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\begin{document}

\title{On triviality of the functor $\text{Coker}(K_1(F) \rightarrow K_1(D))$ 
for division algebras}

\author{Roozbeh Hazrat}
\date{\today}



\begin{abstract}
Let $D$ be a cyclic division algebra over its centre $F$ of index
  $n$. Consider the group $CK_1(D)=D^*/F^*D'$ where $D^*$ is the group
  of invertible elements of $D$ and $D'$ is its commutator subgroup. 
In this note we shall show that the group $CK_1(D)$ is trivial if and only
  if $D$ is an ordinary
  quaternion division algebra over a real Pythagorean field $F$. We construct
  a division algebra $D$ and a division subalgebra $A \subset D$ such
  that $CK_1(A)\cong CK_1(D)$. Using valuation theory, the group
  $CK_1(D)$ is computed for some
  valued division algebras. 
\end{abstract}



\address{
Centre for Mathematics and its Applications\\ 
John Dedman Building (\# 27)\\ 
Australian National University\\
Canberra ACT 0200\\
Australia
}

\email{hazrat@maths.anu.edu.au}


\maketitle
              




\section{Introduction}\label{intro}

Let $A$ be an Azumaya algebra over a commutative ring $R$. Consider
the functor $CK_1(A)=\text{Coker}(K_1(R) \stackrel{i}{\rightarrow}
K_1(A))$ where $i$ is the inclusion map. 
If $R$ is semilocal, since the stable rank of $R$ and $A$ are one, 
$CK_1(A)= A^*/R^*A'$ where $A^*$ and $R^*$ are the group of invertible elements
of $A$ and $R$ respectively and $A'$ is the derived subgroup of
$A^*$. A study of this group in the case of central simple
algebras is initiated in \cite{behrooz} and further in \cite{sk1}. It has 
been established that 
despite of a ``different nature''
of this group from the reduced Whitehead group $SK_1$, these two groups
have the same functorial properties. In \cite{behrooz} this functor is 
determined
for totally ramified division algebras and in particular for any
finite abelian group $G$, a division algebra $D$ such that
$CK_1(D)=G\times G$ is constructed. Further in \cite{sk1}, this functor is
studied in more
cases and examples of cyclic $CK_1$ (even over non local fields) 
constructed.
It was conjectured in \cite{behrooz} that if
the functor $CK_1$ is trivial then the index of the division algebra is
2. 
In this note we characterise quaternion division algebras $D$ such
that $CK_1(D)$ is trivial. In fact triviality of $CK_1$ characterises the 
ground field. We shall show that if $CK_1(D)$, for a quaternion
division 
algebra $D$, is trivial then $D$ is an ordinary quaternion and the
centre of $D$ is a real Pythagorean field. In fact, this is the only class of
cyclic division algebras such that $CK_1$ is trivial (Theorem \ref{c6}).  
We shall show that some of the notions from the theory of 
quadratic forms, like rigidity of an element, can be formulated as a property 
of the group 
$CK_1$.
It seems that this group is highly sensitive to the arithmetic property 
of the 
ground field. For example we will observe that 
$CK_1(\frac{x,x}{\Bbb R((x))})\cong \Bbb Z_2$ whereas
$CK_1(\frac{x,x}{\Bbb F_3((x))})\cong \Bbb Z_2 \oplus \Bbb Z_2$.


\section{Triviality of $CK_1$}\label{quat}

Let $F$ be a field with characteristic not 2. Let $\qt$ be a
quaternion division algebra over $F$. The elements of $D=\qt$ have the
form $c_0+c_1i+c_2j+c_3ij$ and $Nrd_D(c_0+c_1i+c_2j+c_3ij)=
c_0^2-ac_1^2-bc_2^2+abc_3^2$ where $Nrd_D$ is the reduced norm map for $D$. 
Note that $CK_1(D)=D^*/F^*D'$ where $D^*$ and $F^*$ are the group
of invertible elements of $D$ and $F$ respectively and $D'$ is the 
commutator subgroup of $D^*$. If $CK_1(D)=1$ then $Nrd_D(D)={F^*}^2$.
Thus $Nrd_D(i)=-a \in F^2$ and $Nrd_D(j)=-b \in F^2$. Therefore 
$\qt \cong (\frac{-1,-1}{F})$.

Our first theorem shows that triviality of $\qt$ forces that $F$ be a
real Pythagorean field. In fact this property characterises this field.
Recall that $F$ is a real Pythagorean if $-1 \not \in F^{*2}$ and
sum of any two square elements is a square in $F$. It follows immediately 
that $F$ is an ordered field. $F$ is called Euclidean if $F^{*2}$ is an 
ordering of $F$

\begin{theorem}\label{pyth}
Let $F$ be a field of characteristic not 2. Then the following are
equivalent.

1) $\oqt$ is a division ring and $CK_1\oqt$ is trivial.

2) $F$ is the real Pythagorean field.

3) $\oqt$ is a division ring and every maximal subfield of $\oqt$ is
$F$-isomorphic to $F(\sqrt{-1})$.
\end{theorem}
\begin{proof}
We shall show that 1 and 2 are equivalent. The equivalency of 2 and 3 are
known (see \cite{feinsch}).

$1)\Rightarrow 2)$ Since $D=\oqt$ is a division ring, $-1 \not\in {F^*}^2$.
By induction, if $-1={f_1}^2 + \cdots +{f_n}^2$, then considering
${f_1}^2+{f_2}^2=Nrd_D(f_1+f_2i)=f^2$, a contradiction follows.
This shows that $-1$ is not sum of squares plus sum of any two squares
is a square

$2)\Rightarrow 1)$ Suppose $F$ is a real Pythagorean. It is easy to see that 
$D=\oqt$ is a division ring.
Now for any $x\in D^*$, $Nrd_D(x)$ is a sum of four squares, thus 
$Nrd_D(D^*)={F^*}^2$. Since $SK_1(D)$ is trivial, 
this forces that $CK_1$ be trivial.
\end{proof}

Over Euclidean fields, the only quaternion division algebra is the
ordinary quaternion division algebra, and from the above Theorem it
follows that its $CK_1$ is trivial. 

We shall show that an ordinary quaternion division algebra
over a real Pythagorean field is in fact the only type of cyclic division
algebras such that its $CK_1$ is trivial. 
Before we state the theorem, note that there are examples of infinite
dimensional division algebras $D$ such that $D^*$ coincide with
$D'$ \cite{kegel}. In the finite dimensional case, it is not hard to see that
$D^*\not = D'$, infact $K_1(D)=D^*/D'$ is not torsion. In the case of 
algebraic division algebras, everything remains as a mystery.

We need the following theorem which says that the functor $CK_1$ has a
decomposition property similar to one for the reduced Whitehead group.

\begin{theorem}{\cite{sk1}}\label{dec}
Let $A$ and $B$ be division algebras with centre $F$ such that
$(i(A),i(B))=1$. Then $CK_1(A \otimes_F B)\cong CK_1(A) \times CK_1(B)$.
\end{theorem}

 
\begin{theorem}\label{c6}
Let $D$ be a cyclic division algebra finite dimension over its centre $F$. 
Then $CK_1(D)=1$ if and only if $D$ is an ordinary quaternion division algebra
over a real Pythagorean field $F$.
\end{theorem}
\begin{proof}
The only if part is already proved in Theorem \ref{pyth}. For the if
part, we shall show that the index of $D$ is 2. Then the claim follows
again from Theorem \ref{pyth}.
First suppose that $i(D)=n$ is odd. Let $\alpha \in D$ such that
$F(\alpha)$ is a maximal cyclic subfield of $D$ with a mininal
polynomial of the form $x^n-f$ where $f\in F$. 
Since $D^*=F^*D'$, we can
choose $\alpha \in D'$. Now
$\alpha^n=N_{F(\alpha)/F}(\alpha)=Nrd_D(\alpha)=1$.
From this it follows that the degree of the minimal polynomial of $\alpha$
is less than $n$ which can not happen.
Since $CK_1(D)$ has a decomposition property (Theorem \ref{dec}), the
only case which remains, is $n=2^k$, for some integer $k$. 
We shall show that $k=1$.
Again consider $\alpha$ as above. This time $-\alpha^{2^k}=N_{F(\alpha)/F}
(\alpha)=Nrd_D(\alpha)=1$. On the other hand since 
$1+N_{F(\alpha)/F}(\alpha)=N_{F(\alpha)/F}(\alpha+1)\in F^{*2^k}$, it follows that 
$\sqrt{2} \in F$. Thus if $k >1$ then 
$\alpha^{2^k}+1=(\alpha^{2^{k-1}}+1)^2-2\alpha^{2^{k-1}}$ can be
decomposed further which leads to a contradiction that the minimal
polynomial of $\alpha$ has degree less than $n$. Therefore $n=2$ and
the claim follows from Theorem \ref{pyth}
\end{proof}

Note that in the above theorem we could replace $CK_1$ with a smaller 
group $D^*/F^*D^{(1)}$, where $D^{(1)}$ is set of elements with reduced norm
1.

The above theorem states that if in the case of division algebra $D$
of odd prime index, $CK_1(D)$ happens to be trivial, then this forces
the well-known conjecture that any division algebra of prime index is
cyclic would be false!
 
It is a celeberated result of Wedderburn that a division algebra of
index 3 is cyclic. This together with Theorem \ref{dec} and Theorem
\ref{c6} implies that

\begin{theorem}
If $D$ be a division algebra of index $3n$ where $ 3 \nmid n$, then  
$CK_1(D) \not = 1$.
\end{theorem}


\begin{remark}\label{badrem} Let $D$ be a finite dimensional division algebra with 
centre $F$, of index $n$. Consider the sequence 
\begin{equation}\label{nr}
K_1(D) \stackrel{Nrd}{\longrightarrow} K_1(F) \stackrel{i}{\longrightarrow}
 K_1(D)
\end{equation}
Where $Nrd_D$ is the reduced norm map and 
$i$ is the inclusion map. One can
see that the composition $i\,{Nrd}_{D/F}=\eta_n$ where $\eta_n(a)=a^n$. 
(See for example the proof of Lemma 4, p. 157 \cite{draxl}).
From this the formula $a^n={Nrd}_{D/F}(a)c_a$ where 
$a\in D^*$ and $c_a \in D'$ follows. In particular this implies that
the exponent of the abelian group $CK_1(D)$ divides $n$, the index of
$D$.
\end{remark}

It is not known if $exp(CK_1(D)) < i(D)$ what 
would be imposed on the algebraic structure of $D$. We mention that 
if $D$ is a totally ramified division algebra then $exp(CK_1(D)) = i(D)$ 
if and only if $D$ is cyclic \cite{sk1}.
In fact from the above theorem it follows that if $D$ is a cyclic division 
algebra of index $p$, an odd prime, then the exponent of $CK_1(D)$ is 
exactly $p$. The converse is not true, as the following example
shows. This is an
example of a cyclic decomposable $F$-division algebra 
$D$ of index $2p$, $p$ an odd prime, such that it has a proper $F$-division 
subalgebra $A \subset D$, where
$CK_1(A) \cong CK_1(D)$. This in 
particular shows that the exponent of $CK_1(D)$ is less
than the index of $D$, but $D$ is cyclic, thats is, the phenomena 
of cyclicity of
$D$ and $exp(CK_1(D))$ does not follow the same pattern as in the case
of 
totally ramified.
For this we need
the Fein-Schacher-Wadsworth example of a division algebra of 
index $2p$ over a Pythagorean field $F$ \cite{fsw}. We briefly recall the
construction. Let $p$ be an odd prime and $K/F$ be a pair of real 
Pythagorean fields such that $K$ is cyclic over $F$ with the generating automorphism 
$\sigma$. Now consider 
$$D=\Big(\frac{-1,-1}{F((x))}\Big) \otimes_{F((x))} \Big( K((x))/F((x)),\sigma, x\Big).$$ 
      $K((x))/F((x))$ is a cyclic extension where $K((x))$ and
      $F((x))$ are Laurant power 
series fields of $K$ and $F$ respectively. $D$ is a division algebra of 
index $2p$. Since $F$ is real Pythagorean, so does $F((x))$. 
Now by Theorem \ref{dec} the primary decomposition of the division 
algebra $D$ induces a corresponding decomposition for $CK_1(D)$. 
Thus $CK_1(D) \cong CK_1(\frac{-1,-1}{F((x))}) \times 
CK_1(A)$ where $A=\big( K((x))/F((x)),\sigma, x\big)$. By Theorem \ref{pyth}, 
$CK_1(\frac{-1,-1}{F((x))})=1$, thus $CK_1(D)\cong CK_1(A)$.  

\medskip\noindent
{\bf Question.} If $exp(CK_1(D))$ is less than the index $i(D)$, what
can be said about the algebraic structure of $D$.

\medskip
Recall that $a \in F$ is called {\it rigid} if $a \not \in \pm F^{*2}$ and 
$F^{*2}+aF^{*2}=F^{*2} \cup aF^{*2}$. The rigidity concept plays a role in the
study of the extension of Pythagorean fields. In particular if 
$K=F(\sqrt{a})$ is a quadratic extension of $F$, then $K$ is real 
Pythagorean if 
and only if $F$ is real Pythagorean and $a$ is rigid \cite{lam}.



\begin{theorem}\label{rigid}
Let $F$ be a real Pythagorean field. Let $a \not \in \pm F^{*2}$. 
Then $a$ is rigid
if and only if $CK_1(\frac{-1,-a}{F})=\Bbb Z_2$.
\end{theorem}
\begin{proof}
Suppose $a$ is rigid. Since $-a \not \in F^{*2}$, then $D=(\frac{-1,-a}{F})$
is a division ring. Since $F$ is Pythagorean, $Nrd_D(D^*)=c_0^2+ac_1^2$, where
$c_0,c_1 \in F^*$. From rigidity of $a$ it follows that
$Nrd_D(D^*) = F^{*2} \cup aF^{*2}$.
Because $ a \not \in F^{*2}$, then $F^{*2}$ and $F^{*2}a$
are two distinct elements in the group $Nrd_D(D^*)/F^{*2}$. Since $SK_1(D)$ is trivial, 
$CK_1(D)\cong Nrd_D(D^*)/F^{*2}\cong \Bbb Z_2$. 
The if part follows the same way.
\end{proof}

\begin{remark} Let $F$ be a real Pythagorean field and $a \in F^*$
  such that $-a \not \in F^{*2}$. 
Then $CK_1(\frac{-1,-a}{F})\cong
L^*/F^*L^{*2}$ where $L=F(\sqrt{-a})$. For consider the square class
exact sequence $$ 1 \longrightarrow \{F^{*2},F^{*2}a\}\longrightarrow F^*/F^{*2} \longrightarrow L^*/L^{*2} \stackrel{N}
{\longrightarrow} N_{L/F}(L^*)/F^{*2}\longrightarrow 1.$$ In this setting, it is
easy to see that $ N_{L/F}(L^*)= Nrd_{D/F}(D^*)$ thus the claim.
\end{remark}



\section{$CK_1$ for valued quaternion division algebras}

In this section we study the valued 
quaternion division algebras. Valuation theory for division algebras 
is the main tool for computation of $CK_1$. Recall that if $D$ is equipped 
with a valuation, then $\bar D$ and $\bar F$ denote the residue division 
algebra and the residue field of $D$ and $F$ respectively. 
$\Gamma_D$ and $\Gamma_F$ are the value groups of the valuation.

A tame valued quaternion division algebra could be unramified, 
semiramified or totally ramified. This follows from the fact that $D$ is 
defectless. Recall that $D$ is {\it tame} if $char(\bar F)$ does not divide the 
index of $D$. Thus $D$ is {\it defectless}, namely $[\bar D:\bar F][\Gamma_D:\Gamma_F]
=[D:F]$. $D$ is called {\it unramified} if $[\bar D:\bar F]=[D:F]$ and 
{\it totally ramified} if $[\Gamma_D:\Gamma_F]=[D:F]$. $D$ is called 
{\it semiramified}
if $\bar D$ is a field and $[\bar D:\bar F]=[\Gamma_D:\Gamma_F]=i(D)$. 
For an excellent survey of the valuation theory of division algebras see 
\cite{wadval}. 

We will see that there are quaternion division 
algebras of any type mentioned above.

$\bullet$ $D=(\frac{-1,-1}{\Bbb R((x))})$ is unramified, and $CK_1(D)=1$ 
(see Theorem \ref{pyth}).

$\bullet$ $D=(\frac{-1,x}{\Bbb R((x))})$ is semiramified, 
and $CK_1(D)=\Bbb Z_2$ (see Theorem \ref{rigidv}). 

$\bullet$ $D=(\frac{x,y}{\Bbb R((x,y))})$ is totally ramified, and 
$CK_1(D)=\Bbb Z_2\oplus \Bbb Z_2$ (see Remark \ref{xy}).

Although all the examples that we are dealing with here have finite $CK_1$, 
let us mention that $CK_1$ of a quaternion division algebra could be
infinite. It is known that if $p$ is a rational prime such that 
$p \equiv 3 (mod 4)$ then $(\frac{-1,p}{\Bbb Q})$ is a division
algebra. It is not hard to see that $CK_1(\frac{-1,p}{\Bbb Q})=
\bigoplus \Bbb Z_2$ is infinite.  

In order to compute the functor $CK_1$ in the case of 
a valued division algebra, we need the following Theorem.

\begin{theorem}{\cite{sk1}}\label {val}
Let $D$ be a Henselian division algebra tame over its centre with index $n$. 
Then

1) If $D$ is unramified, then $CK_1(D) \cong CK_1(\bar D ).$

2) If $D$ is totally ramified, then $CK_1(D)=\Gamma_D/\Gamma_F.$

3) If $D$ is semiramified, and $\bar{D}$ is cyclic over $\bar F$, then
there is an exact sequence
$$1\longrightarrow N_{\bar D/ \bar F}(\bar D^*)/\bar F^{*n} \longrightarrow CK_1(D) 
\longrightarrow \Gamma_D/\Gamma_F \longrightarrow 1.$$
\end{theorem}

It is known that if $Z(\bar D)$ is separable over $\bar F$, 
then the fundamental homomorphism 
$\Phi: \Gamma_D /\Gamma_F \rightarrow Gal(Z(\bar D)/\bar F)$ is 
epimorphism \cite{wadval}. Since in the above theorem, part 3) $D$ is semiramified over $F$, 
$\Phi$ is an isomorphism. On the other hand $\Bar D$ is cyclic over $\bar F$, 
Thus $\Gamma_D /\Gamma_F \cong \Bbb Z_n$. So we can rewrite the above exact 
sequence as

{\it 3$'$)  If $D$ is semiramified, and $\bar{D}$ is cyclic over $\bar F$, then
there is an exact sequence}
$$1\longrightarrow N_{\bar D/ \bar F}(\bar D^*)/\bar F^{*n} \longrightarrow CK_1(D) 
\longrightarrow \Bbb Z_n \longrightarrow 1.$$

Theorem \ref{val} is our main tool to compute the functor $CK_1$ for
a valued quaternion division algebra. 

Let $F$ be a real Pythagorean field. Let $N$ be the set of 
non rigid elements of $F$. Recall that if $N= \pm F^{*2}$ then $F$ is called
{\it super-Pythagorean} \cite{lam}. The following theorem is a 
demonstration of how 
valuation theory enables us to compute $CK_1$.

\begin{theorem} \label{rigidv}
Let $F$ be a real Pythagorean field and $N$ the 
set of non rigid elements of $F$.
Then for any 
$t \in F((x)) \backslash \pm F((x))^2 \cup N$, 
$CK_1(\frac{-1,-t}{F((x))})=\Bbb Z_2$.
\end{theorem}
\begin{proof}
The valuation of $F((x))$ extends to $D=(\frac{-1,-1}{F((x))})$ as
follows; $v(x)=\frac{1}{2}v(Nrd_D(x))$ where $v$ denotes the
valuation map for both $F$ and $D$.
Suppose first that $t\in F$. It is easy to see that 
$$\big(\frac{-1,-t}{F((x))}\big) \cong \big(\frac{-1,-t}{F}\big ) \otimes_F F((x))$$ is unramified. 
Thus by \ref{val} 1), $CK_1(\frac{-1,-t}{F((x))}) \cong CK_1(\frac{-1,-t}{F})$.
Since $t$ is rigid in $F$ by Theorem \ref{rigid}, $
CK_1(\frac{-1,-t}{F})\cong \Bbb Z_2$ and we are done.

Now if $t \in F((x))\backslash F$, then it is not hard to see that $D$ is semiramified,
$\bar D = F(\sqrt -1)$ and $N_{\bar D}(F(\sqrt -1))=F^{*2}$. Now from Theorem
\ref{val} 3$'$) it follows that $CK_1(D)=\Bbb Z_2$.
\end{proof}

Theorem above is another way to observe that if $F$ is super-Pythagorean then 
so is $F((x))$. 

\begin{remark} \label{xy}
Consider the division algebra $D=(\frac{x,y}{\Bbb R((x,y))})$.
Since $Nrd_D(i)=-x$ and $Nrd_D(j)=-y$, it can be seen
that the extension of the valuation from $\Bbb R((x,y))$ to $D$ is
totally ramified. From Theorem \ref{val} 2) follows that $CK_1(D)=
\Bbb Z_2 \oplus \Bbb Z_2$.
\end{remark}

\subsection{$CK_1$ for local quaternion division algebras}


Let $F$ be a local field and $D$ a quaternion division algebra over $D$. 
We shall use Theorem \ref{val} to show that $CK_1(D)\cong \Bbb Z_2 \oplus 
\Bbb Z_2$. Recall that characteristic of $F$ is not 2. We further assume that
$char(\bar F)$ is not 2.

\begin{theorem}\label{local}
Let $D$ be a quaternion division algebra over a local field $F$. Then
$CK_1(D)\cong \Bbb Z_2 \oplus 
\Bbb Z_2$.
\end{theorem}
\begin{proof}
The valuation of $F$ extends to $D$. Furthermore one can see that with 
this valuation $D$ is semiramified over $F$ and $\bar D$ is cyclic over 
$\bar F$. Since $\bar D$ is finite, $N_{\bar D}(\bar D^*)=\bar F^*$ and the 
exact sequence of Theorem \ref{val} 3$'$ has the look
$$1\longrightarrow \bar F^* /\bar F^{2*} \longrightarrow CK_1(D) 
\longrightarrow \Bbb Z_2 \longrightarrow 1.$$
Since $char (\bar F)$ is not 2, the quotient $\bar F^* /\bar F^{2*}$ is not trivial. 
Because $\bar F^*$ is a cyclic  group 
therefore $\bar F^* /\bar F^{2*} \cong \Bbb Z_2$. Since $exp(CK_1(D))$ divides 2, from the
above exact sequence it follows that $CK_1(D)\cong \Bbb Z_2 \oplus 
\Bbb Z_2$.
\end{proof}
The above theorem shows that 
$CK_1(\frac{-1,x}{\Bbb F_3((x))})=\Bbb Z_2 \oplus \Bbb Z_2$ as mentioned 
before.

We shall mention that this result was not unknown. In fact, since $SK_1(D)$
is trivial and the reduced norm is surjective for division algebras over local fields, 
$CK_1(D) \cong F^*/F^{*2}$. From the theory of quadratic forms for local 
fields we can thus obtain the same 
result. Also if $char \bar F =2$, it is known that $F^*/F^{*2} \cong 
\bigoplus_{2^{n+1}} \Bbb Z_2$ (See P.217  \cite{sch})


 
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