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}
62\newenvironment{defn}[1]
63{\renewcommand\theinnerdefn{#1}\innerdefn}
64{\endinnerdefn}
65
66\newtheorem{innerthm}{Theorem}
67\newenvironment{thm}[1]
68{\renewcommand\theinnerthm{#1}\innerthm}
69{\endinnerthm}
70
71\newtheorem{innerlemma}{Lemma}
72\newenvironment{lemma}[1]
73{\renewcommand\theinnerlemma{#1}\innerlemma}
74{\endinnerlemma}
75
76\newtheorem{innerprop}{Proposition}
77\newenvironment{prop}[1]
78{\renewcommand\theinnerprop{#1}\innerprop}
79{\endinnerprop}
80
81\newtheorem{innercor}{Corollary}
82\newenvironment{cor}[1]
83{\renewcommand\theinnercor{#1}\innercor}
84{\endinnercor}
85
86\newtheorem{innereg}{Example}
87\newenvironment{eg}[1]
88{\renewcommand\theinnereg{#1}\innereg}
89{\endinnereg}
90
91\newtheorem{innerex}{Exercise}
92\newenvironment{ex}[1]
93{\renewcommand\theinnerex{#1}\innerex}
94{\endinnerex}
95
96
97\title{Galois Theory notes}
98\author{arnaucube}
99\date{2025}
100
101\begin{document}
102
103\maketitle
104
105\begin{abstract}
106 Notes taken while studying Galois Theory, mostly from Ian Stewart's book "Galois Theory" \cite{ianstewart}.
107
108 Usually while reading books and papers I take handwritten notes in a notebook, this document contains some of them re-written to $LaTeX$.
109
110 The notes are not complete, don't include all the steps neither all the proofs.
111\end{abstract}
112
113\tableofcontents
114
115\section{Galois Theory notes}
116\subsection{Chapters 4-12}
117(Definitions, theorems, lemmas, corollaries and examples enumeration follows from Ian Stewart's book \cite{ianstewart}).
118
119\begin{defn}{4.10}
120 A \emph{simple extension} is $L:K$ such that $L=K(\alpha)$ for some $\alpha \in L$.
121\end{defn}
122\begin{eg}{4.11}
123 Beware, $L=\mathbb{Q}(i, -i, \sqrt{5}, -\sqrt{5}) = \mathbb{Q}(i, \sqrt{5}) = \mathbb{Q}(i+\sqrt{5})$.
124\end{eg}
125
126\begin{defn}{5.5}[Minimal polynomial]
127 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 $m(\alpha)=0$.
128 \\
129 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 $m(i)=0$.
130\end{defn}
131
132\begin{lemma}{5.9}
133 Every polynomial $a \in K[t]$ is congruent modulo $m$ to a unique polynomial of degree $< \delta m$.
134\end{lemma}
135\begin{proof}
136 Divide $a / m$ with remainder, $a= qm +r$, with $q,r \in K[t]$ and $\delta r < \delta m$.
137 Then, $a-r=qm$, so $a \equiv r \pmod{m}$.
138
139 It remains to prove uniqueness.
140
141 Suppose $\exists~ r \equiv s \pmod{m}$, with $\delta r, \delta s < \delta m$.
142 Then, $r-s$ is divisible by $m$, but has smaller degree than $m$.
143
144 Therefore, $r-s=0$, so $r=s$, proving uniqueness.
145\end{proof}
146
147\begin{thm}{5.10}
148 $\forall 0 \neq f \in \frac{K[t]}{<m>},~~ \exists f^{-1}$ iff $m$ is irreducible in $K[t]$.\\
149 Then $\frac{K[t]}{<m>}$ is a field.
150\end{thm}
151\begin{thm}{5.12} \label{5.12}
152 Let $K(\alpha):K$ simple algebraic extension, let $m$ minimal polynomial of $\alpha$ over $K$.\\
153 $K(\alpha):K$ is isomorphic to $\frac{K[t]}{<m>}$.\\
154 The isomorphism $\frac{K[t]}{<m>} \longrightarrow K(\alpha)$ can be chosen to map $t$ to $\alpha$.
155
156\end{thm}
157\begin{cor}{5.13} \label{5.13}
158 Let $K(\alpha):K$ and $K(\beta):K$ be simple algebraic extensions.\\
159 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 $\beta$.
160\end{cor}
161\begin{proof}
162 By \ref{5.12}, both extensions are isomorphic to $\frac{K[t]}{<m>}$.
163\end{proof}
164
165\begin{lemma}{5.14}
166 Let $K(\alpha):K$ be a simple algebraic extension, let $m$ be the minimal polynomial of $\alpha$ over $K$, let $\delta m =n$.
167
168 Then $\{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}\}$ is a basis for $K(\alpha)$ over $K$.
169 In particular, $[K(\alpha):K]=n$.
170\end{lemma}
171
172\begin{defn}{6.2}
173 The degree $[L:K]$ of a field extension $L:K$ is the dimension of L considered as a vector space over $K$.
174
175 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 $K$.
176\end{defn}
177
178\begin{eg}{6.3}
179 \begin{enumerate}
180 \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 $[\mathbb{C}:\mathbb{R}]=2$.
181 \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 $\mathbb{Q}$.
182 \end{enumerate}
183\end{eg}
184
185\begin{thm}{6.4}\emph{(Short Tower Law)} \label{shorttowerlaw}
186 If $K, L, M \subseteq \mathbb{C}$, and $K \subseteq L \subseteq M$, then $[M:K]=[M:L]\cdot [L:K]$.
187\end{thm}
188\begin{proof}
189 Let $(x_i)_{i \in I}$ be a basis for $L$ over $K$,
190 let $(y_j)_{j \in J}$ be a basis for $M$ over $L$.\\
191 $\forall i \in I, j \in J$, we have $x_i \in L, u_j \in M$.
192 \\
193 Want to show that $(x_i y_j)_{i\in I, j\in J}$ is a basis for $M$ over $K$.
194 \begin{enumerate}[i.]
195 \item prove linear independence:\\
196 Suppose that
197 $$\sum_{ij} k_{ij} x_i y_j = 0 ~(k_{ij} \in K)$$
198 rearrange
199 $$\sum_j (\underbrace{\sum_i k_{ij} x_i}_{\in L}) y_j = 0 ~(k_{ij} \in K)$$
200 Since $\sum_i k_{ij} x_i \in L$, and the $y_j \in M$ are linearly independent over $L$, then $\sum_i k_{ij} x_i = 0$.
201 \\
202 Repeating the argument inside $L$ $\longrightarrow$ $k_{ij}=0 ~~\forall i\in I, j\in J$.
203 \\
204 So the elements $x_i y_j$ are linearly independent over $K$.
205
206 \item prove that $x_i y_j$ span $M$ over $K$:\\
207 Any $x \in M$ can be written
208 $$x=\sum_j \lambda_j y_j$$
209 for $\lambda_j \in L$, because $y_j$ spans $M$ over $L$.
210 Similarly,
211 $$\forall j\in J,~ \lambda_j = \sum_i \lambda_{ij} x_i y_j$$
212 for $\lambda_{ij} \in K$.\\
213 Putting the pieces together,
214 $$x=\sum_{ij} \lambda_{ij} x_i y_j$$
215 as required.
216 \end{enumerate}
217\end{proof}
218
219\begin{cor}{6.6}\emph{(Tower Law)}\\ \label{towerlaw}
220 If $K_0 \subseteq K_1 \subseteq \ldots \subseteq K_n$ are subfields of $\mathbb{C}$, then
221 $$[K_n:K_0] = [K_n:K_{n-1}] \cdot [K_{n-1}:K_{n-2}] \cdot \ldots \cdot [K_1: K_0]$$
222\end{cor}
223\begin{proof}
224 From \ref{shorttowerlaw}.
225\end{proof}
226
227\begin{thm}{6.7}
228 if $K(\alpha):K$
229 \begin{itemize}
230 \item transcendental $\Longrightarrow~~[K(\alpha):K] = \inf$
231 \item algebraic $\Longrightarrow~~[K(\alpha):K] = \delta m$
232 \end{itemize}
233 (where $m$ is the minimal polynomial of $\alpha$ over $K$).
234\end{thm}
235
236\begin{defn}{8.1}
237 $L:K$, a \emph{$K$-automorphism} of $L$ is an automorphism $\alpha$ of $L$ such that $\alpha(k)=k ~~ \forall k \in K$.\\
238 ie. $\alpha$ \emph{fixes} $k$.
239\end{defn}
240
241\begin{thm}{8.2, 8.3}
242 The set of all $K$-automorphisms of $L$ forms a group, $\Gamma(L:K)$, the Galois group of $L:K$.
243\end{thm}
244
245\begin{defn}{8.12}(Radical Extension) \label{8.12}
246 $L:K$ is radical if $L=K(\alpha_1, \ldots, \alpha_m)$ where for each $j=1, \ldots, m$, $\exists~ n_j$ such that $\alpha_j^{n_j} \in K(\alpha_1, \ldots, \alpha_{j-1})~~(j\geq 1)$
247\end{defn}
248
249\begin{lemma}{8.18}
250 Let $q \in L$. The minimal polynomial of $q$ over $K$ \emph{splits} into linear factors over L.
251\end{lemma}
252
253\begin{ex}{E.8.7}
254 TODO
255\end{ex}
256
257
258\begin{defn}{9.1}
259 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
260 $$f(t) = k \cdot (t- \alpha_1) \cdot \ldots \cdot (t - \alpha_n)$$
261 where $k, \alpha_i \in K$.
262
263 $\Longrightarrow$ (Thm 9.3) if $f$ splits over $\Sigma$, $\Sigma$ is the \emph{splitting field}.\\
264 If $K \subseteq \Sigma' \subseteq \Sigma$ and $f$ splits over $\Sigma'$, then $\Sigma' = \Sigma$.
265\end{defn}
266
267\begin{thm}{9.6} \label{9.6}
268 TODO
269\end{thm}
270
271\begin{defn}{9.8}
272 $L:K$ is \emph{normal} if every irreducible polynomial $f \in K[t]$ that has at least one zero in $L$, splits in $L$.
273\end{defn}
274
275\begin{thm}{9.9} \label{9.9}
276 TODO
277\end{thm}
278
279\begin{thm}{9.10}
280 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 equivalently, simple zeros in its splitting field.
281\end{thm}
282
283\begin{lemma}{9.13}
284 $f \in K[t]$ 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 $\Sigma[t]$.\\
285 More details at Rolle's theorem (\ref{rolle}) section.
286\end{lemma}
287
288\begin{thm}{10.5} \label{10.5}
289 $|\Gamma(K:K_0)| = [K:K_0]$, where $K_0$ is the fixed field of $\Gamma(K:K_0)$.
290\end{thm}
291
292
293\begin{defn}{11.1}
294 $K \subseteq L$, $K \subseteq L$. A $K$-monomorphism of $M$ into $L$ is a field monomorphism
295 $$\phi: M \longrightarrow L$$
296 such that $\phi(k)=k ~~ \forall k \in K$.
297\end{defn}
298
299\begin{thm}{11.3} \label{11.3}
300 $L:K$ normal, $K \subseteq M \subseteq L$. Let $\tau$ any $K$-monomorphism $\tau: M \longrightarrow L$.\\
301 Then, $\exists$ a $K$-automorphism $\sigma$ of $L$ such that $\sigma\biggr\vert_M=\tau$.
302\end{thm}
303\begin{proof}
304 $L:K$ normal $\Longrightarrow$ by Thm \ref{9.9}, $L$ splitting field for some poly $f \in K[t]$.
305
306 Hence, $L$ is splitting field over $M$ for $f$ and over $\tau(M)$ for $\tau(f)$.
307
308 Since $\tau \biggr\vert_K$ is the identity, $\tau(f)=f$.
309
310 \vspace{0.5cm}
311 We have
312
313 \begin{tikzpicture}[node distance=1.5cm, auto]
314 \node (M) {$M$};
315 \node (L) [right of=M] {$L$};
316 \node (t) [below of=M] {$\tau(M)$};
317 \node (L2) [below of=L] {$L$};
318
319 \draw[->] (M) to node {$ $} (L);
320 \draw[->] (M) to node {$\tau$} (t);
321 \draw[->] (t) to node {$ $} (L2);
322 \draw[->] (L) to node {$ $} (L2);
323 \end{tikzpicture}
324
325 with $\sigma$ yet to be formed.
326
327 By Theorem \ref{9.6}, $\exists$ isomorphism $\sigma: L \longrightarrow L$ such that $\sigma \biggr\vert_M = \tau$.\\
328 Therefore, $\sigma$ is an automorphism of $L$, and since $\sigma\biggr\vert_K = \tau\biggr\vert_K=id$, $\sigma$ is a $K$-automorphism of $L$.
329\end{proof}
330
331\begin{prop}{11.4} \label{11.4}
332 $L:K$ finite normal, $\alpha,~\beta$ are zeros in $L$ of the irreducible polynomial $p \in K[t]$.
333
334 Then, $\exists$ a $K$-automorphism $\sigma$ of $L$ such that $\sigma(\alpha)=\beta$.
335\end{prop}
336\begin{proof}
337 By Corollary \ref{5.13}, $\exists$ isomorphism $\tau: K(\alpha) \longrightarrow K(\beta)$ such that $\tau \biggr\vert_K$ is the identity, and $\tau(\alpha)=\beta$.
338
339 By Theorem \ref{11.3}, $\tau$ extends to a $K$-automorphism $\sigma$ of $L$.
340\end{proof}
341
342\begin{lemma}{11.8} \label{11.8}
343 $K \subseteq L \subseteq N \subseteq M$, $L:K$ finite, $N$ normal closure of $L:K$.\\
344 Let $\tau$ any $K$-monomorphism $\tau: L \longrightarrow M$.\\
345 Then $\tau(L) \subseteq N$.
346\end{lemma}
347\begin{proof}
348 $\alpha \in L$, $m$ minimal polynomial of $\alpha$ over $K$.\\
349 $\Longrightarrow ~~ m(\alpha)=0$, so $\tau(m(\alpha))=0$
350
351 (since $\tau$ is a $K$-automorphism, ie. maps the zeros of $m(t)$).\\
352
353 Since $\tau$ is a $K$-monomorphism, $\tau(m(\alpha))=m(\tau(\alpha))=0$
354
355 $\Longrightarrow~~ \tau(\alpha)$ is a zero of $m$.\\
356
357 Therefore, $\tau(\alpha)$ lies in $N$, since $N:K$ is normal.\\
358 Henceforth, $\tau(L) \subseteq N$.
359\end{proof}
360
361\begin{thm}{11.9}
362 The following are equivalent:
363 \begin{enumerate}
364 \item $L:K$ normal
365 \item $\exists$ finite normal extension $N$ of $K$ containing $L$,\\
366 such that every $K$-monomorphism $\tau: L \longrightarrow N$ is a $K$-automorphism of $L$.
367 \item for every finite extension $M$ of $K$ containing $L$,\\
368 every $K$-monomorphism $\tau: L \longrightarrow M$ is a $K$-automorphism of $L$.
369 \end{enumerate}
370\end{thm}
371
372\begin{thm}{11.10}
373 $[L:N]=1,~ N$ normal closure of $L:K$. Then,
374
375 $\exists~ n~ K$-monomorphisms $L \longrightarrow N$.\\
376 (the ones proven by Lemma \ref{11.8}).
377\end{thm}
378
379\begin{cor}{11.11} \label{11.11}
380 $|\Gamma(L:K)| = [L:K]$ (if $L:K$ is normal).
381
382 ie. there are precisely $[L:K]$ distinct $K$-automorphisms of $L$.
383\end{cor}
384
385\begin{thm}{11.12}
386 $\Gamma(L:K) = G$. If $L:K$ normal, then $K$ is the fixed field of $G$.
387\end{thm}
388\begin{proof}
389 let $K_0$ be the fixed field of $G$. Let $[L:K]=n$.\\
390 By \ref{11.11}, $|G| = [L:K] = n$.\\
391 By \ref{10.5}, $[L:K_0]=n$ ($K_0$ fixed field).\\
392 Since $K \subseteq K_0$, we must have $K=K_0$.
393
394 $\Longrightarrow$ thus $K$ is the fixed field of $G$.
395\end{proof}
396
397\begin{thm}{11.14}
398 if $L$ any field, $G$ any finite group of automorphisms of $L$, and $K$ its fixed field,
399
400 then $L:K$ is \emph{finite} and \emph{normal}, with Galois group $G$.
401\end{thm}
402
403
404\begin{thm}{12.2}(Fundamental Theorem of Galois Theory)
405 if $L:K$ finite and normal inside $\mathbb{C}$, with $\Gamma(L:K)=G$, then:
406 \begin{enumerate}
407 \item $|\Gamma(L:K)| = [L:K]$
408 (by Corollary \ref{11.11})
409 \item the maps * and $\dagger$ are mutual inverses, and setup an order-reversing one-to-one correspondence between $\mathcal{F}$ and $\mathcal{G}$.
410 \item if $M$ an intermediate field, then
411 $$[L:M] = |M^*|~~~~~~~ [M:K]=\frac{|G|}{|M^*|}$$
412 \item for $M$ an intermediate field, $M:K$ normal iff
413 $$\underbrace{\Gamma(M:K)}_{=M^*} \lhd \underbrace{\Gamma(L:K)}_{=G}$$
414 \item for $M$ intermediate, if $M:K$ normal, then
415 $$\Gamma(M:K) \cong \frac{G}{M^*}$$
416 ie.
417 $$\Gamma(M:K) \cong \frac{\Gamma(L:K)}{\Gamma(L:M)}$$
418 \end{enumerate}
419\end{thm}
420\begin{proof}
421 TODO
422\end{proof}
423
424\subsection{Chapter 13 - Full example}
425
426(Chapter 13 is basically a full example. More examples can be found at section \ref{ex:galoisgroups})
427
428
429\subsection{Detour: Isomorphism Theorems}
430\begin{thm}{i.1}(\emph{First Isomorphism Theorem}) \label{1stisothm}
431
432 \begin{minipage}{0.75\textwidth}
433 If $\psi: G \longrightarrow H$ a group homomorphism, then $ker(\psi) \triangleleft G$.\\
434 Let $\phi: G \longrightarrow G/ker(\psi)$ be the canonical homomorphism.\\
435 Then $\exists$ unique isomorphism $\eta: G/ker(\psi) \longrightarrow \psi(G)$ such that $\psi = \eta \phi$.\\
436 $\Longleftrightarrow$ ie. $G/ker(\psi) \cong \psi(G)$.
437 \end{minipage}
438\hfill
439\begin{minipage}{0.2\textwidth}
440 \begin{tikzpicture}[node distance=2.5cm, auto]
441 \node (G) {$G$};
442 \node (H) [right of=G] {$H$};
443 \node (GmodK) [below of=G, xshift=1.25cm] {$G/\ker(\psi)$};
444
445 \draw[->] (G) to node {$\psi$} (H);
446 \draw[->] (G) to node [swap] {$\phi$} (GmodK);
447 \draw[->] (GmodK) to node [swap] {$\eta$} (H);
448 \end{tikzpicture}
449\end{minipage}
450\end{thm}
451\begin{proof}
452 \emph{(proof from Thomas W. Judson book "Abstract Algebra" \cite{judson})}\\
453
454 Let $K=ker(\psi)$.
455 Since
456 $$\eta: G/K \longrightarrow \psi(G)$$
457 let
458 $$\eta: gK \longrightarrow \psi(g)$$
459 ie. $\eta(gK)=\psi(g)$.
460
461 \begin{enumerate}[i.]
462 \item show that $\eta$ is a \emph{well defined} map:
463
464 if we have two representatives of the same coset, ie. $g_1 K=g_2 K$, we want to show that $\eta(g_1 K) = \eta(g_2 K)$, so that $\eta$ is a well-defined map.
465
466 \vspace{0.3cm}
467 By the coset properties for some $k \in K$, $g_1=g_2 k$, so
468
469 $$\eta(g_1K)=\psi(g_1) = \psi(g_2 k) = \eta(g_2 k K) = \eta(g_2 K)$$
470
471 Thus, $\eta$ does not depend on the choice of coset representatives, and
472 the map $\eta: G/ker(\psi) \longrightarrow \psi(G)$ is uniquely defined
473 since $\psi=\eta\phi$.
474
475 \item show that $\eta$ is a homomorphism:
476
477 Observe:
478 $$\eta(g_1 K g_2 K) = \eta(g_1 g_2 K) = \psi(g_1 g_2) = \psi(g_1) \psi(g_2) = \eta(g_1 K) \eta(g_2 K)$$
479 $\Longrightarrow$ so $\eta$ is a homomorphism.
480
481 \item show that $\eta$ is an isomorphism:
482
483 Since each element of $H=\psi(G)$ has at least a preimage, then $\eta$ is \emph{surjective} (onto $\psi(G)$).
484
485 Show that it is also \emph{injective} (onet-to-one):
486
487 Suppose 2 different preimatges lead to the same image in $\psi(G)$, ie.
488 $\eta(g_1 K) = \eta(g_2 K)$
489
490 then,
491 $$\psi(g_1) = \psi(g_2)$$
492 which implies $\psi(g_1^{-1} g_2) = e$, ie. $g_1^{-1} g_2 \in ker(\psi)$,
493 hence
494 $$g_1^{-1} g_2 K = K$$
495 $$g_1 K = g_2 K$$
496 so $\eta$ is injective.
497 \end{enumerate}
498
499 Since $\eta$ is injective and surjective $\Longrightarrow$ $\eta$ is a bijective homomorphism,\\
500 ie. $\eta$ is an \emph{isomorphism}.
501\end{proof}
502
503
504\begin{thm}{i.2}(\emph{Second Isomorphism Theorem}) \label{2ndisothm}
505 Let $H \subseteq G$, $N \triangleleft G$. Then
506 \begin{enumerate}[i.]
507 \item $HN \subseteq G$
508 \item $H \cap N \triangleleft H$
509 \item $\frac{H}{H \cap N} \cong \frac{HN}{N}$
510 \end{enumerate}
511\end{thm}
512\begin{proof}
513 \emph{(proof from Thomas W. Judson book "Abstract Algebra" \cite{judson})}\\
514
515 \begin{enumerate}[i.]
516 \item show $HN \subseteq G$:\\
517 Note that $HN = \{ hn : h\in H, n\in N \}$. Let $h_1 n_1, h_2 n_2 \in HN$.
518
519 Since $N$ normal $\Longrightarrow~ h_2^{-1} n_1 h_2 \in N$, so
520 $$(h_1 n_1)(h_2 n_2) = h_1 h_2 (h_2^{-1} n_1 h_2) \cdot n_2 \in HN$$
521
522 [Recall: since $N \triangleleft G$, $gN=Ng ~\forall g \in G$ $\Longrightarrow gn=n'g$ for some $n' \in N$.]
523
524 To see that $(hn)^{-1} \in HN$:\\
525 since $(hn)^{-1} = h^{-1} n^{-1} = h^{-1} (h n^{-1} h^{-1})$, thus $(hn)^{-1} \in HN$.
526
527 Thus $HN \subseteq G$.
528
529
530 In fact, $$HN = \bigcup_{h \in H} hN$$
531
532 (TODO: diagram)
533
534 \item show that $H \cap N \triangleleft H$:\\
535 Let $h \in H,~ n \in H \cap N$ (recall: $H \cap N \subseteq H$).\\
536 Then $h^{-1}nh \in H$ $\longleftarrow$ since $h^{-1}, n, h \in H$.\\
537 Since $N \triangleleft G$, $h^{-1} n h \in N$.\\
538 Therefore, $h^{-1}nh \in H \cap N$ $\Longrightarrow~ H \cap N \triangleleft H$
539
540 \item show that $\frac{H}{H \cap N} \cong \frac{HN}{N}$:\\
541 Define a map
542 \begin{align*}
543 \phi: &H \longrightarrow \frac{HN}{N}\\
544 \text{by}~ \phi: &h \longmapsto hN
545 \end{align*}
546
547 $\phi$ is surjective (onto), since any coset $hnN=hN$ is the image of $h \in H$, ie. $\phi(h)$
548
549 $\phi$ is a homomorphism, since
550 $$\phi(h h') = h h' N = hN h'N = \phi(h)\phi(h')$$
551
552 By the First Isomorphism Theorem \ref{1stisothm},
553 $$\frac{HN}{N} \cong \frac{H}{ker(\phi)}$$
554
555 and since
556 \begin{align*}
557 &ker(\phi) = \{ h \in H : h \in N\}\\
558 &\text{then}~~ker(\phi) = H \cap N
559 \end{align*}
560
561 so then,
562 $$\frac{HN}{N} = \phi(H) \cong \frac{H}{ker(\phi)} = \frac{H}{H \cap N}$$
563 thus
564 $$\frac{HN}{N} \cong \frac{H}{H \cap N}$$
565
566 \end{enumerate}
567\end{proof}
568
569
570\begin{thm}{i.3}(\emph{Third Isomorphism Theorem}) \label{3rdisothm}\\
571 Let $H \subseteq K$ and $K \triangleleft G,~ H \triangleleft G$.\\
572 Then
573 $\frac{K}{H} \triangleleft \frac{G}{H}$
574 and
575 $$\frac{ G/H }{ K/H } \cong \frac{G}{K}$$
576\end{thm}
577\begin{proof}
578 \emph{(proof from Dummit and Foote book ``Abstract Algebra" \cite{dummitfoote})}\\
579
580 Easy to see that $\frac{K}{H} \triangleleft \frac{G}{H}$.
581
582 Define
583 \begin{align*}
584 \psi: &\frac{G}{H} \longrightarrow \frac{G}{K}\\
585 \text{by}~ \psi: &gH \longmapsto gK
586 \end{align*}
587
588 To show that $\psi$ is \emph{well defined}:\\
589 suppose $g_1 H = g_2 H$, then $g_1 = g_2 h$ for some $h \in H$.\\
590 Since $H \subseteq K \Longrightarrow~ h \in K$, hence $g_1 K = g_2 K$,\\
591 ie. $\psi(g_1 H) = \psi(g_2 H)$, which shows that $\psi$ is well defined.
592
593 Since $g \in G$ may be chosen arbitrarily in $G$, $\psi$ is a surjective homomorphism.
594
595
596 Finally,
597 \begin{align*}
598 ker(\psi) &= \{ gH \in \frac{G}{H} \mid \psi(gH)=1K \}\\
599 &= \{ gH \in \frac{G}{H} \mid gK =1K \}\\
600 &= \{ gH \in \frac{G}{H} \mid g \in K \}\\
601 &= \frac{K}{H}
602 \end{align*}
603
604
605 By the First Isomorphism Theorem (\ref{1stisothm}),
606
607 \begin{tikzpicture}[node distance=2.5cm, auto]
608 \node (GmodH) {$\frac{G}{H}$};
609 \node (GmodK) [right of=GmodH] {$\frac{G}{K}$};
610 \node (GmodHmodKer) [below of=G, xshift=1.25cm] {$\frac{G/H}{\ker(\psi)} = \frac{G/H}{K/H}$};
611
612 \draw[->] (GmodH) to node {$\psi$} (GmodK);
613 \draw[->] (GmodH) to node [swap] {$\phi$} (GmodHmodKer);
614 \draw[->] (GmodHmodKer) to node [swap] {$\eta$} (GmodK);
615 \end{tikzpicture}
616
617 So, by
618 $$\eta: \frac{G/H}{K/H} \longrightarrow \frac{G}{K}$$
619 since $\eta$ is
620 bijective (we know it by the First Isomorphism Theorem), $\eta$ it is the isomorphism:
621 $$\frac{ G/H }{ K/H } \cong \frac{G}{K}$$
622
623
624\end{proof}
625
626\subsection{Chapter 14}
627
628\begin{defn}{14.1} \label{14.1}
629 a group $G$ is soluble if it has a finite series of subgroups
630 $$1=G_0 \subseteq G_1 \subseteq \ldots \subseteq G_n = G$$
631 such that
632 \begin{enumerate}[i.]
633 \item $G_i \lhd G_{i+1}$ for $i=0,\ldots,n-1$
634 \item $\frac{G_{i+1}}{G_{i+1}}$ is Abelian for for $i=0,\ldots,n-1$
635 \end{enumerate}
636
637 (Note: $G_i \lhd G_{i+1} \lhd G_{i+2}$ does not imply $G_i \lhd G_{i+2}$)
638\end{defn}
639
640\begin{thm}{14.4}
641 $H \subseteq G,~~ N \triangleleft G$, then
642 \begin{enumerate}
643 \item if $G$ soluble $\Longrightarrow H$ soluble
644 \item if $G$ soluble $\Longrightarrow G/N$ soluble
645 \item if $N$ and $G/N$ soluble $\Longrightarrow G$ soluble
646 \end{enumerate}
647\end{thm}
648\begin{proof}
649 \begin{enumerate}
650 \item Since $G$ soluble, by definition: $\exists~~ 1=G_0 \triangleleft G_1 \triangleleft \ldots \triangleleft G_r = G$ with Abelian quotients $\frac{G_{i+1}}{G_i}$.
651
652 Let $H_i = G_i \cap H$, then $H$ has a series $1=H_0 \triangleleft H_1
653 \triangleleft \ldots \triangleleft H_r = H$, next we show that the
654 quotients $\frac{H_{i+1}}{H_i}$ are Abelian (so that H is soluble):
655
656 $$\frac{H_{i+1}}{H_i} = \frac{G_{i+1} \cap H}{G_i \cap H} \stackrel{(*)}{=} \frac{G_{i+1} \cap H}{G_i \cap (G_{i+1}\cap H)}
657 \stackrel{(**)}{\cong} \frac{G_i(G_{i+1} \cap H)}{G_i} \subseteq \frac{G_{i+1}}{G_i}
658 $$
659
660 (*): to see why, $H_i = G_i \cap H = G_i \cap H_i = G_i \cap H_{i+1} = G_i \cap (G_{i+1} \cap H)$.
661 (**): by the 2nd Isomorphism Theorem (\ref{2ndisothm}).
662
663 [TODO: diagram of subgroups] % TODO %
664
665 Notice that $\frac{G_{i+1}}{G_i}$ is Abelian, thus the left-hand-side of the congruence is also Abelian.
666 Therefore, $\frac{H_{i+1}}{H_i}$ is Abelian, thus $H$ is soluble.
667
668 \item For $G/N$ to be soluble, (by definition) it would have the series $\frac{N}{N} = G_0 \frac{N}{N} \triangleleft G_1 \frac{N}{N} \triangleleft \ldots \triangleleft G_r \frac{N}{N} = \frac{G}{N}$,
669 and any quotient being $\frac{G_{i+1}\frac{N}{N}}{G_i \frac{N}{N}}$.
670
671 The series clearly exists, so now we show that the quotients are Abelian, so that $G/N$ is soluble:
672
673 $$
674 \frac{G_{i+1} N}{G_i N} = \frac{G_{i+1}(G_i N)}{G_i N} \stackrel{(*)}{\cong}
675 \frac{G_{i+1}}{G_{i+1} \cap (G_i N)} \cong \frac{G_{i+1}/G_i}{(G_{i+1}
676 \cap (G_i N))/G_i}
677 $$
678
679 (*): by the 2nd Isomorphism Theorem (\ref{2ndisothm}).
680
681 The last quotient is a quotient of the Abelian group $G_{i+1}/G_i$, so it is Abelian.
682
683 Hence, $\frac{G_{i+1}N}{G_i N}$ is also Abelian; so $\frac{G}{N}$ is soluble.
684
685 \item
686 By the definition of $N$ and $G/N$ being soluble,
687 \begin{align*}
688 N \text{soluble} \Longrightarrow~~ 1 &= N_0 \triangleleft N_1 \triangleleft \ldots \triangleleft N_r = N\\
689 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}
690 \end{align*}
691 both with Abelian quotients.
692
693 Consider the series of $G$ given by combining the two previous series:
694
695 $$
696 1 = N_0 \triangleleft N_1 \triangleleft \ldots \triangleleft N_r = N = G_0 \triangleleft G_1 \triangleleft \ldots \triangleleft G_r = G
697 $$
698 the quotients are either
699 \begin{itemize}
700 \item $\frac{N_{i+1}}{N_i}$, Abelian
701 \item $\frac{G_{i+1}}{G_i}$, isomorphic to $\frac{G_{i+1}/N}{G_i/N}$,
702 which is Abelian.
703 \end{itemize}
704 \end{enumerate}
705 Therefore, the quotients are always Abelian; hence $G$ is soluble.
706\end{proof}
707
708
709\begin{defn}{14.5}
710 $G$ is \emph{simple} if it's nontrivial and it's only normal subgroups are $1$ and $G$.
711\end{defn}
712
713\begin{thm}{14.6} \label{14.6}
714 A \emph{soluble} group is \emph{simple} iff it is cyclic of prime order.
715\end{thm}
716\begin{thm}{14.7} \label{14.7}
717 if $n \geq 5$, then $\mathbb{A}_n$ is simple.
718\end{thm}
719\begin{cor}{14.8}
720 $\mathbb{S}_n$ is not soluble if $n \geq 5$.
721\end{cor}
722\begin{proof}
723 if $\mathbb{S}_n$ were soluble, then $\mathbb{A}_n$ would be soluble by Theorem \ref{14.1}(i), and simple by Theorem \ref{14.7}, hence of prime order by Theorem \ref{14.6}.
724
725 But observe: $|\mathbb{A}_n|=\frac{1}{2}(n!)$ is not prime if $n \geq 5$.\\
726 Thus $\mathbb{S}_n$ is not soluble if $n \geq 5$.
727\end{proof}
728
729\begin{lemma}{14.14} \label{14.14}
730 if $A$ finite and abelian group with $p \biggr\vert |A|$ ($p$ prime), then $A$ has an element of order $p$.
731\end{lemma}
732\begin{proof}
733 \begin{enumerate}[i.]
734 \item if $|A|$ prime and Abelian $\Longrightarrow$ then $A$ is cyclic.
735
736 Since $p \vert |A| ~~ \Longrightarrow ~~ \exists! ~ B \subseteq A$ such that $|B|=p$, where $B = <b>$ with $ord(b)=p$. So the lemma is proven.
737
738 \item if $|A|$ non-prime:
739
740 take $M \subseteq A$ with $|M|=m$, $m$ maximal. Then
741
742 \begin{enumerate}[a.]
743 \item if $p|m ~~ \Longrightarrow ~~ \exists! ~ B'=<b'>,~ b' \in A$ with $|B'|=p$ and $ord(b')=p$.
744 \item if $p \nmid m$:
745 Let $b \in A \not M$ and $B=<b>$.\\
746 Then $MB \supseteq M$, and by maximility must be $MB=A$.
747
748 By the 1st Isomorphism Theorem (\ref{1stisothm}),
749 $$|A| = |MB| = \frac{|M| \cdot |B|}{|M \cap B|}$$
750 both $|A|$ and $|B|$ are divisible by $p$ (but recall that $p \nmid m=|M|$), since $B$ is cyclic and $p \vert |B|$
751
752 $\Longrightarrow$ thus, $B$ has an element of order $p$.
753
754 So, if $|B|=r$, and $p|r ~~ \Longrightarrow~~ ord(b^{r/p}) = p$.\\
755
756 Hence, in all cases i, ii.a, ii.b, $A$ contains an element of order $p$.
757 \end{enumerate}
758 \end{enumerate}
759\end{proof}
760
761\begin{thm}{14.15}(Cauchy's Theorem)
762 if $p \biggr\vert |G|$ ($p$ prime), then $\exists ~ x \in G$ such that $ord(x)=p$.
763\end{thm}
764\begin{proof}
765 (induction on $|G|$)
766
767 For $|G|=1,2,3$, trivial.
768
769 Induction step: class equation
770 $$|G|=1+|C_2|+ \ldots + |C_r|$$
771 since $p \biggr\vert |G|$, must have $p \nmid |C_j|$ for some $j \geq 2$.
772
773 If $x \in C_j ~~\Longrightarrow p \biggr\vert |C_G(x)|$ (since $|C_j|=|G|/|C_G(x)|$, recall $p\biggr\vert |G|$).
774
775 \begin{enumerate}[i.]
776 \item if $C_G(x) \neq G$:\\
777 (by induction) since $p \biggr\vert |C_G(x)|$,
778
779 $\exists a \in C_G(x)$ with $ord(a)=p$, and $a \in G$ (since $C_G(x) \subset G$).
780 \item otherwise, $C_G(x)=G$:\\
781 implies $x \in Z(G)$, by choice $x\neq 1$, so $Z(G)\neq 1$.
782
783 Then either
784 \begin{enumerate}[I.]
785 \item $p \biggr\vert |Z(G)| ~~ \longrightarrow$ Abelian case, Lemma \ref{14.14}.
786 \item $p \not\biggr\vert |Z(G)|$:
787 by induction, $\exists x \in G$ such that $\hat{x} \in G/Z(G)$, with $ord(\hat{x})=p$. (where $\hat{x}$ is the image of $x$).
788
789 $\Longrightarrow~~ x^p \in Z(G)$, but $x \not\in Z(G)$.
790
791 Let $X=<x>$, cyclic.\\
792 $XZ(G)$ is Abelian, and $p \biggr\vert |XZ(G))|$\\
793 $\Longrightarrow$ by Lemma \ref{14.14}, it has an element of order $p$, and this element belongs to $G$.
794 \end{enumerate}
795
796 \end{enumerate}
797\end{proof}
798
799\begin{defn}{15.1}(Soluble by radicals)
800 let $f \in K[t],~ K \subseteq \mathbb{C}$, and $\Sigma$ a splitting field of $f$ over $K$.
801
802 $f$ is \emph{soluble by radicals} if\\
803 $\exists$ a field $M$ with $\Sigma \subseteq M$ such that $M:K$ is a radical extension (\ref{8.12}).
804
805 Note: not required $\Sigma:K$ to be radical.
806\end{defn}
807
808\begin{lemma}{15.3}
809 $L:K$ radical extension $\mathbb{C}$, and $M$ normal enclosure of $L:K$, then $M:K$ is radical.
810\end{lemma}
811\begin{proof}
812 let $L=K(\alpha_1, \ldots, \alpha_r)$ with $\alpha_i^{n_i} \in K(\alpha_1, \ldots, \alpha_{j-1})$ (by definition of $L:K$ being a radical extension).
813
814 Let $f_i$ be the minimal polynimal of $\alpha_i$ over $K$.
815
816 Then, $M \supseteq L$ is splitting field of $\prod_{i=1}^r f_i$, since $M$ is normal enclosure of $L:K$.
817
818 For every zero $\beta_{ij}$ of $f_i$ in $M$,\\
819 $\exists$ an isomorphism $\sigma: K(\alpha_i) \longrightarrow K(\beta_{ij})$ by Corollary \ref{5.13}.
820
821 By Proposition \ref{11.4}, since $K(\alpha_i),~K(\beta_{ij}) \subset M$,
822 $\sigma$ extends to a $K$-automorphism
823 $$\tau: M \longrightarrow M$$
824 since $M$ is splitting field (ie. contains the zeros of $\prod f_i$).
825
826 Since $\alpha$ is a member of radical sequence for a subfield of $M$, so it is $\beta_{ij}$.
827
828 By combining the sequences for $M$, $M:K$ is a radical extension.
829\end{proof}
830
831
832\vspace{0.5cm}
833
834The next two lemmas show that certain Galois groups are Abelian.
835
836\begin{lemma}{15.4}
837\end{lemma}
838\begin{proof}
839\end{proof}
840
841\newpage
842
843\subsection{Characterization of finite fields}
844
845\begin{defn}{16.8}
846The prime subfield of a field $K$ is the intersection of all subfields of $K$.
847\end{defn}
848
849\begin{thm}{16.10}
850 Every prime subfield $P$ is isomorphic either to
851 $$P \cong \mathbb{Z}_p~ (p~ \text{prime})~~~~\text{or}~~~~ P \cong \mathbb{Q}$$
852\end{thm}
853\begin{proof}
854Let $K$ a field, $P$ its prime subfield. Then $P$ contains $0$ and $1$ and,
855therefore, contains the elements $n^*$ (for $n \in \mathbb{Z}$) defined by
856
857$$n^* =
858\begin{cases}
859\overbrace{1 + 1 + \cdots + 1}^{n \text{ times}} & \text{if } n > 0\\
8600 & \text{if } n = 0\\
861-(-n)^* & \text{if } n < 0
862\end{cases}$$
863
864The map $* : \mathbb{Z} \to P$ is a ring homomorphism.\\
865Two cases arise:
866\begin{enumerate}[i.]
867 \item $P \cong \mathbb{Z}_p$\\
868 $n^* = 0$ for some $n \neq 0$. Since also $(-n)^* = 0$, there exists a smallest
869 positive integer $p$ such that $p^* = 0$. If $p$ is composite, say $p = rs$ where $r$ and $s$ are
870 smaller positive integers, then $r^* s^* = p^* = 0$, so either $r^* = 0$ or $s^* = 0$, contrary
871 to the definition of $p$. Therefore, $p$ is prime. The elements $n^*$ form a ring isomorphic
872 to $\mathbb{Z}_p$, which is a field by Theorem 16.7. This must be the whole of $P$, since $P$ is the
873 smallest subfield of $K$.
874
875 \item $P \cong \mathbb{Q}$\\
876 $n^* \neq 0$ if $n \neq 0$. Then $P$ must contain all the elements $m^*/n^*$ where $m, n$ are integers and $n \neq 0$. These form a subfield isomorphic to $\mathbb{Q}$ (by the map which sends $m^*/n^*$ to $m/n$), which is necessarily the whole of $P$.
877\end{enumerate}
878\end{proof}
879
880
881\begin{lemma}{16.11}
882 If $K$ is a subfield of $L$, then $K$ and $L$ have the same characteristic.
883\end{lemma}
884\begin{proof}
885 In fact, $K$ and $L$ have the same prime subfield.
886\end{proof}
887
888\vspace{1.0cm}
889
890Finite fields have order $p^n$ with $p$ prime and $n > 0$. In the following theorems we prove it, first proving the case $n=1$ with a specific case proof which works for integers modulo $p$, and then proving the general case for $n > 0$.
891
892\begin{thm}{$(\mathbb{Z}_p)$}
893 The ring $\mathbb{Z}_n$ is a field iff $n=p$, $p$ prime.
894\end{thm}
895\begin{proof}
896 One of the axioms of a field is $\exists$ multiplicative inverse.\\
897 If $\mathbb{Z}_n$ with $n$ no prime, then $n= k \cdot l$ for some $1 \leq k,~l \leq n-1$.\\
898 Then in $\mathbb{Z}_n$, $k \cdot l = 0$, but if $k \cdot l=0$ means that either $k=0$ or $l=0$ (otherwise, we could multiply by (eg) $k^{-1}$ and get $k^{-1} \cdot k \cdot l = k^{-1} \cdot 0$, which leads to $1 \cdot l = 0$).\\
899 which is a contradiction here (since $1 \leq k,~l \leq n-1$).\\
900 Thus $\mathbb{Z}_n$ with $n$ not prime can not be a field.\\
901 Conversely, if $n = p$ prime,\\
902 for $0 \neq x \in \mathbb{Z}_p$, $gcd(x, p)=1$, so Extended Euclidean Algorithm gives $u, v \in \mathbb{Z}$ such that $u x + v p = 1$.\\
903 Then, $ux=1 \pmod p$, so $u=x^{-1}$, so inverses exist.\\
904 Thus $\mathbb{Z}_p$ is a field.
905\end{proof}
906
907\vspace{0.5cm}
908More generally, using Galois theory, we can show that every finite field is isomorphic to $F_{p^n}$:
909
910\begin{thm}{$(F_p^n)$}
911 Every finite field has $p^n$ elements and hence is isomorphic to $F_{p^n}$,
912 for $p$ prime and $n>0$. Denoted as $F_{p^n} = GF(q^n)$.
913\end{thm}
914\begin{proof}
915 Let $K$ be a finite field. Since $K$ finite, $char(K) \neq 0$, ie. $char(K)=p >0$.
916
917 In any field, characteristic must be prime, since if $char(K)=n$ ($n$ not prime), then $n=a \cdot b$ for $1 <a, b < char(K)$, and then, in $K~~ n=0 ~\Longrightarrow~ a \cdot b = 0$; either $a$ or $b$ are zero, a contradiction.
918
919 Thus, $char(K)=p$ prime,\\
920 and the prime subfield of $K$ is $P=F_p$.
921
922 $K$ is a vector space over $P$, with finitely many elements\\
923 \hspace*{2em} $\Longrightarrow~ [K:P]=n < \infty$.
924
925 Let $x_1, \ldots, x_n$ be a basis for $K$ over $P$,
926 $$\forall a \in K, ~ a= \lambda_1 x_1 + \ldots + \lambda_n x_n, ~~w/ \lambda_i \in P$$
927
928 each $\lambda_i$ can be chosen in $p$ ways, since $|P|=p$.
929
930 hence, there are $p^n$ possible $\lambda_1 x_1 + \ldots + \lambda_n x_n$ expressions in $K$.
931
932 Therefore, $|K|=q=p^n$.
933
934 \vspace{0.4cm}
935 Now, the set $K \setminus \{0\}$ forms a group under multiplication, of order $q-1$
936
937 $$\Longrightarrow~ \forall~ 0 \neq x \in K,~ x^{q-1}=1$$
938 $$\text{thus}~ x^q=x~ \Longrightarrow~\text{hence}~ x^q-x=0$$
939
940 Notice that $0^q-0=0$, so that we have that every element of $K$ (including zero) is a zero of $f(t)=t^q-t$.
941
942 Thus $f(t)$ splits in $K$ $~\Longrightarrow~$ ie. $K$ is the splitting field of $f(t)=t^q-t$ over $P$.
943
944 By uniqueness of splitting fields, $F$ is unique up to isomorphism, denoted $F_{p^n}$.
945
946 \vspace{0.4cm}
947 We can construct $F_{p^n}$ as:\\
948 let $h(x) \in F_p[x]=P$ of degree $n$. Then $|F_p[x] / (h)| = p^n$.
949\end{proof}
950
951
952
953
954
955\newpage
956
957\section{Tools}
958This section contains tools that I found useful to solve Galois Theory related problems, and that don't appear in Stewart's book.
959
960\subsection{De Moivre's Theorem and Euler's formula}\label{demoivre}
961Useful for finding all the roots of a polynomial.
962
963Euler's formula:
964$$e^{i \psi} = cos \psi + i \cdot sin \psi$$
965
966The n-th roots of a complex number $z=x + i y = r (cos \theta + i \cdot sin \theta)$ are given by
967
968$$z_k = \sqrt[n]{r} \cdot \left(cos(\frac{\theta + 2k \pi}{n}) + i \cdot sin(\frac{\theta + 2k \pi}{n}) \right)$$
969for $k=0, \ldots, n-1$.
970
971So, by Euler's formula:
972$$z_k = \sqrt[n]{r} \cdot e^{i (\frac{\theta + 2 k \pi}{n})}$$
973
974Usually we will set $\alpha=\sqrt[n]{r}$ and $\zeta = e^{\frac{2 \pi i}{n}}$,
975and find the $\mathbb{Q}$-automorphisms from there (see \ref{ex:galoisgroups} for examples).
976
977\subsection{Einsenstein's Criterion} \label{einsenstein}
978\emph{reference: Stewart's book}
979
980Let $f(t) = a_0 + a_1 t + \ldots + a_n t^n$, suppose there is a prime $q$ such that
981\begin{enumerate}
982 \item $q \nmid a_n$
983 \item $q | a_i$ for $i=0, \ldots, n-1$
984 \item $q^2 \nmid a_0$
985\end{enumerate}
986Then, $f$ is irreducible over $\mathbb{Q}$.
987
988\emph{TODO proof \& Gauss lemma.}
989
990
991\subsection{Elementary symmetric polynomials}
992\emph{TODO from orange notebook, page 36}
993
994\subsection{Cyclotomic polynomials} \label{cyclotomicpoly}
995\subsubsection{From Elmyn Berlekamp's "Algebraic Coding Theory" book}
996The notes in this section are from the book "Algebraic Coding Theory" by Elmyn
997Berlekamp \cite{berlekamp}.
998
999\vspace{0.3cm}
1000
1001Some 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.
1002
1003Observe that in a finite field of order $q$, factoring $x^q - x$ gives
1004$$x^q-x = x(x^{q-1} -1)$$
1005
1006The factor $x^{q-1} -1$ is a special case of $x^n -1$: if we assume that the
1007field contains an element $\alpha$ of order $n$, then the roots of $x^n-1=0$ are
1008$$1, \alpha, \alpha^2, \alpha^3, \ldots, \alpha^{n-1}$$
1009and $\deg(x^n-1)=n$, thus $x^n-1$ has at most $n$ roots in any field, henceforth
1010the powers of $\alpha$ must include all the $n$-th roots of unity.
1011
1012There fore, in any field which contains a primitive $n$-th root of unity we have:
1013
1014\begin{thm}{4.31}
1015 $$x^n -1 = \prod_{i=0}^{n-1} (x - \alpha^i) = \prod_{i=1}^n (x-\alpha^i)$$
1016\end{thm}
1017
1018If $n=k \cdot d$, then $\alpha^k, \alpha^{2k}, \alpha^{3k}, \ldots, \alpha^{dk}$ are all roots of $x^d -1 =0$
1019
1020Every element with order dividing $n$, must be a power of $\alpha$, since an
1021element of order $d$ is a $d$-th root of unity.
1022
1023Every power of $\alpha$ has order which divides $n$, and every field element
1024whose order divides $n$ is a power of $\alpha$. This suggests that we partition
1025the powers of $\alpha$ according to their orders:
1026$$x^n -1 = \prod_{\stackrel{d,}{d|n}} \prod_{\beta} (x- \beta)$$
1027where at each iteration, $\beta$ is a field element of order $d$ for each $d$.
1028
1029The polynomial whose roots are the field elements of order $d$ is called the
1030\emph{cyclotomic polynomial}, denoted by $Q^{(d)}(x)$.
1031
1032\begin{thm}{4.32}
1033 $$x^n -1 = \prod_{\stackrel{d,}{d|n}} Q^{(d)}(x)$$
1034\end{thm}
1035
1036
1037\subsubsection{From Ian Stewart's ``Galois Theory'' book}
1038Notes from Ian Stewart's book \cite{ianstewart}.
1039
1040Consider the case $n=12$, let $\zeta=e^{\pi i /6}$ be a primitive $12$-th root of unity.
1041Classify its powers ($\zeta^j$) according to their minimal power $d$ such that
1042$(\zeta^j)^d = 1$ (ie. when they are primitive $d$-th roots of unity).
1043
1044\begin{enumerate}[]
1045 \item $d=1,~~~ 1$
1046 \item $d=2,~~~ \zeta^6$
1047 \item $d=3,~~~ \zeta^4, \zeta^8$
1048 \item $d=4,~~~ \zeta^3, \zeta^9$
1049 \item $d=6,~~~ \zeta^2, \zeta^{10}$
1050 \item $d=12,~~~ \zeta, \zeta^5, \zeta^7, \zeta^{11}$
1051\end{enumerate}
1052
1053Observe that we can factorize $t^{12} -1$ by grouping the corresponding zeros:
1054\begin{align*}
1055 t^{12}-1 = &(t-1) \times\\
1056 &(t-\zeta^6) \times\\
1057 &(t-\zeta^4) (t-\zeta^8) \times\\
1058 &(t-\zeta^3) (t-\zeta^9) \times\\
1059 &(t-\zeta^2) (t-\zeta^{10}) \times\\
1060 &(t-\zeta) (t-\zeta^5)(t-\zeta^7) (t-\zeta^{11})
1061\end{align*}
1062which simplifies to
1063$$t^{12}-1=(t-1)(t+1)(t^2+t+1)(t^2+1)(t^2-t+1)F(t)$$
1064where $F(t) = (t-\zeta) (t-\zeta^5)(t-\zeta^7) (t-\zeta^{11}) = t^4 -t^2 + 1$ (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).
1065
1066
1067Let $\Phi_d(t)$ be the factor corresponding to primitive $d$-th roots of unity, then we have proved that
1068$$t^{12}-1 = \Phi_1 \Phi_2 \Phi_3 \Phi_4 \Phi_6 \Phi_{12}$$
1069
1070
1071\begin{defn}{21.5}
1072 The polynomial $\Phi_d(t)$ defined by
1073 $$\Phi_n(t) = \prod_{a\in \mathbb{Z}_n,(a,n)=1} (t- \zeta^a)$$
1074 is the $n$-th \emph{cyclotomic polynomial} over $\mathbb{C}$.
1075\end{defn}
1076
1077\begin{cor}{21.6}
1078 $\forall n \in \mathbb{N}$, the polynomial $\Phi_n(t)$ lies in $\mathbb{Z}[t]$ and is monic and irreducible.
1079\end{cor}
1080
1081\begin{thm}{21.9}
1082 \begin{enumerate}
1083 \item The Galois group $\Gamma(\mathbb{Q}(\zeta):\mathbb{Q})$ consists of the
1084 $\mathbb{Q}$-automorphisms $\psi_j$ defined by
1085 $$\psi_j(\zeta)=\zeta^j$$
1086 where $0 \leq j \leq n-1$ and $j$ is prime to $n$.
1087
1088 \item $\Gamma(\mathbb{Q}(\zeta):\mathbb{Q}) \stackrel{iso}{\cong} \mathbb{Z}_n^*$, and is an abelian group.
1089 \item its order is $\phi(n)$
1090 \item if $n$ is prime, $\mathbb{Z}_n^*$ is cyclic
1091 \end{enumerate}
1092\end{thm}
1093
1094
1095
1096\vspace{1cm}
1097
1098\subsubsection{Examples}
1099Examples of cyclotomic polynomials:
1100
1101\begin{align*}
1102 \Phi_n(x) &= x^{n-1} + x^{n-2} + \ldots + x^2 + x + 1 = \sum_{i=0}^{n-1} x^i\\
1103 \Phi_{2p}(x) &= x^{p-1} + \ldots + x^2 - x + 1 = \sum_{i=0}^{p-1} (-x)^i\\
1104 \Phi_m(x) &= x^{m/2} + 1, ~~\text{when $m$ is a power of $2$}
1105\end{align*}
1106
1107
1108\subsection{Lemma 1.42 from J.S.Milne's book}
1109Lemma from J.S.Milne's book \cite{milneFT}.
1110
1111Useful for when dealing with $x^p - 1$ with $p$ prime.
1112
1113Observe that
1114
1115$$x^p -1 = (x-1)(x^{p-1} + x^{p-2} + \ldots + 1)$$
1116
1117Notice that
1118$$\Phi_p(x) = x^{p-1} + x^{p-2} + \ldots + 1$$
1119is the $p$-th Cyclotomic polynomial.
1120
1121\begin{lemma}{1.42}
1122 If $p$ prime, then $x^{p-1} + \ldots + 1$ is irreducible; hence $\mathbb{Q}[e^{2 \pi i /p}]$ has degree $p-1$ over $\mathbb{Q}$.
1123\end{lemma}
1124\begin{proof}
1125 Let $f(x) = (x^p - 1)/(x-1) = x^{p-1} + \ldots + 1$
1126 then
1127 $$
1128 f(x+1) = \frac{(x+1)^p -1}{x+1-1} = \frac{(x+1)^p -1}{x} = x^{p-1} + \ldots + a_i x^i + \ldots + p
1129 $$
1130
1131 with $a_i = \left( \stackrel{p}{i+1} \right)$.
1132
1133 We know that $p | a_i$ for $i= 1, \ldots, p-2$, therefore $f(x+1)$ is irreducibe by Einsenstein's Criterion.
1134
1135 This implies that $f(x)$ is irreducible.
1136\end{proof}
1137
1138
1139\subsection{Dihedral groups - Groups of symmetries} \label{dihedral}
1140Source: Wikipedia and \cite{dihedral}.
1141
1142Dihedral groups ($\mathbb{D}_n$) represent the symmetries of a regular $n$-gon.
1143
1144Properties:
1145\begin{itemize}
1146 \item are non-abelian (for $n>2$), ie. $rs \neq sr$
1147 \item order $2n$
1148 \item generated by a rotation $r$ and a reflection $s$
1149 \item $r^n = s^2 = id,~~~(rs)^2=id$
1150\end{itemize}
1151Subgroups of $\mathbb{D}_n$:
1152\begin{itemize}
1153 \item rotation form a cyclic subgroup of order $n$, denoted as $<r>$
1154 \item for each $d$ such that $d|n$, $\exists~ \mathbb{D}_d$ with order $2d$
1155 \item normal subgroups
1156 \begin{itemize}
1157 \item for $n$ odd: $\mathbb{D}_n$ and $<r^d>$ for every $d|n$
1158 \item for $n$ even: $2$ additional normal subgroups
1159 \end{itemize}
1160 \item Klein four-groups: $\mathbb{Z}_2 \times \mathbb{Z}_2$, of order 4
1161\end{itemize}
1162
1163\vspace{0.3cm}
1164Total 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.
1165
1166\begin{eg}{}
1167For $\mathbb{D}_6$, we have $\{1,2,3,6\} | 6$, so $d(n) = d(6) = 4$, and
1168$s(6) = 1+2+3+6 = 12$; henceforth, the total amount of subgroups is $d(n)+s(n) = 4+12 = 16$.
1169\end{eg}
1170
1171\vspace{0.3cm}
1172For $n \geq 3, ~~\mathbb{D}_n \subseteq \mathbb{S}_n$ (subgroup of the Symmetry group).
1173
1174\subsection{Rolle's theorem} \label{rolle}
1175
1176\begin{thm}{}(Rolle's Theorem)
1177 if a real-valued function $f$ is
1178 \begin{itemize}
1179 \item \emph{continuous} on a proper closed interval $[a,b]$
1180 \item \emph{differentiable} on the open interval $(a,b)$
1181 \item $f(a)=f(b)$
1182 \end{itemize}
1183 then, $\exists$ at least one $c$ in $(a,b)$ such that $Df(c)=0$.
1184\end{thm}
1185\begin{proof} (proof source: cue math website)
1186 Notice that when $Df(x_i)=0$ occurs, is a maximum or minimum (extrema) value of $f$.
1187 $\Longrightarrow$ if a function is continuous, it is guaranteed to have both a maximum and a minimum point in the interval.\\
1188
1189 Two possibilities:
1190 \begin{enumerate}[i.]
1191 \item $f$ is constant\\
1192 $\Longrightarrow$ ie. a horizontal line ($f(a)=f(b)$), ie. no slope $\Longrightarrow$ $Df=0$ everywhere in $[a,b]$.
1193 \item $f$ is not constant:\\
1194 since $f$ not constant, must change directions in ordder to start and end at the same $y$-value ($f(a)=f(b)$).\\
1195 Thus at some point between $a$ and $b$ it will either have a minimum, maximum or both.
1196 \begin{enumerate}[a.]
1197 \item does the maximum occur at a point where $Df > 0$?\\
1198 No, because if $Df > 0$, then $f$ is increasing, but it can not increase since we're at its maximum point.
1199 \item does the maximum occur at a point where $Df < 0$?\\
1200 No, because if $Df < 0$, then $f$ is deccreasing, which means that at our left it was larger, but we're at a maximum point, so a contradiction.
1201 \end{enumerate}
1202 Same with minimus.\\
1203 $\Longrightarrow$ Hence, since $Df \nless 0$ and $Df \ngtr 0$, and $Df$ exists, then $Df=0$.
1204 \end{enumerate}
1205 Thus, $f$ must have extrema (either max or min or both), and at that extrema $Df$ must be zero.
1206\end{proof}
1207
1208
1209Consequence of Rolle's Theorem:
1210
1211\begin{cor}{}(Zero separation property)
1212 Between any two distinct consecutive zeros of $f$, there lies at least one zero of $Df$.
1213\end{cor}
1214\begin{eg}{}
1215 If $f(t)$ has zeros $t_1, t_2$, with $t_1 < t_2$, and $f$ is derivable, then by Rolle'z theorem:
1216
1217 $\exists c \in (t_1, t_2)$ such that $Df(c)=0$.
1218
1219 Hence, the zeros of $Df$ \emph{separate} the zeros of $f$.
1220\end{eg}
1221
1222\newpage
1223
1224\section{Exercises}
1225
1226\subsection{Galois groups}\label{ex:galoisgroups}
1227
1228\subsubsection[t6-7]{$t^6-7 \in \mathbb{Q}$}
1229
1230This exercise comes from a combination of exercises 12.4 and 13.7 from \cite{ianstewart}.
1231
1232First let's find the roots. By De Moivre's Theorem (\ref{demoivre}), $t_k =
1233\sqrt[6]{7} \cdot e^{i \frac{2 \pi k}{6}}$.
1234
1235From which we denote $\alpha = \sqrt[6]{7}$, and $\zeta = e^{\frac{2 \pi i}{6}}$, so that the
1236roots of the polynomial are $\{ \alpha, \alpha \zeta, \alpha \zeta^2, \alpha \zeta^3, \alpha \zeta^4, \alpha \zeta^5\}$, ie.
1237$\{ \alpha \zeta^k \}_0^5$.
1238
1239Hence the \emph{splitting field} is $\mathbb{Q}(\alpha, \zeta)$.
1240
1241\emph{Degree of the extension}
1242
1243In order to find $[\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}$, we're going to split it in tow
1244parts. By the Tower Law (\ref{towerlaw}),
1245
1246$$[\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}] = [\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}(\alpha)] \cdot [\mathbb{Q}(\alpha) : \mathbb{Q}]$$
1247
1248To find each degree, we will find the minimal polynomial of the adjoined term over the base field of the extension:
1249
1250\begin{enumerate}[i.]
1251 \item minimal polynomial of $\alpha$ over $\mathbb{Q}$\\
1252 By Einsenstein's Criterion (\ref{einsenstein}), with $q=7$ we have that $q
1253 \nmid 1$, $7 | {-7,0,0,\ldots}$, and $7^2 \nmid -7$, hence $f(t)$ is
1254 irreducibe over $\mathbb{Q}$, thus is the minimal polynomial
1255 $$m_i(t)= f(t) =t^6-7$$
1256 which has roots $\{ \alpha \zeta^k \}_0^5$.
1257 \item minimal polynomial of $\zeta$ over $\mathbb{Q}(\alpha)$\\
1258 Since $\zeta$ is the primitive $6$th root of unity, we know that the minimal
1259 polynomial will be the $6$th cyclotomic polynomial (\ref{cyclotomicpoly}):
1260 $$m_{ii}(t) = \Phi_6(t) = t^2 - t + 1$$
1261 which has roots $\zeta, -\zeta$.
1262
1263 Since $\mathbb{Q}(\alpha) \subseteq \mathbb{R}$, and the roots of
1264 $\Phi_6(t)=t^2 - t +1$ are in $\mathbb{C}$, $\Phi_6(t)$ remains irreducible
1265 over $\mathbb{Q}(\alpha)$.
1266\end{enumerate}
1267
1268\vspace{0.5cm}
1269Therefore, by the tower of law,
1270$$[\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}] = \deg{\Phi_6(t)} \cdot \deg{f(t)} = 2 \cdot 6 = 12$$
1271and by the Fundamental Theorem of Galois Theory, we know that
1272$$|\Gamma( \mathbb{Q}(\alpha, \zeta) : \mathbb{Q} )| = [\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}] = 12$$
1273which tells us that there exist $12$ $\mathbb{Q}$-automorphisms of the Galois group.
1274
1275
1276\vspace{0.5cm}
1277Let's find the $12$ $\mathbb{Q}$-automorphisms. Start by defining $\sigma$ which
1278fixes $\zeta$ and acts on $\alpha$, sending it to another of the roots of the
1279minimal polynomial of $\alpha$ over $\mathbb{Q}$, $f(t)$, choose $\alpha \zeta$.
1280
1281Now define $\tau$ which fixes $\alpha$ and acts on $\zeta$, sending it into
1282another root of the minimal polynomial of $\zeta$ over $\mathbb{Q}(\alpha)$,
1283choose $-\zeta$.
1284
1285\vspace{0.3cm}
1286\begin{tabular}{@{}l l@{}}
1287 $\begin{aligned}
1288 \sigma: \alpha &\mapsto \alpha \zeta \\
1289 \zeta &\mapsto \zeta
1290 \end{aligned}$
1291 &
1292 $\begin{aligned}
1293 \tau: \alpha &\mapsto \alpha\\
1294 \zeta &\mapsto -\zeta = \zeta^{-1}
1295 \end{aligned}$
1296\end{tabular}
1297
1298In other words, we have $12$ $\mathbb{Q}$-automorphisms, which are the
1299combination of $\sigma$ and $\tau$:
1300
1301$$\begin{aligned}
1302 \sigma^k \tau^j:~~&\alpha \mapsto \alpha \zeta^k\\
1303 &\zeta \mapsto \zeta^j
1304\end{aligned}$$
1305
1306for $0 \leq k \leq 5$ and $j = \pm 1$.
1307
1308\vspace{0.5cm}
1309
1310NOTE: WIP diagram.
1311\begin{tikzpicture}[node distance=2cm]
1312 \def \radius{2}
1313 \draw (0,0) circle (\radius);
1314
1315 \foreach \k in {0,...,5} {
1316 % \node (a\k) at ({360/6 * \k}:\radius) {$\alpha \zeta^{\k}$};
1317 \node (a\k) at ({360/6 * \k}:\radius+0.5) {$\alpha \zeta^{\k}$};
1318 \fill ({360/6 * \k}:\radius) circle (2pt);
1319 }
1320 % real & im axis
1321 \draw[->] (-2.5,0) -- (2.5,0) node[right] {};
1322 \draw[->] (0,-2.5) -- (0,2.5) node[above] {};
1323
1324
1325 % tau:
1326 \draw[<->] (3,1) -- (3,-1) node[right] {$\tau$};
1327 % sigma:
1328 % \foreach \k [evaluate=\k as \next using int(mod(\k+1,6))] in {0,...,5} {
1329 % \coordinate (p\k) at ({360/6 * \k}:\radius);
1330 % \coordinate (p\next) at ({360/6 * \next}:\radius);
1331 %
1332 % \draw[->, bend left=30] (p\k.center) -- node[above] {$\sigma$} (p\next.center);
1333 % }
1334 \foreach \k in {0,...,5} {
1335 \coordinate (p\k) at ({360/6 * \k}:\radius);
1336 }
1337 \foreach \k [evaluate=\k as \next using int(mod(\k+1,6))] in {0,...,5} {
1338 \draw[->, bend left=30] (p\k) -- node[above] {$\sigma$} (p\next);
1339 }
1340\end{tikzpicture}
1341
1342\vspace{0.5cm}
1343
1344Observe, that $\Gamma$ is generated by the combination of $\sigma$ and $\tau$,
1345and it is isomorphic to the group of symmetries of order 12, the dihedral
1346group (\ref{dihedral}) of order 12, $\mathbb{D}_6$, ie. $\Gamma \cong \mathbb{D}_6$.
1347
1348\vspace{0.5cm}
1349
1350Let's find the subgroups of $\Gamma$, and the fixed fields of $\mathbb{Q}(\alpha, \zeta)$.
1351
1352We know that $\Gamma \cong \mathbb{D}_6$, and we know from the properties
1353of the dihedral group (\ref{dihedral}) that the number of subgroups of
1354$\mathbb{D}_6$ will be $d(6) + s(6) = 4 + 12 = 16$ subgroups.
1355
1356
1357\vspace{0.4cm}
1358
1359\hspace*{-3.5cm}
1360\begin{tabular}{ c c c c | p{7.5cm} }
1361 \hline
1362 generators & order & group & fixed field & notes (check fixed field)\\
1363\hline
1364 $\langle \rangle = \langle \sigma^6 \rangle=\langle \tau^2 \rangle$ & 1 & id & $\mathbb{Q}(\alpha,\zeta)$ & \\
1365 $\langle \sigma \rangle = \langle \sigma^5 \rangle$ & 6 & $\mathbb{Z}_6$ & $\mathbb{Q}(\zeta)$ & \\
1366 $\langle \sigma^2 \rangle=\langle \sigma^4 \rangle$ & 3 & $\mathbb{Z}_3$ & $\mathbb{Q}(\alpha^3, \zeta)$ & $\sigma^2(\alpha^3)=\alpha^3 \zeta^{3\cdot 2}=\alpha^3 \zeta^6 = \alpha^3 \cdot 1 = \alpha^3$\\
1367 $\langle \sigma^3 \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha^2,\zeta)$ & $\sigma^3(\alpha^2)=(\alpha\zeta^3)^2=\alpha^2\zeta^6=\alpha^2$\\
1368 \hline
1369 $\langle \tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha)$ & \\
1370 \hline
1371 $\langle \sigma\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta)$ &
1372 $\sigma\zeta(\alpha+\alpha\zeta)=\sigma(\alpha+\alpha\zeta^{-1}) = \alpha\zeta + \alpha\zeta^{-1}\zeta=\alpha\zeta+\alpha$\\
1373 $\langle \sigma^2\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^2), \mathbb{Q}(\alpha\zeta)$ &
1374 $\sigma^2\tau(\alpha+\alpha\zeta^2) = \sigma(\alpha+\alpha\zeta^{-2})=\alpha\zeta^2+ \alpha\zeta^{-2}\zeta^2=\alpha\zeta^2+\alpha$\\
1375 $\langle \sigma^3\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^3)$ &
1376 $\sigma^3\tau(\alpha+\alpha\zeta^3) = \sigma(\alpha+\alpha\zeta^{-3})=\alpha\zeta^3+ \alpha\zeta^{-3}\zeta^3=\alpha\zeta^3+\alpha$\\
1377 $\langle \sigma^4\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^4), \mathbb{Q}(\alpha\zeta^2)$ &
1378 $\sigma^4\tau(\alpha+\alpha\zeta^4) = \sigma(\alpha+\alpha\zeta^{-4})=\alpha\zeta^4+ \alpha\zeta^{-4}\zeta^4=\alpha\zeta^4+\alpha$\\
1379 $\langle \sigma^5\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^5)$ &
1380 $\sigma^5\tau(\alpha+\alpha\zeta^5) = \sigma(\alpha+\alpha\zeta^{-5})=\alpha\zeta^5+ \alpha\zeta^{-5}\zeta^5=\alpha\zeta^5+\alpha$\\
1381 \hline
1382 $\langle \sigma, \tau \rangle = \langle \sigma^5,\tau \rangle$ & $6\cdot2=12$ & $\mathbb{D}_6$ & $\mathbb{Q}$ & \\
1383 $\langle \sigma^2, \tau \rangle = \langle \sigma^4,\tau \rangle$ & $3\cdot2=6$ & $\mathbb{D}_3$ & $\mathbb{Q}(\alpha^3)$ &
1384 $\sigma^2(\alpha^3)=\alpha^3\zeta^{3\cdot 2}=\alpha^3$ and $\tau(\alpha^3)=\alpha^3$\\
1385 $\langle \sigma^3, \tau \rangle$ & $2\cdot2=4$ & $\mathbb{D}_2$ & $\mathbb{Q}(\alpha^2)$ &
1386 $\sigma^3(\alpha^2)=\alpha^2\zeta^{2\cdot 2}=\alpha^2$ and $\tau(\alpha^2)=\alpha^2$\\
1387 \hline
1388 $\langle \sigma^2, \sigma\tau \rangle$ & $3\cdot 2=6$ & $\mathbb{D}_3$ & $\mathbb{Q}(\alpha^3+\alpha^3\zeta^3)$ &
1389 $\sigma^2(\alpha^3 + \alpha^3 \zeta^3) = \alpha^3\zeta^3 + \alpha^3 \zeta^3\zeta^3 = \alpha^3\zeta^3 + \alpha^3\zeta^6 = \alpha^3\zeta^3+\alpha^3$\\
1390 $\langle \sigma^3, \sigma\tau \rangle$ & $2\cdot2=4$ & $\mathbb{Z}_2 \times \mathbb{Z}_2$ & $\mathbb{Q}(\alpha^2\zeta^2),\mathbb{Q}(\alpha^2+\alpha^2\zeta^2)$ &
1391 $\sigma^3(\alpha^2+\alpha^2\zeta^2)=\alpha^2\zeta^{2\cdot3}+\alpha^2\zeta^{2\cdot3}\zeta^2=\alpha^2+\alpha^2\zeta^2$
1392 and
1393 $\sigma\tau(\alpha^2+\alpha^2\zeta^2)=\alpha^2\zeta^2+\alpha^2\zeta^{-2}\zeta^2 = \alpha^2\zeta^2+\alpha^2$\\
1394 $\langle \sigma^3, \sigma^2\tau\rangle$ & $2\cdot2=4$ & $\mathbb{Z}_2 \times \mathbb{Z}_2$ & $\mathbb{Q}(\alpha^2\zeta^4),\mathbb{Q}(\alpha^2+\alpha^2\zeta^4)$ &
1395 $\sigma^2\zeta(\alpha^2\zeta^4)=\alpha^2\zeta^2\zeta^{-4}=\alpha^2\zeta^{-2}=\alpha^2\zeta^4$
1396 and $\sigma^3(\alpha^2\zeta^4)=\alpha^2\zeta^{2\cdot3}\zeta^4=\alpha^2\zeta^4$
1397\end{tabular}
1398
1399
1400
1401\bibliographystyle{unsrt}
1402\bibliography{galois-theory-notes.bib}
1403
1404\end{document}