Algebra II: Lecture 11. Complete Wedderburn proof

Add label to definition of Euler's totient function.

source-II/lecture-09-2019-05-03.tex

@@ -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}

source-II/lecture-11-2019-05-17.tex

@@ -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}