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\newcommand\NK[1][1]{\operatorname{NK}_{#1}} % See def in the paper.
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\begin{document}

\title[Nontriviality of $\NK(D)$]{Nontriviality of $\NK[1](D)$ for division algebras}

\author{Roozbeh Hazrat}

%\forget

\address{
Department of Pure Mathematics\\
Queen's University\\
Belfast BT7 1NN\\
U.K.} \email{r.hazrat@qub.ac.uk}



\author{Adrian Wadsworth} 

\address{
Department of Mathematics\\
University of California at San Diego\\
La Jolla, California 92093-0112\\
U.S.A.}
 \email{arwadsworth@ucsd.edu}

%\forgotten

\date{31 jan 06}


\begin{abstract}
A field $E$ is said to be NKNT if for any noncommutative division
algebra $D$ finite dimensional over $E\subseteq Z(D)=F$ with index
$\ind(D)$, $\Nrd(D^*)/F^{*\ind(D)}$ is nontrivial. It is proved that if
$E$ is a field finitely generated but not algebraic over some
subfield then $E$ is NKNT. As a consequence, if $F=Z(D)$ is
finitely generated over its prime subfield or 
over an algebraically closed field, then
$\CK(D)=\text{Coker}(K_1 F \rightarrow K_1 D)$ is nontrivial.
\end{abstract}




\maketitle



 Let $D$ be a division algebra over its center $F$ of index
$n$. Denote by $D^*$ and $F^*$ the multiplicative group of $D$ and
$F$ respectively. Let $\Nrd_D\!\!:D^* \rightarrow F^*$ be the reduced
norm map, $D^{(1)}$ the kernel of this map and $D'$ the commutator
subgroup of $D^*$. The inclusion map $F \hookrightarrow D$ induces 
a homomorphism $K_1(F) = F^* \rightarrow K_1(D) = D^*/D'$.
Consider the group
$$
\CK(D)=\text{Coker}(K_1F\rightarrow K_1D)  \cong D^* /F^*D'\, . 
$$ 
%which arises from
%the inclusion map $F \hookrightarrow D$.
%  Clearly $\CK(D)\cong D^*/F^* D'$. 
Since $x^{-n}\Nrd(x) \in D^{(1)}$ and the reduced
Whitehead group $\SK(D)=D^{(1)}/D'$ is $n$-torsion (by \cite{draxl}, 
p.~157, Lemma 2), it follows
that $\CK(D)$ is an abelian group of bounded exponent $n^2$ (In
fact one can show that the bound is $n$, see the proof of Lemma 4,
p.~154 in \cite{draxl} or pp.~579--580 in \cite{sk1}). Thus, by 
the Pr\"ufer-Baer
theorem $\CK(D)\cong \bigoplus \Bbb Z_{k_i}$ where each $k_i \mid n$
(see \cite{robinson}, p.~105). Therefore if $\CK(D)$ is nontrivial
then $D^*$ has a (normal) maximal subgroup. The question of
whether $D^*$ has a maximal subgroup seems to remain open and thus
by the above observation is limited to the case when $\CK(D)$ is
trivial. 
 In
\cite{vishne}, Th.~2.12  it was proved that if $D$ is a tensor product of
cyclic algebras then $\CK(D)$ is trivial if and only if $D$ is a
quaternion division algebra $\oqt$ where $F$ is a real Pythagorean
field (see also \cite{kesh}).
It has been conjectured in \cite{hmm} that if $\CK(D)$ is trivial then
$D$ is a quaternion division algebra. 

The group $\CK$ has been computed in \cite{sk1} for certain
division algebras, and its connection with $\SK$ was also studied.
But, $\CK(D)$ is often difficult to work with. We will focus here 
on a related invariant, $\NK(D)$, which is sometimes more tractable,
and can yield information about $\CK(D)$.  Define 
$$
\NK(D) \ = \ {D^*}\big/ {F^*D^{(1)}}  \ \cong
 \ {\Nrd(D^*)}\big/{F^{*\ind(D)}}
$$ 
(with the isomorphism given by the reduced norm map). 
Observe that $\NK(D)$ is a homomorphic image of $\CK(D)$ and that 
whenever $\SK(D) = 1$, we have  $\CK(D) = \NK(D)$.  (Recall that 
$\SK(D) = 1$ whenever $\ind(D)$ is square-free, or the center $F$ of 
$D$ is a local or a global field, by \cite{draxl}, p.~164, Cor.~4, Th.~3, p.~165, (17),
p.~166, (18).)  For example, if 
$Q = \qt$ is a quaternion division algebra with $\mychar(F) \ne 2$,
we have 
$$
\CK(Q) = \NK(Q) \cong \big(\{r^2 - as^2 - bt^2 + abu^2 \,|\ r,s,t,u \in
F\}\backslash \{0\}\big)\big/F^{*2}.
$$ 
%%% FROM PROBLEM:
%>From this formula, it is immediate that $\CK(Q)$ is trivial iff 
From this formula, it is immediate that $\CK(Q)$ is trivial iff 
$F$ is a real Pythagorean field and $Q\cong \oqt$. 


Observe that the condition that $\NK(D)$ be  trivial for 
a noncommutative division ring~$D$ is an extremely strong one.  
Indeed, if $\ind(D) = d$ then $\NK(D) = 1$ iff $\Nrd_D(D^*) = F^{*d}
= \Nrd_D(F^*)$, which holds iff for every maximal subfield $L$ of 
$F$, $N_{L/F}(L^*) = F^{*d} = N_{L/F}(F^*)$.  It was shown in \cite{vishne}
that if $C$ is a noncommutative cyclic algebra with $\NK(C) =1$,
then $C\cong \oqt$ with $F$ a Pythagorean field. 
We will show here that $\NK(D)$ is nontrivial for a great  many other
noncommutative division algebras~$D$. Of course, 
whenever $\NK(D) \ne 1$, we also have $\CK(D) \ne 1$.



%% Define $$G(D) \ = \ {D^*}\big/{\Nrd(D^*)D'}.$$
%% Clearly $\CK(D)$ is a quotient of $G(D)$ and $\NK(D)$ is a
%% quotient of $\CK(D)$. Thus if $G(D)$ is trivial, so is $\CK(D)$
%% (and $\NK(D)$). But the converse is not true (consider $\oqt$
%% where $F$ is a real Pythagorean field which is not Euclidean
%% \cite{vishne}).


\begin{define}
A field $E$ is said to be NKNT (for $\NK$ nontrivial) if for any
noncommutative division algebra $D$ finite dimensional over $E$
(and not necessarily central over $E$), $\NK(D)$ is nontrivial.
\end{define}

It is clear from the definition that if a field $E$ is NKNT then so is
every finite degree field extension of $E$. Here are some examples of
NKNT fields: Clearly a
finite field is NKNT; so is any
algebraically closed field; so also is any field of transcendence degree 
$1$ over an algebraically closed field, by Tsen's Theorem. 
 Since every division algebra  over 
a global field   is a cyclic algebra, the result quoted above
 shows that every  global field is NKNT. Likewise, every nonreal
local field is NKNT.  
However the field of real numbers $\Bbb R$ is not NKNT,
since $\Bbb R$ is a real Pythagorean and thus $\CK(\Bbb H_{\Bbb
R})=1$, but the rational function field $\Bbb R(t)$ is NKNT. 
This is a consequence of our main theorem below.  But we can see 
it directly as follows:
If $L$~is a finite degree extension of $\Bbb R(t)$ and $D$ is an
$L$-central noncommutative division algebra, then by Tsen's
Theorem, $D$ is split by $L(\sqrt{-1})$, so $D$ is a quaternion
algebra; but $\NK(D)$ is then nontrivial because $L$ is not
Pythagorean. 
%% Also, any local and global field is NKNT
%% (\cite{vishne}, \S17 in \cite{pierce}).


In this note our main result is:
% \\

%%\noindent {\bf Theorem.} {\it 

\begin{main}
Let $F$ be a field which is finitely
generated but not algebraic over some subfield~$F_0$. Then, $F$ is
{\rm{NKNT}}.
\end{main}
% \\



Let $D$ be a division algebra with center $F$. In showing 
that $\NK(D)$ is nontrivial, Lemma~\ref{primarydecomplemma} below
allows us to reduce to the case where $\ind(D)$ is a prime power.
Our arguments then  
divide into two cases depending on whether   
 $\ind(D)$ is a power of
$\text{char}(F)$.  Lemma~\ref{cycliclemma} and its
Corollary handle the first case:

\begin{lemma} \label{primarydecomplemma}
Let $D_1, \ldots, D_k$ be division algebras with center $F$ such that
${\text{gcd}(\ind(D_i),\ind(D_j)) = 1}$ whenever $i\ne j$.Then,
$$
\NK(D_1\otimes_F D_2\otimes_F \ldots\otimes_F  D_k)\ \cong\ 
\NK(D_1)\times\ldots\times \NK(D_k)\,.
$$
\end{lemma}

\begin{proof}  It suffices by induction to prove the result for 
$k = 2$.  This can be done the same way as the corresponding result for 
$\CK$ was proved in \cite{sk1}, Th.~2.8.  
\end{proof}

\smallskip

\begin{lemma} \label{cycliclemma}
Let $D$ be a noncommutative division algebra similar to a  cyclic
algebra $A$. In each of the following cases $\NK(D)$ is nontrivial:
\begin{enumerate}
\item $F$ contains a square root of $-1$;
 \item The characteristic
of $F$ is $2$;
 \item The degree of $A$ is odd.
\end{enumerate}
\end{lemma}

\begin{proof}
Since the primary components of $D$ are similar to tensor powers of $A$
which are similar to cyclic algebras, 
 it suffices by Lemma~\ref{primarydecomplemma} to consider the case when $\ind(D)$ is a
power of a prime. Thus, assume $\ind(D) = p^c$, where $p$ is a prime
number and $c\ge 1$ and $D$ is similar to a cyclic algebra $A =
(E/F, \sigma, a)$. Choose $A$ of minimal degree. Then,
$\text{deg}(A) = p^e$ and   $a\notin 
F^{*p}$; for, if $a = b^p$, then $A$ is Brauer equivalent to
$(E_0/F, \sigma, b)$, where $[E : E_0] = p$, contradicting the
minimality of $\text{deg}(A)$.

% We shall show that if $F$ contains a square root of $-1$ or $p$ is
% odd or if $p =2$ and $\text{char}(F) = 2$, then $\NK(D)$ is
% nontrivial.

 Let $d = p^e = \text{deg}(A)$.  Let $\alpha$ be the standard generator of $A$ with
$\alpha^d = a$.  Since the powers of $\alpha$ up to the $d$-th are
part of a base of $A$ over $F$, they are $F$-linearly independent.
Therefore, the minimal polynomial of $\alpha$ over $F$ is $x^d-a$.
If $M$ is any splitting field of $A$, then  for $\alpha \tensor 1 \in A
\tensor_F M$, the minimal polynomial of $\alpha \tensor 1$ over
$M$ is again $x^d - a$. Since the characteristic polynomial of
$\alpha \tensor 1$ has degree $d$, this polynomial is also
$x^d-a$. Hence, $\Nrd_A(\alpha) = \det(\alpha \tensor 1) =
(-1)^{d-1} a$. 
% (Notice that, it is irrelevant whether $x^d-a$ is
% irreducible.) 
So, if $p$ is odd, or $\mychar(F) = 2$, or $F$
contains a square root of $-1$, then $\Nrd_A(\alpha) \notin
F^{*p}$. But thanks to the Dieudonn\'e determinant, $\Nrd_A(A^*)
= \Nrd_D(D^*)$. Thus $\Nrd_A(\alpha)\in\Nrd_D(D^*)\backslash F^{*p^e}$, so
$\NK(D)$ is nontrivial.
\end{proof}

Recall that a $p$-algebra is a central simple algebra of degree a power
of the prime $p$ over a field of characteristic $p$.
Albert's main theorem in the theory of $p$-algebras states that
every $p$-algebra is similar to a cyclic $p$-algebra (see
\cite{albert}, p.~109, Th.~31). Combining this with the 
Lemma above, we obtain:

\begin{cor} \label{palgebra}
Let $D$ be a noncommutative $p$-division algebra. Then $\NK(D)$ is nontrivial.
\end{cor}

\begin{remark}
Let $G(D) \ = \ {D^*}\big/{\Nrd(D^*)D'}$, which is a bigger group than 
$\CK(D)$ in general.
 It is much easier to see that $G(D) \ne 1$ for every noncommutative 
%% the
%% assumption $G(D)=1$ leads to a contradiction for 
$p$-division algebra~$D$.
Indeed, if $G(D)=1$ then $\Nrd(D^*)=\Nrd(D)^{*p^n}$ where
$\ind(D)=p^n$. So, for $F = Z(D)$,
$$
F^{*p^n} \subseteq\  \Nrd(D^*) \ =\  \Nrd(D)^{*p^n} =\ \Nrd(D)^{*p^{2n}}
\subseteq  \ F^{*p^{2n}}. 
$$
Hence,  ${F^*}^{p^n}={F^*}^{p^{2n}}$. Since
$\text{char}(F)=p$, this implies
$F^*={F^*}^p$, i.e., $F$ has no proper purely inseparable extensions.
But one knows  by  \cite{albert}, p.~104, Th.~21, that any $p$-algebra has a purely inseparable
splitting field; hence, $D = F$, a contradiction.
%  which is a contradiction to $F$ being perfect 
% andtherefore $G(D) \not =1$ 
(Compare this argument with \cite{kesh},~Th.~2).
\end{remark}
%\\


In order to prove the main Theorem, we need two propositions.


\begin{proposition}\label{normprop}
Let $F \subseteq L$ be fields with $\dimcol L F=d < \infty$ such that
$N_{L/F}(L^*)=F^{*d}$. If $V$~is a discrete valuation ring of $F$
with residue field $\ol V$, and if the integral closure of $V$ in
$L$ is a finite $V$-module, 
% or $L$ is separable extension of $F$,
then $V$ has a unique extension to a DVR $W$ of $L$, and 
$\dimcol{\ol W}{
\ol V}=\dimcol L F$. 
\end{proposition}

\begin{proof}
Let $v$ be the normalized discrete valuation on $F$ corresponding to the
valuation ring $V$. \lq\lq Normalized" means that the value group
$v(F^*) = \mathbb Z$.
Let $v_1,\ldots,v_s$ be all the 
(inequivalent) extensions of $v$ to $L$. Since $v$ is discrete and
% either 
the integral closure of $V$ in $L$ is a finite $V$\nobreakdash-module,
%or $L/F$ is separable, 
we have 
% it follows that 
$\sum_{i=1}^s e_if_i=\dimcol L F$, 
where $e_i$ 
is the ramification index of $v_i/v$ and $f_i$ is the residue degree
of $v_i/v$ (\cite{bourbaki}, VI, \S 8.3, Cor.~3).
Since $v_i$ extends $v$, the value group of $v_i$ is 
$\frac 1{e_i}\mathbb Z$.
% Let $m_i=e_i f_i$. 
For any $x\in L$, we have 
\begin{equation}\label{formula}
v(N_{L/F}(x))=\textstyle\sum\limits_{i=1}^s e_if_i v_i(x), 
\end{equation}
by, \cite{bourbaki}, VI, \S 8.5, Cor.~2.
% where $x \in L$ and
% $\sum_{i=1}^s m_i=\dimcol L F$.   Here $m_i=e_i f_i$ where $e_i$ 
% is the ramification index of $v_i/v$ and $f_i$ is the residue degree
% of $v_i/v$ (\cite{bourbaki}, VI, \S 8.3, Cor.~3). 
Now 
% if $s\geq 2$, then 
by the Approximation Theorem (\cite{bourbaki}, VI,
\S 7.2, Cor.~~1) one can choose $x\in L$ such that $v_1(x)=1/e_1$
and $v_i(x)=0$ for all $i >1$. Thus by (\ref{formula}),
$v(N_{L/F}(x))=f_1$. But since $N_{L/F}(x) \in F^{*d}$, we must have 
$d\mid v(N_{L/F}(x)) = f_1 \le d$.  This, combined with 
$\sum_{i=1}^s e_if_i= d$ with all $e_i\ge 1$ and $f_i\ge 1$, forces 
$f_1 = d=\dimcol LF$ and $s = 1$, as desired.
%% 
%% But $m_1 <d$ so $N_{L/F}(x) \not \in
%% \mul{F}^d$ which is a contradiction. Thus $s=1$ and $v$ extends
%% uniquely to $L$. On the other hand if the extension is not
%% ramified (i.e., $e_1 \not =1$ ), then $v(N_{L/F}(x))=e_1 f_1
%% v_1(x)$ where $e_1\not = 1$. Choosing $x \in L$ such that
%% $v_1(x)=1/e_1$ implies that $N_{L/F}(x) \not \in \mul{F}^d$ which
%% is again a contradiction. This finishes the proof.
\end{proof}

Using Proposition \ref{normprop}, we obtain the following Theorem
which provides a further class of fields with the NKNT property 
which is not
covered by the main Theorem \ref{thm1}. For example, it shows that
if $F$ is NKNT, then so is the Laurent power series field
$F((x))$.


\begin{theorem}\label{my}
Let $F$ be a discrete valued field with  residue field
$\overline F$ such that $\mychar{(F)}=\mychar {(\overline F)}$. If
$\overline F$ is {\rm NKNT}, then so is $F$.
\end{theorem}

\begin{proof} Suppose there is a   noncommutative division algebra 
~$D$ finite dimensional over~$F$ with
center~$K$ such that $\NK(D) = 1$. 
% We shall show that $\NK(D)$ is nontrivial. 
We can
assume $K=F$. Since by Lemma~\ref{primarydecomplemma},
$\NK$ respects the primary decomposition of
$D$, it is enough to consider the case when 
 $\ind(D)=p^k$, where $p$ is prime and $k\ge 1$. 
If $\mychar(F)=p$, then $D$ is a $p$-algebra and by
Corollary \ref{palgebra}, $\NK(D)$ is nontrivial. Thus we may assume
that $\text{char}(F) \ne p$. 
Hence, every subfield of $D$ containing $F$ is separable over~$F$.

%% We will show that, if there is a division algebra $D$ with
%% $\NK(D)$ trivial, then this forces the discrete valuation $v$ of
%% $F$ has a unique extension to any subfield of $D$ and hence an
%% extension to $D$. Then we observe that $\NK(\overline D)$ is
%% trivial which is a contradiction to the assumption of $\overline
%% F$ being NKNT.
 

% Assume that there is a $F$-division algebra $D$ such
% that $\NK(D)=1$.
 Let $d = p^k = \ind(D)$. Since  $\NK(D)=1$, we have 
 $N_{L/F}(L^*)=\mul{F}^d$ for every maximal subfield $L$
of $D$. Since $L$ is separable over $F$, the integral closure in $L$
of the discrete valuation ring $V_F$ of $v$ 
on $F$ is a finitely generated $V_F$-module. 
Thus by Proposition \ref{normprop}, $v$~extends uniquely to
any maximal subfield of $D$, with no ramification. So, $v$ extends
uniquely to any subfield of $D$. By the theorem of
Ershov-Wadsworth (see \cite{wadval}, Th.~2.1 or \cite{wad87}),
it follows that $v$ extends to a valuation on $D$, which is
denoted again by $v$. Furthermore $D$  is not ramified over $F$, i.e. the
value group $\Gamma_D$ of D coincides with the value
group $\Gamma_F$ of $F$. Let $\ol D$ and $\ol F$ be
the residue division algebra and the residue field of the
valuations on $D$ and~$F$. Since $\text{char}(\ol F)=\text{char}(F)$ 
does not
divide $\ind(D)$, the Ostrowski theorem for valued division algebras,
\cite{morandi} Th.~3, yields $\dimcol D F = 
\dimcol {\ol D}{\ol F}\, \big|\Gamma_D\!:\!\Gamma_F\big|
=\dimcol {\ol D}{\ol F} $.
% . This together with $v$ having
% ramification index $1$ (i.e., $\Gamma_D=\Gamma_F$) give
% $[D:F]=[\ol D: \ol F]$ (see \cite{wadval}, (2.10) or
% \cite{morandi}).
Note also that $\dimcol {Z(\ol D)}{\ol F} \,\big|\,\dimcol D F$; hence,
$Z(\ol D)$ is separable over $\ol F$.  
Thus, the surjectivity of the   fundamental homomorphism
$\Gamma_D / \Gamma_F \rightarrow \Gal(Z(\ol D)/\ol F)$, together
with the fact that $Z(\ol D)$ is normal and separable over $\ol F$ and
$\Gamma_D/\Gamma_F=1$ force that $Z(\ol D)=\ol F$ (see
\cite{wadval}, Prop.~ 2.5 or \cite{jacob}, Prop.~1.7). Hence, $\ind(\ol
D)=\ind(D)$.

Since $\NK(\ol D) \ne 1$ and $\ind(\ol D) = \ind (D) = d$, there is 
$\ol a\in \ol D^*$ with $\Nrd_{\ol D} (\ol a) \notin \ol F^d$.
Let $a$~be any inverse image of $\ol a$ in the valuation ring
$V_D$ of $D$, and let $L$ be any maximal subfield of $D$ 
containing $a$.  Let $V_L$ be the valuation ring of the restriction of 
$v$ to $L$, and let 
$\ol L$ be the residue field of $V_L$. Because $V_L$ is the unique extension
of $V_F$ to $L$, $V_L$ is the integral closure of $V_F$ in $L$;  hence, it is
a finitely-generated $V_F$-module.
Since $\dimcol {\ol D}{\ol F} = \dimcol D F$, we must
have $\dimcol {\ol L}{\ol F} = \dimcol L F$, 
showing that  $\ol L$ is a maximal subfield of $\ol D$.
  If $\ol b_1, \ldots, 
\ol b_d$ are any $\ol F$-vector space base of $\ol L$, then any inverse images
$b_1, \ldots, b_d$ of the $\ol b_i$ in the valuation ring $V_L$ form a base of 
$V_L$ as a free $V_F$-module.  (The $b_i$ generate $V_L$ over $V_F$
by Nakayama's Lemma, and they are $V_F$-independent because 
$V_F$ is a valuation ring and the $\ol b_i$ are $\ol F$-independent.)
By computing the norm $N_{L/F}(a)$  as the determinant of 
the $F$-linear map multiplication by~$a$ using the base $b_1, \ldots,
b_d$, we obtain $N_{L/F}(a) \in V_F$ and 
$$
\ol{N_{L/F}(a)} = N_{\ol L/\ol F}(\ol a) \text { in } \ol F\,.
$$
Because we have assumed $\NK(D) = 1$, we have
$$
N_{L/F}(a) = \Nrd_D(a) \in F^{*d} \cap V_F = V_F^d\,.
$$
Hence, 
$$
\Nrd_{\ol D}(\ol a) = N_{\ol L/\ol F}(\ol a) = \ol {N_{L/F}(a)} \in 
\ol{V_F^d} = \ol F^d\,,
$$
contradicting the choice of $\ol a$.  So, $\NK(D) \ne 1$,
contradicting the choice of $D$.  Thus, $F$ is NKNT. 
\end{proof}

%  Now consider the central simple algebra $D\otimes_F F^h$
% where $F^h$ is the Henselization of $F$. By the theorem of Morandi
% (\cite{wadval}, Theorem 2.3 or \cite{morandi}) $D\otimes_F F^h$ is
% a division algebra and $\overline{D\otimes_F F^h}=\ol D$. Thus the
% valued division algebra $D\otimes_F F^h$ is unramified. Now the
% formula relating the reduced norm of a division algebra to the
% reduced norm of its residue division algebra over a henselian
% field (see \cite{wadval} (6.17) or \cite{ershov}) reduces to the
% simple form of
% $$\overline{\Nrd_{D\otimes_F F^h}(a)}=\Nrd_{\overline{D\otimes_F
% F^h}}(\ol a).$$

% Therefore
% \begin{equation}\label{nrd}
% \overline{\Nrd_{D}(a)}= \overline{\Nrd_{D\otimes_F F^h}(a\otimes
% 1)}=\Nrd_{\overline{D\otimes_F F^h}}(\overline{a\otimes
% 1})=\Nrd_{\overline{D}}(\ol a).
% \end{equation}

% Since $\NK(D)=1$, $\Nrd_{D}(U_D)=U_F ^d$ where $U_D$ and $U_F$ are
% the units of the valuations (i.e., $U_D=\{x\in D^*~ |~ v(x)=0\}$).
% Thus by (\ref{nrd})
% $$\Nrd_{\ol D}(\ol D ^*)=\overline{\Nrd_{D}(U_D)}=\overline{U_F
% ^d}=\mul{\overline{F}}^d.$$

% So $\NK(\ol D)=1$. But by the assumption $\ol F$ is NKNT, hence a
% contradiction. Thus $\NK(D)$ is nontrivial and this finishes the
% proof of the theorem.
% \end{proof}


%\begin{remark} The proof of the theorem also shows the following. Let $F$ be a
%field such that any division algebra finite dimensional over $F$
%is either commutative or a tensor product of cyclic algebras. Then
%if the center of a noncommutative division algebra $D$ is finitely
%generated over $F$ and $D$ is not a quaternion then $\NK(D)\not =
%1$.
%\end{remark}

% \begin{remark}
% One does not need to invoke the Henselisation technique at the
% last part of the proof if one only wants to prove that $\CK(D)$
% (rather than $\NK(D)$) is notrivial. For, assuming $\CK(\ol D)$
% is nontrivial then there is $\ol a \in \ol D$ such that the order
% of $\ol a$ in $\CK(\ol D)$ is not 1. This implies that the order
% of $a$ in $\CK(D)$ is not one, which is a contradiction with the
% assumption that $\CK(D)=1$.
% \end{remark}





\begin{prop} \label{simpleext}
Let $F \subseteq F(t) \subseteq L$ be fields with $t$
transcendental over $F$ and $\dimcol L{F(t)}< \infty$. If
$L=F(t)(\alpha)$ for some $\alpha$, then there is a discrete
valuation ring $V$ of $F(t)$ with $F \subseteq V$ such that $V$
has an extension to a DVR $W$ of $L$ such that $\ol W= \ol V$. $($In
fact, there are infinitely many such $V$.$)$
\end{prop}

%\noindent {\it Proof of Proposition \ref{simpleext}.} 
\begin{proof}
Let
$R=F[t]$. We can assume that $\alpha$ is integral over $R$. Let
$f=x^n+c_{n-1}x^{n-1}+\ldots +c_0$ be the minimal polynomial of
$\alpha$ over $F(t)$. The integrality of $\alpha$ over $R$ (with
$R$ integrally closed) assures that $f \in R[x]$. Let
\begin{eqnarray}\nonumber
\mathcal P_f\ =\ \{ \pi \in F[t]~ |~ \pi \text{~is irreducible and monic in~}
F[t]
%% \\
\text{~and~} \pi | f(r) \text{~for some~} r \in R \}.\nonumber
\end{eqnarray}
We will  show that $|\mathcal P_f|= \infty$. Assume first that 
$c_0=1$, and
write $f=x h(x)+1$ with $h \in R[x]$ and $\text{deg}(h)=n-1$.
Suppose $|\mathcal P_f|=\{ \pi_1,\ldots,\pi_k \}$. Let $s= t\pi_1 \ldots
\pi_k$. Since $h$ has only finitely many roots in $R$, there is a
natural number $\ell$ with $h(s^\ell) \not =0$. Then
$f(s^\ell)=s^\ell h(s^\ell)+1$ has positive degree in $t$, so is not a unit
of $R$. If $p$ is an irreducible monic factor of $f(s^\ell)$, then $p
\in \mathcal P_f$, but $p \nmid s$, so $p \not \in \{ \pi_1,\ldots,\pi_k
\}$, a contradiction. Hence $\mathcal P_f$ cannot be finite if $c_0 = 1$.

Now assume $c_0\not =1$. Let $f(c_0 x)=c_0 g(x)$. So $g \in R[x]$
with $\text{deg}(g)=\text{deg}(f) \geq 1$, and $g$ has constant
term $1$. By the previous case, $|\mathcal P_g|=\infty$. But since$f(c_0
r)=c_0g(r)$, we have $\mathcal P_g \subseteq \mathcal P_f$. 
So, $|\mathcal P_f|=\infty$, as
claimed.

Now, take  any $\pi \in P_f$, and let $V$ be the DVR
$R_{(\pi)}$ which is the  localization of $R$ at its prime ideal $(\pi)$.
 Let $M=\pi
V$, which is the maximal ideal of $V$; so $\ol V = V/M$. Assume
first that the ring $V[\alpha]$ is integrally closed.

Since $\pi | f(r)$ for some $r \in R$, the image $\ol f$ of $f$ in
$\ol V[x]$ has a root $\ol r$ in $\ol V$. Note that $f F(t)[x]
\cap V[x]=f V[x]$, by the division algorithm as $f$ is monic in
$V[x]$. Hence, $V[\alpha] \cong V[x]/f V[x]$ and
\begin{equation}\label{star}
% \frac{V[\alpha]}{MV[\alpha]} \cong \frac{V[x]}{(f,M)} \cong
% \frac{\ol V[x]}{(\ol f)}.
{V[\alpha]}\big/{MV[\alpha]} \ \cong \ {V[x]} \big/{(f,M)} 
\ \cong \ 
{\ol V[x]}\big/{(\ol f)}.
\end{equation}
Because $\ol f(\ol r)=0, x-\ol r$ is an irreducible factor of $\ol
f$ in $\ol V[x]$. Let $N$ be the maximal ideal of $V[\alpha]$
containing $MV[\alpha]$ corresponding to $(x-\ol r)/(f)$ in $\ol
V[x]/(\ol f)$ in the isomorphism given by (\ref{star}). Let $W$ be
the localization $V[\alpha]_N$. Then $W$ is a DVR, as $V[\alpha]$
is the integral closure of $V$ in $L$. Furthermore, $W \cap F(t) =
V$ and $\ol W \cong V[\alpha]/N \cong \ol V[x]/(x-\ol r) \cong \ol
V$.  Thus, the desired $W$ exists for $V = R_{(\pi)}$ whenever 
$\pi \in \mathcal P_f$ and $R_{(\pi)}[\alpha]$ is integrally closed.
 
To complete the proof we show that the 
needed integral closure property of $R_{(\pi)}[\alpha]$ occurs 
for all but finitely many $\pi\in \mathcal P_f$. 
Let $T$ be the integral closure of $R$ in $L$; so $T$ is a
finitely generated $R$-module (\cite{bourbaki}, V, \S 3.2, Th.~2).  
We have $R[\alpha] \subseteq T$,
and $T$ and $R[\alpha]$ each have  quotient field $L$. So,
$T/R[\alpha]$ is a finitely generated torsion $R$-module; hence it
has nonzero annihilator in $R$. Therefore, there is $b \in R$ with
$b \not = 0$ and $b T \subseteq R[\alpha]$. Hence,
$R[\alpha][1/b]=T[1/b]$, which is integrally closed. For any monic
irreducible $\pi \in R$, if $\pi \nmid b$ then the DVR $R_{(\pi)}$
is a localization of $R[1/b]$. Hence, $R_{(\pi)}[\alpha]$ is a
localization of $R[1/b][\alpha]$, so $R_{(\pi)}[\alpha]$ is
integrally closed. There are only finitely many monic irreducibles of
 $R$ dividing $b$.  For all other $\pi$ in the infinite set 
$\mathcal P_f$, we have $R_{(\pi)}(\alpha)$ is integrally closed.
% Since $|\mathcal P_f|=\infty$, we can choose $\pi \in
% \mathcal P_f$ with $\pi \nmid b$. Then the preceding argument shows that
% $V=R_{(\pi)}$ has the desired properties.
\end{proof} 
%% $\Box$

\begin{remark}
For the result of Prop.~\ref{simpleext}, it is not sufficient to assume that 
$\dimcol L{F(t)} <\infty$.  For example, suppose $\mychar(F) = p\ne 0$
and ${\dimcol{F^{1/p}}F\ge p^2}$.  Take any field $K$ with 
$F \subseteq K\subseteq F^{1/p}$ and $p^2 \le \dimcol K F <\infty$,
and let $L = K(t)$.  Take any discrete valuation ring V of $F(t)$ and 
any extension of $V$ to a DVR $W$ of $L$.  Identify $\ol V$ and $K$ 
with their canonical images in $\ol W$.  Since $\ol V = F(\beta)$ for 
some $\beta$, the Theorem of the Primitive Element shows that 
$\ol V\cap K = F(\gamma)$, for some $\gamma\in K$.  Since 
$\gamma^p \in F$, we have $\dimcol {F(\gamma)}F \le p < \dimcol KF$, 
so $\ol V$ doesn't contain all of $K$.  Because $K\subseteq \ol W$,
this shows that $\ol W \ne \ol V$.
\end{remark}





\begin{theorem}\label{thm1}
Let $F$ be a field which is finitely generated but not algebraic
over some subfield~$F_0$. Then, $F$ is NKNT.
\end{theorem}

\begin{proof}
We need to show that for each finite degree extension field $K$ of
$F$ and each noncommutative finite dimensional division algebra
$D$ with center $K$, we have $\NK(D)$ is nontrivial. As in the
proof of Theorem \ref{my}, we can assume that $K=F$ and that $F$
is a finite degree extension of $F_0(t)$, with $t$ transcendental
over $F_0$. Since $\NK$ respects the primary decomposition of $D$,
by Corollary \ref{palgebra} it suffices to consider the
case where $\ind(D)=p^k$, where $p$ is a prime number with $p \not =
\text{char}(F_0)$.

Let $L$ be any maximal subfield of $D$ and let $S$ be the separable
closure of $F_0(t)$ in $L$. Then, $S=F_0(t)(\alpha)$ for some
$\alpha$. By Proposition \ref{simpleext}, applied to the field
extension $F_0(t) \subseteq S$, there is a DVR $V$ of $F_0(t)$
(with $F_0 \subseteq V$) which has an extension to a DVR $W$ of
$S$ with $\ol V =\ol W$. Because $L$ is purely inseparable over
$S$, $W$ has a unique extension to a DVR $Y$ of~$L$, and $\ol Y$
is purely inseparable over $\ol W$. Let $Z=Y \cap F$, which is a
DVR of $F$. Since $\ol W=\ol V \subseteq \ol Z \subseteq \ol Y$,
we have $\ol Y$ is purely inseparable over $\ol Z$. If
$\text{char}(F_0)=0$, it follows that $\ol Y=\ol Z$; hence $[\ol Y
:\ol Z]=1 \not = p^k = [L:F]$. If $\text{char}(F_0)=q \not = 0$,
then $[\ol Y : \ol Z]=q^\ell$ for some $\ell \geq 0$. Since $q \not = p$
by hypothesis, we again have $[\ol Y :\ol Z] \not =[L:F]$.

Let $V_F$ (resp. $V_L$) be the integral closure of $V$ in $F$
(resp.~ $L$), and let $Z_L$ be the integral closure of $Z$ in $L$.
Because the integral closure of $F_0[t]$ in $F$ (resp.~ in $L$) is a
finitely generated $F_0[t]$-module,  by 
\cite{bourbaki}, V, \S 3.2, Th.~2, and $V$ is a localization of
$F_0[t]$ (or $F_0[t^{-1}]$), $V_F$ and $V_L$ are finitely
generated $V$-modules, so $V_L$ is a finitely generated
$V_F$-module. Then, as $Z$ is a localization of $V_F$, $Z_L$ is a
finitely generated $Z$-module. Since the conclusion of Proposition
\ref{normprop} fails for $Z \subseteq Y$ in the field extension $F
\subseteq L$, we have ${F^*}^{p^k} \subsetneqq N_{L/K}(L^*)
\subseteq \Nrd(D^*)$, showing that $\NK(D)$ is nontrivial.
\end{proof}

\begin{cor}
If $D$ is a noncommutative division algebra whose center is 
finitely generated over its prime field or over an algebraically 
closed field, then $\NK (D) \ne 1$.  Hence, ${\CK (D) \ne 1}$ 
and $D^*$ contains a maximal proper normal subgroup.
\end{cor} 

\begin{proof} This is immediate from the Theorem and the comments 
in the introduction.
\end{proof}

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\end{document}

