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\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}