\documentclass[12pt]{article} \usepackage{fancyhdr,fancybox,amssymb,epsfig, amsmath, latexsym} \usepackage[all]{xy} \usepackage{tikz} \usetikzlibrary{arrows} \usetikzlibrary{decorations.markings} \usetikzlibrary{decorations.pathreplacing} \usetikzlibrary{patterns} \usetikzlibrary{shapes.geometric} \usepackage{tkz-graph} \usepackage{tikz-cd} \usepackage{enumerate} \usepackage{multicol} \usepackage{url} \usepackage{bm} \topmargin=-.5in \headsep=0.2in \oddsidemargin=0in \textwidth=6.7in \textheight=9.2in \footskip=.5in % Useful math macros \def\ds{\displaystyle} \def\bold{\bm} % Change ``bm'' to ''mathbf'' for upright boldface. \newcommand{\vv}[2]{\begin{bmatrix} #1 \\ #2 \end{bmatrix}} \newcommand{\vvv}[3]{\begin{bmatrix} #1 \\ #2 \\ #3 \end{bmatrix}} \newcommand{\vvvv}[4]{\begin{bmatrix} #1 \\ #2 \\ #3 \\ #4 \end{bmatrix}} \newcommand{\vvvvv}[5]{\begin{bmatrix} #1 \\ #2 \\ #3 \\ #4 \\ #5 \end{bmatrix}} \def\<{\langle} \def\>{\rangle} \newcommand{\ceil}[1] {\left\lceil #1 \right\rceil} 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\colorlet{lightred}{red!30!white} \colorlet{lightblue}{blue!30!white} \colorlet{lightyellow}{yellow!30!white} \colorlet{keylime}{green!10!white} \colorlet{darkgreen}{green!50!black} \colorlet{midgreen}{green!70!black} \colorlet{lightgreen}{darkgreen!30!white} \colorlet{lightpurple}{violet!60!white} \colorlet{darkgrey}{black!70} \colorlet{midgrey}{black!50} \definecolor{Red}{rgb}{.9,0,0} \definecolor{Blue}{rgb}{0,0,.9} \definecolor{Green}{HTML}{7EC636} \colorlet{lightgreen}{Green!30!white} \definecolor{Gold}{rgb}{.83,.67,.08} \definecolor{conj-purple}{HTML}{E4B7FF} \definecolor{conj-orange}{HTML}{FFD4B7} \colorlet{lred}{red!50!white} \definecolor{lblue}{rgb}{0.60, 0.75, 1} \colorlet{lgreen}{green!40!white} \colorlet{lyellow}{yellow!30!white} %\definecolor{lpurple}{rgb}{0.6 0.35 0.8} \definecolor{lpurple}{rgb}{0.7 0.45 0.9} \colorlet{lorange}{orange!50!white} \definecolor{lightgrey}{rgb}{.85, .85, .85} \definecolor{faded}{rgb}{.75,.75,.75} \definecolor{midgrey}{rgb}{.5,.5,.5} %\definecolor{midgrey}{rgb}{.35, .35, .35} %\definecolor{amethyst}{rgb}{0.6 0.4 0.8} \definecolor{amethyst}{rgb}{0.54 0.17 0.89} \definecolor{amber}{rgb}{1.0 0.49 0.0} \definecolor{turq}{rgb}{0.12 0.69 0.71} \colorlet{lightestgray}{gray!15!white} \colorlet{lightestblue}{blue!15!white} \colorlet{midblue}{blue!45!white} \colorlet{lightestred}{red!15!white} \colorlet{midred}{red!45!white} \colorlet{lightestgreen}{darkgreen!15!white} \colorlet{lightestpurple}{amethyst!20!white} % COLORS FOR MULTIPLICATION TABLES \colorlet{tred}{red!70!white} \colorlet{tblue}{blue!60!white} %(higher number is more faded) \colorlet{tgreen}{green!75!black} %(higher number is brighter) \colorlet{tyellow}{yellow!60!white} \colorlet{tlime}{green!80!white} \colorlet{tpink}{pink} \colorlet{tpurple}{lpurple} \colorlet{torange}{orange} \newcommand*{\n}{5} % counter for Cayley tables %% CAYLEY DIAGRAM GLOBAL DEFAULTS \tikzstyle{v} = [circle, draw, fill=lightgrey,inner sep=0pt, minimum size=6mm] \tikzstyle{r} = [draw, very thick, Red, -stealth] % Red ---> \tikzstyle{b} = [draw, very thick, Blue, -stealth] % Blue ---> \tikzstyle{g} = [draw, very thick, darkgreen, -stealth] % Green ---> \tikzstyle{p} = [draw, very thick, amethyst, -stealth] % Purple ---> \tikzstyle{rr} = [draw, very thick, Red] % Red ---- \tikzstyle{bb} = [draw, very thick, Blue] % Blue ---- \tikzstyle{gg} = [draw, very thick, darkgreen] % Green ---- \tikzstyle{oo} = [draw, very thick, orange] % Purple ---- \tikzstyle{pp} = [draw, very thick, amethyst] % Purple ---- \tikzstyle{tt} = [draw, very thick, turq] % Turquoise ---- \tikzstyle{yy} = [draw, very thick, Gold] % Gold ---- \tikzstyle{R} = [draw, very thick, Red, -stealth'] % Red ---> \tikzstyle{B} = [draw, very thick, Blue, -stealth'] % Blue ---> \tikzstyle{G} = [draw, very thick, darkgreen, -stealth'] % Green ---> \tikzset{r/.style={draw, very thick, Red, -stealth}} % Red ---> \tikzset{R/.style={draw, very thick, Red, -stealth'}} % Red ---> \tikzset{v/.style={circle, draw, fill=lightgrey,inner sep=0pt, minimum size=6mm}} \renewcommand\labelitemi{\rule[0.12em]{0.4em}{0.4em}} \renewcommand\labelitemii{\normalfont\bfseries \rule[0.15em]{0.3em}{0.3em}} \renewcommand\labelitemiii{\textasteriskcentered} \renewcommand\labelitemiv{\textperiodcentered} \renewcommand{\footrulewidth}{1pt} \newenvironment{proof}[1][Proof.]{\smallskip\begin{trivlist} \item[\hskip \labelsep {\sffamily #1}]}{\qed\end{trivlist}\bigskip} \newenvironment{sol}[1][Solution.]{\smallskip\begin{trivlist} \item[\hskip \labelsep {\sffamily #1}]}{\qed\end{trivlist}\bigskip} \newenvironment{fminipage}% {\setlength{\fboxsep}{15pt}\begin{Sbox}\begin{minipage}}% {\end{minipage}\end{Sbox}\fbox{\TheSbox}} \newcommand{\qed}{\hfill \ensuremath{\Box}} \newcommand{\disp}{\displaystyle} \pagestyle{fancy} \lhead{{\sf Homework 10 $|$ Due April 21 (Friday) }} \rhead{\thepage} \cfoot{{\sf Math 4130 $|$ Visual Algebra $|$ Spring 2023 $|$ M.~Macauley}} \begin{document} $\;$ \vspace{-6mm} \begin{enumerate} \item The Galois group of $x^n-1$ naturally acts on the $n^{\rm th}$ roots of unity; this is shown below for $n=3,\dots,8$. The primitive roots are highlighted. \[ \hspace*{-5mm} \tikzstyle{v} = [circle, draw, fill=lightgrey,inner sep=0pt, minimum size=4.5mm] \tikzstyle{v-f} = [circle, draw, grey, fill=lightgrey,inner sep=0pt, minimum size=4.5mm] \tikzstyle{b} = [draw, ultra thick,Blue,-stealth] \tikzstyle{bb} = [draw, ultra thick,Blue] \tikzstyle{b-f} = [draw, ultra thick, lightblue,-stealth] \tikzstyle{bb-f} = [draw, ultra thick, lightblue] \tikzstyle{gg} = [draw, ultra thick, Green] \tikzstyle{gg-f} = [draw, ultra thick, lightgreen] \tikzstyle{g-f} = [draw, ultra thick, lightgreen] \tikzstyle{every node}=[font=\scriptsize] \tikzset{every loop/.style={min distance=5mm,looseness=10}} \begin{tikzpicture}[scale=1.5] %% \begin{scope}[shift={(0,0)}] \draw [faded] (0,-1.3) to (0,1.3); \draw [faded] (-1.55,0) to (1.65,0); \draw[faded] (0,0) circle [radius=1cm]; \node (1) at (0:1) [v-f] {1ドル$}; \node (z) at (120:1) [v] {$\zeta$}; \node (z2) at (240:1) [v] {$\zeta^2$}; %% \path (1) edge [b-f,loop right,-stealth] (1); \draw[bb] (z) to (z2); \end{scope} %% \begin{scope}[shift={(3.5,0)}] \draw [faded] (0,-1.3) to (0,1.3); \draw [faded] (-1.65,0) to (1.65,0); \draw[faded] (0,0) circle [radius=1cm]; \node (1) at (0:1) [v-f] {1ドル$}; \node (z) at (90:1) [v] {$\zeta$}; \node (z2) at (180:1) [v-f] {$\zeta^2$}; \node (z3) at (270:1) [v] {$\zeta^3$}; %% \path (1) edge [b-f,loop right,-stealth] (1); \path (z2) edge [b-f,loop left,-stealth] (z2); \draw[bb] (z) to (z3); \end{scope} %% \begin{scope}[shift={(7,0)}] \draw [faded] (0,-1.3) to (0,1.3); \draw [faded] (-1.55,0) to (1.6,0); \draw[faded] (0,0) circle [radius=1cm]; \node (1) at (0:1) [v-f] {\small 1ドル$}; \node (z) at (72:1) [v] {$\zeta$}; \node (z2) at (144:1) [v] {$\zeta^2$}; \node (z3) at (216:1) [v] {$\zeta^3$}; \node (z4) at (288:1) [v] {$\zeta^4$}; %% \path (1) edge [b-f,loop right,-stealth] (1); \draw[b] (z) to (z3); \draw[b] (z3) to (z4); \draw[b] (z4) to (z2); \draw[b] (z2) to (z); \end{scope} %% \begin{scope}[shift={(0,-3)}] \draw [faded] (0,-1.3) to (0,1.3); \draw [faded] (-1.65,0) to (1.65,0); \draw[faded] (0,0) circle [radius=1cm]; \node (1) at (0:1) [v-f] {\small 1ドル$}; \node (z) at (60:1) [v] {$\zeta$}; \node (z2) at (120:1) [v-f] {$\zeta^2$}; \node (z3) at (180:1) [v-f] {$\zeta^3$}; \node (z4) at (240:1) [v-f] {$\zeta^4$}; \node (z5) at (300:1) [v] {$\zeta^5$}; %% \path (1) edge [b-f,loop right,-stealth] (1); \path (z3) edge [b-f,loop left,-stealth] (z3); \draw[bb] (z) to (z5); \draw[bb-f] (z2) to (z4); \end{scope} %% \begin{scope}[shift={(3.5,-3)}] \draw [faded,very thin] (0,-1.3) to (0,1.3); \draw [faded,very thin] (-1.55,0) to (1.65,0); \draw[faded,very thin] (0,0) circle [radius=1cm]; \node (1) at (0:1) [v-f] {\small 1ドル$}; \node (z) at (51.4:1) [v] {$\zeta$}; \node (z2) at (102.8:1) [v] {$\zeta^2$}; \node (z3) at (154.3:1) [v] {$\zeta^3$}; \node (z4) at (205.7:1) [v] {$\zeta^4$}; \node (z5) at (257.1:1) [v] {$\zeta^5$}; \node (z6) at (308.6:1) [v] {$\zeta^6$}; %% \path (1) edge [b-f,loop right,-stealth] (1); \draw[b,ultra thick] (z) to (z3); \draw[b,ultra thick] (z3) to (z2); \draw[b,ultra thick] (z2) to (z6); \draw[b,ultra thick] (z6) to (z4); \draw[b,ultra thick] (z4) to (z5); \draw[b,ultra thick] (z5) to (z); \end{scope} %% \begin{scope}[shift={(7,-3)}] \draw [faded,very thin] (0,-1.3) to (0,1.3); \draw [faded,very thin] (-1.65,0) to (1.65,0); \draw[faded,very thin] (0,0) circle [radius=1cm]; \node (1) at (0:1) [v-f] {\small 1ドル$}; \node (z) at (45:1) [v] {$\zeta$}; \node (z2) at (90:1) [v-f] {$\zeta^2$}; \node (z3) at (135:1) [v] {$\zeta^3$}; \node (z4) at (180:1) [v-f] {$\zeta^4$}; \node (z5) at (225:1) [v] {$\zeta^5$}; \node (z6) at (270:1) [v-f] {$\zeta^6$}; \node (z7) at (313:1) [v] {$\zeta^7$}; %% \path (1) edge [b-f,loop below,-stealth] (1); \path (z4) edge [b-f,loop below,-stealth] (z4); \path (z2) edge [g-f,loop above,-stealth] (z2); \path (z6) edge [g-f,loop below,-stealth] (z6); \path (1) edge [g-f,loop above,-stealth] (1); \path (z4) edge [g-f,loop above,-stealth] (z4); %% \draw[bb,ultra thick] (z) to (z7); \draw[bb-f] (z2) to (z6); \draw[bb,ultra thick] (z3) to (z5); \draw[gg,ultra thick] (z) to (z5); \draw[gg,ultra thick] (z3) to (z7); \end{scope} \end{tikzpicture} \] \vspace{-9mm} \begin{enumerate} \item Construct analogous diagrams for $n=9,ドル 10ドル,ドル and 16ドル$. \item For each of these three, write the splitting field of $x^n-1$ with elements that do not involve $\zeta$. If possible, write the generating automorphism(s) of $\Gal(x^n-1)$ in terms of them. For example, $\Q(\zeta_8)=\Q(\sqrt{2},i),ドル and $\Gal(x^8-1)=\<\sigma,\tau\>\cong \Z_8^\times\cong V_4,ドル where \[ \left\{\begin{array}{ll} \sigma\colon\sqrt{2}\longmapsto\sqrt{2} \\ \sigma\colon\;\;\;i\longmapsto -i \end{array}\right.\qquad\qquad %% \left\{\begin{array}{ll} \tau\colon\sqrt{2}\longmapsto-\sqrt{2} \\ \tau\colon\;\;\;i\longmapsto i \end{array}\right. \] \vspace{-3.5mm} \item Draw the subgroup lattice of $\Gal(x^n-1)$ and the subfield lattice of $\Q(\zeta)$. \end{enumerate} \item Consider the following polynomial that is irreducible over $\Q$: \[ f(x)=x^4-x^2-5=\Big(x^2-\sqrt{\tfrac{1}{2}+\tfrac{\sqrt{21}}{2}}\Big) \Big(x^2-\sqrt{\tfrac{1}{2}-\tfrac{\sqrt{21}}{2}}\Big). \] Denote its roots by $r_1,r_2,r_3,r_4,ドル its splitting field by $K=\Q(r_1,r_2,r_3,r_4),ドル and its Galois group over $\Q$ by $G=\Gal(f(x))$. \begin{enumerate} \item Since $f(x)$ is irreducible, $[\Q(r_i):\Q]=4$. Use this to find a lower bound on $[K:\Q]$. \item By the tower law, $[K:\Q]=[\Q(r_1,r_2,r_3,r_4):\Q]$ is equal to \[ [\Q(r_1,r_2,r_3,r_4):\Q(r_1,r_2,r_3)]\cdot [\Q(r_1,r_2,r_3):\Q(r_1,r_2)]\cdot [\Q(r_1,r_2):\Q(r_1)]\cdot [\Q(r_1):\Q]. \] Use this to find an upper bound on $[K:\Q]$. \item Find $[K:\Q]=|\Gal(f(x)|,ドル and then the Galois group by process of elimination. \end{enumerate} \item Suppose $\alpha\neq 0$ is algebraic over $\Q$ and $[\Q(\alpha):\Q]$ is odd. \begin{enumerate} \item Show that $\Q(\alpha)=\Q(\alpha+\alpha^{-1})$. \item Give an example to show how this can fail if $[\Q(\alpha):\Q]$ is even. \item Repeat the previous two parts, but for the subfield $\Q(\alpha^2-\alpha)$. \end{enumerate} \end{enumerate} \end{document}