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#+TITLE: Assignment Nine
#+AUTHOR: Lizzy Hunt
#+STARTUP: entitiespretty fold inlineimages
#+LATEX_HEADER: \notindent \notag \usepackage{ dsfont } \usepackage{amsmath} \usepackage[a4paper,margin=1in,landscape]{geometry} \usepackage{polynom} \usepackage{wasysym}
#+LATEX: \setlength\parindent{0pt}
#+OPTIONS: toc:nil
* Section 5.1
** Question One
*** b
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*}
*** c
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*}
** Question Three
$|${ $0, 1, x, x + 1, x^2, x^2 + 1, x^2 + x, x^2 + x + 1$ }$|$ = 8
** Question Four
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. $3^3 = 27$
** Question Six
By Corollary 5.5, all the congruence classes in $F[x]$ are $c \ni c \in F$.
** Question Eight
\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 $k(x)$, $p(x) | f(x) - g(x) \Rightarrow f(x) \equiv_{p(x)} g(x)$
** Question Eleven
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_{p(x)} 0_F$.
Still, $f(x)g(x) \equiv_{}_{p(x)} p(x) \equiv_{p(x)} 0_F$ since $p(x) | p(x)$.
* Section 5.2
** Question One
The congruence classes are those in Section 5.1, Question Three as above.
| + | [0] | [1] | [x] | [x + 1] | [x^2] | [x^2 + 1] | [x^2 + x] | [x^2 + x + 1] |
| [0] | [0] | [1] | [x] | [x + 1] | [x^2] | [x^2 + 1] | [x^2 + x] | [x^2 + x + 1] |
| [1] | [1] | [0] | [x+1] | [x] | [x^2 + 1] | [x^2] | [x^2 + x + 1] | [x^2 + x] |
| [x] | [x] | [x + 1] | [0] | [1] | [x^2 + x] | [x^2 + x + 1] | [x^2] | [x^2 + 1] |
| [x + 1] | [x + 1] | [x] | [1] | [0] | [x^2 + x + 1] | [x^2 + x] | [x^2 + 1] | [x^2] |
| [x^2] | [x^2] | [x^2 + 1] | [x^2 + x] | [x^2 + x + 1] | [0] | [1] | [x] | [x + 1] |
| [x^2 + 1] | [x^2 + 1] | [x^2] | [x^2 + x + 1] | [x^2 + x] | [1] | [0] | [x + 1] | [x] |
| [x^2 + x] | [x^2 + x] | [x^2 + x + 1] | [x^2] | [x^2 + 1] | [x] | [x+1] | [0] | [1] |
| [x^2 + x + 1] | [x^2 + x + 1] | [x^2 + x] | [x^2 + 1] | [x^2] | [x+1] | [x] | [1] | [0] |
| \cdot | [0] | [1] | [x] | [x + 1] | [x^2] | [x^2 + 1] | [x^2 + x] | [x^2 + x + 1] |
| [0] | [0] | [0] | [0] | [0] | [0] | [0] | [0] | [0] |
| [1] | [0] | [1] | [x] | [x + 1] | [x^2] | [x^2 + 1] | [x^2 + x] | [x^2 + x + 1] |
| [x] | [0] | [x] | [x^2] | [x^2+x] | [x+1] | [1] | [x^2+x+1] | [x^2+1] |
| [x + 1] | [0] | [x + 1] | [x^2 + x] | [x^2+1] | [x^2+x+1] | [x^2] | [1] | [x] |
| [x^2] | [0] | [x^2] | [x+1] | [x^2+x+1] | [x^2+x] | [x] | [x^2+1] | [1] |
| [x^2 + 1] | [0] | [x^2 + 1] | [1] | [x^2] | [x] | [x^2+x+1] | [x+1] | [x^2+x] |
| [x^2 + x] | [0] | [x^2 + x] | [x^2+x+1] | [1] | [x^2+1] | [x+1] | [x] | [x+1] |
| [x^2 + x + 1] | [0] | [x^2 + x + 1] | [x^2+1] | [x] | [1] | [x^2+x] | [x^2] | [x+1] |
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).
** Question Two
| + | [0] | [1] | [2] | [x] | [x+1] | [x+2] | [2x] | [2x+1] | [2x+2] |
| [0] | [0] | [1] | [2] | [x] | [x+1] | [x+2] | [2x] | [2x+1] | [2x+2] |
| [1] | [1] | [2] | [0] | [x+1] | [x+2] | [x] | [2x+1] | [2x+2] | [2x] |
| [2] | [2] | [0] | [1] | [x+2] | [x] | [x+1] | [2x+2] | [2x] | [2x+1] |
| [x] | [x] | [x+1] | [x+2] | [2x] | [2x+1] | [2x+2] | [0] | [1] | [2] |
| [x+1] | [x+1] | [x+2] | [x] | [2x+1] | [2x+2] | [2x] | [1] | [2] | [0] |
| [x+2] | [x+2] | [x] | [x+1] | [2x+2] | [2x] | [2x+1] | [2] | [0] | [1] |
| [2x] | [2x] | [2x+1] | [2x+2] | [0] | [1] | [2] | [x] | [x+1] | [1] |
| [2x+1] | [2x+1] | [2x+2] | [2x] | [1] | [2] | [0] | [x+1] | [x+2] | [x] |
| [2x+2] | [2x+2] | [2x] | [2x+1] | [2] | [0] | [1] | [x+2] | [x] | [x+1] |
| \cdot | [0] | [1] | [2] | [x] | [x+1] | [x+2] | [2x] | [2x+1] | [2x+2] |
| [0] | [0] | [0] | [0] | [0] | [0] | [0] | [0] | [0] | [0] |
| [1] | [0] | [1] | [2] | [x] | [x+1] | [x+2] | [2x] | [2x+1] | [2x+2] |
| [2] | [0] | [2] | [1] | [2x] | [2x+2] | [2x+1] | [x] | [x+2] | [x+1] |
| [x] | [0] | [x] | [2x] | [2] | [x+2] | [2x+2] | [1] | [x+1] | [2x+1] |
| [x+1] | [0] | [x+1] | [2x+2] | [x+2] | [2x] | [1] | [2x+1] | [2] | [x] |
| [x+2] | [0] | [x+2] | [2x+1] | [2x+2] | [1] | [x] | [x+1] | [2x] | [2] |
| [2x] | [0] | [2x] | [x] | [1] | [2x+1] | [x+1] | [2] | [2x+2] | [x+2] |
| [2x+1] | [0] | [2x+1] | [x+2] | [x+1] | [2] | [2x] | [2x+2] | [x] | [1] |
| [2x+2] | [0] | [2x+2] | [x+1] | [2x+1] | [x] | [2] | [x+2] | [1] | [2x] |
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).
** Question Three
| + | [0] | [1] | [x] | [x+1] |
| [0] | [0] | [1] | [x] | [x+1] |
| [1] | [1] | [0] | [x+1] | [x] |
| [x] | [x] | [x+1] | [0] | [1] |
| [x+1] | [x+1] | [x] | [1] | [0] |
| \cdot | [0] | [1] | [x] | [x+1] |
| [0] | [0] | [0] | [0] | [0] |
| [1] | [0] | [1] | [x] | [x+1] |
| [x] | [0] | [x] | [1] | [x+1] |
| [x+1] | [0] | [x+1] | [x+1] | [0] |
Not, this is _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).
** Question Six
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 $[ax + b] \cdot [cx + d] = [(ad + bc)x + (2ac + bd)]$
** Question Nine
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 5.3
** Question One
*** a
$x^3 + 2x^2 + x + 1$ does not have any roots in \mathds{Z}_3, so by Corollary 4.19 it must be irreducible,
and thus a field by 5.10
*** b
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.
*** c
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.
** Question Two
*** a
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*}
*** b
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 \smiley):
+ $f$ is injective since $f([ax + b]) = f([cx + d]) \Rightarrow a + b\sqrt{2} = c + d\sqrt{2} \Rightarrow a = c \wedge b = d$
+ $f$ is surjective since each $a + b\sqrt{2}$ is uniquely mapped to $[ax + b]$
+ $f([ax + b] + [cx + d]) = f([(a + c)x + (b + d)]) = (a + c) + (b + d)\sqrt{2} = (a + b\sqrt{2}) + (c + d\sqrt{2}) = f([ax + b]) + f([cx + d])$
+ And via Question Six from section 5.2 above,
$f([ax + b] \cdot [cx + d]) = f([(ad + bc)x + (2ac + bd)]) = (ad + bc) + (2ac + bd)\sqrt{2} = (a + b\sqrt{2})(c + d\sqrt{2}) = f([ax + b])f([cx + d])$
** Question Five
*** a
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*}
*** b
**** Quick Lemma
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 $[ax + b] \cdot [cx + d] = [(ad + bc)x + (3ac + bd)]$
**** Yeah, it's an isomorphism
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 \smiley):
+ $f$ is injective since $f([ax + b]) = f([cx + d]) \Rightarrow a + b\sqrt{3} = c + d\sqrt{3} \Rightarrow a = c \wedge b = d$
+ $f$ is surjective since each $a + b\sqrt{3}$ is uniquely mapped to $[ax + b]$
+ $f([ax + b] + [cx + d]) = f([(a + c)x + (b + d)]) = (a + c) + (b + d)\sqrt{3} = (a + b\sqrt{3}) + (c + d\sqrt{3}) = f([ax + b]) + f([cx + d])$
+ And from the lemma, $f([ax + b] \cdot [cx + d]) = f([(ad + bc)x + (3ac + bd)]) = (ad + bc) + (3ac + bd)\sqrt{3} = (a + b\sqrt{3})(c + d\sqrt{3}) = f([ax + b])f([cx + d])$
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