank of f b sigma for even case }, $\mathcal{E}_i$ is an $(n-2)$-subspace. \newline Similarly let $ \mathcal{V}_j$ be the subspace of $ A^1$ corresponding to $V_j$. Then by Lemmas~\ref{degeneracy for V1 ,V2} and~\ref{rank of f b sigma for even case }, $\mathcal{V}_j$ is an $(n-2)$-subspace. \end{proof} \let\qed\relax \end{proof} \begin{rema} In Theorem~\ref{th for finite field} when $ \alpha > a+1$ and $ l > 1$ then $ \mathcal{E}_i$ is neither an $ n$-subspace nor an $( n-2)$-subspace for $a+1 \leq i \leq \alpha -1$. Indeed, by the definition of $E_i$ \begin{align*} E_{i} = \{ b\in L : \sigma_f^{n/2^i}(b) = -b \} = \{ b\in L : b^{q^{n/2^i} - 1} = -1 \}. \end{align*} Let $C\coloneqq \{b\in L^{\times}: b^{2( q^{n/2^i}-1)}=1\}$. Then $ C $ is a cyclic subgroup of $L^{\times}$. Clearly, $ C = L_{n/2^i}^{\times} \cup ( E_{i} \setminus \{0\} )$. Let $u$ be a generator of $ C$. It is clear that $b_i = u^s \in E_i$ if and only if $s$ is odd. We claim that $f_{b_i, \sigma_f}$ is degenerate if and only if $s$ is an odd multiple of $l$. Indeed, let $ w_i$ and $ \eta_i$ be as in Lemma~\ref{ degeneracy for E_i}. Then \[ w_i = b_i \sigma_f^2(b_i) \cdots \sigma_f^{n/2^i -2}(b_i) = b_i b_i^{q^2}\cdots b_i^{q^{n/2^i -2}} = b_i^{\frac{q^{n/2^{i}} -1}{q^{2} -1}}, \] and \[ \eta_i = w_i^{q-1} = b_i^{\frac{q^{n/2^{i}} -1}{q +1}} = b_i^t,\] where $ t\coloneqq {\frac{q^{n/2^{i}} -1}{q +1}} $. By Lemma~\ref{ degeneracy for E_i}, $f_{b_i, \sigma}$ is degenerate if and only if $\eta_i^{2^i}=1$. Now from the proof of Theorem~\ref{th for finite field}, $\eta_i^{2^{a+1}l}=\eta_i^{2(q+1)}=1$ and $\eta_i^{2^{a}l}=\eta^{(q+1)}\neq1$. Consequently $ f_{b_i,\sigma_f}$ is degenerate if and only if $ \eta_i$ is a primitive $ 2^{a+1}$-th root of unity, that is, if and only if, \begin{equation} 2^{a+ 1} = \ord(\eta_i) = \ord(u^{st}) = \frac{\ord(u)}{\gc(\ord(u), st)}\\ = \frac{2(q+1)t}{\gc(2(q+1)t,st)} = \frac{2^{a+1}l}{\gc (2^{a+1}l, s)}, \end{equation} or, $\gcd(2^{a+1}l,s) = l$. In other words, for $b_i = u^s \in E_i $, $ f_{b_i,\sigma_f}$ is degenerate if and only if $s$ is an odd multiple of $l$. Thus, for example, $ f_{u^l, \sigma_f}$ is degenerate while $ f_{u, \sigma_f}$ is non-degenerate. \end{rema} \subsection{Proof of Theorem\texorpdfstring{\ref{theoD}}{D}} \begin{proof} By~\cite[Proposition~5.4.11]{pg} for every $ n $ there exists exactly one unramified extension $ L $ of $ K=\mathbb{Q}_p $ of degree $ n $ obtained by adjoining a primitive $ (p^n -1)$-th root of unity, say $ \theta$. Moreover according to~\cite[Corollary~2]{number}, the extension $L/K$ constitutes a cyclic extension such that $\gal(L/K) = \langle\sigma \rangle$ where $\sigma$ is defined by $\sigma(\theta) = \theta^{p} $. Since $ -1$ is not a square in $K$ so $p=2^al-1 \equiv 3~ (\mo 4 )$ by~\cite[Proposition~3.4.2]{pg} and thus $a\geq 2$. Let $ E_{i}\coloneqq \{ b\in L : \sigma^{\frac{n}{2^i}}(b) = -b\}$ where $1 \leq i \leq \alpha - 1$. The hypothesis $ \alpha \leq a+1$ means that $1 \leq i \leq a$. Let $w_i,\eta_i$ be as in Lemma~\ref{ degeneracy for E_i}. Again by Lemma~\ref{ degeneracy for E_i}, $f_{b_i,\sigma} \ (b_i \in E_i) $ is degenerate if and only if $\eta_i$ is $ 2^i$-th root of unity such that $\sigma(\eta_i)=-\eta_i^{-1}$. As $2^i \mid 2^a \mid p + 1 \mid p^n -1$, this would mean that $\langle \eta_i \rangle \le \langle \theta \rangle$ and consequently, $\sigma(\eta_i)= \eta_i^p$. But then \[\sigma(\eta_i) \eta_i = \eta_i^{p+1} = \eta_i^{2^{a}l}=1.\] It follows that $ f_{b_i ,\sigma } $ is non-degenerate. Hence $ \mathcal{E}_i$ is an $n$-subspace, where $ \mathcal{E}_i$ is the subspace of $ A^1$ corresponding to $E_i$. \newline Similarly let $ \mathcal{V}_j$ ( $1 \leq j \leq 2$ ) be the subspace of $ A^1$ corresponding to $V_j$. Then by Lemmas~\ref{degeneracy for V1 ,V2} and~\ref{rank of f b sigma for even case }, $\mathcal{V}_j$ is an $(n-2)$-subspace. The theorem now follows. \begin{rema} In the situation of Theorem~\ref{theoD} for $p=2$ the decomposition~\eqref{algebraic dcmpsn Alt L} holds true in view of Remark~\ref{theorem B rema}.\qedhere \end{rema} \end{proof} \section{ A \texorpdfstring{$3$}{3}-dimensional \texorpdfstring{$4$}{4}-subspace in \texorpdfstring{$\alt_4(\mathbb Q)$}{Alt4(Q)}}\label{Sect6} Let $K \coloneqq \mathbb{Q}$ and $L$ be the cyclotomic field $\mathbb{Q} (\eta)$ where $\eta$ is a primitive $5$-th root of unity in $\mathbb C$. Then $L/K$ is a cyclic extension of degree $4$. We will show that the maximum dimension of a $4$-subspace inside $ A^1 $ is $ 3$. Let $ b = x + y\eta +z\eta^{2} + w\eta^{3} \in L $, where $ x,y,z,w \in \mathbb Q$. We take the automorphism $ \sigma$ defined by $ \sigma(\eta)= \eta^3$ as a generator of $\gal(L/K)$. Using the theory of Gauss periods we may find the basis, namely, $ \{ 1, \eta^2 +\eta^3 \} $ for $L_{2}/\mathbb{Q}$. By Proposition~\ref{GMdegen-crit}, $ f_{b,\sigma} $ is degenerate if and only if $ N_{L/L_{2}}(b) \in \mathbb{Q}$, that is, the coefficient of $ \eta^2 +\eta^3 $ in $ N_{L/L_{2}}(b) $ is zero. It is straight-forward to check that this coefficient is $ -xy+ xz +xw - yz + yw - zw $. In this situation we thus obtain the following. \begin{prop} \label{totally isotropic} The maximum dimension of a $ 4 $-subspace inside $ A^{1} $ equals to a maximum dimension of a totally anisotropic subspace of $ L $ with respect to the following quadratic form \[ \mathcal{Q}(x,y,z,w)=xy -xz - xw+ yz - yw + zw .\] \end{prop} \begin{proof} Clear. \end{proof} \begin{theo}[{Legendre's Theorem (\cite[Theorem~1, Chapter~5]{EG})}]\label{legendre} Suppose $ a,b,c \in \mathbb{Z}$ are such that $ abc$ is a non-zero square-free integer. Then the equation $ a X^2 + b Y^2 + c Z^2 = 0 $ has a non-trivial $Z$-solution if and only if \begin{enumerate}[label=(ii\alph*)] \item[(i)] $a,b,c$ do not all have the same sign; \item $-bc $ is a square modulo $|{a}|$, \item $-ac $ is a square modulo $|{b}|$ and \item $-ab $ is a square modulo $|{c}|$. \end{enumerate} \end{theo} \begin{theo}\label{counter} The maximum dimension of a $ 4 $-subspace in $ A^{1} $ is $ 3 $. \end{theo} \begin{proof} Let $ U $ be the $\mathbb Q$-subspace of $ L $ spanned by $ \{\eta + \eta^{2} , -1 + \eta^{3},1+\eta\} $. Let $ b = c_{1} (\eta + \eta^{2}) + c_2(-1 + \eta^{3}) + c_3(1+\eta) $. We claim that $\mathcal W \coloneqq \{f_{b,\sigma} : b\in U \} \le A^1$ is the desired $4$-subspace. Indeed, according to proposition~\ref{totally isotropic} we need to show that the quadratic form $$ \mathcal{Q}(c_1,c_2,c_3)= c_{1}^2 +c_{2}^{2} + c_{3}^{2} +c_1 c_{3} - 3 c_{2}c_{3} $$ has no non-trivial integer solution. It can be checked that $\mathcal{Q}$ reduces to it's diagonal form $$ \mathcal{Q'}= c_{1}^{2} + c_{2}^{2} - 6 c_{3}^{2} .$$ To complete the proof, it suffices to show that $\mathcal{Q'}$ has no non-trivial integer solutions. Based on Theorem~\ref{legendre} it is evident that $\mathcal{Q'}$ has no non-trivial integer solutions since $-ab= -1$ is not square modulo $|c|=6$. \end{proof} \section{Conclusion} Eigenspaces of the elements of the Galois group yield constant rank subspaces in $\alt_K(L)$. We can always find an $ n $-subspace of dimension $ n/2$ in $A^i$ for an arbitrary field $K$ (Remark~\ref{n-subs-exists-in}). However, this may not be the maximum possible dimension of an $n$-subspace in $ A^1$ (as is evident from the example in Section~\ref{Sect5}) unless $n =2k$ with $k$ odd (Theorem~\ref{odd dcmposn for any field}) or $ K$ is finite (or more generally $C^1$~{\cite[Lemma~3]{GQ06}}). Moreover unless $K$ is finite it is not clear that we get an $n$-subspace of maximum dimension of $\alt_{n}(K)$ in this way. The question of the maximum dimension of an $ n$-subspace in $\alt_n(K)$ is closely related to other invariants for skew-forms including $d(K,n,1)$ and $s_n(K)$ defined in~\cite{BGH1987} and~\cite{GQ06} respectively. In particular, it is unknown to the authors if there is a $6$-subspace in $\alt_6(\mathbb{Q})$ of dimension four. \section*{Declaration of Interest} The author do not work for, consult, own shares in or receive funding from any company or organization that would benefit from this article, and have declared no affiliation other than their research organisations. \printbibliography \end{document}