% !TEX TS-program = lualatex \documentclass[notoc,notitlepage,nobib]{tufte-book} % \nonstopmode % uncomment to enable nonstopmode \usepackage{classnotetitle} \title{PMATH348 --- Fields and Galois Theory} \author{Johnson Ng} \subtitle{Classnotes for Winter 2019} \credentials{BMath (Hons), Pure Mathematics major, Actuarial Science Minor} \institution{University of Waterloo} \usepackage{fontspec-luatex} \input{latex-classnotes-preamble.tex} \setmainfont[Ligatures=TeX]{Times Newer Roman} \setsansfont[Ligatures=TeX]{Helvetica Neue LT Std} \makeatletter \providerobustcmd*{\bigcupdot}{% \mathop{% \mathpalette\bigop@dot\bigcup }% } \newrobustcmd*{\bigop@dot}[2]{% \setbox0=\hbox{$\m@th#1#2$}% \vbox{% \lineskiplimit=\maxdimen \lineskip=-0.7\dimexpr\ht0+\dp0\relax \ialign{% \hfil##\hfil\cr $\m@th\cdot$\cr \box0\cr }% }% } \makeatother \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\disc}{disc} \DeclareMathOperator{\Gal}{Gal} \DeclareMathOperator{\Fix}{Fix} \DeclareMathOperator{\stab}{stab} \DeclareMathOperator{\orb}{orb} \DeclareMathOperator{\ch}{ch} \DeclareMathOperator{\Span}{span} \DeclareMathOperator{\lcm}{lcm} \begin{document} \input{latex-classnotes-header.tex} \chapter*{Preface}% \label{chp:preface} \addcontentsline{toc}{chapter}{Preface} % chapter preface This is a 3 part course; it is separated into \begin{enumerate} \item \textbf{Sylow's Theorem} which is a leftover from group theory (\href{https://tex.japorized.ink/PMATH347S18/classnotes.pdf}{PMATH 347}). It has little to do with the rest of the course, but PMATH 347 was a course that is already content-rich to a point where Sylow's Theorem gets pushed into the later course that is this course. \item \textbf{Field Theory} is a somewhat understood concept from ring theory, where we learned that it is a special case of a ring where all of its elements have an inverse. \item \textbf{Galois Theory} is the beautiful theory from the French mathematican Évariste Galois that ties field theory back to group theory. This allows us to reduce certain field theory problems into group theory, which, in some sense, is easier and better understood. \end{enumerate} % chapter preface (end) \tuftepart{Sylow's Theorem}\label{part:sylow_s_theorem} \chapter{Lecture 1 Jan 07th}% \label{chp:lecture_1_jan_07th} % chapter lecture_1_jan_07th \section{Cauchy's Theorem}% \label{sec:cauchy_s_theorem} % section cauchy_s_theorem Recall Lagrange's Theorem. \begin{thm}[Lagrange's Theorem]\index{Lagrange's Theorem}\label{thm:lagrange_s_theorem} If $G$ is a finite group and $H$ is a subgroup of $G$ \sidenote{I shall write this as $H \leq G$ from hereon.}, then $\abs{H} \mid \abs{G}$ \sidenote{This just means $\abs{H}$ divides $\abs{G}$.}. \end{thm} The full converse is not true. \begin{eg} Let $G = A_4$, the \hlnotea{alternating group} of $4$ elements. Then $\abs{G} = 12$ \sidenote{Recall that the symmetric group of $4$ elements $S_4$ has order $4! = 24$, and an alternating group has half of its elements.}. We have that $6 \mid 12$. We shall show that $G$ has no subgroup of order $6$. Suppose to the contrary that $H \leq G$ such that $\abs{H} = 6$. Let $a \in G$ such that $\abs{a} = 3$ \sidenote{i.e. the order of $a$ is $3$. This is a \hlimpo{trick}.} There are $8$ such elements in $G$ \sidenote{This shall be left as an exercise. \begin{ex} Prove that there are $8$ elements in $G$ that have order $3$. \end{ex}}. Note that the \hlnotea{index}\sidenote{The index of a subgroup is the number of unique cosets generated by $H$.} of $H$, $\abs{G : H}$, is $\frac{\abs{G}}{\abs{H}} = 2$. Now consider the \hlnotea{cosets} $H$, $aH$ and $a^2 H$. Since $\abs{G : H} = 2$, we must have either \begin{itemize} \item $aH = H \implies a \in H$; \item $aH = a^{-1}H \overset{\text{`multiply' } a^{-1}}{\implies} H = aH \implies a \in H$; or \item $a^2H = H \overset{\text{`multiply' } a}{\implies} H = aH \implies a \in H$. \end{itemize} Thus all $8$ elements of order $3$ are in $H$ but $\abs{H} = 6$, a contradiction. Therefore, no such subgroup (of order $6$) exists. \end{eg} Our goal now is to establish a partial converse of Lagrange's Theorem. To that end, we shall first lay down some definitions. \begin{defn}[$p$-Group]\index{$p$-Group}\label{defn:_p_group} Let $p$ be prime. We say that a group $G$ is a \hlnoteb{$p$-group} if $\abs{G} = p^k$ for some $k \in \mathbb{N}$. For $H \leq G$, we say that $H$ is a $p$-subgroup of $G$ if $H$ is a $p$-group. \end{defn} \begin{defn}[Sylow $p$-Subgroup]\index{Sylow $p$-Subgroup}\label{defn:sylow_p_subgroup} Let $G$ be a group such that $\abs{G} = p^n m$ for some $n, m \in \mathbb{N}$, such that $p \nmid m$. If $H \leq G$ with order $p^n$, we call $H$ a \hlnoteb{Sylow $p$-subgroup}. \end{defn} Recall Cauchy's Theorem for abelian groups\sidenote{In the course I was in, we were introduced only to the full theorem and actually went through this entire part. See notes on \href{https://tex.japorized.ink/PMATH347S18/classnotes.pdf}{PMATH 347}.}. \begin{thm}[Cauchy's Theorem for Abelian Groups]\index{Cauchy's Theorem for Abelian Groups}\label{thm:cauchy_s_theorem_for_abelian_groups} If $G$ is a finite abelian group, and $p$ is prime such that $p \mid \abs{G}$, then $\abs{G}$ has an element of order $p$. \end{thm} \begin{defn}[Stabilizers and Orbits]\index{Stabilizers}\index{Orbits}\label{defn:stabilizers_and_orbits} Let $G$ be a finite group which acts on a finite set $X$ \sidenote{Recall that a group action is a function $\cdot : G \times X \to X$ such that \begin{enumerate} \item $g(hx) = (gh)x$; and \item $ex = x$. \end{enumerate}}. For $x \in X$, the \hlnoteb{stabilizers} of $x$ is the set \begin{equation*} \stab(x) := \{ g \in G : g x = x \} \leq G. \end{equation*} The orbits of $x$ is a set \begin{equation*} \orb(x) := \{ gx : g \in G \}. \end{equation*} \end{defn} \begin{note} One can verify that the function $G / \stab(x) \to \orb(x)$ such that \begin{equation*} g \stab(x) \mapsto gx \end{equation*} is a bijection. \end{note} \begin{thm}[Orbit-Stabilizer Theorem]\index{Orbit-Stabilizer Theorem}\label{thm:orbit_stabilizer_theorem} Let $G$ be a group acting on a set $X$, and for each $x \in X$, $\stab(x)$ and $\orb(x)$ are the stabilizers and orbits of $x$, respectively. Then \begin{equation*} \abs{G} = \abs{\stab(x)} \cdot \abs{\orb(x)}. \end{equation*} Moreover, if $x, y \in X$, then either $\orb(x) \cap \orb(y) = \emptyset$ or $\orb(x) = \orb(y)$. \end{thm} The theorem is actually equivalent to \href{https://tex.japorized.ink/PMATH347S18/classnotes.pdf#thm.45}{Proposition 45} in the notes for PMATH 347. However, feel free to... \begin{ex} prove \cref{thm:orbit_stabilizer_theorem} as an exercise. \end{ex} Consequently, we have that \begin{equation*} \abs{X} = \sum \abs{\orb(a_i)}, \end{equation*} where $a_i$ are the distinct orbit representatives. Letting \begin{equation*} X_G := \{ x \in X : gx = x, g \in G \}, \end{equation*} we have... \begin{thm}[Orbit Decomposition Theorem]\index{Orbit Decomposition Theorem}\label{thm:orbit_decomposition_theorem} \begin{equation*} \abs{X} = \abs{X_G} + \sum_{a_i \notin X_G} \abs{\orb(a_i)}. \end{equation*} \end{thm} % section cauchy_s_theorem (end) % chapter lecture_1_jan_07th (end) \chapter{Lecture 2 Jan 09th}% \label{chp:lecture_2_jan_09th} % chapter lecture_2_jan_09th \section{Sylow Theory}% \label{sec:sylow_theory} % section sylow_theory From the \hyperref[thm:orbit_decomposition_theorem]{Orbit Decomposition Theorem}, one special case is when $G$ acts on $X = G$ by conjugation. \begin{crly}[Class Equation]\index{Class Equation}\label{crly:class_equation} From \cref{thm:orbit_decomposition_theorem}, if $X = G$, we have \begin{align*} \abs{G} &= \abs{Z(G)} + \sum \overarrow{\abs{ \orb(a_i) }}{\text{non-central}} \\ &= \abs{Z(G)} + \sum [ G : \stab(a_i) ] \text{ by } \hyperref[thm:orbit_stabilizer_theorem]{Orbit-Stabilizer} \\ &= \abs{Z(G)} + \sum [ G : C(a_i) ], \end{align*} where $C(a_i)$ is called the \hldefn{centralizers} of $G$. \end{crly} \begin{thm}[First Sylow Theorem]\index{First Sylow Theorem}\label{thm:first_sylow_theorem} Let $G$ be a finite group, and let $p \mid \abs{G}$ such that $p$ is prime. Then $G$ contains a Sylow $p$-subgroup. \end{thm} \begin{proof} We proceed by induction on the size of $G$. If $\abs{G} = 2$, then $p = 2$, and so $G$ is its own Sylow $p$-subgroup \sidenote{A $2$-cycle is a Sylow $p$-group.}. Consider a finite group $G$ with $\abs{G} \geq 2$. Let $p$ be a prime that divides $\abs{G}$, and assume that the desired result holds for smaller groups. Let $\abs{G} = p^n m$, where $n, m \in \mathbb{N}$, and $p \nmid m$. \noindent \hlbnotea{Case 1: $p \mid \abs{Z(G)}$} By \cref{thm:cauchy_s_theorem_for_abelian_groups}, $\exists a \in Z(G)$ such that $\abs{a} = p$. Since $\langle a \rangle \subsetneq Z(G)$, we have that \begin{equation*} \langle a \rangle \triangleleft G \text{ and } \abs{ \langle a \rangle } = p. \end{equation*} \sidenote{This feels like a struck of genius. Let's break it down and find some way that makes it easier to remember. We want to find $H \leq G$ such that $\abs{H} = p^n$. We have $\abs{ \langle a \rangle } = p$. We want to be able to use the \hlimpo{Correspondence Theorem}, so we should adjust our materials to fit that mold: since $\abs{\langle a \rangle} = p$, notice that \begin{equation*} \frac{\abs{G}}{\abs{\langle a \rangle}} = p^{n - 1} m. \end{equation*} This is a smaller group than $G$, and so IH tells us that it has a Sylow $p$-subgroup, say $\bar{H}$. By the Correspondence Theorem, we may retrieve $H$.} Notice that the group $G / \langle a \rangle$ is a group that has a lower order than $G$, and so by IH, $\exists \bar{H} \leq G / \langle a \rangle$ such that $\bar{H}$ is a Sylow $p$-subgroup of $G / \langle a \rangle$. Note that if $n = 1$. then $\langle a \rangle$ itself is the Sylow $p$-subgroup. WMA $n > 1$. We have that $\abs{H} = p^{n - 1}$. By \hyperref[thm:correspondence_theorem]{correspondence}, \begin{equation*} \bar{H} = H / \langle a \rangle, \end{equation*} where $H \leq G$. By comparing the orders, we have \begin{equation*} p^{n - 1} = \frac{\abs{H}}{p} \implies \abs{H} = p^n. \end{equation*} Therefore $H$ is a Sylow $p$-subgroup of $G$. \noindent \hlbnotea{Case 2: $p \nmid Z(G)$} By the \hyperref[crly:class_equation]{class equation}, notice that \begin{equation}\label{eq:first_sylow_theorem_eq1} p^n m = \abs{G} = \abs{Z(G)} + \sum [ G : C(a_i) ], \end{equation} and the summation cannot be $0$ or $p$ would otherwise divide $Z(G)$. Since $p$ divides the LHS of \cref{eq:first_sylow_theorem_eq1} and not $\abs{Z(G)}$, and the sum is nonzero, we must have that $\exists a_i \in G$ such that $p \nmid [ G : C(a_i) ]$, since only then would $p \mid \abs{G}$ \sidenote{This is after having this term `neutralizing' $\abs{G}$ so that the entire RHS is also divisible by $p$. If $p$ already divides everything, and does not divide $\abs{Z(G)}$, then $p$ would not divide $\abs{Z(G)}$.}. Since $p \mid \abs{G}$ but not $\abs{ G : C(a_i)}$, it must be that $p^n \mid \abs{C(a_i)}$ by Lagrange \sidenote{Having $p^n \mid \abs{C(a_i)}$ would cancel out all the $p$'s in $\abs{G}$, thus rendering $p$ unable to divide $\abs{G : C(a_i)}$.}. Note that we have $\abs{C(a_i)} \leq \abs{G}$. Thus by IH, $C(a_i)$ has a Sylow $p$-subgroup, which is also a Sylow $p$-subgroup of $G$. \end{proof} \begin{crly}[Cauchy's Theorem]\index{Cauchy's Theorem}\label{crly:cauchy_s_theorem} If $p$ is prime and $p \mid \abs{G}$, then $G$ has an element of order $p$. \end{crly} \begin{proof} WLOG, WMA $\abs{G} = p^n m$, where $n, m \in \mathbb{N}$ and $p \nmid m$. By \cref{thm:first_sylow_theorem}, $\exists H \leq G$ such that $H$ is a Sylow $p$-subgroup. Take $a \in H \setminus \left\{ e \right\}$. Then $\abs{a} = p^k$ for some $k \leq n$. Let $b = a^{p^{k - 1}}$. Notice that $b \neq e$, or it would contradict the definition of an order (for $a$). Then $b^p = \left( a^{p^{k - 1}} \right)^p = a^p = e$. Therefore $\abs{b} = p$ and $b \in G$. \end{proof} \begin{defn}[Normalizer]\index{Normalizer}\label{defn:normalizer} Let $G$ be a group, and $H \leq G$. The set \begin{equation*} N_G(H) = \left\{ g \in G \mmid gHg^{-1} = H \right\} \end{equation*} is called the \hlnoteb{normalizer} of $H$ in $G$. \end{defn} \begin{ex} Verify that $N_G(H)$ is the largest subgroup of $G$ that contains $H$ as a normal subgroup. \end{ex} \begin{proof} It is clear by definition of a normalizer that $H \triangleleft N_G(H)$. Suppose there exists $N_G(H) < \tilde{H} \leq G$ such that $H \triangleleft \tilde{H}$. Let $h \in \tilde{H} \setminus N_G(H)$. But since $H \triangleleft \tilde{H}$, we have \begin{equation*} hHh^{-1} = H, \end{equation*} which implies that $h \in N_G(H)$, a contradiction. Therefore $N_G(H)$ is the largest subgroup that contains $H$ as a normal subgroup. \end{proof} Before proceeding with the Sylow's next theorem, we require two lemmas. \marginnote{\cref{lemma:intersection_of_a_sylow_p_subgroup_with_any_other_p_subgroups} tells us that if we can find a $p$-subgroup $Q$ of $G$, then the elements in $Q$ that serves as the stabilizers of $P$ are precisely the elements that $Q$ shares with $P$. This is uninteresting if $P$ is either abelian or normal, but it would highlight what $Z(P)$ is.} \begin{lemma}[Intersection of a Sylow $p$-subgroup with any other $p$-subgroups]\label{lemma:intersection_of_a_sylow_p_subgroup_with_any_other_p_subgroups} Let $G$ be a finite group and $p$ a prime such that $p \mid \abs{G}$. Let $P, Q \leq G$ be a Sylow $p$-subgroup and a (regular) $p$-subgroup, respectively. Then \begin{equation}\label{eq:intersection_of_a_sylow_p_subgroup_with_any_other_p_subgroups} Q \cap N_G(P) = Q \cap P. \end{equation} \end{lemma} \begin{proof} Since $P \subseteq N_G(P)$, $\subseteq$ of \cref{eq:intersection_of_a_sylow_p_subgroup_with_any_other_p_subgroups} is done. Let $N = N_G(P)$, and let $H = Q \cap N$. WTS $H \subseteq Q \cap P$. Since $H = Q \cap N \subseteq Q$, it suffices to show that $H \subseteq P$. Since $P$ is a Sylow $p$-subgroup, let $\abs{P} = p^n$. By Lagrange, we have that $\abs{H} = p^m$ for some $m \leq n$. Since $P \triangleleft N$, we have that $HP \leq N$ \sidenote{See \href{https://tex.japorized.ink/PMATH347S18/classnotes.pdf\#thm.30}{PMATH 347}}. Moreover, we have that \begin{equation*} \abs{HP} = \frac{\abs{H}\abs{P}}{\abs{H \cap P}} = p^k \end{equation*} for some $k \leq n$. Also, $P \subset HP$, and so $n \leq k$, implying that $k = n$. Thus $P = HP$, and thus \begin{equation*} H \subseteq HP = P, \end{equation*} as required. \end{proof} % section sylow_theory (end) % chapter lecture_2_jan_09th (end) \chapter{Lecture 3 Jan 11th}% \label{chp:lecture_3_jan_11th} % chapter lecture_3_jan_11th \section{Sylow Theory (Continued)}% \label{sec:sylow_theory_continued} % section sylow_theory_continued \begin{lemma}[Counting The Conjugates of a Sylow $p$-Subgroup]\label{lemma:counting_the_conjugates_of_a_sylow_p_subgroup} Let $G$ be a finite group, and $p$ a prime such that $p \mid \abs{G}$. Let \begin{itemize} \item $P$ be a Sylow $p$-subgroup; \item $Q$ be a $p$-subgroup; \item $K = \left\{ gPg^{-1} \mmid g \in G \right\}$; \item $Q$ act on $K$ by conjugation; and \item $P = P_1, P_2, \ldots, P_r$ be the distinct orbit representatives from the action of $Q$ on $K$. \end{itemize} Then \begin{equation*} \abs{K} = \sum_{i=1}^{r} \left[ Q : Q \cap P_i \right]. \end{equation*} \end{lemma} \begin{proof} From the definition of $K$, and the fact that $Q$ acts on $K$, we have \begin{align*} \abs{K} &= \sum_{i=1}^{r} \abs{ \orb(P_i) } \\ &= \sum_{i=1}^{r} \abs{ Q } / \abs{ \stab(P_i) } \quad \hyperref[thm:orbit_stabilizer_theorem]{\text{orbit-stabilizer}} \\ &= \sum_{i=1}^{r} \abs{ Q } / \abs{ N_G(P_i) \cap Q } \quad \text{ by the action } \\ &= \sum_{i=1}^{r} [ Q : N_G(P_i) \cap Q ] \quad \text{ by definition } \\ &= \sum_{i=1}^{r} [ Q : Q \cap P_i ] \quad \hyperref[lemma:intersection_of_a_sylow_p_subgroup_with_any_other_p_subgroups]{\text{the last lemma}}. \end{align*} % TODO : Question for prof \sidenote{Why can we use \cref{lemma:intersection_of_a_sylow_p_subgroup_with_any_other_p_subgroups}? Are the $P_i$'s Sylow $p$-subgroups?} \end{proof} \begin{thm}[Second Sylow Theorem]\index{Second Sylow Theorem}\label{thm:second_sylow_theorem} If $P$ and $Q$ are Sylow $p$-subgroups of $G$, then $\exists g \in G$ such that $P = gQg^{-1}$. \end{thm} \begin{proof} Let $K = \left\{ qPq^{-1} \mmid q \in G \right\}$. WTS $Q \in K$. We shall also note that $\abs{P} = p^k$ for some $k \in \mathbb{N}$. Let $P$ act on $K$ by conjugation. Let the orbit representatives be \begin{equation*} P = P_1, P_2, \ldots, P_r. \end{equation*} By \cref{lemma:counting_the_conjugates_of_a_sylow_p_subgroup}, we have \begin{equation*} \abs{K} = \sum_{i=1}^{r} [ P : P \cap P_i ] = [ P : P ] + \sum_{i=2}^{r} [ P : P \cap P_i ] = 1 + \sum_{i=2}^{r} [ P : P \cap P_i ]. \end{equation*} Thus \begin{equation*} \abs{K} \equiv 1 \mod p. \end{equation*} Now let $Q$ act on $K$ by conjugation. Reordering if necessary, the orbit representatives are \begin{equation*} P = P_1, P_2, \ldots, P_s, \end{equation*} where $s$ is not necessarily $r$. From here, it suffices to show that $Q = P_i$ for some $i \in \left\{ 1, 2, \ldots, s \right\}$. Suppose not. Then by \cref{lemma:counting_the_conjugates_of_a_sylow_p_subgroup}, \begin{equation*} \abs{K} = \sum_{i=1}^{s} [ Q : P_i \cap Q ]. \end{equation*} Note that it must be the case that $[ Q : P_i \cap Q ] > 1$, for some if not all $i$, for otherwise it would imply that $Q \cap P_i$ and that would be a contradiction. Then by Lagrange, \begin{equation*} \abs{K} \equiv 0 \mod p. \end{equation*} This contradicts the fact that $\abs{K} \equiv 1 \mod p$. This shows that $Q = P_i$ for some $i \in \left\{ 1, 2, \ldots, s \right\}$, and so $Q$ is a conjugate of $P$. \end{proof} \begin{note}[Notation] We shall denote $n_p$ as the number of Sylow $p$-subgroups in $G$. \end{note} \begin{thm}[Third Sylow Theorem]\index{Third Sylow Theorem}\label{thm:third_sylow_theorem} Let $p$ be a prime, and that it divides $\abs{G}$, where $G$ is a group. Suppose $\abs{G} = p^n m$, where $n, m \in \mathbb{N}$ and $p \nmid m$. Then \begin{enumerate} \item $n_p \equiv 1 \mod p$; and \item $n_p \mid m$. \end{enumerate} \end{thm} \begin{proof} Let $P$ be a Sylow $p$-subgroup of $G$, and let \begin{equation*} K = \left\{ gPg^{-1} \mmid g \in G \right\}. \end{equation*} By \hyperref[thm:second_sylow_theorem]{Sylow's second theorem}, $n_p = \abs{K}$ as all the conjugates are exactly the Sylow $p$-subgroups. And by our last proof, we saw that $n_p \equiv 1 \mod p$. Let $G$ act on $K$ by conjugation. Then by the \hyperref[thm:orbit_stabilizer_theorem]{Orbit-Stabilizer Theorem}, \begin{equation*} \abs{G} = \abs{\stab(P)}\abs{\orb(P)}. \end{equation*} Thus \begin{equation}\label{eq:third_sylow_theorem_eq1} p^n m = \abs{ N_G(P) } n_p. \end{equation} Thus $n_p \mid p^n m$. Since $n_p \equiv 1 \not\equiv 0 \mod p$, we must have $n_p \mid m$. \end{proof} \begin{remark} \begin{enumerate} \item From \cref{eq:third_sylow_theorem_eq1}, we have that \begin{equation*} n_p = [ G : N_G(P) ]. \end{equation*} \item \imponote\ Note that \begin{equation*} n_p = 1 \iff \forall g \in G \; gPg^{-1} = P \iff P \triangleleft G. \end{equation*} However, note that $P$ \hlimpo{may be trivial}! This means that if $G$ is simple, it does not imply that $n_p = 1$. \end{enumerate} \end{remark} \begin{defn}[Simple Group]\index{Simple Group}\label{defn:simple_group} A group is said to be \hlnoteb{simple} if it has no non-trivial\sidenote{By non-trivial, we mean that the normal subgroup is not the group with only the identity element.} normal subgroups. \end{defn} \marginnote{ \begin{procedure}[No simple subgroup of order $n$]\label{procedure:no_simple_subgroup_of_order_n} The approach to showing that there are no simple groups of a certain order is as follows: \begin{itemize} \item we make use of the fact that each group has a Sylow subgroup, and there are usually not many such subgroups; \item using each of the possiblities as cases, we find out if a group of the given order will have a normal subgroup. \end{itemize} \end{procedure} } \begin{eg} Prove that there is no simple group of order $56$. \end{eg} \begin{proof} Let $G$ be a group. Note that $56 = 2^3 \cdot 7$. Then $n_7 \equiv 1 \mod 7$ and $n_7 \mid 8 = 2^3$. Thus \begin{equation*} n_7 = 1 \text{ or } n_7 = 8. \end{equation*} \hlbnotea{$n_7 = 1$} By the remark above, $G$ has a normal Sylow $7$-subgroup. Thus $G$ is not simple. \hlbnotea{$n_7 = 8$} By Lagrange, since $7$ is prime \sidenote{This makes use of the fact that the Sylow $7$-subgroup has a prime order, not just because $7$ itself is prime. We say this here because if the order of the Sylow $p$-subgroup is prime, then by \hyperref[thm:lagrange_s_theorem]{Lagrange}, $\abs{P \cap Q}$, where $P$ and $Q$ are distinct Sylow $p$-subgroups, is a subgroup of $P$ (and $Q$), and must hence either be $1$ or $p$. But this intersection cannot have order $p$, since $P$ and $Q$ are distinct. Thus $\abs{P \cap Q} = 1$. It is also important to note that this is only true if the order of the Sylow $p$-subgroups are prime, i.e. simply $p$ itself. If their orders are $p^n$ for some $n > 1$, this is not necessarily true. }, the distinct Sylow $7$-subgroups of $G$ intersect trivially. Therefore, there are $8 \times 6 = 48$ elements of order $7$ in $G$. But this implies that $56 - 48 = 8$ elements that are not of order $7$. One of them is the identity, thus the remaining $7$ elements must have order $2$ \sidenote{They cannot be of any other order as that would create a cyclic group that is not of order $2$ or $7$, which is impossible.}. This implies that \begin{equation*} n_2 = 7 \equiv 1 \mod 2, \end{equation*} which by our remark means that $G$ has a normal Sylow $2$-subgroup. Thus $G$ is not simple by both accounts. \end{proof} % section sylow_theory_continued (end) % chapter lecture_3_jan_11th (end) \chapter{Lecture 4 Jan 14th}% \label{chp:lecture_4_jan_14th} % chapter lecture_4_jan_14th \section{Sylow Theory (Continued 2)}% \label{sec:sylow_theory_continued_2} % section sylow_theory_continued_2 \begin{remark} \begin{enumerate} \item Let $p \neq q$ both be primes, and $p, q \mid \abs{ G }$. Let $H_p$ and $H_q$ be a Sylow $p$-subgroup and a Sylow $q$-subgroup of $G$, respectively. By Lagrange's Theorem, we must have that $H_p \cap H_q = \left\{ e \right\}$. Then \begin{equation*} \abs{ H_p \cup H_q } = \abs{ H_p } + \abs{ H_q } - 1. \end{equation*} \item Let $\abs{ G } = pm$ and $p \nmid m$, where $p$ is prime. If $H, K$ are Sylow $p$-subgroups of $G$ with $H \neq K$, then $H \cap K = \left\{ e \right\}$. \end{enumerate} \end{remark} \begin{eg} Let $G = D_6$. Notice that \begin{equation*} H = \langle 1, s \rangle, \quad K = \langle 1, rs \rangle \end{equation*} are both Sylow $2$-subgroups of $D_6$ and $H \neq K$, and their intersection is trivial. \end{eg} \begin{eg} Let $\abs{ G } = pq$ where $p, q$ are primes with $p < q$ and $p \nmid q - 1$. Then $\abs{G}$ is cyclic. \end{eg} \begin{proof} By the Third Sylow Theorem, $n_p \equiv 1 \mod p$ and $n_p \mid q$. Notice that $n_p = 1$, since if $n_p = q$, then $n_p \equiv 1 \mod p \implies p \mid q - 1$, contradicting our assumption. By our remark last lecture, $G$ has a normal Sylow $p$-subgroup, which we shall call $H_p$. On the other hand, $n_q \equiv 1 \mod q$ and $n_q \mid p$. Since $p < q$, $q \nmid p - 1$, and so the same argument as before holds. Hence $n_q = 1$, and so $G$ has a normal Sylow $q$-subgroup. Since $H_p \triangleleft G$, we know that $H_p H_q \leq G$, and we notice that \begin{equation*} \abs{ H_p H_q } = \frac{\abs{ H_p } \abs{ H_q }}{\abs{ H_p \cap H_q }} = pq = \abs{ G }. \end{equation*} Thus $G = H_p H_q$. Let $a, b \in G$. If $a, b$ is either both in $H_p$ or both in $H_q$, then $ab = ba$ \sidenote{Note: $H_p$ and $H_q$ are normal subgroups.}. WMA $a \in H_p$ and $b \in H_q$. By our first remark today, note that $H_p \cap H_q = \left\{ e \right\}$. Then, observe that \begin{equation*} \underbrace{aba^{-1}}_{H_q} \tikzmark{a}b^{-1} \in H_q \qquad \tikzmark{b}a \underbrace{ba^{-1}b^{-1}}_{H_p} \in H_p \end{equation*} \begin{tikzpicture}[remember picture,overlay] \draw[latex'-] ([shift={(1pt,-2pt)}]pic cs:a) |- ([shift={(1pt,-10pt)}]pic cs:a) node[anchor=north] {$H_q$}; \draw[latex'-] ([shift={(1pt,-2pt)}]pic cs:b) |- ([shift={(1pt,-10pt)}]pic cs:b) node[anchor=north] {$H_p$}; \end{tikzpicture} Thus $aba^{-1}b^{-1} = e \implies ab = ba$. So $G$ is abelian. By the Fundamental Theorem of Finite Abelian Groups \begin{equation*} G \simeq \mathbb{Z}_p \times \mathbb{Z}_q \simeq \mathbb{Z}_{pq}, \end{equation*} which is cyclic. \end{proof} \begin{eg} By the Fundamental Theorem of Finite Abelian Groups \begin{equation*} S_3 \simeq \mathbb{Z}_2 \times \mathbb{Z}_3, \end{equation*} and $\abs{ S_3 } = 6 = 2 \cdot 3$, is not cyclic. Notice that $S_3$ does not fulfill the requirements for the last example since $2 \mid 3 - 1 = 2$. \end{eg} \begin{eg} If $\abs{ G } = 30$, then $G$ has a subgroup isomorphic to $\mathbb{Z}_{15}$. Note that $\abs{ G } = 2 \cdot 3 \cdot 5$. By the Third Sylow Theorem, \begin{equation*} n_5 \equiv 1 \mod 5 \text{ and } n_5 \mid 6 \implies n_5 = 1 \text{ or } 6 \end{equation*} and \begin{equation*} n_3 \equiv 1 \mod 3 \text{ and } n_3 \mid 10 \implies n_3 = 1 \text{ or } 10. \end{equation*} Suppose $n_5 = 6$ and $n_3 = 10$. Since the Sylow $3$-subgroups and Sylow $5$-subgroups intersect trivially, this accounts for $(6 \times 4) + ( 10 \times 2 ) = 44$ elements but $\abs{G} = 30 < 44$. Thus we must have $n_5 = 1$ or $n_3 = 1$. Thus $G$ is not simple. Let $H_3$ and $H_5$ be Sylow $3$- and $5$-subgroups, respectively. WLOG, suppose $H_3 \triangleleft G$. Then $H_3 H_5 \leq G$, and notice that $\abs{H_3 H_5} = 15$. Since $15 = 3 \cdot 5$ and $3 \nmid 4 = 5 - 1$, we know that $H_3 H_5 \simeq \mathbb{Z}_{15}$ by an earlier example. \end{eg} \begin{eg}\label{eg:g_60_is_simple} Let $\abs{ G } = 60$ with $n_5 > 1$. Then $G$ is simple. \end{eg} This is an important example for it is with this that we can prove the following: \begin{crly}[$A_5$ is Simple]\label{crly:_a_5_is_simple} $A_5$ is simple. \end{crly} \begin{proof} Note that $\abs{ A_5 } = \frac{5!}{2} = 60$, and \begin{equation*} \lra{\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \end{pmatrix}} \text{ and } \lra{\begin{pmatrix} 1 & 3 & 2 & 4 & 5 \end{pmatrix}} \end{equation*} are both Sylow $5$-subgroups that are distinct (one has odd \hlnotea{parity} while the other has even). \end{proof} \begin{proof}[For \cref{eg:g_60_is_simple}] Suppose $n_5 > 1$. Notice that $60 = 2^2 \cdot 3 \cdot 5$. By \cref{thm:third_sylow_theorem}, $n_5 \equiv 1 \mod 5$ and $n_5 \mid 12$, and thus $n_5 = 6$. This accounts for $6 \times 4 + 1 = 25$ elements. Now suppose $H \triangleleft G$ is proper and non-trivial. If $5 \mid \abs{ H }$, then $H$ contains a Sylow $5$-subgroup of $G$. Since $H \triangleleft G$, $H$ contains all the conjugates of this Sylow $5$-subgroup. Thus by our argument above, we have that $\abs{ H } \geq 25$ \sidenote{These are the $25$ elements that were found in the last paragraph.}. Also, $H \mid 60$. Thus it must be that $\abs{ H } = 30$. But then by the last example, $n_5 = 1$, a contradiction. So $5 \nmid \abs{ H }$. By Lagrange, it remains that \begin{equation*} \abs{ H } = 2, \, 3, \, 4, \, 6\, \text{ or } 12. \end{equation*} \hlbnoted{Case A} $\abs{ H } = 12 = 2^2 \cdot 3$.\sidenote{ \begin{ex} Prove that either $n_2 = 1$ or $n_3 = 1$. \end{ex}} So $H$ contains a normal Sylow $2$- or $3$-subgroup that is normal in $G$. \end{proof} The proof shall be continued next lecture. % section sylow_theory_continued_2 (end) % chapter lecture_4_jan_14th (end) \tuftepart{Fields} \chapter{Lecture 5 Jan 14th}% \label{chp:lecture_5_jan_14th} % chapter lecture_5_jan_14th \section{Sylow Theory (Continued 3)}% \label{sec:sylow_theory_continued_3} % section sylow_theory_continued_3 We shall continue with the last proof from where we left off. \begin{proof}[\cref{eg:g_60_is_simple} continued] \hlbnoted{Case A} $\abs{H} = 12$. WLOG, let $K$ be a normal Sylow $3$-subgroup of $H$, which is also normal in $G$ \sidenote{In Sylow Theory, normality is transitive: \begin{mproof} If $P$ is a normal Sylow $p$-subgroup of $G$, and $Q$ is a normal subgroup of $P$, then $\forall q \in Q$, we have $q \in P$ and so $gqg^{-1} = q$ by normality of $P$. It follows that $gQg^{-1} = Q$ and so $Q$ is also normal in $G$. \end{mproof} }. \noindent \hlbnoted{Case B} $\abs{H} = 6$. $H$ would then have a normal Sylow $3$-subgroup, which is normal in $G$. We shall also call this subgroup $K$. \noindent By replacing $H$ with $K$ if necessary, wma $\abs{ H } \in \left\{ 2, 3, 4 \right\}$. Consider $\bar{G} = G / H$. Then $\abs{ \bar{G} } \in \left\{ 15, 20, 30 \right\}$. \sidenote{ \begin{ex} Prove that $\bar{G}$ has a normal Sylow $5$-subgroup in all the three possible orders of $\bar{G}$. \end{ex}} In any case, $\bar{G}$ has a normal Sylow $5$-subgroup. Call this normal subgroup $\bar{P}$. By \hyperref[thm:correspondence_theorem]{correspondence}, $\bar{P} = P / H$ where $P$ is a normal subgroup of $G$ \sidenote{Note: correspondence works for the normal case as well.}. Thus $P$ is a proper non-trivial normal subgroup of $G$. Also, \begin{equation*} \abs{ P } = \abs{ \bar{P} } \cdot \abs{ H } = 5 \cdot \abs{ H }. \end{equation*} Thus $5 \mid \abs{ P }$, putting us back to the case where $5 \mid \abs{ H }$. Thus $G$ does not have a non-trivial normal subgroup, i.e. $G$ is simple. \end{proof} % section sylow_theory_continued_3 (end) \section{Review of Ring Theory}% \label{sec:review_of_ring_theory} % section review_of_ring_theory Let $F$ be a \hlnotea{field}, and $I$ be an \hlnotea{ideal} of $F[x]$, its \hlnotea{polynomial ring}. Since $F[x]$ is a PID, we have $I = \langle p(x) \rangle$ for some $p(x) \in F[x]$. Moreover, $I$ is \hlnotea{maximal} iff $p(x)$ is \hlnotea{irreducible}. Thus we observe that \begin{center} $F[x]/I$ is a field iff $I = \langle p(x) \rangle$ is maximal iff $p(x) \in F[x]$ is irreducible. \end{center} Therefore, to talk about fields, we need to understand irreducibles. % section review_of_ring_theory (end) \section{Irreducibles}% \label{sec:irreducibles} % section irreducibles \begin{defn}[Irreducible]\index{Irreducible}\label{defn:irreducible} Let $R$ be an integral domain (ID) \sidenote{\hldefn{Integral domains} are commutative rings that has no zero divisors.}. We say that $f(x) \in R[x]$ is \hlnoteb{irreducible} (over $R$) if \begin{enumerate} \item $f(x) \neq 0$; \item $f(x) \notin R^\times$, where $R^\times$ is the set of units of $R$; \item whenever $f(x) = g(x) h(x)$, where $g(x), h(x) \in R[x]$, then either $g(x) \in R^\times$ or $h(x) \in R^\times$. \end{enumerate} If $f(x) \neq 0$, $f(x) \notin R^\times$ and $f(x)$ is not irreducible, we say that $f(x)$ is \hldefn{reducible} (over $R$). \end{defn} \begin{eg} $f(x) = x^2 - 2$ is irreducible over $\mathbb{Q}$ but reducible over $\mathbb{R}$ as \begin{equation*} f(x) = \left(x - \sqrt{2}\right)\left(x + \sqrt{2}\right). \end{equation*} \end{eg} Let $F$ be a field, $f(x) \in F[x]$ and $a \in F$. By the \hlnotea{Division Algorithm}, we can write \begin{equation*} f(x) = (x - a) q(x) + r(x), \end{equation*} where $q(x), r(x) \in F[x]$. Note that we either have $r(x) = 0$ or $\deg r < \deg (x - a) = 1$. In the latter case, $r \in F$, and so \begin{equation*} f(x) = (x - a) q(x) + r. \end{equation*} Then $f(a) = 0 + r = r$, and so $f(x) = (x - a) q(x) + f(a)$. \begin{equation*} \therefore (x - a) \mid f(x) \iff f(a) = 0. \end{equation*} \begin{propo}[Polynomials with Roots are Reducible]\label{propo:polynomials_with_roots_are_reducible} Let $F$ be a field. If $f(x) \in F[x]$ with $\deg f > 1$, and $f$ has a root in $F$, then $f$ is reducible (over $F$). \end{propo} \begin{eg} Let $f(x) = x^6 + x^3 + x^4 + x^3 + 3 \in \mathbb{Z}_7[x]$. Then $f(1) = 0$. Therefore \begin{equation*} f(x) = (x - 1) g(x) \text{ where } g(x) \in \mathbb{Z}_7[x]. \end{equation*} Thus $f(x)$ is reducible over $\mathbb{Z}_7$. \end{eg} \begin{propo}[Irreducible Rootless Polynomials]\label{propo:irreducible_rootless_polynomials} Let $F$ be a field\sidenote{Note that this does not work in an ID. For example, $2x^2 + 2$. }. If $f(x) \in F[x]$ with $\deg f \in \{ 2, 3 \}$, then $f(x)$ is irreducible over $F$ iff $f(x)$ has no roots in $F$. \end{propo} \begin{warning} $(x^2 + 1)^2 \in \mathbb{R}[x]$ is reducible but has no root in $\mathbb{R}$. Note that the degree of the polynomial is $4$. \end{warning} \begin{eg} Let $f(x) = x^3 + x + 1 \in \mathbb{Z}_2[x]$. Note that $f(0) = 1$ and $f(1) = 3 \equiv 1 \mod 2$. Since $\deg f = 3$ and $f$ has no roots in $\mathbb{Z}_2$, $f(x)$ is irreducible over $\mathbb{Z}_2$. \end{eg} \begin{thm}[Gauss' Lemma]\index{Gauss' Lemma}\label{thm:gauss_lemma} Let $R$ be a Unique Factorization Domain (UFD), with field of fractions $F$. Let $p(x) \in R[x]$. If \begin{equation*} p(x) = A(x) B(x), \end{equation*} where $A(x), B(x)$ are non-constant in $F[x]$, then $\exists r, s \in F^\times$ non-zero such that \begin{equation*} p(x) = a(x) b(x), \end{equation*} where $a(x) = rA(x)$ and $b(x) = sB(x)$. \end{thm} \begin{note} If $p(x) \in R[x]$ is reducible over $F$, then $p(x)$ is reducible over $R$. \end{note} \begin{note} If $R = \mathbb{Z}$ and $F = \mathbb{Q}$, then if $p(x)$ is irreducible over $\mathbb{Z}$, then $p(x)$ is irreducible over $\mathbb{Q}$. \end{note} % section irreducibles (end) % chapter lecture_5_jan_14th (end) \chapter{Lecture 6 Jan 18th}% \label{chp:lecture_6_jan_18th} % chapter lecture_6_jan_18th \section{Irreducibles (Continued)}% \label{sec:irreducibles_continued} % section irreducibles_continued Our goal in this section is to develop methods to test for the irreducibility of polynomials. \begin{warning} Note that $f(x) = 2x + 4 = 2(x + 2)$ is reducible over $\mathbb{Z}$ \sidenote{This is interesting over $\mathbb{Z}$, since $2 \notin \mathbb{Z}^\times$.} but irreducible over $\mathbb{Q}$. \end{warning} \begin{propo}[Mod-$p$ Irreducibility Test]\index{Mod-$p$ Irreducibility Test}\label{propo:mod_p_irreducibility_test} Let $f(x) \in \mathbb{Z}[x]$ with $\deg f \geq 1$. Let $p \in \mathbb{Z}$ be prime. If $\bar{f}(x)$ is the corresponding polynomial in $\mathbb{Z}_p[x]$ such that \begin{itemize} \item the coefficients of $\bar{f}(x)$ are coefficients of $f(x)$ in mod $p$, \item $\deg f = \deg \bar{f}$ \sidenote{This means that the leading coefficient of $f$ is not killed off.}, and \item $\bar{f}$ is irreducible over $\mathbb{Z}_p$, \end{itemize} then $f(x)$ is irreducible over $\mathbb{Q}$. \end{propo} \begin{proof} Suppose $\deg f = \deg \bar{f}$, and $\bar{f}(x) \in \mathbb{Z}_p$ is irreducible over $\mathbb{Z}_p$. Suppose to the contrary that $f(x)$ is reducible over $\mathbb{Q}$. Then for some $g(x), h(x) \in \mathbb{Q}[x]$ with $\deg g, \deg h < \deg f$, we have \begin{equation*} f(x) = g(x) h(x). \end{equation*} By \hyperref[thm:gauss_lemma]{Gauss' Lemma}, wma $g(x), h(x) \in \mathbb{Z}[x]$. Then we have \begin{equation*} \bar{f}(x) = \bar{g}(x) \bar{h}(x) \in \mathbb{Z}_p[x]. \end{equation*} By assumption, $\bar{f}$ is irreducible over $\mathbb{Z}_p$, either \begin{equation*} \deg \bar{g} = 0 \text{ or } \deg \bar{h} = 0. \end{equation*} Wlog, $\deg \bar{g} = 0$. Then \begin{equation*} \deg h \leq \deg f = \deg \bar{f} = \deg \bar{h} \leq \deg h, \end{equation*} which implies that $\deg f = \deg h$ but $\deg h < \deg f$. Thus $f$ is irreducible over $\mathbb{Q}$. \end{proof} \begin{eg} Consider the polynomial \begin{equation*} f(x) = 3x^3 + 22x^2 + 17x + 471. \end{equation*} Then consider \begin{equation*} \bar{f}(x) = x^3 + x + 1 \in \mathbb{Z}_2[x]. \end{equation*} Since $\bar{f}(0) \neq 0$ and $\bar{f}(1) \neq 0$, and $\deg f = 3$, by \cref{propo:irreducible_rootless_polynomials}, $\bar{f}(x)$ is irreducible over $\mathbb{Z}_2$. Since $\deg f = \deg \bar{f}$, $f$ is irreducible over $\mathbb{Q}$ by the \hyperref[propo:mod_p_irreducibility_test]{Mod-$2$ irreducible test}. \end{eg} \begin{warning} Consider $f(x) = 2x^2 + x \in \mathbb{Q}[x]$, which is reducible over $\mathbb{Q}$. However, $\bar{f}(x) = x \in \mathbb{Z}_2[x]$ is \hlimpo{irreducible} over $\mathbb{Z}_2$. Notice here that $\deg \bar{f} \neq \deg f$. \end{warning} More generally so... \begin{propo}[Polynomials that Cannot be Factored Over the Ideals is Irreducible]\label{propo:polynomials_that_cannot_be_factored_over_the_ideals_is_irreducible} Let $I$ be a proper ideal of an ID $R$. Let $p(x) \in R[x]$ be monic and non-const. If $p(x)$ cannot be factored in $\left( R / I \right)[x]$ \sidenote{Note that $\left( R / I \right)$ may not be an ID even if $R$ is one.} into polynomials of lesser degree, then $p(x)$ is irreducible over $R$. \end{propo} \begin{proof} Sps to the contrary that $p(x)$ is reducible over $R$. Then \begin{equation*} p(x) = f(x) g(x) \end{equation*} for some $f(x), g(x) \notin R^\times$. Since $p(x)$ is monic, and $\deg f, \deg g < \deg p$, wma $f(x)$ and $g(x)$ are also monic. Then \begin{equation*} \bar{p}(x) = \bar{f}(x) \bar{g}(x) \in \left( R/I \right)[x]. \end{equation*} Since $I \subsetneq R$, we have that $1 \notin I$, and so \begin{equation*} \deg \bar{f}, \deg \bar{g} < \deg \bar{p} \end{equation*} but that implies that $p(x)$ can be factored in $\left( R/I \right)[x]$. \end{proof} \begin{propo}[Eisenstein's Criterion]\index{Eisenstein's Criterion}\label{propo:eisenstein_s_criterion} Let $R$ be an ID. Let $P$ be a prime ideal of $R$. Let \begin{equation*} f(x) = x^n + a_{n - 1} x^{n - 1} + \hdots + a_1 x + a_0 \in R[x] \end{equation*} with $n \geq 1$. Note that $f$ is monic. Now if \begin{equation*} a_{n - 1}, a_{n - 2}, \ldots, a_1, a_0 \in P \text{ and } a_0 \notin P^2, \end{equation*} then $f$ is irreducible over $R$. \end{propo} \begin{proof} Sps to the contrary that $f$ is reducible over $R$. Since $f(x)$ is monic, \begin{equation*} f(x) = g(x) h(x) \end{equation*} where $g(x), h(x) \in R[x]$ and $\deg g, \deg h < \deg f$. Then \begin{equation*} \bar{f}(x) = \bar{g}(x) \bar{h}(x) = x^n \in (R/P)[x] \end{equation*} since $a_{n - 1}, a_{n - 2}, \ldots, a_1, a_0 \in P$. Since $P$ is prime, $R/P$ is an ID, we have that either $\bar{g}(0) = 0$ or $\bar{h}(0) = 0$. Wlog, $\bar{g}(0) = 0 \in P$. But that implies that $a_0 = \bar{g}(0)\bar{h}(0) = 0 \in P^2$, a contradiction. \end{proof} % section irreducibles_continued (end) % chapter lecture_6_jan_18th (end) \chapter{Lecture 7 Jan 21st}% \label{chp:lecture_7_jan_21st} % chapter lecture_7_jan_21st \section{Irreducibles (Continued 2)}% \label{sec:irreducibles_continued_2} % section irreducibles_continued_2 \begin{eg} Prove that $f(x, y) = x^2 + y^2 - 1$ is irreducible in $\mathbb{Q}[x, y] = (\mathbb{Q}[x])[y]$. \end{eg} \begin{proof} Consider $f(x, y) = y^2 + (x^2 + 1)$. Since $x - 1$ is irreducible, let $P = \langle x - 1 \rangle$, which is therefore a prime ideal of $\mathbb{Q}[x]$. Moreover, then since $x + 1 \in \mathbb{Q}[x]$, notice that \begin{equation*} x^2 - 1 = ( x + 1 )( x - 1 ) \in P. \end{equation*} Since $(x - 1)^2 \nmid \left(x^2 - 1\right)$, \sidenote{The only polynomial of degree 2 in $P$ is $(x - 1)^2$.} we have that $x^2 - 1 \notin P^2$. Then by \hyperref[propo:eisenstein_s_criterion]{Eisenstein}, we have that $f(x, y)$ is irreducible. \end{proof} \begin{crly}[Eisenstein + Gauss]\label{crly:eisenstein_gauss} Let $p \in \mathbb{Z}$ be a prime, and let \begin{equation*} f(x) = x^n + a_{n - 1} x^{n - 1} + \hdots + a_1 x + a_0 \end{equation*} be non-const in $\mathbb{Z}[x]$. If $p \mid a_i$ for all $i \in \{0, \ldots, n - 1\}$, and $p^2 \nmid a_0$, then $f$ is irreducible over $\mathbb{Q}$. \end{crly} \marginnote{Recall that the prime ideals of $\mathbb{Z}$ are $\mathbb{Z}_p$ where $p$ is prime.} \begin{proof} Let $P = \langle p \rangle$. It follows from Eisenstein that $f$ is irreducible over $\mathbb{Z}$, and then from Gauss that $f$ is irreducible over $\mathbb{Q}$. \end{proof} \begin{eg} Let $f(x) = x^n - d \in \mathbb{Z}[x]$ where $\exists p \in \mathbb{Z}$ prime such that $p^2 \nmid d$ and $p \mid d$. Let $P = \langle p \rangle$ and so by \cref{crly:eisenstein_gauss}, $f$ is irreducible over $\mathbb{Q}$. \end{eg} \begin{eg} It is not always true that such a prime $p$ can be found. For example, consider the polynomial \begin{equation*} x^2 - 72 = x^2 - \left( 2^3 3^2 \right) \in \mathbb{Q}[x], \end{equation*} which has roots $\pm 6 \sqrt{2} \notin \mathbb{Q}$. Thus $x^2 - 72$ is irreducible in $\mathbb{Q}[x]$. \end{eg} \begin{note} The above example is noteworthy since it will appear rather often throughout this course. Notice that if we have polynomials of the above form, then we immediately have that the polynomial is irreducible. \end{note} \begin{eg}\label{eg:examples_of_irreducible_polynomials} Are the following irreducible over $\mathbb{Q}$? \begin{enumerate} \item $f(x) = x^7 + 21 x^5 + 15x^2 + 9x + 6$ Yes. Notice that all the non-leading coefficients have a factor of $3$, and so if we let $p = 3$, since $3^2 = 9 \nmid 6$, it follows from Eisenstein that $f$ is irreducible over $\mathbb{Q}$. \item $f(x) = x^3 + 2x + 16$ Eisenstein can't help us here, at least, not right now, since $\gcd(2, 16) = 2$ and $2^2 = 4 \mid 16$. Consider $\tilde{f}(x) = x^3 + 2x + 1 \in \mathbb{Z}_3[x]$. Notice that $\tilde{f}(0) = 1 = \tilde{f}(2)$ and $\tilde{f}(1) = 4$. Since $\deg \tilde{f} = 3$, it follows from \cref{propo:irreducible_rootless_polynomials} that $\tilde{f}$ is irreducible over $\mathbb{Z}_3$. Since $\deg f = \deg \tilde{f}$, it follows from the \hyperref[propo:mod_p_irreducibility_test]{Mod-$3$ irreducible test} that $f$ is irreducible over $\mathbb{Q}$. \sidenote{ \begin{marginwarning} Note that we cannot use Eisenstein on $\tilde{f}$ since while $\mathbb{Z}_p$ is an integral domain, the only proper prime ideal is $\{0\}$, which would make the tool useless. In addition to the above, note that $\mathbb{Z}_p$ cannot have other proper prime ideals. If $1 < k < p$, then by Fermat's Little Theorem, $k^{p-1} \equiv 1 \mod p$, i.e. $1 \in \langle k \rangle$, which means $\langle k \rangle = \mathbb{Z}_p$. \end{marginwarning} } \item $f(x) = x^4 + 5x^3 + 6x^2 - 1$ Again, Eisenstein can't help us here, since $5 \coprime 6 \coprime 1$ \sidenote{$\coprime$ is a common notation for coprimeness.}. Consider \begin{equation*} \bar{f}(x) = x^4 + x^3 + 1 \in \mathbb{Z}_2[x]. \end{equation*} We know that $\bar{f}(0) = 1 = \bar{f}(1)$, and so $\bar{f}$ has no roots in $\mathbb{Z}_2$. \sidenote{Note that we cannot use \cref{propo:irreducible_rootless_polynomials} here as $\deg \bar{f} = 4 > 3$.} Consider the quadratics\sidenote{\hlwarn{Why did we only check for the quadratics and not others?} We did so as we have already checked for the linear factors by checking for roots, which also checks for the cubic factors, since if we can factor out a linear factor, we are left with a cubic factor. Ruling out linear factors in turn rules out cubic factors.} of $\mathbb{Z}_2[x]$: we have \begin{equation*} x^2, \quad x^2 + x, \quad x^2 + 1, \quad x^2 + x + 1, \end{equation*} all, but the last, of which are reducible. However, notice that \begin{equation*} \left( x^2 + x + 1 \right)^2 = x^4 + x^2 + 1 \neq \bar{f}(x) \end{equation*} (by the Freshman's Dream). Thus $\bar{f}$ is irreducible in $\mathbb{Z}_2$. Since $\deg f = \deg \bar{f}$, by \hyperref[propo:mod_p_irreducibility_test]{Mod-$2$ irreducible test}. \item \label{eg:polynomial_with_prime_minus_1_degree}\imponote Let $p$ be a prime, and let \begin{equation*} f(x) = x^{p - 1} + x^{p - 2} + \hdots + x^2 + x + 1. \end{equation*} Note that $f(x)(x - 1) = x^p - 1$, and so $f(x) = \frac{x^p - 1}{x - 1}$. Furthermore, notice that \begin{align*} f(x + 1) &= \frac{(x + 1)^p - 1}{x} = \sum_{k = 0}^{p} \binom{p}{k} x^{p - k} - \frac{1}{x} \\ &= x^{p - 1} + \binom{p}{p - 1} x^{p - 2} + \hdots + \binom{p}{2} x + \binom{p}{1}. \end{align*} By setting $P = \langle p \rangle$, we have that $f(x + 1)$ is irreducible by Eisenstein. It follows from A3Q2 that $f(x)$ is also irreducible. \end{enumerate} \end{eg} \subsection{Summary for Proving Irreducibility}% \label{sub:summary_for_proving_irreducibility} % subsection summary_for_proving_irreducibility \marginnote{ \begin{procedure}[Summary for Proving Irreducibility]\label{procedure:summary_for_proving_irreducibility} This subsection is dedicated to summarize the ways that are available to us to finding out if a given polynomial is reducible or irreducible. This includes common heuristics and/or techniques. \end{procedure} } \paragraph{Showing reducibility} Let $f(x) \in F[x]$. $f(x)$ is reducible if \begin{itemize} \item \textbf{$f(x)$ has a root $\alpha$ in $F[x]$}, because that means $x - \alpha \in F[x]$, and thus $x - \alpha \mid f(x)$. \item $f(x) = g(x) h(x)$ for some non-constant $g(x), h(x) \in F[x]$, which is just by \hyperref[defn:irreducible]{definition}. \end{itemize} \paragraph{Showing irreducibility} Let $f(x) \in F[x]$. Note that it is possible to mix and match these tools to achieve our goal. Our trusty tools include: \begin{itemize} \item \textbf{\hyperref[propo:eisenstein_s_criterion]{Eisenstein}}: find a prime or irreducible polynomial (prime ideals) such that \begin{itemize} \item \hlimpo{each of the coefficients, except for the leading coefficient, is divisible by the prime (or irreducible)}, and \item \hlimpo{the square of the last coefficient is not divisible by the prime (or irreducible)}. \end{itemize} \item If $\deg f \in \{ 2, 3 \}$, then by \cref{propo:irreducible_rootless_polynomials}, we simply need to check that the polynomial does not have roots. This sounds crazy in an infinite field, but we can combine this with... \item \textbf{\hyperref[propo:mod_p_irreducibility_test]{Mod-$p$ irreducibility}}: find the equivalent $\tilde{f}(x) \in \mathbb{Z}_p[x]$: that is, let $\tilde{f}(x) \in \mathbb{Z}_p[x]$ be such that we replace each of the coefficients of $f$ are replaced with their counterparts in $\mathbb{Z}_p$. Then use the other methods here to check for irreducibility. \item By \hyperref[item:a3q2a]{A3Q2(a)}, we have that $f(x) \in F[x]$ is irreducible over $F[x]$ if $f(x + a)$ is irreducible over $F[x]$ for some $a \in F^\times$. \end{itemize} Some somewhat helpful heuristics that can reduce our work and/or make a problem much easier include: \begin{itemize} \item If our polynomial is of the form \begin{equation*} f(x) = x^{2n} + a_{2n-2}x^{2n-2} + \hdots + a_{2}x^2 + a_0, \end{equation*} then we can let $y = x^2$ and consider the polynomial \begin{equation*} g(y) = y^n + a_{2n-2}y^{n-1} + \hdots + a_2y + a_0. \end{equation*} Note that if $g(y)$ is reducible, then so is $f(x)$. Conversely, if we want $f(x)$ to be irreducible, it had better be the case that $g(y)$ is irreducible. \item Depending on where our polynomial comes from, the \hlimpo{graph} of the function may be helpful. For instance, the function $x^4 + 3$ looks like a quadratic graph that does not intersect the $x$-axis, so we know that it does not have real roots. \begin{itemize} \item We can do this even for polynomials that look like $x^4 - 2x^2 + 9$, since \begin{equation*} x^4 - 2x^2 + 9 = (x^2 - 1)^2 + 8 > 0. \end{equation*} \end{itemize} \end{itemize} % subsection summary_for_proving_irreducibility (end) % section irreducibles_continued_2 (end) \section{Field Extensions}% \label{sec:field_extensions} % section field_extensions Let $K$ be a field. Recall that a non-empty subset $F \subseteq K$ is called a \hldefn{subfield} of $K$ if $F$ is a field under the same operations. \begin{eg} $\mathbb{Q}(\sqrt{2}) := \left\{ a + b\sqrt{2} \mmid a, b \in \mathbb{Q} \right\}$ is a subfield of $\mathbb{C}$. We call this field $\mathbb{Q}$ `\hldefn{adjoin}' $\sqrt{2}$. \end{eg} \begin{note} We did not actually show that $\mathbb{Q}(\sqrt{2})$ is indeed a field but note the following: let $a + b \sqrt{2} \neq 0 \in \mathbb{Q}(\sqrt{2})$. Then \begin{equation*} \frac{1}{a + b\sqrt{2}} \cdot \frac{(a - b\sqrt{2})}{(a - b \sqrt{2})} = \frac{a - b \sqrt{2}}{a^2 - 2b^2} \in \mathbb{Q}(\sqrt{2}), \end{equation*} and note that \begin{equation*} a^2 - 2b^2 \neq 0 \iff \frac{a}{b} = \sqrt{2}, \end{equation*} which does not happen in $\mathbb{Q}$ itself. \end{note} \begin{defn}[Field Extension]\index{Field Extension}\label{defn:field_extension} Let $F$ be a field. A \hlnoteb{field extension} (or an \hlnoteb{extension}) of $F$ is a field $K$ which contains an \hlimpo{isomorphic} copy of $F$ as a subfield. We denote this notion of $K/F$. \end{defn} \begin{eg} \begin{itemize} \item We have that $\mathbb{C} / \mathbb{R}$ and $\mathbb{Q}(\sqrt{2}) / \mathbb{Q}$. \item For a prime $p$, if \begin{equation*} \mathbb{Z}_p(x) = \left\{ \frac{f(x)}{g(x)} \mmid f(x), g(x) \in \mathbb{Z}_p[x], g \neq 0 \right\}, \end{equation*} then $\mathbb{Z}_p(x) / \mathbb{Z}_p$. We call $\mathbb{Z}_p(x)$ the fraction field of $\mathbb{Z}_p$. \item Let $F$ be a field, and $f(x) \in F[x]$ be irreducible. Then let $K = F[x]/\langle f(x) \rangle$. Then $K / F$. \item Note that $\mathbb{Q}$ is not an extension of $\mathbb{Z}_p$ for any prime $p$, because they have \hlimpo{different characteristics}. In particular, $\mathbb{Q}$ subfields of $\mathbb{Q}$ must have the same characteristic as $\mathbb{Q}$. \end{itemize} \end{eg} \begin{note} Note that in the last example, $K$ is not a `direct' extension of $F$, but it contains an isomorphic copy of $F$. This allows us to have more flexibility in what we can do. \end{note} \begin{warning} If given $\mathbb{Z}_p = \{ 0, 1, 2, \ldots, p - 1 \}$, then $\mathbb{Q}$ is not an extension of $\mathbb{Z}_p$ since the two use different operations. \end{warning} % section field_extensions (end) % chapter lecture_7_jan_21st (end) \chapter{Lecture 8 Jan 23rd}% \label{chp:lecture_8_jan_23rd} % chapter lecture_8_jan_23rd \section{Field Extensions (Continued)}% \label{sec:field_extensions_continued} % section field_extensions_continued \begin{eg} Let $F$ be a field. \begin{itemize} \item If the characteristic $ch(F) = p > 0$ is a prime, then \begin{equation*} F \supset \{ 0, 1, 2, \ldots, p - 1 \} \simeq \mathbb{Z}_p. \end{equation*} Thus $F / \mathbb{Z}_p$. \item If $\ch(F) = 0$, then $F / \mathbb{Q}$. \end{itemize} In either of these cases, we call $\mathbb{Z}_p$ and/or $\mathbb{Q}$ the \hldefn{prime subfield} of $F$. \end{eg} \begin{defn}[Generated Field Extension]\index{Generated Field Extension}\label{defn:generated_field_extension} Let $K / F$, and $\alpha_1, \ldots, \alpha_n \in K$. The \hlnoteb{field extension of $F$ generated by $\{a_i\}_{i = 1}^{n}$} is \begin{equation*} F(\alpha_1, \ldots, \alpha_n) := \left\{ \frac{f(\alpha_1, \ldots, \alpha_n)}{g(\alpha_1, \ldots, \alpha_n)} \mmid f, g \in F[x_1, \ldots, x_n], g \neq 0 \right\}, \end{equation*} of which we call as $F$ \hldefn{adjoin} $\alpha_1, \ldots, \alpha_n$. \end{defn} \begin{note} We have that $F(\alpha_1, \ldots, \alpha_n) / F$, and in turn $K / F(\alpha_1, \ldots, \alpha_n)$. \end{note} \begin{remark}[Minimality]\label{remark:minimality_of_an_extension} Let $K / F$, and $\alpha_1, \ldots, \alpha_n \in K$. If we have $E / F$ such that $K / E$ and $\alpha_i \in E$ for all $i$, then \begin{equation*} F(\alpha_1, \ldots, \alpha_n) \subseteq E, \end{equation*} i.e. $F(\alpha_1, \ldots, \alpha_n)$ is the smallest extension of $F$ that contains the $\alpha_i$'s. \end{remark} \begin{eg}[A classical example of field extensions] Show that $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$. \end{eg} \begin{proof} Since $\sqrt{2}, \sqrt{3} \in \mathbb{Q}(\sqrt{2}, \sqrt{3})$, by closure, we have that $\sqrt{2} + \sqrt{3} \in \mathbb{Q}(\sqrt{2}, \sqrt{3})$, and so $\mathbb{Q}(\sqrt{2} + \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$. For the other direction, we have that $\sqrt{2} + \sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$. Then in particular $\frac{1}{\sqrt{2} + \sqrt{3}} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$. Notice that \begin{equation*} \frac{1}{\sqrt{2} + \sqrt{3}} \cdot \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}} = \sqrt{3} - \sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3}). \end{equation*} So $2 \sqrt{3}, 2 \sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ \sidenote{$2 \sqrt{2}$ follows from a similar argument by using $1 = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}$.}, and in turn $\sqrt{2}, \sqrt{3} \in \mathbb{Q}(\sqrt{2}, \sqrt{3})$. Then by minimality, $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2} + \sqrt{3})$. \end{proof} \begin{remark} Notice that $F(\alpha, \beta) = \left[ F(\alpha) \right](\beta)$. We have that $F(\alpha) \subseteq F(\alpha, \beta), \beta \in F(\alpha, \beta)$, which implies that $F(\alpha)(\beta) \subseteq F(\alpha, \beta)$ by minimality. Also, since $F \subseteq F(\alpha, \beta)$, and $\alpha, \beta \in F(\alpha, \beta)$, we have, by minimality (again), that $F(\alpha, \beta) \subseteq F(\alpha)(\beta)$. \end{remark} \begin{propo}[Span of the Extension]\label{propo:span_of_the_extension} Let $K / F$ and $\alpha \in K$. If $\alpha$ is a root of some non-zero $f(x) \in F[x]$ irreducible over $F$, then $F(\alpha) \simeq F[x] / \langle f(x) \rangle$. Moreover, if $\deg f = n$, then \begin{equation*} F(\alpha) = \Span_F \{ 1, \alpha, \ldots, \alpha^{n - 1} \}. \end{equation*} \end{propo} \begin{proof} Sps $\alpha \in K$ is a root of an irreducible $f(x) \in F[x]$ over $F$. Let $\deg f = n \in \mathbb{N}$. Define $\phi : F[x] \to F(\alpha)$ by $\phi(g(x)) = g(\alpha)$. Note that this is a ring homomorphism. Let \begin{equation*} I = \{ g(x) \in F[x] \mid g(\alpha) = 0 \} = \ker \phi, \end{equation*} which is an ideal. Since $F[x]$ is a PID \sidenote{See \href{https://tex.japorized.ink/PMATH347S18/classnotes.tex}{PMATH347}.}, $\exists g(x) \in F[x]$ such that $I = \langle g(x) \rangle$. Since $\alpha$ is a root of $f(x)$, $f(x) \in I$, and so $f(x) = g(x) h(x)$ for some $h(x) \in F[x]$. Since $I \neq F[x]$ and $f$ is irreducible, $h(x) \in F^\times$. Thus $\langle g(x) \rangle = \langle g(x) \rangle$. Then by the \hlnotea{First Isomorphism Theorem}, \begin{equation*} F[x] / \langle f(x) \rangle \simeq \phi(F[x]). \end{equation*} By construction, $\phi(F[x]) \subseteq F(\alpha)$. Since $\phi(F[x])$ is a field (by isomorphism) which contains $\alpha = \phi(x)$ and $F$, and so by minimality $F(\alpha) \subseteq \phi(F[x])$. Therefore \begin{equation*} F[x]/ \langle f(x) \rangle \simeq F(\alpha), \end{equation*} as required. Through the isomorphism, for any $h(x) \in F[x]$, we have \begin{equation*} h(x) + \langle f(x) \rangle \mapsto h(\alpha). \end{equation*} So \begin{equation*} F[x] / \langle f(x) \rangle = \left\{ c_{n - 1} x^{n - 1} + \hdots + c_1 x + c_0 + \langle f(x) \rangle \mmid c_i \in F \right\} \end{equation*} and thus \begin{align*} F(\alpha) &= \left\{ c_{n - 1} \alpha^{n - 1} + \hdots + c_1 \alpha + c_0 + \langle f(x)\rangle \mmid c_i \in F \right\} \\ &= \Span_F \left\{ 1, \alpha, \ldots, \alpha^{n - 1} \right\}, \end{align*} as claimed. \end{proof} % section field_extensions_continued (end) % chapter lecture_8_jan_23rd (end) \chapter{Lecture 9 Jan 25th}% \label{chp:lecture_9_jan_25th} % chapter lecture_9_jan_25th \section{Field Extensions (Continued 2)}% \label{sec:field_extensions_continued_2} % section field_extensions_continued_2 Let $K / F$, and $0 \neq g(x) \in F[x]$, and $\alpha \in K$ such that $g(\alpha) = 0$. Since $F[x]$ is an ID, $g(x)$ must have an irreducible factor $f(x) \in F[x]$ such that $f(\alpha) = 0$. By the proof of \cref{propo:span_of_the_extension}, \begin{equation*} \langle f(x) \rangle = \ker \phi = I = \{ h(x) \in F[x] \mid h(\alpha) = 0 \}. \end{equation*} In particular, \begin{itemize} \item If $h(x) \in F[x]$ such that $h(\alpha) = 0$, then $h(x) \in \langle f(x) \rangle$. In particular, $f(x) \mid h(x)$. \item $\langle f(x) \rangle$ contains a unique, monic, irreducible polynomial: for any $g(x) \in \langle f(x) \rangle$ that is irreducible, we know that $g(x) = uf(x)$, where $0 \neq u \in F^\times$, and so we can just divide the polynomial $g$ by $u$ to make it monic. \end{itemize} \begin{defn}[Minimal Polynomial]\index{Minimal Polynomial}\label{defn:minimal_polynomial} Let $K / F$, and $\alpha \in K$ be a root of a non-zero polynomial in $F[x]$. Then there exists a unique irreducible monic polynomial $f(x) \in F[x]$ such that $f(\alpha) = 0$. We call this $f(x)$ the \hlnoteb{minimal polynomial} for $\alpha$ over $F$. If $\deg f = n$, we call $n$ the \hldefn{degree} of $\alpha$ over $F$, denoted $\deg_F(\alpha)$. \end{defn} \begin{note} For an $\alpha \in K$, its minimal polynomial is unique, but a minimal polynomial need not have only one root. \end{note} \begin{propo}[Span of an Extension if Linearly Independent]\label{propo:span_of_an_extension_if_linearly_independent} Let $K / F$, and $\alpha \in K$ with minimal polynomial $f(x) \in F[x]$, with $\deg_F(\alpha) = n$. Then the span $F(\alpha) = \Span_F \{ 1, \alpha, \ldots, \alpha^{n - 1} \}$ is linearly independent over $F$. \end{propo} \begin{proof} Sps to the contrary that \begin{equation*} c_{n - 1} \alpha^{n - 1} + c_{n - 2} \alpha^{n - 2} + \hdots + c_1 \alpha + c_0 = 0, \; c_i \in F, \end{equation*} has a non-trivial solution, i.e. not all $c_i$'s are $0$ (i.e. we assume that the $\alpha$'s are linearly dependent). Consider \begin{equation*} g(x) = c_{n - 1} x^{n - 1} + \hdots + c_1 x + c_0, \end{equation*} and so $g \neq 0$. However, $g(\alpha) = 0$, so $g(x) \in \langle f(x) \rangle$, i.e. $f(x) \mid g(x)$. However, that contradicts the fact that $\deg f = n > n - 1 \geq \deg g$. \end{proof} \begin{eg} Consider $K / F$, and $\alpha \in K$. Then \begin{equation*} \deg_F(\alpha) = 1 \iff \text{ min. polym } f(x) = x - \alpha \in F[x] \iff \alpha \in F. \end{equation*} \end{eg} \begin{eg} Consider $\mathbb{Q}(\sqrt{2}) / \mathbb{Q}$. Let $\alpha = \sqrt{2}$. Note that $f(\alpha) = 0$ for $f(x) = x^2 - 2$, which is irreducible by Eisenstein by $P = \langle 2 \rangle$. Thus $\deg_F(\alpha) = 2$, and so \begin{equation*} \mathbb{Q}(\sqrt{2}) = \Span_{\mathbb{Q}} \{ 1, \alpha \} = \{ a + b \sqrt{2} \mid a, b \in \mathbb{Q} \}. \end{equation*} \end{eg} \marginnote[-150pt]{ \begin{procedure}[Computing Degree of an Algebraic Number]\label{procedure:computing_degree_of_an_algebraic_number} You are usually given some algebraic number whose degree you are told to compute. \begin{enumerate} \item If it looks like the given number can be brought back into the base field after taking powers and some algebraic manipulation, then let $\alpha$ be that element. Do what you noticed until all the instances of elements not in the base field are gone. Try to take a little steps as possible. \item Use that formula while replacing each instance of $\alpha$ with $x$. This will give you a polynomial where the given number is a root. \item Check if this polynomial is minimal. If not, recompute. \end{enumerate} \end{procedure} } \begin{eg} Let $\alpha = \sqrt{1 + \sqrt{3}}$. Notice that $\alpha^2 = 1 + \sqrt{3}$, and so $(\alpha^2 - 1)^2 = 3$. Thus \begin{equation*} \alpha^4 - 2 \alpha^2 + 1 - 3 = 0. \end{equation*} Let $f(x) = x^4 - 2x^2 - 2 \in \mathbb{Q}[x]$. Note that $f$ is monic and $f(\alpha) = 0$. By Eisenstein, $f$ is irreducible if we pick $P = \langle 2 \rangle$ \sidenote{From hereon, we shall just say, for instance, that ``the polynomial is irreducible by $2$-Eisenstein.''}. Thus $f$ is a minimal polynomial for $\alpha$. We have that \begin{equation*} \deg_{\mathbb{Q}}(\alpha) = \deg f = 4. \end{equation*} \end{eg} \marginnote{ \begin{procedure}[Computing Finite Fields of Given Size]\label{procedure:computing_finite_fields_of_given_size} We will usually be given a number $m$ that is the power of a single prime number $p$. \begin{enumerate} \item Determine $n$ for $p^n = m$. \item Find an \hyperref[defn:irreducible]{irreducible polynomial} of degree $n$ in $\mathbb{Z}_p[x]$. \item Pick one of the roots $\alpha$ and adjoin it to $\mathbb{Z}_p$, i.e. consider \begin{equation*} \mathbb{Z}_p(\alpha) = \Span_{\mathbb{Z}_p} \{1, \alpha, \alpha^2, \ldots, \alpha^{n-1} \}. \end{equation*} This field will have order $m$. \end{enumerate} \end{procedure} } \begin{eg}\label{eg:computing_finite_fields} Let $f(x) = x^3 + x + 1 \in \mathbb{Z}_2[x]$. Let $\alpha$ be a root of $f(x)$ in some extension of $\mathbb{Z}_2$. Compute the size of $\mathbb{Z}_2(\alpha)$. \end{eg} \begin{solution} We showed in one of our previous examples that such an $f$ is irreducible in $\mathbb{Z}_2$. Thus $\deg_{\mathbb{Z}_2}(\alpha) = 3$. Then \begin{equation*} \mathbb{Z}_2(\alpha) = \Span_{\mathbb{Z}_2} \{ 1, \alpha, \alpha^2 \}, \end{equation*} where $\{ 1, \alpha, \alpha^2 \}$ is linearly independent over $\mathbb{Z}_2$. Thus \begin{equation*} \abs{\mathbb{Z}_2(\alpha)} = 2 \times 2 \times 2 = 8, \end{equation*} since each of these elements have agree $2$. \end{solution} \begin{note} Up till now, we never guaranteed that such a root exists, but it does, which is a theorem that we shall prove later. (See \hyperref[thm:kronecker_s_theorem]{Kronecker's Theorem}) \end{note} \begin{crly}[Isomorphism between Extensions]\label{crly:isomorphism_between_extensions} Let $K / F$ and $\alpha, \beta \in K$ have the same minimal polynomial $f(x) \in F[x]$. Then $F(\alpha) \simeq F(\beta)$. \end{crly} \begin{proof} From \cref{propo:span_of_the_extension}, we have that \begin{equation*} F(\alpha) \simeq F[x] / \langle f(x) \rangle \simeq F(\beta). \end{equation*} \end{proof} % section field_extensions_continued_2 (end) % chapter lecture_9_jan_25th (end) \chapter{Lecture 10 Jan 28th}% \label{chp:lecture_10_jan_28th} % chapter lecture_10_jan_28th \section{Field Extensions (Continued 3)}% \label{sec:field_extensions_continued_3} % section field_extensions_continued_3 How can we work with field extensions algebraically? \subsection{Linear Algebra on Field Extensions}% \label{sub:linear_algebra_on_field_extensions} % subsection linear_algebra_on_field_extensions We can look at $K / F$ as $K$ being an $F$-vector space. \begin{defn}[Finite Extension]\index{Finite Extension}\label{defn:finite_extension} We say $K / F$ is a \hlnoteb{finite extension} if $K$ is a finite dimensional $F$-vector space. We call the dimension, $\dim_F K$, the \hldefn{degree} of $K / F$, and denote this dimension as \begin{equation*} [ K : F ]. \end{equation*} \end{defn} \begin{eg} We have $[ \mathbb{C} : \mathbb{R} ] = \abs{ \{ 1, i \} } = 2$. \end{eg} \begin{eg} $[ \mathbb{R} : \mathbb{Q} ] = \infty$. \end{eg} \begin{eg} Let $K / F$ and $\alpha \in K$ with the minimal polynomial $f(x) \in F[x]$. Then $[ F(\alpha) : F ] = \abs{ \{ 1, \alpha, \ldots, \alpha^{n - 1} \} } = n$, where $n = \deg f = \deg_F(\alpha)$.\sidenote{This is why we call the dimension of $K/F$ as a degree.} \end{eg} \begin{defn}[Tower of Fields]\index{Tower of Fields}\label{defn:tower_of_fields} We say $F_1 / F_2 / F_3 / \hdots / F_n$ is a \hlnoteb{tower of fields} if each $F_i / F_{i + 1}$ is a field extension. \end{defn} \begin{thm}[Tower Theorem]\index{Tower Theorem}\label{thm:tower_theorem} If $K / E$ and $E / F$ are finite extensions, then \begin{equation*} [K : F] = [K : E] [E : F]. \end{equation*} \end{thm} \begin{proof} Let $\mathcal{B}_v = \left\{ v_1, \ldots, v_n \right\}$ be a basis for $K / E$ and $\mathcal{B}_w = \left\{ w_1, \ldots, w_m \right\}$ be a basis for $E / F$. \noindent \hlbnoted{Claim} The set $\left\{ v_i w_j : 1 \leq i \leq n, 1 \leq j \leq m \right\}$ is a basis for $K / F$. \noindent \hlbnotea{Linear Independence} Assume \begin{equation}\label{eq:tower_of_fields_eq1} \sum_{i, j} c_{i, j} w_j v_i = 0. \end{equation} Notice that we may write \cref{eq:tower_of_fields_eq1} as \begin{equation*} \sum_{i} \left( \sum_{j} c_{i, j} w_j \right) v_i = 0. \end{equation*} Since $\mathcal{B}_v$ is a basis of $K / E$, for each $i$, we have \begin{equation*} \sum_{j} c_{i, j} w_j = 0. \end{equation*} Since $\mathcal{B}_w$ is a basis for $E / F$, for each $j$, we have \begin{equation*} c_{i, j} = 0. \end{equation*} It follows that the $w_j v_i$'s are linearly independent of each other. \noindent \hlbnotea{Span} Let $u \in K$. Then \begin{equation*} u = \sum_{i=1}^{n} c_i v_i, \end{equation*} where $c_i \in E$ is given by \begin{equation*} c_i = \sum_{j=1}^{m} d_{i, j} w_j. \end{equation*} Then \begin{equation*} u = \sum_{i, j} d_{i, j} w_j v_i. \end{equation*} Thus $\{ v_i, w_j \}$ is a basis for $K / F$. \end{proof} \begin{eg} Compute $[\mathbb{Q}(\sqrt[3]{5}, i) : \mathbb{Q}]$. \end{eg} \begin{solution} By the \hyperref[thm:tower_theorem]{Tower Theorem}, we have that \begin{equation*} [\mathbb{Q}(\sqrt[3]{5}, i) : \mathbb{Q}] = [\mathbb{Q}(\sqrt[3]{5})(i) : \mathbb{Q}(\sqrt[3]{5})] \cdot [\mathbb{Q}(\sqrt[3]{5}) : \mathbb{Q}]. \end{equation*} Notice that \begin{equation*} [\mathbb{Q}(\sqrt[3]{5}) : \mathbb{Q}] = \deg ( x^3 - 5 ) = 3. \end{equation*} For $[\mathbb{Q}(\sqrt[3]{5})(i) : \mathbb{Q}(\sqrt[3]{5})]$, let $p(x)$ be the minimal polynomial for $i$ over $\mathbb{Q}(\sqrt[3]{5})$. Since $i^2 + 1 = 0$, we know that $i$ is a root of $x^2 + 1 = 0$. Then in particular, we must have $p(x) \mid x^2 + 1$. So $\deg p \in \{ 1, 2 \}$. Now since $\mathbb{Q}(\sqrt[3]{5}) \subseteq \mathbb{R}$ and $i \notin \mathbb{Q}(\sqrt[3]{5})$, we observe that $\deg p \neq 1$. Thus $\deg p = 2$. It follows that \begin{equation*} [\mathbb{Q}(\sqrt[3]{5})(i) : \mathbb{Q}(\sqrt[3]{5})] = 2. \end{equation*} Therefore \begin{equation*} [\mathbb{Q}(\sqrt[3]{5}, i) : \mathbb{Q}] = 2 \cdot 3 = 6. \end{equation*} \end{solution} % subsection linear_algebra_on_field_extensions (end) \subsection{Polynomials on Field Extensions}% \label{sub:polynomials_on_field_extensions} % subsection polynomials_on_field_extensions \begin{defn}[Algebraic and Transcendental]\index{Algebraic}\index{Transcendental}\label{defn:algebraic_and_transcendental} Let $K / F$. We say that $\alpha \in K$ is \hlnoteb{algebraic} over $F$ if $\exists 0 \neq f(x) \in F[x]$ such that $f(\alpha) = 0$. Otherwise, we say that $\alpha$ is \hlnoteb{transcendental} over $F$; that is, there is no non-zero polynomial over $F$ such that $\alpha$ is a root. We say that $K / F$ is algebraic if every $\alpha \in K$ is \hlnoteb{ algebraic } over $F$. Otherwise, we say that $K / F$ is \hlnoteb{transcendental}. \end{defn} \begin{eg} $\pi$ is transcendental over $\mathbb{Q}$ \sidenote{The proof of this statement is beyond our power at this point.}. However, $\pi$ is algebraic over $\mathbb{R}$ (note that $x - \pi \in \mathbb{R}[x]$.). \end{eg} \begin{eg} As a direct consequence of the above example, we have that $\mathbb{R}/\mathbb{Q}$ is transcendental. \end{eg} \begin{eg} As we have seen numerous times, $\mathbb{Q}(\sqrt{2}) / \mathbb{Q}$ is algebraic. \end{eg} \begin{remark} If $\alpha \in K$ is algebraic over $F$, then $\alpha$ has a minimal polynomial in $F[x]$. \end{remark} \begin{thm}[Finite Extensions are Algebraic]\label{thm:finite_extensions_are_algebraic} If $K / F$ is finite, then $K / F$ is algebraic. \end{thm} \marginnote{ \begin{strategy} The idea is to make use of the fact that the extension will at least have the algebraic number as a span up to some degree $n$, and instead of working with the spanning set, we work with one $\alpha$ away. There will be two cases, each of which can be dealt with at relative ease. \end{strategy} } \begin{proof} Suppose $[K : F] = n < \infty$. Let $\alpha \in K$. Consider \begin{equation*} \alpha, \alpha^2, \ldots, \alpha^n, \alpha^{n + 1}. \end{equation*} \noindent \hlbnotea{Case 1} Suppose $\alpha^i = \alpha^j$ for some $i \neq j \in \{ 1, \ldots, n + 1\}$. Then $\alpha$ is certainly a root of $f(x) = x^i - x^j$. \noindent \hlbnotea{Case 2} Suppose $\alpha^i \neq \alpha^j$ for all $i \neq j$. Then we must have that \begin{equation*} \{\alpha, \alpha^2, \ldots, \alpha^n, \alpha^{n + 1}\} \end{equation*} is linearly dependent over $F$. Thus we may have \begin{equation*} c_1 \alpha + c_2 \alpha^2 + \hdots + c_{n + 1} \alpha^{n + 1} = 0 \end{equation*} where not all $c_i$'s are $0$. Then $\alpha$ is a root of \begin{equation*} f(x) = c_{n+1} x^{n+1} + \hdots + c_1 x, \end{equation*} which is a non-zero polynomial. In either case, we observe that $\alpha$ is algebraic over $F$. Therefore $K / F$ is algebraic. \end{proof} % subsection polynomials_on_field_extensions (end) % section field_extensions_continued_3 (end) % chapter lecture_10_jan_28th (end) \chapter{Lecture 11 Jan 30th}% \label{chp:lecture_11_jan_30th} % chapter lecture_11_jan_30th \section{Field Extensions (Continued 4)}% \label{sec:field_extensions_continued_4} % section field_extensions_continued_4 \subsection{Polynomials on Field Extensions (Continued)}% \label{sub:polynomials_on_field_extensions_continued} % subsection polynomials_on_field_extensions_continued \begin{note} Recall that given $K / F$, \begin{itemize} \item Finite (defn): $\dim_F K = [ K : F ] < \infty$ \item Algebraic (defn) :$\forall \alpha \in K, \exists 0 \neq f \in F[x]$, such that $f(\alpha) = 0$ \item Finite $\implies$ Algebraic \end{itemize} \end{note} \begin{defn}[Finitely Generated Extension]\index{Finitely Generated Extension}\label{defn:finitely_generated_extension} We say $K$ is a \hlnoteb{finitely generated extension} of $F$ if $\exists \alpha_1, \alpha_2, \ldots, \alpha_n \in K$ such that $K = F(\alpha_1, \ldots, a_n)$. \end{defn} \begin{propo}[Finitely Generated Algebraic Extensions are Finite]\label{propo:finitely_generated_algebraic_extensions_are_finite} If $K$ is a finitely generated algebraic extension of $F$, then $K / F$ is finite.\sidenote{This proposition is actually an \hlnotea{iff} statement in disguise.} \end{propo} \begin{proof} Sps $K / F$ is algebraic, where $K = F(\alpha_1, \ldots, \alpha_n)$. We shall proceed by performing induction on $n$. If $n = 1$, then $[F(\alpha_1) : F] = \deg_F(\alpha_1) < \infty$. Now suppose that the result holds for $n$. Consider \begin{equation*} K = F(\alpha_1, \ldots, \alpha_n, \alpha_{n + 1}). \end{equation*} Then by the \hyperref[thm:tower_theorem]{Tower Theorem}, \begin{align*} &[F(\alpha_1, \ldots, \alpha_n, \alpha_{n + 1}) : F] \\ &= [F(\alpha_1, \ldots, \alpha_{n})(\alpha_{n + 1}) : F(\alpha_1, \ldots, \alpha_n)] \cdot [F(\alpha_1, \ldots, \alpha_n) : F]. \end{align*} It follows from the base case and the induction hypothesis that \\ \noindent $[F(\alpha_1, \ldots, \alpha_{n+1}):F]$ is finite. \end{proof} \begin{note} Finite extensions are, therefore, finitely generated. \end{note} \begin{warning} The field $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{4}, \ldots)$ is an algebraic extension of $\mathbb{Q}$ but it is not a finite extension. \end{warning} \begin{propo}[Greater Algebraic Extensions]\label{propo:greater_algebraic_extensions} If $K / E$ and $E / F$ are algebraic extensions, then $K / F$ is an algebraic extension. \end{propo} \begin{proof} Let $\alpha \in K$. Since $K / E$ is algebraic, $\alpha$ has a minimal polynomial in $E[x]$, say it is \begin{equation*} p(x) = x^n + c_{n - 1} x^{n - 1} + \hdots + c_1 x + c_0. \end{equation*} It is clear that $p(x) \in F(c_{n-1}, \ldots, c_0)[x] \subseteq E[x]$. Further, we see that $\alpha$ is algebraic over $F(c_{n - 1}, \ldots, c_1, c_0)$. Consider $F(c_{n-1}, \ldots, c_0, \alpha)$. Since $F(c_{n-1}, \ldots, c_0, \alpha) / F(c_{n-1}, \ldots, c_0)$ is finitely generated, by \cref{propo:finitely_generated_algebraic_extensions_are_finite}, \begin{equation*} [F(c_{n - 1}, \ldots, c_1, c_0, \alpha) : F(c_{n - 1}, \ldots, c_0)] \text{ is finite}. \end{equation*} It is clear that $F(c_{n-1}, \ldots, c_0, \alpha) / F$ is finitely generated, i.e. \begin{equation*} [F(c_{n - 1}, \ldots, c_0, \alpha) : F] < \infty. \end{equation*} Then $\alpha$ has a minimal polynomial $q(x) \in F[x]$, i.e. $\alpha$ is algebraic over $F$. \end{proof} \begin{propo}[Algebraic Numbers Form a Subfield]\label{propo:algebraic_numbers_form_a_subfield} Let $K / F$. The set of elements of $K$ algebraic over $F$ form a subfield of $K$. \end{propo} \begin{proof}[Sketch proof] Let $L = \{ \alpha \in K : \alpha \text{ is alg. over } F \}$. Let $\alpha, \beta \in L$ and $\beta \neq 0$. Then \begin{equation*} \alpha, \beta, \alpha + \beta, \alpha \beta, \beta^{-1} \in F(\alpha, \beta). \end{equation*} Then $[F(\alpha, \beta) : F] < \infty$ implies that $L$ is finitely generated, which is thus algebraic, and is hence a subfield of $K$. \end{proof} % subsection polynomials_on_field_extensions_continued (end) % section field_extensions_continued_4 (end) \section{Splitting Fields}% \label{sec:splitting_fields} % section splitting_fields From various examples in the past, we notice that many of the roots that we have come across live in $\mathbb{C}$. We shall see why later on, but we can ask ourselves if we can generalize this notion and make use of properties from this notion. \begin{defn}[Splits]\index{Splits}\label{defn:splits} Let $f(x) \in F[x]$ be non-constant. We say $f(x)$ \hlnoteb{splits} in an extension $K/F$ if there exists $\exists u \in F$, and $\exists \alpha_1, \ldots, \alpha_n \in K$ such that \begin{equation*} f(x) = u(x - \alpha_1) \hdots (x - \alpha_n). \end{equation*} \end{defn} \begin{eg} Every non-constant polynomial in $\mathbb{R}[x]$ splits in $\mathbb{C}$. \end{eg} \marginnote{Kronecker's Theorem is much more profound than what meets the eye here. See \href{http://feyzioglu.boun.edu.tr/book/chapter5/ch5(51).pdf}{this PDF}. \hlwarn{Notes will be added at a later time.}} % TODO : expand on Kronecker's Theorem \begin{thm}[Kronecker's Theorem]\index{Kronecker's Theorem}\label{thm:kronecker_s_theorem} Let $f(x) \in F[x]$ be non-constant. There exists an extension $K / F$ such that $f(x)$ has a root in $K$. \end{thm} \begin{proof} Let $f(x) \in F[x]$ be non-constant. Then let $p(x) \in F[x]$ be an irreducible factor of $f(x)$. Then consider $K = F[t] /
$. which we know is a field. Then \begin{equation*} \bar{t} = t + p(t) \in K \end{equation*} is a root of $p(x)$, which means that $\bar{t}$ is also a root for $f(x)$. \end{proof} \begin{thm}[Repeated Kronecker's Theorem]\label{thm:repeated_kronecker_s_theorem} Let $f(x) \in F[x]$ be non-constant. Then there exists an extension $K / F$ such that $f(x)$ splits over $K$. \end{thm} \begin{proof} By the Fundamental Theorem of Algebra, if we suppose that $\deg f = n < \infty$, then $f$ has $n$ roots. Consequently, we need only to apply \cref{thm:kronecker_s_theorem} for at most $n$-many times to get to an extension where $f(x)$ splits. \end{proof} % section splitting_fields (end) % chapter lecture_11_jan_30th (end) \chapter{Lecture 12 Feb 01st}% \label{chp:lecture_12_feb_01st} % chapter lecture_12_feb_01st \section{Splitting Fields (Continued)}% \label{sec:splitting_fields_continued} % section splitting_fields_continued \begin{defn}[Splitting Field]\index{Splitting Field}\label{defn:splitting_field} Let $f(x) \in F[x]$ be non-constant. A \hlnotea{minimal extension} $K$ of $F$ with the property that $f(x)$ splits over $K$ is called a \hlnoteb{splitting field} for $f(x)$ over $F$. \end{defn} The following result is a direct consequence of \cref{thm:repeated_kronecker_s_theorem}. \begin{propo}[A Splitting Field is Generated]\label{propo:a_splitting_field_is_generated} Let $f(x) \in F[x]$ be non-constant, and let $K / F$ be such that $f(x)$ splits over $K$. Suppose \begin{equation*} f(x) = u(x - \alpha_1) \hdots (x - \alpha_n), \end{equation*} where $u \in F$ and $\alpha_1, \ldots, \alpha_n \in K$. Then a splitting field for $f(x)$ over $F$ is $F(\alpha_1, \ldots, \alpha_n)$. \end{propo} \begin{eg} Find a splitting field for \begin{equation*} f(x) = x^4 + x^2 - 6 \end{equation*} over $\mathbb{Q}$. \end{eg} \begin{solution} Notice that \begin{equation*} f(x) = (x^2 + 3)(x^2 - 2) = (x + \sqrt{3} i)(x - \sqrt{3}i)(x - \sqrt{2})(x + \sqrt{2}) \end{equation*} in $\mathbb{C}[x]$. Then a splitting field of $f(x)$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt{2}, \sqrt{3}i)$. \end{solution} \newthought{Now what if} we had two differing extensions at which $f(x)$ splits, say $K$ and $E$, and $K$ and $E$ are not the same field extension? In particular, $K$ and $E$ would contain some subfield, say $F(\alpha_1, \ldots, \alpha_n)$ and $F(\beta_1, \ldots, \beta_n)$ respectively, which may not be the same spliting field. How are these splitting fields related? \begin{marginfigure} \centering \resizebox{\marginparwidth}{!}{ \begin{tikzpicture} \node[label={270:{$f(x) \in F[x]$}}] at (0, 0) {}; \node[label={90:{$f(x)$ splits in $K$}}] at (-2, 2) {}; \node[label={90:{$f(x)$ splits in $E$}}] at (2, 2) {}; \draw (-1, 0) -- (-2, 2) node[midway,color=foreground,fill=background] {$F(\alpha_1, \ldots, \alpha_n)$}; \draw (1, 0) -- (2, 2) node[midway,color=foreground,fill=background] {$F(\beta_1, \ldots, \beta_n)$}; \draw[latex-latex] (-0.5, 1.3) to[bend left] (0.5, 1.3) node[midway,above=1.5] {relation?}; \end{tikzpicture}} \caption{Differing Splitting Fields} \label{fig:differing_splitting_fields} \end{marginfigure} \begin{lemma}[Isomorphic Fields have Isomorphic Polynomial Rings]\label{lemma:isomorphic_fields_have_isomorphic_polynomial_rings} Let $F$ ad $F'$ be fields. If $\phi : F \to F'$ is an isomorphism, there exists a map $\tilde{\phi} : F[x] \to F'[x]$ that is also an isomorphism. \end{lemma} \begin{proof} The map $\tilde{\phi} : F[x] \to F'[x]$ given by \begin{equation*} \tilde{\phi}(\alpha_n x^n + \hdots + \alpha_1 x + \alpha_0) = \tilde{\alpha}_n x^n \hdots + \tilde{\alpha}_1 x + \tilde{\alpha}_0 \end{equation*} is clearly an isomorphism between $F[x]$ and $F'[x]$. \end{proof} \begin{note} Since there is no difference between talking about $\phi$ and $\tilde{\phi}$, we shall freely write $\tilde{\phi}$ as $\phi$ without remorse. \end{note} \begin{lemma}[Isomorphism Extension Lemma]\index{Isomorphism Extension Lemma}\label{lemma:isomorphism_extension_lemma} Let $F$ and $F'$ be fields, $\phi : F[x] \to F'[x]$ be an isomorphism, $f(x) \in F[x]$ be irreducible, $\alpha$ be a root of $f(x)$ in an extension of $F$, and $\beta$ be a root of $f(x)$ in an extension of $F'$. Then there exists an isomorphism $\psi : F(\alpha) \to F'(\beta)$ such that $\psi \restriction_F = \phi$. Moreover, $\psi(\alpha) = \beta$. \end{lemma} \begin{proof}[Sketch] Using the \hlnotea{First Isomorphism Theorem} to find $\rho_1$ and $\rho_2$, we have \begin{equation*} F(\alpha) \overset{\rho_1}{\to} \quotient{F[x]}{\langle f(x) \rangle} \overset{\sigma}{\to} \quotient{F'[x]}{\langle \phi(f(x)) \rangle} \overset{\rho_2}{\to} F'(\beta), \end{equation*} \sidenote{ \begin{ex} Prove that $\phi(f(x))$ is irreducible. \end{ex} } where $\sigma(\bar{g(x)}) = \bar{\phi(g(x))}$. \sidenote{ \begin{ex} Prove that $\sigma$ is an isomorphism. \end{ex} } Then $\psi = \rho_2 \circ \sigma \rho_1 : F(\alpha) \to F'(\beta)$ is an isomorphism. Let $a \in F$. Then \begin{equation*} \psi(a) = \rho_2 \circ \sigma \circ \rho_1 (a) = \rho_2 \circ \sigma(\bar{a}) = \rho_2(\bar{\phi(a)}) = \phi(a). \end{equation*} Also, \begin{equation*} \psi(\alpha) = \rho_2 \circ \sigma \circ \rho_1(\alpha) = \rho_2 \circ \sigma(\bar{x}) = \rho_2(\bar{\phi(x)}) = \rho_2(\bar{x}) = \beta. \end{equation*} \end{proof} It follows from induction that \begin{lemma}[Extended Isomorphism Extension Lemma]\label{lemma:extended_isomorphism_extension_lemma} Let $F$ be a field, $f(x) \in F[x]$ non-constant, $K$ a splitting field for $f(x)$ over $F$, $F'$ a field, $\phi : F \to F'$ an isomorphism, and $K'$ a splitting field for $\phi(f(x))$ over $F'$. Then there is an isomorphism $\psi : K \to K'$ such that $\psi \restriction_F = \phi$. \end{lemma} \begin{crly}[Splitting Fields are Unique up to Isomorphism]\label{crly:splitting_fields_are_unique_up_to_isomorphism} Let $f(x) \in F[x]$ be non-constant. If $K$ and $K'$ are splitting fields for $f(x)$ over $F$, then $K \cong K'$. \end{crly} \begin{proof} Consider $\phi = \id$ and use \cref{lemma:extended_isomorphism_extension_lemma}. \end{proof} % section splitting_fields_continued (end) \section{Algebraic Closures}% \label{sec:algebraic_closures} % section algebraic_closures We talked about algebraicity, and it makes sense asking about where exactly `upstairs' that we will be able to find all of the algebraic numbers over our given field. A lot of the machinery has been taken care of with the introduction of \hyperref[sec:splitting_fields]{splitting fields}. \begin{defn}[Algebraic Closures]\index{Algebraic Closures}\label{defn:algebraic_closures} A field $\bar{F}$ is an \hlnoteb{algebraic closure} of a field $F$ if \begin{enumerate} \item $\bar{F} / F$ is algebraic; and \item it is the smallest extension such that every non-constant $f(x) \in F[x]$ splits over $\bar{F}$. \end{enumerate} \end{defn} \begin{eg} $\mathbb{C}$ is an algebraic closure for $\mathbb{R}$. \end{eg} \begin{eg} $\mathbb{C}$ is \hlimpo{not} an algebraic closure for $\mathbb{Q}$. This is because $\mathbb{C}$ contains $\pi$, which is transcendental to $\mathbb{Q}$. Similarly so, $\mathbb{R}$ is \hlimpo{not} an algebraic closure for $\mathbb{Q}$. \end{eg} \begin{defn}[Algebraically Closed]\index{Algebraically Closed}\label{defn:algebraically_closed} A field $F$ is \hlnoteb{algebraically closed} if every non-constant $f(x) \in F[x]$ has a root in $F$. \end{defn} \begin{remark} If $F$ is algebraically closed, then every non-constant $f(x) \in F[x]$ splits over $F$. \end{remark} \begin{eg}[Fundamental Theorem of Algebra]\label{eg:fundamental_theorem_of_algebra} $\mathbb{C}$ is algebraically closed. This is also called the \hlnoteb{Fundamental Theorem of Algebra}. It is equivalent to saying that for any polynomial $f(x) \in \mathbb{C}[x]$, if $\deg f(x) = n \in \mathbb{N}$, then $\exists \alpha_1, \ldots, \alpha_n \in \mathbb{C}$ such that $f(\alpha_i) = 0$ for all $i \in \{1, \ldots, n\}$. \end{eg} \begin{note} It is common to see the Fundamental Theorem of Algebra proven in \hlnotea{Complex Analysis}. It is possible to prove it using almost solely concepts from Abstract Algebra. However, we are nowhere close to being able to prove this fundamental theorem at this point of the course. (See \hyperref[item:a10q3]{A10Q3}) \end{note} % section algebraic_closures (end) % chapter lecture_12_feb_01st (end) \chapter{Lecture 13 Feb 04th}% \label{chp:lecture_13_feb_04th} % chapter lecture_13_feb_04th \section{Algebraic Closures (Continued)}% \label{sec:algebraic_closures_continued} % section algebraic_closures_continued This may seem obvious from the names (closure, closed?), but it is actually not immediately clear that algebraic closures are algebraically closed. \begin{propo}[Algebraic Closures are Algebraically Closed]\label{propo:algebraic_closures_are_algebraically_closed} If $\bar{F}$ is an algebraic closure for $F$, then $\bar{F}$ is algebraically closed. \end{propo} \begin{proof} Let $f(x) \in \bar{F}[x]$ be non-constant. Then by \hyperref[thm:kronecker_s_theorem]{Kronecker's Theorem}, $f(x)$ has a root $\alpha$ in some extension of $\bar{F}$. Since $\bar{F}(\alpha) / \bar{F}$ is algebraic and $\bar{F} / F$ is also algebraic, we have that $\bar{F}(\alpha) / F$ is algebraic. Thus $\alpha$ is a root of some $p(x) \in F[x]$. Since $\bar{F}$ is the algebraic closure of $\bar{F}$, $p(x)$ splits over $\bar{F}[x]$, and so it follows that $\alpha \in \bar{F}$. Therefore, $\bar{F}$ is algebraically closed. \end{proof} \marginnote{ \begin{mnote} \cref{thm:every_field_has_an_algebraic_closure} is an exercise in A5. \end{mnote} } \begin{thm}[Every Field has an Algebraic Closure]\label{thm:every_field_has_an_algebraic_closure} For every field $F$, there exists an algebraically closed field that contains $F$. \end{thm} \begin{thm}[Adding All Algebraic Elements Algebraically Closes the Field]\label{thm:adding_all_algebraic_elements_algebraically_closes_the field} Let $K$ be an algebraically closed field that contains $F$. The collection of elements in $K$ which are algebraic over $F$ is an algebraic closure of $F$. \end{thm} \begin{proof} Let \begin{equation*} L := \left\{ \alpha \in K \mmid \alpha \text{ is algebraic over } F \right\}. \end{equation*} As given in the statement, we want to show that $L$ is an algebraic closure of $F$. It is clear that $L / F$, since every $\beta \in F$ is algebraic over $F$ and is hence in $L$. Let $f(x) \in F[x]$ with $\deg f \geq 1$. Since $f(x)$ splits over $K$, we have \begin{equation*} f(x) = u(x - \alpha_1)(x - \alpha_2) \hdots (x - \alpha_n), \end{equation*} where $u \in F^\times$ and $\alpha_i \in K$ for $i \in \{ 1, \ldots, n \}$. Then since $f(\alpha_i) = 0$ for all $i$, it follows that each of the $\alpha_i \in L$. In other words, $f(x)$ splits over $L$. \end{proof} % section algebraic_closures_continued (end) \section{Cyclotomic Extensions}% \label{sec:cyclotomic_extensions} % section cyclotomic_extensions We look into a specific class of field extensions, which is rather important to us. Consider the following question: \begin{quotebox}{green}{white} What is the splitting field of the polynomial $f(x) = x^n - 1$ over $\mathbb{Q}$? \end{quotebox} The following definition should remind one of MATH 135. \begin{defn}[$n$\textsuperscript{th} Roots of Unity]\index{$n$\textsuperscript{th} Roots of Unity}\label{defn:_n_th_roots_of_unity} We call the roots of $x^n - 1$ (over $\mathbb{C}$) the \hlnoteb{$n$\textsuperscript{th} roots of unity}. \end{defn} \begin{eg} We can obtain all the $n$\textsuperscript{th} roots of unity using Euler's identity \begin{equation*} \zeta_n = \cos \left( \frac{2\pi}{n} \right) + i \sin \left( \frac{2 \pi}{n} \right), \end{equation*} which we label these roots as $1 = \zeta_n^0,\, \zeta_n^1,\, \zeta_n^2,\, \zeta_n^3,\, \ldots,\, \zeta_n^{n - 1}$. \end{eg} Following the various results that we have proven in the last few lectures, we know that the splitting field of $x^n - 1$ over $\mathbb{Q}$ is therefore $\mathbb{Q}(\zeta_n)$. \newthought{We can then ask} ourselves what is the degree of $\mathbb{Q}(\zeta_n)$ over $\mathbb{Q}$, i.e. what is $[ \mathbb{Q}(\zeta_n) : \mathbb{Q} ]$? If $n = p$ where $p$ is prime, then since we may write \begin{equation*} x^p - 1 = (x - 1)(x^{p - 1} + x^{p - 2} + \hdots + x + 1), \end{equation*} by \cref{eg:polynomial_with_prime_minus_1_degree} in \cref{eg:examples_of_irreducible_polynomials}, we know that \begin{equation*} \Phi_p(x) = x^{p - 1} + \hdots + x + 1 \end{equation*} is irreducible over $\mathbb{Q}$. So $\Phi_p(x)$ is the minimal polynomial for $\zeta_p$ over $\mathbb{Q}$. It thus follows that $[ \mathbb{Q}(\zeta_p) : \mathbb{Q} ] = p - 1$. \begin{eg} We shall calculate $[ \mathbb{Q}(\zeta_6) : \mathbb{Q} ]$. Note that \begin{equation*} \zeta_6 = \cos \left( \frac{2 \pi}{6} \right) + i \sin \left( \frac{2 \pi}{6} \right) = \frac{1}{2} + i \frac{\sqrt{3}}{2}. \end{equation*} Since $1, 2 \in \mathbb{Q}$, we have that $\mathbb{Q}(\zeta_6) = \mathbb{Q}(i \sqrt{3})$. By \hyperref[propo:eisenstein_s_criterion]{$3$-Eisenstein}, the polynomial $x^2 + 3$ is irreducible and is a polynomial where $i \sqrt{3}$ is a root. Thus \begin{equation*} [ \mathbb{Q}(\zeta_6) : \mathbb{Q} ] = [ \mathbb{Q}(i \sqrt{3}) : \mathbb{Q} ] = \deg (x^2 + 3) = 2. \end{equation*} \end{eg} \begin{remark} The $n$\textsuperscript{th} roots of unity form a cyclic group. A generator of this group is called a \hldefn{primitive $n$\textsuperscript{th} root of unity}. In other words, $\zeta_n^k$ is an primitive $n$\textsuperscript{th} root of unity iff $(\zeta_n^k)^m \neq 1$ for $m = 1, 2, \ldots, n - 1$. From Group Theory, $\zeta_n^k$ is a primitive $n$\textsuperscript{th} root of unity iff $\gcd(n, k) = 1$. Thus, there are \begin{equation*} \phi(n) = \abs{ \{ 1 \leq k \leq n : \gcd(k, n) = 1 \} } \end{equation*} primitive $n$\textsuperscript{th} root of unity\sidenote{Explanation required.}. \end{remark} \begin{defn}[$n$\textsuperscript{th} Cyclotomic Polynomial]\index{$n$\textsuperscript{th} Cyclotomic Polynomial}\label{defn:_n_th_cyclotomic_polynomial} For $n \geq 1$, the \hlnoteb{$n$\textsuperscript{th} cyclotomic polynomial} is \begin{equation*} \Phi_n(x) = \prod_{\substack{1 \leq k \leq n \\ \gcd(k, n) = 1}} \left( x - e^{2 \pi i \frac{k}{n}} \right) = (x - \alpha_1)(x - \alpha_2) \hdots (x - \alpha_{\phi(n)}), \end{equation*} where the $\alpha_i$'s are the primitive $n$\textsuperscript{th} roots of unity. \end{defn} \begin{remark} Since $\Phi_n(x)$ has rational coefficients, we know that $\Phi_n(x) \in \mathbb{C}[x]$. \end{remark} In fact, $\Phi_n(x)$ is the minimal polynomial for $\zeta_n$ over $\mathbb{Q}$, which then gives us that $[ \mathbb{Q}(\zeta_n) : \mathbb{Q} ] = \phi(n)$. However, we are not yet ready to show this. \begin{eg} The following are $n$\textsuperscript{th} cyclotomic polynomials, where $n = 1, 2, 3$ and $4$: \marginnote{See the first 30 cyclotomic polynomials on \href{https://en.wikipedia.org/wiki/Cyclotomic_polynomial\#Examples}{Wikipedia}.} \begin{itemize} \item $\Phi_1(x) = x - 1$ \item $\Phi_2(x) = \left(x - e^{2 \pi i \frac{1}{2}}\right) = (x + 1)$ \item $\Phi_3(x) = \left(x + e^{2 \pi i \frac{1}{3}}\right)\left(x - e^{2 \pi i \frac{2}{3}}\right) = x^2 + x + 1$ \item $\Phi_4(x) = (x + i)(x - i) = x^2 + 1$ \end{itemize} \end{eg} \begin{eg} Let $n = p$ be prime. Then the $p$\textsuperscript{th} roots of unity are \begin{equation*} 1, \zeta_p^2, \zeta_p^3, \ldots, \zeta_p^{p - 1} \end{equation*} and the primitives are \begin{equation*} \zeta_p^2, \zeta_p^3, \ldots, \zeta_p^{p - 1}. \end{equation*} Thus \begin{equation*} x^p - 1 = (x - 1)(x^{p - 1} + x^{p - 2} + \hdots + x^2 + x + 1) = (x - 1) \Phi_p(x). \end{equation*} \end{eg} A good question to ask here is: \begin{quotebox}{green}{foreground} Is there an easier way to compute $\Phi_n(x)$ for all $n$? \end{quotebox} % section cyclotomic_extensions (end) % chapter lecture_13_feb_04th (end) \chapter{Lecture 14 Feb 08th}% \label{chp:lecture_14_feb_08th} % chapter lecture_14_feb_08th \section{Cyclotomic Extensions (Continued)}% \label{sec:cyclotomic_extensions_continued} % section cyclotomic_extensions_continued \begin{remark} Note that $Z := \{ z \in \mathbb{C} : z^n = 1 \}$ is a group. We may write \begin{equation*} \bigcupdot_{d \mid n} \left\{ \text{ primitive } d\text{\textsuperscript{th} roots of unity } \right\}. \end{equation*} \end{remark} \begin{lemma}[$x^n - 1 = \prod_{d \mid n} \Phi_d(x)$]\label{lemma:_x_n_1_d mid n_phi_d_x_} We have \begin{equation*} x^n - 1 = \prod_{d \mid n} \Phi_d(x). \end{equation*} \end{lemma} \begin{eg} \begin{align*} \Phi_6(x) &= \frac{x^6 - 1}{\Phi_1(x) \Phi_2(x) \Phi_3(x)} \\ &= \frac{x^6 - 1}{(x - 1)(x + 1)(x^2 + x + 1)} = x^2 - x + 1. \end{align*} \end{eg} \begin{propo}[Cyclotomic Polynomials have Integer Coefficients]\label{propo:cyclotomic_polynomials_have_integer_coefficients} For every $n \geq 1$, $\Phi_n(x) \in \mathbb{Z}[x]$. \end{propo} \marginnote{ \begin{strategy} We shall use strong induction here. \end{strategy} } \begin{proof} We proceed by induction on $n$. If $n = 1$, then $\Phi_1(x) = x - 1 \in \mathbb{Z}[x]$. Suppose the results holds for all $l < n$. By \cref{lemma:_x_n_1_d mid n_phi_d_x_}, we have \begin{equation*} x^n - 1 = f(x) \Phi_n(x) \end{equation*} where \begin{equation*} f(x) = \prod_{\substack{d \mid n \\ d < n}} \Phi_d(x). \end{equation*} By the induction hypothesis, $f(x) \in \mathbb{Z}[x]$. Let $F = \mathbb{Q}(\zeta_n)$ so that $\Phi_n(x) \in F[x]$. By the division algorithm, $\exists ! q(x), r(x) \in F[x]$ such that \begin{equation*} x^n - 1 = f(x) q(x) + r(x). \end{equation*} Similarly, $\exists ! \tilde{q}(x), \tilde{r}(x) \in \mathbb{Q}[x] \supset \mathbb{Z}[x]$ such that \begin{equation*} x^n - 1 f(x) \tilde{q}(x) + \tilde{r}(x). \end{equation*} Since $\mathbb{Q} \subseteq F \subseteq \mathbb{C}$, by uniqueness\sidenote{This part should be thought of in the following way: we know that there is some $q(x) \in F[x]$, which is an extension of $\mathbb{Q}[x]$, and we also found that there is some $\tilde{q}(x) \in \mathbb{Q}[x]$, and so uniqueness tells us that the two must be the same.}, \begin{equation*} \Phi_n(x) = q(x) = \tilde{q}(x) \in \mathbb{Q}[x]. \end{equation*} It follows by \hyperref[thm:gauss_lemma]{Gauss' Lemnma} that $\Phi_n(x) \in \mathbb{Z}[x]$. \end{proof} \marginnote{The proof for \cref{thm:cyclotomic_polynomials_are_irreducible_over_q_} is provided over two separate lectures, in particular it is provided at the end of this lecture and the beginning of Lecture 16. For sanity, the entire proof will be provided here.} \begin{thm}[Cyclotomic Polynomials are Irreducible over $\mathbb{Q}$]\label{thm:cyclotomic_polynomials_are_irreducible_over_q_} For $n \geq 1$, $\Phi_n(x)$ is irreducible over $\mathbb{Q}$. \end{thm} \marginnote{ \begin{strategy} We will show that $\Phi_n(x)$ is a minimal polynomial. If $g(x) \in \mathbb{Q}[x]$ is a minimal polynomial for $\zeta_n$, then since $\zeta_n$ is also a root of $\Phi_n(x)$, we must have $g(x) \mid \Phi_n(x)$. So to show that $g(x)$ is actually $\Phi_n(x)$, it suffices to show that $\Phi_n(x) \mid g(x)$. \end{strategy} } \begin{proof} Let $g(x) \in \mathbb{Q}[x]$ be a minimal polynomial for $\zeta_n$. It suffices for us to show that $\Phi_n(x) \mid g(x)$. To that end, we can show that every root of $\Phi_n(x)$ is a root of $g(x)$ (in $\mathbb{C}$). Let $\alpha$ be a root of $\Phi_n(x)$. Then by \cref{defn:_n_th_cyclotomic_polynomial}, $\alpha = \zeta_n^k$ for some $k \in \{ 1, \ldots, n - 1 \}$ such that $\gcd(k, n) = 1$. Then let $k = p_1 p_2 \hdots p_N$, where each $p_i$ is a prime and $p_i \nmid n$ \sidenote{Note that this must be the case since $\gcd(k, n) = 1$.}. Thus, the statement which we wish to prove becomes the following: $\zeta_n^{p_1}, \zeta_n^{p_1 p_2}, \ldots, \zeta_n^{p_1 p_2 \hdots p_N} = \alpha$ are roots of $g(x)$. To prove the above, it suffices for us to show that if $\zeta \in \mathbb{C}$ is a root of $g(x)$, then $\zeta^p$, where $p$ is prime and $p \nmid n$, is also a root of $g(x)$. Suppose not \faDizzy, i.e. that $g(\zeta) = 0$ but $g\left(\zeta^p\right) \neq 0$, where $p$ is prime and $p \nmid n$. Now since $g(x) \mid \Phi_n(x)$, we have $\Phi_n(\zeta) = 0$. Since $p \nmid n$, it follows that $\zeta^p$ is also a primitive $n$\textsuperscript{th} root of unity, i.e. $\Phi_n(\zeta_n^p) = 0$. Now since $g(x) \mid \Phi_n(x)$, $\exists h(x) \in \mathbb{Q}[x]$ such that $\Phi_n(x) =g(x) h(x)$. By \hyperref[thm:gauss_lemma]{Gauss}, WMA $h(x) \in \mathbb{Z}[x]$. Since $\mathbb{Z}[x$ is an integral domain, $\Phi_n(\zeta^p) = 0$ and $g(\zeta^p) \neq 0 \implies h(\zeta^p) = 0$. Let $f(x) = h(x^p) \in \mathbb{Z}[x]$. Then $f(\zeta) = 0$. Moreover, we have $g(x) \mid f(x)$ in $\mathbb{Q}[x]$. Thus $f(x) = g(x) k(x)$ for some $k(x) \in \mathbb{Z}[x]$ (again, through Gauss). Suppose $h(x) = \sum b_j x^j$, which then implies that $f(x) = \sum b_j x^{pj}$. Consider $\bar{f}(x) \in \mathbb{Z}_p[x]$, i.e. \begin{equation*} \bar{f}(x) = \sum \bar{b}_j x^{pj}, \quad \bar{b}_j \equiv b_j \mod p. \end{equation*} Then \begin{align*} \bar{f}(x) &= \sum \bar{b}_j^p x^{pj} \quad \because \text{ \hlnotea{Fermat's Little Theorem} } \\ &= \left( \sum \bar{b}_j x^j \right)^p \quad \because \text{ \hlnotea{Freshman's Dream} } \\ &= \left( \bar{h}(x) \right)^p. \end{align*} It follows that \begin{equation*} \left( \bar{h}(x) \right)^p = \bar{f}(x) = \bar{g}(x) \bar{k}(x) \in \mathbb{Z}_p[x]. \end{equation*} Now let $\bar{l}(x)$ be an irreducible factor of $\bar{g}(x)$ over $\mathbb{Z}_p[x]$ \sidenote{Note that this $\bar{l}(x)$ may be $\bar{g}(x)$ itself if $\bar{g}(x)$ is still irreducible over $\mathbb{Z}_p[x]$.}. Since $\bar{l}(x) \mid \bar{h}(x)^p$, we have that $\bar{l}(x) \mid \bar{h}(x)$ \sidenote{\hlwarn{Why?}}. % TODO : Find out why On the other hand, in $\mathbb{Z}_p[x]$, we have that $\bar{\Phi}_n(x) = \bar{g}(x) \bar{h}(x)$. It follows that $\bar{l}(x)^2 \mid \bar{\Phi}_n(x)$ \sidenote{\hlwarn{Why?}}. % TODO : Find out why Since $\bar{\Phi}_n(x) \mid x^n - 1$, we have that \begin{equation*} x^n - 1 = \bar{l}(x)^2 \bar{q}(x) \in \mathbb{Z}_p[x]. \end{equation*} By \hlnotea{taking derivatives} on both sides, we have \begin{align*} \bar{n} x^{n - 1} &= 2 \bar{l}(x) \bar{l}'(x) \bar{q}(x) + \bar{l}(x)^2 \bar{q}'(x) \\ &= \bar{l}(x) [ \faEllipsisH ] \in \mathbb{Z}_p[x], \end{align*} where $\faEllipsisH$ is an irrelevant factor. Since $\bar{n} \neq 0$, we have that the only root of LHS is $\bar{0}$, and so the only root of $\bar{l}(x)$ is some extension of $\mathbb{Z}_p$ is $\bar{0}$. Since $\bar{l}(x) \mid x^n - \bar{1}$, we have that $\bar{0}^n - \bar{1} = 0$ but that mean $0 = 1 \in \mathbb{Z}_p$, a contradiction. Tracing back our long convoluted line of thought, we have that \faDizzy is not true, and so we must have $g(\zeta^p) = 0$, which \begin{itemize} \item[$\implies$] all the $\zeta_n^{p_1}, \zeta_n^{p_1 p_2}, \ldots, \alpha$ are all roots of $g(x)$; \item[$\implies$] $\Phi_n(x) \mid g(x)$; \item[$\implies$] $\Phi_n(x) = g(x)$, \end{itemize} which is what we want to show. \end{proof} \begin{crly}[Cyclotomic Polynomials are Minimal Polynomials of Its Roots over $\mathbb{Q}$]\label{crly:cyclotomic_polynomials_are_minimal_polynomials_of_its_roots_over_q_} $\Phi_n(x)$ is the minimal polynomial for $\zeta_n$ over $\mathbb{Q}$. In particular, $[ \mathbb{Q}(\zeta_n) : \mathbb{Q} ] = \phi(n)$. \end{crly} \begin{eg} Let $f(x) = x^5 - 3$. Describe the splitting field of $f(x)$ over $\mathbb{Q}$. We shall find a basis for this splitting field over $\mathbb{Q}$. The roots of $f(x)$ are \begin{equation*} \sqrt[5]{3}, \, \zeta_5 \sqrt[5]{3}, \zeta_5^2 \sqrt[5]{3}, \zeta_5^3 \sqrt[5]{3}, \zeta_5^4 \sqrt[5]{3}. \end{equation*} It follows that the splitting field for $f$ is $F = \mathbb{Q}(\sqrt[5]{3}, \zeta_5)$. Note that since \begin{equation*} \deg_{\mathbb{Q}}(\zeta_5) = \phi(5) = 4 \text{ and } \deg_{\mathbb{Q}}(\sqrt[5]{3}) = 5, \end{equation*} it follows from A4Q2 that \begin{equation*} [ \mathbb{Q}(\sqrt[5]{3}, \zeta_5) : \mathbb{Q} ] = [ \mathbb{Q}(\sqrt[5]{3}) : \mathbb{Q} ][ \mathbb{Q}(\zeta_5) : \mathbb{Q} ] = 4 \cdot 5 = 20. \end{equation*} Now a \hlbnotea{basis for $\mathbb{Q}(\zeta_5)(\sqrt[5]{3}) / \mathbb{Q}(\zeta_5)$} is \begin{equation*} \left\{ 1, \sqrt[5]{3}, \left( \sqrt[5]{3} \right)^2, \left( \sqrt[5]{3} \right)^3, \left( \sqrt[5]{3} \right)^4 \right\}, \end{equation*} while a \hlbnotea{basis for $\mathbb{Q}(\zeta_5) / \mathbb{Q}$} is \begin{equation*} \left\{ 1, \zeta_5, \zeta_5^2, \zeta_5^3, \zeta_5^4 \right\}. \end{equation*} Following the \hyperref[thm:tower_theorem]{Tower Theorem}, a basis for the splitting field $F$ is \begin{equation} \left\{ \sqrt[5]{3} \left( \zeta_5 \right)^j \mmid 0 \leq j \leq 4 \right\}. \end{equation} \end{eg} % section cyclotomic_extensions_continued (end) % chapter lecture_14_feb_08th (end) \chapter{Lecture 15 Feb 11th}% \label{chp:lecture_15_feb_11th} % chapter lecture_15_feb_11th \section{Finite Fields}% \label{sec:finite_fields} % section finite_fields Finite fields are very easy to work with and grasp. The nice thing about finite fields is that, up to isomorphism, there is only one field that has order prime to some power, which we shall show in this section. \begin{lemma}[Units of a Finite Field Form a Finite Cyclic Group]\label{lemma:units_of_a_finite_field_form_a_finite_cyclic_group} Let $F$ be a finite field. Then $G = F^\times$ is a finite cyclic group. \end{lemma} \begin{proof} Since $G$ is the set of units of $F$, we know that $G$ is an abelian group by its construction, and it is finite since $F$ is finite. Then, by the \hlnotea{Finite Abelian Group Structure}, $\exists n_1, \ldots, n_m \in \mathbb{Z}$ such that \begin{equation}\label{eq:lemma_units_of_finite_fields_eq1} G \simeq \mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2} \times \hdots \times \mathbb{Z}_{n_m}, \end{equation} and each $n_i$ is a prime power. Let \begin{align*} N &:= n_1 n_2 \hdots n_m \text{ and } \\ M &:= \lcm(n_1, \ldots, n_m). \end{align*} By construction, $M \leq N$. Now $\forall a \in G$, we have that $a$ is a root of $x^M - 1 \in F[x]$ due to \cref{eq:lemma_units_of_finite_fields_eq1}\sidenote{$a$ is of one of the orders $n_1, n_2, \ldots. n_m$, so it is a root of $x^M - 1$.}. Note that $N = \abs{G}$, and the polynomial $x^M - 1$ has at most $M$ roots. Therefore, $N \leq M$. Thus we must have $N = M$, thus forcing the $n_i$'s to be coprimes, and so we have \begin{equation*} G \simeq \mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2} \times \hdots \times \mathbb{Z}_{n_m} = \mathbb{Z}_N. \end{equation*} \end{proof} \begin{propo}[Order of Finite Fields are Powers of Its Primal Characteristic]\label{propo:order_of_finite_fields_are_powers_of_its_primal_characteristic} Let $F$ be a finite field. Then \begin{enumerate} \item $\abs{F} = p^n$, where $p$ is the \hlnotea{characteristic}\sidenote{Recall from PMATH 347 that the definition of the characteristic is the order of $1$ under addition. We shall use $\Char(F)$ to mean the characteristic of the field $F$.} of $F$ and $n = [F : \mathbb{Z}_p]$. \item $F = \mathbb{Z}_p(\alpha)$ for some $\alpha$ such that $\deg_{\mathbb{Z}_p}(\alpha) = n$. \end{enumerate} \end{propo} \begin{proof} Let $F$ be a finite field with characteristic $p$. Then $\mathbb{Z}_p$ is a prime subfield of $F$, and in particular $F / \mathbb{Z}_p$. Let $n = [ F : \mathbb{Z}_p ]$. By \cref{lemma:units_of_a_finite_field_form_a_finite_cyclic_group}, let $\alpha \in G = F^\times$ be such that $G = \langle \alpha \rangle$. By adding a unit of $F$ to $\mathbb{Z}_p$, since $\mathbb{Z}_p$ is a prime subfield, we have that $\mathbb{Z}_p(\alpha) = F$. Now since $n = [F : \mathbb{Z}_p]$, we have that \begin{equation*} F = \Span_{\mathbb{Z}_p} \{ 1, \alpha, \alpha^2, \ldots, \alpha^{n - 1} \}. \end{equation*} It follows that $\abs{F} = p^n$. \end{proof} \marginnote{\cref{thm:finite_fields_as_splitting_fields} is the important theorem that tells us that there is only one finite field for every $p^n$ up to isomorphism, and this follows from the \hyperref[crly:splitting_fields_are_unique_up_to_isomorphism]{uniqueness of splitting fields}.} \begin{thm}[Finite Fields as Splitting Fields]\label{thm:finite_fields_as_splitting_fields} Let $p$ be a prime and $n \in \mathbb{N}$. Then $F$ is a finite field of order $p^n$ iff $F$ is the splitting field of $f(x) = x^{p^n} - x$ over $\mathbb{Z}_p[x]$. \end{thm} \begin{proof} Suppose $\abs{F} = p^n$. By \hyperref[thm:lagrange_s_theorem]{Lagrange}\sidenote{Is it really Lagrange?}, $a^{p^n - 1} - 1 = 0$ for every $a \in F^\times$. Then in particular, \begin{equation*} a(a^{p^n} - 1) = a^{p^n} - a = 0. \end{equation*} It follows that every $a \in F$ is a root of $x^{p^n} - x$. Since $x^{p^n} - x$ has at most $p^n$ roots, $F$ must thus contain al roots of $x^{p^n} - x$, and so $x^{p^n} - x$ splits over $F[x]$. Any proper subfield of $F$ would not have enough elements to be a splitting field for $x^{p^n} - x$. Thus $F$ is a splitting field of $x^{p^n} - x$. For the \hlbnoted{$\impliedby$} direction, let $F$ be the splitting field of $f(x) = x^{p^n} - x$. Let \begin{equation*} K = \{ \alpha \in F : f(\alpha) = 0 \}. \end{equation*} \begin{ex} $K$ is a field. \end{ex} \marginnote{ \begin{solution}[to the ex. in the proof] For $\alpha, \beta \in K$, we have that \begin{equation*} \alpha^{p^n} - \alpha = 0 \text{ and } \beta^{p^n} - \beta = 0. \end{equation*} It then follows by the Freshman's Dream that \begin{gather*} \left( \alpha^{p^n} + \beta^{p^n} \right) - \alpha - \beta = 0 \\ \implies \\ (\alpha + \beta)^{p^n} - (\alpha + \beta) = 0. \end{gather*} \end{solution}} Then $K \leq F$. However, we also have that $F \leq K$, since all roots of $f$ are in $F$ since $F$ is a splitting field, and $f$ also splits over $K$. Also, note that $f'(x) = -1$ since $\Char F = p$, and so $f$ has no repeated roots since it is a decreasing function. \end{proof} % section finite_fields (end) % chapter lecture_15_feb_11th (end) \chapter{Lecture 16 Feb 13th}% \label{chp:lecture_16_feb_13th} % chapter lecture_16_feb_13th \section{Finite Fields (Continued)}% \label{sec:finite_fields_continued} % section finite_fields_continued \marginnote{Since I moved the `second half' of the proof of \cref{thm:cyclotomic_polynomials_are_irreducible_over_q_} over to \cref{chp:lecture_14_feb_08th}, not too much content is left here.} By \cref{lemma:units_of_a_finite_field_form_a_finite_cyclic_group}, \cref{propo:order_of_finite_fields_are_powers_of_its_primal_characteristic} and \cref{thm:finite_fields_as_splitting_fields}, we have the following result. \begin{thm}[Classification of Finite Fields]\label{thm:classification_of_finite_fields} For any prime $p$ and $n \in \mathbb{N}$, we have \begin{itemize} \item there exists a field $F$ such that $\abs{F} = p^n$; and \item any 2 fields of order $p^n$ are isomorphic to one another. \end{itemize} \end{thm} \begin{note}[Notation] We denote the field of order $p^n$ by $\mathbb{F}_{p^n}$, i.e. \begin{equation*} \mathbb{F}_{p^n} := \left\{ x \mmid f(x) = x^{p^n} - x = 0 \right\}. \end{equation*} \end{note} In the next lecture, we shall prove the following theorem. \begin{thmnonum}[Subfields of Finite Fields]\label{thmnonum:subfields_of_finite_fields} If $E$ is a subfield of $\mathbb{F}_{p^n}$, then $E \simeq \mathbb{F}_{p^r}$, where $r \mid n$. Moreover, if $r \mid n$, then $\mathbb{F}_{p^n}$ has a unique\sidenote{This is truly unique, not unique up to isomorphism, which is \hlimpo{rare}.} subfield of order $p^r$. \end{thmnonum} The above theorem gives us the following example. \begin{eg} Given the finite field $\mathbb{F}_{2^{12}}$, we know that the divisors of $12$ are \begin{equation*} 1, \, 2, \, 3, \, 4, \, 6, \, 12. \end{equation*} By the above theorem, we have the following \hlnotea{lattice structure}. \begin{figure}[ht] \centering \begin{tikzcd} & \mathbb{F}_{2^{12}} \arrow[rd, no head] \arrow[ld, no head] & & \\ \mathbb{F}_{2^4} \arrow[rd, no head] & & \mathbb{F}_{2^6} \arrow[ld, no head] \arrow[rd, no head] & \\ & \mathbb{F}_{2^2} \arrow[rd, no head] & & \mathbb{F}_{2^3} \arrow[ld, no head] \\ & & \mathbb{F}_{2^1} & \end{tikzcd} \caption{Lattice of $\mathbb{F}_{2^12}$} \label{fig:lattice_of_f__2_12_} \end{figure} \end{eg} % section finite_fields_continued (end) % chapter lecture_16_feb_13th (end) \tuftepart{Galois Theory} \chapter{Lecture 17 Feb 15th}% \label{chp:lecture_17_feb_15th} % chapter lecture_17_feb_15th \section{Finite Fields (Continued 2)}% \label{sec:finite_fields_continued_2} % section finite_fields_continued_2 We shall now prove the last theorem that we stated. \begin{thm}[Subfields of Finite Fields]\label{thm:subfields_of_finite_fields} If $E$ is a subfield of $\mathbb{F}_{p^n}$, then $E \simeq \mathbb{F}_{p^r}$, where $r \mid n$. Moreover, if $r \mid n$, then $\mathbb{F}_{p^n}$ has a unique\sidenote{This is truly unique, not unique up to isomorphism, which is \hlimpo{rare}.} subfield of order $p^r$. \end{thm} \begin{proof} \hlbnoted{Part 1} Let $E < \mathbb{F}_{p^n}$. By \cref{propo:order_of_finite_fields_are_powers_of_its_primal_characteristic} and \hyperref[thm:tower_theorem]{the Tower Theorem}, we have \begin{equation*} n = [ \mathbb{F}_{p^n} : \mathbb{F}_p ] = [ \mathbb{F}_{p^n} : E ][ E : \mathbb{F}_p ]. \end{equation*} Then by letting $r = [ E : \mathbb{F}_p ]$, we have that $r \mid n$ and $\abs{E} = p^r$. \noindent \hlbnoted{Part 2} Suppose $r \mid n$, i.e. $\exists k \in \mathbb{Z}$ such that $n = rk$. Consider \begin{equation*} \mathbb{F}_{p^n} = \left\{ \alpha \in \bar{\mathbb{F}}_p \mmid \alpha^{p^{rk}} - \alpha = 0 \right\}, \end{equation*} \sidenote{Note that we consider the closure just so that we contain all the roots. \hlwarn{Should $\mathbb{F}_{p^n}$ not already have everything?}}which we see is the splitting field of $x^{p^n} - x$, i.e. it is the set of roots of $x^{p^n} - x$. Since $r \mid n$, we have \begin{equation*} p^n - 1 = (p^r - 1)(p^{n - r} + p^{n - 2r} + \hdots + p^r + 1). \end{equation*} Then, let \begin{align*} E &:= \left\{ \alpha \in \bar{\mathbb{F}}_p \mmid \alpha^{p^r} - \alpha = 0 \right\} \\ &= \left\{ \alpha \in \bar{\mathbb{F}}_p \mmid \alpha^{p^r - 1} - 1 = 0 \right\} \cup \{ 0 \} \\ &\subseteq \bar{\mathbb{F}}_{p^n}. \end{align*} Moreover, we have that $\abs{E} = p^r$. For uniqueness, suppose if there exists $K < \mathbb{F}_{p^n}$ with order $p^r$. Then $\forall \alpha \in K$, \begin{equation*} \alpha^{p^r} - \alpha = 0 \implies \alpha \in E. \end{equation*} Thus $K = E$. \end{proof} % section finite_fields_continued_2 (end) \section{Introduction to Galois Theory}% \label{sec:introduction_to_galois_theory} % section introduction_to_galois_theory Let $f(x) \in F[x]$ be non-constant, and $\alpha_1, \ldots, \alpha_n$ be the roots of $f(x)$ in its splitting field $K$. Our goal is to study these roots by permuting them under automorphisms of the splitting field $K$. \begin{defn}[Galois Group]\index{Galois Group}\label{defn:galois_group} Let $K / F$. We define the \hlnoteb{Galois Group} of $K / F$, by \begin{equation*} \Gal(K / F) := \left\{ \phi \in \Aut(K) \mmid \phi \restriction_{F} = \id \right\} \leq \Aut(K), \end{equation*} where $\Aut(K)$ is the \hlnotea{group of automorphisms} of $K$. \end{defn} \begin{lemma}[The Galois Group permutes algebraic roots]\label{lemma:the_galois_group_permutes_algebraic_roots} Let $K / F$. If $\alpha \in K$ is a root of $f(x) \in F[x]$ and $\phi \in \Gal(K / F)$, then $\phi(\alpha)$ is also a root of $f(x)$. \sidenote{It is important to note that the Galois automorphisms only permute the \hlimpo{algebraic} roots, since it would fix $\alpha$ if $\alpha \in F$.} \end{lemma} \begin{proof} Let $f(x) \in F[x]$. Then $f(x) = \sum a_i x^i$. Since $\alpha$ is a root, we have $f(\alpha) = \sum a_i \alpha^i = 0$. Since $\phi$ is an automorphism, we must therefore have $0 = \phi(0)$. Since $\phi \in \Gal(K / F)$, we have that \begin{align*} 0 = \phi(0) = \phi ( \sum a_i \alpha^i ) = \sum \phi(a_i) \phi(\alpha^i) \overset{(*)}{=} \sum a_i \phi(\alpha)^i = f(\phi(\alpha)), \end{align*} where $(*)$ is since $\phi$ fixes $F$. \end{proof} \begin{crly}[Elements of the Galois Group permutes roots of the same minimal polynomial]\label{crly:elements_of_the_galois_group_permutes_roots_of_the_same_minimal_polynomial} Let $K / F$. If $\alpha \in K$ is algebraic over $F$, and $\phi \in \Gal(K / F)$, then $\phi(\alpha)$ is algebraic over $F$, and $\alpha$ and $\phi(\alpha)$ has the same minimal polynomial in $F[x]$. \end{crly} \begin{eg} Let $F = \mathbb{Q}$ and $K = F(\sqrt{2})$. Then $\Gal( \mathbb{Q}(\sqrt{2}) / \mathbb{Q}) = \Aut \mathbb{Q}(\sqrt{2})$ \sidenote{\hlwarn{Why}? Is it cause there is very little room for us to wiggle around $\phi \restriction_F = \id$?}. Note that the minimal polynomial of $\sqrt{2}$ is $x^2 - 2 = (x - \sqrt{2})(x + \sqrt{2}) \in K[x]$. Thus if $\phi \in \Gal(\mathbb{Q}(\sqrt{2})/\mathbb{Q})$, then $\phi(\sqrt{2}) = \sqrt{2}$ or $-\sqrt{2}$ \sidenote{Note that we must fix everything else, by definition of a Galois group.}. It follows that the only two maps in $\Gal(K / F)$ are \begin{gather*} \phi_1 : a + b \sqrt{2} \mapsto a + b \sqrt{2} \\ \phi_2 : a + b \sqrt{2} \mapsto a - b \sqrt{2}. \end{gather*} Thus $\Gal(K / F) = \{ \phi_1, \phi_2 \} \simeq \mathbb{Z}_2$. \end{eg} % section introduction_to_galois_theory (end) % chapter lecture_17_feb_15th (end) \chapter{Lecture 18 Feb 25th}% \label{chp:lecture_18_feb_25th} % chapter lecture_18_feb_25th \section{Introduction to Galois Theory (Continued)}% \label{sec:introduction_to_galois_theory_continued} % section introduction_to_galois_theory_continued \marginnote{ \begin{procedure}[Computing Galois Groups]\label{procedure:computing_galois_groups} We will only deal with finite extensions for Galois theory. Given a base field and its extension: \begin{enumerate} \item Figure out what's being adjoined. \item Look for their minimal polynomial(s). \item Look at what sort of automorphisms are permitted to be in the group, i.e. look at the possible ways that the roots can be permuted. \begin{itemize} \item Note \cref{lemma:the_galois_group_permutes_algebraic_roots} and \cref{crly:elements_of_the_galois_group_permutes_roots_of_the_same_minimal_polynomial}. \end{itemize} \end{enumerate} \end{procedure} } \begin{eg} Consider the Galois group $\Gal( \mathbb{Q}(\sqrt{2}, \sqrt{3}) / \mathbb{Q} )$. Now the minimal polynomial for $\sqrt{2}$ and $\sqrt{3}$ are \begin{equation*} x^2 - 2, \text{ and } x^2 - 3, \end{equation*} respectively. Then we can only have $\phi(\sqrt{2}) = \pm \sqrt{2}$ and $\phi(\sqrt{3}) = \pm \sqrt{3}$, i.e. \begin{table}[ht] \centering \caption[][20pt]{All possible elements of $\Gal(\mathbb{Q}(\sqrt{2}, \sqrt{3}) / \mathbb{Q})$} \label{table:all_possible_elements_of_gal_q_sqrt_2_sqrt_3_q} \begin{tabular}{c | c c c c | c} & $\sqrt{2}$ & $-\sqrt{2}$ & $\sqrt{3}$ & $-\sqrt{3}$ & $S_4$ \\ \hline $\phi_1$ & $\sqrt{2}$ & $-\sqrt{2}$ & $\sqrt{3}$ & $-\sqrt{3}$ & $\epsilon$ \\ $\phi_2$ & $\sqrt{2}$ & $-\sqrt{2}$ & $-\sqrt{3}$ & $\sqrt{3}$ & $(3 \; 4)$ \\ $\phi_3$ & $-\sqrt{2}$ & $\sqrt{2}$ & $\sqrt{3}$ & $-\sqrt{3}$ & $(1 \; 2)$ \\ $\phi_4$ & $-\sqrt{2}$ & $\sqrt{2}$ & $-\sqrt{3}$ & $\sqrt{3}$ & $(1 \; 2)(3 \; 4)$ \end{tabular} \end{table} So $\Gal( \mathbb{Q}(\sqrt{2}, \sqrt{3}) / \mathbb{Q} ) = \{ \phi_i : i = 1, 2, 3, 4 \}$. Note that $\abs{\phi_i} = 2$ for $i = 2, 3, 4$. It follows that $\Gal( \mathbb{Q}(\sqrt{2}, \sqrt{3}) / \mathbb{Q} )$ is abelian and has order $4$. Therefore \begin{equation*} \Gal \left( \quotient{\mathbb{Q}(\sqrt{2}, \sqrt{3})}{\mathbb{Q}} \right) \simeq \mathbb{Z}_2 \times \mathbb{Z}_2, \end{equation*} where the first $\mathbb{Z}_2$ comes from $\langle (1 \; 2) \rangle$ while the other comes from $\langle (3 \; 4) \rangle$. \end{eg} \marginnote{ Notice that in \cref{eg:galois_elements_can_only_permute_roots_in_the_same_field}, the field where the roots lie in is important; we see that the Galois group ended up being the trivial group because the other roots of the minimal polynomial of $\sqrt[3]{2}$ live in a higher extension. } \begin{eg}\label{eg:galois_elements_can_only_permute_roots_in_the_same_field} Consider $G = \Gal(\mathbb{Q}(\sqrt[3]{2}) / \mathbb{Q})$. Let $\phi \in G$. Since $\phi(\sqrt[3]{2})$ is a root of $x^3 - 2$, we must have that \begin{equation*} \phi(\sqrt[3]{2}) \in \left\{ \sqrt[3]{2}, \, \sqrt[3]{2} \zeta_3, \, \sqrt[3]{2} \zeta_3^2 \right\}. \end{equation*} However, $\sqrt[3]{2} \zeta_3, \, \sqrt[3]{2} \zeta_3^2 \notin \mathbb{Q}(\sqrt[3]{2})$. Therefore we must have $\phi(\sqrt[3]{2}) = \sqrt[3]{2}$, i.e. $\phi = \id$. It follows that $G = \{ 1 \}$. \end{eg} % section introduction_to_galois_theory_continued (end) \section{The Galois Group as a Permutation Group}% \label{sec:the_galois_group_as_a_permutation_group} % section the_galois_group_as_a_permutation_group Let $F$ be a field, $f(x) \in F[x]$, $\deg f = n \geq 1$, and $K$ a splitting field of $f(x)$ over $F$. Let $\alpha_1, \ldots, \alpha_n \in K$ be the roots of $f(x)$, and let $G = \Gal(K / F)$. From the last few examples, we notice that for any $\phi \in G$, $\phi(\alpha_i) = \alpha_j$. In this section, we will show that $G$ is actually a \hlnotea{permutation group} of the roots, as a subgroup of $S_n$ in the case of permuting the roots of $f(x)$, the degree $n$ polynomial. In fact, more is true, but we shall see that down the road. \newthought{From} the last two examples, one cannot help but notice a possible problem: \begin{quotebox}{green}{foreground} what if there are repeated roots? \end{quotebox} If there are, indeed, repeated roots, say $\alpha_1 = \alpha_2$ among the roots $\alpha_1, \alpha_2, \alpha_3, \alpha_4$, where $\alpha_4 \neq \alpha_3 \neq \alpha_1 \neq \alpha_4$, then the identity element would be indistinguishable from $\phi$ that is defined as \begin{equation*} \phi(\alpha_1) = \alpha_2, \, \phi(\alpha_2) = \alpha_1, \, \phi(\alpha_3) = \alpha_3, \, \phi(\alpha_4) = \alpha_4. \end{equation*} So it suffices for us to consider for the case where $f(x)$ does not have multiple roots of the same value, i.e. the \hlnotea{multiplicity} of all roots is $1$. Such polynomials are called \hlnotea{separable} polynomials. \begin{defn}[Separable Polynomials]\index{Separable Polynomials}\label{defn:separable_polynomials} A polynomial $f(x) \in F[x]$ is said to be \hlnoteb{separable} if all of its roots have multiplicity $1$. \end{defn} \newthought{Let} $f(x) \in F[x]$ be separable with $\deg f = n \geq 1$, and suppose $K$ is the splitting field of $f(x)$ over $F$. Let $\alpha_1, \ldots, \alpha_n$ be the roots of $f$ in $K$. From our discussion above, we want to show that $\Gal(K / F) \simeq P \leq S_n$. In other words, we want to see that given $\phi \in \Gal(K / F)$, $\exists \pi \in P \leq S_n$ such that $\phi(\alpha_i) = \alpha_{\pi(i)}$. \begin{notation} Given $f(x) \in F[x]$, and $K$ the splitting field of $f(x)$, we sometimes write \begin{equation*} \Gal(f(x)) := \Gal(K / F). \end{equation*} In other words, when we write $\Gal(f(x))$, we are talking about the Galois group over the splitting field of $f(x)$ over $F$. \end{notation} \begin{eg} Recall an earlier example of ours where $f(x) = (x^2 - 2)(x^2 - 3) \in \mathbb{Q}[x]$, where we showed that the Galois group $\Gal(f(x)) = \mathbb{Z}_2 \times \mathbb{Z}_2$. Let \begin{equation*} \alpha_1 = \sqrt{2}, \, \alpha_2 = - \sqrt{2}, \, \alpha_3 = \sqrt{3}, \, \alpha_4 = - \sqrt{3}. \end{equation*} Then \begin{equation*} \Gal(f(x)) \simeq \{ \epsilon, \, (3 \; 4), \, (1 \; 2), \, (1 \; 2)(3 \; 4) \}. \end{equation*} \end{eg} \begin{eg} Let $x^2 + 1 \in \mathbb{Q}[x]$. Then \sidenote{The adjoined elements are $\pm i$.} \begin{equation*} \Gal(x^2 + 1) \simeq \mathbb{Z}_2. \end{equation*} However, if we consider $x^2 + 1 \in \mathbb{Z}_2[x]$, then \begin{equation*} \Gal(x^2 + 1) = \Gal((x + 1)^2) = \{ 1 \}. \end{equation*} \end{eg} The following is a quick corollary of from our discussion and observation. \begin{crly}[The Galois Group completely captures all permutation of the roots]\label{crly:the_galois_group_completely_captures_all_permutation_of_the_roots} Let $F$ be a field, $f(x) \in F[x]$ a non-constant and irreducible, $K$ a splitting field of $f(x)$ over $F$. Then $\forall \alpha, \, \beta \in K$ such that $f(\alpha) = 0 = f(\beta)$, $\exists \phi \in \Gal(K / F) = \Gal(f(x))$ such that $\phi(\alpha) = \beta$. \end{crly} \begin{proof} We shall use the \hyperref[lemma:isomorphism_extension_lemma]{Isomorphism Extension Lemma} to prove this. \begin{marginfigure} \centering \begin{tikzcd} K \arrow[rr, "\phi"] & & K \arrow[dd, no head] \\ & & \\ F(\alpha) \arrow[uu, no head] & \overset{\alpha \mapsto \beta}{\simeq} & F(\beta) \arrow[dd, no head] \\ & & \\ F \arrow[rr, "id"] \arrow[uu, no head] & & F \end{tikzcd} \caption{Constructing elements of the Galois Group}\label{fig:constructing_elements_of_the_galois_group} \end{marginfigure} Consider the identity as our isomorphism $\id : F \to F$. The Isomorphism Extension Lemma gives us the isomorphism that goes from $F(\alpha)$ to $F(\beta)$ by mapping $\alpha$ to $\beta$. We may thus define $\phi$ such that $\phi$ fixes $F$ and $\phi(\alpha) = \beta$, in $K$. \end{proof} \newthought{Permutation groups} that allows one to traverse all around the indices, such as the Galois group, have a special name. \begin{defn}[Transitive Subgroup]\index{Transitive Subgroup}\label{defn:transitive_subgroup} A subgroup $H \leq S_n$ is \hlnoteb{transitive} if $\forall i, j \in \{ 1, \ldots, n \}$, $\exists \pi \in H$ such that $\pi(i) = j$. \end{defn} \begin{crly}[The Galois Group of a Separable, Irreducible Polynomial is Transitive]\label{crly:the_galois_group_of_a_separable_irreducible_polynomial_is_transitive} Let $f(x) \in F[x]$, with $\deg f = n \geq 1$, be separable and irreducible. Then $\Gal(f(x)) \simeq H \leq S_n$ that is transitive. \end{crly} \begin{eg} Consider $G = \Gal(x^3 - 2)$ over $\mathbb{Q}[x]$. Since $f(x) = x^3 - 2$ is irreducible (by \hyperref[propo:eisenstein_s_criterion]{$2$-Eisenstein}) and $\Char \mathbb{Q} = 0$, $f(x)$ is separable \sidenote{See A5Q3(d).}. It follows from \cref{crly:the_galois_group_of_a_separable_irreducible_polynomial_is_transitive} that $G \simeq H \leq S_3$ transitive. Let $\alpha_1, \, \alpha_2, \, \alpha_3$ be the roots of $f(x)$. Let $X = \{ \alpha_1, \, \alpha_2, \, \alpha_3 \}$, and $G$ act on $X$ via $\phi \cdot \alpha_i = \phi(\alpha_i)$. By the \hyperref[thm:orbit_stabilizer_theorem]{Orbit-Stabilizer Theorem}, we have \begin{equation*} \abs{ G } = \abs{ \orb(\alpha_1) } \cdot \abs{ \stab(\alpha_1) } = 3 \cdot \abs{ \stab(\alpha_1) }, \end{equation*} where we note that $\abs{ \orb(\alpha_1) }$ since all the orbits of $\alpha_1$ are exactly elements of $X$. It follows that $3 \mid \abs{G}$. Since the only subgroups of $S_3$ that are divisible by $3$ are $A_3$ and $S_3$, we either have \begin{equation*} G \simeq A_3 \text{ or } G \simeq S_3. \end{equation*} \end{eg} We shall finish the rest of this example in the next lecture. % section the_galois_group_as_a_permutation_group (end) % chapter lecture_18_feb_25th (end) \chapter{Lecture 19 Feb 27th}% \label{chp:lecture_19_feb_27th} % chapter lecture_19_feb_27th \section{The Galois Group as a Permutation Group (Continued)}% \label{sec:the_galois_group_as_a_permutation_group_continued} % section the_galois_group_as_a_permutation_group_continued We shall continue with the last example of the last lecture. \begin{eg} We considered $G = \Gal(x^3 - 2)$ over $\mathbb{Q}[x]$, and showed that we either have \begin{equation*} G \simeq A_3 \text{ or } G \simeq S_3. \end{equation*} Recall that the roots of $f(x) = x^3 - 2$ are \begin{equation*} \alpha_1 = \sqrt[3]{2}, \, \alpha_2 = \alpha_1 \zeta_3, \, \alpha_3 = \alpha_1 \zeta_3^2. \end{equation*} \sidenote{We want to find an element in $G$ that is an odd permutation, which is something that $A_3$ does not have. To do this, notice that $\zeta_3$ and $\zeta_3^2$ belong to a different minimal polynomial which is inaccessible by $\sqrt[3]{2}$.} Note that $f(x)$ is irreducible over $\mathbb{Q}(\zeta_3)$ \sidenote{Well, none of the roots of $f$ are in $\mathbb{Q}(\zeta_3)$, after all. Also, note that since $\alpha_1 \notin \mathbb{Q}(\zeta_3)$, $f$ remains the minimal polynomial of $\alpha_1$ over $\mathbb{Q}(\zeta_3)$, and so $\deg_{\mathbb{Q}(\zeta_3)} (\alpha_1) = 3$, but $[ \mathbb{Q}(\zeta_3) : \mathbb{Q} ] = 2$.}. By the \cref{crly:the_galois_group_completely_captures_all_permutation_of_the_roots}, $\exists \phi \in G$ such that we have the relation as shown in \cref{fig:crly_galois_group_has_all_permutations_in_action}. \begin{figure}[ht] \centering \begin{tikzcd} \mathbb{Q}(\sqrt[3]{2}, \zeta_3) \arrow[rr, "\phi"] & & \mathbb{Q}(\sqrt[3]{2}, \zeta_3) \arrow[dd, no head] \\ & & \\ \mathbb{Q}(\zeta_3) \arrow[rr, "\zeta_3 \mapsto \zeta_3^2"] \arrow[uu,no head] & & \mathbb{Q}(\zeta_3) \arrow[dd, no head] \\ & & \\ \mathbb{Q} \arrow[rr, "id"] \arrow[uu, no head] & & \mathbb{Q} \end{tikzcd} \caption[][80pt]{\cref{crly:the_galois_group_completely_captures_all_permutation_of_the_roots} in action} \label{fig:crly_galois_group_has_all_permutations_in_action} \end{figure} \sidenote{Note that $\zeta_3 \mapsto \zeta_3^2$ is a valid isomorphism, especially since they have the same minimal polynomial.} Note that \begin{gather*} \phi(\alpha_1) = \alpha_1 \\ \phi(\alpha_2) = \phi(\alpha_1 \zeta_3) = \alpha_1 \zeta_3^2 = \alpha_3 \\ \phi(\alpha_3) = \phi(\alpha_1 \zeta_3^2) = \alpha_1 \zeta_3 = \alpha_2 \end{gather*} It thus follows that $\phi \sim (2 \; 3)$, a $2$-cycle, in $G$. Thus $\phi$ is an element of order $2$, which is an element that $A_3$ does not have. Thus $G \simeq S_3$. \end{eg} From the above example, we notice the following helpful observation. \begin{remark} When computing $G = \Gal(K / F)$, it is often helpful to first know $\abs{G}$. \end{remark} Fortunately, in the finite dimensional world, $\abs{G}$ has an upper bound. \begin{defn}[$F$-map]\index{$F$-map}\label{defn:_f_map} Let $K / F$ and $E / F$. Any \hlnotea{homomorphism} $\phi : K \to E$ which fixes $F$, i.e. $\phi \restriction_F = \id_F$, is called an \hlnoteb{$F$-map}. \end{defn} \begin{remark} Suppose $K / F$ and $E / F$, and $\phi : K \to E$ an $F$-map. \begin{enumerate} \item Since $\ker \phi \neq K$, we have $\ker \Phi = 0$ \sidenote{Note that in finite fields, $\ker \phi \in \{ \{ 0 \}, K \}$.}. Thus $\phi$ is \hlnotea{injective}. \item For any $\alpha \in F$, $v \in K$, $\phi(av) = \phi(a) \phi(v) = a \phi(v)$ since $\phi$ is a homomorphism. It follows that $\phi$ is a \hlnotea{linear transformation}. \item Let $\phi : K \to K$ be an $F$-map, and suppose $K$ is a finite-dimensional $F$-vector space with $[ K : F ] < \infty$. Then $\phi $ is surjective. It follows that $\phi : K \to K$ ($[K : F] < \infty$) is an $F$-map $\iff \phi \in \Gal(K / F)$. \end{enumerate} \end{remark} \begin{lemma}[Number of Distinct $F$-maps]\label{lemma:number_of_distinct_f_maps} Let $K / F$ and $E / F$, and suppose $K / F$ is a finite extension. The number of distinct $F$-maps from $K$ to $E$ is at most $[K : F]$. \end{lemma} \begin{proof} We shall do induction on the number of generators of $K / F$, which is also $[ K : F ] = n$, which is what we can iterate on. When $n = 1$, we have $K = F(\alpha_1)$ and $\phi : K \to E$ an $F$-map. Then the roots $\alpha_1$ and $\phi(\alpha_1)$ have the same minimal polynomial \sidenote{\hlwarn{Why?}} over $F$. Thus, there are at most $[ K : F ]$-many choices for $\phi(\alpha_1)$, meaning that there are at most $[ K : F ]$-many such $F$-maps. \marginnote{\WTS that the number of $F$-maps is at most $[K : F] = [ K : L ][ L : F]$. We can get $[ L : F ]$ from the \hlnotea{induction hypothesis} and $[ K : L ]$ from an argument similar to the base case.} Continuing with this inductive line of thought, suppose that the statement is true for $K = F(\alpha_1, \ldots, \alpha_n)$ for some $n > 1$. Now let \begin{equation*} L = F(\alpha_1, \ldots, \alpha_{n - 1}), \text{ so that } K = L(\alpha_n). \end{equation*} Let $\phi : K \to E$ be an $F$-map. Note that $\phi \restriction_L : L \to E$ is still an $F$-map. By the induction hypothesis, the number of choices for $\phi \restriction_L$ is at most $[ L : F ]$. Since $\phi$ is completely determined by $\phi \restriction_L$ and $\phi(\alpha_n)$, there are, therefore, at most \begin{equation*} [L : F][ L(\alpha_n) : L ] = [ K : F ]\text{-many} \end{equation*} choices for $\phi$, following the Tower Theorem. \end{proof} The following corollary follows immediately from the realization that $F$-maps going from $K \to K$ are exactly the elements of the Galois group $\Gal(K / F)$. \begin{crly}[Upper Bound for the Galois Group of Finite Extensions]\label{crly:upper_bound_for_the_galois_group_of_finite_extensions} If $K / F$ is finite, then \begin{equation*} \abs{ \Gal(K / F) } \leq [ K : F ]. \end{equation*} \end{crly} \begin{warning} There are extensions $K$ of a field $F$ such that $\abs{\Gal(K / F)} < [ K : F ]$. \begin{enumerate} \item We saw in an earlier example that $G = \Gal( \mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q} ) = \{ 1 \}$, but $[ \mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q} ] = 3$, and $\sqrt[3]{2} \zeta_3, \, \sqrt[3]{2} \zeta_3^2 \notin \mathbb{Q}(\sqrt[3]{2})$. In this case, the Galois group is too tiny. \item Consider $G = \Gal( \mathbb{Z}_2(x) / \mathbb{Z}_2(t^2) )$. Note that $[\mathbb{Z}_2(x) : \mathbb{Z}_2(t^2)] = 2$, since the minimal polynomial of $t$ in $\mathbb{Z}_2(t^2)[x]$ is \begin{equation*} x^2 - t^2 = (x - t)^2 \in \mathbb{Z}_2(t)[x]. \end{equation*} Thus if $\phi \in G$, then it is necesssary that $\phi(t) = t$, implying that $G = \{ 1 \}$. In this case, it is because $t$ is a root with multiplicity $> 1$. \end{enumerate} \end{warning} % section the_galois_group_as_a_permutation_group_continued (end) % chapter lecture_19_feb_27th (end) \chapter{Lecture 20 Mar 01st}% \label{chp:lecture_20_mar_01st} % chapter lecture_20_mar_01st \section{Galois Group of Separable Fields}% \label{sec:galois_group_of_separable_fields} % section galois_group_of_separable_fields \begin{quotebox}{green}{foreground} So when exactly does $\abs{ \Gal(K / F) } = [ K : F ]$? \end{quotebox} \begin{defn}[Separable Elements and Separable Extensions]\index{Separable Elements}\index{Separable Extensions}\label{defn:separable_elements_and_separable_extensions} Let $K / F$ \sidenote{This need not be a finite extension.}. We say that $\alpha \in K$ is \hlnoteb{separable} if $\alpha$ is \hlnotea{algebraic} over $F$ and its minimal polynomial is \hyperref[defn:separable_polynomials]{separable} (over $F$) \sidenote{This also means that the root is unique.}. We say that the extension $K / F$ is \hlnoteb{separable} if $K / F$ is \hyperref[defn:algebraic_and_transcendental]{algebraic} and $\forall \alpha \in K$, $\alpha$ is separable over $F$. \end{defn} \marginnote{ \begin{remark} \cref{defn:perfect_fields} means that all polynomials over the field are separable, i.e. they do not have repeated roots. \end{remark}} \begin{defn}[Perfect Fields]\index{Perfect Fields}\label{defn:perfect_fields} We say that a field $F$ is \hlnoteb{perfect} if every algebraic extension of $F$ is separable. \end{defn} \begin{note} Recall from A5, we showed that given an irreducible $f(x) \in F[x]$, \begin{equation*} f(x) \text{ is separable } \iff f'(x) \neq 0. \end{equation*} \end{note} \begin{propo}[Separability and the Characteristic of a Field]\label{propo:separability_and_the_characteristic_of_a_field} Let $f(x) \in F[x]$ be irreducible. \begin{enumerate} \item If $\Char F = 0$, then $f(x)$ is separable. \sidenote{This is proven in A5.} \item If $\Char F = p$ prime, then $f(x)$ is not separable iff $f(x) = g(x^p)$ for some $g(x) \in F[x]$. \end{enumerate} \end{propo} \begin{proof} \begin{enumerate} \setcounter{enumi}{1} \item Let \begin{equation*} f(x) \in a_n x^n + a_{n - 1} x^{n - 1} + \hdots + a_1 x + a_0 \end{equation*} Then $f(x)$ is not separable \\ $\iff$ $f'(x) = 0$ \\ $\iff$ $na_n x^{n - 1} + (n - 1)a_{n - 1} x^{n - 2} + \hdots + a_1 = 0$ \\ $\iff$ $ka_k = 0$ for $k \in \{ 1, \, \ldots, \, n \}$ \\ $\iff$ $ka_k = pm_k a_k$ where $m_k \in \mathbb{N}$, $k \in \{ 1, \ldots, n \}$ since either $p \mid k$ or $a_k = 0$ \\ $\iff$ $f(x) = a_n x^{m_n p} + a_{n - 1}x^{m_{n - 1} p} + \hdots + a_1 x^{m_1 p} + a_0$ \\ $\iff$ $f(x) = g(x^p)$ where \begin{equation*} g(x) = a_n x^{m_n} + a_{n - 1} x^{m_{n - 1}} + \hdots + a_1 x^{m_1} + a_0. \end{equation*} \end{enumerate} \end{proof} \begin{crly}[Fields of Characteristic Zero are Perfect]\label{crly:fields_of_characteristic_zero_are_perfect} If $\Char F = 0$, then $F$ is perfect. \end{crly} \begin{eg} Note that in $\mathbb{Z}_2(t) / \mathbb{Z}_2(t^2)$, we have that \begin{equation*} x^2 - t^2 = (x - t)^2, \end{equation*} i.e. $t$ is a root with multiplicity $2$. Thus $\mathbb{Z}_2(t^2)$ is not perfect. \end{eg} \begin{crly}[Every Finite Field is Perfect]\label{crly:every_finite_field_is_perfect} Every finite field $F$ is perfect. \end{crly} \marginnote{ \begin{strategy} Of course, we want to use \cref{propo:separability_and_the_characteristic_of_a_field}. We can do so by supposing that $f(x)$ is irreducible but not separable, which then forces $f(x) = g(x^p)$. The important point here is to notice that in a finite field, all elements of the field will eventually cycle back as we add or multiply them. Then, by using the fact that $\Char F = p$ is prime, in particular by the \hlnotea{Freshman's Dream}, we can use \hldefn{Frobenius's Homomorphism} $\phi(a) = a^p$, and we end up showing that every element in $F$ is some other element of $F$ with power $p$. This will cause $f(x)$ to become reducible due to the Freshman's Dream. \end{strategy}} \begin{proof} Let $F$ be finite with $\Char F = p > 0$ \sidenote{Note that fields of characteristic $0$ must be infinite, so this is a valid assumption.}. Suppose to the contrary that $\exists f(x) \in F[x]$ such that $f(x)$ is irreducible but not separable. Then $\exists g(x) \in F[x]$ such that $f(x) = g(x^p)$. In particular, we have \begin{equation*} f(x) = a_n x^{p m_n} + a_{n - 1} x^{p m_{n - 1}} + \hdots + a_1 x^{p m_1} + a_0. \end{equation*} Now consider $\phi : F \to F$ given by $\phi(a) = a^p$. By the \hlnotea{Freshman's Dream}, $\phi$ is a homomorphism. It is clear that it is injective since if $a \neq b$, then $a^p \neq b^p$, for otherwise \begin{equation*} 0 = a^p - b^p = (a - b)^p \iff 0 = a - b \iff a = b. \end{equation*} Also, since $F$ is finite, injectivity of $\phi$ guarantees that it is surjective. This means that $\forall a_k \in F$, $\exists b_k \in F$ such that \begin{equation*} a_k = b_k^p = \phi(b_k). \end{equation*} Then we have \begin{align*} f(x) &= b_n^p x^{p m_n} + b_{n - 1}^p x^{p m_{n - 1}} + \hdots + b_1^p x^{p m_1} + b_0^p \\ &= ( b_n x^{m_n} + b_{n - 1} x^{m_{n - 1}} + \hdots + b_1 x^{m_1} + b_0 )^p, \end{align*} again, by the \hlnotea{Freshman's Dream}. Therefore $f(x)$ is reducible, contradicting our assumption. \end{proof} \begin{thm}[Galois Group of a Splitting Field of a Separable Polynomial has Order the Degree of the Extension]\label{thm:galois_group_of_a_splitting_field_of_a_separable_polynomial_has_order_the_degree_of_the_extension} Let $f(x) \in F[x]$ be non-constant and separable. Let $K$ be the splitting field of $f(x)$ over $F$. Then \begin{equation*} \abs{ \Gal(K / F) } = \abs{ \Gal(f(x)) } = [ K : F ]. \end{equation*} \end{thm} \begin{proof} We shall perform induction on $[ K : F ] = n$. \noindent \hlbnoted{$n = 1$} We have \begin{equation*} 1 \leq \abs{ \Gal(K / F) } \leq [ K : F ] \leq 1, \end{equation*} since we always have $\epsilon \in \Gal(K / F)$. Proceeding inductively... \noindent \hlbnoted{$n = k + 1$} Let $p(x) \in F[x]$ be an irreducible factor of $f(x)$ \sidenote{Note that it suffices for us to show for irreducible polynomials, since we can always factor a polynomial into irreducible terms.}. Note that $p(x)$ is also separable over $F$. Let \begin{equation*} \alpha_1, \ldots, \alpha_m \in K \end{equation*} be the roots of $p(x)$, where $m = \deg p(x)$, and we note that $\alpha_i \neq \alpha_j$ for all $i \neq j$ since $p(x)$ is separable. Now since $[ K : F ] > 1$, wma $\alpha_1 \notin F$. Then consider $E = F(\alpha_1)$. Since $p(x)$ is irreducible in $F[x]$, it follows that $[ E : F ] = m$. Thus by the \hyperref[thm:tower_theorem]{Tower Theorem}, we have \begin{equation*} [ K : E ] = \frac{[ K : F ]}{[ E : F ]} = \frac{n}{m} < n. \end{equation*} Note that we still have $K$ as the splitting field of $f(x)$ over $E$. It follows from induction that \begin{equation}\label{eq:galois_group_of_splitting_field_of_separable_polym_1} \abs{ \Gal( K / E ) } = [ K : E ] = \frac{n}{m}. \end{equation} Since $p(x)$ is irreducible, by the \hyperref[lemma:isomorphism_extension_lemma]{Isomorphism Extension Lemma}, $\forall j$, $\exists \phi_j \in \Gal(K / F)$ such that $\phi_j(\alpha_1) = \alpha_j$. Since the roots are distinct, it follows that each of the $\phi_j$'s are distinct in $\Gal (K / F)$, and there are $m$-many such automorphisms. Furthermore, we have that $\phi_j^{-1} \phi_i (\alpha_1) \neq \alpha_1 \in E$, and so $\phi_j^{-1} \phi_i \notin \Gal(K / E)$. This means that \begin{equation*} \phi_j \Gal(K / E) \neq \phi_i \Gal(K / E), \end{equation*} and so we have that there must be \begin{equation*} \abs{ \Gal(K / F) / \Gal(K / E) } \geq m. \end{equation*} By \hyperref[thm:lagrange_s_theorem]{Lagrange}, we have from \cref{eq:galois_group_of_splitting_field_of_separable_polym_1} that \begin{equation*} \abs{ \Gal(K / F) } \geq m \cdot \abs{ \Gal(K / E) } = m \cdot \frac{n}{m} = n, \end{equation*} as desired. \end{proof} % section galois_group_of_separable_fields (end) % chapter lecture_20_mar_01st (end) \chapter{Lecture 21 Mar 04th}% \label{chp:lecture_21_mar_04th} % chapter lecture_21_mar_04th \section{The Primitive Element Theorem}% \label{sec:the_primitive_element_theorem} % section the_primitive_element_theorem We shall now look at a rather `simple' case of splitting fields of separable polynomials. \begin{defn}[Simple Extension and Primitive Elements]\index{Simple Extension}\index{Primitive Element}\label{defn:simple_extension_and_primitive_elements} We say that $K / F$ is \hlnoteb{simple} if $\exists \alpha \in K$ such that $K = F(\alpha)$. We call $\alpha$ a \hlnoteb{primitive element} for $K / F$. \end{defn} \marginnote{ You may want to look at \nameref{strategy:analysis_of_proof_of_primitive_element_theorem} first before diving into the proof for \cref{thm:primitive_element_theorem}. The proof provided in class makes the proof look as if it is a struck of genius, but it is actually through some frolicking around with finding out what we need, that one can realize why we chose to pick such a specifically defined $S$. } \begin{thm}[Primitive Element Theorem]\index{Primitive Element Theorem}\label{thm:primitive_element_theorem} If $K / F$ is finite and \hyperref[defn:separable_elements_and_separable_extensions]{separable}, then $K / F$ is simple. \end{thm} This is an important result to us, since it would imply the following. \begin{crly}[Finite Extensions of Perfect Fields are Simple]\label{crly:finite_extensions_of_perfect_fields_are_simple} If $F$ is perfect and $K / F$ is finite, then $K / F$ is simple. \end{crly} \begin{eg} Fields of characteristic $0$ and finite fields have simple extensions. \end{eg} \begin{proof}[\cref{thm:primitive_element_theorem}] \hlbnoted{$F$ is finite} Then $K$ is necessarily finite since it is a finite extension, and thus $K^\times = \langle \alpha \rangle$ for some $\alpha \in K$. Hence $K = F(\alpha)$. \marginnote{ \begin{strategy} Note that \cref{thm:primitive_element_theorem} does not assume if $F$ is finite or infinite, and so we must deal with either cases separately. \end{strategy} } \noindent \hlbnoted{$F$ is infinite} Since $K / F$ is finite, we may assume \begin{equation*} K = F(\pi_1, \ldots, \pi_n) \end{equation*} for some $\pi_i \in K$. It suffices for us to show that $\forall \alpha, \beta \in K$, $\exists \gamma \in K$ such that such that \begin{equation*} K = F(\alpha, \beta) = F(\gamma), \end{equation*} since our desired result will simply follow by arguing repeatedly. Let $p(x)$ and $q(x)$ be the minimal polynomials of $\alpha$ and $\beta$, respectively, over $F$. Now let $L$ be the splitting field of $p(x) q(x)$ over $K$. Let the roots of $p(x)$ be \begin{equation*} \alpha = \alpha_1, \, \alpha_2, \, \ldots, \, \alpha_n, \end{equation*} and the roots of $q(x)$ be \begin{equation*} \beta = \beta_1, \, \beta_2, \, \ldots, \, \beta_m. \end{equation*} By separability, $\alpha_i \neq \alpha_j$ and $\beta_i \neq \beta_j$ for all $i \neq j$. Now let \sidenote{Note that had we wanted to start with $\gamma = \beta + u \alpha$, we would have declared $S$ with elements like $\frac{\beta_j - \beta_1}{\alpha_1 - \alpha_i}$.} \begin{equation*} S := \left\{ \frac{\alpha_i - \alpha_1}{\beta_1 - \beta_j} \mmid 1 < i \leq n, \, 1 < j \leq m \right\}. \end{equation*} Since $S$ is finite while $F$ is infinite, $\exists u \in F^\times$ such that $u \notin S$. Let $\gamma = \alpha + u \beta$. \noindent \hlbnotea{Claim: $F(\alpha, \beta) = F(\gamma)$} Clearly, $F(\gamma) \subseteq F(\alpha, \beta)$ since $\gamma \in F(\alpha, \beta)$. Let $h(x)$ be the minimal polynomial of $\beta$ over $F(\gamma)$. Since $q(\beta) = 0$, we have that $h(x) \mid q(x)$, and consequently if $h(\faLock) = 0$, then $\faLock = \beta_j$ for some $j \in \{ 1, \ldots, m \}$. Now let $k(x) = p(\gamma - ux) \in F(\gamma)[x]$. Notice that \begin{equation*} k(\beta) = p(\gamma - u \beta) = p(\alpha) = 0. \end{equation*} Thus $h(x) \mid k(x)$. Notice that for $j > 1$, we have \begin{align*} k(\beta_j) = 0 &\iff p(\gamma - u \beta_j) = 0 \\ &\iff \gamma - u \beta_j = \alpha_i \text{ for some } i \\ &\iff \alpha_1 + u \beta_1 - u \beta_j = \alpha_i \\ &\iff u = \frac{\alpha_i - \alpha_1}{\beta_1 - \beta_j} \in S. \end{align*} Thus we know that for these $j$'s, $k(\beta_j) \neq 0$ since we chose $u \notin S$. It follows that $h(\beta_j) \neq 0$ for $j > 1$, and so $h(x) = x - \beta \in F(\gamma)[x]$, implying that $\beta \in F(\gamma)$. Using the same argument, we can show that $\alpha \in F(\gamma)$. This completes the proof, \end{proof} \marginnote{This is indeed a very profound result, making use of relatively simple notions such as minimal polynomials and splitting fields, and then proving for us a theorem that helps us narrow down the choice of the algebraic number to a single number that extends the base field to the extension.} \begin{strategy}[Analysis of the proof for the Primitive Element Theorem]\label{strategy:analysis_of_proof_of_primitive_element_theorem} If we want $F(\alpha, \beta) = F(\gamma)$ for some $\gamma$, one naive choice is to go with $\gamma = \alpha + u \beta$ for some $u \in F^\times$ and hope that this will force $\alpha, \beta \in F(\gamma)$. The argument is similar for either $\alpha$ or $\beta$ (by switching variables), so let's think about only one of them. Now $q(x)$ is not necessarily a minimal polynomial of $\beta$ over $F(\gamma)$, so let's make use of that. If $\beta \in F(\gamma)$, then we must have $x - \beta \mid q(x)$. So let's \hlbnotec{consider the minimal polynomial} $h(x)$ of $\beta$ in $F(\gamma)$, which would divide $q(x)$. Of course, ideally, we want $h(x) = x - \beta$. Then let's suppose that $h(x)$ has some root other than $\beta$. Then \hlbnotec{in the splitting field} of $q(x)$, where $h(x)$ must then also split, since $q(x)$ is separable, we have that $h(x)$ must therefore be able to split into linear terms, where each linear term has a root of $q(x)$ as its constant value. In other words, \hlbnotec{all roots of $h(x)$ are roots of $q(x)$.} Then we notice another possible polynomial that such an $h(x)$ can divide: we know that $\alpha = \gamma - u \beta$, and $\alpha$ is a root of $p(x)$. Then if we let $k(x) = p(\gamma - u x)$, we have \begin{equation*} k(\beta) = p(\gamma - u \beta) = p(\alpha) = 0. \end{equation*} So $h(x) \mid k(x)$. Now since all the roots of $h(x)$ are roots of $q(x)$, these roots must also be roots of $k(x)$. Let these other roots of $q(x)$ be labelled $\beta_j$'s. Then picking $\beta_j \neq \beta$, we have \begin{equation*} k(\beta_j) = 0 \iff p(\gamma - u \beta_j) = 0. \end{equation*} We already know what the roots of $p(x)$ are so let's label those as $\alpha_i$. Then \begin{equation*} \gamma - u \beta_j = \alpha_i. \end{equation*} Note that $\alpha_i \neq \alpha$ since the roots are unique. Following that, \begin{equation*} \alpha + u \beta - u \beta_j = \alpha_i, \end{equation*} which then \begin{equation}\label{eq:pet_strat_choice_of_units} u = \frac{\alpha_i - \alpha}{\beta - \beta_j}. \end{equation} We notice that there are only finitely many such $u$'s in $F^\times$ since there are only as many as the roots $\alpha_i$'s and $\beta_j$'s can allow. However, $F^\times$ is infinite by our assumption, i.e. there are always units of $F$ that cannot be expressed as in \cref{eq:pet_strat_choice_of_units}. So by picking a $u \in F^\times$ that is not determined by \cref{eq:pet_strat_choice_of_units}, we rule out the possibility that $k(x)$ has these other $\beta_j$'s as roots, and hence forcing $h(x)$ to be what we want: that is $h(x) = x - \beta$. Thus our job is done for showing that $\beta \in F(\gamma)$! We can then apply the same argument to showing that $\alpha \in F(\gamma)$, by letting $\gamma = \beta + u' \alpha$ by choosing $u'$ in a similar fashion as above. In this case, we would have to extend our working field to the splitting field of $p(x)$. Then to put the two together, we could have then started working with an extension where both $p(x)$ and $q(x)$ splits, and the splitting field of $p(x) q(x)$ is exactly where we should be working in. \end{strategy} % section the_primitive_element_theorem (end) % chapter lecture_21_mar_04th (end) \chapter{Lecture 22 Mar 06th}% \label{chp:lecture_22_mar_06th} % chapter lecture_22_mar_06th \section{Normal Extensions}% \label{sec:normal_extensions} % section normal_extensions \begin{note} Given $f(x) \in F[x]$ irreducible, $K$ the splitting field of $f(x)$ over $F$, $\alpha, \beta \in K$ the roots of $f(x)$, the \hyperref[lemma:isomorphism_extension_lemma]{Isomorphism Extension Lemma} \sidenote{See also \cref{crly:the_galois_group_completely_captures_all_permutation_of_the_roots}.} tells us that $\exists \phi \in \Gal(K / F)$ such that $\phi(\alpha) = \beta$. \end{note} There is a slightly more general result which we shall use today, whose proof shall be left as an exercise. \begin{ex}\label{ex:galois_permutations_work_even_with_reducible_polyms} Let $f(x) \in F[x]$ be non-constant, $K$ the splitting field of $f(x)$ over $F$, and $\alpha, \beta \in K$ have the same minimal polynomial in $F[x]$. Then $\exists \phi \in \Gal(K / F)$ such that $\phi(\alpha) = \beta$. \end{ex} \begin{eg} Recall our `favorite' example $\Gal( \mathbb{Q}(\sqrt[3]{2}) / \mathbb{Q} ) = \{ 1 \}$, where we had that the other roots of $x^3 - 2$, which are $\sqrt[3]{2} \zeta_3$ and $\sqrt[3]{2} \zeta_3^2$, are not in $\mathbb{Q}(\sqrt[3]{2})$. Notice that $x^3 - 2$ does not split in $\mathbb{Q}(\sqrt[3]{2})$. \end{eg} \begin{defn}[Normal Extension]\index{Normal Extension}\label{defn:normal_extension} Let $[ K : F ] < \infty$. We say that $K / F$ is \hlnoteb{normal} if $K$ is the \hyperref[defn:splitting_field]{splitting field} of some non-constant $f(x) \in F[x]$. \end{defn} \begin{defn}[$F$-conjugates]\index{$F$-conjugates}\label{defn:_f_conjugates} Let $[ K : F ] < \infty$. Let $\alpha \in K$ with minimal polynomial $p(x) \in F[x]$. The roots of $p(x)$ in its splitting field are called the \hlnoteb{$F$-conjugates} (or just conjugates) of $\alpha$. \end{defn} We have the following theorem that characterizes normality. \begin{thm}[Normality Theorem]\index{Normality Theorem}\label{thm:normality_theorem} Let $[ K : F ] < \infty$. TFAE: \begin{enumerate} \item $K / F$ is normal. \item For all extensions $L$ over $K$, if $\phi$ is an \hyperref[defn:_f_map]{$F$-map} from $L \to L$, then $\phi \restriction_K \in \Gal(K / F)$. \item If $\alpha \in K$, then all $F$-conjugates of $\alpha$ are also in $K$. \item If $\alpha \in K$, then its minimal polynomial splits over $K$. \end{enumerate} \end{thm} \begin{proof} It is clear that $(3) \implies (4)$. \noindent \hlbnoted{$(1) \implies (2)$} Suppose $K / F$ is normal. Then $K$ is the splitting field of some non-constant $f(x) \in F[x]$. Let $\phi : L \to L$ be an $F$-map. Since $[ K : F ] < \infty$, wma \begin{equation*} K = F(\alpha_1, \ldots, \alpha_n), \quad \alpha_i \in K, \, f(\alpha_i) = 0. \end{equation*} Then $\forall i$, $\exists j$ such that $\phi \restriction_K (\alpha_i) = \alpha_j \in K$, since $\phi$ is an $F$-map. We see that $\phi \restriction_K \in \Gal(K / F)$ both fixes $F$ and is an automorphism of $K$. \noindent \hlbnoted{$(2) \implies (3)$} Let $\alpha \in K$ with minimal polynomial $f(x) \in F[x]$. Since $K / F$ is finite, once again, let \begin{equation*} K = F(\alpha_1, \ldots, \alpha_n) \end{equation*} for $\alpha_i \in K$. For each $i$, let $h_i(x)$ be the minimal polynomial of $\alpha_i$ over $F$. Now define \sidenote{We want to consider such a polynomial because it will contain all of the $\alpha_i$'s and their conjugates, and we hope to see that the splitting field of $p(x)$, where we can then apply our assumption.} \begin{equation*} p(x) = f(x) h_1(x) h_2(x) \hdots h_n(x). \end{equation*} Let $L$ be the splitting field of $p(x)$ over $F$. By construction, $L / K / F$ is a tower of fields. Let $\beta \in L$ be a root of $f(x)$ \sidenote{i.e. we consider a conjugate of $\alpha_i$, for one of the $i$'s.}. By \cref{ex:galois_permutations_work_even_with_reducible_polyms}, $\exists \phi \in \Gal(K / F)$ such that $\phi(\alpha_i) = \beta$. Since $\phi$ is an $F$-map, our assumption tells us that $\phi \restriction_K \in \Gal(K / F)$. Since $\alpha_i \in K$, we have that $\beta \in K$. This proves what is required. \noindent \hlbnoted{$(4) \implies (1)$} Suppose $K / F$ is finite, and write \begin{equation*} K = F(\alpha_1, \ldots, \alpha_n) \end{equation*} for $\alpha_i \in K$. Let $h_i(x)$ be the minimal polynomial of $\alpha_i$ over $F$. Let $f(x) = \prod_{i=1}^{n} h_i(x)$. It follows from (3) that the splitting field of $f(x)$ over $F$ is $F(\alpha_1, \ldots, \alpha_n) = K$ since $K$ contains all of the $F$-conjugates of each $\alpha_i$. Thus $K / F$ is normal. \end{proof} \begin{eg} As we've just seen in this lecture, $\mathbb{Q}(\sqrt[3]{2}) / \mathbb{Q}$ is not normal since $\sqrt[3]{2} \zeta_3 \notin \mathbb{Q}(\sqrt[3]{2})$ while $\sqrt[3]{2} \in \mathbb{Q}(\sqrt[3]{2})$. \end{eg} \begin{eg} $\mathbb{F}_{p^n} / \mathbb{F}_p$ is normal since $\mathbb{F}_{p^n}$ is the splitting field of $x^{p^n} - x \in \mathbb{F}_p[x]$. \end{eg} \begin{eg} \hlnotea{Cyclotomic extensions} $\mathbb{Q}(\zeta_n) / \mathbb{Q}$ are normal since it is the splitting field of $\Phi_n(x)$. \end{eg} \begin{eg}[\imponote] $\mathbb{Z}_p(t) / \mathbb{Z}(t^p)$ is normal since it is the splitting field of $x^p - t^p = (x - t)^p$. This is an example of a normal extension that is \hlimpo{not separable}. \end{eg} % section normal_extensions (end) \section{Galois Extensions}% \label{sec:galois_extensions} % section galois_extensions \begin{defn}[Galois Extension]\index{Galois Extension}\label{defn:galois_extension} Suppose $[ K : F ] < \infty$. We say that $K / F$ is \hlnoteb{Galois} if $K / F$ is \hyperref[defn:normal_extension]{normal} and \hyperref[defn:separable_elements_and_separable_extensions]{separable}. \end{defn} % section galois_extensions (end) % chapter lecture_22_mar_06th (end) \chapter{Lecture 23 Mar 08th}% \label{chp:lecture_23_mar_08th} % chapter lecture_23_mar_08th \section{Galois Extensions (Continued)}% \label{sec:galois_extensions_continued} % section galois_extensions_continued \begin{defn}[Fixed Field]\index{Fixed Field}\label{defn:fixed_field} Let $K$ be a field. If $G \leq \Aut(K)$, the \hlnoteb{fixed field} of $G$ is defined as \begin{equation*} \Fix(G) := \left\{ a \in K \mmid \phi(a) = a, \, \phi \in G \right\}. \end{equation*} \end{defn} \begin{ex} Check that $\Fix(G) \leq K$ is indeed a field. \end{ex} \begin{remark} The following is noteworthy: \begin{equation*} \Fix(\Gal(K / F)) \supseteq F. \end{equation*} This follows immediately from the \hyperref[defn:galois_group]{definition of a Galois group}. \end{remark} \begin{eg} Consider \begin{equation*} f(x) = (x - \sqrt{2})^2 (x + \sqrt{2})^2 = x^4 - 4x^2 + 4. \end{equation*} Then there are $\phi \neq \epsilon \in G = \Gal(f(x))$ such that $\phi(\sqrt{2}) = \sqrt{2}$ (and $\phi(-\sqrt{2}) = \sqrt{2}$). In this case, \begin{equation*} \Fix(G) \supsetneq F. \end{equation*} \end{eg} \begin{thm}[Characterization of Galois Extensions]\index{Characterization of Galois Extensions}\label{thm:characterization_of_galois_extensions} Suppose $[ K : F ] < \infty$. TFAE: \begin{enumerate} \item $K$ is the splitting field of a non-constant separable $f(x) \in F[x]$ over $F$. \item $\abs{\Gal(K / F)} = [ K : F ]$. \item $\Fix(\Gal(K / F)) = F$. \item $K / F$ is Galois. \end{enumerate} \end{thm} \begin{proof} \hlbnoted{$(1) \implies (2)$} See \cref{thm:galois_group_of_a_splitting_field_of_a_separable_polynomial_has_order_the_degree_of_the_extension}. \noindent \hlbnoted{$(2) \implies (3)$} Suppose $\abs{ \Gal(K / F) } = [ K : F ]$. By our earlier remark, we have that $\Fix(\Gal(K / F)) \supseteq F$. So it suffices to show that $\Fix(\Gal(K / F)) \subseteq F$. Let $E = \Fix(\Gal(K / F))$. Then we have a tower $K / E / F$. By the \hyperref[thm:tower_theorem]{Tower Theorem}, \begin{equation*} [K : F] = [K : E][E : F]. \end{equation*} It suffices to show that $[ E : F ] = 1$. Now, note that we have $\Gal(K / E) \leq \Gal(K / F)$ \sidenote{Note that the former fixes more things, and so there is less `space' for the permutations to move around.}. By assumption, we have \begin{equation*} \abs{ \Gal(K / E) } \leq \abs{ \Gal(K / F) } = [K : F]. \end{equation*} Let $\alpha \in E$ and $\phi \in \Gal(K / F)$. By the construction of $E$, we have that $\phi(\alpha) = \alpha$. It follows that $\phi$ also fixes $E$, and so $\phi \in \Gal(K / E)$. Thus $\Gal(K / F) \leq \Gal(K / E)$. Hence $\Gal(K / E) = \Gal(K / F)$. This shows that \begin{equation*} [ K : F ] = \abs{ \Gal(K / E) } \leq [ K : E ] \leq [ K : F ], \end{equation*} which implies that $[E : F] = 1$, as we desired. \noindent \hlbnoted{$(3) \implies (4)$} Suppose $\Fix(\Gal(K / F)) = F$. Let $\alpha \in K$ with minimal polynomial $p(x)$ over $F$. WTS $p(x)$ splits over $K$ with no repeated roots. Let $G = \Gal(K / F)$. Consider \sidenote{$\Delta$ gives us all the roots of $p(x)$.} \begin{equation*} \Delta = \left\{ \phi(\alpha) : \phi \in G \right\} \subseteq K. \end{equation*} Let $\alpha_1, \ldots, \alpha_n \in \Delta$ be distinct, and wlog wma $\alpha = \alpha_1$. Now consider \begin{equation*} h(x) = (x - \alpha_1)(x - \alpha_2) \hdots (x - \alpha_n) \in K[x]. \end{equation*} Then in $K[x]$, we have that $h(x) \mid p(x)$, since all roots of $h(x)$ are roots of $p(x)$. Now for any $\phi \in G$, we have that $\phi(\alpha_i) = \alpha_j$, and so \begin{equation*} \phi(h(x)) = h(x), \end{equation*} which means $h(x) \in \Fix(G)[x] = F[x]$ by assumption. Since $p(x)$ is the minimal polynomial of $\alpha$ in $F[x]$, it follows that $p(x) \mid h(x)$ and so $p(x) = h(x)$, which splits over $K$ and has no repeated roots, just as we wanted. \noindent \hlbnoted{$(4) \implies (1)$} Suppose $K / F$ is Galois \sidenote{By \hyperref[defn:galois_extension]{definition}, $K / F$ is normal and separable, which means \begin{itemize} \item (\textbf{normal}) $K$ is the splitting field of some non-constant $f(x) \in F[x]$; and \item (\textbf{separable}) for each $\alpha \in K$, the minimal polynomial $p(x)$ of $\alpha$ over $F$ is separable in $K$. \end{itemize} So there is actually something to prove because we only know that $K$ is the splitting field of some polynomial that may or may not be a minimal polynomial of any $\alpha \in K$, and $K$ may not be the splitting field of any of the minimal polynomial of its elements.}. Since $[ K : F ] < \infty$, let \begin{equation*} K = F(\alpha_1, \ldots, \alpha_n), \end{equation*} where $\alpha_i \in K$. Let $q_i(x) \in F[x]$ be the minimal polynomial of each $\alpha_i$. Let $p_1(x), \ldots, p_m(x)$ be the distinct $q_i(x)$'s. Since $K / F$ is separable, we know that the $q_i(x)$'s are separable and hence so are the $p_i(x)$'s. Let \begin{equation*} f(x) = p_1(x) p_2(x) \hdots p_n(x) \in F[x]. \end{equation*} By A6Q3(b), we have that $f(x)$ is separable over $F$ in $K$. Also, $K / F$ is normal since the splitting field of $f(x)$ over $F$ is exactly $K$. \end{proof} \begin{eg} Let's look at our non-Galois extension of $\mathbb{Q}$, $\mathbb{Q}(\sqrt[3]{2})$. We observe that \begin{equation*} \Fix ( \Gal( \mathbb{Q}(\sqrt[3]{2}) / \mathbb{Q} ) ) = \mathbb{Q}(\sqrt[3]{2}), \end{equation*} since $\sqrt[3]{2}$ fixes even $\sqrt[3]{2}$ itself. \end{eg} \begin{eg} Let $\alpha = \sqrt{2 + \sqrt{3}} \in \mathbb{C}$. Note that \begin{align*} \alpha^2 & = 2 + \sqrt{3} \\ \alpha^2 - 2 & = \sqrt{3} \\ (\alpha^2 - 2)^2 & = 3 \\ \alpha^4 - 4 \alpha^2 + 4 & = 3 \\ \alpha^4 - 4 \alpha^2 + 1 & = 0 \end{align*} Consider the polynomial \begin{equation*} f(x) = x^4 - 4x^2 + 1 \in \mathbb{Q}[x], \end{equation*} and $\alpha$ is a root of $f(x)$. Now consider the polynomial \begin{equation*} g(y) = y^2 - 4y + 1. \end{equation*} Note that in $\mathbb{Z}_5$, under $\tilde{g}(y) = y^2 + y + 1$, we have \begin{gather*} 0 \mapsto 1 \quad 1 \mapsto 3 \quad 2 \mapsto 2 \\ 3 \mapsto 3 \quad 4 \mapsto 1. \end{gather*} By \cref{propo:irreducible_rootless_polynomials}, $\tilde{g}(y)$ is irreducible, and by \hyperref[propo:mod_p_irreducibility_test]{Mod-$5$ irreducibility}, $g(y)$ is irreducible. Thus $f(x)$ itself is irreducible. Since $f$ is monic, it is the minimal polynomial of $\alpha$. Note that by construction, all the roots of $f(x)$ are \begin{equation*} \pm \sqrt{2 \pm \sqrt{3}}. \end{equation*} Thus $f(x)$ is separable. Note that \begin{equation*} \sqrt{2 + \sqrt{3}} \sqrt{2 - \sqrt{3}} = 1, \end{equation*} and so $\sqrt{2 - \sqrt{3}}$ is the inverse of $\sqrt{2 + \sqrt{3}}$, and so it is necessarily in $\mathbb{Q}(\alpha)$. It follows that all the $\mathbb{Q}$-conjugates of $\alpha$ are in $\mathbb{Q}(\alpha)$. Thus $\mathbb{Q}(\alpha) / \mathbb{Q}$ is normal. It follows that $\mathbb{Q}(\alpha) / \mathbb{Q}$ is a Galois extension. By \hyperref[thm:characterization_of_galois_extensions]{the characterization of Galois extensions}, we have \begin{equation*} \abs{\Gal(\mathbb{Q}(\alpha)/\mathbb{Q})} = [\mathbb{Q}(\alpha) : \mathbb{Q}] = 4. \end{equation*} Let $\beta = \sqrt{2 - \sqrt{3}}$. By the note that $\alpha \beta = 1$, we have that the following table describes all the possible actions of $G$ on the $\mathbb{Q}$-conjugates of $\alpha$. \begin{table}[ht] \centering \caption{Table of elements of $\Gal(\mathbb{Q}(\alpha)/\mathbb{Q})$.} \label{table:table_of_elements_of_gal_q_alpha_q} \begin{tabular}{c | c c c c | c} & $\alpha$ & $-\alpha$ & $\beta$ & $-\beta$ & $S_4$ \\ \hline $\phi_1$ & $\alpha$ & $-\alpha$ & $\beta$ & $-\beta$ & $\epsilon$ \\ $\phi_2$ & $-\alpha$ & $\alpha$ & $-\beta$ & $\beta$ & $(1 \; 2)(3 \; 4)$ \\ $\phi_3$ & $\beta$ & $-\beta$ & $\alpha$ & $-\alpha$ & $(1 \; 3)(2 \; 4)$ \\ $\phi_4$ & $-\beta$ & $\beta$ & $-\alpha$ & $\alpha$ & $(1 \; 4)(2 \; 3)$ \end{tabular} \end{table} By \cref{table:table_of_elements_of_gal_q_alpha_q}, it follows that \begin{equation*} G \simeq K_4 \simeq \mathbb{Z}_2 \times \mathbb{Z}_2. \end{equation*} \end{eg} % section galois_extensions_continued (end) % chapter lecture_23_mar_08th (end) \chapter{Lecture 24 Mar 11th}% \label{chp:lecture_24_mar_11th} % chapter lecture_24_mar_11th \section{Fundamental Theorem of Galois Theory}% \label{sec:fundamental_theorem_of_galois_theory} % section fundamental_theorem_of_galois_theory We shall first prove the following theorem, which will take away the bulk of the work required to prove the fundamental theorem later on. \begin{thm}[Artin's Theorem]\index{Artin's Theorem}\label{thm:artin_s_theorem} Let $K$ be a field, and $H \leq \Aut(K)$ a finite subgroup. Let $F = \Fix(H)$. Then \begin{enumerate} \item $K / F$ is Galois. \item $\Gal(K / F) = H$. \item $\abs{H} = \abs{\Gal(K / F)} = [K : F]$. \end{enumerate} \end{thm} \marginnote{ \begin{strategy} One sensible approach is to see how large is this extension. Since $[K : F] \leq \abs{H}$ is what we want, let's go against that. One may resort to try to use technique that were used lately, particularly, considering minimal polynomials of elements in $K$ algebraic over $F$, and then try to get to the splitting field of the polynomial of the product of those unique minimal polynomials. This, however, is not helpful, for we would go up into an extension where there is more `freedom', and so making it difficult to find a contradiction. \end{strategy}} \begin{proof} First, note that since $F = \Fix(H)$, we have that $H \subseteq \Gal(K / F)$ since, in a rather dumb way of putting it, $H$ only moves the elements in $K$ that it can, and it cannot move the elements in $F$ since $F$ is exactly the elements that $H$ cannot move. Then we have one part of (3), i.e. \begin{equation*} \abs{H} \leq \abs{\Gal(K / F)} \leq [K : F], \end{equation*} where the second inequality is because we do not know if $K$ is the splitting field of some non-constant separable polynomial in $F[x]$. Observe that if $[K : F] \leq \abs{H}$, then all of our results will follow. In particular, (3) would then be true. Then, $\abs{\Gal(K / F)} = [K : F]$ implies that $K / F$ is Galois by \hyperref[thm:characterization_of_galois_extensions]{the characterization of Galois extensions}. Also, since $H \subseteq \Gal(K / F)$, $\abs{H} = \abs{\Gal(K / F)}$ would give us (2). Let $\abs{H} = m$. Let $\beta_1, \ldots, \beta_n \in K^\times$, and suppose to the contrary that $n > m$.\sidenote{It is logical to make the next step provided that one is being conscious of knowledge from linear algebra.} \noindent \hlbnoted{Claim} $\{\beta_1, \ldots, \beta_n\}$ is linearly dependent on $F$. Consider the system of linear equations \begin{equation*} \phi(\beta_1) x_1 + \phi(\beta_2) x_2 + \hdots + \phi(\beta_n) x_n = 0, \end{equation*} where we let $\phi$ range over $H$. Since $\abs{H} = m$, we have $m$-many such linear equations. Notice that there are more variables than there are equations, and so we must have a non-trivial solution $(x_1, x_2, \ldots, x_n) \in K^n$. Note that if $\psi \in H$, for any $\phi \in H$, since $\psi \in H$ is a homomorphism we have \begin{align*} &\phi(\beta_1) \psi(x_1) + \phi(\beta_2) \psi(x_2) + \hdots \phi(\beta_n) \psi_n \\ &= \psi\left(\psi^{-1} \phi(\beta_1) x_1 + \psi^{-1} \phi(\beta_2) x_2 + \hdots \psi^{-1} \phi(\beta_n) x_n\right) \\ &= \psi^{-1}(0) = 0, \end{align*} where the second last equality is because the term in the parenthesis is one of the linear equations from before. It follows that $(\psi(x_1), \ldots, \psi(x_n)) \in K^n$ is also a non-trivial solution. From this, we know that there are $m$-many such non-trivial solutions. \sidenote{This is a non-trivial step. I can't seem to figure out how this can come by though.} Let $(x_1, \ldots, x_n)$ be a non-trivial solution with a minimal number of non-zero entries \sidenote{We want this guy to have as many zeroes as possible.}. Reordering if necessary, wma \begin{equation*} (x_1, \ldots, x_n) = (x_1, \ldots, x_r, 0, \ldots, 0), \end{equation*} where for $i = 1,\,2,\,\ldots,\,r$, we have $x_i \neq 0$. Note that $r > 1$, for otherwise $\phi(\beta_1) x_1 = 0 \implies x_1 = 0$, a contradiction. Notice that \begin{equation*} \left( 1, \frac{x_2}{x_1}, \frac{x_3}{x_1}, \ldots, \frac{x_n}{x_1} \right) \end{equation*} is also a solution. Thus wma $x_1 = 1$, i.e. we assume \begin{equation}\label{eq:key_to_artin} (x_1, x_2, \ldots, x_r, 0, \ldots, 0) = (1, x_2, x_3, \ldots, x_r, 0, \ldots, 0). \end{equation} Now, notice that if $x_2, x_3, \ldots, x_r \in F$, then we have \begin{equation*} \beta_1 + \beta_2 x_2 + \hdots \beta_n x_n = 0, \end{equation*} then $\beta_1$ is dependent on the others, and that would complete the proof. \noindent \hlbnoted{Sub-claim} $x_2, x_3, \ldots, x_r \in F$. Suppose not, i.e. suppose $x_i \notin F$. Then since $F = \Fix(H)$, we know that $\exists \psi \in H$ such that $\psi(x_i) \neq x_i$. Wlog, sps $x_i = x_2$. We know that \begin{equation*} (1, \psi(x_2), \psi(x_3), \ldots, \psi(x_r), 0, \ldots, 0) \end{equation*} is also a non-trivial solution to the earlier system of equations. Then consider \begin{align*} &(1, x_2, \ldots, x_r) - (1, \psi(x_2), \psi(x_3), \ldots, \psi(x_r), 0, \ldots, 0) \\ &= (0, x_2 - \psi(x_2), 0, \ldots, 0), \end{align*} which is also a non-trivial solution to the system of equations. However, this contradicts the minimality of \cref{eq:key_to_artin}. Thus $x_2, x_3, \ldots, x_r \in F$. This completes our proof. \end{proof} \begin{defn}[Galois Correspondences]\index{Galois Correspondences}\label{defn:galois_correspondences} Let $K$ be an extension of $F$. Let $\mathcal{E}$ denote the set of intermediate subfields of $K / F$, i.e. \begin{equation*} \mathcal{E} := \{ E : F \subseteq E \subseteq K \}, \end{equation*} and $\mathcal{H}$ denote the subgroups of $\Gal(K / F)$, i.e. \begin{equation*} \mathcal{H} := \{ H : H \leq \Gal(K / F) \}. \end{equation*} We define the \hlnoteb{Galois correspondences} by \begin{equation*} \Gal(K / -) : \mathcal{E} \to \mathcal{H} \text{ as } E \mapsto \Gal(K / E) \end{equation*} and \begin{equation*} \Fix : \mathcal{H} \to \mathcal{E} \text{ and } H \mapsto \Fix H. \end{equation*} \end{defn} \begin{note} Notice that $\Gal(K / -)$ bring subfields to subgroups, while $\Fix$ brings subgroups to subfields. \end{note} % section fundamental_theorem_of_galois_theory (end) % chapter lecture_24_mar_11th (end) \chapter{Lecture 25 Mar 13th}% \label{chp:lecture_25_mar_13th} % chapter lecture_25_mar_13th \section{Fundamental Theorem of Galois Theory (Continued)}% \label{sec:fundamental_theorem_of_galois_theory_continued} % section fundamental_theorem_of_galois_theory_continued \begin{remark} The \hyperref[defn:galois_correspondences]{Galois correspondences} are \hldefn{inclusion reversing}: notice that \begin{enumerate} \item $E_1 \subseteq E_2 \in \mathcal{E} \implies \Gal(K / E_2) \subseteq \Gal(K / E_1)$ since the later Galois group has less elements to fix. \item $H_1 \subseteq H_2 \in \mathcal{H} \implies \Fix H_2 \subseteq \Fix H_1$ since $H_2$ has more elements, and so it will move more things around, making the fix smaller. \end{enumerate} \end{remark} \begin{thm}[Fundamental Theorem of Galois Theory]\index{Fundamental Theorem of Galois Theory}\label{thm:fundamental_theorem_of_galois_theory} Let $K / F$ be a finite \hyperref[defn:galois_extension]{Galois extension}. Then the Galois correspondences give an inclusion reversing bijection between $\mathcal{E}$ and $\mathcal{H}$, i.e. \begin{enumerate} \item if $E \in \mathcal{E}$, then $\Fix (\Gal(K / E)) = E$. In particular, $K / E$ is Galois; \item if $H \in \mathcal{H}$, then $\Gal(K / \Fix H) = H$. \end{enumerate} \end{thm} \begin{proof} \begin{enumerate} \item Since $K / F$ is normal and separable, by A7, we have that $K / E$ is also normal and separable. Thus it follows from \hyperref[thm:characterization_of_galois_extensions]{the characterization of Galois extensions} that \begin{equation*} \Fix(\Gal(K / E)) = E. \end{equation*} \item This is exactly \hyperref[thm:artin_s_theorem]{Artin's Theorem}! \end{enumerate} \end{proof} \begin{crly}[Relation between Index and Degree]\label{crly:relation_between_index_and_degree} Let $K / F$ be a finite Galois extension. If $H_1 \subseteq H_2 \in \mathcal{H}$, then \begin{equation*} \abs{ H_2 : H_1 } = [ \Fix H_1 : \Fix H_2 ]. \end{equation*} If $K / E_1 / E_2 / F$ is a tower of fields, then \begin{equation*} [ E_1 : E_2 ] = \abs{ \Gal(K / E_2) : \Gal(K / E_1) }. \end{equation*} \end{crly} \begin{proof} Notice that by the \hyperref[thm:tower_theorem]{Tower Theorem}, we have \begin{align*} [ \Fix H_1 : \Fix H_2 ] &= \frac{[ K : \Fix H_2 ]}{[ K : \Fix H_1 ]} \\ &= \frac{\abs{\Gal(K / \Fix H_2)}}{\abs{\Gal(K / \Fix H_1)}} \\ &= \frac{\abs{H_2}}{\abs{H_1}} = \abs{ H_2 : H_1 }. \end{align*} On the other hand, we have \begin{align*} [ E_1 : E_2 ] &= \frac{[ K : E_2 ]}{[ K : E_1 ]} \\ &= \frac{\abs{\Gal(K / E_2)}}{\abs{\Gal(K / E_1)}} \\ &= \abs{\Gal(K / E_2) : \Gal(K / E_1)}. \end{align*} \end{proof} \cref{fig:visual_representation_of_the_fundamental_theorem_of_galois_theory} illustrates what the \hyperref[thm:fundamental_theorem_of_galois_theory]{fundamental theorem} does. \begin{figure*}[ht] \centering \begin{tikzpicture}[baseline= (a).base] \node[scale=0.8] (a) at (0, 0) { \begin{tikzcd} & & G = \Gal(K/F) \arrow[ld, "n_1"', no head] \arrow[d, "n_2", no head] \arrow[rd, "n_3", no head] & & & & & & & \Fix(\{1\}) = K & & \\ & H_1 \arrow[d, no head] \arrow[ld, no head] & H_2 \arrow[d, no head] \arrow[ld, no head] \arrow[rd, no head] & H_3 \arrow[d, no head] \arrow[rd, no head] & \arrow[rr, "\Fix"] & & & \vdots \arrow[rd, no head] \arrow[rru, no head] & \vdots \arrow[d, no head] \arrow[ru, no head] & \vdots \arrow[d, no head] \arrow[rd, no head] \arrow[ld, no head] \arrow[u, no head] & \vdots \arrow[d, no head] \arrow[lu, no head] & \vdots \arrow[ld, no head] \arrow[llu, no head] \\ \vdots \arrow[rrd, no head] & \vdots \arrow[rd, no head] & \vdots \arrow[d, no head] & \vdots \arrow[ld, no head] & \vdots \arrow[lld, no head] & & & & \Fix H_1 \arrow[rd, "n_1"', no head] & \Fix H_2 \arrow[d, "n_2", no head] & \Fix H_3 \arrow[ld, "n_3", no head] & \\ & & \{1\} & & & & & & & \Fix(G) = F & & \end{tikzcd} }; \end{tikzpicture} \caption[][20pt]{Visual Representation of the Fundamental Theorem of Galois Theory} \label{fig:visual_representation_of_the_fundamental_theorem_of_galois_theory} \end{figure*} \begin{note}[\imponote] Notice that the index of each of the subgroups with respect to the base group becomes the degree of which the fix of the subgroup extends the base field. \end{note} The following is a typical problem for finding out about a Galois extension. \begin{eg} Consider our `favorite' extension \begin{equation*} G = \Gal( \mathbb{Q}(\alpha, \zeta_3) / \mathbb{Q} ) = \Gal(x^3 - 2), \end{equation*} where $\alpha = \sqrt[3]{2}$. \begin{enumerate} \item Prove that $\mathbb{Q}(\alpha, \zeta_3) / \mathbb{Q}$ is Galois. \item Find $\abs{G}$. \item Find $G$. \item Draw the subfield lattice. \end{enumerate} \end{eg} \begin{solution} \begin{enumerate} \item Note that $\mathbb{Q}$ is perfect, so $x^3 - 2$ is separable. Also it is irreducible by \hyperref[propo:eisenstein_s_criterion]{$2$-Eisenstein}. And indeed, $\mathbb{Q}(\alpha, \zeta_3)$ is the splitting field of $x^3 - 2 \in \mathbb{Q}[x]$. Thus $\mathbb{Q}(\alpha, \zeta_3) / \mathbb{Q}$ is Galois. \item Observe that \begin{equation*} \abs{G} = [ \mathbb{Q}(\alpha, \zeta_3) : \mathbb{Q} ] = 2 \cdot 3 = 6, \end{equation*} where the second equality follows from one of our assignment problems. \item Since we have $G \leq S_3$ and $\abs{G} = 6$, it follows that $G \simeq S_3$. \item Since $G \simeq S_3$, we have the following \hldefn{subgroup lattice}: \begin{figure*}[ht] \centering \begin{tikzcd} & & S_3 \arrow[lld, "2"', no head] \arrow[ld, "3", no head] \arrow[rd, "3"', no head] \arrow[rrd, "3", no head] & & \\ <(1 \; 2 \; 3)> \arrow[rrd, "3"', no head] & <(1 \; 2)> \arrow[rd, "2", no head] & & <(1 \; 3)> \arrow[ld, "2"', no head] & <(2\; 3)> \arrow[lld, "2", no head] \\ & & \{ \epsilon \} & & \end{tikzcd} \caption{Subgroup Lattice of $\Gal(\mathbb{Q}(\sqrt[3]{2}, \zeta_3) / \mathbb{Q})$} \label{fig:subgroup_lattice_of_gal_q_sqrt_3_2_zeta_3_q_} \end{figure*} One can indeed check that all $6$ of the possible automorphisms on the roots must occur, of which its result is shown in \cref{table:structure_of_gal_q_sqrt_3_2_zeta_3_q}. \begin{table}[hb] \centering \caption{Structure of $\Gal(\mathbb{Q}(\sqrt[3]{2}, \zeta_3) / \mathbb{Q})$} \label{table:structure_of_gal_q_sqrt_3_2_zeta_3_q} \begin{tabular}{c | c c c | c} & $\alpha$ & $\alpha \zeta_3$ & $\alpha \zeta_3^2$ & $S_3$ \\ \hline $\phi_1$ & $\alpha$ & $\alpha \zeta_3$ & $\alpha \zeta_3^2$ & $\epsilon$ \\ $\phi_2$ & $\alpha$ & $\alpha \zeta_3^2$ & $\alpha \zeta_3$ & $(2 \; 3)$ \\ $\phi_3$ & $\alpha \zeta_3^2$ & $\alpha$ & $\alpha \zeta_3$ & $(1 \; 2 \; 3)$ \\ $\phi_4$ & $\alpha \zeta_3^2$ & $\alpha \zeta_3$ & $\alpha$ & $(1 \; 3)$ \\ $\phi_5$ & $\alpha \zeta_3$ & $\alpha \zeta_3^2$ & $\alpha$ & $(1 \; 3 \; 2)$ \\ $\phi_6$ & $\alpha \zeta_3$ & $\alpha$ & $\alpha \zeta_3^2$ & $(1 \; 2)$ \end{tabular} \end{table} By the \hyperref[thm:fundamental_theorem_of_galois_theory]{fundamental theorem}, we have that the subfield lattice is as shown in \cref{fig:subfield_lattice_of_gal_q_sqrt_3_2_zeta_3_q_}. \begin{figure*}[ht] \centering \begin{tikzpicture}[baseline= (a).base] \node[scale=0.6] (a) at (0, 0) { \begin{tikzcd} & & & {\Fix \{\epsilon\} = \mathbb{Q}(\alpha, \zeta_3)} \arrow[llld, "3"', no head] \arrow[ld, "2", no head] \arrow[rd, "2"', no head] \arrow[rrrd, "2", no head] & & & \\ \Fix <(1 \; 2 \; 3)> = \mathbb{Q}(\zeta_3) \arrow[rrrd, "2", no head] & & \Fix <(1 \; 2)> = \mathbb{Q}(\alpha\zeta_3^2) \arrow[rd, "3", no head] & & \Fix <(1 \; 3)> = \mathbb{Q}(\alpha\zeta_3) \arrow[ld, "3"', no head] & & \Fix <(2 \; 3)> = \mathbb{Q}(\alpha) \arrow[llld, "3", no head] \\ & & & \Fix(G) = \mathbb{Q} & & & \end{tikzcd} }; \end{tikzpicture} \caption{Subfield Lattice of $\Gal(\mathbb{Q}(\sqrt[3]{2}, \zeta_3) / \mathbb{Q})$} \label{fig:subfield_lattice_of_gal_q_sqrt_3_2_zeta_3_q_} \end{figure*} \end{enumerate} \end{solution} % section fundamental_theorem_of_galois_theory_continued (end) % chapter lecture_25_mar_13th (end) \chapter{Lecture 26 Mar 15th}% \label{chp:lecture_26_mar_15th} % chapter lecture_26_mar_15th \section{Fundamental Theorem of Galois Theory (Continued 2)}% \label{sec:fundamental_theorem_of_galois_theory_continued_2} % section fundamental_theorem_of_galois_theory_continued_2 \begin{note} It is important to note that given a finite Galois extension, the \hyperref[thm:fundamental_theorem_of_galois_theory]{Fundamental Theorem of Galois Theory} tells us that there are only finitely many intermediary subfields up to isomorphism. \end{note} \begin{propo}[Other Subfields Through Group Normality]\label{propo:other_subfields_through_group_normality} Let $K / F$ be a finite Galois extension. Let $E$ be an intermediate subfield of $K / F$. Then for any $\phi \in \Gal(K / F)$, we have \begin{equation*} \phi \Gal(K / E) \phi^{-1} = \Gal(K / \phi(E)). \end{equation*} \end{propo} \begin{proof} We have the following chain of iffs: \begin{align*} & \forall \psi \in \Aut(K) \; \psi \in \Gal(K / E) \\ &\iff \forall \alpha \in E \; \psi(\alpha) = \alpha \\ &\iff \forall \beta \in E \; \psi \phi^{-1} (\beta) = \phi^{-1} (\beta) \\ &\iff \forall \beta \in E \; \phi \psi \phi^{-1} (\beta) = \beta \\ &\iff \phi \psi \phi^{-1} \in \Gal(K / \phi(E)). \end{align*} \end{proof} \begin{defn}[$H$-invariant]\index{$H$-invariant}\label{defn:_h_invariant} Let $K / E / F$ be a tower of fields, and $H \leq \Aut(K)$. We say $E$ is \hlnoteb{invariant} under $H$ (or \hlnoteb{$H$-invariant}) if $\forall \phi \in H$, $\phi(E) = E$. \end{defn} \begin{propo}[Intermediate Subfields and Normal Subfields]\label{propo:intermediate_subfields_and_normal_subfields} Let $K / F$ be a finite Galois extension. If $E$ is an intermediate subfield of $K / F$, then TFAE: \begin{enumerate} \item $E / F$ is Galois. \item $E$ is $\Gal(K / F)$-invariant. \item $\Gal(K / E) \trianglelefteq \Gal(K / F)$. \end{enumerate} \end{propo} \begin{proof} By definition and \cref{propo:other_subfields_through_group_normality}, we immediately have $(2) \iff (3)$. \noindent \hlbnoted{$(1) \implies (2)$} Suppose $E / F$ is Galois. Let $\phi \in G \coloneqq \Gal(K / F)$. Since $E / F$ is Galois, in particular, normal, we know that $\phi \restriction_{E} \in \Gal(E / F)$. Thus we have $\phi \restriction_E (E) = \phi(E) = E$ by the surjectivity of $\phi$ \sidenote{Note that $\phi$ is a automorphism, in particular, bijective.}. It follows that $E$ is $G$-invariant, simply by definition. \noindent \hlbnoted{$(2) \implies (1)$} Suppose $E$ is $G$-invariant, where again we let $G \coloneqq \Gal(K / F)$. Now by A7, since $K / E / F$ is a tower of fields, we have that $E / F$ is separable. Thus it suffices for us to show that $E / F$ is normal. Let $\alpha \in E$ with minimal polynomial $f(x) \in F[x]$. \sidenote{\st{I think there was a slight oversight during lectures. The original reasoning here was that since $K / F$ is normal, we have $f(x)$ splits over $K$. But that is not necessarily true. Being a Galois extension only requires $K$ to be the splitting field of some non-constant polynomial (and separability of the extension), but it does not necessitate that $K$ is where all polynomials in $F[x]$ split. This is only true if $K$ is the algebraic closure of $F[x]$.} My mistake! The reasoning is correct by \hyperref[thm:normality_theorem]{the normality theorem}.} Since $K / F$ is normal, since $\alpha \in E \subseteq K$, we have that its minimal polynomial $f(x)$ must split in $K$. Then let $\beta \in K$ be an $F$-conjugate of $\alpha$. Since $f(x) \in F[x]$ is irreducible, we have that $G$ is \hyperref[defn:transitive_subgroup]{transitive} by \cref{crly:the_galois_group_of_a_separable_irreducible_polynomial_is_transitive}, and so $\exists \phi \in G$ such that $\phi(\alpha) = \beta$. Then by assumption, we have that \begin{equation*} \beta = \phi(\alpha) \in \phi(E) = E. \end{equation*} It follows that $E / F$ is indeed normal. \end{proof} \newthought{A natural} question to ask: \begin{quotebox}{green}{foreground} In what way is $\Gal(E / F)$ related to $\Gal(K / F)$? \end{quotebox} \begin{propo}[First Isomorphism Theorem on Galois Groups]\label{propo:first_isomorphism_theorem_on_galois_groups} Suppose we have a tower of fields $K / E / F$, and $K / F$ is a finite Galois extension. If $E / F$ is Galois, then \begin{equation*} \Gal(E / F) \simeq \quotient{\Gal(K / F)}{\Gal(K / E)}. \end{equation*} \end{propo} \begin{proof} Consider the function $\psi : \Gal(K / F) \to \Gal(E / F)$ given by $\psi(\phi) = \phi \restriction_E$. Note that this is a homomorphism: for $\phi_1, \phi_2 \in \Gal(K / F)$, observe that \begin{equation*} \psi(\phi_1 \phi_2) = (\phi_1 \phi_2) \restriction_E = \phi_1 \restriction_{\phi_2 \restriction_E} \phi_2 \restriction_E = \phi_1 \restriction_E \phi_2 \restriction_E = \psi(\phi_1) \psi(\phi_2). \end{equation*} It is clear that $\ker \psi = \Gal(K / E)$. It follows from the \hlnotea{First Isomorphism Theorem} that \begin{equation*} \Gal(E / F) \simeq \quotient{\Gal(K / F)}{\Gal(K / E)}, \end{equation*} which is what we want. \end{proof} % section fundamental_theorem_of_galois_theory_continued_2 (end) \section{Special Galois Groups}% \label{sec:special_galois_groups} % section special_galois_groups \begin{eg} Compute $\Gal\left(\quotient{\mathbb{Q}(\zeta_n)}{\mathbb{Q}}\right)$. \begin{solution} Note that $\quotient{\mathbb{Q}(\zeta_n)}{\mathbb{Q}}$ is the splitting field for the separable polynomial $\Phi_n(x)$ over $\mathbb{Q}$. Thus $\quotient{\mathbb{Q}(\zeta_n)}{\mathbb{Q}}$ is Galois. We also know that \begin{equation*} \abs{\Gal \left( \quotient{\mathbb{Q}(\zeta_n)}{\mathbb{Q}} \right)} = \phi(n), \end{equation*} where we note that $\phi(n)$ here is the \hlnotea{Euler $\phi$-function}, for the sake of clarity. \noindent \hlbnoted{Claim: $G = \Gal \left( \quotient{\mathbb{Q}(\zeta_n)}{\mathbb{Q}} \right) \simeq \mathbb{Z}_n^\times$} Consider the map \begin{equation*} \psi : \mathbb{Z}_n^\times \to G \text{ given by } \psi(k) = \phi_k, \end{equation*} where $\phi_k(\zeta_n) = \zeta_n^k$. \noindent \hlbnotea{$\psi$ is a homomorphism} Observe that \begin{equation*} \phi_k \circ \phi_l (\zeta_n) = \phi_k (\zeta_n^l) = \zeta_n^{kl} = \phi_{kl} (\zeta_n). \end{equation*} Thus we have $\psi(k)\psi(l) = \psi(kl)$. \noindent \hlbnotea{$\psi$ is injective} Observe that \begin{equation*} k \in \ker \psi \iff \psi(k) = 1 \iff \zeta_n^k = \zeta_n \iff k = 1. \end{equation*} \noindent \hlbnotea{$\psi$ is surjective} Notice that $\abs{\mathbb{Z}_n^\times} = \phi(n) = \abs{G}$, where $\phi$ is the Euler $\phi$-function. It follows from the injectivity of $\psi$ that $\psi$ is surjective. Therefore, $\psi$ is an isomorphism, which is what is required. \end{solution} \end{eg} \begin{eg} Compute $G = \Gal \left( \quotient{\mathbb{F}_{p^n}}{\mathbb{F}_p} \right)$. \begin{solution} First, note that $\mathbb{F}_{p^n}$ is the splitting field of the separable polynomial $x^{p^n} - x$ over $\mathbb{F}_{p}$. So $\quotient{\mathbb{F}_{p^n}}{\mathbb{F}_p}$ is indeed Galois and we are not making a fool out of ourselves. Consequently, we have that \begin{equation*} \abs{\Gal \left( \quotient{\mathbb{F}_{p^n}}{\mathbb{F}_p} \right)} = [ \mathbb{F}_{p^n} : \mathbb{F}_p ] = n. \end{equation*} Consider the \hldefn{Forbenius map} \begin{equation*} \phi : \mathbb{F}_{p^n} \to \mathbb{F}_{p^n} \text{ given by } \phi(a) = a^p. \end{equation*} By \hlnotea{Fermat's Little Theorem}, $a^p = a$, and so we have that $\phi \in G$. Let $j = \abs{\phi}$. Note that $j \leq n$ \sidenote{\hlwarn{Why?}}. Now by definition, every element of $\mathbb{F}_{p^n}$ is a root of $x^{p^j} - x$, i.e. $p^j \geq p^n$, which implies $j \geq n$. Thus $\abs{\phi} = j = n$. It follows that \begin{equation*} \Gal \left( \quotient{\mathbb{F}_{p^n}}{\mathbb{F}_p} \right) = \langle \phi \rangle \simeq \mathbb{Z}_n. \end{equation*} \end{solution} \end{eg} % section special_galois_groups (end) % chapter lecture_26_mar_15th (end) \chapter{Lecture 27th Mar 18th}% \label{chp:lecture_27th_mar_18th} % chapter lecture_27th_mar_18th \section{Galois Groups of Polynomials}% \label{sec:galois_groups_of_polynomials} % section galois_groups_of_polynomials We know how hard it is to find roots of a polynomial. It would be ideal if we can find ways to work around them when trying to construct Galois groups. \begin{defn}[Discriminant]\index{Discriminant}\label{defn:discriminant} Let $f(x) \in F[x]$ be non-constant, with $K$ being its splitting field. We can write \begin{equation*} f(x) = u(x - \alpha_1)(x - \alpha_2) \hdots (x - \alpha_n) \in K[x]. \end{equation*} We define the \hlnoteb{discriminant} of $f(x)$ as \begin{equation*} \disc f(x) \coloneqq \prod_{i < j} (\alpha_i - \alpha_j)^2. \end{equation*} \end{defn} \begin{remark} \begin{enumerate} \item Note that \begin{equation*} \disc f(x) \neq 0 \iff f(x) \text{ is separable } \end{equation*} \item For a \hldefn{quadratic} \begin{equation*} f(x) = x^2 + bx + c = (x - \alpha)(x - \beta) = x^2 - (\alpha + \beta) + \alpha\beta, \end{equation*} we have that its discriminant is \begin{align*} \disc f(x) &= (\beta - \alpha)^2 = \alpha^2 + \beta^2 - 2 \alpha\beta \\ &= (\alpha + \beta)^2 - 4 \alpha \beta \\ &= (-b)^2 - 4c. \end{align*} One may note that the \hyperref[defn:discriminant]{discriminant} is like a generalization from this. \end{enumerate} \end{remark} \begin{lemma}[The Discriminant Lives in the Base Field]\label{lemma:the_discriminant_lives_in_the_base_field} Let $f(x) \in F[x]$ be non-constant. Then $\disc f(x) \in F$. \end{lemma} \begin{proof} We break this into two cases: \noindent \hlbnoted{$f(x)$ is not separable} In this case $\disc f(x) = 0 \in F$. \noindent \hlbnoted{$f(x)$ is separable} Notice that for any $\phi \in G = \Gal f(x)$, we have \begin{equation*} \phi(\disc f(x)) = \disc f(x), \end{equation*} since $\phi$ is an automorphism, in particular, a homomorphism. Thus we have that $\disc f(x) \in \Fix G = F$. \end{proof} \begin{propo}[Galois Group of Finite Extensions]\label{propo:galois_group_of_finite_extensions} Suppose $\Char F \neq 2$, $f(x)$ a separable polynomial with $\deg f = n \geq 2$, and $G = \Gal f(x)$. Let \begin{equation*} d = \prod_{i < j} (\alpha_i - \alpha_j), \end{equation*} where the $\alpha_i$'s are the roots of $f(x)$ in its splitting field $K$. If $\phi \in G \subseteq S_n$, then $\phi(d) = \pm d$. Moreover, $\phi(d) = d \iff \phi \in A_n$, and $\Gal \left( \quotient{K}{F(d)} \right) = G \cap A_n$. Also, $G \subseteq A_n \iff d \in \Fix(G) = F \iff \disc f(x)$ is a square in $F$. \end{propo} \begin{proof} Let $\phi \in G$. Since we have $\Char F \neq 2$, we have that $d$ and $\phi(d)$ are roots of \begin{equation*} x^2 - d^2 = x^2 - \disc f(x) \in F[x]. \end{equation*} It follows that $\phi(d) = \pm d$, as we want. Observe that $S_n$ acts on $X = \{d, -d\}$ by \begin{equation*} \sigma \cdot \prod (\alpha_i - \alpha_j) = \prod (\alpha_{\sigma(i)} - \alpha_{\sigma(j)}). \end{equation*} Moreover, observe that \begin{equation*} \epsilon \cdot d = d \text{ and } (n \; (n - 1)) \cdot d = -d. \end{equation*} It follows that the group action is transitive. By the \hyperref[thm:orbit_stabilizer_theorem]{Orbit-Stabilizer Theorem}, we have that \begin{equation*} n! = \abs{S_n} = \abs{\stab(d)} \abs{\orb(d)} = 2 \abs{\stab(d)}. \end{equation*} Thus $\abs{\stab(d)} = \frac{n!}{2}$, and since $\stab(d) \leq S_n$, it follows that $\stab(d) = A_n$. This also means that \begin{equation*} \phi(d) = d \iff \phi \in A_n = \stab(d). \end{equation*} The rest of the statement follows immediately by the way they are introduced. \end{proof} \subsection{Quadratics}% \label{sub:quadratics} % subsection quadratics Note that $\disc f(x)$ is not a square in $F$ \\ $\iff \Gal f(x) \not\subseteq A_2 = \{1\}$ \\ $\iff \Gal f(x) = S_2 \simeq \mathbb{Z}_2$ \\ $\iff f(x)$ is irreducible. We mostly know how to deal with quadratics, especially since we have a formula for finding roots of quadratic polynomials, in particular the \hlnotea{quadratic formula}: given $f(x) = ax^2 + bx + c$, we have \begin{equation*} x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \end{equation*} % subsection quadratics (end) \subsection{Cubics}% \label{sub:cubics} % subsection cubics Recall that if $f(x) \in F[x]$ is irreducible and separable, then $\Gal f(x) \leq S_3$ is \hyperref[crly:the_galois_group_of_a_separable_irreducible_polynomial_is_transitive]{transitive}, and so either \begin{equation*} \Gal f(x) \simeq S_3 \text{ or } \Gal f(x) \simeq A_3. \end{equation*} One may find it rather difficult to find roots for cubics, especially when the \hlnotea{rational roots theorem} does not give us any results. \begin{defn}[Depressed Cubic]\index{Depressed Cubic}\label{defn:depressed_cubic} Suppose $\Char F \notin \{2, 3\}$. Let \begin{equation*} g(x) = x^3 + \alpha x^2 + \beta x + \gamma \in F[x] \end{equation*} be irreducible and separable. Let \begin{equation*} f(x) = g \left( x - \frac{\alpha}{3} \right) = x^3 + bx + c \in F[x]. \end{equation*} $f(x)$ is called a \hlnoteb{depressed cubic}. \end{defn} \begin{note} We can derive a formula for $b$ and $c$ in the depressed cubic, in particular they are \begin{equation*} b = \beta - \frac{1}{3} \alpha^2 \text{ and } c = \gamma + \frac{2}{27} \alpha^3 - \frac{1}{3} \alpha\beta. \end{equation*} It is important to note that $f(x)$ is still irreducible and separable, and most importantly \begin{equation*} \Gal f(x) = \Gal g(x). \end{equation*} This is exactly why we want to consider depressed cubics, since they are easier to deal with, especially when we need to do \hlnotea{long division}. Furthermore, if we let $\alpha_1, \alpha_2, \alpha_3$ be the roots of $f(x)$, then one can derive that \begin{equation*} \disc f(x) = -4b^3 - 27c^2. \end{equation*} Then by \cref{propo:galois_group_of_finite_extensions}, we have that \begin{equation*} \Gal f(x) = \begin{cases} A_3 & \disc f(x) = d^2, \, d \in F \\ S_3 & \text{otherwise } \end{cases} \end{equation*} \end{note} \begin{eg} Consider the polynomial $f(x) = x^3 - 3x + 1 \in \mathbb{Q}[x]$. If we use the \hyperref[propo:mod_p_irreducibility_test]{Mod-2 irreducibility test}, one can immedaitely see that it is an irreducible polynomial that we have seen before. Thus $f(x)$ itself is irreducible. We have that \begin{equation*} \disc f(x) = -4(-3)^3 - 27(1)^2 = 3 \cdot 27 = 9^2, \end{equation*} and $9 \in \mathbb{Q}$. It follows that $\Gal f(x) \simeq A_3$. \end{eg} % subsection cubics (end) % section galois_groups_of_polynomials (end) % chapter lecture_27th_mar_18th (end) \chapter{Lecture 28 Mar 20th}% \label{chp:lecture_28_mar_20th} % chapter lecture_28_mar_20th \section{Galois Groups of Polynomials (Continued)}% \label{sec:galois_groups_of_polynomials_continued} % section galois_groups_of_polynomials_continued \subsection{Quartics}% \label{sub:quartics} % subsection quartics \begin{defn}[Depressed Quartic]\index{Depressed Quartic}\label{defn:depressed_quartic} Suppose $\Char F \neq 2$. Given \begin{equation*} f(x) = x^4 + \alpha x^3 + \beta x^2 + \gamma x + \delta \in F[x], \end{equation*} we shall consider \begin{equation*} g(x) = f\left(x - \frac{\alpha}{4}\right) = x^4 + bx^2 + cx + d \in F[x]. \end{equation*} This is similarly called a \hldefn{depressed quartic}. \end{defn} \begin{note} Furthermore, $\Gal(f(x)) = \Gal(g(x))$. \marginnote{$g(x)$ is irreducible and separable iff $f(x)$ is irreducible and separable.} \end{note} \newthought{Let} \begin{equation*} f(x) = x^4 + b x^2 + c x + d \in F[x], \end{equation*} If $G = \Gal(f(x))$, then $G$ is a \hyperref[defn:transitive_subgroup]{transitive} of $S_4$, with $4 \mid \abs{G}$. Thus our options are: \begin{equation*} S_4, A_4, D_4, V, \mathbb{Z}_4, \end{equation*} where $V$ is the Klein-$4$ group \sidenote{This is very \hlimpo{important}.}, \begin{equation*} V = \left\{ \epsilon, (1 \; 2)(3 \; 4), (1 \; 3)(2 \; 4), (1\; 4)(2 \; 3) \right\} \simeq \mathbb{Z}_2 \times \mathbb{Z}_2. \end{equation*} Suppose the roots of $f(x)$ are $\alpha_1, \alpha_2, \alpha_3, \alpha_4$ which are distinct, and let $K = F(\alpha_1, \alpha_2, \alpha_3, \alpha_4)$. Also, let \begin{align*} u &= \alpha_1 \alpha_2 + \alpha_3 \alpha_4 \\ v &= \alpha_1 \alpha_3 + \alpha_2 \alpha_4 \\ w &= \alpha_1 \alpha_4 + \alpha_2 \alpha_3. \end{align*} \begin{defn}[Resolvent Cubic]\index{Resolvent Cubic}\label{defn:resolvent_cubic} The \hlnoteb{resolvent cubic} of $f(x)$ is defined by \begin{align*} \Res f(x) &= (x - u)(x - v)(x - w) \\ &= x^3 - b x^2 - 4d x + 4bd - c^2 \in F[x]. \end{align*} \end{defn} Let $L = F(u, v, w)$. Then $K / L / F$ is a tower of fields \sidenote{$K$ is the splitting field of $f(x)$ and $L$ is the splitting field of $\Res f(x)$.}. Now let $G = \Gal(f(x))$. Note that $K / F$ is Galois. Thus $K / L$ is Galois \sidenote{By A7.}. Also \begin{equation*} \Gal(\Res f(x)) = \Gal( L / F ), \end{equation*} where $L / F$ is also Galois. Since $\Gal( K / L ) = G \cap V$ and $L / F$ is Galois, we have that \begin{equation*} \Gal( K / L ) \trianglelefteq \Gal(K / F), \end{equation*} and \begin{quotebox}{red}{foreground} \begin{equation*} \Gal(\Res f(x)) = \Gal(L / F) = G / (G \cap V). \end{equation*} \end{quotebox} Let $m = \abs{\Gal(\Res f(x))} = \abs{\Gal(L / F)}$. Then we have the following possibilities: \begin{table}[ht] \centering \caption{Galois group $G$} \label{table:galois_group_g} \begin{tabular}{c | c c c c c} $G$ & $S_4$ & $A_4$ & $D_4$ & $V$ & $\mathbb{Z}_4$ \\ \hline $G \cap V$ & $V$ & $V$ & $V$ & $V$ & $\mathbb{Z}_2$ \\ $G / G \cap V$ & $S_3$ & $\mathbb{Z}_3$ & $\mathbb{Z}_2$ & $\{ 1 \}$ & $\mathbb{Z}_2$ \\ $m$ & $6$ & $3$ & $2$ & $1$ & $2$ \end{tabular} \end{table} Note that the $m = 2$ can be a problem. \begin{remark} $G$ is uniquely determined when $m \in \{ 1, 3, 6 \}$. \end{remark} For the rest of our discussion, we shall focus on when $m = 2$. \newthought{We know that} $G \simeq D_4$ or $G \simeq \mathbb{Z}_4$. Fortunately, the two groups have different order. Since $\deg \Res f(x) = 3$, and $m = 2$, thus $f(x)$ factors into a linear and quadratic terms. Thus exactly one of either $u, v,$ or $w$ is in $F$. Wlog, $u \in F$. Either option for $G$ has a $4$-cycle which fixes $u$. Therefore \begin{equation*} \sigma = (1 \; 3 \; 2 \; 4) \in G. \end{equation*} Note that $\sigma^2 = (1 \; 2)(3 \; 4) \in G$. Consider: \begin{equation*} (x - \alpha_1 \alpha_2)(x - \alpha_3 \alpha_4) = x^2 - ux + d \end{equation*} and \begin{equation*} (x - ( \alpha_1 + \alpha_2 ))(x - (\alpha_3 + \alpha_4)) = x^2 + (b - u). \end{equation*} \hlbnoted{Claim} $G = \langle \sigma \rangle \simeq \mathbb{Z}_4$ iff both of these polynomials split over $L$. \begin{proof} \hlbnoted{$(\implies)$} Suppose $G = \langle \sigma \rangle$. Then \begin{equation*} \Gal(K / L) = G \cap V = \langle \sigma^2 \rangle. \end{equation*} Thus $\alpha_1 \alpha_2, \, \alpha_3 \alpha_4, \, \alpha_1 + \alpha_2, \, \alpha_3 + \alpha_4 \in \Fix \langle \sigma^2 \rangle = L$. \noindent \hlbnoted{$(\impliedby)$} Suppose $\alpha_1 \alpha_2, \alpha_3 \alpha_4, \alpha_1 + \alpha_2, \alpha_3 + \alpha_4 \in L$. We need only to show that $\langle \sigma \rangle$ has at most order $4$. Note that $\alpha_1 \alpha_2 \in L(\alpha_1) \implies \alpha_1, \alpha_2 \in L(\alpha_1)$. Further, $v - w = (\alpha_1 - \alpha_2) (\alpha_3 - \alpha_4) \in L$. Thus $\alpha_3 - \alpha_4 \in L(\alpha_1)$, which means $\alpha_3 \in L(\alpha_1) \implies \alpha_4 \in L(\alpha_1)$. \sidenote{Note that we needed $\Char F \neq 2$.}. Therefore $K = F(\alpha_1, \alpha_2, \alpha_3, \alpha_4) = L(\alpha_1)$. Hence, $[ K : L ] = [ L(\alpha_1) : L ] = \abs{ \Gal(K / L) }$. This means that if we consider \begin{equation*} p(x) = x^2 - (\alpha_1 + \alpha_2) x + \alpha_1 \alpha_2 \in L[x], \end{equation*} then $p(\alpha_1) = 0$. Therefore, $[ K : L ] \leq 2$. This implies that $[ K : F ] = [ K : L ][ L : F ] = 2 [ K : L ] \leq 4$. Since $G \simeq \mathbb{Z}_4$ or $G \simeq D_4$ with $\abs{G} \leq 4$, we have that \begin{equation*} G = \langle \sigma \rangle \simeq \mathbb{Z}_4. \end{equation*} \end{proof} \begin{note} Rework this entire lecture to work out the reasoning. \end{note} % subsection quartics (end) % section galois_groups_of_polynomials_continued (end) \tuftepart{Solvability by Radicals} % chapter lecture_28_mar_20th (end) \chapter{Lecture 29 Mar 22nd}% \label{chp:lecture_29_mar_22nd} % chapter lecture_29_mar_22nd \section{Galois Groups of Polynomials (Continued 2)}% \label{sec:galois_groups_of_polynomials_continued_2} % section galois_groups_of_polynomials_continued_2 \subsection{Quartics (Continued)}% \label{sub:quartics_continued} % subsection quartics_continued \begin{eg} Let $f(x) = x^4 - 2x - 2 \in \mathbb{Q}[x]$. This is irreducible by \hyperref[propo:eisenstein_s_criterion]{$2$-Eisenstein} and $\mathbb{Q}$ is perfect. The resolvent of $f(x)$ is \begin{equation*} \Res f(x) = x^3 + 8x - 4. \end{equation*} This guy has no rational roots \sidenote{Check!}, and so $\Res f(x)$ is irreducible. Now \begin{equation*} \disc \Res f(x) = -4(8^3) - 27((-4)^2) < 0, \end{equation*} thus $\disc \Res f(x)$ is not a square in $\mathbb{Q}$. THus \begin{equation*} \Gal \Res f(x) = S_3. \end{equation*} It follows that \begin{equation*} \Gal f(x) = S_4. \end{equation*} \end{eg} \begin{eg} Let $g(x) = x^4 + 5x + 5 \in \mathbb{Q}[x]$. $g(x)$ is irreducible by \hyperref[propo:eisenstein_s_criterion]{$5$-Eisenstein} and separable. We have \begin{equation*} \Res g(x) = x^3 - 20x - 25. \end{equation*} Note that $\Res g(5) = 0$, and so \begin{equation*} \Res g(x) = (x - 5)(x^2 + 5x + 5), \end{equation*} and the second guy is irreducible, again, by $5$-Eisenstein. Thus \begin{equation*} \Gal \Res{g(x)} = \mathbb{Z}_2. \end{equation*} Thus $m = 2$. Let $u = 5 \in \mathbb{Q}$, and consider \begin{equation*} x^2 - 5x + 5 \text{ and } x^2 - 5. \end{equation*} Notice that the roots of $x^2 + 5x + 5$ are \begin{equation*} \frac{-5 \pm \sqrt{25 - 20}}{2} = \frac{-5 \pm \sqrt{5}}{2}. \end{equation*} Thus $L = \mathbb{Q}(\sqrt{5})$. The roots of $x^2 - 5x + 5$ are \begin{equation*} \frac{5 \pm \sqrt{5}}{2} \in L, \end{equation*} and the roots of $x^2 - 5$ are: \begin{equation*} \pm \sqrt{5} \in L. \end{equation*} Thus both the polynomial splits in $L$ and so \begin{equation*} \Gal f(x) = \mathbb{Z}_4. \end{equation*} \end{eg} % subsection quartics_continued (end) \subsection{Summary of Galois Groups for Polynomials}% \label{sub:summary_of_galois_groups_for_polynomials} % subsection summary_of_galois_groups_for_polynomials Throughout this section, suppose $\Char(F) \neq 2, 3$. \begin{enumerate} \item $\disc(x^3 + bx + c) = -4b^3 - 27c^2$ \item If $f(x) = x^3 + bx + c \in F[x]$ is irreducible and separable then \begin{equation*} \Gal(f(x)) \simeq \begin{cases} A_3 & \text{ if } \disc(f(x)) \text{ is a square in } F \\ S_3 & \text{ otherwise } \end{cases} \end{equation*} \item If $f(x) = x^4 + bx^2 + cx + d \in F[x]$ is separable and irreducible and $m = \abs{\Gal(\Res(f(x)))}$ then \begin{equation*} \Gal(f(x)) \simeq \begin{cases} S_4 & m = 6 \\ A_4 & m = 3 \\ V & m = 1 \\ D_4 \text{ or } \mathbb{Z}_4 & m = 2 \end{cases}. \end{equation*} Moreover, if $m = 2$, $u$ is the root of $\Res(f(x))$ in $F$, and $L$ is the splitting field for $\Res(f(x))$ over $F$, then $\Gal(f(x)) \simeq \mathbb{Z}_4$ iff both $x^2 - ux + d$ and $x^2 + (b - u)$ split over $L$. \end{enumerate} % subsection summary_of_galois_groups_for_polynomials (end) % section galois_groups_of_polynomials_continued_2 (end) \section{Solvability by Radicals}% \label{sec:solvability_by_radicals} % section solvability_by_radicals \subsection{Solvable Groups}% \label{sub:solvable_groups} % subsection solvable_groups \begin{defn}[Solvable Groups]\index{Solvable Groups}\label{defn:solvable_groups} A group $G$ is \hlnoteb{solvable} if there exists a finite chain of subgroups starting at $G = G_0$: \begin{equation*} G = G_0 \supseteq G_1 \supseteq G_2 \supseteq \hdots \supseteq G_n = \{1\}, \end{equation*} where each $G_{i + 1} \trianglelefteq G_i$, and $G_i / G_{i+1}$ is abelian. \end{defn} \begin{note} It is not necessary that $G_{i + 1} \trianglelefteq G_{i-1}$. \end{note} \begin{eg} Abelian $\implies$ solvable $\because G \supseteq \{1\}$. \end{eg} \begin{eg} $S_4$ is solvable: \begin{equation*} S_4 \supseteq A_4 \supseteq V \supseteq \{1\}. \end{equation*} \end{eg} \begin{eg} $D_{2n}$ is always solvable since it always have a $\mathbb{Z}_n$ as a normal subgroup. \end{eg} \begin{eg} Suppose $G$ is simple (no non-trivial normal subgroup). Then $G$ is solvable iff $G$ is abelian. \end{eg} \begin{eg} Since $A_5$ is simple (by Sylow) and non-abelian, we have that $A_5$ is \hlimpo{not solvable}. \end{eg} \begin{propo}[Subgroups of Solvable Groups are Solvable]\label{propo:subgroups_of_solvable_groups_are_solvable} Let $G$ be a solvable group. If $N \leq G$, then $N$ is solvable. If $N \trianglelefteq G$, then $G / N$ is solvable. \end{propo} \begin{proof}[sketch] We know \begin{equation*} G = G_0 \supseteq G_1 \supseteq \hdots \supseteq G_n =\{1\}. \end{equation*} Then \begin{equation*} N = N \cap G_0 \supseteq N \cap G_1 \supseteq \hdots \supseteq N \cap G_n = \{ 1 \}. \end{equation*} We know that $N \cap G_{i+1} \trianglelefteq N \cap G_i$. Also, by the \hlnotea{Second Isomorphism Theorem} \sidenote{$AB / B \simeq A / A \cap B$}, \begin{equation*} N \cap G_i / N \cap G_{i+1} \simeq (N \cap G_i) G_{i+1} / G_{i+1} \subseteq G_i / G_{i+1}, \end{equation*} where the last guy is abelian. Thus $N$ is solvable. We have \begin{equation*} G / N = \overline{G}_0 \supseteq \overline{G}_1 \supseteq \hdots \supseteq \overline{G}_n = \overline{\{1\}}. \end{equation*} Note that $\overline{G}_i / \overline{G}_{i+1} = \left( G_i N / N \right) / \left( G_{i+1} N / N \right) \simeq G_i N / G_{i+1} N$ by the \hlnotea{Third Isomorphism Theorem}. Go check that the last guy is abelian. \end{proof} % subsection solvable_groups (end) % section solvability_by_radicals (end) % chapter lecture_29_mar_22nd (end) \chapter{Lecture 30 Mar 25th}% \label{chp:lecture_30_mar_25th} % chapter lecture_30_mar_25th \section{Solvability by Radicals (Continued)}% \label{sec:solvability_by_radicals_continued} % section solvability_by_radicals_continued \begin{propo}[Converse of \cref{propo:subgroups_of_solvable_groups_are_solvable}]\label{propo:converse_of_subgroups_of_solvable_groups_are_solvable} Let $N \trianglelefteq G$. Then $G$ is solvable iff $N$ and $\quotient{G}{N}$ are solvable. \end{propo} \begin{proof} \hlbnoted{$(\implies)$} This is done by \cref{propo:subgroups_of_solvable_groups_are_solvable}. \noindent \hlbnoted{$(\impliedby)$} Given the assumptions, consider the chains \begin{equation}\label{eq:converse_of_subgroups_of_solvable_groups_eq_1} N = N_0 \supseteq N_1 \supseteq \hdots \supseteq N_m = \{1\} \end{equation} and \begin{equation*} \overline{G} = \overline{G}_0 \supseteq \overline{G}_1 \supseteq \hdots \supseteq \overline{G}_l = \overline{N} = \{N\}, \end{equation*} where for convenience, we write \begin{equation*} \overline{G}_i = \quotient{G_i}{N}, \text{ for } G_i \leq G, \, N \subseteq G_i. \end{equation*} By the \href{https://tex.japorized.ink/PMATH347S18/classnotes.pdf\#thm.41}{Third Isomorphism Theorem}, we have that \begin{equation*} \quotient{\overline{G}_i}{\overline{G}_{i+1}} \simeq \quotient{G_i}{G_{i+1}}, \end{equation*} and in particular \begin{equation*} G_{i+1} \trianglelefteq G_i. \end{equation*} Thus, we have that \begin{equation*} G = G_0 \supseteq G_1 \supseteq \hdots \supseteq N \end{equation*} and then by \cref{eq:converse_of_subgroups_of_solvable_groups_eq_1}, we can continue the chain down to the trivial subgroup, i.e. \begin{equation*} G = G_0 \supseteq G_1 \supseteq \hdots \supseteq N \supseteq N_1 \supseteq \hdots \supseteq \{1\}. \end{equation*} \end{proof} \begin{note} Suppose $G$ is a finite solvable group. By \hlnotea{refining} \sidenote{By refining, we mean to make the chain of normal subgroups as long as possible.} the chain as much as possible, wma \begin{equation*} G = G_0 \supseteq G_1 \supseteq \hdots \supseteq G_n = \{1\}, \end{equation*} such that $\quotient{G_i}{G_{i+1}}$ is abelian and $G_{i+1} \trianglelefteq G_i$. The refinement process means that $\not\exists H_i \leq G$ such that \begin{equation*} G_{i+1} \subsetneq H_i \subsetneq G_i \text{ and } G_{i+1} \trianglelefteq H_i \trianglelefteq G_i. \end{equation*} Note that if such $H$'s exists, then we would automatically have \begin{equation*} \quotient{G_i}{H_i} \text{ and } \quotient{H_i}{G_{i+1}} \text{ both being abelian, } \end{equation*} but this would mean that the refinement process was not sufficient in the first place. Following that, we know that after refinement, it is necessary that each $\quotient{G_i}{G_{i+1}}$ is abelian and \hyperref[defn:simple_extension_and_primitive_elements]{simple}. This means that each $\quotient{G_i}{G_{i+1}}$ has prime order \sidenote{\hlwarn{Why}?}. % TODO : Why do each of the quotient have prime order? We never mentioned % anything about primes. \end{note} \begin{defn}[Simple Radical Extension]\index{Simple Radical Extension}\label{defn:simple_radical_extension} We say that $K / F$ is a \hlnoteb{simple radical extension} if $K = F(\alpha)$ for some $\alpha \in K$ such that $\alpha^n \in F$ for some $n \in \mathbb{N}$. \end{defn} \begin{defn}[Radical Tower]\index{Radical Tower}\label{defn:radical_tower} A \hlnoteb{radical tower} of $F$ is a tower of fields \begin{equation}\label{eq:defn_radical_tower} K_m / K_{m-1} / \hdots / K_2 / K_1 / F \end{equation} such that $K_1 / F$ and each $K_{i+1} / K_i$ is a simple radical extension. \end{defn} \begin{defn}[Radical Extension]\index{Radical Extension}\label{defn:radical_extension} We say that $K / F$ is a \hlnoteb{radical extension} (or just \hlnoteb{radical}) if there exists a radical tower over $F$ starting at $K$, i.e. \begin{equation*} K_m = K \end{equation*} in \cref{eq:defn_radical_tower}. \end{defn} \begin{defn}[Solvable by Radicals]\index{Solvable by Radicals}\label{defn:solvable_by_radicals} We say that $f(x) \in F[x]$ is \hlnoteb{solvable by radicals} over $F$ if its splitting field is contained in a radical extension of $F$. \end{defn} \begin{eg} Consider $f(x) = x^3 - 5 \in \mathbb{Q}[x]$. Then we have \begin{equation*} \mathbb{Q}(\sqrt[3]{5}, \zeta_3) \supseteq \mathbb{Q}(\sqrt[3]{5}) \supseteq \mathbb{Q}. \end{equation*} Note that $\zeta_3^3 = 1$ and $\left( \sqrt[3]{5} \right)^3 = 5$, both of which are in $\mathbb{Q}$. It follows that $\mathbb{Q}(\sqrt[3]{5}, \zeta_3)$ is radical and so $f(x)$ is solvable by radicals. \end{eg} \begin{eg} Consider $f(x) = x^4 - 4x^2 + 2 \in \mathbb{Q}[x]$. Then we have \begin{equation*} \mathbb{Q}(\sqrt{2 + \sqrt{2}}) \supseteq \mathbb{Q}(\sqrt{2}) \supseteq \mathbb{Q}. \end{equation*} Once again, we see that $f(x)$ is solvable by radicals. \end{eg} An uninteresting non-example would be $K = F(\pi)$, which is not a radical extension, and particularly because $\pi$ is \hyperref[defn:algebraic_and_transcendental]{transcendental} over $\mathbb{Q}$. We shall see a more interesting example in A10Q2. % TODO: provide assignment problem for the example \begin{defn}[Cyclic Extension]\index{Cyclic Extension}\label{defn:cyclic_extension} We say that $K / F$ is a \hlnoteb{cyclic extension} if $K / F$ is finite and Galois, and its Galois group is cyclic.\sidenote{In other words, we say that a finite Galois extension is cyclic if its Galois group is cyclic.} \end{defn} \begin{note} For the rest of the course, we shall always assume that we are working in a field of characteristic 0. \end{note} \begin{propo}[Simple Primitive Extensions are Cyclic]\label{propo:simple_primitive_extensions_are_cyclic} If $F$ contains a primitive $n$\textsuperscript{th} root of unity and $K = F(\alpha)$ with $\alpha^n \in F$, then $K / F$ is cyclic. \end{propo} \marginnote{ \begin{strategy} We can show that the Galois group $G$ to be isomorphic to a cyclic group, and we know that we can show that by showing how $G$ permutes its roots. Since our extension contains a primitive $n$\textsuperscript{th} of unity, if we make use of that, we will have a way of moving around all $n$ of the roots. The polynomial which captures all of the roots we want is exactly $f(x) = x^n - \alpha^n$. \end{strategy}} \begin{proof} Consider the polynomial \begin{equation*} f(x) = x^n - \alpha^n \in F[x]. \end{equation*} Let $\zeta \in F$ we a primitive $n$\textsuperscript{th} root of unity. Now the roots of $f(x)$ in $K$ are \begin{equation*} \alpha,\,\alpha\zeta,\,\alpha\zeta^2,\,\ldots,\,\alpha\zeta^{n-1}. \end{equation*} We see that $K$ is the splitting field of $f(x)$ over $F$. Since $F$ is perfect (this is our assumption for the rest of the course!), we know that $K / F$ is therefore Galois. Now for each $\phi \in G = \Gal(K / F)$, $\exists! 0 \leq i \leq n-1$ such that $\phi(\alpha) = \alpha\zeta^i$. Let $\Gamma : G \to \mathbb{Z}_n$ such that $\Gamma(\phi) = i$. It is clear that $\Gamma$ is a homomorphism: taking two elements is the same as just adding the powers. It is also injective, which is rather clear, since each of the $\phi \in G$ has a uniquely associated $0 \leq i \leq n-1$. It follows that $G \simeq \mathbb{Z}_n$ by the First Isomorphism Theorem, and so $G$ is cyclic. \end{proof} % section solvability_by_radicals_continued (end) % chapter lecture_30_mar_25th (end) \chapter{Lecture 31 Mar 27th}% \label{chp:lecture_31_mar_27th} % chapter lecture_31_mar_27th \section{Solvability by Radicals (Continued 2)}% \label{sec:solvability_by_radicals_continued_2} % section solvability_by_radicals_continued_2 \begin{defn}[Linearly Dependent and Independent]\index{Linearly Dependent}\index{Linearly Independent}\label{defn:linearly_dependent_and_independent} We say that $\{ \sigma_1, \sigma_2, \ldots, \sigma_n \} \subseteq \Aut K$ is \hlnoteb{linearly dependent} over $K$ if $\exists a_i \in K$, with not all $a_i \neq 0$ such that \begin{equation*} a_1 \sigma_1(\alpha) + \hdots a_n \sigma_n(\alpha) = 0 \end{equation*} for all $\alpha \in K$. Otherwise, we say that $\{ \sigma_1, \sigma_2, \ldots, \sigma_n \}$ is \hlnoteb{linearly independent}. \end{defn} \begin{lemma}[The Galois Group is Linearly Independent]\label{lemma:the_galois_group_is_linearly_independent} Suppose $[K : F] < \infty$, Then $\Gal( K / F )$ is linearly independent over $K$. \end{lemma} \begin{proof} Suppose not. Let $\{ \sigma1, \ldots, \sigma_r \}$ be a minimal linearly dependent subset of $G = \Gal(K / F)$. Notice that since $a_1 \in K^\times$ m we have that $a_1 \sigma_1(\alpha) = 0$ for any $\alpha \in K$ implies that $\sigma_1 = 0$, which is impossible. Thus we must have $r > 1$. Now by assumption, $\exists \alpha_i \in K$ such that $\forall \alpha \in K$, we have \begin{equation*} a_1 \sigma_1(\alpha) + a_2 \sigma_2(\alpha) + \hdots + a_r \sigma_r(\alpha) = 0. \end{equation*} Let $\beta \in K$ such that wlog, $\sigma_1(\beta) \neq \sigma_2(\beta)$. Then for any $\alpha \in K$, we have \begin{equation}\label{eq:linearly_indep_galois_eq_1} a_1\sigma_1(\alpha)\sigma_1(\beta) + a_2\sigma_2(\alpha)\sigma_2(\beta) + \hdots + a_r\sigma_r(\alpha)\sigma_r(\beta) = 0 \end{equation} and \begin{equation}\label{eq:linearly_indep_galois_eq_2} a_1\sigma_1(\alpha)\sigma_1(\beta) + a_2\sigma_2(\alpha)\sigma_1 + \hdots + a_r\sigma_r(\alpha)\sigma_1(\beta) = 0. \end{equation} Subtracting \cref{eq:linearly_indep_galois_eq_2} from \cref{eq:linearly_indep_galois_eq_1}, we have \begin{equation*} [a_2\sigma_2(\beta) - a_2\sigma_1(\beta)]\sigma_2(\alpha) + \hdots + [a_r\sigma_r(\beta) - a_r\sigma_1(\beta)]\sigma_r(\alpha) = 0. \end{equation*} Notice that the cofficient of first term is non-zero. Thus $\{ \sigma_2, \ldots, \sigma_r \}$ is also a linearly dependent subset of $G$. This contradicts minimality, and so it is impossible that $G$ is linearly dependent. \end{proof} \begin{propo}[Cyclic Extensions over Base with Primitive Roots are Simple Radical]\label{propo:cyclic_extensions_over_base_with_primitive_roots_are_simple_radical} Let $F$ be a field, which contains a primitive $n$\textsuperscript{th} root of unity $\zeta$. If $K / F$ is cyclic with $[K : F] = n$, then $K / F$ is simple radical. \end{propo} \begin{proof} \sidenote{This is a proof that involves a lot of non-trivial steps. Each of these steps are given a \faLightbulb mark.}By assumption, we may assume that \begin{equation*} G = \Gal(K / F) = \langle \sigma \rangle, \text{ where } \abs{\sigma} = n = [K : F] = \abs{G}, \end{equation*} for some $\sigma \in G$. \noindent \hlbnoted{Finding an element that will serve as the adjoined element} \faLightbulb For any $\alpha \in K$, consider \begin{equation*} g(\alpha) = \alpha + \zeta\sigma(\alpha) + \zeta^2\sigma^2(\alpha) + \hdots + \zeta^n\sigma^n(\alpha). \end{equation*} \faLightbulb Notice that since $\zeta$ is a primitive $n$\textsuperscript{th} of unity, we have that \begin{equation*} \zeta\sigma(g(\alpha)) = g(\alpha) \implies \sigma(g(\alpha)) = \zeta^{-1} g(\alpha) = \zeta^{n-1} g(\alpha). \end{equation*} Since $\sigma$ is automorphism, in particular a homomorphism, we have that \begin{equation*} \sigma(g(\alpha)^n) = \sigma(g(\alpha))^n = (\zeta^{n-1} g_(\alpha))^n = g(\alpha). \end{equation*} It follows that $g(\alpha) \in \Fix G = F$. Note that since $\alpha \in K$ was arbitrary, this process works for any $\alpha$. \noindent \hlbnoted{Showing that $g(\alpha)$ cannot live in any of the intermediate extensions} Now since $G$ is linearly independent over $K$ (by \cref{lemma:the_galois_group_is_linearly_independent}), we know that $\exists \alpha \in K$ such that $g(\alpha) \neq 0$. Let's consider such an $\alpha$. By our argument just slightly before this, we know that \begin{equation*} \sigma^i(g(\alpha)) \neq g(\alpha) \text{ for } 1 \leq i \leq n - 1. \end{equation*} This means that $g(\alpha) \notin \Fix H$ for any $\{1\} \neq H \leq G$. It follows from the \hyperref[thm:fundamental_theorem_of_galois_theory]{Fundamental Theorem of Galois Theory} that $g(\alpha) \notin E$ for any $F \subseteq E \subsetneq K$. It thus follows that $F(g(\alpha)) = K$. Since $g(\alpha)^n \in F$, it follows by definition that $K / F$ is a simple radical. \end{proof} The following proposition is important for us to move forward, but we shall prove this in \cref{chp:lecture_32_mar_29th}. \begin{propononum}[A Condition for Solvable Galois Groups] Suppose we have a tower of fields $K / E / F$, $K / E$ is radical, and $E / F$ is Galois. Then there exists $L / K$ such that \begin{itemize} \item $L / F$ is Galois; \item $L / E$ is radical; and \item $\Gal(L / E)$ is solvable. \end{itemize} \end{propononum} \begin{crly}[Radical Extensions have Solvable Extensions]\label{crly:radical_extensions_have_solvable_extensions} If $K / F$ is radical, then there exists $L / K$ such that $L / F$ is radical and Galois, with $\Gal(L / F)$ being solvable. \end{crly} \begin{proof} We simply need to use the last proposition and set $E = F$. \end{proof} \begin{thm}[Galois Theorem]\index{Galois Theorem}\label{thm:galois_theorem} Let $f(x) \in F[x]$. Then $f(x)$ is solvable by radicals over $F$ iff $\Gal(f(x))$ is solvable. \end{thm} \begin{proof} \sidenote{We shall proof only the $(\implies)$ direction, which is what we will be using for the rest of the course. Please read $(\impliedby)$ direction in the recommended text of the course (shall be added to an appendix page if I come to read it).} Suppose $f(x)$ is solvable by radicals over $F$. WMA \begin{equation*} f(x) = p_1(x)^{i_1} p_2(x)^{i_2} \hdots p_l(x)^{i_l}, \end{equation*} where each $p_i(x)$ is irreducible and distinct from one another. Now by replacing $f(x)$ with $p_1(x)p_2(x) \hdots p_l(x)$, \sidenote{We may do this because $\Gal(f(x)) = \Gal(p_1(x)p_2(x)\hdots p_l(x))$.} wma $f(x)$ is separable. Let $E$ be the splitting field $f(x)$ over $F$. Then $E / F$ is Galois \sidenote{cf. \hyperref[thm:fundamental_theorem_of_galois_theory]{the fundmental theorem}.}. Moreover, since $f(x)$ is, still, solvable by radicals, we have that $E \subseteq K$ where $K / F$ is a \hyperref[defn:radical_extension]{radical extension}. By \cref{crly:radical_extensions_have_solvable_extensions}, $\exists L / K$ such that $L / F$ is Galois and radical, and in particular $\Gal(L / F)$ is solvable. Since $E / F$ is Galois, we have that $\Gal(L / E) \trianglelefteq \Gal(L / F)$ by \cref{propo:intermediate_subfields_and_normal_subfields}, and by \cref{propo:first_isomorphism_theorem_on_galois_groups}, we have \begin{equation*} \Gal(f(x)) = \Gal(E / F) \simeq \quotient{\Gal(L / F)}{\Gal(L / E)}. \end{equation*} Since $\quotient{\Gal(L / F)}{\Gal(L / E)}$ is abelian, it follows that $\Gal(f(x))$ is indeed solvable. \end{proof} \begin{note} The contrapositive of \cref{thm:galois_theorem} is particularly useful; the statement is \begin{quotebox}{magenta}{foreground} If $\Gal(f(x))$ is not solvable, then $f(x)$ is not solvable by radicals. \end{quotebox} \end{note} \begin{warning} Going back to \cref{chp:lecture_30_mar_25th}, recall \hyperref[defn:solvable_by_radicals]{the definition of solvability by radicals}, and notice that Galois' Theorem does not tell us that the splitting field of $f(x)$ need to be a radical extension, despite the splitting field being contained in a radical extension. \end{warning} % section solvability_by_radicals_continued_2 (end) % chapter lecture_31_mar_27th (end) \chapter{Lecture 32 Mar 29th}% \label{chp:lecture_32_mar_29th} % chapter lecture_32_mar_29th \section{Solvability by Radicals (Continued 3)}% \label{sec:solvability_by_radicals_continued_3} % section solvability_by_radicals_continued_3 Recall that $f(x) \in F[x]$ is solvable by radicals iff $\Gal f(x)$ is solvable (cf. \hyperref[thm:galois_theorem]{Galois' Theorem}). Note that we still assume that $\Char F = 0$. \begin{eg} Let $f(x) \in \mathbb{Q}[x]$ such that $1 \leq \deg f(x) < 5$. Then $f(x)$ is solvable by radicals. \begin{proof} We can get $f(x) \mapsto g(x)$ by deleting repeated factors. Then $g(x)$ is separable. Then $\Gal f(x) = \Gal g(x) \leq S_4$. Recall that $S_4$ is solvable. \end{proof} \end{eg} We require the following proposition from group theory. You can prove this for yourself. \begin{note} $S_n = <(1 \; 2), (1 \; 2 \; 3 \; \hdots \; n)>$. If $p$ is prime, then $S_p = <\tau, \sigma>$, where $\tau$ is any transposition and $\sigma$ is any $p$-cycle. \end{note} \begin{lemma}[Galois groups of Polynomials with Non-Real Roots] Let $f(x) \in \mathbb{Q}[x]$ be irreducible with prime degree $p$. If $f(x)$ has exactly $2$ non-real roots, then $\Gal f(x) \simeq S_p$. \end{lemma} \begin{proof} Let $\alpha$ be a root of $f(x)$ in its splitting field $K$. Then $[ \mathbb{Q}(\alpha) : \mathbb{Q} ] = \deg f = p$. By the Tower Theorem, $p \mid [ K : \mathbb{Q} ] = \abs{\Gal f(x)}$. This means that $\exists \sigma \in \Gal f(x), \, \abs{\sigma} = p$. Wlog, wma $\sigma = (1 \; 2 \; 3 \; \hdots \; p)$. Let $\phi : \mathbb{C} \to \mathbb{C}$ be given by $\phi(z) = \overline{z}$, which is a $\mathbb{Q}$-map \sidenote{Indeed, $\phi$ fixes $\mathbb{Q}$}. By the \hyperref[thm:normality_theorem]{Normality Theorem}, $\phi \restriction_{K} \in \Gal f(x)$. Since $f(x)$ has only two non-real roots, it follows that $\phi \restriction_{K} = (i \; j)$. By the note above, we have that $\Gal f(x) \simeq S_p$. \end{proof} \begin{eg} Consider $f(x) = x^5 + 2x^3 - 24x - 2 \in \mathbb{Q}[x]$. Note that $f(x)$ is irreducible by $2$-Eisenstein. Also, \begin{table}[ht] \centering \caption{Some values of $f(x)$} \label{table:some_values_of_f_x} \begin{tabular}{c c c c} $f(-100)$ & $f(-1)$ & $f(1)$ & $f(100)$ \\ \hline $< 0$ & $> 0$ & $< 0$ & $> 0$ \end{tabular} \end{table} We see that there are at least 3 real roots between all of the above values by the \hlnotea{Intermediate Value Theorem}. Say the roots of $f(x)$ are $\alpha_i$, $1 \leq i \leq 5$. Then, $\sum \alpha_i = - [x^4] f(x) = 0$, where $[x^4] f(x)$ is the coefficient of $x^4$ in $f(x)$, and \begin{equation*} \sum_{i < j} \alpha_i \alpha_j = [x^3] f(x) = 2. \end{equation*} Therefore, $\sum \alpha_i^2 = \left( \sum \alpha_i \right)^2 - 2 \sum_{i < j} \alpha_i \alpha_j = -4$. Thus, not all roots of $f(x)$ are real. Since non-real roots of $f(x)$ appear in conjugate pairs, it follows that $f(x)$ has exactly two non-real roots. By the lemma, $\Gal f(x) \simeq S_5$. Since $S_5$ is not solvable (cause $A_5 \leq S_5$), $f(x)$ is not solvable by radicals. \end{eg} \begin{procedure}[Showing Insolvability of a Quintic]\label{procedure:showing_insolvability_of_a_quintic} There are several ways we can do this. Again, one can mix and match these methods to show that a quintic is not solvable by radicals. \begin{itemize} \item The above example gives us a heuristical method to check if any of the roots are non-real, and the key to the above method is to pay attention to \begin{equation*} \sum \alpha_i = - [x^4]f(x) \text{ and } \sum_{i < j} \alpha_i \alpha_j = [x^3]f(x), \end{equation*} and there will be non-real roots if \begin{equation*} \sum \alpha_i^2 = \left( \sum \alpha_i \right)^2 - 2 \sum_{i < j} \alpha_i \alpha_j < 0, \end{equation*} since that is not a value that we can expect coming from the reals. \item We can also check the \hlnotea{derivative} of $f(x)$, and inspect from there where the `turning points' of $f(x)$ on $\mathbb{R}^2$ are and how many are there. \end{itemize} \end{procedure} \begin{thm}[Insolvability of the Quintics]\index{Insolvability of the Quintics}\label{thm:insolvability_of_the_quintics} Not every quintic $f(x) \in \mathbb{Q}[x]$ is solvable by radicals. \end{thm} \begin{eg} An example of a quintic that is solvable by radicals is $x^5 - 1$. \end{eg} \newthought{Going back} to that black box that we have yet to prove: recall the proposition. \begin{propo}[A Higher Extension that is Both Galois and Radical]\label{propo:a_higher_extension_that_is_both_galois_and_radical} Suppose $K / E / F$ is a tower of fields, $E / F$ is Galois and $K / E$ is radical. Then $\exists L / K$ such that $L / F$ is Galois and $L / E$ is radical, and $\Gal(L / E)$ is solvable. \end{propo} \begin{proof} We prove the result when $K / E$ is simple radical. The more general case follows by `an' \sidenote{There is some work to do, but this is not important. Still a nice exercise I reckon.} induction. Say $K = E(\alpha)$ where $\alpha^n = \beta \in E$. Also, suppose $G = \Gal(E / F) = \{ \sigma_1, \sigma_2, ..., \sigma_r \}$. Consider \sidenote{We want to grab the $n$ roots of unity. Notice that $\prod_{i=1}^{r} (x^n - \sigma_i(\beta))$ stays in $\Fix G$ if we apply any of the $\sigma$'s. In particular, notice that for any $\sigma_j$, when applied to $f(x)$, we know that $\Phi_n(x)$ is definitely fixed, but the other requires some care. However, it is not difficult: notice that we end up with $(\sigma_j \sigma_i) (\beta)$, over all $i$, and for each $i$, we get another $\sigma_{i'}$ that is different from other $i$'s. It is this subtle observation that eventually keeps the polynomial in the fix field.} \begin{equation*} f(x) = \Phi_n(x) \prod_{i=1}^{r} (x^n - \sigma_i(\beta)) \in \Fix G[x] = F[x]. \end{equation*} Let $L$ be the splitting field for $f(x)$ over $K$. \noindent \hlbnoted{$L / F$ is Galois} Notice that \begin{equation*} K = K(\text{roots of } f(x)) = K(\alpha, \text{ others}) = E(\alpha, \text{ others}). \end{equation*} Thus $L$ is the splitting field for $f(x)$ over $E$ (\hlbnoted{$L / E$} is radical). Since $E / F$ is Galois, $E$ is the splitting field of some separable polynomial $h(x) \in F[x]$. Thus $L$ is the splitting field for $h(x) f(x)$ over $F$. Since $\Char F = 0$, $L / F$ is Galois. \hlbnoted{$L / E$ is radical} Let $\zeta$ be a root of $\Phi_n(x)$, and we know that $\zeta \in L$. Extend each $\sigma_i \in G$ to $\sigma^*_i \in \Gal( E / F )$ by the \hyperref[lemma:isomorphism_extension_lemma]{Isomorphism Extension Lemma}. Then the roots of $f(x)$ are either $\zeta^i$ or $\zeta^i \sigma^*_i(\alpha)$. This means that \begin{equation*} L = E(\zeta, \sigma_1^*(\alpha), \sigma_2^*(\alpha), \ldots, \sigma_r^*(\alpha)). \end{equation*} Note that we have \begin{equation*} (\sigma_i^*(\alpha))^n = \sigma_i^*(\beta) = \sigma_i(\beta) \in E, \end{equation*} since $\beta \in E$. Then we have the chain \begin{equation*} E \subseteq E(\zeta) \subseteq E(\zeta, \sigma_1^*(\alpha)) \subseteq \hdots \subseteq L. \end{equation*} Since each of the adjoined element at each stage is radical, it follows that $L / E$ is indeed radical as claimed. \hlbnoted{$\Gal(L / E)$ is solvable} Let $E_0 = E(\zeta)$, and for $i \leq i \leq r$, \begin{equation*} E_i = E(\zeta, \sigma_1^*(\alpha), \ldots, \sigma_i^*(\alpha)) = E_{i-1}(\sigma_i^*(\alpha)). \end{equation*} Note that $E_r = L$. Let $G_i = \Gal(L / E_i)$. By the \hyperref[thm:fundamental_theorem_of_galois_theory]{Fundamental Theorem of Galois Theory} \sidenote{We want to reverse the inclusion, and the fundamental theorem gives us exactly this.}, we have that \begin{equation*} \{1\} = G_r \leq G_{r-1} \leq \hdots \leq G_2 \leq G_1 \leq G_0 = \Gal(L / E(\zeta)). \end{equation*} Moreover, \begin{equation*} G_0 \leq G' \coloneqq \Gal(L / E). \end{equation*} Thus, we have that \begin{equation*} G_0 = \Gal(L / E(\zeta)) \trianglelefteq \Gal(L / E) = G' \end{equation*} since $E(\zeta) / E$ is Galois, since $E(\zeta)$ is the splitting field of $\Phi_n(x)$. Furthermore, \begin{equation*} G' / G_0 \simeq \Gal(E(\zeta) / E), \end{equation*} which is abelian \sidenote{Recall that $\phi(\zeta) = \zeta^i$ and $\phi \mapsto i$.}. Now $G_{i+1} \trianglelefteq G_i$ since $E_{i+1} / E_i$ is Galois ($\because$ \cref{propo:simple_primitive_extensions_are_cyclic}; $\zeta \in E_i$ and $\sigma_{i+1}^*(\alpha)^n \in E_i$ and so $E_{i+1} / E_i$ is cyclic). Also we have that \begin{equation*} G_i / G_{i+1} \simeq \Gal(E_{i+1} / E_i) \end{equation*} is also cyclic and abelian. \end{proof} % section solvability_by_radicals_continued_3 (end) % chapter lecture_32_mar_29th (end) \chapter{Lecture 35 Apr 05th}% \label{chp:lecture_35_apr_05th} % chapter lecture_35_apr_05th Lecture 33 was given by a guest lecturer on a topic not required by the course. \noindent Lecture 34 was just to complete \cref{propo:a_higher_extension_that_is_both_galois_and_radical}. \section{Final Examination Information}% \label{sec:final_examination_information} % section final_examination_information \paragraph{Logistics} \begin{itemize} \item (Mon) Apr 22 0900 \item STC 0040 \textbf{(do double check before the exam)} \item Assigned seating (CHECK) \end{itemize} \paragraph{Office Hours} \begin{itemize} \item Normal office hours are cancelled \item (Tue) Apr 16 1300 - 1500 \item (Thu) Apr 18 1300 - 1500 \item Appointment \end{itemize} \paragraph{Exam Info} \begin{itemize} \item 8Q $\times$ 10m $\implies$ 80 marks total \item ordinary exam (no scaling) - easier than midterm \begin{itemize} \item Read back on assignments and examples, etc. \end{itemize} \item Materials \begin{enumerate} \item minimal polynomials, field extensions \item show an ext $K / F$ is Galois and find $\Gal(K / F)$ \item $\Gal f(x)$ \item 3 assignment-related questions (parts) \item 2 proofs from lecture (2 part) \begin{itemize} \item Second half of the course (post-midterm, after finite fields) \end{itemize} \item \begin{enumerate} \item New proof \item Assignment proof (from assignments) \end{enumerate} \item Solvability by radicals stuff \item Give examples or DNE (10 parts) \end{enumerate} \end{itemize} Important to read assignments for 1, 2, 3, 4, 6(\faStar), 7, 8 % section final_examination_information (end) \section{Example Problem Practice}% \label{sec:example_problem_practice} % section example_problem_practice \begin{enumerate} \item A Galois $K / F$ such that $[K : F] = 36$. \textbf{A:} $\mathbb{Q}(\zeta_{37}) / \mathbb{Q}$ (note 37 is prime) \item An irreducible $f(x) \in \mathbb{Q}[x]$ such that $\deg f(x) = 7$ and $\abs{\Gal f(x)} = 20$. \textbf{A:} No, $\Gal(f(x))$ is a transitive subgroup but $7 \nmid 20$. \item A field $E$ such that $\mathbb{F}_p \subseteq E \subseteq \mathbb{F}_{p^{12}}$ such that $E / \mathbb{F}_p$ is NOT Galois. \textbf{A:} No. $\Gal( \mathbb{F}_{p^{12}} / \mathbb{F}_p ) \simeq \mathbb{Z}_{12}$ is abelian. and \begin{equation*} \Gal( \mathbb{F}_{p^{12}} / E ) \trianglelefteq \Gal( \mathbb{F}_{p^{12}} / \mathbb{F}_p) \end{equation*} \item An irreducible quintic in $\mathbb{Q}[x]$ which is solvable by radicals. \textbf{A:} $x^5 - 2$ \item An infinite field of characteristic $7$. \textbf{A:} $\overline{\mathbb{F}_7}, \, \mathbb{Z}_7(x)$ \item A Galois ext of $\mathbb{Z}_2(t^2)$ . \textbf{A:} $\mathbb{Z}_2(t^2)$ \item A finite ext of $\mathbb{C}(t)$ which is not simple. \textbf{A:} $\mathbb{C}$ is a perfect field, and primitive element theorem says that all of the extensions of $\mathbb{C}$ are simple. \item An irreducible polynomial in $\mathbb{F}_{3^{10}}[x]$ which is not separable. \textbf{A:} DNE. Finite fields are perfect. \end{enumerate} % section example_problem_practice (end) % chapter lecture_35_apr_05th (end) \appendix \chapter{Asides and Prior Knowledge}% \label{chp:asides_and_prior_knowledge} % chapter asides_and_prior_knowledge \section{Correspondence Theorem}% \label{sec:correspondence_theorem} % section correspondence_theorem \hlnotea{The Correspondence Theorem} is somewhat widely known as the Fourth Isomorphism Theorem, although some authors associates the name with a proposition known as \href{https://en.wikipedia.org/wiki/Zassenhaus_lemma}{Zaessenhaus Lemma}. \begin{thm}[Correspondence Theorem]\index{Correspondence Theorem}\label{thm:correspondence_theorem} Let $G$ be a group, and $N \triangleleft G$ \sidenote{Recall that this symbol means that $N$ is a normal subgroup of $G$.}. Then there exists a bijection between the set of all subgroups $A \leq G$ such that $A \supseteq N$ and the set of subgroups $A / N$ of $G / N$. \end{thm} % TODO add Rational roots theorem % section correspondence_theorem (end) % chapter asides_and_prior_knowledge (end) \chapter{Assignment Problems}% \label{chp:assignment_problems} % chapter assignment_problems \section{Assignment 1}% \label{sec:assignment_1} % section assignment_1 \begin{enumerate} \item \begin{enumerate} \item Let $G$ be a group and let $Z(G)$ denote the centre of $G$. Prove that if $G / Z(G)$ is cyclic then $G$ is abelian. \item Prove that any group of order $p^2$, where $p$ is a prime, is abelian. (Hint: Class Equation) \end{enumerate} \item Prove that there is no simple group of order $132$. \item \begin{enumerate} \item Let $G$ be a simple group and let $H$ be a subgroup of $G$ such that $\abs{ G : H } = n > 1$. Prove that there exists an injective group homomorphism $\phi : G \to S_n$. \item Let $p$ be a prime dividing the order of a simple group $G$, where $G$ is not of order $p$. Let $n_p$ denote the number of Sylow $p$-subgroups of $G$. Prove that the order of $G$ divides $n_p !$. \item Prove that there is no simple group of order $48$. \end{enumerate} \item Let $p$ be a prime which divides the order of a finite group $G$. Say $\abs{G} = p^n m$, with $n \in \mathbb{N}$ and $p \nmid m$. Prove that $G$ has a subgroup of order $p^k$, for $k = 1, 2, \ldots, n$. \item Let $p$ be a prime which divides the order of a finite group $G$. Show that any $p$-subgroup of $G$ is contained in a Sylow $p$-subgroup of $G$. \end{enumerate} % section assignment_1 (end) \section{Assignment 2}% \label{sec:assignment_2} % section assignment_2 \begin{enumerate} \item Find all Sylow $3$-subgroups of $GL_2(\mathbb{Z}_3)$. Be sure to justify that \item The purpose of this exercise is to ensure all students see a proof of the Correspondence Theorem at least once! \begin{enumerate} \item (Correspondence Theorem) Let $G$ be a group and let $N$ be a normal subgroup of $G$. Prove that every subgroup of $G / N$ is of the form $H / N$, where $H$ is a subgroup of $G$ which contains $N$. \item (Normal Correspondence Theorem) Let $G$ be a group and let $N$ be a normal subgroup of $G$. Prove that every normal subgroup of $G / N$ is of the form $H / N$, where $H$ is a normal subgroup of $G$ which contains $N$. \item Let $G$ be a finite group and let $p$ be a prime which divides the order of $G$. Let $N$ be a normal subgroup of $G$. Prove that the number of Sylow $p$-subgroups of $G / N$ is less than or equal to the number of Sylow $p$-subgroups of $G$. (Note: if $p$ does not divide the order of $G / N$, we will say $G / N$ has $0$ Sylow $p$-subgroups.) \end{enumerate} you have found them all. \item Consider the following group-theoretic proposition. \begin{quotebox}{magenta}{foreground} Let $G$ be a group and let $H_1, H_2, \ldots, H_n$ be a collection of normal subgroups of $G$. If $(H_1 H_2 \hdots H_{i - 1}) \cap H_i = \{ e \}$ for $i = 2, 3, \ldots, n$ then $H_1 H_2 \hdots H_n \cong H_1 \times H_2 \times \hdots \times H_n$. \end{quotebox} \begin{enumerate} \item Prove the proposition when $n = 2$. \item Let $G$ be a finite group. Prove that every Sylow subgroup of $G$ is normal in $G$ if and only if $G$ is an external direct product of its Sylow subgroups. (You may freely use the above proposition.) \end{enumerate} \item Use Sylow Theory to classify all groups of order $175$ up to isomorphism. \end{enumerate} % section assignment_2 (end) \section{Assignment 3}% \label{sec:assignment_3} % section assignment_3 \begin{enumerate} \item Determine whether or not the following polynomials are irreducible over the indicated ring. \begin{enumerate} \item $f(x) = 8x^7 + 3x^4 + x^3 + 2x^2 + 5x + 9 \in \mathbb{Z}_{11}[x]$. \item $f(x) = x^4 + 3 \in \mathbb{R}[x]$. \item $f(x) = x^{n - 1} + x^{n - 2} + \hdots + x^2 + x + 1 \in \mathbb{Q}[x]$, where $n > 1$ is a composite number. \item $f(x) = x^3 + 5x^2 - 2x + 12 \in \mathbb{Z}[x]$. \item $f(x, y) = x^4 + y^5 x^3 + y^3 x + y \in \mathbb{Q}[y, x]$. \end{enumerate} \item \begin{enumerate} \item Let $F$ be a field and let $a \in F$, Prove that $f(x) \in F[x]$ is irreducible if and only if $f(x + a) \in F[x]$ is irreducible. \label{item:a3q2a} \item Prove that $f(x) = x^4 + 1$ is irreducible over $\mathbb{Q}$. \end{enumerate} \item \begin{enumerate} \item Compute $\deg_{\mathbb{Q}}(i + \sqrt{2})$. \item Compute $\deg_{\mathbb{Q}}(\cos(\pi / 9))$ (Hint: De Moivre's Theorem involving $\cos(3 \theta)$). \end{enumerate} \item Prove that $\mathbb{Q}(\sqrt{2},, \sqrt[3]{2}) = \mathbb{Q}(\sqrt[6]{2})$. \item Construct a field of order $125$. \end{enumerate} % section assignment_3 (end) \section{Assignment 4}% \label{sec:assignment_4} % section assignment_4 \begin{enumerate} \item Let $K / E$ be an extension and let $E / F$ be a finite extension. Suppose $\alpha \in K$ is algebraic over $F$. \begin{enumerate} \item Prove that $[E(\alpha) : E] \leq [F(\alpha) : F]$. \item Prove that $[E(\alpha) : F(\alpha)] \leq [E : F]$. \end{enumerate} \item Let $K / F$ be an extension and let $\alpha, \beta \in K$ be algebraic over $F$. Prove that if $\deg_F(\alpha)$ and $\deg_F(\beta)$ are coprime, then $[F(\alpha, \beta) : F] = [F(\alpha) : F][F(\beta) : F]$. \item Compute the following degrees: \begin{enumerate} \item $[\mathbb{Q}(\alpha, \beta) : \mathbb{Q}]$, where $\alpha = \cos(\pi / 9)$ and $\beta = i + \sqrt{2}$ (Hint : A3) \item $[\mathbb{Q}(\sqrt{p}, \sqrt{q}) : \mathbb{Q}]$, where $p$ and $q$ are distinct primes. \item $[\mathbb{Z}_3(\alpha) : \mathbb{Z}_3]$, where $\alpha$ is a root of $f(x) = x^3 + x + 2 \in \mathbb{Z}_3[x]$. \item $[\mathbb{R}(t) : \mathbb{R}(t^2)]$. \end{enumerate} \item Let $K / F$ be an extension and let $\alpha, \beta \in K$ be transcendental over $F$. Prove that $F(\alpha) \cong F(x)$, the rational function field over $F$. Conclude that $F(\alpha) \cong F(\beta)$. \item Suppose $f(x)$ is an irreducible polynomial in $\mathbb{Q}[x]$ of degree $2n$. Prove that if $E$ is a field extension of $\mathbb{Q}$ of degree $2$, then $f(x)$ is either irreducible in $E[x]$ or $f(x)$ factors in $E[x]$ as a product of two irreducible factors each of degree $n$. \end{enumerate} % section assignment_4 (end) \section{Assignment 5}% \label{sec:assignment_5} % section assignment_5 \begin{enumerate} \item Find the splitting field $K$ of the following polynomials over $\mathbb{Q}$ and determine $[ K : \mathbb{Q} ]$. \begin{enumerate} \item $f(x) = x^{11} - 2$. \item $f(x) = x^4 - x^2 + 4$ (Optional hint: $f(x) = (x^2 + 2)^2 - 5x^2$) \item $f(x) = x^4 + 2$. \end{enumerate} \item \begin{enumerate} \item Prove that for any field $F$ there exists an algebraically closed field $K$ containing $F$. Hint: Use the following steps: \begin{enumerate} \item For every nonconstant monic polynomial $f(x) \in F[x]$, let $x_f$ be a symbol/indeterminate. Consider the polynomial ring $R = F[\ldots, x_f, \ldots]$. Let $I$ be the ideal of $R$ generated by elements of the form $f(x_f)$. Show that $I \neq R$. \item Since $I$ is a proper ideal of $R$, $I$ is contained in some maximal ideal $M$ of $R$. Consider the field \begin{equation*} K_1 := R / M, \end{equation*} which is a field extension of $F$. Show that any non-constant polynomial in $F[x]$ has a root in $K_1$. \item Similarly, you could construct $K_2$ so that any non-constant polynomial in $K_1[x]$ has a root in $K_2$. (You don't need to do this.). Continuing in this way we get a chain of field extensions \begin{equation*} F = K_0 \subseteq K_1 \subseteq K_2 \subseteq \hdots. \end{equation*} Let \begin{equation*} K := \bigcup_{i=1}^{\infty} K_i. \end{equation*} Show that $K$ is an algebraically closed field containing $F$. \end{enumerate} \item Prove that $\bar{\mathbb{Q}} \neq \mathbb{C}$. \end{enumerate} \item Let $F$ be a field and let \begin{equation*} f(x) = a_n x^n + \hdots + a_1 x + a_0 \in F[x] \end{equation*} be a polynomial. The formal \textit{derivative} of $f(x)$, denoted by $f'(x)$, is defined to be the polynomial \begin{equation*} f'(x) = na_n x^{n - 1} + (n - 1) a_{n - 1} x^{n - 2} + \hdots + a_1 \in F[x]. \end{equation*} Of course, all the usual derivative properties (product rule, quotient rule, chain rule etc.) apply. Now let $\alpha$ be a root of $f(x)$ in some extension $K$ of $F$. Then we say that $\alpha$ is a root of $f(x)$ in $K$ of \textit{multiplicity} $n$ if $f(x) = q(x)(x - \alpha)^n$, where $q(x) \in K[x]$ and $q(\alpha) \neq 0$. If $\alpha$ is a root of $f(x)$ of multiplicity greater than $1$ it is said to be a \textit{multiple root} in $K$. \begin{enumerate} \item Let $F$ be a field and let $0 \neq f(x) \in F[x]$. Prove that $f(x)$ has a multiple root in some extension of $F$ if and only if $f(x)$ and $f'(x)$ have a common factor in $F[x]$ of positive degree. (Hint: If $f(x)$ and $f'(x)$ have no common factor of positive degree then there exists $a(x), b(x) \in F[x]$ such that $f(x) a(x) + f'(x) b(x) = 1$. This follows from the division algorithm for polynomial rings.) \item Prove that $f(x) = x^6 - 2 \in \mathbb{Z}_3[x]$ has a multiple root in an extension of $\mathbb{Z}_3$. \item A polynomial $f(x) \in F[x]$ is said to be \textit{separable} if $f(x)$ has no multiple roots in any extension of $F$. Prove that an irreducible polynomial $f(x) \in F[x]$ is separable if and only if $f'(x) \neq 0$. \item Let $F$ be a field of characteristic $0$. Prove that every irreducible polynomial in $F[x]$ is separable. \end{enumerate} \item Let $K$ be a the splitting field of $x^n - c$ over $\mathbb{Z}_p$, where $p$ is prime, $n$ is a positive integer, and $c \in \mathbb{Z}_p^\times$. (Note: $\mathbb{Z}_p^\times = \mathbb{Z}_p \setminus \{ 0 \}$.) \begin{enumerate} \item Prove that if $p$ does not divide $n$ then $x^n - c$ has $n$ distinct roots in $K$. \item Suppose $p \mid n$. Find an expression in terms of $p$ and $n$ for the number of distinct roots of $x^n - c$ in $K$. \end{enumerate} \end{enumerate} % section assignment_5 (end) \section{Assignment 6}% \label{sec:assignment_6} % section assignment_6 \begin{enumerate} \item Let $F$ be a field. \begin{enumerate} \item Let $f(x) \in F[x]$ be irreducible, separable, and of degree $n$. Prove that if $G = \Gal(f(x))$ then $n$ divides $\abs{G}$ and $\abs{G}$ divides $n!$. \item Prove that if $f(x) \in F[x]$ is an irreducible and separable quadratic, then $\Gal(f(x)) \simeq \mathbb{Z}_2$. \item Prove that there is no irreducible and separable quartic $f(x) \in F[x]$ such that $\Gal(f(x)) \simeq S_3$. \item Give an example of a separable quartic $f(x) \in \mathbb{Q}[x]$ such that \\ $\Gal(f(x)) \simeq S_3$. \end{enumerate} \item Let $F$ be a field. \begin{enumerate} \item Prove the following converse to a result from lecture: Let $f(x) \in F[x]$ be separable and let $K$ be the splitting field of $f(x)$ over $F$. Prove that if for all roots $\alpha, \beta \in K$ of $f(x)$, there exists $\phi \in \Gal(K / F)$ such that $\phi(\alpha) = \beta$, then $f(x)$ is irreducible over $F$. \item Let $f(x) \in F[x]$ be separable with $f(0) \neq 0$ and $\deg(f(x)) = n \geq 1$. Now let $g(x) = x^n f \left( \frac{1}{x} \right) \in F[x]$. Use part (a) to prove that $f(x)$ is irreducible over $F$ if and only if $g(x)$ is irreducible over $F$. \item Let $f(x) = 3x^4 + 9x^3 - 21 x^2 + 81x + 1 \in \mathbb{Q}[x]$. Find $g(x)$, as in part (b). What pattern do you notice? Prove that $f(x)$ is irreducible over $\mathbb{Q}$. \end{enumerate} \item \begin{enumerate} \item Let $f(x), g(x) \in F[x]$ be distinct monic, separable, and irreducible polynomials in $F[x]$. Let $K$ be the splitting field of $f(x) g(x)$ over $F$. Prove that $f(x)$ and $g(x)$ do not have any roots in common in $K$. \item Let $K / F$ be an algebraic field extension. Prove that if $\alpha_1, \alpha_2, \ldots, \alpha_n \in K$ have distinct, separable minimal polynomials $p_1(x), p_2(x), \ldots, p_n(x) \in F[x]$, respectively, then \begin{equation*} f(x) = p_1(x) p_2(x) \hdots p_n(x) \in F[x] \end{equation*} is separable. \end{enumerate} \item Consider $f(x) = x^4 - 2 \in \mathbb{Q}[x]$. The roots of $f(x)$ in $\mathbb{C}$ are \begin{equation*} \alpha_1 = \sqrt[4]{2}, \, \alpha_2 = -\alpha_1, \, \alpha_3 = i \alpha_1, \, \alpha_4 = -i \alpha_1. \end{equation*} Let $G = \Gal(f(x))$, viewed as a subgroup of $S_4$ in the usual way. \begin{enumerate} \item Prove that $(1324), (14)(23) \in G$. \item Sketch the roots of $f(x)$ in the complex plane. By considering the actions of the permutations from part (a) on the roots of $f(x)$, which well-known group do you suspect $G$ is isomorphic to? You do not need to prove your answer. \end{enumerate} \end{enumerate} % section assignment_6 (end) \section{Assignment 7}% \label{sec:assignment_7} % section assignment_7 \begin{enumerate} \item Let $K / E / F$ be a tower of fields. Prove or disprove the following: \begin{enumerate} \item If $K / F$ is separable then $K / E$ is separable. \item If $K / F$ is separable then $E / F$ is separable. \item An algebraic extension of a perfect field is a perfect field. \end{enumerate} \item Let $K / E / F$ be a tower of fields such that $[ K : F ] < \infty$. Prove or disprove the following: \begin{enumerate} \item If $E / F$ is normal and $K / E$ is normal then $K / F$ is normal. \item If $K / F$ is normal then $K / E$ is normal. \item If $K / F$ is normal then $E / F$ is normal. \item If $[ K : F ] = 2$ then $K / F$ is normal. \end{enumerate} \item Let $f(x) = x^p - x - a \in \mathbb{Z}_p[x]$, where $p$ is prime and $a \in \mathbb{Z}_p^\times$. \begin{enumerate} \item Prove that $f(x)$ is irreducible over $\mathbb{Z}_p$. (Hint: If you have found one root of $f(x)$ you know them all!) \item Let $\alpha$ be a root of $f(x)$ over $\mathbb{Z}_p$. Show that $\mathbb{Z}_p(\alpha) / \mathbb{Z}_p$ is Galois. \item Find $\Gal(\mathbb{Z}_p(\alpha) / \mathbb{Z}_p)$. \end{enumerate} \item Let $c = \sqrt{2 + \sqrt{2}} \in \mathbb{C}$. \begin{enumerate} \item Find $[ \mathbb{Q}(c) : \mathbb{Q} ]$. \item Show $\mathbb{Q} \subseteq \mathbb{Q}(c)$ is a Galois extension. \item Let $G = \Gal(\mathbb{Q}(c) / \mathbb{Q})$. Make a table describing the actions of $G$ on the $\mathbb{Q}$-conjugates of $c$. Write each element of $G$ as a permutation, in the usual way. \item Which well-known group is $G$ isomorphic to? \end{enumerate} \end{enumerate} % section assignment_7 (end) \section{Assignment 8}% \label{sec:assignment_8} % section assignment_8 \begin{enumerate} \item \begin{enumerate} \item Let $\alpha_1, \alpha_2, \, \ldots, \, \alpha_n \in \mathbb{C}$ be the $n > 1$ distinct roots of an irreducible $f(x) \in \mathbb{Q}[x]$. Prove that \begin{equation*} \frac{3 + \alpha_1}{\alpha_1^7} + \frac{3 + \alpha_2}{\alpha_2^7} + \hdots + \frac{3 + \alpha_n}{\alpha_n^7} \in \mathbb{Q}. \end{equation*} \item Let $K / F$ be a finite Galois extension such that $p \mid [ K : F ]$, where $p$ is prime. Prove that there exists an intermediate subfield $E$ such that $[K : E] = p$. \item Let $F$ be a subfield of $\mathbb{C}$ such that $F / \mathbb{Q}$ is a finite Galois extension. Prove that if $\abs{ \Gal(F / \mathbb{Q}) }$ is odd then $F \subseteq \mathbb{R}$. \end{enumerate} \item Let $F$ be a field and let $F(\alpha_1, \ldots, \alpha_n)$ be a finitely generated field extension of $F$, where each $\alpha_i$ is separable over $F$. \begin{enumerate} \item Let $q_i(x)$ be the minimal polynomial for $\alpha_i$ over $F$, for $1 \leq i \leq n$. Let $p_1(x), p_2(x), \ldots, p_m(x)$ be the list of the distinct $q_i(x)$'s. If $K$ is the splitting field of $p_1(x) p_2(x) \hdots p_m(x)$ over $F$, prove that $K / F$ is Galois. \\ \textbf{Note}: The field $K$ is called the \textit{Galois closure} of $F(\alpha_1, \ldots, \alpha_n)$ over $F$. \item Conclude that $F(\alpha_1, \ldots, \alpha_n)$ is a separable extension of $F$. \item Write down the Galois closure of $\mathbb{Q}(\sqrt[3]{2}, \sqrt[7]{3})$ over $\mathbb{Q}$. \end{enumerate} \item Let $K$ be the splitting field of $f(x) = (x^2 - 7)(x^4 + x^3 + x^2 + x + 1)$ over $\mathbb{Q}$. \begin{enumerate} \item Explain why $K / \mathbb{Q}$ is Galois. \item Make a table, in the usual way, to describe the elements of $\Gal(K/\mathbb{Q})$ as permutations in $S_6$. \item Which well-known group is $\Gal(K/\mathbb{Q})$ isomorphic to? \item With full justification, draw the subfield lattice of $K/\mathbb{Q}$. Be sure to include all degress in your diagram. \end{enumerate} \item How many intermediate fields $F$ are there such that $\mathbb{Q}(i) \subseteq F \subseteq \mathbb{Q}(\sqrt[8]{2}, i)$? \end{enumerate} % section assignment_8 (end) \section{Assignment 9}% \label{sec:assignment_9} % section assignment_9 \begin{enumerate} \item Find the Galois groups of the following polynomials in $\mathbb{Q}[x]$. \begin{enumerate} \item $x^3 + 3x^2 - 2x + 1$ \item $x^4 + 3x + 3$ \item $x^4 + 4x^2 + 1$ \end{enumerate} \item Let $p(x) = x^2 + x + 1 \in \mathbb{Q}[x]$ and $q(x) = x^2 + 3 \in \mathbb{Q}[x]$. Prove that \begin{equation*} \Gal(p(x)q(x)) \not\simeq \Gal(p(x)) \times \Gal(q(x)). \end{equation*} \item Suppose $K / F$ is a finite Galois extension, where $G = \Gal(K / F)$ is abelian. Prove that if $K$ is the spltting field of an irreducible $f(x) \in F[x]$, then $\deg(f(x)) = \abs{G}$. \item Suppose $K / F$ is a finite Galois extension with $\Gal(K / F) \simeq S_3$. Prove that $K$ is the splitting field of an irreducible cubic in $F[x]$. \item Let $F$ be a finite field with $\Char(F) \notin \{2, 3\}$. Prove that if $f(x)$ is an irreducible cubic in $F[x]$, then its discriminant is a square in $F$. \end{enumerate} % section assignment_9 (end) \section{Assignment 10}% \label{sec:assignment_10} % section assignment_10 \begin{enumerate} \item \begin{enumerate} \item Prove that $f(x) = x^5 - 64x + 2 \in \mathbb{Q}[x]$ is not solvable by radicals over $\mathbb{Q}$. \item Let $K$ be the splitting field of an irreducible $f(x) \in \mathbb{Q}[x]$ over $\mathbb{Q}$. Prove that if $[ K : \mathbb{Q} ] = p^n$, for a prime $p$ and $n \in \mathbb{N}$, then $f(x)$ is solvable by radicals over $F$. \end{enumerate} \item Let $K$ be the splitting field of $f(x) = x^3 - 3x - 1 \in \mathbb{Q}[x]$ over $\mathbb{Q}$. Prove that $K$ is not a radical extension of $F$, even though $f(x)$ is solvable by radicals over $F$. \item The goal of this exercise is to prove the Fundamental Theorem of Algebra using Sylow Theory and Galois Theory (and no complex analysis). \label{item:a10q3} \begin{enumerate} \item For the sake of contradiction, suppose $L / \mathbb{C}$ is an algebraic extension with $L \neq \mathbb{C}$. Take $\alpha \in L, \alpha \notin \mathbb{C}$, and let $g(x)$ be the minimal polynomial for $\alpha$ over $\mathbb{R}$. Let $K$ be the splitting field for $g(x)$ over $\mathbb{C}$. Prove that $K / \mathbb{R}$ is a finite Galois extension. \item Let $G = \Gal(K / \mathbb{R})$ so that $\abs{G} = 2^j m$ for some $j \geq 0$ and odd $m \in \mathbb{N}$. Suppose $H$ is a Sylow-2 subgroup of $G$. If $j = 0$, take $H = \{1\}$. If $E = \Fix(H)$, compute $[K : E]$ and $[E : \mathbb{R}]$ in terms of $j$ and $m$. \item Prove that $[E : \mathbb{R}] = 1$. (Hint: Be sure not to use anything about conjugate roots, which relies on the Fundamental Theorem of Algebra.) \item Prove that there exists a subfield $F$ of $K$ such that $[F : \mathbb{C}] = 2$. \item Obtain a contradiction to finish off the proof. \end{enumerate} \end{enumerate} % section assignment_10 (end) % chapter assignment_problems (end) \backmatter \fancyhead[LE]{\thepage \enspace \textsl{\leftmark}} % \nobibliography* % \bibliography{references} \printindex \end{document} % vim:tw=80:fdm=syntax