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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, $f([ax + b] \cdot [cx + d]) = f([(ad + bc)x + (3ac + bd)]) = (ad + bc) + (3ac + bd)\sqrt{3} = (a + b\sqrt{3})(c + d\sqrt{3}) = f([ax + b])f([cx + d])$