\documentclass{article} \usepackage[utf8]{inputenc} \usepackage{amsfonts} \usepackage{amsthm} \usepackage{amsmath} \usepackage{enumerate} \usepackage{hyperref} \usepackage{amssymb} \begin{filecontents}[overwrite]{galois-theory-notes.bib} @misc{ianstewart, author = {Ian Stewart}, title = {{Galois Theory, Third Edition}}, year = {2004} } @misc{dihedral, author = {Gaurab Bardhan and Palash Nath and Himangshu Chakraborty} title = {Subgroups and normal subgroups of dihedral group up to isomorphism} year = {2010}, note = {\url{https://scipp.ucsc.edu/~haber/ph251/Dn_subgroups.pdf}}, url = {https://scipp.ucsc.edu/~haber/ph251/Dn_subgroups.pdf} } \end{filecontents} \nocite{*} \theoremstyle{definition} \newtheorem{innerdefn}{Definition} \newenvironment{defn}[1] {\renewcommand\theinnerdefn{#1}\innerdefn} {\endinnerdefn} \newtheorem{innerthm}{Theorem} \newenvironment{thm}[1] {\renewcommand\theinnerthm{#1}\innerthm} {\endinnerthm} \newtheorem{innerlemma}{Lemma} \newenvironment{lemma}[1] {\renewcommand\theinnerlemma{#1}\innerlemma} {\endinnerlemma} \newtheorem{innercor}{Lemma} \newenvironment{cor}[1] {\renewcommand\theinnercor{#1}\innercor} {\endinnercor} \newtheorem{innereg}{Example} \newenvironment{eg}[1] {\renewcommand\theinnereg{#1}\innereg} {\endinnereg} \title{Galois Theory notes} \author{arnaucube} \date{2025} \begin{document} \maketitle \begin{abstract} Notes taken while studying Galois Theory, mostyly from Ian Stewart's book "Galois Theory" \cite{ianstewart}. Usually while reading books and papers I take handwritten notes in a notebook, this document contains some of them re-written to $LaTeX$. The notes are not complete, don't include all the steps neither all the proofs. \end{abstract} \tableofcontents \section{Recap on the degree of field extensions} (Definitions, theorems, lemmas, corollaries and examples enumeration follows from Ian Stewart's book \cite{ianstewart}). \begin{defn}{4.10} A \emph{simple extension} is $L:K$ such that $L=K(\alpha)$ for some $\alpha \in L$. \end{defn} \begin{eg}{4.11} Beware, $L=\mathbb{Q}(i, -i, \sqrt{5}, -\sqrt{5}) = \mathbb{Q}(i, \sqrt{5}) = \mathbb{Q}(i+\sqrt{5})$. \end{eg} \begin{defn}{5.5} 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 $m(\alpha)=0$. \\ 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 $m(i)=0$. \end{defn} \begin{lemma}{5.9} Every polynomial $a \in K[t]$ is congruent modulo $m$ to a unique polynomial of degree $< \delta m$. \end{lemma} \begin{proof} Divide $a / m$ with remainder, $a= qm +r$, with $q,r \in K[t]$ and $\delta r < \delta m$. Then, $a-r=qm$, so $a \equiv r \pmod{m}$. It remains to prove uniqueness. Suppose $\exists~ r \equiv s \pmod{m}$, with $\delta r, \delta s < \delta m$. Then, $r-s$ is divisible by $m$, but has smaller degree than $m$. Therefore, $r-s=0$, so $r=s$, proving uniqueness. \end{proof} \begin{lemma}{5.14} Let $K(\alpha):K$ be a simple algebraic extension, let $m$ be the minimal polynomial of $\alpha$ over $K$, let $\delta m =n$. Then $\{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}\}$ is a basis for $K(\alpha)$ over $K$. In particular, $[K(\alpha):K]=n$. \end{lemma} \begin{defn}{6.2} The degree $[L:K]$ of a field extension $L:K$ is the dimension of L considered as a vector space over $K$. 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 $K$. \end{defn} \begin{eg}{6.3} \begin{enumerate} \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 $[\mathbb{C}:\mathbb{R}]=2$. \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 $\mathbb{Q}$. \end{enumerate} \end{eg} \begin{thm}{6.4}\emph{(Short Tower Law)} \label{shorttowerlaw} If $K, L, M \subseteq \mathbb{C}$, and $K \subseteq L \subseteq M$, then $[M:K]=[M:L]\cdot [L:K]$. \end{thm} \begin{proof} Let $(x_i)_{i \in I}$ be a basis for $L$ over $K$, let $(y_j)_{j \in J}$ be a basis for $M$ over $L$.\\ $\forall i \in I, j \in J$, we have $x_i \in L, u_j \in M$. \\ Want to show that $(x_i y_j)_{i\in I, j\in J}$ is a basis for $M$ over $K$. \begin{enumerate}[i.] \item prove linear independence:\\ Suppose that $$\sum_{ij} k_{ij} x_i y_j = 0 ~(k_{ij} \in K)$$ rearrange $$\sum_j (\underbrace{\sum_i k_{ij} x_i}_{\in L}) y_j = 0 ~(k_{ij} \in K)$$ Since $\sum_i k_{ij} x_i \in L$, and the $y_j \in M$ are linearly independent over $L$, then $\sum_i k_{ij} x_i = 0$. \\ Repeating the argument inside $L$ $\longrightarrow$ $k_{ij}=0 ~~\forall i\in I, j\in J$. \\ So the elements $x_i y_j$ are linearly independent over $K$. \item prove that $x_i y_j$ span $M$ over $K$:\\ 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 $L$. Similarly, $\forall j\in J,~ \lambda_j = \sum_i \lambda_{ij} x_i y_j$ for $\lambda_{ij} \in K$.\\ Putting the pieces together, $x=\sum_{ij} \lambda_{ij} x_i y_j$ as required. \end{enumerate} \end{proof} \begin{cor}{6.6}\emph{(Tower Law)}\\ \label{towerlaw} If $K_0 \subseteq K_1 \subseteq \ldots \subseteq K_n$ are subfields of $\mathbb{C}$, then $$[K_n:K_0] = [K_n:K_{n-1}] \cdot [K_{n-1}:K_{n-2}] \cdot \ldots \cdot [K_1: K_0]$$ \end{cor} \begin{proof} From \ref{shorttowerlaw}. \end{proof} [...] \newpage \section{Tools} This section contains tools that I found useful to solve Galois Theory related problems, and that don't appear in Stewart's book. \subsection{De Moivre's Theorem and Euler's formula}\label{demoivre} Useful for finding all the roots of a polynomial. Euler's formula: $$e^{i \psi} = cos \psi + i \cdot sin \psi$$ The n-th roots of a complex number $z=x + i y = r (cos \theta + i \cdot sin \theta)$ are given by $$z_k = \sqrt[n]{r} \cdot \left(cos(\frac{\theta + 2k \pi}{n}) + i \cdot sin(\frac{\theta + 2k \pi}{n}) \right)$$ for $k=0, \ldots, n-1$. So, by Euler's formula: $$z_k = \sqrt[n]{r} \cdot e^{i (\frac{\theta + 2 k \pi}{n})}$$ \subsection{Einsenstein's Criterion} \label{einsenstein} \emph{reference: Stewart's book} Let $f(t) = a_0 + a_1 t + \ldots + a_n t^n$, suppose there is a prime $q$ such that \begin{enumerate} \item $q \nmid a_n$ \item $q | a_i$ for $i=0, \ldots, n-1$ \item $q^2 \nmid a_0$ \end{enumerate} Then, $f$ is irreducible over $\mathbb{Q}$. \emph{TODO proof \& Gauss lemma.} \subsection{Elementary symmetric polynomials} \emph{TODO from orange notebook, page 36} \subsection{Cyclotomic polynomials} \label{cyclotomicpoly} \emph{TODO theory from brown muji notebook, page 82} Examples: \begin{align*} \Phi_n(x) &= x^{n-1} + x^{n-2} + \ldots + x^2 + x + 1 = \sum_{i=0}^{n-1} x^i\\ \Phi_{2p}(x) &= x^{p-1} + \ldots + x^2 - x + 1 = \sum_{i=0}^{p-1} (-x)^i\\ \Phi_m(x) &= x^{m/2} + 1, ~~\text{when $m$ is a power of $2$} \end{align*} \subsection{Lemma 1.42 from J.S.Milne's book} \emph{TODO add reference to Milne's book} Useful for when dealing with $x^p - 1$ with $p$ prime. Observe that $$x^p -1 = (x-1)(x^{p-1} + x^{p-2} + \ldots + 1)$$ Notice that $$\Phi_p(x) = x^{p-1} + x^{p-2} + \ldots + 1$$ is the $p$-th Cyclotomic polynomial. \begin{lemma}{1.42} If $p$ prime, then $x^{p-1} + \ldots + 1$ is irreducible; hence $\mathbb{Q}[e^{2 \pi i /p}]$ has degree $p-1$ over $\mathbb{Q}$. \end{lemma} \begin{proof} Let $f(x) = (x^p - 1)/(x-1) = x^{p-1} + \ldots + 1$ then $$ f(x+1) = \frac{(x+1)^p -1}{x+1-1} = \frac{(x+1)^p -1}{x} = x^{p-1} + \ldots + a_i x^i + \ldots + p $$ with $a_i = \left( \stackrel{p}{i+1} \right)$. We know that $p | a_i$ for $i= 1, \ldots, p-2$, therefore $f(x+1)$ is irreducibe by Einsenstein's Criterion. This implies that $f(x)$ is irreducible. \end{proof} \subsection{Dihedral groups - Groups of symmetries} \label{dihedral} Source: Wikipedia and \cite{dihedral}. Dihedral groups ($\mathbb{D}_n$) represent the symmetries of a regular $n$-gon. Properties: \begin{itemize} \item are non-abelian (for $n>2$), ie. $rs \neq sr$ \item order $2n$ \item generated by a rotation $r$ and a reflextion $s$ \item $r^n = s^2 = id,~~~(rs)^2=id$ \end{itemize} Subgroups of $\mathbb{D}_n$: \begin{itemize} \item rotation form a cyclic subgroup of order $n$, denoted as $$ \item for each $d$ such that $d|n$, $\exists~ \mathbb{D}_d$ with order $2d$ \item normal subgroups \begin{itemize} \item for $n$ odd: $\mathbb{D}_n$ and $$ for every $d|n$ \item for $n$ even: $2$ additional normal subgroups \end{itemize} \item Klein four-groups: $\mathbb{Z}_2 \times \mathbb{Z}_2$, of order 4 \end{itemize} \vspace{0.3cm} Total 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. \begin{eg}{} For $\mathbb{D}_6$, we have $\{1,2,3,6\} | 6$, so $d(n) = d(6) = 4$, and $s(6) = 1+2+3+6 = 12$; henceforth, the total amount of subgroups is $d(n)+s(n) = 4+12 = 16$. \end{eg} \vspace{0.3cm} For $n \geq 3, ~~\mathbb{D}_n \subseteq \mathbb{S}_n$ (subgroup of the Symmetry group). \newpage \section{Exercises} \subsection{Galois groups} \subsubsection[t6-7]{$t^6-7 \in \mathbb{Q}$} This exercise comes from a combination of exercises 12.4 and 13.7 from \cite{ianstewart}. First let's find the roots. By De Moivre's Theorem (\ref{demoivre}), $t_k = \sqrt[6]{7} \cdot e^{i \frac{2 \pi k}{6}}$. From which we denote $\alpha = \sqrt[6]{7}$, and $\zeta = e^{\frac{2 \pi i}{6}}$, so that the roots of the polynomial are $\{ \alpha, \alpha \zeta, \alpha \zeta^2, \alpha \zeta^3, \alpha \zeta^4, \alpha \zeta^5\}$, ie. $\{ \alpha \zeta^k \}_0^5$. Hence the \emph{splitting field} is $\mathbb{Q}(\alpha, \zeta)$. \emph{Degree of the extension} In order to find $[\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}$, we're going to split it in tow parts. By the Tower Law (\ref{towerlaw}), $$[\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}] = [\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}(\alpha)] \cdot [\mathbb{Q}(\alpha) : \mathbb{Q}]$$ To find each degree, we will find the minimal polynomial of the adjoined term over the base field of the extension: \begin{enumerate}[i.] \item minimal polynomial of $\alpha$ over $\mathbb{Q}$\\ By Einsenstein's Criterion (\ref{einsenstein}), with $q=7$ we have that $q \nmid 1$, $7 | {-7,0,0,\ldots}$, and $7^2 \nmid -7$, hence $f(t)$ is irreducibe over $\mathbb{Q}$, thus is the minimal polynomial $$m_i(t)= f(t) =t^6-7$$ which has roots $\{ \alpha \zeta^k \}_0^5$. \item minimal polynomial of $\zeta$ over $\mathbb{Q}(\alpha)$\\ Since $\zeta$ is the primitive $6$th root of unity, we know that the minimal polynomial will be the $6$th cyclotomic polynomial (\ref{cyclotomicpoly}): $$m_{ii}(t) = \Phi_6(t) = t^2 - t + 1$$ which has roots $\zeta, -\zeta$. Since $\mathbb{Q}(\alpha) \subseteq \mathbb{R}$, and the roots of $\Phi_6(t)=t^2 - t +1$ are in $\mathbb{C}$, $\Phi_6(t)$ remains irreducible over $\mathbb{Q}(\alpha)$. \end{enumerate} \vspace{0.5cm} Therefore, by the tower of law, $$[\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}] = \deg{\Phi_6(t)} \cdot \deg{f(t)} = 2 \cdot 6 = 12$$ and by the Fundamental Theorem of Galois Theory, we know that $$|\Gamma( \mathbb{Q}(\alpha, \zeta) : \mathbb{Q} )| = [\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}] = 12$$ which tells us that there exist $12$ $\mathbb{Q}$-automorphisms of the Galois group. \vspace{0.5cm} Let's find the $12$ $\mathbb{Q}$-automorphisms. Start by defining $\sigma$ which fixes $\zeta$ and acts on $\alpha$, sending it to another of the roots of the minimal polynomial of $\alpha$ over $\mathbb{Q}$, $f(t)$, choose $\alpha \zeta$. Now define $\tau$ which fixes $\alpha$ and acts on $\zeta$, sending it into another root of the minimal polynomial of $\zeta$ over $\mathbb{Q}(\alpha)$, choose $-\zeta$. \vspace{0.3cm} \begin{tabular}{@{}l l@{}} $\begin{aligned} \sigma: \alpha &\mapsto \alpha \zeta \\ \zeta &\mapsto \zeta \end{aligned}$ & $\begin{aligned} \tau: \alpha &\mapsto \alpha\\ \zeta &\mapsto -\zeta = \zeta^{-1} \end{aligned}$ \end{tabular} In other words, we have $12$ $\mathbb{Q}$-automorphisms, which are the combination of $\sigma$ and $\tau$: $$\begin{aligned} \sigma^k \tau^j:~~&\alpha \mapsto \alpha \zeta^k\\ &\zeta \mapsto \zeta^j \end{aligned}$$ for $0 \leq k \leq 5$ and $j = \pm 1$. \vspace{0.5cm} \emph{TODO diagram} \vspace{0.5cm} Observe, that $\Gamma$ is generated by the combination of $\sigma$ and $\tau$, and it is isomorphic to the group of symmetries of order 12, the dihedral group (\ref{dihedral}) of order 12, $\mathbb{D}_6$, ie. $\Gamma \cong \mathbb{D}_6$. \vspace{0.5cm} Let's find the subgroups of $\Gamma$, and the fixed fields of $\mathbb{Q}(\alpha, \zeta)$. We know that $\Gamma \cong \mathbb{D}_6$, and we know from the properties of the dihedral group (\ref{dihedral}) that the number of subgroups of $\mathbb{D}_6$ will be $d(6) + s(6) = 4 + 12 = 16$ subgroups. \vspace{0.4cm} \hspace*{-3.5cm} \begin{tabular}{ c c c c | p{7.5cm} } \hline generators & order & group & fixed field & notes (check fixed field)\\ \hline $\langle \rangle = \langle \sigma^6 \rangle=\langle \tau^2 \rangle$ & 1 & id & $\mathbb{Q}(\alpha,\zeta)$ & \\ $\langle \sigma \rangle = \langle \sigma^5 \rangle$ & 6 & $\mathbb{Z}_6$ & $\mathbb{Q}(\zeta)$ & \\ $\langle \sigma^2 \rangle=\langle \sigma^4 \rangle$ & 3 & $\mathbb{Z}_3$ & $\mathbb{Q}(\alpha^3, \zeta)$ & $\sigma^2(\alpha^3)=\alpha^3 \zeta^{3\cdot 2}=\alpha^3 \zeta^6 = \alpha^3 \cdot 1 = \alpha^3$\\ $\langle \sigma^3 \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha^2,\zeta)$ & $\sigma^3(\alpha^2)=(\alpha\zeta^3)^2=\alpha^2\zeta^6=\alpha^2$\\ \hline $\langle \tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha)$ & \\ \hline $\langle \sigma\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta)$ & $\sigma\zeta(\alpha+\alpha\zeta)=\sigma(\alpha+\alpha\zeta^{-1}) = \alpha\zeta + \alpha\zeta^{-1}\zeta=\alpha\zeta+\alpha$\\ $\langle \sigma^2\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^2), \mathbb{Q}(\alpha\zeta)$ & $\sigma^2\tau(\alpha+\alpha\zeta^2) = \sigma(\alpha+\alpha\zeta^{-2})=\alpha\zeta^2+ \alpha\zeta^{-2}\zeta^2=\alpha\zeta^2+\alpha$\\ $\langle \sigma^3\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^3)$ & $\sigma^3\tau(\alpha+\alpha\zeta^3) = \sigma(\alpha+\alpha\zeta^{-3})=\alpha\zeta^3+ \alpha\zeta^{-3}\zeta^3=\alpha\zeta^3+\alpha$\\ $\langle \sigma^4\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^4), \mathbb{Q}(\alpha\zeta^2)$ & $\sigma^4\tau(\alpha+\alpha\zeta^4) = \sigma(\alpha+\alpha\zeta^{-4})=\alpha\zeta^4+ \alpha\zeta^{-4}\zeta^4=\alpha\zeta^4+\alpha$\\ $\langle \sigma^5\tau \rangle$ & 2 & $\mathbb{Z}_2$ & $\mathbb{Q}(\alpha+\alpha\zeta^5)$ & $\sigma^5\tau(\alpha+\alpha\zeta^5) = \sigma(\alpha+\alpha\zeta^{-5})=\alpha\zeta^5+ \alpha\zeta^{-5}\zeta^5=\alpha\zeta^5+\alpha$\\ \hline $\langle \sigma, \tau \rangle = \langle \sigma^5,\tau \rangle$ & $6\cdot2=12$ & $\mathbb{D}_6$ & $\mathbb{Q}$ & \\ $\langle \sigma^2, \tau \rangle = \langle \sigma^4,\tau \rangle$ & $3\cdot2=6$ & $\mathbb{D}_3$ & $\mathbb{Q}(\alpha^3)$ & $\sigma^2(\alpha^3)=\alpha^3\zeta^{3\cdot 2}=\alpha^3$ and $\tau(\alpha^3)=\alpha^3$\\ $\langle \sigma^3, \tau \rangle$ & $2\cdot2=4$ & $\mathbb{D}_2$ & $\mathbb{Q}(\alpha^2)$ & $\sigma^3(\alpha^2)=\alpha^2\zeta^{2\cdot 2}=\alpha^2$ and $\tau(\alpha^2)=\alpha^2$\\ \hline $\langle \sigma^2, \sigma\tau \rangle$ & $3\cdot 2=6$ & $\mathbb{D}_3$ & $\mathbb{Q}(\alpha^3+\alpha^3\zeta^3)$ & $\sigma^2(\alpha^3 + \alpha^3 \zeta^3) = \alpha^3\zeta^3 + \alpha^3 \zeta^3\zeta^3 = \alpha^3\zeta^3 + \alpha^3\zeta^6 = \alpha^3\zeta^3+\alpha^3$\\ $\langle \sigma^3, \sigma\tau \rangle$ & $2\cdot2=4$ & $\mathbb{Z}_2 \times \mathbb{Z}_2$ & $\mathbb{Q}(\alpha^2\zeta^2),\mathbb{Q}(\alpha^2+\alpha^2\zeta^2)$ & $\sigma^3(\alpha^2+\alpha^2\zeta^2)=\alpha^2\zeta^{2\cdot3}+\alpha^2\zeta^{2\cdot3}\zeta^2=\alpha^2+\alpha^2\zeta^2$ and $\sigma\tau(\alpha^2+\alpha^2\zeta^2)=\alpha^2\zeta^2+\alpha^2\zeta^{-2}\zeta^2 = \alpha^2\zeta^2+\alpha^2$\\ $\langle \sigma^3, \sigma^2\tau\rangle$ & $2\cdot2=4$ & $\mathbb{Z}_2 \times \mathbb{Z}_2$ & $\mathbb{Q}(\alpha^2\zeta^4),\mathbb{Q}(\alpha^2+\alpha^2\zeta^4)$ & $\sigma^2\zeta(\alpha^2\zeta^4)=\alpha^2\zeta^2\zeta^{-4}=\alpha^2\zeta^{-2}=\alpha^2\zeta^4$ and $\sigma^3(\alpha^2\zeta^4)=\alpha^2\zeta^{2\cdot3}\zeta^4=\alpha^2\zeta^4$ \end{tabular} \bibliographystyle{unsrt} \bibliography{galois-theory-notes.bib} \end{document}