math/galois-theory-notes.tex
galois-theory-notes.texRaw
1\documentclass{article}
2\usepackage[utf8]{inputenc}
3\usepackage{amsfonts}
4\usepackage{amsthm}
5\usepackage{amsmath}
6\usepackage{enumerate}
7\usepackage{hyperref}
8\usepackage{amssymb}
9\usepackage{tikz} % diagram
10
11\begin{filecontents}[overwrite]{galois-theory-notes.bib}
12@misc{ianstewart,
13 author = {Ian Stewart},
14 title = {{Galois Theory, Third Edition}},
15 year = {2004}
16}
17
18@misc{milneFT,
19 author={Milne, James S.},
20 title={Fields and Galois Theory (v5.10)},
21 year={2022},
22 note={Available at \url{https://jmilne.org/math/} },
23 pages={144}
24}
25
26@misc{berlekamp,
27 author={Elmyn Berlekamp},
28 title={Algebraic Coding Theory},
29 year={1984},
30 note={Revised Edition from 1984}
31}
32
33@misc{dihedral,
34 author = {Gaurab Bardhan and Palash Nath and Himangshu Chakraborty},
35 title = {Subgroups and normal subgroups of dihedral group up to isomorphism},
36 year = {2010},
37 note = {\url{https://scipp.ucsc.edu/~haber/ph251/Dn_subgroups.pdf}},
38 url = {https://scipp.ucsc.edu/~haber/ph251/Dn_subgroups.pdf}
39}
40
41@misc{judson,
42 author={Thomas W. Judson},
43 title={Abstract Algebra: theory and applications},
44 year={1994},
45 note={Available at \url{http://abstract.ups.edu/download.html} },
46 pages={438}
47}
48
49@misc{dummitfoote,
50 author={David S. Dummit and Richard M. Foote},
51 title={Abstract Algebra (Third Edition)},
52 year={2004},
53 pages={945}
54}
55\end{filecontents}
56\nocite{*}
57
58
59\theoremstyle{definition}
60
61\newtheorem{innerdefn}{Definition}
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70
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75
76\newtheorem{innercor}{Corollary}
77\newenvironment{cor}[1]
78{\renewcommand\theinnercor{#1}\innercor}
79{\endinnercor}
80
81\newtheorem{innereg}{Example}
82\newenvironment{eg}[1]
83{\renewcommand\theinnereg{#1}\innereg}
84{\endinnereg}
85
86
87\title{Galois Theory notes}
88\author{arnaucube}
89\date{2025}
90
91\begin{document}
92
93\maketitle
94
95\begin{abstract}
96 Notes taken while studying Galois Theory, mostly from Ian Stewart's book "Galois Theory" \cite{ianstewart}.
97
98 Usually while reading books and papers I take handwritten notes in a notebook, this document contains some of them re-written to $LaTeX$.
99
100 The notes are not complete, don't include all the steps neither all the proofs.
101\end{abstract}
102
103\tableofcontents
104
105\section{Galois Theory notes}
106\subsection{Chapters 4-6}
107(Definitions, theorems, lemmas, corollaries and examples enumeration follows from Ian Stewart's book \cite{ianstewart}).
108
109\begin{defn}{4.10}
110 A \emph{simple extension} is $L:K$ such that $L=K(\alpha)$ for some $\alpha \in L$.
111\end{defn}
112\begin{eg}{4.11}
113 Beware, $L=\mathbb{Q}(i, -i, \sqrt{5}, -\sqrt{5}) = \mathbb{Q}(i, \sqrt{5}) = \mathbb{Q}(i+\sqrt{5})$.
114\end{eg}
115
116\begin{defn}{5.5}
117 Let $L:K$, suppose $\alpha \in L$ is algebraic over $K$. Then, the \emph{minimal polynomial} of $\alpha$ over $K$ is the unique monic polynomial $m$ over $K$, $m(t) \in K[t]$, of smallest degree such that $m(\alpha)=0$.
118 \\
119 eg.: $i \in \mathbb{C}$ is algebraic over $\mathbb{R}$. The minimal polynomial of $i$ over $\mathbb{R}$ is $m(t)=t^2 +1$, so that $m(i)=0$.
120\end{defn}
121
122\begin{lemma}{5.9}
123 Every polynomial $a \in K[t]$ is congruent modulo $m$ to a unique polynomial of degree $< \delta m$.
124\end{lemma}
125\begin{proof}
126 Divide $a / m$ with remainder, $a= qm +r$, with $q,r \in K[t]$ and $\delta r < \delta m$.
127 Then, $a-r=qm$, so $a \equiv r \pmod{m}$.
128
129 It remains to prove uniqueness.
130
131 Suppose $\exists~ r \equiv s \pmod{m}$, with $\delta r, \delta s < \delta m$.
132 Then, $r-s$ is divisible by $m$, but has smaller degree than $m$.
133
134 Therefore, $r-s=0$, so $r=s$, proving uniqueness.
135\end{proof}
136
137\begin{thm}{5.10}
138 $\forall 0 \neq f \in \frac{K[t]}{<m>},~~ \exists f^{-1}$ iff $m$ is irreducible in $K[t]$.\\
139 Then $\frac{K[t]}{<m>}$ is a field.
140\end{thm}
141\begin{thm}{5.12} \label{5.12}
142 Let $K(\alpha):K$ simple algebraic extension, let $m$ minimal polynomial of $\alpha$ over $K$.\\
143 $K(\alpha):K$ is isomorphic to $\frac{K[t]}{<m>}$.\\
144 The isomorphism $\frac{K[t]}{<m>} \longrightarrow K(\alpha)$ can be chosen to map $t$ to $\alpha$.
145
146\end{thm}
147\begin{cor}{5.13}
148 Let $K(\alpha):K$ and $K(\beta):K$ be simple algebraic extensions.\\
149 If $\alpha,~ \beta$ have same minimal polynomial $m$ over $K$, then the two extensions are isomorphic, and the isomorphism of the larger fields map $\alpha$ to $\beta$.
150\end{cor}
151\begin{proof}
152 By \ref{5.12}, both extensions are isomorphic to $\frac{K[t]}{<m>}$.
153\end{proof}
154
155\begin{lemma}{5.14}
156 Let $K(\alpha):K$ be a simple algebraic extension, let $m$ be the minimal polynomial of $\alpha$ over $K$, let $\delta m =n$.
157
158 Then $\{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}\}$ is a basis for $K(\alpha)$ over $K$.
159 In particular, $[K(\alpha):K]=n$.
160\end{lemma}
161
162\begin{defn}{6.2}
163 The degree $[L:K]$ of a field extension $L:K$ is the dimension of L considered as a vector space over $K$.
164
165 Equivalently, the dimension of $L$ as a vector space over $K$ is the number of terms in the expression for a general element of $L$ using coefficients from $K$.
166\end{defn}
167
168\begin{eg}{6.3}
169 \begin{enumerate}
170 \item $\mathbb{C}$ elements are 2-dimensional over $\mathbb{R}$ ($p+qi \in \mathbb{C}$, with $p,q \in \mathbb{R}$), because a basis is $\{1, i\}$, hence $[\mathbb{C}:\mathbb{R}]=2$.
171 \item $[ \mathbb{Q}(i, \sqrt{5}) : \mathbb{Q}]=4$, since the elements $\{1, \sqrt{5}, i, i\sqrt{5}\}$ form a basis for $\mathbb{Q}(i, \sqrt{5})$ over $\mathbb{Q}$.
172 \end{enumerate}
173\end{eg}
174
175\begin{thm}{6.4}\emph{(Short Tower Law)} \label{shorttowerlaw}
176 If $K, L, M \subseteq \mathbb{C}$, and $K \subseteq L \subseteq M$, then $[M:K]=[M:L]\cdot [L:K]$.
177\end{thm}
178\begin{proof}
179 Let $(x_i)_{i \in I}$ be a basis for $L$ over $K$,
180 let $(y_j)_{j \in J}$ be a basis for $M$ over $L$.\\
181 $\forall i \in I, j \in J$, we have $x_i \in L, u_j \in M$.
182 \\
183 Want to show that $(x_i y_j)_{i\in I, j\in J}$ is a basis for $M$ over $K$.
184 \begin{enumerate}[i.]
185 \item prove linear independence:\\
186 Suppose that
187 $$\sum_{ij} k_{ij} x_i y_j = 0 ~(k_{ij} \in K)$$
188 rearrange
189 $$\sum_j (\underbrace{\sum_i k_{ij} x_i}_{\in L}) y_j = 0 ~(k_{ij} \in K)$$
190 Since $\sum_i k_{ij} x_i \in L$, and the $y_j \in M$ are linearly independent over $L$, then $\sum_i k_{ij} x_i = 0$.
191 \\
192 Repeating the argument inside $L$ $\longrightarrow$ $k_{ij}=0 ~~\forall i\in I, j\in J$.
193 \\
194 So the elements $x_i y_j$ are linearly independent over $K$.
195
196 \item prove that $x_i y_j$ span $M$ over $K$:\\
197 Any $x \in M$ can be written
198 $$x=\sum_j \lambda_j y_j$$
199 for $\lambda_j \in L$, because $y_j$ spans $M$ over $L$.
200 Similarly,
201 $$\forall j\in J,~ \lambda_j = \sum_i \lambda_{ij} x_i y_j$$
202 for $\lambda_{ij} \in K$.\\
203 Putting the pieces together,
204 $$x=\sum_{ij} \lambda_{ij} x_i y_j$$
205 as required.
206 \end{enumerate}
207\end{proof}
208
209\begin{cor}{6.6}\emph{(Tower Law)}\\ \label{towerlaw}
210 If $K_0 \subseteq K_1 \subseteq \ldots \subseteq K_n$ are subfields of $\mathbb{C}$, then
211 $$[K_n:K_0] = [K_n:K_{n-1}] \cdot [K_{n-1}:K_{n-2}] \cdot \ldots \cdot [K_1: K_0]$$
212\end{cor}
213\begin{proof}
214 From \ref{shorttowerlaw}.
215\end{proof}
216
217\begin{thm}{6.7}
218 if $K(\alpha):K$
219 \begin{itemize}
220 \item transcendental $\Longrightarrow~~[K(\alpha):K] = \inf$
221 \item algebraic $\Longrightarrow~~[K(\alpha):K] = \delta m$
222 \end{itemize}
223 (where $m$ is the minimal polynomial of $\alpha$ over $K$).
224\end{thm}
225
226\begin{defn}{8.1}
227 $L:K$, a \emph{$K$-automorphism} of $L$ is an automorphism $\alpha$ of $L$ such that $\alpha(k)=k ~~ \forall k \in K$.\\
228 ie. $\alpha$ \emph{fixes} $k$.
229\end{defn}
230
231\begin{thm}{8.2, 8.3}
232 The set of all $K$-automorphisms of $L$ forms a group, $\Gamma(L:K)$, the Galois group of $L:K$.
233\end{thm}
234\begin{lemma}{8.18}
235 Let $q \in L$. The minimal polynomial of $q$ over $K$ \emph{splits} into linear factors over L.
236\end{lemma}
237
238\begin{defn}{9.1}
239 For $K \subseteq \mathbb{C}$, and $f \in K[t]$, $f$ \emph{splits} over $K$ if it can be expressed as a product of linear factors
240 $$f(t) = k \cdot (t- \alpha_1) \cdot \ldots \cdot (t - \alpha_n)$$
241 where $k, \alpha_i \in K$.
242
243 $\Longrightarrow$ (Thm 9.3) if $f$ splits over $\Sigma$, $\Sigma$ is the \emph{splitting field}.\\
244 If $K \subseteq \Sigma' \subseteq \Sigma$ and $f$ splits over $\Sigma'$, then $\Sigma' = \Sigma$.
245\end{defn}
246
247\begin{thm}{9.6} \label{9.6}
248 TODO
249\end{thm}
250
251\begin{defn}{9.8}
252 $L:K$ is \emph{normal} if every irreducible polynomial $f \in K[t]$ that has at least one zero in $L$, splits in $L$.
253\end{defn}
254
255\begin{thm}{9.9} \label{9.9}
256 TODO
257\end{thm}
258
259\begin{thm}{9.10}
260 An irreducible polynomial $f \in K[t]$ ($K \subseteq \mathbb{C}$) is \emph{separable over} $K$ if it has simple zeros in $\mathbb{C}$, or equivelently, simple zeros in its splitting field.
261\end{thm}
262
263\begin{lemma}{9.13}
264 $f \in K[t]$ with splitting field $\Sigma$. $f$ has multiple zeros (in $\Sigma$ or $\mathbb{C}$) iff $f$ and $Df$ have a common factor of degree $\geq 1$ in $\Sigma[t]$.\\
265 More details at Rolle's theorem (\ref{rolle}) section.
266\end{lemma}
267
268\begin{thm}{10.5} \label{10.5}
269 $|\Gamma(K:K_0)| = [K:K_0]$, where $K_0$ is the fixed field of $\Gamma(K:K_0)$.
270\end{thm}
271
272
273\begin{defn}{11.1}
274 $K \subseteq L$, $K \subseteq L$. A $K$-monomorphism of $M$ into $L$ is a field monomorphism
275 $$\phi: M \longrightarrow L$$
276 such that $\phi(k)=k ~~ \forall k \in K$.
277\end{defn}
278
279\begin{thm}{11.3}
280 $L:K$ normal, $K \subseteq M \subseteq L$. Let $\tau$ any $K$-monomorphism $\tau: M \longarrow L$.\\
281 Then, $\exists$ a $K$-automorphism $\sigma$ of $L$ such that $\sigma\biggr\vert_M=\tau$.
282\end{thm}
283\begin{proof}
284 $L:K$ normal $\Longrightarrow$ by Thm \ref{9.9}, $L$ splitting field for some poly $f \in K[t]$.
285
286 Hence, $L$ is splitting field over $M$ for $f$ and over $\tau(M)$ for $\tau(f)$.
287
288 Since $\tau \biggr\vert_K$ is the identity, $\tau(f)=f$.
289
290 \vspace{0.5cm}
291 We have
292
293 \begin{tikzpicture}[node distance=1.5cm, auto]
294 \node (M) {$M$};
295 \node (L) [right of=M] {$L$};
296 \node (t) [below of=M] {$\tau(M)$};
297 \node (L2) [below of=L] {$L$};
298
299 \draw[->] (M) to node {$ $} (L);
300 \draw[->] (M) to node {$\tau$} (t);
301 \draw[->] (t) to node {$ $} (L2);
302 \draw[->] (L) to node {$ $} (L2);
303 \end{tikzpicture}
304
305 with $\sigma$ yet to be formed.
306
307 By Theorem \ref{9.6}, $\exists$ isomorphism $\sigma: L \longrightarrow L$ such that $\sigma \biggr\vert_M = \tau$.\\
308 Therefore, $\sigma$ is an automorphism of $L$, and since $\sigma\biggr\vert_K = \tau\biggr\vert_K=id$, $\sigma$ is a $K$-automorphism of $L$.
309\end{proof}
310
311\begin{lemma}{11.8} \label{11.8}
312 $K \subseteq L \subseteq N \subseteq M$, $L:K$ finite, $N$ normal closure of $L:K$.\\
313 Let $\tau$ any $K$-monomorphism $\tau: L \longrightarrow M$.\\
314 Then $\tau(L) \subseteq N$.
315\end{lemma}
316\begin{proof}
317 $\alpha \in L$, $m$ minimal polynomial of $\alpha$ over $K$.\\
318 $\Longrightarrow ~~ m(\alpha)=0$, so $\tau(m(\alpha))=0$
319
320 (since $\tau$ is a $K$-automorphism, ie. maps the zeros of $m(t)$).\\
321
322 Since $\tau$ is a $K$-monomorphism, $\tau(m(\alpha))=m(\tau(\alpha))=0$
323
324 $\Longrightarrow~~ \tau(\alpha)$ is a zero of $m$.\\
325
326 Therefore, $\tau(\alpha)$ lies in $N$, since $N:K$ is normal.\\
327 Henceforth, $\tau(L) \subseteq N$.
328\end{proof}
329
330\begin{thm}{11.9}
331 The following are equivalent:
332 \begin{enumerate}
333 \item $L:K$ normal
334 \item $\exists$ finite normal extension $N$ of $K$ containing $L$,\\
335 such that every $K$-monomorphism $\tau: L \longrightarrow N$ is a $K$-automorphism of $L$.
336 \item for every finite extension $M$ of $K$ containing $L$,\\
337 every $K$-monomorphism $\tau: L \longrightarrow M$ is a $K$-automorphism of $L$.
338 \end{enumerate}
339\end{thm}
340
341\begin{thm}{11.10}
342 $[L:N]=1,~ N$ normal closure of $L:K$. Then,
343
344 $\exists~ n~ K$-monomorphisms $L \longrightarrow N$.\\
345 (the ones proven by Lemma \ref{11.8}).
346\end{thm}
347
348\begin{cor}{11.11} \label{11.11}
349 $|\Gamma(L:K)| = [L:K]$ (if $L:K$ is normal).
350
351 ie. there are precisely $[L:K]$ distinct $K$-automorphisms of $L$.
352\end{cor}
353
354\begin{thm}{11.12}
355 $\Gamma(L:K) = G$. If $L:K$ normal, then $K$ is the fixed field of $G$.
356\end{thm}
357\begin{proof}
358 let $K_0$ be the fixed field of $G$. Let $[L:K]=n$.\\
359 By \ref{11.11}, $|G| = [L:K] = n$.\\
360 By \ref{10.5}, $[L:K_0]=n$ ($K_0$ fixed field).\\
361 Since $K \subseteq K_0$, we must have $K=K_0$.
362
363 $\Longrightarrow$ thus $K$ is the fixed field of $G$.
364\end{proof}
365
366\begin{thm}{11.14}
367 if $L$ any field, $G$ any finite group of automorphisms of $L$, and $K$ its fixed field,
368
369 then $L:K$ is \emph{finite} and \emph{normal}, with Galois group $G$.
370\end{thm}
371
372
373\begin{thm}{12.2}(Fundamental Theorem of Galois Theory)
374 if $L:K$ finite and normal inside $\mathbb{C}$, with $\Gamma(L:K)=G$, then:
375 \begin{enumerate}
376 \item $|\Gamma(L:K)| = [L:K]$
377 (by Corollary \ref{11.11})
378 \item the maps * and $\dagger$ are mutual inverses, and setup an order-reversing one-to-one correspondence between $\mathcal{F}$ and $\mathcal{G}$.
379 \item if $M$ an intermediate field, then
380 $$[L:M] = |M^*|~~~~~~~ [M:K]=\frac{|G|}{|M^*|}$$
381 \item for $M$ an intermediate field, $M:K$ normal iff
382 $$\underbrace{\Gamma(M:K)}_{=M^*} \lhd \underbrace{\Gamma(L:K)}_{=G}$$
383 \item for $M$ intermediate, if $M:K$ normal, then
384 $$\Gamma(M:K) \cong \frac{G}{M^*}$$
385 ie.
386 $$\Gamma(M:K) \cong \frac{\Gamma(L:K)}{\Gamma(L:M)}$$
387 \end{enumerate}
388\end{thm}
389\begin{proof}
390 TODO
391\end{proof}
392
393[Chapter 13 is basically a full example. More examples can be found at section \ref{ex:galoisgroups}]
394
395
396\subsection{Detour: Isomorphism Theorems}
397\begin{thm}{i.1}(\emph{First Isomorphism Theorem}) \label{1stisothm}
398
399 \begin{minipage}{0.75\textwidth}
400 If $\psi: G \longrightarrow H$ a group homomorphism, then $ker(\psi) \triangleleft G$.\\
401 Let $\phi: G \longrightarrow G/ker(\psi)$ be the canonical homomorphism.\\
402 Then $\exists$ unique isomorphism $\eta: G/ker(\psi) \longrightarrow \psi(G)$ such that $\psi = \eta \phi$.\\
403 $\Longleftrightarrow$ ie. $G/ker(\psi) \cong \psi(G)$.
404 \end{minipage}
405\hfill
406\begin{minipage}{0.2\textwidth}
407 \begin{tikzpicture}[node distance=2.5cm, auto]
408 \node (G) {$G$};
409 \node (H) [right of=G] {$H$};
410 \node (GmodK) [below of=G, xshift=1.25cm] {$G/\ker(\psi)$};
411
412 \draw[->] (G) to node {$\psi$} (H);
413 \draw[->] (G) to node [swap] {$\phi$} (GmodK);
414 \draw[->] (GmodK) to node [swap] {$\eta$} (H);
415 \end{tikzpicture}
416\end{minipage}
417\end{thm}
418\begin{proof}
419 \emph{(proof from Thomas W. Judson book "Abstract Algebra" \cite{judson})}\\
420
421 Let $K=ker(\psi)$.
422 Since
423 $$\eta: G/K \longrightarrow \psi(G)$$
424 let
425 $$\eta: gK \longrightarrow \psi(g)$$
426 ie. $\eta(gK)=\psi(g)$.
427
428 \begin{enumerate}[i.]
429 \item show that $\eta$ is a \emph{well defined} map:
430
431 if $g_1 K=g_2 K$, then for some $k \in K$, $g_1 k =g_2$, so
432 $$\eta(g_1K)=\psi(g_1) = \psi(g_1)\psi(k) = \psi(g_1 k) = \psi(g_2) = \eta(g_2 k)$$
433
434 Thus, $\eta$ does not depend on the choice of coset representatives, and
435 the map $\eta: G/ker(\psi) \longrightarrow \psi(G)$ is uniquely defined
436 since $\psi=\eta\phi$.
437
438 \item show that $\eta$ is a homomorphism:
439
440 Observe:
441 $$\eta(g_1 K g_2 K) = \eta(g_1 g_2 K) = \psi(g_1 g_2) = \psi(g_1) \psi(g_2) = \eta(g_1 K) \eta(g_2 K)$$
442 $\Longrightarrow$ so $\eta$ is a homomorphism.
443
444 \item show that $\eta$ is an isomorphism:
445
446 Since each element of $H=\psi(G)$ has at least a preimage, then $\eta$ is \emph{surjective} (onto $\psi(G)$).
447
448 Show that it is also \emph{injective} (onet-to-one):
449
450 Suppose 2 different preimatges lead to the same image in $\psi(G)$, ie.
451 $\eta(g_1 K) = \eta(g_2 K)$
452
453 then,
454 $$\psi(g_1) = \psi(g_2)$$
455 which implies $\psi(g_1^{-1} g_2) = e$, ie. $g_1^{-1} g_2 \in ker(\psi)$,
456 hence
457 $$g_1^{-1} g_2 K = K$$
458 $$g_1 K = g_2 K$$
459 so $\eta$ is injective.
460 \end{enumerate}
461
462 Since $\eta$ is injective and surjective $\Longrightarrow$ $\eta$ is a bijective homomorphism,\\
463 ie. $\eta$ is an \emph{isomorphism}.
464\end{proof}
465
466
467\begin{thm}{i.2}(\emph{Second Isomorphism Theorem}) \label{2ndisothm}
468 Let $H \subseteq G$, $N \triangleleft G$. Then
469 \begin{enumerate}[i.]
470 \item $HN \subseteq G$
471 \item $H \cap N \triangleleft H$
472 \item $\frac{H}{H \cap N} \cong \frac{HN}{N}$
473 \end{enumerate}
474\end{thm}
475\begin{proof}
476 \emph{(proof from Thomas W. Judson book "Abstract Algebra" \cite{judson})}\\
477
478 \begin{enumerate}[i.]
479 \item show $HN \subseteq G$:\\
480 Note that $HN = \{ hn : h\in H, n\in N \}$. Let $h_1 n_1, h_2 n_2 \in HN$.
481
482 Since $N$ normal $\Longrightarrow~ h_2^{-1} n_1 h_2 \in N$, so
483 $$(h_1 n_1)(h_2 n_2) = h_1 h_2 (h_2^{-1} n_1 h_2) \in HN$$
484
485 [Recall: since $N \triangleleft G$, $gN=Ng ~\forall g \in G$ $\Longrightarrow gn=n'g$ for some $n' \in N$.]
486
487 To see that $(hn)^{-1} \in HN$:\\
488 since $(hn)^{-1} = n^{-1} h^{-1} = h^{-1} (h n^{-1} h^{-1})$, thus $(hn)^{-1} \in HN$.
489
490 Thus $HN \subseteq G$.
491
492
493 In fact, $$HN = \bigcup_{h \in H} hN$$
494
495 (TODO: diagram)
496
497 \item show that $H \cap N \triangleleft H$:\\
498 Let $h \in H,~ n \in H \cap N$ (recall: $H \cap N \subseteq H$).\\
499 Then $h^{-1}nh \in H$ $\longleftarrow$ since $h^{-1}, n, h \in H$.\\
500 Since $N \triangleleft G$, $h^{-1} n h \in N$.\\
501 Therefore, $h^{-1}nh \in H \cap N$ $\Longrightarrow~ H \cap N \triangleleft H$
502
503 \item show that $\frac{H}{H \cap N} \cong \frac{HN}{N}$:\\
504 Define a map
505 \begin{align*}
506 \phi: &H \longrightarrow \frac{HN}{N}\\
507 \text{by}~ \phi: &h \longmapsto hN
508 \end{align*}
509
510 $\phi$ is surjective (onto), since any coset $hnN=hN$ is the image of $h \in H$, ie. $\phi(h)$
511
512 $\phi$ is a homomorphism, since
513 $$\phi(h h') = h h' N = hN h'N = \phi(h)\phi(h')$$
514
515 By the First Isomorphism Theorem \ref{1stisothm},
516 $$\frac{HN}{N} \cong \frac{H}{ker(\phi)}$$
517
518 and since
519 \begin{align*}
520 &ker(\phi) = \{ h \in H : h \in N\}\\
521 &\text{then}~~ker(\phi) = H \cap N
522 \end{align*}
523
524 so then,
525 $$\frac{HN}{N} = \phi(H) \cong \frac{H}{ker(\phi)} = \frac{H}{H \cap N}$$
526 thus
527 $$\frac{HN}{N} \cong \frac{H}{H \cap N}$$
528
529 \end{enumerate}
530\end{proof}
531
532
533\begin{thm}{i.3}(\emph{Third Isomorphism Theorem}) \label{2ndisothm}\\
534 Let $H \subseteq K$ and $K \triangleleft G,~ H \triangleleft G$.\\
535 Then
536 $\frac{K}{H} \triangleleft \frac{G}{H}$
537 and
538 $$\frac{ G/H }{ K/H } \cong \frac{G}{K}$$
539\end{thm}
540\begin{proof}
541 \emph{(proof from Dummit and Foote book ``Abstract Algebra" \cite{dummitfoote})}\\
542
543 Easy to see that $\frac{K}{H} \triangleleft \frac{G}{H}$.
544
545 Define
546 \begin{align*}
547 \psi: &\frac{G}{H} \longrightarrow \frac{G}{K}\\
548 \text{by}~ \psi: &gH \longmapsto gK
549 \end{align*}
550
551 To show that $\psi$ is \emph{well defined}:\\
552 suppose $g_1 H = g_2 H$, then $g_1 = g_2 h$ for some $h \in H$.\\
553 Since $H \subseteq K \Longrightarrow~ h \in K$, hence $g_1 K = g_2 K$,\\
554 ie. $\psi(g_1 H) = \psi(g_2 H)$, which shows that $\psi$ is well defined.
555
556 Since $g \in G$ may be chosen arbitrarily in $G$, $\psi$ is a surjective homomorphism.
557
558
559 Finally,
560 \begin{align*}
561 ker(\psi) &= \{ gH \in \frac{G}{H} \mid \psi(gH)=1K \}\\
562 &= \{ gH \in \frac{G}{H} \mid gK =1K \}\\
563 &= \{ gH \in \frac{G}{H} \mid g \in K \}\\
564 &= \frac{K}{H}
565 \end{align*}
566
567
568 By the First Isomorphism Theorem (\ref{1stisothm}),
569
570 \begin{tikzpicture}[node distance=2.5cm, auto]
571 \node (GmodH) {$\frac{G}{H}$};
572 \node (GmodK) [right of=GmodH] {$\frac{G}{K}$};
573 \node (GmodHmodKer) [below of=G, xshift=1.25cm] {$\frac{G/H}{\ker(\psi)} = \frac{G/H}{K/H}$};
574
575 \draw[->] (GmodH) to node {$\psi$} (GmodK);
576 \draw[->] (GmodH) to node [swap] {$\phi$} (GmodHmodKer);
577 \draw[->] (GmodHmodKer) to node [swap] {$\eta$} (GmodK);
578 \end{tikzpicture}
579
580 So, by
581 $$\eta: \frac{G/H}{K/H} \longrightarrow \frac{G}{K}$$
582 since $\eta$ is
583 bijective (we know it by the First Isomorphism Theorem), $\eta$ it is the isomorphism:
584 $$\frac{ G/H }{ K/H } \cong \frac{G}{K}$$
585
586
587\end{proof}
588
589\subsection{Chapter 14}
590
591\begin{defn}{14.1}
592 a group $G$ is soluble if it has a finite series of subgroups
593 $$1=G_0 \subseteq G_1 \subseteq \ldots \subseteq G_n = G$$
594 such that
595 \begin{enumerate}[i.]
596 \item $G_i \lhd G_{i+1}$ for $i=0,\ldots,n-1$
597 \item $\frac{G_{i+1}}{G_{i+1}}$ is Abelian for for $i=0,\ldots,n-1$
598 \end{enumerate}
599
600 (Note: $G_i \lhd G_{i+1} \lhd G_{i+2}$ does not imply $G_i \lhd G_{i+2}$)
601\end{defn}
602
603\begin{thm}{14.4}
604 $H \subseteq G,~~ N \triangleleft G$, then
605 \begin{enumerate}
606 \item if $G$ soluble $\Longrightarrow H$ soluble
607 \item if $G$ soluble $\Longrightarrow G/N$ soluble
608 \item if $N$ and $G/N$ soluble $\Longrightarrow G$ soluble
609 \end{enumerate}
610\end{thm}
611\begin{proof}
612 \begin{enumerate}
613 \item Since $G$ soluble, by definition: $\exists~~ 1=G_0 \triangleleft G_1 \triangleleft \ldots \triangleleft G_r = G$ with Abelian quotients $\frac{G_{i+1}}{G_i}$.
614
615 Let $H_i = G_i \cap H$, then $H$ has a series $1=H_0 \triangleleft H_1
616 \triangleleft \ldots \triangleleft H_r = H$, next we show that the
617 quotients $\frac{H_{i+1}}{H_i}$ are Abelian (so that H is soluble):
618
619 $$\frac{H_{i+1}}{H_i} = \frac{G_{i+1} \cap H}{G_i \cap H} \stackrel{(*)}{=} \frac{G_{i+1} \cap H}{G_i \cap (G_{i+1}\cap H)}
620 \stackrel{(**)}{\cong} \frac{G_i(G_{i+1} \cap H)}{G_i} \subseteq \frac{G_{i+1}}{G_i}
621 $$
622
623 (*): to see why, $H_i = G_i \cap H = G_i \cap H_i = G_i \cap H_{i+1} = G_i \cap (G_{i+1} \cap H)$.
624 (**): by the 2nd Isomorphism Theorem (\ref{2ndisothm}).
625
626 [TODO: diagram of subgroups] % TODO %
627
628 Notice that $\frac{G_{i+1}}{G_i}$ is Abelian, thus the left-hand-side of the congruence is also Abelian.
629 Therefore, $\frac{H_{i+1}}{H_i}$ is Abelian, thus $H$ is soluble.
630
631 \item For $G/N$ to be soluble, (by definition) it would have the series $\frac{N}{N} = G_0 \frac{N}{N} \triangleleft G_1 \frac{N}{N} \triangleleft \ldots \triangleleft G_r \frac{N}{N} = \frac{G}{N}$,
632 and any quotient being $\frac{G_{i+1}\frac{N}{N}}{G_i \frac{N}{N}}$.
633
634 The series clearly exists, so now we show that the quotients are Abelian, so that $G/N$ is soluble:
635
636 $$
637 \frac{G_{i+1} N}{G_i N} = \frac{G_{i+1}(G_i N)}{G_i N} \stackrel{(*)}{\cong}
638 \frac{G_{i+1}}{G_{i+1} \cap (G_i N)} \cong \frac{G_{i+1}/G_i}{(G_{i+1}
639 \cap (G_i N))/G_i}
640 $$
641
642 (*): by the 2nd Isomorphism Theorem (\ref{2ndisothm}).
643
644 The last quotient is a quotient of the Abelian group $G_{i+1}/G_i$, so it is Abelian.
645
646 Hence, $\frac{G_{i+1}N}{G_i N}$ is also Abelian; so $\frac{G}{N}$ is soluble.
647
648 \item
649 By the definition of $N$ and $G/N$ being soluble,
650 \begin{align*}
651 N \text{soluble} \Longrightarrow~~ 1 &= N_0 \triangleleft N_1 \triangleleft \ldots \triangleleft N_r = N\\
652 G/N \text{soluble} \Longrightarrow~~ 1= \frac{N}{N} &= \frac{G_0}{N} \triangleleft \frac{G_1}{N} \triangleleft \ldots \triangleleft \frac{G_r}{N} = \frac{G}{N}
653 \end{align*}
654 both with Abelian quotients.
655
656 Consider the series of $G$ given by combining the two previous series:
657
658 $$
659 1 &= N_0 \triangleleft N_1 \triangleleft \ldots \triangleleft N_r = N = G_0 \triangleleft G_1 \triangleleft \ldots \triangleleft G_r = G
660 $$
661 the quotients are either
662 \begin{itemize}
663 \item $\frac{N_{i+1}}{N_i}$, Abelian
664 \item $\frac{G_{i+1}}{G_i}$, isomorphic to $\frac{G_{i+1}/N}{G_i/N}$,
665 which is Abelian.
666 \end{itemize}
667 \end{enumerate}
668 Therefore, the quotients are always Abelian; hence $G$ is soluble.
669\end{proof}
670
671\newpage
672
673\section{Tools}
674This section contains tools that I found useful to solve Galois Theory related problems, and that don't appear in Stewart's book.
675
676\subsection{De Moivre's Theorem and Euler's formula}\label{demoivre}
677Useful for finding all the roots of a polynomial.
678
679Euler's formula:
680$$e^{i \psi} = cos \psi + i \cdot sin \psi$$
681
682The n-th roots of a complex number $z=x + i y = r (cos \theta + i \cdot sin \theta)$ are given by
683
684$$z_k = \sqrt[n]{r} \cdot \left(cos(\frac{\theta + 2k \pi}{n}) + i \cdot sin(\frac{\theta + 2k \pi}{n}) \right)$$
685for $k=0, \ldots, n-1$.
686
687So, by Euler's formula:
688$$z_k = \sqrt[n]{r} \cdot e^{i (\frac{\theta + 2 k \pi}{n})}$$
689
690Usually we will set $\alpha=\sqrt[n]{r}$ and $\zeta = e^{\frac{2 \pi i}{n}}$,
691and find the $\mathbb{Q}$-automorphisms from there (see \ref{ex:galoisgroups} for examples).
692
693\subsection{Einsenstein's Criterion} \label{einsenstein}
694\emph{reference: Stewart's book}
695
696Let $f(t) = a_0 + a_1 t + \ldots + a_n t^n$, suppose there is a prime $q$ such that
697\begin{enumerate}
698 \item $q \nmid a_n$
699 \item $q | a_i$ for $i=0, \ldots, n-1$
700 \item $q^2 \nmid a_0$
701\end{enumerate}
702Then, $f$ is irreducible over $\mathbb{Q}$.
703
704\emph{TODO proof \& Gauss lemma.}
705
706
707\subsection{Elementary symmetric polynomials}
708\emph{TODO from orange notebook, page 36}
709
710\subsection{Cyclotomic polynomials} \label{cyclotomicpoly}
711\subsubsection{From Elmyn Berlekamp's "Algebraic Coding Theory" book}
712The notes in this section are from the book "Algebraic Coding Theory" by Elmyn
713Berlekamp \cite{berlekamp}.
714
715\vspace{0.3cm}
716
717Some times we might find polynomials that have the shape of $t^n - 1$, those are \emph{cyclotomic polynomials}, and have some properties that might be useful.
718
719Observe that in a finite field of order $q$, factoring $x^q - x$ gives
720$$x^q-x = x(x^{q-1} -1)$$
721
722The factor $x^{q-1} -1$ is a special case of $x^n -1$: if we assume that the
723field contains an element $\alpha$ of order $n$, then the roots of $x^n-1=0$ are
724$$1, \alpha, \alpha^2, \alpha^3, \ldots, \alpha^{n-1}$$
725and $\deg(x^n-1)=n$, thus $x^n-1$ has at most $n$ roots in any field, henceforth
726the powers of $\alpha$ must include all the $n$-th roots of unity.
727
728There fore, in any field which contains a primitive $n$-th root of unity we have:
729
730\begin{thm}{4.31}
731 $$x^n -1 = \prod_{i=0}^{n-1} (x - \alpha^i) = \prod_{i=1}^n (x-\alpha^i)$$
732\end{thm}
733
734If $n=k \cdot d$, then $\alpha^k, \alpha^{2k}, \alpha^{3k}, \ldots, \alpha^{dk}$ are all roots of $x^d -1 =0$
735
736Every element with order dividing $n$, must be a power of $\alpha$, since an
737element of order $d$ is a $d$-th root of unity.
738
739Every power of $\alpha$ has order which divides $n$, and every field element
740whose order divides $n$ is a power of $\alpha$. This suggests that we partition
741the powers of $\alpha$ according to their orders:
742$$x^n -1 = \prod_{\stackrel{d,}{d|n}} \prod_{\beta} (x- \beta)$$
743where at each iteration, $\beta$ is a field element of order $d$ for each $d$.
744
745The polynomial whose roots are the field elements of order $d$ is called the
746\emph{cyclotomic polynomial}, denoted by $Q^{(d)}(x)$.
747
748\begin{thm}{4.32}
749 $$x^n -1 = \prod_{\stackrel{d,}{d|n}} Q^{(d)}(x)$$
750\end{thm}
751
752
753\subsubsection{From Ian Stewart's ``Galois Theory'' book}
754Notes from Ian Stewart's book \cite{ianstewart}.
755
756Consider the case $n=12$, let $\zeta=e^{\pi i /6}$ be a primitive $12$-th root of unity.
757Classify its powers ($\zeta^j$) according to their minimal power $d$ such that
758$(\zeta^j)^d = 1$ (ie. when they are primitive $d$-th roots of unity).
759
760\begin{enumerate}[]
761 \item $d=1,~~~ 1$
762 \item $d=2,~~~ \zeta^6$
763 \item $d=3,~~~ \zeta^4, \zeta^8$
764 \item $d=4,~~~ \zeta^3, \zeta^9$
765 \item $d=6,~~~ \zeta^2, \zeta^{10}$
766 \item $d=12,~~~ \zeta, \zeta^5, \zeta^7, \zeta^{11}$
767\end{enumerate}
768
769Observe that we can factorize $t^{12} -1$ by grouping the corresponding zeros:
770\begin{align*}
771 t^{12}-1 = &(t-1) \times\\
772 &(t-\zeta^6) \times\\
773 &(t-\zeta^4) (t-\zeta^8) \times\\
774 &(t-\zeta^3) (t-\zeta^9) \times\\
775 &(t-\zeta^2) (t-\zeta^{10}) \times\\
776 &(t-\zeta) (t-\zeta^5)(t-\zeta^7) (t-\zeta^{11})
777\end{align*}
778which simplifies to
779$$t^{12}-1=(t-1)(t+1)(t^2+t+1)(t^2+1)(t^2-t+1)F(t)$$
780where $F(t) = (t-\zeta) (t-\zeta^5)(t-\zeta^7) (t-\zeta^{11}) = t^4 -t^2 + 1$ (this last step can be obtained either by multiplying $(t-\zeta)(t-\zeta^5)(t-\zeta^7) (t-\zeta^{11})$ together, or by dividing $t^{12}-1$ by all the other factors).
781
782
783Let $\Phi_d(t)$ be the factor corresponding to primitive $d$-th roots of unity, then we have proved that
784$$t^{12}-1 = \Phi_1 \Phi_2 \Phi_3 \Phi_4 \Phi_6 \Phi_{12}$$
785
786
787\begin{defn}{21.5}
788 The polynomial $\Phi_d(t)$ defined by
789 $$\Phi_n(t) = \prod_{a\in \mathbb{Z}_n,(a,n)=1} (t- \zeta^a)$$
790 is the $n$-th \emph{cyclotomic polynomial} over $\mathbb{C}$.
791\end{defn}
792
793\begin{cor}{21.6}
794 $\forall n \in \mathbb{N}$, the polynomial $\Phi_n(t)$ lies in $\mathbb{Z}[t]$ and is monic and irreducible.
795\end{cor}
796
797\begin{thm}{21.9}
798 \begin{enumerate}
799 \item The Galois group $\Gamma(\mathbb{Q}(\zeta):\mathbb{Q})$ consists of the
800 $\mathbb{Q}$-automorphisms $\psi_j$ defined by
801 $$\psi_j(\zeta)=\zeta^j$$
802 where $0 \leq j \leq n-1$ and $j$ is prime to $n$.
803
804 \item $\Gamma(\mathbb{Q}(\zeta):\mathbb{Q}) \stackrel{iso}{\cong} \mathbb{Z}_n^*$, and is an abelian group.
805 \item its order is $\phi(n)$
806 \item if $n$ is prime, $\mathbb{Z}_n^*$ is cyclic
807 \end{enumerate}
808\end{thm}
809
810
811
812\vspace{1cm}
813
814\subsubsection{Examples}
815Examples of cyclotomic polynomials:
816
817\begin{align*}
818 \Phi_n(x) &= x^{n-1} + x^{n-2} + \ldots + x^2 + x + 1 = \sum_{i=0}^{n-1} x^i\\
819 \Phi_{2p}(x) &= x^{p-1} + \ldots + x^2 - x + 1 = \sum_{i=0}^{p-1} (-x)^i\\
820 \Phi_m(x) &= x^{m/2} + 1, ~~\text{when $m$ is a power of $2$}
821\end{align*}
822
823
824\subsection{Lemma 1.42 from J.S.Milne's book}
825Lemma from J.S.Milne's book \cite{milneFT}.
826
827Useful for when dealing with $x^p - 1$ with $p$ prime.
828
829Observe that
830
831$$x^p -1 = (x-1)(x^{p-1} + x^{p-2} + \ldots + 1)$$
832
833Notice that
834$$\Phi_p(x) = x^{p-1} + x^{p-2} + \ldots + 1$$
835is the $p$-th Cyclotomic polynomial.
836
837\begin{lemma}{1.42}
838 If $p$ prime, then $x^{p-1} + \ldots + 1$ is irreducible; hence $\mathbb{Q}[e^{2 \pi i /p}]$ has degree $p-1$ over $\mathbb{Q}$.
839\end{lemma}
840\begin{proof}
841 Let $f(x) = (x^p - 1)/(x-1) = x^{p-1} + \ldots + 1$
842 then
843 $$
844 f(x+1) = \frac{(x+1)^p -1}{x+1-1} = \frac{(x+1)^p -1}{x} = x^{p-1} + \ldots + a_i x^i + \ldots + p
845 $$
846
847 with $a_i = \left( \stackrel{p}{i+1} \right)$.
848
849 We know that $p | a_i$ for $i= 1, \ldots, p-2$, therefore $f(x+1)$ is irreducibe by Einsenstein's Criterion.
850
851 This implies that $f(x)$ is irreducible.
852\end{proof}
853
854
855\subsection{Dihedral groups - Groups of symmetries} \label{dihedral}
856Source: Wikipedia and \cite{dihedral}.
857
858Dihedral groups ($\mathbb{D}_n$) represent the symmetries of a regular $n$-gon.
859
860Properties:
861\begin{itemize}
862 \item are non-abelian (for $n>2$), ie. $rs \neq sr$
863 \item order $2n$
864 \item generated by a rotation $r$ and a reflection $s$
865 \item $r^n = s^2 = id,~~~(rs)^2=id$
866\end{itemize}
867Subgroups of $\mathbb{D}_n$:
868\begin{itemize}
869 \item rotation form a cyclic subgroup of order $n$, denoted as $<r>$
870 \item for each $d$ such that $d|n$, $\exists~ \mathbb{D}_d$ with order $2d$
871 \item normal subgroups
872 \begin{itemize}
873 \item for $n$ odd: $\mathbb{D}_n$ and $<r^d>$ for every $d|n$
874 \item for $n$ even: $2$ additional normal subgroups
875 \end{itemize}
876 \item Klein four-groups: $\mathbb{Z}_2 \times \mathbb{Z}_2$, of order 4
877\end{itemize}
878
879\vspace{0.3cm}
880Total number of subgroups in $\mathbb{D}_n$: $d(n) + s(n)$, where $d(n)$ is the number of positive disivors of $n$, and $s(n)$ is the sum of those divisors.
881
882\begin{eg}{}
883For $\mathbb{D}_6$, we have $\{1,2,3,6\} | 6$, so $d(n) = d(6) = 4$, and
884$s(6) = 1+2+3+6 = 12$; henceforth, the total amount of subgroups is $d(n)+s(n) = 4+12 = 16$.
885\end{eg}
886
887\vspace{0.3cm}
888For $n \geq 3, ~~\mathbb{D}_n \subseteq \mathbb{S}_n$ (subgroup of the Symmetry group).
889
890\subsection{Rolle's theorem} \label{rolle}
891TODO
892
893
894\newpage
895
896\section{Exercises}
897
898\subsection{Galois groups}\label{ex:galoisgroups}
899
900\subsubsection[t6-7]{$t^6-7 \in \mathbb{Q}$}
901
902This exercise comes from a combination of exercises 12.4 and 13.7 from \cite{ianstewart}.
903
904First let's find the roots. By De Moivre's Theorem (\ref{demoivre}), $t_k =
905\sqrt[6]{7} \cdot e^{i \frac{2 \pi k}{6}}$.
906
907From which we denote $\alpha = \sqrt[6]{7}$, and $\zeta = e^{\frac{2 \pi i}{6}}$, so that the
908roots of the polynomial are $\{ \alpha, \alpha \zeta, \alpha \zeta^2, \alpha \zeta^3, \alpha \zeta^4, \alpha \zeta^5\}$, ie.
909$\{ \alpha \zeta^k \}_0^5$.
910
911Hence the \emph{splitting field} is $\mathbb{Q}(\alpha, \zeta)$.
912
913\emph{Degree of the extension}
914
915In order to find $[\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}$, we're going to split it in tow
916parts. By the Tower Law (\ref{towerlaw}),
917
918$$[\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}] = [\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}(\alpha)] \cdot [\mathbb{Q}(\alpha) : \mathbb{Q}]$$
919
920To find each degree, we will find the minimal polynomial of the adjoined term over the base field of the extension:
921
922\begin{enumerate}[i.]
923 \item minimal polynomial of $\alpha$ over $\mathbb{Q}$\\
924 By Einsenstein's Criterion (\ref{einsenstein}), with $q=7$ we have that $q
925 \nmid 1$, $7 | {-7,0,0,\ldots}$, and $7^2 \nmid -7$, hence $f(t)$ is
926 irreducibe over $\mathbb{Q}$, thus is the minimal polynomial
927 $$m_i(t)= f(t) =t^6-7$$
928 which has roots $\{ \alpha \zeta^k \}_0^5$.
929 \item minimal polynomial of $\zeta$ over $\mathbb{Q}(\alpha)$\\
930 Since $\zeta$ is the primitive $6$th root of unity, we know that the minimal
931 polynomial will be the $6$th cyclotomic polynomial (\ref{cyclotomicpoly}):
932 $$m_{ii}(t) = \Phi_6(t) = t^2 - t + 1$$
933 which has roots $\zeta, -\zeta$.
934
935 Since $\mathbb{Q}(\alpha) \subseteq \mathbb{R}$, and the roots of
936 $\Phi_6(t)=t^2 - t +1$ are in $\mathbb{C}$, $\Phi_6(t)$ remains irreducible
937 over $\mathbb{Q}(\alpha)$.
938\end{enumerate}
939
940\vspace{0.5cm}
941Therefore, by the tower of law,
942$$[\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}] = \deg{\Phi_6(t)} \cdot \deg{f(t)} = 2 \cdot 6 = 12$$
943and by the Fundamental Theorem of Galois Theory, we know that
944$$|\Gamma( \mathbb{Q}(\alpha, \zeta) : \mathbb{Q} )| = [\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}] = 12$$
945which tells us that there exist $12$ $\mathbb{Q}$-automorphisms of the Galois group.
946
947
948\vspace{0.5cm}
949Let's find the $12$ $\mathbb{Q}$-automorphisms. Start by defining $\sigma$ which
950fixes $\zeta$ and acts on $\alpha$, sending it to another of the roots of the
951minimal polynomial of $\alpha$ over $\mathbb{Q}$, $f(t)$, choose $\alpha \zeta$.
952
953Now define $\tau$ which fixes $\alpha$ and acts on $\zeta$, sending it into
954another root of the minimal polynomial of $\zeta$ over $\mathbb{Q}(\alpha)$,
955choose $-\zeta$.
956
957\vspace{0.3cm}
958\begin{tabular}{@{}l l@{}}
959 $\begin{aligned}
960 \sigma: \alpha &\mapsto \alpha \zeta \\
961 \zeta &\mapsto \zeta
962 \end{aligned}$
963 &
964 $\begin{aligned}
965 \tau: \alpha &\mapsto \alpha\\
966 \zeta &\mapsto -\zeta = \zeta^{-1}
967 \end{aligned}$
968\end{tabular}
969
970In other words, we have $12$ $\mathbb{Q}$-automorphisms, which are the
971combination of $\sigma$ and $\tau$:
972
973$$\begin{aligned}
974 \sigma^k \tau^j:~~&\alpha \mapsto \alpha \zeta^k\\
975 &\zeta \mapsto \zeta^j
976\end{aligned}$$
977
978for $0 \leq k \leq 5$ and $j = \pm 1$.
979
980\vspace{0.5cm}
981
982NOTE: WIP diagram.
983\begin{tikzpicture}[node distance=2cm]
984 \def \radius{2}
985 \draw (0,0) circle (\radius);
986
987 \foreach \k in {0,...,5} {
988 % \node (a\k) at ({360/6 * \k}:\radius) {$\alpha \zeta^{\k}$};
989 \node (a\k) at ({360/6 * \k}:\radius+0.5) {$\alpha \zeta^{\k}$};
990 \fill ({360/6 * \k}:\radius) circle (2pt);
991 }
992 % real & im axis
993 \draw[->] (-2.5,0) -- (2.5,0) node[right] {};
994 \draw[->] (0,-2.5) -- (0,2.5) node[above] {};
995
996
997 % tau:
998 \draw[<->] (3,1) -- (3,-1) node[right] {$\tau$};
999 % sigma:
1000 % \foreach \k [evaluate=\k as \next using int(mod(\k+1,6))] in {0,...,5} {
1001 % \coordinate (p\k) at ({360/6 * \k}:\radius);
1002 % \coordinate (p\next) at ({360/6 * \next}:\radius);
1003 %
1004 % \draw[->, bend left=30] (p\k.center) -- node[above] {$\sigma$} (p\next.center);
1005 % }
1006 \foreach \k in {0,...,5} {
1007 \coordinate (p\k) at ({360/6 * \k}:\radius);
1008 }
1009 \foreach \k [evaluate=\k as \next using int(mod(\k+1,6))] in {0,...,5} {
1010 \draw[->, bend left=30] (p\k) -- node[above] {$\sigma$} (p\next);
1011 }
1012\end{tikzpicture}
1013
1014\vspace{0.5cm}
1015
1016Observe, that $\Gamma$ is generated by the combination of $\sigma$ and $\tau$,
1017and it is isomorphic to the group of symmetries of order 12, the dihedral
1018group (\ref{dihedral}) of order 12, $\mathbb{D}_6$, ie. $\Gamma \cong \mathbb{D}_6$.
1019
1020\vspace{0.5cm}
1021
1022Let's find the subgroups of $\Gamma$, and the fixed fields of $\mathbb{Q}(\alpha, \zeta)$.
1023
1024We know that $\Gamma \cong \mathbb{D}_6$, and we know from the properties
1025of the dihedral group (\ref{dihedral}) that the number of subgroups of
1026$\mathbb{D}_6$ will be $d(6) + s(6) = 4 + 12 = 16$ subgroups.
1027
1028
1029\vspace{0.4cm}
1030
1031\hspace*{-3.5cm}
1032\begin{tabular}{ c c c c | p{7.5cm} }
1033 \hline
1034 generators & order & group & fixed field & notes (check fixed field)\\
1035\hline
1036 $\langle \rangle = \langle \sigma^6 \rangle=\langle \tau^2 \rangle$ & 1 & id & $\mathbb{Q}(\alpha,\zeta)$ & \\
1037 $\langle \sigma \rangle = \langle \sigma^5 \rangle$ & 6 & $\mathbb{Z}_6$ & $\mathbb{Q}(\zeta)$ & \\
1038 $\langle \sigma^2 \rangle=\langle \sigma^4 \rangle$ & 3 & $\mathbb{Z}_3$ & $\mathbb{Q}(\alpha^3, \zeta)$ & $\sigma^2(\alpha^3)=\alpha^3 \zeta^{3\cdot 2}=\alpha^3 \zeta^6 = \alpha^3 \cdot 1 = \alpha^3$\\
1039 $\langle \sigma^3 \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha^2,\zeta)$ & $\sigma^3(\alpha^2)=(\alpha\zeta^3)^2=\alpha^2\zeta^6=\alpha^2$\\
1040 \hline
1041 $\langle \tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha)$ & \\
1042 \hline
1043 $\langle \sigma\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta)$ &
1044 $\sigma\zeta(\alpha+\alpha\zeta)=\sigma(\alpha+\alpha\zeta^{-1}) = \alpha\zeta + \alpha\zeta^{-1}\zeta=\alpha\zeta+\alpha$\\
1045 $\langle \sigma^2\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^2), \mathbb{Q}(\alpha\zeta)$ &
1046 $\sigma^2\tau(\alpha+\alpha\zeta^2) = \sigma(\alpha+\alpha\zeta^{-2})=\alpha\zeta^2+ \alpha\zeta^{-2}\zeta^2=\alpha\zeta^2+\alpha$\\
1047 $\langle \sigma^3\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^3)$ &
1048 $\sigma^3\tau(\alpha+\alpha\zeta^3) = \sigma(\alpha+\alpha\zeta^{-3})=\alpha\zeta^3+ \alpha\zeta^{-3}\zeta^3=\alpha\zeta^3+\alpha$\\
1049 $\langle \sigma^4\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^4), \mathbb{Q}(\alpha\zeta^2)$ &
1050 $\sigma^4\tau(\alpha+\alpha\zeta^4) = \sigma(\alpha+\alpha\zeta^{-4})=\alpha\zeta^4+ \alpha\zeta^{-4}\zeta^4=\alpha\zeta^4+\alpha$\\
1051 $\langle \sigma^5\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^5)$ &
1052 $\sigma^5\tau(\alpha+\alpha\zeta^5) = \sigma(\alpha+\alpha\zeta^{-5})=\alpha\zeta^5+ \alpha\zeta^{-5}\zeta^5=\alpha\zeta^5+\alpha$\\
1053 \hline
1054 $\langle \sigma, \tau \rangle = \langle \sigma^5,\tau \rangle$ & $6\cdot2=12$ & $\mathbb{D}_6$ & $\mathbb{Q}$ & \\
1055 $\langle \sigma^2, \tau \rangle = \langle \sigma^4,\tau \rangle$ & $3\cdot2=6$ & $\mathbb{D}_3$ & $\mathbb{Q}(\alpha^3)$ &
1056 $\sigma^2(\alpha^3)=\alpha^3\zeta^{3\cdot 2}=\alpha^3$ and $\tau(\alpha^3)=\alpha^3$\\
1057 $\langle \sigma^3, \tau \rangle$ & $2\cdot2=4$ & $\mathbb{D}_2$ & $\mathbb{Q}(\alpha^2)$ &
1058 $\sigma^3(\alpha^2)=\alpha^2\zeta^{2\cdot 2}=\alpha^2$ and $\tau(\alpha^2)=\alpha^2$\\
1059 \hline
1060 $\langle \sigma^2, \sigma\tau \rangle$ & $3\cdot 2=6$ & $\mathbb{D}_3$ & $\mathbb{Q}(\alpha^3+\alpha^3\zeta^3)$ &
1061 $\sigma^2(\alpha^3 + \alpha^3 \zeta^3) = \alpha^3\zeta^3 + \alpha^3 \zeta^3\zeta^3 = \alpha^3\zeta^3 + \alpha^3\zeta^6 = \alpha^3\zeta^3+\alpha^3$\\
1062 $\langle \sigma^3, \sigma\tau \rangle$ & $2\cdot2=4$ & $\mathbb{Z}_2 \times \mathbb{Z}_2$ & $\mathbb{Q}(\alpha^2\zeta^2),\mathbb{Q}(\alpha^2+\alpha^2\zeta^2)$ &
1063 $\sigma^3(\alpha^2+\alpha^2\zeta^2)=\alpha^2\zeta^{2\cdot3}+\alpha^2\zeta^{2\cdot3}\zeta^2=\alpha^2+\alpha^2\zeta^2$
1064 and
1065 $\sigma\tau(\alpha^2+\alpha^2\zeta^2)=\alpha^2\zeta^2+\alpha^2\zeta^{-2}\zeta^2 = \alpha^2\zeta^2+\alpha^2$\\
1066 $\langle \sigma^3, \sigma^2\tau\rangle$ & $2\cdot2=4$ & $\mathbb{Z}_2 \times \mathbb{Z}_2$ & $\mathbb{Q}(\alpha^2\zeta^4),\mathbb{Q}(\alpha^2+\alpha^2\zeta^4)$ &
1067 $\sigma^2\zeta(\alpha^2\zeta^4)=\alpha^2\zeta^2\zeta^{-4}=\alpha^2\zeta^{-2}=\alpha^2\zeta^4$
1068 and $\sigma^3(\alpha^2\zeta^4)=\alpha^2\zeta^{2\cdot3}\zeta^4=\alpha^2\zeta^4$
1069\end{tabular}
1070
1071
1072
1073\bibliographystyle{unsrt}
1074\bibliography{galois-theory-notes.bib}
1075
1076\end{document}