math/galois-theory-notes.tex
galois-theory-notes.texRaw
1\documentclass{article}
2\usepackage[utf8]{inputenc}
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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
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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{lemma}{5.14}
138 Let $K(\alpha):K$ be a simple algebraic extension, let $m$ be the minimal polynomial of $\alpha$ over $K$, let $\delta m =n$.
139
140 Then $\{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}\}$ is a basis for $K(\alpha)$ over $K$.
141 In particular, $[K(\alpha):K]=n$.
142\end{lemma}
143
144\begin{defn}{6.2}
145 The degree $[L:K]$ of a field extension $L:K$ is the dimension of L considered as a vector space over $K$.
146
147 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 $K$.
148\end{defn}
149
150\begin{eg}{6.3}
151 \begin{enumerate}
152 \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 $[\mathbb{C}:\mathbb{R}]=2$.
153 \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 $\mathbb{Q}$.
154 \end{enumerate}
155\end{eg}
156
157\begin{thm}{6.4}\emph{(Short Tower Law)} \label{shorttowerlaw}
158 If $K, L, M \subseteq \mathbb{C}$, and $K \subseteq L \subseteq M$, then $[M:K]=[M:L]\cdot [L:K]$.
159\end{thm}
160\begin{proof}
161 Let $(x_i)_{i \in I}$ be a basis for $L$ over $K$,
162 let $(y_j)_{j \in J}$ be a basis for $M$ over $L$.\\
163 $\forall i \in I, j \in J$, we have $x_i \in L, u_j \in M$.
164 \\
165 Want to show that $(x_i y_j)_{i\in I, j\in J}$ is a basis for $M$ over $K$.
166 \begin{enumerate}[i.]
167 \item prove linear independence:\\
168 Suppose that
169 $$\sum_{ij} k_{ij} x_i y_j = 0 ~(k_{ij} \in K)$$
170 rearrange
171 $$\sum_j (\underbrace{\sum_i k_{ij} x_i}_{\in L}) y_j = 0 ~(k_{ij} \in K)$$
172 Since $\sum_i k_{ij} x_i \in L$, and the $y_j \in M$ are linearly independent over $L$, then $\sum_i k_{ij} x_i = 0$.
173 \\
174 Repeating the argument inside $L$ $\longrightarrow$ $k_{ij}=0 ~~\forall i\in I, j\in J$.
175 \\
176 So the elements $x_i y_j$ are linearly independent over $K$.
177
178 \item prove that $x_i y_j$ span $M$ over $K$:\\
179 Any $x \in M$ can be written $x=\sum_j \lambda_j y_j$ for $\lambda_j \in L$, because $y_j$ spans $M$ over $L$.
180 Similarly, $\forall j\in J,~ \lambda_j = \sum_i \lambda_{ij} x_i y_j$ for $\lambda_{ij} \in K$.\\
181 Putting the pieces together, $x=\sum_{ij} \lambda_{ij} x_i y_j$ as required.
182 \end{enumerate}
183\end{proof}
184
185\begin{cor}{6.6}\emph{(Tower Law)}\\ \label{towerlaw}
186 If $K_0 \subseteq K_1 \subseteq \ldots \subseteq K_n$ are subfields of $\mathbb{C}$, then
187 $$[K_n:K_0] = [K_n:K_{n-1}] \cdot [K_{n-1}:K_{n-2}] \cdot \ldots \cdot [K_1: K_0]$$
188\end{cor}
189\begin{proof}
190 From \ref{shorttowerlaw}.
191\end{proof}
192
193[...] TODO: pending to add key parts up to Chapter 15.
194
195\subsection{Detour: Isomorphism Theorems}
196\begin{thm}{i.1}(\emph{First Isomorphism Theorem}) \label{1stisothm}
197
198 \begin{minipage}{0.75\textwidth}
199 If $\psi: G \longrightarrow H$ a group homomorphism, then $ker(\psi) \triangleleft G$.\\
200 Let $\phi: G \longrightarrow G/ker(\psi)$ be the canonical homomorphism.\\
201 Then $\exists$ unique isomorphism $\eta: G/ker(\psi) \longrightarrow \psi(G)$ such that $\psi = \eta \phi$.\\
202 $\Longleftrightarrow$ ie. $G/ker(\psi) \cong \psi(G)$.
203 \end{minipage}
204\hfill
205\begin{minipage}{0.2\textwidth}
206 \begin{tikzpicture}[node distance=2.5cm, auto]
207 \node (G) {$G$};
208 \node (H) [right of=G] {$H$};
209 \node (GmodK) [below of=G, xshift=1.25cm] {$G/\ker(\psi)$};
210
211 \draw[->] (G) to node {$\psi$} (H);
212 \draw[->] (G) to node [swap] {$\phi$} (GmodK);
213 \draw[->] (GmodK) to node [swap] {$\eta$} (H);
214 \end{tikzpicture}
215\end{minipage}
216\end{thm}
217\begin{proof}
218 \emph{(proof from Thomas W. Judson book "Abstract Algebra" \cite{judson})}\\
219
220 Let $K=ker(\psi)$.
221 Since
222 $$\eta: G/K \longrightarrow \psi(G)$$
223 let
224 $$\eta: gK \longrightarrow \psi(g)$$
225 ie. $\eta(gK)=\psi(g)$.
226
227 \begin{enumerate}[i.]
228 \item show that $\eta$ is a \emph{well defined} map:
229
230 if $g_1 K=g_2 K$, then for some $k \in K$, $g_1 k =g_2$, so
231 $$\eta(g_1K)=\psi(g_1) = \psi(g_1)\psi(k) = \psi(g_1 k) = \psi(g_2) = \eta(g_2 k)$$
232
233 Thus, $\eta$ does not depend on the choice of coset representatives, and
234 the map $\eta: G/ker(\psi) \longrightarrow \psi(G)$ is uniquely defined
235 since $\psi=\eta\phi$.
236
237 \item show that $\eta$ is a homomorphism:
238
239 Observe:
240 $$\eta(g_1 K g_2 K) = \eta(g_1 g_2 K) = \psi(g_1 g_2) = \psi(g_1) \psi(g_2) = \eta(g_1 K) \eta(g_2 K)$$
241 $\Longrightarrow$ so $\eta$ is a homomorphism.
242
243 \item show that $\eta$ is an isomorphism:
244
245 Since each element of $H=\psi(G)$ has at least a preimage, then $\eta$ is \emph{surjective} (onto $\psi(G)$).
246
247 Show that it is also \emph{injective} (onet-to-one):
248
249 Suppose 2 different preimatges lead to the same image in $\psi(G)$, ie.
250 $\eta(g_1 K) = \eta(g_2 K)$
251
252 then,
253 $$\psi(g_1) = \psi(g_2)$$
254 which implies $\psi(g_1^{-1} g_2) = e$, ie. $g_1^{-1} g_2 \in ker(\psi)$,
255 hence
256 $$g_1^{-1} g_2 K = K$$
257 $$g_1 K = g_2 K$$
258 so $\eta$ is injective.
259 \end{enumerate}
260
261 Since $\eta$ is injective and surjective $\Longrightarrow$ $\eta$ is a bijective homomorphism,\\
262 ie. $\eta$ is an \emph{isomorphism}.
263\end{proof}
264
265
266\begin{thm}{i.2}(\emph{Second Isomorphism Theorem}) \label{2ndisothm}
267 Let $H \subseteq G$, $N \triangleleft G$. Then
268 \begin{enumerate}[i.]
269 \item $HN \subseteq G$
270 \item $H \cap N \triangleleft H$
271 \item $\frac{H}{H \cap N} \cong \frac{HN}{N}$
272 \end{enumerate}
273\end{thm}
274\begin{proof}
275 \emph{(proof from Thomas W. Judson book "Abstract Algebra" \cite{judson})}\\
276
277 \begin{enumerate}[i.]
278 \item show $HN \subseteq G$:\\
279 Note that $HN = \{ hn : h\in H, n\in N \}$. Let $h_1 n_1, h_2 n_2 \in HN$.
280
281 Since $N$ normal $\Longrightarrow~ h_2^{-1} n_1 h_2 \in N$, so
282 $$(h_1 n_1)(h_2 n_2) = h_1 h_2 (h_2^{-1} n_1 h_2) \in HN$$
283
284 [Recall: since $N \triangleleft G$, $gN=Ng ~\forall g \in G$ $\Longrightarrow gn=n'g$ for some $n' \in N$.]
285
286 To see that $(hn)^{-1} \in HN$:\\
287 since $(hn)^{-1} = n^{-1} h^{-1} = h^{-1} (h n^{-1} h^{-1})$, thus $(hn)^{-1} \in HN$.
288
289 Thus $HN \subseteq G$.
290
291
292 In fact, $$HN = \bigcup_{h \in H} hN$$
293
294 (TODO: diagram)
295
296 \item show that $H \cap N \triangleleft H$:\\
297 Let $h \in H,~ n \in H \cap N$ (recall: $H \cap N \subseteq H$).\\
298 Then $h^{-1}nh \in H$ $\longleftarrow$ since $h^{-1}, n, h \in H$.\\
299 Since $N \triangleleft G$, $h^{-1} n h \in N$.\\
300 Therefore, $h^{-1}nh \in H \cap N$ $\Longrightarrow~ H \cap N \triangleleft H$
301
302 \item show that $\frac{H}{H \cap N} \cong \frac{HN}{N}$:\\
303 Define a map
304 \begin{align*}
305 \phi: &H \longrightarrow \frac{HN}{N}\\
306 \text{by}~ \phi: &h \longmapsto hN
307 \end{align*}
308
309 $\phi$ is surjective (onto), since any coset $hnN=hN$ is the image of $h \in H$, ie. $\phi(h)$
310
311 $\phi$ is a homomorphism, since
312 $$\phi(h h') = h h' N = hN h'N = \phi(h)\phi(h')$$
313
314 By the First Isomorphism Theorem \ref{1stisothm},
315 $$\frac{HN}{N} \cong \frac{H}{ker(\phi)}$$
316
317 and since
318 \begin{align*}
319 &ker(\phi) = \{ h \in H : h \in N\}\\
320 &\text{then}~~ker(\phi) = H \cap N
321 \end{align*}
322
323 so then,
324 $$\frac{HN}{N} = \phi(H) \cong \frac{H}{ker(\phi)} = \frac{H}{H \cap N}$$
325 thus
326 $$\frac{HN}{N} \cong \frac{H}{H \cap N}$$
327
328 \end{enumerate}
329\end{proof}
330
331
332\begin{thm}{i.3}(\emph{Third Isomorphism Theorem}) \label{2ndisothm}\\
333 Let $H \subseteq K$ and $K \triangleleft G,~ H \triangleleft G$.\\
334 Then
335 $\frac{K}{H} \triangleleft \frac{G}{H}$
336 and
337 $$\frac{ G/H }{ K/H } \cong \frac{G}{K}$$
338\end{thm}
339\begin{proof}
340 \emph{(proof from Dummit and Foote book ``Abstract Algebra" \cite{dummitfoote})}\\
341
342 Easy to see that $\frac{K}{H} \triangleleft \frac{G}{H}$.
343
344 Define
345 \begin{align*}
346 \psi: &\frac{G}{H} \longrightarrow \frac{G}{K}\\
347 \text{by}~ \psi: &gH \longmapsto gK
348 \end{align*}
349
350 To show that $\psi$ is \emph{well defined}:\\
351 suppose $g_1 H = g_2 H$, then $g_1 = g_2 h$ for some $h \in H$.\\
352 Since $H \subseteq K \Longrightarrow~ h \in K$, hence $g_1 K = g_2 K$,\\
353 ie. $\psi(g_1 H) = \psi(g_2 H)$, which shows that $\psi$ is well defined.
354
355 Since $g \in G$ may be chosen arbitrarily in $G$, $\psi$ is a surjective homomorphism.
356
357
358 Finally,
359 \begin{align*}
360 ker(\psi) &= \{ gH \in \frac{G}{H} \mid \psi(gH)=1K \}\\
361 &= \{ gH \in \frac{G}{H} \mid gK =1K \}\\
362 &= \{ gH \in \frac{G}{H} \mid g \in K \}\\
363 &= \frac{K}{H}
364 \end{align*}
365
366
367 By the First Isomorphism Theorem (\ref{1stisothm}),
368
369 \begin{tikzpicture}[node distance=2.5cm, auto]
370 \node (GmodH) {$\frac{G}{H}$};
371 \node (GmodK) [right of=GmodH] {$\frac{G}{K}$};
372 \node (GmodHmodKer) [below of=G, xshift=1.25cm] {$\frac{G/H}{\ker(\psi)} = \frac{G/H}{K/H}$};
373
374 \draw[->] (GmodH) to node {$\psi$} (GmodK);
375 \draw[->] (GmodH) to node [swap] {$\phi$} (GmodHmodKer);
376 \draw[->] (GmodHmodKer) to node [swap] {$\eta$} (GmodK);
377 \end{tikzpicture}
378
379 So, by
380 $$\eta: \frac{G/H}{K/H} \longrightarrow \frac{G}{K}$$
381 since $\eta$ is
382 bijective (we know it by the First Isomorphism Theorem), $\eta$ it is the isomorphism:
383 $$\frac{ G/H }{ K/H } \cong \frac{G}{K}$$
384
385
386\end{proof}
387
388\subsection{Chapter 14}
389
390
391\begin{thm}{14.4}
392 $H \subseteq G,~~ N \triangleleft G$, then
393 \begin{enumerate}
394 \item if $G$ soluble $\Longrightarrow H$ soluble
395 \item if $G$ soluble $\Longrightarrow G/N$ soluble
396 \item if $N$ and $G/N$ soluble $\Longrightarrow G$ soluble
397 \end{enumerate}
398\end{thm}
399\begin{proof}
400 \begin{enumerate}
401 \item Since $G$ soluble, by definition: $\exists~~ 1=G_0 \triangleleft G_1 \triangleleft \ldots \triangleleft G_r = G$ with Abelian quotients $\frac{G_{i+1}}{G_i}$.
402
403 Let $H_i = G_i \cap H$, then $H$ has a series $1=H_0 \triangleleft H_1
404 \triangleleft \ldots \triangleleft H_r = H$, next we show that the
405 quotients $\frac{H_{i+1}}{H_i}$ are Abelian (so that H is soluble):
406
407 $$\frac{H_{i+1}}{H_i} = \frac{G_{i+1} \cap H}{G_i \cap H} \stackrel{(*)}{=} \frac{G_{i+1} \cap H}{G_i \cap (G_{i+1}\cap H)}
408 \stackrel{(**)}{\cong} \frac{G_i(G_{i+1} \cap H)}{G_i} \subseteq \frac{G_{i+1}}{G_i}
409 $$
410
411 (*): to see why, $H_i = G_i \cap H = G_i \cap H_i = G_i \cap H_{i+1} = G_i \cap (G_{i+1} \cap H)$.
412 (**): by the 2nd Isomorphism Theorem (\ref{2ndisothm}).
413
414 [TODO: diagram of subgroups]
415
416 Notice that $\frac{G_{i+1}}{G_i}$ is Abelian, thus the left-hand-side of the congruence is also Abelian.
417 Therefore, $\frac{H_{i+1}}{H_i}$ is Abelian, thus $H$ is soluble.
418
419 \item For $G/N$ to be soluble, (by definition) it would have the series $\frac{N}{N} = G_0 \frac{N}{N} \triangleleft G_1 \frac{N}{N} \triangleleft \ldots \triangleleft G_r \frac{N}{N} = \frac{G}{N}$,
420 and any quotient being $\frac{G_{i+1}\frac{N}{N}}{G_i \frac{N}{N}}$.
421
422 The series clearly exists, so now we show that the quotients are Abelian, so that $G/N$ is soluble:
423
424 $$
425 \frac{G_{i+1} N}{G_i N} = \frac{G_{i+1}(G_i N)}{G_i N} \stackrel{*}{\cong}
426 \frac{G_{i+1}}{G_{i+1} \cap (G_i N)} \cong \frac{G_{i+1}/G_i}{(G_{i+1}
427 \cap (G_i N))/G_i}
428 $$
429
430 (*): by the 2nd Isomorphism Theorem (\ref{2ndisothm}).
431
432 The last quotient is a quotient of the Abelian group $G_{i+1}/G_i$, so it is Abelian.
433
434 Hence, $\frac{G_{i+1}N}{G_i N}$ is also Abelian; so $\frac{G}{N}$ is soluble.
435
436 \item
437 By the definition of $N$ and $G/N$ being soluble,
438 \begin{align*}
439 N \text{soluble} \Longrightarrow~~ 1 &= N_0 \triangleleft N_1 \triangleleft \ldots \triangleleft N_r = N\\
440 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}
441 \end{align*}
442 both with Abelian quotients.
443
444 Consider the series of $G$ given by combining the two previous series:
445
446 $$
447 1 &= N_0 \triangleleft N_1 \triangleleft \ldots \triangleleft N_r = N = G_0 \triangleleft G_1 \triangleleft \ldots \triangleleft G_r = G
448 $$
449 the quotients are either
450 \begin{itemize}
451 \item $\frac{N_{i+1}}{N_i}$, Abelian
452 \item $\frac{G_{i+1}}{G_i}$, isomorphic to $\frac{G_{i+1}/N}{G_i/N}$,
453 which is Abelian.
454 \end{itemize}
455 \end{enumerate}
456 Therefore, the quotients are always Abelian; hence $G$ is soluble.
457\end{proof}
458
459\newpage
460
461\section{Tools}
462This section contains tools that I found useful to solve Galois Theory related problems, and that don't appear in Stewart's book.
463
464\subsection{De Moivre's Theorem and Euler's formula}\label{demoivre}
465Useful for finding all the roots of a polynomial.
466
467Euler's formula:
468$$e^{i \psi} = cos \psi + i \cdot sin \psi$$
469
470The n-th roots of a complex number $z=x + i y = r (cos \theta + i \cdot sin \theta)$ are given by
471
472$$z_k = \sqrt[n]{r} \cdot \left(cos(\frac{\theta + 2k \pi}{n}) + i \cdot sin(\frac{\theta + 2k \pi}{n}) \right)$$
473for $k=0, \ldots, n-1$.
474
475So, by Euler's formula:
476$$z_k = \sqrt[n]{r} \cdot e^{i (\frac{\theta + 2 k \pi}{n})}$$
477
478Usually we will set $\alpha=\sqrt[n]{r}$ and $\zeta = e^{\frac{2 \pi i}{n}}$,
479and find the $\mathbb{Q}$-automorphisms from there (see \ref{ex:galoisgroups} for examples).
480
481\subsection{Einsenstein's Criterion} \label{einsenstein}
482\emph{reference: Stewart's book}
483
484Let $f(t) = a_0 + a_1 t + \ldots + a_n t^n$, suppose there is a prime $q$ such that
485\begin{enumerate}
486 \item $q \nmid a_n$
487 \item $q | a_i$ for $i=0, \ldots, n-1$
488 \item $q^2 \nmid a_0$
489\end{enumerate}
490Then, $f$ is irreducible over $\mathbb{Q}$.
491
492\emph{TODO proof \& Gauss lemma.}
493
494
495\subsection{Elementary symmetric polynomials}
496\emph{TODO from orange notebook, page 36}
497
498\subsection{Cyclotomic polynomials} \label{cyclotomicpoly}
499\subsubsection{From Elmyn Berlekamp's "Algebraic Coding Theory" book}
500The notes in this section are from the book "Algebraic Coding Theory" by Elmyn
501Berlekamp \cite{berlekamp}.
502
503\vspace{0.3cm}
504
505Some 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.
506
507Observe that in a finite field of order $q$, factoring $x^q - x$ gives
508$$x^q-x = x(x^{q-1} -1)$$
509
510The factor $x^{q-1} -1$ is a special case of $x^n -1$: if we assume that the
511field contains an element $\alpha$ of order $n$, then the roots of $x^n-1=0$ are
512$$1, \alpha, \alpha^2, \alpha^3, \ldots, \alpha^{n-1}$$
513and $\deg(x^n-1)=n$, thus $x^n-1$ has at most $n$ roots in any field, henceforth
514the powers of $\alpha$ must include all the $n$-th roots of unity.
515
516There fore, in any field which contains a primitive $n$-th root of unity we have:
517
518\begin{thm}{4.31}
519 $$x^n -1 = \prod_{i=0}^{n-1} (x - \alpha^i) = \prod_{i=1}^n (x-\alpha^i)$$
520\end{thm}
521
522If $n=k \cdot d$, then $\alpha^k, \alpha^{2k}, \alpha^{3k}, \ldots, \alpha^{dk}$ are all roots of $x^d -1 =0$
523
524Every element with order dividing $n$, must be a power of $\alpha$, since an
525element of order $d$ is a $d$-th root of unity.
526
527Every power of $\alpha$ has order which divides $n$, and every field element
528whose order divides $n$ is a power of $\alpha$. This suggests that we partition
529the powers of $\alpha$ according to their orders:
530$$x^n -1 = \prod_{\stackrel{d,}{d|n}} \prod_{\beta} (x- \beta)$$
531where at each iteration, $\beta$ is a field element of order $d$ for each $d$.
532
533The polynomial whose roots are the field elements of order $d$ is called the
534\emph{cyclotomic polynomial}, denoted by $Q^{(d)}(x)$.
535
536\begin{thm}{4.32}
537 $$x^n -1 = \prod_{\stackrel{d,}{d|n}} Q^{(d)}(x)$$
538\end{thm}
539
540
541\subsubsection{From Ian Stewart's ``Galois Theory'' book}
542Notes from Ian Stewart's book \cite{ianstewart}.
543
544Consider the case $n=12$, let $\zeta=e^{\pi i /6}$ be a primitive $12$-th root of unity.
545Classify its powers ($\zeta^j$) according to their minimal power $d$ such that
546$(\zeta^j)^d = 1$ (ie. when they are primitive $d$-th roots of unity).
547
548\begin{enumerate}[]
549 \item $d=1,~~~ 1$
550 \item $d=2,~~~ \zeta^6$
551 \item $d=3,~~~ \zeta^4, \zeta^8$
552 \item $d=4,~~~ \zeta^3, \zeta^9$
553 \item $d=6,~~~ \zeta^2, \zeta^{10}$
554 \item $d=12,~~~ \zeta, \zeta^5, \zeta^7, \zeta^{11}$
555\end{enumerate}
556
557Observe that we can factorize $t^{12} -1$ by grouping the corresponding zeros:
558\begin{align*}
559 t^{12}-1 = &(t-1) \times\\
560 &(t-\zeta^6) \times\\
561 &(t-\zeta^4) (t-\zeta^8) \times\\
562 &(t-\zeta^3) (t-\zeta^9) \times\\
563 &(t-\zeta^2) (t-\zeta^{10}) \times\\
564 &(t-\zeta) (t-\zeta^5)(t-\zeta^7) (t-\zeta^{11})
565\end{align*}
566which simplifies to
567$$t^{12}-1=(t-1)(t+1)(t^2+t+1)(t^2+1)(t^2-t+1)F(t)$$
568where $F(t) = (t-\zeta) (t-\zeta^5)(t-\zeta^7) (t-\zeta^{11}) = t^4 -t^2 + 1$ (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).
569
570
571Let $\Phi_d(t)$ be the factor corresponding to primitive $d$-th roots of unity, then we have proved that
572$$t^{12}-1 = \Phi_1 \Phi_2 \Phi_3 \Phi_4 \Phi_6 \Phi_{12}$$
573
574
575\begin{defn}{21.5}
576 The polynomial $\Phi_d(t)$ defined by
577 $$\Phi_n(t) = \prod_{a\in \mathbb{Z}_n,(a,n)=1} (t- \zeta^a)$$
578 is the $n$-th \emph{cyclotomic polynomial} over $\mathbb{C}$.
579\end{defn}
580
581\begin{cor}{21.6}
582 $\forall n \in \mathbb{N}$, the polynomial $\Phi_n(t)$ lies in $\mathbb{Z}[t]$ and is monic and irreducible.
583\end{cor}
584
585\begin{thm}{21.9}
586 \begin{enumerate}
587 \item The Galois group $\Gamma(\mathbb{Q}(\zeta):\mathbb{Q})$ consists of the
588 $\mathbb{Q}$-automorphisms $\psi_j$ defined by
589 $$\psi_j(\zeta)=\zeta^j$$
590 where $0 \leq j \leq n-1$ and $j$ is prime to $n$.
591
592 \item $\Gamma(\mathbb{Q}(\zeta):\mathbb{Q}) \stackrel{iso}{\cong} \mathbb{Z}_n^*$, and is an abelian group.
593 \item its order is $\phi(n)$
594 \item if $n$ is prime, $\mathbb{Z}_n^*$ is cyclic
595 \end{enumerate}
596\end{thm}
597
598
599
600\vspace{1cm}
601
602\subsubsection{Examples}
603Examples of cyclotomic polynomials:
604
605\begin{align*}
606 \Phi_n(x) &= x^{n-1} + x^{n-2} + \ldots + x^2 + x + 1 = \sum_{i=0}^{n-1} x^i\\
607 \Phi_{2p}(x) &= x^{p-1} + \ldots + x^2 - x + 1 = \sum_{i=0}^{p-1} (-x)^i\\
608 \Phi_m(x) &= x^{m/2} + 1, ~~\text{when $m$ is a power of $2$}
609\end{align*}
610
611
612\subsection{Lemma 1.42 from J.S.Milne's book}
613Lemma from J.S.Milne's book \cite{milneFT}.
614
615Useful for when dealing with $x^p - 1$ with $p$ prime.
616
617Observe that
618
619$$x^p -1 = (x-1)(x^{p-1} + x^{p-2} + \ldots + 1)$$
620
621Notice that
622$$\Phi_p(x) = x^{p-1} + x^{p-2} + \ldots + 1$$
623is the $p$-th Cyclotomic polynomial.
624
625\begin{lemma}{1.42}
626 If $p$ prime, then $x^{p-1} + \ldots + 1$ is irreducible; hence $\mathbb{Q}[e^{2 \pi i /p}]$ has degree $p-1$ over $\mathbb{Q}$.
627\end{lemma}
628\begin{proof}
629 Let $f(x) = (x^p - 1)/(x-1) = x^{p-1} + \ldots + 1$
630 then
631 $$
632 f(x+1) = \frac{(x+1)^p -1}{x+1-1} = \frac{(x+1)^p -1}{x} = x^{p-1} + \ldots + a_i x^i + \ldots + p
633 $$
634
635 with $a_i = \left( \stackrel{p}{i+1} \right)$.
636
637 We know that $p | a_i$ for $i= 1, \ldots, p-2$, therefore $f(x+1)$ is irreducibe by Einsenstein's Criterion.
638
639 This implies that $f(x)$ is irreducible.
640\end{proof}
641
642
643\subsection{Dihedral groups - Groups of symmetries} \label{dihedral}
644Source: Wikipedia and \cite{dihedral}.
645
646Dihedral groups ($\mathbb{D}_n$) represent the symmetries of a regular $n$-gon.
647
648Properties:
649\begin{itemize}
650 \item are non-abelian (for $n>2$), ie. $rs \neq sr$
651 \item order $2n$
652 \item generated by a rotation $r$ and a reflection $s$
653 \item $r^n = s^2 = id,~~~(rs)^2=id$
654\end{itemize}
655Subgroups of $\mathbb{D}_n$:
656\begin{itemize}
657 \item rotation form a cyclic subgroup of order $n$, denoted as $<r>$
658 \item for each $d$ such that $d|n$, $\exists~ \mathbb{D}_d$ with order $2d$
659 \item normal subgroups
660 \begin{itemize}
661 \item for $n$ odd: $\mathbb{D}_n$ and $<r^d>$ for every $d|n$
662 \item for $n$ even: $2$ additional normal subgroups
663 \end{itemize}
664 \item Klein four-groups: $\mathbb{Z}_2 \times \mathbb{Z}_2$, of order 4
665\end{itemize}
666
667\vspace{0.3cm}
668Total 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.
669
670\begin{eg}{}
671For $\mathbb{D}_6$, we have $\{1,2,3,6\} | 6$, so $d(n) = d(6) = 4$, and
672$s(6) = 1+2+3+6 = 12$; henceforth, the total amount of subgroups is $d(n)+s(n) = 4+12 = 16$.
673\end{eg}
674
675\vspace{0.3cm}
676For $n \geq 3, ~~\mathbb{D}_n \subseteq \mathbb{S}_n$ (subgroup of the Symmetry group).
677
678
679
680\newpage
681
682\section{Exercises}
683
684\subsection{Galois groups}\label{ex:galoisgroups}
685
686\subsubsection[t6-7]{$t^6-7 \in \mathbb{Q}$}
687
688This exercise comes from a combination of exercises 12.4 and 13.7 from \cite{ianstewart}.
689
690First let's find the roots. By De Moivre's Theorem (\ref{demoivre}), $t_k =
691\sqrt[6]{7} \cdot e^{i \frac{2 \pi k}{6}}$.
692
693From which we denote $\alpha = \sqrt[6]{7}$, and $\zeta = e^{\frac{2 \pi i}{6}}$, so that the
694roots of the polynomial are $\{ \alpha, \alpha \zeta, \alpha \zeta^2, \alpha \zeta^3, \alpha \zeta^4, \alpha \zeta^5\}$, ie.
695$\{ \alpha \zeta^k \}_0^5$.
696
697Hence the \emph{splitting field} is $\mathbb{Q}(\alpha, \zeta)$.
698
699\emph{Degree of the extension}
700
701In order to find $[\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}$, we're going to split it in tow
702parts. By the Tower Law (\ref{towerlaw}),
703
704$$[\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}] = [\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}(\alpha)] \cdot [\mathbb{Q}(\alpha) : \mathbb{Q}]$$
705
706To find each degree, we will find the minimal polynomial of the adjoined term over the base field of the extension:
707
708\begin{enumerate}[i.]
709 \item minimal polynomial of $\alpha$ over $\mathbb{Q}$\\
710 By Einsenstein's Criterion (\ref{einsenstein}), with $q=7$ we have that $q
711 \nmid 1$, $7 | {-7,0,0,\ldots}$, and $7^2 \nmid -7$, hence $f(t)$ is
712 irreducibe over $\mathbb{Q}$, thus is the minimal polynomial
713 $$m_i(t)= f(t) =t^6-7$$
714 which has roots $\{ \alpha \zeta^k \}_0^5$.
715 \item minimal polynomial of $\zeta$ over $\mathbb{Q}(\alpha)$\\
716 Since $\zeta$ is the primitive $6$th root of unity, we know that the minimal
717 polynomial will be the $6$th cyclotomic polynomial (\ref{cyclotomicpoly}):
718 $$m_{ii}(t) = \Phi_6(t) = t^2 - t + 1$$
719 which has roots $\zeta, -\zeta$.
720
721 Since $\mathbb{Q}(\alpha) \subseteq \mathbb{R}$, and the roots of
722 $\Phi_6(t)=t^2 - t +1$ are in $\mathbb{C}$, $\Phi_6(t)$ remains irreducible
723 over $\mathbb{Q}(\alpha)$.
724\end{enumerate}
725
726\vspace{0.5cm}
727Therefore, by the tower of law,
728$$[\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}] = \deg{\Phi_6(t)} \cdot \deg{f(t)} = 2 \cdot 6 = 12$$
729and by the Fundamental Theorem of Galois Theory, we know that
730$$|\Gamma( \mathbb{Q}(\alpha, \zeta) : \mathbb{Q} )| = [\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}] = 12$$
731which tells us that there exist $12$ $\mathbb{Q}$-automorphisms of the Galois group.
732
733
734\vspace{0.5cm}
735Let's find the $12$ $\mathbb{Q}$-automorphisms. Start by defining $\sigma$ which
736fixes $\zeta$ and acts on $\alpha$, sending it to another of the roots of the
737minimal polynomial of $\alpha$ over $\mathbb{Q}$, $f(t)$, choose $\alpha \zeta$.
738
739Now define $\tau$ which fixes $\alpha$ and acts on $\zeta$, sending it into
740another root of the minimal polynomial of $\zeta$ over $\mathbb{Q}(\alpha)$,
741choose $-\zeta$.
742
743\vspace{0.3cm}
744\begin{tabular}{@{}l l@{}}
745 $\begin{aligned}
746 \sigma: \alpha &\mapsto \alpha \zeta \\
747 \zeta &\mapsto \zeta
748 \end{aligned}$
749 &
750 $\begin{aligned}
751 \tau: \alpha &\mapsto \alpha\\
752 \zeta &\mapsto -\zeta = \zeta^{-1}
753 \end{aligned}$
754\end{tabular}
755
756In other words, we have $12$ $\mathbb{Q}$-automorphisms, which are the
757combination of $\sigma$ and $\tau$:
758
759$$\begin{aligned}
760 \sigma^k \tau^j:~~&\alpha \mapsto \alpha \zeta^k\\
761 &\zeta \mapsto \zeta^j
762\end{aligned}$$
763
764for $0 \leq k \leq 5$ and $j = \pm 1$.
765
766\vspace{0.5cm}
767
768NOTE: WIP diagram.
769\begin{tikzpicture}[node distance=2cm]
770 \def \radius{2}
771 \draw (0,0) circle (\radius);
772
773 \foreach \k in {0,...,5} {
774 % \node (a\k) at ({360/6 * \k}:\radius) {$\alpha \zeta^{\k}$};
775 \node (a\k) at ({360/6 * \k}:\radius+0.5) {$\alpha \zeta^{\k}$};
776 \fill ({360/6 * \k}:\radius) circle (2pt);
777 }
778 % real & im axis
779 \draw[->] (-2.5,0) -- (2.5,0) node[right] {};
780 \draw[->] (0,-2.5) -- (0,2.5) node[above] {};
781
782
783 % tau:
784 \draw[<->] (3,1) -- (3,-1) node[right] {$\tau$};
785 % sigma:
786 % \foreach \k [evaluate=\k as \next using int(mod(\k+1,6))] in {0,...,5} {
787 % \coordinate (p\k) at ({360/6 * \k}:\radius);
788 % \coordinate (p\next) at ({360/6 * \next}:\radius);
789 %
790 % \draw[->, bend left=30] (p\k.center) -- node[above] {$\sigma$} (p\next.center);
791 % }
792 \foreach \k in {0,...,5} {
793 \coordinate (p\k) at ({360/6 * \k}:\radius);
794 }
795 \foreach \k [evaluate=\k as \next using int(mod(\k+1,6))] in {0,...,5} {
796 \draw[->, bend left=30] (p\k) -- node[above] {$\sigma$} (p\next);
797 }
798\end{tikzpicture}
799
800\vspace{0.5cm}
801
802Observe, that $\Gamma$ is generated by the combination of $\sigma$ and $\tau$,
803and it is isomorphic to the group of symmetries of order 12, the dihedral
804group (\ref{dihedral}) of order 12, $\mathbb{D}_6$, ie. $\Gamma \cong \mathbb{D}_6$.
805
806\vspace{0.5cm}
807
808Let's find the subgroups of $\Gamma$, and the fixed fields of $\mathbb{Q}(\alpha, \zeta)$.
809
810We know that $\Gamma \cong \mathbb{D}_6$, and we know from the properties
811of the dihedral group (\ref{dihedral}) that the number of subgroups of
812$\mathbb{D}_6$ will be $d(6) + s(6) = 4 + 12 = 16$ subgroups.
813
814
815\vspace{0.4cm}
816
817\hspace*{-3.5cm}
818\begin{tabular}{ c c c c | p{7.5cm} }
819 \hline
820 generators & order & group & fixed field & notes (check fixed field)\\
821\hline
822 $\langle \rangle = \langle \sigma^6 \rangle=\langle \tau^2 \rangle$ & 1 & id & $\mathbb{Q}(\alpha,\zeta)$ & \\
823 $\langle \sigma \rangle = \langle \sigma^5 \rangle$ & 6 & $\mathbb{Z}_6$ & $\mathbb{Q}(\zeta)$ & \\
824 $\langle \sigma^2 \rangle=\langle \sigma^4 \rangle$ & 3 & $\mathbb{Z}_3$ & $\mathbb{Q}(\alpha^3, \zeta)$ & $\sigma^2(\alpha^3)=\alpha^3 \zeta^{3\cdot 2}=\alpha^3 \zeta^6 = \alpha^3 \cdot 1 = \alpha^3$\\
825 $\langle \sigma^3 \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha^2,\zeta)$ & $\sigma^3(\alpha^2)=(\alpha\zeta^3)^2=\alpha^2\zeta^6=\alpha^2$\\
826 \hline
827 $\langle \tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha)$ & \\
828 \hline
829 $\langle \sigma\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta)$ &
830 $\sigma\zeta(\alpha+\alpha\zeta)=\sigma(\alpha+\alpha\zeta^{-1}) = \alpha\zeta + \alpha\zeta^{-1}\zeta=\alpha\zeta+\alpha$\\
831 $\langle \sigma^2\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^2), \mathbb{Q}(\alpha\zeta)$ &
832 $\sigma^2\tau(\alpha+\alpha\zeta^2) = \sigma(\alpha+\alpha\zeta^{-2})=\alpha\zeta^2+ \alpha\zeta^{-2}\zeta^2=\alpha\zeta^2+\alpha$\\
833 $\langle \sigma^3\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^3)$ &
834 $\sigma^3\tau(\alpha+\alpha\zeta^3) = \sigma(\alpha+\alpha\zeta^{-3})=\alpha\zeta^3+ \alpha\zeta^{-3}\zeta^3=\alpha\zeta^3+\alpha$\\
835 $\langle \sigma^4\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^4), \mathbb{Q}(\alpha\zeta^2)$ &
836 $\sigma^4\tau(\alpha+\alpha\zeta^4) = \sigma(\alpha+\alpha\zeta^{-4})=\alpha\zeta^4+ \alpha\zeta^{-4}\zeta^4=\alpha\zeta^4+\alpha$\\
837 $\langle \sigma^5\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^5)$ &
838 $\sigma^5\tau(\alpha+\alpha\zeta^5) = \sigma(\alpha+\alpha\zeta^{-5})=\alpha\zeta^5+ \alpha\zeta^{-5}\zeta^5=\alpha\zeta^5+\alpha$\\
839 \hline
840 $\langle \sigma, \tau \rangle = \langle \sigma^5,\tau \rangle$ & $6\cdot2=12$ & $\mathbb{D}_6$ & $\mathbb{Q}$ & \\
841 $\langle \sigma^2, \tau \rangle = \langle \sigma^4,\tau \rangle$ & $3\cdot2=6$ & $\mathbb{D}_3$ & $\mathbb{Q}(\alpha^3)$ &
842 $\sigma^2(\alpha^3)=\alpha^3\zeta^{3\cdot 2}=\alpha^3$ and $\tau(\alpha^3)=\alpha^3$\\
843 $\langle \sigma^3, \tau \rangle$ & $2\cdot2=4$ & $\mathbb{D}_2$ & $\mathbb{Q}(\alpha^2)$ &
844 $\sigma^3(\alpha^2)=\alpha^2\zeta^{2\cdot 2}=\alpha^2$ and $\tau(\alpha^2)=\alpha^2$\\
845 \hline
846 $\langle \sigma^2, \sigma\tau \rangle$ & $3\cdot 2=6$ & $\mathbb{D}_3$ & $\mathbb{Q}(\alpha^3+\alpha^3\zeta^3)$ &
847 $\sigma^2(\alpha^3 + \alpha^3 \zeta^3) = \alpha^3\zeta^3 + \alpha^3 \zeta^3\zeta^3 = \alpha^3\zeta^3 + \alpha^3\zeta^6 = \alpha^3\zeta^3+\alpha^3$\\
848 $\langle \sigma^3, \sigma\tau \rangle$ & $2\cdot2=4$ & $\mathbb{Z}_2 \times \mathbb{Z}_2$ & $\mathbb{Q}(\alpha^2\zeta^2),\mathbb{Q}(\alpha^2+\alpha^2\zeta^2)$ &
849 $\sigma^3(\alpha^2+\alpha^2\zeta^2)=\alpha^2\zeta^{2\cdot3}+\alpha^2\zeta^{2\cdot3}\zeta^2=\alpha^2+\alpha^2\zeta^2$
850 and
851 $\sigma\tau(\alpha^2+\alpha^2\zeta^2)=\alpha^2\zeta^2+\alpha^2\zeta^{-2}\zeta^2 = \alpha^2\zeta^2+\alpha^2$\\
852 $\langle \sigma^3, \sigma^2\tau\rangle$ & $2\cdot2=4$ & $\mathbb{Z}_2 \times \mathbb{Z}_2$ & $\mathbb{Q}(\alpha^2\zeta^4),\mathbb{Q}(\alpha^2+\alpha^2\zeta^4)$ &
853 $\sigma^2\zeta(\alpha^2\zeta^4)=\alpha^2\zeta^2\zeta^{-4}=\alpha^2\zeta^{-2}=\alpha^2\zeta^4$
854 and $\sigma^3(\alpha^2\zeta^4)=\alpha^2\zeta^{2\cdot3}\zeta^4=\alpha^2\zeta^4$
855\end{tabular}
856
857
858
859\bibliographystyle{unsrt}
860\bibliography{galois-theory-notes.bib}
861
862\end{document}