### 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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