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%%\fancyhead[CE]{{\footnotesize MICHAEL ADAM}}
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%%\fancyhead[C]{\scshape Seminar "uber Algebra}
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%%  {\footnotesize Prof. Dr. Ina Kersten\\ Dipl.-Math. Michael Adam}}
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\title{On Wedderburn's theorem about finite division algebras}

\subjclass[2000]{16K20, 12E20; 16K50}
\keywords{finite dimensional division algebras, finite fields, Brauer groups}

\author{Michael Adam}
\address{Michael Adam, Mathematisches Institut\\ 
  Universit\"at G\"ottingen\\
  Bunsenstra\ss e 3--5\\
  D-37073 G\"ottingen\\ Germany}
  \email{mad@uni-math.gwdg.de}

\author{Birte Julia Mutschler}
\address{Birte Julia Mutschler, Mathematisches Institut\\ 
  Universit\"at G\"ottingen\\
  Bunsenstra\ss e 3--5\\
  D-37073 G\"ottingen\\ Germany}
\email{Birte.Julia@web.de}

%\date{\today}
\date{23 April, 2003}

\begin{abstract}
  Wedderburn's first proof of his theorem on finite division algebras
  contains a gap. We analyse the gap, give a variant of
  Wedderburn's proof that goes completely without the gap-producing
  statement, and we show how to fill the gap in a way
  Wedderburn could probably have done it.
\end{abstract}

\maketitle



\numberwithin{equation}{section}



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\tableofcontents


%%
%% -----------------------------------------------------
%%



\section{Introduction}


In the famous paper ``A theorem on finite
algebras''~\cite{Wedderburn:finite-algebras} from the year 1905,
Wedderburn\footnote{Why is he called ``Joseph H.\ M.\ Wedderburn''
  nowadays while according to his papers, his name appears to have
  been ``J.\ H.\ Maclagan-Wedderburn''?} first stated his theorem that
any finite division algebra is commutative and he gave three different
proofs for it.
%%
In 1927, Emil Artin~\cite{Artin:ueber-Wedderburn} remarked that
Wedderburn's first proof is not valid. Artin did however not explicate
what the gap or flaw is, not even whether it is a gap that can be
closed or a serious error that vitiates the whole argument. Instead,
he gave a new proof in the spirit of Wedderburn's original ideas.
%%
Karen H.~Parshall, in her article~\cite{Parshall:pursuit} about the
history of Wedderburn's finite division algebra theorem, also
discussed the first proof. She explained that it is in fact a gap that
can be filled today (but she did not say how) and judged that this
would probably have been beyond Wedderburn's scope.


In this paper, we first describe and analyse the gap.  
%%
Then we present a variant of Wedderburn's first proof which makes no
use at all of the statement that produced the gap. 
%%
We also demonstrate how to fill the gap in a manner that Wedderburn
might have done had he been aware of it.
%%
Finally, we give a (hopefully rather complete) chronological list of
proofs of Wedderburn's theorem in literature.


%%\subsection*{Acknowledgements}
\medskip\noindent
\textbf{Acknowledgements. }
%%
We would like to express our gratitude to Ina Kersten for her kind
support and encouragement.
%%
We further thank Keith Dennis for pointing out Kasch's proof of 
Wedderburn's theorem to us.


%%
%% -----------------------------------------------------
%%

\section{The gap}

We first describe the gap in Wedderburn's proof. This incorporates a
good amount of interpretation, for what Wedderburn writes down in the
decisive section is rather vague and misleading at first sight,
which is probably due to the fact that the strict modern language and
perception of algebra was just emerging at that time. We therefore go
into this in rather detail in the beginning.

Wedderburn considers a finite division ring $A$. It is an algebra of finite
dimension $s$ over its center $\F_{p^m} = GF[p^m]$, where $p$ is a prime
number.  He chooses a basis $x_1,\dots,x_s$ of $A$ over $\F_{p^m}$ and hence
for each $x\in A$ a representation $x=\sum_{i=1}^s \xi_i x_i$ with
coefficients $\xi_i \in \F_{p^m}$ (called the co\"{o}rdinates of $x$ by
Wedderburn).  The questionable section in Wedderburn's article (pages 350
and 351 in \cite{Wedderburn:finite-algebras}) 
%%reads
starts as follows\footnote{Like Parshall, we have corrected a misprint in the
quotation: The constant coefficient in equation $(2)$ was called $a_{r-1}$
instead of $a_r$.}:
%%
\begin{quote}
  It follows from the theory of hypercomplex numbers, that there is an
  equation of lowest degree,
  %%
  \[ (2) \text{\ \ \ \ \ } f(x) \equiv x^r + a_1
  x^{r-1} + a_2 x^{r-2} + \dots + a_r = 0,
  \]
  %%
  with coefficients in $GF[p^m]$, which is satisfied identically by any
  given number $x$ of the algebra, irrespective of any special relation
  between the co\"{o}rdinates of $x$, except the condition that they lie in
  $GF[p^m]$. Further, there is at least one element of the algebra which
  satisfies no similar equation of lower degree.
\end{quote}
%%
At first sight, Wedderburn really seems to consider the normed
polynomial $f\in \F_{p^m}[X]$ of lowest degree which is satisfied by
each element of the algebra\footnote{Note that such a polynomial
  exists if and only if the algebra is finite.} and then to claim that
it is the minimal polynomial of some specific element $x$ of the
algebra. This is of course impossible since under these circumstances,
$f$ would be on the one hand irreducible, and would on the other hand
contain a linear factor $(X-a)$ for each $a\in \F_{p^m}$. So in an
attempt to rescue the statement one could consider the polynomial
which is satisfied only by all $a\in A\setminus \F_{p^m}$. But this
$f$ cannot be the minimal polynomial of an element of the algebra
either, which can be seen as
%%
  follows\footnote{A word of warning: One cannot simply divide $f$ by each
  linear factor $(X-a)$, concluding that $f$ would have degree at least
  $|A\setminus\F_{p^m}| = p^{ms}-p^m$, while a minimal polynomial of an
  element $a\in A$ can have degree at most $[A:\F_{p^m}]=s$. The reason is
  that evaluation of polynomials with non-commutative coefficients is not
  compatible with multiplication, and this leads to strange phenomena. Here
  is an example of a polynomial of degree $2$ which has at least $6$
  different roots:
  %%
  Let $D= F \oplus F u \oplus F v \oplus F w$ be the quaternion algebra over
  some field $F$, i.e.\ $u^2=v^2=w^2=-1$ and $uv=w$. Then the polynomial
  $X^2+1$ has $\pm u$, $\pm v$ and $\pm w$ as roots.}:
%%
For each root $a$, the polynomial $f$ can be divided (from the right)
by the linear polynomial $(X-a)$. In general, if an irreducible
polynomial $f \in \F_{p^m}[X]$ is divisible in $A[X]$ from the right
by two linear factors $(X-a)$ and $(X-b)$, then $a$ and $b$ are
conjugate in $A$ (cf.\ Artin~\cite{Artin:ueber-Wedderburn}). Thus,
because each element of $A\setminus \F_{p^m}$ is a root of $f$, it
follows that $A\setminus \F_{p^m}$ would consist of one single
conjugacy class. So we would obtain for any $a\in A\setminus \F_{p^m}$
with centralizer $Z(a)$
\[
p^{ms} - p^m 
=  | A\setminus \F_{p^m} | 
= |A^*| / |Z(a)^*| 
= (p^{ms} -1) / (p^{mt} -1)
\]
where $[Z(a):\F_{p^m}]=t$. In any case this would yield the contradiction
that $p$ divides one.
%%
But even if there was such an element with $f$ as minimal polynomial,
this polynomial $f$ would not fulfill the requirements for the rest of
the proof.  See section~\ref{sec:wedderburns-proof}.


This indicates that Wedderburn must have meant something else than he
apparently wrote down. A first hint is given by the argument for the
claim cited above:
%%
\begin{quote}
  Indeed, $(2)$ states that $x^{r-1}, x^{r-2}, \dots, x^0$ are linearely
  independent with respect to $GF[p^m]$, and the condition of independence
  can evidently be put in a form which states that certain determinants,
  whose elements are rational integral functions of the co\"{o}rdinates of
  $x$, do not all vanish identically. Hence there must be some set of values
  of the co\"{o}rdinates for which $x^{r-1}, x^{r-2}, \dots, x^0$ are
  independent and hence the particular $x$ so obtained can satisfy no
  equation of lower degree than $r$. $(2)$ is called the characteristic or
  identical equation, while the equation of lowest degree satisfied by a
  given $x$ is called its reduced equation.
\end{quote}
%%
Furthermore, comparing with the section ``The identical equation'' of
his article ``On hypercomplex
numbers''~\cite{Wedderburn:hypercomplex-numbers} from 1907, one
sees\footnote{Parshall~\cite[p.~285]{Parshall:pursuit} also made
  this---to our judgement non-obvious---interpretation but gave no
  hint as to how she arrived at this viewpoint.} that Wedderburn must
actually be talking about the \emph{generic minimal polynomial}
$M_{X,A}=M_X$ of $A$ (using modern terminology): Let $X_1,\dots,X_s$
be algebraically independent over $\F_{p^m}$. The \emph{generic element}
of $A$ is defined to be
\[
X  = \sum_{i=1}^s X_i \cdot x_i \ \in\  
     \F_{p^m}(X_1,\dots,X_s)\tensor_{\F_{p^m}} A \text{.}
\]
Its minimal polynomial $M_{X,A}$ over the function field
$\F_{p^m}(X_1,\dots,X_s)$ is called the generic minimal polynomial of
$A$.
%%
We will see in section~\ref{sec:wedderburns-proof} that this
polynomial does the job in the rest of the proof, which leaves no
doubt as to which polynomial Wedderburn was talking about.
%%
The coefficients of the generic minimal polynomial are in fact
polynomials in the $X_i$, and so we can evaluate the coefficients of
$M_{X,A}$ at the coordinates $\xi_i$ of an element $x = \sum_i \xi_i
x_i$ of $A$, thus obtaining a polynomial $M_{x,A} \in \F_{p^m}[T]$,
called \emph{the generic minimal polynomial of $A$ evaluated at $x$}.
It has the property that $M_{x,A}(x) =0$.
%%
In this terminology, Wedderburn's claim reads as follows:
%%
\begin{claim}\label{claim}
  For a finite dimensional central division algebra $A$ over
  $\F_{p^m}$, there exists an element $x\in A$ such that $M_{x,A}$
  equals the minimal polynomial of $x$ over $\F_{p^m}$.
\end{claim}
%%
Wedderburn argues as follows: Write $X^j = \sum_i
f_{ij}(X_1,\dots,X_s) x_i$ with $f_{ij} \in \F_{p^m}[X_1,\dots,X_s]$,
and let $M_{X,A}$ be of degree $r$. Then from the
$\F_{p^m}(X_1,\dots,X_s)$-linear independence of the powers $1, X,
X^2,\dots,X^{r-1}$ it follows that some $r$-minor of the matrix
$(f_{ij})_{ij}$ is a nonzero polynomial in $\F_{p^m}[X_1,\dots,X_s]$.
Now Wedderburn concludes without further reasoning that there are
elements in $\F_{p^m}$ for which this polynomial does not vanish.


Wedderburn really does not seem to have had any further argument for
this step in mind because two years later, he made the same statement
as claim~\ref{claim} for an arbitrary finite dimensional (associative)
algebra instead of a division algebra, saying it was obvious.
%%
While for a division algebra this can be proven, as shown in
section~\ref{sec:filling} below, it is false for an arbitrary algebra.
The following counterexample is tributetd to K.\ McCrimmon by Parshall
(\cite{Parshall:pursuit}, p.\ 285): Let $A=\F_{p^m}e_1 \oplus \dots
\oplus \F_{p^m}e_n$ with orthogonal idempotents $e_i$. Then the
generic minimal polynomial of $X$ has degree $n$ while each element of
the algebra satisfies the equation $x^{p^m} -x=0$. So one just has to
choose $n$ large enough to obtain an algebra for which the generic
minimal polynomial has degree strictly larger than any of the minimal
polynomials of elements of $A$.

\begin{conclusion}
  We consider the gap in Wedderburn's first proof to be the insufficient
  proof for statement~$\ref{claim}$. \bdende
\end{conclusion}



%%
%% -----------------------------------------------------
%%


\section{The generic minimal polynomial}\label{sec:generic-mipo}

In this section, we present some lemmas around the generic minimal
polynomial of a finite dimensional algebra, specifically a division
algebra. Note that even if the proofs are given in a modern language,
they use only methods that were available to Wedderburn.

Let us fix some notations. We consider finite dimensional algebras $A$
over some field $F$. These are always meant to be associative and to
have a unit. Then for an element $a\in A$ we denote the minmal
polynomial of $a$ over $F$ by $m_a$ or $m_{a/F}$ if the reference to
the ground field has to be made clear. Similarly, the characteristic
polynomial of $a$ in $A$ is denoted by $p_{a} = p_{a,A/F}$. If $A$ is
a division algebra, we have $p_{a,A/F} = m_{a/F}^\alpha$ for some
integer $\alpha$.

Let $x_1,\dots,x_s$ be a $F$-basis of $A$, and let $X=\sum X_i \cdot
x_i \in F(X_1,\dots,X_s)\tensor_F A$ be the \emph{generic element} of
$A$. We have already defined the generic minimal polynomial of $A$ as
the minimal polynomial $M_{X,A}:= m_{X/F(X_1,\dots,X_s)}$ of $X$. Now
we define the \emph{generic characteristic polynomial} of $A$ as the
characteristic polynomial $P_{X,A} := p_{X,F(X_1,\dots,X_s)\tensor A}$
of $X$. As with the generic minimal polynomial, the coefficients of
$P_{X,A}$ are polynomials in the $X_i$ and hence for any $a = \sum a_i
x_i \in A$, we can \emph{evaluate} (the coefficients of) $P_{X,A}$ at
$a$ to obtain $P_{a,A}\in F[T]$. We always have $P_{a,A} = p_{a,A/F}$.
In particular, $\deg P_{X,A} = [A:F]$.

\medskip
\begin{lemma}\label{lem:deg-gen-mipo-smaller}
  Let $F$ be any field, and let $A$ be a finite dimensional
  non-commutative algebra over $F$ with $F\subset C(A)$, the center of
  $A$.  Then $\deg M_{X,A} < [A:F]$.
\end{lemma}

\begin{proof}
  Let $x_1,\dots,x_s$ be an $F$-basis for $A$, and let
  $L=F(X_1,\dots,X_s)$.  The $L$-subalgebra $L[X]$ of $L\tensor_F A$
  generated by the generic element $X = \sum_i X_i\cdot x_i$ is of
  dimension $\deg M_{X,A}$. Now $L\subset C(L\tensor_F A) = L\tensor_F
  C(A)$ since $F \subset C(A)$, and so $L[X]$ is commutative. Hence
  $L[X] \subsetneq L\tensor_F A$, because $L\tensor_F A$ is
  non-commutative.
\end{proof}


\begin{lemma}\label{lem:division-tensor-transzendent}
  Let $A$ be a finite dimensional division algebra over some field
  $F$. Let $T_1,\dots,T_n$ be indeterminates over $F$. Then
  $F(T_1,\dots,T_n)\tensor_F A$ is a division algebra.
\end{lemma}

This lemma is well known. We we reproduce the following proof from
Jacobson~\cite[p.\ 33]{Jacobson:division-algebras} in order to show
that the result was within Wedderburn's reach.

\begin{proof}
  The polynomial ring $A[T_1] = F[T_1]\tensor_F A$ is a domain. Hence
  by induction, $F[T_1,\dots,T_n] \tensor_F A = A[T_1,\dots,T_n]$ is a
  domain.  Since every element of $F(T_1,\dots,T_n)\tensor_F A$ can be
  written in the form $f\cdot g^{-1}$ with $f \in A[T_1,\dots,T_n]$
  and $g\in F[T_1,\dots,T_n]\setminus \{0\}$, it follows that
  $F(T_1,\dots,T_n)\tensor_F A$ is a domain, too. As a finite
  dimensional algebra over the field $F(T_1,\dots,T_n)$ it is thus a
  division algebra.
\end{proof}

\begin{corollary}
  For a finite dimensional division algebra $A$ over a field $F$, we
  have $P_{X,A} = M_{X,A}^\alpha$ for a suitable integer $\alpha$.
\end{corollary}

\begin{proof}
  $P_{X,A}$ and $M_{X,A}$ are, respectively, the characteristic and
  minimal polynomial of the same element $X$ of the division algebra
  $F(X_1\dots,X_s)\tensor_F A$.
\end{proof}

\begin{lemma}\label{lem:deg-mipo-divides-deg-generic-mipo}
  Let $F$ be any field, and let $A$ be a finite dimensional division
  algebra over $F$. Let $a\in A$. Then $M_{a,A}$ is a power of the
  minimal polynomial $m_{a/F}$ of $a$ over $F$. In particular, the
  degree of $m_a$ divides the degree of the generic minimal
  polynomial.
%%  $M_{X,A}$.
\end{lemma}

\begin{proof}
  By the above corollary, $P_{X,A} = M_{X,A}^\alpha$ for some integer
  $\alpha$. Hence also for any $a\in A$, we have $P_{a,A} =
  M_{a,A}^\alpha$. Now $P_{a,A} = p_{a,A/F} = m_{a/F}^\beta$ for some
  integer $\beta$. From the irreducibility of $m_a$, we conclude that
  $\beta$ is divisible by $\alpha$ and that $M_{a,A} = m_a^\gamma$ for
  $\gamma = \beta / \alpha$.
\end{proof}



%%
%% -----------------------------------------------------
%%




\section{A variant of Wedderburn's first proof}\label{sec:wedderburns-proof}

In this section, we present a variant of Wedderburn's first proof in a
slightly modernised language, which does not use the gap-producing
claim~\ref{claim} at all.


\begin{proof}[Variant of Wedderburn's proof]
  Let $A$ be a finite division ring. It is of some dimension $s$ over
  its center $C(A) \cong \F_{p^m} = \F_q$, where $p$ is a prime
  number. We divide the proof into two steps:
  \begin{enumerate}
  \item We choose representatives $a_1,\dots,a_t$ for the conjugacy
    classes of elements of $A \setminus \F_{p^m}$. Denoting $s_i :=
    [C(a_i):\F_q]$, we then have by the class equation
    \begin{equation*}
      q^{s} - 1 = |A^*| = |\F_{q}^*| + \sum_{i=1}^t \frac{|A^*|}{|C(a_i)|}
      =  q - 1 + \sum \frac{q^{s}-1}{q^{s_i}-1}
    \end{equation*}
    It follows, that $q-1$ is divisible by
    \begin{eqnarray*}
      \gcd\biggl(\frac{q^{s}-1}{q^{s_i}-1} : i=1,\dots t\biggr) 
      & = & \frac{q^{s}-1}{\lcm(q^{s_i}-1 : i=1,\dots,t)} \\
      & = & \frac{q^{s}-1}{q^{\lcm(s_1,\dots,s_t)}-1}
    \end{eqnarray*}
    Hence, denoting $s'=\lcm(s_1,\dots,s_t)$, we have 
    \begin{equation*}
    (q - 1) \cdot (q^{s'}-1) = l \cdot (q^{s}-1)
    \end{equation*}
    for some integer $l$. It follows that $l\equiv -1 \pmod {q}$,
    hence $l = k\cdot q - 1$ for some postive $k$.  Using $s'\leq s$,
    we see that $k=1$ and $s' = s$.
    
  \item Now we suppose by induction that all division rings of order
    strictly less than $|A|$ are commutative.  Then in particular, the
    centralizers $C(a)$ of elements $a\in A\setminus \F_q$ are
    commutative, so these are precisely the maximal subfields of $A$.
    %%
    Since $\F_{q}$ is perfect, each centralizer $C$ is, as a finite
    field extension of $\F_{q}$, generated by one element $a$, hence
    $[C:\F_{q}]$ equals the degree of the minimal polynomial of $a$.
    But by lemma~\ref{lem:deg-mipo-divides-deg-generic-mipo},
    the degree of the minimal polynomial always divides the degree $r$
    of the generic minimal polynomial.  So we have
    \begin{equation}\label{eq}
      s = \lcm\bigl([C(a):\F_{q}] \ : \ a\in A\setminus \F_q \bigr)
        \ \mid \  r \ \leq \ s 
    \end{equation}
    which implies $[A:F]=s=r=\deg M_{X,A}$. But by
    lemma~\ref{lem:deg-gen-mipo-smaller}, this yields $A=F$.
    \qedhere
 \end{enumerate}
    %%
\end{proof}




\begin{remark}
  Step one above is taken more or less directly from Wedderburn's
  proof. In the second step, Wedderburn argues that in addition by
  claim~\ref{claim}, there is one $b\in A\setminus \F_q$, for which
  $m_{b/\F_q} = M_{b,A}$, whence equation~\eqref{eq} simplifies to
  \begin{equation*}
    %%\label{eq}
    s = \lcm\bigl([C(a):\F_{q}] \ : \ a\in A\setminus \F_q \bigr)
      = [C(b):\F_q] = r \text{.}
  \end{equation*}
  Then he concludes that $C(b) = A$, contradicting $b\in A\setminus
  C(A)$.  \bdende
\end{remark}



%%
%% -----------------------------------------------------
%%


\section{Filling the gap}\label{sec:filling}

In this section we demonstrate how to fill the gap in Wedderburn's
proof. We do not rescue his insufficient argument but instead give a
new proof of claim~\ref{claim}.
%%
The proof is based upon the following three statements:
%%
\begin{lemma}\label{rettungs-lemma}
  Let $F$ be any field, and let $A$ be finite dimensional central
  division algebra over $F$.
  \begin{enumerate}
  \item\label{item:dim-square} 
    %%
    The dimension of $A$ over $F$ is a square.
    %%
  \item\label{item:max-subfield} 
    %%
    Each maximal subfield $L$ of $A$ has degree $[L:F] =
    \sqrt{[A:F]}$.
    %%
  \item\label{item:degree-generic-min-pol} 
    %%
    The degree of the generic minimal polynomial of $A$ is
    $\sqrt{[A:F]}$.
    %%
  \end{enumerate}
\end{lemma}

\medskip
\noindent
We first show how to prove claim~\ref{claim} using these statements.

\begin{proof}[Proof of claim~\ref{claim}]
  For a finite ground field $F=\F_{p^m}$ and a central division algebra $A$
  of dimension $s = r^2$ over $F$, take a maximal subfield $L$ and choose a
  generator $a$ (note that $\F_{p^m}$ is perfect and hence $L$ is separable
  over $\F_{p^m}$). Then the minimal polynomial $m_a$ of $a$ has degree $r$.
  The degree of the generic minimal polynomial also equals $r$ and, since
  $M_a(a)=0$, we conclude $m_a = M_a$.
\end{proof}


We now investigate to what extent Wedderburn knew or could have proven the
above statements.

\medskip
\noindent
\emph{Statement $\eqref{item:dim-square}$:}
%%
The first statement has surely been within Wedderburn's reach.  He
could have argued as follows using his famous structure
theorem~\cite{Wedderburn:hypercomplex-numbers} on simple algebras:
Central simple algebras stays simple under extension of the base
field.  So in particular for any algebraically closed field extension
$F'$ of $F$, the $F'$-algebra $F'\tensor_F A$ is isomorphic to a
matrix algebra over a finite dimensional division algebra over $F'$ by
the structure theorem.  But over an algebraically closed field, there
is no finite dimensional division algebra except for $F'$ itself since
any bigger algebra would contain a proper finite commutative field
extension of $F'$ which is impossible.
%%
So as a matrix algebra, $F'\tensor_F A$ has square dimension over
$F'$, and this dimension equals the dimension of $A$ over $F$.


\medskip
\noindent
\emph{Statement $\eqref{item:max-subfield}$:} 
%%
Statement \eqref{item:max-subfield} (and therefore also
\eqref{item:dim-square}) can be proven completely elementary for a
finite base field, cf.\ e.g.\ Artin~\cite{Artin:ueber-Wedderburn} or
Schue~\cite{Schue:Wedderburn}.  Artin's proof uses polynomials with
coefficients in a skew field, conjugation of subfields, some linear
algebra and counting arguments.  Schue's proof employs normalizers,
some Galois theory and a lot of linear algebra. All of these means
have been available to Wedderburn.

\begin{remark} 
  Although Wedderburn could have proven
  statement~\eqref{item:max-subfield}, he was surely not aware of this
  fact. If he had been, he would also have noticed that this
  immediately yields another, much shorter proof of his theorem, since
  it stongly simplifies the class equation also employed in
  Wedderburn's proof:
  
  \smallskip
  \parbox{\indentedpar}{Let $A$ be a central division algebra of
    dimension $s=r^2$ over $F=\F_q$.  Assume by induction, that all
    division algebras of lower dimension are commutative.  In
    particular all centralizers $C_1,\dots,C_t$ of elements of
    $A\setminus F$ are maximal subfields, hence $[C_i:F] = r$. Then
    the class equation for conjugacy classes of $A^*$ reads
  \[
    q^{r^2} -1 = |A^*| = |F^*| + \sum_{i=1}^t \frac{q^{r^2} - 1}{q^r -1}
    \text{,}
  \] 
  which implies that $\frac{q^{r^2}-1}{q^r-1}$ divides $q-1$ and hence
  that $r=1$. (This is also how Schue completes his proof of
  Wedderburn's theorem.)}

  \smallskip
  \noindent  
  Furthermore, statement \eqref{item:max-subfield} also abbreviates
  step $(2)$ of Wedderburn's original proof, if one wants to use the
  class equation in his way:
  
  
  \smallskip \parbox{\indentedpar}{As before, we have by induction 
    that the centralizers
    are exactly the maximal subfields. Combined with the statement
    that the maximal subfields all have degree $r$ with $r^2=s$, this
    yields $r^2=s=\lcm\bigl([C(a):\F_{q}] \ : \ a\in A\setminus \F_q
    \bigr) =r$, and thus $r=1$.  \bdende}
  %%
\end{remark}



\medskip
\noindent
\emph{Statement $\eqref{item:degree-generic-min-pol}$:} 
So we are left with the task of proving the third statement, using only
means available to Wedderburn.
%%
%%
%%
\begin{proof}[Proof of \eqref{item:degree-generic-min-pol}]
  Let $A$ be a central division algebra of dimension $s=r^2$ over a
  field $F$.
  %%
  We first prove that the degree of the generic minimal polynomial is
  at least $r$.
  %%
  Choose an algebraically extension field $F'$ of $F$. Then
  $F'\tensor_F A = A_{F'} \cong \Mat_{r}(F')$, the $F'$-algebra of
  $r\times r$ matrices with entries in $F'$, as shown in the argument
  for statement~\eqref{item:dim-square}.
  %%
  We choose an $F$-basis $e_{i,j}$ ($i, j = 1,\dots r$) of $A$ such
  that the $1\tensor e_{i,j}$ correspond to the standard basis of
  $\Mat_r(F')$. Since $M_{X,A_{F'}}$ divides $M_{X,A}$, it suffices to
  show that $\deg M_{X,A_{F'}} \geq r$. Over $F'$, the generic element
  $X=\sum X_{ij} e_{ij}$ is thus the ``generic matrix'' $(X_{ij})$. If
  we specialise $(X_{ij})$ to a diagonal matrix in
  $A_{F'}\cong\Mat_{r}(F')$ with pairwise distinct non-vanishing
  diagonal entries, the minimal polynomial of this matrix equals its
  characteristic polynomial and hence has degree $r$. (Note that for
  such a matrix to exist, $F'$ must contain at least $r+1$ distinct
  elements; since we choose $F'$ to be algebraically closed, we are on
  the safe side.) We conclude $\deg M_{X,A_{F'}} \geq r$, as
  required\footnote{By the theorem of Cayley-Hamilton, the $r\times
    r$-matrix $(X_{ij})$ satisfies its characteristic polynomial,
    which is of degree $r$. Thus, we do in fact have $\deg
    M_{X,A_{F'}} = r$.}.
  %%
  
  To show that the degree of the generic minimal polynomial is at most
  $r$, we note that by lemma~\ref{lem:division-tensor-transzendent}
  above, $F(X_{11},\dots,X_{rr}) \tensor_F A$ is again a central
  division algebra of dimension $r^2$, and thus the subfield generated
  by $X$ is of degree at most $r$ over $F(X_{11},\dots,X_{rr})$ by
  statement \eqref{item:max-subfield} of lemma~\ref{rettungs-lemma};
  but the degree of this subfield is just the degree of the generic
  minimal polynomial, hence we are done.
\end{proof}

\begin{remark}
  The statement of \eqref{item:degree-generic-min-pol} is true more
  generally for any finite dimensional central simple algebra. One can
  also prove the more precise statement that the generic minimal
  polynomial equals the generic reduced characteristic polynomial. The
  proof is very similar to the proof given above, but one of course
  has to systematically introduce the reduced characteristic
  polynomial of a central simple algebra---which was implicitly used
  above in case of a matrix algebra---and show that it is well defined
  using the theorem of Skolem-Noether. As we need the statement only
  in the simpler form of \eqref{item:degree-generic-min-pol}, we skip this
  proof. \bdende
\end{remark}

\medskip

\begin{conclusion}
  Our final conclusion is that Wedderburn's proof did in fact contain
  a gap he was not aware of, but that he could well have provided the
  necessary arguments to fill it with. \bdende
\end{conclusion}

%%
%% -----------------------------------------------------
%%

%%\appendix

\section*{Appendix. Chronological list of proofs}

All over the past century, new proofs of Wedderburn's theorem on
finite division algebras have been given by numerous mathematicians.
The proofs vary enormously with respect to length and depth of the
means employed. In her graduation paper~\cite{Birte-examen}, the
second author has searched through the literature for proofs and gave
a detailed exposition of each proof found in a thematically ordered
compilation as well as a comparison of the different proofs. As a
result of this effort, we present here a chronologically ordered and
hopefully rather complete list of (original) proofs of Wedderburn's
theorem present in literature with short comments on each proof.


\begin{description}
  
\item[1905, Wedderburn] There are three different proofs in ``A
  theorem on finite algebras''~\cite{Wedderburn:finite-algebras}.  The
  first proof was discussed in the preceding sections, the two others
  are more group theoretic, using Zsigmondy's theorem and the
  automorphism group of the algebra.

\item[1905, Dickson] Dickson published a proof in the same
  year~\cite{Dickson:finite-algebras}. This proof is very similar to
  Wedderburn's second and third proof. Dickson remarked that
  Wedderburn found these proofs only after having seen Dickson's
  proof.
  
\item[1927, Artin] In~\cite{Artin:ueber-Wedderburn}, after remarking
  that Wedderburn's first proof from~\cite{Wedderburn:finite-algebras}
  was not valid, Artin gave a new proof close to Wedderburn's main
  ideas, using some group theory, divisibility properties of
  polynomials, proving that all maximal subfields of the algebra are
  conjugate.
  
\item[1928, Noether] This proof from lectures by E.\ Noether, written
  down in 1931 by van der Waerden~\cite{Waerden:Algebra-2}, is very
  similar to that of Artin, using maximal subfields and group
  theoretic conjugacy arguments.
  
\item[1929, Brauer] In the article~\cite{Brauer:hyperkomplexe-zahlen}
  concerned with the study of algebras via group theoretic means using
  factor systems (certain Galois cocycles), Brauer derives
  Wedderburn's theorem from the more general theorem that the Schur
  index of a finite simple algebra equals one.
  
\item[1931, Witt] In this one-page
  paper~\cite{Witt:endliche-schiefkoerper}, Witt uses only cyclotomic
  polynomials and elementary group theory to prove Wedderburn's
  theorem.
  
\item[1932, Hasse-Noether] Noether~\cite{Noether:Hyperkomplex} and
  Hasse~\cite{Hasse:Struktur-Brauergruppe} remark that the relative
  Brauer group $\Br(L/K)$ is isomorphic to $K^*/N_{L/K}(L)$ for a
  finite cyclic extension $L/K$.
  %%
  Combined with the fact that the norm is surjective for finite
  extensions of finite fields, this yields Wedderburn's
  theorem.\footnote{Although the authors do not mention this
    consequence explicitly, there is no doubt that they were aware of
    it. Furthermore, since the two articles given above do neither
    contain a proof of the result nor a reference, it is not clear to
    us to whom the proof should be tributed, and so we have cited
    both.}
  
\item[1935, Chevalley] Inspired by Tsens
  work~\cite{Tsen:divisionsalgebren} about function fields of
  transcendence degree one over an algebraically closed field, and a
  remark of Artin, Chevalley proved
  in~\cite{Chevalley:endliche-koerper-c_1} that finite fields are
  $C_1$ and that over a $C_1$-field, there are no finite dimensional
  central division algebras except for the field itself, obtaining
  Wedderburn's theorem as a corollary.
  
\item[1951, Serre] %%\cite{Sem-Henri-Cartan:50-51}
  In his report \cite{Serre:applications-cohomologie-groupes} on
  Galois cohomology and the theory of simple algebras, Serre proves
  Wedderburn's theorem using (and proving) the facts\footnote{Each of
    these statements was known before, cf.\ e.g.\ 
    Eilenberg~\cite{Eilenberg:cohomology-of-groups}, but we have not
    found them put together like this to give a proof of Wedderburn's
    theorem, any earlier.}  that the relative Brauer group $\Br(L/K)
  \cong H^2\bigl(\Gal(L/K),L^*\bigr)$ for a finite Galois extension
  $L/K$, that $H^2\bigl(\Gal(L/K),L^*\bigr) \cong K^*/N_{L/K}(L^*)$
  for a cyclic extension $L/K$, and the classical fact that the norm
  is onto for finite fields.
  
  
\item[1952, Zassenhaus] In his rather long group theoretic
  proof~\cite{Zassenhaus:Wedderburn}, Zassenhaus derives Wedderburn's
  theorem from the theorem (proved in the same paper) that a finite
  group is abelian if for each abelian subgroup, the normalizer
  conincides with the centralizer.
  
\item[1961, Herstein] Herstein~\cite{Herstein:Wedderburn-Jacobson}
  gives a completely elementary proof by longish calculations, using
  only some theory of finite fields and group theory.
  
\item[1964, Kaczynski] Using deep results from the theory of finite
  groups, Kac\-zyn\-ski~\cite{Kaczynski:Wedderburn} shows that the group
  of units of a finite division algebra is simple and solvable, which
  is only possible for an abelian group.
  
\item[1964, Scott] In this group theoretical proof~\cite[p.\ 
  427]{Scott:Group-Theory}, Scott shows that all $q$-Sylow subgroups
  of the group of units of the division algebra are cyclic for $q\neq
  2$ and cyclic or generalised quaternion groups for $q=2$, and
  derives a contradiction studying the order of suitable elements and
  subgroups.

\item[1969, Ebey and Sitaram] The
  authors~\cite{Ebey-Sitaram:Wedderburn} use projective geometries to
  describe the group of units as the Frobenius complement of a
  Frobenius group and then employ deep results from the theory of
  finite groups to prove Wedderburn's theorem.
  
\item[1970, Burn and Maduram] Burn and
  Maduram~\cite{Burn-Maduram:Wedderburn} give a variant
  of~\cite{Ebey-Sitaram:Wedderburn}, avoiding the use of geometry.
  
\item[1971, Rogers] Rogers' proof~\cite{Rogers:Jacobson} strongly
  resembles Herstein's proof~\cite{Herstein:Wedderburn-Jacobson} from
  1961.  Being hardly less elementary, the calculations are more
  structured and make use of conjugates of subfields.
  
\item[1974, Nagahara and Tominaga] Proving a theorem of Jacobson
  in~\cite{Nagahara-Tominaga:Wedderburn}, the authors give two proofs
  of Wedderburn's theorem, applying elementary ring and group theory
  and several dimension arguments.
  
\item[1988, Schue] Schue~\cite{Schue:Wedderburn} calculates the
  dimension of centralizers as the square root of the dimension of the
  algebra by means of linear algebra and Galois theory, and then
  applies the class equation.

\item[1989, Meixner] 
  For a finite division algebra,
  Meixner~\cite{Meixner:Kreisteilungspolynome} considers a Sylow
  subgroup for the maximal prime divisor of the order of the group of
  units.  Using elementary group theory, he concludes that the
  corresponding cyclotomic number must be a power of $2$, deriving a
  contradiction.
  
\item[1990, Lorenz] Lorenz' textbook~\cite{Lorenz:Algebra-II-BI} on
  algebra contains a proof of Wedderburn's theorem on page 269, that
  we did not find earlier in literature. It uses an explicit formula
  for the behaviour of the presentation of central division algebras
  as crossed products under cyclic extensions.

\item[1994, Kasch] In his booklet~\cite{Kasch:Galoistheorie-Schiefkoerper}, 
  Kasch gives a simple proof of the main theorem of Galois theory for
  skew fields. As an application, this theorem is used in a proof of
  Wedderburn's theorem (pages 22--23, loc.\ cit.) to calculate the
  degree of a maximal subfield of a finite central division algebra
  and to obtain an upper bound for the number of maximal subfields.
  
  

\item[1998, Grundh\"ofer] Grundh\"ofer's proof~\cite{Grundhoefer:Wedderburn}
  resembles Witt's proof in that he uses cyclotomic polynomials, group
  theory and elementary number theory, but the main argument is
  transferred to obtain a purely group theoretical result at first.

\end{description}

\begin{remark*}
  We have not taken into account any of the huge number of
  generalizations of Wedderburn's theorem but only listed publications
  which contain a ``direct'' proof that does not merely obtain the
  theorem as a special case of a more general result.
\end{remark*}



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\begin{thebibliography}{10}

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\bibitem{Scott:Group-Theory}
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\end{thebibliography}

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%%\noindent
%%\rule{3cm}{0.4pt}\\[2ex]
%%Michael Adam\\
%%Mathematisches Institut\\
%%Bunsenstra\ss e 3--5\\
%%D--37073 G\"{o}ttingen\\
%%email: \texttt{mad@uni-math.gwdg.de}

%%\noindent\rule{3cm}{0.4pt}
%%
%%\medskip
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%%Michael Adam
%%
%%Mathematisches Institut
%%
%%Bunsenstra"se 3--5
%%
%%D-37073 G"ottingen
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%%email: \texttt{mad@uni-math.gwdg.de}
 
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