1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
|
% Created 2023-03-16 Thu 18:22
% Intended LaTeX compiler: pdflatex
\documentclass[11pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{graphicx}
\usepackage{longtable}
\usepackage{wrapfig}
\usepackage{rotating}
\usepackage[normalem]{ulem}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{capt-of}
\usepackage{hyperref}
\notindent \notag \usepackage{ dsfont } \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry}
\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}
|
