Unverified Commit 581bfac2 authored by rrueger's avatar rrueger
Browse files

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 $n$.
......@@ -125,3 +125,209 @@ meaning the extension is not Galois by the very first definition of a Galois
\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.
Finite skew fields are fields.
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.
When the multiplicative group of field acts on itself by conjugation it holds
Z(x) = \Stab _{F^\times} (x) \sqcup \set{0}
where $Z(x) = \set{y \in F : xy = yx}$ denotes the centraliser of $x$ in $F$,
the field.
By definition of a field $F = F^\times \sqcup \set{0}$. Then
& = \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}
\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
= |A| - 1
= q^n -1
= \sum _{\substack{x \\ \text{~repr.}}} |\Orb _{A^\times} (x)|
In fact, we know that all elements in $F$ have a trivial conjugacy class, so
we can further write
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)|
``$x \notin F \text{~repr.}$'' 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
|\Orb _{A^\times} (x)|
= \frac{|A^\times|}{|\Stab _{A^\times} (x)|}
= \frac{q^n -1}{|\Stab _{A^\times} (x)|}.
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
|\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}
and in summary we now have
q^n - 1
= q - 1 + \sum _{x \notin F \text{~repr.}} \frac{q^n -1}{q^{m_x} - 1}.
Hearkening back to our work with cyclotomic polynomials, it seems sensible to
consider this as a polynomial in $\Z[q]$, and write
q^n - 1 = \prod _{d | n} \Phi _d (q), \quad
q^{m_x} - 1 = \prod _{d | m_x} \Phi _{d} (q)
Since each $m_x$ divides $n$, each $d$ that divides $m_x$ also divides $n$;
\frac{q^n - 1}{q^{m_x} - 1} \in \Z [q].
still contains all primitive roots. In turn, we then have
\Phi _n (q) \,\, | \,\, \left ( \frac{q^n - 1}{q^{m_x} - 1} \right )
and, by rearranging Equation~\ref{eq:main}, it follows that
\Phi _n (q) \,\, | \,\, (q - 1)
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
|\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
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
|q - 1| < |q - \zeta|
for $q \in \N$. This is clearest, when seen geometrically
point/.style = {draw, circle, fill = black, inner sep = 1pt},
dot/.style = {draw, circle, fill = black, inner sep = 0.2pt},
% The circle
\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);
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.
Supports Markdown
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment