% 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}