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+% Created 2023-03-27 Mon 22:00
+% Intended LaTeX compiler: pdflatex
+\documentclass[11pt]{article}
+\usepackage[utf8]{inputenc}
+\usepackage[T1]{fontenc}
+\usepackage{graphicx}
+\usepackage{longtable}
+\usepackage{wrapfig}
+\usepackage{rotating}
+\usepackage[normalem]{ulem}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{capt-of}
+\usepackage{hyperref}
+\notindent \notag \usepackage{ dsfont } \usepackage{amsmath} \usepackage[a4paper,margin=1in,landscape]{geometry} \usepackage{polynom} \usepackage{wasysym}
+\author{Lizzy Hunt}
+\date{\today}
+\title{Assignment Nine}
+\hypersetup{
+ pdfauthor={Lizzy Hunt},
+ pdftitle={Assignment Nine},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.6.1)}, 
+ pdflang={English}}
+\begin{document}
+
+\maketitle
+\setlength\parindent{0pt}
+
+\section{Section 5.1}
+\label{sec:orga03a7b4}
+\subsection{Question One}
+\label{sec:org41cc8bf}
+\subsubsection{b}
+\label{sec:org39f646c}
+Yes. In \(F\), \(f(x) - g(x) = -x^3 + x = x^3 + x\).
+
+\begin{equation*}
+\polylongdiv[style=A]{x^3 + x}{x^2+1}
+\end{equation*}
+
+\subsubsection{c}
+\label{sec:orge019c41}
+No. \(f(x) - g(x) = x^5 - 2x^4 + 4x^3 - 8x^2 + 3x - 2\)
+
+\begin{equation*}
+\polylongdiv[style=A]{x^5 - 2x^4 + 4x^3 - 8x^2 + 3x - 2}{x^3 - x^2 + x - 1}
+\end{equation*}
+
+\subsection{Question Three}
+\label{sec:org1718036}
+\(|\)\{ \(0, 1, x, x + 1, x^2, x^2 + 1, x^2 + x, x^2 + x + 1\) \}\(|\) = 8
+
+\subsection{Question Four}
+\label{sec:orgd724559}
+For every \(a,b,c \in \mathds{Z}_3\) we can generate a polynomial \(ax^2 + bx + c\), by part two of Corollary 5.5. \(3^3 = 27\)
+
+\subsection{Question Six}
+\label{sec:orgb081ff5}
+By Corollary 5.5, all the congruence classes in \(F[x]\) are \(c \ni c \in F\).
+
+\subsection{Question Eight}
+\label{sec:org4c8f836}
+\begin{align*}
+f(x)k(x) &\equiv_{p(x)} g(x)k(x) \\
+& \Rightarrow p(x) | f(x)k(x) - g(x)k(x) \\
+& \Rightarrow p(x) | (f(x) - g(x))(k(x))
+\end{align*}
+
+By Theorem 4.10, since \(p(x)\) is relatively prime to \(k(x)\), \(p(x) | f(x) - g(x) \Rightarrow f(x) \equiv_{p(x)} g(x)\)
+
+\subsection{Question Eleven}
+\label{sec:org98dcd2c}
+Since \(p(x)\) is reducible, it can be rewritten as \(p(x) = f(x)g(x)\) with \(f(x), g(x) \in F[x]\) with each \(f(x)\) and \(g(x)\) having a degree greater than 0, summing to the
+degree of \(p(x)\).
+
+Then, it is impossible for \(p(x)\) to divide \(f(x)\) or \(g(x)\) since \(p(x)\) has a higher degree. So, neither \(f(x)\) or \(g(x)\) can \$ \(\equiv\)\textsubscript{p(x)} 0\textsubscript{F}\$.
+
+Still, \(f(x)g(x) \equiv_{}_{p(x)} p(x) \equiv_{p(x)} 0_F\) since \(p(x) | p(x)\).
+
+\section{Section 5.2}
+\label{sec:org9a01661}
+\subsection{Question One}
+\label{sec:org8ab076f}
+The congruence classes are those in Section 5.1, Question Three as above.
+
+\begin{center}
+\begin{tabular}{lllllllll}
++ & [0] & [1] & [x] & [x + 1] & [x\textsuperscript{2}] & [x\textsuperscript{2} + 1] & [x\textsuperscript{2} + x] & [x\textsuperscript{2} + x + 1]\\[0pt]
+[0] & [0] & [1] & [x] & [x + 1] & [x\textsuperscript{2}] & [x\textsuperscript{2} + 1] & [x\textsuperscript{2} + x] & [x\textsuperscript{2} + x + 1]\\[0pt]
+[1] & [1] & [0] & [x+1] & [x] & [x\textsuperscript{2} + 1] & [x\textsuperscript{2}] & [x\textsuperscript{2} + x + 1] & [x\textsuperscript{2} + x]\\[0pt]
+[x] & [x] & [x + 1] & [0] & [1] & [x\textsuperscript{2} + x] & [x\textsuperscript{2} + x + 1] & [x\textsuperscript{2}] & [x\textsuperscript{2} + 1]\\[0pt]
+[x + 1] & [x + 1] & [x] & [1] & [0] & [x\textsuperscript{2} + x + 1] & [x\textsuperscript{2} + x] & [x\textsuperscript{2} + 1] & [x\textsuperscript{2}]\\[0pt]
+[x\textsuperscript{2}] & [x\textsuperscript{2}] & [x\textsuperscript{2} + 1] & [x\textsuperscript{2} + x] & [x\textsuperscript{2} + x + 1] & [0] & [1] & [x] & [x + 1]\\[0pt]
+[x\textsuperscript{2} + 1] & [x\textsuperscript{2} + 1] & [x\textsuperscript{2}] & [x\textsuperscript{2} + x + 1] & [x\textsuperscript{2} + x] & [1] & [0] & [x + 1] & [x]\\[0pt]
+[x\textsuperscript{2} + x] & [x\textsuperscript{2} + x] & [x\textsuperscript{2} + x + 1] & [x\textsuperscript{2}] & [x\textsuperscript{2} + 1] & [x] & [x+1] & [0] & [1]\\[0pt]
+[x\textsuperscript{2} + x + 1] & [x\textsuperscript{2} + x + 1] & [x\textsuperscript{2} + x] & [x\textsuperscript{2} + 1] & [x\textsuperscript{2}] & [x+1] & [x] & [1] & [0]\\[0pt]
+\end{tabular}
+\end{center}
+
+\begin{center}
+\begin{tabular}{lllllllll}
+\(\cdot\) & [0] & [1] & [x] & [x + 1] & [x\textsuperscript{2}] & [x\textsuperscript{2} + 1] & [x\textsuperscript{2} + x] & [x\textsuperscript{2} + x + 1]\\[0pt]
+[0] & [0] & [0] & [0] & [0] & [0] & [0] & [0] & [0]\\[0pt]
+[1] & [0] & [1] & [x] & [x + 1] & [x\textsuperscript{2}] & [x\textsuperscript{2} + 1] & [x\textsuperscript{2} + x] & [x\textsuperscript{2} + x + 1]\\[0pt]
+[x] & [0] & [x] & [x\textsuperscript{2}] & [x\textsuperscript{2}+x] & [x+1] & [1] & [x\textsuperscript{2}+x+1] & [x\textsuperscript{2}+1]\\[0pt]
+[x + 1] & [0] & [x + 1] & [x\textsuperscript{2} + x] & [x\textsuperscript{2}+1] & [x\textsuperscript{2}+x+1] & [x\textsuperscript{2}] & [1] & [x]\\[0pt]
+[x\textsuperscript{2}] & [0] & [x\textsuperscript{2}] & [x+1] & [x\textsuperscript{2}+x+1] & [x\textsuperscript{2}+x] & [x] & [x\textsuperscript{2}+1] & [1]\\[0pt]
+[x\textsuperscript{2} + 1] & [0] & [x\textsuperscript{2} + 1] & [1] & [x\textsuperscript{2}] & [x] & [x\textsuperscript{2}+x+1] & [x+1] & [x\textsuperscript{2}+x]\\[0pt]
+[x\textsuperscript{2} + x] & [0] & [x\textsuperscript{2} + x] & [x\textsuperscript{2}+x+1] & [1] & [x\textsuperscript{2}+1] & [x+1] & [x] & [x+1]\\[0pt]
+[x\textsuperscript{2} + x + 1] & [0] & [x\textsuperscript{2} + x + 1] & [x\textsuperscript{2}+1] & [x] & [1] & [x\textsuperscript{2}+x] & [x\textsuperscript{2}] & [x+1]\\[0pt]
+\end{tabular}
+\end{center}
+
+Yes, this is a field since by Theorem 5.7, it is a commutative ring with identity, and each
+non-zero row in the multiplication table contains the multiplicative identity (each is a unit).
+
+\subsection{Question Two}
+\label{sec:org9db49ef}
+\begin{center}
+\begin{tabular}{llllllllll}
++ & [0] & [1] & [2] & [x] & [x+1] & [x+2] & [2x] & [2x+1] & [2x+2]\\[0pt]
+[0] & [0] & [1] & [2] & [x] & [x+1] & [x+2] & [2x] & [2x+1] & [2x+2]\\[0pt]
+[1] & [1] & [2] & [0] & [x+1] & [x+2] & [x] & [2x+1] & [2x+2] & [2x]\\[0pt]
+[2] & [2] & [0] & [1] & [x+2] & [x] & [x+1] & [2x+2] & [2x] & [2x+1]\\[0pt]
+[x] & [x] & [x+1] & [x+2] & [2x] & [2x+1] & [2x+2] & [0] & [1] & [2]\\[0pt]
+[x+1] & [x+1] & [x+2] & [x] & [2x+1] & [2x+2] & [2x] & [1] & [2] & [0]\\[0pt]
+[x+2] & [x+2] & [x] & [x+1] & [2x+2] & [2x] & [2x+1] & [2] & [0] & [1]\\[0pt]
+[2x] & [2x] & [2x+1] & [2x+2] & [0] & [1] & [2] & [x] & [x+1] & [1]\\[0pt]
+[2x+1] & [2x+1] & [2x+2] & [2x] & [1] & [2] & [0] & [x+1] & [x+2] & [x]\\[0pt]
+[2x+2] & [2x+2] & [2x] & [2x+1] & [2] & [0] & [1] & [x+2] & [x] & [x+1]\\[0pt]
+\end{tabular}
+\end{center}
+
+
+\begin{center}
+\begin{tabular}{llllllllll}
+\(\cdot\) & [0] & [1] & [2] & [x] & [x+1] & [x+2] & [2x] & [2x+1] & [2x+2]\\[0pt]
+[0] & [0] & [0] & [0] & [0] & [0] & [0] & [0] & [0] & [0]\\[0pt]
+[1] & [0] & [1] & [2] & [x] & [x+1] & [x+2] & [2x] & [2x+1] & [2x+2]\\[0pt]
+[2] & [0] & [2] & [1] & [2x] & [2x+2] & [2x+1] & [x] & [x+2] & [x+1]\\[0pt]
+[x] & [0] & [x] & [2x] & [2] & [x+2] & [2x+2] & [1] & [x+1] & [2x+1]\\[0pt]
+[x+1] & [0] & [x+1] & [2x+2] & [x+2] & [2x] & [1] & [2x+1] & [2] & [x]\\[0pt]
+[x+2] & [0] & [x+2] & [2x+1] & [2x+2] & [1] & [x] & [x+1] & [2x] & [2]\\[0pt]
+[2x] & [0] & [2x] & [x] & [1] & [2x+1] & [x+1] & [2] & [2x+2] & [x+2]\\[0pt]
+[2x+1] & [0] & [2x+1] & [x+2] & [x+1] & [2] & [2x] & [2x+2] & [x] & [1]\\[0pt]
+[2x+2] & [0] & [2x+2] & [x+1] & [2x+1] & [x] & [2] & [x+2] & [1] & [2x]\\[0pt]
+\end{tabular}
+\end{center}
+
+Yes, this is a field since by Theorem 5.7, it is a commutative ring with identity, and each
+non-zero row in the multiplication table contains the multiplicative identity (each is a unit).
+
+\subsection{Question Three}
+\label{sec:orgc80a66e}
+
+\begin{center}
+\begin{tabular}{lllll}
++ & [0] & [1] & [x] & [x+1]\\[0pt]
+[0] & [0] & [1] & [x] & [x+1]\\[0pt]
+[1] & [1] & [0] & [x+1] & [x]\\[0pt]
+[x] & [x] & [x+1] & [0] & [1]\\[0pt]
+[x+1] & [x+1] & [x] & [1] & [0]\\[0pt]
+\end{tabular}
+\end{center}
+
+\begin{center}
+\begin{tabular}{lllll}
+\(\cdot\) & [0] & [1] & [x] & [x+1]\\[0pt]
+[0] & [0] & [0] & [0] & [0]\\[0pt]
+[1] & [0] & [1] & [x] & [x+1]\\[0pt]
+[x] & [0] & [x] & [1] & [x+1]\\[0pt]
+[x+1] & [0] & [x+1] & [x+1] & [0]\\[0pt]
+\end{tabular}
+\end{center}
+
+Not, this is \uline{not} a field since by Theorem 5.7, it is a commutative ring with identity, but
+not every non-zero row in the multiplication table contains the multiplicative identity (\(x+1\) is
+not a unit).
+
+\subsection{Question Six}
+\label{sec:orga040020}
+By Corollary 5.5, each congruence class can be rewritten with \(a,b \in \mathds{Q}\): \([ax + b]\).
+
+Addition is defined as \([ax + b] + [cx + d] = [(a + c)x + bd]\).
+
+\((ax + b)(cx + d) = acx^2 + dax + bcx + bd = acx^2 + (da + bc)x + bd\)
+
+\begin{equation*}
+\polylongdiv[style=A]{acx^2 + (da + bc)x + bd}{x^2 - 2}
+\end{equation*}
+
+Multiplication is thusly defined as \([ax + b] \cdot [cx + d] = [(ad + bc)x + (2ac + bd)]\)
+
+\subsection{Question Nine}
+\label{sec:org3bd28c4}
+Given that \([a + bx]\) is a nonzero congruence class, either
+\(a > 0\) or \(b > 0\). Then let \(c = \frac{-a}{a^2 + b^2}\) and \(d = \frac{b}{a^2 + b^2}\).
+
+\begin{equation*}
+\polylongdiv[style=A]{acx^2 + (da + bc)x + bd}{x^2 + 1}
+\end{equation*}
+
+\begin{align*}
+[ax + b][cx + d] & = [(ad + bc)x + (bd - ac)] \\
+&= [(\frac{ab}{a^2 + b^2} + \frac{-ab}{a^2 + b^2})x + \frac{b^2}{a^2 + b^2} - \frac{-a^2}{a^2 + b^2}] \\
+&= [0x + \frac{b^2 + a^2}{a^2 + b^2}] \\
+&= [1]
+\end{align*}
+
+\section{Section 5.3}
+\label{sec:org6b8ca6a}
+\subsection{Question One}
+\label{sec:org09f321d}
+\subsubsection{a}
+\label{sec:org00adeb0}
+\(x^3 + 2x^2 + x + 1\) does not have any roots in \mathds{Z}\textsubscript{3}, so by Corollary 4.19 it must be irreducible,
+and thus a field by 5.10
+
+\subsubsection{b}
+\label{sec:orge4a6bab}
+This is not a field by Theorem 5.10 since 2 is a root in \(Z_5\), so by Corollary 4.19 it must be reducible.
+
+\subsubsection{c}
+\label{sec:org1cc5f89}
+This is not a field by Theorem 5.10 since \(x^4 + x^2 + 1 = (x^2 - x + 1)(x^2 + x + 1) \equiv_2 (x^2 + x + 1)^2\) shows \(x^4 + x^2 + 1\) is reducible.
+
+\subsection{Question Two}
+\label{sec:org1cd3926}
+\subsubsection{a}
+\label{sec:org265177d}
+Since \(\mathds{Q} (\sqrt{2})\) is a subset of \(\mathds{R}\), multiplication and addition are associative, commutative, and distributive.
+
+The additive identity of \$\mathds{Q} (\sqrt{2}) is \(0 + 0\sqrt{2}\) and the multiplicative identity is \(1 + 0\sqrt{2}\).
+
+It must be a field since every non-zero element \(a + b \sqrt{2}\) is a unit:
+
+\begin{align*}
+(a + b\sqrt{2})x = 1 & \Rightarrow x = \frac{1}{a + b\sqrt{2}} \\
+& \Rightarrow x = \frac{a - b\sqrt{2}}{(a + b\sqrt{2})(a - b \sqrt{2})} \\
+& \Rightarrow x = \frac{a}{a^2 - 2b^2} - \frac{b}{a^2 - 2b^2}\sqrt{2} \\
+\end{align*}
+
+\subsubsection{b}
+\label{sec:orgef6853e}
+Every element in \(\mathds{Q} / (x^2 - 2)\) can be rewritten as a member of the congruence class \([ax + b]\) with \(a, b \in \mathds{Q}\) by Corollary 5.5.
+
+Then, we can define a function \(f\) such that \(f([ax + b]) = a + b\sqrt{2}\) so that \(f(x) \in \mathds{Q}(\sqrt{2})\).
+
+\(f\) is thus an isomorphism since (a chance at redemption from my midterm \(\ddot\smile\)):
+
+\begin{itemize}
+\item \(f\) is injective since \(f([ax + b]) = f([cx + d]) \Rightarrow a + b\sqrt{2} = c + d\sqrt{2} \Rightarrow a = c \wedge b = d\)
+\item \(f\) is surjective since each \(a + b\sqrt{2}\) is uniquely mapped to \([ax + b]\)
+\item \(f([ax + b] + [cx + d]) = f([(a + c)x + (b + d)]) = (a + c) + (b + d)\sqrt{2} = (a + b\sqrt{2}) + (c + d\sqrt{2}) = f([ax + b]) + f([cx + d])\)
+
+\item And via Question Six from section 5.2 above,
+\(f([ax + b] \cdot [cx + d]) = f([(ad + bc)x + (2ac + bd)]) = (ad + bc) + (2ac + bd)\sqrt{2} = (a + b\sqrt{2})(c + d\sqrt{2}) = f([ax + b])f([cx + d])\)
+\end{itemize}
+
+\subsection{Question Five}
+\label{sec:orgc53207a}
+\subsubsection{a}
+\label{sec:orgf8f7862}
+Since \(\mathds{Q} (\sqrt{3})\) is a subset of \(\mathds{R}\), multiplication and addition are associative, commutative, and distributive.
+
+The additive identity of \$\mathds{Q} (\sqrt{3}) is \(0 + 0\sqrt{3}\) and the multiplicative identity is \(1 + 0\sqrt{3}\).
+
+It must be a field since every non-zero element \(r + s \sqrt{3}\) is a unit (by first assuming that the inverse of \(r + s\sqrt{3}\) from the
+back of the book, is \(\frac{r}{t} - \frac{s}{t}\sqrt{3}\) with \(t=r^2 - 3s^2\)):
+
+\begin{align*}
+1 &= (r + s\sqrt{3})(\frac{r}{t} - \frac{s}{t}\sqrt{3}) \\
+  &= \frac{r^2}{t} - \frac{sr}{t}\sqrt{3} + \frac{sr}{t}\sqrt{3} - \frac{3s^2}{t} \\
+  &= \frac{r^2 - 3s^2}{t} \\
+  &= 1
+\end{align*}
+
+\subsubsection{b}
+\label{sec:org1d5d4ca}
+\begin{enumerate}
+\item Quick Lemma
+\label{sec:org6e61eec}
+In \(\mathds{Q}{x^2 - 3}\), by Corollary 5.5, each congruence class can be rewritten with \(a,b \in \mathds{Q}\): \([ax + b]\).
+
+\((ax + b)(cx + d) = acx^2 + dax + bcx + bd = acx^2 + (da + bc)x + bd\)
+
+\begin{equation*}
+\polylongdiv[style=A]{acx^2 + (da + bc)x + bd}{x^2 - 3}
+\end{equation*}
+
+Multiplication is thusly defined as \([ax + b] \cdot [cx + d] = [(ad + bc)x + (3ac + bd)]\)
+
+\item Yeah, it's an isomorphism
+\label{sec:org454b3a6}
+
+Every element in \(\mathds{Q} / (x^2 - 3)\) can be rewritten as a member of the congruence class \([ax + b]\) with \(a, b \in \mathds{Q}\) by Corollary 5.5.
+
+Then, we can define a function \(f\) such that \(f([rx + s]) = r + s\sqrt{3}\) so that \(f(x) \in \mathds{Q}(\sqrt{3})\).
+
+\(f\) is thus an isomorphism since (a chance of redemption from my midterm \(\ddot\smile\)):
+
+\begin{itemize}
+\item \(f\) is injective since \(f([ax + b]) = f([cx + d]) \Rightarrow a + b\sqrt{3} = c + d\sqrt{3} \Rightarrow a = c \wedge b = d\)
+\item \(f\) is surjective since each \(a + b\sqrt{3}\) is uniquely mapped to \([ax + b]\)
+\item \(f([ax + b] + [cx + d]) = f([(a + c)x + (b + d)]) = (a + c) + (b + d)\sqrt{3} = (a + b\sqrt{3}) + (c + d\sqrt{3}) = f([ax + b]) + f([cx + d])\)
+\item And from the lemma, \(f([ax + b] \cdot [cx + d]) = f([(ad + bc)x + (3ac + bd)]) = (ad + bc) + (3ac + bd)\sqrt{3} = (a + b\sqrt{3})(c + d\sqrt{3}) = f([ax + b])f([cx + d])\)
+\end{itemize}
+\end{enumerate}
+\end{document}
\ No newline at end of file