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+% Created 2023-01-18 Wed 12:13
+% 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}
+\noindent \notag \usepackage{ dsfont }
+\author{Logan Hunt}
+\date{\today}
+\title{Assignment One}
+\hypersetup{
+ pdfauthor={Logan Hunt},
+ pdftitle={Assignment One},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.5.5)}, 
+ pdflang={English}}
+\begin{document}
+
+\maketitle
+\tableofcontents
+
+
+\section{Section 1.1}
+\label{sec:org35f7eb6}
+\subsection{Question 5}
+\label{sec:orge503259}
+By reforming the given expression to show \(ca\) as a multiple of \(q\) and some remainder:
+
+\begin{align}
+a = bq + r \\
+ca = (cb)q + (c)r
+\end{align}
+
+and that \(0 \leq r < b \Rightarrow 0 \leq cr < cb\), Theorem 1.1 tells us that there is only one unique quotient:
+which is \(q\), and the remainder is \((c)r\).
+
+\subsection{Question 7}
+\label{sec:org6afde38}
+By Theorem 1.1, \(a = bq + r, 0 \leq r \lt b\) implies that when b = 3, \(a = 3q + r\) where \(0 \leq r \lt b\),
+thus \(a \in \mathds{Z} \Rightarrow a = 3q + r\) where \(r \in {0,1,2}\).
+
+By squaring \(a\), we have three options which simplify to the form \(3k\) or \(3k + 1\):
+\begin{enumerate}
+\item \(a^2 = (3q)^2 = 9q^2\) which is \(3k \ni k = 3q^2\)
+\item \(a^2 = (3q + 1)^2 = 9q^2 + 3q + 1 = 3(3q^2 + q) + 1\) which is \(3k + 1 \ni k = (3q^2 + q)\)
+\item \(a^2 = (3q + 2)^2 = 9q^2 + 6q + 4 = 3(3q^2 + 2q) + 4\) which is
+\(3l + 4 \ni l = (3q^2 + 2q)\) which is also \(3l + 1 + 3 = 3(l+1) + 1\),
+which again is \(3k + 1 \ni k = l+1\)
+\end{enumerate}
+
+\subsection{Question 8}
+\label{sec:orgfe8a512}
+By Theorem 1.1, \(a = bq + r, 0 \leq r \lt b\) implies that when b = 4, \(a = 4q + r\) where \(0 \leq r \lt b\),
+and thus \(a = 4q + r\) and \(r \in {0,1,2,3}\).
+
+Therefore we have four cases:
+
+\begin{enumerate}
+\item \(a = 4q\), and \(a\) must be even which is invalid
+\item \(a = 4q + 1\), and \(4q + 1 = 2k + 1 \ni k = 2q\) which fits the definition of an odd number
+\item \(a = 4q + 2\), thus \(a\) must be even as \(a = 2k \ni k = 2q + 1\)
+\item \(a = 4q + 3\), can be rewritten to \(4q + 1 + 2\), which is odd: \(2k + 1 \ni k = 2q + 1\)
+\end{enumerate}
+
+And thus any odd number can be rewritten as \(4q + 1\) or \(4q + 3\).
+
+\subsection{Question 10}
+\label{sec:org3c66372}
+The division of \(a\) and \(c\) by \(n\) can each be represented by
+
+\begin{align}
+a = q_{a}n + r_a \\
+c = q_c_{}n + r_c
+\end{align}
+
+where \(0 \le r_a < n\) and \(0 \le r_c < n\) by Theorem 1.1, and we can subtract each side of the
+second equation from the first:
+
+\begin{align}
+a - c = (q_{a}n + r_a) - (q_c_{}n + r_c)
+\end{align}
+
+With this work now in hand, we will prove by showing that the conjecture is true both ways:
+
+For the first, we will suppose that \(r_a = r_c\). Then, we find that
+\begin{align}
+a - c = n(q_a - q_c) \\
+a - c = nk
+\end{align}
+for some integer \(k\).
+
+For the second, we will prove that if \(a - c = nk\), when \(a\) and \(c\) are divided by \(n\),
+they leave the same remainder \(r\).
+
+From the work we did previously, and by substituting \(a-c = nk\), we find that
+\begin{align}
+a - c = (q_{a}n + r_a) - (q_c_{}n + r_c) \\
+nk = (q_{a}n + r_a) - (q_{c}n + r_c) \\
+nk = n(q_a - q_c) + (r_a - r_c) \\
+n(k - q_a + q_c) = r_a - r_c
+\end{align}
+and thus \(r_a - r_c\) is a multiple of \(n\).
+
+From the inequalities produced previously by Theorem 1.1, if \(r_a \geq r_c\) then \(0 \le r_a - r_c < n\)
+and \(-n < r_c - r_a \leq 0\). Else if \(r_c \geq r_a\) then \(0 \leq r_c - r_a < n\) and \(-n < r_a - r_c \leq 0\).
+
+By combining the two cases, it must be that \$ -n < r\textsubscript{a} - r\textsubscript{c} < n \$, and from the above fact that
+\(r_a - r_c\) is a multiple of n, the only multiple of n in \((-n, n)\) is 0. Thus \(r_a = r_c\).
+
+\section{Section 1.2}
+\label{sec:org4226083}
+\subsection{Question 1}
+\label{sec:org4ccc206}
+\subsubsection{c}
+\label{sec:org802af33}
+1,  57 and 112 are co-prime
+\subsection{Question 3}
+\label{sec:orgb341781}
+\begin{align}
+a | b \Rightarrow \exists n \in \mathds{Z} \ni an = b \\
+b | c \Rightarrow \exists m \in \mathds{Z} \ni bm = c \\
+(an)m = c \Rightarrow a | c
+\end{align}
+\subsection{Question 4}
+\label{sec:orge55e932}
+\subsubsection{a}
+\label{sec:org7540edc}
+\begin{align}
+a | b \Rightarrow \exists n \in \mathds{Z} \ni an = b \\
+a | c \Rightarrow \exists m \in \mathds{Z} \ni am = c \\
+a | (b + c) \Rightarrow \exists l \in \mathds{Z} \ni al = b + c \\
+al = an + am \\
+b + c = a(n + m)
+\Rightarrow a | b+c
+\end{align}
+\subsubsection{b}
+\label{sec:orge72f7c4}
+By continuing from "a":
+
+\begin{align}
+br + ct = (an)r + (am)t \\
+\Rightarrow br + ct = a(nr + mt) \\
+\Rightarrow a | (br + ct)
+\end{align}
+
+
+\subsection{Question 7}
+\label{sec:org7273a90}
+\(|a|\) since \(|a| \textbar a\) and \(|a| \textbar 0\), and there cannot be an integer larger than
+\(|a|\) that divides \(a\).
+
+\subsection{Question 9}
+\label{sec:org6b4815b}
+No, not every multiple of two factors of an integer divides that integer.
+
+For example, \(3 | 9\) and \(9 | 9\) but \(27 \nmid 9\).
+
+\subsection{Question 15}
+\label{sec:orgf96c6cd}
+\subsubsection{c}
+\label{sec:orgad14e2b}
+\begin{align}
+1003 = (2)(456) + 91 \\
+456 = (5)(91) + 1 \\
+91 = (91)(1) + 0 \\
+\Rightarrow (1003, 456) = 1
+\end{align}
+\subsubsection{d}
+\label{sec:orgf89fea4}
+\begin{align}
+322 = (2)(148) + 26 \\
+148 = (5)(26) + 18 \\
+26 = (1)(18) + 8 \\
+18 = (2)(8) + 2 \\
+8 = (4)(2) + 0 \\
+\Rightarrow (322, 148) = 2
+\end{align}
+
+\subsection{Question 17}
+\label{sec:org66d011a}
+
+\begin{align}
+a | c \Rightarrow \exists n \in \mathds{Z} \ni an = c \\
+b | c \Rightarrow b | an
+\end{align}
+
+And by Theorem 1.4, \((a, b) = 1 \Rightarrow b | n \Rightarrow ab | an \Rightarrow ab | c\):
+
+\subsection{Question 19}
+\label{sec:org905b244}
+Given some \(d\) such that \(d | a\) and \(d | b\), then \(a = nd\) and \(b = md\).
+\begin{align}
+a | (b + c) \Rightarrow \exists p \in \mathds{Z} \ni ap = b + c \\
+c = ap - md \Rightarrow c = (nd)p - md \Rightarrow c = d(np - md) \Rightarrow d | c
+\end{align}
+
+Since we're given that \((b, c) = 1 \Rightarrow (md, c) = 1\) and from the above fact that \(d | c\), we can conclude
+that \(md = 1\). Therefore \((a, b) = (a, md) = (a, 1) = 1\).
+
+Given some \(e\) such that \(e | a\) and \(e | c\), then \(a = ue\) and \(c = we\).
+\begin{align}
+a | (b + c) \Rightarrow \exists o \in \mathds{Z} \ni ao = b + c \\
+b = ao - we \Rightarrow b = (ue)o - we \Rightarrow b = e(uo - w) \Rightarrow e | b
+\end{align}
+
+And from similar reasoning we can conclude that \((a, c) = (a, we) = (a, 1) = 1\).
+
+\subsection{Question 31}
+\label{sec:orgc12b666}
+\subsubsection{a}
+\label{sec:org4312735}
+\([6, 10] = 60\)
+
+\([4, 5, 6, 10] = 60\)
+
+\([20, 42] = 840\)
+
+\([2, 3, 14, 36, 42] = 252\)
+
+\subsubsection{b}
+\label{sec:orgb964c2c}
+Let \(m = [a_1, a_2, \ldots, a_k]\), then by the Division Algorithm, \(t = mq + r\) for integers \(q\) and \(r\),
+with \(0 \leq r < m\).
+
+As given that all \(a_i | t\) \(\Rightarrow\) \(a_i | mq + r\), and as \(mq\) is a multiple of the LCM including
+a\textsubscript{i}, \(a_i\) must divide \(mq\), and thus \(a_i | r\).
+
+Because \(a_i | r\), \(r = na_i\) and is thus a multiple of all \(a_i\), but the above statement that
+\(0 \leq r < m\) shows that \(m\) - the LCM by definition - is greater than \(r\). Therefore, \(r = 0\) and
+\(t = mq \Rightarrow m | t\).
+\end{document}
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diff --git a/Homework/math4310/328e45ffe10876ae6e22f4d5a990f625-emacs-elixir-format.ex b/Homework/math4310/328e45ffe10876ae6e22f4d5a990f625-emacs-elixir-format.ex
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+File alg_structures_assn_4.pdf changed on disk.  Reread from disk? (yes or no) 
\ No newline at end of file
diff --git a/Homework/math4310/abstract_algebra_assn_10.org b/Homework/math4310/abstract_algebra_assn_10.org
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+#+TITLE: Assignment Ten
+#+AUTHOR: Lizzy Hunt
+#+STARTUP: entitiespretty fold inlineimages
+#+LATEX_HEADER: \notindent \notag \usepackage{ dsfont } \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} \usepackage{polynom} \usepackage{wasysym}
+#+LATEX: \setlength\parindent{0pt}
+#+OPTIONS: toc:nil
+
+* Section 6.1
+** Question Four
+No. As a counterexample, suppose $n = \big(\begin{smallmatrix}
+  0 & 0\\
+  0 & 1
+\end{smallmatrix}\big) \in J$, and $m = \big(\begin{smallmatrix}
+  1 & 1  \\
+  1 & 1
+\end{smallmatrix}\big) \in M(\mathds{R})$, $mn = \big(\begin{smallmatrix}
+  0 & 1  \\
+  0 & 1
+\end{smallmatrix}\big) \notin J$
+
+** Question Five
+By Theorem 3.6, $K$ is a subring since it is closed under subtraction: $\big(\begin{smallmatrix}
+  a & b  \\
+  0 & 0
+\end{smallmatrix}\big) -
+(\begin{smallmatrix}
+  c & d  \\
+  0 & 0
+\end{smallmatrix}\big) =
+(\begin{smallmatrix}
+  a - c & b - d  \\
+  0 & 0
+\end{smallmatrix}\big)$, and absorbs products of $M(\mathds{R})$ on the right (which is a superset of $K$, so $K$ is closed under this multiplication):
+$(\begin{smallmatrix}
+  a & b  \\
+  0 & 0
+\end{smallmatrix}\big) \cdot
+(\begin{smallmatrix}
+  c & d  \\
+  e & f
+\end{smallmatrix}\big)
+=
+(\begin{smallmatrix}
+ ac + be & ad + bf  \\
+  0 & 0
+\end{smallmatrix}\big)$
+
+But from the left, as a counterexample $n = (\begin{smallmatrix}
+1 & 1  \\
+0 & 0
+\end{smallmatrix}\big) \in K, m =(\begin{smallmatrix}
+  2 & 1  \\
+  3 & 0
+\end{smallmatrix}\big) \in M(\mathds{R})$, then $mn = (\begin{smallmatrix}
+  2 & 2  \\
+  3 & 3
+\end{smallmatrix}\big) \notin K$
+
+** Question Eleven
+*** a
+$(1) = (2) = (3) = (4) = \mathds{Z}_5$ and $(0) =$ { 0 }
+
+*** b
+$(1) = (2) = (4) = (5) = (7) = (8) = \mathds{Z}_9$, $(0) =$ { 0 }, and $(3) = (6) =$ { 0, 3, 6 }
+
+*** c
+$(1) = (5) = (7) = (11) = \mathds{Z}_{12}$, $(2) = (6) = (10) =$ { 0, 2, 4, 6, 8, 10 }, $(4) = (8) =$ { 0, 4, 8 },
+$(3) = (9) =$ { 0, 3, 6, 9 }, and $(6) =$ { 0, 6 }
+
+** Question Thirteen
+No, $\mathds{Z}_5$ is a commutative ring, but in question above, $(1) = (2)$ but $1 \neq 2$.
+
+** Question Sixteen
+*** a
+If we can show that $(4, 6) \sube (2)$ and $(2) \sube (4, 6)$, then we can conclude that $(4, 6) = (2)$
+
+Each element $x \in (4, 6) \Rightarrow x = 4m + 6n$ for $m,n \in \mathds{Z}$.
+
+Thus $x = 2(2m + 3n)$ which shows that $x$ is a multiple of two, and thus, $(4, 6) \sube (2)$.
+
+Each element $y \in (2) \Rightarrow y = 2p$ for $p \in \mathds{Z}$. When $p$ is itself a multiple of two, this can be rewritten as $y = 2(2o) = 4o$ and thus $y \in (4, 6)$.
+
+When $y$ is odd, $y = 2(2(q - 1) + 3) = 4q + 6$ for some $q \in \mathds{Z}$. Therefore, $y$ is in $(4, 6)$.
+
+*** b
+We'll take the same approach as a:
+
+Each element $x \in (6, 9, 15) = 6m + 9n + 15o$ with $m, n, o \in \mathds{Z}$.
+
+Thus $x = 3(2m + 3n + 5o)$ is a multiple of three, so $x \in (3)$.
+
+Each element $y \in (3) \Rightarrow y = 3p$ for $p \in \mathds{Z}$. When $p$ is a multiple of two, then $y = 3(2q) = 6q \in (6, 9, 15)$.
+
+When $p$ is odd, $y = 3(2(u - 1) + 7)$ for some $u \in \mathds{Z}$, $y = 6u + 15 \in (6, 9, 15)$.
+
+** Question Seventeen
+*** a
+If $a \in I \cap J$, and $b \in I \cap J$, then $a \in I$, $b \in I$, $a \in J$, and $b \in J$. Then, $a - b \in I$ and $a - b \in J$, so
+$a - b \in I \cap J$.
+
+If $a \in I \cap J$ and $r \in R$, then $ra \in I$, $ra \in J$, $ar \in J$, and $ar \in I$ by definition of I and J.
+
+Therefore, $I \cap J$ is an ideal in $R$.
+
diff --git a/Homework/math4310/abstract_algebra_assn_10.pdf b/Homework/math4310/abstract_algebra_assn_10.pdf
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--- /dev/null
+++ b/Homework/math4310/abstract_algebra_assn_10.pdf
diff --git a/Homework/math4310/abstract_algebra_assn_10.tex b/Homework/math4310/abstract_algebra_assn_10.tex
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+% Created 2023-04-16 Sun 21:55
+% 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{ dsfont } \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} \usepackage{polynom} \usepackage{wasysym}
+\author{Lizzy Hunt}
+\date{\today}
+\title{Assignment Ten}
+\hypersetup{
+ pdfauthor={Lizzy Hunt},
+ pdftitle={Assignment Ten},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.6.1)}, 
+ pdflang={English}}
+\begin{document}
+
+\maketitle
+\setlength\parindent{0pt}
+
+\section{Section 6.1}
+\label{sec:org1d97e2b}
+\subsection{Question Four}
+\label{sec:orgd166f06}
+No. As a counterexample, suppose \(n = \big(\begin{smallmatrix}
+  0 & 0\\
+  0 & 1
+\end{smallmatrix}\big) \in J\), and \(m = \big(\begin{smallmatrix}
+  1 & 1  \\
+  1 & 1
+\end{smallmatrix}\big) \in M(\mathds{R})\), \(mn = \big(\begin{smallmatrix}
+  0 & 1  \\
+  0 & 1
+\end{smallmatrix}\big) \notin J\)
+
+\subsection{Question Five}
+\label{sec:orgd391af3}
+By Theorem 3.6, \(K\) is a subring since it is closed under subtraction: \(\big(\begin{smallmatrix}
+  a & b  \\
+  0 & 0
+\end{smallmatrix}\big) -
+(\begin{smallmatrix}
+  c & d  \\
+  0 & 0
+\end{smallmatrix}\big) =
+(\begin{smallmatrix}
+  a - c & b - d  \\
+  0 & 0
+\end{smallmatrix}\big)\), and absorbs products of \(M(\mathds{R})\) on the right (which is a superset of \(K\), so \(K\) is closed under this multiplication):
+\((\begin{smallmatrix}
+  a & b  \\
+  0 & 0
+\end{smallmatrix}\big) \cdot
+(\begin{smallmatrix}
+  c & d  \\
+  e & f
+\end{smallmatrix}\big)
+=
+(\begin{smallmatrix}
+ ac + be & ad + bf  \\
+  0 & 0
+\end{smallmatrix}\big)\)
+
+But from the left, as a counterexample \(n = (\begin{smallmatrix}
+1 & 1  \\
+0 & 0
+\end{smallmatrix}\big) \in K, m =(\begin{smallmatrix}
+  2 & 1  \\
+  3 & 0
+\end{smallmatrix}\big) \in M(\mathds{R})\), then \(mn = (\begin{smallmatrix}
+  2 & 2  \\
+  3 & 3
+\end{smallmatrix}\big) \notin K\)
+
+\subsection{Question Eleven}
+\label{sec:orgd8049eb}
+\subsubsection{a}
+\label{sec:org51730ce}
+\((1) = (2) = (3) = (4) = \mathds{Z}_5\) and \((0) =\) \{ 0 \}
+
+\subsubsection{b}
+\label{sec:org3030fdd}
+\((1) = (2) = (4) = (5) = (7) = (8) = \mathds{Z}_9\), \((0) =\) \{ 0 \}, and \((3) = (6) =\) \{ 0, 3, 6 \}
+
+\subsubsection{c}
+\label{sec:orgac8d99a}
+\((1) = (5) = (7) = (11) = \mathds{Z}_{12}\), \((2) = (6) = (10) =\) \{ 0, 2, 4, 6, 8, 10 \}, \((4) = (8) =\) \{ 0, 4, 8 \},
+\((3) = (9) =\) \{ 0, 3, 6, 9 \}, and \((6) =\) \{ 0, 6 \}
+
+\subsection{Question Thirteen}
+\label{sec:org8889fae}
+No, \(\mathds{Z}_5\) is a commutative ring, but in question above, \((1) = (2)\) but \(1 \neq 2\).
+
+\subsection{Question Sixteen}
+\label{sec:org132058d}
+\subsubsection{a}
+\label{sec:org121e6c5}
+If we can show that \((4, 6) \sube (2)\) and \((2) \sube (4, 6)\), then we can conclude that \((4, 6) = (2)\)
+
+Each element \(x \in (4, 6) \Rightarrow x = 4m + 6n\) for \(m,n \in \mathds{Z}\).
+
+Thus \(x = 2(2m + 3n)\) which shows that \(x\) is a multiple of two, and thus, \((4, 6) \sube (2)\).
+
+Each element \(y \in (2) \Rightarrow y = 2p\) for \(p \in \mathds{Z}\). When \(p\) is itself a multiple of two, this can be rewritten as \(y = 2(2o) = 4o\) and thus \(y \in (4, 6)\).
+
+When \(y\) is odd, \(y = 2(2(q - 1) + 3) = 4q + 6\) for some \(q \in \mathds{Z}\). Therefore, \(y\) is in \((4, 6)\).
+
+\subsubsection{b}
+\label{sec:orgc2b77e9}
+We'll take the same approach as a:
+
+Each element \(x \in (6, 9, 15) = 6m + 9n + 15o\) with \(m, n, o \in \mathds{Z}\).
+
+Thus \(x = 3(2m + 3n + 5o)\) is a multiple of three, so \(x \in (3)\).
+
+Each element \(y \in (3) \Rightarrow y = 3p\) for \(p \in \mathds{Z}\). When \(p\) is a multiple of two, then \(y = 3(2q) = 6q \in (6, 9, 15)\).
+
+When \(p\) is odd, \(y = 3(2(u - 1) + 7)\) for some \(u \in \mathds{Z}\), \(y = 6u + 15 \in (6, 9, 15)\).
+
+\subsection{Question Seventeen}
+\label{sec:org35871b0}
+\subsubsection{a}
+\label{sec:orgcdd82ef}
+If \(a \in I \cap J\), and \(b \in I \cap J\), then \(a \in I\), \(b \in I\), \(a \in J\), and \(b \in J\). Then, \(a - b \in I\) and \(a - b \in J\), so
+\(a - b \in I \cap J\).
+
+If \(a \in I \cap J\) and \(r \in R\), then \(ra \in I\), \(ra \in J\), \(ar \in J\), and \(ar \in I\) by definition of I and J.
+
+Therefore, \(I \cap J\) is an ideal in \(R\).
+\end{document}
\ No newline at end of file
diff --git a/Homework/math4310/abstract_algebra_assn_11.org b/Homework/math4310/abstract_algebra_assn_11.org
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+#+TITLE: Assignment Eleven
+#+AUTHOR: Lizzy Hunt
+#+STARTUP: entitiespretty fold inlineimages
+#+LATEX_HEADER: \notindent \notag \usepackage{ dsfont } \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} \usepackage{polynom} \usepackage{wasysym}
+#+LATEX: \setlength\parindent{0pt}
+#+OPTIONS: toc:nil
+
+* Section 6.1
+** Question Twenty-One
+The ideal $(a) + (b)$ is generated by all linear combinations of $a$ and $b$, $x \in (a) + (b) \Rightarrow x = a c_1 + b c_2$
+
+By definition, $d | a$ and $d | b$, so $a = d a_1$ and $b = d b_1$. Then, $x \in (a) + (b) \Rightarrow x = d a_1 c_1 + d b_1 c_2 = d(a_1 c_1 + b_1 c_2)$,
+so $(a) + (b) \sube (c)$.
+
+By Theorem 1.2, $d = au + bv$ which is a linear combination of $a$ and $b$, so $d \in (a) + (b)$. Since
+every multiple of $d$ is also a multiple of $a$ and $b$, $(d) \sube (a) + (b)$.
+
+So, $(d) = (a) + (b)$.
+
+* Section 6.2
+** Question Two
+Let $f : F \rightarrow R$ with $R$ being the image, so the kernel of $f$ is an ideal in the field $F$ by Theorem 6.10.
+Then assuming from the hint, that question ten in 6.1 is true, the only ideals in $F$ are either $(0_F)$ or $F$
+itself. So, $\text{ker}(f) = (0_F)$ or $F$.
+
+When $\text{ker}(f) = (0_F)$ then $f$ is injective by Theorem 6.11, and is thus an isomorphism.
+
+When $\text{ker}(f) = F$ then $R = \{ 0_F \}$.
+
+** Question Four
+*** a
+$f$ is consistent, as when
+$[a]_{12}_{} = [b]_{12}_{}_{} (a \equiv b \text{ mod } 12 \Rightarrow a - b = 12n)$, then $[a]_4 = [b]_4$ as
+$a - b \equiv 20n \text{ mod } 4 = a - b \equiv 0 \text{ mod } 4 \Rightarrow a \equiv b \text{ mod } 4$.
+
+*** b
+The kernel of $f$ is the ideal $(4)$ as $f(x) \ni x \in (4) \Rightarrow f(x) = [4n]_{4} = [0]_4$.
+
+** Question Six
+The kernel of \phi are polynomials in $\mathds{R}[x]$ with a root of 2. By Theorem 4.16, $x - 2$ must be a factor, so
+$\text{ker}(\phi)$ is the set of all multiples of $x-2$ in $\mathds{R}[x] = (x - 2)$.
+
+** Question Nine
+*** a
+There is surjectivity as $x \in \mathds{Z}$, then the matrix $\big(\begin{smallmatrix}
+  x & 0\\
+  0 & 0
+\end{smallmatrix}\big)$ maps to $x$.
+
+It is a homomorphism since $f(x + y) = \big(\begin{smallmatrix}
+  a & 0\\
+  c & d
+\end{smallmatrix}\big) + \big(\begin{smallmatrix}
+  e & 0\\
+  g & h
+\end{smallmatrix}\big) = \big(\begin{smallmatrix}
+  a + e & 0 \\
+  c + g & d + h
+\end{smallmatrix}\big) = (a + e) = f(x) + f(y)$,
+$f(xy) = \big(\begin{smallmatrix}
+  a & 0\\
+  c & d
+\end{smallmatrix}\big) \cdot \big(\begin{smallmatrix}
+  e & 0\\
+  g & h
+3\end{smallmatrix}\big) = \big(\begin{smallmatrix}
+  ae & 0 \\
+  ce + dc & dh
+\end{smallmatrix}\big) = ae = f(x)f(y)$, and the identity matrix is mapped to $1$.
+
+*** b
+$\{ \big(\begin{smallmatrix}
+  0 & 0 \\
+  c & d
+\end{smallmatrix}\big) | c, d \in \mathds{Z} \}$
+
+** Question Ten 
+*** a
+By definition, $x, y \in f(I) \Rightarrow \exists z, t \in I \ni f(z) = x, f(t) = b$, and $f(I)$ is also a subset of $S$.
+
+By hormomorphism, $f(z) - f(t) = x - y \Rightarrow f(z - t) = x - y \in I$.
+
+For $s \in S \Rightarrow \exists r \in R \ni f(r) = s$ by surjection. Then, $x f(r) = s x = f(z) f(r) = f(zr) \in f(I)$
+
+*** b
+Let $R = \mathds{R}$, $S = \mathds{C}$, and $f : R \rightarrow S \ni f(x) = x + 0i$. $f$ is not a surjective
+homomorphism ($i$ has no associated element in $\mathds{R}$), and $\mathds{R}$ is an ideal in $\mathds{R}$.
+
+However, $f(\mathds{R})$ is not an ideal by the trivial example that
+$f(1)(i) = (1 + 0i)(0 + i) = i \notin \mathds{R}$.
+
+** Question Seventeen
+*** a
+$f(a + b) = ((a + b) + I, (a + b) + J) = ((a + I) + (b + I), (a + J) + (b + J)) = (a + I, a + J) + (b + I, b + J) = f(a) + f(b)$.
+
+Similarly, $f(a)f(b) = f(ab)$: $f(ab) = (ab + I, ab + J)$ and $f(a)f(b) = (a + I, a + J)(b + I, b + J) = ((a + I)(b + I), (a + J)(b + J)) = (ab + I, ab + J)$.
+
+*** b
+If we consider $R = \mathds{Z}, I = (2)$, and $J = (4)$, the elements in \mathds{Z}/(2) x \mathds{Z}/(4) are:
+
+{(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3)}
+
+There's at least one domain element that is impossible to get to: $(1, 0)$ - an integer
+can't be equivalent to $1$ mod $2$ and also $0$ mod $4$. So, $f$ is not necessarily surjective.
+
+** Question Twenty-One
+Let $f : \mathds{Z}_{20} \rightarrow \mathds{Z}_5$ such that $f([a]_{20}) = [a]_5$. $f$ is consistent because when
+$[a]_{20} = [b]_{20}_{} (a \equiv b \text{ mod } 20 \Rightarrow a - b = 20n)$, then $[a]_5 = [b]_5$ as
+$a - b \equiv 20n \text{ mod } 5 = a - b \equiv 0 \text{ mod } 5 \Rightarrow a \equiv b \text{ mod } 5$.
+
+Also, $f$ is a surjective homomorphism:
++ $f([a]_{20} + [b]_{20}) = f([a + b]_{20}) = [(a + b)]_5 = [a]_5 + [b]_5 = f([a]_{20}) + f([b]_{20})$
++ $f([a]_{20} * [b]_{20}) = f([a * b]_{20}) = [(a * b)]_5 = [a]_5 * [b]_5 = f([a]_{20}) * f([b]_{20})$
++ For each $[x]_5$ there exists $[y]_{20}$ such that $f([y]_{20}) = [x]_5$. The trivial solution is
+  that $x = y$.
+
+Finally, the kernel of $f$ is the ideal $(5)$ as $x \in (5) \Rightarrow f(x) = [5n]_5 = [0]_5$. Thus,
+$\mathds{Z}_{20} / (5)$ is isomorphic to $\mathds{Z}_5$ by the First Isomorphism Theorem.
diff --git a/Homework/math4310/abstract_algebra_assn_11.pdf b/Homework/math4310/abstract_algebra_assn_11.pdf
new file mode 100644
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--- /dev/null
+++ b/Homework/math4310/abstract_algebra_assn_11.pdf
diff --git a/Homework/math4310/abstract_algebra_assn_11.tex b/Homework/math4310/abstract_algebra_assn_11.tex
new file mode 100644
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--- /dev/null
+++ b/Homework/math4310/abstract_algebra_assn_11.tex
@@ -0,0 +1,162 @@
+% Created 2023-04-17 Mon 12:52
+% 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,portrait]{geometry} \usepackage{polynom} \usepackage{wasysym}
+\author{Lizzy Hunt}
+\date{\today}
+\title{Assignment Eleven}
+\hypersetup{
+ pdfauthor={Lizzy Hunt},
+ pdftitle={Assignment Eleven},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.6.1)}, 
+ pdflang={English}}
+\begin{document}
+
+\maketitle
+\setlength\parindent{0pt}
+
+\section{Section 6.1}
+\label{sec:org7b8fe71}
+\subsection{Question Twenty-One}
+\label{sec:orga85b892}
+The ideal \((a) + (b)\) is generated by all linear combinations of \(a\) and \(b\), \(x \in (a) + (b) \Rightarrow x = a c_1 + b c_2\)
+
+By definition, \(d | a\) and \(d | b\), so \(a = d a_1\) and \(b = d b_1\). Then, \(x \in (a) + (b) \Rightarrow x = d a_1 c_1 + d b_1 c_2 = d(a_1 c_1 + b_1 c_2)\),
+so \((a) + (b) \sube (c)\).
+
+By Theorem 1.2, \(d = au + bv\) which is a linear combination of \(a\) and \(b\), so \(d \in (a) + (b)\). Since
+every multiple of \(d\) is also a multiple of \(a\) and \(b\), \((d) \sube (a) + (b)\).
+
+So, \((d) = (a) + (b)\).
+
+\section{Section 6.2}
+\label{sec:orgf462b61}
+\subsection{Question Two}
+\label{sec:org6ec012f}
+Let \(f : F \rightarrow R\) with \(R\) being the image, so the kernel of \(f\) is an ideal in the field \(F\) by Theorem 6.10.
+Then assuming from the hint, that question ten in 6.1 is true, the only ideals in \(F\) are either \((0_F)\) or \(F\)
+itself. So, \(\text{ker}(f) = (0_F)\) or \(F\).
+
+When \(\text{ker}(f) = (0_F)\) then \(f\) is injective by Theorem 6.11, and is thus an isomorphism.
+
+When \(\text{ker}(f) = F\) then \(R = \{ 0_F \}\).
+
+\subsection{Question Four}
+\label{sec:orgb116fb8}
+\subsubsection{a}
+\label{sec:org3bb1cfa}
+\(f\) is consistent, as when
+\([a]_{12}_{} = [b]_{12}_{}_{} (a \equiv b \text{ mod } 12 \Rightarrow a - b = 12n)\), then \([a]_4 = [b]_4\) as
+\(a - b \equiv 20n \text{ mod } 4 = a - b \equiv 0 \text{ mod } 4 \Rightarrow a \equiv b \text{ mod } 4\).
+
+\subsubsection{b}
+\label{sec:orgf598f8c}
+The kernel of \(f\) is the ideal \((4)\) as \(f(x) \ni x \in (4) \Rightarrow f(x) = [4n]_{4} = [0]_4\).
+
+\subsection{Question Six}
+\label{sec:org7e819f2}
+The kernel of \(\phi\) are polynomials in \(\mathds{R}[x]\) with a root of 2. By Theorem 4.16, \(x - 2\) must be a factor, so
+\(\text{ker}(\phi)\) is the set of all multiples of \(x-2\) in \(\mathds{R}[x] = (x - 2)\).
+
+\subsection{Question Nine}
+\label{sec:orgff395bd}
+\subsubsection{a}
+\label{sec:org2a1ef62}
+There is surjectivity as \(x \in \mathds{Z}\), then the matrix \(\big(\begin{smallmatrix}
+  x & 0\\
+  0 & 0
+\end{smallmatrix}\big)\) maps to \(x\).
+
+It is a homomorphism since \(f(x + y) = \big(\begin{smallmatrix}
+  a & 0\\
+  c & d
+\end{smallmatrix}\big) + \big(\begin{smallmatrix}
+  e & 0\\
+  g & h
+\end{smallmatrix}\big) = \big(\begin{smallmatrix}
+  a + e & 0 \\
+  c + g & d + h
+\end{smallmatrix}\big) = (a + e) = f(x) + f(y)\),
+\(f(xy) = \big(\begin{smallmatrix}
+  a & 0\\
+  c & d
+\end{smallmatrix}\big) \cdot \big(\begin{smallmatrix}
+  e & 0\\
+  g & h
+3\end{smallmatrix}\big) = \big(\begin{smallmatrix}
+  ae & 0 \\
+  ce + dc & dh
+\end{smallmatrix}\big) = ae = f(x)f(y)\), and the identity matrix is mapped to \(1\).
+
+\subsubsection{b}
+\label{sec:org036e3e8}
+\(\{ \big(\begin{smallmatrix}
+  0 & 0 \\
+  c & d
+\end{smallmatrix}\big) | c, d \in \mathds{Z} \}\)
+
+\subsection{Question Ten}
+\label{sec:org3604c40}
+\subsubsection{a}
+\label{sec:org626043a}
+By definition, \(x, y \in f(I) \Rightarrow \exists z, t \in I \ni f(z) = x, f(t) = b\), and \(f(I)\) is also a subset of \(S\).
+
+By hormomorphism, \(f(z) - f(t) = x - y \Rightarrow f(z - t) = x - y \in I\).
+
+For \(s \in S \Rightarrow \exists r \in R \ni f(r) = s\) by surjection. Then, \(x f(r) = s x = f(z) f(r) = f(zr) \in f(I)\)
+
+\subsubsection{b}
+\label{sec:org2946823}
+Let \(R = \mathds{R}\), \(S = \mathds{C}\), and \(f : R \rightarrow S \ni f(x) = x + 0i\). \(f\) is not a surjective
+homomorphism (\(i\) has no associated element in \(\mathds{R}\)), and \(\mathds{R}\) is an ideal in \(\mathds{R}\).
+
+However, \(f(\mathds{R})\) is not an ideal by the trivial example that
+\(f(1)(i) = (1 + 0i)(0 + i) = i \notin \mathds{R}\).
+
+\subsection{Question Seventeen}
+\label{sec:orgb24a5cb}
+\subsubsection{a}
+\label{sec:org15e7223}
+\(f(a + b) = ((a + b) + I, (a + b) + J) = ((a + I) + (b + I), (a + J) + (b + J)) = (a + I, a + J) + (b + I, b + J) = f(a) + f(b)\).
+
+Similarly, \(f(a)f(b) = f(ab)\): \(f(ab) = (ab + I, ab + J)\) and \(f(a)f(b) = (a + I, a + J)(b + I, b + J) = ((a + I)(b + I), (a + J)(b + J)) = (ab + I, ab + J)\).
+
+\subsubsection{b}
+\label{sec:org2bba32c}
+If we consider \(R = \mathds{Z}, I = (2)\), and \(J = (4)\), the elements in \mathds{Z}/(2) x \mathds{Z}/(4) are:
+
+\{(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3)\}
+
+There's at least one domain element that is impossible to get to: \((1, 0)\) - an integer
+can't be equivalent to \(1\) mod \(2\) and also \(0\) mod \(4\). So, \(f\) is not necessarily surjective.
+
+\subsection{Question Twenty-One}
+\label{sec:org567c5e4}
+Let \(f : \mathds{Z}_{20} \rightarrow \mathds{Z}_5\) such that \(f([a]_{20}) = [a]_5\). \(f\) is consistent because when
+\([a]_{20} = [b]_{20}_{} (a \equiv b \text{ mod } 20 \Rightarrow a - b = 20n)\), then \([a]_5 = [b]_5\) as
+\(a - b \equiv 20n \text{ mod } 5 = a - b \equiv 0 \text{ mod } 5 \Rightarrow a \equiv b \text{ mod } 5\).
+
+Also, \(f\) is a surjective homomorphism:
+\begin{itemize}
+\item \(f([a]_{20} + [b]_{20}) = f([a + b]_{20}) = [(a + b)]_5 = [a]_5 + [b]_5 = f([a]_{20}) + f([b]_{20})\)
+\item \(f([a]_{20} * [b]_{20}) = f([a * b]_{20}) = [(a * b)]_5 = [a]_5 * [b]_5 = f([a]_{20}) * f([b]_{20})\)
+\item For each \([x]_5\) there exists \([y]_{20}\) such that \(f([y]_{20}) = [x]_5\). The trivial solution is
+that \(x = y\).
+\end{itemize}
+
+Finally, the kernel of \(f\) is the ideal \((5)\) as \(x \in (5) \Rightarrow f(x) = [5n]_5 = [0]_5\). Thus,
+\(\mathds{Z}_{20} / (5)\) is isomorphic to \(\mathds{Z}_5\) by the First Isomorphism Theorem.
+\end{document}
\ No newline at end of file
diff --git a/Homework/math4310/abstract_algebra_assn_12.org b/Homework/math4310/abstract_algebra_assn_12.org
new file mode 100644
index 0000000..6102c0a
--- /dev/null
+++ b/Homework/math4310/abstract_algebra_assn_12.org
@@ -0,0 +1,26 @@
+#+TITLE: Assignment Twelve
+#+AUTHOR: Lizzy Hunt
+#+STARTUP: entitiespretty fold inlineimages
+#+LATEX_HEADER: \notindent \notag \usepackage{ dsfont } \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} \usepackage{polynom} \usepackage{wasysym}
+#+LATEX: \setlength\parindent{0pt}
+#+OPTIONS: toc:nil
+
+* Section 6.3
+** Question One
+$n = cd$ with some $1 < |c| < |n|$ and $1 < |d| < |n|$ since $n$ is composite, so $c$ and $d$ are not multiples of $n$. Therefore as $cd \in (n)$ but $c \notin (n)$ and $d \notin (n)$ then $(n)$ is not a prime ideal
+by definition.
+
+** Question Five
+Both $\mathds{Z}_6$ and $\mathds{Z}_{12}$'s maximal ideals are $(2)$ and $(3)
+** Question Six
+*** a
+The only maximal ideal of $\mathds{Z}_8$ is $(2)$ since it is its prime divisor.
+
+Similarly, the only maximal ideal of $\mathds{Z}_9$ is $(3)$.
+
+*** b
+In $\mathds{Z}_{10}$ the maximal ideals are $(2)$ and $(5)$, similarly for $\mathds{Z}_{15}$: $(3)$ and $(5)$.
+
+** Question Eight
+Consider $(2) \cap (3)$ which generates $(6)$, and is not prime in $\mathds{Z}$; $3 \cdot 2 \in (6)$ but $3 \notin (6)$ and $2 \notin (6)$.
+
diff --git a/Homework/math4310/abstract_algebra_assn_12.pdf b/Homework/math4310/abstract_algebra_assn_12.pdf
new file mode 100644
index 0000000..1a6b5c8
--- /dev/null
+++ b/Homework/math4310/abstract_algebra_assn_12.pdf
diff --git a/Homework/math4310/abstract_algebra_assn_12.tex b/Homework/math4310/abstract_algebra_assn_12.tex
new file mode 100644
index 0000000..0e39503
--- /dev/null
+++ b/Homework/math4310/abstract_algebra_assn_12.tex
@@ -0,0 +1,56 @@
+% Created 2023-04-23 Sun 13:45
+% 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,portrait]{geometry} \usepackage{polynom} \usepackage{wasysym}
+\author{Lizzy Hunt}
+\date{\today}
+\title{Assignment Twelve}
+\hypersetup{
+ pdfauthor={Lizzy Hunt},
+ pdftitle={Assignment Twelve},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.6.1)}, 
+ pdflang={English}}
+\begin{document}
+
+\maketitle
+\setlength\parindent{0pt}
+
+\section{Section 6.3}
+\label{sec:org86da123}
+\subsection{Question One}
+\label{sec:org600275c}
+\(n = cd\) with some \(1 < |c| < |n|\) and \(1 < |d| < |n|\) since \(n\) is composite, so \(c\) and \(d\) are not multiples of \(n\). Therefore as \(cd \in (n)\) but \(c \notin (n)\) and \(d \notin (n)\) then \((n)\) is not a prime ideal
+by definition.
+
+\subsection{Question Five}
+\label{sec:org37fce42}
+Both \(\mathds{Z}_6\) and \(\mathds{Z}_{12}\)'s maximal ideals are \((2)\) and \$(3)
+\subsection{Question Six}
+\label{sec:org18d9056}
+\subsubsection{a}
+\label{sec:org0890b14}
+The only maximal ideal of \(\mathds{Z}_8\) is \((2)\) since it is its prime divisor.
+
+Similarly, the only maximal ideal of \(\mathds{Z}_9\) is \((3)\).
+
+\subsubsection{b}
+\label{sec:org77d60ec}
+In \(\mathds{Z}_{10}\) the maximal ideals are \((2)\) and \((5)\), similarly for \(\mathds{Z}_{15}\): \((3)\) and \((5)\).
+
+\subsection{Question Eight}
+\label{sec:org38c7c8c}
+Consider \((2) \cap (3)\) which generates \((6)\), and is not prime in \(\mathds{Z}\); \(3 \cdot 2 \in (6)\) but \(3 \notin (6)\) and \(2 \notin (6)\).
+\end{document}
\ No newline at end of file
diff --git a/Homework/math4310/abstract_algebra_assn_9.org b/Homework/math4310/abstract_algebra_assn_9.org
new file mode 100644
index 0000000..41fb8a0
--- /dev/null
+++ b/Homework/math4310/abstract_algebra_assn_9.org
@@ -0,0 +1,229 @@
+#+TITLE: Assignment Nine
+#+AUTHOR: Lizzy Hunt
+#+STARTUP: entitiespretty fold inlineimages
+#+LATEX_HEADER: \notindent \notag \usepackage{ dsfont } \usepackage{amsmath} \usepackage[a4paper,margin=1in,landscape]{geometry} \usepackage{polynom} \usepackage{wasysym}
+#+LATEX: \setlength\parindent{0pt}
+#+OPTIONS: toc:nil
+
+* Section 5.1
+** Question One
+*** b
+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*}
+
+*** c
+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*}
+
+** Question Three
+$|${ $0, 1, x, x + 1, x^2, x^2 + 1, x^2 + x, x^2 + x + 1$ }$|$ = 8
+
+** Question Four
+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. $3^3 = 27$
+
+** Question Six
+By Corollary 5.5, all the congruence classes in $F[x]$ are $c \ni c \in F$.
+
+** Question Eight
+\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 $k(x)$, $p(x) | f(x) - g(x) \Rightarrow f(x) \equiv_{p(x)} g(x)$
+
+** Question Eleven
+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_{p(x)} 0_F$.
+
+Still, $f(x)g(x) \equiv_{}_{p(x)} p(x) \equiv_{p(x)} 0_F$ since $p(x) | p(x)$.
+
+* Section 5.2
+** Question One
+The congruence classes are those in Section 5.1, Question Three as above.
+
+| +            | [0]          | [1]          | [x]          | [x + 1]      | [x^2]         | [x^2 + 1]     | [x^2 + x]     | [x^2 + x + 1] |
+| [0]          | [0]          | [1]          | [x]          | [x + 1]      | [x^2]         | [x^2 + 1]     | [x^2 + x]     | [x^2 + x + 1] |
+| [1]          | [1]          | [0]          | [x+1]        | [x]          | [x^2 + 1]     | [x^2]         | [x^2 + x + 1] | [x^2 + x]     |
+| [x]          | [x]          | [x + 1]      | [0]          | [1]          | [x^2 + x]     | [x^2 + x + 1] | [x^2]         | [x^2 + 1]     |
+| [x + 1]      | [x + 1]      | [x]          | [1]          | [0]          | [x^2 + x + 1] | [x^2 + x]     | [x^2 + 1]     | [x^2]         |
+| [x^2]         | [x^2]         | [x^2 + 1]     | [x^2 + x]     | [x^2 + x + 1] | [0]          | [1]          | [x]          | [x + 1]      |
+| [x^2 + 1]     | [x^2 + 1]     | [x^2]         | [x^2 + x + 1] | [x^2 + x]     | [1]          | [0]          | [x + 1]      | [x]          |
+| [x^2 + x]     | [x^2 + x]     | [x^2 + x + 1] | [x^2]         | [x^2 + 1]     | [x]          | [x+1]        | [0]          | [1]          |
+| [x^2 + x + 1] | [x^2 + x + 1] | [x^2 + x]     | [x^2 + 1]     | [x^2]         | [x+1]        | [x]          | [1]          | [0]          |
+
+| \cdot            | [0] | [1]          | [x]      | [x + 1]  | [x^2]     | [x^2 + 1] | [x^2 + x] | [x^2 + x + 1] |
+| [0]          | [0] | [0]          | [0]      | [0]      | [0]      | [0]      | [0]      | [0]          |
+| [1]          | [0] | [1]          | [x]      | [x + 1]  | [x^2]     | [x^2 + 1] | [x^2 + x] | [x^2 + x + 1] |
+| [x]          | [0] | [x]          | [x^2]     | [x^2+x]   | [x+1]    | [1]      | [x^2+x+1] | [x^2+1]       |
+| [x + 1]      | [0] | [x + 1]      | [x^2 + x] | [x^2+1]   | [x^2+x+1] | [x^2]     | [1]      | [x]          |
+| [x^2]         | [0] | [x^2]         | [x+1]    | [x^2+x+1] | [x^2+x]   | [x]      | [x^2+1]   | [1]          |
+| [x^2 + 1]     | [0] | [x^2 + 1]     | [1]      | [x^2]     | [x]      | [x^2+x+1] | [x+1]    | [x^2+x]       |
+| [x^2 + x]     | [0] | [x^2 + x]     | [x^2+x+1] | [1]      | [x^2+1]   | [x+1]    | [x]      | [x+1]        |
+| [x^2 + x + 1] | [0] | [x^2 + x + 1] | [x^2+1]   | [x]      | [1]      | [x^2+x]   | [x^2]     | [x+1]        |
+
+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).
+  
+** Question Two
+| +      | [0]    | [1]    | [2]    | [x]    | [x+1]  | [x+2]  | [2x]   | [2x+1] | [2x+2] |
+| [0]    | [0]    | [1]    | [2]    | [x]    | [x+1]  | [x+2]  | [2x]   | [2x+1] | [2x+2] |
+| [1]    | [1]    | [2]    | [0]    | [x+1]  | [x+2]  | [x]    | [2x+1] | [2x+2] | [2x]   |
+| [2]    | [2]    | [0]    | [1]    | [x+2]  | [x]    | [x+1]  | [2x+2] | [2x]   | [2x+1] |
+| [x]    | [x]    | [x+1]  | [x+2]  | [2x]   | [2x+1] | [2x+2] | [0]    | [1]    | [2]    |
+| [x+1]  | [x+1]  | [x+2]  | [x]    | [2x+1] | [2x+2] | [2x]   | [1]    | [2]    | [0]    |
+| [x+2]  | [x+2]  | [x]    | [x+1]  | [2x+2] | [2x]   | [2x+1] | [2]    | [0]    | [1]    |
+| [2x]   | [2x]   | [2x+1] | [2x+2] | [0]    | [1]    | [2]    | [x]    | [x+1]  | [1]    |
+| [2x+1] | [2x+1] | [2x+2] | [2x]   | [1]    | [2]    | [0]    | [x+1]  | [x+2]  | [x]    |
+| [2x+2] | [2x+2] | [2x]   | [2x+1] | [2]    | [0]    | [1]    | [x+2]  | [x]    | [x+1]  |
+
+
+| \cdot      | [0] | [1]    | [2]    | [x]    | [x+1]  | [x+2]  | [2x]   | [2x+1] | [2x+2] |
+| [0]    | [0] | [0]    | [0]    | [0]    | [0]    | [0]    | [0]    | [0]    | [0]    |
+| [1]    | [0] | [1]    | [2]    | [x]    | [x+1]  | [x+2]  | [2x]   | [2x+1] | [2x+2] |
+| [2]    | [0] | [2]    | [1]    | [2x]   | [2x+2] | [2x+1] | [x]    | [x+2]  | [x+1]  |
+| [x]    | [0] | [x]    | [2x]   | [2]    | [x+2]  | [2x+2] | [1]    | [x+1]  | [2x+1] |
+| [x+1]  | [0] | [x+1]  | [2x+2] | [x+2]  | [2x]   | [1]    | [2x+1] | [2]    | [x]    |
+| [x+2]  | [0] | [x+2]  | [2x+1] | [2x+2] | [1]    | [x]    | [x+1]  | [2x]   | [2]    |
+| [2x]   | [0] | [2x]   | [x]    | [1]    | [2x+1] | [x+1]  | [2]    | [2x+2] | [x+2]  |
+| [2x+1] | [0] | [2x+1] | [x+2]  | [x+1]  | [2]    | [2x]   | [2x+2] | [x]    | [1]    |
+| [2x+2] | [0] | [2x+2] | [x+1]  | [2x+1] | [x]    | [2]    | [x+2]  | [1]    | [2x]   |
+
+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).
+
+** Question Three
+
+| +     | [0]   | [1]   | [x]   | [x+1] |
+| [0]   | [0]   | [1]   | [x]   | [x+1] |
+| [1]   | [1]   | [0]   | [x+1] | [x]   |
+| [x]   | [x]   | [x+1] | [0]   | [1]   |
+| [x+1] | [x+1] | [x]   | [1]   | [0]   |
+
+| \cdot     | [0] | [1]   | [x]   | [x+1] |
+| [0]   | [0] | [0]   | [0]   | [0]   |
+| [1]   | [0] | [1]   | [x]   | [x+1] |
+| [x]   | [0] | [x]   | [1]   | [x+1] |
+| [x+1] | [0] | [x+1] | [x+1] | [0]   |
+
+Not, this is _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).
+
+** Question Six
+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 $[ax + b] \cdot [cx + d] = [(ad + bc)x + (2ac + bd)]$
+
+** Question Nine
+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 5.3
+** Question One
+*** a
+$x^3 + 2x^2 + x + 1$ does not have any roots in \mathds{Z}_3, so by Corollary 4.19 it must be irreducible,
+and thus a field by 5.10
+
+*** b
+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.
+
+*** c
+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.
+
+** Question Two
+*** a
+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*}
+
+*** b
+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 \smiley):
+
++ $f$ is injective since $f([ax + b]) = f([cx + d]) \Rightarrow a + b\sqrt{2} = c + d\sqrt{2} \Rightarrow a = c \wedge b = d$
++ $f$ is surjective since each $a + b\sqrt{2}$ is uniquely mapped to $[ax + b]$
++ $f([ax + b] + [cx + d]) = f([(a + c)x + (b + d)]) = (a + c) + (b + d)\sqrt{2} = (a + b\sqrt{2}) + (c + d\sqrt{2}) = f([ax + b]) + f([cx + d])$
+  
++ And via Question Six from section 5.2 above,
+   $f([ax + b] \cdot [cx + d]) = f([(ad + bc)x + (2ac + bd)]) = (ad + bc) + (2ac + bd)\sqrt{2} = (a + b\sqrt{2})(c + d\sqrt{2}) = f([ax + b])f([cx + d])$
+
+** Question Five
+*** a
+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*}
+
+*** b
+**** Quick Lemma
+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 $[ax + b] \cdot [cx + d] = [(ad + bc)x + (3ac + bd)]$
+
+**** Yeah, it's an isomorphism 
+
+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 \smiley):
+
++ $f$ is injective since $f([ax + b]) = f([cx + d]) \Rightarrow a + b\sqrt{3} = c + d\sqrt{3} \Rightarrow a = c \wedge b = d$
++ $f$ is surjective since each $a + b\sqrt{3}$ is uniquely mapped to $[ax + b]$
++ $f([ax + b] + [cx + d]) = f([(a + c)x + (b + d)]) = (a + c) + (b + d)\sqrt{3} = (a + b\sqrt{3}) + (c + d\sqrt{3}) = f([ax + b]) + f([cx + d])$
++ And from the lemma, $f([ax + b] \cdot [cx + d]) = f([(ad + bc)x + (3ac + bd)]) = (ad + bc) + (3ac + bd)\sqrt{3} = (a + b\sqrt{3})(c + d\sqrt{3}) = f([ax + b])f([cx + d])$
diff --git a/Homework/math4310/abstract_algebra_assn_9.pdf b/Homework/math4310/abstract_algebra_assn_9.pdf
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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
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+            {Lizzy Hunt}^^M
+*** (cannot \read from terminal in nonstop modes)
+
+ 
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diff --git a/Homework/math4310/abstract_algebra_midterm_2.org b/Homework/math4310/abstract_algebra_midterm_2.org
new file mode 100644
index 0000000..b4ba650
--- /dev/null
+++ b/Homework/math4310/abstract_algebra_midterm_2.org
@@ -0,0 +1,141 @@
+#+TITLE: Midterm 2
+#+AUTHOR: Lizzy Hunt
+#+STARTUP: entitiespretty fold inlineimages
+#+LATEX_HEADER: \notindent \notag \usepackage{ dsfont } \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} \usepackage{polynom} \usepackage{wasysym}
+#+LATEX: \setlength\parindent{0pt}
+#+OPTIONS: toc:nil
+
+* Question One
+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, $a(x) = b(x)q(x) + r(x) = (2x^2 + 1)(2x) + (2x + 1)$
+
+* Question Two
+\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 $\mathds{Z}_3[x]$
+
+* Question Three
+We will use the Euclidean algorithm to find the GCD:
+** GCD
+\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)
+
+** (supplement) Division Algorithm Work
+\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 $\frac{45}{16} \equiv_5 0$.
+
+* Question Four
+** a
+$\mathds{Z}_3[x]$ 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.
+
+** b
+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 $\mathds{Z}[3]$.
+
+* Question Five
+Firstly, $\mathds{Z}_5[x]$ is a field by Theorem 2.8.
+
+** a
+There are no roots of $x^2 + 2x + 3$ in $\mathds{Z}_5$, so by Theorem 5.10 it is irreducible.
+
+$f_1 : (0, 1, 2, 3, 4) \rightarrow (3, 1, 1, 3, 2)$
+
+** b
+This is reducible since $1$ is a root.
+
+\begin{equation*}
+\polylongdiv[style=A]{x^3 + x + 3}{x - 1}
+\end{equation*}
+
+Thus, $f_2(x) = (x + 4)(x^2 + x + 2)$
+
+
+* Question Six
+Following pages 137 - 138 of the book...
+
+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.
+
diff --git a/Homework/math4310/abstract_algebra_midterm_2.pdf b/Homework/math4310/abstract_algebra_midterm_2.pdf
new file mode 100644
index 0000000..a2f8126
--- /dev/null
+++ b/Homework/math4310/abstract_algebra_midterm_2.pdf
diff --git a/Homework/math4310/abstract_algebra_midterm_2.tex b/Homework/math4310/abstract_algebra_midterm_2.tex
new file mode 100644
index 0000000..a3f0a2f
--- /dev/null
+++ b/Homework/math4310/abstract_algebra_midterm_2.tex
@@ -0,0 +1,176 @@
+% 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}
\ No newline at end of file
diff --git a/Homework/math4310/alg_structures_assn_2.org b/Homework/math4310/alg_structures_assn_2.org
new file mode 100644
index 0000000..d7dee4c
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_2.org
@@ -0,0 +1,135 @@
+#+TITLE: Assignment Two
+#+AUTHOR: Logan Hunt
+#+STARTUP: entitiespretty fold inlineimages
+#+LATEX_HEADER: \notindent \notag \usepackage{ dsfont } \usepackage{amsmath}
+#+LATEX: \setlength\parindent{0pt}
+
+
+* Section 1.3
+** Question 3
+701, 1009, 1949, 1951 are all prime
+** Question 7
+\begin{align*}
+p | a \Rightarrow np &= a \\
+p | a + bc \Rightarrow mp &= a + bc \\
+mp &= np + bc \\
+mp - np &= bc \\
+p(m - n) &= bc 
+\end{align*}
+
+Since $bc$ is a multiple of $p$, and $p$ is prime, by Theorem 1.5, $p | b$ or $p | c$.
+** Question 15
+If $p | a^n \Rightarrow p | a \cdot a \dotsc a$ then by corollary 1.6, $p | a$. Then, $a = pm$ and
+$a^n = p^n \cdot m^n$. $p^n$ is a factor of $a^n$ and thus $p^n | a^n$.
+
+** Question 17
+
+From the Fundamental Theorem of Arithmetic, both $a$ and $b$ must be a product of primes such that
+$a = (p \cdot p_1 \cdot p_2 \dotsc p_i)$ and $b = (p \cdot q_1 \cdot q_2 \dotsc q_j)$ with each $p_i$ and $q_j$ being prime.
+
+Then, $a^2 = (p \cdot p_1 \cdot p_2 \dotsc p_r) \cdot (p \cdot p_1 \cdot p_2 \dots p_r)$ and $b^2 = (p \cdot q_1 \cdot q_2 \dotsc q_j) \cdot (p \cdot q_1 \cdot q_2 \dotsc q_j)$.
+$p^2$ is then a common factor of $a^2$ and $b^2$, and since each other factor ($p_i^2$, $q_i^2$) is a square of a prime, there
+can be no other greater common divisor than $p^2$.
+
+Therefore, $(a, b) = p \Rightarrow (a^2, b^2) = p^2$ when p is prime.
+
+** Question 30
+*** a
+Firstly assume that there are $a,b \ni a^2 = 2b^2$. Then, $a^2 = p_1 p_2 \dotsc p_r$ and $2b^2 = q_1 q_2 \dotsc q_s$,
+and by the Fundamental Theorem of Arithmetic, every $p_i = \pm q_j$. Since $a^2$ is even as it is equal
+to $2b^2$, let $p_1$ be the factor corresponding to a power of $2$. Then $p_1 = 2^n$ and $n$ must be a
+multiple of $2$ since $a^2 = 2^n \dotsc \Rightarrow a = 2^{\frac{n}{2}} \dotsc$
+
+However, $2b^2 = \pm 2^n \dotsc$ implies that $b^2 = \pm 2^{n-1} \dotsc$ and from similar reasoning $n-1$ must also
+be a multiple of $2$.
+
+This is a contradiction - not both $n$ and $n-1$ can be multiples of $2$!
+
+*** b
+Just reformat it:
+
+\begin{equation*}
+\sqrt{2} = \frac{a}{b} \Rightarrow 2 = \frac{a^2}{b^2} \Rightarrow 2b^2 = a^2
+\end{equation*}
+
+
+* Section 2.1
+** Question 2
+*** a
+\begin{equation*}
+6k + 5 \equiv_4 6 + 5 \equiv_4 11 \equiv_4 3
+\end{equation*}
+
+*** b
+\begin{equation*}
+2r + 3s \equiv_{10} 2(3) + 3(-7) \equiv_{10} 6 - 21 \equiv_{10} -15 \equiv_{10} 5
+\end{equation*}
+
+** Question 3b
+\begin{align*}
+& 10(0) + 9(0) + 8(3) + 7(1) + 6(1) + 5(0) + 4(5) + 3(5) + 2(9) + 5 \\
+&\equiv_{11 }24 + 7 + 6 + 20 + 15 + 18 + 5 \\
+&\equiv_{11} 95 \\
+&\equiv_{11} 7
+\end{align*}
+
+Invalid ISBN
+
+** Question 5
+*** a
+Theorem 2.2 states that if $a \equiv_4 b$, and $c \equiv_4 d$, then $ac \equiv_4 bd$.
+
+Since $5 \equiv_4 1$ and $5 \cdot 5 \equiv_4 1 \cdot 1$, then $5 \cdot 5 \cdot 5 \equiv_4 1 \cdot 1 \cdot 1$.
+
+We can continue chaining these together until we find $5^{2000} \equiv_4 1^{2000}$, and thus [5^{2000}] = [1] in $\mathds{Z}_4$.
+*** b
+By repeating the same process as in a, $4 \equiv_5 4$ and $4^2 \equiv_5 1$. Then, $4^3 \equiv_5 4 \cdot 1$. Then, $4^4 \equiv_5 4^2 \Rightarrow 4^4 \equiv_5 1$.
+
+In general as we keep chaining on and because of Theorem 2.2, even powers of 4 will be equivalent mod 5 to 1,
+and odd powers of 4 will be equivalent mod 5 to 4.
+
+Therefore, $4^{2001} \equiv_5 1$ and $[4^{2001}] = [1]$ in $\mathds{Z}_5$.
+** Question 7
+If $a \in \mathds{Z}$ then $a \equiv_4 m$ with $m \in {1,2,3,4}$. Then, by Theorem 2.2 again, $a^2 \equiv_4 m^2$.
+
+$[m^2] \in \mathds{Z}_4$ must be equivalent to any ${[1^2], [2^2], [3^2], [4^2]} = {[1], [0], [1], [0]}$.
+
+Therefore, $[a^2]$ cannot be in ${[3], [4]} \subset {[2], [3], [4]}$
+
+** Question 14
+*** a
+A simple python script will prove this is false:
+
+\begin{verbatim}
+seen = set()
+for n in range(1, 10):
+  for a in range(0, 10):
+    for b in range(0, 10):
+      seenfoo = (a, b, n) in seen or (b, a, n) in seen
+      if (a * b) % n == 0 and a % n != 0 and b % n != 0 and not seenfoo:
+        seen.add((a, b, n))
+        print(f"a={a}, b={b}, n={n}")
+\end{verbatim}
+
+And we receive several counterexamples:
+
+\begin{verbatim}
+a=2, b=2, n=4
+a=2, b=6, n=4
+a=6, b=6, n=4
+a=2, b=3, n=6
+a=2, b=9, n=6
+...
+a=2, b=4, n=8
+a=4, b=6, n=8
+a=3, b=3, n=9
+a=3, b=6, n=9
+a=6, b=6, n=9
+\end{verbatim}
+
+*** b
+If $ab \equiv_n 0$, then $ab = mn + 0$ for some $m \in \mathds{Z}$ by definition.
+
+Then, $ab = mn$ implies that $ab$ is a multiple of $n$, and since $n$ is prime, then by Theorem 1.8, $n | ab$ implies that $n | a$ or $n | b$.
+
+
diff --git a/Homework/math4310/alg_structures_assn_2.pdf b/Homework/math4310/alg_structures_assn_2.pdf
new file mode 100644
index 0000000..cd405a0
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_2.pdf
diff --git a/Homework/math4310/alg_structures_assn_2.tex b/Homework/math4310/alg_structures_assn_2.tex
new file mode 100644
index 0000000..fa7ea3e
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_2.tex
@@ -0,0 +1,180 @@
+% Created 2023-01-25 Wed 08:50
+% 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 \notga \usepackage{ dsfont } \usepackage{amsmath}
+\author{Logan Hunt}
+\date{\today}
+\title{Assignment Two}
+\hypersetup{
+ pdfauthor={Logan Hunt},
+ pdftitle={Assignment Two},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.5.5)}, 
+ pdflang={English}}
+\begin{document}
+
+\maketitle
+\tableofcontents
+
+\setlength\parindent{0pt}
+
+
+\section{Section 1.3}
+\label{sec:orgb642b1a}
+\subsection{Question 3}
+\label{sec:orgefa83b9}
+701, 1009, 1949, 1951 are all prime
+\subsection{Question 7}
+\label{sec:org19f664b}
+\begin{align*}
+p | a \Rightarrow np &= a \\
+p | a + bc \Rightarrow mp &= a + bc \\
+mp &= np + bc \\
+mp - np &= bc \\
+p(m - n) &= bc 
+\end{align*}
+
+Since \(bc\) is a multiple of \(p\), and \(p\) is prime, by Theorem 1.5, \(p | b\) or \(p | c\).
+\subsection{Question 15}
+\label{sec:orgec25485}
+If \(p | a^n \Rightarrow p | a \cdot a \dotsc a\) then by corollary 1.6, \(p | a\). Then, \(a = pm\) and
+\(a^n = p^n \cdot m^n\). \(p^n\) is a factor of \(a^n\) and thus \(p^n | a^n\).
+
+\subsection{Question 17}
+\label{sec:orgf0ee3ce}
+
+From the Fundamental Theorem of Arithmetic, both \(a\) and \(b\) must be a product of primes such that
+\(a = (p \cdot p_1 \cdot p_2 \dotsc p_i)\) and \(b = (p \cdot q_1 \cdot q_2 \dotsc q_j)\) with each \(p_i\) and \(q_j\) being prime.
+
+Then, \(a^2 = (p \cdot p_1 \cdot p_2 \dotsc p_r) \cdot (p \cdot p_1 \cdot p_2 \dotsc p_r)\) and \(b^2 = (p \cdot q_1 \cdot q_2 \dotsc q_j) \cdot (p \cdot q_1 \cdot q_2 \dotsc q_j)\).
+\(p^2\) is then a common factor of \(a^2\) and \(b^2\), and since each other factor (\(p_i^2\), \(q_i^2\)) is a square of a prime, there
+can be no other greater common divisor than \(p^2\).
+
+Therefore, \((a, b) = p \Rightarrow (a^2, b^2) = p^2\) when p is prime.
+
+\subsection{Question 30}
+\label{sec:org711c1fd}
+\subsubsection{a}
+\label{sec:org622b53c}
+Firstly assume that there are \(a,b \ni a^2 = 2b^2\). Then, \(a^2 = p_1 p_2 \dotsc p_r\) and \(2b^2 = q_1 q_2 \dotsc q_s\),
+and by the Fundamental Theorem of Arithmetic, every \(p_i = \pm q_j\). Since \(a^2\) is even as it is equal
+to \(2b^2\), let \(p_1\) be the factor corresponding to a power of \(2\). Then \(p_1 = 2^n\) and \(n\) must be a
+multiple of \(2\) since \(a^2 = 2^n \dotsc \Rightarrow a = 2^{\frac{n}{2}} \dotsc\)
+
+However, \(2b^2 = \pm 2^n \dotsc\) implies that \(b^2 = \pm 2^{n-1} \dotsc\) and from similar reasoning \(n-1\) must also
+be a multiple of \(2\).
+
+This is a contradiction - not both \(n\) and \(n-1\) can be multiples of \(2\)!
+
+\subsubsection{b}
+\label{sec:orgfba7c43}
+Just reformat it:
+
+\begin{equation*}
+\sqrt{2} = \frac{a}{b} \Rightarrow 2 = \frac{a^2}{b^2} \Rightarrow 2b^2 = a^2
+\end{equation*}
+
+
+\section{Section 2.1}
+\label{sec:orgd42fe0f}
+\subsection{Question 2}
+\label{sec:org8c3b581}
+\subsubsection{a}
+\label{sec:orgc7b35b9}
+\begin{equation*}
+6k + 5 \equiv_4 6 + 5 \equiv_4 11 \equiv_4 3
+\end{equation*}
+
+\subsubsection{b}
+\label{sec:org07ec8aa}
+\begin{equation*}
+2r + 3s \equiv_{10} 2(3) + 3(-7) \equiv_{10} 6 - 21 \equiv_{10} -15 \equiv_{10} 5
+\end{equation*}
+
+\subsection{Question 3b}
+\label{sec:org29b1c99}
+\begin{align*}
+& 10(0) + 9(0) + 8(3) + 7(1) + 6(1) + 5(0) + 4(5) + 3(5) + 2(9) + 5 \\
+&\equiv_{11 }24 + 7 + 6 + 20 + 15 + 18 + 5 \\
+&\equiv_{11} 95 \\
+&\equiv_{11} 7
+\end{align*}
+
+Invalid ISBN
+
+\subsection{Question 5}
+\label{sec:org37d83f3}
+\subsubsection{a}
+\label{sec:org770db68}
+Theorem 2.2 states that if \(a \equiv_4 b\), and \(c \equiv_4 d\), then \(ac \equiv_4 bd\).
+
+Since \(5 \equiv_4 1\) and \(5 \cdot 5 \equiv_4 1 \cdot 1\), then \(5 \cdot 5 \cdot 5 \equiv_4 1 \cdot 1 \cdot 1\).
+
+We can continue chaining these together until we find \(5^{2000} \equiv_4 1^{2000}\), and thus [5\textsuperscript{2000}] = [1] in \(\mathds{Z}_4\).
+\subsubsection{b}
+\label{sec:org12acb79}
+By repeating the same process as in a, \(4 \equiv_5 4\) and \(4^2 \equiv_5 1\). Then, \(4^3 \equiv_5 4 \cdot 1\). Then, \(4^4 \equiv_5 4^2 \Rightarrow 4^4 \equiv_5 1\).
+
+In general as we keep chaining on and because of Theorem 2.2, even powers of 4 will be equivalent mod 5 to 1,
+and odd powers of 4 will be equivalent mod 5 to 4.
+
+Therefore, \(4^{2001} \equiv_5 1\) and \([4^{2001}] = [1]\) in \(\mathds{Z}_5\).
+\subsection{Question 7}
+\label{sec:orgc6a9940}
+If \(a \in \mathds{Z}\) then \(a \equiv_4 m\) with \(m \in {1,2,3,4}\). Then, by Theorem 2.2 again, \(a^2 \equiv_4 m^2\).
+
+\([m^2] \in \mathds{Z}_4\) must be equivalent to any \({[1^2], [2^2], [3^2], [4^2]} = {[1], [0], [1], [0]}\).
+
+Therefore, \([a^2]\) cannot be in \({[3], [4]} \subset {[2], [3], [4]}\)
+
+\subsection{Question 14}
+\label{sec:org0f658a6}
+\subsubsection{a}
+\label{sec:org9ee67bf}
+A simple python script will prove this is false:
+
+\begin{verbatim}
+seen = set()
+for n in range(1, 10):
+  for a in range(0, 10):
+    for b in range(0, 10):
+      seenfoo = (a, b, n) in seen or (b, a, n) in seen
+      if (a * b) % n == 0 and a % n != 0 and b % n != 0 and not seenfoo:
+        seen.add((a, b, n))
+        print(f"a={a}, b={b}, n={n}")
+\end{verbatim}
+
+And we receive several counterexamples:
+
+\begin{verbatim}
+a=2, b=2, n=4
+a=2, b=6, n=4
+a=6, b=6, n=4
+a=2, b=3, n=6
+a=2, b=9, n=6
+...
+a=2, b=4, n=8
+a=4, b=6, n=8
+a=3, b=3, n=9
+a=3, b=6, n=9
+a=6, b=6, n=9
+\end{verbatim}
+
+\subsubsection{b}
+\label{sec:org767f9d7}
+If \(ab \equiv_n 0\), then \(ab = mn + 0\) for some \(m \in \mathds{Z}\) by definition.
+
+Then, \(ab = mn\) implies that \(ab\) is a multiple of \(n\), and since \(n\) is prime, then by Theorem 1.8, \(n | ab\) implies that \(n | a\) or \(n | b\).
+\end{document}
\ No newline at end of file
diff --git a/Homework/math4310/alg_structures_assn_3.org b/Homework/math4310/alg_structures_assn_3.org
new file mode 100644
index 0000000..c0165a0
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_3.org
@@ -0,0 +1,115 @@
+#+TITLE: Assignment Three
+#+AUTHOR: Logan Hunt
+#+STARTUP: entitiespretty fold inlineimages
+#+LATEX_HEADER: \notindent \notag \usepackage{ dsfont } \usepackage{amsmath}
+#+LATEX: \setlength\parindent{0pt}
+#+OPTIONS: toc:nil
+
+* Section 2.2
+** Question One
+*** b
+| \oplus | 0 | 1 | 2 | 3 |
+|  0 | 0 | 1 | 2 | 3 |
+|  1 | 1 | 2 | 3 | 0 |
+|  2 | 2 | 3 | 0 | 1 |
+|  3 | 3 | 0 | 1 | 2 |
+
+| \odot | 0 | 1 | 2 | 3 |
+| 0 | 0 | 0 | 0 | 0 |
+| 1 | 0 | 1 | 2 | 3 |
+| 2 | 0 | 2 | 0 | 2 |
+| 3 | 0 | 3 | 2 | 1 |
+
+*** c
+| \oplus | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
+|  0 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
+|  1 | 1 | 2 | 3 | 4 | 5 | 6 | 0 |
+|  2 | 2 | 3 | 4 | 5 | 6 | 0 | 1 |
+|  3 | 3 | 4 | 5 | 6 | 0 | 1 | 2 |
+|  4 | 4 | 5 | 6 | 0 | 1 | 2 | 3 |
+|  5 | 5 | 6 | 0 | 1 | 2 | 3 | 4 |
+|  6 | 6 | 0 | 1 | 2 | 3 | 4 | 5 |
+
+| \odot | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
+| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
+| 1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
+| 2 | 0 | 2 | 4 | 6 | 1 | 3 | 5 |
+| 3 | 0 | 3 | 6 | 2 | 5 | 1 | 4 |
+| 4 | 0 | 4 | 1 | 5 | 2 | 6 | 3 |
+| 5 | 0 | 5 | 3 | 1 | 6 | 4 | 2 |
+| 6 | 0 | 6 | 5 | 4 | 3 | 2 | 1 |
+
+** Question Three
+$x = [1], [3], [5], [7]$
+
+** Question Five
+$x = [1], [2], [4], [5]$
+
+** Question Eight
+$x = [1], [2], [6], [7]$
+
+** Question Eleven
+*** b
+$x = [0], [1], [2], [3]$
+** Question Fifteen
+*** c
+From the Binomial Theorem,
+
+\begin{align*}
+(a + b)^5 &= a^5 + 5a^4 b + 10a^3 b^2 + 10a^2 b^3 + 5ab^4 + b^5
+\end{align*}
+
+Then,
+\begin{equation*}
+(a + b)^5 &\equiv_5 (a^5 + 5a^4 b + 10a^3 b^2 + 10a^2 b^3 + 5ab^4 + b^5) \\
+&\equiv_5 (a^5 + b^5) \\
+\end{equation*}
+
+since each of the terms $5a^4 b, 10a^3 b^2, 10a^2 b^3, 5ab^4$ are zero as $5 \equiv_5 0$ and $10 \equiv_5 0$.
+
+
+
+* Section 2.3
+** Question One
+*** b
+$[1], [3], [5], [7]$ since $[7 * 7] = [49] = [1]$, $[5 * 5] = [25] = [1]$, $[3 * 3] = [9] = [1]$, and
+$[1 * 1] = [1]$.
+** Question Two
+*** b
+$[2], [4], [6]$ since $[2 * 4] = [8] = [0]$, $[4 * 4] = [16] = [0]$, and $[6 * 4] = [24] = [0]$.
+** Question Eight
+*** a
+1. $2x = 1$
+2. $2x = 3$
+3. $2x = 5$
+*** b
+Yes, each one is equivalent to 0 when $x = 6$.
+** Question Nine
+*** a
+By definition, there exists $b$, the inverse of $a$, such that $ab = 1$.
+Assume that $a$ is a zero divisor, then there exists $c \neq 0$
+such that $ac = 0$. Then, $(ab)c = 0b \Rightarrow (1)(c) = 0$ implies that $c = 0$, which is a contradiction.
+*** b
+By definition, there exists $b$, with $b \neq 0$ such that $ab = 0$.
+Assume that $a$ is a unit, then there exists $c$ such that $ac = 1$.
+Then, $(b)ac = 1b \Rightarrow 0c = b$ implies that $b = 0$, which is a contradiction.
+
+** Question Eleven
+By definition of $a$ being a unit, there exists $ay = 1$ with $y$ being an inverse of $a$.
+By multiplying our target $(y)ax = (y)b \Rightarrow x = yb$.
+
+To prove this is unique, assume that $k$ and $l$ are solutions of $ax = b$. Then, $ak = b$ and $al = b$. Since $a$ is a unit, by using our
+previous strategy, $(y)ak = (y)b$ and $(y)al = (y)b$, so $k = yb$ and $l = yb$ and thus $k = l$.
+
+
+* Chapter 13
+** A2
+As $p | c \Rightarrow c = pk$, and $p | c \Rightarrow c = ql$ then $pk = ql$ and thus by Theorem 1.5 since $p$ and $q$ are prime $p | l$
+or $p | q$ and $q | p$ or $q | k$, but since $p$ and $q$ are prime, then it must be that only $p | l$ and $q | k$.
+
+Since $p | l$ then $l = mp$ and $c = ql \Rightarrow c = qmp$ and $qp$ is a factor of $c$.
+
+** A3 (a)
+GO = 0715
+
+715^3 (mod 2773) = 107
diff --git a/Homework/math4310/alg_structures_assn_3.pdf b/Homework/math4310/alg_structures_assn_3.pdf
new file mode 100644
index 0000000..99e66c1
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_3.pdf
diff --git a/Homework/math4310/alg_structures_assn_3.tex b/Homework/math4310/alg_structures_assn_3.tex
new file mode 100644
index 0000000..bb9c222
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_3.tex
@@ -0,0 +1,183 @@
+% Created 2023-01-31 Tue 22:49
+% 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}
+\author{Logan Hunt}
+\date{\today}
+\title{Assignment Three}
+\hypersetup{
+ pdfauthor={Logan Hunt},
+ pdftitle={Assignment Three},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.6.1)}, 
+ pdflang={English}}
+\begin{document}
+
+\maketitle
+\setlength\parindent{0pt}
+
+\section{Section 2.2}
+\label{sec:orgfecc6e5}
+\subsection{Question One}
+\label{sec:orgd22e42b}
+\subsubsection{b}
+\label{sec:org757c583}
+\begin{center}
+\begin{tabular}{rrrrr}
+\(\oplus\) & 0 & 1 & 2 & 3\\[0pt]
+0 & 0 & 1 & 2 & 3\\[0pt]
+1 & 1 & 2 & 3 & 0\\[0pt]
+2 & 2 & 3 & 0 & 1\\[0pt]
+3 & 3 & 0 & 1 & 2\\[0pt]
+\end{tabular}
+\end{center}
+
+\begin{center}
+\begin{tabular}{rrrrr}
+\(\odot\) & 0 & 1 & 2 & 3\\[0pt]
+0 & 0 & 0 & 0 & 0\\[0pt]
+1 & 0 & 1 & 2 & 3\\[0pt]
+2 & 0 & 2 & 0 & 2\\[0pt]
+3 & 0 & 3 & 2 & 1\\[0pt]
+\end{tabular}
+\end{center}
+
+\subsubsection{c}
+\label{sec:org603554a}
+\begin{center}
+\begin{tabular}{rrrrrrrr}
+\(\oplus\) & 0 & 1 & 2 & 3 & 4 & 5 & 6\\[0pt]
+0 & 0 & 1 & 2 & 3 & 4 & 5 & 6\\[0pt]
+1 & 1 & 2 & 3 & 4 & 5 & 6 & 0\\[0pt]
+2 & 2 & 3 & 4 & 5 & 6 & 0 & 1\\[0pt]
+3 & 3 & 4 & 5 & 6 & 0 & 1 & 2\\[0pt]
+4 & 4 & 5 & 6 & 0 & 1 & 2 & 3\\[0pt]
+5 & 5 & 6 & 0 & 1 & 2 & 3 & 4\\[0pt]
+6 & 6 & 0 & 1 & 2 & 3 & 4 & 5\\[0pt]
+\end{tabular}
+\end{center}
+
+\begin{center}
+\begin{tabular}{rrrrrrrr}
+\(\odot\) & 0 & 1 & 2 & 3 & 4 & 5 & 6\\[0pt]
+0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\[0pt]
+1 & 0 & 1 & 2 & 3 & 4 & 5 & 6\\[0pt]
+2 & 0 & 2 & 4 & 6 & 1 & 3 & 5\\[0pt]
+3 & 0 & 3 & 6 & 2 & 5 & 1 & 4\\[0pt]
+4 & 0 & 4 & 1 & 5 & 2 & 6 & 3\\[0pt]
+5 & 0 & 5 & 3 & 1 & 6 & 4 & 2\\[0pt]
+6 & 0 & 6 & 5 & 4 & 3 & 2 & 1\\[0pt]
+\end{tabular}
+\end{center}
+
+\subsection{Question Three}
+\label{sec:org6ac99a4}
+\(x = [1], [3], [5], [7]\)
+
+\subsection{Question Five}
+\label{sec:org69e894a}
+\(x = [1], [2], [4], [5]\)
+
+\subsection{Question Eight}
+\label{sec:org26350fa}
+\(x = [1], [2], [6], [7]\)
+
+\subsection{Question Eleven}
+\label{sec:orgcd603cc}
+\subsubsection{b}
+\label{sec:org2896606}
+\(x = [0], [1], [2], [3]\)
+\subsection{Question Fifteen}
+\label{sec:org5bca8ba}
+\subsubsection{c}
+\label{sec:org830aeb3}
+From the Binomial Theorem,
+
+\begin{align*}
+(a + b)^5 &= a^5 + 5a^4 b + 10a^3 b^2 + 10a^2 b^3 + 5ab^4 + b^5
+\end{align*}
+
+Then,
+\begin{equation*}
+(a + b)^5 &\equiv_5 (a^5 + 5a^4 b + 10a^3 b^2 + 10a^2 b^3 + 5ab^4 + b^5) \\
+&\equiv_5 (a^5 + b^5) \\
+\end{equation*}
+
+since each of the terms \(5a^4 b, 10a^3 b^2, 10a^2 b^3, 5ab^4\) are zero as \(5 \equiv_5 0\) and \(10 \equiv_5 0\).
+
+
+
+\section{Section 2.3}
+\label{sec:org7ebf66c}
+\subsection{Question One}
+\label{sec:orge906711}
+\subsubsection{b}
+\label{sec:orgddb4d23}
+\([1], [3], [5], [7]\) since \([7 * 7] = [49] = [1]\), \([5 * 5] = [25] = [1]\), \([3 * 3] = [9] = [1]\), and
+\([1 * 1] = [1]\).
+\subsection{Question Two}
+\label{sec:org4efa77f}
+\subsubsection{b}
+\label{sec:org41d490d}
+\([2], [4], [6]\) since \([2 * 4] = [8] = [0]\), \([4 * 4] = [16] = [0]\), and \([6 * 4] = [24] = [0]\).
+\subsection{Question Eight}
+\label{sec:org27e3ed4}
+\subsubsection{a}
+\label{sec:orgeac97d5}
+\begin{enumerate}
+\item \(2x = 1\)
+\item \(2x = 3\)
+\item \(2x = 5\)
+\end{enumerate}
+\subsubsection{b}
+\label{sec:orgb99f90c}
+Yes, each one is equivalent to 0 when \(x = 6\).
+\subsection{Question Nine}
+\label{sec:orge8d0de5}
+\subsubsection{a}
+\label{sec:org8045c18}
+By definition, there exists \(b\), the inverse of \(a\), such that \(ab = 1\).
+Assume that \(a\) is a zero divisor, then there exists \(c \neq 0\)
+such that \(ac = 0\). Then, \((ab)c = 0b \Rightarrow (1)(c) = 0\) implies that \(c = 0\), which is a contradiction.
+\subsubsection{b}
+\label{sec:orgc5282b2}
+By definition, there exists \(b\), with \(b \neq 0\) such that \(ab = 0\).
+Assume that \(a\) is a unit, then there exists \(c\) such that \(ac = 1\).
+Then, \((b)ac = 1b \Rightarrow 0c = b\) implies that \(b = 0\), which is a contradiction.
+
+\subsection{Question Eleven}
+\label{sec:org6981e66}
+By definition of \(a\) being a unit, there exists \(ay = 1\) with \(y\) being an inverse of \(a\).
+By multiplying our target \((y)ax = (y)b \Rightarrow x = yb\).
+
+To prove this is unique, assume that \(k\) and \(l\) are solutions of \(ax = b\). Then, \(ak = b\) and \(al = b\). Since \(a\) is a unit, by using our
+previous strategy, \((y)ak = (y)b\) and \((y)al = (y)b\), so \(k = yb\) and \(l = yb\) and thus \(k = l\).
+
+
+\section{Chapter 13}
+\label{sec:org62ae8b0}
+\subsection{A2}
+\label{sec:org0b1499e}
+As \(p | c \Rightarrow c = pk\), and \(p | c \Rightarrow c = ql\) then \(pk = ql\) and thus by Theorem 1.5 since \(p\) and \(q\) are prime \(p | l\)
+or \(p | q\) and \(q | p\) or \(q | k\), but since \(p\) and \(q\) are prime, then it must be that only \(p | l\) and \(q | k\).
+
+Since \(p | l\) then \(l = mp\) and \(c = ql \Rightarrow c = qmp\) and \(qp\) is a factor of \(c\).
+
+\subsection{A3 (a)}
+\label{sec:orge4f13f2}
+GO = 0715
+
+715\textsuperscript{3} (mod 2773) = 107
+\end{document}
\ No newline at end of file
diff --git a/Homework/math4310/alg_structures_assn_4.org b/Homework/math4310/alg_structures_assn_4.org
new file mode 100644
index 0000000..cc35190
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_4.org
@@ -0,0 +1,176 @@
+#+TITLE: Assignment Four
+#+AUTHOR: Lizzy Hunt
+#+STARTUP: entitiespretty fold inlineimages
+#+LATEX_HEADER: \notindent \notag \usepackage{ dsfont } \usepackage{amsmath}
+#+LATEX: \setlength\parindent{0pt}
+#+OPTIONS: toc:nil
+
+* Section 3.1
+** Question One
+*** a
+$a \in R, b \in R \Rightarrow a + b \in R$
+*** b
+$a \in R \Rightarow x \in R \ni a + x = 0_R$
+** Question Three
+1. All operations are closed since only the elements in ${0, e, a, b}$ appear in the tables.
+2. From the second row and columns in the multiplication table we see that $e$ is the multiplicative identity.
+3. From the first row and columns in the addition table we see that $0$ is the zero element.
+4. In this field, each element is its own additive inverse.
+5. There is commutativity as the transpose of each table is identical to the original (symmetry along the diagonals).
+** Question Six
+*** a
+Since our addition and multiplication operators are the same in $\mathds{Z}$, we have
+associativity, commutativity, and distributivity.
+
+Sums of multiples of 3 are also multiples of 3: $3n + 3m = 3(n + m)$, so this set is closed under addition.
+
+Products of multiples of 3 are also multiples of 3: $(3n)(3m) = (3)(3nm)$, so this set is closed under multiplication.
+
+The additive inverse exists in the set for every element: $3n + x = 0 \Rightarrow x = 3 \cdot (-n)$.
+
+$0$ is the zero element and is a multiple of 3 since $3 \cdot 0 = 0$.
+
+Therefore ${x : x = 3n \ni n \in \mathds{Z}}$ is a subring of $\mathds{Z}$
+
+*** b
+Since our addition and multiplication operators are the same in $\mathds{Z}$, we have
+associativity, commutativity, and distributivity.
+
+Sums of multiples of $k$ are also multiples of $k$: $kn + km = k(n + m)$, so this set is closed under addition.
+
+Products of multiples of $k$ are also multiples of $k$: $(kn)(km) = (k)(knm)$, so this set is closed under multiplication.
+
+The additive inverse exists exists in the set for every element: $kn + x = 0 \Rightarrow k = k \cdot (-n)$.
+
+$0$ is the zero element and is a multiple of k since $k \cdot 0 = 0$.
+
+Therefore ${x : x = kn \ni n \in \mathds{Z}}$ is a subring of $\mathds{Z}$
+
+** Question Nine
+*** a
+${(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5)}$
+*** b
+Since our addition and multiplication operators are the same as in $R$, then we have
+associativity, commutativity, and distributivity.
+
+Sums of elements in $R*$ are also in $R*$ since any element $(r,r) \in R*, (j, j) \in R* = (r + j, r + j) \in R*$.
+
+Products of elements in $R*$ are also in $R*$ since any element $(r, r) \in R*, (j, j) \in R* = (rj, rj) \in R*$.
+
+The additive inverse exists in the set for every element: $(r, r) + x = 0 \Rightarrow x = (-r, -r) \in R*$.
+
+$(0, 0) \in R*$ is the zero element.
+
+Therefore ${(r, r) : (r, r) \in R*}$ is a subring of $R*$.
+
+** Question Ten
+Consider $a \in S \ni a = (10, -10)$ and $b \in S \ni b = (10, -10)$, then $ab \notin S$ since
+$ab = (100, 100)$ which does not follow the rule that $100 + 100 = 0$.
+** Question Eleven
+*** a
+Addition is closed since
+$\begin{smallmatrix}
+a & a \\
+b & b 
+\end{smallmatrix}$ + $\begin{smallmatrix}
+c & c \\
+d & d 
+\end{smallmatrix}$ = $\begin{smallmatrix}
+(a + c) & (a + c) \\
+(b + d) & (b + d)
+\end{smallmatrix}$ which of the form of the given rule.
+
+It is also closed under multiplication since
+$\begin{smallmatrix}
+a & a \\
+b & b 
+\end{smallmatrix}$ $\cdot$ $\begin{smallmatrix}
+c & c \\
+d & d 
+\end{smallmatrix}$ = $\begin{smallmatrix}
+(ac + ad) & (ac + ad) \\
+(bc + bd) & (bc + bd)
+\end{smallmatrix}$ which is also of the form of the given rule.
+
+The zero element is the zero matrix $\begin{smallmatrix}
+0 & 0 \\
+0 & 0
+\end{smallmatrix}$, trivially.
+
+Associativity and commutivity (of addition) come from these operations existing for 2x2 matrices in $M(\mathds{R})$.
+
+*** b
+$\begin{smallmatrix}
+a & a \\
+b & b 
+\end{smallmatrix}$ $\cdot$ $\begin{smallmatrix}
+1 & 1 \\
+0 & 0 
+\end{smallmatrix}$ = $\begin{smallmatrix}
+(a(1) + a(0)) & (a(1) + a(0)) \\
+(b(1) + b(0)) & (b(1) + b(0))
+\end{smallmatrix}$
+which is equivalent to $\begin{smallmatrix}
+a & a \\
+b & b 
+\end{smallmatrix}$
+
+*** c
+$\begin{smallmatrix}
+1 & 1 \\
+0 & 0 
+\end{smallmatrix}$ $\cdot$ $\begin{smallmatrix}
+1 & 1 \\
+2 & 2
+\end{smallmatrix}$ = $\begin{smallmatrix}
+((1)(1) + (1)(2)) & ((1)(1) + (1)(2)) \\
+((0)(1) + (0)(2)) & ((0)(1) + (0)(2))
+\end{smallmatrix}$ = $\begin{smallmatrix}
+3 & 3 \\
+0 & 0
+\end{smallmatrix}$.
+** Question Twelve
+$\mathds{Z}[i]$ is closed under addition: $(a + bi) + (c + di) = (a + c) + (b + d)i$ and since
+$\mathds{Z}$ is a ring itself, $(a + c) + (b + d)i$ is also in $\mathds{Z}[i]$.
+
+$\mathds{Z}[i]$ is closed under multiplication: $(a + bi) \cdot (c + di) = ac + adi + cbi + bdi^2 = (ac - bd) + (ad + bd)i$
+by similar logic.
+
+The additive inverse always exists in the set: $(a + bi) + x = 0 \Rightarrow x = -a - bi$.
+
+Finally, the zero element is trivially $0 + 0i$ since $(a + bi) + (0 + 0i) = a + bi$.
+
+** Question Fourteen
+$S$ is closed under addition. For example given some $f, g \in S$ then $(f + g)(x) = h$ and $h$ will still satisfy the rule that
+$h(2) = 0$ since $(f + g)(2) = f(2) + g(2) = 0 + 0$, and addition was already closed in the domain given that question 8 is a ring
+itself.
+
+Similarly, $S$ is closed under multiplication. $(f \cdot g)(x) = h$ and h will still satisfy the rule as $h(2) = f(2) \cdot g(2) = 0 \cdot 0$.
+
+The zero element is $f \ni f(x) = 0$.
+
+Finally, the additive inverse $g$ exists in the set for each $f$ since $f + g = 0$ implies that $g$ is just $-f$ (the rule still
+stands here).
+
+** Question Fifteen
+*** b
+| \odot      | (0, 0) | (0, 1) | (0, 2) | (1, 0) | (1, 1) | (1, 2) |
+| (0, 0) | (0, 0) | (0, 0) | (0, 0) | (0, 0) | (0, 0) | (0, 0) |
+| (0, 1) | (0, 0) | (0, 1) | (0, 2) | (0, 0) | (0, 1) | (0, 2) |
+| (0, 2) | (0, 0) | (0, 2) | (0, 1) | (0, 0) | (0, 2) | (0, 1) |
+| (1, 0) | (0, 0) | (0, 0) | (0, 0) | (1, 0) | (1, 0) | (1, 0) |
+| (1, 1) | (0, 0) | (0, 1) | (0, 2) | (1, 0) | (1, 1) | (1, 2) |
+| (1, 2) | (0, 0) | (0, 2) | (0, 1) | (1, 0) | (1, 2) | (1, 1) |
+
+| \oplus     | (0, 0) | (0, 1) | (0, 2) | (1, 0) | (1, 1) | (1, 2) |
+| (0, 0) | (0, 0) | (0, 1) | (0, 2) | (1, 0) | (1, 1) | (1, 2) |
+| (0, 1) | (0, 1) | (0, 2) | (0, 0) | (1, 1) | (1, 2) | (1, 0) |
+| (0, 2) | (0, 2) | (0, 0) | (0, 1) | (1, 2) | (1, 0) | (1, 1) |
+| (1, 0) | (1, 0) | (1, 1) | (1, 2) | (2, 0) | (2, 1) | (2, 2) |
+| (1, 1) | (1, 1) | (1, 2) | (1, 0) | (2, 1) | (2, 2) | (2, 0) |
+| (1, 2) | (1, 2) | (1, 0) | (1, 1) | (2, 2) | (2, 0) | (2, 1) |
+
+** Question Eighteen
+No, by definition, as the distributive axiom is violated for this to be a ring.
+
+For example, $1(1 + 1) = (1)(1) + (1)(1) = 2$ but $1(1 + 1) = 1$.
diff --git a/Homework/math4310/alg_structures_assn_4.pdf b/Homework/math4310/alg_structures_assn_4.pdf
new file mode 100644
index 0000000..3e6a16d
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_4.pdf
diff --git a/Homework/math4310/alg_structures_assn_4.tex b/Homework/math4310/alg_structures_assn_4.tex
new file mode 100644
index 0000000..bf588dd
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_4.tex
@@ -0,0 +1,233 @@
+% Created 2023-02-08 Wed 09:17
+% 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}
+\author{Lizzy Hunt}
+\date{\today}
+\title{Assignment Four}
+\hypersetup{
+ pdfauthor={Lizzy Hunt},
+ pdftitle={Assignment Four},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.6.1)}, 
+ pdflang={English}}
+\begin{document}
+
+\maketitle
+\setlength\parindent{0pt}
+
+\section{Section 3.1}
+\label{sec:org5ad4880}
+\subsection{Question One}
+\label{sec:org4dd5421}
+\subsubsection{a}
+\label{sec:orgfe8dc40}
+\(a \in R, b \in R \Rightarrow a + b \in R\)
+\subsubsection{b}
+\label{sec:orgccbb819}
+\(a \in R \Rightarow x \in R \ni a + x = 0_R\)
+\subsection{Question Three}
+\label{sec:orgc539ab7}
+\begin{enumerate}
+\item All operations are closed since only the elements in \({0, e, a, b}\) appear in the tables.
+\item From the second row and columns in the multiplication table we see that \(e\) is the multiplicative identity.
+\item From the first row and columns in the addition table we see that \(0\) is the zero element.
+\item In this field, each element is its own additive inverse.
+\item There is commutativity as the transpose of each table is identical to the original (symmetry along the diagonals).
+\end{enumerate}
+\subsection{Question Six}
+\label{sec:orge1c7f2c}
+\subsubsection{a}
+\label{sec:orgb356ff9}
+Since our addition and multiplication operators are the same in \(\mathds{Z}\), we have
+associativity, commutativity, and distributivity.
+
+Sums of multiples of 3 are also multiples of 3: \(3n + 3m = 3(n + m)\), so this set is closed under addition.
+
+Products of multiples of 3 are also multiples of 3: \((3n)(3m) = (3)(3nm)\), so this set is closed under multiplication.
+
+The additive inverse exists in the set for every element: \(3n + x = 0 \Rightarrow x = 3 \cdot (-n)\).
+
+\(0\) is the zero element and is a multiple of 3 since \(3 \cdot 0 = 0\).
+
+Therefore \({x : x = 3n \ni n \in \mathds{Z}}\) is a subring of \(\mathds{Z}\)
+
+\subsubsection{b}
+\label{sec:orga2b6add}
+Since our addition and multiplication operators are the same in \(\mathds{Z}\), we have
+associativity, commutativity, and distributivity.
+
+Sums of multiples of \(k\) are also multiples of \(k\): \(kn + km = k(n + m)\), so this set is closed under addition.
+
+Products of multiples of \(k\) are also multiples of \(k\): \((kn)(km) = (k)(knm)\), so this set is closed under multiplication.
+
+The additive inverse exists exists in the set for every element: \(kn + x = 0 \Rightarrow k = k \cdot (-n)\).
+
+\(0\) is the zero element and is a multiple of k since \(k \cdot 0 = 0\).
+
+Therefore \({x : x = kn \ni n \in \mathds{Z}}\) is a subring of \(\mathds{Z}\)
+
+\subsection{Question Nine}
+\label{sec:org4f6749e}
+\subsubsection{a}
+\label{sec:orgc523c2e}
+\({(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5)}\)
+\subsubsection{b}
+\label{sec:org970c88d}
+Since our addition and multiplication operators are the same as in \(R\), then we have
+associativity, commutativity, and distributivity.
+
+Sums of elements in \(R*\) are also in \(R*\) since any element \((r,r) \in R*, (j, j) \in R* = (r + j, r + j) \in R*\).
+
+Products of elements in \(R*\) are also in \(R*\) since any element \((r, r) \in R*, (j, j) \in R* = (rj, rj) \in R*\).
+
+The additive inverse exists in the set for every element: \((r, r) + x = 0 \Rightarrow x = (-r, -r) \in R*\).
+
+\((0, 0) \in R*\) is the zero element.
+
+Therefore \({(r, r) : (r, r) \in R*}\) is a subring of \(R*\).
+
+\subsection{Question Ten}
+\label{sec:orgfdaf968}
+Consider \(a \in S \ni a = (10, -10)\) and \(b \in S \ni b = (10, -10)\), then \(ab \notin S\) since
+\(ab = (100, 100)\) which does not follow the rule that \(100 + 100 = 0\).
+\subsection{Question Eleven}
+\label{sec:org3b794ec}
+\subsubsection{a}
+\label{sec:orgef5a91b}
+Addition is closed since
+\(\begin{smallmatrix}
+a & a \\
+b & b 
+\end{smallmatrix}\) + \(\begin{smallmatrix}
+c & c \\
+d & d 
+\end{smallmatrix}\) = \(\begin{smallmatrix}
+(a + c) & (a + c) \\
+(b + d) & (b + d)
+\end{smallmatrix}\) which of the form of the given rule.
+
+It is also closed under multiplication since
+\(\begin{smallmatrix}
+a & a \\
+b & b 
+\end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix}
+c & c \\
+d & d 
+\end{smallmatrix}\) = \(\begin{smallmatrix}
+(ac + ad) & (ac + ad) \\
+(bc + bd) & (bc + bd)
+\end{smallmatrix}\) which is also of the form of the given rule.
+
+The zero element is the zero matrix \(\begin{smallmatrix}
+0 & 0 \\
+0 & 0
+\end{smallmatrix}\), trivially.
+
+Associativity and commutivity (of addition) come from these operations existing for 2x2 matrices in \(M(\mathds{R})\).
+
+\subsubsection{b}
+\label{sec:orgb4b3acf}
+\(\begin{smallmatrix}
+a & a \\
+b & b 
+\end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix}
+1 & 1 \\
+0 & 0 
+\end{smallmatrix}\) = \(\begin{smallmatrix}
+(a(1) + a(0)) & (a(1) + a(0)) \\
+(b(1) + b(0)) & (b(1) + b(0))
+\end{smallmatrix}\)
+which is equivalent to \(\begin{smallmatrix}
+a & a \\
+b & b 
+\end{smallmatrix}\)
+
+\subsubsection{c}
+\label{sec:orga515fb6}
+\(\begin{smallmatrix}
+1 & 1 \\
+0 & 0 
+\end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix}
+1 & 1 \\
+2 & 2
+\end{smallmatrix}\) = \(\begin{smallmatrix}
+((1)(1) + (1)(2)) & ((1)(1) + (1)(2)) \\
+((0)(1) + (0)(2)) & ((0)(1) + (0)(2))
+\end{smallmatrix}\) = \(\begin{smallmatrix}
+3 & 3 \\
+0 & 0
+\end{smallmatrix}\).
+\subsection{Question Twelve}
+\label{sec:orgb1926b2}
+\(\mathds{Z}[i]\) is closed under addition: \((a + bi) + (c + di) = (a + c) + (b + d)i\) and since
+\(\mathds{Z}\) is a ring itself, \((a + c) + (b + d)i\) is also in \(\mathds{Z}[i]\).
+
+\(\mathds{Z}[i]\) is closed under multiplication: \((a + bi) \cdot (c + di) = ac + adi + cbi + bdi^2 = (ac - bd) + (ad + bd)i\)
+by similar logic.
+
+The additive inverse always exists in the set: \((a + bi) + x = 0 \Rightarrow x = -a - bi\).
+
+Finally, the zero element is trivially \(0 + 0i\) since \((a + bi) + (0 + 0i) = a + bi\).
+
+\subsection{Question Fourteen}
+\label{sec:org31f60fc}
+The zero element is given as 2.
+
+\(S\) is closed under addition. For example given some \(f, g \in S\) then \((f + g)(x) = h\) and \(h\) will still satisfy the rule that
+\(h(2) = 0\) since \((f + g)(2) = f(2) + g(2) = 0 + 0\), and addition was already closed in the domain given that question 8 is a ring
+itself.
+
+Similarly, \(S\) is closed under multiplication. \((f \cdot g)(x) = h\) and h will still satisfy the rule as \(h(2) = f(2) \cdot g(2) = 0 \cdot 0\).
+
+The zero element is \(f \ni f(x) = 0\).
+
+Finally, the additive inverse \(g\) exists in the set for each \(f\) since \(f + g = 0\) implies that \(g\) is just \(-f\) (the rule still
+stands here).
+
+\subsection{Question Fifteen}
+\label{sec:org04d3ba1}
+\subsubsection{b}
+\label{sec:orgae94898}
+\begin{center}
+\begin{tabular}{lllllll}
+\(\odot\) & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2)\\[0pt]
+(0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0)\\[0pt]
+(0, 1) & (0, 0) & (0, 1) & (0, 2) & (0, 0) & (0, 1) & (0, 2)\\[0pt]
+(0, 2) & (0, 0) & (0, 2) & (0, 1) & (0, 0) & (0, 2) & (0, 1)\\[0pt]
+(1, 0) & (0, 0) & (0, 0) & (0, 0) & (1, 0) & (1, 0) & (1, 0)\\[0pt]
+(1, 1) & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2)\\[0pt]
+(1, 2) & (0, 0) & (0, 2) & (0, 1) & (1, 0) & (1, 2) & (1, 1)\\[0pt]
+\end{tabular}
+\end{center}
+
+\begin{center}
+\begin{tabular}{lllllll}
+\(\oplus\) & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2)\\[0pt]
+(0, 0) & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2)\\[0pt]
+(0, 1) & (0, 1) & (0, 2) & (0, 0) & (1, 1) & (1, 2) & (1, 0)\\[0pt]
+(0, 2) & (0, 2) & (0, 0) & (0, 1) & (1, 2) & (1, 0) & (1, 1)\\[0pt]
+(1, 0) & (1, 0) & (1, 1) & (1, 2) & (2, 0) & (2, 1) & (2, 2)\\[0pt]
+(1, 1) & (1, 1) & (1, 2) & (1, 0) & (2, 1) & (2, 2) & (2, 0)\\[0pt]
+(1, 2) & (1, 2) & (1, 0) & (1, 1) & (2, 2) & (2, 0) & (2, 1)\\[0pt]
+\end{tabular}
+\end{center}
+
+\subsection{Question Eighteen}
+\label{sec:orga487be8}
+No, by definition, as the distributive axiom is violated for this to be a ring.
+
+For example, \(1(1 + 1) = (1)(1) + (1)(1) = 2\) but \(1(1 + 1) = 1\).
+\end{document}
\ No newline at end of file
diff --git a/Homework/math4310/alg_structures_assn_5.org b/Homework/math4310/alg_structures_assn_5.org
new file mode 100644
index 0000000..78b0957
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_5.org
@@ -0,0 +1,315 @@
+#+TITLE: Assignment Five
+#+AUTHOR: Lizzy Hunt
+#+STARTUP: entitiespretty fold inlineimages
+#+LATEX_HEADER: \notindent \notag \usepackage{ dsfont } \usepackage{amsmath} \usepackage[a4paper,margin=1in,landscape]{geometry}
+#+LATEX: \setlength\parindent{0pt}
+#+OPTIONS: toc:nil
+
+* Section 3.2
+** Question One
+*** a
+$a^2 - ab + ba - b^2$
+*** b
+$a^3 + ba^2 + 2a^2b + 2ab^2 + ab^2 + b^3$
+*** c
+(a) could be $a^2 - b^2$
+
+(b) could be $a^3 + 3a^2b + 3ab^2 + b^3$
+
+** Question Three
+*** a
+$\begin{smallmatrix}
+0 & 0 \\
+0 & 0 \\
+\end{smallmatrix}$, $\begin{smallmatrix}
+1 & 0 \\
+0 & 1 \\
+\end{smallmatrix}$, $\begin{smallmatrix}
+1 & 0 & 0 \\
+0 & 1 & 0 \\
+0 & 0 & 1 \\
+\end{smallmatrix}$, $\begin{smallmatrix}
+0 & 0 & 0 \\
+0 & 0 & 0 \\
+0 & 0 & 0 \\
+\end{smallmatrix}$
+*** b
+\begin{verbatim}
+>>> set(filter(lambda y: all([y**2 % 12 == y for x in range(12)]), range(12)))
+{0, 1, 4, 9}
+\end{verbatim}
+
+** Question Seven
+S is closed under multiplication:
+$i \in S, j \in S \Rightarrow i \cdot j = n1_R \cdot m1_R = (nm)1_R$.
+
+S is closed under subtraction:
+$i \in S, j \in S \Rightarrow i - j = n1_R - m1_R = (n-m)1_R$
+
+** Question Eight
+Considering $n,m \in T$ then $n = xb, m = yb$ with $x,y \in R$:
+
+1. T is closed under multiplication: $n \cdot m = xb \cdot yb = (x \cdot y)b$, which follows the rule.
+2. T is closed under subtraction: $n - m = xb - yb = (x - y)b$, which also follows the rule.
+
+We also know $T$ is not empty since it must have at least 0_R.
+
+** Question Ten
+*** a
+$\bar{R} = {(0, 0), (1, 0), (2, 0)}$
+
+$\bar{S} = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4)}$
+
+*** b
+Considering $n, m \in \bar{R}$, then $n = (x, 0_S), n = (y, 0_S)$ with $x,y \in R$.
+
+1. $\bar{R}$ is closed under multiplication: $n \cdot m = (x, 0_S) \cdot (y, 0_S) = (x \cdot y, 0_S)$ and $x \cdot y \in R$, so thus $n \cdot m \in R \times S$.
+2. $\bar{R}$ is closed under subtraction: $n - m = (x, 0_S) - (y, 0_S) = (x - y, 0_S)$ and $x - y \in R$, so thus $n - m \in R \times S$.
+
+We also know that $\bar{R}_0 = (0_R, 0_S) \in R \times S$ so $\bar{R}$ is not empty.
+
+*** c
+Considering $n, m \in \bar{S}$, then $n = (0_R, x), n = (0_R, y)$ with $x,y \in S$.
+
+1. $\bar{S}$ is closed under multiplication: $n \cdot m = (0_R, x) \cdot (0_R, y) = (x \cdot y, 0_S)$ and $x \cdot y \in S$, so thus $n \cdot m \in R \times S$.
+2. $\bar{S}$ is closed under subtraction: $n - m = (0_R, x) - (0_R, y) = (0_R, x - y)$ and $x - y \in R$, so thus $n - m \in R \times S$.
+
+We also know that $\bar{S}_0 = (0_R, 0_S) \in R \times S$ so $\bar{S}$ is not empty.
+
+** Question Thirteen
+*** a
+Considering $n, m \in S \cap T$, then $n \in S$ and $n \in T$, $m \in S$ and $m \in T$, therefore:
+
+1. $S \cap T$ is closed under multiplication as $m \cdot n$ must also be in $S \cap T$
+2. $S \cap T$ is closed under addition as $m + n$ must also be in $S \cap T$
+
+Since $S$ and $T$ are both subrings of $R$, $0_R \in S \cap T$.
+
+*** b
+No, consider $S$ being the integer multiples of 8 and $T$ being the integer multiples of 3 being subrings of $\mathds{Z}$ (proof of these being subrings is in
+Question Six of Section 3.1 in Assignment Four), then $n \in S$ with $n = 8$ and $m \in T$ with $m = 3$ then $n + m = 11 \notin S \cup T$.
+
+** Question Fifteen - TODO
+
+** Question Twenty-One
+*** a
+From $ab = ac \Rightarrow ab - ac = 0_R \Rightarrow a(b - c) = 0_R$, we know $b-c$ is equivalent to $0_R$ since we're given that $a$ is a non-zero element.
+
+$b-c = 0_R \Rightarrow b = c$
+
+*** b
+From $ba = ca \Rightarrow ba - ca = 0_R \Rightarrow (b - c)(a) = 0_R$ we come to the same conclusion as (a)
+
+$b-c = 0_R \Rightarrow b = c$
+
+* Section 3.3
+** Question One
+In the true nature of being a computer science student, automate something for 2 hours that you could've done in 20 minutes!
+#+BEGIN_SRC python :session "cct" :results none
+  cartesian_prod = lambda zn, zm: list([(i, j) for i in range(zn) for j in range(zm)])
+  mult_tup = lambda a, b, zn, zm: ((a[0] * b[0]) % zn, (a[1] * b[1]) % zm)
+  add_tup = lambda a, b, zn, zm: ((a[0] + b[0]) % zn, (a[1] + b[1]) % zm)
+  empty_table = lambda col, row, symbol: [[symbol] + [str(x) for x in col]] + [[str(x)] + [0 for i in range(len(col))] for x in row]
+
+  def make_cartesian_congruence_table(zn, zm, op, mapping, symbol):
+    cp = cartesian_prod(zn, zm)
+    mapped_cp = [cp[mapping[i]] for i in range(len(cp))]
+    table = empty_table(mapped_cp, mapped_cp, symbol)
+    for i in range(len(cp)):
+      for j in range(len(cp)):
+        table[i+1][j+1] = str(op(mapped_cp[i], mapped_cp[j], zn, zm))
+    return table
+
+  def make_normal_table(n, op, symbol):
+    table = empty_table(list(range(n)), list(range(n)), symbol)
+    for i in range(n):
+      for j in range(n):
+        table[i+1][j+1] = str(op(i, j) % n)
+    return table
+#+END_SRC
+
+*** Z_2 \times Z_3 with bijection
+#+BEGIN_SRC python :session "cct" :results none
+  bijection = [0, 4, 2, 3, 1, 5]
+#+END_SRC 
+
+**** Multiplication Tables
+#+BEGIN_SRC python :session "cct" :results table
+  make_cartesian_congruence_table(2, 3, mult_tup, bijection, '\odot')
+#+END_SRC 
+
+| \odot      | (0, 0) | (1, 1) | (0, 2) | (1, 0) | (0, 1) | (1, 2) |
+| (0, 0) | (0, 0) | (0, 0) | (0, 0) | (0, 0) | (0, 0) | (0, 0) |
+| (1, 1) | (0, 0) | (1, 1) | (0, 2) | (1, 0) | (0, 1) | (1, 2) |
+| (0, 2) | (0, 0) | (0, 2) | (0, 1) | (0, 0) | (0, 2) | (0, 1) |
+| (1, 0) | (0, 0) | (1, 0) | (0, 0) | (1, 0) | (0, 0) | (1, 0) |
+| (0, 1) | (0, 0) | (0, 1) | (0, 2) | (0, 0) | (0, 1) | (0, 2) |
+| (1, 2) | (0, 0) | (1, 2) | (0, 1) | (1, 0) | (0, 2) | (1, 1) |
+
+**** Addition Tables
+#+BEGIN_SRC python :session "cct" :results table
+  make_cartesian_congruence_table(2, 3, add_tup, bijection, '\oplus')
+#+END_SRC 
+
+| \oplus     | (0, 0) | (1, 1) | (0, 2) | (1, 0) | (0, 1) | (1, 2) |
+| (0, 0) | (0, 0) | (1, 1) | (0, 2) | (1, 0) | (0, 1) | (1, 2) |
+| (1, 1) | (1, 1) | (0, 2) | (1, 0) | (0, 1) | (1, 2) | (0, 0) |
+| (0, 2) | (0, 2) | (1, 0) | (0, 1) | (1, 2) | (0, 0) | (1, 1) |
+| (1, 0) | (1, 0) | (0, 1) | (1, 2) | (0, 0) | (1, 1) | (0, 2) |
+| (0, 1) | (0, 1) | (1, 2) | (0, 0) | (1, 1) | (0, 2) | (1, 0) |
+| (1, 2) | (1, 2) | (0, 0) | (1, 1) | (0, 2) | (1, 0) | (0, 1) |
+
+*** Z_6
+**** Multiplication Tables
+#+BEGIN_SRC python :session "cct" :results table
+  make_normal_table(6, lambda a, b: a * b, '\odot')
+#+END_SRC 
+
+| \odot | 0 | 1 | 2 | 3 | 4 | 5 |
+| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
+| 1 | 0 | 1 | 2 | 3 | 4 | 5 |
+| 2 | 0 | 2 | 4 | 0 | 2 | 4 |
+| 3 | 0 | 3 | 0 | 3 | 0 | 3 |
+| 4 | 0 | 4 | 2 | 0 | 4 | 2 |
+| 5 | 0 | 5 | 4 | 3 | 2 | 1 |
+
+**** Addition Tables
+#+BEGIN_SRC python :session "cct" :results table
+  make_normal_table(6, lambda a, b: a + b, '\oplus')
+#+END_SRC 
+
+| \oplus | 0 | 1 | 2 | 3 | 4 | 5 |
+|  0 | 0 | 1 | 2 | 3 | 4 | 5 |
+|  1 | 1 | 2 | 3 | 4 | 5 | 0 |
+|  2 | 2 | 3 | 4 | 5 | 0 | 1 |
+|  3 | 3 | 4 | 5 | 0 | 1 | 2 |
+|  4 | 4 | 5 | 0 | 1 | 2 | 3 |
+|  5 | 5 | 0 | 1 | 2 | 3 | 4 |
+
+** Question Two
+#+BEGIN_SRC python :session "cct" :results table
+  bijection = [0, 3, 2, 1]
+  make_cartesian_congruence_table(2, 2, mult_tup, bijection, '\odot')
+#+END_SRC
+
+| \odot      | (0, 0) | (1, 1) | (1, 0) | (0, 1) |
+| (0, 0) | (0, 0) | (0, 0) | (0, 0) | (0, 0) |
+| (1, 1) | (0, 0) | (1, 1) | (1, 0) | (0, 1) |
+| (1, 0) | (0, 0) | (1, 0) | (1, 0) | (0, 0) |
+| (0, 1) | (0, 0) | (0, 1) | (0, 0) | (0, 1) |
+
+#+BEGIN_SRC python :session "cct" :results table
+  bijection = [0, 3, 2, 1]
+  make_cartesian_congruence_table(2, 2, add_tup, bijection, '\oplus')
+#+END_SRC
+
+| \oplus     | (0, 0) | (1, 1) | (1, 0) | (0, 1) |
+| (0, 0) | (0, 0) | (1, 1) | (1, 0) | (0, 1) |
+| (1, 1) | (1, 1) | (0, 0) | (0, 1) | (1, 0) |
+| (1, 0) | (1, 0) | (0, 1) | (0, 0) | (1, 1) |
+| (0, 1) | (0, 1) | (1, 0) | (1, 1) | (0, 0) |
+
+** Question Three
+1. $f$ is injective, since $f(a) = f(b) \Rightarrow (a,a) = (b,b) \Rightarrow a = b$
+2. $f$ is surjective since every range element in $R^*$, $(a,a)$ is mapped to $a \in R$ by definition
+3. $f(a) + f(b) = (a, a) + (b, b) = (a + b, a + b) = f(a + b)$ and $f(a)f(b) = (a, a) \cdot (b, b) = (ab, ab) = f(ab)$
+
+** Question Four
+$f(1)f(3) = (2)(6) \equiv_{10} 2$ but $f(3) = 6$ which doesn't hold the properties of homomorphism.
+
+** Question Five
+Consider the given function:
+1. $f$ is injective, since $f(a) = f(b) \Rightarrow$ $\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}$ = $\begin{smallmatrix} 0 & 0 \\ 0 & b \end{smallmatrix}$ $\Rightarrow a = b$
+2. $f$ is surjective, since every range element $\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}$ has an element in $\mathds{R}$, $a$, such that $f(a) =$ $\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}$ by definition
+3. $f(a) + f(b) =$ $\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}$ + $\begin{smallmatrix} 0 & 0 \\ 0 & b \end{smallmatrix}$ = $\begin{smallmatrix} 0 & 0 \\ 0 & (a + b) \end{smallmatrix}$ = $f(a + b)$,
+   and $f(a) \cdot f(b) =$ $\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}$ \cdot $\begin{smallmatrix} 0 & 0 \\ 0 & b \end{smallmatrix}$ = $\begin{smallmatrix} 0 & 0 \\ 0 & (a \cdot b) \end{smallmatrix}$ = $f(a \cdot b)$,
+   
+** Question Nine - TODO
+
+** Question Eleven
+*** b
+\begin{verbatim}
+>>> f = lambda x: 3 * x
+>>> list(filter(lambda x: f(x * x) != (f(x) * f(x)), range(0, 20, 2)))
+[2, 4, 6, 8, 10, 12, 14, 16, 18]
+\end{verbatim}
+*** d
+\begin{verbatim}
+>>> k = lambda x: 0 if x == 0 else (x ** -1)
+>>> list(filter(lambda x: k(x * x) != (k(x) * k(x)), range(0, 20)))
+[5, 10, 13, 19]
+\end{verbatim}
+** Question Twelve
+*** c
+Not a homomorphism. Just from reducing $f(x + x)$ we find $f(x + x) \neq f(x) + f(x)$:
+
+\begin{equation*}
+f(x + x) = \dfrac{1}{(x + x)^2 + 1} = \dfrac{1}{4x^2 + 1}
+\end{equation*}
+
+\begin{equation*}
+f(x) + f(x) = \dfrac{1}{x^2 + 1} + \dfrac{1}{x^2 + 1} = \dfrac{2}{x^2 + 1}
+\end{equation*}
+
+Thus $f(x + x) \neq f(x) + f(x)$.
+*** d
+\begin{verbatim}
+>>> import numpy as np
+>>> h = lambda x: [[-x, 0], [x, 0]]
+>>> list(filter(lambda x: not np.array_equiv(h(x * x), np.dot(h(x), h(x))), range(0, 20))
+[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
+\end{verbatim}
+** Question Thirteen
+*** a
+If $r \in R$ then $(r, 0_S) \in R \times S$, and $f((r, 0_S)) = r$, so every range element has a domain element mapping to it.
+
+For some $a = (r, m) \in R \times S$ and $b = (t, v) \in R \times S$ then $f((r, m)) + f((t, v)) = r + t = f((r + t, v + m))$, and
+$f((r, m)) \cdot f((t, v)) = r \cdot t = f((r \cdot t, v \cdot m))$.
+
+*** b
+If $s \in S$ then $(0_R, s) \in R \times S$, and $f((0_R, s)) = s$, so every range element has a domain element mapping to it.
+
+For some $a = (r, m) \in R \times S$ and $b = (t, v) \in R \times S$ then $f((r, m)) + f((t, v)) = m + v = f((r + t, m + v))$, and
+$f((r, m)) \cdot f((t, v)) = m \cdot v = f((r \cdot t, m \cdot v))$.
+
+** Question Fifteen
+Consider $f: Z_4 \rightarrow Z_4 \ni f(x) = 0$, then 2 is a zero divisor, but 0 is not by definition.
+
+** Question Twenty One
+*** Lemma
+We assume a multiplicative identity $x$ exists in $\mathds{Z}^{*}$:
+
+$b \odot x = b \Rightarrow b + x - bx = b \Rightarrow x - bx = 0 \Rightarrow x(1-b) = 0$ and $1-b \neq 0 \forall b \in \mathds{Z}^{*}$ so $x = 0$.
+
+which is verified by:
+
+$0 \odot b = 0 + b - 0 \cdot b = b$ and $b \odot 0 = b + 0 - b \cdot 0 = b$.
+
+*** Proof
+
+Consider $f : \mathds{Z}^{} \rightarrow \mathds{Z}^{*}$ to be the isomorphism we so desire, then by Theorem 3.10,
+f(1 \in \mathds{Z}) = 0 \in \mathds{Z}^{*}.
+
+Therefore, f(2) = f(1 + 1) = f(1) \oplus f(1) = -1, f(3) = f(2 + 1) = -1 \oplus f(1) = -2.
+
+$f : \mathds{Z}^{} \rightarrow \mathds{Z}^{*} \ni f(x) = 1 - x$ seems to fit the bill nicely.
+
+1. $f$ is a bijection since we have an inverse $x = 1 - f(x)$
+2. $f(a \oplus b) = 1 - (a \oplus b) = 1 - (a + b - 1) = (1 - a) + (1 - b) = f(a) + f(b)$ and
+   $f(a \odot b) = 1 - (a \odot b) = 1 - (a + b - ab) = (1-a) \cdot (1-b) = f(a) \cdot f(b)$
+
+** Question Twenty Four
+*** a
+1. By the usual coordinate addition we know that $a, b \in R \Rightarrow a + b \in R$ and is associative, commutative, and the additive identity is $(0, 0)$.
+2. $R$ is closed under multiplication: $a, b \in R \Rightarrow a = (c, d) \wedge b = (e, f) \Rightarrow a \cdot b = (ce, de)$ and since $c, d, e, f \in \mathds{R}$ then $(ce, de) \in \mathds{R} \times \mathds{R}$ by definition.
+3. Multiplication is associative: $a, b, c \in R \Rightarrow a = (d, e) \wedge b = (f, g) \wedge c = (h, i) \Rightarrow (a \cdot b) \cdot c = (df, ef) \cdot (h, i) = (dfh, efh)$ and $a \cdot (b \cdot c) = (d, e) \cdot (fh, gh) = (dfh, efh)$
+4. Multiplication is distributive: $a, b, c \in R \Rightarrow a = (d, e) \wedge b = (f, g) \wedge c = (h, i) \Rightarrow a(b + c) = (e, f) \cdot (f + h, g + i) = (ef + eh, ff + fh) = (ef, ff) + (eh, fh) = a \cdot b + a \cdot c$
+   and $a, b, c \in R \Rightarrow a = (d, e) \wedge b = (f, g) \wedge c = (h, i) \Rightarrow (a + b) \cdot c = (d + f, e + g) \cdot (h, i) = (dh + fh, eh + gh) = (dh, eh) + (fh, gh) = a \cdot c + b \cdot c$
+*** b
+Consider the function $f: R \rightarrow M(\mathds{R})$ is an isomorphism, such that $f((a, b)) =$ $\begin{smallmatrix} a & 0 \\ b & 0 \end{smallmatrix}$, then:
+1. $f$ is surjective since every element in the range has a domain element $\begin{smallmatrix} a & 0 \\ b & 0 \end{smallmatrix} = y \ni f((a, b)) = y$
+2. $f$ is injective since $f((c, d) = x) = f((e, f) = y) \Rightarrow$ $\begin{smallmatrix} c & 0 \\ d & 0 \end{smallmatrix}$ = $\begin{smallmatrix} e & 0 \\ f & 0 \end{smallmatrix}$ $\Rightarrow c = e \wedge d = f \Rightarrow x = y$
+3. $f((c, d) = x) + f((e, f) = y) \Rightarrow$ $\begin{smallmatrix} c & 0 \\ d & 0 \end{smallmatrix}$ + $\begin{smallmatrix} e & 0 \\ f & 0 \end{smallmatrix}$ = $\begin{smallmatrix} c + e & 0 \\ d + f & 0 \end{smallmatrix}$
+   $= f(x + y)$, and $f((c, d) = x) \cdot f((e, f) = y) \Rightarrow$ $\begin{smallmatrix} c & 0 \\ d & 0 \end{smallmatrix}$ $\cdot$ $\begin{smallmatrix} e & 0 \\ f & 0 \end{smallmatrix}$ = $\begin{smallmatrix} ce & 0 \\ de & 0 \end{smallmatrix}$
+   $= f(x \cdot y)$
+
diff --git a/Homework/math4310/alg_structures_assn_5.pdf b/Homework/math4310/alg_structures_assn_5.pdf
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@@ -0,0 +1,434 @@
+% Created 2023-02-17 Fri 12:59
+% 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}
+\author{Lizzy Hunt}
+\date{\today}
+\title{Assignment Five}
+\hypersetup{
+ pdfauthor={Lizzy Hunt},
+ pdftitle={Assignment Five},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.6.1)}, 
+ pdflang={English}}
+\begin{document}
+
+\maketitle
+\setlength\parindent{0pt}
+
+\section{Section 3.2}
+\label{sec:org258154b}
+\subsection{Question One}
+\label{sec:orgf02bb85}
+\subsubsection{a}
+\label{sec:orgccb1c0c}
+\(a^2 - ab + ba - b^2\)
+\subsubsection{b}
+\label{sec:org9164b85}
+\(a^3 + ba^2 + 2a^2b + 2ab^2 + ab^2 + b^3\)
+\subsubsection{c}
+\label{sec:orgdaba485}
+(a) could be \(a^2 - b^2\)
+
+(b) could be \(a^3 + 3a^2b + 3ab^2 + b^3\)
+
+\subsection{Question Three}
+\label{sec:org4d02a16}
+\subsubsection{a}
+\label{sec:orgb743fca}
+\(\begin{smallmatrix}
+0 & 0 \\
+0 & 0 \\
+\end{smallmatrix}\), \(\begin{smallmatrix}
+1 & 0 \\
+0 & 1 \\
+\end{smallmatrix}\), \(\begin{smallmatrix}
+1 & 0 & 0 \\
+0 & 1 & 0 \\
+0 & 0 & 1 \\
+\end{smallmatrix}\), \(\begin{smallmatrix}
+0 & 0 & 0 \\
+0 & 0 & 0 \\
+0 & 0 & 0 \\
+\end{smallmatrix}\)
+\subsubsection{b}
+\label{sec:org8300d7e}
+\begin{verbatim}
+>>> set(filter(lambda y: all([y**2 % 12 == y for x in range(12)]), range(12)))
+{0, 1, 4, 9}
+\end{verbatim}
+
+\subsection{Question Seven}
+\label{sec:org9152891}
+S is closed under multiplication:
+\(i \in S, j \in S \Rightarrow i \cdot j = n1_R \cdot m1_R = (nm)1_R\).
+
+S is closed under subtraction:
+\(i \in S, j \in S \Rightarrow i - j = n1_R - m1_R = (n-m)1_R\)
+
+\subsection{Question Eight}
+\label{sec:org3ed31b0}
+Considering \(n,m \in T\) then \(n = xb, m = yb\) with \(x,y \in R\):
+
+\begin{enumerate}
+\item T is closed under multiplication: \(n \cdot m = xb \cdot yb = (x \cdot y)b\), which follows the rule.
+\item T is closed under subtraction: \(n - m = xb - yb = (x - y)b\), which also follows the rule.
+\end{enumerate}
+
+We also know \(T\) is not empty since it must have at least 0\textsubscript{R}.
+
+\subsection{Question Ten}
+\label{sec:orgc4efb36}
+\subsubsection{a}
+\label{sec:orgbc2ab9d}
+\(\bar{R} = {(0, 0), (1, 0), (2, 0)}\)
+
+\(\bar{S} = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4)}\)
+
+\subsubsection{b}
+\label{sec:org564675e}
+Considering \(n, m \in \bar{R}\), then \(n = (x, 0_S), n = (y, 0_S)\) with \(x,y \in R\).
+
+\begin{enumerate}
+\item \(\bar{R}\) is closed under multiplication: \(n \cdot m = (x, 0_S) \cdot (y, 0_S) = (x \cdot y, 0_S)\) and \(x \cdot y \in R\), so thus \(n \cdot m \in R \times S\).
+\item \(\bar{R}\) is closed under subtraction: \(n - m = (x, 0_S) - (y, 0_S) = (x - y, 0_S)\) and \(x - y \in R\), so thus \(n - m \in R \times S\).
+\end{enumerate}
+
+We also know that \(\bar{R}_0 = (0_R, 0_S) \in R \times S\) so \(\bar{R}\) is not empty.
+
+\subsubsection{c}
+\label{sec:org5905763}
+Considering \(n, m \in \bar{S}\), then \(n = (0_R, x), n = (0_R, y)\) with \(x,y \in S\).
+
+\begin{enumerate}
+\item \(\bar{S}\) is closed under multiplication: \(n \cdot m = (0_R, x) \cdot (0_R, y) = (x \cdot y, 0_S)\) and \(x \cdot y \in S\), so thus \(n \cdot m \in R \times S\).
+\item \(\bar{S}\) is closed under subtraction: \(n - m = (0_R, x) - (0_R, y) = (0_R, x - y)\) and \(x - y \in R\), so thus \(n - m \in R \times S\).
+\end{enumerate}
+
+We also know that \(\bar{S}_0 = (0_R, 0_S) \in R \times S\) so \(\bar{S}\) is not empty.
+
+\subsection{Question Thirteen}
+\label{sec:orgfc1192c}
+\subsubsection{a}
+\label{sec:orgdc9f5c9}
+Considering \(n, m \in S \cap T\), then \(n \in S\) and \(n \in T\), \(m \in S\) and \(m \in T\), therefore:
+
+\begin{enumerate}
+\item \(S \cap T\) is closed under multiplication as \(m \cdot n\) must also be in \(S \cap T\)
+\item \(S \cap T\) is closed under addition as \(m + n\) must also be in \(S \cap T\)
+\end{enumerate}
+
+Since \(S\) and \(T\) are both subrings of \(R\), \(0_R \in S \cap T\).
+
+\subsubsection{b}
+\label{sec:orgeb2e7d6}
+No, consider \(S\) being the integer multiples of 8 and \(T\) being the integer multiples of 3 being subrings of \(\mathds{Z}\) (proof of these being subrings is in
+Question Six of Section 3.1 in Assignment Four), then \(n \in S\) with \(n = 8\) and \(m \in T\) with \(m = 3\) then \(n + m = 11 \notin S \cup T\).
+
+\subsection{Question Fifteen - TODO}
+\label{sec:org675f8a8}
+
+\subsection{Question Twenty-One}
+\label{sec:org0c713ff}
+\subsubsection{a}
+\label{sec:org6158430}
+From \(ab = ac \Rightarrow ab - ac = 0_R \Rightarrow a(b - c) = 0_R\), we know \(b-c\) is equivalent to \(0_R\) since we're given that \(a\) is a non-zero element.
+
+\(b-c = 0_R \Rightarrow b = c\)
+
+\subsubsection{b}
+\label{sec:org4391edc}
+From \(ba = ca \Rightarrow ba - ca = 0_R \Rightarrow (b - c)(a) = 0_R\) we come to the same conclusion as (a)
+
+\(b-c = 0_R \Rightarrow b = c\)
+
+\section{Section 3.3}
+\label{sec:orgff5f7f5}
+\subsection{Question One}
+\label{sec:org08aff62}
+In the true nature of being a computer science student, automate something for 2 hours that you could've done in 20 minutes!
+\begin{verbatim}
+cartesian_prod = lambda zn, zm: list([(i, j) for i in range(zn) for j in range(zm)])
+mult_tup = lambda a, b, zn, zm: ((a[0] * b[0]) % zn, (a[1] * b[1]) % zm)
+add_tup = lambda a, b, zn, zm: ((a[0] + b[0]) % zn, (a[1] + b[1]) % zm)
+empty_table = lambda col, row, symbol: [[symbol] + [str(x) for x in col]] + [[str(x)] + [0 for i in range(len(col))] for x in row]
+
+def make_cartesian_congruence_table(zn, zm, op, mapping, symbol):
+  cp = cartesian_prod(zn, zm)
+  mapped_cp = [cp[mapping[i]] for i in range(len(cp))]
+  table = empty_table(mapped_cp, mapped_cp, symbol)
+  for i in range(len(cp)):
+    for j in range(len(cp)):
+      table[i+1][j+1] = str(op(mapped_cp[i], mapped_cp[j], zn, zm))
+  return table
+
+def make_normal_table(n, op, symbol):
+  table = empty_table(list(range(n)), list(range(n)), symbol)
+  for i in range(n):
+    for j in range(n):
+      table[i+1][j+1] = str(op(i, j) % n)
+  return table
+\end{verbatim}
+
+\subsubsection{Z\textsubscript{2} \texttimes{} Z\textsubscript{3} with bijection}
+\label{sec:orgcac27c3}
+\begin{verbatim}
+bijection = [0, 4, 2, 3, 1, 5]
+\end{verbatim}
+
+\begin{enumerate}
+\item Multiplication Tables
+\label{sec:org1f4aa9f}
+\begin{verbatim}
+make_cartesian_congruence_table(2, 3, mult_tup, bijection, '\odot')
+\end{verbatim}
+
+\begin{center}
+\begin{tabular}{lllllll}
+\(\odot\) & (0, 0) & (1, 1) & (0, 2) & (1, 0) & (0, 1) & (1, 2)\\[0pt]
+(0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0)\\[0pt]
+(1, 1) & (0, 0) & (1, 1) & (0, 2) & (1, 0) & (0, 1) & (1, 2)\\[0pt]
+(0, 2) & (0, 0) & (0, 2) & (0, 1) & (0, 0) & (0, 2) & (0, 1)\\[0pt]
+(1, 0) & (0, 0) & (1, 0) & (0, 0) & (1, 0) & (0, 0) & (1, 0)\\[0pt]
+(0, 1) & (0, 0) & (0, 1) & (0, 2) & (0, 0) & (0, 1) & (0, 2)\\[0pt]
+(1, 2) & (0, 0) & (1, 2) & (0, 1) & (1, 0) & (0, 2) & (1, 1)\\[0pt]
+\end{tabular}
+\end{center}
+
+\item Addition Tables
+\label{sec:org807c598}
+\begin{verbatim}
+make_cartesian_congruence_table(2, 3, add_tup, bijection, '\oplus')
+\end{verbatim}
+
+\begin{center}
+\begin{tabular}{lllllll}
+\(\oplus\) & (0, 0) & (1, 1) & (0, 2) & (1, 0) & (0, 1) & (1, 2)\\[0pt]
+(0, 0) & (0, 0) & (1, 1) & (0, 2) & (1, 0) & (0, 1) & (1, 2)\\[0pt]
+(1, 1) & (1, 1) & (0, 2) & (1, 0) & (0, 1) & (1, 2) & (0, 0)\\[0pt]
+(0, 2) & (0, 2) & (1, 0) & (0, 1) & (1, 2) & (0, 0) & (1, 1)\\[0pt]
+(1, 0) & (1, 0) & (0, 1) & (1, 2) & (0, 0) & (1, 1) & (0, 2)\\[0pt]
+(0, 1) & (0, 1) & (1, 2) & (0, 0) & (1, 1) & (0, 2) & (1, 0)\\[0pt]
+(1, 2) & (1, 2) & (0, 0) & (1, 1) & (0, 2) & (1, 0) & (0, 1)\\[0pt]
+\end{tabular}
+\end{center}
+\end{enumerate}
+
+\subsubsection{Z\textsubscript{6}}
+\label{sec:orgbe8adf1}
+\begin{enumerate}
+\item Multiplication Tables
+\label{sec:orga98b12b}
+\begin{verbatim}
+make_normal_table(6, lambda a, b: a * b, '\odot')
+\end{verbatim}
+
+\begin{center}
+\begin{tabular}{rrrrrrr}
+\(\odot\) & 0 & 1 & 2 & 3 & 4 & 5\\[0pt]
+0 & 0 & 0 & 0 & 0 & 0 & 0\\[0pt]
+1 & 0 & 1 & 2 & 3 & 4 & 5\\[0pt]
+2 & 0 & 2 & 4 & 0 & 2 & 4\\[0pt]
+3 & 0 & 3 & 0 & 3 & 0 & 3\\[0pt]
+4 & 0 & 4 & 2 & 0 & 4 & 2\\[0pt]
+5 & 0 & 5 & 4 & 3 & 2 & 1\\[0pt]
+\end{tabular}
+\end{center}
+
+\item Addition Tables
+\label{sec:orgfcf7bde}
+\begin{verbatim}
+make_normal_table(6, lambda a, b: a + b, '\oplus')
+\end{verbatim}
+
+\begin{center}
+\begin{tabular}{rrrrrrr}
+\(\oplus\) & 0 & 1 & 2 & 3 & 4 & 5\\[0pt]
+0 & 0 & 1 & 2 & 3 & 4 & 5\\[0pt]
+1 & 1 & 2 & 3 & 4 & 5 & 0\\[0pt]
+2 & 2 & 3 & 4 & 5 & 0 & 1\\[0pt]
+3 & 3 & 4 & 5 & 0 & 1 & 2\\[0pt]
+4 & 4 & 5 & 0 & 1 & 2 & 3\\[0pt]
+5 & 5 & 0 & 1 & 2 & 3 & 4\\[0pt]
+\end{tabular}
+\end{center}
+\end{enumerate}
+
+\subsection{Question Two}
+\label{sec:org9540aed}
+\begin{verbatim}
+bijection = [0, 3, 2, 1]
+make_cartesian_congruence_table(2, 2, mult_tup, bijection, '\odot')
+\end{verbatim}
+
+\begin{center}
+\begin{tabular}{lllll}
+\(\odot\) & (0, 0) & (1, 1) & (1, 0) & (0, 1)\\[0pt]
+(0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0)\\[0pt]
+(1, 1) & (0, 0) & (1, 1) & (1, 0) & (0, 1)\\[0pt]
+(1, 0) & (0, 0) & (1, 0) & (1, 0) & (0, 0)\\[0pt]
+(0, 1) & (0, 0) & (0, 1) & (0, 0) & (0, 1)\\[0pt]
+\end{tabular}
+\end{center}
+
+\begin{verbatim}
+bijection = [0, 3, 2, 1]
+make_cartesian_congruence_table(2, 2, add_tup, bijection, '\oplus')
+\end{verbatim}
+
+\begin{center}
+\begin{tabular}{lllll}
+\(\oplus\) & (0, 0) & (1, 1) & (1, 0) & (0, 1)\\[0pt]
+(0, 0) & (0, 0) & (1, 1) & (1, 0) & (0, 1)\\[0pt]
+(1, 1) & (1, 1) & (0, 0) & (0, 1) & (1, 0)\\[0pt]
+(1, 0) & (1, 0) & (0, 1) & (0, 0) & (1, 1)\\[0pt]
+(0, 1) & (0, 1) & (1, 0) & (1, 1) & (0, 0)\\[0pt]
+\end{tabular}
+\end{center}
+
+\subsection{Question Three}
+\label{sec:org421e770}
+\begin{enumerate}
+\item \(f\) is injective, since \(f(a) = f(b) \Rightarrow (a,a) = (b,b) \Rightarrow a = b\)
+\item \(f\) is surjective since every range element in \(R^*\), \((a,a)\) is mapped to \(a \in R\) by definition
+\item \(f(a) + f(b) = (a, a) + (b, b) = (a + b, a + b) = f(a + b)\) and \(f(a)f(b) = (a, a) \cdot (b, b) = (ab, ab) = f(ab)\)
+\end{enumerate}
+
+\subsection{Question Four}
+\label{sec:org1f8479a}
+\(f(1)f(3) = (2)(6) \equiv_{10} 2\) but \(f(3) = 6\) which doesn't hold the properties of homomorphism.
+
+\subsection{Question Five}
+\label{sec:orgeb656c3}
+Consider the given function:
+\begin{enumerate}
+\item \(f\) is injective, since \(f(a) = f(b) \Rightarrow\) \(\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}\) = \(\begin{smallmatrix} 0 & 0 \\ 0 & b \end{smallmatrix}\) \(\Rightarrow a = b\)
+\item \(f\) is surjective, since every range element \(\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}\) has an element in \(\mathds{R}\), \(a\), such that \(f(a) =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}\) by definition
+\item \(f(a) + f(b) =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}\) + \(\begin{smallmatrix} 0 & 0 \\ 0 & b \end{smallmatrix}\) = \(\begin{smallmatrix} 0 & 0 \\ 0 & (a + b) \end{smallmatrix}\) = \(f(a + b)\),
+and \(f(a) \cdot f(b) =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix} 0 & 0 \\ 0 & b \end{smallmatrix}\) = \(\begin{smallmatrix} 0 & 0 \\ 0 & (a \cdot b) \end{smallmatrix}\) = \(f(a \cdot b)\),
+\end{enumerate}
+
+\subsection{Question Nine - TODO}
+\label{sec:org2f29e8e}
+
+\subsection{Question Eleven}
+\label{sec:org3e455ed}
+\subsubsection{b}
+\label{sec:org24734c1}
+\begin{verbatim}
+>>> f = lambda x: 3 * x
+>>> list(filter(lambda x: f(x * x) != (f(x) * f(x)), range(0, 20, 2)))
+[2, 4, 6, 8, 10, 12, 14, 16, 18]
+\end{verbatim}
+\subsubsection{d}
+\label{sec:orgc2ba394}
+\begin{verbatim}
+>>> k = lambda x: 0 if x == 0 else (x ** -1)
+>>> list(filter(lambda x: k(x * x) != (k(x) * k(x)), range(0, 20)))
+[5, 10, 13, 19]
+\end{verbatim}
+\subsection{Question Twelve}
+\label{sec:org95ce95e}
+\subsubsection{c}
+\label{sec:org77d2e96}
+Not a homomorphism. Just from reducing \(f(x + x)\) we find \(f(x + x) \neq f(x) + f(x)\):
+
+\begin{equation*}
+f(x + x) = \dfrac{1}{(x + x)^2 + 1} = \dfrac{1}{4x^2 + 1}
+\end{equation*}
+
+\begin{equation*}
+f(x) + f(x) = \dfrac{1}{x^2 + 1} + \dfrac{1}{x^2 + 1} = \dfrac{2}{x^2 + 1}
+\end{equation*}
+
+Thus \(f(x + x) \neq f(x) + f(x)\).
+\subsubsection{d}
+\label{sec:org1c2b201}
+\begin{verbatim}
+>>> import numpy as np
+>>> h = lambda x: [[-x, 0], [x, 0]]
+>>> list(filter(lambda x: not np.array_equiv(h(x * x), np.dot(h(x), h(x))), range(0, 20))
+[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
+\end{verbatim}
+\subsection{Question Thirteen}
+\label{sec:org57e2c18}
+\subsubsection{a}
+\label{sec:org7515001}
+If \(r \in R\) then \((r, 0_S) \in R \times S\), and \(f((r, 0_S)) = r\), so every range element has a domain element mapping to it.
+
+For some \(a = (r, m) \in R \times S\) and \(b = (t, v) \in R \times S\) then \(f((r, m)) + f((t, v)) = r + t = f((r + t, v + m))\), and
+\(f((r, m)) \cdot f((t, v)) = r \cdot t = f((r \cdot t, v \cdot m))\).
+
+\subsubsection{b}
+\label{sec:orgfb2a3ab}
+If \(s \in S\) then \((0_R, s) \in R \times S\), and \(f((0_R, s)) = s\), so every range element has a domain element mapping to it.
+
+For some \(a = (r, m) \in R \times S\) and \(b = (t, v) \in R \times S\) then \(f((r, m)) + f((t, v)) = m + v = f((r + t, m + v))\), and
+\(f((r, m)) \cdot f((t, v)) = m \cdot v = f((r \cdot t, m \cdot v))\).
+
+\subsection{Question Fifteen}
+\label{sec:org4046d12}
+Consider \(f: Z_4 \rightarrow Z_4 \ni f(x) = 0\), then 2 is a zero divisor, but 0 is not by definition.
+
+\subsection{Question Twenty One}
+\label{sec:org48d8cc7}
+\subsubsection{Lemma}
+\label{sec:org3598e64}
+We assume a multiplicative identity \(x\) exists in \(\mathds{Z}^{*}\):
+
+\(b \odot x = b \Rightarrow b + x - bx = b \Rightarrow x - bx = 0 \Rightarrow x(1-b) = 0\) and \(1-b \neq 0 \forall b \in \mathds{Z}^{*}\) so \(x = 0\).
+
+which is verified by:
+
+\(0 \odot b = 0 + b - 0 \cdot b = b\) and \(b \odot 0 = b + 0 - b \cdot 0 = b\).
+
+\subsubsection{Proof}
+\label{sec:orgee74248}
+
+Consider \(f : \mathds{Z}^{} \rightarrow \mathds{Z}^{*}\) to be the isomorphism we so desire, then by Theorem 3.10,
+f(1 \(\in\) \mathds{Z}) = 0 \(\in\) \mathds{Z}\textsuperscript{*}.
+
+Therefore, f(2) = f(1 + 1) = f(1) \(\oplus\) f(1) = -1, f(3) = f(2 + 1) = -1 \(\oplus\) f(1) = -2.
+
+\(f : \mathds{Z}^{} \rightarrow \mathds{Z}^{*} \ni f(x) = 1 - x\) seems to fit the bill nicely.
+
+\begin{enumerate}
+\item \(f\) is a bijection since we have an inverse \(x = 1 - f(x)\)
+\item \(f(a \oplus b) = 1 - (a \oplus b) = 1 - (a + b - 1) = (1 - a) + (1 - b) = f(a) + f(b)\) and
+\(f(a \odot b) = 1 - (a \odot b) = 1 - (a + b - ab) = (1-a) \cdot (1-b) = f(a) \cdot f(b)\)
+\end{enumerate}
+
+\subsection{Question Twenty Four}
+\label{sec:org07a0b7a}
+\subsubsection{a}
+\label{sec:org897ff21}
+\begin{enumerate}
+\item By the usual coordinate addition we know that \(a, b \in R \Rightarrow a + b \in R\) and is associative, commutative, and the additive identity is \((0, 0)\).
+\item \(R\) is closed under multiplication: \(a, b \in R \Rightarrow a = (c, d) \wedge b = (e, f) \Rightarrow a \cdot b = (ce, de)\) and since \(c, d, e, f \in \mathds{R}\) then \((ce, de) \in \mathds{R} \times \mathds{R}\) by definition.
+\item Multiplication is associative: \(a, b, c \in R \Rightarrow a = (d, e) \wedge b = (f, g) \wedge c = (h, i) \Rightarrow (a \cdot b) \cdot c = (df, ef) \cdot (h, i) = (dfh, efh)\) and \(a \cdot (b \cdot c) = (d, e) \cdot (fh, gh) = (dfh, efh)\)
+\item Multiplication is distributive: \(a, b, c \in R \Rightarrow a = (d, e) \wedge b = (f, g) \wedge c = (h, i) \Rightarrow a(b + c) = (e, f) \cdot (f + h, g + i) = (ef + eh, ff + fh) = (ef, ff) + (eh, fh) = a \cdot b + a \cdot c\)
+and \(a, b, c \in R \Rightarrow a = (d, e) \wedge b = (f, g) \wedge c = (h, i) \Rightarrow (a + b) \cdot c = (d + f, e + g) \cdot (h, i) = (dh + fh, eh + gh) = (dh, eh) + (fh, gh) = a \cdot c + b \cdot c\)
+\end{enumerate}
+\subsubsection{b}
+\label{sec:org92e2744}
+Consider the function \(f: R \rightarrow M(\mathds{R})\) is an isomorphism, such that \(f((a, b)) =\) \(\begin{smallmatrix} a & 0 \\ b & 0 \end{smallmatrix}\), then:
+\begin{enumerate}
+\item \(f\) is surjective since every element in the range has a domain element \(\begin{smallmatrix} a & 0 \\ b & 0 \end{smallmatrix} = y \ni f((a, b)) = y\)
+\item \(f\) is injective since \(f((c, d) = x) = f((e, f) = y) \Rightarrow\) \(\begin{smallmatrix} c & 0 \\ d & 0 \end{smallmatrix}\) = \(\begin{smallmatrix} e & 0 \\ f & 0 \end{smallmatrix}\) \(\Rightarrow c = e \wedge d = f \Rightarrow x = y\)
+\item \(f((c, d) = x) + f((e, f) = y) \Rightarrow\) \(\begin{smallmatrix} c & 0 \\ d & 0 \end{smallmatrix}\) + \(\begin{smallmatrix} e & 0 \\ f & 0 \end{smallmatrix}\) = \(\begin{smallmatrix} c + e & 0 \\ d + f & 0 \end{smallmatrix}\)
+\(= f(x + y)\), and \(f((c, d) = x) \cdot f((e, f) = y) \Rightarrow\) \(\begin{smallmatrix} c & 0 \\ d & 0 \end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix} e & 0 \\ f & 0 \end{smallmatrix}\) = \(\begin{smallmatrix} ce & 0 \\ de & 0 \end{smallmatrix}\)
+\(= f(x \cdot y)\)
+\end{enumerate}
+\end{document}
\ No newline at end of file
diff --git a/Homework/math4310/alg_structures_assn_6.org b/Homework/math4310/alg_structures_assn_6.org
new file mode 100644
index 0000000..501130d
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_6.org
@@ -0,0 +1,75 @@
+#+TITLE: Assignment Five
+#+AUTHOR: Lizzy Hunt
+#+STARTUP: entitiespretty fold inlineimages
+#+LATEX_HEADER: \notindent \notag \usepackage{ dsfont } \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry}
+#+LATEX: \setlength\parindent{0pt}
+#+OPTIONS: toc:nil
+
+* Section 4.1
+** Question One
+*** d
+$2x^5 + x^4 + 6x^2 + 3x + 2$
+
+** Question Three
+*** b
+\begin{verbatim}
+>>> list(filter(lambda x: x, \
+    [f"{a}x^2 + {b}x + {c}" if not a == 0 or not b == 0 else "" \
+    for a in range(3) for b in range(3) for c in range(3)]))
+\end{verbatim}
+
+{'0x^2 + 1x + 0',
+ '0x^2 + 1x + 1',
+ '0x^2 + 1x + 2',
+ '0x^2 + 2x + 0',
+ '0x^2 + 2x + 1',
+ '0x^2 + 2x + 2',
+ '1x^2 + 0x + 0',
+ '1x^2 + 0x + 1',
+ '1x^2 + 0x + 2',
+ '1x^2 + 1x + 0',
+ '1x^2 + 1x + 1',
+ '1x^2 + 1x + 2',
+ '1x^2 + 2x + 0',
+ '1x^2 + 2x + 1',
+ '1x^2 + 2x + 2',
+ '2x^2 + 0x + 0',
+ '2x^2 + 0x + 1',
+ '2x^2 + 0x + 2',
+ '2x^2 + 1x + 0',
+ '2x^2 + 1x + 1',
+ '2x^2 + 1x + 2',
+ '2x^2 + 2x + 0',
+ '2x^2 + 2x + 1',
+ '2x^2 + 2x + 2'}
+
+** Question Five
+Unfortunately, latex'ing long division is not trivial. Sorry in advance.
+
+#+attr_latex: :width 400px
+[[./q5.jpeg]]
+
+** Question Six
+*** c
+No, since two polynomials of degree $\leq k$ with $k=2$, under multiplication, could produce
+a polynomial of degree 4.
+*** d
++ Closed under addition (two polynomials with even degrees can only add to polynomials with even degrees)
++ Closed under multiplication (two polynomials with even degrees can only multiply to polynomials with even degrees)
++ 0_R exists in the set
++ For a polynomial $a$ with only even powers, then $a + x = 0_R$ implies that $a = -x$ which will only change coefficients
+
+Seems like a subring (although not rigorously shown) to me!
+
+*** e
+Not a subring, since $x^3 \cdot x = x^4$, with $x^3$ and $x$ both elements in the described set.
+** Question Eleven
+
+\begin{verbatim}
+>>> list(filter(lambda x: ((3 + x) % 9 == 0) and ((3 * x) % 9 == 0), range(9)))
+[6]
+\end{verbatim}
+
+$(1 + 3x)(1 + 6x) = 1 + 6x + 3x + 18x^2 = 1 + 9x + 18x^2 = 1$
+
+
diff --git a/Homework/math4310/alg_structures_assn_6.pdf b/Homework/math4310/alg_structures_assn_6.pdf
new file mode 100644
index 0000000..dc9d43e
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_6.pdf
diff --git a/Homework/math4310/alg_structures_assn_6.tex b/Homework/math4310/alg_structures_assn_6.tex
new file mode 100644
index 0000000..3ca16ba
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_6.tex
@@ -0,0 +1,111 @@
+% Created 2023-03-01 Wed 08:52
+% 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,portrait]{geometry}
+\author{Lizzy Hunt}
+\date{\today}
+\title{Assignment Five}
+\hypersetup{
+ pdfauthor={Lizzy Hunt},
+ pdftitle={Assignment Five},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.6.1)}, 
+ pdflang={English}}
+\begin{document}
+
+\maketitle
+\setlength\parindent{0pt}
+
+\section{Section 4.1}
+\label{sec:orge1c762a}
+\subsection{Question One}
+\label{sec:org6594d20}
+\subsubsection{d}
+\label{sec:orgf9d77b6}
+\(2x^5 + x^4 + 6x^2 + 3x + 2\)
+
+\subsection{Question Three}
+\label{sec:org8e6afc5}
+\subsubsection{b}
+\label{sec:org7f9f78d}
+\begin{verbatim}
+>>> list(filter(lambda x: x, \
+    [f"{a}x^2 + {b}x + {c}" if not a == 0 or not b == 0 else "" \
+    for a in range(3) for b in range(3) for c in range(3)]))
+\end{verbatim}
+
+\{'0x\textsuperscript{2} + 1x + 0',
+ '0x\textsuperscript{2} + 1x + 1',
+ '0x\textsuperscript{2} + 1x + 2',
+ '0x\textsuperscript{2} + 2x + 0',
+ '0x\textsuperscript{2} + 2x + 1',
+ '0x\textsuperscript{2} + 2x + 2',
+ '1x\textsuperscript{2} + 0x + 0',
+ '1x\textsuperscript{2} + 0x + 1',
+ '1x\textsuperscript{2} + 0x + 2',
+ '1x\textsuperscript{2} + 1x + 0',
+ '1x\textsuperscript{2} + 1x + 1',
+ '1x\textsuperscript{2} + 1x + 2',
+ '1x\textsuperscript{2} + 2x + 0',
+ '1x\textsuperscript{2} + 2x + 1',
+ '1x\textsuperscript{2} + 2x + 2',
+ '2x\textsuperscript{2} + 0x + 0',
+ '2x\textsuperscript{2} + 0x + 1',
+ '2x\textsuperscript{2} + 0x + 2',
+ '2x\textsuperscript{2} + 1x + 0',
+ '2x\textsuperscript{2} + 1x + 1',
+ '2x\textsuperscript{2} + 1x + 2',
+ '2x\textsuperscript{2} + 2x + 0',
+ '2x\textsuperscript{2} + 2x + 1',
+ '2x\textsuperscript{2} + 2x + 2'\}
+
+\subsection{Question Five}
+\label{sec:orgf501bb3}
+Unfortunately, latex'ing long division is not trivial. Sorry in advance.
+
+\begin{center}
+\includegraphics[width=400px]{./q5.jpeg}
+\end{center}
+
+\subsection{Question Six}
+\label{sec:orgf9d6188}
+\subsubsection{c}
+\label{sec:org0feefbc}
+No, since two polynomials of degree \(\leq k\) with \(k=2\), under multiplication, could produce
+a polynomial of degree 4.
+\subsubsection{d}
+\label{sec:org18fd125}
+\begin{itemize}
+\item Closed under addition (two polynomials with even degrees can only add to polynomials with even degrees)
+\item Closed under multiplication (two polynomials with even degrees can only multiply to polynomials with even degrees)
+\item 0\textsubscript{R} exists in the set
+\item For a polynomial \(a\) with only even powers, then \(a + x = 0_R\) implies that \(a = -x\) which will only change coefficients
+\end{itemize}
+
+Seems like a subring (although not rigorously shown) to me!
+
+\subsubsection{e}
+\label{sec:org6de4719}
+Not a subring, since \(x^3 \cdot x = x^4\), with \(x^3\) and \(x\) both elements in the described set.
+\subsection{Question Eleven}
+\label{sec:orgb3179d8}
+
+\begin{verbatim}
+>>> list(filter(lambda x: ((3 + x) % 9 == 0) and ((3 * x) % 9 == 0), range(9)))
+[6]
+\end{verbatim}
+
+\((1 + 3x)(1 + 6x) = 1 + 6x + 3x + 18x^2 = 1 + 9x + 18x^2 = 1\)
+\end{document}
\ No newline at end of file
diff --git a/Homework/math4310/alg_structures_assn_7.org b/Homework/math4310/alg_structures_assn_7.org
new file mode 100644
index 0000000..d683a34
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_7.org
@@ -0,0 +1,175 @@
+#+TITLE: Assignment Seven
+#+AUTHOR: Lizzy Hunt
+#+STARTUP: entitiespretty fold inlineimages
+#+LATEX_HEADER: \notindent \notag \usepackage{ dsfont } \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry}
+#+LATEX: \setlength\parindent{0pt}
+#+OPTIONS: toc:nil
+
+
+* Section 4.2
+** Question Five
+*** b
+\begin{align*}
+x^5 + x^4 + 2x^3 - x^2 - x - 2 &= (x - 1)(x^4 + 2x^3 + 5x^2 + 4x + 4) + (-x^3 - x + 2) \\
+x^4 + 2x^3 + 5x^2 + 4x + 4 &= (-x-2)(-x^3 - x + 2) + (4x^2 + 4x + 8)\\
+-x^3 - x + 2 &= (\frac{-x}{4})(4x^2 + 4x + 8) + 0
+\end{align*}
+
+$(4x^2 + 4x + 8) = 4(x^2 + x + 2)$
+
+$x^2 + x + 2$
+*** c
+$deg(d) = 2$ is the greatest degree of a possible common divisor, so we'll stick with $x^2 - 1$.
+*** g
+\begin{align*}
+2x^4 + 5x^3 - 5x - 2 &= (x + 4)(2x^3 - 3x^2 - 2x) + (14x^2 + 3x - 2) \\
+2x^3 - 3x^2 - 2x &= (\frac{x}{7} - \frac{12}{49})(14x^2 + 3x - 2) + (\frac{-48x - 24}{49}) \\
+14x^2 + 3x - 2 &= (\frac{-7x}{24})(\frac{-48x - 24}{49}) + 0
+\end{align*}
+
+$\frac{-48x - 24}{49} = \left(\frac{49}{-48}\right)\left(\frac{\left(-48x-24\right)}{49}\right) = x + \frac{1}{2}$
+
+$x + \frac{1}{2}$
+** Question Ten
+$x^3 - 3abx + a^3 + b^3$ can be factored to $(a + b + x) (a^2 - a b - a x + b^2 - b x + x^2)$, so $a + b + x$ is the gcd.
+
+* Section 4.3
+** Question Three
+*** a
+{ $x^2 + x + 1$, $2x^2 + 2x + 2$, $3x^2 + 3x + 3$, $4x^2 + 4x + 4$ }
+*** b
+{ $3x + 2$, $6x + 4$, $2x + 6$, $5x + 1$, $x + 3$, $4x + 5$ }
+
+** Question Six
+Assume that $x^2 + 1$ is in fact reducible. Then, $x^2 + 1 = (ax + b)(cx + d)$. Thus, $ax \cdot cx = x^2 \Rightarrow ac = 1$, $axd + bcx = 0 \Rightarrow ad + bc = 0$, and $bd = 1$ with $a,b,c,d$ all being nonzero.
+
+Then, $a = \frac{1}{c}$ and $d = \frac{1}{b}$, so $ad = -bc \Rightarrow \frac{1}{cb} = - bc$, which is impossible.
+
+** Question Nine
+*** a
+$x^2 + x + 1$
+
+*** b
+$x^3 + x^2 + 1$ and $x^3 + x + 1$
+
+** Question Ten
+*** a
+In $\mathds{Q}[x]$, no. In $\mathds{R}[x]$, yes: $x^2 - 3 = (x - \sqrt{3})(x + \sqrt{3})$.
+
+*** b
+Yes, in both fields: $x^2 + x - 2 = (x + 2)(x - 1)$
+
+** Question Eleven
+Assume that $x^3 - 3$ were reducible in $\mathds{Z}_7 [x]$. Then, there must be a monic factor of degree one, as
+$x^3 - 3 = g(x)h(x) \Rightarrow deg(x^3 - 3) = deg(g(x)h(x)) \Rightarrow 3 = deg(g(x)) + deg(h(x))$ by Theorem 4.2, and either $deg(g(x))$ or $deg(h(x))$ must be one, with the other, two.
+
+So, one of the factors must be of the form $x + a, a \in \mathds{Z}_7$. None such factor exists since polynomial division always returns a non-zero remainder:
+
+$x \nmid x^3 - 3$, trivially.
+
+$\frac{x^3 - 3}{x + 1}$ gives a remainder of $4$.
+
+$\frac{x^3 - 3}{x + 2}$ gives a remainder of $-11 \equiv_7 3$.
+
+$\frac{x^3 - 3}{x + 3}$ gives a remainder of $-30 \equiv_7 5$.
+
+$\frac{x^3 - 3}{x + 4}$ gives a remainder of $-67 \equiv_7 3$.
+
+$\frac{x^3 - 3}{x + 5}$ gives a remainder of $-128 \equiv_7 5$.
+
+$\frac{x^3 - 3}{x + 6}$ gives a remainder of $-219 \equiv_7 5$.
+
+** Question Twelve
+In $\mathds{Q}[x]$, $x^4 - 4 = (x^2 - 2)(x^2 + 2)$
+
+In $\mathds{R}[x]$, $x^4 - 4 = (x - \sqrt{2})(x + \sqrt{2})(x^2 + 2)$
+
+In $\mathds{C}[x]$, $x^4 - 4 = (x - \sqrt{2})(x + \sqrt{2})(x - i \sqrt{2})(x + i \sqrt{2})$
+
+** Question Fourteen
+$x^2 + x \equiv_6 (x + 4)(x + 3)$
+
+$x^2 + x \equiv_6 x(x + 1)$
+
+* Section 4.4
+** Question Two
+*** c
+By Theorem 4.15, the remainder is $f(-1) = 5$
+
+** Question Three
+*** c
+If the remainder of $\frac{f(x)}{h(x)}$ is 0, then $h$ is a factor of $f$, so using
+Theorem 4.15 we can find that $f(-2) = -55 \equiv_5 0$, so $h$ is indeed a factor.
+
+** Question Four
+*** b
+We need to find k such that $f(-1) = 0 (mod 5)$
+
+\begin{verbatim}
+>>> f = lambda k,x: (x**4 + 2 * x**3 - 3 * x**2 + k * x + 1) % 5
+>>> for i in range(5):
+...   if f(i, -1) == 0:
+...     print(i)
+...
+2
+\end{verbatim}
+
+$k=2$ works nicely!
+
+** Question Seven
+\begin{verbatim}
+>>> set(filter(lambda x: (x**7 - x) % 7 == 0, range(7)))
+{0, 1, 2, 3, 4, 5, 6}
+\end{verbatim}
+
+Shows that each element is a root, so the factoring is correct by the Factor Theorem.
+
+** Question Eight
+*** b
+The polynomial is irreducible since its only roots are $\pm \sqrt{7} \notin \mathds{Q}$, by Corollary 4.19
+
+*** d
+\begin{verbatim}
+>>> set(filter(lambda x: (2 * x**3 + x**2 + 2 * x + 2) % 5 == 0, range(5)))
+set()
+\end{verbatim}
+
+It's irreducible since there are no roots in $\mathds{Z}_5$.
+
+** Question Nine
+\begin{verbatim}
+>>> def find_irr_mon_polys(deg, mod):
+...    z_s = range(mod)
+...    polys = set()
+...    for b in z_s:
+...       for c z_s:
+...          f_repr = f"x^2 + {b}x + {c}"
+...          f = lambda x: (x**2 + b*x + c) % mod
+...          if not any(map(lambda x: f(x) == 0, z_s)):
+...              polys.add(f_repr)
+...    return polys
+
+>>> find_irr_mon_polys(2, 5)
+\end{verbatim}
+
+{ $x^2 + 0x + 2,
+ x^2 + 0x + 3,
+ x^2 + 1x + 1,
+ x^2 + 1x + 2,
+ x^2 + 2x + 3,
+ x^2 + 2x + 4,
+ x^2 + 3x + 3,
+ x^2 + 3x + 4,
+ x^2 + 4x + 1,
+ x^2 + 4x + 2$ }
+ 
+\begin{verbatim}
+>>> find_irr_mon_polys(2, 3)
+\end{verbatim}
+{ $x^2 + 0x + 1, x^2 + 1x + 2, x^2 + 2x + 2$ }
+
+** Question Thirteen
+*** a
+If $f(x) = cg(x)$ with $c \neq 0_F$, then $g(x) = c^{-1}f(x)$ and $f(x) = c^{-1}g(x)$. Then, $g(y) = 0_F \Leftrightarrow f(y) = 0_F$.
+*** b
+No, consider $f(x) = x$ and $g(x) = x^2$ in $\mathds{Z}$, then $f$ and $g$ share $0$ as their only root, but they are not associates.
diff --git a/Homework/math4310/alg_structures_assn_7.pdf b/Homework/math4310/alg_structures_assn_7.pdf
new file mode 100644
index 0000000..535c3c8
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_7.pdf
diff --git a/Homework/math4310/alg_structures_assn_7.tex b/Homework/math4310/alg_structures_assn_7.tex
new file mode 100644
index 0000000..d38f7d9
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_7.tex
@@ -0,0 +1,234 @@
+% Created 2023-03-16 Thu 18:22
+% 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,portrait]{geometry}
+\author{Lizzy Hunt}
+\date{\today}
+\title{Assignment Seven}
+\hypersetup{
+ pdfauthor={Lizzy Hunt},
+ pdftitle={Assignment Seven},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.6.1)}, 
+ pdflang={English}}
+\begin{document}
+
+\maketitle
+\setlength\parindent{0pt}
+
+
+\section{Section 4.2}
+\label{sec:org39ef51e}
+\subsection{Question Five}
+\label{sec:orgafa157f}
+\subsubsection{b}
+\label{sec:org072214f}
+\begin{align*}
+x^5 + x^4 + 2x^3 - x^2 - x - 2 &= (x - 1)(x^4 + 2x^3 + 5x^2 + 4x + 4) + (-x^3 - x + 2) \\
+x^4 + 2x^3 + 5x^2 + 4x + 4 &= (-x-2)(-x^3 - x + 2) + (4x^2 + 4x + 8)\\
+-x^3 - x + 2 &= (\frac{-x}{4})(4x^2 + 4x + 8) + 0
+\end{align*}
+
+\((4x^2 + 4x + 8) = 4(x^2 + x + 2)\)
+
+\(x^2 + x + 2\)
+\subsubsection{c}
+\label{sec:orgb01ae88}
+\(deg(d) = 2\) is the greatest degree of a possible common divisor, so we'll stick with \(x^2 - 1\).
+\subsubsection{g}
+\label{sec:org4766198}
+\begin{align*}
+2x^4 + 5x^3 - 5x - 2 &= (x + 4)(2x^3 - 3x^2 - 2x) + (14x^2 + 3x - 2) \\
+2x^3 - 3x^2 - 2x &= (\frac{x}{7} - \frac{12}{49})(14x^2 + 3x - 2) + (\frac{-48x - 24}{49}) \\
+14x^2 + 3x - 2 &= (\frac{-7x}{24})(\frac{-48x - 24}{49}) + 0
+\end{align*}
+
+\(\frac{-48x - 24}{49} = \left(\frac{49}{-48}\right)\left(\frac{\left(-48x-24\right)}{49}\right) = x + \frac{1}{2}\)
+
+\(x + \frac{1}{2}\)
+\subsection{Question Ten}
+\label{sec:org6ab06e1}
+\(x^3 - 3abx + a^3 + b^3\) can be factored to \((a + b + x) (a^2 - a b - a x + b^2 - b x + x^2)\), so \(a + b + x\) is the gcd.
+
+\section{Section 4.3}
+\label{sec:org9759eff}
+\subsection{Question Three}
+\label{sec:orgf67495a}
+\subsubsection{a}
+\label{sec:org3a3c23d}
+\{ \(x^2 + x + 1\), \(2x^2 + 2x + 2\), \(3x^2 + 3x + 3\), \(4x^2 + 4x + 4\) \}
+\subsubsection{b}
+\label{sec:org0eff32f}
+\{ \(3x + 2\), \(6x + 4\), \(2x + 6\), \(5x + 1\), \(x + 3\), \(4x + 5\) \}
+
+\subsection{Question Six}
+\label{sec:org58db881}
+Assume that \(x^2 + 1\) is in fact reducible. Then, \(x^2 + 1 = (ax + b)(cx + d)\). Thus, \(ax \cdot cx = x^2 \Rightarrow ac = 1\), \(axd + bcx = 0 \Rightarrow ad + bc = 0\), and \(bd = 1\) with \(a,b,c,d\) all being nonzero.
+
+Then, \(a = \frac{1}{c}\) and \(d = \frac{1}{b}\), so \(ad = -bc \Rightarrow \frac{1}{cb} = - bc\), which is impossible.
+
+\subsection{Question Nine}
+\label{sec:orge1fd195}
+\subsubsection{a}
+\label{sec:org33e3a63}
+\(x^2 + x + 1\)
+
+\subsubsection{b}
+\label{sec:orgf9e0897}
+\(x^3 + x^2 + 1\) and \(x^3 + x + 1\)
+
+\subsection{Question Ten}
+\label{sec:org4da4dbc}
+\subsubsection{a}
+\label{sec:orgc043f8c}
+In \(\mathds{Q}[x]\), no. In \(\mathds{R}[x]\), yes: \(x^2 - 3 = (x - \sqrt{3})(x + \sqrt{3})\).
+
+\subsubsection{b}
+\label{sec:org68f2a14}
+Yes, in both fields: \(x^2 + x - 2 = (x + 2)(x - 1)\)
+
+\subsection{Question Eleven}
+\label{sec:org957dc65}
+Assume that \(x^3 - 3\) were reducible in \(\mathds{Z}_7 [x]\). Then, there must be a monic factor of degree one, as
+\(x^3 - 3 = g(x)h(x) \Rightarrow deg(x^3 - 3) = deg(g(x)h(x)) \Rightarrow 3 = deg(g(x)) + deg(h(x))\) by Theorem 4.2, and either \(deg(g(x))\) or \(deg(h(x))\) must be one, with the other, two.
+
+So, one of the factors must be of the form \(x + a, a \in \mathds{Z}_7\). None such factor exists since polynomial division always returns a non-zero remainder:
+
+\(x \nmid x^3 - 3\), trivially.
+
+\(\frac{x^3 - 3}{x + 1}\) gives a remainder of \(4\).
+
+\(\frac{x^3 - 3}{x + 2}\) gives a remainder of \(-11 \equiv_7 3\).
+
+\(\frac{x^3 - 3}{x + 3}\) gives a remainder of \(-30 \equiv_7 5\).
+
+\(\frac{x^3 - 3}{x + 4}\) gives a remainder of \(-67 \equiv_7 3\).
+
+\(\frac{x^3 - 3}{x + 5}\) gives a remainder of \(-128 \equiv_7 5\).
+
+\(\frac{x^3 - 3}{x + 6}\) gives a remainder of \(-219 \equiv_7 5\).
+
+\subsection{Question Twelve}
+\label{sec:orge9f1b7e}
+In \(\mathds{Q}[x]\), \(x^4 - 4 = (x^2 - 2)(x^2 + 2)\)
+
+In \(\mathds{R}[x]\), \(x^4 - 4 = (x - \sqrt{2})(x + \sqrt{2})(x^2 + 2)\)
+
+In \(\mathds{C}[x]\), \(x^4 - 4 = (x - \sqrt{2})(x + \sqrt{2})(x - i \sqrt{2})(x + i \sqrt{2})\)
+
+\subsection{Question Fourteen}
+\label{sec:orge490368}
+\(x^2 + x \equiv_6 (x + 4)(x + 3)\)
+
+\(x^2 + x \equiv_6 x(x + 1)\)
+
+\section{Section 4.4}
+\label{sec:orgced6b81}
+\subsection{Question Two}
+\label{sec:org80309c5}
+\subsubsection{c}
+\label{sec:org684714e}
+By Theorem 4.15, the remainder is \(f(-1) = 5\)
+
+\subsection{Question Three}
+\label{sec:org4950bb3}
+\subsubsection{c}
+\label{sec:org38b222f}
+If the remainder of \(\frac{f(x)}{h(x)}\) is 0, then \(h\) is a factor of \(f\), so using
+Theorem 4.15 we can find that \(f(-2) = -55 \equiv_5 0\), so \(h\) is indeed a factor.
+
+\subsection{Question Four}
+\label{sec:org5aa1b14}
+\subsubsection{b}
+\label{sec:orge7c045e}
+We need to find k such that \(f(-1) = 0 (mod 5)\)
+
+\begin{verbatim}
+>>> f = lambda k,x: (x**4 + 2 * x**3 - 3 * x**2 + k * x + 1) % 5
+>>> for i in range(5):
+...   if f(i, -1) == 0:
+...     print(i)
+...
+2
+\end{verbatim}
+
+\(k=2\) works nicely!
+
+\subsection{Question Seven}
+\label{sec:orgecff464}
+\begin{verbatim}
+>>> set(filter(lambda x: (x**7 - x) % 7 == 0, range(7)))
+{0, 1, 2, 3, 4, 5, 6}
+\end{verbatim}
+
+Shows that each element is a root, so the factoring is correct by the Factor Theorem.
+
+\subsection{Question Eight}
+\label{sec:orge5e9620}
+\subsubsection{b}
+\label{sec:org65737b6}
+The polynomial is irreducible since its only roots are \(\pm \sqrt{7} \notin \mathds{Q}\), by Corollary 4.19
+
+\subsubsection{d}
+\label{sec:orga790f86}
+\begin{verbatim}
+>>> set(filter(lambda x: (2 * x**3 + x**2 + 2 * x + 2) % 5 == 0, range(5)))
+set()
+\end{verbatim}
+
+It's irreducible since there are no roots in \(\mathds{Z}_5\).
+
+\subsection{Question Nine}
+\label{sec:org2cf5693}
+\begin{verbatim}
+>>> def find_irr_mon_polys(deg, mod):
+...    z_s = range(mod)
+...    polys = set()
+...    for b in z_s:
+...       for c z_s:
+...          f_repr = f"x^2 + {b}x + {c}"
+...          f = lambda x: (x**2 + b*x + c) % mod
+...          if not any(map(lambda x: f(x) == 0, z_s)):
+...              polys.add(f_repr)
+...    return polys
+
+>>> find_irr_mon_polys(2, 5)
+\end{verbatim}
+
+\{ \(x^2 + 0x + 2,
+ x^2 + 0x + 3,
+ x^2 + 1x + 1,
+ x^2 + 1x + 2,
+ x^2 + 2x + 3,
+ x^2 + 2x + 4,
+ x^2 + 3x + 3,
+ x^2 + 3x + 4,
+ x^2 + 4x + 1,
+ x^2 + 4x + 2\) \}
+
+\begin{verbatim}
+>>> find_irr_mon_polys(2, 3)
+\end{verbatim}
+\{ \(x^2 + 0x + 1, x^2 + 1x + 2, x^2 + 2x + 2\) \}
+
+\subsection{Question Thirteen}
+\label{sec:org4f87227}
+\subsubsection{a}
+\label{sec:org32ba550}
+If \(f(x) = cg(x)\) with \(c \neq 0_F\), then \(g(x) = c^{-1}f(x)\) and \(f(x) = c^{-1}g(x)\). Then, \(g(y) = 0_F \Leftrightarrow f(y) = 0_F\).
+\subsubsection{b}
+\label{sec:org84283d0}
+No, consider \(f(x) = x\) and \(g(x) = x^2\) in \(\mathds{Z}\), then \(f\) and \(g\) share \(0\) as their only root, but they are not associates.
+\end{document}
\ No newline at end of file
diff --git a/Homework/math4310/alg_structures_assn_9.odt b/Homework/math4310/alg_structures_assn_9.odt
new file mode 100644
index 0000000..b874bc3
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_9.odt
diff --git a/Homework/math4310/alg_structures_assn_9.pdf b/Homework/math4310/alg_structures_assn_9.pdf
new file mode 100644
index 0000000..88a2686
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_9.pdf
diff --git a/Homework/math4310/alg_structures_assn_9.tex b/Homework/math4310/alg_structures_assn_9.tex
new file mode 100644
index 0000000..f118c8b
--- /dev/null
+++ b/Homework/math4310/alg_structures_assn_9.tex
@@ -0,0 +1,213 @@
+% Created 2023-03-26 Sun 19:34
+% 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}
+\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:org93c6765}
+\subsection{Question One}
+\label{sec:org1b933f1}
+\subsubsection{b}
+\label{sec:org9c622bd}
+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:org3b14848}
+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:org4de1470}
+\(|\)\{ \(0, 1, x, x + 1, x^2, x^2 + 1, x^2 + x, x^2 + x + 1\) \}\(|\) = 8
+
+\subsection{Question Four}
+\label{sec:org7437458}
+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:orgf9878df}
+By Corollary 5.5, all the congruence classes in \(F[x]\) are \(c \ni c \in F\).
+
+\subsection{Question Eight}
+\label{sec:org7f80a1c}
+\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 - TODO}
+\label{sec:orgbb6f81a}
+
+
+\section{Section 5.2}
+\label{sec:org105955c}
+\subsection{Question One}
+\label{sec:org2696380}
+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 - TODO}
+\label{sec:org55dc879}
+
+\subsection{Question Three - TODO}
+\label{sec:orga648a70}
+
+\subsection{Question Six}
+\label{sec:orgc033877}
+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:org76f0944}
+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:org82d2650}
+\subsection{Question One}
+\label{sec:org2711f62}
+\subsubsection{a}
+\label{sec:org805523d}
+\(x^3 + 2x^2 + x + 1\) 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:orgf91e167}
+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:orgaf8d3bb}
+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:org09929ee}
+\subsubsection{a}
+\label{sec:org889a4d4}
+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:orgbfbb6ba}
+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 of 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])
+\end{itemize}
+
+\subsection{Question Five}
+\label{sec:org246fde8}
+\subsubsection{a}
+\label{sec:org748a4c5}
+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, \(\frac{r}{r^2 - 3s^2} s \frac{s}{r^2 - 3s^2}\sqrt{3}\):
+
+\begin{align*}
+(a + b\sqrt{3})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*}
+\end{document}
\ No newline at end of file
diff --git a/Homework/math4310/alg_structures_midterm_1.org b/Homework/math4310/alg_structures_midterm_1.org
new file mode 100644
index 0000000..db6dc8e
--- /dev/null
+++ b/Homework/math4310/alg_structures_midterm_1.org
@@ -0,0 +1,119 @@
+#+TITLE: Midterm One
+#+AUTHOR: Lizzy Hunt
+#+STARTUP: entitiespretty fold inlineimages
+#+LATEX_HEADER: \notindent \notag \usepackage{ dsfont } \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry}
+#+LATEX: \setlength\parindent{0pt}
+#+OPTIONS: toc:nil
+
+* Question One
+\begin{verbatim}
+In [1]: sols = lambda n: set(filter(lambda x: (x**2 + x) % n == 0, range(n)))
+
+In [2]: sols(5)
+Out[2]: {0, 4}
+
+In [3]: sols(6)
+Out[3]: {0, 2, 3, 5}
+\end{verbatim}
+
+* Question Two
+$p | a \Rightarrow np = a$
+
+$p | a + bc \Rightarrow mp = a + bc \Rightarrow mp = np + bc \Rightarrow (m - n)p = bc$ which implies $p$ is a factor of $bc$ ($p | bc$), so by Theorem 1.5, $p | b$ or $p | c$.
+
+* Question Three
+From the multiplication table we can see that $e$ is the identity element $1_R$, the zero
+element $0_R$ is $z$. From the addition table we find that every element is its own
+additive inverse.
+
+** a
+The units in this ring are ${e, g}$, and the zero divisors are ${a, b, c, f}$.
+
+** b
+$3bd = 3(bd) = 3z = b$
+
+$2ac = 2(ac) = 2b = b + b = z$
+
+$g^{-1}f^3 = (g)(fff) (gf)(ff) = (d)(f) = d$
+
+Thus, $3bd - 2ac + g^{-1}f^3 = b - z + d = b + d = f$.
+
+* Question Four
+To get a hunch,
+
+#+BEGIN_SRC python
+In [1]: import numpy as np
+
+In [2]: def find_counter(matrix_generator, in_ring, search_space=range(-100, 100)):
+...:     for a1 in search_space:
+...:         n = matrix_generator(a1)
+...:         for a2 in search_space:
+...:             m = matrix_generator(a2)
+...:             dot = np.dot(n, m)
+...:             add = np.add(n, m)
+...:
+...:             if not in_ring(dot):
+...:                 return (n, m, dot, '*')
+...:             if not in_ring(add):
+...:                 return (n, m, add, '+')
+...:
+...:     return None
+...:
+
+In [3]: find_counter(lambda a: [[0, a], [0, -a]], \
+                     lambda m: m[0][1] == -m[1][1] and m[0][0] == 0 \
+                     and m[1][0] == 0)
+
+In [4]: find_counter(lambda a: [[0, a], [a, 0]], \
+                     lambda m: m[0][1] == m[1][0] and m[1][1] == 0 \
+                     and m[0][0] == 0)
+Out[4]:
+([[0, -100], [-100, 0]],
+ [[0, -100], [-100, 0]],
+ array([[10000,     0],
+        [    0, 10000]]),
+ '*')
+#+END_SRC
+** a
+Using Theorem 3.2, this is a ring:
+
+1. S_1 is closed under addition: $x, y \in S_1 \Rightarrow x + y =$ $\begin{smallmatrix} 0 & a \\ 0 & -a \end{smallmatrix}$ $+$ $\begin{smallmatrix} 0 & b \\ 0 & -b \end{smallmatrix}$ $\Rightarrow$ $x + y = \begin{smallmatrix} 0 & a + b \\ 0 & (-a) + (-b) \end{smallmatrix} = \begin{smallmatrix} 0 & a + b \\ 0 & -(a + b) \end{smallmatrix}$ which satisfies the rule
+2. S_1 is closed under multiplication: $x, y \in S_1 \Rightarrow x \cdot y =$ $\begin{smallmatrix} 0 & a \\ 0 & -a \end{smallmatrix}$ $\cdot$ $\begin{smallmatrix} 0 & b \\ 0 & -b \end{smallmatrix}$ $\Rightarrow$ $x \cdot y =$ $\begin{smallmatrix} 0 + (a)(0) & 0(b) + a(-b) \\ 0 + (-a)(0) & 0b + (-a)(-b) \end{smallmatrix}$ $=$ $\begin{smallmatrix} 0 & -ab \\ 0 & ab \end{smallmatrix}$ which satisfies the rule
+3. $0_R_{} = 0_{S_1} =$ $\begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix}$
+4. $a \in S_1 \Rightarrow a + x = 0_R \wedge x \in S_1$ since $\begin{smallmatrix} 0 & a \\ 0 & -a \end{smallmatrix}$ $+$ $x = 0_{S_1} =$ $\begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix}$ $\Rightarrow x =$ $\begin{smallmatrix} 0 & -a \\ 0 & a \end{smallmatrix}$ $\in S_1$
+
+** b
+$\begin{smallmatrix} 0 & -100 \\ -100 & 0 \end{smallmatrix}$ $\cdot$ $\begin{smallmatrix} 0 & -100 \\ -100 & 0 \end{smallmatrix}$ = $\begin{smallmatrix} 10000 & 0 \\ 0 & 10000 \end{smallmatrix}$ is not in the ring.
+
+* Question Five
+No, it is not a homomorphism from the definition in 3.3.
+
+Consider $n =$ $\begin{smallmatrix} 0 & 0 \\ 0 & 1 \end{smallmatrix}$ and $m =$ $\begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix}$.
+
+Then $nm =$ $\begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix}$ and thus $Trace(n) \cdot Trace(m) = 1 \wedge Trace(nm) = 0 \Rightarrow f(n)f(m) \neq f(nm)$.
+
+* Question Six
+No, I don't believe $f$ to be an isomorphism.
+** Lemma
+We need to show that $x \in Q \wedge x \sqrt{2} = y \in Q \Rightarrow x = 0$.
+
+If $x \neq 0$ then $y$ can be represented as $\frac{a}{b}$ with $a, b \neq 0 \wedge a,b \in \mathds{Z}$ and $(a, b) = 1$.
+
+So, $x \sqrt{2} = \frac{a}{b} \Rightarrow x \sqrt{2} b = a \Rightarrow x^2(2b^2) = a^2$ and thus $2 | aa$, and by applying Corollary 1.6, $2 | a$.
+
+By substituting $2c = a$ (a is divisible by 2), $x^2(2b^2) = 4c^2 \Rightarrow x^2 b^2 = 2c^2$ which implies $2 | (x^2 b^2) \Rightarrow 2 | x$ or $2 | b$ (again by Corollary 1.6).
+
+But 2 cannot divide $b$ as well as $a$, since $(a, b) = 1$ and $b \neq 0$.
+
+Thus $2 | x$. So $x = 2d \Rightarrow 4d^2(2b^2) = a^2 \Rightarrow 4 | a^2$, so $4e = a$, and by similar logic for $b$ above, $4 | x$.
+Then, subsituting $x = 4f \Rightarrow 16f^2(2b^2) = a^2 \Rightarrow 16 | a^2$. Again, $16 | x$.
+
+In fact, all range elements $R$ of the recursive function on the natural numbers $f \ni f(1) = 2, f(n) = f(n-1)^2 \text{\{} n > 1 \text{\}}$ must divide $x$.
+
+Thus, x must be sup($R$), or 0 (everything divides zero). Obviously, though, $R$ is unbounded on the right, so it follows $x = 0$.
+
+** Proof
+$a, b \in Q(\sqrt{2}) \Rightarrow f(a) = n + m \sqrt{2} \wedge b = x + y \sqrt{2} \Rightarrow f(a) = n - m \sqrt{2} \vee f(b) = x - y \sqrt{2}$.
+
+$f$ is not injective; if we assume $f(a) = f(b) \Rightarrow a = b$ then $f(a) - f(b) = 0 \Rightarrow n - m \sqrt{2} - x - y \sqrt{2} = 0 \Rightarrow 0 = (n-x) - (m + y)\sqrt{2} \Rightarrow n-x = (m+y)\sqrt{2}$ and, as $n-x \in Q$ since
+Q is a ring and by definition, $(m + y)\sqrt{2} \in Q$ implies $(m + y) = 0$ by the lemma, thus $m = -y$. But $a = b \Rightarrow m = y$, a contradiction.
diff --git a/Homework/math4310/alg_structures_midterm_1.pdf b/Homework/math4310/alg_structures_midterm_1.pdf
new file mode 100644
index 0000000..b57c5c6
--- /dev/null
+++ b/Homework/math4310/alg_structures_midterm_1.pdf
diff --git a/Homework/math4310/alg_structures_midterm_1.tex b/Homework/math4310/alg_structures_midterm_1.tex
new file mode 100644
index 0000000..0babb7a
--- /dev/null
+++ b/Homework/math4310/alg_structures_midterm_1.tex
@@ -0,0 +1,157 @@
+% Created 2023-02-21 Tue 22:21
+% 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,portrait]{geometry}
+\author{Lizzy Hunt}
+\date{\today}
+\title{Midterm One}
+\hypersetup{
+ pdfauthor={Lizzy Hunt},
+ pdftitle={Midterm One},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.6.1)}, 
+ pdflang={English}}
+\begin{document}
+
+\maketitle
+\setlength\parindent{0pt}
+
+\section{Question One}
+\label{sec:org0a6f4f9}
+\begin{verbatim}
+In [1]: sols = lambda n: set(filter(lambda x: (x**2 + x) % n == 0, range(n)))
+
+In [2]: sols(5)
+Out[2]: {0, 4}
+
+In [3]: sols(6)
+Out[3]: {0, 2, 3, 5}
+\end{verbatim}
+
+\section{Question Two}
+\label{sec:orgc3f2004}
+\(p | a \Rightarrow np = a\)
+
+\(p | a + bc \Rightarrow mp = a + bc \Rightarrow mp = np + bc \Rightarrow (m - n)p = bc\) which implies \(p\) is a factor of \(bc\) (\(p | bc\)), so by Theorem 1.5, \(p | b\) or \(p | c\).
+
+\section{Question Three}
+\label{sec:org1b549c3}
+From the multiplication table we can see that \(e\) is the identity element \(1_R\), the zero
+element \(0_R\) is \(z\). From the addition table we find that every element is its own
+additive inverse.
+
+\subsection{a}
+\label{sec:org692c3d2}
+The units in this ring are \({e, g}\), and the zero divisors are \({a, b, c, f}\).
+
+\subsection{b}
+\label{sec:orgace7d7f}
+\(3bd = 3(bd) = 3z = b\)
+
+\(2ac = 2(ac) = 2b = b + b = z\)
+
+\(g^{-1}f^3 = (g)(fff) (gf)(ff) = (d)(f) = d\)
+
+Thus, \(3bd - 2ac + g^{-1}f^3 = b - z + d = b + d = f\).
+
+\section{Question Four}
+\label{sec:org2300bc4}
+To get a hunch,
+
+\begin{verbatim}
+In [1]: import numpy as np
+
+In [2]: def find_counter(matrix_generator, in_ring, search_space=range(-100, 100)):
+...:     for a1 in search_space:
+...:         n = matrix_generator(a1)
+...:         for a2 in search_space:
+...:             m = matrix_generator(a2)
+...:             dot = np.dot(n, m)
+...:             add = np.add(n, m)
+...:
+...:             if not in_ring(dot):
+...:                 return (n, m, dot, '*')
+...:             if not in_ring(add):
+...:                 return (n, m, add, '+')
+...:
+...:     return None
+...:
+
+In [3]: find_counter(lambda a: [[0, a], [0, -a]], \
+                     lambda m: m[0][1] == -m[1][1] and m[0][0] == 0 \
+                     and m[1][0] == 0)
+
+In [4]: find_counter(lambda a: [[0, a], [a, 0]], \
+                     lambda m: m[0][1] == m[1][0] and m[1][1] == 0 \
+                     and m[0][0] == 0)
+Out[4]:
+([[0, -100], [-100, 0]],
+ [[0, -100], [-100, 0]],
+ array([[10000,     0],
+        [    0, 10000]]),
+ '*')
+\end{verbatim}
+\subsection{a}
+\label{sec:orgd1aade2}
+Using Theorem 3.2, this is a ring:
+
+\begin{enumerate}
+\item S\textsubscript{1} is closed under addition: \(x, y \in S_1 \Rightarrow x + y =\) \(\begin{smallmatrix} 0 & a \\ 0 & -a \end{smallmatrix}\) \(+\) \(\begin{smallmatrix} 0 & b \\ 0 & -b \end{smallmatrix}\) \(\Rightarrow\) \(x + y = \begin{smallmatrix} 0 & a + b \\ 0 & (-a) + (-b) \end{smallmatrix} = \begin{smallmatrix} 0 & a + b \\ 0 & -(a + b) \end{smallmatrix}\) which satisfies the rule
+\item S\textsubscript{1} is closed under multiplication: \(x, y \in S_1 \Rightarrow x \cdot y =\) \(\begin{smallmatrix} 0 & a \\ 0 & -a \end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix} 0 & b \\ 0 & -b \end{smallmatrix}\) \(\Rightarrow\) \(x \cdot y =\) \(\begin{smallmatrix} 0 + (a)(0) & 0(b) + a(-b) \\ 0 + (-a)(0) & 0b + (-a)(-b) \end{smallmatrix}\) \(=\) \(\begin{smallmatrix} 0 & -ab \\ 0 & ab \end{smallmatrix}\) which satisfies the rule
+\item \(0_R_{} = 0_{S_1} =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix}\)
+\item \(a \in S_1 \Rightarrow a + x = 0_R \wedge x \in S_1\) since \(\begin{smallmatrix} 0 & a \\ 0 & -a \end{smallmatrix}\) \(+\) \(x = 0_{S_1} =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix}\) \(\Rightarrow x =\) \(\begin{smallmatrix} 0 & -a \\ 0 & a \end{smallmatrix}\) \(\in S_1\)
+\end{enumerate}
+
+\subsection{b}
+\label{sec:org3c559c3}
+\(\begin{smallmatrix} 0 & -100 \\ -100 & 0 \end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix} 0 & -100 \\ -100 & 0 \end{smallmatrix}\) = \(\begin{smallmatrix} 10000 & 0 \\ 0 & 10000 \end{smallmatrix}\) is not in the ring.
+
+\section{Question Five}
+\label{sec:orgb488460}
+No, it is not a homomorphism from the definition in 3.3.
+
+Consider \(n =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & 1 \end{smallmatrix}\) and \(m =\) \(\begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix}\).
+
+Then \(nm =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix}\) and thus \(Trace(n) \cdot Trace(m) = 1 \wedge Trace(nm) = 0 \Rightarrow f(n)f(m) \neq f(nm)\).
+
+\section{Question Six}
+\label{sec:orgba6a67f}
+No, I don't believe \(f\) to be an isomorphism.
+\subsection{Lemma}
+\label{sec:org72bdcf0}
+We need to show that \(x \in Q \wedge x \sqrt{2} = y \in Q \Rightarrow x = 0\).
+
+If \(x \neq 0\) then \(y\) can be represented as \(\frac{a}{b}\) with \(a, b \neq 0 \wedge a,b \in \mathds{Z}\) and \((a, b) = 1\).
+
+So, \(x \sqrt{2} = \frac{a}{b} \Rightarrow x \sqrt{2} b = a \Rightarrow x^2(2b^2) = a^2\) and thus \(2 | aa\), and by applying Corollary 1.6, \(2 | a\).
+
+By substituting \(2c = a\) (a is divisible by 2), \(x^2(2b^2) = 4c^2 \Rightarrow x^2 b^2 = 2c^2\) which implies \(2 | (x^2 b^2) \Rightarrow 2 | x\) or \(2 | b\) (again by Corollary 1.6).
+
+But 2 cannot divide \(b\) as well as \(a\), since \((a, b) = 1\) and \(b \neq 0\).
+
+Thus \(2 | x\). So \(x = 2d \Rightarrow 4d^2(2b^2) = a^2 \Rightarrow 4 | a^2\), so \(4e = a\), and by similar logic for \(b\) above, \(4 | x\).
+Then, subsituting \(x = 4f \Rightarrow 16f^2(2b^2) = a^2 \Rightarrow 16 | a^2\). Again, \(16 | x\).
+
+In fact, all range elements \(R\) of the recursive function on the natural numbers \(f \ni f(1) = 2, f(n) = f(n-1)^2 \text{\{} n > 1 \text{\}}\) must divide \(x\).
+
+Thus, x must be sup(\(R\)), or 0 (everything divides zero). Obviously, though, \(R\) is unbounded on the right, so it follows \(x = 0\).
+
+\subsection{Proof}
+\label{sec:org69eb679}
+\(a, b \in Q(\sqrt{2}) \Rightarrow f(a) = n + m \sqrt{2} \wedge b = x + y \sqrt{2} \Rightarrow f(a) = n - m \sqrt{2} \vee f(b) = x - y \sqrt{2}\).
+
+\(f\) is not injective; if we assume \(f(a) = f(b) \Rightarrow a = b\) then \(f(a) - f(b) = 0 \Rightarrow n - m \sqrt{2} - x - y \sqrt{2} = 0 \Rightarrow 0 = (n-x) - (m + y)\sqrt{2} \Rightarrow n-x = (m+y)\sqrt{2}\) and, as \(n-x \in Q\) since
+Q is a ring and by definition, \((m + y)\sqrt{2} \in Q\) implies \((m + y) = 0\) by the lemma, thus \(m = -y\). But \(a = b \Rightarrow m = y\), a contradiction.
+\end{document}
\ No newline at end of file
diff --git a/Homework/math4310/alg_structurs_1.org b/Homework/math4310/alg_structurs_1.org
new file mode 100644
index 0000000..ae92f51
--- /dev/null
+++ b/Homework/math4310/alg_structurs_1.org
@@ -0,0 +1,188 @@
+#+TITLE: Assignment One
+#+AUTHOR: Logan Hunt
+#+STARTUP: entitiespretty fold inlineimages latexpreview
+#+LATEX_HEADER: \noindent \notag \usepackage{ dsfont }
+
+* Section 1.1
+** Question 5
+By reforming the given expression to show $ca$ as a multiple of $q$ and some remainder:
+
+\begin{align}
+a = bq + r \\
+ca = (cb)q + (c)r
+\end{align}
+
+and that $0 \leq r < b \Rightarrow 0 \leq cr < cb$, Theorem 1.1 tells us that there is only one unique quotient:
+which is $q$, and the remainder is $(c)r$.
+
+** Question 7
+By Theorem 1.1, $a = bq + r, 0 \leq r \lt b$ implies that when b = 3, $a = 3q + r$ where $0 \leq r \lt b$,
+thus $a \in \mathds{Z} \Rightarrow a = 3q + r$ where $r \in {0,1,2}$.
+
+By squaring $a$, we have three options which simplify to the form $3k$ or $3k + 1$:
+1. $a^2 = (3q)^2 = 9q^2$ which is $3k \ni k = 3q^2$
+2. $a^2 = (3q + 1)^2 = 9q^2 + 3q + 1 = 3(3q^2 + q) + 1$ which is $3k + 1 \ni k = (3q^2 + q)$
+3. $a^2 = (3q + 2)^2 = 9q^2 + 6q + 4 = 3(3q^2 + 2q) + 4$ which is
+   $3l + 4 \ni l = (3q^2 + 2q)$ which is also $3l + 1 + 3 = 3(l+1) + 1$,
+   which again is $3k + 1 \ni k = l+1$
+
+** Question 8
+By Theorem 1.1, $a = bq + r, 0 \leq r \lt b$ implies that when b = 4, $a = 4q + r$ where $0 \leq r \lt b$,
+and thus $a = 4q + r$ and $r \in {0,1,2,3}$.
+
+Therefore we have four cases:
+
+1. $a = 4q$, and $a$ must be even which is invalid
+2. $a = 4q + 1$, and $4q + 1 = 2k + 1 \ni k = 2q$ which fits the definition of an odd number
+3. $a = 4q + 2$, thus $a$ must be even as $a = 2k \ni k = 2q + 1$
+4. $a = 4q + 3$, can be rewritten to $4q + 1 + 2$, which is odd: $2k + 1 \ni k = 2q + 1$
+
+And thus any odd number can be rewritten as $4q + 1$ or $4q + 3$.
+
+** Question 10
+The division of $a$ and $c$ by $n$ can each be represented by
+
+\begin{align}
+a = q_{a}n + r_a \\
+c = q_c_{}n + r_c
+\end{align}
+
+where $0 \le r_a < n$ and $0 \le r_c < n$ by Theorem 1.1, and we can subtract each side of the
+second equation from the first:
+
+\begin{align}
+a - c = (q_{a}n + r_a) - (q_c_{}n + r_c)
+\end{align}
+
+With this work now in hand, we will prove by showing that the conjecture is true both ways:
+
+For the first, we will suppose that $r_a = r_c$. Then, we find that
+\begin{align}
+a - c = n(q_a - q_c) \\
+a - c = nk
+\end{align}
+for some integer $k$.
+
+For the second, we will prove that if $a - c = nk$, when $a$ and $c$ are divided by $n$,
+they leave the same remainder $r$.
+
+From the work we did previously, and by substituting $a-c = nk$, we find that
+\begin{align}
+a - c = (q_{a}n + r_a) - (q_c_{}n + r_c) \\
+nk = (q_{a}n + r_a) - (q_{c}n + r_c) \\
+nk = n(q_a - q_c) + (r_a - r_c) \\
+n(k - q_a + q_c) = r_a - r_c
+\end{align}
+and thus $r_a - r_c$ is a multiple of $n$.
+
+From the inequalities produced previously by Theorem 1.1, if $r_a \geq r_c$ then $0 \le r_a - r_c < n$
+and $-n < r_c - r_a \leq 0$. Else if $r_c \geq r_a$ then $0 \leq r_c - r_a < n$ and $-n < r_a - r_c \leq 0$.
+
+By combining the two cases, it must be that $ -n < r_a - r_c < n $, and from the above fact that
+$r_a - r_c$ is a multiple of n, the only multiple of n in $(-n, n)$ is 0. Thus $r_a = r_c$.
+
+* Section 1.2
+** Question 1
+*** c
+1,  57 and 112 are co-prime
+** Question 3
+\begin{align}
+a | b \Rightarrow \exists n \in \mathds{Z} \ni an = b \\
+b | c \Rightarrow \exists m \in \mathds{Z} \ni bm = c \\
+(an)m = c \Rightarrow a | c
+\end{align}
+** Question 4
+*** a
+\begin{align}
+a | b \Rightarrow \exists n \in \mathds{Z} \ni an = b \\
+a | c \Rightarrow \exists m \in \mathds{Z} \ni am = c \\
+a | (b + c) \Rightarrow \exists l \in \mathds{Z} \ni al = b + c \\
+al = an + am \\
+b + c = a(n + m)
+\Rightarrow a | b+c
+\end{align}
+*** b
+By continuing from "a":
+
+\begin{align}
+br + ct = (an)r + (am)t \\
+\Rightarrow br + ct = a(nr + mt) \\
+\Rightarrow a | (br + ct)
+\end{align}
+
+
+** Question 7
+$|a|$ since $|a| \textbar a$ and $|a| \textbar 0$, and there cannot be an integer larger than
+$|a|$ that divides $a$.
+
+** Question 9
+No, not every multiple of two factors of an integer divides that integer.
+
+For example, $3 | 9$ and $9 | 9$ but $27 \nmid 9$.
+
+** Question 15
+*** c
+\begin{align}
+1003 = (2)(456) + 91 \\
+456 = (5)(91) + 1 \\
+91 = (91)(1) + 0 \\
+\Rightarrow (1003, 456) = 1
+\end{align}
+*** d
+\begin{align}
+322 = (2)(148) + 26 \\
+148 = (5)(26) + 18 \\
+26 = (1)(18) + 8 \\
+18 = (2)(8) + 2 \\
+8 = (4)(2) + 0 \\
+\Rightarrow (322, 148) = 2
+\end{align}
+
+** Question 17
+
+\begin{align}
+a | c \Rightarrow \exists n \in \mathds{Z} \ni an = c \\
+b | c \Rightarrow b | an
+\end{align}
+
+And by Theorem 1.4, $(a, b) = 1 \Rightarrow b | n \Rightarrow ab | an \Rightarrow ab | c$:
+
+** Question 19
+Given some $d$ such that $d | a$ and $d | b$, then $a = nd$ and $b = md$.
+\begin{align}
+a | (b + c) \Rightarrow \exists p \in \mathds{Z} \ni ap = b + c \\
+c = ap - md \Rightarrow c = (nd)p - md \Rightarrow c = d(np - md) \Rightarrow d | c
+\end{align}
+
+Since we're given that $(b, c) = 1 \Rightarrow (md, c) = 1$ and from the above fact that $d | c$, we can conclude
+that $md = 1$. Therefore $(a, b) = (a, md) = (a, 1) = 1$.
+
+Given some $e$ such that $e | a$ and $e | c$, then $a = ue$ and $c = we$.
+\begin{align}
+a | (b + c) \Rightarrow \exists o \in \mathds{Z} \ni ao = b + c \\
+b = ao - we \Rightarrow b = (ue)o - we \Rightarrow b = e(uo - w) \Rightarrow e | b
+\end{align}
+
+And from similar reasoning we can conclude that $(a, c) = (a, we) = (a, 1) = 1$.
+
+** Question 31
+*** a
+$[6, 10] = 60$
+
+$[4, 5, 6, 10] = 60$
+
+$[20, 42] = 840$
+
+$[2, 3, 14, 36, 42] = 252$
+
+*** b
+Let $m = [a_1, a_2, \ldots, a_k]$, then by the Division Algorithm, $t = mq + r$ for integers $q$ and $r$,
+with $0 \leq r < m$.
+
+As given that all $a_i | t$ \Rightarrow $a_i | mq + r$, and as $mq$ is a multiple of the LCM including
+a_i, $a_i$ must divide $mq$, and thus $a_i | r$.
+
+Because $a_i | r$, $r = na_i$ and is thus a multiple of all $a_i$, but the above statement that
+$0 \leq r < m$ shows that $m$ - the LCM by definition - is greater than $r$. Therefore, $r = 0$ and
+$t = mq \Rightarrow m | t$.
+
diff --git a/Homework/math4310/final.org b/Homework/math4310/final.org
new file mode 100644
index 0000000..2189562
--- /dev/null
+++ b/Homework/math4310/final.org
@@ -0,0 +1,31 @@
+#+TITLE: Sample Problems
+
+WELL-DEFINED:
+when x \in domain = y \in domain then f(x) = f(y) and each x \in domain has y \in range
+
+PRINCIPAL IDEAL:
+ideal generated by an element
+
+SURJECTION:
+every y \in range has (non-unique) x such that f(x) = y
+
+* Question Three
+** a
+$f$ is well defined as $f([a]_12)$ = $f([b]_12)$ implies $[a] = [b]$, as $[a]_4 = [b]_4$ implies that $a \equiv_4 b$ so $a - b \equiv_4 0 \Rightarrow a - b \equiv_4 12$
+which implies $a - b = 12n$
+** b
+$f$ is a homomorphism:
+f(a + b) = f(a) + f(b) by f([a]_12 + [b]_12) = f([a + b]_12) = [a + b]_4 = [a]_4 + [b]_4 = f([a]_12) + f([b]_12)
+f(ab) = f(a)f(b) by f([a]_12 [b]_12) = f([ab]_12) = [ab]_4 = [a]_4[b]_4 = f([a]_12)f([b]_12)
+
+$f$ is surjective:
+every element in the range [a]_4 can be mapped to an element in the domain: [a]_12 since f([a]_12) = [a]_4
+** c
+the kernel are all the elements of $Z_12 \equiv_4 0$ (0, 4, 8)
+
+** d
+first isomorphism theorem: it's Z_4
+
+* Question Four
+
+
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diff --git a/Homework/math4310/midterm_1.pdf b/Homework/math4310/midterm_1.pdf
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@@ -0,0 +1,151 @@
+% Created 2023-02-20 Mon 22:30
+% 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}
+\author{Lizzy Hunt}
+\date{\today}
+\title{Midterm One}
+\hypersetup{
+ pdfauthor={Lizzy Hunt},
+ pdftitle={Midterm One},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.6.1)}, 
+ pdflang={English}}
+\begin{document}
+
+\maketitle
+\setlength\parindent{0pt}
+
+\section{Question One}
+\label{sec:org6315799}
+\begin{verbatim}
+In [1]: sols = lambda n: set(filter(lambda x: (x**2 + x) % n == 0, range(n)))
+
+In [2]: sols(5)
+Out[2]: {0, 4}
+
+In [3]: sols(6)
+Out[3]: {0, 2, 3, 5}
+\end{verbatim}
+
+\section{Question Two - TODO}
+\label{sec:org6c24b4c}
+
+
+\section{Question Three}
+\label{sec:org76bcd7b}
+From the multiplication table we can see that \(e\) is the identity element \(1_R\), the zero
+element \(0_R\) is \(z\). From the addition table we find that every element is its own
+additive inverse.
+
+\subsection{a}
+\label{sec:org983fdae}
+The units in this ring are \({e, g}\), and the zero divisors are \({a, b, c, f}\).
+
+\subsection{b}
+\label{sec:org9072029}
+\(3bd = 3(bd) = 3z = b\)
+
+\(2ac = 2(ac) = 2b = b + b = z\)
+
+\(g^{-1}f^3 = (g)(fff) (gf)(ff) = (d)(f) = d\)
+
+Thus, \(3bd - 2ac + g^{-1}f^3 = b - z + d = b + d = f\).
+
+\section{Question Four}
+\label{sec:orge571a7e}
+To get a hunch,
+
+\begin{verbatim}
+In [1]: import numpy as np
+
+In [2]: def find_counter(matrix_generator, in_ring, search_space=range(-100, 100)):
+   ...:     for a1 in search_space:
+   ...:         n = matrix_generator(a1)
+   ...:         for a2 in search_space:
+   ...:             m = matrix_generator(a2)
+   ...:             dot = np.dot(n, m)
+   ...:             add = np.add(n, m)
+   ...:
+   ...:             if not in_ring(dot):
+   ...:                 return (n, m, dot, '*')
+   ...:             if not in_ring(add):
+   ...:                 return (n, m, add, '+')
+   ...:
+   ...:     return None
+   ...:
+
+In [3]: find_counter(lambda a: [[0, a], [0, -a]], lambda m: m[0][1] == -m[1][1] and m[0][0] == 0 and m[1][0] == 0)
+
+In [4]: In [31]: find_counter(lambda a: [[0, a], [a, 0]], lambda m: m[0][1] == m[1][0] and m[1][1] == 0 and m[0][0] == 0)
+Out[4]:
+([[0, -100], [-100, 0]],
+ [[0, -100], [-100, 0]],
+ array([[10000,     0],
+        [    0, 10000]]),
+ '*')
+\end{verbatim}
+\subsection{a}
+\label{sec:orge0ecef4}
+Using Theorem 3.2, this is a ring:
+
+\begin{enumerate}
+\item S\textsubscript{1} is closed under addition: \(x, y \in S_1 \Rightarrow x + y =\) \(\begin{smallmatrix} 0 & a \\ 0 & -a \end{smallmatrix}\) \(+\) \(\begin{smallmatrix} 0 & b \\ 0 & -b \end{smallmatrix}\) \(\Rightarrow\) \(x + y = \begin{smallmatrix} 0 & a + b \\ 0 & (-a) + (-b) \end{smallmatrix} = \begin{smallmatrix} 0 & a + b \\ 0 & -(a + b) \end{smallmatrix}\) which satisfies the rule
+\item S\textsubscript{1} is closed under multiplication: \(x, y \in S_1 \Rightarrow x \cdot y =\) \(\begin{smallmatrix} 0 & a \\ 0 & -a \end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix} 0 & b \\ 0 & -b \end{smallmatrix}\) \(\Rightarrow\) \(x \cdot y =\) \(\begin{smallmatrix} 0 + (a)(0) & 0(b) + a(-b) \\ 0 + (-a)(0) & 0b + (-a)(-b) \end{smallmatrix}\) \(=\) \(\begin{smallmatrix} 0 & -ab \\ 0 & ab \end{smallmatrix}\) which satisfies the rule
+\item \(0_R_{} = 0_{S_1} =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix}\)
+\item \(a \in S_1 \Rightarrow a + x = 0_R \wedge x \in S_1\) since \(\begin{smallmatrix} 0 & a \\ 0 & -a \end{smallmatrix}\) \(+\) \(x = 0_{S_1} =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix}\) \(\Rightarrow x =\) \(\begin{smallmatrix} 0 & -a \\ 0 & a \end{smallmatrix}\) \(\in S_1\)
+\end{enumerate}
+
+\subsection{b}
+\label{sec:org1a8af9b}
+\(\begin{smallmatrix} 0 & -100 \\ -100 & 0 \end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix} 0 & -100 \\ -100 & 0 \end{smallmatrix}\) = \(\begin{smallmatrix} 10000 & 0 \\ 0 & 10000 \end{smallmatrix}\) is not in the ring.
+
+\section{Question Five}
+\label{sec:org85339f8}
+No, it is not a homomorphism from the definition in 3.3.
+
+Consider \(n =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & 1 \end{smallmatrix}\) and \(m =\) \(\begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix}\).
+
+Then \(nm =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix}\) and thus \(Trace(n) \cdot Trace(m) = 1 \wedge Trace(nm) = 0 \Rightarrow f(n)f(m) \neq f(nm)\).
+
+\section{Question Six}
+\label{sec:org405929b}
+No, I don't believe \(f\) to be an isomorphism.
+\subsection{Lemma}
+\label{sec:org323f970}
+We need to show that \(x \in Q \wedge x \sqrt{2} = y \in Q \Rightarrow x = 0\).
+
+If \(x \neq 0\) then \(y\) can be represented as \(\frac{a}{b}\) with \(a, b \neq 0 \wedge a,b \in \mathds{Z}\) and \((a, b) = 1\).
+
+So, \(x \sqrt{2} = \frac{a}{b} \Rightarrow x \sqrt{2} b = a \Rightarrow x^2(2b^2) = a^2\) and thus \(2 | aa\), and by applying Corollary 1.6, \(2 | a\).
+
+By substituting \(2c = a\) (a is divisible by 2), \(x^2(2b^2) = 4c^2 \Rightarrow x^2 b^2 = 2c^2\) which implies \(2 | (x^2 b^2) \Rightarrow 2 | x\) or \(2 | b\) (again by Corollary 1.6).
+
+But 2 cannot divide \(b\) as well as \(a\), since \((a, b) = 1\) and \(b \neq 0\).
+
+Thus \(2 | x\). So \(x = 2d \Rightarrow 4d^2(2b^2) = a^2 \Rightarrow 4 | a^2\), so \(4e = a\), and by similar logic for \(b\) above, \(4 | x\).
+Then, subsituting \(x = 4f \Rightarrow 16f^2(2b^2) = a^2 \Rightarrow 16 | a^2\). Again, \(16 | x\).
+
+In fact, all range elements \(R\) of the function on the natural numbers \$f \(\ni\) \$ \(f(1) = 2, f(n) = f(n-1)^2 { n > 1 }\) must divide \(x\).
+
+Thus, x must be sup(\(R\)) or 0. Obviously \(R\) is unbounded on the right, so it follows \(x = 0\).
+
+\subsection{Proof}
+\label{sec:orge723534}
+\(a, b \in Q(\sqrt{2}) \Rightarrow f(a) = n + m \sqrt{2} \wedge b = x + y \sqrt{2} \Rightarrow f(a) = n - m \sqrt{2} \vee f(b) = x - y \sqrt{2}\).
+
+\(f\) is not injective; if we assume \(f(a) = f(b) \Rightarrow a = b\) then \(f(a) - f(b) = 0 \Rightarrow n - m \sqrt{2} - x - y \sqrt{2} = 0 \Rightarrow 0 = (n-x) - (m + y)\sqrt{2} \Rightarrow n-x = (m+y)\sqrt{2}\) and, as \(n-x \in Q\) since
+Q is a ring and by definition, \((m + y)\sqrt{2} \in Q\) implies \((m + y) = 0\) by the lemma, thus \(m = -y\). But for \(a = b\), \(m = y\), a contradiction.
+\end{document}
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