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% Created 2023-04-16 Sun 21:55
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\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}
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