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