%%This is a standard LaTeX2e article document template. personal version 12/5/200%% \documentclass[11pt,twoside]{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%packages%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \pagestyle{empty} \usepackage{latexsym} \usepackage{amssymb} \usepackage{amsfonts} \usepackage{amsthm} \usepackage{amstext} \usepackage{multicol} \usepackage[all]{xy} \usepackage{amsxtra} \usepackage[pdftex, breaklinks, linktocpage=true, bookmarksopen=true,bookmarksopenlevel=0,bookmarksnumbered=true]{hyperref} \hypersetup{ pdftitle = {Math 371 Assignment 1}, pdfauthor = {\textcopyright\ Bryden Cais}, pdfcreator = {\LaTeX\ with package \flqq hyperref\frqq}, colorlinks = {true} } %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%formatting%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \setlength{\topmargin}{0in} %%% This sets all the spacing stuff to use the page more \setlength{\oddsidemargin}{0in} %%% efficiently than the normal "article" setup would. \setlength{\evensidemargin}{0in} %%% It's OK to play with these some. \setlength{\textheight}{8.5in} %%% \setlength{\textwidth}{6.5in} %%% \setlength{\headsep}{0in} %%% \setlength{\headheight}{0in} %%% %\setlength{\footskip}{0in} %%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \DeclareMathOperator{\ann}{ann} \DeclareMathOperator{\Gal}{Gal} \DeclareMathOperator{\Tor}{Tor} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\End}{End} \newcommand{\Q}{\mathbf{Q}} \newcommand{\R}{\mathbf{R}} \newcommand{\Z}{\mathbf{Z}} \newcommand{\C}{\mathbf{C}} \newcommand{\F}{\mathbf{F}} \newcommand{\p}{\mathfrak{p}} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}{Corollary} \begin{document} \begin{center} {\bf \Large \underline{Honors Algebra 4, MATH 371 Winter 2010}}\\ {\large Assignment 6}\\Due Wednesday, March 24 at 08:35 \end{center} \begin{enumerate} \item Let $K/F$ be a degree $2$ extension of fields. \begin{enumerate} \item If the characteristic of $F$ is not 2, prove that $K=F(a)$ for some $a\in K\setminus F$ with $a^2\in F$.\label{one} \item Give a counterexample to (\ref{one}) if $F$ has characteristic 2. \item Fix $F$ of characteristic not 2 and let $K_1,K_2$ be quadratic extensions of $F$ with $K_1 = F(a_1)$ and $K_2=F(a_2)$ where $a_i^2=b_i\in F$. Prove that $K_1\simeq K_2$ as extensions of $F$ (i.e. that there exists an isomorphism of fields $K_1\simeq K_2$ restricting to the identity on $F$) if and only if $b_1/b_2\in (F^{\times})^2$ is a square.\label{three} Conclude that the isomorphism classes of quadratic extensions of $F$ are in bijection with the group $F^{\times}/(F^{\times})^2$. \item Using (\ref{three}), give a complete list (without repetition) of all isomorphism classes of quadratic extensions of $\Q$. \end{enumerate} \noindent{\bf Solution} : \begin{enumerate} \item Fix $b\in K\setminus F$. Then $\{1,b\}$ is an $F$-basis of $K$, so $b$ satisfies a degree 2 polynomial $b^2+ ub +v=0$ with $u,v\in F$. Since the characteristic of $F$ is not 2, $2\in F^{times}$ so $u/2$ makes sense and we have $(b+u/2)^2 =u^2/4-v$ by completing the square. Thus, $a:=b+u/2\in K\setminus F$ has $a^2\in F$ and clearly $K=F(a)$. \item The extension $\F_2[X]/(X^2+X+1)$ of $\F_2$ gives a counterexample, since $(a+bX)^2 = (a^2+b^2)+ b^2X$ lies in $F$ if and only if $b=0$. \item If $K_1\simeq K_2$ as extensions of $F$, then $b_1$ must be a square in $K_2$, say $b_1= (u+va_2)^2$. This gives $b_1=u^2+v^2b_2 + 2uva_2$ from which it follows (as $2\neq 0$ in $F$) that either $u$ or $v$ must be zero. The second case cannot occur as otherwise $b_1$ would be a square in $F$ and $K_1=F$. Thus ,$b_1=v^2b_2$ for some $v\in F$. Conversely, If $b_1=v^2b_2$ then $b_1=(va_2)^2$ is a square in $K_2$, and the map $F[X]\rightarrow K_2$ sending $X$ to $va_2$ is surjective and yields an isomorphism $F[X]/(X^2-b_1)\simeq K_2$. As the source of this isomorphism is isomorphic to $K_1$, we get $K_1\simeq K_2$ as extensions of $F$. \item These are parameterized by $\Q^{\times}/(\Q^{\times})^2 = \{2,3,5,6,7,10,11,13,14,15,\ldots\}$ (the positive square-free integers). \end{enumerate} \item For $a\in \F_p$, set $$f_a(x):=X^p-X-a\in \F_p[X].$$ \begin{enumerate} \item If $a=0$, show that $f_a(X)=\prod_{u\in \F_p} (X-u)$. \item Suppose that $a\neq 0$ and let $E_a$ be a splitting field of $f_a(X)$. If $r_1,r_2\in E_a$ are roots of $f_a$, prove that $r_1-r_2\in \F_p$. \item Show that $f_a(X)$ is irreducible for all $a\in \F_p^{\times}$. \item Prove that $f_b(X)$ splits completely over $E_a$ for each fixed $a\in \F_p^{\times}$ and all $b\in \F_p^{\times}$. Conclude that $E_a$ is independent of $a$. \end{enumerate} \noindent{Solution:} \begin{enumerate} \item Every $u\in \F_p$ satisfies $X^p-X=0$ and this gives $p$ roots of the degree $p$ polynomial $X^p-X$ in the Euclidean domain $F[X]$, so we get the claimed factorization. \item Observe that $(r_1-r_2)^p-(r_1-r_2) = (r_1^p-r_1)-(r_2^p-r_2) = 0$ so $r_1-r_2$ is a root of $X^p-X$ and hence an element of $\F_p$ by the first part. \item Certainly $f_a$ has no root in $\F_p$ for $a\in \F_p^{\times}$, by part 1. Over $E_a$, we have the factorization $$f_a=\prod_{0\le i