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\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}