math/galois-theory-notes.tex
galois-theory-notes.texRaw
1\documentclass{article}
2\usepackage[utf8]{inputenc}
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4\usepackage{amsthm}
5\usepackage{amsmath}
6\usepackage{enumerate}
7\usepackage{hyperref}
8\usepackage{amssymb}
9
10\begin{filecontents}[overwrite]{galois-theory-notes.bib}
11@misc{ianstewart,
12 author = {Ian Stewart},
13 title = {{Galois Theory, Third Edition}},
14 year = {2004}
15}
16
17@misc{dihedral,
18 author = {Gaurab Bardhan and Palash Nath and Himangshu Chakraborty}
19 title = {Subgroups and normal subgroups of dihedral group up to isomorphism}
20 year = {2010},
21 note = {\url{https://scipp.ucsc.edu/~haber/ph251/Dn_subgroups.pdf}},
22 url = {https://scipp.ucsc.edu/~haber/ph251/Dn_subgroups.pdf}
23}
24\end{filecontents}
25\nocite{*}
26
27
28\theoremstyle{definition}
29
30\newtheorem{innerdefn}{Definition}
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34
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39
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44
45\newtheorem{innercor}{Lemma}
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50\newtheorem{innereg}{Example}
51\newenvironment{eg}[1]
52{\renewcommand\theinnereg{#1}\innereg}
53{\endinnereg}
54
55
56\title{Galois Theory notes}
57\author{arnaucube}
58\date{2025}
59
60\begin{document}
61
62\maketitle
63
64\begin{abstract}
65 Notes taken while studying Galois Theory, mostyly from Ian Stewart's book "Galois Theory" \cite{ianstewart}.
66
67 Usually while reading books and papers I take handwritten notes in a notebook, this document contains some of them re-written to $LaTeX$.
68
69 The notes are not complete, don't include all the steps neither all the proofs.
70\end{abstract}
71
72\tableofcontents
73
74\section{Recap on the degree of field extensions}
75(Definitions, theorems, lemmas, corollaries and examples enumeration follows from Ian Stewart's book \cite{ianstewart}).
76
77\begin{defn}{4.10}
78 A \emph{simple extension} is $L:K$ such that $L=K(\alpha)$ for some $\alpha \in L$.
79\end{defn}
80\begin{eg}{4.11}
81 Beware, $L=\mathbb{Q}(i, -i, \sqrt{5}, -\sqrt{5}) = \mathbb{Q}(i, \sqrt{5}) = \mathbb{Q}(i+\sqrt{5})$.
82\end{eg}
83
84\begin{defn}{5.5}
85 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 $m(\alpha)=0$.
86 \\
87 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 $m(i)=0$.
88\end{defn}
89
90\begin{lemma}{5.9}
91 Every polynomial $a \in K[t]$ is congruent modulo $m$ to a unique polynomial of degree $< \delta m$.
92\end{lemma}
93\begin{proof}
94 Divide $a / m$ with remainder, $a= qm +r$, with $q,r \in K[t]$ and $\delta r < \delta m$.
95 Then, $a-r=qm$, so $a \equiv r \pmod{m}$.
96
97 It remains to prove uniqueness.
98
99 Suppose $\exists~ r \equiv s \pmod{m}$, with $\delta r, \delta s < \delta m$.
100 Then, $r-s$ is divisible by $m$, but has smaller degree than $m$.
101
102 Therefore, $r-s=0$, so $r=s$, proving uniqueness.
103\end{proof}
104
105\begin{lemma}{5.14}
106 Let $K(\alpha):K$ be a simple algebraic extension, let $m$ be the minimal polynomial of $\alpha$ over $K$, let $\delta m =n$.
107
108 Then $\{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}\}$ is a basis for $K(\alpha)$ over $K$.
109 In particular, $[K(\alpha):K]=n$.
110\end{lemma}
111
112\begin{defn}{6.2}
113 The degree $[L:K]$ of a field extension $L:K$ is the dimension of L considered as a vector space over $K$.
114
115 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 $K$.
116\end{defn}
117
118\begin{eg}{6.3}
119 \begin{enumerate}
120 \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 $[\mathbb{C}:\mathbb{R}]=2$.
121 \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 $\mathbb{Q}$.
122 \end{enumerate}
123\end{eg}
124
125\begin{thm}{6.4}\emph{(Short Tower Law)} \label{shorttowerlaw}
126 If $K, L, M \subseteq \mathbb{C}$, and $K \subseteq L \subseteq M$, then $[M:K]=[M:L]\cdot [L:K]$.
127\end{thm}
128\begin{proof}
129 Let $(x_i)_{i \in I}$ be a basis for $L$ over $K$,
130 let $(y_j)_{j \in J}$ be a basis for $M$ over $L$.\\
131 $\forall i \in I, j \in J$, we have $x_i \in L, u_j \in M$.
132 \\
133 Want to show that $(x_i y_j)_{i\in I, j\in J}$ is a basis for $M$ over $K$.
134 \begin{enumerate}[i.]
135 \item prove linear independence:\\
136 Suppose that
137 $$\sum_{ij} k_{ij} x_i y_j = 0 ~(k_{ij} \in K)$$
138 rearrange
139 $$\sum_j (\underbrace{\sum_i k_{ij} x_i}_{\in L}) y_j = 0 ~(k_{ij} \in K)$$
140 Since $\sum_i k_{ij} x_i \in L$, and the $y_j \in M$ are linearly independent over $L$, then $\sum_i k_{ij} x_i = 0$.
141 \\
142 Repeating the argument inside $L$ $\longrightarrow$ $k_{ij}=0 ~~\forall i\in I, j\in J$.
143 \\
144 So the elements $x_i y_j$ are linearly independent over $K$.
145
146 \item prove that $x_i y_j$ span $M$ over $K$:\\
147 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 $L$.
148 Similarly, $\forall j\in J,~ \lambda_j = \sum_i \lambda_{ij} x_i y_j$ for $\lambda_{ij} \in K$.\\
149 Putting the pieces together, $x=\sum_{ij} \lambda_{ij} x_i y_j$ as required.
150 \end{enumerate}
151\end{proof}
152
153\begin{cor}{6.6}\emph{(Tower Law)}\\ \label{towerlaw}
154 If $K_0 \subseteq K_1 \subseteq \ldots \subseteq K_n$ are subfields of $\mathbb{C}$, then
155 $$[K_n:K_0] = [K_n:K_{n-1}] \cdot [K_{n-1}:K_{n-2}] \cdot \ldots \cdot [K_1: K_0]$$
156\end{cor}
157\begin{proof}
158 From \ref{shorttowerlaw}.
159\end{proof}
160
161[...]
162
163
164\newpage
165
166\section{Tools}
167This section contains tools that I found useful to solve Galois Theory related problems, and that don't appear in Stewart's book.
168
169\subsection{De Moivre's Theorem and Euler's formula}\label{demoivre}
170Useful for finding all the roots of a polynomial.
171
172Euler's formula:
173$$e^{i \psi} = cos \psi + i \cdot sin \psi$$
174
175The n-th roots of a complex number $z=x + i y = r (cos \theta + i \cdot sin \theta)$ are given by
176
177$$z_k = \sqrt[n]{r} \cdot \left(cos(\frac{\theta + 2k \pi}{n}) + i \cdot sin(\frac{\theta + 2k \pi}{n}) \right)$$
178for $k=0, \ldots, n-1$.
179
180So, by Euler's formula:
181$$z_k = \sqrt[n]{r} \cdot e^{i (\frac{\theta + 2 k \pi}{n})}$$
182
183\subsection{Einsenstein's Criterion} \label{einsenstein}
184\emph{reference: Stewart's book}
185
186Let $f(t) = a_0 + a_1 t + \ldots + a_n t^n$, suppose there is a prime $q$ such that
187\begin{enumerate}
188 \item $q \nmid a_n$
189 \item $q | a_i$ for $i=0, \ldots, n-1$
190 \item $q^2 \nmid a_0$
191\end{enumerate}
192Then, $f$ is irreducible over $\mathbb{Q}$.
193
194\emph{TODO proof \& Gauss lemma.}
195
196
197\subsection{Elementary symmetric polynomials}
198\emph{TODO from orange notebook, page 36}
199
200\subsection{Cyclotomic polynomials} \label{cyclotomicpoly}
201\emph{TODO theory from brown muji notebook, page 82}
202
203Examples:
204
205\begin{align*}
206 \Phi_n(x) &= x^{n-1} + x^{n-2} + \ldots + x^2 + x + 1 = \sum_{i=0}^{n-1} x^i\\
207 \Phi_{2p}(x) &= x^{p-1} + \ldots + x^2 - x + 1 = \sum_{i=0}^{p-1} (-x)^i\\
208 \Phi_m(x) &= x^{m/2} + 1, ~~\text{when $m$ is a power of $2$}
209\end{align*}
210
211
212\subsection{Lemma 1.42 from J.S.Milne's book}
213\emph{TODO add reference to Milne's book}
214
215Useful for when dealing with $x^p - 1$ with $p$ prime.
216
217Observe that
218
219$$x^p -1 = (x-1)(x^{p-1} + x^{p-2} + \ldots + 1)$$
220
221Notice that
222$$\Phi_p(x) = x^{p-1} + x^{p-2} + \ldots + 1$$
223is the $p$-th Cyclotomic polynomial.
224
225\begin{lemma}{1.42}
226 If $p$ prime, then $x^{p-1} + \ldots + 1$ is irreducible; hence $\mathbb{Q}[e^{2 \pi i /p}]$ has degree $p-1$ over $\mathbb{Q}$.
227\end{lemma}
228\begin{proof}
229 Let $f(x) = (x^p - 1)/(x-1) = x^{p-1} + \ldots + 1$
230 then
231 $$
232 f(x+1) = \frac{(x+1)^p -1}{x+1-1} = \frac{(x+1)^p -1}{x} = x^{p-1} + \ldots + a_i x^i + \ldots + p
233 $$
234
235 with $a_i = \left( \stackrel{p}{i+1} \right)$.
236
237 We know that $p | a_i$ for $i= 1, \ldots, p-2$, therefore $f(x+1)$ is irreducibe by Einsenstein's Criterion.
238
239 This implies that $f(x)$ is irreducible.
240\end{proof}
241
242
243\subsection{Dihedral groups - Groups of symmetries} \label{dihedral}
244Source: Wikipedia and \cite{dihedral}.
245
246Dihedral groups ($\mathbb{D}_n$) represent the symmetries of a regular $n$-gon.
247
248Properties:
249\begin{itemize}
250 \item are non-abelian (for $n>2$), ie. $rs \neq sr$
251 \item order $2n$
252 \item generated by a rotation $r$ and a reflextion $s$
253 \item $r^n = s^2 = id,~~~(rs)^2=id$
254\end{itemize}
255Subgroups of $\mathbb{D}_n$:
256\begin{itemize}
257 \item rotation form a cyclic subgroup of order $n$, denoted as $<r>$
258 \item for each $d$ such that $d|n$, $\exists~ \mathbb{D}_d$ with order $2d$
259 \item normal subgroups
260 \begin{itemize}
261 \item for $n$ odd: $\mathbb{D}_n$ and $<r^d>$ for every $d|n$
262 \item for $n$ even: $2$ additional normal subgroups
263 \end{itemize}
264 \item Klein four-groups: $\mathbb{Z}_2 \times \mathbb{Z}_2$, of order 4
265\end{itemize}
266
267\vspace{0.3cm}
268Total 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.
269
270\begin{eg}{}
271For $\mathbb{D}_6$, we have $\{1,2,3,6\} | 6$, so $d(n) = d(6) = 4$, and
272$s(6) = 1+2+3+6 = 12$; henceforth, the total amount of subgroups is $d(n)+s(n) = 4+12 = 16$.
273\end{eg}
274
275\vspace{0.3cm}
276For $n \geq 3, ~~\mathbb{D}_n \subseteq \mathbb{S}_n$ (subgroup of the Symmetry group).
277
278
279
280\newpage
281
282\section{Exercises}
283
284\subsection{Galois groups}
285
286\subsubsection[t6-7]{$t^6-7 \in \mathbb{Q}$}
287
288This exercise comes from a combination of exercises 12.4 and 13.7 from \cite{ianstewart}.
289
290First let's find the roots. By De Moivre's Theorem (\ref{demoivre}), $t_k =
291\sqrt[6]{7} \cdot e^{i \frac{2 \pi k}{6}}$.
292
293From which we denote $\alpha = \sqrt[6]{7}$, and $\zeta = e^{\frac{2 \pi i}{6}}$, so that the
294roots of the polynomial are $\{ \alpha, \alpha \zeta, \alpha \zeta^2, \alpha \zeta^3, \alpha \zeta^4, \alpha \zeta^5\}$, ie.
295$\{ \alpha \zeta^k \}_0^5$.
296
297Hence the \emph{splitting field} is $\mathbb{Q}(\alpha, \zeta)$.
298
299\emph{Degree of the extension}
300
301In order to find $[\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}$, we're going to split it in tow
302parts. By the Tower Law (\ref{towerlaw}),
303
304$$[\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}] = [\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}(\alpha)] \cdot [\mathbb{Q}(\alpha) : \mathbb{Q}]$$
305
306To find each degree, we will find the minimal polynomial of the adjoined term over the base field of the extension:
307
308\begin{enumerate}[i.]
309 \item minimal polynomial of $\alpha$ over $\mathbb{Q}$\\
310 By Einsenstein's Criterion (\ref{einsenstein}), with $q=7$ we have that $q
311 \nmid 1$, $7 | {-7,0,0,\ldots}$, and $7^2 \nmid -7$, hence $f(t)$ is
312 irreducibe over $\mathbb{Q}$, thus is the minimal polynomial
313 $$m_i(t)= f(t) =t^6-7$$
314 which has roots $\{ \alpha \zeta^k \}_0^5$.
315 \item minimal polynomial of $\zeta$ over $\mathbb{Q}(\alpha)$\\
316 Since $\zeta$ is the primitive $6$th root of unity, we know that the minimal
317 polynomial will be the $6$th cyclotomic polynomial (\ref{cyclotomicpoly}):
318 $$m_{ii}(t) = \Phi_6(t) = t^2 - t + 1$$
319 which has roots $\zeta, -\zeta$.
320
321 Since $\mathbb{Q}(\alpha) \subseteq \mathbb{R}$, and the roots of
322 $\Phi_6(t)=t^2 - t +1$ are in $\mathbb{C}$, $\Phi_6(t)$ remains irreducible
323 over $\mathbb{Q}(\alpha)$.
324\end{enumerate}
325
326\vspace{0.5cm}
327Therefore, by the tower of law,
328$$[\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}] = \deg{\Phi_6(t)} \cdot \deg{f(t)} = 2 \cdot 6 = 12$$
329and by the Fundamental Theorem of Galois Theory, we know that
330$$|\Gamma( \mathbb{Q}(\alpha, \zeta) : \mathbb{Q} )| = [\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}] = 12$$
331which tells us that there exist $12$ $\mathbb{Q}$-automorphisms of the Galois group.
332
333
334\vspace{0.5cm}
335Let's find the $12$ $\mathbb{Q}$-automorphisms. Start by defining $\sigma$ which
336fixes $\zeta$ and acts on $\alpha$, sending it to another of the roots of the
337minimal polynomial of $\alpha$ over $\mathbb{Q}$, $f(t)$, choose $\alpha \zeta$.
338
339Now define $\tau$ which fixes $\alpha$ and acts on $\zeta$, sending it into
340another root of the minimal polynomial of $\zeta$ over $\mathbb{Q}(\alpha)$,
341choose $-\zeta$.
342
343\vspace{0.3cm}
344\begin{tabular}{@{}l l@{}}
345 $\begin{aligned}
346 \sigma: \alpha &\mapsto \alpha \zeta \\
347 \zeta &\mapsto \zeta
348 \end{aligned}$
349 &
350 $\begin{aligned}
351 \tau: \alpha &\mapsto \alpha\\
352 \zeta &\mapsto -\zeta = \zeta^{-1}
353 \end{aligned}$
354\end{tabular}
355
356In other words, we have $12$ $\mathbb{Q}$-automorphisms, which are the
357combination of $\sigma$ and $\tau$:
358
359$$\begin{aligned}
360 \sigma^k \tau^j:~~&\alpha \mapsto \alpha \zeta^k\\
361 &\zeta \mapsto \zeta^j
362\end{aligned}$$
363
364for $0 \leq k \leq 5$ and $j = \pm 1$.
365
366\vspace{0.5cm}
367\emph{TODO diagram}
368\vspace{0.5cm}
369
370Observe, that $\Gamma$ is generated by the combination of $\sigma$ and $\tau$,
371and it is isomorphic to the group of symmetries of order 12, the dihedral
372group (\ref{dihedral}) of order 12, $\mathbb{D}_6$, ie. $\Gamma \cong \mathbb{D}_6$.
373
374\vspace{0.5cm}
375
376Let's find the subgroups of $\Gamma$, and the fixed fields of $\mathbb{Q}(\alpha, \zeta)$.
377
378We know that $\Gamma \cong \mathbb{D}_6$, and we know from the properties
379of the dihedral group (\ref{dihedral}) that the number of subgroups of
380$\mathbb{D}_6$ will be $d(6) + s(6) = 4 + 12 = 16$ subgroups.
381
382
383\vspace{0.4cm}
384
385\hspace*{-3.5cm}
386\begin{tabular}{ c c c c | p{7.5cm} }
387 \hline
388 generators & order & group & fixed field & notes (check fixed field)\\
389\hline
390 $\langle \rangle = \langle \sigma^6 \rangle=\langle \tau^2 \rangle$ & 1 & id & $\mathbb{Q}(\alpha,\zeta)$ & \\
391 $\langle \sigma \rangle = \langle \sigma^5 \rangle$ & 6 & $\mathbb{Z}_6$ & $\mathbb{Q}(\zeta)$ & \\
392 $\langle \sigma^2 \rangle=\langle \sigma^4 \rangle$ & 3 & $\mathbb{Z}_3$ & $\mathbb{Q}(\alpha^3, \zeta)$ & $\sigma^2(\alpha^3)=\alpha^3 \zeta^{3\cdot 2}=\alpha^3 \zeta^6 = \alpha^3 \cdot 1 = \alpha^3$\\
393 $\langle \sigma^3 \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha^2,\zeta)$ & $\sigma^3(\alpha^2)=(\alpha\zeta^3)^2=\alpha^2\zeta^6=\alpha^2$\\
394 \hline
395 $\langle \tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha)$ & \\
396 \hline
397 $\langle \sigma\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta)$ &
398 $\sigma\zeta(\alpha+\alpha\zeta)=\sigma(\alpha+\alpha\zeta^{-1}) = \alpha\zeta + \alpha\zeta^{-1}\zeta=\alpha\zeta+\alpha$\\
399 $\langle \sigma^2\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^2), \mathbb{Q}(\alpha\zeta)$ &
400 $\sigma^2\tau(\alpha+\alpha\zeta^2) = \sigma(\alpha+\alpha\zeta^{-2})=\alpha\zeta^2+ \alpha\zeta^{-2}\zeta^2=\alpha\zeta^2+\alpha$\\
401 $\langle \sigma^3\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^3)$ &
402 $\sigma^3\tau(\alpha+\alpha\zeta^3) = \sigma(\alpha+\alpha\zeta^{-3})=\alpha\zeta^3+ \alpha\zeta^{-3}\zeta^3=\alpha\zeta^3+\alpha$\\
403 $\langle \sigma^4\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^4), \mathbb{Q}(\alpha\zeta^2)$ &
404 $\sigma^4\tau(\alpha+\alpha\zeta^4) = \sigma(\alpha+\alpha\zeta^{-4})=\alpha\zeta^4+ \alpha\zeta^{-4}\zeta^4=\alpha\zeta^4+\alpha$\\
405 $\langle \sigma^5\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^5)$ &
406 $\sigma^5\tau(\alpha+\alpha\zeta^5) = \sigma(\alpha+\alpha\zeta^{-5})=\alpha\zeta^5+ \alpha\zeta^{-5}\zeta^5=\alpha\zeta^5+\alpha$\\
407 \hline
408 $\langle \sigma, \tau \rangle = \langle \sigma^5,\tau \rangle$ & $6\cdot2=12$ & $\mathbb{D}_6$ & $\mathbb{Q}$ & \\
409 $\langle \sigma^2, \tau \rangle = \langle \sigma^4,\tau \rangle$ & $3\cdot2=6$ & $\mathbb{D}_3$ & $\mathbb{Q}(\alpha^3)$ &
410 $\sigma^2(\alpha^3)=\alpha^3\zeta^{3\cdot 2}=\alpha^3$ and $\tau(\alpha^3)=\alpha^3$\\
411 $\langle \sigma^3, \tau \rangle$ & $2\cdot2=4$ & $\mathbb{D}_2$ & $\mathbb{Q}(\alpha^2)$ &
412 $\sigma^3(\alpha^2)=\alpha^2\zeta^{2\cdot 2}=\alpha^2$ and $\tau(\alpha^2)=\alpha^2$\\
413 \hline
414 $\langle \sigma^2, \sigma\tau \rangle$ & $3\cdot 2=6$ & $\mathbb{D}_3$ & $\mathbb{Q}(\alpha^3+\alpha^3\zeta^3)$ &
415 $\sigma^2(\alpha^3 + \alpha^3 \zeta^3) = \alpha^3\zeta^3 + \alpha^3 \zeta^3\zeta^3 = \alpha^3\zeta^3 + \alpha^3\zeta^6 = \alpha^3\zeta^3+\alpha^3$\\
416 $\langle \sigma^3, \sigma\tau \rangle$ & $2\cdot2=4$ & $\mathbb{Z}_2 \times \mathbb{Z}_2$ & $\mathbb{Q}(\alpha^2\zeta^2),\mathbb{Q}(\alpha^2+\alpha^2\zeta^2)$ &
417 $\sigma^3(\alpha^2+\alpha^2\zeta^2)=\alpha^2\zeta^{2\cdot3}+\alpha^2\zeta^{2\cdot3}\zeta^2=\alpha^2+\alpha^2\zeta^2$
418 and
419 $\sigma\tau(\alpha^2+\alpha^2\zeta^2)=\alpha^2\zeta^2+\alpha^2\zeta^{-2}\zeta^2 = \alpha^2\zeta^2+\alpha^2$\\
420 $\langle \sigma^3, \sigma^2\tau\rangle$ & $2\cdot2=4$ & $\mathbb{Z}_2 \times \mathbb{Z}_2$ & $\mathbb{Q}(\alpha^2\zeta^4),\mathbb{Q}(\alpha^2+\alpha^2\zeta^4)$ &
421 $\sigma^2\zeta(\alpha^2\zeta^4)=\alpha^2\zeta^2\zeta^{-4}=\alpha^2\zeta^{-2}=\alpha^2\zeta^4$
422 and $\sigma^3(\alpha^2\zeta^4)=\alpha^2\zeta^{2\cdot3}\zeta^4=\alpha^2\zeta^4$
423\end{tabular}
424
425
426
427\bibliographystyle{unsrt}
428\bibliography{galois-theory-notes.bib}
429
430\end{document}