lecture-09-2019-05-03.tex 11.7 KB
 rrueger committed May 06, 2019 1 2 3 4 5 \chapter{Constructibility} \section{Some Examples} \begin{example}[Non-counter example]  rrueger committed Jun 01, 2019 6  Suppose we want to find a splitting field $E$ to some polynomial $p \in \F[Z]$  rrueger committed May 06, 2019 7 8  such that $E/F$ is \textbf{not} a Galois extension. Naturally, $p$ cannot have distinct roots, otherwise the extension is normal and thus Galois. To find  rrueger committed May 11, 2019 9 10  such an example we can look in $\F _p (X)$, the field of rational functions in $X$ with coefficients in $\F _p$. Let $E$ be the splitting field of $p(Z) =  rrueger committed Jun 01, 2019 11 12 13  Z^p - X$. Suppose $r$ is a root of $p$, then we have $r^p -X = 0$ so $r^p = X$ and then $p(Z) = Z^p - r^p = {(Z - r)}^p$ as we are in characteristic $p$. Clearly this polynomial has repeated roots. So we have that $E = F(r)$.  rrueger committed May 06, 2019 14 15 \end{example}  rrueger committed May 11, 2019 16 17 18 19 20 \begin{exercise} As contrived as it may seem, this is one of the simplest examples of a Non-Galois splitting field. Why? \end{exercise}  rrueger committed May 06, 2019 21 22 23 24 25 26 27 28 29 30 31 32 33 34 \begin{example}[Infinite dimensional extension] We proved at the beginning of this semester, that for a set of $n$ pairwise distinct field automorphisms on some field $E$, which fix a field $H \subseteq E$ we have $\dim (E/H) \geq n$. This gives an upper lower bound on the extension. Perhaps there are cases with $\dim (E/H) = \infty$. Infinite dimensional extensions haven't been the focus of our study yet of course they exist. We can choose as a base field $\F _p$; now clearly we can't choose $\F _{p^n}$ as an extended field, as $\dim (\F _{p^n} / \F _p) = n$, so instead we elect the algebraic closure $\overline{\F _p}$ which is infinite dimensional. We can now consider $\sigma _f (\alpha) = \alpha ^p$ a Frobenius automorphism and let $\set{\sigma}$ be the set of automorphisms in the theorem. Then the field fixed by $\set{\sigma}$ is the field with the property $X^p - X = 0$ which is exactly $\F _p$.  rrueger committed May 11, 2019 35  % @todo: Show that F_p is exactly the field that is fixed.  rrueger committed May 06, 2019 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 \end{example} \section{Fundamental Examples of The Fundamental theorem of Galois Thoery} We have seen that $\F _{p^n} / \F _p$ and $\C (X_1, \ldots, X_n) / \C (e_1, \ldots, e_n)$ are Galois extensions. In this lecture we will investigate more carefully the constructibility of $n$-gons, that is regular polygons with n vertices. % \begin{siderules} % The story of constructibility begins on the sun kissed beaches of icarus... % @todo % Finish the story. % \end{siderules} We will now reason a series of reductions on what it means for a number to be constructible prove one direction of Theorem~\ref{thm:gauss-wantzel} (Gauss-Wantzel). We will show that if an $n$-gon is constructible, then $n$ is the product of a power of two and a Fermat prime. So, we ask ourselves, how do we construct the regular $n$-gon? Visually we have \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{3} \node (origin) at (0,0) [point, label = {below right:$\mathcal{O}$}]{}; \draw[dashed] (origin) circle (\rad); % The Axes \draw[->] (-1.25*\rad,0) -- (1.25*\rad,0); \draw[->] (0,-1.25*\rad) -- (0, 1.25*\rad); % Points \node (t1) at +(0:\rad) [point] {}; \node (t2) at +(120:\rad) [point, label = above:$\omega _3$] {}; \node (t3) at +(-120:\rad) [point] {}; \node (s1) at +(0:\rad) [point] {}; \node (s2) at +(90:\rad) [point, label = above right:$\omega _4$] {}; \node (s3) at +(180:\rad) [point] {}; \node (s4) at +(270:\rad) [point] {}; \node (p1) at +(0:\rad) [point] {}; \node (p2) at +(72:\rad) [point, label = above right:$\omega _5$] {}; \node (p3) at +(144:\rad) [point] {}; \node (p4) at +(-144:\rad) [point] {}; \node (p5) at +(-72:\rad) [point] {}; % Line connecting \draw [red] (t1) -- (t2) -- (t3) -- (t1); \draw [blue] (s1) -- (s2) -- (s3) -- (s4) -- (s1); \draw [green] (p1) -- (p2) -- (p3) -- (p4) -- (p5) -- (p1); \end{tikzpicture} \end{center} %  rrueger committed May 11, 2019 99 100 101 102 and the question becomes, for which $n$ can we construct $\omega _n = e^{\frac{2 \pi}{n} \ii}$? Again, we know that this is only constructible if $\omega _n \in E$ for which $E$ has a complete tower of degree 2 extensions reaching down to $\Q$. This is useful, as this is a completely algebraic description of our  rrueger committed Jun 26, 2019 103 problem. We are now only interested in the extension of $\Q (\omega _n) / \Q$.  rrueger committed May 11, 2019 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239  \subsection{The extension $\Q (\omega _n) / \Q$} Henceforth we'll fix $n \geq 2$ and drop the index $n$ to identify $\omega _n = \omega$. We immediately see that $\omega$ is a root of the polynomial $X^n - 1$ meaning that it is an algebraic number. This is good as it allows us to apply all of our known tools. In particular we note that if $\omega$ is constructible, it lives in some field $E$ which is an extension of degree $2^k$ for some $k$ over $\Q$. In turn the extension $\Q (\omega) / \Q$ must then also have a degree of a power of two. % @todo: Why must it be a power of two? % We know this as $\Q (\omega) \subseteq E$, and thus corresponds to some group % $S \subset \Gal (E/ \Q (\omega))$. % % The problem here is that it uses that the extension is Galois. This is true, % but it isn't proven yet. In essence, we want to answer the two questions \begin{enumerate}[(i)] \item What is $\dim (\Q (\omega) / \Q)$? \item Is $\Q (\omega) / \Q$ Galois? \end{enumerate} To answer (i), we will find the minimal irreducible polynomial with root $\omega$ over $\Q$. It is then one of our results from field theory, that the degree of this polynomial is the degree of the extension. To do this we will take a short excursion concerning itself with the factorisation of $X^n - 1$. We can look at examples for $n=1, 2, 3, 4$: % \begin{itemize} \item (n=2, $\omega = -1$) The equation $X^2 - 1$ factorises to $(X+1)(X-1)$. \item (n=3, $\omega = -\frac{1}{2} + \ii \frac{\sqrt{3}}{2}$) The equation $X^3 - 1$ factorises to $(X-1)(X^2 + X + 1)$. \item (n=4, $\omega = -\ii$) The equation $X^4 - 1$ factorises to $(X-1)(X+1)(X^2 + 1)$. \end{itemize} % A general factorisation of $X^n - 1$ over $\C$ is $\prod _{j=1} ^{n-1} X - \omega ^j$ and we know that $\Sigma := \set{\omega ^0, \omega ^1, \ldots, \omega ^{n-1}}$ forms a group. This last fact allows us to observe an innocent detail stemming from group theory that for $g \in \Sigma$ % \begin{align*} \mathrm{ord}(g) \mid (|\Sigma| = n) \end{align*} % We can see how this manifests itself in an example \begin{example}[n=4] Then $\Sigma = \set{1, \ii, -1, -\ii}$, and $\mathrm{ord}(1) = 1, \mathrm{ord}(\ii) = 4, \mathrm{ord}(-1) = 2, \mathrm{ord}(-\ii) = 4$. Looking at the factorisation $X^4 - 1 = (X + 1)(X - 1)(X^2 + 1)$ we see that there is a factor for each order. \end{example} This gives rise to the idea, that we could group $\Sigma$ by the order of each element. For this we will introduce \begin{definition}[Primitive root of unity] A \textbf{primitive} n\textsuperscript{th} \textbf{root} of unity, is one that generates all the n\textsuperscript{th} roots of unity. This means the primitive roots are exactly those of order $n$. \end{definition} \begin{definition}[n\textsuperscript{th} Cyclotomic Polynomial] Let $J \subseteq \Sigma$ consist of the primitive roots of unity, then we define the n\textsuperscript{th} \begin{align*} \Phi _n (X) = \prod _{z \in J} (X - z) \end{align*} \end{definition} \begin{example} For $n=4$, we have \begin{align*} \Phi _4 (X) = (X - \ii)(X + \ii) = X^2 + 1 \end{align*} \end{example} \begin{theorem} \begin{align*} \Phi _n \in \Q [X] \end{align*} \end{theorem} \begin{proof} We claim that $\Q (\omega)$ is the splitting field of $p(X) = X^n - 1$. This holds, as $\Q (\omega)$ clearly contains all the roots of $p$ and it cannot split in a subfield as a subfield must contain $\omega$, yet $\Q (\omega)$ is by definition the smallest field containing $\Q$ and $\omega$. Now let $\sigma \in \Gal (\Q (\omega) / \Q)$. Since $X^n - 1 = 0$ we have $X^n = 1$ and $\sigma (X^n) = {\sigma (X)} ^n = 1$. We can now assert that $\sigma$ preserves the order of a root. Assume this were not the case: then for a primitive root $z$ there would be some $k < n$ such that ${\sigma (z)}^k = 1 = {\sigma (z)}^n$. We would then have ${\sigma (z)} ^{n-k}= \sigma (z^{n-k}) = 1$. Applying the inverse of $\sigma$ gives $z ^{n-k} = 1$: a contradiction to $z$ being of order n i.e.\ primitive. In particular, this means that $\sigma$ maps primitive roots to primitive roots. We then draw % \begin{align*} \sigma (\Phi _n (X)) = \sigma \left ( \prod _{z~\text{prim.}} (X - z) \right ) = \prod _{z~\text{prim.}} (X - \sigma (z)) = \prod _{z~\text{prim.}} (X - z) = \Phi _n (X) \end{align*} % that $\Phi _n$ is Galois invariant. It follows that the coefficients are fixed by the Galois group, and thus must be rational. $\Phi _n \in \Q[X]$. \end{proof} This theorem delivers an immediate answer to question (ii) \begin{corollary}[$\Q (\omega) / \Q$ is Galois] It is the splitting field of a polynomial in $\Q [X]$ with distinct roots, thus normal and hence Galois. \end{corollary} \begin{theorem} The n\textsuperscript{th} cyclotomic polynomial s irreducible over the rationals. \end{theorem} We will discuss the proof in a later lecture, and focus on what ramifications this entails. Specifically, we will have the degree of the extension $\Q (\omega) / \Q$, namely the degree of the n\textsuperscript{th} cyclotomic polynomial. The degree of $\Phi _n$ is given by the number of primitive roots. We know $\set{\omega ^0, \omega ^1, \ldots, \omega ^{n-1}} \cong \rquot{\Z}{n\Z}$, so the number of roots with order $n$ is equal to the number of elements in $\rquot{\Z}{n\Z}$ with order n. The roots of order $n$ however are exactly the primitive roots, and the elements in $\rquot{\Z}{n\Z}$ with 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  rrueger committed Jun 30, 2019 240 \begin{definition}[Euler's totient function]\label{def:euler_totient}  rrueger committed May 11, 2019 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  We define $\varphi (n){:}\, \N \to \N$ as the number of elements $1 \leq k \leq n$ that are coprime to $n$. \end{definition} \begin{lemma} For $p$ prime, it holds that \begin{align*} \varphi(p^n) = p^{n-1} (p-1) \end{align*} \end{lemma} \begin{proof} The total number of candidates $1 \leq k \leq p^n$ are $p^n$. The only ones that aren't coprime, are those that are multiples of $p$, of which there are $p^{n-1}$ in $\set{1, \ldots, p^n}$. Thus the total number that are coprime $p^n - p^{n-1} = p^{n-1}(p-1)$. \end{proof} \noindent To conclude, we have \begin{theorem} For $n \geq 2, \omega = e^{\frac{2\pi}{n}\ii}$, \begin{align*} \dim (\Q (\omega) / \Q) = \varphi(n) \end{align*} \end{theorem} \noindent We can now return to the original problem of constructing $n$-gons. \subsection{Construction} We have now that if $\omega$ is constructible, then $\dim (\Q (\omega) / \Q) = \varphi(n) = 2^k$ for some $k$. We can now factor $n = 2^a p_1 ^{\lambda _1}  rrueger committed Jun 22, 2019 276 277 \cdots p_n ^{\lambda _n}$ for $p_i$ distinct primes not equal 2 and as a consequence of the Chinese remainder theorem, we have that  rrueger committed May 11, 2019 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 % \begin{align*} \varphi (n) = \varphi (2^a) \varphi (p_1 ^{\lambda _1}) \cdots \varphi (p_n ^{\lambda _n}) = 2^{a-1} p_1 ^{\lambda _1 -1} (p_1 -1) \cdots p_n ^{\lambda _n -1} (p_n -1). \end{align*} % The only chance that $\varphi (n) = 2^k$ is if each $\lambda _j = 1$ and if each $p_j - 1 = 2^{k_j}$ i.e.\ each $p_j = 2^{j_k} + 1$. % @todo: Why is this then a Fermat prime? In conclusion we have that the regular $n$-gon is constructible if $n$ is the product of a power of $2$ and some Fermat primes. The theorem of Gauss is one that goes in both directions, we will prove the converse next lecture.