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+#+TITLE: Midterm 2
+#+AUTHOR: Lizzy Hunt
+#+STARTUP: entitiespretty fold inlineimages
+#+LATEX_HEADER: \notindent \notag \usepackage{ dsfont } \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} \usepackage{polynom} \usepackage{wasysym}
+#+LATEX: \setlength\parindent{0pt}
+#+OPTIONS: toc:nil
+
+* Question One
+In $\mathds{Z}_3[x]$, $4x^3 + 4x + 4 \equiv x^3 + x + 1$
+\begin{equation*}
+\polylongdiv[style=A]{x^3 + x + 1}{2x^2+1}
+\end{equation*}
+
+$\frac{1}{2}x \equiv_3 2x$ and $\frac{1}{2}x + 1 \equiv_3 (2x + 1)$
+
+So, $a(x) = b(x)q(x) + r(x) = (2x^2 + 1)(2x) + (2x + 1)$
+
+* Question Two
+\begin{equation*}
+\polylongdiv[style=A]{x^4 + x^3 + x + 1}{x-2}
+\end{equation*}
+
+Shows that the remainder will be zero when we are in $\mathds{Z}_3[x]$
+
+* Question Three
+We will use the Euclidean algorithm to find the GCD:
+** GCD
+\begin{align*}
+3x^3 + 2x^2 + 2x + 3 &= (3x - 1)(x^2 + x + 3) + (4x + 1) \\
+(x^2 + x + 3) &\equiv_5 (4x + 1)(4x + 3) + 45 \equiv_5 (4x + 1)(4x + 3) + 0 \\
+\end{align*}
+
+So, the GCD is (4x + 1)
+
+** (supplement) Division Algorithm Work
+\begin{equation*}
+\polylongdiv[style=A]{3x^3 + 2x^2 + 2x + 3}{x^2 + x + 3}
+\end{equation*}
+
+\begin{equation*}
+\polylongdiv[style=A]{x^2 + x + 3}{4x + 1}
+\end{equation*}
+
+Check the GCD:
+
+\begin{equation*}
+\polylongdiv[style=A]{3x^3 + 2x^2 + 2x + 3}{4x + 1}
+\end{equation*}
+
+$\frac{3}{4}x^2 + \frac{5}{16}x + \frac{27}{64} \equiv_5 2x^2 + 3$
+and $\frac{165}{64} \equiv_5 0$
+
+\begin{equation*}
+\polylongdiv[style=A]{x^2 + x + 3}{4x + 1}
+\end{equation*}
+
+$\frac{1}{4}x + \frac{3}{16} \equiv_5 4x + 3$
+and $\frac{45}{16} \equiv_5 0$.
+
+* Question Four
+** a
+$\mathds{Z}_3[x]$ is a field by Theorem 2.8.
+
+There are no roots of $2x^3 + x + 1$ in $\mathds{Z}_3$, so by Theorem 5.10 $\mathds{Z}_3[x] / (2x^3 + x + 1)$ is a field, as it is
+irreducible by Corollary 4.19.
+
+** b
+Since we've proven $\mathds{Z}_3[x] / (2x^3 + x + 1)$ is a field, we know that $[2x + 1]$ must be a unit.
+
+By similar reasoning to the proof of Theorem 5.10 (1), $gcd(2x + 1, 2x^3 + x + 1) = 1$
+
+Then,
+
+\begin{equation*}
+\polylongdiv[style=A]{2x^3 + x + 1}{2x + 1}
+\end{equation*}
+
+Shows that:
+\begin{align*}
+2x^3 + x + 1 &= (2x + 1)(x^2 + x) + 1 \\
+1 &= (2x^3 + x + 1) + (2x+1)(-x^2 - x) \\
+1 - (2x^3 + x + 1) &= (2x + 1)(-x^2 - x) \\
+1 &\equiv_{2x^3 + x + 1} (2x+1)(-x^2 - x)
+\end{align*}
+and thus, by similar logic again found in the proof mentioned before, $[-x^2 - x] = [2x^2 + 2x]$ must be the inverse.
+
+As a sanity check,
+
+\begin{equation*}
+\polylongdiv[style=A]{2x^3 + x + 1}{2x^2 + 2x}
+\end{equation*}
+
+shows that the remainder is a unit in $\mathds{Z}[3]$.
+
+* Question Five
+Firstly, $\mathds{Z}_5[x]$ is a field by Theorem 2.8.
+
+** a
+There are no roots of $x^2 + 2x + 3$ in $\mathds{Z}_5$, so by Theorem 5.10 it is irreducible.
+
+$f_1 : (0, 1, 2, 3, 4) \rightarrow (3, 1, 1, 3, 2)$
+
+** b
+This is reducible since $1$ is a root.
+
+\begin{equation*}
+\polylongdiv[style=A]{x^3 + x + 3}{x - 1}
+\end{equation*}
+
+Thus, $f_2(x) = (x + 4)(x^2 + x + 2)$
+
+
+* Question Six
+Following pages 137 - 138 of the book...
+
+By Corollary 4.19 we know $(x^2 + 1)$ is irreducible in $\mathds{R}[x]$ since its only roots are in $\mathds{C}$: $\pm i$.
+
+Because of Corollary 5.12, we know that there is an extension field $K$ that contains a root to $(x^2 + 1)$. Namely,
+some $\alpha = [x]$. By Corollary 5.5, each element can be rewritten as $[ax + b]$ with $a, b \in \mathds{R}$.
+We can create a mapping $f$ such that $[a + bx] \rightarrow [a] + [b][x] = a + b \alpha$ is unique and $f^{-1}(a + b \alpha) \rightarrow [a + bx]$ (bijection).
+
+Additionally, we can show that $f([a + bx]) + f([c + dx]) = (a + b \alpha) + (c + d \alpha) = (a + c) + (b + d) \alpha = f([(a + c)x + (b + d)])$.
+
+\begin{equation*}
+[ax + b][cx + b] = [acx^2 + abx + bcx + bd]
+\end{equation*}
+
+\begin{equation*}
+\polylongdiv[style=A]{acx^2 + abx + bcx + bd}{x^2 + 1}
+\end{equation*}
+
+So, multiplication in $K$ is given:
+
+\begin{equation*}
+[ax + b][cx + b] = [(ab + bc)x + ac - bd]
+\end{equation*}
+
+Then, we can show that by definition of $\alpha$, $f([a + bx]) \cdot f([c + dx]) = (a + b \alpha) \cdot (c + d \alpha) = (ac + ad \alpha + bc \alpha + bd \alpha^2) = (ac - bd + (ad + bc)\alpha) = f([(ab + bc)x + ac - bd]) = f([a + bx][c + dx])$.
+
+Now, replacing our work with $\alpha = i$, $f$ is an isomorphism from $\mathds{R}/(x^2 + 1)$ to $\mathds{C}$ since it is a bijection and satisfies the addition and multiplication rules.
+