3-commutative-rings.tex 29.4 KB
 rrueger committed Jun 02, 2019 1 2 3 4 \chapter{Commutative Rings} \section{Definition of Rings and Domains}  rrueger committed Jun 09, 2019 5 \begin{definition}[Ring]  rrueger committed Jun 02, 2019 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20  A \textbf{ring} $R$ is a set with two binary operations $R \times R \to R$: addition $(a,b) \mapsto a+b$ and multiplication $(a,b) \mapsto ab$, such that \begin{enumerate} \item $R$ is an abelian group under addition; that is, \begin{enumerate} \item $a + (b + c) = (a + b) + c$ for all $a,b,c \in R$; \item there is an element $0 \in R$ with $0 + a = a$ for all $a \in R;$ \item for each $a \in R$, there is $a' \in R$ with $a' + a = 0$; \item $a + b = b + a$. \end{enumerate} \item \textbf{Associativity:} $a(bc) = (ab)c$ for every $a,b,c \in R$;  heldf committed Jun 19, 2019 21 22  \item there is $1 \in R$ with $1 a=a=a 1$ for every $a \in R$;  rrueger committed Jun 02, 2019 23 24 25 26 27 28 29 30 31 32 33 34  \item\textbf{Distributivity:} $a(b+c) = ab+ ac$ and $(b+c)a = ba + ca$ for every $a,b,c \in R$. \end{enumerate} \end{definition} \begin{proposition}[Elementary properties] Let $R$ be a ring. \begin{enumerate}  heldf committed Jun 19, 2019 35  \item $0 \cdot a = 0 = a \cdot 0$ for every $a \in R$.  rrueger committed Jun 02, 2019 36 37  \item If $1 = 0$ then $R$ consists of the single element $0$ (\textbf{zero  38  ring}, or \textbf{trivial ring}).  rrueger committed Jun 02, 2019 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54  \item If $-a$ is the additive inverse of $a$, then $(-1)(-a) = a = (-a)(-1)$, in particular $(-1)(-1) = 1$. \item $(-1)a = -a = a(-1)$ for all $a \in R$. \item If $n \in \N$ and $n \cdot 1 = 0$, then $n \cdot a = 0$ for all $a \in R$, where \begin{align*} n \cdot a = \underbrace{a + a \ldots a}_{n \text{~summands}} \end{align*} \end{enumerate} \end{proposition}  rrueger committed Jun 09, 2019 55 \begin{definition}[Subring, Proper subring]  rrueger committed Jun 02, 2019 56 57 58 59 60 61 62 63  A subset $S$ of a ring $R$ is a \textbf{subring} of $R$ if % \begin{enumerate} \item $1 \in S$, \item if $a,b \in S$, then $a-b \in S$, \item if $a,b \in S$, then $ab \in S$. \end{enumerate} %  zengw committed Jun 21, 2019 64  A \textbf{proper subring} $S \subseteq R$ is a subring of $R$ with $S \neq  rrueger committed Jun 02, 2019 65 66 67 68  R$. \end{definition}  rrueger committed Jun 09, 2019 69 \begin{definition}[Domain, Integral domain]  70 71 72  A \textbf{domain} (or integral domain) is a nonzero commutative ring $R$ that satisfies the \textbf{cancellation law}: for all $a,b,c \in R$, if $ca = cb$ and $c \neq 0$, it follows $a = b$.  rrueger committed Jun 02, 2019 73 74 75 76 77 78 79 80 81 82 83 84 85 86 \end{definition} \begin{proposition} A nonzero commutative ring $R$ is a domain if and only if the product of two nonzero elements of $R$ is nonzero. \end{proposition} \begin{proposition} The commutative ring $\Z_m$ is a domain if and only if $m$ is prime. \end{proposition} \noindent \textbf{\textsc{From here on, all rings are assumed to be commutative.}}  rrueger committed Jun 09, 2019 87 \begin{definition}[Divisior]  rrueger committed Jun 02, 2019 88 89 90 91 92 93 94 95 96 97  Let $a$ and $b$ be elements of a commutative ring $R$. Then $a$ \textbf{divides} $b$ in $\mathbf{R}$, denoted by % \begin{align*} a \mid b, \end{align*} % if there exists an element $c \in R$ with $b = ca$. \end{definition}  rrueger committed Jun 09, 2019 98 \begin{definition}[Unit]  rrueger committed Jun 02, 2019 99 100 101 102 103 104 105 106 107 108 109 110 111  An element $u$ in a commutative ring $R$ is called a \textbf{unit} if $u \mid 1$ in $R$, that is, if there exists $v \in R$ with $uv = 1$. \end{definition} \begin{proposition} If $a$ is an integer, then $[a]$ is a unit in $\Z_m$ if and only if $a$ and $m$ are relatively prime. In fact, if $sa + tm = 1$, then $[a]^{-1} = [s]$. \end{proposition} \begin{corollary} If $p$ is prime, then every nonzero $[a] \in \Z_{p}$ is a unit. \end{corollary}  rrueger committed Jun 09, 2019 112 \begin{definition}[The group of units]  rrueger committed Jun 02, 2019 113 114 115 116 117 118 119 120 121  If $R$ is a nonzero commutative ring, then the \textbf{group of units} of $R$ is % \begin{align*} U(R)= \set{\text{all units in~}R} \end{align*} % \end{definition}  rrueger committed Jun 09, 2019 122 \begin{definition}[Field]  rrueger committed Jun 09, 2019 123 124  A \textbf{field} $F$ is a nontrivial commutative ring in which every nonzero element $a$ is a unit; that is, there is $a^{-1} \in F$ with $a^{-1} a = 1$.  rrueger committed Jun 02, 2019 125 126 127 128 129 130 131 132 133 134 \end{definition} \begin{proposition} The commutative ring $\Z_m$ is a field if and only if $m$ is prime. \end{proposition} \begin{proposition} Every field $F$ is a domain. \end{proposition}  rrueger committed Jun 23, 2019 135 136 137 138 \begin{proposition} Every finite domain is a field. \end{proposition}  rrueger committed Jun 02, 2019 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 \begin{theorem}[Existence of the fraction field] \label{thm:existenceoffractionfield} If $R$ is a domain, then there is a field containing $R$ as a subring. Moreover, such a field $F$ can be chosen so that, for each $f \in F$ there are $a,b \in R$ with $b \neq 0$ and $f = ab^{-1}$. \end{theorem} \begin{idea} To prove this theorem, we constructed the following relation on $R \times R^{\times}$ % \begin{align*} \forall a,c \in R, b,d \in R^{\times}:\, (a,b) \equiv (c,d) \overset{\text{Def.}}{\iff} ad = bc \end{align*} % It is easy to see, that this relation is indeed an equivalence relation. We define $F$ as the set of all equivalence classes $[a,b]$ and we equip $F$ with the following addition and multiplication % \begin{align*} [a,b] + [c,d] = [ad + bc, bd] \text{~and~} [a,b][c,d] = [ac, bd] \end{align*} % It is easy to verify, that these operations are well-defined and that $F$ is a field. Now $R' = \{[a,1]:a \in R\}$ is a subring and we identify $a \in R$ with $[a,1] \in R'$. \end{idea}  rrueger committed Jun 09, 2019 168 \begin{definition}[Fraction field]  rrueger committed Jun 02, 2019 169 170 171 172 173  The field $F$ constructed from $R$ in Theorem~\ref{thm:existenceoffractionfield} is called the \textbf{fraction field} of $R$, denoted by $\mathrm{Frac}(R)$. \end{definition}  rrueger committed Jun 24, 2019 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 \begin{theorem}[Universal property of the fraction field] For $R$ an integral domain, $K$ a field and $\varphi \colon R \to K$ an injective ring homomorphism, there exists a unique $\overline{\varphi} \colon F = \Frac (R) \to K$ such that the following diagram commutes % \begin{figure}[H] \centering \begin{tikzpicture} \matrix (m)[matrix of math nodes,row sep=3em,column sep=2em]{% R & & F \\ & K & \\ }; \path[-stealth] (m-1-1) edge node [above] {$\imath$} (m-1-3) (m-1-1) edge node [below left] {$\varphi$} (m-2-2) (m-1-3) edge node [below right] {$\overline{\varphi}$} (m-2-2); \end{tikzpicture} \end{figure} % \noindent i.e\ that $\varphi = \imath \circ \overline{\varphi}$. Here $\imath \colon R \to \Frac (R)$ is the natural inclusion $r \mapsto \frac{r}{1}$. The unique map is is given by % \begin{align*} \overline{\varphi} \left( \frac{r}{s} \right) := \varphi (r) {\varphi (s)}^{-1}. \end{align*} \end{theorem}  rrueger committed Jun 09, 2019 205 \begin{definition}[Subfield]  rrueger committed Jun 02, 2019 206 207 208 209 210 211 212  A \textbf{subfield} of a field $K$ is a subring $k$ of $K$ that is also a field. \end{definition} \section{Polynomials}  rrueger committed Jun 09, 2019 213 \begin{definition}[Formal power series]  rrueger committed Jun 02, 2019 214 215 216 217 218 219 220 221 222 223  If $R$ is a commutative ring, then a \textbf{formal power series} over $R$ is a sequence of elements $s_i \in R$ for all $i \geq 0$, called the \textbf{coefficients} of $\sigma$: % \begin{align*} \sigma = (s_0, s_1, s_2, \ldots, s_i, \ldots). \end{align*} % \end{definition}  rrueger committed Jun 09, 2019 224 \begin{definition}[Polynomial, Leading Coefficient, Degree]  rrueger committed Jun 02, 2019 225 226 227 228 229 230 231 232  A \textbf{polynomial} over a commutative ring $R$ is a formal power series $\sigma=(s_0,s_1, \ldots, s_i, \ldots)$ over $R$ for which there exists some integer $n \geq 0$ with $s_i = 0$ for all $i > n$; that is, % \begin{align*} \sigma =(s_0, s_1, \ldots, s_n, 0,0,\ldots). \end{align*} %  heldf committed Jun 19, 2019 233  if $s_n \neq 0$ then $s_n$ is called the \textbf{leading coefficient} and $n$ is the  rrueger committed Jun 02, 2019 234 235 236  \textbf{degree}: $n = \deg(\sigma)$. If $s_n = 1$, we call $\sigma$ monic. \end{definition}  rrueger committed Jun 09, 2019 237 \begin{remark}[Definition of degree of the zero polynomial]  heldf committed Jun 19, 2019 238  By this definition, the degree of the zero polynomial $(0, \ldots)$ is not defined.  rrueger committed Jun 09, 2019 239 240 241 242 243 244 245  Some authors choose to define it as being $-\infty$. This makes sense later in the context of the degree formulae (Lemma~\ref{commutative-rings:degree-formulae}), as we will not have to require the polynomials to be nonzero for the product or sum formulae. Ultimately this is down to convention. \end{remark}  rrueger committed Jun 09, 2019 246 \begin{remark}[Notation]  rrueger committed Jun 02, 2019 247 248 249 250 251 252  If $R$ is a commutative ring, then $R[[x]]$ denotes the set of all formal power series over $R$, and $R[x] \subseteq R[[x]]$ denotes the set of all polynomials over $R$. \end{remark} \begin{proposition}  rrueger committed Jun 09, 2019 253 254 255  If $R$ is a commutative ring, then $R[[x]]$ is a commutative ring that contains $R[x]$ and $R'$ as subrings where $R' = \set{(r,0,0,\ldots): r \in R} \subset R[x]$.  rrueger committed Jun 02, 2019 256 257 \end{proposition}  rrueger committed Jun 09, 2019 258 \begin{lemma}\label{commutative-rings:degree-formulae}  rrueger committed Jun 09, 2019 259 260 261 262 263 264 265 266 267 268 269 270  Let $R$ be a commutative ring and let $p,q \in R[x]$ be nonzero polynomials. % \begin{enumerate} \item Either $pq = 0$ or $\deg(pq) \leq \deg(p) + deg(q)$. \item If $R$ is a domain, then $pq \neq 0$ and \begin{align*} \deg(pq) = deg(p) + deg(q). \end{align*} \item If $R$ is a domain, $p,q \neq 0$ and $q \mid p$ in $R[x]$, then $\deg(q) \leq \deg(p)$. \item If $R$ is a domain, then $R[x]$ is a domain. \end{enumerate}  rrueger committed Jun 02, 2019 271 272 \end{lemma}  rrueger committed Jun 09, 2019 273 \begin{definition}[Interdeterminate]  rrueger committed Jun 09, 2019 274 275 276 277 278  The \textbf{indeterminate} $x \in R[x]$ is % \begin{align*} x = (0,1,0,0, \ldots). \end{align*}  rrueger committed Jun 02, 2019 279 280 281 \end{definition} \begin{lemma}  rrueger committed Jun 09, 2019 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297  The indeterminate $x$ in $R[x]$ has the following properties. % \begin{enumerate} \item If $p = (s_0,s_1,\ldots)$, then \begin{align*} x p = (0, s_0, s_1, \ldots); \end{align*} that is, multiplying by $x$ shifts each coefficient one step to the right. \item If $n \geq 0$, then $x^n$ is the polynomial having $0$ everywhere except for $1$ in the nth coordinate. \item If $r \in R$, then \begin{align*} (r,0,0,\ldots)(s_0, s_1, \ldots, s_j, \ldots) = (r s_0, r s_1, \ldots, r s_j, \ldots). \end{align*} \end{enumerate}  rrueger committed Jun 02, 2019 298 299 300 \end{lemma} \begin{corollary}  rrueger committed Jun 09, 2019 301 302  Formal power series $s_0 + s_1 x + s_2 x^2 + \ldots$ and $t_0 + t_1 x + t_2 x^2 + \ldots$ in $R[[x]]$ are equal if and only if $s_i = t_i$ for all $i$.  rrueger committed Jun 02, 2019 303 304 \end{corollary}  rrueger committed Jun 09, 2019 305 \begin{definition}[Field of rational functions]  rrueger committed Jun 09, 2019 306 307  Let $k$ be a field. The fraction field $\Frac(k[x])$ of $k[x]$, denoted by $k(x)$, is called the \textbf{field of rational functions} over $k$.  rrueger committed Jun 02, 2019 308 309 310 \end{definition} \begin{proposition}  rrueger committed Jun 09, 2019 311 312  If $k$ is a field, then the elements of $k(x)$ have the form $f(x) / g(x)$, where $f(x), g(x) \in k[x]$ and $g(x) \neq 0$.  rrueger committed Jun 02, 2019 313 314 315 \end{proposition} \begin{proposition}  rrueger committed Jun 09, 2019 316 317  If $p$ is prime, then the field of rational functions $\mathbb{F}_{p}(x)$ is an infinite field containing $\mathbb{F}_p$ as a subfield.  rrueger committed Jun 02, 2019 318 319 320 321 322 \end{proposition} \section{Homomorphisms}  rrueger committed Jun 09, 2019 323 \begin{definition}[Ring homomorphism, Isomorphism]  rrueger committed Jun 09, 2019 324 325 326 327 328 329 330  If $A$ and $R$ are (not necessarily commutative) rings, a \textbf{ring homomorphism} is a function $\varphi: A \to R$ such that \begin{enumerate} \item $\varphi(1) = 1$ \item $\varphi(a + a') = \varphi(a) + \varphi(a')$ for all $a,a' \in A$, \item $\varphi(aa') = \varphi(a) \varphi(a')$ for all $a,a' \in A$. \end{enumerate}  rrueger committed Jun 09, 2019 331  A ring homomorphism that is also a bijection is called an  rrueger committed Jun 09, 2019 332  \textbf{isomorphism}.  rrueger committed Jun 02, 2019 333 334 \end{definition}  rrueger committed Jun 25, 2019 335 336 337 \begin{theorem}[Universal property of Rings] Let $R$ and $S$ be commutative rings, and let $\varphi \colon R \to S$ be a homomorphism. For fixed $s_1, \ldots, s_n \in S$, there exists a unique  rrueger committed Jun 09, 2019 338 339 340  homomorphism % \begin{align*}  rrueger committed Jun 25, 2019 341 342  \Phi \colon R[x_1, \ldots, x_n] & \to S \\ x_i & \mapsto s_i  rrueger committed Jun 09, 2019 343 344  \end{align*} %  rrueger committed Jun 25, 2019 345 346  with $\Phi(r) = \varphi(r)$ for all $r \in R$. That is, $\Phi$ uniquely extends $\varphi$ from $R$ to $R[x_1, \ldots, x_n]$, sending $x_i$ to $s_i$.  rrueger committed Jun 02, 2019 347 348 \end{theorem}  rrueger committed Jun 09, 2019 349 \begin{definition}[Evaluation map]  rrueger committed Jun 09, 2019 350 351  If $R$ is a commutative ring and $a \in R$, then \textbf{evaluation at} $a$ is the function $e_a: R[x] \to R$, defined by $e_a(f(x)) = f(a)$.  rrueger committed Jun 02, 2019 352 353 354 \end{definition} \begin{corollary}  rrueger committed Jun 09, 2019 355 356  If $R$ is a commutative ring, then evaluation $e_a: R[x] \to R$ is a homomorphism for every $a \in R$.  rrueger committed Jun 02, 2019 357 358 359 \end{corollary} \begin{corollary}  rrueger committed Jun 09, 2019 360 361 362 363 364 365 366 367 368  If $R$ and $S$ are commutative rings and $\varphi: R \to S$ is a homomorphism, then there is a homomorphism $\varphi_{*}: R[x] \to S[x]$ given by % \begin{align*} \varphi_{*}: r_0 + r_1 x + r_2x^2 + \ldots \mapsto \varphi(r_0) + \varphi(r_1) x + \varphi(r_2)x^2 + \ldots \end{align*} % Moreover, $\varphi_{*}$ is an isomorphism if $\varphi$ is.  rrueger committed Jun 02, 2019 369 370 371 \end{corollary} \begin{proposition}  rrueger committed Jun 09, 2019 372 373 374 375 376 377 378 379  Let $\varphi: A \to R$ be a homomorphism. \begin{enumerate} \item $\varphi(a^n) = \varphi(a)^n$ for all $n \geq 0$ and all $a \in A$. \item If $a \in A$ is a unit, then $\varphi(a)$ is a unit and $\varphi(a^{-1}) = \varphi(a)^{-1}$ and so $\varphi(U(A)) \subseteq U(R)$, where $U(A)$ is the group of units of $A$. Moreover, if $\varphi$ is an isomorphism, then $U(A) \cong U(R)$ (as groups). \end{enumerate}  rrueger committed Jun 02, 2019 380 381 \end{proposition}  rrueger committed Jun 09, 2019 382 \begin{definition}[Kernel, Image]  rrueger committed Jun 09, 2019 383 384 385 386 387 388 389 390 391 392 393  If $\varphi: A \to R$ is a homomorphism, then its \textbf{kernel} is % \begin{align*} \ker \varphi = \set{a \in A \text{~with~} \varphi(a) = 0} \end{align*} % and its \textbf{image} is % \begin{align*} \im \varphi = \set{r \in R: r = \varphi(a) \text{~for some~} a \in R}. \end{align*}  rrueger committed Jun 02, 2019 394 395 \end{definition}  rrueger committed Jun 09, 2019 396 \begin{definition}[Ideal, Proper ideal]  rrueger committed Jun 09, 2019 397 398 399 400 401 402 403 404 405 406  An \textbf{ideal} in a commutative ring $R$ is a subset $I$ of $R$ such that % \begin{enumerate} \item $0 \in I$, \item if $a,b \in I$, then $a + b \in I$, \item if $a \in I$ and $r \in R$, then $ra \in I$. \end{enumerate} % The ring $R$ itself and $(0)$, the subset containing only $0$, are always ideals. An ideal $I \neq R$ is called a \textbf{proper ideal}.  rrueger committed Jun 02, 2019 407 408 409 \end{definition} \begin{remark} \mbox{}\\  rrueger committed Jun 09, 2019 410 411 412 413 414  \begin{enumerate} \item If $I \subseteq R$ contains $1$, then $I = R$. \item If $R$ is a field, the only ideals in $R$ are $(0)$ and $R$ itself, i.e.\ the only proper ideal in $R$ is $(0)$. \end{enumerate}  rrueger committed Jun 02, 2019 415 416 417 \end{remark} \begin{proposition}  rrueger committed Jun 09, 2019 418 419  A homomorphism $\varphi: A \to R$ is an injection if and only if $\ker \varphi = (0)$.  rrueger committed Jun 02, 2019 420 421 422 \end{proposition} \begin{corollary}  rrueger committed Jun 09, 2019 423 424  If $k$ is a field and $\varphi: k \to R$ is a homomorphism, where $R$ is not the zero ring, then $\varphi$ is an injection.  rrueger committed Jun 02, 2019 425 426 \end{corollary}  rrueger committed Jun 09, 2019 427 \begin{definition}[Principal ideal]  rrueger committed Jun 09, 2019 428 429 430 431 432 433 434 435 436  If $b_1, b_2, \ldots , b_n$ lie in $R$, then the set of all linear combinations % \begin{align*} I = \set{r_1 b_1 + \ldots r_n b_n: r_i \in R \forall i} \end{align*} % is an ideal in $R$. We write $(b_1, b_2, \ldots, b_n)$. The ideal $(b)$ is called \textbf{principal ideal} generated by $b$.  rrueger committed Jun 02, 2019 437 438 439 \end{definition} \begin{theorem}  rrueger committed Jun 09, 2019 440 441  Every ideal $I$ in $\Z$ is a principal ideal; that is, there is $d \in \Z$ with $I = (d)$.  rrueger committed Jun 02, 2019 442 443 444 \end{theorem} \begin{proposition}  rrueger committed Jun 09, 2019 445 446  Let $R$ be a commutative ring and let $a,b \in R$. If $a \mid b$ and $b \mid a$, then $(a) = (b)$.  rrueger committed Jun 02, 2019 447 448 \end{proposition}  rrueger committed Jun 09, 2019 449 \begin{definition}[Associates]  rrueger committed Jun 09, 2019 450 451  Elements $a$ and $b$ in a commutative ring $R$ are \textbf{associates} if there exists a unit $u \in R$ with $b = ua$.  rrueger committed Jun 02, 2019 452 453 454 \end{definition} \begin{proposition}  rrueger committed Jun 09, 2019 455 456 457 458 459 460 461  Let $R$ be a domain and let $a,b \in R$. % \begin{enumerate} \item $a \mid b$ and $b \mid a$ if and only if $a$ and $b$ are associates. \item The principal ideals $(a)$ and $(b)$ are equal if and only if $a$ and $b$ are associates. \end{enumerate}  rrueger committed Jun 02, 2019 462 463 \end{proposition}  rrueger committed Jun 09, 2019 464   rrueger committed Jun 02, 2019 465 466 467 \section{Quotient Rings} Quotient rings are a generalization of the commutative rings $\Z_m$.  rrueger committed Jun 09, 2019 468 \begin{definition}[Coset]  rrueger committed Jun 09, 2019 469 470 471 472 473 474 475 476 477 478 479 480  Let $I$ be an ideal in a commutative ring $R$. If $a \in R$, then the \textbf{coset} $a + I$ is the subset % \begin{align*} a + I = \{a + i: i \in I\}. \end{align*} % The coset $a + I$ is often called $a \mod I$. % \begin{align*} R/I = \{a + I: a \in R\}. \end{align*}  rrueger committed Jun 02, 2019 481 482 483 \end{definition} \begin{remark}  rrueger committed Jun 09, 2019 484 485 486 487 488 489 490 491 492 493  Given an ideal $I$ in a commutative ring $R$, the relation $\equiv$ on $R$, defined by % \begin{align*} a \equiv b \text{~if~} a - b \in I \end{align*} % is called \textbf{congruence mod I}; it is an equivalence relation on $R$ and its equivalence classes are the cosets. The family of all cosets is a \textbf{partition} of $R$.  rrueger committed Jun 02, 2019 494 495 496 \end{remark} \begin{proposition}  rrueger committed Jun 09, 2019 497 498 499  Let $I$ be an ideal in a commutative ring $R$. If $a,b \in R$, then $a + I = b+ I$ if and only if $a - b \in I$. In particular, $a + I = I$ if and only if $a \in I$.  rrueger committed Jun 02, 2019 500 501 \end{proposition}  rrueger committed Jun 09, 2019 502 \begin{definition}[Addition and multiplication on quotients]  rrueger committed Jun 09, 2019 503 504 505 506 507 508 509 510 511 512 513 514  Let $R$ be a commutative ring and $I$ be an ideal in $R$. Define addition $\alpha: R/I \times R/I \to R/I$ by % \begin{align*} \alpha:(a + I, b + I) \mapsto a+b+I, \end{align*} % and multiplication $\mu: R/I \times R/I \to R/I$ by % \begin{align*} \mu: (a+ I, b+I) \mapsto ab + I \end{align*}  rrueger committed Jun 02, 2019 515 516 517 \end{definition} \begin{theorem}  rrueger committed Jun 09, 2019 518 519  If $I$ is an ideal in a commutative ring $R$, then $R/I$ is a commutative ring and it is called \textbf{quotient ring} of $R$ modulo $I$.  rrueger committed Jun 02, 2019 520 521 \end{theorem}  rrueger committed Jun 09, 2019 522 \begin{definition}[Natural map]  rrueger committed Jun 09, 2019 523 524  Let $I$ be an ideal in a commutative ring $R$. The \textbf{natural map} is the function $\pi: R \to R/I$ given by $a \mapsto a + I$.  rrueger committed Jun 02, 2019 525 526 527 \end{definition} \begin{proposition}  rrueger committed Jun 09, 2019 528 529  If $I$ is an ideal in a commutative ring $R$, then the natural map $\pi: R \to R/I$ is a surjective homomorphism and $\ker \pi = I$.  rrueger committed Jun 02, 2019 530 531 532 \end{proposition} \begin{corollary}  rrueger committed Jun 09, 2019 533 534 535  Given an ideal $I$ in a commutative ring $R$, there exists a commutative ring $A$ and a (surjective) homomorphism $\varphi: R \to A$ with $I = \ker \varphi$.  rrueger committed Jun 02, 2019 536 537 538 \end{corollary} \begin{theorem}[First Isomorphism Theorem]  rrueger committed Jun 09, 2019 539 540 541 542 543 544  Let $R$ and $A$ be commutative rings. If $\varphi: R \to A$ is a homomorphism, then $\ker \varphi$ is an ideal in $R$, $\im \varphi$ is a subring of $A$ and % \begin{align*} R/\ker \varphi \cong \im \varphi \end{align*}  rrueger committed Jun 02, 2019 545 546 \end{theorem}  rrueger committed Jun 09, 2019 547 \begin{definition}[Prime field]  rrueger committed Jun 09, 2019 548 549  If $k$ is a field, the intersection of all the subfields of $k$ is called the \textbf{prime field} of $k$.  rrueger committed Jun 02, 2019 550 551 552 \end{definition} \begin{proposition}  rrueger committed Jun 09, 2019 553 554 555 556 557 558 559 560  Let $k$ be a field with unit $l$ and let $\chi: \Z \to k$ be the homomorphism $\chi: n \mapsto nl$. \begin{enumerate} \item Either $\im \chi \cong \Z$ or $\im \chi \cong \mathbb{F}_p$ for some prime $p$. \item The prime field of $k$ is isomorphic to $\mathbb{Q}$ or to $\mathbb{F}_p$ for some prime $p$. \end{enumerate}  rrueger committed Jun 02, 2019 561 562 \end{proposition}  rrueger committed Jun 09, 2019 563 \begin{definition}[Characteristic]  rrueger committed Jun 09, 2019 564 565 566  A field $k$ has \textbf{characteristic} 0 if its prime field is isomorphic to $\mathbb{Q}$; it has characteristic $p$ if its prime field is isomorphic to $\mathbb{F}_p$ for some prime $p$.  rrueger committed Jun 02, 2019 567 568 569 \end{definition} \begin{proposition}  rrueger committed Jun 09, 2019 570 571  If $K$ is a finite field, then $|K| = p^n$ for some prime $p$ and some $n \geq 1$.  rrueger committed Jun 02, 2019 572 573 574 575 \end{proposition} \section{Maximal Ideals and Prime Ideals}  rrueger committed Jun 09, 2019 576   rrueger committed Jun 09, 2019 577 \begin{definition}[Maximal ideal]  rrueger committed Jun 09, 2019 578 579 580  An ideal $I$ in a commutative ring $R$ is called a \textbf{maximal ideal} if $I$ is a proper ideal for which there is no proper ideal $J$ with $I \subsetneq J$.  rrueger committed Jun 02, 2019 581 582 583 \end{definition} \begin{proposition}  rrueger committed Jun 09, 2019 584 585  A proper ideal $I$ in a commutative ring $R$ is a maximal ideal if and only if $R/I$ is a field.  rrueger committed Jun 02, 2019 586 587 588 \end{proposition} \begin{proposition}  rrueger committed Jun 09, 2019 589 590  If $k$ is a field, then $I = (x_1 - a_1, \ldots, x_n - a_n)$ is a maximal ideal in $k[x_1,\ldots, x_n]$ whenever $a_1, \ldots, a_n \in k$.  rrueger committed Jun 02, 2019 591 592 593 \end{proposition} \begin{theorem}  rrueger committed Jun 09, 2019 594 595  Let $R$ be a commutative nonzero ring. For any proper ideal $I \subset R$, there exists a maximal ideal $J \subset R$, such that $I \subset J$.  rrueger committed Jun 02, 2019 596 597 \end{theorem}  rrueger committed Jun 09, 2019 598 \begin{definition}[Prime ideal]  rrueger committed Jun 09, 2019 599 600  An ideal $I$ in a commutative ring $R$ is called \textbf{prime ideal} if $I$ is a proper ideal such that $ab \in I$ implies $a \in I$ or $b \in I$.  rrueger committed Jun 02, 2019 601 602 603 \end{definition} \begin{proposition}  rrueger committed Jun 09, 2019 604 605  If $I$ is a proper ideal in a commutative ring $R$, then $I$ is a prime ideal if and only if $R/I$ is a domain.  rrueger committed Jun 02, 2019 606 607 608 \end{proposition} \begin{corollary}  rrueger committed Jun 09, 2019 609  Every maximal ideal is a prime ideal.  rrueger committed Jun 02, 2019 610 611 612 \end{corollary} \begin{definition}[Multiplication of ideals]  rrueger committed Jun 09, 2019 613 614 615 616 617  If $I$ and $J$ are ideals in a commutative ring $R$, then % \begin{align*} IJ = \set{\text{all finite sums~} \sum_{l} a_l b_l: a_l \in I, b_l \in J} \end{align*}  rrueger committed Jun 02, 2019 618 619 620 621 622 623 624 625 626 627 628 629 \end{definition} \begin{theorem} If $R$ is a nonzero commutative ring, then $R$ has a maximal ideal. Indeed, every proper ideal $U$ in $R$ is contained in a maximal ideal. \end{theorem} \begin{proposition} Let $P$ be a prime ideal in a commutative ring $R$. If $I$ and $J$ are ideals with $IJ \subseteq P$, then $I \subseteq P$ or $J \subseteq P$. \end{proposition}  rrueger committed Jun 09, 2019 630 \begin{definition}[Coprime ideals]  rrueger committed Jun 02, 2019 631 632 633 634 635 636  Let $R$ be a commutative ring and $I,J$ ideals in $R$. Then $I$ and $J$ are \textbf{coprime ideals} if $I + J = R$, i.e.\ there exists $a \in I$ and $b \in J$ such that $a + b = 1$. \end{definition} \begin{theorem}[Chinese Remainder Theorem]  rrueger committed Jun 09, 2019 637  Let $I_1, \ldots, I_k$ be pairwise coprime ideals in a commutative ring $R$.  rrueger committed Jun 02, 2019 638 639 640 641 642 643 644 645 646 647 648 649 650 651 652 653 654 655 656 657 658 659 660 661 662 663 664 665 666 667 668 669 670 671 672 673 674 675 676 677 678  Then the homomorphism % \begin{align*} \varphi: & R \to R/I_1 \times \cdots \times R/I_k \\ & x \mapsto (x + I_1, \ldots , x+ I_k) \end{align*} % is surjective and $\ker\varphi = I_1 \cap \ldots \cap I_k$. \end{theorem} \section{From Arithmetic to Polynomials} \begin{theorem}[Division Algorithm] If $k$ is a field and $f(x), g(x) \in k[x]$ with $f(x) \neq 0$, then there are unique polynomials $q(x), r(x) \in k[x]$ with % \begin{align*} g = qf + r \end{align*} % where either $r = 0$ or $\deg(r) < \deg(f)$. \end{theorem} \begin{corollary} Let $R$ be a commutative ring, and let $f(x) \in R[x]$ be a monic polynomial. If $g(x) \in R[x]$, then there exist $q(x) , r(x) \in R[x]$ with % \begin{align*} g(x) = q(x) f(x) + r(x), \end{align*} % where either $r(x) = 0$ or $\deg(r) < \deg(f)$. (Now $q,r$ are not unique!) \end{corollary} \begin{theorem} If $k$ is a field, then every ideal $I$ in $k[x]$ is a principal ideal; that is, there is $d \in I$ with $I = (d)$. Moreover, if $I \neq (0)$, then $d$ can be chosen to be monic. \end{theorem}  rrueger committed Jun 09, 2019 679 \begin{definition}[Root of a polynomial]  rrueger committed Jun 02, 2019 680 681 682 683 684 685 686 687 688 689 690 691 692  If $f(x) \in k[x]$, where $k$ is a field, then a \textbf{root} of $f$ in $k$ is an element $a \in k$ with $f(a) = 0$. \end{definition} \begin{lemma} Let $f(x) \in k[x]$, where $k$ is a field, and let $u \in k$. Then there is $q(x) \in k[x]$ with % \begin{align*} f(x) = q(x) (x-u) + f(u) \end{align*} \end{lemma}  rrueger committed Jun 09, 2019 693 \begin{definition}[Irreducible]  rrueger committed Jun 02, 2019 694 695 696 697 698 699 700 701 702 703 704 705 706  An element $p$ in a domain $R$ is \textbf{irreducible} if $p$ is neither $0$ nor a unit and in every factorization $p = uv$ in $R$, either $u$ or $v$ is a unit. \end{definition} \begin{theorem}[Reduction of Coefficients] Let $f(x) = a_0 + a_1 x + \dotsm + a_{n-1}x^{n-1} + x^n \in \Z[x]$ be monic and let $p$ be prime. If $\overline{f}(x) = \overline{a_0} + \overline{a_1} x + \dotsm \overline{a_{n-1}}x^{n-1} + x^n$ is irreducible in $\mathbb{F}_p$, then $f$ is irreducible in $\mathbb{Q}[x]$. \end{theorem} \begin{lemma}[Translation]  rrueger committed Jun 09, 2019 707  Let $g(x) \in \Z[x]$. If there is $c \in \Z$ with $g(x + c)$ irreducible in  rrueger committed Jun 02, 2019 708 709 710 711 712 713 714 715 716 717  $\Z[x]$, then $g$ is irreducible in $\mathbb{Q}[x]$. \end{lemma} \begin{theorem}[Eisenstein Criterion] Let $f(x) = a_0 + a_1 x + \dotsm + a_{n-1}x^{n-1} + a_n x^n \in \Z[x]$. If there is a prime $p$ with $p \mid a_i$ for all $i < n$ but $p \nmid a_n$ and $p^2 \nmid a_0$, then $f$ is irreducible in $\mathbb{Q}[x]$. \end{theorem}  rrueger committed Jun 09, 2019 718 \section{Euclidean Rings and Principal Ideal Domains*}  rrueger committed Jun 02, 2019 719   rrueger committed Jun 09, 2019 720 \begin{definition}[Greatest common divisor]  rrueger committed Jun 02, 2019 721 722 723 724 725  If $a,b$ lie in a commutative ring $R$, then a \textbf{greatest common divisor} of $a,b$ is a common divisor $d \in R$ which is divisible by every common divisor; that is, if $c \mid a$ and $c \mid b$, then $c \mid d$. \end{definition}  rrueger committed Jun 09, 2019 726 727 \begin{definition}[Euclidean ring, Degree function] A \textbf{euclidean ring} is a domain $R$ that is equipped with a function  rrueger committed Jun 02, 2019 728 729 730 731 732 733 734 735 736 737 738 739 740 741 742 743 744 745 746 747 748 749 750 751  % \begin{align*} \partial: R \setminus \set{0} \to \N, \end{align*} % called a \textbf{degree function}, such that % \begin{enumerate} \item $\partial(f) \leq \partial(fg)$ for all $f,g \in R$ with $f,g \neq 0$; \item \textbf{Division Algorithm:} for all $f,g \in R$ with $f \neq 0$, there exist $q,r \in R$ with % \begin{align*} g = qf + r, \end{align*} % where either $r = 0$ or $\partial(r) < \partial(f)$. \end{enumerate} \end{definition} \begin{theorem}  rrueger committed Jun 09, 2019 752  Let $R$ be a euclidean ring.  rrueger committed Jun 02, 2019 753 754 755 756 757 758 759 760 761 762 763 764 765 766 767 768 769 770  \begin{enumerate} \item Every ideal $I$ in $R$ is a principal ideal. \item Every pair $a,b \in R$ has a $\gcd$, say $d$, that is a linear combination of $a$ and $b$. \item \textbf{Euclid's Lemma}: If an irreducible element $p \in R$ divides a product $ab$, then either $p \mid a$ or $p \mid b$. \item \textbf{Unique Factorization}: If $a \in R$ and $a = p_1 \hdots p_m$, where the $p_i$ are irreducible elements, then this factorization is unique in the following sense: if $a = q_1 \hdots q_k$, where the $q_j$ are irreducible, then $k = m$ and the $q$'s can be reindexed so that $p_i$ and $q_i$ are associates for all $i$. \end{enumerate} \end{theorem}  rrueger committed Jun 09, 2019 771 \begin{definition}[Principal ideal domain]  rrueger committed Jun 02, 2019 772 773 774 775 776 777 778  A \textbf{principal ideal domain} (PID) is a domain $R$ in which every ideal is a principal ideal. \end{definition} \section{Unique Factorization Domains}  rrueger committed Jun 09, 2019 779 \begin{definition}[Unique factorization domain, Factorial ring]  rrueger committed Jun 02, 2019 780 781 782 783 784 785 786 787 788 789 790 791 792 793 794 795 796 797 798 799 800 801 802 803 804 805 806 807 808 809 810 811 812 813 814 815 816 817 818 819 820 821 822 823 824 825 826 827  A domain $R$ is a \textbf{UFD} (unique factorization domain or factorial ring) if % \begin{enumerate} \item every $r \in R$, neither 0 nor a unit, is a product of irreducibles; \item if $p_1 \hdots p_m = q_1 \hdots q_n$, where all $p_i$ and $q_j$ are irreducible, then $m = n$ and there is a permutation $\sigma \in S_n$ with $p_i$ and $q_{\sigma(i)}$ associates for all $i$. \end{enumerate} \end{definition} \begin{proposition} Let $R$ be a domain in which every $r \in R$, neither 0 nor a unit, is a product of irreducibles. Then $R$ is a UFD if and only if $(p)$ is a prime ideal in $R$ for every irreducible element $p \in R$. \end{proposition} \begin{lemma}\mbox{} \begin{enumerate} \item If $R$ is a commutative ring and % \begin{align*} I_1 \subseteq I_2 \subseteq \ldots \subseteq I_n \subseteq I_{n+1} \subseteq \ldots \end{align*} % is an ascending chain of ideals in $R$, then $J = \bigcup_{n \geq 1} I_n$ is an ideal in $R$. \item If $R$ is a PID, then it has no infinite strictly ascending chain of ideals % \begin{align*} I_1 \subsetneq I_2 \subsetneq \ldots \subsetneq I_n \subsetneq I_{n+1} \subsetneq \ldots \end{align*} \item If $R$ is a PID and $r \in R$ is neither 0 nor a unit, then $r$ is a product of irreducibles. \end{enumerate} \end{lemma} \begin{theorem} Every PID is a UFD. \end{theorem}  rrueger committed Jun 09, 2019 828 \begin{proposition}\label{prop:existenceofgcdinufd}  rrueger committed Jun 02, 2019 829 830 831 832 833 834 835 836 837  If $R$ is a UFD, then a $\gcd(a_1, \ldots, a_n)$ of any finite set of elements $a_1, \ldots, a_n$ in $R$ exists. \end{proposition} \begin{remark} Proposition~\ref{prop:existenceofgcdinufd} does not say that the $\gcd$ is a linear combination of the elements. In general, this is not true. \end{remark}  rrueger committed Jun 09, 2019 838 \begin{definition}[Relatively prime]  rrueger committed Jun 02, 2019 839 840 841 842 843 844  Elements $a_1, \ldots, a_n$ in a UFD $R$ are called \textbf{relatively prime} if their $\gcd$ is a unit. \end{definition} Recall, that if $R$ is a domain, then the units in $R[x]$ are the units in $R$.  rrueger committed Jun 09, 2019 845 \begin{definition}[Primitive]  rrueger committed Jun 02, 2019 846 847 848 849 850 851  A polynomial $f(x) = a_n x^n + \cdots + a_1 x + a_0 \in R[x]$, where $R$ is a UFD, is called \textbf{primitive} if its coefficients are relatively prime; that is, the only common divisors of $a_n, \ldots, a_1, a_0$ are units. \end{definition} \begin{remark}  rrueger committed Jun 12, 2019 852  If $R$ is a UFD, then every irreducible $p(x) \in R[x]$ of positive degree is  rrueger committed Jun 02, 2019 853 854 855 856 857 858 859 860 861 862 863 864 865 866  primitive. Otherwise, there is an irreducible $q \in R$ with $p(x) = q g(x)$; Note that $\deg(q) = 0$. Since $p$ is irreducible, its only factors are units and associates; since $q$ is not a unit, it must be an associate of $p$. Units have degree 0 in $R[x]$. Hence, associates in $R[x]$ have the same degree. As $p$ has positive degree, $q$ cannot be an associate. Contradiction. \end{remark} \begin{lemma}[Gauss] If $R$ is a UFD and $f(x), g(x) \in R[x]$ are both primitive, then their product $fg$ is also primitive. \end{lemma} \begin{lemma} Let $R$ be a UFD, let $Q = \Frac(R)$ and let $f(x) \in Q[x]$ be nonzero.  rrueger committed Jun 09, 2019 867  %  rrueger committed Jun 02, 2019 868 869 870 871 872 873 874 875 876 877 878 879 880 881 882 883 884 885 886 887 888 889  \begin{enumerate} \item There is a factorization % \begin{align*} f(x) = c(f) f^{*}(x), \end{align*} % where $c(f) \in Q$ and $f^{*} \in R[x]$ is primitive. This factorization is unique up to multiplication with a unit. \item If $f(x), g(x) \in R[x]$ , then $c(fg), c(f)c(g)$ are associates in $R$ and $(fg)^{*}, f^{*}g^{*}$ are associates in $R[x]$. \item Let $f(x) \in Q[x]$ have a factorization $f = qg^{*}$, where $q \in Q$ and $g^{*} \in R[x]$ is primitive. Then $f \in R[x]$ if and only if $q\in R$. \item Let $g^{*},f \in R[x]$. If $g^{*}$ is primitive and $g^{*} \mid bf$, where $b \in R$ and $b \neq 0$, then $g^{*} \mid f$. \end{enumerate} \end{lemma}  rrueger committed Jun 09, 2019 890 \begin{definition}[Associated primitive content, Content]  rrueger committed Jun 02, 2019 891 892 893 894 895 896 897 898 899 900 901 902 903 904 905 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920 921 922 923  Let $R$ be UFD with $Q = \Frac(R)$. If $f(x) \in Q[x]$, there is a factorization $f = c(f) f^{*}$, where $c(f) \in Q$ and $f^{*} \in R[x]$ is primitive. We call $c(f)$ the \textbf{content} of $f$ and $f^{*}$ the \textbf{associated primitive polynomial}. \end{definition} \begin{theorem}[Gauss] If $R$ is a UFD, then $R[x]$ is also a UFD. \end{theorem} \begin{corollary} If $k$ is a field, then $k[x_1, \ldots, x_n]$ is a UFD. \end{corollary} \begin{corollary} If $k$ is a field, then $p = p(x_1, \ldots, x_n) \in k[x_1, \ldots, x_n]$ is irreducible if and only if $p$ generates a prime ideal in $k[x_1, \ldots, x_n]$. \end{corollary} \begin{corollary}[Gauss' Lemma] Let $R$ be a UFD, let $Q = \Frac(R)$, and let $f(x) \in R[x]$. If $f = GH$ in $Q[x]$, then there is a factorization % \begin{align*} f = gh \text{~in~} R[x], \end{align*} % where $\deg(g) = \deg(G)$ and $\deg(h) = \deg(H)$; in fact, $G$ is a constant multiple of $g$ and $H$ is a constant multiple of $h$. Therefore, if $f$ does not factor into polynomials of smaller degree in $R[x]$, then $f$ is irreducible in $Q[x]$. \end{corollary}