From 6bf4b90c90f15f4ab60833bddf5b5756d1a6b1f6 Mon Sep 17 00:00:00 2001
From: Elizabeth Alexander Hunt <me@liz.coffee>
Date: Thu, 2 Jul 2026 11:55:17 -0700
Subject: Init

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+% Created 2023-10-07 Sat 14:55
+% 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{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} \usepackage{polynom} \usepackage{wasysym}
+\author{Lizzy Hunt}
+\date{\today}
+\title{Midterm 2}
+\hypersetup{
+ pdfauthor={Lizzy Hunt},
+ pdftitle={Midterm 2},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.7-pre)}, 
+ pdflang={English}}
+\begin{document}
+
+\maketitle
+\setlength\parindent{0pt}
+
+\section{Question One}
+\label{sec:orgd83a31d}
+In \(\mathds{Z}_3[x]\), \(4x^3 + 4x + 4 \equiv x^3 + x + 1\)
+\begin{equation*}
+\polylongdiv[style=A]{x^3 + x + 1}{2x^2+1}
+\end{equation*}
+
+\(\frac{1}{2}x \equiv_3 2x\) and \(\frac{1}{2}x + 1 \equiv_3 (2x + 1)\)
+
+So, \(a(x) = b(x)q(x) + r(x) = (2x^2 + 1)(2x) + (2x + 1)\)
+
+\section{Question Two}
+\label{sec:org245f338}
+\begin{equation*}
+\polylongdiv[style=A]{x^4 + x^3 + x + 1}{x-2}
+\end{equation*}
+
+Shows that the remainder will be zero when we are in \(\mathds{Z}_3[x]\)
+
+\section{Question Three}
+\label{sec:org154bff9}
+We will use the Euclidean algorithm to find the GCD:
+\subsection{GCD}
+\label{sec:orga7765bb}
+\begin{align*}
+3x^3 + 2x^2 + 2x + 3 &= (3x - 1)(x^2 + x + 3) + (4x + 1) \\
+(x^2 + x + 3) &\equiv_5 (4x + 1)(4x + 3) + 45 \equiv_5 (4x + 1)(4x + 3) + 0 \\
+\end{align*}
+
+So, the GCD is (4x + 1)
+
+\subsection{(supplement) Division Algorithm Work}
+\label{sec:orgdafc72b}
+\begin{equation*}
+\polylongdiv[style=A]{3x^3 + 2x^2 + 2x + 3}{x^2 + x + 3}
+\end{equation*}
+
+\begin{equation*}
+\polylongdiv[style=A]{x^2 + x + 3}{4x + 1}
+\end{equation*}
+
+Check the GCD:
+
+\begin{equation*}
+\polylongdiv[style=A]{3x^3 + 2x^2 + 2x + 3}{4x + 1}
+\end{equation*}
+
+\(\frac{3}{4}x^2 + \frac{5}{16}x + \frac{27}{64} \equiv_5 2x^2 + 3\)
+and \(\frac{165}{64} \equiv_5 0\)
+
+\begin{equation*}
+\polylongdiv[style=A]{x^2 + x + 3}{4x + 1}
+\end{equation*}
+
+\(\frac{1}{4}x + \frac{3}{16} \equiv_5 4x + 3\)
+and \(\frac{45}{16} \equiv_5 0\).
+
+\section{Question Four}
+\label{sec:org9492a2d}
+\subsection{a}
+\label{sec:orgc2e7831}
+\(\mathds{Z}_3[x]\) is a field by Theorem 2.8.
+
+There are no roots of \(2x^3 + x + 1\) in \(\mathds{Z}_3\), so by Theorem 5.10 \(\mathds{Z}_3[x] / (2x^3 + x + 1)\) is a field, as it is
+irreducible by Corollary 4.19.
+
+\subsection{b}
+\label{sec:org68f838e}
+Since we've proven \(\mathds{Z}_3[x] / (2x^3 + x + 1)\) is a field, we know that \([2x + 1]\) must be a unit.
+
+By similar reasoning to the proof of Theorem 5.10 (1), \(gcd(2x + 1, 2x^3 + x + 1) = 1\)
+
+Then,
+
+\begin{equation*}
+\polylongdiv[style=A]{2x^3 + x + 1}{2x + 1}
+\end{equation*}
+
+Shows that:
+\begin{align*}
+2x^3 + x + 1 &= (2x + 1)(x^2 + x) + 1 \\
+1 &= (2x^3 + x + 1) + (2x+1)(-x^2 - x) \\
+1 - (2x^3 + x + 1) &= (2x + 1)(-x^2 - x) \\
+1 &\equiv_{2x^3 + x + 1} (2x+1)(-x^2 - x)
+\end{align*}
+and thus, by similar logic again found in the proof mentioned before, \([-x^2 - x] = [2x^2 + 2x]\) must be the inverse.
+
+As a sanity check,
+
+\begin{equation*}
+\polylongdiv[style=A]{2x^3 + x + 1}{2x^2 + 2x}
+\end{equation*}
+
+shows that the remainder is a unit in \(\mathds{Z}[3]\).
+
+\section{Question Five}
+\label{sec:orgaf8ce04}
+Firstly, \(\mathds{Z}_5[x]\) is a field by Theorem 2.8.
+
+\subsection{a}
+\label{sec:orgf579bc7}
+There are no roots of \(x^2 + 2x + 3\) in \(\mathds{Z}_5\), so by Theorem 5.10 it is irreducible.
+
+\(f_1 : (0, 1, 2, 3, 4) \rightarrow (3, 1, 1, 3, 2)\)
+
+\subsection{b}
+\label{sec:orgb0b2569}
+This is reducible since \(1\) is a root.
+
+\begin{equation*}
+\polylongdiv[style=A]{x^3 + x + 3}{x - 1}
+\end{equation*}
+
+Thus, \(f_2(x) = (x + 4)(x^2 + x + 2)\)
+
+
+\section{Question Six}
+\label{sec:org009bcbb}
+Following pages 137 - 138 of the book\ldots{}
+
+By Corollary 4.19 we know \((x^2 + 1)\) is irreducible in \(\mathds{R}[x]\) since its only roots are in \(\mathds{C}\): \(\pm i\).
+
+Because of Corollary 5.12, we know that there is an extension field \(K\) that contains a root to \((x^2 + 1)\). Namely,
+some \(\alpha = [x]\). By Corollary 5.5, each element can be rewritten as \([ax + b]\) with \(a, b \in \mathds{R}\).
+We can create a mapping \(f\) such that \([a + bx] \rightarrow [a] + [b][x] = a + b \alpha\) is unique and \(f^{-1}(a + b \alpha) \rightarrow [a + bx]\) (bijection).
+
+Additionally, we can show that \(f([a + bx]) + f([c + dx]) = (a + b \alpha) + (c + d \alpha) = (a + c) + (b + d) \alpha = f([(a + c)x + (b + d)])\).
+
+\begin{equation*}
+[ax + b][cx + b] = [acx^2 + abx + bcx + bd]
+\end{equation*}
+
+\begin{equation*}
+\polylongdiv[style=A]{acx^2 + abx + bcx + bd}{x^2 + 1}
+\end{equation*}
+
+So, multiplication in \(K\) is given:
+
+\begin{equation*}
+[ax + b][cx + b] = [(ab + bc)x + ac - bd]
+\end{equation*}
+
+Then, we can show that by definition of \(\alpha\), \(f([a + bx]) \cdot f([c + dx]) = (a + b \alpha) \cdot (c + d \alpha) = (ac + ad \alpha + bc \alpha + bd \alpha^2) = (ac - bd + (ad + bc)\alpha) = f([(ab + bc)x + ac - bd]) = f([a + bx][c + dx])\).
+
+Now, replacing our work with \(\alpha = i\), \(f\) is an isomorphism from \(\mathds{R}/(x^2 + 1)\) to \(\mathds{C}\) since it is a bijection and satisfies the addition and multiplication rules.
+\end{document}
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