path: root/Homework/math4310/abstract_algebra_assn_9.tex

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\author{Lizzy Hunt}
\date{\today}
\title{Assignment Nine}
\hypersetup{
 pdfauthor={Lizzy Hunt},
 pdftitle={Assignment Nine},
 pdfkeywords={},
 pdfsubject={},
 pdfcreator={Emacs 28.2 (Org mode 9.6.1)}, 
 pdflang={English}}
\begin{document}

\maketitle
\setlength\parindent{0pt}

\section{Section 5.1}
\label{sec:orga03a7b4}
\subsection{Question One}
\label{sec:org41cc8bf}
\subsubsection{b}
\label{sec:org39f646c}
Yes. In \(F\), \(f(x) - g(x) = -x^3 + x = x^3 + x\).

\begin{equation*}
\polylongdiv[style=A]{x^3 + x}{x^2+1}
\end{equation*}

\subsubsection{c}
\label{sec:orge019c41}
No. \(f(x) - g(x) = x^5 - 2x^4 + 4x^3 - 8x^2 + 3x - 2\)

\begin{equation*}
\polylongdiv[style=A]{x^5 - 2x^4 + 4x^3 - 8x^2 + 3x - 2}{x^3 - x^2 + x - 1}
\end{equation*}

\subsection{Question Three}
\label{sec:org1718036}
\(|\)\{ \(0, 1, x, x + 1, x^2, x^2 + 1, x^2 + x, x^2 + x + 1\) \}\(|\) = 8

\subsection{Question Four}
\label{sec:orgd724559}
For every \(a,b,c \in \mathds{Z}_3\) we can generate a polynomial \(ax^2 + bx + c\), by part two of Corollary 5.5. \(3^3 = 27\)

\subsection{Question Six}
\label{sec:orgb081ff5}
By Corollary 5.5, all the congruence classes in \(F[x]\) are \(c \ni c \in F\).

\subsection{Question Eight}
\label{sec:org4c8f836}
\begin{align*}
f(x)k(x) &\equiv_{p(x)} g(x)k(x) \\
& \Rightarrow p(x) | f(x)k(x) - g(x)k(x) \\
& \Rightarrow p(x) | (f(x) - g(x))(k(x))
\end{align*}

By Theorem 4.10, since \(p(x)\) is relatively prime to \(k(x)\), \(p(x) | f(x) - g(x) \Rightarrow f(x) \equiv_{p(x)} g(x)\)

\subsection{Question Eleven}
\label{sec:org98dcd2c}
Since \(p(x)\) is reducible, it can be rewritten as \(p(x) = f(x)g(x)\) with \(f(x), g(x) \in F[x]\) with each \(f(x)\) and \(g(x)\) having a degree greater than 0, summing to the
degree of \(p(x)\).

Then, it is impossible for \(p(x)\) to divide \(f(x)\) or \(g(x)\) since \(p(x)\) has a higher degree. So, neither \(f(x)\) or \(g(x)\) can \$ \(\equiv\)\textsubscript{p(x)} 0\textsubscript{F}\$.

Still, \(f(x)g(x) \equiv_{}_{p(x)} p(x) \equiv_{p(x)} 0_F\) since \(p(x) | p(x)\).

\section{Section 5.2}
\label{sec:org9a01661}
\subsection{Question One}
\label{sec:org8ab076f}
The congruence classes are those in Section 5.1, Question Three as above.

\begin{center}
\begin{tabular}{lllllllll}
+ & [0] & [1] & [x] & [x + 1] & [x\textsuperscript{2}] & [x\textsuperscript{2} + 1] & [x\textsuperscript{2} + x] & [x\textsuperscript{2} + x + 1]\\[0pt]
[0] & [0] & [1] & [x] & [x + 1] & [x\textsuperscript{2}] & [x\textsuperscript{2} + 1] & [x\textsuperscript{2} + x] & [x\textsuperscript{2} + x + 1]\\[0pt]
[1] & [1] & [0] & [x+1] & [x] & [x\textsuperscript{2} + 1] & [x\textsuperscript{2}] & [x\textsuperscript{2} + x + 1] & [x\textsuperscript{2} + x]\\[0pt]
[x] & [x] & [x + 1] & [0] & [1] & [x\textsuperscript{2} + x] & [x\textsuperscript{2} + x + 1] & [x\textsuperscript{2}] & [x\textsuperscript{2} + 1]\\[0pt]
[x + 1] & [x + 1] & [x] & [1] & [0] & [x\textsuperscript{2} + x + 1] & [x\textsuperscript{2} + x] & [x\textsuperscript{2} + 1] & [x\textsuperscript{2}]\\[0pt]
[x\textsuperscript{2}] & [x\textsuperscript{2}] & [x\textsuperscript{2} + 1] & [x\textsuperscript{2} + x] & [x\textsuperscript{2} + x + 1] & [0] & [1] & [x] & [x + 1]\\[0pt]
[x\textsuperscript{2} + 1] & [x\textsuperscript{2} + 1] & [x\textsuperscript{2}] & [x\textsuperscript{2} + x + 1] & [x\textsuperscript{2} + x] & [1] & [0] & [x + 1] & [x]\\[0pt]
[x\textsuperscript{2} + x] & [x\textsuperscript{2} + x] & [x\textsuperscript{2} + x + 1] & [x\textsuperscript{2}] & [x\textsuperscript{2} + 1] & [x] & [x+1] & [0] & [1]\\[0pt]
[x\textsuperscript{2} + x + 1] & [x\textsuperscript{2} + x + 1] & [x\textsuperscript{2} + x] & [x\textsuperscript{2} + 1] & [x\textsuperscript{2}] & [x+1] & [x] & [1] & [0]\\[0pt]
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{lllllllll}
\(\cdot\) & [0] & [1] & [x] & [x + 1] & [x\textsuperscript{2}] & [x\textsuperscript{2} + 1] & [x\textsuperscript{2} + x] & [x\textsuperscript{2} + x + 1]\\[0pt]
[0] & [0] & [0] & [0] & [0] & [0] & [0] & [0] & [0]\\[0pt]
[1] & [0] & [1] & [x] & [x + 1] & [x\textsuperscript{2}] & [x\textsuperscript{2} + 1] & [x\textsuperscript{2} + x] & [x\textsuperscript{2} + x + 1]\\[0pt]
[x] & [0] & [x] & [x\textsuperscript{2}] & [x\textsuperscript{2}+x] & [x+1] & [1] & [x\textsuperscript{2}+x+1] & [x\textsuperscript{2}+1]\\[0pt]
[x + 1] & [0] & [x + 1] & [x\textsuperscript{2} + x] & [x\textsuperscript{2}+1] & [x\textsuperscript{2}+x+1] & [x\textsuperscript{2}] & [1] & [x]\\[0pt]
[x\textsuperscript{2}] & [0] & [x\textsuperscript{2}] & [x+1] & [x\textsuperscript{2}+x+1] & [x\textsuperscript{2}+x] & [x] & [x\textsuperscript{2}+1] & [1]\\[0pt]
[x\textsuperscript{2} + 1] & [0] & [x\textsuperscript{2} + 1] & [1] & [x\textsuperscript{2}] & [x] & [x\textsuperscript{2}+x+1] & [x+1] & [x\textsuperscript{2}+x]\\[0pt]
[x\textsuperscript{2} + x] & [0] & [x\textsuperscript{2} + x] & [x\textsuperscript{2}+x+1] & [1] & [x\textsuperscript{2}+1] & [x+1] & [x] & [x+1]\\[0pt]
[x\textsuperscript{2} + x + 1] & [0] & [x\textsuperscript{2} + x + 1] & [x\textsuperscript{2}+1] & [x] & [1] & [x\textsuperscript{2}+x] & [x\textsuperscript{2}] & [x+1]\\[0pt]
\end{tabular}
\end{center}

Yes, this is a field since by Theorem 5.7, it is a commutative ring with identity, and each
non-zero row in the multiplication table contains the multiplicative identity (each is a unit).

\subsection{Question Two}
\label{sec:org9db49ef}
\begin{center}
\begin{tabular}{llllllllll}
+ & [0] & [1] & [2] & [x] & [x+1] & [x+2] & [2x] & [2x+1] & [2x+2]\\[0pt]
[0] & [0] & [1] & [2] & [x] & [x+1] & [x+2] & [2x] & [2x+1] & [2x+2]\\[0pt]
[1] & [1] & [2] & [0] & [x+1] & [x+2] & [x] & [2x+1] & [2x+2] & [2x]\\[0pt]
[2] & [2] & [0] & [1] & [x+2] & [x] & [x+1] & [2x+2] & [2x] & [2x+1]\\[0pt]
[x] & [x] & [x+1] & [x+2] & [2x] & [2x+1] & [2x+2] & [0] & [1] & [2]\\[0pt]
[x+1] & [x+1] & [x+2] & [x] & [2x+1] & [2x+2] & [2x] & [1] & [2] & [0]\\[0pt]
[x+2] & [x+2] & [x] & [x+1] & [2x+2] & [2x] & [2x+1] & [2] & [0] & [1]\\[0pt]
[2x] & [2x] & [2x+1] & [2x+2] & [0] & [1] & [2] & [x] & [x+1] & [1]\\[0pt]
[2x+1] & [2x+1] & [2x+2] & [2x] & [1] & [2] & [0] & [x+1] & [x+2] & [x]\\[0pt]
[2x+2] & [2x+2] & [2x] & [2x+1] & [2] & [0] & [1] & [x+2] & [x] & [x+1]\\[0pt]
\end{tabular}
\end{center}


\begin{center}
\begin{tabular}{llllllllll}
\(\cdot\) & [0] & [1] & [2] & [x] & [x+1] & [x+2] & [2x] & [2x+1] & [2x+2]\\[0pt]
[0] & [0] & [0] & [0] & [0] & [0] & [0] & [0] & [0] & [0]\\[0pt]
[1] & [0] & [1] & [2] & [x] & [x+1] & [x+2] & [2x] & [2x+1] & [2x+2]\\[0pt]
[2] & [0] & [2] & [1] & [2x] & [2x+2] & [2x+1] & [x] & [x+2] & [x+1]\\[0pt]
[x] & [0] & [x] & [2x] & [2] & [x+2] & [2x+2] & [1] & [x+1] & [2x+1]\\[0pt]
[x+1] & [0] & [x+1] & [2x+2] & [x+2] & [2x] & [1] & [2x+1] & [2] & [x]\\[0pt]
[x+2] & [0] & [x+2] & [2x+1] & [2x+2] & [1] & [x] & [x+1] & [2x] & [2]\\[0pt]
[2x] & [0] & [2x] & [x] & [1] & [2x+1] & [x+1] & [2] & [2x+2] & [x+2]\\[0pt]
[2x+1] & [0] & [2x+1] & [x+2] & [x+1] & [2] & [2x] & [2x+2] & [x] & [1]\\[0pt]
[2x+2] & [0] & [2x+2] & [x+1] & [2x+1] & [x] & [2] & [x+2] & [1] & [2x]\\[0pt]
\end{tabular}
\end{center}

Yes, this is a field since by Theorem 5.7, it is a commutative ring with identity, and each
non-zero row in the multiplication table contains the multiplicative identity (each is a unit).

\subsection{Question Three}
\label{sec:orgc80a66e}

\begin{center}
\begin{tabular}{lllll}
+ & [0] & [1] & [x] & [x+1]\\[0pt]
[0] & [0] & [1] & [x] & [x+1]\\[0pt]
[1] & [1] & [0] & [x+1] & [x]\\[0pt]
[x] & [x] & [x+1] & [0] & [1]\\[0pt]
[x+1] & [x+1] & [x] & [1] & [0]\\[0pt]
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{lllll}
\(\cdot\) & [0] & [1] & [x] & [x+1]\\[0pt]
[0] & [0] & [0] & [0] & [0]\\[0pt]
[1] & [0] & [1] & [x] & [x+1]\\[0pt]
[x] & [0] & [x] & [1] & [x+1]\\[0pt]
[x+1] & [0] & [x+1] & [x+1] & [0]\\[0pt]
\end{tabular}
\end{center}

Not, this is \uline{not} a field since by Theorem 5.7, it is a commutative ring with identity, but
not every non-zero row in the multiplication table contains the multiplicative identity (\(x+1\) is
not a unit).

\subsection{Question Six}
\label{sec:orga040020}
By Corollary 5.5, each congruence class can be rewritten with \(a,b \in \mathds{Q}\): \([ax + b]\).

Addition is defined as \([ax + b] + [cx + d] = [(a + c)x + bd]\).

\((ax + b)(cx + d) = acx^2 + dax + bcx + bd = acx^2 + (da + bc)x + bd\)

\begin{equation*}
\polylongdiv[style=A]{acx^2 + (da + bc)x + bd}{x^2 - 2}
\end{equation*}

Multiplication is thusly defined as \([ax + b] \cdot [cx + d] = [(ad + bc)x + (2ac + bd)]\)

\subsection{Question Nine}
\label{sec:org3bd28c4}
Given that \([a + bx]\) is a nonzero congruence class, either
\(a > 0\) or \(b > 0\). Then let \(c = \frac{-a}{a^2 + b^2}\) and \(d = \frac{b}{a^2 + b^2}\).

\begin{equation*}
\polylongdiv[style=A]{acx^2 + (da + bc)x + bd}{x^2 + 1}
\end{equation*}

\begin{align*}
[ax + b][cx + d] & = [(ad + bc)x + (bd - ac)] \\
&= [(\frac{ab}{a^2 + b^2} + \frac{-ab}{a^2 + b^2})x + \frac{b^2}{a^2 + b^2} - \frac{-a^2}{a^2 + b^2}] \\
&= [0x + \frac{b^2 + a^2}{a^2 + b^2}] \\
&= [1]
\end{align*}

\section{Section 5.3}
\label{sec:org6b8ca6a}
\subsection{Question One}
\label{sec:org09f321d}
\subsubsection{a}
\label{sec:org00adeb0}
\(x^3 + 2x^2 + x + 1\) does not have any roots in \mathds{Z}\textsubscript{3}, so by Corollary 4.19 it must be irreducible,
and thus a field by 5.10

\subsubsection{b}
\label{sec:orge4a6bab}
This is not a field by Theorem 5.10 since 2 is a root in \(Z_5\), so by Corollary 4.19 it must be reducible.

\subsubsection{c}
\label{sec:org1cc5f89}
This is not a field by Theorem 5.10 since \(x^4 + x^2 + 1 = (x^2 - x + 1)(x^2 + x + 1) \equiv_2 (x^2 + x + 1)^2\) shows \(x^4 + x^2 + 1\) is reducible.

\subsection{Question Two}
\label{sec:org1cd3926}
\subsubsection{a}
\label{sec:org265177d}
Since \(\mathds{Q} (\sqrt{2})\) is a subset of \(\mathds{R}\), multiplication and addition are associative, commutative, and distributive.

The additive identity of \$\mathds{Q} (\sqrt{2}) is \(0 + 0\sqrt{2}\) and the multiplicative identity is \(1 + 0\sqrt{2}\).

It must be a field since every non-zero element \(a + b \sqrt{2}\) is a unit:

\begin{align*}
(a + b\sqrt{2})x = 1 & \Rightarrow x = \frac{1}{a + b\sqrt{2}} \\
& \Rightarrow x = \frac{a - b\sqrt{2}}{(a + b\sqrt{2})(a - b \sqrt{2})} \\
& \Rightarrow x = \frac{a}{a^2 - 2b^2} - \frac{b}{a^2 - 2b^2}\sqrt{2} \\
\end{align*}

\subsubsection{b}
\label{sec:orgef6853e}
Every element in \(\mathds{Q} / (x^2 - 2)\) can be rewritten as a member of the congruence class \([ax + b]\) with \(a, b \in \mathds{Q}\) by Corollary 5.5.

Then, we can define a function \(f\) such that \(f([ax + b]) = a + b\sqrt{2}\) so that \(f(x) \in \mathds{Q}(\sqrt{2})\).

\(f\) is thus an isomorphism since (a chance at redemption from my midterm \(\ddot\smile\)):

\begin{itemize}
\item \(f\) is injective since \(f([ax + b]) = f([cx + d]) \Rightarrow a + b\sqrt{2} = c + d\sqrt{2} \Rightarrow a = c \wedge b = d\)
\item \(f\) is surjective since each \(a + b\sqrt{2}\) is uniquely mapped to \([ax + b]\)
\item \(f([ax + b] + [cx + d]) = f([(a + c)x + (b + d)]) = (a + c) + (b + d)\sqrt{2} = (a + b\sqrt{2}) + (c + d\sqrt{2}) = f([ax + b]) + f([cx + d])\)

\item And via Question Six from section 5.2 above,
\(f([ax + b] \cdot [cx + d]) = f([(ad + bc)x + (2ac + bd)]) = (ad + bc) + (2ac + bd)\sqrt{2} = (a + b\sqrt{2})(c + d\sqrt{2}) = f([ax + b])f([cx + d])\)
\end{itemize}

\subsection{Question Five}
\label{sec:orgc53207a}
\subsubsection{a}
\label{sec:orgf8f7862}
Since \(\mathds{Q} (\sqrt{3})\) is a subset of \(\mathds{R}\), multiplication and addition are associative, commutative, and distributive.

The additive identity of \$\mathds{Q} (\sqrt{3}) is \(0 + 0\sqrt{3}\) and the multiplicative identity is \(1 + 0\sqrt{3}\).

It must be a field since every non-zero element \(r + s \sqrt{3}\) is a unit (by first assuming that the inverse of \(r + s\sqrt{3}\) from the
back of the book, is \(\frac{r}{t} - \frac{s}{t}\sqrt{3}\) with \(t=r^2 - 3s^2\)):

\begin{align*}
1 &= (r + s\sqrt{3})(\frac{r}{t} - \frac{s}{t}\sqrt{3}) \\
  &= \frac{r^2}{t} - \frac{sr}{t}\sqrt{3} + \frac{sr}{t}\sqrt{3} - \frac{3s^2}{t} \\
  &= \frac{r^2 - 3s^2}{t} \\
  &= 1
\end{align*}

\subsubsection{b}
\label{sec:org1d5d4ca}
\begin{enumerate}
\item Quick Lemma
\label{sec:org6e61eec}
In \(\mathds{Q}{x^2 - 3}\), by Corollary 5.5, each congruence class can be rewritten with \(a,b \in \mathds{Q}\): \([ax + b]\).

\((ax + b)(cx + d) = acx^2 + dax + bcx + bd = acx^2 + (da + bc)x + bd\)

\begin{equation*}
\polylongdiv[style=A]{acx^2 + (da + bc)x + bd}{x^2 - 3}
\end{equation*}

Multiplication is thusly defined as \([ax + b] \cdot [cx + d] = [(ad + bc)x + (3ac + bd)]\)

\item Yeah, it's an isomorphism
\label{sec:org454b3a6}

Every element in \(\mathds{Q} / (x^2 - 3)\) can be rewritten as a member of the congruence class \([ax + b]\) with \(a, b \in \mathds{Q}\) by Corollary 5.5.

Then, we can define a function \(f\) such that \(f([rx + s]) = r + s\sqrt{3}\) so that \(f(x) \in \mathds{Q}(\sqrt{3})\).

\(f\) is thus an isomorphism since (a chance of redemption from my midterm \(\ddot\smile\)):

\begin{itemize}
\item \(f\) is injective since \(f([ax + b]) = f([cx + d]) \Rightarrow a + b\sqrt{3} = c + d\sqrt{3} \Rightarrow a = c \wedge b = d\)
\item \(f\) is surjective since each \(a + b\sqrt{3}\) is uniquely mapped to \([ax + b]\)
\item \(f([ax + b] + [cx + d]) = f([(a + c)x + (b + d)]) = (a + c) + (b + d)\sqrt{3} = (a + b\sqrt{3}) + (c + d\sqrt{3}) = f([ax + b]) + f([cx + d])\)
\item And from the lemma, \(f([ax + b] \cdot [cx + d]) = f([(ad + bc)x + (3ac + bd)]) = (ad + bc) + (3ac + bd)\sqrt{3} = (a + b\sqrt{3})(c + d\sqrt{3}) = f([ax + b])f([cx + d])\)
\end{itemize}
\end{enumerate}
\end{document}