lecture-11-2019-05-17.tex 13.4 KB
 rrueger committed Jun 01, 2019 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 \chapter{Galois Theory without Galois Extensions} When dealing with Galois extensions, it at some point becomes inevitable to ask: \textit{What do we do when an extension is \textbf{not} Galois?}. Will \textit{none} of our tools work? It turns out, that when we consider some finite extension $E/F$ not Galois, but $E$ is contained in some field $\hat{E}$ such that $\hat{E}/F$ \textit{is} Galois we can apply some known results to arrive at the \begin{theorem}[Primitve Element Theorem] Let $E/F$ be a finite extension, with F a field with characteristic 0. Then there exists some $\alpha$ in $E$ such that $E = F(\alpha)$. \end{theorem} However, before we prove this, we will take a step back and investigate in which situations such an $\hat{E}$ exists. \section{Characteristic 0} \begin{siderules} \begin{definition}[Reminder Algebra I, Characteristic]  rrueger committed Jun 28, 2019 22  The characteristic, $p$, of a field is given by the number of times the  rrueger committed Jun 01, 2019 23  multiplicative neutral element must be added to itself before we arrive at  rrueger committed Jun 28, 2019 24  the additive neutral element. If no such $p$ exists, we say the field has  rrueger committed Jun 01, 2019 25 26  characteristic 0. \begin{align*}  rrueger committed Jun 28, 2019 27  \underset{p \text{~times}}{\underbrace{1+1+1+\cdots}} = 0  rrueger committed Jun 01, 2019 28 29 30 31 32 33 34 35 36  \end{align*} \end{definition} \begin{example}~ \begin{itemize} \item The characteristic of $\F _{p^n}$ is $p$. \item The characteristics of $\R, \C, \Q$ are $0$. \end{itemize} \end{example}  rrueger committed Jun 28, 2019 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56  This definition becomes more natural when considering it in the following manner: let $\chi \colon \Z \to k$ be a homomorphism, mapping $n \mapsto nl$ for $l$ some unit of $k$. Then $\ker (\chi) = (p)$ for $p$ zero or a prime (This is a result from Algebra I. It is a direct consequence of $\Z$ being a PID and $k$ an integral domain). If $p \neq 0$, then $p$ is clearly the characteristic as defined before, since if $x \in (p) = \ker (\chi)$, then $x = pl$ as $(p)$ is a prime ideal. Then % \begin{align*} pl = 0 = pll^{-1} = p1 = \underset{p \text{~times}}{\underbrace{1+1+1+\cdots}} \end{align*} % Assigning a field characteristic zero, when no such $p$ exists at first seemed somewhat arbitrary, but now is quite canonical, as it simply is the $p$ that generates the ideal that is $\ker (\chi)$.  rrueger committed Jun 01, 2019 57 58 59 60 61 62 63 64 \end{siderules} \begin{theorem} Consider a finite field extension $E/F$ and suppose $F$ has characteristic 0. Then there exists some $\hat{E}$ that contains $E$ with $\hat{E}/F$ being Galois. \end{theorem}  rrueger committed Jun 29, 2019 65 66 67 \begin{proof} \end{proof}  rrueger committed Jun 21, 2019 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 %\begin{proof} % We will utilise the splitting field of a polynomial with distinct roots in % this proof as by the characterisation of Galois extensions over normal % extensions we will have our $\hat{E}$. So we must seek out a polynomial $P \in % F[X]$ with distinct roots whose splitting field will contain $E$. The % extension is finite, so we can write $E = F(\alpha _1, \ldots, \alpha _m)$. It % seems then close at hand then to look for a polynomial with roots $\alpha _1, % \ldots, \alpha _m$. We already know of some polynomials with some $\alpha _i$ % as roots, namely the irreducible minimal polynomials of each $\alpha _i$, $p_i % \in F[X]$ - the existence of these is guaranteed by the fact that our % extension is finite, and thus algebraic. We can guarantee that $P$ has % \textit{all} the roots $\alpha_ i$ if we define % \begin{align*} % \quad \quad \quad \quad % P = \prod _{i=1} ^m p_i \quad \in F[X] % \end{align*} % However, we have no idea whether $P$ has distinct roots. In fact, we know that % in general it will not. We may have to omit some $p_i$'s for any hope for this % to work. Even then, it may be that a $p_i$ has repeated roots. So let us % investigate the $p_i$'s.  rrueger committed Jun 01, 2019 88   rrueger committed Jun 21, 2019 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 % We will show that each $p_i$ has distinct roots, by showing that % % % \begin{align*} % \gcd (p_i, p_i ') = 1 % \end{align*} % % % Immediately we see that $\deg (p_i) \geq 2$, otherwise $\alpha_i$ would lie in % $F$. % Crucially, also, we have that $p_i ' \neq 0$ as if $p_i = X^d + \cdots$, then % $p_i ' = d X^{d-1} + \cdots$ and in characteristic 0, we have guaranteed that % $d \neq 0$ and thus $p' \neq 0$. We now suppose by contradiction, that $\gcd % (p_i, p_i') \neq 1$. It then follows that $p_i ' | p_i$ - a contradiction to % $p_i$ being irreducible. We are closer to our goal. The $p_i$'s have distinct % roots. What about $p_i$ and $p_j$? We know they can't share a subset of their % roots, as otherwise, by previous argumentation, some factor would divide them. % However, they could be the same. If so omit one of them in the product. %\end{proof}  rrueger committed Jun 29, 2019 106 107 108 109 110 111  \section{Characteristic p} In the case of $F$ having characteristic $p \neq 0$, there is no such $\hat E$ in general. To show this, we pose the following counterexample:  rrueger committed Jun 29, 2019 112 113 114 115 116 117 118 119 120 121 122 123 124 125 Let $F = \F _p(X)$. Moreover, let $E = F(r)$ for $r$ a root of $p(Z) = Z^p - X$. We immediately now notice that $p'(Z) = 0$, so our previous line of argument doesn't work. Since $r$ is a root, $0 = p(r) = r^p - X$ implying $X = r^p$ so $P(Z) = Z^p - r^p$; as we are in characteristic $p$ we have that $p(Z) = Z^p - r^p = {(Z - r)}^p$. This means that $E$ is the splitting field over a polynomial with a single repeated root. We know that the automorphism group of $E/F$ can only permute the roots, and as there is only one root, $\Aut (E/F)$ is trivial. The extension is not Galois. The question is now, \textit{is there are larger $\hat E \supseteq E$ such that $\hat E /F$ is Galois?} Assume there was such an $\hat E$, then for any automorphism $\sigma \in \Aut (\hat E /F)$ it holds that $X \in F = \F _p (X)$ is fixed, so is $X^p$ and thus also $r$. It follows that $F(r) = E$ is also fixed by $\sigma$ and thus not \textbf{only} $F$ is fixed, meaning the extension is not Galois by the very first definition of a Galois extension.  rrueger committed Jun 29, 2019 126 127  \section{Skew Fields and Wedderburn's Theorem}  rrueger committed Jun 30, 2019 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333  \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}