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Unverified Commit 581bfac2 authored by rrueger's avatar rrueger
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Algebra II: Lecture 11. Complete Wedderburn proof

Add label to definition of Euler's totient function.
parent 9fcdce19
......@@ -237,7 +237,7 @@ order $n$ are exactly those $\overline{k}$ with $k$ coprime to $n$. So we may
hunt the elements $k \in \set{1, \ldots, n}$ with $(n, k) = 1$. In general this
is difficult so we simply give this a name
\begin{definition}[Euler's totient function]
\begin{definition}[Euler's totient function]\label{def:euler_totient}
We define $\varphi (n){:}\, \N \to \N$ as the number of elements $1 \leq k \leq
n$ that are coprime to $n$.
\end{definition}
......
......@@ -125,3 +125,209 @@ meaning the extension is not Galois by the very first definition of a Galois
extension.
\section{Skew Fields and Wedderburn's Theorem}
\begin{definition}[Skew Field, Division Algebra]
A \textbf{skew field}, or \textbf{division algebra} is \textit{almost} a
field. A skew field is defined like a field, with the exception that the
multiplicative group must not be abelian. Formally, there is an abelian group
$(A, +, 0)$, a multiplicative group $(A, \cdot, 1)$ and the same compatibility
laws as in the definition of a field.
\end{definition}
\begin{theorem}[Wedderburn]\label{thm:weddeburn}
Finite skew fields are fields.
\end{theorem}
\begin{remark}
Here the real utility of the definition of a skew field becomes apparent: one
can show that a finite structure is a field without having to show
commutativity of multiplication. It reminds of a similar result that
we have seen in Algebra I, namely that finite domains are fields. Once we have
proven Theorem~\ref{thm:weddeburn}, we will have shown that there is no
distinction between finite fields, finite domains and finite skew fields.
\end{remark}
\begin{lemma}\label{lemma:stab_central}
When the multiplicative group of field acts on itself by conjugation it holds
that
%
\begin{align*}
Z(x) = \Stab _{F^\times} (x) \sqcup \set{0}
\end{align*}
%
where $Z(x) = \set{y \in F : xy = yx}$ denotes the centraliser of $x$ in $F$,
the field.
\end{lemma}
\begin{proof}
By definition of a field $F = F^\times \sqcup \set{0}$. Then
\begin{align*}
Z(x)
& = \set{y \in F: xy = yx} \\
& = \set{y \in F^\times: xy = yx} \cup \set{0} \\
& = \set{y \in F^\times: x = yxy^{-1}} \cup \set{0} \\
& = \Stab _{F^\times} (x) \cup \set{0}
\end{align*}
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:weddeburn}]
To show this, it only remains to show that all elements commute. To that end
let $A$ be a skew field, and $F = Z(A) = \set{x \in A : \forall a \in A: ax =
xa}$ its centre. We immediately note that $F$ is a field, and as it is finite
(as a subset of a finite skew field), it is isomorphic to some $\F _q$. So $A$
is some $F$ vector space with dimension $n$ meaning $|A| = |F|^n = q^{n}$.
Our goal is to show that $n=1$ and thus $A = F$.
We will show this by contradiction, but first, we note some numerical results:
As we have done many times before, we'll look at the action of $A^\times$ on
itself, acting by conjugation. As orbits form a partition, we immediately see
that
%
\begin{align*}
|A^\times|
= |A| - 1
= q^n -1
= \sum _{\substack{x \\ \text{~repr.}}} |\Orb _{A^\times} (x)|
\end{align*}
%
In fact, we know that all elements in $F$ have a trivial conjugacy class, so
we can further write
%
\begin{align*}
q^n - 1
= |F| + \sum _{\substack{x \notin F \\ \text{~repr.}}} |\Orb _{A^\times}(x)|
= q-1 + \sum _{\substack{x \notin F \\ \text{~repr.}}} |\Orb _{A^\times}(x)|
\end{align*}
%
``$x \notin F \text{~repr.}$'' being representatives of orbits not contained
in $F$. This is well defined, as if any element of a conjugacy class lies in
$F$, then that conjugacy class is trivial. Using the orbit-stabiliser theorem,
we know that each
%
\begin{align*}
|\Orb _{A^\times} (x)|
= \frac{|A^\times|}{|\Stab _{A^\times} (x)|}
= \frac{q^n -1}{|\Stab _{A^\times} (x)|}.
\end{align*}
%
We can now investigate the stabilisers $\Stab _{A^\times} (x)$. From
Lemma~\ref{lemma:stab_central}, we see that working with $Z(x)$ gives us
almost identical results, but being a field makes it much easier it much
easier to do that work. In fact, on inspection, we see that $F \subset Z(x)$.
This inclusion is clear, when framed slightly differently: $Z(x)$ is the field
of elements that commute with $x$, whereas $F$ is the field of elements that
commute with \textit{all} $a \in A$. Specifically, we now have that $F
\subseteq Z(x) \subseteq A$, so $Z(x)$ is a vector space over $F$, with its
dimension $m_x$ a divisor of $n$. We can now write
%
\begin{align*}
|\Orb _{A^\times} (x)|
= \frac{q^n -1}{|\Stab _{A^\times} (x)|}
= \frac{q^n -1}{|Z(x) \setminus \set{0}|}
= \frac{q^n -1}{q^{m_x} - 1}
\end{align*}
%
and in summary we now have
%
\begin{align}\label{eq:main}
q^n - 1
= q - 1 + \sum _{x \notin F \text{~repr.}} \frac{q^n -1}{q^{m_x} - 1}.
\end{align}
%
Hearkening back to our work with cyclotomic polynomials, it seems sensible to
consider this as a polynomial in $\Z[q]$, and write
%
\begin{align*}
q^n - 1 = \prod _{d | n} \Phi _d (q), \quad
q^{m_x} - 1 = \prod _{d | m_x} \Phi _{d} (q)
\end{align*}
%
Since each $m_x$ divides $n$, each $d$ that divides $m_x$ also divides $n$;
consequently
%
\begin{align*}
\frac{q^n - 1}{q^{m_x} - 1} \in \Z [q].
\end{align*}
%
still contains all primitive roots. In turn, we then have
%
\begin{align*}
\Phi _n (q) \,\, | \,\, \left ( \frac{q^n - 1}{q^{m_x} - 1} \right )
\end{align*}
%
and, by rearranging Equation~\ref{eq:main}, it follows that
%
\begin{align*}
\Phi _n (q) \,\, | \,\, (q - 1)
\end{align*}
%
Importantly, this means that when considering this polynomial over $\C$, that
$|\Phi _n(q)| \leq | q - 1 | = q - 1$; the latter equality arising from the
fact, that $q \in \N$. We now have everything that we need to coerce a
contradiction if $n > 1$.
Firstly, we write $\Phi _n (q)$ in its product form $\Phi _n (q) = \prod _i (q
- \zeta_i)$ where $\zeta_i$ are the primitive $n$-th roots of unity. We
then note
%
\begin{align}\label{eq:inequality}
|\Phi _n (q)|
= \left| \prod _{i=1} ^{\varphi(n)} (q - \zeta _i) \right|
= \prod _{i=1} ^{\varphi(n)} |q - \zeta _i|
\overset{(*)}{\geq} \prod _{i=1} ^{\varphi(n)} \left| |q| - |\zeta _i| \right|
= \prod _{i=1} ^{\varphi(n)} |q - 1|
= (q - 1) ^{\varphi(n)}
\geq q - 1
\end{align}
%
where $\varphi$ is Euler's totient Function
(Definition~\ref{def:euler_totient}). Now we are posed a problem, we have on
the one hand that $\Phi _n (q) | (q^n - 1) \implies |\Phi _n (q)| \leq q -1$,
yet on the other hand $|\Phi _n (q)| \geq q - 1$ by the previous calculation.
So we conclude $|\Phi _n (q)| = q - 1$ which happens if and only if both
inequalities in~\ref{eq:inequality} are in fact equalities. We can demonstrate
that $(*)$ is a strict inequality when $n > 1$: For an $n$-th root of unity
$\zeta \neq 1$, it holds true that
%
\begin{align*}
|q - 1| < |q - \zeta|
\end{align*}
%
for $q \in \N$. This is clearest, when seen geometrically
%
\begin{center}
\begin{tikzpicture}
[
scale=1,
>=stealth,
point/.style = {draw, circle, fill = black, inner sep = 1pt},
dot/.style = {draw, circle, fill = black, inner sep = 0.2pt},
]
% The circle
\def\rad{1.5}
\node (origin) at (0,0) [point, label = {below right:$\mathcal{O}$}]{};
\draw[dashed] (origin) circle (\rad);
% The Axes
\draw[->] (-1.25*\rad,0) -- (4.5*\rad,0);
\draw[->] (0,-1.25*\rad) -- (0, 1.25*\rad);
% Points
\node (w) at +(60:\rad) [point, label = {above right:$\omega_n$}] {};
\node (q) at (3*\rad,0) [point, label = {below:$q$}] {};
\node (u) at (\rad,0) [point, label = {below right:$1$}] {};
\node at (1*\rad,0) [point, label = {below right:$1$}] {};
\node at (2*\rad,0) [point, label = {below:$2$}] {};
\node at (4*\rad,0) [point, label = {below:$4$}] {};
% Line connecting
\draw [] (w) -- (q) -- (u) -- (w);
\end{tikzpicture}
\end{center}
%
So when $n > 1$, there is a primitive root of unity $\neq 1$, and thus, the
inequality strict. This is cause of our contradiction, so $n=1$ which in turn
dictates $F=A$, hence all elements in $A$ commute and $A$ is a field.
\end{proof}
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