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+% 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}
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