├── .gitignore ├── README.md ├── main.fls ├── main.pdf ├── main.tex ├── src ├── fields.tex ├── group.tex ├── images │ ├── Dihedral2.pdf │ ├── Dihedral2_120.pdf │ ├── Dihedral2_240.pdf │ ├── Dihedral2_360.pdf │ ├── Dihedral3.pdf │ ├── cube.tex │ ├── first_iso.pdf │ ├── second_iso.pdf │ └── wave.png └── rings.tex └── styles.sty /.gitignore: -------------------------------------------------------------------------------- 1 | main.synctex.gz 2 | *.fls 3 | *.log 4 | main.fdb_latexmk 5 | main.thm 6 | main.aux -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | # LaTeX source code for Algebra I 2 | 3 | See the published PDF at [published PDF](https://murphyshark.vercel.app/static/notes/abstract_algebra.pdf) 4 | 5 | Or play with the project in the [Crixet sandbox here](https://app.crixet.com/?github=https%3A%2F%2Fgithub.com%2Fpehcy%2Falgebra-notes-latex%2Ftree%2Fmain&mode=sandbox) 6 | -------------------------------------------------------------------------------- /main.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/pehcy/algebra-notes-latex/d7939b136bed4b3995914aa403c720a14c345990/main.pdf -------------------------------------------------------------------------------- /main.tex: -------------------------------------------------------------------------------- 1 | \documentclass[11pt,twoside]{report} 2 | \usepackage[a4paper, margin=22.4mm,bottom=1in,top=1in,footskip=31.5pt]{geometry} 3 | \usepackage{styles} 4 | \usepackage{import} 5 | \usepackage{fancyhdr} 6 | \usepackage{fontawesome} 7 | 8 | \definecolor{fireEngRed}{RGB}{205, 32, 41} 9 | \definecolor{capri}{RGB}{0, 189, 254} 10 | \definecolor{water}{RGB}{216,242,255} 11 | \definecolor{extinctGreen}{RGB}{85,163,56} 12 | \definecolor{darkGoldrod}{RGB}{184,134,11} 13 | 14 | \pagestyle{fancy} 15 | \fancyhf{} 16 | \renewcommand{\headrulewidth}{0pt} 17 | \renewcommand{\footrulewidth}{0.5pt} 18 | \renewcommand{\footrule}{\hbox to\headwidth{\color{fireEngRed}\leaders\hrule height \footrulewidth\hfill}} 19 | \fancyfoot[LE]{\sffamily \thepage \quad $\vert$ \hspace{1em} \leftmark} 20 | \fancyfoot[RO]{\sffamily \leftmark \hspace{1em} $\vert$ \quad \thepage} 21 | 22 | \title{Abstract Algebra} 23 | \begin{document} 24 | \begin{titlepage} 25 | \centering 26 | \vfill 27 | {\Large 28 | \textbf{Abstract Algebra}\\ 29 | \vskip2cm 30 | pehcy (MurphyShark) \quad \textsf{https://github.com/pehcy} \\ 31 | } 32 | \vfill 33 | \includegraphics[width=0.55\paperwidth]{./src/images/wave.png} % also works with logo.pdf 34 | \vskip2cm 35 | { 36 | Based on lectures UECM 3363 Abstract Algebra in Universiti Tunku Abdul Rahman in 2020\\ 37 | Notes taken by MurphyShark. 38 | } 39 | \vfill 40 | \vfill 41 | \end{titlepage} 42 | \newpage 43 | \import{src/}{group.tex} 44 | \import{src/}{rings.tex} 45 | \import{src/}{fields.tex} 46 | %\import{src/}{modelValidation.tex} 47 | \end{document} -------------------------------------------------------------------------------- /src/fields.tex: -------------------------------------------------------------------------------- 1 | \chapter{Fields} 2 | 3 | \section{Extenstion Fields} 4 | 5 | \subsection{Simple Extension} 6 | 7 | \begin{definition} 8 | Let $E$ be an extension field of a field $F$ and let $\mathfrak{a} \in E$. We call an element 9 | $\mathfrak{a}$ \bred{algebraic} over $F$ if $\mathfrak{a}$ is the zero of some nonzero 10 | polynomial in $F[x]$. If $\mathfrak{a}$ is not algebraic over $F$, it is called 11 | \bred{transcendental} over $F$. 12 | \end{definition} 13 | 14 | \begin{example} 15 | $\mathbb{C}$ is an extension field of $\mathbb{R}$. 16 | \[ 17 | \begin{tikzcd}[row sep=1.0em] 18 | \mathbb{C} \arrow[dash]{d}\\ 19 | \mathbb{R} 20 | \end{tikzcd} 21 | \] 22 | The imaginary number $i = \sqrt{-1}$ is said to be algebraic since $x^2 + 1 = 0 \in \mathbb{R}[x]$. While 23 | $\pi$ is transcendental since it is not a zero in $\mathbb{R}[x]$. 24 | \end{example} 25 | 26 | \begin{theorem} 27 | Let $K$ be an extension field of $F$, and $u \in K$ is an algebraic element over $F$. Then there exists 28 | a unique monic irreducible polynomial $p(x)$ in $F[x]$ that has $u$ as a root. 29 | Furthermore if $u$ is a root of $g(x) \in F[x]$, then $p(x)$ divide $g(x)$. 30 | \end{theorem} 31 | \begin{proof} 32 | Notice that $u \in K$ is algebraic over $F$ if and only if there is a nonzero polynomial 33 | $f(x) \in F[x]$ such that $f(u) = 0_K$. 34 | 35 | Let $S$ be the set of all nonzero polynomials in $F[x]$ that have $u$ as a root, then $S$ is nonempty set. 36 | By well-ordering principle, $\exists p(x) \in S$ such that $p(x)$ has the smallest degree in $S$. 37 | 38 | Suppose that $f(x) \in F[x]$ with $f(u) = 0_K$. By division algorithm, 39 | \[ 40 | f(x) = p(x) q(x) + r(x) 41 | \] 42 | with $\deg p(x) > \deg r(x)$ or $r(x) = 0$. 43 | 44 | If $r(x) \neq 0$, 45 | \[ 46 | f(x) - p(x) q(x) = r(x) \> \Longrightarrow f(u) - p(u) q(u) = r(u) = 0. 47 | \] 48 | This contradicting the fact that $p(x)$ is the smallest polynomial. Thus $r(x) = 0$ and 49 | $p(x)$ divide $f(x)$. And now if we let $p(x)$ and $q(x)$ be the smallest polynomial. Then, 50 | $p(x) | q(x)$ and $q(x) | p(x)$ implies that $p(x) = q(x)$. 51 | \end{proof} 52 | 53 | \begin{remark} 54 | The $p(x)$ is called the "minimal polynomial of $u$ over $F$". 55 | \end{remark} 56 | 57 | \begin{example} 58 | $\mathbb{R}$ is an extension field of $\mathbb{Q}$ and $\sqrt{3} \in \mathbb{R}$ is algebraic, 59 | then 60 | \[ 61 | p(x) = x^2 - 3 \in \mathbb{Q}[x] 62 | \] 63 | \end{example} 64 | 65 | \begin{example} 66 | $\mathbb{R}$ is an extension field of $\mathbb{Q}$ and $\sqrt{3} + \sqrt{5} \in \mathbb{R}$ is algebraic, 67 | Then 68 | \[ 69 | p(x) = x^4 - 16x^2 + 4 \in \mathbb{Q}[x]. 70 | \] 71 | \end{example} 72 | \begin{solution} 73 | Let $x = \sqrt{3} + \sqrt{5} \in \mathbb{R}$, then 74 | \begin{align*} 75 | x^2 = 3 + 2\sqrt{15} + 5 &\Rightarrow x^2 - 8 = 2\sqrt{15}\\ 76 | &\Rightarrow (x^2 - 8)^2 = 4 \cdot 15\\ 77 | &\Rightarrow x^4 - 16x^2 + 4 = 0 78 | \end{align*} 79 | Thus $p(x) = x^4 - 16x^2 + 4 \in \mathbb{Q}[x]$. 80 | \end{solution} 81 | 82 | \begin{theorem} 83 | Let $K$ be an extension field of $F$ and $u \in K$ is an algebraic element 84 | over $F$ with minimal polynomial $p(x)$ of degree $n$, then 85 | \begin{enumerate} 86 | \item $F[u]$ is a field isomorphism of $F[x]/p(x)$. 87 | \item $\{ 1, u, u^2, \ldots, u^{n-1}\}$ is a basis of the 88 | vector space $F(u)$ over $F$. 89 | \item $|F(u):F|=n$. 90 | \end{enumerate} 91 | \end{theorem} 92 | \begin{proof} 93 | \begin{enumerate} 94 | \item Since $F(u)$ contains $F$ and $u$, so $F(u)$ contains every element of the form 95 | \[ 96 | b_0 + b_1u + b_2u^2 + \cdots + b_tu^t \quad \forall b_i \in F. 97 | \] 98 | We again define a function $\varphi : F[x] \to F(u)$ by 99 | \[ 100 | \varphi(f(x)) = f(u). 101 | \] 102 | Then $\varphi$ is ring homomorphism. 103 | 104 | Note that $ker\, \varphi = \langle p(x) \rangle$ where $p(x)$ is the minimal polynomial 105 | of $u$ over $F$. By the first isomorphism theorem, $F[x] / p(x) \cong \Ima \varphi$. 106 | Since $p(x)$ is irreducible, the quotient ring $F[x] / \langle p(x) \rangle$ is a field and 107 | $\Ima \varphi$ is also a field. 108 | 109 | Note that $\varphi(c) = c \> \forall c \in F$ and $\varphi (x) = u$. Thus $F \subset \Ima \varphi$ and 110 | $u \in \Ima \varphi$. By definition of simple extension, $F(u) = \Ima \varphi$. 111 | 112 | \item Since $F(u) = \Ima \varphi, \> \forall w \in F(u), \exists f(x) \in F \> s.t. \> 113 | f(u) = w$. If $\deg f(x) > n$. By division algorithm, we have 114 | \[ 115 | f(x) = p(x) q(x) + r(x) 116 | \] 117 | If $r(x) = 0$, $f(u) = p(u)q(u) = 0 = w$. 118 | 119 | Otherwise, if $r(x) \neq 0$, then 120 | \[ 121 | r(x) = f(x) - p(x)q(x) \Longrightarrow r(u) = f(u) - p(u)q(u) = w 122 | \] 123 | with $\deg r(x) < n$. Thus 124 | \[ 125 | F(u) = \text{Span}\{1 , u, u^2, \ldots, u^{n-1} \}. 126 | \] 127 | 128 | \item Let $c_0 +c_1u + c_2u^2 + \cdots + c_{n-1}u^{n-1} = 0$. If 129 | $\exists c_k \in F \> s.t. \> c_k \neq 0$. Then $p(x)$ is not the minimal polynomial of 130 | $u$ over $F$. Thus $c_i = 0 \forall c_i \in F$. Hence, we can say that 131 | $\{1, u, u^2, \ldots, u^{n-1} \}$ is linearly independent and it is also 132 | a basis of $F(u)$ over $F \Longrightarrow |F(u):F| = n$. 133 | \end{enumerate} 134 | \end{proof} 135 | 136 | \begin{example} 137 | $\mathbb{Q}[\sqrt{3}]$ is isomorphic to $\mathbb{Q}[\sqrt{3}]/\langle x^2 - 3 \rangle$ 138 | \end{example} 139 | 140 | \begin{example} 141 | If $u$ and $v$ have the same minimal polynomial $p(x)$ in $F[x]$, then $F(u)$ is 142 | isomorphic to $F(v)$. For instance, 143 | \[ 144 | \mathbb{Q}[\sqrt{3}] \cong \mathbb{Q}[-\sqrt{3}]. 145 | \] 146 | \end{example} 147 | 148 | Let $\sigma: F \to E$ be an isomorphism then we again define 149 | $\olsi{\sigma} : F[x] \to E[x]$ by for $a_0 + a_1x + \cdots + a_n x^n \in F[x]$. We can write 150 | \begin{equation} 151 | \sigma (a_0 + a_1x + \cdots + a_n x^n) = \sigma(a_0) + \sigma(a_1x) + \cdots + \sigma(a_n x^n) 152 | \end{equation} 153 | and $\olsi{\sigma}$ is also isomorphism. 154 | 155 | \begin{corollary} 156 | Let $\sigma : F \to E$ be an isomorphism of fields. Let $u$ be an algebraic element in 157 | "some" extension field of $F$ with minimal polynomial $p(x) \in F[x]$. Again we let 158 | $v$ be an algebraic element in some extension field of $E$ with minimal polynomial 159 | $\sigma p(x) \in E[x]$. Then $\sigma$ extends to an isomorphism of fields 160 | $\olsi{\sigma} : F(u) \to E(v)$ such that 161 | \[ 162 | \olsi{\sigma}(u) = v \> \text{ and } \> \olsi{\sigma}(c) = \sigma(c) \quad \forall c \in F. 163 | \] 164 | \end{corollary} 165 | \begin{proof} 166 | By previous theorem, $\varphi:F[x]/\left( p(x) \right) \to F(u)$ and 167 | $\olsi{\varphi}:E[x]/\left( \sigma p(x) \right) \to E(v)$ are isomorphism where 168 | $\varphi( \left[ f(x) \right]) = f(u)$ and $\olsi{\varphi}( \left[ g(x) \right]) = g(v)$. 169 | 170 | Furthermore, we let $\xi $ be the surjective isomorphism 171 | \[ 172 | \olsi{\xi}: E[x] \to E[x]/(\sigma p(x)) 173 | \] 174 | defined by $\olsi{\xi}(g(x)) = |g(x)|$. 175 | 176 | Note that 177 | \[ 178 | \begin{tikzcd}[row sep=2.5em] 179 | F[x] \arrow[r, "\sigma"] \arrow[dash]{d} & E[x] \arrow[r, "\olsi{\xi}"] \arrow[dash]{d} & E[x] / \left( \sigma p(x)\right) \arrow[r, "\olsi{\varphi}"] \arrow[dash]{d}& E[v] \arrow[dash]{d}\\ 180 | f(x) \arrow[r] & \sigma f(x) \arrow[r] & \left[ \sigma f(x) \right] \arrow[r] & \sigma f(v)\\ 181 | \end{tikzcd} 182 | \] 183 | Since $\sigma, \olsi{\varphi}$ and $\olsi{xi}$ are surjective, so is the composite function. 184 | \begin{align*} 185 | ker\, \olsi{\phi}(\olsi{\xi}(\sigma)) &= 186 | \{ f(x) \in F[x] \> | \> \sigma f(v) = 0 \}\\ 187 | &= \{ f(x) \in F[x] \> | \> \sigma f(x) \in \langle \sigma f(x) \rangle \}\\ 188 | &= \langle p(x) \rangle 189 | \end{align*} 190 | By First isomorphism theorem, 191 | \[ 192 | F(u) \cong F[x] / \langle p(x) \rangle \cong^{\theta} E(v) 193 | \] 194 | Since $\theta([f(x)]) = \sigma f(v)$. Note that 195 | \[ 196 | \theta([x]) = \sigma \cdot 1_F \cdot v = 1_V \cdot v = v 197 | \] 198 | 199 | so we have the following situation 200 | \[ 201 | \begin{tikzcd}[row sep=1.0em] 202 | F(u) & F[x] / \langle p(x) \rangle \arrow[l, "\varphi"] \arrow[r, "\theta"] & E(v)\\ 203 | f(u) & \left[ f(x) \right] \arrow[l, "\varphi"] \arrow[r, "\theta"] & \sigma f(v)\\ 204 | c & \left[ c \right] \arrow[l, "\varphi"] \arrow[r, "\theta"] & \sigma(c)\\ 205 | \end{tikzcd} 206 | \] 207 | 208 | The composite function $\theta \circ \varphi^{-1} : F(u) \to E(v)$ is an isomorphism 209 | that extends $\sigma$ and maps $u$ to $v$. 210 | \[ 211 | \tikz[overlay]{ 212 | \draw (8.5,-2.25) node[anchor=south, color=cyan!70!black] {\textsf{By First isomorphism thm.}}; 213 | \draw[ultra thick, cyan!70!black, fill=cyan!90!black, opacity=0.25, rounded corners] (5.35,-1.65) -- 214 | (10,-1.65) -- (10,0.01) -- (7.35,0.01) -- (7.35,1.225) -- (2.82,1.225) -- (2.82,0.02) -- (5.35,0.02) -- cycle; 215 | } 216 | \begin{tikzcd}[row sep=2.5em] 217 | & F \arrow[r, "\sigma"] \arrow[dash, "\subseteq"]{d} & E \arrow[dash, "\subseteq"]{d} & \\ 218 | & F[x] \arrow[r, "\sigma"] \arrow[d, "\xi"] & E[x] \arrow[d, "\olsi{\xi}"] & \\ 219 | F(u) \arrow[r, "\varphi"] & F[x] / \langle p(x) \rangle \arrow[r, "\sigma"] & E[x] / \langle \sigma p(x)\rangle \arrow[r, "\olsi{\varphi}"] & E(v)\\ 220 | \end{tikzcd} 221 | \] 222 | \end{proof} 223 | 224 | \begin{example} 225 | $x^3 - 2$ is irreducible in $\mathbb{Q}[x]$ by Eisenstein's criterion. $\sqrt[3]{2} \in \mathbb{R}$ is a root of it. 226 | Verify that $\sqrt[3]{2} \omega$ is also a root of $x^3 - 2$ in $\mathbb{C}$ where 227 | \[ 228 | \omega = \frac{-1 + \sqrt{3}i}{2} 229 | \] 230 | is a complex cube root of $1$. 231 | \end{example} 232 | \begin{solution} 233 | Let $\sigma$ be the identity function from $\mathbb{Q}$ to $\mathbb{Q}$. By applying the previous corollary, we have 234 | \[ 235 | \mathbb{Q}(\sqrt[3]{2}) \cong^\theta \mathbb{Q}(\sqrt[3]{2} \omega) 236 | \] 237 | such that $\olsi{\sigma}(\sqrt[3]{2}) = \sqrt[3]{2} \omega$. And now $(\sqrt[3]{2} \omega)^3 = 2 \omega^3 = 2$. 238 | \end{solution} 239 | 240 | 241 | \subsection{Algebraic Extension} 242 | 243 | \begin{definition}[Algebraic extension] 244 | An extension field $K$ of a field $F$ is said to be an algebraic extension of $F$ if every element 245 | of $K$ is algebraic over $F$. 246 | \end{definition} 247 | 248 | \begin{example} 249 | $\mathbb{C}$ is an algebraic extension of $\mathbb{R}$. $\forall a+bi \in \mathbb{C}$, where 250 | $a,b \in \mathbb{R}$ and $i = \sqrt{-1}$. We have 251 | \[ 252 | (x+a+bi)(x+a-bi) = x^2 +2ax + a^2 + b^2. 253 | \] 254 | Thus $a+bi$ is a root of $x^2 +2ax + a^2 + b^2 = 0$. 255 | \end{example} 256 | 257 | \begin{theorem} 258 | If $K$ is a finite-dimensional extension field of $F$, then $K$ is an algebraic extension of $F$. 259 | \end{theorem} 260 | \begin{proof} 261 | Let $\{ V_1, V_2, \ldots, V_n \}$ be the basis of $K$ over $F$. For all $u \in K$, 262 | $\{ 1, u, u^2, \ldots, u^n \}$ is linearly dependent. That is, 263 | \[ 264 | \exists u^k \in K \> s.t. \> u^k = \text{Span} \{1, u, u^2, \ldots, u^n\} = c_0 + c_1 u + c_2 u^2 + 265 | \cdots + c_{k-1} u^{k-1} (k \geq 1). 266 | \] 267 | Thus $u$ is a root of $f(x) = x^k - c_{k-1} u^{k-1} - \cdots - c_0$, this implies $K$ is an algebraic extension. 268 | \end{proof} 269 | 270 | \hrule 271 | \vspace{10pt} 272 | 273 | In fact, a simple extension is an algebraic extension if $u$ is algebraic. If extension field $K$ contains a 274 | transcental element $u$, then $K$ must be infinite dimensional over $F$. 275 | \[ 276 | \textsf{Non algebraic} \Longrightarrow \textsf{Infinite dimension} 277 | \] 278 | Note that $F(u)$ denote the intersection of all subfields of $K$ that contains both $F$ and $u$. It said to be a simple 279 | extension of $F$. If $u_1, u_2, \ldots, u_n$ are elements of an extension field $K$ of $F$. Let 280 | $F(u_1, \ldots, u_n)$ denote the intersection of all the subfields of $K$ that contain $F$ and every $u$ (known as 281 | generalized simple extension); $F(u, u_1, \ldots, u_n)$ is said to be a finitely generated extension of $F$. 282 | 283 | \begin{theorem} 284 | If $K = F(u_1, u_2, \ldots, u_n)$ is a finitely generated extension field of 285 | $F$ and each $u_i$ is algebraic over $F$, then $K$ is a finite-dimensional algebraic extension of $F$. 286 | \end{theorem} 287 | \begin{proof} 288 | Note that if $u, v$ is algebraic over $F$, then $v$ is algebraic over $F(u)$. Thus 289 | \[ 290 | |F(u,v) : F(u)|\cdot |F(u) : F| < \infty \Longrightarrow 291 | |F(u,v) : F| = |F(u,v) : F(u)| \cdot |F(u) : F| < \infty. 292 | \] 293 | By mathematical induction, we have 294 | \[ 295 | |F(u_1, u_2, \ldots, u_n) : F(u_1, u_2, \ldots, u_{n-1})| \ldots 296 | |F(u_1) : F| < \infty 297 | \] 298 | which is also finite. 299 | \end{proof} 300 | 301 | \begin{corollary} 302 | If $L$ is algebraic extension of $K$ and $K$ is an algebraic extension of $L$. Then 303 | $L$ is an algebraic extension field of $F$. 304 | \end{corollary} 305 | \begin{proof} 306 | $\forall \omega \in L, \> \exists f(x) \in K[x] \quad s.t. \> f(\omega) = a_0 + a_1 \omega 307 | + \cdots + a_n \omega^n$. 308 | 309 | Note that $F(a_0, a_1, \ldots, a_n)$ is finitely generated extension of $F$ and all $a_i$'s 310 | \\are algebraic. Thus it is finite dimensional algebraic extension of $F$. Since 311 | $\omega$ is algebraic over $F(a_0, a_1, \ldots, a_n)$. So $F(a_0, a_1, \ldots, a_n)$ is finite 312 | dimensional extension of $F \Longrightarrow \omega$ is algebraic over $F$. 313 | Thus $L$ is an algebraic extension of $F$. 314 | \end{proof} 315 | 316 | \begin{remark} 317 | Algebraic subfield $E$ of $\mathbb{C}$ over $\mathbb{Q}$ is called the \textbf{field of algebraic numbers}. 318 | Where $E$ is an finite-dimensional algebraic extension over $K$. 319 | \[ 320 | \begin{tikzcd}[row sep=tiny] 321 | \mathbb{C} &\\ 322 | & E \arrow[dash]{ul} 323 | \\ 324 | \mathbb{Q} \arrow[uu, "\pi"] \arrow[ur, "\mu"]&\\ 325 | \end{tikzcd} 326 | \] 327 | 328 | \begin{itemize} 329 | \item $\mu$ denote algebraic extension over $\mathbb{Q}$, e.g.: $\sqrt{2}, \sqrt{3}, i, \ldots$. 330 | \item $\pi$ denote non-algebraic extension. 331 | \end{itemize} 332 | \end{remark} 333 | 334 | \begin{corollary} 335 | Let $K$ be an extension field of $F$ and let $E$ be the set of all elements of $K$ that are 336 | algebraic over $F$. Then $E$ is a subfield of $K$ and an algebraic extension field of $F$. 337 | \end{corollary} 338 | \begin{proof} 339 | We only need to show that $E$ is a field. Let $u,v \in F$, note that $F(u,v)$ 340 | is finitely generated extension of $F$, so $E$ is algebraic extension. 341 | $E$ is closed under subtraction and multiplication. Moreover $u^{-1}$ is 342 | algebraic over $F$. Thus $E$ is a subfield of $K$. 343 | \end{proof} 344 | 345 | 346 | \begin{example} 347 | \[ 348 | \mathbb{Q}(i, -i) = \mathbb{Q}(i) 349 | \] 350 | \end{example} 351 | 352 | \begin{example} 353 | $$\mathbb{Q}(\sqrt{3}, i) = \mathbb{Q}(\sqrt{3})(i)$$ 354 | \end{example} 355 | \begin{solution} 356 | \begin{align*} 357 | |\mathbb{Q}(\sqrt{3}, i)| &= |\mathbb{Q}(\sqrt{3})(i): \mathbb{Q}|\\ 358 | &= |\mathbb{Q}(\sqrt{3})(i): \mathbb{Q}(\sqrt{3})| \, \cdot |\mathbb{Q}(\sqrt{3}): \mathbb{Q}|\\ 359 | &= 2 \cdot 2\\ 360 | &= 4 361 | \end{align*} 362 | \end{solution} 363 | 364 | \begin{example} 365 | Every finite-dimensional extension is also finitely generated. If $\{ u_1, u_2, \ldots, u_n \}$ 366 | is a basis of $K$ over $F$. This implies $F(u_1, u_2, \ldots, u_n) \subseteq K$ and 367 | $K \subseteq F(u_1, u_2, \ldots, u_n)$. 368 | Thus, 369 | \[ 370 | K = F(u_1, u_2, \ldots, u_n) = \text{Span}\{ u_1, u_2, \ldots, u_n \}. 371 | \] 372 | \end{example} 373 | 374 | \begin{example}[Non-example] 375 | $$\mathbb{Q}(\sqrt{3}, \sqrt{5}) \neq \mathbb{Q}(\sqrt{3})$$ 376 | \end{example} 377 | \begin{solution} 378 | For the sake of contradiction, consider $\mathbb{Q}(\sqrt{3}, \sqrt{5}) = \mathbb{Q}(\sqrt{3})$, then 379 | \[ 380 | \sqrt{5} = a+ b\sqrt{3}, \quad \forall a,b \in \mathbb{Q} 381 | \] 382 | Altering this equation by moving $a$ to left-hand side, then squaring both sides. We obtain 383 | \begin{align*} 384 | (\sqrt{5} - a)^2 = (b\sqrt{3})^2 \quad &\Rightarrow 5 - 2\sqrt{5}a + a^2 = 3b^2\\ 385 | &\Rightarrow \frac{5 + a^2 - 3b^2}{2a} = \sqrt{5} \quad (a\neq 0) 386 | \end{align*} 387 | However, when $a=0$, we have $5 = 3b^2$. Which is a contradiction. 388 | \end{solution} 389 | 390 | \section{Splitting Field} 391 | 392 | In last chapter we had discussed about the integral domain. Suppose 393 | polynomial $f(x)$ has degree $n$. Then $f(x)$ has at most $n$ roots in \textit{any} field. 394 | Suppose that $K$ contains fewer than $n$ roots of $f(x)$. It might be possible to find an extension field of 395 | $K$ that contains additional roots of $f(x)$. 396 | 397 | \begin{definition}[Splitting field] 398 | If $F$ is a field and $f(x) \in F[x]$, then an extension field $K$ of $F$ is said to 399 | be a \bred{splitting field} (or \bred{root field}) of $f(x)$ over $F$ provided that 400 | \begin{itemize} 401 | \item $f(x)$ splits over $K$, say 402 | \begin{equation} 403 | f(x) = c(x-u_1)\ldots (x-u_n) 404 | \end{equation} 405 | \item and 406 | \begin{equation} 407 | K = \underbrace{F(u_1, u_2, \ldots, u_n).}_{\text{smallest field}} 408 | \end{equation} 409 | \end{itemize} 410 | \end{definition} 411 | 412 | \begin{example} 413 | If $f(x) = x^4 - x^2 - 2 = (x^2 - 2)(x^2 + 1)$ in $\mathbb{Q}[x]$. Then 414 | \[ 415 | \mathbb{Q}(\sqrt{2}, -\sqrt{2}, i, -i) = \mathbb{Q}(\sqrt{2}, i) 416 | \] 417 | is a splitting field of $f(x)$ over $\mathbb{Q}$. 418 | \end{example} 419 | 420 | \newpage 421 | \section{Finite Fields} 422 | 423 | \begin{theorem} 424 | Let $R$ be a ring with identity. Then 425 | \begin{enumerate} 426 | \item The set 427 | \[ 428 | \mathfrak{P} = \{ k \cdot 1_R \> | \> k \in \mathbb{Z} \} 429 | \] 430 | is a subring of $R$. 431 | \item If $R$ has characteristic $0$, then $\mathfrak{P}$ is isomorphic to $\mathbb{Z}$. 432 | \item If $R$ has characteristic $n > 0$, then $\mathfrak{P}$ is isomorphic to $\mathbb{Z}_n$. 433 | \end{enumerate} 434 | \end{theorem} 435 | \begin{proof} 436 | We prove each of the statements listed above. 437 | \begin{enumerate} 438 | \item First of all, we use subring test to check if $\mathfrak{P}$ is a subring of $R$. 439 | \[ 440 | \begin{cases} 441 | a \cdot 1_R - b \cdot 1_R = (a-b) \cdot 1_R \in \mathfrak{P}\\ 442 | a \cdot 1_R \cdot b \cdot 1_R = ab\, 1_R \in \mathfrak{P} 443 | \end{cases} 444 | \] 445 | so $\mathfrak{P}$ is a subring of $R$. 446 | \end{enumerate} 447 | 448 | We now prove (2), (3) at once, We consider a map $f : \mathbb{Z} \to \mathbb{R}$ 449 | defined by 450 | \[ 451 | f(n) = n \cdot 1_R \quad \forall n \in \mathbb{Z}. 452 | \] 453 | Then $f$ is homomorphism because 454 | \[ 455 | f(n + m) = (n + m) \cdot 1_R = f(n) + f(m) 456 | \] 457 | and the kernel is 458 | \[ 459 | ker\, f = \{ n \in \mathbb{Z} \> | \> n \cdot 1_R = 0_R \}. 460 | \] 461 | By the first isomorphism theorem, $\mathbb{Z} / ker\, f$ is isomorphic to $\mathbb{R}$. 462 | \begin{itemize} 463 | \item If $R$ has a characteristic $0$, then $ker\, f = \langle 0 \rangle \Longrightarrow \mathbb{Z} \cong \mathbb{R}$. 464 | \item If $R$ has a characteristic $n$, then $ker\, f = \langle n \rangle \Longrightarrow \mathbb{Z}/\langle n \rangle \cong \mathbb{R}$. 465 | \end{itemize} 466 | \end{proof} 467 | 468 | \begin{corollary} 469 | Every finite field has characteristic $p$ for some prime $p$. 470 | \end{corollary} 471 | \begin{proof} 472 | Since finite field is integral domain, it must has characteristic prime $p$. 473 | \end{proof} 474 | 475 | \begin{remark} 476 | \begin{enumerate} 477 | \item The converse of it is false. There are infinite fields of characteristic $p$ (i.e. $\mathbb{Z}_p[x]$). 478 | \[ 479 | \mathfrak{P} = \{ k \cdot 1_R \> | \> k \in \mathbb{Z} \} 480 | \] 481 | is called the prime subfield of $K$ and is contained in every subfield of $K$. Which means 482 | every field of characteristic $p$ must contains $\mathbb{Z}_p$. 483 | 484 | \item The number of elements in a finite field $K$ is called the order of $K$. 485 | \end{enumerate} 486 | \end{remark} 487 | 488 | \subsection{Order of finite field} 489 | 490 | \begin{theorem} 491 | A finite field $K$ has order $p^n$, where $p$ is the characteristic of $K$ 492 | and $n = |K:\mathbb{Z}_p|$. 493 | \end{theorem} 494 | \begin{proof} 495 | Let $K$ be a finite dimensional extension of $\mathbb{Z}_p$. Let $n = |K:\mathbb{Z}_p|$, then 496 | $\{ u_1, u_2, \ldots, u_n \}$ is a basis of $K$. 497 | 498 | $\forall k \in K$, $k$ is represented uniquely be 499 | \[ 500 | k = c_1 u_1 + c_2 u_2 + \cdots + c_nu_n. 501 | \] 502 | There are precisely $p^n$ distinct linear combinations of the form. Thus 503 | $|K| = p^n$. 504 | \end{proof} 505 | 506 | \begin{lemma}[The Freshman's dream] 507 | Let $R$ be a commutative ring with identity of characteristic $p$, where $p$ is a prime. 508 | Then for every $a, b \in R$ and for all positive integer $n$ we have 509 | \begin{equation} 510 | (a + b)^{p^n} = a^{p^n} + b^{p^n}. 511 | \end{equation} 512 | \end{lemma} 513 | \begin{proof} 514 | We will use the induction on $n$. 515 | 516 | Assume $n = 1$, we expand $(a+b)^p$ with binomial theorem. 517 | \[ 518 | (a+b)^p = \sum^p_{k=0} {p \choose k} a^{n-k}b^k = a^p + 519 | {p \choose 1} a^{p-1}b + \cdots + {p \choose p-1} ab^{p-1} + b^p. 520 | \] 521 | Note that 522 | \[ 523 | {p \choose k} = \frac{p!}{(p-k)!\, k!}, \quad k, p-k < p \> \text{ for } 1 \leq k < p. 524 | \] 525 | This implies that $p$ divide ${p \choose k} \Longrightarrow 526 | {p \choose k}a^{p-k}b^k = 0 \> (\text{mod } p)$. Thus $(a+b)^p = a^p + b^p$. 527 | We are done for base case. 528 | 529 | Assume that it holds for all less than $n$. 530 | \begin{align*} 531 | (a+b)^{p^n} &= \left( (a+b)^p \right)^{p^{n-1}} \\ 532 | &= (a^p + b^p)^{p^{n-1}}\\ 533 | &= (a^p)^{p^{n-1}} + (b^p)^{p^{n-1}}\\ 534 | &= a^{p^n} + b^{p^n}. 535 | \end{align*} 536 | 537 | Therefore the theorem is true for every positive integer $n$. Now we are done. 538 | \end{proof} 539 | 540 | \begin{theorem}[Existence of finite field] 541 | Let $K$ be an extension field $\mathbb{Z}_p$. For all positive integer $n$, 542 | $K$ has order $p^n$ if and only if $K$ is a splitting field of $x^{p^n} - x$ over $\mathbb{Z}_p$. 543 | \end{theorem} 544 | \begin{proof} 545 | $(\Rightarrow)$ Let $f(x) = x^{p^n} - x$ over $\mathbb{Z}_p$. Then 546 | \[ 547 | f(x) = p^n x^{p^n - 1} - 1 = -1 548 | \] 549 | and so $f(x)$ and $f'(x)$ are relatively prime $\Longrightarrow f(x)$ is separable. 550 | Then $f(x)$ has $p^n$ distinct roots in splitting field. Let $c$ be any nonzero 551 | element of $K$. Let $\{c_1, c_2, \ldots, c_{p^n - 1}\}$ be all the nonzero elements of 552 | $K$ and let 553 | \[ 554 | u = c_1 c_2 \cdots c_k. 555 | \] 556 | 557 | Note that $\{cc_1, cc_2, \ldots, cc_k\}$ are all distinct and nonzero elements of $K$. Thus 558 | \[ 559 | \{cc_1, cc_2, \ldots, cc_k\} = \{c_1, c_2, \ldots, c_{p^n - 1}\} 560 | \] 561 | and 562 | \[ 563 | cc_1 \cdot cc_2 \cdots cc_{p^n - 1} = c_1 \cdot c_2 \cdots c_{p^n - 1} \Rightarrow 564 | c^{p^n - 1} = 1_K \Rightarrow c^{p^n} - c = 0. 565 | \] 566 | So all nonzero elements of $K$ consisting of the $p^n$ distinct roots of $x^{p^n} - x$ over 567 | $\mathbb{Z}_p$. 568 | 569 | $(\Leftarrow)$ On the other hand, let $E$ be the subset of $K$ consisting of the 570 | $p^n$ distinct roots of $x^{p^n} - x$. We want to show $E$ is a subfield of $K$. 571 | 572 | For all $a,b$ in $E$, $a^{p^n} = a$ and $b^{p^n} = b$. So 573 | \[ 574 | (a+b)^{p^n} = a^{p^n} + b^{p^n} = a+b \Rightarrow a+b\in E 575 | \] 576 | \[ 577 | (ab)^{p^n} = a^{p^n}\, b^{p^n} = ab \Rightarrow ab\in E 578 | \] 579 | \[ 580 | (-a)^{p^n} = (-1)^{p^n}\, a^{p^n} = -a \Rightarrow a\in E 581 | \] 582 | and 583 | \[ 584 | (a^{-1})^{p^n} = (a^{p^n})^{-1} = a^{-1}. 585 | \] 586 | 587 | Clearly $0,1 \in E$. Since $K$ is a field, the rest of axioms are true. 588 | Since $K$ is splitting field of $x^{p^n} - x, K \subseteq E$. Hence 589 | $K = E$ and $|K| = |E| = p^n$. 590 | \end{proof} 591 | 592 | \section{Galois Theory} 593 | 594 | The classical question of algebra : Whether or not there were formulas for the solution of higher degree polynomial 595 | equations. There are no formula for the solution of all polynomial equations of degree $n$ when $n \geq 5$. 596 | 597 | Galois theory had a profound influence on the development of later mathematics for beyond the scope of 598 | the original solvability problem. Why? Because his theory connecting the field extensions with groups. 599 | 600 | \begin{definition}[Galois group] 601 | Let $K$ be an extension field of $F$. An $F$-automorphism of $K$ is an isomorphism $\sigma : K \to K$ that 602 | fixes $F$ elementwise. That is, $\sigma(c) \to c, \forall c \in F$. 603 | 604 | The set of all $F$-automorphisms of $K$ is denoted $\gal_F K$ and is called the Galois group of $K$ over $F$. 605 | \end{definition} 606 | 607 | \begin{theorem} 608 | If $K$ is an extension field of $F$, then $\gal_F K$ is a group under the operation 609 | of composition of functions. 610 | \end{theorem} 611 | \begin{proof} 612 | We need to show that the function composition from a group: 613 | 614 | [Closedness]: $\forall \sigma, \tau \in \gal_F K, \quad \sigma \circ \tau (c) = c \> \forall c \in F$. 615 | 616 | [Associativity]: Associativity holds because of function composition's property. 617 | 618 | [Existence of identity]: $\forall \sigma \in \gal_F K$, $\sigma \circ \tau = \tau \circ \sigma = \sigma$. 619 | 620 | [Existence of inverse]: $\forall \sigma \in \gal_F K$, $\sigma^{-1}(c) = c \> \forall c \in F$. So 621 | $\sigma^{-1}$ is the inverse in $\gal_F K$. 622 | \end{proof} 623 | 624 | \begin{theorem} 625 | Let $K$ be an extension field of $F$ and $f(x) \in F[x]$. If $u \in K$ is a root of $f(x)$ and $\sigma \in \gal_F K$, 626 | then $\sigma(u)$ is also a root of $f(x)$. 627 | \end{theorem} 628 | \begin{proof} 629 | Take a $u \in K$, $f(u) = 0 \Longrightarrow \sigma(f(u)) = \sigma(0) = 0$. 630 | 631 | Since $\sigma \in \gal_F K, \sigma \left( f(u) \right) = f\left( \sigma(u) \right) = 0$. 632 | \end{proof} 633 | 634 | \begin{remark} 635 | Every root of $p(x)$ in $K$ is the image of $u$ under some automorphism of $\gal_F K$. 636 | \end{remark} 637 | 638 | \begin{theorem} 639 | Let $K$ be the splitting field of some polynomial over $F$ and $u,v \in K$. Then there exists an $\sigma \in \gal_F K$ 640 | such that $\sigma(u) = v$ if and only if $u$ and $v$ have the same minimal polynomial. 641 | \end{theorem} 642 | \begin{proof} 643 | $(\Rightarrow)$ Note that if $K$ is splitting field of $f(x)$ with $\deg f = n$. Then 644 | \[ 645 | |K:F| \leq n! \Longrightarrow K \textsf{ is algebraic over } F. 646 | \] 647 | So $u$ has a minimal polynomial $q(x)$ over $F$. 648 | 649 | By previous theorem, $p(v) = 0$ and $q(u) = 0$. These imply $p(x) | q(x)$ and 650 | $q(x) | p(x)$ and hence $p(x) = q(x)$. We are done for this direction. 651 | 652 | $(\Leftarrow)$ On the other hand, in simple extension, we can extend an isomorphism 653 | $\sigma: F \to F$ to $\olsi{\sigma}: \olsi{F} \to \olsi{F}$ such that 654 | $\olsi{\sigma}(u) = v$ and $\olsi{\sigma}(c) = c$ for all constant $c \in F$. 655 | 656 | Since $K$ is splitting field of some polynomial over $F$, which is also a splitting field of 657 | $F(u)$ and $F(v)$. We can extend an isomorphism $\olsi{\sigma}$ to an isomorphism 658 | $\tilde{\sigma}: K \to K$. If $\sigma = 1$, then $\tilde{\sigma} \in \gal_F K$. 659 | \end{proof} 660 | 661 | \begin{example} 662 | Given that $\sigma \in \gal_\mathbb{R} \mathbb{C}$, find $\sigma(i)$. 663 | \end{example} 664 | \begin{solution} 665 | By previous theorem, 666 | \[ 667 | \sigma(i) = \{ i, - i \}. 668 | \] 669 | Thus $\gal_F K = \{1, \sigma \}$ is a group of order 2 and hence $\gal_\mathbb{R} \mathbb{C}$ 670 | is isomorphic to $\mathbb{Z}_2$. 671 | \end{solution} 672 | 673 | \begin{theorem}[Galois group of finitely generated extension] 674 | Let $K = F(u_1, u_2, \ldots, u_n)$ be an algebraic extension field of $F$. If 675 | $\sigma \circ \tau \in \gal_F K $ and $\sigma(u_i) = \tau(u_i)$ for each $i = 1,2,\ldots, n$. 676 | Then $\sigma = \tau$. 677 | 678 | In other words, an automorphism in $\gal_F K$ is completely determined by its action 679 | on $u_1, u_2, \ldots u_n$. 680 | \end{theorem} 681 | \begin{proof} 682 | Let $\beta = \sigma^{-1} \circ \tau$, then 683 | \[ 684 | \sigma^{-1} \circ \tau (u_i) = \sigma^{-1} \circ \sigma(u_i) = u_i. 685 | \] 686 | Let $v \in F(u_i)$, then there exists a constant $c_i \in F$ such that 687 | \[ 688 | c_0 + c_1u_1 + \cdots + c_{n-1}u^{n-1}_1 = w \Longrightarrow \beta(v) = v. 689 | \] 690 | Again we let $w \in F(u_1, u_2)$, then there exists a constant $c_i \in F$ such that 691 | \[ 692 | c_0 + c_1u_2 + \cdots + c_{n-1}u^{n-1}_2 = w \Longrightarrow \beta(w) = w. 693 | \] 694 | Repeating this process, we conclude that for all $v \in K$, $\beta(v) = v \Longrightarrow \beta = 1$. 695 | Thus $\tau = \sigma$. 696 | \end{proof} 697 | 698 | \begin{example} 699 | Consider $\sigma \in \gal_\mathbb{Q} \mathbb{Q}[\sqrt{3}, \sqrt{5}]$. We let two actions 700 | \[ 701 | \tau(\sqrt{3}) = -\sqrt{3}, \quad \tau(\sqrt{5}) = \sqrt{5} 702 | \] 703 | and 704 | \[ 705 | \alpha(\sqrt{3}) = \sqrt{3}, \quad \alpha(\sqrt{5}) = -\sqrt{5} 706 | \] 707 | and defined $\beta = \alpha \circ \tau$. 708 | 709 | Then $\gal_\mathbb{Q} \mathbb{Q}[\sqrt{3}, \sqrt{5}] = \{ 1, \tau, \alpha, \beta \}$ 710 | such that 711 | 712 | \begin{center} 713 | \begin{tabular}{|c|c|c|c|c|} 714 | \hline 715 | & $1$ & $\tau$ & $\alpha$ & $\beta$\\ 716 | \hline 717 | $\sqrt{3}$ & $\sqrt{3}$ & $-\sqrt{3}$ & $\sqrt{3}$ & $-\sqrt{3}$\\[0.225em] 718 | \hline 719 | $\sqrt{5}$ & $\sqrt{5}$ & $\sqrt{5}$ & $-\sqrt{5}$ & $-\sqrt{5}$\\ 720 | \hline 721 | \end{tabular} 722 | \end{center} 723 | 724 | Note that $\tau, \alpha, \beta$ both have an order 2. Thus $\gal_\mathbb{Q} \mathbb{Q}[\sqrt{3}, \sqrt{5}] \cong 725 | \mathbb{Z}_2 \times \mathbb{Z}_2$. 726 | \end{example} 727 | 728 | \begin{corollary} 729 | If $K$ is splitting field of a separable polynomial $f(x)$ of degree $n$ in $F[x]$, then $\gal_F K$ is isomorphic 730 | to a subgroup of $S_n$. 731 | \end{corollary} 732 | \begin{proof} 733 | Let $u_1, u_2, \ldots, u_n$ be distinct roots of $f(x)$. Then $K = F(u_1, u_2, \ldots, u_n)$ and let 734 | \[\mathcal{U} = \{ u_1, u_2, \ldots, u_n \}.\] For all permutation $\sigma \in \gal_F K$, $\forall 1 \leq i \leq n$, 735 | $\sigma(u_i) = u_j$ for some $j$. 736 | 737 | Now define a function $\theta : \gal_F K \to S_n$ defined by $\theta: \sigma \mapsto \sigma \circ 1_\mathcal{U}$. $\theta$ 738 | is well defined. Note that 739 | \[ 740 | \sigma \circ 1_\mathcal{U} = \tau \circ 1_\mathcal{U} \Longrightarrow \sigma = \tau 741 | \] 742 | by previous theorem. Thus $\theta$ is injective, and $\theta$ is homomorphism. 743 | 744 | By the first isomorphism theorem, we say that 745 | \[ 746 | \gal_F K \cong \Ima \theta = \textsf{a subgroup of } S_n. 747 | \] 748 | \end{proof} 749 | 750 | \begin{definition}[Intermediate field] 751 | Let $K$ be an extension field of $F$. A field $E$ such that 752 | \[ 753 | F \subseteq E \subseteq K 754 | \] 755 | is called an \bred{intermediate field} of the extension. Clearly 756 | that $\gal_E K \cong \gal_F K$. 757 | \end{definition} 758 | 759 | \begin{theorem} 760 | Let $K$ be an extension field of $F$. If $H$ is a subgroup of $\gal_F K$, let 761 | \[ 762 | E_H = \{ k \in K \> | \> \sigma(k)=k \textsf{ for every } \sigma \in H \}. 763 | \] 764 | Then $E_H$ is an intermediate field of the extension $K$. The field $E_H$ is called 765 | the "fixed field" of subgroup $H$. 766 | \end{theorem} 767 | \begin{proof} 768 | It is clear that $F \subseteq E_H \subseteq K$. So we only want to show that $E_H$ is a subfield. 769 | 770 | Let $a,b \in H$. For all $\sigma \in H$, we have 771 | \[ 772 | \sigma(a + b) = \sigma(a) + \sigma(b) = a+b 773 | \] 774 | \[ 775 | \sigma(ab) = \sigma(a) \sigma(b) = ab 776 | \] 777 | \[ 778 | \sigma(0_K) = 0_K, \sigma(1_K) = 1_K 779 | \] 780 | \[ 781 | \sigma(-a) = - \sigma(a) = -a 782 | \] 783 | \[ 784 | \sigma(a^{-1}) =\left( \sigma(a)\right)^{-1} = a^{-1} 785 | \] 786 | Thus $E_H$ is a subfield of $K$. 787 | \end{proof} 788 | 789 | \begin{example} 790 | From previous example, 791 | \[ 792 | \mathbb{Q} \subseteq \mathbb{Q}[\sqrt{3}] \subseteq \mathbb{Q}[\sqrt{3}, \sqrt{5}]. 793 | \] 794 | And $\gal_{\mathbb{Q}[\sqrt{3}]} \mathbb{Q}[\sqrt{3}, \sqrt{5}] = \{ 1, \alpha \}$, 795 | and $\gal_{\mathbb{Q}} \mathbb{Q}[\sqrt{3}, \sqrt{5}] = \{1, \tau, \alpha, \beta \}$. 796 | \end{example} 797 | 798 | \begin{example} 799 | $\gal_{\mathbb{Q}[\sqrt{3}]} \mathbb{Q}[\sqrt{3}, \sqrt{5}] = \{ 1, \alpha \}$ is the fixed 800 | field of $\mathbb{Q}[\sqrt{3}] = \{ 1, \alpha \}$, where 801 | \[ 802 | \alpha(\sqrt{3}) = \sqrt{3}, \alpha(\sqrt{5}) = -\sqrt{5}. 803 | \] 804 | \end{example} 805 | \begin{solution} 806 | For all $a_i \in \mathbb{Q}$. Compute 807 | \begin{align*} 808 | &\alpha (a_0 + a_1 \sqrt{3} + a_2 \sqrt{5} + a_3 \sqrt{15}) = a_0 + a_1 \sqrt{3} + a_2 \sqrt{5} + a_3 \sqrt{15}\\ 809 | &\Longleftrightarrow \> a_0 + a_1 \sqrt{3} - a_2 \sqrt{5} - a_3 \sqrt{15} = a_0 + a_1 \sqrt{3} + a_2 \sqrt{5} + a_3 \sqrt{15}\\ 810 | &\Longleftrightarrow \> a_2 \sqrt{5} + a_3 \sqrt{15} = 0\\ 811 | &\Longleftrightarrow \> a_2 + a_3 \sqrt{3} = 0\\ 812 | \end{align*} 813 | 814 | Since $a_2a^{-1}_3 \in \mathbb{Q} \Longrightarrow -\sqrt{3} \in \mathbb{Q}$, which is a contradiction. 815 | \end{solution} 816 | 817 | \begin{example} 818 | $\gal_{\mathbb{R}} \mathbb{C} = \{ 1, \alpha \}$, and $\alpha(a + bi) \Longrightarrow a - bi = a+ bi \Longrightarrow b =0$. 819 | Thus fixed field of $\gal_{\mathbb{R}} \mathbb{C}$ is the field $\mathbb{R}$. 820 | \end{example} 821 | 822 | \begin{remark} 823 | The ground field $F$ need not always be the fixed field of the group $\gal_F K$. 824 | \end{remark} 825 | 826 | \begin{example} 827 | $\sqrt[3]{2}$ is the root of $x^3 - 2 = 0$. So $\sqrt[3]{2} \mapsto \{ \sqrt[3]{2}, \sqrt[3]{2} \omega, \sqrt[3]{2} \omega^2 \}$, 828 | where $\omega$ is the cube root of unity. 829 | 830 | However, $\sqrt[3]{2} \omega, \sqrt[3]{2} \omega^2 \notin \mathbb{Q}[\sqrt[3]{2}]$. $\forall \sigma \in \gal_\mathbb{Q} \mathbb{Q}[\sqrt[3]{2}]$, 831 | $\sigma(\sqrt[3]{2}) = \sqrt[3]{2}$. Thus $\sigma = 1$. The fixed field of $\gal_\mathbb{Q} \mathbb{Q}[\sqrt[3]{2}]$ is 832 | $\mathbb{Q}[\sqrt[3]{2}]$. 833 | \end{example} 834 | 835 | \subsection{Fundamental Theorem of Galois theory} 836 | 837 | \begin{definition}[Galois correspondence] 838 | Let $K$ be a finite-dimensionalextension field of $F$, and let 839 | $S$ be the set of all intermediate fields. Again we let 840 | $T$ be the set of all subgroups of the Galois group $\gal_F K$. 841 | 842 | Define a map $\phi : T \to S$ by this rule. For each intermediate field $E$, 843 | \begin{equation} 844 | \phi(E) = \gal_E K. 845 | \end{equation} 846 | This function $\phi$ is called the Galois correspondence. 847 | 848 | \begin{center} 849 | $\begin{tikzcd} 850 | \gal_K K \arrow[r] 851 | & K \arrow[d, "S"] \\ 852 | \gal_F K \arrow[u,"T"] \arrow[r] 853 | & F 854 | \end{tikzcd} 855 | $ 856 | \end{center} 857 | \end{definition} 858 | 859 | \begin{example} 860 | \[ 861 | \mathbb{Q} \rightarrow \gal_\mathbb{Q} \mathbb{Q}(\sqrt{3}, \sqrt{5}) = \{1, \tau, \alpha, \beta \} 862 | \] 863 | \end{example} 864 | 865 | \begin{lemma} 866 | Let $K$ be a finite-dimensional extension field of $F$. If $H$ is a subgroup of the Galois group 867 | $\gal_F K$ and $E$ is the \bgold{fixed field} of $H$, then $K$ is 868 | \textit{simple, normal, separable extension} of $E$. 869 | 870 | \begin{center} 871 | $\begin{tikzcd} 872 | \gal_F K 873 | & K \arrow[dash]{d} \\ 874 | H \arrow[dash]{u} \arrow[r] 875 | & E_H 876 | \end{tikzcd} 877 | $ 878 | \end{center} 879 | \end{lemma} 880 | \begin{proof} 881 | Since $K$ is finite-dimensional extension field, so $K$ is algebraic over $F$. Let $\mathfrak{U} \in K$ and 882 | $p(x) \in E[x]$ be minimal polynomial of $\mathfrak{U}$, and $\forall \sigma \in H$, $\sigma(\mathfrak{U})$ is 883 | some root of $p(x)$. 884 | 885 | Therefore, $\mathfrak{U}$ has a finite number of distinct images under automorphisms in $H$, said 886 | \[ 887 | \mathfrak{U} = u_1, u_2, \ldots, u_t \in K, \quad \text{where } t \leq \deg\, p(x) 888 | \] 889 | If $\sigma \in H$ and $u_i = \tau(\mathfrak{U})$ with $\tau \in H$, then $\sigma(u_i) = \sigma \circ \tau (\mathfrak{U})$. 890 | 891 | Since $\sigma$ is injective, so 892 | \[ 893 | \mathfrak{U} \xhookrightarrow{H} \{ u_1, u_2, \ldots, u_t \} 894 | \] 895 | which $\{ u_1, u_2, \ldots u_t\}$ is image of $\mathfrak{U}$. And $u_i = \tau(\mathfrak{U})$ for some $\tau \in H$. 896 | is injective 897 | \[ 898 | \{ u_1, u_2, \ldots, u_t \} \xhookrightarrow[\text{permutation } \sigma]{} \{ u_1, u_2, \ldots, u_t \} 899 | \] 900 | Every automorphism in $H$ permutes $u_1, u_2, \ldots, u_t$. Let 901 | \[ 902 | f(x) = (x-u_1)(x-u_2)\ldots(x-u_t) 903 | \] 904 | Since all $u_i$'s are distinct, $f(x)$ is separable. 905 | 906 | Now we claim that $f(x) \in E[x]$. Note that $\sigma f(x) = f(x)$ for all $\sigma \in H$. All coefficients of 907 | $f(x)$ is fixed by $\sigma \in H$. Thus $f(x) \in E[x]$. Since $u = u_1$ is a root of $f(x) \in E[x]$, $u$ is separable over 908 | $E. \Longrightarrow \> K$ is separable extension of $E$. 909 | 910 | From the previous theorem, $K = E(V)$ for some $V \in K$. Let $g(x) = (x - v)(x - v_2) \ldots (x - v_s)$ 911 | where $\{ v_1 = v, v_2, \ldots, v_s \}$ are images of $\sigma \in H$. Similarly, $g(x) \in E[x]$ and 912 | $K = E(v)$ is splitting field of $g(x)$. Therefore $K$ is normal extension of $E$. 913 | \end{proof} 914 | 915 | \begin{definition}[Galois extension] 916 | If $K$ is a finite dimensional, normal, separable extension field of the field $F$, we say that 917 | $K$ is a Galois extension of $F$ or that $K$ is Galois over $F$. 918 | \end{definition} 919 | 920 | \begin{corollary} 921 | Let $K$ be a finite dimensional extension field of $F$. Then $K$ is Galois over $F$ if and only if 922 | $F$ is fixed field of the Galois group over $F$. 923 | \end{corollary} 924 | \begin{proof} 925 | $(\Rightarrow)$ By previous theorem, $E = F$. Thus $F$ is fixed field of the Galois group over $F$. 926 | 927 | $(\Leftarrow)$ By lemma, $K$ is simple, normal, separable extension of $F$. 928 | \end{proof} 929 | 930 | \begin{theorem}[Fundamental theorem of Galois Theory] 931 | If $K$ is a Galois extension field of $F$, 932 | \begin{enumerate} 933 | \item There is a bijection between the set $S$ of all the intermediate fields of the extension and 934 | the set $T$ of all subgroups of the Galois group $\gal_F K$, given by assigning each intermediate 935 | field $E$ to the subgroup $\gal_E K$. Furthermore 936 | \[ 937 | |K:E| = |\gal_E K| 938 | \] 939 | and 940 | \[ 941 | |E:F| = |\gal_F K : \gal_E K|. 942 | \] 943 | 944 | \item An intermediate field $E$ is normal extension of $F$ if and only if the 945 | corresponding group $\gal_E K$ is a normal subgroup of $\gal_F K$. 946 | \end{enumerate} 947 | \end{theorem} 948 | \begin{proof} 949 | \begin{enumerate} 950 | \item By theorem, fixed field of $\gal_E K$ is $E$. By theorem we have 951 | \[ 952 | |\gal_E K| = |K:E|. 953 | \] 954 | Similarly, 955 | \[ 956 | |\gal_F K| = |K:F|. 957 | \] 958 | In fact $|K:F| = |K:E|\,|E:F|$. Thus 959 | \[ 960 | |\gal_F K| = |\gal_E K| \cdot |E:F| \Longrightarrow |E:F| = |\gal_F K: \gal_E K|. 961 | \] 962 | 963 | \item $(\Leftarrow)$ Assume first that $\gal_E K$ is a normal subgroup of $\gal_F K$. If $p(x)$ 964 | is an irreducible polynomial in $F[x]$ with a root $u \in E$, we must show that $p(x)$ splits in $E[x]$. 965 | 966 | Since $K$ is normal over $F$. We know that $p(x)$ splits in $K[x]$. So we know that $p(x)$ also splits 967 | in $K[x]$. So we need to show that each root $v$ of $p(x)$ in $K$ is actually in $E$. There is an automorphism 968 | $\sigma \in \gal_F K$ such that $\sigma(u) = v$ by theorem. So $\gal_F K = |K:F| = E$. 969 | 970 | \[ 971 | \begin{tikzcd}[row sep=2.5em] 972 | K=F(u) \arrow[r, "\sigma"] \arrow[dash]{d} & F(u) = K \arrow[dash]{d}\\ 973 | F \arrow[r, "1"] & F 974 | \end{tikzcd} 975 | \] 976 | 977 | Now if $\tau$ is any element of $\gal_E K$, then normality implies 978 | \[ 979 | \tau \circ \sigma = \sigma \circ \tau' \quad \textsf{ for some } \tau' \in \gal_E K. 980 | \] 981 | Since $u \in E$, we have 982 | \begin{align*} 983 | \tau(v) &= \tau \left( \sigma(u) \right)\\ 984 | &= \sigma \left( \tau'(u) \right)\\ 985 | &= \sigma(u)\\ 986 | &= v. 987 | \end{align*} 988 | So $\gal_E K$ fixes other roots for all $v \in E$. Thus $E$ is normal extension of $F$. 989 | 990 | $(\Rightarrow)$ Assume that $E$ is normal subgroup of $F$. Then there exists a surjective 991 | homomorphism of groups $\theta: \gal_F K \to \gal_F E$ whose kernel is 992 | $\gal_E K$. Then $\gal_E K$ is normal subgroup of $\gal_F K$. 993 | 994 | Therefore by First isomorphism theorem, $\gal_F E \cong \gal_F K / \gal_E K$. 995 | \end{enumerate} 996 | \end{proof} 997 | 998 | \begin{lemma} 999 | Let $K$ be a finite-dimensional normal extension field of $F$ and $E$ an intermediate field 1000 | which is normal over $F$. Then there is a surjective homomorphism of groups 1001 | \[ 1002 | \theta : \gal_F K \to \gal_F E 1003 | \] 1004 | where kernel of $\theta$ is $\gal_E K$. 1005 | \end{lemma} 1006 | \begin{proof} 1007 | Let $\sigma \in \gal_F K$ and an $u \in E$. Then $u$ is algebraic over $F$ with minimal polynomial 1008 | $p(x)$. 1009 | 1010 | Since $E$ is normal, all roots of $p(x)$ are in $E$. Since $\sigma(u)$ is also 1011 | root of $p(x)$, $\sigma(u) \in E$. Therefore 1012 | \[ 1013 | \sigma(E) \subseteq E \quad \forall \sigma \in \gal_F K. 1014 | \] 1015 | Thus we can restrict $\sigma$ to $E$ and $\sigma \circ 1_E$ is an $F$-isomorphism 1016 | that is $E \cong \sigma(E)$. Hence, 1017 | \[ 1018 | |E:F| = |\sigma(E):F|. 1019 | \] 1020 | 1021 | Since $K$ is splitting field over $F$, $K$ is also splitting field over $E$. $\forall \sigma \in \gal_F E$. 1022 | $\sigma$ can be extended to an $F$-automorphism in $\gal_F K$. Its kernel consists of the automorphisms of $K$ 1023 | whose restriction to $E$ is the identity map as $\gal_E K$. 1024 | \end{proof} 1025 | 1026 | \begin{example} 1027 | Let $K$ be the splitting field of $x^3 - 2$. Note that 1028 | \[ 1029 | \mathbb{Q} \subseteq \mathbb{Q}[\sqrt[3]{2}] \subseteq K 1030 | \] 1031 | and $|\mathbb{Q}[\sqrt[3]{2}]: \mathbb{Q}| = 3$. $K$ is Galois extension of $\mathbb{Q}$ 1032 | such that 1033 | \[ 1034 | | \gal_\mathbb{Q} K| = |K:\mathbb{Q}| 1035 | \] 1036 | and $\gal_\mathbb{Q} K$ is isomorphic to a subgroup of $S_3$. 1037 | 1038 | \begin{align*} 1039 | 3 < |K:\mathbb{Q}| \leq 6 &\Longrightarrow 3 < |K: \mathbb{Q}[\sqrt[3]{2}]| \cdot |\mathbb{Q}[\sqrt[3]{2}] : \mathbb{Q}| \leq 6\\ 1040 | &\Longrightarrow 3 < 3\, |K: \mathbb{Q}[\sqrt[3]{2}] | \leq 6\\ 1041 | &\Longrightarrow 1 < |K: \mathbb{Q}[\sqrt[3]{2}] | \leq 2. 1042 | \end{align*} 1043 | We must have $|K: \mathbb{Q}| = 6$ and $\gal_\mathbb{Q} K \cong S_3$. 1044 | 1045 | $\mathbb{Q}[\omega]$ is the splitting field of $x^2 + x + 1$ and hence $\mathbb{Q}[\omega]$ is 1046 | normal and separable. Thus $K$ is Galois over $\mathbb{Q}$. 1047 | \end{example} 1048 | 1049 | \section{Solvability by Radicals} 1050 | 1051 | We shall assume that all fields have characteristic $0$. A "formula" is a specific procedure 1052 | that starts with coefficients of the polynomial $f(x) \in F[x]$ and arrives at the solutions of equation 1053 | $f(x) = 0_F$ by using only the field operations $(+_F, -_F, \times_F, \div_F)$ and the extraction 1054 | of roots (such as $\sqrt[n]{\cdot}$). 1055 | 1056 | In this context, an $n$-th root of an element $c$ in field $F$ is any root of the polynomial 1057 | $x^n - c$ in some extension field of $F$. If that "formula" really exists, then there exists 1058 | an extension field $K$ of $F$ such that 1059 | \[ 1060 | \begin{tikzcd} 1061 | F \arrow[rr, blue, bend left=50, "\text{splitting field of } x^n - c"] & \supseteq & F' \arrow[rr, red, bend left=50, "\text{splitting field of } x^{n'} - c'"] & \supseteq & \ldots \arrow[rr, bend left=50]& \supseteq & K\\ 1062 | \end{tikzcd} 1063 | \] 1064 | 1065 | \begin{example} 1066 | The solutions of $x^3 + 3x + 2 = 0 \in \mathbb{Q}[x]$ are 1067 | \[ 1068 | \fbox{$\sqrt[3]{-1+\sqrt{2}} + \sqrt[3]{-1-\sqrt{2}}$} \quad \fbox{$\omega \sqrt[3]{-1+\sqrt{2}} + (\omega^2) \sqrt[3]{-1-\sqrt{2}}$} 1069 | \quad \fbox{$(\omega^2) \sqrt[3]{-1+\sqrt{2}} + \omega \sqrt[3]{-1-\sqrt{2}}$} 1070 | \] 1071 | \end{example} 1072 | 1073 | \begin{definition}[Radical extensions] 1074 | A field $K$ is said to be a radical extension of a field $F$ if there is a chain of fields 1075 | \[ 1076 | F = F_0 \subseteq F_1 \subseteq \ldots \subseteq F_t = K 1077 | \] 1078 | such that for each $i = 1, 2, \ldots, t$. $F_i = F_{i - 1}(u_i)$ and 1079 | some power of $u_i$ is in $F_{i-1}$ ($u_i^{K_{i}} \in F_{i-1}$). 1080 | 1081 | The equation $f(x) = 0_F$ is said to be solvable by radicals if there is a radical extension of $F$ that contains 1082 | a splitting field of $f(x)$. 1083 | If not, then $f(x)$ is not solvable by radical. 1084 | \end{definition} 1085 | 1086 | \begin{definition}[Solvable groups] 1087 | A group $G$ is said to be solvable if it has a "chain" of subgroups 1088 | \[ 1089 | G = G_0 \triangleright G_1 \triangleright \ldots \triangleright G_{n-1} \triangleright G_n 1090 | = \langle e \rangle 1091 | \] 1092 | and $G_{i-1} / G_i$ is abelian. 1093 | \end{definition} 1094 | 1095 | \begin{example} 1096 | Every abelian group $G$ is solvable. $G \supseteq \langle e \rangle$ and $G/\langle e \rangle \cong G$ 1097 | is abelian. 1098 | \end{example} 1099 | \begin{solution} 1100 | Since $S_3 \triangleright \langle (1\, 2\, 3) \rangle$, and 1101 | \[ 1102 | (1\, 2) \, \langle (1\, 2\, 3) \rangle = \{ (1 \, 2), (1 \, 3), (2\, 3)\} 1103 | \] 1104 | \[ 1105 | \langle (1\, 2\, 3) \rangle\, (1\, 2) = \{ (1 \, 2), (2 \, 3), (1\, 3)\} 1106 | \] 1107 | \end{solution} 1108 | 1109 | \begin{theorem} 1110 | For $n \geq 5$ the group $S_n$ is not solvable. 1111 | \end{theorem} 1112 | \begin{proof} 1113 | For the sake of contradiction, suppose that 1114 | $S_n$ is solvable and that 1115 | \[ 1116 | S_n = G_0 \supseteq G_1 \supseteq \cdots \supseteq G_t = \langle 1 \rangle. 1117 | \] 1118 | Let $(r\, s\, t)$ be any 3-cycle in $S_n$ and let $\alpha, \beta$ be any 1119 | element of $\{1, 2, \ldots, n\}$ other than $r, s, t$ (they always exist since $n \geq 5$). 1120 | Since $S_n / G_1$ is abelian, by theorem of dihedral group, 1121 | \begin{align*} 1122 | (t\, \alpha\, s)(s\, r\, \beta)(t\, \alpha\, s)^{-1} (s\, r\, \beta)^{-1} &= 1123 | (t\, \alpha\, s)(s\, r\, \beta)(t\, s\, \alpha)(s\, \beta\, r)\\ 1124 | &= (r\, s\, t) \in G_1 1125 | \end{align*} 1126 | Note that the cycle $\langle (r\, s\, t) \rangle \subseteq G_1$, and $G_1$ definitely contains all the 3-cycles 1127 | since $G_1/G_2$ is abelian, repeating upper process, $G_2$ also contains $\langle (r\, s\, t) \rangle$. 1128 | 1129 | In conclusion, $\forall i \in \{1,2,\ldots, n\}$, $G_i$ contains all the 3-cycles. This contradicting the fact 1130 | that $S_n$ is solvable. 1131 | \end{proof} 1132 | 1133 | \begin{theorem} 1134 | Every homomorphic image of a solvable group $G$ is solvable. 1135 | \end{theorem} 1136 | \begin{proof} 1137 | Let $G$ be a solvable group. Then there exists a chain of groups 1138 | \[ 1139 | G = G_0 \supseteq G_1 \ldots \supseteq G_n = \langle e \rangle 1140 | \] 1141 | such that for all $i$, $G_{i-1} \triangleright G_i$ and $G_{i-1}/G_i$ is abelian. 1142 | 1143 | Consider $f$ is the homomorphism of $G$. Then $f(G_i)$ is also a group, and the chain of group is now 1144 | \[ 1145 | f(G) = f(G_0) \supseteq f(G_1) \ldots \supseteq f(G_n) = \langle e \rangle 1146 | \] 1147 | and $aba^{-1}b^{-1} \in G_i$ whenever $a,b$ in $G_{i-1}$. This implies 1148 | \[ 1149 | f(aba^{-1}b^{-1}) = f(a)f(b) f(a)^{-1}f(b)^{-1} \in f(G_i) \quad \forall a,b \in G_{i-1} 1150 | \] 1151 | and 1152 | \[ 1153 | \forall c, d \in f(G_{i-1}), \exists a,b \in G_{i-1} \quad s.t. \> f(a) = c \textsf{ and } f(b) =d. 1154 | \] 1155 | So we have 1156 | \[ 1157 | f(a)f(b) f(a)^{-1}f(b)^{-1} = cdc^{-1}d^{-1} \in f(G_i). 1158 | \] 1159 | Therefore $f(G_{i-1})/f(G_i)$ is abelian. Hence $f(G)$ is solvable group. 1160 | \end{proof} 1161 | 1162 | \begin{definition} 1163 | A generator of this cyclic group of $n$-th roots of unity in $K$ is called a primitive $n$-th root of unity. 1164 | \end{definition} 1165 | 1166 | This definition states that $\zeta$ is a primitive $n$-th roots of unity iff $\zeta, \zeta^2, \ldots, \zeta^n$ 1167 | are the $n$ distinct $n$-th roots of unity. 1168 | 1169 | \begin{example} 1170 | Consider $x^4 - 1 \in \mathbb{Q}[x]$. The 4-th roots of unity in $\mathbb{C} = \{1, -1, i, -1\} = \langle i \rangle$. 1171 | $i$ and $-i$ are primitive 4-th root of unity in $\mathbb{C}$. 1172 | \end{example} 1173 | 1174 | \begin{example} 1175 | According to De Moivre's theorem, 1176 | \[ 1177 | \cos \left(\frac{2 \pi}{n}\right) + i \sin \left(\frac{2 \pi}{n}\right) 1178 | \] 1179 | is a primitive $n$-th root of unity in $\mathbb{C}$. 1180 | \end{example} 1181 | 1182 | \begin{theorem} 1183 | Every homomorphic image of a solvable group $G$ is solvable. 1184 | \end{theorem} 1185 | \begin{proof} 1186 | Let $G$ be a solvable group. Then there exists a chain of groups 1187 | \[ 1188 | G = G_0 \supseteq G_1 \supseteq \ldots \supseteq G_n = \langle e \rangle 1189 | \] 1190 | such that $G_{i-1} \triangleright G_i$ and $G_{i-1} / G_i$ is abelian $\forall i$. 1191 | 1192 | Let $f$ be the homomorphism of $G$, then $f(G_i)$ is also a group. The chain of group is now 1193 | \[ 1194 | f(G) = f(G_0) \supseteq f(G_1) \supseteq \ldots \supseteq f(G_n) = \langle e \rangle 1195 | \] 1196 | and $\forall a, b \in G_{i-1}$, we have $aba^{-1}b^{-1} \in G_i$. And 1197 | \[ 1198 | f(aba^{-1}b^{-1}) = f(a)f(b)f(a)^{-1}f(b)^{-1} \in f(G_i). 1199 | \] 1200 | And for all $c, d$ in $f(G_i)$, there exists some $a, b \in G_i$ such that $f(a) = c$ and 1201 | $f(b) = d$. So 1202 | \[ 1203 | f(a)f(b)f(a)^{-1}f(b)^{-1} = cdc^{-1}d^{-1} \in f(G_i). 1204 | \] 1205 | Thus $f(G_{i-1})/f(G_i)$ is abelian. Thus $f(G)$ is solvable group. 1206 | \end{proof} 1207 | 1208 | \begin{lemma} 1209 | Let $F$ be a field and $\zeta$ a primitive $n$-th root of unity of $F$. Then $F$ contains a primity $d$-th root of unity 1210 | for every positive divisor $d$ of $n$. 1211 | \end{lemma} 1212 | \begin{proof} 1213 | Because $\zeta$ is a primitive $n$-th root of unity of $F$, that is, $\zeta^n = 1_F$. If $\zeta$ 1214 | has order $n$ and $n = dk$, $\zeta^k$ has order $d$. 1215 | 1216 | Note that $\{ \zeta^k, \zeta^{2k}, \ldots, \zeta^{nk} \}$ are all distinct and roots of $x^d - 1_F$. Thus 1217 | $\zeta^k$ is a primitive $d$-th root of unity. 1218 | \end{proof} 1219 | 1220 | \begin{example} 1221 | There is no formula (involving only field operations and extraction of roots) for the 1222 | solution of all 5th-degree polynomial equations. 1223 | \end{example} 1224 | \begin{solution} 1225 | Consider $f(x) = 2x^5 - 10x + 5 \in \mathbb{Q}[x]$. We check the zeros of the derivatives. 1226 | \[ 1227 | f'(x) = \frac{d}{dx} 2x^5 - 10x + 5 = 10x^4 - 10 \quad \Longrightarrow \textsf{roots are } \pm 1, \pm i 1228 | \] 1229 | \[ 1230 | f''(x) = \frac{d^2}{dx^2} 2x^5 - 10x + 5 = 40x^3 \quad \Longrightarrow \textsf{the only root is } 0. 1231 | \] 1232 | 1233 | \begin{figure}[ht] 1234 | \pgfplotsset{compat=1.16} 1235 | \begin{center} 1236 | \begin{tikzpicture} 1237 | \begin{axis}[xmin=-2, xmax=2, ymin=-15, ymax=19, axis x line=middle, axis y line=middle, legend pos=outer north east] 1238 | \addplot[domain=-2:2, color=red, samples=200, line width=0.75pt]{2*x^5-10*x+5}; 1239 | %\node[left,red] at (1.65,15) {$2x^5-10x+5$}; 1240 | \legend{\(2x^5 - 10x +5\)}; 1241 | \addplot[color=red, mark=*] coordinates {(-1.601,0)}; 1242 | \addplot[color=red, mark=*] coordinates {(1.329,0)}; 1243 | \addplot[color=red, mark=*] coordinates {(0.507,0)}; 1244 | \end{axis} 1245 | \end{tikzpicture} 1246 | \end{center} 1247 | \caption{By graph, $f(x) = 2x^5 - 10x + 5$ has 3 real roots.} 1248 | \end{figure} 1249 | 1250 | Note that by Eisenstein's criterion, $5|10$, $5|4$, but $5 \nmid 2$ and $5^2 \nmid 5$. So $f(x)$ 1251 | is irreducible in $\mathbb{Q}[x]$. 1252 | 1253 | If $K$ is splitting field of $f(x)$ in $\mathbb{C}$. Then 1254 | \[ 1255 | |\gal_\mathbb{Q} K| = [K:\mathbb{Q}] 1256 | \] 1257 | Since $K$ is Galois field of $\mathbb{Q}$. 1258 | 1259 | If $r$ is any root of $f(x)$, then 1260 | \[ 1261 | |K:\mathbb{Q}| = |K: \mathbb{Q}(r)| \cdot |\mathbb{Q}(r) : \mathbb{Q}| = 5\, |K: \mathbb{Q}(r)|. 1262 | \] 1263 | Thus $|K : \mathbb{Q}|$ is divisible by $5$. By Cauchy's theorem, there exists $sigma \in \gal_\mathbb{Q} K$ with 1264 | $\ord(\sigma) = 5$. Note that $\gal_\mathbb{Q} K \subseteq S_5$ and the only elements of order 1265 | $5$ are (5-cycles), the 5-cycle is in $\gal_\mathbb{Q} K$. The complex conjugation $a + bi \mapsto a -bi$ 1266 | induces an automorphism on $K$ since $\sigma \in \gal_\mathbb{Q} \mathbb{C}, \sigma \vert_K \in \gal_\mathbb{Q} K$. 1267 | 1268 | Since $\sigma \vert_K$ interchanges the two nonreal roots of $f(x)$, implies that 1269 | $\gal_{\mathbb{Q}} K$ contains a transposition (2-cyle). Note that the only 1270 | subgroup of $S_5$ that contains both a 5-cycle and a transposition is $S_5$ itself. 1271 | So $\gal_\mathbb{Q} K = S_5 \Longrightarrow f(x)$ is not solvable by radicals. 1272 | 1273 | \end{solution} 1274 | 1275 | \section*{Tutorials} 1276 | 1277 | \begin{mdframed} 1278 | \vspace{-0.25cm} 1279 | \hspace{-0.25cm} 1280 | \begin{Exercise} 1281 | Let $E$ be an extension field of a field $F$ and let $\alpha \in E$ be transcendental over $F$. Show that every element of $F(\alpha)$ is not in $F$ 1282 | is also transcendental over $F$. 1283 | \end{Exercise} 1284 | 1285 | \vspace{0.752cm} 1286 | \begin{Exercise} 1287 | Let $F$ be a finite field of characteristic $p$. Show that every element of $F$ is algebraic over the 1288 | prime field $\mathbb{Z}_p \leq F$. 1289 | \end{Exercise} 1290 | \end{mdframed} -------------------------------------------------------------------------------- /src/images/Dihedral2.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/pehcy/algebra-notes-latex/d7939b136bed4b3995914aa403c720a14c345990/src/images/Dihedral2.pdf -------------------------------------------------------------------------------- /src/images/Dihedral2_120.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/pehcy/algebra-notes-latex/d7939b136bed4b3995914aa403c720a14c345990/src/images/Dihedral2_120.pdf -------------------------------------------------------------------------------- /src/images/Dihedral2_240.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/pehcy/algebra-notes-latex/d7939b136bed4b3995914aa403c720a14c345990/src/images/Dihedral2_240.pdf -------------------------------------------------------------------------------- /src/images/Dihedral2_360.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/pehcy/algebra-notes-latex/d7939b136bed4b3995914aa403c720a14c345990/src/images/Dihedral2_360.pdf -------------------------------------------------------------------------------- /src/images/Dihedral3.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/pehcy/algebra-notes-latex/d7939b136bed4b3995914aa403c720a14c345990/src/images/Dihedral3.pdf -------------------------------------------------------------------------------- /src/images/cube.tex: -------------------------------------------------------------------------------- 1 | \begin{tikzpicture} 2 | 3 | \begin{scope} 4 | \begin{scope}[scale=3,y={(0.2cm,0.3cm)},x={(1cm,0cm)}, z={(0cm,1cm)}] 5 | \coordinate (O) at (0, 0, 0); 6 | 7 | \draw[ very thick,Bcolor, dashed ] (0,0,1)--(0,0,-1); 8 | \draw[ very thick,Bcolor ] (0,0,-1.4)--(0,0,-1); 9 | \draw[ very thick,Bcolor ] (0,0,1.4)--(0,0,1); 10 | 11 | \draw[dotted, very thick,medgrey] (1,1,-1)--(-1,1,-1)--(-1,-1,-1); 12 | \draw[dotted, very thick,medgrey] (-1,1,-1)--(-1,1,1); 13 | 14 | \draw[thick,opacity=0.2,fill=Ccolor] (1,1,1)--(1,-1,1)--(-1,-1,1)--(-1,1,1)--cycle; % top 15 | \draw[thick,opacity=0.2,fill=Ccolor] (1,-1,1)--(1,-1,-1)--(-1,-1,-1)--(-1,-1,1)--cycle; % front 16 | \draw[thick,opacity=0.2,fill=Acolor] (1,1,-1)--(1,1,1)--(1,-1,1)--(1,-1,-1)--cycle; 17 | 18 | \end{scope} 19 | \node[inner sep = 0cm,text width=7cm, align=center] at (0,-6.5) () {One of three possible axes of rotation through the centers of opposite faces.\\ Each rotation could be $0^\circ$, $90^\circ$, $180^\circ$, or $270^\circ$, for a total of $3\cdot 4 = 12$ rotations of this type. But three in this count are the trivial identity rotation, which we only count once, so there are really 10 unique rotations along these axes. }; 20 | \end{scope} 21 | 22 | \begin{scope}[xshift=10cm] 23 | \begin{scope}[scale=3,y={(0.2cm,0.3cm)},x={(1cm,0cm)}, z={(0cm,1cm)}] 24 | \coordinate (O) at (0, 0, 0); 25 | 26 | \draw[ very thick,Bcolor, dashed ] (-1,1,1)--(1,-1,-1); 27 | \draw[ very thick,Bcolor ] (1.2,-1.2,-1.2)--(1,-1,-1); 28 | \draw[ very thick,Bcolor ] (-1.2,1.2,1.2)--(-1,1,1); 29 | 30 | 31 | \draw[ dotted, very thick,medgrey ] (1,1,-1)--(-1,1,-1)--(-1,-1,-1); 32 | \draw[dotted, very thick ,medgrey] (-1,1,-1)--(-1,1,1); 33 | 34 | \draw[thick,opacity=0.2,fill=Ccolor] (1,1,1)--(1,-1,1)--(-1,-1,1)--(-1,1,1)--cycle; % top 35 | \draw[thick,opacity=0.2,fill=Ccolor] (1,-1,1)--(1,-1,-1)--(-1,-1,-1)--(-1,-1,1)--cycle; % front 36 | \draw[thick,opacity=0.2,fill=Acolor] (1,1,-1)--(1,1,1)--(1,-1,1)--(1,-1,-1)--cycle; 37 | 38 | \end{scope} 39 | \node[inner sep = 0cm,text width=5cm, align=center] at (0,-6.5) () {One of four possible axes of rotation through opposite vertices. Each could be either $120^\circ$ or $240^\circ$, so there are $4 \cdot 2 = 8$ rotations of this type.}; 40 | \end{scope} 41 | 42 | \begin{scope}[xshift=20cm] 43 | \begin{scope}[scale=3,y={(0.2cm,0.3cm)},x={(1cm,0cm)}, z={(0cm,1cm)}] 44 | \coordinate (O) at (0, 0, 0); 45 | 46 | \draw[ very thick,Bcolor, dashed ] (0,1,1)--(0,-1,-1); 47 | \draw[ very thick,Bcolor ] (0,-1.2,-1.2)--(0,-1,-1); 48 | \draw[ very thick,Bcolor ] (0,1.2,1.2)--(0,1,1); 49 | 50 | \draw[ dotted, very thick,medgrey ] (1,1,-1)--(-1,1,-1)--(-1,-1,-1); 51 | \draw[dotted, very thick ,medgrey] (-1,1,-1)--(-1,1,1); 52 | 53 | \draw[thick,opacity=0.2,fill=Ccolor] (1,1,1)--(1,-1,1)--(-1,-1,1)--(-1,1,1)--cycle; % top 54 | \draw[thick,opacity=0.2,fill=Ccolor] (1,-1,1)--(1,-1,-1)--(-1,-1,-1)--(-1,-1,1)--cycle; % front 55 | \draw[thick,opacity=0.2,fill=Acolor] (1,1,-1)--(1,1,1)--(1,-1,1)--(1,-1,-1)--cycle; 56 | 57 | \end{scope} 58 | \node[inner sep = 0cm,text width=7cm, align=center] at (0,-4.5) () {One of six possible axes of rotation through the centers of opposite edges. Only a $180^\circ$ rotation around these axes would preserve the shape, so we have only 6 rotations possible.}; 59 | \end{scope} 60 | 61 | \node[inner sep = 0cm,text width=20cm, align=center] at (10,6) () {\Huge{The Symmetries of a Cube}}; 62 | \end{tikzpicture} -------------------------------------------------------------------------------- /src/images/first_iso.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/pehcy/algebra-notes-latex/d7939b136bed4b3995914aa403c720a14c345990/src/images/first_iso.pdf -------------------------------------------------------------------------------- /src/images/second_iso.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/pehcy/algebra-notes-latex/d7939b136bed4b3995914aa403c720a14c345990/src/images/second_iso.pdf -------------------------------------------------------------------------------- /src/images/wave.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/pehcy/algebra-notes-latex/d7939b136bed4b3995914aa403c720a14c345990/src/images/wave.png -------------------------------------------------------------------------------- /src/rings.tex: -------------------------------------------------------------------------------- 1 | \chapter{Rings} 2 | 3 | Ring is an algebraic structure which is a set of elements with two operations: addition and multiplication. 4 | 5 | \begin{axiom}[Rings] 6 | A ring $R$ is a set with two binary operation, addition (usually denoted by $+$) and multiplication 7 | (usually denoted by $ab$), such that for all $a,b,c \in R$. 8 | \begin{enumerate} 9 | \item Addition is commutative, $a + b = b + a$. 10 | \item Associativity holds in addition, $(a + b) + c = a + (b + c)$. 11 | \item There is an additive identity $0_R$. That is, there is an element $0_R \in R$ such that 12 | \[ 13 | a + 0_R = a = 0_R + a 14 | \] 15 | for all $a \in R$. 16 | \item There is an additive inverse $-a \in R$ such that 17 | \[ 18 | a + (-a) = 0_R = -a + a. 19 | \] 20 | \item Associativity holds in multiplication, $a(bc)=(ab)c$. 21 | \item Distributive law holds in $R$, 22 | \[ 23 | a(b+c) = ab + ac 24 | \] 25 | and 26 | \[ 27 | (b+c)a = ba + ca. 28 | \] 29 | \end{enumerate} 30 | \end{axiom} 31 | 32 | Here are a few things you should take notes: 33 | \begin{enumerate} 34 | \item A ring is an Abelian group under addition, also having an associative 35 | multiplication that is left and right distributive over addition. 36 | 37 | \item Note that multiplication need not be commutative. When it is, we say that the 38 | ring is commutative. 39 | 40 | \item A ring need not have an identity under multiplication. A unity (or identity) in a 41 | ring is a nonzero element that is an identity under multiplication. 42 | 43 | \item A nonzero element of a commutative ring with unity need not have 44 | multiplicative inverse. When it does, we say that it is a unit of the ring. Thus, 45 | $x$ is a unit if $x^{-1}$ exists. 46 | 47 | \item We follow the following terminology and notation. If $x$ and $y$ belong to a 48 | commutative ring $R$ and $x$ is nonzero, we say that $x$ divides $y$ and write $x | y$, 49 | if there exists an element $c$ in $R$ such that $y = xc$. 50 | 51 | \item If $x$ is an element from a group under the operation of addition and $n$ is a 52 | positive integer, $nx$ means $\underbrace{x + x + \cdots + x}_{n \text{ times}}$, where there are $n$ summands. 53 | \end{enumerate} 54 | 55 | \begin{example} 56 | The set $\mathbb{Z}$ of integers under ordinary addition and multiplication is a commutative ring with unity $1$. The units 57 | of $\mathbb{Z}$ are $1$ and $-1$. 58 | \end{example} 59 | 60 | \begin{example} 61 | The set $M_2(\mathbb{Z})$ of $2 \times 2$ matrices with integer entries is a noncommutative ring with unity 62 | $\begin{bmatrix} 63 | 1 & 0\\ 0 & 1 64 | \end{bmatrix}$. 65 | \end{example} 66 | 67 | \begin{example} 68 | The set of all continuous real-valued functions of a real variable whose graphs pass through 69 | the point $(1, 0)$ is a commutative ring without unity under the operations of pointwise addition and multiplication, that is, 70 | \[ 71 | (f + g)(x) = f(x) + g(x) 72 | \] 73 | and 74 | \[ 75 | (fg)(x) = f(x)g(x). 76 | \] 77 | \end{example} 78 | 79 | \begin{theorem} 80 | Let $a,b$ and $c$ belong to a ring $R$. Then 81 | \begin{enumerate} 82 | \item $a0 = 0a = 0$. 83 | \item $a(-b) = (-a)b = -(ab)$. 84 | \item $(-a)(-b) = ab$. 85 | \item $a(b-c) = ab - ac$ and $(b-c)a = ba-ca$. 86 | \item Furthermore, if $R$ has a unity element 1, then $(-1)a = -a$. 87 | \item $(-1)(-1) = (-1)^2 = 1$. 88 | \end{enumerate} 89 | \end{theorem} 90 | \begin{proof} 91 | We are going to prove each of them: 92 | 93 | \begin{enumerate} 94 | \item \begin{align*} 95 | a(0 + 0) = a0 &\Rightarrow a0 + a0 = a0\\ 96 | &\Rightarrow {\color{red}{} (-a0)} + a0 + a0 = {\color{red}{} (-a0)} + a0\\ 97 | &\Rightarrow 0 + a0 = 0\\ 98 | &\Rightarrow a0 = 0. 99 | \end{align*} 100 | 101 | \item \begin{align*} 102 | a0 = 0 &\Rightarrow a(b + (-b)) = 0\\ 103 | &\Rightarrow ab + a(-b) = 0\\ 104 | &\Rightarrow {\color{red}{} -(ab)} + ab + a(-b) = {\color{red}{} -(ab)} + 0\\ 105 | &\Rightarrow a(-b) = -(ab). 106 | \end{align*} 107 | 108 | \item \begin{align*} 109 | (-a)(-b) &= -(a(-b))\\ 110 | &= -(a(-b))\\ 111 | &= -(-(ab))\\ 112 | &= ab. 113 | \end{align*} 114 | 115 | \item \[ 116 | a(b-c) = a(b+(-c)) = ab + a(-c) = ab - ac. 117 | \] 118 | 119 | \item \begin{align*} 120 | 0a = 0 &\Rightarrow (1+(-1))a = 0\\ 121 | &\Rightarrow 1a + (-1)a = 0\\ 122 | &\Rightarrow a + (-1)a = 0\\ 123 | &\Rightarrow {\color{red}{} (-a)} + a + (-1)a = {\color{red}{} (-a)} + 0\\ 124 | &\Rightarrow 0 + (-1)a = {\color{red}{} (-a)}\\ 125 | &\Rightarrow (-1)a = -a. 126 | \end{align*} 127 | 128 | \item \[ 129 | (-1)(-1) = -(1(-1)) = -(-(1 \cdot 1)) = 1. 130 | \] 131 | \end{enumerate} 132 | \end{proof} 133 | 134 | \begin{lemma} 135 | If a ring has a unity, it is unique. 136 | \end{lemma} 137 | \begin{proof} 138 | Suppose $1_R$ and $1'_R$ are both unity of ring $R$. Then 139 | For all $a \in R$, we must have 140 | \[ 141 | a \cdot 1_R = a = 1_R \cdot a 142 | \] 143 | and 144 | \[ 145 | a \cdot 1'_R = a = 1'_R \cdot a. 146 | \] 147 | Thus, we say that $a \cdot 1_R = a = a \cdot 1'_R$. By cancellation law this yields 148 | \[ 149 | 1_R = 1'_R. 150 | \] 151 | So ring $R$ has unique unity. 152 | \end{proof} 153 | 154 | \section{subrings} 155 | 156 | \begin{theorem}[Subring test] 157 | A nonempty subset $S$ of the ring $R$ is a subring if $S$ is closed under subtraction and multiplication, that is, 158 | \begin{enumerate} 159 | \item $S \neq \varnothing$. 160 | \item $a - b \in S \quad \forall a, b \in S$. 161 | \item $ab \in S \quad \forall a, b \in S$. 162 | \end{enumerate} 163 | \end{theorem} 164 | \begin{proof} 165 | Since addition in $R$ is commutative and $S$ closed under subtraction, we know that by 166 | the \textit{one-step subgroup test} that $S$ is an Abelian group under addition. 167 | 168 | Also, since multiplication in $R$ is asscoiative as well as distributive over addition, the 169 | same is true for multiplication in $S$. 170 | 171 | Therefore, the only condition remaining to be checked is that multiplication is a binary 172 | operation on $S$. But this is exactly what closure means. 173 | \end{proof} 174 | 175 | \begin{example} 176 | $S = \{ 0, 2, 4 \}$ is a subring of $\mathbb{Z}_6$. 177 | \end{example} 178 | \begin{solution} 179 | Using subring test, the subtraction and multiplication form a group in $S$. 180 | 181 | \begin{center} 182 | \begin{minipage}{.25\textwidth} 183 | { 184 | \arrayrulewidth=1pt 185 | \renewcommand{\arraystretch}{1.5} 186 | \begin{tabular}{c|*{3}{>{\columncolor{white}}c}} 187 | $-$ & \cellcolor{myred}0 & \cellcolor{mygreen}2 & \cellcolor{mypurple}4 \\ 188 | \hline 189 | \cellcolor{myred}0 & \cellcolor{myred}0 & \cellcolor{mypurple}4 & \cellcolor{mygreen}2 \\ 190 | \cellcolor{mygreen}2 & \cellcolor{mygreen}2 & \cellcolor{myred}0 & \cellcolor{mypurple}4\\ 191 | \cellcolor{mypurple}4 & \cellcolor{mypurple}4 & \cellcolor{mygreen}2 & \cellcolor{myred}0\\ 192 | \end{tabular} 193 | } 194 | \end{minipage}% This must go next to `\end{minipage}` 195 | \begin{minipage}{.25\textwidth} 196 | { 197 | \arrayrulewidth=1pt 198 | \renewcommand{\arraystretch}{1.5} 199 | \begin{tabular}{c|*{3}{>{\columncolor{white}}c}} 200 | $\cdot$ & \cellcolor{myred}0 & \cellcolor{mygreen}2 & \cellcolor{mypurple}4 \\ 201 | \hline 202 | \cellcolor{myred}0 & \cellcolor{myred}0 & \cellcolor{myred}0 & \cellcolor{myred}0 \\ 203 | \cellcolor{mygreen}2 & \cellcolor{myred}0 & \cellcolor{mypurple}4 & \cellcolor{mygreen}2\\ 204 | \cellcolor{mypurple}4 & \cellcolor{myred}0 & \cellcolor{mygreen}2 & \cellcolor{mypurple}4\\ 205 | \end{tabular} 206 | } 207 | \end{minipage} 208 | \end{center} 209 | 210 | Thus $S \leq \mathbb{Z}_6$. 211 | \end{solution} 212 | 213 | \section{Quotient rings, Ideals} 214 | 215 | \subsection{Ideals} 216 | 217 | \begin{definition}[Ideals] 218 | A subring $A$ of a ring $R$ is called an ideal if for every $r \in R$ and every $a \in A$ both $ra$ and 219 | $ar$ are in $A$. 220 | \end{definition} 221 | 222 | So, a subring $A$ of a ring $R$ is an ideal of $R$ if 223 | \[ 224 | rA = \{ ra \> | \> a \in A \} \subseteq A 225 | \] 226 | and 227 | \[ 228 | Ar = \{ ar \> | \> a \in A \} \subseteq A 229 | \] 230 | for all $r \in R$. An ideal $A$ of $R$ is called a \bred{proper ideal} of $R$ if $A$ is a proper subset of $R$. 231 | 232 | \begin{lemma} 233 | Let $I$ be a subring of a ring $R$. Then $I$ is an ideal in $R$ if and only if multiplication 234 | \[ 235 | (a + I)(b + I) = (ab + I) 236 | \] 237 | is a well-defined operation on the cosets of $I$ in $R$. 238 | \end{lemma} 239 | \begin{proof} 240 | $(\Rightarrow)$ Assume that $I$ is an ideal in $R$, and suppose 241 | that $a_1 + I = a_2 + I$ and $b_1 + I = b_2 + I$. This implies that $a_1 = a_2 + k$ and 242 | $b_1 = b_2 + j$ for some $i, j \in I$. Then we have 243 | \[ 244 | a_1 b_1 = a_2 b_2 + a_2 j + kb_2 + kj. 245 | \] 246 | Since $I$ is a subring of $R$, and therefore it closed under multiplication, as well as addition, and $kj \in I$. 247 | Since $I$ is an ideal, $a_2 j \in I$ and $kb_2 \in I$, and so $a_2 j + kb_2 + jk \in I$. 248 | 249 | Therefore, $a_1b_1 \in a_2b_2 + I$ and $a_1b_1 + I = a_2b_2 + I$. Thus, the multiplication on the set 250 | of cosets of $I$ is well-defined. 251 | 252 | $(\Leftarrow)$ Assume that the indicated operation is well-defined. We need to show that for all $r \in R$, and 253 | for all $x \in I$, we have $rx \in I$ and $xr \in I$. So we have $x + I = 0 + I = I$. Hence 254 | \[ 255 | rx + I = (r + I)(x + I) = (r + I)(0 + I) = 0 + I = I. 256 | \] 257 | Again we have $xr \in I$. Thus $I$ is an ideal in $R$. 258 | \end{proof} 259 | 260 | \begin{theorem}[Ideal test] 261 | A nonempty subset $A$ of a ring $R$ is an ideal of $R$ provided 262 | \begin{enumerate} 263 | \item $A \neq \varnothing$. 264 | \item $a - b \in A$ whenever $a, b \in A$. 265 | \item $ra$ and $ar$ are in $A$ for all $a \in A$ and $r \in R$. 266 | \end{enumerate} 267 | \end{theorem} 268 | 269 | \begin{example} 270 | For any ring $R$, $\{0 \}$ and $R$ are ideals of $R$. The ideal $\{ 0 \}$ is called the trivial ideal. 271 | \end{example} 272 | 273 | \begin{example} 274 | For any positive integer $n$, the set 275 | \[ 276 | n\mathbb{Z} = \{ 0, \pm n, \pm 2n, \pm 3n, \ldots \} 277 | \] 278 | is an ideal of $\mathbb{Z}$. 279 | \end{example} 280 | \begin{solution} 281 | We can show $n\mathbb{Z} \unlhd \mathbb{Z}$ using ideal test. 282 | \begin{enumerate} 283 | \item Since $0 \in \mathbb{Z}$, then $n \cdot 0 = 0 \in n\mathbb{Z} \neq \varnothing$. 284 | \item For all $a, b \in n\mathbb{Z}$, we let $a=nt_1$ and $b=nt_2$ for $t_1, t_2 \in \mathbb{Z}$. We have 285 | \[ 286 | a - b = nt_1 - nt_2 = n(t_1 - t_2) \in n\mathbb{Z} 287 | \] 288 | since $t_1 - t_2$ is also an integer. 289 | 290 | \item Whenever $a \in A$ and $r \in R$, let $a = nt', \> t' \in \mathbb{Z}$. We have 291 | \[ 292 | ar = (nt')r = n(t'r) \in n\mathbb{Z}, \quad t'r \in \mathbb{Z}. 293 | \] 294 | and 295 | \[ 296 | ra = r(nt') = nrt' = n(rt') = n(t'r) = ar \in n\mathbb{Z}. 297 | \] 298 | 299 | Therefore $n\mathbb{Z} \unlhd \mathbb{Z}$. 300 | \end{enumerate} 301 | \end{solution} 302 | 303 | \begin{example} 304 | Let $R$ be a commutative ring with unity and let $a \in R$. The set 305 | \[ 306 | \langle \mathfrak{a} \rangle = \{ r\mathfrak{a} \> | \> r \in R \} 307 | \] 308 | is an ideal of $R$ called the \bred{principal ideal} generated by $\mathfrak{a}$. 309 | \end{example} 310 | \begin{solution} 311 | Using ideal test to check 312 | \begin{enumerate} 313 | \item Since $R$ is a ring, then $0_R \in R$ and so $0_R = 0_R \cdot \mathfrak{a} \in \langle \mathfrak{a} \rangle \neq \varnothing$. 314 | \item For all $b, c \in \langle \mathfrak{a} \rangle$, let $b = r_1 \mathfrak{a}$, $c = r_2 \mathfrak{a}$, where 315 | $r_1, r_2 \in R$. Then 316 | \[ 317 | b - c = r_1 \mathfrak{a} - r_2 \mathfrak{a} = (r_1 - r_2) \mathfrak{a} \in \langle \mathfrak{a} \rangle. 318 | \] 319 | and $r_1 - r_2$ is in $R$. 320 | 321 | \item For all $\olsi{a} \in \langle \mathfrak{a} \rangle$, $r \in R$, we let $\olsi{a} = r' \mathfrak{a}$ and 322 | \[ 323 | \olsi{a}r = (r' \mathfrak{a}) r = \mathfrak{a}(r'r) \in \langle \mathfrak{a} \rangle. 324 | \] 325 | where $r'r \in R$. 326 | 327 | On the other hand, 328 | \[ 329 | r\olsi{a} = r(r' \mathfrak{a}) = \mathfrak{a}(rr') = \mathfrak{a}(r'r) = \olsi{a} r \in \langle \mathfrak{a} \rangle. 330 | \] 331 | 332 | Therefore $\langle \mathfrak{a} \rangle$ is a subring of $R$. 333 | \end{enumerate} 334 | \end{solution} 335 | 336 | \subsection{Quotient Rings} 337 | 338 | \begin{theorem} 339 | Let $R$ be a ring and let $A$ be a subring of $R$. The set of cosets 340 | \[ 341 | R/A = \{ r+A \> | \> r \in A \} 342 | \] 343 | is a ring under the operations 344 | \begin{itemize} 345 | \item $(s+A) + (t+A) = s + t + A$ 346 | \item $(s+A)(t+A) = st + A$ 347 | \end{itemize} 348 | if and only if $A$ is an ideal of $R$. 349 | \end{theorem} 350 | \begin{proof} 351 | Let $R$ to be a ring, and let $A \trianglelefteq R$. For all $s+A, t+A$ in $R/A$ we define 352 | addition as 353 | \[ 354 | \boxplus (s+A, t+A) := (s+A) \boxplus (t + A) = s+t+A 355 | \] 356 | and multiplication as 357 | \[ 358 | \odot(s+A, t+A) := (s+A) \odot (t + A) = st+A. 359 | \] 360 | We want to show $(R/A, \boxplus, \odot)$ is a ring. 361 | 362 | \begin{enumerate} 363 | \item (Closureness) Suppose that $s + A = s' + A$ and $t+A = t' + A$ for all $s, s', t, t' \in \mathbb{R}$. First we 364 | need to show 365 | \[ 366 | (s+t) + A = (s' + t')+A. 367 | \] 368 | We are going to express $s, t$ in term of $s', t'$ respectively. 369 | \begin{align*} 370 | s + A = s' + A \> &\Rightarrow s - s' \in A\\ 371 | &\Rightarrow s - s' = a_1 \in A\\ 372 | &\Rightarrow \fbox{$s = a_1 + s'$}\>, \quad a_1 \in A. \label{eq:r2.31} \tag{{\color{red} $\heartsuit$}} 373 | \end{align*} 374 | and 375 | \begin{align*} 376 | t + A = t' + A \> &\Rightarrow t - t' \in A\\ 377 | &\Rightarrow t - t' = a_2 \in A\\ 378 | &\Rightarrow \fbox{$t = a_2 + t'$}\>, \quad a_2 \in A. \label{eq:r2.32} \tag{{\color{gray} $\clubsuit $}} 379 | \end{align*} 380 | 381 | Summing up \eqref{eq:r2.31} together with \eqref{eq:r2.32} we have 382 | \[ 383 | s + t = (a_1 + s') + (a_2 + t') = a_1 + a_2 + s' + t', \quad a_1 + a_2 \in A. 384 | \] 385 | Subtracting $s'+t'$ on both side of the equation yields 386 | \[ 387 | s+t-(s'+t') = a_1 + a_2, \Longrightarrow s+t-(s'+t') \in A. 388 | \] 389 | 390 | We have shown $R/A$ closed under addition $\boxplus$, we continue to proof 391 | \[ 392 | st+A = s't' + A. 393 | \] 394 | Which is equivalent to show $R/A$ closed under the multiplication $\odot$. Applying 395 | the results that we found from \eqref{eq:r2.31}, \eqref{eq:r2.32} we have 396 | \[ 397 | st -s't' = a_1 a_2 + a_1t' + s'a_2 398 | \] 399 | Is $a_1t'$ in $A$? Of course, since $a_1 \in A, t' \in R \Longrightarrow a_1t' \in A \triangleleft R$. 400 | So as $s'a_2 \in A \triangleleft R$. 401 | 402 | \item (Existence of additive identity) For all $s+A \in R/A$, there exists $e+A \in R/A$ such that 403 | \begin{align*} 404 | (s + A) \boxplus (e+A) = s + A &\Rightarrow (s+e) + A = 0 + s + A\\ 405 | &\Rightarrow s+e = s\\ 406 | &\Rightarrow e = 0_A. 407 | \end{align*} 408 | Thus the additive identity is $0_A + A$. 409 | 410 | \item (Existence of additive inverse) For all $s+A \in R/A$, there exists $r+A \in R/A$ such that 411 | \begin{align*} 412 | (s + A) \boxplus (r+A) = 0 + A &\Rightarrow (s+r) + A = 0 + A\\ 413 | &\Rightarrow s+r = 0\\ 414 | &\Rightarrow r =-s. 415 | \end{align*} 416 | Thus the additive inverse of $s+A$ is $-s+A$ in $R/A$. 417 | 418 | \item (Associativity of multiplication) For all $s+A, t+A, u+A$ in $R/A$, compute 419 | \begin{align*} 420 | (s+A) \odot [(t+A) \odot (u+A)] &= (s+A) \odot (tu + A)\\ 421 | &= s(tu) +A\\ 422 | &= (st)u + A\\ 423 | &= [(st)+A] \odot (u+A)\\ 424 | &= [(s+A) \odot (t+A)] \odot (u+A). 425 | \end{align*} 426 | Associativity in $\odot$ holds. 427 | 428 | \item (Existence of unity) For all $s+A \in R/A$, there exists $e'+A \in R/A$ such that 429 | \begin{align*} 430 | (s + A) \odot (e'+A) = s + A &\Rightarrow se'+A = s + A\\ 431 | &\Rightarrow se' = s\\ 432 | &\Rightarrow e' = 1_A \in R. 433 | \end{align*} 434 | The multiplicative identity is $1_A + A$ in $R/A$. 435 | 436 | \item (Existence of multiplicative inverse) For all $s+A \in R/A$, there exists $r+A \in R/A$ such that 437 | \begin{align*} 438 | (s + A) \odot (r+A) = 1_A + A &\Rightarrow sr+A = 1_A + A\\ 439 | &\Rightarrow sr = 1_A\\ 440 | &\Rightarrow r = s^{-1}. 441 | \end{align*} 442 | The multiplicative inverse of $s+A$ is $s^{-1} + A$ in $R/A$, provided $s^{-1}$ exists in $R$. 443 | 444 | \item (Distributive Law) For all $s+A, t+A, u+A$ in $R/A$, compute 445 | \begin{align*} 446 | (s+A) \odot [(t+A) \boxplus (u+A)] &= (s+A) \odot [(t+u) + A]\\ 447 | &= s(t+u) + A\\ 448 | &= st + su + A\\ 449 | &= (st + A) + (su + A)\\ 450 | &= (st+A) \boxplus (su+A)\\ 451 | &= (s+A) \odot (t+A) \boxplus (s+A) \odot (u+A) 452 | \end{align*} 453 | Distributive law holds in $(R/A, \boxplus, \odot)$. 454 | \end{enumerate} 455 | 456 | Therefore $(R/A, \boxplus, \odot)$ is a ring. 457 | \end{proof} 458 | 459 | \begin{example} 460 | $\mathbb{Z}/4\mathbb{Z} = \{ 4\mathbb{Z},\> 1 + 4\mathbb{Z},\> 2+4\mathbb{Z},\> 3+4\mathbb{Z} \}$ 461 | \end{example} 462 | \begin{solution} 463 | The integers with multiple of $4$ is 464 | \[ 465 | 4\mathbb{Z} = \{\ldots, -8, -4, 0, 4, 8, 12, \ldots \}. 466 | \] 467 | The left ideals are 468 | \begin{align*} 469 | 1+_4 4\mathbb{Z} &= \{ \ldots, -7, -3, 1, 5, 9, 13, \ldots \}\\ 470 | 2+_4 4\mathbb{Z} &= \{ \ldots, -6, -2, 2, 6, 10, 14, \ldots \}\\ 471 | 3+_4 4\mathbb{Z} &= \{ \ldots, -5, -1, 3, 7, 11, 15, \ldots \}\\ 472 | 4+_4 4\mathbb{Z} &= \{ \ldots, -4, 0, 4, 8, 12, 16, \ldots \} = 4\mathbb{Z} 473 | \end{align*} 474 | \end{solution} 475 | 476 | \begin{definition}[characteristic of ring] 477 | The characteristic of a ring $R$ is the least positive integer $n$ such that $nx = 0$ for all $x \in R$. If no such integer exists, we say that 478 | $R$ has characteristic 0. The characteristic of $R$ is denoted by $\text{char}(R)$. 479 | \end{definition} 480 | 481 | \begin{example} 482 | The characteristic of ring $\mathbb{Z}_7$ is 7. 483 | \end{example} 484 | 485 | \begin{example} 486 | Find the characteristic of $R = \mathbb{Z}[i] / \langle 2 - i \rangle$. 487 | \end{example} 488 | \begin{solution} 489 | Given that 490 | \[ 491 | R = \mathbb{Z}[i] / \langle 2 - i \rangle = \{ a + bi + \langle 2-i \rangle \}. 492 | \] 493 | Observe that 494 | \[ 495 | 2 - i + \langle 2-i \rangle = 0 + \langle 2-i \rangle 496 | \] 497 | and so we treat $2 - i$ as zero, and $i = 2$ and so 498 | \[ 499 | -1 = i^2 = 4 \Longrightarrow 5 = 0 \in R. 500 | \] 501 | Thus $R = \{ 0+\langle 2-i \rangle, 1 + \langle 2-i \rangle, 2 + \langle 2-i \rangle, 502 | 3 + \langle 2-i \rangle, 4 + \langle 2-i \rangle\}$. So $\text{Char}(R) = 5$. 503 | \end{solution} 504 | 505 | \begin{definition}[Prime ideal] 506 | An ideal $I$ in a commutative ring $R$ is said to be prime if $I \neq R$ and whenever 507 | $ab \in I$, then either $a \in I$ or $b \in I$. 508 | \end{definition} 509 | 510 | \begin{lemma} 511 | Let $R$ be a commutative ring with unity, and $I$ be an ideal in $R$. Then $I$ is 512 | a prime ideal in $R$ if and only if $R/I$ is an integral domain. 513 | \end{lemma} 514 | \begin{proof} 515 | $R/I$ will therefore be an integral domain and only if it has no zero divisors. This condition 516 | is equivalent to the condition that 517 | \[ 518 | (a+I)(b+I) = I \iff a + I = I \> \textsf{ or } \> b + I = I. 519 | \] 520 | Thus $R/I$ is an integral domain if and only if $ab + I = I$ implies that 521 | $a + I = I$ or $b+I = I$ or, in other words, if and only if 522 | $ab \in I$ implies that $a \in I$ or $b \in I$, which is to say that $I$ is a prime ideal 523 | in $R$. 524 | \end{proof} 525 | 526 | \begin{definition}[Maximal ideal] 527 | An ideal $I$ in a ring $R$ is said to be maximal if $I \neq R$ and whenever $J$ is an ideal such that 528 | \[ 529 | I \subset J \subset R 530 | \] 531 | then $I = J$ or $J = R$. 532 | \end{definition} 533 | 534 | \begin{lemma} 535 | Consider $R$ is a ring with nonzero unity, and $M$ is an ideal such that $M \neq R$. If $R/M$ is a 536 | division ring, then $M$ is a maximal ideal. 537 | \end{lemma} 538 | \begin{proof} 539 | Suppose $I$ is an ideal such that $M \subsetneq I \subseteq R$. Then $\exists a \in I \> s.t. \> a \notin M$. 540 | Then $a + M \neq 0 + M$ and there exists $b + M \in R/M$ such that 541 | \[ 542 | (a+M)(b+M) = 1_R + M \Longrightarrow (1_R - ab) \in M \Longrightarrow ab+m = 1_R 543 | \] 544 | for some $m \in M$. Since $ab \in I$ and $m \in M \subset I$. Also $1_R \in I \Longrightarrow I = R$. Thus $M$ is 545 | a maximal ideal. 546 | \end{proof} 547 | 548 | \begin{theorem} 549 | Let $M$ be an ideal in a commutative ring $R$ with identity. Then $M$ is a maximal ideal if and only if 550 | the quotient ring $R/M$ is a field. 551 | \end{theorem} 552 | \begin{proof} 553 | $(\Leftarrow)$ If $R/M$ is a field, then $M$ is a maximal ideal by previous lemma. 554 | 555 | $(\Rightarrow)$ Since $M \neq R$, $R/I$ is a commutative ring with 556 | $1_R + R \neq 0_R + M$. Take any nonzero $a + M \in R/M, a \notin M$ and put 557 | \[ 558 | N := Ra + M = \{ ra+m \> | \> r \in R, m \in M \}. 559 | \] 560 | Note that $Ra$ is an ideal and $M$ is also an ideal ($Ra = \langle a \rangle$). 561 | Thus $Ra + M$ is ideal that include $M$. 562 | 563 | Since $M$ is maximal, this implies that $N = R \Longrightarrow 1_R \in N$. 564 | $ra + m = 1_R$ for some $r \in R, m \in M$. Compute 565 | \begin{align*} 566 | ra + m = 1_R &\Rightarrow ra + M = 1_R + M & \text{Since } (ra-1_R) \in M\\ 567 | &\Rightarrow (a+M)(r+M) = 1_R + M. 568 | \end{align*} 569 | We can now see that $a+M$ is actually a unit in $R/M$. Hence $R/M$ is a field. 570 | \end{proof} 571 | 572 | \begin{corollary} 573 | In a commutative ring $R$ with unity, every maximal ideal is a prime ideal. 574 | \end{corollary} 575 | \begin{proof} 576 | If $I$ is a maximal ideal in $R$, then $R/I$ is a field. Every field is an integral domain, so $R/I$ is 577 | also an integral domain, and $I$ is a prime ideal. 578 | \end{proof} 579 | 580 | \begin{example} 581 | The ideal $\langle x^2 + 1 \rangle$ is maximal in $\mathbb{R}[x]$. 582 | \end{example} 583 | \begin{solution} 584 | Suppose that $A$ is an ideal of $\mathbb{R}[x]$ that properly 585 | contains $\langle x^2 + 1 \rangle$. We will 586 | prove that $A = \mathbb{R}[x]$ by showing that $A$ contains some nonzero real number 587 | $c$. Then 588 | \[ 589 | 1 = \frac{1}{c} c \in A 590 | \] 591 | and therefore $A = \mathbb{R}[x]$. 592 | 593 | Let a polynomial $f(x) \in A$, but $f(x) \notin \langle x^2 + 1 \rangle$. Then 594 | \[ 595 | f(x) = q(x) (x^2 + 1) + r(x), 596 | \] 597 | where $r(x) = ax + b$, where $a,b$ both are nonzero, and hence 598 | \[ 599 | a^2 x^2 - b^2 = (ax+b)(ax-b) \in A \textsf{ and } a^2(x^2 + 1) \in A. 600 | \] 601 | So that 602 | \[ 603 | 0 \neq a^2 + b^2 = (a^2x^2 + a^2) - (a^2x^2 - b^2) \in A. 604 | \] 605 | \end{solution} 606 | 607 | \begin{example} 608 | The ideal $\langle x^2 + 1 \rangle$ is not prime in $\mathbb{Z}_2[x]$. 609 | \end{example} 610 | \begin{solution} 611 | (Counterexample) Take 612 | \[ 613 | (x+1)^2 = x^2 + 2x + 1 = x^2 + 1 \in \langle x^2 + 1 \rangle 614 | \] 615 | but $1 + x \notin \langle x^2 + 1 \rangle$. 616 | \end{solution} 617 | 618 | \section{Ring homomorphism} 619 | 620 | \begin{definition}[Ring homomorphism] 621 | A ring homomorphism $f$ from a ring $(R, {\color{red} \oplus}, {\color{red} \odot})$ to a ring 622 | $(S, {\color{blue} \boxplus}, {\color{blue} \boxdot})$ is a mapping from $R$ to $S$ that preserves the 623 | ring additions (${\color{red} \oplus}, {\color{blue} \boxplus}$) and multiplications (${\color{red} \odot}, {\color{blue} \boxdot}$); that is, 624 | \begin{equation*} 625 | f(a \> {\color{red} \oplus} \> b) = f(a) \> {\color{blue} \boxplus} \> f(b) 626 | \end{equation*} 627 | and 628 | \begin{equation*} 629 | f(a \> {\color{red} \odot} \> b) = f(a) \> {\color{blue} \boxdot} \> f(b) 630 | \end{equation*} 631 | A ring homomorphism that is one-to-one and onto is called the \bred{ring isomorphism}. 632 | \end{definition} 633 | 634 | \begin{example} 635 | The map $\phi: \mathbb{Z} \to \mathbb{Z}_3$ defined by 636 | \[ 637 | \phi(x) = x \> (\text{mod }3) \quad \forall x \in \mathbb{Z} 638 | \] 639 | is a ring homomorphism. 640 | \end{example} 641 | \begin{solution} 642 | Clearly, for all $x, y \in \mathbb{Z}$ 643 | \begin{align*} 644 | \phi(x + y) = (x + y) \> (\text{mod }3) &= (x \> \text{mod }3) + (y \> (\text{mod }3))\\ 645 | &= \phi(x) +_3 \phi(y) 646 | \end{align*} 647 | 648 | and 649 | \begin{align*} 650 | \phi(xy) = (xy) \> (\text{mod }3) &= (x \> \text{mod }3) \cdot (y \> (\text{mod }3))\\ 651 | &= \phi(x) \cdot_3 \phi(y) 652 | \end{align*} 653 | 654 | This is an example of a map that respects both operations. 655 | \end{solution} 656 | 657 | \begin{example} 658 | Consider the map $\phi: \mathbb{Z}_4 \to \mathbb{Z}_6, \phi(x) = 3x$ for all $x$ in 659 | $\mathbb{Z}_4$. $\phi$ is a ring homomorphism. 660 | \end{example} 661 | \begin{solution} 662 | For all $x, y \in \mathbb{Z}$, we check that 663 | \begin{align*} 664 | \phi(x + y) = 3(x + y) \> (\text{mod }6) &= (3x \> \text{mod }6) + (3y \> (\text{mod }6))\\ 665 | &= \phi(x) +_6 \phi(y) 666 | \end{align*} 667 | 668 | and 669 | \begin{align*} 670 | \phi(xy) = 3(xy) \> (\text{mod }6) = 9(xy) \> (\text{mod }6) &= (3x \> \text{mod }6) \cdot (3y \> (\text{mod }6))\\ 671 | &= \phi(x) \cdot_6 \phi(y) 672 | \end{align*} 673 | 674 | this map preserves both operations. So $\phi$ is a ring homomorphism. 675 | 676 | In our calculation, we can have used the fact that $3 = 9 \> (\text{mod }6)$. The jump from 3 to 9 modulo 6 677 | can be better seem in 678 | \[ 679 | 3 \> (\text{mod }6) = \phi(1) = \phi(1 \cdot 1) = \phi(1)\, \phi(1) = 3 \cdot 3 = 9 \> (\text{mod }6). 680 | \] 681 | \end{solution} 682 | 683 | \begin{example} 684 | For $a, b \in \mathbb{R}$, let $A(a,b) = M_2(\mathbb{R})$ be defined by 685 | \[ 686 | A(a, b) = \begin{bmatrix} 687 | a & b \\ -b & a 688 | \end{bmatrix}. 689 | \] 690 | Let $R = \{ A(a,b) \> | \> a, b \in \mathbb{R} \} \subseteq M_2(\mathbb{R})$. Then $R \cong \mathbb{C}$. 691 | \end{example} 692 | \begin{solution} 693 | Let $\phi: R \to \mathbb{C}$ be defined by 694 | \[ 695 | \phi(A(a,b)) = a + bi \in \mathbb{C}. 696 | \] 697 | We show firstly that $\phi$ is a ring homomorphism. 698 | 699 | For addition, we have 700 | \begin{align*} 701 | \phi(A(a,b) + A(c,d)) &= \phi \left( A(a+c, b+ d)\right) \\ 702 | &= (a + c) + (b+d)i\\ 703 | &= (a+bi) + (c + di)\\ 704 | &= \phi\left( A(a,b) \right) + \phi\left( A(c,d) \right). 705 | \end{align*} 706 | 707 | For multiplication, we have 708 | \begin{align*} 709 | \phi(A(a,b) + A(c,d)) &= \phi \left( \begin{bmatrix} 710 | a & b \\ -b & a 711 | \end{bmatrix} 712 | \begin{bmatrix} 713 | c & d \\ -d & c 714 | \end{bmatrix} \right) \\ 715 | &= \phi \left( \begin{bmatrix} 716 | ac-bd & ad + bc \\ -(ad+bc) & ac - bd 717 | \end{bmatrix} \right)\\ 718 | &= \phi \left( A(ac-bd, ad + bc )\right)\\ 719 | &= (a+bi)(c+di)\\ 720 | &= \phi \left( A(a,b)\right)\, \phi \left( A(c,d)\right). 721 | \end{align*} 722 | 723 | Now, $\phi$ is one-to-one and onto since $\phi \left( A(a,b) \right) = a+bi = 0$ if and only if 724 | $a = b = 0$, and $Ker\, \phi =\{ A(0,0) \}$ is trivial. 725 | \end{solution} 726 | 727 | \begin{example} 728 | Show that the equation $2x^3 - 5x^2 + 7x - 8 = 0$ has no solutions in $\mathbb{Z}$. 729 | \end{example} 730 | \begin{solution} 731 | Let $\phi: \mathbb{Z} \to \mathbb{Z}_3$ be the natural homomorphism $\phi(x) = x \text{ mod } 3$. Suppose that there 732 | is an integer $a \in \mathbb{Z}$ such that 733 | \[ 734 | 2a^3 - 5a^2 + 7a - 8 = 0. 735 | \] 736 | Then 737 | \[ 738 | 0 = \phi(0) = \phi(2a^3 - 5a^2 + 7a - 8) = 2\phi(a)^3 - 5\phi(a)^2 + 7\phi(a) - 8. 739 | \] 740 | Since $-5 = 7 = -8 = 1 \, (\text{mod }3)$ in $\mathbb{Z}_3$, we have 741 | \[ 742 | 2\phi(a)^3 - 5\phi(a)^2 + 7\phi(a) - 8 \> = \> 2\phi(a)^3 + \phi(a)^2 + \phi(a) + 1 743 | \] 744 | and thus $2b^3 + b^2 + b + 1 = 0$, where $b = \phi(a)$ in $\mathbb{Z}_3$. 745 | 746 | However, one can easily check that no element $b \in \{ 0, 1, 2 \}$ in $\mathbb{Z}_3$ is a solution to this equation. 747 | Therefore there is no such integer $a \in \mathbb{Z}$ to the original equation. 748 | \end{solution} 749 | 750 | \begin{example}[Tutorial] 751 | Show that the rings $2 \mathbb{Z}$ and $3 \mathbb{Z}$ are not isomorphic. 752 | \end{example} 753 | \begin{solution} 754 | Assume the contrary and let $\phi: 2 \mathbb{Z} \to 3\mathbb{Z}$ to be an isomorphism. Let us examine 755 | $\phi(2)$. Note that for some $k \in \mathbb{Z}, \phi(2) = 3k$. Since $\phi$ is a homomorphism, 756 | \[ 757 | \phi(k) = \phi(2 + 2) = \phi(2) + \phi(2) = 3k + 3k = 6k. 758 | \] 759 | But $\phi$ is a ring homomorphism and 760 | \[ 761 | \phi(k) = \phi(2 \cdot 2) = \phi(2)\, \phi(2) = (3k)(3k) = 9k^2. 762 | \] 763 | This implies that $6k = 9k^2 \Longrightarrow k = 0$ or $k = \frac{2}{3}$. 764 | 765 | For $k = 0 \Longrightarrow \phi(x) = 0$ is not one-to-one and not onto. Also, $k = \frac{2}{3} \notin \mathbb{Z}$, and thus 766 | $\phi$ cannot be an isomorphism. 767 | \end{solution} 768 | 769 | \begin{example} 770 | Determine all ring homomorphism from $\mathbb{Z}$ to $\mathbb{Z}_6$. 771 | \end{example} 772 | \begin{solution} 773 | Since $\mathbb{Z}$ is generated from $1$ by addition and subtraction, if a ring homomorphism $f: \mathbb{Z} \to \mathbb{Z}_6$, then for any 774 | $a \in \mathbb{Z}$, we have 775 | \[ 776 | f(a) = am 777 | \] 778 | where $m = f(1)$. Then $f$ is linear, so 779 | \[ 780 | f(a) + f(b) = am + bm = (a+b)m = f(a+b) \quad \forall a, b \in \mathbb{Z}. 781 | \] 782 | So $f$ is a ring homomorphism if and only if 783 | \[ 784 | 0 = f(ab) - f(a)f(b) = abm - (am)(bm) = ab(m - m^2) 785 | \] 786 | for any $a,b \in \mathbb{Z}$. 787 | 788 | In particular, taking $a = b = 1$, we need to find $m$ such that $0 = m - m^2 \> (\text{mod} 6)$. Working modulo 6 one by one 789 | \begin{align*} 790 | 0 - 0^2 = 0-0=0, &1-1^2=1-1=0, &2 - 2^2 = 2 - 4 = 4 \neq 0\\ 791 | 3 - 3^2 = 3 - 9 = -6 = 0, &4-4^2 = 4 -16 = 0, &5 - 5^2 = 5 - 25 = 2\neq 0 792 | \end{align*} 793 | 794 | The possible values of $m$ are $0, 1, 3$ and $4$. So the homomorphisms are as follow 795 | \begin{itemize} 796 | \item $f(a) = 0 \> (\text{mod} 6), \quad \forall a \in \mathbb{Z}$. 797 | \item $f(a) = a \> (\text{mod} 6), \quad \forall a \in \mathbb{Z}$. 798 | \item $f(a) = 3a \> (\text{mod} 6), \quad \forall a \in \mathbb{Z}$. 799 | \item $f(a) = 4a \> (\text{mod} 6), \quad \forall a \in \mathbb{Z}$. 800 | \end{itemize} 801 | \end{solution} 802 | 803 | \begin{theorem}[The first isomorphism for rings] 804 | Let $f$ be a ring homomorphism from ring $R$ ro $S$. Then the mapping from $R / ker(f)$ to $f(R)$, given by 805 | \[ 806 | r + ker(f) \to f(r) 807 | \] 808 | is an isomorphism. In symbols, $R/ker(f) \cong f(R)$. 809 | \end{theorem} 810 | \begin{proof} 811 | Define a map $f: R/K \to \Ima \olsi{f}$ by 812 | \[ 813 | f(a + K) = \olsi{f}(a)\quad \forall a \in R, a+ K \in R/K. 814 | \] 815 | \begin{enumerate} 816 | \item Since $f$ is well defined, so 817 | \begin{align*} 818 | a + K = b + K &\Rightarrow a - b \in K\\ 819 | &\Rightarrow \olsi{f}(a - b) = 0_S\\ 820 | &\Rightarrow \olsi{f}(a) - \olsi{f}(b) = 0_S\\ 821 | &\Rightarrow \olsi{f}(a) = \olsi{f}(b) 822 | \end{align*} 823 | 824 | \item $f$ is injective since 825 | \begin{align*} 826 | f(a +K) = f(b+K) &\Rightarrow \olsi{f}(a) = \olsi{f}(b)\\ 827 | &\Rightarrow \olsi{f}(a-b) = 0_S\\ 828 | &\Rightarrow a - b \in K\\ 829 | &\Rightarrow a + K = b + K. 830 | \end{align*} 831 | 832 | \item $f$ is surjective. For all $f(a) \in \Ima \olsi{f}, \exists a+K \in R/K$ 833 | such that $\olsi{f}(a+K) = f(a)$. 834 | 835 | \item $f$ is homomorphism, 836 | \begin{align*} 837 | f(a + K + b + K) &\Rightarrow f((a+b) + K)\\ 838 | &\Rightarrow \olsi{f}(a+b)\\ 839 | &\Rightarrow \olsi{f}(a) + \olsi{f}(b)\\ 840 | &\Rightarrow f(a+K) + f(b+K). 841 | \end{align*} 842 | 843 | \begin{align*} 844 | f(a + K) \cdot f(b + K) &\Rightarrow f(ab + K)\\ 845 | &\Rightarrow \olsi{f}(ab)\\ 846 | &\Rightarrow \olsi{f}(a) \cdot \olsi{f}(b)\\ 847 | &\Rightarrow f(a+K) \cdot f(b+K). 848 | \end{align*} 849 | \end{enumerate} 850 | Thus $f: R/K \cong \Ima f$ as rings. 851 | \end{proof} 852 | 853 | \begin{example} 854 | Let $\phi: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}_3$ be the ring homomorphism defined by 855 | \[ 856 | \phi((a,b)) = b \text{ mod }3. 857 | \] 858 | Then $ker(\phi) = \mathbb{Z} \times 3 \mathbb{Z}$ and $(\mathbb{Z} \times \mathbb{Z}) / (\mathbb{Z} \times 3 \mathbb{Z})$ 859 | is isomorphic to $\mathbb{Z}_3$, which is a field. Thus $\mathbb{Z} \times 3\mathbb{Z}$ is a maximal ideal of 860 | $\mathbb{Z} \times \mathbb{Z}$. 861 | \end{example} 862 | 863 | \section{Polynomial rings} 864 | 865 | \begin{definition} 866 | Let $R$ be a commutative ring. We define 867 | \begin{equation} 868 | R[x] = \{ r_nx^n + r_{n-1}x^{n-1} + \cdots + r_1 x + r_0 \> | \> r_i \in R \}. 869 | \end{equation} 870 | \end{definition} 871 | 872 | The letter $x$ here can be thought of a variable or just a placeholder. Either way the familiar structure 873 | allows us to add, subtract and multiply these as we do traditional polynomials even if the ring 874 | were some strange abstract entity. 875 | 876 | \section{Factorization of polynomials} 877 | 878 | \begin{theorem}[Division algorithm] 879 | Let $R$ be a ring with identity and $f(x),g(x) \in R[x]$ with 880 | $g(x) \neq 0$. Then there exists unique polynomials 881 | $q(x)$ and $r(x)$ in $R[x]$ such that 882 | \begin{equation} 883 | f(x) = q(x) g(x) + r(x) 884 | \end{equation} 885 | and $\text{deg}(r) < \text{deg}(g)$. $r(x) = 0$ if there is no remainder. 886 | \end{theorem} 887 | 888 | \begin{proof} 889 | The basic idea is to formalize the process of long division in an inductive sense. We omit the details here. 890 | They're boring here. 891 | \end{proof} 892 | 893 | \begin{example} 894 | In $\mathbb{Z}_3$ we can divide $2x^2 + 1$ into $x^4 + 2x^3 + 2x + 1$. Then we have 895 | \[ 896 | x^4 + 2x^3 + 2x + 1 = (2x^2 + 1)(2x^2 + x + 2) 897 | \] 898 | \end{example} 899 | 900 | \begin{theorem}[Factor theorem] 901 | Let $F$ be a field, $a \in F$ and $f(x) \in F[x]$. Then $a$ is a \bgreen{root} (or \bgreen{zero}) of 902 | $f(x)$ if and only if $x - a$ is a factor of $f(x)$. 903 | \end{theorem} 904 | \begin{proof} 905 | $(\Rightarrow)$ Assume that $a \in F$ is a zero of $f(x) \in F[x]$. We wish to show that 906 | $x - a$ is a factor of $f(x)$. To do so, apply the division algorithm. By division algorithm, 907 | $\exists \,$ unique polynomials $q(x)$ and $r(x)$ such that 908 | \[ 909 | f(x) = (x-a)q(x) + r(x) 910 | \] 911 | and the $deg(r) < deg(x-a) = 1$, so $r(x) = c \in F$, where $c$ is a constant. Also, the fact that 912 | $a$ is a zero of $f(x)$ implies $f(a) = 0$. So 913 | \[ 914 | f(x) = (x-a)q(x) + c \> \Longrightarrow 0 = f(a) = (a-a)q(a) + c. 915 | \] 916 | Thus $c = 0$, and $x - a$ is a factor of $f(x)$. 917 | 918 | $(\Leftarrow)$ On the other way, we want to show 919 | \end{proof} 920 | 921 | \begin{definition}[Algebraically closed] 922 | Given $F$ a field, we call $F$ \bred{algebraically closed} if every $f \in F[x]$ such that 923 | $deg(F) > 0$ has a root in $F$. 924 | \end{definition} 925 | 926 | \begin{example} 927 | Show that $x^2 + 3x - 4 \in \mathbb{Z}_{12}[x]$ has 4 roots. 928 | \end{example} 929 | \begin{solution} 930 | We list down all the values of $f(x) = x^2 + 3x - 4$ for $x = 0, 1, \ldots, 11$. 931 | 932 | \begin{center} 933 | \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|} 934 | \hline 935 | $x$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11\\ 936 | \hline 937 | $x^2+3x-4\> (\text{mod } 12)$ & 8 & {\color{red} 0} & 6 & 2 & {\color{red} 0} & {\color{red} 0} & 938 | 2 & 6 & {\color{red} 0} & 8 & 6 & 6\\ 939 | \hline 940 | \end{tabular} 941 | \end{center} 942 | 943 | which now we can see: $x^2 + 3x - 4$ has 4 zeros in $\mathbb{Z}_{12}[x]$. Thus, a polynomial of degree $n$ can have more than $n$ roots 944 | in a ring. The problem is that $\mathbb{Z}_{12}$ is not a domain: $(x+4)(x-1) = 0$ 945 | does not imply one of the factors must be zero. 946 | \end{solution} 947 | 948 | \begin{example} 949 | Show that the polynomial $2x^3 + 3x^2 -7x - 5$ can be factored into linear factors in $Z_{11}[x]$. 950 | \end{example} 951 | \begin{solution} 952 | We can use synthetic division, 953 | 954 | \begin{center} 955 | \begin{tabular}{cccc|c} 956 | $2$ & $3 = -8$ & $-7=4$ & $6$ & \\ 957 | & $-2$ & $-10$ & $-6$ & $-1$\\ 958 | \cmidrule{1-4} 959 | $2$ & $-10 = 1$ & $-6$ & & \\ 960 | & $-4$ & $6$ & & $-2$\\ 961 | \cmidrule{1-4} 962 | $2$ & $-3$ & & & \\ 963 | \end{tabular} 964 | \end{center} 965 | 966 | Thus, $2x^3 + 3x^2 -7x - 5 = (x+1)(x+2)(2x-3)$ in $\mathbb{Z}_{11}[x]$. 967 | \end{solution} 968 | 969 | \subsection{Irreducibility tests} 970 | 971 | There are various methods to check if a polynomial in $\mathbb{Z}[x]$ is irreducible in $\mathbb{Q}[x]$. 972 | 973 | \begin{theorem}[Rational root test] 974 | Let 975 | \[ f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \label{eq:r2.2} \tag{{\color{orange} $\bigstar $}} \] 976 | be a polynomial with integers coefficients. If $r \neq 0$ and 977 | the rational number $r/s$ (in lowest terms) is a root of $f(x)$, then 978 | $r|a_0$ and $s|a_n$. 979 | \end{theorem} 980 | \begin{proof} 981 | Plug $x = r/s$ into \eqref{eq:r2.2} and equating with zero. The equation is now 982 | \[ 983 | a_n \left( \frac{r}{s}\right)^n + a_{n-1}\left( \frac{r}{s}\right)^{n-1} + \cdots + a_1\left( \frac{r}{s}\right) + a_0 984 | = 0. 985 | \] 986 | Again multiplying $s^n$ on both sides 987 | \[ 988 | a_nr^n + a_{n-1}r^{n-1}s + \cdots + a_1rs^{n-1}x + a_0s^n = 0. 989 | \] 990 | Factoring $r$ out and moving $a_0s^n$ to the right-hand side. We obtained 991 | \[ 992 | r(a_nr^{n-1} + a_{n-1}r^{n-2}s + \cdots + a_1s^{n-1}x) = -a_0s^n. 993 | \] 994 | Since $\gcd (r,s) = 1$, thus $r|a_0$ and similarly $s|a_n$. 995 | \end{proof} 996 | 997 | \begin{example} 998 | The polynomial $f(x) = 2x^4 + x^3 - 21x^2 - 14x + 12$ is reducible in $\mathbb{Q}[x]$. 999 | \end{example} 1000 | \begin{solution} 1001 | If $r/s$ is a root of $f(x)$, where $r|12$ and $s|2$. Thus the possible roots are 1002 | \[ 1003 | \pm 1,\> \pm 2,\> \pm 3,\> \pm 4,\> \pm 6,\> \pm 12,\> \pm \frac{1}{2},\> \pm \frac{3}{2}. 1004 | \] 1005 | In fact, $f(x) = (x+3) \left(x - \frac{1}{2} \right)(2x^2 - 4x - 8) \in \mathbb{Q}[x]$. 1006 | \end{solution} 1007 | 1008 | \begin{example} 1009 | The polynomial $g(x) = x^3 + 4x^2 + x - 1$ is irreducible in $\mathbb{Q}[x]$. 1010 | \end{example} 1011 | \begin{solution} 1012 | The possible roots are $\{-1, 1\}$. However 1013 | \[ 1014 | g(1) = 1 + 4 + 1 - 1 = 5 \quad \text{and }\quad g(-1) = -1 + 4 - 1 - 1 = 1 1015 | \] 1016 | So $g(x)$ has no root and $\deg g(x) = 3$. Thus $g(x)$ is irreducible over $\mathbb{Q}[x]$. 1017 | \end{solution} 1018 | 1019 | \begin{theorem}[Mod $p$ Irreducibility test] 1020 | Let $p$ be a prime and let $f(x) \in \mathbb{Z}[x]$ with degree 1 or greater. Let 1021 | $\overline{f} \in \mathbb{Z}_p[x]$ obtained by reducing all of $f(x)$'s coefficients 1022 | mod $p$. Then if 1023 | \begin{equation} 1024 | \deg(\overline{f}) = \deg(f) 1025 | \end{equation} 1026 | and $\overline{f}$ is irreducible over $\mathbb{Z}_p$ then $f(x)$ is irreducible over 1027 | $\mathbb{Q}$. 1028 | \end{theorem} 1029 | \begin{proof} 1030 | Assume that $f(x) = p(x) q(x)$ in $\mathbb{Z}[x]$. Since $\phi: \mathbb{Z}[x] \to \mathbb{Z}_p[x]$ 1031 | defined by $\phi f(x) = \overline{f}(x)$ is a ring homomorphism. So 1032 | \[ 1033 | \overline{f}(x) = \overline{p(x) q(x)} = \overline{p}(x) \overline{q}(x). 1034 | \] 1035 | If $p \nmid a_k$, then $p$ does not divide the leading coefficients of $p(x)$ and $q(x)$. Thus 1036 | $\deg \overline{p}(x) = \deg p(x)$ and $\deg \overline{q}(x) = \deg q(x)$. 1037 | \end{proof} 1038 | 1039 | \begin{example} 1040 | The polynomial $f(x) = x^5 + 8x^4 + 3x^2 + 4x + 7$ is irreducible in $\mathbb{Q}$. 1041 | \end{example} 1042 | \begin{solution} 1043 | We define 1044 | \[ 1045 | \olsi{f}(x) = x^5 + x^2 + 1 \in \mathbb{Z}_2[x]. 1046 | \] 1047 | By rational root test, the only possible root is $0.1$ from $\mathbb{R}$ but it is not an inetger. 1048 | There are several quadratic polynomials in $\mathbb{Z}_2[x]$ such as 1049 | \[ 1050 | x^2 + x + 1, \quad x^2 + 1,\quad x^2 + x, \quad x^2. 1051 | \] 1052 | Since $x^2 + 1, x^2 + x, x^2$ both have roots, they cannot be factor of $\olsi{f}$. The only possible 1053 | factor of $\olsi{f}$ is $x^2 + x + 1$. Thus 1054 | \[ 1055 | x^5 + x^2 + 1 = (x^2 + x+ 1)(x^3 + ax^2 + bx + c). 1056 | \] 1057 | Equating coefficients of both sides, we have 1058 | \[ 1059 | \begin{cases} 1060 | 1 + a = 0\\ 1061 | 1 + a + b = 0\\ 1062 | a + b + c = 0\\ 1063 | b + c = 0\\ 1064 | c = 1 1065 | \end{cases}. 1066 | \] 1067 | On solving yields $a = -1 = 1 (\text{mod } 2)$,\> $b = 0$ and $c = 1$ but $b + c \neq 0$ and is contradiction. 1068 | So $f(x)$ does not has a quadratic factor. It means that $f(x)$ is irreducible in both 1069 | $\mathbb{Z}_2[x]$ and $\mathbb{Z}$. So $f(x)$ is also irreducible in $\mathbb{Q}[x]$. 1070 | \end{solution} 1071 | 1072 | \begin{theorem}[Eisenstein's criterion] 1073 | Let $f(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n \in \mathbb{Z}[x] \setminus \{ 0 \}$. 1074 | If there is a prime number $p$ such that 1075 | $p \nmid a_n$, but $p | a_{n-1}, \ldots p | a_{2}$ and $p^2 | a_0$. Then $f(x)$ is 1076 | irreducible over $\mathbb{Q}$. 1077 | \end{theorem} 1078 | \begin{proof} 1079 | Suppose that $f(x)$ is reducible over $\mathbb{Q}$ then 1080 | \[ 1081 | f(x) = g(x) h(x) 1082 | \] 1083 | and $g(x), h(x)$ are nonconstant polynomials. 1084 | 1085 | Let 1086 | \[ 1087 | f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0, 1088 | \] 1089 | \[ 1090 | g(x) = b_rx^r + b_{r-1}x^{r-1} + \cdots + b_1x + b_0, 1091 | \] 1092 | \[ 1093 | h(x) = c_sx^s + c_{s-1}x^{s-1} + \cdots + c_1x + c_0. 1094 | \] 1095 | Since $p|a_0 =b_0c_0 \Longrightarrow p | b_0$ or $p|c_0$, and $p^2 \nmid a_0$. This implies that 1096 | $p$ divides only one of them. Assume that $p | b_0$ and $p \nmid c_0$, then 1097 | \[ 1098 | p|a_0 = b_0c_1 + b_1c_0. 1099 | \] 1100 | Since $p|b_0c_1$ and $p \nmid c_0 \Longrightarrow p|b_1$. Assume that $p|b_i \> \forall 0 \leq i < m$ 1101 | for some $m \leq r$. Then 1102 | \[ 1103 | p|a_m = \sum_{\substack{i+j=m \\ j \leq s}} b_ic_j \Longrightarrow p|b_mc_0 \Longrightarrow p|b_m. 1104 | \] 1105 | By mathematical induction, $p|b_r$. Thus $p|a_n = b_rc_s$. This contradicting the fact that $f(x)$ 1106 | is reducible. 1107 | \end{proof} 1108 | 1109 | \begin{example} 1110 | $x^9 + 5$ is irreducible in $\mathbb{Q}[x]$ with $p=5$. 1111 | \end{example} 1112 | 1113 | \begin{example} 1114 | $x^{17} + 6x^{13} - 15x^{4} + 3x^2 - 9x + 12$ is irreducible in $\mathbb{Q}[x]$ with $p=3$. 1115 | \end{example} 1116 | 1117 | \begin{example} 1118 | $x^n + 5$ is irreducible in $\mathbb{Q}[x]$ for all $n \geq 1$. There are irreducible polynomials of 1119 | every degree in $\mathbb{Q}[x]$. 1120 | \end{example} 1121 | 1122 | \vspace*{6cm} 1123 | \begin{corollary} 1124 | For any prime $p$, the $p$-th cyclotomic polynomial 1125 | \[ 1126 | \Phi_p(x) = x^{p-1} + x^{p-2} + \cdots + x + 1 1127 | \] 1128 | is irreducible over $\mathbb{Q}$. 1129 | \end{corollary} 1130 | \begin{proof} 1131 | Let $\zeta = e^{2\pi i / n}$. Then $\zeta, \zeta^2, \ldots, \zeta^n$ are the $n$-th roots of unity. They form 1132 | the vertices of a regular $n$-gon in the complex plane. If $\gcd(a,n) > 1$ then $\zeta^a$ is a root of unity 1133 | of order $n / \gcd(a,n) < n$, but if $\gcd(a,n) = 1$ then $\zeta$ is not a root of lower order, and in this case we call 1134 | $\zeta^a$ a primitive $n$-th root of unity. We define the $n$-th cyclotomic polynomial $\Phi_n(x)$ to be the monic polynomial 1135 | of degree $\phi(n)$ whose roots are the primitive $n$-th root of unity: 1136 | \begin{equation} 1137 | \Phi_n(x) = \prod_{\substack{a = 1\\ \gcd(a,n)=1}}^{n}(x - \zeta^a). 1138 | \end{equation} 1139 | 1140 | The first few cyclotomic polynomials are as follows: 1141 | 1142 | \begin{center} 1143 | \begin{tabular}{c|c} 1144 | $n = $ & \\ 1145 | \hline 1146 | $1$ & $\Phi_1(x) = x-1$\\[0.225em] 1147 | $2$ & $\Phi_2(x) = x + 1$\\[0.225em] 1148 | $3$ & $\Phi_3(x) = x^2 + x + 1$\\[0.225em] 1149 | $4$ & $\Phi_4(x) = x^2 + 1$\\[0.225em] 1150 | $5$ & $\Phi_5(x) = x^4 + x^3 + x^2 + x + 1$\\[0.225em] 1151 | $6$ & $\Phi_6(x) = x^2 - x + 1$\\[0.225em] 1152 | $7$ & $\Phi_7(x) = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$\\[0.225em] 1153 | $8$ & $\Phi_8(x) = x^4 + 1$\\[0.225em] 1154 | $9$ & $\Phi_9(x) = x^6 + x^3 + 1$\\[0.225em] 1155 | $10$ & $\Phi_{10}(x) = x^4 - x^3 + x^2 - x + 1$\\[0.225em] 1156 | $11$ & $\Phi_{11}(x) = x^{10} + x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$\\[0.225em] 1157 | $12$ & $\Phi_{12}(x) = x^4 - x^2 + 1$\\[0.225em] 1158 | $13$ & $\Phi_{13}(x) = x^{12} + x^{11} + x^{10} + x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$\\[0.225em] 1159 | $14$ & $\Phi_{14}(x) = x^6 - x^5 + x^4 - x^3 + x^2 -x + 1$\\[0.225em] 1160 | $15$ & $\Phi_{15}(x) = x^8 - x^7 + x^5 - x^4 + x^3 -x + 1$\\[0.225em] 1161 | $16$ & $\Phi_{16}(x) = x^8 + 1$\\[0.225em] 1162 | \end{tabular} 1163 | \end{center} 1164 | 1165 | Let $p$ denote a given prime number. For any polynomial $f(x)$ with integral coefficients let 1166 | $\overline{f}(x)$ be the polynomial whose coefficients are the residue classes (mod $p$) determined by the coefficients 1167 | of $f(x)$. Thus the assertion $\overline{f} = \overline{g}$ means that there is a polynomial $h(x)$ with 1168 | integral coefficients such that $f(x) = ph(x)$. 1169 | 1170 | \begin{lemma} 1171 | (Sch\"{o}nemann, 1846) Let $A(x)$ be a monic polynomial with integral coefficients, for instance 1172 | \[ 1173 | A(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0 = \prod^{n}_{i = 1} (x - \alpha_i). 1174 | \] 1175 | Let $p$ be a prime, and put 1176 | \[ 1177 | C(x) = \prod_{i=1}^{n} (x - \alpha^p_i). 1178 | \] 1179 | Then $\overline{C} = \overline{A}$. 1180 | \end{lemma} 1181 | \begin{proof} 1182 | Let $\sigma_k(\alpha)$ denote the $k$-th symmetric function of $\alpha_i$. When $\sigma_k(\alpha)^p$ is expanded 1183 | by the multinomial theorem, all coefficients except the extreme ones are divisible by $p$. That is, 1184 | \[ 1185 | \frac{\sigma_k(\alpha_1, \alpha_2, \ldots, \alpha_n)^p - \sigma_k(\alpha^p_1, \alpha^p_2, \ldots, \alpha_n^p)}{p} 1186 | \] 1187 | is a symmetric polynomial in the $\alpha$, with integral coefficients, and hence by the symmetric function theorem the 1188 | quantity must be a rational integer. 1189 | \end{proof} 1190 | 1191 | \begin{lemma} 1192 | Put $f(x) = x^n - 1$. Then $\overline{f}$ is a squarefree if and only if $p \nmid n$. 1193 | \end{lemma} 1194 | \begin{proof} 1195 | By previous lemma we can see that if $p \nmid n$. Then $\gcd(\overline{f}, \overline{f}') = 1$, and hence 1196 | that $\overline{f}$ is squarefree. On the other hand, if $p|n$, say $n = mp$ for some integer $m$, then 1197 | \[ 1198 | \overline{f} = \overline{(x^m - 1)^p} 1199 | \] 1200 | and hence $\overline{f}$ is not squarefree. 1201 | 1202 | Let $\Phi_n(x)$ denote the $n$-th cyclotomic polynomial. Since $\Phi_n | f$, it follows from the above 1203 | that if $p \nmid n$, then $\overline{\Phi_n}$ is also squarefree. 1204 | \end{proof} 1205 | 1206 | \begin{theorem} 1207 | (Kronecker, 1854) The polynomial $\Phi_n(x)$ is irreducible over $\mathbb{Q}$. 1208 | \end{theorem} 1209 | \begin{proof} 1210 | Suppose that $A$ and $B$ are monic polynomials with rational coefficients such that 1211 | $\Phi_n = AB$, and suppose also that $\deg A > 0$. We know that A and B 1212 | have integral coefficients. Let $Z$ denote the roots of $A$. Let $C$ be the monic polynomial 1213 | whose roots are the numbers $\zeta^p$ for $\zeta \in Z$. Here $p$ is an arbitrary prime not dividing $n$. 1214 | Our first step is to show that $A = C$. 1215 | 1216 | Since the map $\zeta \mapsto \zeta^p$ merely permutes the roots of 1217 | $\Phi_n$, we know that $C|\Phi_n$. Let $G = \gcd(B,C)$. Then $\overline{G}|B$ and $\overline{G}|C$. But $\overline{A} = \overline{C}$ by previous lemma, 1218 | and hence $\overline{G}^2 | \olsi{A} \olsi{B}$. But $\Phi_n$ is squarefree, by previous lemma. Hence $G = \olsi{1}$, so $G = 1$, 1219 | and consequently $C |A$. But $C$ and $A$ have the same degree, so in fact $A = C$. 1220 | 1221 | Now let $\zeta$ be a root of $A$, and $\zeta'$ a root of $\Phi_n$. Then there exists a positive integer $a$, 1222 | $\gcd(a, n) = 1$, such that $\zeta' = \zeta^a$. We factor $a$, $a = p_1p_2 \ldots p_k$. Since $\zeta$ is a root of $A$, it follows 1223 | from the argument above that $\zeta^{p_1}$ is also a root of A. Then by a second application of 1224 | the above argument, we see that $\zeta^{p_1p_2}$ is also a root of $A$. Continuing in this manner, we 1225 | deduce that $\zeta'$ is a root of $A$. Since this is valid for every root $\zeta'$ of $\Phi_n$, we conclude that 1226 | $A = \Phi_n$. Hence $\Phi_n$ is irreducible. 1227 | \end{proof} 1228 | \end{proof} 1229 | 1230 | 1231 | 1232 | \begin{theorem} 1233 | Let $F$ be a field and let $p(x) \in F[x]$. Then $\langle p(x) \rangle$ is a maximal ideal in $F[x]$ 1234 | if and only if $p(x)$ is irreducible over $F$. 1235 | \end{theorem} 1236 | \begin{proof} 1237 | Suppose $\langle p(x) \rangle$ is a maximal ideal in $F[x]$. We know that $p(x) \neq 0$ and 1238 | $p(x)$ is not a unit since neither $\{ 0 \}$ nor $\langle 1_F \rangle = F[x]$ is a 1239 | maximal ideal in $F[x]$. Let 1240 | \[ 1241 | p(x) = g(x) h(x) 1242 | \] 1243 | be a factorization. Then $\langle p(x) \rangle \subseteq \langle g(x) \rangle 1244 | \subseteq \langle F[x] \rangle$ and since $\langle p(x) \rangle$ is maximal we either have 1245 | $\langle g(x) \rangle = \langle p(x) \rangle$ or $\langle g(x) \rangle = F[x]$. In the 1246 | first case we get 1247 | \end{proof} 1248 | 1249 | \begin{theorem}[Fundamental Theorem of Algebra] 1250 | Every nonconstant polynomial in $\mathbb{C}[x]$ has a root in $\mathbb{C}$. 1251 | \end{theorem} 1252 | \begin{remark} 1253 | The field $\mathbb{C}$ is algebraically closed. 1254 | \end{remark} 1255 | 1256 | \begin{corollary} 1257 | A polynomial is irreducible in $\mathbb{C}[x]$ if and only if it has a degree 1. 1258 | \end{corollary} 1259 | \begin{proof} 1260 | All linear equation with degree 1 only have one root in $\mathbb{R}$. 1261 | \end{proof} 1262 | 1263 | \begin{corollary} 1264 | Every nonconstant polynomial $f(x)$ of degree $n$ can be written in the form 1265 | \[ 1266 | c(x-a_1)(x-a_2) \ldots (x-a_n) 1267 | \] 1268 | for some $c, a_1, a_2, \ldots, a_n \in \mathbb{C}$. This factorization is unique except for the order 1269 | of the factors. 1270 | \end{corollary} 1271 | \begin{proof} 1272 | By the fundamental theorem of algebra, 1273 | \begin{align*} 1274 | f(x) &= (r_1x + s_1)(r_2x + s_2) \ldots (r_nx + s_n)\\ 1275 | &= r_1\, r_2 \ldots r_n (x + s_1r_1^{-1})(x + s_2r^{-1}_2) \ldots (x + s_nr^{-1}_n). 1276 | \end{align*} 1277 | Since $f(x)$ has $n$ unique roots, factorization is also unique. 1278 | \end{proof} 1279 | 1280 | \begin{lemma} 1281 | If $f(x)$ is a polynomial in $\mathbb{R}[x]$ and $a + bi$ is a root of $f(x)$ in $\mathbb{C}$, 1282 | then $a - bi$ is also a root of $f(x)$. 1283 | \end{lemma} 1284 | \begin{proof} 1285 | Let $z = a+bi$ and the conjugate $\olsi{z} = a - bi$. Define a map 1286 | $\varphi: \mathbb{C}[x] \to \mathbb{C}[x]$ by $\varphi f(x) = \olsi{f}$. 1287 | Bijective is trivial in $\varphi$, e.g. 1288 | $\varphi(f + g) = \olsi{f} + \olsi{g}$ 1289 | and $\varphi(fg)= \olsi{f} \, \olsi{g}$ since 1290 | \[ 1291 | \overline{(a+bi)(c+di)} = \overline{(a+bi)}\> \overline{(c+di)}. 1292 | \] 1293 | If $f(x)$ has a root $z$ then $\olsi{f}(x)$ will has a root $\olsi{z}$. 1294 | If coefficients of $f(x)$ are all real numbers, then $f(x) = \olsi{f}(x)$. 1295 | Thus $f(x)$ has a root $\olsi{z}$. 1296 | \end{proof} 1297 | 1298 | \begin{theorem} 1299 | A polynomial $f(x)$ is irreducible in $\mathbb{R}[x]$ \text{if and only if} $f(x)$ is a 1300 | first-degree polynomial or $f(x) = ax^2 + bx + c$ with $b^2 - 4ac < 0$. 1301 | \end{theorem} 1302 | \begin{proof} 1303 | In $\mathbb{C}[x]$, 1304 | \[ 1305 | f(x) = c(x - a_1)(x-a_2)\ldots (x-a_n). 1306 | \] 1307 | If $a_i = c + di, a_j = c-di$ for some $1 \leq j \leq n$. The product of the conjugates are 1308 | \[ 1309 | (x - a_i)(x - a_j) = (x - c -di)(x - c + di) = x^2 -2cx + c^2 + d^2 \in \mathbb{R}[x]. 1310 | \] 1311 | Thus we can pair them and so $f(x)$ can be split by irreducible polynomials whose degree is either 1312 | 1 or 2. 1313 | 1314 | Now we knew every irreducible polynomial has a degree 1 or 2. When its degree is 2, then 1315 | \[ 1316 | f(x) = ax^2 + bx + c \quad \forall a,b,c \in \mathbb{R} \label{eq:r2.1} \tag{{\color{myred} $\clubsuit$}} 1317 | \] 1318 | We now continue to work on the "formula" to solve $x$. Completing the square on \eqref{eq:r2.1} 1319 | \begin{align*} 1320 | ax^2 + bx + c = 0 &\Rightarrow a \left[ x^2 + \frac{b}{a}x + \left(\frac{b}{2a} \right)^2 \right] - \left(\frac{b}{2a} \right)^2 = 0\\ 1321 | &\Rightarrow \left(x + \frac{b}{2a} \right)^2 = \frac{b^2 - 4ac}{4a^2}\\[0.35em] 1322 | &\Rightarrow x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}\\[0.35em] 1323 | &\Rightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a \neq 0\\ 1324 | \end{align*} 1325 | Now we can take a look on determinant $\Delta = b^2 - 4ac$. If $\Delta < 0$, the two roots 1326 | will be in $\mathbb{C} \setminus \mathbb{R}$, else the two roots are in $\mathbb{R}$. (Either $\Delta > 0$ or 1327 | $\Delta = 0$). Hence the first-degree polynomial or quadratic polynomial is irreducible in $\mathbb{R}[x]$. 1328 | \end{proof} 1329 | 1330 | \begin{corollary} 1331 | Every polynomial $f(x)$ of odd degree in $\mathbb{R}[x]$ has a root in $\mathbb{C}$. 1332 | \end{corollary} 1333 | \begin{proof} 1334 | Consequently, we can tell if a polynomial in $\mathbb{R}[x]$ or $\mathbb{C}[x]$ is irreducible without any elaborate 1335 | tests. 1336 | \end{proof} 1337 | 1338 | \section{Integral Domains} 1339 | 1340 | Let $R$ be a commutative ring. A \bred{zero divisor} is a nonzero element $a \in R$ such that 1341 | \begin{equation} 1342 | ab=0 1343 | \end{equation} 1344 | for some nonzero $b \in R$. The most familiar integral domain is $\mathbb{Z}$. It is a 1345 | commutative ring with unity one. If $a, b \in \mathbb{Z}$ and $ab=0$, then either $a=0$ or 1346 | $b=0$. 1347 | 1348 | \begin{definition} 1349 | A ring with unity $1$ having no zero divisors is an integral domain. 1350 | \end{definition} 1351 | 1352 | \begin{lemma} 1353 | Fields are integral domain 1354 | \end{lemma} 1355 | \begin{proof} 1356 | Let $F$ be a field. We want to show that $F$ has no zero divisors. Suppose $ab =0$ and 1357 | $a \neq 0$. Then $a$ must has an inverse $a^{-1}$ such that $a^{-1}\, a b = a^{-1} \cdot 0 \Longrightarrow b = 0$. 1358 | Therefore, $F$ has no zero divisors, and so $F$ is an integral domain. 1359 | \end{proof} 1360 | 1361 | \begin{corollary} 1362 | $\mathbb{Z}_p$ is a field. 1363 | \end{corollary} 1364 | \begin{proof} 1365 | According to previous theorem, we need only prove that $\mathbb{Z}_p$ has no zero 1366 | divisors. 1367 | 1368 | Suppose that $a, b \in \mathbb{Z}_p$ and $ab = 0$. Then $ab = pk$ for some 1369 | integer $k$. But then we have either $p \, | \, a$ or $p p \, | \, b$ and in 1370 | $\mathbb{Z}_p$ is equivalent to either $a = 0$ or $b = 0$. 1371 | \end{proof} 1372 | 1373 | \begin{lemma} 1374 | If $R$ is an integral domain, then the characteristic of $R$ is either $0$ or a positive prime. 1375 | \end{lemma} 1376 | \begin{solution} 1377 | Suppose not, suppose $R$ has characteristic $n=ab$ with $1 < a < b < n$. Then 1378 | \[ 1379 | (\underbrace{1_R + \cdots + 1_R}_{a \text{ times}})(\underbrace{1_R + \cdots + 1_R}_{b \text{ times}}) 1380 | = ab1_R = n1_R = 0. 1381 | \] 1382 | Since $R$ is an integral domain, it is either $a \cdot 1_R = 0$ or $b \cdot 1_R = 0$. This is contradicts 1383 | with our assumption that $n$ is not prime. 1384 | \end{solution} 1385 | 1386 | \begin{definition} 1387 | If $F$ is a field, then the only ideals are $\{0\}$ and $F$ itself. 1388 | \end{definition} 1389 | \begin{proof} 1390 | Let $F$ be a field, and let $I \subset F$ be an ideal. Assume $I \neq \{0 \}$, and find $x \neq 0 \in I$. 1391 | Since $F$ is a field, $x$ is invertible; Since $I$ is an ideal, $1 = x^{-1} \cdot x \in I$. Therefore $I = F$. 1392 | \end{proof} 1393 | 1394 | \begin{example} 1395 | The extended ring 1396 | \[ 1397 | \mathbb{Q}[\sqrt{2}] = \{ a+b\sqrt{2} \> | \> a,b \in \mathbb{Q} \} 1398 | \] 1399 | is a field and that every nonzero element has a multiplicative inverse. 1400 | \end{example} 1401 | \begin{solution} 1402 | This is clearly a ring. To show that every nonzero element has a multiplicative inverse. 1403 | Consider $a + b\sqrt{2} \neq 0 \in \mathbb{Q}[\sqrt{2}]$. The multiplicative inverse is 1404 | \[ 1405 | \frac{1}{a + b\sqrt{2}} 1406 | \] 1407 | Then multiplying top and bottom by conjugate, we have 1408 | \[ 1409 | \frac{a - b\sqrt{2}}{(a + b\sqrt{2})(a - b\sqrt{2})} = \frac{a - b\sqrt{2}}{a^2 - 2b^2}. 1410 | \] 1411 | Now we want to show $a^2 - 2b^2 \neq 0$. 1412 | 1413 | If $a = 0$ and $b \neq 0$ or if $a \neq 0$ and $b=0$, then $a^2 - 2b^2 \neq 0$. Since 1414 | $a^2 - 2b^2 \neq 0$, the only other possibility is $a, b \neq 0$. 1415 | 1416 | Thus, $a^2 = 2b^2$ with $a,b \neq 0$. We may assume that $a$ and $b$ are integers -- 1417 | in fact, now we can see $2$ divides $2b^2$, so $2 \,| \, a^2 \Longrightarrow \> 2 \, | \, a$. So 1418 | $a = 2c$ for some integer $c$. Plugging in gives $4c^2 = 2b^2 \Longrightarrow 2c^2 = b^2$. 1419 | 1420 | It follows that every nonzero element of $\mathbb{Q}[\sqrt{2}]$ is invertible, so 1421 | $\mathbb{Q}[\sqrt{2}]$ is a field. 1422 | \end{solution} 1423 | 1424 | \begin{example}[Non-example] 1425 | $\displaystyle M_2(\mathbb{Z}) = \left\{ \begin{bmatrix} 1426 | a & b\\ c & d 1427 | \end{bmatrix} \> \bigg \vert \> a,b,c,d \in \mathbb{Z} \right\}$ is not an integral domain. 1428 | \end{example} 1429 | \begin{solution} 1430 | Choose 1431 | \[ 1432 | A = \begin{bmatrix} 1433 | 1 & 0\\ 0 & 0 1434 | \end{bmatrix}, \quad 1435 | B = \begin{bmatrix} 1436 | 0 & 0\\ 1 & 0 1437 | \end{bmatrix} 1438 | \] 1439 | from $M_2(\mathbb{Z})$, and compute the matrix product 1440 | \[ 1441 | AB = \begin{bmatrix} 1442 | 1 & 0\\ 0 & 0 1443 | \end{bmatrix} 1444 | \begin{bmatrix} 1445 | 0 & 0\\ 1 & 0 1446 | \end{bmatrix} = \begin{bmatrix} 1447 | 0 & 0\\ 0 & 0 1448 | \end{bmatrix} = \mathbf{0}. 1449 | \] 1450 | $A, B$ are zero divisors but none of them are zero. Thus $M_2(\mathbb{Z})$ is not an integral domain. 1451 | \end{solution} 1452 | 1453 | \begin{theorem} 1454 | A finite integral domain is a field 1455 | \end{theorem} 1456 | \begin{proof} 1457 | Let $D$ be a finite integral domain. Since $D$ is an integral domain, then $D$ is a commutative 1458 | ring with unity, and hence we need to show that $D$ is a field. In order to do this, we 1459 | want to show $\forall a \neq 0 \in D, \> \exists a^{-1} \in D$ such that 1460 | \[ 1461 | a \cdot a^{-1} = 1_D = a^{-1} \cdot a. 1462 | \] 1463 | Without loss of generality, we let 1464 | \[ 1465 | D = \{ a, a^2, a^3, \ldots, a^t \} 1466 | \] 1467 | where $a \neq 0$ for some $t \in \mathbb{N}$. Consider two elements $a^i, a^j$ from $D$, we have 1468 | \begin{align*} 1469 | a^i = a^j &\Rightarrow a^{i-j} = 1_D\\ 1470 | &\Rightarrow a\, a^{i-j-1} = 1_D\\ 1471 | &\Rightarrow a^{-1} = a^{i-j-1}\\ 1472 | &\Rightarrow a \cdot a^{-1} = a^{i-j} = 1_D. 1473 | \end{align*} 1474 | and the multiplication is commutative, therefore $D$ is a field. 1475 | \end{proof} 1476 | \begin{remark} 1477 | \begin{align*} 1478 | \textsf{if } \mathbb{Z}_p \textsf{ is a field} &\Longrightarrow \mathbb{Z}_p \textsf{ has no zero divisors.}\\ 1479 | &\Longrightarrow \mathbb{Z}_p \textsf{ is an integral domain.}\\ 1480 | &\Longrightarrow |\mathbb{Z}_p| = p \textsf{ is prime and is finite.} 1481 | \end{align*} 1482 | \end{remark} 1483 | 1484 | \begin{theorem} 1485 | Let $R$ be a commutative ring with unity and let $A$ be an ideal of $R$. Then 1486 | $R/A$ is an integral domain if and only if $A$ is prime. 1487 | \end{theorem} 1488 | \begin{proof} 1489 | We are going to prove in two directions: 1490 | 1491 | $(\Rightarrow)$ Suppose $R/A$ is an integral domain and $ab \in A$. Then 1492 | \[ 1493 | (a + A)(b + A) = ab + A = A 1494 | \] 1495 | is the zero element of the ring $R/A$. So it is either $a + A = A$ 1496 | or $b + A = A$; that is, either $a \in A$ or $b = A$. Thus $A$ is prime. 1497 | 1498 | $(\Leftarrow)$ On the other hand, observe that $R/A$ is a commutative ring with 1499 | unity for any proper ideal $A$. Thus, we only need to show that when $A$ 1500 | is prime, $R/A$ has no zero divisors. 1501 | 1502 | Now suppose that $A$ is prime and 1503 | \[ 1504 | (a + A) (b+A) = 0 + A = A. 1505 | \] 1506 | Then $ab \in A$, and therefore, it is either $a \in A$ or 1507 | $b \in A$. Thus, one of 1508 | $a + A$ or $b + A$ is the zero coset in $R/A$. Now we are done. 1509 | \end{proof} 1510 | 1511 | \section{Principal Ideal Domain} 1512 | 1513 | \begin{definition} 1514 | An integral domain $R$ is called a \bred{principal ideal domain} (or \bred{PID}) if every ideal in $R$ is principal. 1515 | \end{definition} 1516 | 1517 | \begin{example} 1518 | The integers $\mathbb{Z}$ and polynomial rings over fields are principal ideal domains. 1519 | \end{example} 1520 | 1521 | \begin{theorem} 1522 | If $F$ is a field then $F[x]$ is a PID. 1523 | \end{theorem} 1524 | \begin{proof} 1525 | We know $F[x]$ is integral domain since $F$ is an integral domain. Let $I$ be an ideal of 1526 | $F[x]$. 1527 | 1528 | \textbf{Case 1}: If $I = \{ 0 \}$ then $I = \langle 0 \rangle$ and we are done. 1529 | 1530 | \textbf{Case 2}: If $I \neq \{ 0 \}$ let $g(x)$ be a nonzero polynomial of minimal degree in 1531 | $I$ (which exists by well-ordering). If $g(x)$ is constant then $g(x) = \alpha \in F$ and then 1532 | $I = F = \langle \alpha \rangle$ because for any $r \in F$ we have 1533 | \[ 1534 | r = r\alpha^{-1} \alpha \in \langle \alpha \rangle. 1535 | \] 1536 | Suppose then that $g(x)$ is not constant, we claim $I = \langle g(x) \rangle$. Since 1537 | $g(x) \in I$ we have $\langle g(x) \rangle \subseteq I$. We claim $I \subseteq \langle g(x) \rangle$. 1538 | Let $f(x) \in I$. By the \textit{division alogrithm}, we can write 1539 | \[ 1540 | f(x) = q(x)g(x) + r(x) 1541 | \] 1542 | with $0 \leq deg(r(x)) < deg(g(x))$. Since 1543 | \[ 1544 | r(x) = f(x) - q(x)\, g(x) 1545 | \] 1546 | we have $r(x) \in I$ and the fact that $g(x)$ is a nonzero polynomial of minimal degree implies 1547 | that $r(x) = 0$ and so $f(x) = q(x)\, g(x) \Longrightarrow f(x) \in \langle g(x) \rangle$. 1548 | \end{proof} 1549 | 1550 | \newpage 1551 | \section{Unique Factorization Domain} 1552 | 1553 | \begin{definition} 1554 | An integral domain $D$ is a \bred{unique factorization domain} (\bred{UFD} in short) 1555 | if 1556 | \begin{enumerate} 1557 | \item Every nonzero element of $D$ that is not a unit can be written as a product of 1558 | irreducibles of $D$, and 1559 | \item The factorization into irreducibles is unique up to associates and the order in which 1560 | the factors appear. 1561 | \end{enumerate} 1562 | \end{definition} 1563 | 1564 | \begin{theorem} 1565 | Every PID is a UFD. 1566 | \end{theorem} 1567 | \begin{proof} 1568 | Let $R$ be a PID and suppose that a nonzero element $a$ of $R$ can be express in two different ways 1569 | as a product of irreducibles. Suppose 1570 | \[ 1571 | a = p_1 p_2 \cdots p_r \quad \text{and } a = q_1 q_2 \cdots q_s 1572 | \] 1573 | where each $p_i$ and $q_j$ is irreducible in $R$, and $s \geq r$. Then $p_1$ divides the 1574 | product $q_1, q_2, \cdots, q_s $ and so $p_1 | q_j$ for some $j$, as $p_1$ is prime. 1575 | After reordering the $q_j$ we can consider $p_1|q_1$. Then $q_1 = u_1\, p_1$ for some unit $u_1$ of $R$, 1576 | since $q_1$ and $p_1$ are both irreducible. Thus 1577 | \[ 1578 | p_1 p_2 \cdots p_r = u_1 p_1 q_2 \cdots q_s 1579 | \] 1580 | and cancelling $p_1$ on both side 1581 | \[ 1582 | p_2 \cdots p_r = u_1 q_2 \cdots q_s. 1583 | \] 1584 | Continuing this process we reach 1585 | \[ 1586 | 1 = u_1 u_2 \ldots u_r\, q_{r+1} \ldots q_s. 1587 | \] 1588 | Since none of the $q_j$ is a unit, this means that $r=s$ and $p_1 p_2 \cdots p_r$ are associates of 1589 | $q_1 q_2 \cdots q_r$ in some order. Thus $R$ is a unique factorization domain. 1590 | \end{proof} 1591 | 1592 | \begin{theorem} 1593 | Every field is a UFD. 1594 | \end{theorem} 1595 | \begin{proof} 1596 | Every field $F$ is a UFD because it is PID $\Rightarrow$ it is an integral domain and every nonzero 1597 | is a unit, and it contains no prime. 1598 | \end{proof} 1599 | 1600 | \section{Euclidean Domains} 1601 | 1602 | \begin{definition} 1603 | An integral domain $D$ is called a \bred{Euclidean Domain} if there is a function $d$ 1604 | (called the measure) from the nonzero elements of $D$ to the positive integers such that 1605 | 1606 | \begin{enumerate} 1607 | \item $d(a) \leq d(ab)$ for all nonzero $a,b \in D$; and 1608 | \item If $a,b \in D, b \neq 0$, then there exists elements $q, r \in D$ such that 1609 | \[ 1610 | a = bq + r 1611 | \] 1612 | where $r = 0$ or $d(r) < d(b)$. 1613 | \end{enumerate} 1614 | \end{definition} 1615 | 1616 | \begin{example} 1617 | The ring $\mathbb{Z}$ is a Euclidean domain with $d(a) = |a|$. 1618 | \end{example} 1619 | 1620 | \begin{example} 1621 | Let $F$ be a field. Then $F[x]$ is a Euclidean domain with $d(f(x)) = \deg\, f(x)$. 1622 | \end{example} 1623 | 1624 | \begin{example} 1625 | Gaussian integers $\mathbb{Z}[i]$ is a Euclidean domain with 1626 | \[ 1627 | d(a+bi) = a^2 + b^2. 1628 | \] 1629 | \end{example} 1630 | 1631 | \begin{theorem} 1632 | Every Euclidean domain is a PID. 1633 | \end{theorem} 1634 | \begin{proof} 1635 | Let $E$ be a Euclidean domain. Consider an ideal $I$ of $E$. If $I = \{ 0 \}$, then 1636 | $I = \langle 0 \rangle$. 1637 | 1638 | Let $I \neq \{ 0 \}$. Then $N = \{ d(x) \> | \> x \in I, x \neq 0 \}$ is a nonempty 1639 | set of nonnegative integers; and so, by the well-ordering principle 1640 | it has the least element. 1641 | 1642 | Let $a \in I, a \neq 0$ such that $d(a)$ is the least element of $N$. i.e. $d(a) \leq d(x)$ for all 1643 | nonzero $x$ in $I$. We want to show $I = Ea$. Since $a \in I$, it follows that 1644 | \[ 1645 | Ea \subseteq I. 1646 | \] 1647 | Let $b \in I$. Since $E$ is a Euclidean domain, there exist $q, r \in E$ such that 1648 | \[ 1649 | b = aq+r, \quad \text{where } r = 0 \text{ or } d(r) < d(a). 1650 | \] 1651 | If $r \neq 0$, then $r = b - aq \in I$ shows that $d(r) \in N$; and since 1652 | $d(r) < d(a)$, this contradicts the minimality of $d(a)$ in $N$. Therefore, $r = 0$ and so $b = aq \in Ea$. 1653 | Thus $I \subseteq Ea$ and hence $I = Ea$. 1654 | \end{proof} 1655 | 1656 | \begin{remark} 1657 | If $D$ is Euclidean domain, then it is both PID and UFD. 1658 | \end{remark} 1659 | 1660 | \begin{theorem} 1661 | If $D$ is UFD, then $D[x]$ is UFD. 1662 | \end{theorem} 1663 | \begin{proof} 1664 | Let $D$ be an Euclidean domain and $\varnothing \neq I \triangleleft D$. 1665 | 1666 | Among all the nonzero elements of $I$, let $a \in I$ be such that $d(a)$ is a minimum. 1667 | Then 1668 | \[ 1669 | I = \langle a \rangle. 1670 | \] 1671 | For if $b \in I$, there exist some $q,r$ such that 1672 | \[ 1673 | b = aq+r, 1674 | \] 1675 | where $r=0$ or $d(r) < d(a)$. But $r = b-aq \in I$, so $d(r)$ cannot be less than $d(a)$. 1676 | Thus, $r = 0$ and $b = aq \in \langle a \rangle$ 1677 | \end{proof} 1678 | 1679 | \section*{Tutorials} 1680 | 1681 | \begin{mdframed} 1682 | \vspace{-0.25cm} 1683 | \hspace{-0.25cm} 1684 | \begin{Exercise} 1685 | For each of the following, decide whether the indicated operations on the set will form a ring. If a ring is not formed, 1686 | state the reason why this is the case. If a ring is formed state whether the ring is commutative, whether it has unity, 1687 | and whether it is a field. 1688 | \begin{enumerate} 1689 | \item $n\mathbb{Z}$, under the usual addition and multiplication. 1690 | \item $n\mathbb{R}^+$, under the usual addition and multiplication. 1691 | \item $n\mathbb{Z} \times \mathbb{Z}$ with addition and multiplication by components. 1692 | \item $n\mathbb{Z} \times 2\mathbb{Z}$ with addition and multiplication by components. 1693 | \item $\{ a + b\sqrt{5} \> | \> a,b \in \mathbb{Q} \}$ with the usual addition and multiplication. 1694 | \item $\{ ri \> | \> r \in \mathbb{R} \}$ with the usual addition and multiplication where $i^2 = -1$. 1695 | \end{enumerate} 1696 | \end{Exercise} 1697 | 1698 | \vspace{0.752cm} 1699 | \begin{Exercise} 1700 | Let $\alpha = \sqrt[3]{5}$ and $\mathbb{Z}[\alpha] = \{ a + b\alpha + c\alpha^2 \> | \> a,b,c \in \mathbb{Z} \}$. Prove whether 1701 | $\mathbb{Z}[\alpha]$ is a subring of $\mathbb{R}$. 1702 | \end{Exercise} 1703 | 1704 | \vspace{0.752cm} 1705 | \begin{Exercise} 1706 | Let $X$ be some arbitrary set, and $P(X)$ be the set of all subsets of $X$. Define operators on $P(X)$ as follows, 1707 | where $a,b$ in $P(X)$: 1708 | \[ 1709 | a + b = (a \cup b) \setminus (a \cap b) 1710 | \] 1711 | and 1712 | \[ 1713 | ab = a \cap b. 1714 | \] 1715 | Show that $P(X)$ is a commutative ring. 1716 | \end{Exercise} 1717 | 1718 | \vspace{0.752cm} 1719 | \begin{Exercise} 1720 | Let $\mathbb{A}$ be the set $\mathbb{A} = \{ a + bi \> | \> a, b \in \mathbb{Q} \}$ 1721 | where $i^2 = -1$. Here, 1722 | \[ 1723 | (a + bi) + (c + di) = (a + c) + (b+d)i 1724 | \] 1725 | and 1726 | \[ 1727 | (a+bi)(c+di) = (ac-bd) + (ad - bc)i. 1728 | \] 1729 | Show that $\mathbb{A}$ is a field. 1730 | \end{Exercise} 1731 | 1732 | \vspace{0.752cm} 1733 | \begin{Exercise} 1734 | Show that the rings $2\mathbb{Z}$ and $3\mathbb{Z}$ are not isomorphic. 1735 | \end{Exercise} 1736 | 1737 | \vspace{0.752cm} 1738 | \begin{Exercise} 1739 | Show that a ring $R$ has no nonzero nilpotent element if and only if $0$ is the only solution of $x^2 = 0$ in $R$. 1740 | \end{Exercise} 1741 | 1742 | \vspace{0.752cm} 1743 | \begin{Exercise} 1744 | Show that if $R$ is a ring with unity and $N$ is an ideal of $R$ such that $N \neq R$, then $R/N$ is a ring with unity. 1745 | \end{Exercise} 1746 | 1747 | \vspace{0.752cm} 1748 | \begin{Exercise} 1749 | If $F$ is a field, show that $(F \setminus \{0\}, \cdot)$ is a group. 1750 | \end{Exercise} 1751 | 1752 | \vspace{0.752cm} 1753 | \begin{Exercise} 1754 | Show that in a field $F$, the only ideals are $F$ and $\{ 0 \}$. 1755 | \end{Exercise} 1756 | 1757 | \vspace{0.752cm} 1758 | \begin{Exercise} 1759 | Show that each homomorphism from a field to a ring is either one to one or maps everything onto 0. 1760 | \end{Exercise} 1761 | 1762 | \vspace{0.752cm} 1763 | \begin{Exercise} 1764 | Find the characteristic of the following rings: 1765 | \begin{enumerate} 1766 | \item $2 \mathbb{Z}$. 1767 | \item $\mathbb{Z}_3 \times 3 \mathbb{Z}$. 1768 | \item $\mathbb{Z}_5 \times \mathbb{Z}_5$. 1769 | \end{enumerate} 1770 | \end{Exercise} 1771 | 1772 | \vspace{0.752cm} 1773 | \begin{Exercise} 1774 | Show that the matrix $\begin{bmatrix} 1775 | 1 & 2 \\ 2 & 4 1776 | \end{bmatrix}$ is a zero divisor in $M_2(\mathbb{Z})$. 1777 | \end{Exercise} 1778 | 1779 | \vspace{0.752cm} 1780 | \begin{Exercise} 1781 | An element $\mathfrak{a}$ of a ring $R$ is idempotent if $\mathfrak{a}^2 = \mathfrak{a}$. Show that a division ring 1782 | contains exactly two idempotent elements. 1783 | \end{Exercise} 1784 | 1785 | \vspace{0.752cm} 1786 | \begin{Exercise} 1787 | If $A$ and $B$ are ideals of a ring $R$, then the sum $A + B$ of $A$ and $B$ is defined by 1788 | \[ 1789 | A + B = \{ a + b \> | \> a \in A, b \in B \}. 1790 | \] 1791 | \begin{enumerate} 1792 | \item Show that $A + B$ is an ideal of $R$. 1793 | \item Show that $A \subseteq A + B$. 1794 | \end{enumerate} 1795 | \end{Exercise} 1796 | 1797 | \vspace{0.752cm} 1798 | \begin{Exercise} 1799 | If $A$ and $B$ are ideals of a ring $R$, then the product $AB$ of $A$ and $B$ is defined by 1800 | \[ 1801 | AB = \left\{ \sum^n_{i=1} a_ib_i \> \bigg | \> a_i \in A, b_i \in B, n \in \mathbb{Z}^+ \right\}. 1802 | \] 1803 | \begin{enumerate} 1804 | \item Show that $AB$ is an ideal of $R$. 1805 | \item Show that $AB \subseteq (A \cap B)$. 1806 | \end{enumerate} 1807 | \end{Exercise} 1808 | 1809 | \vspace{0.752cm} 1810 | \begin{Exercise} 1811 | Find $q(x)$ and remainder $r(x)$ as described by the division algorithm so that 1812 | \[ 1813 | f(x) = g(x)q(x) + r(x) 1814 | \] 1815 | with $r(x) = 0$ or of degree less than the degree of $g(x)$. 1816 | \begin{enumerate} 1817 | \item $f(x) = x^6 + 3x^5 + 4x^2 - 3x + 2$ and $g(x) = x^2 + 2x - 3$ in $\mathbb{Z}_7[x]$. 1818 | \item $f(x) = x^5 - 2x^4 + 3x - 5$ and $g(x) = 2x + 1$ in $\mathbb{Z}_{11}[x]$. 1819 | \end{enumerate} 1820 | \end{Exercise} 1821 | \end{mdframed} -------------------------------------------------------------------------------- /styles.sty: -------------------------------------------------------------------------------- 1 | \ProvidesPackage{styles} 2 | \usepackage{amsmath, nccmath} 3 | \usepackage{amssymb} 4 | \usepackage{amsbsy} 5 | \usepackage{bbm} 6 | \usepackage{graphicx} 7 | \usepackage[export]{adjustbox} 8 | \newcommand\sbullet[1][.5]{\mathbin{\vcenter{\hbox{\scalebox{#1}{$\bullet$}}}}} 9 | \usepackage{amsthm} 10 | \usepackage{thmtools} 11 | \usepackage{mdframed} 12 | \usepackage{mathtools, cuted} 13 | \usepackage{actuarialsymbol} 14 | \usepackage{tabularx} 15 | \usepackage{tabu} 16 | \usepackage{colortbl} 17 | \usepackage{cancel} 18 | \usepackage[table]{xcolor} 19 | \usepackage{booktabs} 20 | \usepackage{titlesec} 21 | %\usepackage{algorithm2e} 22 | %\usepackage{algorithm} 23 | %\usepackage{algorithmic} 24 | \usepackage{array} 25 | \usepackage[most]{tcolorbox} 26 | \usepackage{cmupint} 27 | \DeclareMathOperator*{\argmax}{arg\,max} 28 | \DeclareMathOperator*{\argmin}{arg\,min} 29 | \DeclareMathOperator{\aut}{Aut} 30 | \DeclareMathOperator{\gal}{Gal} 31 | \DeclareMathOperator{\ord}{ord} 32 | \DeclareMathOperator{\stab}{Stab} 33 | \DeclareMathOperator{\orb}{Orbit} 34 | \DeclareMathOperator{\Ima}{Im} 35 | \newcommand\perm[2][^n]{\prescript{#1\mkern-2.5mu}{}P_{#2}} 36 | \newcommand\comb[2][^n]{\prescript{#1\mkern-0.5mu}{}C_{#2}} 37 | \newcommand{\Cancel}[2][black]{{\color{#1}\cancel{\color{black}#2}}} 38 | 39 | \DeclareRobustCommand{\bxseries}{\fontseries{bx}\selectfont} 40 | \DeclareTextFontCommand{\textbx}{\fontseries{bx}} 41 | \DeclareTextFontCommand{\bred}{\color{fireEngRed}\bfseries} 42 | \DeclareTextFontCommand{\bgreen}{\color{extinctGreen}\bfseries} 43 | \DeclareTextFontCommand{\bgold}{\color{darkGoldrod}\bfseries} 44 | 45 | \definecolor{Acolor}{RGB}{104,129,174} 46 | \definecolor{Bcolor}{RGB}{227,118,59} 47 | \definecolor{Ccolor}{RGB}{169,183,209} 48 | \definecolor{medgrey}{RGB}{150,150,150} 49 | 50 | \definecolor{amaranth}{rgb}{0.9, 0.17, 0.31} 51 | \colorlet{codegreen}{green!20} 52 | \colorlet{codeyellow}{yellow!60} 53 | \colorlet{codered}{red!30} 54 | 55 | \definecolor{myred}{RGB}{240,150,150} 56 | \definecolor{mygreen}{RGB}{150,240,150} 57 | \definecolor{myblue}{RGB}{150,150,255} 58 | \definecolor{mypurple}{RGB}{210,150,255} 59 | \definecolor{myyellow}{RGB}{255,195,0} 60 | \definecolor{mygrey}{RGB}{230,230,230} 61 | \definecolor{lightgrey}{rgb}{.85, .85, .85} 62 | 63 | \colorlet{lightred}{red!30!white} 64 | \colorlet{lightblue}{blue!30!white} 65 | \colorlet{lightyellow}{yellow!30!white} 66 | \colorlet{keylime}{green!10!white} 67 | \colorlet{darkgreen}{green!50!black} 68 | \colorlet{midgreen}{green!70!black} 69 | \colorlet{lightgreen}{darkgreen!30!white} 70 | \colorlet{lightpurple}{violet!60!white} 71 | \colorlet{darkgrey}{black!70} 72 | \colorlet{midgrey}{black!50} 73 | \definecolor{Red}{rgb}{.9,0,0} 74 | \definecolor{Blue}{rgb}{0,0,.9} 75 | \definecolor{Green}{rgb}{0,.9,0} 76 | \definecolor{Gold}{rgb}{.83,.67,.08} 77 | 78 | \definecolor{conj-purple}{HTML}{E4B7FF} 79 | \definecolor{conj-orange}{HTML}{FFD4B7} 80 | 81 | \definecolor{grey}{rgb}{.7, .7, .7} 82 | \colorlet{lred}{red!50!white} 83 | \definecolor{lblue}{rgb}{0.60, 0.75, 1} 84 | \colorlet{lgreen}{green!40!white} 85 | \colorlet{lyellow}{yellow!30!white} 86 | %\definecolor{lpurple}{rgb}{0.6 0.35 0.8} 87 | \definecolor{lpurple}{rgb}{0.7 0.45 0.9} 88 | \colorlet{lorange}{orange!50!white} 89 | \definecolor{lightgrey}{rgb}{.85, .85, .85} 90 | \definecolor{faded}{rgb}{.75,.75,.75} 91 | \definecolor{midgrey}{rgb}{.5,.5,.5} 92 | %\definecolor{midgrey}{rgb}{.35, .35, .35} 93 | %\definecolor{amethyst}{rgb}{0.6 0.4 0.8} 94 | \definecolor{amethyst}{rgb}{0.54 0.17 0.89} 95 | \definecolor{amber}{rgb}{1.0 0.49 0.0} 96 | \definecolor{turq}{rgb}{0.12 0.69 0.71} 97 | 98 | \newcommand{\olsi}[1]{\,\overline{\!{#1}}} % overline short italic 99 | %\usepackage{mathrsfs} 100 | %\usepackage{titletoc} 101 | 102 | % pseudocode formatting 103 | \usepackage{setspace} 104 | \usepackage[linesnumbered,ruled,vlined]{algorithm2e} 105 | \renewcommand{\KwSty}[1]{\textnormal{\textcolor{amaranth!90!black}{\bfseries #1}}\unskip} 106 | \renewcommand{\ArgSty}[1]{\textnormal{\ttfamily #1}\unskip} 107 | \SetKwComment{Comment}{\color{codegreen!80!black!190}// }{} 108 | \renewcommand{\CommentSty}[1]{\textnormal{\ttfamily\color{codegreen!80!black!190}#1}\unskip} 109 | \newcommand{\var}{\texttt} 110 | \newcommand{\FuncCall}[2]{\texttt{\bfseries #1(#2)}} 111 | \SetKwProg{Function}{function}{}{} 112 | \SetKw{Continue}{continue} 113 | \renewcommand{\ProgSty}[1]{\texttt{\bfseries #1}} 114 | \DontPrintSemicolon 115 | \SetAlFnt{\small} 116 | \SetAlgorithmName{Pseudocode}{algorithmautorefname} 117 | 118 | % pseudocode highlighting 119 | \usepackage{tikz} 120 | \usetikzlibrary{fit,calc} 121 | % two slightly different boxit commands, to ensure the inner boxes end at the same spot 122 | \newcommand{\boxit}[2]{ 123 | \tikz[remember picture,overlay] \node (A) {};\ignorespaces 124 | \tikz[remember picture,overlay]{\node[yshift=3pt,fill=#1,opacity=.25,fit={($(A)+(0,0.15\baselineskip)$)($(A)+(.9\linewidth,-{#2}\baselineskip - 0.25\baselineskip)$)}] {};}\ignorespaces 125 | } 126 | \newcommand{\boxitt}[2]{ 127 | \tikz[remember picture,overlay] \node (A) {};\ignorespaces 128 | \tikz[remember picture,overlay]{\node[yshift=3pt,fill=#1,opacity=.25,fit={($(A)+(0,0.15\baselineskip)$)($(A)+(.858\linewidth,-{#2}\baselineskip - 0.25\baselineskip)$)}] {};}\ignorespaces 129 | } 130 | 131 | \usepackage{algpseudocode} 132 | 133 | \usepackage{tikz} 134 | \usepackage{tikz-3dplot} 135 | \usetikzlibrary{arrows} 136 | \tikzstyle{every picture}+=[remember picture] 137 | \usepackage{pgfplots} 138 | \pgfplotsset{compat=1.16} 139 | \usetikzlibrary{arrows,shapes,positioning,calc,chains,matrix,positioning,scopes} 140 | \usetikzlibrary{positioning, arrows.meta, graphs} 141 | \usetikzlibrary{decorations.markings} 142 | \usetikzlibrary{arrows.meta, positioning, bending} 143 | \makeatletter 144 | \usepackage{tikz-cd} 145 | 146 | \usepackage{enumerate} 147 | \usepackage[inline]{enumitem} 148 | \usepackage{setspace} 149 | \singlespacing 150 | \setlength{\parindent}{2em} 151 | \setlength{\parskip}{0.5em} 152 | 153 | \usepackage{fancybox} 154 | \usepackage{framed} 155 | \usepackage{ragged2e} 156 | 157 | %\theoremstyle{definition} 158 | %\newtheorem{definition}{Definition}[section] 159 | % \newtheorem{theorem}{Theorem}[section] 160 | % \newtheorem{axiom}{Axiom}[section] 161 | \newtheorem*{remark}{Remark} 162 | %\newmdtheoremenv{definition}{Definition}[section] 163 | 164 | \newcommand\norm[1]{\left\lVert#1\right\rVert} 165 | 166 | \newtheoremstyle{break}{0pt}{0pt}{\normalfont}{0pt}{\bfseries}{}{\newline}{\thmname{#1} \thmnumber{#2} \quad \thmnote{\textbf#3}} 167 | \declaretheorem[numberwithin=chapter,style=break,name=Theorem]{theorem} 168 | 169 | \tcolorboxenvironment{theorem}{ 170 | breakable, 171 | colback=green!5, 172 | colframe={green!55!black}, 173 | } 174 | 175 | \newtheoremstyle{break}{0pt}{0pt}{\normalfont}{0pt}{\bfseries}{}{\newline}{\thmname{#1} \thmnumber{#2} \quad \thmnote{\textbf#3}} 176 | \declaretheorem[numberwithin=chapter,style=break,name=Definition]{definition} 177 | 178 | \tcolorboxenvironment{definition}{ 179 | breakable, 180 | colback=red!5, 181 | colframe={red!55!black}, 182 | } 183 | 184 | \newtheoremstyle{break}{0pt}{0pt}{\normalfont}{0pt}{\bfseries}{}{\newline}{\thmname{#1} \thmnumber{#2} \quad \thmnote{\textbf#3}} 185 | \declaretheorem[numberwithin=chapter,style=break,name=Lemma]{lemma} 186 | 187 | \tcolorboxenvironment{lemma}{ 188 | breakable, 189 | colback=orange!5, 190 | colframe={orange!55!black}, 191 | } 192 | 193 | \newtheoremstyle{break}{0pt}{0pt}{\normalfont}{0pt}{\bfseries}{}{\newline}{\thmname{#1} \thmnumber{#2} \quad \thmnote{\textbf#3}} 194 | \declaretheorem[numberwithin=chapter,style=break,name=Axiom]{axiom} 195 | 196 | \tcolorboxenvironment{axiom}{ 197 | breakable, 198 | colback=orange!5, 199 | colframe={orange!55!black}, 200 | } 201 | 202 | \newtheoremstyle{break}{0pt}{0pt}{\normalfont}{0pt}{\bfseries}{}{\newline}{\thmname{#1} \thmnumber{#2} \quad \thmnote{\textbf#3}} 203 | \declaretheorem[numberwithin=chapter,style=break,name=Corollary]{corollary} 204 | 205 | \tcolorboxenvironment{corollary}{ 206 | breakable, 207 | colback=blue!5, 208 | colframe={blue!55!black}, 209 | } 210 | 211 | \usepackage{thmtools} 212 | 213 | \declaretheoremstyle[ 214 | spaceabove=20pt, spacebelow=6pt, 215 | headfont=\normalfont\bfseries, 216 | notefont=\mdseries, notebraces={(}{)}, 217 | bodyfont=\normalfont, 218 | postheadspace=1em, 219 | numberwithin=section 220 | ]{exstyle} 221 | \declaretheoremstyle[ 222 | spaceabove=6pt, spacebelow=20pt, 223 | headfont=\normalfont\bfseries, 224 | notefont=\mdseries, notebraces={(}{)}, 225 | bodyfont=\normalfont, 226 | postheadspace=1em, 227 | headpunct={}, 228 | qed=$\blacktriangleleft$, 229 | numbered=no 230 | ]{solstyle} 231 | \declaretheorem[style=exstyle]{example} 232 | \declaretheorem[style=solstyle]{solution} 233 | 234 | %\usepackage[most]{tcolorbox} 235 | %\newtcbtheorem{policy}{Policy Type}{enhanced,frame empty,interior empty,colframe=ForestGreen!50!white,coltitle=ForestGreen!50!black,fonttitle=\bfseries,colbacktitle=ForestGreen!15!white,borderline={0.5mm}{0mm}{ForestGreen!15!white},borderline={0.5mm}{0mm}{ForestGreen!50!white,dashed},attach boxed title to top center={yshift=-2mm},boxed title style={boxrule=0.4pt},varwidth boxed title}{th} 236 | %\newtcbtheorem{policy}{Policy Type}{fonttitle=\bfseries\upshape,fontupper=\itshape,colframe=blue!0!black,colback=white!99,colbacktitle=blue!20!white,coltitle=blue!75!black}{th} 237 | 238 | \usepackage[lastexercise,answerdelayed]{exercise} 239 | \counterwithin{Exercise}{section} 240 | \counterwithin{Answer}{section} 241 | \newcommand{\AtBeginExerciseHeader}{\hspace{-0.75cm}} 242 | \renewcommand{\ExerciseHeader}{\AtBeginExerciseHeader \textbf{\ExerciseHeaderDifficulty \ExerciseName~\ExerciseHeaderNB} \quad } 243 | %\renewcommand{\ExerciseHeader}{\textbf{\ExerciseName~\ExerciseHeaderNB}\quad} 244 | \renewcommand{\AnswerHeader}{\AtBeginExerciseHeader \textbf{\AnswerName~\ExerciseHeaderNB}\smallskip\newline} 245 | 246 | \titleformat{\chapter}[display] 247 | {\sffamily\bfseries\filright} 248 | {\hrule height 0pt \Huge\color{fireEngRed}\thechapter\hrule} 249 | {-\baselineskip} 250 | {\Huge\strut} 251 | \titlespacing*{\chapter} 252 | {0pt} 253 | {-\topskip} 254 | {2\normalbaselineskip} 255 | 256 | %\titleformat{\chapter} 257 | % {\normalfont\LARGE\bfseries}{\thechapter}{1em}{} 258 | %\titlespacing*{\chapter}{0pt}{3.5ex plus 1ex minus .2ex}{2.3ex plus .2ex} 259 | 260 | % \usepackage{txfonts} 261 | \usepackage{tgpagella} 262 | %\usepackage{mathptmx, newtxmath} 263 | %\usepackage{libertine} 264 | %\usepackage[T1]{fontenc} 265 | %\usepackage[libertine]{newtxmath} 266 | 267 | % \usepackage[scaled=0.85]{DejaVuSansMono} 268 | \usepackage{minted} 269 | 270 | %% Section/Subsection Font Family 271 | \titleformat{\section} 272 | {\fontfamily{qhv}\selectfont\bfseries\Large} 273 | {\thesection} 274 | {1em} 275 | {} 276 | \titleformat{\subsection} 277 | {\fontfamily{qhv}\selectfont\bfseries\large} 278 | {\thesubsection} 279 | {1em} 280 | {} 281 | 282 | \usepackage[most]{tcolorbox} 283 | \tcbuselibrary{minted,skins} 284 | 285 | \newtcblisting{pycode}[1][]{ 286 | listing engine=minted, 287 | colback=pybg, 288 | colframe=white, 289 | listing only, 290 | minted style=manni, 291 | minted language=python, 292 | minted options={linenos=true,numbersep=3mm,texcl=true,baselinestretch=1.1, #1} 293 | } 294 | 295 | \newtcblisting{rcode}{ 296 | listing engine=minted, 297 | colback=codeblockbg, 298 | colframe=black!50, 299 | listing only, 300 | minted style=pastie, 301 | minted language=r, 302 | minted options={linenos=true,numbersep=3mm,texcl=true,baselinestretch=1.1}, 303 | left=5mm,enhanced, 304 | overlay={\begin{tcbclipinterior}\fill[black!25] (frame.south west) 305 | rectangle ([xshift=5mm]frame.north west);\end{tcbclipinterior}} 306 | } 307 | 308 | \definecolor{codeblockbg}{rgb}{0.97,0.97,0.97} 309 | 310 | \definecolor{pybg}{rgb}{0.98,0.98,0.98} 311 | %\newminted{pycon}{bgcolor=pybg, linenos=true, tabsize=4} 312 | %\usemintedstyle{vs} 313 | --------------------------------------------------------------------------------