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

\maketitle
\setlength\parindent{0pt}


\section{Section 4.2}
\label{sec:org39ef51e}
\subsection{Question Five}
\label{sec:orgafa157f}
\subsubsection{b}
\label{sec:org072214f}
\begin{align*}
x^5 + x^4 + 2x^3 - x^2 - x - 2 &= (x - 1)(x^4 + 2x^3 + 5x^2 + 4x + 4) + (-x^3 - x + 2) \\
x^4 + 2x^3 + 5x^2 + 4x + 4 &= (-x-2)(-x^3 - x + 2) + (4x^2 + 4x + 8)\\
-x^3 - x + 2 &= (\frac{-x}{4})(4x^2 + 4x + 8) + 0
\end{align*}

\((4x^2 + 4x + 8) = 4(x^2 + x + 2)\)

\(x^2 + x + 2\)
\subsubsection{c}
\label{sec:orgb01ae88}
\(deg(d) = 2\) is the greatest degree of a possible common divisor, so we'll stick with \(x^2 - 1\).
\subsubsection{g}
\label{sec:org4766198}
\begin{align*}
2x^4 + 5x^3 - 5x - 2 &= (x + 4)(2x^3 - 3x^2 - 2x) + (14x^2 + 3x - 2) \\
2x^3 - 3x^2 - 2x &= (\frac{x}{7} - \frac{12}{49})(14x^2 + 3x - 2) + (\frac{-48x - 24}{49}) \\
14x^2 + 3x - 2 &= (\frac{-7x}{24})(\frac{-48x - 24}{49}) + 0
\end{align*}

\(\frac{-48x - 24}{49} = \left(\frac{49}{-48}\right)\left(\frac{\left(-48x-24\right)}{49}\right) = x + \frac{1}{2}\)

\(x + \frac{1}{2}\)
\subsection{Question Ten}
\label{sec:org6ab06e1}
\(x^3 - 3abx + a^3 + b^3\) can be factored to \((a + b + x) (a^2 - a b - a x + b^2 - b x + x^2)\), so \(a + b + x\) is the gcd.

\section{Section 4.3}
\label{sec:org9759eff}
\subsection{Question Three}
\label{sec:orgf67495a}
\subsubsection{a}
\label{sec:org3a3c23d}
\{ \(x^2 + x + 1\), \(2x^2 + 2x + 2\), \(3x^2 + 3x + 3\), \(4x^2 + 4x + 4\) \}
\subsubsection{b}
\label{sec:org0eff32f}
\{ \(3x + 2\), \(6x + 4\), \(2x + 6\), \(5x + 1\), \(x + 3\), \(4x + 5\) \}

\subsection{Question Six}
\label{sec:org58db881}
Assume that \(x^2 + 1\) is in fact reducible. Then, \(x^2 + 1 = (ax + b)(cx + d)\). Thus, \(ax \cdot cx = x^2 \Rightarrow ac = 1\), \(axd + bcx = 0 \Rightarrow ad + bc = 0\), and \(bd = 1\) with \(a,b,c,d\) all being nonzero.

Then, \(a = \frac{1}{c}\) and \(d = \frac{1}{b}\), so \(ad = -bc \Rightarrow \frac{1}{cb} = - bc\), which is impossible.

\subsection{Question Nine}
\label{sec:orge1fd195}
\subsubsection{a}
\label{sec:org33e3a63}
\(x^2 + x + 1\)

\subsubsection{b}
\label{sec:orgf9e0897}
\(x^3 + x^2 + 1\) and \(x^3 + x + 1\)

\subsection{Question Ten}
\label{sec:org4da4dbc}
\subsubsection{a}
\label{sec:orgc043f8c}
In \(\mathds{Q}[x]\), no. In \(\mathds{R}[x]\), yes: \(x^2 - 3 = (x - \sqrt{3})(x + \sqrt{3})\).

\subsubsection{b}
\label{sec:org68f2a14}
Yes, in both fields: \(x^2 + x - 2 = (x + 2)(x - 1)\)

\subsection{Question Eleven}
\label{sec:org957dc65}
Assume that \(x^3 - 3\) were reducible in \(\mathds{Z}_7 [x]\). Then, there must be a monic factor of degree one, as
\(x^3 - 3 = g(x)h(x) \Rightarrow deg(x^3 - 3) = deg(g(x)h(x)) \Rightarrow 3 = deg(g(x)) + deg(h(x))\) by Theorem 4.2, and either \(deg(g(x))\) or \(deg(h(x))\) must be one, with the other, two.

So, one of the factors must be of the form \(x + a, a \in \mathds{Z}_7\). None such factor exists since polynomial division always returns a non-zero remainder:

\(x \nmid x^3 - 3\), trivially.

\(\frac{x^3 - 3}{x + 1}\) gives a remainder of \(4\).

\(\frac{x^3 - 3}{x + 2}\) gives a remainder of \(-11 \equiv_7 3\).

\(\frac{x^3 - 3}{x + 3}\) gives a remainder of \(-30 \equiv_7 5\).

\(\frac{x^3 - 3}{x + 4}\) gives a remainder of \(-67 \equiv_7 3\).

\(\frac{x^3 - 3}{x + 5}\) gives a remainder of \(-128 \equiv_7 5\).

\(\frac{x^3 - 3}{x + 6}\) gives a remainder of \(-219 \equiv_7 5\).

\subsection{Question Twelve}
\label{sec:orge9f1b7e}
In \(\mathds{Q}[x]\), \(x^4 - 4 = (x^2 - 2)(x^2 + 2)\)

In \(\mathds{R}[x]\), \(x^4 - 4 = (x - \sqrt{2})(x + \sqrt{2})(x^2 + 2)\)

In \(\mathds{C}[x]\), \(x^4 - 4 = (x - \sqrt{2})(x + \sqrt{2})(x - i \sqrt{2})(x + i \sqrt{2})\)

\subsection{Question Fourteen}
\label{sec:orge490368}
\(x^2 + x \equiv_6 (x + 4)(x + 3)\)

\(x^2 + x \equiv_6 x(x + 1)\)

\section{Section 4.4}
\label{sec:orgced6b81}
\subsection{Question Two}
\label{sec:org80309c5}
\subsubsection{c}
\label{sec:org684714e}
By Theorem 4.15, the remainder is \(f(-1) = 5\)

\subsection{Question Three}
\label{sec:org4950bb3}
\subsubsection{c}
\label{sec:org38b222f}
If the remainder of \(\frac{f(x)}{h(x)}\) is 0, then \(h\) is a factor of \(f\), so using
Theorem 4.15 we can find that \(f(-2) = -55 \equiv_5 0\), so \(h\) is indeed a factor.

\subsection{Question Four}
\label{sec:org5aa1b14}
\subsubsection{b}
\label{sec:orge7c045e}
We need to find k such that \(f(-1) = 0 (mod 5)\)

\begin{verbatim}
>>> f = lambda k,x: (x**4 + 2 * x**3 - 3 * x**2 + k * x + 1) % 5
>>> for i in range(5):
...   if f(i, -1) == 0:
...     print(i)
...
2
\end{verbatim}

\(k=2\) works nicely!

\subsection{Question Seven}
\label{sec:orgecff464}
\begin{verbatim}
>>> set(filter(lambda x: (x**7 - x) % 7 == 0, range(7)))
{0, 1, 2, 3, 4, 5, 6}
\end{verbatim}

Shows that each element is a root, so the factoring is correct by the Factor Theorem.

\subsection{Question Eight}
\label{sec:orge5e9620}
\subsubsection{b}
\label{sec:org65737b6}
The polynomial is irreducible since its only roots are \(\pm \sqrt{7} \notin \mathds{Q}\), by Corollary 4.19

\subsubsection{d}
\label{sec:orga790f86}
\begin{verbatim}
>>> set(filter(lambda x: (2 * x**3 + x**2 + 2 * x + 2) % 5 == 0, range(5)))
set()
\end{verbatim}

It's irreducible since there are no roots in \(\mathds{Z}_5\).

\subsection{Question Nine}
\label{sec:org2cf5693}
\begin{verbatim}
>>> def find_irr_mon_polys(deg, mod):
...    z_s = range(mod)
...    polys = set()
...    for b in z_s:
...       for c z_s:
...          f_repr = f"x^2 + {b}x + {c}"
...          f = lambda x: (x**2 + b*x + c) % mod
...          if not any(map(lambda x: f(x) == 0, z_s)):
...              polys.add(f_repr)
...    return polys

>>> find_irr_mon_polys(2, 5)
\end{verbatim}

\{ \(x^2 + 0x + 2,
 x^2 + 0x + 3,
 x^2 + 1x + 1,
 x^2 + 1x + 2,
 x^2 + 2x + 3,
 x^2 + 2x + 4,
 x^2 + 3x + 3,
 x^2 + 3x + 4,
 x^2 + 4x + 1,
 x^2 + 4x + 2\) \}

\begin{verbatim}
>>> find_irr_mon_polys(2, 3)
\end{verbatim}
\{ \(x^2 + 0x + 1, x^2 + 1x + 2, x^2 + 2x + 2\) \}

\subsection{Question Thirteen}
\label{sec:org4f87227}
\subsubsection{a}
\label{sec:org32ba550}
If \(f(x) = cg(x)\) with \(c \neq 0_F\), then \(g(x) = c^{-1}f(x)\) and \(f(x) = c^{-1}g(x)\). Then, \(g(y) = 0_F \Leftrightarrow f(y) = 0_F\).
\subsubsection{b}
\label{sec:org84283d0}
No, consider \(f(x) = x\) and \(g(x) = x^2\) in \(\mathds{Z}\), then \(f\) and \(g\) share \(0\) as their only root, but they are not associates.
\end{document}