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\title{\bf On symbol lengths\\ in Milnor $K$-theory}
\author{Karim Johannes Becher, Detlev Hoffmann
\footnote{This work was supported by the TMR Network `Algebraic $K$-Theory,
Linear Algebraic Groups and Related Structures' (ERB FMRX CT 97-0107) and by the subsequent RTN Network (HPRN CT-2002-0287).}
}

\date{}
\begin{document}
\maketitle


\begin{abstract}
Let $F$ be a field and $p$ a prime number. The $p$-symbol length of $F$,
denoted by $\lambda_p(F)$, is the least integer $l$ such that every element of
the group $K_2 F/p K_2F$ can be written as a sum of $\leq l$ symbols
(with the convention that $\lambda_p(F)=\infty$ if no such integer exists).
In this article, we obtain an upper bound for $\lambda_p(F)$ in 
the case where the group $\mg{F}/{\mg{F}}^p$ is finite of order $p^m$.
This bound is $\lambda_p(F)\leq \frac{m}{2}$, except for the case where
$p=2$ and $F$ is real, when the bound is $\lambda_2(F)\leq \frac{m+1}{2}$.
We further give examples showing that these bounds are sharp.
\end{abstract}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}       %%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Let $F$ be a commutative field. 
Let $\mg{F}$ denote the multiplicative group of $F$.
We recall the definition of the groups $K_n F$\, $(n\geq 1)$ of Milnor $K$-theory, as defined in \cite{Mil}.
$K_1 F$ stands just for the multiplicative group $\mg{F}$ in additive notation.
In order to distinguish between addition in $K_1 F$ and in $F$, 
we write $\{x\}\in K_1 F$ for $x\in\mg{F}$, so that 
we have $\{x\}+\{y\}=\{xy\}$ in $K_1 F$  for $x,y\in\mg{F}$.
For $n\geq 2$, the group $K_n F$ is defined as the quotient of the group $(K_1 F)^{\otimes n}$ modulo the subgroup
generated by the elementary tensors $\{x_1\}\otimes\dots\otimes\{x_n\}\in(K_1 F)^{\otimes n}$ where $x_1,\dots,x_n\in\mg{F}$ and $x_i+x_{i+1}=1$ for some $i<n$.

Let now be $n\geq 1$ and let $H$ denote the group $K_n F$ or some quotient of it.
For $x_1,\dots,x_n\in\mg{F}$, we denote the canonical image of $\{x_1\}\otimes\dots\otimes\{x_n\}\in(K_1 F)^{\otimes n}$
in $H$ by $\{x_1,\dots,x_n\}$
and we call such an element a \emph{symbol in $H$}.
This notation differs from Milnor's original one where the same symbol is denoted by $l(x_1)\cdots l(x_n)$.
Obviously, $H$ is generated by its symbols. Note further that the zero element of $H$ is a symbol.
We may then ask whether there exists an integer $l\geq 0$ such that every element of $H$ can be written as a sum of $l$ symbols.
If this is the case then we denote by $\lambda(H)$ the least such integer $l$; 
otherwise we set $\lambda(H):=\infty$.
We call the value $\lambda(H)\in \nat_0\cup\{\infty\}$ the \emph{symbol length of $H$}.

Quotients of $K_n F$ of particular interest are $K_n F/l\, K_n F$
where $l$ is a positive integer; 
we shall abbreviate this quotient by $K_n^{(l)} F$.
In the case where $l=2$, we may also use Milnor's notation $k_n F$ for $K_n F/ 2\, K_n F$.

Let now $p$ be a prime number. We define
$\lambda_p(F):=\lambda (K_2^{(p)} F)$ and call this the \emph{$p$-symbol
length} of the field $F$. 
% The purpose of this article is to study these invariants.

Lenstra showed that if $F$ is a global field, then
$K_2 F$ consists of symbols (cf. \cite{Len}); 
it follows then that $\lambda_p(F)=1$ for all primes $p$. 
The same is true if $F$ is a local field. 
Whether  the rational function field in two
variables over the complex numbers $F=\cc(X,Y)$ also satisfies
$\lambda_p(F)=1$ for all primes $p$ is a striking open question; at
least for $p\leq 3$ the answer is positive, by a theorem of Artin (cf. \cite{MArt}).
A recent result of Saltman implies that $\lambda_2(F)=2$ if
$F$ is the function field of a curve over a $q$-adic local field
where $q\neq 2$ (cf. \cite{Salt}). 

%Suppose now that $F$ is a function field of an algebraic surface
%over some algebraically closed field. 
%Very recently, it has been shown by J. A. de Jong that for any
%central simple algebra over such $F$, index and exponent coincide.
%Identifying $K_2^{(p)} F$ with ${\rm Br}_p(F)$, the
%$p$-torsion part of the Brauer group of $F$, via the Merkurjev--Suslin
%theorem, the question whether $\lambda_p(F)=1$ is equivalent to
%the question whether any central simple algebra of degree $p$ of such
%a field $F$ is cyclic.
%At least for $p\leq 3$ this is true,
%and actually it was known before that
%$\lambda_p(F)=1$ for $p\leq 3$, by the Tsen--Lang theorem together
%with an argument of M. Artin for $p=3$. 

\medskip


In this article, we  establish an upper bound for
$\lambda_p(F)$ under the assumption  that
$\mg{F}/F^{\times p}$ is finite and we show further that it is generally the best possible.
\smallskip

\begin{thm}\label{T:estimate}
Suppose that $|\mg{F}/{\mg{F}}^p|= p^m$ where $m\in\nat$. Then
$$
\renewcommand{\arraystretch}{1.5}
\lambda_p (F)\,\leq\,\, 
\left\{
\begin{array}{cl}
\left[\frac{m}{2}\right] & \mbox{if $p\neq 2$ or if $F$ is nonreal,} \\
\left[\frac{m+1}{2}\right]  & \mbox{if $F$ is real and $p=2$.}
\end{array}
\right.
$$
\end{thm}
\vspace{3mm}

Here we use the notation $[x]$ for the integral part of $x\in\rr$.
Recall that the field $F$ is nonreal if and only if $-1$ is a sum of squares in $F$.
\medskip

\noindent
The proof of Theorem~\ref{T:estimate} is divided into several parts, 
which will occupy the next two sections.
In Section~2, we relate the $K$-groups modulo $p$ to certain exterior power spaces and, under the condition that $p\neq 2$ or that $-1$ is a square in $F$,
we deduce the estimate $\lambda_p(F)\leq[\frac{m}{2}]$  from a known
fact on alternating spaces (\ref{P:p^m}). 
In a similar way we get the weaker bound $\lambda_p(F)\leq[\frac{m+1}{2}]$  for $p=2$ (\ref{C:m+1/2}).
We will therefore be left with the task to exclude that $\lambda_2 (F)$ be equal to $\frac{m+1}{2}$ unless $F$ is a real field.
This will be done in the third section by an argument involving  quadratic form theory (\ref{C:specialcase}).

Finally we show in the fourth section that in all cases, according to whether $F$ is real or not and whether $p$ equals $2$ or not, the estimate
stated above is best possible (\ref{T:opt}). 
To do so, we give examples where $F$ is a field containing a primitive $p$-th root of unity, such that $|\mg{F}/{\mg{F}}^p|= p^m$
 and such that there is a simple $F$-division algebra which is a
 product of $l$ symbol algebras of degree $p$, where
 $l=[\frac{m+1}{2}]$ if $p=2$ and if $F$ is real and where 
$l=[\frac{m}{2}]$ otherwise; the existence of such a division algebra implies indeed that $\lambda_p (F)\geq l$.
We will further apply a similar argument to show that for the rational 
function field in  $m+1$ variables over a field $F$ containing a primitive $p$-th root of unity,
we always have $\lambda_p(F(X_0,\dots,X_m)\,)\geq m$ (\ref{P:imp}). 
This improves the bound $\lambda_p(F(X_0,\dots,X_m)\,)\geq [\frac{m+1}{2}]$, shown in \cite[Proposition~3]{Kahn00} (under a stronger hypothesis).


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{$K$-groups modulo $p$ and exterior powers} %%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Let $k$ denote a commutative field, $V$ a vector space over $k$ and $n$ a positive integer.
Let $\Lambda^n V$  denote the exterior power of degree $n$ over $V$.
This is a vector space over $k$ generated by elements $v_1\wedge\dots\wedge v_n$, where $v_1,\dots,v_n\in V$, subject to the relations of $k$-multilinearity
as well as to the relation that $v_1\wedge\dots\wedge v_n=0$ whenever $v_i=v_{i+1}$ for some $i<n$. 
An element of $\Lambda^n V$ which is of the shape $v_1\wedge\dots\wedge v_n$ is called a 
\emph{pure $n$-vector}.

Suppose now that $V$ has finite dimension~$m$. 
There exists a least integer $N$ such that every element of $\Lambda^n V$ is a sum of $N$ pure $n$-vectors.
If the dimension of $V$ is $m$ then we denote this integer by $N(k;m,n)$
(since it depends only on $k$, $n$ and $m$).
For $n=2$ a classical result tells us that $N(k;m,2)=[\frac{m}{2}]$, independently of the field $k$ (cf. \cite[\S 5, Corollaire~2]{B-A9}).
\medskip

Let $p$ be a prime number and $n$ a positive integer. Let $\ff_p$ denote the finite field with $p$ elements.
The group $K_n^{(p)} F$, associated to the field $F$, is endowed with a natural structure as a vector space over $\ff_p$.
Note that for $m\in\nat$, equality $|\mg{F}/{\mg{F}}^p|=p^m$ means
that $K_1^{(p)} F$ as a vector space over $\ff_p$ has dimension~$m$. 


Let $x_1,\dots, x_n$ be elements of $\mg{F}$.
If $x_i= x_{i+1}$ for some $i<n$ then $\{x_1,\dots,
x_n\}=\{x_1,\dots,x_i,-1,x_{i+2},\dots, x_n\}$ in $K^{(p)}_2 F$ by
\cite[Lemma~1.2]{Mil}; in particular, this symbol is of order at most $2$. 
Suppose now that $-1\in{\mg{F}}^p$, that is either $p\neq 2$ or
$p=2$ and $-1\in\sq{F}$.
This implies that any symbol in $K_n^{(p)} F$
with a repeated coefficient is zero.
Therefore, if $-1\in F^{\times p}$ then the natural $\ff_p$-homomorphism $(K_1^{(p)} F)^n \longrightarrow K_n^{(p)} F$
induces a natural \mbox{$\ff_p$-homomorphism} $\Lambda^n (K_1^{(p)} F)\longrightarrow K_n^{(p)} F$
which maps a pure $n$-vector \mbox{$\{x_1\}\wedge\dots\wedge\{x_n\}$} to the
symbol $\{x_1,\dots,x_n\}$ and which is therefore
surjective.
From this we conclude:


\begin{lem}\label{L:gen}
Suppose that $p\neq 2$ or that $-1\in\sq{F}$. If  $|\mg{F}/{\mg{F}}^p|=p^m$ then $\lambda(K_n^{(p)}F)\leq N(\ff_p;m,n)$.\hfill$\Box$
\end{lem}

Unfortunately, the exact value of $N(\ff_p;m,n)$ is not known in general when $n\geq 3$ (cf. \cite{Kahn92} on this problem and approximative results).
%\begin{cor}\label{C:m/2}
%Suppose that $|\mg{F}/{\mg{F}}^p|=p^m$ where $p\neq 2$ or $-1\in\sq{F}$.
%Then every element of $K_2^{(p)} F$ can be written as a sum of\, $[\frac{m}{2}]$ symbols.\hfill$\Box$
%\end{cor}
We may further observe that the bound in (\ref{L:gen}) is best possible for all $m,n$ and $p$.
Indeed, one will have $\lambda(K_n^{(p)} F)= N(\ff_p;m,n)$ if
$-1\in{\mg{F}}^p$, $|\mg{F}/{\mg{F}}^p|=p^m$ 
and if the natural $\ff_p$-homomorphism  $\Lambda^n (K_1^{(p)}
F)\longrightarrow K_n^{(p)} F$ is bijective, and examples where these
conditions hold will be given in (\ref{E:p^m}).

\begin{prop}\label{P:eqcond}
Suppose that $p\neq 2$ or that $-1\in\sq{F}$.
The following are equivalent:
\begin{itemize}
\item[$(i)$] for any $n\geq 2$, the  natural $\ff_p$-homomorphism
  $\Lambda^n (K_1^{(p)} F)\longrightarrow K_n^{(p)} F$ is an isomorphism;
\item[$(ii)$] the natural  $\ff_p$-homomorphism
  $\Lambda^2 (K_1^{(p)} F)\longrightarrow K_2^{(p)} F$ is an isomorphism;
\item[$(iii)$] whenever $a,b\in\mg{F}$ are such that the symbol
  $\{a,b\}\in K_2^{(p)} F$ is zero then the elements $\{a\}$ and $\{b\}$ in
  $K_1^{(p)} F$ are $\ff_p$-linearily dependent;
\item[$(iv)$] for any $a\in \mg{F}\setminus {\mg{F}}^p$ there exists
  $r\geq 0$ such that $a^r(1-a)\in{\mg{F}}^p$.
\end{itemize}
\end{prop}
\begin{proof} 
The  implications $(i\Rightarrow ii\Rightarrow iii\Rightarrow iv)$ are
obvious. 

Assume now that $(iv)$ holds.
Then it follows for $n\geq 2$ that a pure $n$-vector $\{x_1\}\wedge\dots\wedge\{x_n\}\in\Lambda^n
(K_1^{(p)} F)$ is zero as soon as $x_i+x_{i+1}=1$ for some $i<n$. 
Since this corresponds to the defining relation for $K_n^{(p)} F$ as a quotient of $(K_1^{(p)} F)^{\otimes
  n}$, we conclude that the natural $\ff_p$-homomorphism  $\Lambda^n
(K_1^{(p)} F)\longrightarrow K_n^{(p)} F$ is bijective. This shows that $(iv)$
implies $(i)$.
\end{proof}

In the case where $p\neq 2$ and $F$ contains a primitive $p$-th root
of unity, the conditions $(i)$--$(iv)$ are satisfied if and only if
$F$ is a \emph{$p$-rigid field}, as defined in \cite[p.~772]{Ware92}.

\begin{prop}\label{P:p^m}
Suppose that $|\mg{F}/{\mg{F}}^p|=p^m$, and further that $p\neq 2$ or that \mbox{$-1\in\sq{F}$}. Then
\begin{itemize}
\item[$(a)$] every element of $K_2^{(p)} F$ is a sum of\, $[\frac{m}{2}]$ symbols,
\item[$(b)$] every element of $K_{m-1}^{(p)} F$ is a symbol,
\item[$(c)$] $K_n^{(p)} F =0$ for any $n>m$,
\item[$(d)$] the equivalent conditions  in $(\ref{P:eqcond})$ hold if and only if
$K_m^{(p)}F\neq 0$, and in this case one has $K_m^{(p)} F\, \cong\, \zz/ p\,\zz$.
\end{itemize}
\end{prop}
\begin{proof}
(a)\,\, By (\ref{L:gen})  every element of $K_2^{(p)} F$ can be
written as a sum of\,
$N(\ff_p;m,2)$ symbols, and this number is known to be equal to
$[\frac{m}{2}]$.

(b)\,\, It is easy to see that $N(k;m,m\!-\!1)=1$ for any field $k$. 
Therefore $\lambda(K_{m-1}^{(p)}F)\leq 1$ by  (\ref{L:gen}).

(c)\,\, $K_n^{(p)} F$ can be considered as a quotient of the space
$\Lambda^n (K_1^{(p)} F)$, which vanishes as
soon as $n$ exceedes $m$, the dimension of $K_1^{(p)} F$.

(d)\,\, Since $K_1^{(p)}F$ is of dimension $m$, one has $\Lambda^m
(K_1^{(p)} F)\cong \zz/p\,\zz$. Therefore the first condition in Proposition~\ref{P:eqcond} 
implies that $K_m^{(p)} F\cong \zz/p\,\zz$.

Assume now that the equivalent conditions in Proposition
(\ref{P:eqcond}) do not hold for $F$. In particular, since $(iii)$
does not hold, there are $a_1,a_2\in\mg{F}$\, $\ff_p$-linearly
independent modulo $p$-th powers such that $\{a_1,a_2\}=0$ in
$K_2^{(p)}F$. We may choose $a_3,\dots,a_m\in\mg{F}$ such that 
$\{a_1\},\dots,\{a_m\}$ form an $\ff_p$-basis of $K_1^{(p)} F$.
Then the pure $n$-vector $\{a_1\}\wedge\dots\wedge\{a_m\}$ in
$\Lambda^m (K_1^{(p)}F)$ is nonzero, while the symbol $\{a_1,\dots,a_m\}$
in $K_m^{(p)}F$ is zero. Hence the natural surjection $\Lambda^m (K_1^{(p)}
F)\longrightarrow K_m^{(p)} F$ is not injective. 
As $\Lambda^m (K_1^{(p)} F)\cong \zz/p\,\zz$, it follows that
$K_m^{(p)} F=0$.
\end{proof}

\begin{lem}
Suppose $p\neq 2$ or that $-1\in\sq{F}$, and that $F$ is $p$-henselian with respect to a discrete valuation
$v$ with
residue field $\bar{F}$ with ${\rm char} (\bar{F})\neq p$. 
If the equivalent conditions in $(\ref{P:eqcond})$ hold for $\bar{F}$
then they hold for $F$ as well.
\end{lem}  
\begin{proof} 
Since $F$ is $p$-henselian and $p\neq {\rm char} (\bar{F})$, we know that
for any $u\in F$ with $v(u)=0$  we have $u\in{\mg{F}}^p$ if and only if
$\ovl{u}\in {\bar{F}}^{\times p}$.
We show  that Condition~$(iv)$ in $(\ref{P:eqcond})$ holds for the field $F$
if it holds for $\bar{F}$.

Let $a$ be an element of $F$ which is not a $p$-th power in $F$.
We are looking for an integer $r$ such that
$a^r-a^{r+1}$ is a $p$-th power in $F$.
Using that $-1\in{\mg{F}}^p$ and possibly replacing $a$ by $-a^{-1}$,
we may assume that $v(a)\geq 0$.
If $v(a)>0$ then $a^r-a^{r+1}$ is a $p$-th power if and only of $a^r$ is
a $p$-th power, hence we are done with $r=p$.
On the other hand, if $v(a)=0$ then we saw that $\ovl{a}$ cannot be a $p$-th power in $\bar{F}$;
hence, by hypothesis, there exists $r\in\zz$ such that
$\ovl{a^r (1-a)}$ is a $p$-th power in $\bar{F}$ and
it follows that $a^r(1-a)$ is a $p$-th power in $F$.
\end{proof}

\begin{exs}\label{E:p^m}
In each of the following cases, the field $F$ satisfies the equi\-valent conditions of $(\ref{P:eqcond})$:
\begin{itemize}
\item $F=C(\!(X_1)\!)\dots(\!(X_m)\!)$ where $C$ is an algebraically
  closed field,
\item $F=R(\!(X_1)\!)\dots(\!(X_m)\!)$ where $p\neq 2$ and $R$ is a
  real closed field,
\item $F=F$ is a \emph{non-$p$-adic $m$-dimensional local field}, i.e. there
exists a sequence of fields $F_0,\dots,F_m=F$ where $F_0$ is finite
with ${\rm char}(F_0)\neq p$ and, for $1<i\leq m$, the field $F_i$ is complete with respect 
to a discrete valuation with residue field $F_{i-1}$. 
\end{itemize}
Indeed, this follows from the basic case $m=0$, using the above lemma.
Note further that $|\mg{F}/{\mg{F}}^p|=p^m$ in the first two cases.
In the third case one has $|\mg{F}/{\mg{F}}^p|=p^{m+1}$ if $p$ divides
$({\rm char}(F_0)-1)$ and $|\mg{F}/{\mg{F}}^p|=p^{m}$ otherwise.
\end{exs}
\medskip

The remainder of this section together with the following section are
devoted to study the case where $p=2$. Then, without the hypothesis
that $-1\in\sq{F}$, we cannot relate $K_n^{(p)} F$ to the exterior
power space $\Lambda^n(K_1^{(p)} F)$ over $\ff_p$, as we did before.
We therefore are going to define a kind of "twisted exterior power
space" over the field with two elements.
  
In the sequel, the letter $k$ will only be used to
denote the Milnor $K$-groups modulo $2$ of a field, that is $k_n F=
K_n^{(2)} F$ ($n\geq 0$).

Let $V$ be a vector space over the field $\ff_2$, $\varepsilon$ a fixed element of $V$ and $n$ a positive integer. 
We shall write $\Lambda_{\varepsilon}^n V$ for the vector space over $\ff_2$ generated by elements $v_1\wedge\dots\wedge v_n$, with $v_1,\dots,v_n\in V$,
subject to the relations of multilinearity and symmetry (i.e. $v_1\wedge\dots\wedge v_n$ does not depend on the order of the coefficients $v_i$) and
further to the relation that $v_1\wedge\dots\wedge v_n=0$ whenever $v_{i+1}=v_i+\varepsilon$ for some $i<n$.
Note that if $\varepsilon=0$, then
$\Lambda_{\varepsilon}^n V$ is just the exterior power space $\Lambda^n V$.

We call \emph{pure $n$-vectors in $\Lambda_{\varepsilon}^n V$} the elements of the form $v_1\wedge\dots\wedge v_n$.
From now on we restrict to the case where $n=2$. For given $x\in V$, we shall say that a pure $2$-vector is {of the shape $x\wedge \ast$} if it can
be written as $x\wedge y$ for some $y\in V$.

\begin{lem}\label{L:transform}
Let $V$ be a vector space over $\ff_2$ with a special element $\varepsilon\in V$.
Let $\xi$ be an element of $\Lambda_{\varepsilon}^2 V$ and let $v_1,\dots,v_l,w_1,\dots,w_l$ $\in V$ be such that $\xi=v_1\wedge w_1 +\dots+ v_l\wedge w_l$.
\begin{itemize}
\item[$(a)$] 
If $v$ is a nontrivial sum of some of the elements $v_1,\dots,v_l$ then $\xi$ 
can be written as a sum of $l$ pure $2$-vectors where the first is of the shape $v\wedge\ast$.
\item[$(b)$] If $v$ is a nontrivial sum of some of the elements $v_1,\dots,v_l,w_1,\dots,w_l$ then 
$\xi$ can be written as a sum of $l$ pure $2$-vectors where the first is of the shape $v\wedge \ast$ or $(v+\varepsilon)\wedge \ast$.
\item[$(c)$] If  $v_1,\dots,v_l,w_1,\dots,w_l,\ \varepsilon$ are linearly dependent then
$\xi$ can be written as a sum of $l$ pure $2$-vectors where the first is of the shape $\varepsilon\wedge\ast$.
\end{itemize}
\end{lem}
\begin{proof}
(a)\,\,
If $l=1$ this is trivial. In the case $l=2$ one uses the equality $v_1\wedge w_1 + v_2\wedge w_2 =(v_1+v_2)\wedge w_1 + v_2\wedge (w_1+w_2)$.
 The general case follows from this by induction on $l$.

(b)\,\, Using the relations
$v_i\wedge w_i = w_i\wedge v_i = (v_i + w_i +\varepsilon)\wedge w_i$, one readily sees that the statement follows from (a).

(c)\,\, 
Suppose that $v_1,\dots,v_l,w_1,\dots,w_l,\ \varepsilon$ are linearly dependent.
Then one of the elements $0$ and $\varepsilon$ can be written as a nontrivial sum of some of the elements $v_1,\dots,v_l,w_1,\dots,w_l$.
Observing that a pure $2$-vector of the shape $0\wedge \ast$ is zero, hence equal to $\varepsilon\wedge 0$, we conclude by $(b)$ that
$\xi$ can be written as a sum of  $l$ pure $2$-vectors where the first is of the shape $\varepsilon\wedge\ast$. 
\end{proof}

\begin{prop}\label{P:eps-alt-space}
Let $V$ be a vector space over $\ff_2$ of dimension $m$ and let $\varepsilon\in V$.
Then any element of $\Lambda_{\varepsilon}^2 V$ is a sum of $[\frac{m+1}{2}]$ pure $2$-vectors. 
If \mbox{$\xi\in \Lambda_{\varepsilon}^2 V$} is not a sum of $[\frac{m}{2}]$ pure $2$-vectors, then
$\xi=\varepsilon\wedge\varepsilon + v_1\wedge w_1 + \dots + v_n\wedge w_n$,
where $v_1,w_1,\dots,v_n,w_n\in V$ and $2n+1=m$;
moreover, under these circumstances, $(v_1,w_1,\dots,v_n,w_n,\varepsilon)$ is a basis of $V$.
\end{prop}

\begin{proof}
Suppose that $\xi\in \Lambda_{\varepsilon}^2 V$ is a sum of $l$ but not of $l-1$ pure $2$-vectors where $l>\frac{m}{2}$.  
In a representation $\xi$ as a sum of $l$ pure $2$-vectors, the $2l$ coefficients are  necessarily \mbox{linearly} dependent.
By (\ref{L:transform}(c)), we may write $\xi=\varepsilon\wedge w + \xi'$ with $w\in V$ and with $\xi'\in \Lambda_{\varepsilon}^2 V$ equal to a sum of $l-1$ symbols.
Assume that $w$ is different from $\varepsilon$. 
Then the subspace of $V$ generated by the coefficients of any representation of $\xi'$ as a sum of $l-1$ 
pure $2$-vectors contains one of the elements $0$, $\varepsilon$, $w$ and $w+\varepsilon$. 
Using the fact that $0\wedge \ast=\varepsilon\wedge 0$, It follows from (\ref{L:transform}(b)) 
that $\xi'$ can be written as a sum of $l-1$ pure $2$-vectors where the first one is of the shape $x\wedge y$ with $y\in V$ and
$x$ equal to one of the elements $w$, $\varepsilon$ and $w+\varepsilon$. 
Now $\varepsilon\wedge w + x\wedge y$ cannot be equal to a pure $2$-vector, since otherwise $\xi$ would be equal to a sum of $l-1$ pure $2$-vectors.
Hence, by the relations in $\Lambda_{\varepsilon}^2 V$, we must have $x=w+\varepsilon$ and therefore
 $\varepsilon\wedge w + x\wedge y=\varepsilon \wedge (x+\varepsilon) + y\wedge x = \varepsilon\wedge\varepsilon + (y+\varepsilon)\wedge x$.

From this we conclude that we can write $\xi=\varepsilon\wedge\varepsilon + v_1\wedge w_1 + \dots + v_{l-1}\wedge w_{l-1}$
with  $v_1,w_1,\dots,v_{l-1},w_{l-1}\in V$. 
Since $\xi$ is not a sum of less than $l$ pure $2$-vectors, we conclude from (\ref{L:transform}(c)) 
that $v_1,w_1,\dots,v_{l-1},w_{l-1}, \varepsilon$ are linearly independent.
Then, since $2l-1\geq m$, these elements form a basis of $V$, in particular, if we put  $n:=l-1$, then 
$2n+1=m$.
\end{proof}

Since $k_1 F$ has an obvious structure of a vector space over $\ff_2$, 
we may apply the above results to $k_1 F$ together with the special element $\varepsilon:=\{-1\}$. 
By the relations defining $k_n F$, there exists a surjective $\ff_2$-homomorphism $\Lambda_{_{\{-1\}}}^n (k_1 F) \longrightarrow k_n F$ which maps 
a pure $n$-vector $\{x_1\}\wedge\dots\wedge\{x_n\}$ to a symbol $\{x_1,\dots,x_n\}$. From this together with  the last proposition we obtain immediately:


\begin{cor}\label{C:m+1/2}
Suppose that $|\scg{F}|=2^m$.
Then every element of $k_2 F$ can be written as a sum of\, $[\frac{m+1}{2}]$ symbols.\hfill$\Box$
\end{cor}



\begin{rem}\label{R:proof-of-thm}
Note that (\ref{L:gen}) and (\ref{C:m+1/2}) immediately imply
following weaker version of Theorem~\ref{T:estimate}:
if $|\mg{F}/{\mg{F}}^p|= p^m$ then
$$
\renewcommand{\arraystretch}{1.5}
\lambda_p (F)\,\leq\,\, 
\left\{
\begin{array}{cl}
\left[\frac{m}{2}\right] & \mbox{if $p\neq 2$ or $-1\in\sq{F}$,} \\
\left[\frac{m+1}{2}\right]  & \mbox{if $p=2$ and $-1\notin\sq{F}$.}
\end{array}
\right.
$$
In order to establish Theorem~\ref{T:estimate} entirely,
it remains to show that we have \mbox{$\lambda_2(F)\leq [\frac{m}{2}]$} for any
nonreal field $F$ with $|\mg{F}/{\mg{F}}^2|= 2^m$.
This will be accomplished with (\ref{C:specialcase}). 
\end{rem}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{$K$-groups modulo $2$ and quadratic forms} %%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

For the following definitions and facts, we suppose that the characteristic of $F$ is different from $2$.
We shall use the standard notations in quadratic form theory as established in \cite{Lam} and \cite{Scharlau}.
However, we use a different convention for Pfister forms:
if $a_1,\dots,a_n\in\mg{F}$ then $\lla a_1,\dots, a_n\rra$ will denote the $n$-fold Pfister form $\la 1,- a_1\ra\otimes\dots\otimes \la 1,- a_n\ra$.
Let $W(F)$ denote the Witt ring of $F$ and  $I F$ the fundamental ideal consisting of the classes
of quadratic forms of even dimension.
Let further $I^n F := (I F)^n$ and $\bar{I}^n F:= I^n F/ I^{n+1} F$ for any $n\geq 0$.
The ideal $I^n F$ is additively generated by the $n$-fold Pfister forms over $F$.

Milnor has defined for any $n\geq 1$ a homomorphism $s_n : k_n F\longrightarrow \bar{I}^n F$,
mapping a symbol $\{x_1,\dots,x_n\}$ to the class of the Pfister form $\lla x_1,\dots, x_n\rra$
and determined by this property.
Milnor  showed that  $s_2 : k_2 F \rightarrow \bar{I}^2 F$ is an isomorphism \cite[Theorem~4.1.]{Mil}.
It has been proven recently that $s_n : k_n F\longrightarrow \bar{I}^n F$ is an isomorphism for any $n\geq 1$
\cite[Theorem 4.1]{OVV}.


We write $\pm\sq{F}:=\sq{F}\cup -\sq{F}$.
Recall that $F$ is a \emph{real euclidean field}, if $\sq{F}$ is an ordering of $F$, i.e. if $F$ is real and $\mg{F}=\pm\sq{F}$.

\begin{prop} \label{P:euclidean}
Suppose that $|\scg{F}|=2$ and $n\geq 2$. 
If  $F$ is nonreal then $k_n F=0$, otherwise $F$ is real euclidean and the unique nonzero element in $k_n F$ is $\{-1,\dots,-1\}$.
\end{prop}

\begin{proof}
By the hypothesis we have $\mg{F}=\sq{F}\cup a\sq{F}$ for some $a\in\mg{F}$.
Suppose that $k_n F$ is nontrivial. It is obvious that then the unique non\-zero element of $k_n F$ is the symbol $\{a,\dots,a\}$, which can
also be written as $\{-1,\dots,-1, a\}$. As this symbol is nontrivial, $-1$ cannot be a square and $a$ cannot be a sum of two squares in $F$.
Hence $a\sq{F}=-\sq{F}$ and $-1$ is not a sum of two squares.
Therefore, $\mg{F}=\pm\sq{F}$ and every sum of two squares is again a square in $F$.
We conclude that $F$ is a real euclidean field.
To complete the proof we remind that, if $F$ is any real field then the symbol $\{-1,\dots,-1\}\in k_n F$ is nontrivial.
\end{proof}

In the sequel we will focus on the study of $k_2 F$. If $\varphi$ is a quadratic form in $I^2 F$ then we write $\bar{\varphi}$ for its class in $\bar{I}^2 F$.
The following statement is folklore, at least in the context where $k_2 F$ is replaced by $\brd{F}$, the $2$-torsion of the Brauer group of $F$, and $s_2: k_2 F\rightarrow \bar{I}^2 F$
by the inverse of the homomorphism $c:\bar{I}^2 F\rightarrow \brd{F}$, induced by the Clifford invariant (cf. \cite[Lemma 2.2.]{H01} and the references given there). 
We wish to give a self-contained proof here in our setting, where the statement does not depend on Merkurjev's result \cite{Mer81} that  $c:\bar{I}^2 F\rightarrow \brd{F}$ is an isomorphism.

\begin{lem}\label{L:folklore} Suppose that ${\rm char}(F)\neq 2$. Let $m\geq 1$ and $\xi\in k_2 F$. 
A~necessary and sufficient condition that $\xi$ be a sum of $m$ symbols is that the class
$s_2(\xi)\in \bar{I}^2 F$ contain a quadratic form of dimension $2m+2$.
\end{lem}
\begin{proof}
We first prove  by induction on $m$ that the condition is necessary.
We write $\xi=\xi'+\{a,b\}$ where $\xi'\in k_2 F$ is a sum of $m-1$ symbols and $a,b\in\mg{F}$.
Now, if $m=1$ then we have $\xi'=0$ and $\xi=\{a,b\}$, hence $s_2(\xi)=\ovl{\lla a,b\rra}$, and the dimension of $\lla a,b\rra$ is $4=2m+2$.
Suppose now that $m>1$.
By induction hypothesis we have $s_2(\xi')=\ovl{\varphi'}$ for a quadratic form $\varphi'$ of dimension $2(m\!-\!1)+2=2m$. 
We decompose $\varphi'$ into $\varphi''\perp\la r\ra$, where $\varphi''$ is a quadratic form of dimension $2m-1$ and $r\in\mg{F}$. 
As $s_2(\{a,b\})=\ovl{-r\lla a,b\rra}$ we obtain
$s_2(\xi)=s_2(\xi')+s_2(\{a,b\})=\ovl{\varphi'\perp -r\lla a,b\rra}$.
But $\varphi'\perp -r\lla a,b\rra$ is Witt equivalent to  $\varphi:=\varphi''\perp\la ra,rb,-rab\ra$.
Therefore we have also $s_2(\xi)=\ovl{\varphi}$, with $\dim \varphi=2m+2$.

To show that the condition is sufficient, we use again induction~on~$m$.
Let $\psi$ be a quadratic form in $I^2 F$ of dimension $2m+2$ such that $s_2(\xi)=\ovl{\psi}$.
We decompose $\psi$ into $\psi'\perp\la c,d,e\ra$ with $\psi'$ a quadratic form of dimension $2m-1$ and $c,d,e\in\mg{F}$. 
Since $\psi$ is Witt equivalent to $\psi'\perp\la -cde\ra\perp e\lla -ce,-de\rra$, we have
$$
s_2(\xi)\,\,  =\,\,  \ovl{\psi'\perp\la -cde\ra}+\ovl{e\lla -ce,-de\rra}
	\,\,  =\,\,  \ovl{\psi'\perp\la -cde\ra} +s_2(\{-ce,-de\})\,.$$
We write $\xi=\xi'+\{-ce,-de\}$, with $\xi'\in k_2 F$. Since $s_2$ is an isomorphism, 
it follows that $s_2(\xi')=\ovl{\psi'\perp\la -cde\ra}$,
where the quadratic form $\psi'\perp\la -cde\ra$ has dimension $2m=2(m-1)+2$ and trivial discriminant.
Hence, if $m=1$ then $\psi'\perp\la -cde\ra$ must be the hyperbolic plane, thus $\xi'=0$ by injectivity of $s_2$ and $\xi=\{-ce,-de\}$.
If $m>1$ then by induction hypothesis $\xi'$ is a sum of $m-1$ symbols, which implies that $\xi$ is a sum of $m$ symbols.
\end{proof}

\begin{lem}\label{L:3squares}
Suppose that ${\rm char}(F)\neq 2$. Let $a\in \mg{F}$. If $a$ can be written in $F$ as a sum of  three squares (or less), then there exists $b\in\mg{F}$ such that $\{-1,-1\}\,=\,\{-a,-b\}$.
\end{lem}
\begin{proof}
If $a$ is a sum of three squares in $F$, then there is some  $b\in\mg{F}$ such that $\lla -1,-1\rra=\la 1,1,1,1\ra=\la 1, a,b,ab\ra=\lla -a,-b\rra$;
hence $s_2(\{-1,-1\})\,=\,s_2(\{-a,-b\})$  and thus $\{-1,-1\}\,=\,\{-a,-b\}$, by injectivity of $s_2$.
\end{proof}

\begin{prop}\label{P:specialcase}
Suppose that $|\scg{F}|= 2^m$ where $m>1$ and that there is an element $\xi\in k_2 F$ which cannot be written as a sum of $[\frac{m}{2}]$ symbols.
Then there exist $a_1,b_1,\dots,a_n,b_n\in \mg{F}$, where $2n+1=m$, such that
$$\xi =\{-1,-1\}+\{-a_1,-b_1\}+\dots+\{-a_n,-b_n\} .$$
Furthermore, given any such representation of $\xi$ as a sum of $n$ symbols, the elements $\{-1\},\{a_1\},\{b_1\},\dots,\{a_n\},\{b_n\}$ 
form an $\ff_2$-basis of $k_1 F$ and none of the elements $a_1,b_1,\dots,a_n,b_n$ is a sum of squares in $F$.
\end{prop}

\begin{proof}
By (\ref{P:p^m}, a) and by the hypothesis on $\xi$, the element $-1$ cannot be a square in $F$, in particular, ${\rm char}(F)\neq 2$.

Let $\theta$ be an element of $\Lambda_{_{\{-1\}}}^2 (k_1 F)$ which maps to $\xi$ under the cano\-nical homomorphism 
 $\Lambda_{_{\{-1\}}}^2 (k_1 F)\longrightarrow k_2 F$. The hypothesis implies that $\theta$ is not a
sum of $[\frac{m}{2}]$ pure $2$-vectors in  $\Lambda_{_{\{-1\}}}^2 (k_1 F)$.
Since $k_1 F$ has dimension $m$, we conclude by (\ref{P:eps-alt-space}) that $m$ is odd and that $\theta$ is a sum of $\frac{m+1}{2}$
pure $2$-vectors where the first one is $\{-1\}\wedge\{-1\}$.
Hence we may choose elements $a_1,b_1,\dots,a_n,b_n\in \mg{F}$, where $2n+1=m$, such that 
$$\theta =\{-1\}\wedge\{-1\}+\{-a_1\}\wedge\{-b_1\}+\dots+\{-a_n\}\wedge\{-b_n\}.$$
Further, using $\{-1\}+\{-x\}=\{x\}$ in $k_1 F$ for any $x\in\mg{F}$, 
we conclude from the second part of (\ref{P:eps-alt-space}) that an $\ff_2$-basis of $k_1 F$ 
is given by the elements $\{-1\},\{a_1\},\{b_1\},\dots,\{a_n\},\{b_n\}$.

From the above representation of $\theta$ we obtain immediately that $$\xi =\{-1,-1\}+\{-a_1,-b_1\}+\dots+\{-a_n,-b_n\} .$$
The proof will be complete if we can show for any $l\in\nat$ and any representation of $\xi$ as above, 
that none of the elements $a_1,b_1,\dots,a_n,b_n$ can be a sum of $l$ squares in $F$.
Suppose that this statement is true for a certain positive integer $l$.
To show that it still holds after replacing $l$ by $l+1$, we may assume that 
$a_1$ is a sum of $l+1$ squares in $F$, in order to derive a contradiction.

Since $\{-1,-1\}+\{-a_1,-b_1\}$ cannot be equal to a symbol, $a_1$ cannot be a sum of three squares in $F$
by Lemma~\ref{L:3squares}, hence $l\geq 3$.
By our assumption we may write $a_1=c+e$ with elements $c,e\in\mg{F}$, where $c$ is a sum of two squares and  $e$ is a sum of $l-1$ squares in $F$.


As $l-1\geq 2$, the element $e$ is not a square in $F$.
Since there exists an element of $F$ which is a sum of $l+1$ but not of $l$ squares in $F$, 
the form  $l\times \la 1\ra$ over $F$ is necessarily anisotropic.
It follows that $e\notin -\sq{F}$.
As $a_1$ is not a sum of $l$ squares in $F$, we also obtain that $e\notin a_1\sq{F}$.
Further, by the equality $a_1=(a_1^2-e a_1)/c$ and since $c$ is a sum of $2$ squares in $F$ while this is not the case for $a_1$,
we see that $-ea_1$ is not a square in $F$, i.e. $e\notin -a_1\sq{F}$.
Until here we have shown that $e\notin \pm \sq{F}\cup \pm a_1\sq{F}$.

In $k_2 F$ we have the equality $\{-a_1, e\}= \{-c,a_1 e\}$.
Since $c$ and $e$ are sums of $l$ squares in $F$, we get from the induction hypothesis that the symbol $\{-a_1,-b_1\}$ cannot be equal neither
to $\{-a_1, -e\}$ nor to $\{-a_1,e\}$. 
Hence the pure $2$-vectors $\{-a_1\}\wedge\{-eb_1\}$ and $\{-a_1\}\wedge\{eb_1\}$ in $\Lambda_{_{\{-1\}}}^2 (k_1 F)$ are both nonzero.
From this we conclude that $e\notin \pm b_1\sq{F}\cup \pm a_1 b_1\sq{F}$.

Since  $\{-1\},\{a_1\},\{b_1\},\dots,\{a_n\},\{b_n\}$ form an $\ff_2$-basis of $k_1 F$,
we may write $e=xy$ where $x\in  \pm\sq{F}\cup\pm a_1\sq{F} \cup\pm b_1\sq{F}\cup \pm a_1b_1\sq{F}$ and where $y$ is equal to $1$ or to a nontrivial product
of some of the elements $a_2,b_2,\dots,a_n,b_n$.
Since $e\notin  \pm\sq{F}\cup\pm a_1\sq{F} \cup\pm b_1\sq{F}\cup \pm a_1b_1\sq{F}$ as we have shown above, we have $y\neq 1$.
Therefore (\ref{L:transform},b) shows that in $\Lambda_{_{\{-1\}}}^2 (k_1 F)$ the element 
$\{-a_2\}\wedge\{-b_2\}+\dots+\{-a_n\}\wedge\{-b_n\}$ can be rewritten as a sum of $(n-1)$ pure $2$-vectors where the first is
of the shape $\{\pm y\}\wedge \{\ast\}$.
Hence we may suppose that $a_2=\pm y$.
By the relations in $\Lambda_{_{\{-1\}}}^2 (k_1 F)$ and since $e=xy$ and  $x\in \pm\sq{F}\cup\pm a_1\sq{F} \cup\pm b_1\sq{F}\cup \pm a_1b_1\sq{F}$,
we can rewrite $\{-a_1\}\wedge\{-b_1\}+\{-a_2\}\wedge\{-b_2\}$ as a sum of two pure $2$-vectors 
which is either of the shape $\{-a_1\}\wedge \{\ast\} + \{\pm e\}\wedge \{\ast\}$ or  
of the shape  $\{-a_1\}\wedge\{\pm e\}+ \{\ast\}\wedge \{\ast\}$.
So we may actually suppose that $e$ is equal to one of $\pm b_1$ and $\pm a_2$.
However, by the induction hypothesis, neither $b_1$ nor $a_2$ can be a sum of $l$ squares in $F$.
Hence $e$ is equal either to $-a_2$ or to $-b_1$. In particular, since $a_1= c+e$ where $c$ is a sum of $2$ squares in $F$, the quadratic form 
$\la 1,1, -a_1,-b_1,-a_2\ra$ is isotropic.


We put $\zeta:=\{-1,-1\}+\{-a_1,-b_1\}+\{-a_2,-b_2\}\in k_2 F$.
In $\bar{I}^2 F$ we compute
\begin{eqnarray*}
s_2(\zeta) & = & \ovl{\lla -1,-1\rra} + \ovl{(-1)\cdot\lla -a_1,-b_1\rra}+\ovl{(-1)\cdot\lla -a_2,-b_2\rra}\\
	   & = & \ovl{\la  1,1,-a_1,-b_1,a_1b_1,-a_2,-b_2,a_2b_2\ra}\,.
\end{eqnarray*}
By the above, the $8$-dimensional form $\la 1,1,-a_1,-b_1,a_1b_1,-a_2,-b_2,a_2b_2\ra$  is isotropic, hence Witt equivalent to a $6$-dimensional form $\varphi$.
Then $s_2(\zeta)=\ovl{\varphi}$, thus $\zeta$ is equal to a sum of two symbols by (\ref{L:folklore}).
But then $\xi$ can be written as a sum of $n<\frac{m}{2}$ symbols, in contradiction to the hypothesis.
\end{proof}

\begin{cor}\label{C:specialcase}
If $F$ is a nonreal field such that $|\scg{F}|= 2^m$ then every element of $k_2 F$ can be written as a sum of $[\frac{m}{2}]$ symbols.
\end{cor}
\begin{proof}
For $m=1$ this is clear by (\ref{P:euclidean}).
Assume that $m>1$ and that there exists an element in $k_2 F$
which cannot be written as a sum of $[\frac{m}{2}]$ symbols.
Then we must have $-1\notin\sq{F}$ by (\ref{R:proof-of-thm}), 
and by the last proposition there exist elements in $F$ which are not sums of squares.
Therefore $F$ is real.
\end{proof}

The last corollary completes the proof of Theorem~\ref{T:estimate} (see \ref{R:proof-of-thm}).


\medskip
If the field $F$ is nonreal, then one denotes by $s(F)$ the least positive integer $s$ such that $-1$ is a sum of $s$ squares over $F$,
otherwise one puts $s(F):=\infty$. The invariant $s(F)$ is  called the \emph{level of $F$}.
By a famous result due to Pfister, the integers occurring as the level of some field are precisely the powers of $2$. However, it is still an unsolved problem
whether there is a nonreal field $F$ of level greater than $4$ which has finite square class group $\scg{F}$.
(For further information on this problem, see \cite{B01} and the references given there.)
If the answer to this question turned out to be negative then one
could simplify the proof of Proposition~\ref{P:specialcase}.

We will see in the next section that, 
whenever there exist fields of finite level $s$ and with finite
square class group, then for $m$ sufficiently large there is such a field $L$
such that, in addition to $s(L)=s$, one has $|\scg{L}|=2^m$ and $\lambda_2(L)= [\frac{m}{2}]$ 
(see Corollary~\ref{C:level}).
This shows in particular that the estimate given in Corollary~\ref{C:specialcase} is generally
the best possible.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Symbols and central simple algebras} %%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Let $p$ be a prime number.
We suppose for the moment that the field $F$ contains a primitive $p$-th root of unity $\omega$;
in particular, ${\rm char}(F)\neq p$.
Given $a,b\in\mg{F}$ we denote by $(a,b)_{F,\omega}$ the central simple $F$-algebra of degree $p$
generated by two elements $\alpha$ and $\beta$ which are subject to the relations $\alpha^p=a$, $\beta^p=b$ and $\beta\alpha=\omega\alpha\beta$.
We call this a  \emph{symbol algebra of degree $p$ over $F$}. 
In the case $p=2$, we have $\omega=-1$ and, hence, we recover the definition of an $F$-quaternion algebra;
we then write $(a,b)_F$  instead of $(a,b)_{F,-1}$.

Let $\br{F}$ denote the Brauer group of $F$ and by ${\rm Br}_p (F)$ its $p$-torsion subgroup.
There exists a canonical group homomorphism $K_2^{(p)} F\longrightarrow {\rm Br}_p (F)$ which maps a symbol $\{a,b\}\in K_2^{(p)} F$, with $a,b\in\mg{F}$, to the
class of $(a,b)_{F,\omega}$ in ${\rm Br}_p (F)$.
By the Merkurjev--Suslin theorem (cf. \cite{MerSus} or \cite{Mer86}), this is actually an isomorphism, but we shall not use this fact in the sequel.

\begin{lem}\label{L:criterion}
Suppose that $F$  contains a primitive  $p$-th root of unity. 
Let $a_1,b_1,\dots,a_n,b_n\in \mg{F}$.
If $(a_1,b_1)_{F,\omega}\otimes_F\dots\otimes_F (a_n,b_n)_{F,\omega}$ is a division algebra then
the element $\{a_1,b_1\}+\dots+\{a_n,b_n\}$ in $K_2^{(p)} F$ cannot be written as a sum of less than $n$ symbols.
\end{lem}
\begin{proof}
We put $\xi:=\{a_1,b_1\}+\dots+\{a_n,b_n\}$. Let $D$ denote the central $F$-division algebra
whose class in $\br{F}$ is the image of $\xi$ under the canonical homomorphism
$K_2^{(p)}\longrightarrow {\rm Br}_p(F)$.
If $\xi$ is a sum of $m$ symbols in $K_2^{(p)} F$, then the degree of $D$ is at most $p^m$.
On the other hand, if $(a_1,b_1)_{F,\omega}\otimes_F\dots\otimes_F (a_n,b_n)_{F,\omega}$
is a division algebra, then it is $F$-isomorphic to $D$, which then must be of degree $p^n$.
These facts together imply the statement.
\end{proof}

\begin{thm}\label{T:opt}
Suppose that ${\rm char}(F)\neq p$ and that $|\mg{F}/{\mg{F}}^p|=p^n$.
Let $m\geq 2n-1$ and \mbox{$L:=F(\!(X_1)\!)\dots(\!(X_{m-n})\!)$}.
Then $|\mg{L}/{\mg{L}}^p|=p^m$ and
$$
\renewcommand{\arraystretch}{1.5}
\lambda_p (L)\,=\,\, 
\left\{
\begin{array}{cl}
\left[\frac{m}{2}\right]  & \mbox{if $p\neq 2$ or if $F$ is nonreal,}\\
\left[\frac{m+1}{2}\right] & \mbox{if $p=2$ and $F$ is real.}
\end{array}
\right.
$$
\end{thm}

\begin{proof}
It is well-known that $|\mg{L}/{\mg{L}}^p|=p^{m-n}\cdot |\mg{F}/{\mg{F}}^p|=p^n$
To prove the remaining claims we may assume that $m$ is  equal  either to $2n-1$ or to $2n$.
Indeed, if $m>2n$ then we replace $n$ by $n':=[\frac{m}{2}]$ and $F$ by $F':=F(\!(X_1)\!)\dots(\!(X_{n'-n})\!)$, observing that
$|\mg{{F'}}/{{F'}}^{\times p}|=p^{n'}$ and that $L$ is the iterated power series field in $m-n'$ variables over $F'$.

Let $\omega$ denote a primitive $p$-th root of unity. Recall that $F(\omega)/F$ is a field extension of degree dividing $p-1$.
Therefore any irreducible polynomial of degree $p$ over $F$ will stay irreducible over $F(\omega)$.
For any $a\in\mg{F}\setminus{\mg{F}}^p$, the polynomial $X^p-a$ is irreducible over $F$ \cite[Chapter~VIII, Theorem~9.1.]{LangAlg}, hence also over $F(\omega)$.
This shows that the canonical homomorphism $\mg{F}/{\mg{F}}^p\longrightarrow \mg{F(\omega)}/{\mg{F(\omega)}}^p$ is injective.
The hypothesis therefore implies $|{\mg{F(\omega)}}/{\mg{F(\omega)}}^p|\geq p^n$.

By Kummer theory (cf. \cite[Chapter~VIII, \S~8]{LangAlg}, for example),
we may choose elements $a_1,\dots,a_n\in \mg{F}$ such that $F(\omega,a_1^{1/p},\dots,a_n^{1/p})$ is an extension of $F(\omega)$ of degree $p^n$.
Note that  $L(\omega)=F(\omega)(\!(X_1)\!)\dots(\!(X_{m-n})\!)$.

We assume now that $m=2n$.
By \cite[Proposition 2.10]{T87a}, the central simple $L(\omega)$-algebra $(a_1,X_1)_{L(\omega),\omega}\otimes_{{L(\omega)}}\dots\otimes_{{L(\omega)}} (a_n,X_n)_{L(\omega),\omega}$
is a division algebra. Therefore, the element $\{a_1,X_1\}+\dots+\{a_n,X_n\}$ in $K_2^{(p)} L(\omega)$
is not a sum of less than $n$ symbols.
It follows that in $K_2^{(p)} L$ this element cannot be a sum of less than $n$ symbols either.
This together with Theorem~\ref{T:estimate} shows that $\lambda_p (L)=n=[\frac{m}{2}]=[\frac{m+1}{2}]$.

Assume next that $m=2n-1$ and that $p\neq 2$ or  $F$ is nonreal.
The same argument as before shows that the element $\{a_1,X_1\}+\dots+\{a_{n-1},X_{n-1}\}\in K_2^{(p)} L$ is not a sum of less than $n-1= [\frac{m}{2}] $ symbols.
This together with~(\ref{T:estimate}) yields $\lambda_p (L)= [\frac{m}{2}]$.

In the remaining case, $F$ is a real field,  $p=2$ and \mbox{$m=2n-1$}. We may then choose the elements $a_1,\dots, a_n\in\mg{F}$ as above in such a way that
 $a_1,\dots, a_{n-1}$ become squares in some real closure $E$ of $F$.
The $F$-quaternion algebra $(-1,-1)_{F}$ does not split over the field  $F(\sqrt{a_1},\dots,\sqrt{a_{n-1}})$, which is contained in $E$.
Hence, by \cite[Proposition 2.10]{T87a}, the product of quaternion algebras 
$(-1,-1)_L\otimes_{L} (a_1,X_1)_{L}\otimes_{L}\dots\otimes_{L} (a_{n-1},X_{n-1})_{L}$ is a division algebra.
So, by (\ref{L:criterion}), the element $\{-1,-1\}+ \{a_1,X_1\}+\dots+\{a_{n-1},X_{n-1}\}\in K_2^{(p)} L$ cannot be written as a sum of less than $n$ symbols.
By (\ref{T:estimate}) we conclude that  $\lambda_2 (L)= n=[\frac{m+1}{2}]$.
\end{proof}

\begin{cor}\label{C:level}
Suppose that ${\rm char}(F)\neq 2$ and that $|\scg{F}|=2^n$.
Let  $m\geq 2n-1$ and \mbox{$L:=F(\!(X_1)\!)\dots(\!(X_{m-n})\!)$}.
Then $|\scg{L}|=2^m$, $s(L)=s(F)$ and
$$
\renewcommand{\arraystretch}{1.5}
\lambda_2 (L)\,=\,\, 
\left\{
\begin{array}{cl}
\left[\frac{m}{2}\right]  & \mbox{if $F$ is nonreal,}\\
\left[\frac{m+1}{2}\right] & \mbox{otherwise.}
\end{array}
\right.
$$
\end{cor}
\begin{proof}
It is well-known that $L$ has the same level as $F$; the rest of the statement is contained in the theorem.
\end{proof}

\begin{exs} 
In each of the following situations, $L$ is a nonreal field with 
$|\scg{L}|=2^m$ and $\lambda_2 (L)\,=\,\left[\frac{m}{2}\right]$.

(1)\,\, $L=\cc(\!(X_1)\!)\dots(\!(X_m)\!)$, where $m\geq 0$. The level of $L$ is $1$ in this case.

(2)\,\, $L=\ff_3(\!(X_1)\!)\dots(\!(X_{m-1})\!)$, where $m\geq 1$ and where $\ff_3$ denotes the field with
three elements. In this case, $s(L)=2$.

(3)\,\, $L=\qq_2(\!(X_1)\!)\dots (\!(X_{m-3})\!)$, where $m\geq 5$ and where $\qq_2$ is the field of dyadic numbers.
On the other hand, a direct but tedious calculation shows that one has
$\lambda_2(F)=1$ for any field $F$ of level $4$ with $|\scg{F}|\leq 2^4$.
(This can also be seen from \cite{Szcz}.)
\end{exs}

The key idea in the proof of the following statement goes back to Naka\-yama
\cite{Nak35}.
\vspace{-0.5cm}

\begin{prop}\label{P:imp}\hphantom{xxxx}
Suppose that $F$ contains a primitive $p$-th root of unity $\omega$.
Let $L$ be an extension of $F(X_0,\dots,X_m)$\, contained in $F(X_0)(\!(X_1)\!)\!\dots\!(\!(X_m)\!)$.
There is a tensor product of $m$ symbol algebras of degree $p$ over $L$ which is a division algebra.
In particular, one has $\lambda_p (L)\geq m$. 
\end{prop}
\begin{proof}
$\!\!$ We choose distinct monic irreducible polynomials \mbox{$f_1,\dots,f_m\in F\left[X_0\right]$.}
They represent $\ff_p$-linearly independent classes 
in  $\mg{F(X_0)}/{\mg{F(X_0)}}^p$.
Hence, \cite[Proposition 2.10]{T87a} implies that 
\mbox{$(f_1,X_1)_{L,\omega}\otimes_{_L}\dots\otimes_{_L}(f_m,X_m)_{L,\omega}$} is an central division algebra
over $L=F(X_0)(\!(X_1)\!)\!\dots\!(\!(X_m)\!)$.
It is now clear that the same holds if $L$ is any subfield of $F(X_0)(\!(X_1)\!)\!\cdots\!(\!(X_m)\!)$ containing $F(X_0,\dots,X_m)$; in particular,
 $\lambda_p (L)\geq m$, by (\ref{L:criterion}).
\end{proof}

In view of the inequality of the proposition, it seems natural to ask:

\begin{qu} Let $m$ be a positive integer, $p$ a prime, 
and $C$ an algebraically closed field with ${\rm char}(F)\neq p$.
Let $F=C(X_0,\dots,X_m)$ be the rational function field in $m$ variables over $C$.
Do we have $\lambda_p(F)= m$?
\end{qu} 

Assume now that $m=1$.
Already here, the problem is unsolved and very interesting.
Showing that $\lambda_p(F)=1$ would imply, via the
Merkurjev--Suslin Theorem, that any central division
algebra over $F=C(X_0,X_1)$ of exponent $p$ is cyclic (equivalently, a symbol
algebra of degree $p$).
This is true at least for  $p\leq 3$ (cf. \cite{MArt}). 
By a very recent result of de Jong, any central division
algebra of exponent $p$ over $F$ is of degree $p$ (cf. \cite{AJDJ}), but it is 
not presently known whether such an algebra is necessarily cyclic when $p\geq 5$.

An important and rather open problem in the study of symbol lengths is
their behaviour under field extensions. First of all, it would be
interesting to know whether $\lambda_p(L)<\lambda_p(F)$ is possible
for a prime $p$ and a finite extension $L/F$ of degree prime to $p$
(cf. \cite[Conjecture~3]{Kahn00} for $p=2$).
Furthermore, no upper bound is known for $\lambda_p(L)$ in terms of
$\lambda_p (F)$ and the degree of the (finite) extension $L/F$.

To finish, let us indicate what can happen with
respect to the latter problem in the case $p=2$.
In our examples, the base field $F$ will have finitely many square classes.
A far more systematic discussion of the behaviour of the $2$-symbol
length $\lambda_2$ under field extensions $L/F$ of degree $2$ and $3$
can be found in \cite{MinWad}.

\begin{exs}
(1)\,\, Let $F_0$ be the quadratic closure of $\qq$ and $L_0/F_0$ an extension of degree $3$. 
For given $m\geq 0$, let
  $F=F_0(\!(X_1)\!)\dots(\!(X_m)\!)$ and
  $L=L_0(\!(X_1)\!)\dots(\!(X_m)\!)$. Then one has $[L:F]=[L_0/F_0]=3$
and it turns out that $\lambda_2(F)=\left[\frac{m}{2}\right]$, by
(\ref{T:opt}), whereas $\lambda_2(L)=m$.

As for the last claim, one may argue as follows.
By \cite[Chapter~VII, Appendix, Corollary~3]{Lam}, $L_0$ has infinitely
many square classes. 
Let $a_1,\dots,a_m\in\mg{L_0}$ be
$\ff_2$-independent in $\scg{L_0}$.
Then \cite[Proposition 2.10]{T87a} shows that 
$(a_1,X_1)_L\otimes_{_L}\cdots\otimes_{_L} (a_n,X_n)_L$ is a division algebra. 
Using (\ref{L:criterion}), it follows that $\lambda_2(L)\geq m$.
On the other hand, any sum of $m+1$ symbols in $k_2 L$
can be written as a sum of one symbol defined over $L_0$ and with $m$ other
symbols; however, $k_1 F_0=0$ and $[L_0:F_0]=3$ imply that $k_2 L_0=0$, by \cite[Lemma~2]{Mer83},
because $k_1 F_0=0$ and $[L_0:F_0]=3$. One conludes that $\lambda_2(L)=m$.

(2)\,\, 
Let $F_0$ be a nonreal field such that $|\scg{F}|=2^m$ for given $m\geq 1$, and such that $k_2 F_0=0$. 
Let $L_0/F_0$ be a quadratic extension. Then it follows from \cite[Chapter~VII, Theorem 3.4.]{Lam} that
  $|\scg{L_0}|=2^{2m-1}$.
We put 
$F=F_0(\!(X_1)\!)\dots(\!(X_{n-m})\!)$ and
  $L=L_0(\!(X_1)\!)\dots(\!(X_{n-m})\!)$ for given $n\geq 3m-1$.
Then $[L:F]=[L_0:F_0]=2$, and (\ref{T:opt}) yields
$\lambda_2(F)=[\frac{n}{2}]$ and $\lambda_2(L)= [\frac{n+m-1}{2}]$.
In particular, for $n= 3m-1$ we get $\lambda_2(F)=[\frac{3m-1}{2}]$
and $\lambda_2(L)=2m-1$, so $\lambda_2(L)=\frac{4\lambda_2(F)+(-1)^{m}}{3}$.
\end{exs}


 

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\bigskip\noindent
Karim Johannes Becher, Universit\"at Konstanz, Fachbereich Mathematik und Statistik, D204,\ 78467 Konstanz, Germany; e-mail: {\tt becher@maths.ucd.ie}

\bigskip\noindent
Detlev Hoffmann, Universit\'e de Franche-Comt\'e, UFR des Sciences et Techniques, 
D\'epartement de Math\'ematiques de Besan\c{c}on,
16 Route de Gray, 
25030 Besan\c{c}on Cedex, France; e-mail: {\tt detlev@math.univ-fcomte.fr}

\bigskip\noindent
\emph{May 14, 2003}
\end{document}
