\documentclass[a4paper,11pt,twoside,reqno]{amsart} \usepackage{typearea} \usepackage{graphicx} \title{Groups and Rings} \author{David Pierce} \date{\today} \address{Mathematics Dept\\Middle East Tech.\ Univ.\\Ankara 06531, Turkey} \email{dpierce@metu.edu.tr} \urladdr{http://metu.edu.tr/~dpierce/} %\usepackage[notref,notcite]{showkeys} \usepackage{url} \usepackage{amsmath,amssymb,amsthm,amscd} \usepackage[mathscr]{euscript} \usepackage{upgreek} \usepackage{multicol} \usepackage{stmaryrd} % \triangle{left,right}eqslant \usepackage[matrix,arrow]{xy} \usepackage{hfoldsty} \usepackage{verbatim} \usepackage{paralist} %%%%%%%%%%%%%%%%%%%%%%%%%% % % Theorems % %%%%%%%%%%%%%%%%%%%%%%%% %\swapnumbers \newtheorem{theorem}{Theorem} \newtheorem{lemma}{Lemma} \newtheorem*{porism}{Porism} \newtheorem*{corollary}{Corollary} \theoremstyle{definition} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} %%%%%%%%%%%%%%%%%%%%%%%%%%%%% \makeindex \usepackage{bm} \newcommand{\tuple}[1]{\bm{#1}} \newcommand{\vscr}[1]{#1'} % successor ordinal (v for von Neumann) \newcommand{\vnn}{\upomega} % my usual preference for this \newcommand{\included}{\subseteq} % [the name suggests the meaning here] \newcommand{\pincluded}{\subset} % [the name suggests the meaning here] \renewcommand{\leq}{\leqslant} \renewcommand{\geq}{\geqslant} \renewcommand{\emptyset}{\varnothing} \renewcommand{\land}{\mathrel{\&}} \renewcommand{\setminus}{\smallsetminus} \renewcommand{\phi}{\varphi} \newcommand{\stnd}[1]{\mathbb{#1}} \newcommand{\N}{\stnd{N}} \newcommand{\Z}{\stnd{Z}} % integers \newcommand{\Q}{\stnd{Q}} % rationals \newcommand{\C}{\stnd{C}} % complex numbers \newcommand{\R}{\stnd{R}} % real numbers \newcommand{\F}{\stnd{F}} % \newcommand{\Ham}{\stnd{H}} % quaternions \newcommand{\Oct}{\stnd{O}} % octonions \newcommand{\id}{\operatorname{id}} % identity-map \newcommand{\gid}{\operatorname{e}} % identity of group \newcommand{\inv}{^{-1}} % mult. inverse \newcommand{\Mat}[2][n]{\operatorname M_{#1}(#2)} \newcommand{\MatR}[1][n]{\Mat[#1]{R}} \newcommand{\MatZ}[1][n]{\Mat[#1]{\Z}} \newcommand{\GL}[2][n]{\operatorname{GL}_{#1}(#2)} \newcommand{\GLZ}[1][n]{\GL[#1]{\Z}} \newcommand{\GLR}[1][n]{\GL[#1]{R}} \newcommand{\Kfg}{\mathrm V} % Klein four group \newcommand{\quat}{\mathrm Q_8} % Quaternion group \newcommand{\str}[1]{\mathfrak{#1}} % structure \newcommand{\qsep}{\;} % follows a quantified variable \newcommand{\Forall}[1]{\forall{#1}\qsep } \newcommand{\Exists}[1]{\exists{#1}\qsep } \newcommand{\modsim}{/\mathord{\sim}} % modulo the eq-ren \sim \newcommand{\eqc}[1]{[#1]} % equivalence-class \newcommand{\divides}{\mathrel{|}} \newcommand{\ndivides}{\mathrel{\nmid}} \newcommand{\order}[1]{\lvert#1\rvert} \newcommand{\gpgen}[1]{\langle#1\rangle}% subgroup generated by #1 \newcommand{\unordered}[2]{[#2]^{#1}} % unordered #1-tuples from #2 \newcommand{\free}[1]{\operatorname{F}(#1)} % free group on #1 \newcommand{\fggen}{I} % generating set of a free group \newcommand{\gprels}{B} % relations \newcommand{\setactedon}{A} % set acted on by a group \newcommand{\setcolon}{\colon} \newcommand{\subgp}{<} % subgroup \newcommand{\nsubgp}{\vartriangleleft} % normal subgroup \newcommand{\nsupgp}{\vartriangleright} % normal supergroup \newcommand{\psubgp}{\lneqq} \newcommand{\congruence}{\equiv} \newcommand{\siml}{\congruence_{\ell}^H} \newcommand{\simr}{\congruence_{\mathrm r}^H} \newcommand{\weakprod}{\sideset{}{^{\mathrm{w}}}\prod} \newcommand{\textweakprod}{\prod^{\mathrm w}} \newcommand{\freeprod}{\sideset{}{^*}\prod} \newcommand{\textfreeprod}{\prod^*} \newcommand{\gpres}[2]{\gpgen{#1\mid#2}}% group on #1 with rel'ns #2 \newcommand{\centr}[1]{\operatorname{C}(#1)} % center \newcommand{\cseries}[2]{\operatorname{C}_{#1}(#2)} % central series \newcommand{\cseriesplain}[1]{\operatorname{C}_{#1}} % central series \newcommand{\centralizer}[2]{\operatorname{C}_{#2}(#1)} % centralizer \newcommand{\normalizer}[2]{\operatorname{N}_{#2}(#1)} \newcommand{\dsubgp}[2]{#2^{(#1)}} % n-th derived subgroup of #2, where n=#1. \newcommand{\tsubgp}[1]{#1_{\mathrm{t}}} % torsion sub-group \DeclareMathOperator{\im}{im} % image \newcommand{\family}[1]{\mathcal{#1}} % family (of sets) \newcommand{\class}[1]{\mathbf{#1}} % class \newcommand{\unit}[1]{{#1}^{\times}} % group of units of a ring \newcommand{\Zmod}[1]{\Z_{#1}} \newcommand{\Zmodu}[1]{\unit{\Zmod{#1}}} \DeclareMathOperator{\lcm}{lcm} \newcommand{\rest}[1]{\restriction{#1}}% restriction of function to #1 \newcommand{\modulo}{\emph{modulo}} \newcommand{\bracket}{\operatorname b} % (Lie) bracket % Concerning permutations: \newcommand{\sgn}[1]{\operatorname{sgn}(#1)} \newcommand{\sq}[1]{q_{\sigma}(#1)} % used to define sgn \newcommand{\Sym}[1]{\operatorname{Sym}(#1)} \newcommand{\Alt}[1]{\operatorname{Alt}(#1)} % alternating group \newcommand{\Dih}[1]{D_{#1}} % dihedral group %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\mi}{\mathrm i} \newcommand{\mj}{\mathrm j} \newcommand{\mk}{\mathrm k} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\setimb}[1]{[#1]} % image of a set, using brackets \newcommand{\abs}[1]{\left\lvert#1\right\rvert} % absolute value \newcommand{\size}[1]{\lvert#1\rvert} % cardinality \newcommand{\so}[1]{\operatorname{E}(#1)} \newcommand{\End}[1]{\operatorname{End}(#1)} \newcommand{\Aut}[1]{\operatorname{Aut}(#1)} \newcommand{\Hom}[1]{\operatorname{Hom}(#1)} \newcommand{\Inn}[1]{\operatorname{Inn}(#1)} \newcommand{\Der}[1]{\operatorname{Der}(#1)} %\newcommand{\pid}{\textsc{pid}} \newcommand{\pid}{PID} \newcommand{\ufd}{UFD} \newcommand{\ed}{ED} \newcommand{\primei}{\mathfrak{p}} % a prime ideal \newcommand{\maxi}{\mathfrak{m}} % a maximal ideal \newcommand{\supp}[1]{\operatorname{supp}(#1)} \newcommand{\Supp}[1]{\operatorname{supp}[#1]} \newcommand{\symdiff}{\mathbin{\triangle}} %\newcommand{\lang}{\mathcal{L}} % a language or signature \newcommand{\pow}[1]{\mathscr{P}(#1)} % power set \let\oldsqrt\sqrt \renewcommand{\sqrt}[2][1]{\oldsqrt{\vphantom{#1}}{#2}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \renewcommand{\theequation}{\roman{equation}} \renewcommand{\thepart}{\Roman{part}} \begin{document} %\frontmatter \maketitle \section*{Preface} These are lecture notes for the first semester of a one-year graduate course in algebra. The main reference for the course is Hungerford's \emph{Algebra} \cite{MR600654}. The present notes are mostly derived from the notes I kept while teaching Math 503 at METU in the fall of 2008. However, \S~\ref{sect:N} has a different source: a course called Non-standard Analysis, which I gave at the Nesin Mathematics Village, \c Sirince, in the summer of 2009. I have built up Part~\ref{part:N} around this section. I edit these notes as I teach Math 503 in the fall of 2009. I do some reorganizing according to how I should like to do things in the future. %\newpage \tableofcontents \newpage \part{Foundations}\label{part:N} %\setcounter{section}{-1} \section{Functions and relations}\label{sect:f} %I start with a brief set-theoretic review of \emph{functions.} If $A$ and $B$ are sets, then their \textbf{cartesian product,}\index{cartesian product} denoted by \begin{equation*} A\times B, \end{equation*} is the set $\{(x,y)\colon x\in A\land y\in B\}$. Here the \textbf{ordered pair}\index{ordered pair} $(x,y)$ is defined so that \begin{equation*} (a,b)=(x,y)\iff a=x\land b=y. \end{equation*} One definition that accomplishes this is $(x,y)=\{\{x\},\{x,y\}\}$, but we never actually need the precise definition. An \textbf{ordered triple}\index{ordered triple} $(x,y,z)$ can be defined as $((x,y),z)$, and so forth. A \textbf{function}\index{function} or \textbf{map}\index{map} from $B$ to $A$ is a subset $f$ of $B\times A$ such that, for each $b$ in $B$, there is exactly one $a$ in $A$ such that $(b,a)\in f$. Then instead of $(b,a)\in f$, we write \begin{equation}\label{eqn:f} f(b)=a. \end{equation} I assume the reader is familiar with the \emph{kinds} of functions from $B$ to $A$: injective, surjective, and so forth. A \textbf{singulary operation}\index{singulary}\footnote{The word \textbf{unary}\index{unary} is more common, but less etymologically correct.} on $A$ is a function from $A$ to itself; a \textbf{binary}\index{binary} operation on $A$ is a function from $A\times A$ to $A$. A \textbf{binary relation} on $A$ is a subset of $A\times A$; if $R$ is such, and $(a,b)\in R$, we often write \begin{equation*} a\mathrel Rb. \end{equation*} However, a singulary operation on $A$ is a particular kind of binary relation on $A$: a kind of relation for which we already have the special notation in~\eqref{eqn:f}. I assume the reader is familiar with other kinds of binary relations, namely orderings. \section{An axiomatic development of the natural numbers}\label{sect:N} The set of natural numbers, commonly denoted by \begin{equation*} \N, \end{equation*} can be understood as having \begin{enumerate} \item a distinguished \textbf{initial element,}\index{initial element} denoted by \begin{equation*} 0 \end{equation*} and called \textbf{zero,}\index{zero} and \item a distinguished singulary operation of \textbf{succession,}\index{succession, successor} denoted by \begin{equation*} n\mapsto n+1, \end{equation*} where $n+1$ is called the \textbf{successor} of $n$. \end{enumerate} I propose to refer to the ordered triple $(\N,0,n\mapsto n+1)$ as an \emph{iterative structure.} In general, by an \textbf{iterative structure,}\index{iterative} I mean any set that has a distinuished element and a distinguished singulary operation. Here the underlying set is sometimes called the \textbf{universe}\index{universe} of the structure. If one wants a simple notational distinction between a structure and its universe, and the universe is $A$, then the structure might be denoted by $\str A$. (Here $\str A$ is the Fraktur version of $A$. See Appendix~\ref{app:German}.) The iterative structure $(\N,0,n\mapsto n+1)$ is distinguished among iterative structures for satisfying the following axioms. \begin{enumerate} \item\label{ax:0} $0$ is not a successor: $0\neq n+1$. \item\label{ax:inj} Succession is injective: if $m+1=n+1$, then $m=n$. \item\label{ax:ind} the structure admits \textbf{proof by induction,}\index{induction} in the sense that, of all subsets of the universe, the only subset $A$ with the following two closure properties is the whole universe: \begin{enumerate} \item $0\in A$; \item for all $n$, if $n\in A$, then $n+1\in A$. \end{enumerate} \end{enumerate} These axioms seem to have been discovered originally by Dedekind~\cite[II, VI (71), p.~67]{MR0159773}, although they were also written down by Peano~\cite{Peano} and are often known as the \textbf{Peano axioms.}\index{Peano} \begin{theorem}[Recursion] For every iterative structure $(A,b,f)$, there is a unique \textbf{homomorphism}\index{homomorphism} to this structure from $(\N,0,n\mapsto n+1)$: that is, there is a unique function $h$ from $\N$ to $A$ such that \begin{enumerate} \item $h(0)=b$, \item $h(n+1)=f(h(n))$ for all $n$ in $\N$. \end{enumerate} \end{theorem} \begin{proof} We seek $h$ as a particular subset of $\N\times A$. Let $B$ be the set whose elements are the subsets $C$ of $\N\times A$ such that, if $(x,y)\in C$, then either \begin{enumerate} \item $(x,y)=(0,b)$ or else \item $C$ has an element $(u,v)$ such that $(x,y)=(u+1,f(v))$. \end{enumerate} Let $R=\bigcup B$; so $R$ is a subset of $\N\times A$. We may say $R$ is a \emph{relation} from $\N$ to $A$. If $(x,y)\in R$, we may write also \begin{equation*} x\mathrel Ry. \end{equation*} Since ${(0,b)}\in B$, we have $0\mathrel Rb$. If $n\mathrel Ry$, then $(n,y)\in C$ for some $C$ in $B$, but then $C\cup\{(n+1,f(y))\}\in B$ by definition of $B$, so $(n+1)\mathrel R f(y)$. Therefore $R$ is the desired function $h$, provided it is a \emph{function} from $\N$ to $A$. Proving this has two stages. \begin{asparaenum}[1.] \item For all $n$ in $\N$, there is $y$ in $A$ such that $n\mathrel Ry$. Indeed, let $D$ be the set of such $n$. Then we have just seen that $0\in D$, and if $n\in D$, then $n+1\in D$. By induction, $D=\N$. \item For all $n$ in $\N$, if $n\mathrel Ry$ and $n\mathrel Rz$, then $y=z$. Indeed, let $E$ be the set of such $n$. Suppose $0\mathrel R y$. Then $(0,y)\in C$ for some $C$ in $B$. Since $0$ is not a successor, we must have $y=b$, by definition of $B$. Therefore $0\in E$. Suppose $n\in E$, and $(n+1)\mathrel Ry$. Then $(n+1,y)\in C$ for some $C$ in $B$. Again since $0$ is not a successor, we must have $(n+1,y)=(m+1,f(v))$ for some $(m,v)$ in $C$. Since succession is injective, we must have $m=n$. Since $n\in E$, we know $v$ is \emph{unique} such that $n\mathrel Rv$. Since $y=f(v)$, therefore $y$ is unique such that $(n+1)\mathrel Ry$. Thus $n+1\in E$. By induction, $E=\N$. \end{asparaenum} So $R$ is the desired function $h$. Finally, $h$ is unique by induction. \end{proof} \begin{corollary} For every set $A$ with a distinguished element $b$, and for every function $F$ from $\N\times B$ to $B$, there is a unique function $H$ from $\N$ to $A$ such that \begin{enumerate} \item $H(0)=b$, \item $H(n+1)=F(n,H(n))$ for all $n$ in $\N$. \end{enumerate} \end{corollary} \begin{proof} Let $h$ be the unique homomorphism from $(\N,0,n\mapsto n+1)$ to $(\N\times A,(0,b),f)$, where $f$ is the operation $(n,x)\mapsto(n+1,F(n,x)))$. In particular, $h(n)$ is always an ordered pair. By induction, the first entry of $h(n)$ is always $n$; so there is a function $H$ from $\N$ to $A$ such that $h(n)=(n,H(n))$. Then $H$ is as desired. By induction, $H$ is unique. \end{proof} We can now use recursion to \emph{define} the binary operation $(x,y)\mapsto x+y$ of \textbf{addition,}\index{addition} along with the binary operation $(x,y)\mapsto x\cdot y$ or $(x,y)\mapsto xy$ of \textbf{multiplication,}\index{multiplication} on $\N$. The definitions are: \begin{align*} n+0&=n,& n+(m+1)&=(n+m)+1,& n\cdot0&=0,& n\cdot(m+1)&=n\cdot m+n. \end{align*} \begin{lemma}\label{lem:+} For all $n$ and $m$ in $\N$, \begin{align*} 0+n&=n,&(m+1)+n&=(m+n)+1. \end{align*} \end{lemma} \begin{proof} Induction. \end{proof} \begin{theorem}\label{thm:N-comm} Addition on $\N$ is \begin{enumerate} \item \textbf{commutative:}\index{commutative} $n+m=m+n$; and \item \textbf{associative:}\index{associative} $n+(m+k)=(n+m)+k$. \end{enumerate} \end{theorem} \begin{proof} Induction and the lemma. \end{proof} \begin{theorem}\label{thm:cancel} Addition on $\N$ allows \textbf{cancellation:}\index{cancellation} if $n+x=n+y$, then $x=y$. \end{theorem} \begin{proof} Induction, and injectivity of succession. \end{proof} \begin{lemma}\label{lem:.} For all $n$ and $m$ in $\N$, \begin{align*} 0\cdot n&=0,&(m+1)\cdot n&=m\cdot n+n. \end{align*} \end{lemma} \begin{proof} Induction. \end{proof} \begin{theorem}\label{thm:mult-comm} Multiplication on $\N$ is \begin{enumerate} \item commutative: $nm=mn$; \item \textbf{distributive}\index{distributive} over addition: $n(m+k)=nm+nk$; and \item associative: $n(mk)=(nm)k$. \end{enumerate} \end{theorem} \begin{proof} Induction and the lemma. \end{proof} Landau \cite{MR12:397m} proves \emph{using induction alone} that $+$ and $\cdot$ exist as given by the recursive definitions above. However, Theorem~\ref{thm:cancel} needs more than induction. Also, the existence of \textbf{exponentiation,}\index{exponentiation} as an operation $(x,y)\mapsto x^y$ such that \begin{align*} n^0&=1,& n^{m+1}&=n^m\cdot n, \end{align*} requires more than induction. The usual ordering $<$ of $\N$ is defined recursively as follows. First note that $m\leq n$ means simply $mgp} In every group, the following equations are identities: \begin{gather}\label{def-1} x\gid=x=\gid{}x,\\\label{def-inv} xx\inv=\gid{}=x\inv x,\\\label{def-ass} (xy)z=x(yz). \end{gather} \end{theorem} \begin{proof} With $\phi$ as above, we have $\phi(x\gid{})=\phi(x)\circ\phi(\gid{})=\phi(x)\circ\id_A=\phi(x)$, so $x\gid{}=x$ since $\phi$ is injective. The remaining identities are established likewise. \end{proof} Any element $a$ of a group determines a singulary operation $\lambda_a$ on the group, given by \begin{equation*} \lambda_a(x)=ax. \end{equation*} \begin{theorem}\label{thm:Cay} The function $x\mapsto\lambda_x$ embeds a group in its symmetry group. \end{theorem} \begin{proof} Let $G$ be a group, and $a\in G$. We have \begin{equation*} \lambda_{a\inv}(\lambda_a(x))=a\inv(ax)=(a\inv a)x=\gid{}x=x \end{equation*} by Theorem~\ref{thm:sym=>gp}, so $\lambda_{a\inv}\circ\lambda_a=\id_G$. Likewise \begin{equation*} \lambda_{a}(\lambda_{a\inv}(x))=a(a\inv x)=(aa\inv)x=\gid{}x=x, \end{equation*} so $\lambda_a\circ\lambda_{a\inv}=\id_G$. Thus $\lambda_a$ is invertible and therefore belongs to $\Sym G$, and \begin{equation*} \lambda_{a}{}\inv=\lambda_{a\inv}. \end{equation*} We have also \begin{equation*} \lambda_{\gid}(x)={\gid}x=x=\id_G(x), \end{equation*} so \begin{equation*} \lambda_{\gid}=\id_G, \end{equation*} and \begin{equation*} \lambda_{ab}(x)=(ab)x=a(bx)=\lambda_a(\lambda_b)(x)=(\lambda_a\circ\lambda_b)(x), \end{equation*} so \begin{equation*} \lambda_{ab}=\lambda_a\circ\lambda_b. \end{equation*} Finally, if $\lambda_a=\lambda_b$, then \begin{gather*} ax=bx,\\ (ax)x\inv=(bx)x\inv,\\ a(xx\inv)=b(xx\inv),\\ a{\gid}=b{\gid},\\ a=b.\qedhere \end{gather*} %So $x\mapsto\lambda_x$ is as claimed. \end{proof} The following is known as \textbf{Cayley's Theorem.}\index{Cayley's Theorem} \index{Cayley's Theorem} \index{theorem!Cayley's Th---} \begin{porism} The converse of Theorem~\ref{thm:sym=>gp} holds. \end{porism} The binary operation of a group is often referred to as \textbf{multiplication;} singulary, \textbf{inversion;} nullary, the \textbf{identity}\index{identity} or the \textbf{neutral element.}\index{neutral} The identity is sometimes denoted by $1$ rather than $\gid$. \section{Simplifications}\label{sect:simp} A \textbf{monoid}\index{monoid} is a structure $(G,{\gid},\cdot)$ satisfying~\eqref{def-1} and~\eqref{def-ass} above; a \textbf{semigroup}\index{semigroup} is a structure $(G,\cdot)$ satisfying~\eqref{def-ass}. Given a set $A$, let us denote by \begin{equation*} \so A \end{equation*} the set of functions from $A$ to itself (that is, the set of singulary operations on $A$). Then $(\so A,\id_A,\circ)$ is a monoid. If $(G,\gid,\cdot)$ is a monoid, then by the proof of Theorem~\ref{thm:Cay}, $x\mapsto\lambda_x$ is a homomorphism from $(G,\gid,\cdot)$ to $(\so G,\id_G,\circ)$; however, it might not be an embedding. The following will be used in Theorem~\ref{thm:solutions}. \begin{theorem}\label{thm:left} Any structure that satisfies \begin{gather*} {\gid}x=x,\\ x\inv x=\gid,\\ x(yz)=(xy)z \end{gather*} is a group. In other words, any semigroup with a left-identity and with left-inverses is a group. \end{theorem} \begin{proof} Using the given identies, we have \begin{equation*} (xx\inv)(xx\inv)=x(x\inv x)x\inv=x{\gid}x\inv=xx\inv, \end{equation*} and so \begin{equation*} xx\inv={\gid}xx\inv=(xx\inv)\inv(xx\inv)(xx\inv)=(xx\inv)\inv(xx\inv)={\gid}. \end{equation*} Hence also \begin{equation*} x{\gid}=x(x\inv x)=(xx\inv)x={\gid}x=x.\qedhere \end{equation*} \end{proof} %The lemma has an obvious dual. A semigroup \textbf{expands}\index{expands} to a group if it can be given an identity and an inversion so as to become a group (while the underlying set remains the same). \begin{theorem}\label{thm:solutions} Let $G$ be a nonempty semigroup. The following are equivalent. \begin{enumerate} \item $G$ expands to a group. \item $G$ expands uniquely to a group. \item Each equation $ax=b$ and $ya=b$ with coefficients from $G$ has a solution in $G$. \item Each equation $ax=b$ and $ya=b$ with coefficients from $G$ has a unique solution in $G$. \end{enumerate} \end{theorem} \begin{proof} In a group, the equation $b=ax$ implies $a\inv b=a\inv(ax)$, and \begin{equation*} a\inv(ax) =(a\inv a)x ={\gid}x =x; \end{equation*} so the equation has at most one solution. It has at least one solution, since indeed $a(a\inv b)=(aa\inv)b={\gid}b=b$. Likewise for the equation $b=ya$. Conversely, suppose $G$ is a nonempty semigroup in which all of the given equations have solutions. If $c\in G$, let $\gid$ be a solution to $yc=c$. If $b\in G$, let $d$ be a solution to $cx=b$. Then \begin{equation*} {\gid}b={\gid}(cd)=({\gid}c)d=cd=b. \end{equation*} Also the equation $yc={\gid}$ has a solution: call it $c\inv$. Now use Theorem~\ref{thm:left}. \end{proof} By the theorem, we can characterize groups as those semigroups that satisfy the axiom \begin{equation*} \Forall x\Forall y\Exists z\Exists w(xz=y\land wx=y). \end{equation*} More is true: \begin{theorem} A map from one group to another is a homomorphism, provided it is a homomorphism of semigroups. \end{theorem} \begin{proof} In a group, if $a$ is an element, then the identity is the unique solution of $xa=a$, and $a\inv$ is the unique solution of $yaa=a$. A semigroup homomorphism $\phi$, where $\phi(a)=b$, takes solutions of these equations to solutions of $xb=b$ and $ybb=b$. \end{proof} \section{The integers} A group or monoid or semigroup is \textbf{abelian}\index{abelian} if it satisfies the identity \begin{equation*} xy=yx. \end{equation*} Multiplication on an abelian group is often (though not always) called \textbf{addition}\index{addition} and denoted by $+$; in this case, the identity may be denoted by $0$. Let \begin{equation*} \N^+=\vnn\setminus\{0\}. \end{equation*} \begin{theorem} $(\vnn,1,\cdot)$ and $(\N^+,1,\cdot)$ are abelian monoids. \end{theorem} \begin{proof} The claim follows from the definition of addition on $\vnn$ and from Theorem~\ref{thm:mult-comm}. \end{proof} If an abelian semigroup $(G,+)$ also has a total ordering such that \begin{equation*} x1$; but $\order a\divides p$, so $\order a=p$, that is, $G=\gpgen a$. \end{proof} \begin{corollary} If $G$ is finite and $a\in G$, then $a^{\order G}=\gid$. \end{corollary} \begin{proof} $a^{\order a}=\gid$ and $\order a$ divides $\order G$. \end{proof} An application is the theorems of Fermat and Euler (Theorems~\ref{thm:Fermat} and~\ref{thm:Euler}). The first Sylow Theorem (Theorem~\ref{thm:Sylow-1}) is a partial converse. \begin{theorem} $\Zmodu n=\{[x]\in\Zmod n\setcolon \gcd(x,n)=1\}$. \end{theorem} \begin{proof} $\gcd(m,n)=1$ if and only if $am+bn=1$ for some integers $a$ and $b$; but this just means $[a][m]=1$ for some $a$. \end{proof} \begin{theorem}[Fermat]\label{thm:Fermat} If the prime $p$ is not a factor of $a$, then \begin{equation*} a^{p-1}\equiv 1\pmod p. \end{equation*} Hence $a^p\equiv a\pmod p$ for any integer $a$. \end{theorem} \begin{proof} The order of $\Zmodu p$ is $p-1$, and $[a]\in\Zmodu p$. This proves the first claim, and the second if $p\ndivides a$; the second is trivial if $p\divides a$. \end{proof} If $n\neq0$, let the order of $\Zmodu n$ be denoted by \begin{equation*} \phi(n). \end{equation*} \begin{theorem}[Euler]\label{thm:Euler}\index{Euler's Theorem} \index{theorem!Euler's Th---} If $\gcd(a,n)=1$, then $a^{\phi(n)}\equiv 1\pmod n$. \end{theorem} \section{Normal subgroups} If $H\subgp G$, then there are equivalences $\siml$ and $\simr$ on $G$ given by \begin{equation*} x\siml y\iff xH=yH;\qquad\qquad x\simr y\iff Hx=Hy. \end{equation*} \begin{theorem}\label{thm:n} Suppose $H\subgp G$. The following are equivalent: \begin{enumerate} \item $G/H$ is a group. \item $\siml$ and $\simr$ are the same. \item $aH=Ha$ for all $a$ in $G$. \item $a\inv Ha=H$ for all $g$ in $G$. \end{enumerate} \end{theorem} \begin{proof} Suppose $G/H$ is a group, that is, $\siml$ is a congruence relation. This means \begin{equation*} xH=x'H\land yH=y'H\implies xyH=x'y'H. \end{equation*} As a special case, we have that, if $h\in H$, so that $hH=H$, then $hyH=yH$, so $y\inv hyH=H$. Thus \begin{equation*} y\inv Hy=H, \end{equation*} equivalently, $Hy=yH$. Therefore $\siml$ and $\simr$ are the same. Conversely, suppose these relations are the same. Then every $x$ has the same congruence class with respect to either one: $xH=Hx$. If $xH=x'H$ and $yH=y'H$, then $xyH=xy'H=xHy'=x'Hy'=x'y'H$. Thus $G/H$ is a group. \end{proof} A subgroup $H$ of $G$ meeting any of these equivalent conditions is called \textbf{normal,}\index{normal!--- subgroup} and we write \begin{equation*} H\nsubgp G. \end{equation*} Of abelian groups, all subgroups are normal. In general, if $N\nsubgp G$, then the group $G/N$ is the \textbf{quotient-group}% \index{quotient!--- group}\index{group!quotient ---} of $G$ by $N$. \begin{theorem}\label{thm:NGHG} If $N\nsubgp G$ and $H\subgp G$, then $N\cap H\nsubgp H$. (That is, normality is preserved in subgroups.) \end{theorem} \begin{proof} The defining property of normal subgroups is universal, that is, $N\nsubgp G$ means $(G,N)\models\Forall x\Forall y(x\in N\to yxy\inv\in N)$. \end{proof} \begin{theorem} If $N\nsubgp G$ and $H\subgp G$, then $\gpgen{N\cup H}=NH$. \end{theorem} \begin{proof} Suppose $n\in N$ and $h\in H$. Then $nh=hh\inv nh$. Since $N\nsubgp{N\cup H}$, we have $h\inv nh\in N$, so $nh\in HN$. Thus $NH\included HN$, so by symmetry $NH=HN$. Therefore \begin{equation*} NH(NH)\inv=NHH\inv N\inv=NHHN\included NHN=NNH\included NH, \end{equation*} that is, $NH$ is closed under $(x,y)\mapsto xy\inv$. Since $NH$ also contains $\gid$, it is a subgroup of $G$ by Theorem~\ref{thm:subgp}. \end{proof} \begin{theorem}\label{thm:isdp} Suppose $N\nsubgp G$ and $H\subgp G$ and $N\cap H=\gpgen{\gid}$. Then the surjection $(x,y)\mapsto xy$ from $N\times H$ to $NH$ is a bijection. \end{theorem} \begin{proof} If $g$ and $h$ are in $H$, and $m$ and $n$ are in $N$, and $gm=hn$, then \begin{equation*} h\inv g=nm\inv, \end{equation*} so each side must be $\gid$, and hence $g=h$ and $m=n$. \end{proof} In the theorem, $NH$ is the \textbf{internal semidirect product}\index{internal semidirect product} of $N$ and $H$. Note well that the bijection between $N\times H$ and $NH$ need not be an isomorphism, since we have, in $N\times H$, \begin{equation*} (m,g)(n,h)=(mn,gh), \end{equation*} while, in $NH$, \begin{equation}\label{eqn:sdp} (mg)(nh)=(mgng\inv)(gh). \end{equation} Theorem~\ref{thm:wdp} below establishes conditions under which the bijection \emph{is} an isomorphism. Semidirect products in general are treated in \S~\ref{sect:semidirect}. \begin{theorem} The normal subgroups of a group are precisely the kernels of homomorphisms on the group. \end{theorem} \begin{proof} If $f$ is a homomorphism from $G$ to $H$, then $f(ana\inv)=f(a)f(n)f(a)\inv=\gid$ for all $n$ in $\ker f$, so $a(\ker f)a\inv\included \ker f$; thus $\ker f\nsubgp G$. Conversely, if $N\nsubgp G$, then the map $x\mapsto xN$ from $G$ to $G/N$ is a homomorphism with kernel~$N$. \end{proof} In the proof, the map $x\mapsto xN$ is the \textbf{canonical projection}% \index{projection}\index{canonical!--- projection} or the \textbf{quotient map}\index{quotient map} of $G$ onto $G/N$; it may be denoted by $\pi$. \begin{theorem}\label{thm:hom-n} If $f$ is a homomorphism from $G$ to $H$, and $N$ is a normal subgroup of $G$ such that $N\subgp \ker f$, then there is a unique homomorphism $\tilde f$ from $G/N$ to $H$ such that $f=\tilde f\circ \pi$, that is, the following diagram \textbf{commutes}\index{commutes} (all directed paths from one node to another represent the same function). \begin{equation*} \xymatrix{ N \ar[r] \ar[dr]_1 & G \ar[r]^{\pi} \ar[d]^f & G/N \ar[dl]^{\tilde f}\\ &H& } \end{equation*} \end{theorem} \begin{proof} If $\tilde f$ exists, it must satisfy $\tilde f(xN)=f(x)$ for all $x$ in $G$. Such $\tilde f$ does exist, since if $xN=yN$, then $xy\inv\in N\subgp \ker f$, so $f(xy\inv)=\gid$ and $f(x)=f(y)$. \end{proof} \begin{corollary}[First Isomorphism Theorem]% \index{isomorphism!I--- Theorems|(}\index{theorem!Isomorphism Th---s|(} $G/\ker f\cong \im f$ for any homomorphism $f$ on~$G$. \end{corollary} \begin{proof} Let $N=\ker f$; then $\tilde f$ is the desired homomorphism. \end{proof} \begin{corollary} If $f$ is a homomorphism from $G$ to $H$, and $N$ is a normal subgroup of $G$, and $M\nsubgp H$, and $f[N]\subgp M$, then there is a homomorphism $\tilde f$ from $G/N$ to $H/M$ such that the following diagram commutes: \begin{equation*} \xymatrix{ N\ar[d]\ar[r] & G \ar[r]^{\pi}\ar[d]_f & G/N\ar[d]^{\tilde f}\\ M \ar[r] & H \ar[r] & H/M } \end{equation*} \end{corollary} \begin{proof} The induced homomorphism from $N$ to $H/M$ is trivial. \end{proof} \begin{theorem}[Second Isomorphism] If $H\subgp G$ and $N\nsubgp G$, then \begin{equation*} % H/(H\cap N)\cong NH/N. \frac H{H\cap N}\cong\frac{HN}N. \end{equation*} \end{theorem} \begin{proof} The map $h\mapsto hN$ from $H$ to $HN/N$ is surjective with kernel $H\cap N$. So the claim follows by the First Isomorphism Theorem (a corollary to Theorem~\ref{thm:hom-n}). \end{proof} For example, In $\Z$, since $\gpgen n\cap\gpgen m=\gpgen{\lcm(n,m)}$ and $\gpgen n+\gpgen m=\gpgen{\gcd(n,m)}$, we have \begin{equation*} \frac{\gpgen n}{\gpgen{\lcm(n,m)}} \cong\frac{\gpgen{\gcd(n,m)}}{\gpgen m}. \end{equation*} \begin{theorem}[Third Isomorphism] If $N$ and $K$ are normal subgroups of $G$ and $N\subgp K$, then $K/N\nsubgp G/N$ and \begin{equation*} \frac{G/N}{K/N}\cong G/K. \end{equation*} \end{theorem} \begin{proof} By Theorem~\ref{thm:hom-n}, the map $xN\mapsto xK$ from $G/N$ to $G/K$ is a well-defined epimorphism. The kernel contains $xN$ if and only if $x\in K$, that is, $xN\in K/N$. Again the claim now follows by the First Isomorphism Theorem (a corollary to Theorem~\ref{thm:hom-n}). \end{proof} \index{isomorphism!I--- Theorems|)}\index{theorem!Isomorphism Th---s|)} Theorem~\ref{thm:hom-n} will also be used to prove von Dyck's Theorem (Theorem~\ref{thm:vD}). \begin{lemma} If $f$ is an epimorphism from $G$ onto $H$, then there is a one-to-one correspondence $K\mapsto f[K]$ between subgroups of $G$ that include $\ker f$ and subgroups of $H$; under this, normal subgroups correspond. \begin{equation*} \xymatrix{ \ker f \ar[r]\ar[d] & K \ar[r]\ar[d] & G \ar@{>>}[d]^f\\ \{e\}\ar[r] & f[K] \ar[r] & H } \end{equation*} \end{lemma} \begin{theorem} If $N\nsubgp G$, then every subgroup of $G/N$ is $K/N$ for some subgroup $K$ of $G$ that includes $N$, and moreover $K/N$ is normal in $G/N$ if and only if $K$ is normal in~$G$. \begin{equation*} \xymatrix{ N \ar[r]\ar[d] & K \ar[r]\ar[d] & G \ar@{>>}[d]^f\\ \{e\}\ar[r] & K/N \ar[r] & G/N } \end{equation*} \end{theorem} \section{Finite groups} Since every group can be considered as a symmetry group of \emph{itself,} every \emph{finite} group $G$ can be considered as a symmetry group of finite set. In particular, $G$ can be considered as a subgroup of $\Sym n$ for some $n$ in $\vnn$. An element $\sigma$ of $\Sym n$ can be denoted by \begin{equation*} \begin{pmatrix} 0 & 1 & \cdots & n-1\\ \sigma(0) & \sigma(1) & \cdots & \sigma(n-1) \end{pmatrix}. \end{equation*} In particular, the permutation \begin{equation*} \begin{pmatrix} 0 & 1 & \cdots & n-2 & n-1\\ 1 & 2 & \cdots & n-1 & 0 \end{pmatrix}. \end{equation*} can be called a \emph{cycle.}\index{cycle} More generally, if $m\leq n$, then the permutation \begin{equation*} \begin{pmatrix} 0 & 1 & \cdots & m-2 & m-1 & m & \cdots & n-1\\ 1 & 2 & \cdots & m-1 & 0 & m & \cdots & n-1 \end{pmatrix} \end{equation*} is a cycle too, or more precisely an \emph{$m$-cycle.} For the moment, let us call this $\sigma_m$. In the most general sense, an \textbf{$m$-cycle,} or a cycle of \textbf{length}\index{length} $m$, in $\Sym n$ is an element of the form \begin{equation*} \begin{pmatrix} \tau(0) &\tau(1) &\cdots &\tau(m-2) &\tau(m-1) &\tau(m) &\cdots &\tau(n-1)\\ \tau(1) &\tau(2) &\cdots &\tau(m-1) & \tau(0) &\tau(m) &\cdots &\tau(n-1) \end{pmatrix} \end{equation*} where $\tau\in\Sym n$. Let this $m$-cycle be called $\sigma$. Then $\sigma(\tau(k))=\tau(\sigma_m(k))$, so \begin{equation*} \sigma=\tau\sigma_m\tau\inv. \end{equation*} In general, the length of a cycle is its order. The $m$-cycle $\sigma$ above can be written more neatly as \begin{equation*} \begin{pmatrix} \tau(0) & \tau(1) & \tau(m-1) \end{pmatrix}. \end{equation*} In this notation, the same cycle $\sigma$ can be written in $m$ different ways, as \begin{equation*} \begin{pmatrix} \tau(i) & \tau(i+1) & \cdots & \tau(m-1) & \tau(0) & \cdots & \tau(i-1) \end{pmatrix} \end{equation*} for any $i$ in $m$. Two elements $\sigma$ and $\tau$ of $\Sym n$ are \textbf{disjoint}\index{disjoint} if, for all $x$ in $n$, \begin{equation*} \sigma(x)\neq x\implies\tau(x)=x. \end{equation*} In this case, $\sigma\tau=\tau\sigma$. \begin{theorem} Every element of $\Sym n$ is a product of disjoint cycles of length at least $2$, uniquely up to order of factors. \end{theorem} \begin{proof} Let $\sigma\in\Sym n$. If $k\in n$, let \begin{equation*} [k]=\{\sigma^{\ell}(k)\colon \ell\in\Z\}. \end{equation*} Then the sets $[k]$ partition $n$: we have \begin{equation*} n=[k_0]\cup\cdots\cup[k_{\ell-1}] \end{equation*} for some $\ell$, the union being disjoint. If $i\in\ell$, define $\sigma_i$ by \begin{equation*} \sigma_i(x)= \begin{cases} \sigma(x),&\text{ if }x\in[k_i],\\ x,&\text{ otherwise.} \end{cases} \end{equation*} If $[k_i]$ has size $\ell_i$, then $\sigma_i$ is the $\ell_i$-cycle $ \begin{pmatrix} k & \sigma(k) & \cdots & \sigma^{\ell_i-1}(k) \end{pmatrix} $. Finally, $\sigma$ is the product (that is, the composition) of all of the $\sigma_i$ such that $\ell_i>1$. \end{proof} \begin{theorem} The order of a finite permutation is the least common multiple of the orders of its disjoint cyclic factors. \end{theorem} A $2$-cycle is also called a \textbf{transposition.}\index{transposition} \begin{corollary} Every finite permutation is a product of transpositions. \end{corollary} \begin{proof} $\begin{pmatrix} 0 & 1 & \cdots & m-1 \end{pmatrix} = \begin{pmatrix} 0 & m-1 \end{pmatrix} \dotsm \begin{pmatrix} 0 & 2 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix}$. \end{proof} Let the set of $2$-element subsets of $n$ by denoted by \begin{equation*} \unordered{2}{n}. \end{equation*} If $\sigma\in\Sym n$, and $\{i,j\}\in\unordered 2n$, then we can define \begin{equation*} \sigma(\{i,j\})=\{\sigma(i),\sigma(j)\}. \end{equation*} Thus we have a homomorphism from $\Sym n$ to $\Sym{\unordered 2n}$. Understanding $n$ as the subset $\{0,\dots,n-1\}$ of $\Q$, we have a function $X\mapsto\sq X$ from $\unordered 2n$ to $\unit{\Q}$ given by \begin{equation*} \sq{\{i,j\}}=\frac{\sigma(i)-\sigma(j)}{i-j}. \end{equation*} Then we can define the function $\sigma\mapsto\sgn{\sigma}$ from $\Sym n$ into $\unit{\Q}$ by \begin{equation*} \sgn{\sigma}= \prod_{X\in \unordered{2}{n}} \sq X. \end{equation*} \begin{theorem} The function $\sigma\mapsto\sgn{\sigma}$ is an homomorphism from $\Sym n$ onto the subgroup $\gpgen{-1}$ of $\unit{\Q}$; it takes every transposition to $-1$. \end{theorem} \begin{proof} If $\sigma= \begin{pmatrix} k&\ell \end{pmatrix}$, then \begin{equation*} \sgn{\sigma}=\sq{\{k,\ell\}}\prod_{i\in n\setminus\{k,\ell\}}(\sq{\{i,\ell\}}\sq{\{k,i\}}) =\frac{\ell-k}{k-\ell}\cdot\prod_{i\in n\setminus\{k,\ell\}}\Bigl(\frac{i-k}{i-\ell}\cdot\frac{\ell-i}{k-i}\Bigr) =-1. \end{equation*} If $\sigma$ and $\tau$ are arbitrary elements of $\Sym n$, then \begin{align*} \sgn{\sigma\tau} &=\prod_{\{i,j\}\in \unordered{2}{n}}\frac{\sigma(\tau(i))-\sigma(\tau(j))}{i-j}\\ &=\prod_{\{i,j\}\in \unordered{2}{n}}\left(\frac{\sigma(\tau(i))-\sigma(\tau(j))} {\tau(i)-\tau(j)}\cdot \frac{\tau(i)-\tau(j)}{i-j}\right)\\ &=\prod_{X\in \unordered{2}{n}}\sq{\tau(X)}\cdot\sgn{\tau}\\ &=\sgn{\sigma}\sgn{\tau} \end{align*} since $\tau$ permutes $\unordered 2n$. \end{proof} The value $\sgn{\sigma}$ can now be called the \textbf{signum}\index{signum} of $\sigma$; it is $1$ if and only if $\sigma$ is the product of an even number of transpositions. Such a product is itself called \textbf{even;}\index{even} the other permutations, with signum $-1$, are called \textbf{odd.}\index{odd} The \textbf{alternating group}\index{alternating}\index{group!alternating ---} of degree $n$ is the kernel of $\sigma\mapsto\sgn \sigma$ on $\Sym n$ and is denoted by \begin{equation*} \Alt n. \end{equation*} Hence $\Alt n\nsubgp\Sym n$ and $[\Sym n:\Alt n]=2$. A group is \textbf{simple}\index{simple group}% \index{group!simple ---} if it has no proper nontrivial normal subgroups. For example, $\Zmod n$ is simple just in case $\abs n$ is prime. Hence the only simple abelian groups are the $\Zmod p$, where $p$ is prime. \begin{lemma} $\Alt n$ is generated by the $3$-cycles in $\Sym n$. \end{lemma} \begin{proof} The group $\Alt n$ is generated by the products $\begin{pmatrix} a&b \end{pmatrix} \begin{pmatrix} a&c \end{pmatrix}$ and $\begin{pmatrix} a&b \end{pmatrix} \begin{pmatrix} c&d \end{pmatrix}$, where $a$, $b$, $c$, and $d$ are distinct elements of $n$. But \begin{gather*} \begin{pmatrix} a & b \end{pmatrix} \begin{pmatrix} a & c \end{pmatrix}= \begin{pmatrix} a & c & b \end{pmatrix},\\ \begin{pmatrix} a & b \end{pmatrix} \begin{pmatrix} c & d \end{pmatrix} = \begin{pmatrix} b & c & a \end{pmatrix} \begin{pmatrix} c & d & b \end{pmatrix}. \end{gather*} Hence all $3$-cycles belong to $\Alt n$, and this group is generated by these cycles. \end{proof} \begin{lemma} $\Alt n$ is generated by the $3$-cycles $\begin{pmatrix} 0 & 1 & k \end{pmatrix}$, where $14$, then a normal subgroup of $\Alt n$ contains a $3$-cycle, provided it has a nontrivial element whose factorization into disjoint cycles contains one of the following: \begin{enumerate} \item a cycle of length at least $4$; \item two cycles of length $3$; \item transpositions, only one $3$-cycle, and no other cycles; or \item only transpositions. \end{enumerate} \end{lemma} \begin{proof} \begin{asparaenum} \item If $k\geq 4$, and $\sigma$ is disjoint from $\begin{pmatrix} 0 & 1 & \dots & k-1 \end{pmatrix}$, then \begin{equation*} \begin{pmatrix} 0 & 1 & 2 \end{pmatrix} \begin{pmatrix} 0 & 1 & \dots & k-1 \end{pmatrix}\sigma \begin{pmatrix} 2 & 1 & 0 \end{pmatrix} \sigma\inv \begin{pmatrix} k-1&\dots & 1 & 0 \end{pmatrix}= \begin{pmatrix} 0 & 1 & 3 \end{pmatrix}. \end{equation*} \item If $\sigma$ is disjoint from $\begin{pmatrix} 0 & 1 & 2 \end{pmatrix} \begin{pmatrix} 3 & 4 & 5 \end{pmatrix}$, then we reduce to the previous case: \begin{equation*} \begin{pmatrix} 0 & 1 & 3 \end{pmatrix} \underbrace{ \begin{pmatrix} 0 & 1 & 2 \end{pmatrix} \begin{pmatrix} 3 & 4 & 5 \end{pmatrix}}\sigma \begin{pmatrix} 3 & 1 & 0 \end{pmatrix} \sigma\inv \underbrace{ \begin{pmatrix} 5 & 4 & 3 \end{pmatrix} \begin{pmatrix} 2 & 1 & 0 \end{pmatrix}}= \begin{pmatrix} 0 & 1 & 4 & 2 & 3 \end{pmatrix}. \end{equation*} \item If $\sigma$ is disjoint from $\begin{pmatrix} 0 & 1 & 2 \end{pmatrix}$ and is the product of transpositions, then \begin{equation*} \left[\begin{pmatrix} 0 & 1 & 2 \end{pmatrix}\sigma\right]^2= \begin{pmatrix} 2 & 1 & 0 \end{pmatrix}. \end{equation*} \item If $\sigma$ is a product of transpositions disjoint from $\begin{pmatrix} 0&1 \end{pmatrix}$ and $\begin{pmatrix} 2&3 \end{pmatrix}$, then \begin{gather*} \begin{pmatrix} 0 & 1 & 2 \end{pmatrix} \underbrace{ \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 3 \end{pmatrix} \sigma} \begin{pmatrix} 2 & 1 & 0 \end{pmatrix} \underbrace{ \sigma \begin{pmatrix} 3 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix}} = \begin{pmatrix} 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 3 \end{pmatrix},\\ \begin{pmatrix} 0 & 2 & 4 \end{pmatrix} \underbrace{ \begin{pmatrix} 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 3 \end{pmatrix}} \begin{pmatrix} 4 & 2 & 0 \end{pmatrix} \underbrace{ \begin{pmatrix} 3 & 1 \end{pmatrix} \begin{pmatrix} 2 & 0 \end{pmatrix}} = \begin{pmatrix} 0 & 4 & 2 \end{pmatrix}.\qedhere \end{gather*} \end{asparaenum} \end{proof} % END OF DAY 6 (October 13, 2008) \begin{theorem} $\Alt n$ is simple if and only if $n\neq 4$. \end{theorem} \begin{proof} $\Alt 1$ and $\Alt 2$ are trivial, and $\Alt 3\cong\Zmod 3$. The case when $n>4$ is handled by the previous lemmas. Finally, every element of $\Alt 4$ (in fact, of $\Sym 4$) can be considered as a permutation of the set \begin{equation*} \Bigl\{ \bigl\{\{0,1\},\{2,3\}\bigr\}, \bigl\{\{0,2\},\{1,3\}\bigr\}, \bigl\{\{0,3\},\{1,2\}\bigr\}\Bigr\}. \end{equation*} Thus we get an epimorphism from $\Alt 4$ to $\Sym 3$ whose kernel is therefore a proper nontrivial normal subgroup. \end{proof} The normal subgroup of $\Alt 4$ found in the proof is \begin{equation*} \gpgen{ \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 3 \end{pmatrix}, \begin{pmatrix} 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 3 \end{pmatrix}, \begin{pmatrix} 0 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix}}. \end{equation*} We can obtain it by considering $\Alt 4$ as the group of rotational symmetries of the regular tetrahedron. The vertices of this tetrahedron can be taken as $4$ of the $8$ vertices of a cube: say, the vertices with coordinates $(1,1,1)$, $(1,-1,-1)$, $(-1,1,-1)$, and $(-1,-1,1)$. Then a symmetry of the tetrahedron determines a permutation of the $3$ coordinate axes, hence an element of $\Sym 3$. \section{Determinants}\label{sect:det} Let $R$ be a commutative ring. We define the function $X\mapsto\det(X)$ from $\MatR$ to $R$ by \begin{equation*} \det((a^i_j)^{i2$, and $G=\gpgen{a,b}$, where $\order a=n$ and $\order b=2=\order{ab}$, then $G\cong\Dih n$. \end{theorem} \begin{proof} Assume $n\geq 2$. Since $abab=\gid$ and $b\inv=b$, we have \begin{align*} ba&=a\inv b,& ba\inv&=ab. \end{align*} Therefore $ba^k=a^{-k}b$ for all integers $k$. This shows \begin{equation*} G=\{a^ib^j\setcolon(i,j)\in n\times 2\}. \end{equation*} It remains to show $\order G=2n$. Suppose \begin{equation*} a^ib^j=a^kb^{\ell}, \end{equation*} where $(i,j)$ and $(k,\ell)$ are in $n\times 2$. Then \begin{equation*} a^{i-k}=b^{\ell-j}. \end{equation*} If $b^{\ell-j}=\gid$, then $\ell=j$ and $i=k$. The alternative is that $b^{\ell-j}=b$. In this case, \begin{equation*} n\divides2(i-k). \end{equation*} If $n\divides i-k$, then $i=k$ and hence $j=\ell$. The only other possibility is that $n=2m$ for some $m$, and $i-k=\pm m$, so that $a^m=b$. But then $aa^maa^m=a^2$, while $abab=\gid$, so $n=2$. \end{proof} \section{Products and sums}\label{sect:prod-sum} \begin{theorem}\label{thm:prod} Let $G_0$, $G_1$ and $H$ be groups. For each $i$ in $2$, let $\pi_i$ be the homomorphism $(x_0,x_1)\mapsto x_i$ from $G_0\times G_1$ to $G_i$, and let $f_i$ be a homomorphism from $H$ to $G_i$. Then there is a homomorphism \begin{equation*} x\mapsto(f_0(x),f_1(x)) \end{equation*} from $H$ to $G_0\times G_1$, and this is the unique homomorphism $f$ from $H$ to $G_0\times G_1$ such that, for each $i$ in~$2$, \begin{equation*} \pi_if=f_i \end{equation*} ---that is, the following diagram commutes: \begin{equation*} \xymatrix{ G_0 & \ar[l]_-{\pi_0} G_0\times G_1 \ar[r]^-{\pi_1} & G_1\\ & \ar[ul]^{f_0} \ar[u]_f H \ar[ur]_{f_1} & } \end{equation*} \end{theorem} \begin{proof} If $u\in G_0\times G_1$, then $u=(\pi_0(u),\pi_1(u))$. Hence, if $f\colon H\to G_0\times G_1$, then $f(x)=(\pi_0f(x),\pi_1f(x))$. In particular then, $f$ is as desired if and only if $f(x)=(f_0(x),f_1(x))$. \end{proof} We can generalize this theorem by considering an indexed family $(G_i\colon i\in I)$ of groups. The \textbf{direct product}\index{direct product} of this family is denoted by \begin{equation*} \prod_{i\in I}G_i. \end{equation*} This is, first of all, the set whose elements are $(x_i\colon i\in I)$ (that is, functions $i\mapsto x_i$ on~$I$) such that $x_i\in G_i$ for each $i$ in $I$. An operation of multiplication on this set is given by \begin{equation*} (x_i\colon i\in I)(y_i\colon i\in I)=(x_iy_i\colon i\in I). \end{equation*} Under this multiplication, $\prod_{i\in I}G_i$ becomes a group. If $i\in I$, we define a homomorphism $\pi_i$ from $\prod_{i\in I}G_i$ to $G_i$ by \begin{equation*} \pi_i(x_j\colon j\in I)=x_i. \end{equation*} In case $I=n$, we may write $\prod_{i\in I}G_i$ also as \begin{equation*} G_0\times\cdots\times G_{n-1}, \end{equation*} and a typical element of this as \begin{equation*} (x_0,\dots,x_{n-1}). \end{equation*} To the previous theorem we have: \begin{porism} Suppose $(G_i\colon i\in I)$ is an indexed family of groups, and $H$ is a group, and for each $i$ in $I$ there is a homomorphism from $H$ to $G_i$. Then there is a homomorphism \begin{equation*} x\mapsto(f_i(x)\colon i\in I) \end{equation*} from $H$ to $\prod_{i\in I}G_i$, and this is the unique homomorphism $f$ from $H$ to $\prod_{i\in I}G_i$ such that, for each $i$ in~$I$, \begin{equation*} \pi_if=f_i. \end{equation*} \end{porism} The direct product of a family of abelian groups is an abelian group. When we restrict attention to abelian groups, then we can reverse the arrows in Theorem~\ref{thm:prod}: \begin{theorem}\label{thm:oplus} Let $G_0$, $G_1$ and $H$ be abelian groups. Let $\iota_0$ be the homomorphism $x\mapsto(x,0)$ from $G_0$ to $G_0\oplus G_1$, and let $\iota_1$ be $x\mapsto(0,x)$ from $G_1$ to $G_0\oplus G_1$. For each $i$ in $2$, let $f_i$ be a homomorphism from $G_i$ to $H$. Then there is a homomorphism \begin{equation*} (x_0,x_1)\mapsto f_0(x_0)+f_1(x_1) \end{equation*} from $G_0\oplus G_1$ to $H$, and this is the unique homomorphism $f$ from $G_0\oplus G_1$ to $H$ such that, for each $i$ in~$2$, \begin{equation*} f\iota_i=f_i \end{equation*} ---that is, the following diagram commutes: \begin{equation*} \xymatrix{ G_0 \ar[r]^-{\iota_0} \ar[dr]_{f_0} & G_0\oplus G_1 \ar[d]^f & \ar[l]_-{\iota_1} \ar[dl]^{f_1} G_1\\ & H & } \end{equation*} \end{theorem} \begin{proof} Every element $(x_0,x_1)$ of $G_0\oplus G_1$ is $\iota_0(x_0)+\iota_1(x_1)$, so that, if $f$ is a homomorphism on $G_0\oplus G_1$, then \begin{equation}\label{eqn:f+} f(x_0,x_1)=f\iota_0(x_0)+f\iota_1(x_1). \end{equation} Hence $f$ is as desired if and only if $f(x_0,x_1)=f_0(x_0)+f_1(x_1)$. The function so defined is indeed a homomorphism, since \begin{multline*} f((x_0,x_1)+(u_0,u_1)) =f(x_0+u_0,x_1+u_1) =f_0(x_0+u_0)+f_1(x_1+u_1)\\ \begin{aligned} &=f_0(x_0)+f_0(u_0)+f_1(x_1)+f_1(u_1)\\ &=f_0(x_0)+f_1(x_1)+f_0(u_0)+f_1(u_1) =f(x_0,x_1)+f(u_0,u_1), \end{aligned} \end{multline*} because $H$ is abelian. \end{proof} In the proof, the definition of $f$ in~\eqref{eqn:f+} relies on the \emph{finiteness} of the family $(G_i\colon i\in 2)$; more precisely, it relies on the finiteness of $\{i\in 2\colon x_i\neq e)$. Of an arbitrary indexed family $(G_i\colon i\in I)$ of groups, we define the \textbf{\emph{weak} direct product}\index{weak direct product} to be the subgroup, denoted by \begin{equation*} \weakprod_{i\in I}G_i, \end{equation*} of $\prod_{i\in I}G_i$ comprising those elements $(x_i\colon i\in I)$ such that $\{i\in I\colon x_i\neq\gid\}$ is finite. We define a homomorphism $\iota_i$ from each $G_i$ to $\textweakprod_{j\in I}G_j$ by \begin{equation*} \iota_i(x)=(x_j\colon j\in I), \end{equation*} where \begin{equation*} x_j= \begin{cases} x,&\text{ if }j=i;\\ \gid,&\text{ otherwise.} \end{cases} \end{equation*} If $I$ is finite, then the weak direct product is the same as the (full) direct product. Proving that $f$ as in~\eqref{eqn:f+} is a \emph{homomorphism} uses that $H$ is abelian. The weak direct product of a family $(G_i\colon i\in I)$ of abelian groups is called the \textbf{direct sum}\index{direct sum} and is denoted by \begin{equation*} \sum_{i\in I}G_i. \end{equation*} In case $I=n$, we may write $\sum_{i\in I}G_i$ also as \begin{equation*} G_0\oplus\cdots\oplus G_{n-1}. \end{equation*} To the previous theorem we have: \begin{porism} Suppose $(G_i\colon i\in I)$ is an indexed family of abelian groups, and $H$ is an abelian group, and for each $i$ in $I$ there is a homomorphism $f_i$ from $G_i$ to $H$. Then there is a homomorphism \begin{equation*} x\mapsto\sum_{i\in I}f_i(x_i) \end{equation*} from $\sum_{i\in I}G_i$ to $H$, and this is the unique homomorphism $f$ from $\sum_{i\in I}G_i$ to $H$ such that, for each $i$ in~$I$, \begin{equation*} f\iota_i=f_i. \end{equation*} \end{porism} Now we can provide an example promised in \S~\ref{sect:rings}. Let $E$ be the abelian group $\sum_{n\in\vnn}\Z$. Suppose $f$ is a singulary operation on $\vnn$. An element $f^*$ of $\End E$ is induced, given by \begin{equation*} f^*(x_n\colon n\in\vnn)=(x_{f(n)}\colon n\in\vnn). \end{equation*} Then $f^*\iota_{f(n)}=\iota_n$. Let $f$ be the operation $x\mapsto x+1$ on $\vnn$, and let $g$ be the operation given by \begin{equation*} g(x)= \begin{cases} y,&\text{ if }f(y)=x,\\ 0,&\text{ if }x=0. \end{cases} \end{equation*} Then $gf(x)=x$, so $f^*g^*=(gf)^*$, the identity in $\End E$; but $g^*f^*$ is not the identity, since it is $(fg)^*$, and $fg(0)=1=fg(1)$. We have two kinds of products so far, related as follows. \begin{theorem} Let $(G_i\colon i\in I)$ be an indexed family of groups. Then \begin{align*} \iota_j[G_j]&\nsubgp\weakprod_{i\in I}G_i,& \weakprod_{i\in I}G_i&\nsubgp\prod_{i\in I}G_i,& \iota_j[G_j]&\nsubgp\prod_{i\in I}G_i.\qedhere \end{align*} \end{theorem} Theorem~\ref{thm:oplus} and its porism can be generalized to some cases of arbitrary groups: \begin{theorem} Suppose $(G_i\colon i\in I)$ is an indexed family of groups, and $H$ is a group, and for each $i$ in $I$ there is a homomorphism $f_i$ from $G_i$ to $H$. Suppose further that, for all $i$ and $j$ in $I$, \begin{equation*} f_i(x)f_j(y)=f_j(y)f_i(x). \end{equation*} Then there is a homomorphism \begin{equation*} x\mapsto\prod_{i\in I}f_i(x_i) \end{equation*} from $\textweakprod_{i\in I}G_i$ to $H$, and this is the unique homomorphism $f$ from $\textweakprod_{i\in I}G_i$ to $H$ such that, for each $i$ in~$I$, \begin{equation*} f\iota_i=f_i. \end{equation*} \end{theorem} As a special case of this theorem, we have the next theorem below, by means of the following: \begin{lemma} If $M$ and $N$ are normal subgroups of $G$, and \begin{equation*} M\cap N=\gpgen{\gid}, \end{equation*} then each element $m$ of $M$ commutes with each element $n$ of $N$, that is, \begin{equation*} mn=nm. \end{equation*} \end{lemma} \begin{proof} We can analyze $mnm\inv n\inv$ both as the element $(mnm\inv)n\inv$ of $N$ and as the element $m(nm\inv n\inv)$ in $M$; so the element is $\gid$, and therefore $mn=(m\inv n\inv)\inv=nm$. \end{proof} \begin{theorem}\label{thm:wdp} If $(N_i\colon i\in I)$ is an indexed family of normal subgroups of a group, and for each $j$ in $I$, \begin{equation} N_j\cap\Bigl\langle\bigcup_{i\in I\setminus\{j\}}N_i\Bigr\rangle=\gpgen{\gid}, \end{equation} then \begin{equation*} \Bigl\langle\bigcup_{i\in I}N_i\Bigr\rangle \cong\weakprod_{i\in I}N_i. \end{equation*} \end{theorem} \begin{proof} Say the $N_i$ are normal subgroups of $G$. Since $N_i\cap N_j=\gpgen{\gid}$ whenever $i\neq j$, the last theorem and the lemma guarantee that there is a homomorphism $h$ from $\textweakprod_{i\in I}N_i$ into $G$ such that, for each $i$ in $I$, the composition $h\iota_i$ is just the inclusion of $N_i$ in $G$. Then the range of $h$ is $\Bigl\langle\bigcup_{i\in I}N_i\Bigr\rangle$. To show that $h$ is injective, note that, if $n\in\textweakprod_{i\in I}N_i$ and $h(n)=\gid$, then, for each $j$ in $I$, we have \begin{equation*} n_j{}\inv=\prod_{i\in I\setminus\{j\}}n_i. \end{equation*} The left member is in $N_j$, the right in $\Bigl\langle\bigcup_{i\in I\setminus\{j\}}N_i\Bigr\rangle$, so each side is $\gid$; in particular, $n_j=\gid$. Therefore $n=\gid$. \end{proof} In the conclusion of the theorem, $G$ is the \textbf{\emph{internal} weak direct product}\index{internal weak direct product} of the~$N_i$. \section{Free groups} The direct sum $\sum_{i\in I}\Z$ has elements $\gid^i$, namely $\iota_i(1)$ or $(\delta_j^i\colon j\in I)$, where \begin{equation*} \delta_j^i= \begin{cases} 1,&\text{ if }j=i,\\ 0,&\text{ otherwise.} \end{cases} \end{equation*} An arbitrary element of $\sum_{i\in I}$ is a \textbf{`formal sum,'}\index{formal sum} \begin{equation*} \sum_{i\in I}x_i\gid^i. \end{equation*} \begin{theorem}\label{thm:free-ab} Suppose $G$ is an abelian group, $I$ is a set, and $f$ is a function from $I$ to $G$. Then there is a homomorphism \begin{equation*} \sum_{i\in I}x_i\gid^i\mapsto\sum_{i\in I}x_if(i) \end{equation*} from $\sum_{i\in I}\Z$ to $G$, and this is the unique homomorphism $\tilde f$ from $\sum_{i\in I}$ to $G$ such that, for each $i$ in~$I$, \begin{equation*} \tilde f(\gid^i)=f(i) \end{equation*} ---that is, the following diagram commutes, where $\iota$ is the map $i \mapsto\gid^i$: \begin{equation*} \xymatrix{ I \ar[r]^-{\iota} \ar[d]_f & \ar[dl]^{\tilde f} \displaystyle\sum_{i\in I}\Z \\ G & } \end{equation*} \end{theorem} The direct sum $\sum_{i\in I}\Z$ in the theorem is the \textbf{free abelian group}\index{free abelian group} on $I$ with respect to the map $i\mapsto\gid^i$. There is also a \textbf{free group}\index{free group} on $I$, which we may denoted by \begin{equation*} \free I. \end{equation*} This is the group of \emph{reduced words} on $I$. A \textbf{word}\index{word} on $I$ is a finite nonempty string $t_0t_1\cdots t_n$, where each entry $t_k$ is either $\gid$, or else $a$ or $a\inv$ for some $a$ in $I$. A word is \textbf{reduced}\index{reduced} if $a$ and $a\inv$ are never adjacent in it, and $\gid$ is never adjacent to any other entry (so $\gid$ can appear only in the string $\gid$). We make $\free I$ into a group when the multiplication is defined as juxtaposition followed by \textbf{reduction,}\index{reduction} namely, replacement of each occurrence of $aa\inv$ or $a\inv a$ with $\gid$, and replacement of each occurrence of $x\gid$ or $\gid x$ with $x$. Thus, when an element $a$ of $I$ is written as $a^{+1}$, we have \begin{equation*} (a_{m}^{\epsilon(m)}\cdots a_{0}^{\epsilon(0)}) (b_{0}^{\zeta(0)}\cdots b_{n}^{\zeta(n)})= a_{m}^{\epsilon(m)}\cdots a_{j}^{\epsilon(j)}b_j^{\zeta(j)}\cdots b_{n}^{\zeta(n)}, \end{equation*} where $j$ is maximal such that, if $i2$, then $\Dih n$ has the presentation $\gpres{a,b}{a^n,b^2,abab}$. \end{theorem} \begin{proof} Let $G=\gpres{a,b}{a^n,b^2,abab}$. Then the order of (the image of) $a$ in $G$ divides $n$, and the order of $b$ divides $2$. But by von Dyck's Theorem and Theorem~\ref{thm:Dn}, $G$ maps onto $\Dih n$, and hence $n$ divides the order of $a$ in $G$, and $2$ divides the order of $b$. Therefore $\Dih n\cong G$. \end{proof} \begin{theorem} The group $\gpres{\mi,\mj}{\mi^4,\mi^2\mj^2,\mi\mj\mi^3\mj}$ has order $8$, and its elements are (the images of) $\pm 1$, $\pm\mi$, $\pm\mj$, $\pm\mk$, where $1=\gid$ and $\mk=\mi\mj$ and $-x=\mi^2x$. \end{theorem} \begin{proof} Let the group be called $G$. In $G$, we have $\mj^2=\mi^{-2}=\mi^2$, so $\mj^4=1$. Hence also $\mk=\mi\mj=\mj^3\mi$, so $\mi^3\mj=\mj\mi$. This shows that every element of $G$ can be written as $\mi^n\mj^m$, where $n\in4$ and $m\in 2$; hence it is one of the given elements. \end{proof} \section[Fin.~gen.~ab.~groups]{Finitely generated abelian groups} To \textbf{classify}\index{classify} a collection of groups is to find a function $f$ such that \begin{equation*} f(G)=f(H)\iff G\cong H \end{equation*} for all groups $G$ and $H$ in the collection. We do this now with the finitely generated abelian groups, and in particular with the finite abelian groups. The next theorem will be needed for Theorem~\ref{thm:Zp-cross}. \begin{theorem}\label{thm:fin-gen-ab} For every abelian group $G$ on $n$ generators, there is a unique element $k$ of $n$, along with positive integers $d_0$, \dots, $d_{k-1}$, where \begin{equation}\label{d} d_0\divides\dotsm\divides d_{k-1}, \end{equation} such that \begin{equation}\label{FH} G\cong \Zmod{d_0}\oplus\dotsb\oplus\Zmod{d_{k-1}}\oplus \underbrace{\Z\oplus\dotsb\oplus\Z}_{n-k}. \end{equation} \end{theorem} \begin{proof} %Suppose $G$ is an abelian group with a generating set of size $n$. Let $F$ be the free abelian group $\sum_{i\in n}\Z$. Then \begin{equation*} G\cong F/N, \end{equation*} where $N$ is the kernel of the induced epimorphism from $F$ onto $G$. As before, each element of $F$ can be understood as a formal sum $\sum_{i\in n}x_i\gid^i$. Then $F$ itself is $\gpgen{\gid^0,\dots,\gid^{n-1}}$. If $N=\gpgen{d_0\gid^0,\dots,d_{k-1}\gid^{k-1}}$, then $G$ is as in~\eqref{FH}. Not every subgroup of $F$ is given to us so neatly, but we can use linear algebra to put it into this form. Every element of $F$, considered as a formal sum, can be written also as a matrix product: \begin{equation*} x_0a^0+\dotsb+x_{n-1}a^{n-1}= \begin{pmatrix} x_0&\cdots&x_{n-1} \end{pmatrix} \begin{pmatrix} {\gid}^0\\\vdots\\{\gid}^{n-1} \end{pmatrix}=\tuple x\mathbf e. \end{equation*} %for some \emph{unique} $n$-tuple $(x_0,\dots,x_{n-1})$ of integers. The generators of a (finitely generated) subgroup of $F$ can be considered as the entries of a column vector, and this column can be considered as the product of a matrix over $\Z$ with $\mathbf e$: %Considering finite columns of such elements, we have \begin{equation*} \begin{pmatrix} x_0^0{\gid}^0+\dotsb+x_{n-1}^0{\gid}^{n-1}\\ \vdots\\ x_0^{m-1}{\gid}^0+\dotsb+x_{n-1}^{m-1}{\gid}^{n-1} \end{pmatrix} =\begin{pmatrix} x_0^0&\dots&x_{n-1}^0\\ \vdots&\ddots&\vdots\\ x_0^{m-1}&\dots&x_{n-1}^{m-1} \end{pmatrix} \begin{pmatrix} {\gid}^0\\\vdots\\{\gid}^{n-1} \end{pmatrix} =X\mathbf e. \end{equation*} The subgroup of $F$ generated by the rows of $X\mathbf e$ can be denoted by $\gpgen {X\mathbf e}$. If $P$ is an $m\times m$ matrix with integer entries, then \begin{equation*} \gpgen{PX\mathbf e}\included\gpgen{X\mathbf e}. \end{equation*} If also $P$ is \emph{invertible}---that is, $\det(P)=\pm1$---then \begin{equation*} \gpgen{PX\mathbf e}=\gpgen{X\mathbf e}. \end{equation*} We can therefore perform the following row-operations on $X$, without changing the group $\gpgen{X\mathbf e}$. We can \begin{enumerate} \item interchange two rows, \item multiply a row by $-1$, \item add an integer multiple of one row to another. \end{enumerate} These operations allow us to perform Gaussian elimination. Adding rows of zeros as necessary, we may also assume that $m\geq n$. Then for some invertible integer matrix $P$, we have \begin{equation*} PX= \begin{pmatrix} T\\\hline0 \end{pmatrix}, \end{equation*} where $T$ is an $n\times n$ upper-triangular matrix, \begin{equation*} T= \begin{pmatrix} *&\cdots&*\\ &\ddots&\vdots\\ 0& &* \end{pmatrix}. \end{equation*} By using also invertible \emph{column}-operations, we can diagonalize $T$. That is, there are invertible integer matrices $P$ and $Q$ such that \begin{equation*} PXQ= \begin{pmatrix} D\\\hline0 \end{pmatrix}, \end{equation*} where \begin{equation*} D= \begin{pmatrix} d_0& &0\\ &\ddots&\\ 0& &d_{n-1} \end{pmatrix}. \end{equation*} We now have \begin{equation*} \gpgen{X\mathbf e}=\gpgen{PXQQ\inv\mathbf e}=\gpgen{DQ\inv\mathbf e}\cong\gpgen{D\mathbf e}. \end{equation*} Working further on $D$ with invertible row- and column- operations, we may assume~\eqref{d} holds, while $d_k=\cdots=d_{n-1}=0$. Indeed, suppose $b,c\in\Z$ and $\gcd(b,c)=d$. By invertible operations, from \begin{equation*} \begin{pmatrix} b&0\\0&c \end{pmatrix} \end{equation*} we obtain $\begin{pmatrix} b&0\\c&c \end{pmatrix}$ and then $\begin{pmatrix} d&e\\0&f \end{pmatrix}$, where $e$ and $f$ are multiples of $c$ and hence of $d$; hence, with an invertible column-operation, we get \begin{equation*} \begin{pmatrix} d&0\\0&f \end{pmatrix}. \end{equation*} where again $d\divides f$. Applying such transformations as needed to pairs of entries in $D$ yields~\eqref{d}. \end{proof} \begin{porism} Every subgroup of a free abelian group on $n$ generators is free abelian on $n$ generators or fewer. \end{porism} We can show uniqueness of the numbers $d_j$ by an alternative analysis. \begin{theorem}[Chinese Remainder]\label{thm:CRT} If $\gcd(m,n)=1$, then the homomorphism $x\mapsto(x,x)$ from $\Zmod{mn}$ to $\Zmod m\oplus\Zmod n$ is an isomorphism. \end{theorem} \begin{proof} If $x\equiv0\pmod m$ and $x\equiv 0\pmod n$, then $x\equiv0\pmod{mn}$. Hence the given homomorphism is injective. Its surjectivity follows by counting. \end{proof} The Chinese Remainder Theorem will be generalized as Theorem~\ref{thm:CRT-R}. In the usual formulation of the theorem, every system \begin{align*} x&\equiv a\pmod m,&x&\equiv b\pmod n \end{align*} has a unique solution \emph{modulo} $mn$; but this solution is just the inverse image of $(a,b)$ under the isomorphism $x\mapsto(x,x)$. \begin{theorem} For every finite abelian group, there are unique primes $p_0$, \dots, $p_{k-1}$, not necessarily distinct, along with unique positive integers $m(0)$, \dots, $m(k-1)$, such that \begin{equation*} G\cong \Zmod{p_0{}^{m(0)}}\oplus\dotsb\oplus\Zmod{p_{k-1}{}^{m(k-1)}}. \end{equation*} \end{theorem} \begin{proof} To obtain the analysis, apply the Chinese Remainder Theorem to Theorem~\ref{thm:fin-gen-ab}. The analysis is unique, provided it is unique in the case where all of the $p_j$ are the same. But in this case, the analysis is unique, by repeated application of the observation that the order of the group is the highest prime power appearing in the factorization. \end{proof} \section{Semidirect products}\label{sect:semidirect} An isomorphism from a structure to itself is an \textbf{automorphism.}\index{automorphism} \begin{theorem} The automorphisms of a group $G$ compose a subgroup of $\Sym G$. \end{theorem} The subgroup in the theorem is denoted by \begin{equation*} \Aut G. \end{equation*} \begin{theorem} For every group $G$, there is a homomorphism \begin{equation*} g\mapsto(x\mapsto gxg\inv) \end{equation*} from $G$ to $\Aut G$. \end{theorem} An automorphism $x\mapsto gxg\inv$ as in the theorem is \textbf{conjugation}\index{conjugation} by $g$ and is an \textbf{inner automorphism}\index{inner automorphism} of $G$. The kernel of the homomorphism in the theorem is the \textbf{center}\index{center} of $G$, denoted by\footnote{An alternative formulation of the center of a group is given and generalized in \S~\ref{sect:nilpotent}.} \begin{equation*} \centr G. \end{equation*} Then $G$ is \textbf{centerless}\index{centerless} if $\centr G$ is trivial. Repeating the process of forming inner automorphisms, we obtain a chain \begin{equation*} G\to\Aut G\to\Aut{\Aut G}\to\dotsb, \end{equation*} called the \textbf{automorphism tower}\index{automorphism tower} of $G$. The tower reaches a fixed point, perhaps after transfinitely many steps: Simon Thomas~\cite{MR801316} shows this in case $G$ is centerless; Joel Hamkins~\cite{MR1487370}, in the general case. \begin{theorem}\label{thm:GNG} For every group $G$, if $N\nsubgp G$, then there is a homomorphism \begin{equation*} g\mapsto(x\mapsto gxg\inv) \end{equation*} from $G$ to $\Aut N$. \end{theorem} In the theorem, let the homomorphism be $g\mapsto\sigma_g$. Suppose also $H\subgp G$, and $N\cap H=\gpgen{\gid}$. Then the conditions of Theorem~\ref{thm:isdp} are met, and $NH$ is an internal semidirect product. Equation~\eqref{eqn:sdp}, describing multiplication on $NH$, can be rewritten as \begin{equation*} (mg)(nh)=(m\cdot\sigma_g(n))(gh). \end{equation*} \begin{theorem} Suppose $N$ and $H$ are groups, and $g\mapsto\sigma_g$ is a homomorphism from $H$ to $\Aut N$. Then the set $N\times H$ becomes a group when multiplication is defined by \begin{equation*} (m,g)(n,h)=(m\cdot\sigma_g(n),gh). \end{equation*} \end{theorem} \begin{proof} To check that the multiplication is associative means checking that \begin{equation*} \lambda_{(m,g)}\lambda_{(n,h)}=\lambda_{(m,g)(n,h)}. \end{equation*} We can write $\lambda_{(m,g)}$ as $\lambda_m\sigma_g\times \lambda_g$. Then \begin{align*} \lambda_{(m,g)}\lambda_{(n,h)} =(\lambda_m\sigma_g\times\lambda_g)(\lambda_n\sigma_h\times\lambda_h) &=\lambda_m\sigma_g\lambda_n\sigma_h\times\lambda_g\lambda_h\\ &=\lambda_m\lambda_{\sigma_g(n)}\sigma_g\sigma_h\times\lambda_{gh}\\ &=\lambda_{m\cdot\sigma_g(n)}\sigma_{gh}\times\lambda_{gh}\\ &=\lambda_{(m\cdot\sigma_g(n),gh)}\\ &=\lambda_{(m,g)(n,h)}. \end{align*} Finally, $(\gid,\gid)$ is an identity, and $(\sigma_{h\inv}(n\inv),h\inv)$ is an inverse of $(n,h)$. \end{proof} The group given by the theorem is the \textbf{semidirect product}\index{semidirect product} of $N$ and $H$ with respect to $\sigma$; it can be denoted by \begin{equation*} N\rtimes_{\sigma}H. \end{equation*} The bijection in Theorem~\ref{thm:isdp} is an isomorphism from $N\rtimes_{\sigma}H$ to $NH$ when $\sigma$ is as in Theorem~\ref{thm:GNG}. % $g\mapsto(x\mapsto gxg\inv)$. \begin{theorem}\label{thm:Zp-cross} If $p$ is prime, then $\Zmodu p\cong\Zmod{p-1}$. \end{theorem} \begin{proof} The group $\Zmodu p$ has order $p-1$ and, by Theorem~\ref{thm:fin-gen-ab}, is isomorphic to \begin{equation*} \Zmod {d_0}\oplus\Zmod{d_{k-1}}\oplus \Zmod m, \end{equation*} where $d_0\divides\cdots\divides d_{k-1}\divides m$. Hence every element of $\Zmodu p$ is a root of the polynomial $x^m-1$. But this polynomial can have at most $m$ roots in $\Zmod p$, since this is a \emph{field.} Hence $p-1\leq m$, so $m=p-1$, and $k=0$. \end{proof} \begin{theorem} The embedding $x\mapsto\lambda_x$ of a ring $(E,\cdot)$ in $(\End E,\circ)$ restricts to an embedding of $\unit{(E,\cdot)}$ in $\Aut E$. In case $E$ is $\Zmod n$, each embedding is an isomorphism. In particular, if $a$ is an element of $\Zmodu n$ of order $m$, and $m\divides t$, then $\Zmod t$ acts on $\Zmod n$ by $(x,y)\mapsto a^xy$. Conversely, if some $\Zmod t$ acts on $\Zmod n$, then the action is so given for some such $a$. \end{theorem} \begin{theorem}\label{thm:pq} For every odd prime $p$, for every prime divisor $q$ of $p-1$, there is a non-abelian semidirect product $\Zmod p\rtimes_{\sigma}\Zmod q$, which is unique up to isomorphism. \end{theorem} \begin{proof} As $\Zmodu p$ is cyclic, it has a unique subgroup $G$ of order $q$. As $q$ is prime, every nontrivial element of $G$ is a generator. If $a\in G\setminus\{1\}$, let $\sigma$ be the homomorphism $x\mapsto(y\mapsto a^xy)$ from $\Zmod q$ to $\Aut{\Zmod p}$. Then we can form \begin{equation*} \Zmod p\rtimes_{\sigma}\Zmod q. \end{equation*} If $\Zmod p\rtimes_{\tau}\Zmod q$ is some other non-abelian semidirect product, then $\tau_1$ is $x\mapsto b\cdot x$ for some $b$ in $G\setminus\{1\}$. But then $b^n=a$ for some $n$, so there is an isomorphism from $\Zmod p\rtimes_{\sigma}\Zmod q$ to $\Zmod p\rtimes_{\tau}\Zmod q$ that takes $(x,y)$ to $(x,ny)$. \end{proof} Because of its uniqueness, we may refer to the semidirect product of the theorem as \begin{equation*} \Zmod p\rtimes\Zmod q. \end{equation*} In case $q=2$, this group is $\Dih p$. The next section develops the tools used in \S~\ref{sect:class-small} to show that there is no other way to obtain a group of order $pq$ for distinct primes $p$ and $q$. \section{Actions of groups}\label{sect:actions} \begin{theorem} Let $G$ be a group, and $\setactedon$ a set. There is a one-to-one correspondence between \begin{enumerate} \item homomorphisms $g\mapsto(a\mapsto ga)$ from $G$ into $\Sym{\setactedon}$, and \item functions $(g,a)\mapsto ga$ from $G\times A$ into $A$ such that \begin{gather}\label{act:1} \gid a=a,\\\label{act:gha} (gh)a=g(ha). \end{gather} for all $h$ and $h$ in $G$ and $a$ in $A$. \end{enumerate} \end{theorem} \begin{proof} If $g\mapsto(a\mapsto ga)$ maps $G$ homomorphically into $\Sym{\setactedon}$, then~\eqref{act:1} and~\eqref{act:gha} follow. Suppose conversely that these hold. Then, in particular, \begin{equation*} g(g\inv a)=(gg\inv)a=\gid a=a \end{equation*} and likewise $g\inv(ga)=a$, so $a\mapsto g\inv a$ is the inverse of $a\mapsto ga$, and the function $g\mapsto(a\mapsto ga)$ does map $G$ into $\Sym{\setactedon}$, homomorphically by~\eqref{act:gha}. \end{proof} Either of two functions that correspond as in the theorem is a \textbf{(left) action}\index{action}\index{left!--- action} of $G$ on $A$. Examples include the following. \begin{asparaenum}[1.] \item A symmetry group of a set acts on the set in the obvious way, by \begin{equation*} (\sigma,x)\mapsto\sigma(x). \end{equation*} \item An arbitrary group $G$ acts on itself by left multiplication: \begin{equation*} (g,x)\mapsto\lambda_g(x). \end{equation*} \item If $H\subgp G$, then $G$ acts on the set $G/H$ by \begin{equation*} (g,xH)\mapsto gxH. \end{equation*} \item Finally, $G$ acts on itself by conjugation: \begin{equation*} (g,x)\mapsto x\mapsto gxg\inv. \end{equation*} \end{asparaenum} Suppose $(g,x)\mapsto gx$ is an arbitrary action of $G$ on $\setactedon$. If $a\in\setactedon$, then the subset $\{g\setcolon ga=a\}$ of $G$ is the \textbf{stabilizer}\index{stabilizer} of $a$, denoted by \begin{equation*} G_a; \end{equation*} the subset $\{ga\setcolon g\in G\}$ of ${\setactedon}$ is the \textbf{orbit}\index{orbit} of $a$, denoted by \begin{equation*} Ga. \end{equation*} The subset $\{x\setcolon G_x=G\}$ of ${\setactedon}$ can be denoted by \begin{equation*} {\setactedon}_0. \end{equation*} See Appendix~\ref{App:ga} for an alternative development of these notions. \begin{theorem}\label{thm:action} Let $G$ act on ${\setactedon}$ by $(g,x)\mapsto gx$. \begin{enumerate} \item The orbits partition $\setactedon$; \item $G_a\subgp G$; \item\label{GGa} $[G:G_a]=\size {Ga}$; \end{enumerate} \end{theorem} \begin{proof} For~\eqref{GGa}, we establish a bijection between $G/G_a$ and $Ga$ by noting that \begin{equation*} gG_a=hG_a\iff h\inv g\in G_a\iff ga=ha; \end{equation*} so the bijection is $gG_a\mapsto ga$. \end{proof} \begin{corollary} If there are only finitely many orbits in $\setactedon$ under $G$, then \begin{equation}\label{eqn:class} \size {\setactedon}=\size{{\setactedon}_0} +\sum_{a\in X}[G:G_a] \end{equation} for some set $X$ of elements of $\setactedon$ whose orbits are nontrivial. \end{corollary} Equation~\eqref{eqn:class} is the \textbf{class equation.}\index{class equation} For example, suppose $G$ acts on itself by conjugation, and $g\in G$. Then $Gg$ is the \textbf{conjugacy class}\index{conjugacy class} of $g$, while $G_g$ is the \textbf{centralizer}\index{centralizer} of $g$, denoted by\footnote{More generally, if $H}[l]\ar[d]\\ G \ar[d] & \cseries nG' \ar[l]\ar[d] & \cseries{n-1}G'\ar[l]\ar[d] & \ar[l]\ar[d]\cseries{n-2}G'\ar[l] & \centr G' \ar@{.>}[l] \ar[d]\\ \cseries nG & \ar[l] \cseries {n-1}G & \ar[l] \cseries{n-2}G & \ar[l] \cseries{n-3}G & \gpgen{\gid} \ar@{.>}[l] } \end{equation*} That is, we know $\dsubgp 0G\subgp\cseries nG$; and if $\dsubgp k G \subgp \cseries{n-k}G$ for some $k$ in $n$, then \begin{equation*} \dsubgp{k+1}G=(\dsubgp kG)'\subgp \cseries{n-k}G'\subgp \cseries{n-(k+1)}G. \end{equation*} By induction then, $\dsubgp nG\subgp \cseries 0G=\gpgen{\gid}$, so $\dsubgp nG=\gpgen{\gid}$. \end{proof} \begin{theorem} Solubility is preserved in subgroups and quotients. If $N\nsubgp G$, and $N$ and $G/N$ are soluble, then $G$ is soluble. \end{theorem} \begin{proof} Suppose $f\colon G\to H$. By~\eqref{eqn:f([x,y])}, we have $f(\dsubgp nG)\subgp \dsubgp nH$, with equality is $f$ is surjective. The case where $f$ is an inclusion of $G$ in $H$ shows that subgroups of soluble groups are soluble. The case where $f$ is a quotient map shows that quotients of soluble groups are soluble. Finally, if $N\nsubgp G$, then $(G/N)'=G'N/N$. Suppose $\dsubgp n{(G/N)}=\gpgen{\gid}$, and $\dsubgp mN=\gpgen{\gid}$. Then $\dsubgp nG\subgp N$ and so $\dsubgp{n+m}G=\gpgen{\gid}$. \end{proof} \begin{theorem} Groups with non-abelian simple subgroups are not soluble. In particular, $\Sym 5$ is not soluble if $n\geq 5$. \end{theorem} \begin{proof} Suppose $H$ is simple. Since $H'\nsubgp H$, we have either $H'=\gpgen{\gid}$ or $H'=H$. In the former case, $H$ is abelian; in the latter, $H$ is insoluble. \end{proof} The last theorem suggests the origin of the notion of solubility of groups: the general 5th-degree polynomial equation \begin{equation*} a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+x^5=0 \end{equation*} is ``insoluble by radicals'' precisely because $\Sym 5$ is an insoluble group. \section{Normal series} A \textbf{normal series}\index{normal!--- series} for a group $G$ is a sequence $(G_n\colon n\in\vnn)$ of subgroups, where $G_{n+1}\nsubgp G_n$ in each case; the situation can be depicted by \begin{equation*} G=G_0\nsupgp G_1\nsupgp G_2\nsupgp\dotsb \end{equation*} (If one wants to distinguish, one may call this a \textbf{subnormal series}\index{subnormal series}\index{series!subnormal ---}, normal if each $G_i$ is normal in $G$.) The \textbf{factors}\index{factor} of the normal series are the quotients $G_i/G_{i+1}$. If $G_n=\gpgen{\gid}$ for some $n$, then the series is called \begin{enumerate} \item a \textbf{composition series,}\index{composition series}\index{series!composition ---} if the factors are simple; \item a \textbf{soluble series,}\index{soluble series}\index{soluble!--- series}\index{series!soluble ---} if the factors are abelian. \end{enumerate} For example, if $G$ is nilpotent, then the series \begin{equation*} \gpgen{\gid}\nsubgp \centr G\nsubgp \cseries 2G\nsubgp\dotsb \nsubgp G \end{equation*} is a soluble series. \begin{theorem} A group is soluble if and only if it has a soluble series. \end{theorem} \begin{proof} If the series \begin{equation*} G\nsupgp G_1\nsupgp G_2\nsupgp\dotsb\nsupgp G_n=\gpgen{\gid} \end{equation*} is soluble, then, by Theorem~\ref{thm:G'}, we have \begin{align*} G'&1$ and $n>1$, then $m+\gpgen {mn}$ and $n+\gpgen{mn}$ are zero-divisors in $\Zmod {mn}$. The unique element of the trivial ring $\Zmod 1$ is a unit, but not a zero-divisor. A commutative ring is an \textbf{integral domain}\index{integral domain}\index{domain!integral ---}% \index{ring|seealso{domain}} if it has no zero-divisors and $1\neq0$. So fields are integral domains. But $\Z$ is an integral domain that is not a field. If $p$ is prime, then $\Zmod p$ is a field, denoted by $\F_p$. An arbitrary ring $R$ such that $R\setminus\unit R=\{0\}$ is a \textbf{division ring.}\index{division ring} So fields are division rings; but $\Ham$ is a non-commutative division ring. If $R$ is a ring, and $G$ is a group, we can form the direct sum $\sum_{g\in G}R$, which is, first of all, an abelian group; we can give it a multiplication as follows. We write an element $(r_g\colon g\in G)$ of the direct sum as \begin{equation*} \sum_{g\in G}r_gg; \end{equation*} this is a \textbf{formal finite $R$-linear combination}\index{linear combination} of the elements of $G$. Then multiplication is defined as one expects: if $r,s\in R$ and $g,h\in G$, then \begin{equation*} (rg)(sh)=(rs)(gh), \end{equation*} and the definition extends to all of $\sum_{g\in G}R$ by distributivity. The resulting ring can be denoted by \begin{equation*} R(G); \end{equation*} it is the \textbf{group ring}\index{group ring} of $G$ over $R$. We can do the same construction with monoids, rather than groups. For example, if we start with the free monoid generated by a symbol $X$, we get a \textbf{polynomial ring}\index{polynomial ring} in one variable, denoted by \begin{equation*} R[X]; \end{equation*} this is the ring of formal $R$-linear combinations \begin{equation*} \sum_{k=0}^na_kx^k, \end{equation*} where $n\in\vnn$ and $a_k\in R$. We could use a second variable, getting for example $R[X,Y]$. Usually $R$ here is commutative and is in particular a field. \section{Ideals} If $A$ is a sub-ring of $R$, then we can form the abelian group $R/A$. We could try to define a multiplication on this by \begin{equation*} (x+A)(y+A)=xy+A. \end{equation*} However, if $x-x'\in A$, and $y-y'\in A$, we need not have $xy-x'y'\in A$. A \textbf{left ideal}\index{left!--- ideal}\index{ideal!left ---} of $R$ is a sub-ring $I$ such that \begin{equation*} RI\included I, \end{equation*} that is, $rx\in I$ whenever $r\in R$ and $x\in I$. Likewise, \textbf{right}\index{right!--- ideal}\index{ideal!right ---} and \textbf{two-sided} \index{two-sided ideal}\index{ideal!two-sided ---}ideal. For example, the set of matrices \begin{equation*} \begin{bmatrix} * & 0 & \dots & 0\\ \vdots & \vdots & & \vdots\\ * & 0 & \dots & 0 \end{bmatrix} \end{equation*} is a left ideal of $\MatR$, but not a right ideal unless $n=1$. Also, $Rx$ is a left ideal of $R$, while $RxR$ is a two-sided ideal. \begin{theorem} If $I$ is a two-sided ideal of $R$, then $R/I$ is a well-defined ring. The kernel of a ring-homomorphism is a two-sided ideal. \end{theorem} Suppose $(A_i\setcolon i\in I)$ is an indexed family of left ideals of a ring $R$. Let the abelian subgroup of $R$ generated by $\bigcup_{i\in I}A_i$ be denoted by \begin{equation*} \sum_{i\in I}A_i; \end{equation*} this is the \textbf{sum}\index{sum} of the left ideals $A_i$. This must not be confused with the \emph{direct sums} defined in \S~\ref{sect:prod-sum}. If in particular $I=n$, let the abelian subgroup of $R$ generated by \begin{equation*} \{a_0\dotsm a_{n-1}\setcolon a_i\in A_i\} \end{equation*} be denoted by \begin{equation*} A_0\dotsb A_{n-1}; \end{equation*} this is the \textbf{product}\index{product} of the left ideals $A_i$. \begin{theorem} Sums and finite products of left ideals are left ideals; sums and products of two-sided ideals are two-sided ideals. Addition and multiplication of ideals are associative; addition is commutative; multiplication distributes over addition. \end{theorem} \begin{theorem} If $A$ and $B$ are left ideals of a ring, then so is $A\cap B$, and $AB\included A\cap B$. \end{theorem} Usually $AB$ does not include $A\cap B$, since for example $A^2$ might not include $A$; such is the case when $A=2\Z$, since then $A^2=4\Z$. \begin{theorem} If $f\colon R\to S$, a homomorphism of rings, and $I$ is a two-sided ideal of $R$ included in $\ker f$, then there is a unique homomorphism $\tilde f$ from $R/I$ to $S$ such that $f=\tilde f\circ\pi$. \end{theorem} Hence the isomorphism theorems, as for groups. \section{Commutative rings}\label{sect:comm} Henceforth, let all rings be commutative, so all ideals are two-sided. A subset $A$ of a ring $R$ determines the ideal denoted by \begin{equation*} (A), \end{equation*} namely the smallest ideal including $A$. This consists of the \textbf{$R$-linear combinations}\index{linear combination} of elements of $A$, namely the well-defined sums \begin{equation*} \sum_{a\in A}r_aa, \end{equation*} where $r_a\in R$; in particular, $r_a=0$ for all but finitely many $a$. If $A=\{a\}$, then $(A)$ is denoted by \begin{equation*} (a) \end{equation*} or $Ra$ and is called a \textbf{principal ideal.}\index{principal!--- ideal}\index{ideal!principal ---} A \textbf{principal ideal domain}\index{principal!--- ideal domain}\index{domain!principal ideal ---} or \pid\ is an integral domain whose every ideal is principal. For example, $\Z$ is a \pid\ by Theorem~\ref{thm:Z-subg}. But in the polynomial ring $\R[X,Y]$, the ideal $(X,Y)$ is not principal. An ideal is proper if and only if it does not contain a unit. A \emph{proper} ideal $P$ is \textbf{prime}\index{prime} if \begin{equation}\label{eqn:p-ideal} ab\in P\implies a\in P\lor b\in P. \end{equation} So a ring in which $1\neq0$ is an integral domain if and only if $(0)$ is a prime ideal. Compare the definition of prime ideal with the following: a positive integer $p$ is prime if and only if \begin{equation*} p\divides ab\implies p\divides a\lor p\divides b. \end{equation*} We shall address the relation between prime integers and prime ideals in \S~\ref{sect:factor}. Meanwhile, an equivalent formulation of prime ideals is given by the following. \begin{theorem} A proper ideal $P$ of a ring is prime if and only if, for all ideals $I$ and $J$ of the ring, \begin{equation}\label{eqn:IJ} IJ\included P\iff I\included P\lor J\included P. \end{equation} \end{theorem} \begin{proof} The given condition has~\eqref{eqn:p-ideal} as a special case, since the latter can be written as \begin{equation*} (a)(b)\included P\implies(a)\included P\lor (b)\included P. \end{equation*} Also, if~\eqref{eqn:IJ} fails, so that $IJ\included P$, but $I\setminus P$ contains some $a$, and $J\setminus P$ contains some $b$, then $ab\in P$, so~\eqref{eqn:p-ideal} fails. \end{proof} \begin{theorem} A proper ideal $P$ of a ring $R$ is prime if and only if $R/P$ is an integral domain. \end{theorem} \begin{proof} That $I$ is prime means~\eqref{eqn:p-ideal}, which can be written as \begin{equation*} (a+I)(b+I)=I\implies a+I=I\lor b+I=I; \end{equation*} but this means $R/I$ is integral. \end{proof} An ideal is called \textbf{maximal}\index{maximal} if it is maximal as a proper ideal. A ring is a field if and only if $(0)$ is a maximal ideal. (Note that $(0)$ is in fact the ideal with \emph{no} generators, so it could be written as $(\ )$; but it usually is not.) \begin{theorem} A proper ideal $I$ of a ring $R$ is maximal if and only if $R/I$ is a field. \end{theorem} \begin{proof} That $R/I$ is a field means that, if $a\in R\setminus I$, then for some $b$, \begin{equation*} ab\in 1+I. \end{equation*} That $I$ is maximal means that, if $a\in R\setminus I$, then \begin{equation*} I+(a)=R, \end{equation*} equivalently, $1\in I+(a)$, which means that, for some $b$, $ba-1\in I$. \end{proof} \begin{corollary} Maximal ideals are prime. \end{corollary} The converse fails easily, since the prime ideals of $\Z$ are the ideals $(0)$ and $(p)$, where $p$ is prime, and the latter are maximal, but $(0)$ is not. However, it is not even the case that prime ideals other than $(0)$ are always maximal. For example, $\R[X,Y]$ has the prime ideal $(X)$, which is not maximal. A ring is \textbf{Boolean}\index{Boolean} if it respects the identity \begin{equation*} x^2=x. \end{equation*} For example, if $\Omega$ is a set, then $\pow{\Omega}$ is a Boolean ring, where multiplication is intersection, and addition is the taking of \textbf{symmetric differences,}\index{symmetric difference} where the symmetric difference of $x$ and $y$ is $x\setminus y)\cup(y\setminus x)$, denoted by $x\symdiff y$. \begin{theorem}\label{thm:Boole} In Boolean rings, all prime ideals are maximal. \end{theorem} \begin{proof} In a Boolean ring, we have $2x=(2x)^2=4x^2=4x$, %$x+x=(x+x)^2=x^2+2x+x^2=x+2x+x$, so \begin{equation*} 2x=0. \end{equation*} (Thus nontrivial Boolean rings have characteristic $2$.) Hence \begin{equation*} x(1+x)=x+x^2=x+x=0, \end{equation*} so $x$ is a zero-divisor unless it or $1+x$ is $0$, that is, unless $x$ is $0$ or $1$. Therefore there are no Boolean integral domains besides $\F_2$, which is a field. \end{proof} In $\Z$, the ideal $(a,b)$ is the principal ideal generated by $\gcd(a,b)$. So $a$ and $b$ are coprime if $(a,b)=\Z$. This condition can be written as $(a)+(b)=\Z$. Then the following generalizes Theorem~\ref{thm:CRT}. \begin{theorem}[Chinese Remainder]\label{thm:CRT-R}\index{Chinese Remainder Theorem} \index{theorem!Chinese Remainder Th---} Suppose $R$ has an indexed family $(I_i\colon iN(J)>1$. In case $d=-5$, we compute \begin{gather*} (2,1+\tau_{-5})=\gpgen{2,2\tau_{-5},1+\tau_{-5},\tau_{-5}-5} =\gpgen{2,1+\tau_{-5}},\\ (3,1\pm\tau_{-5})=\gpgen{3,3\tau_{-5},1\pm\tau_{-5},\tau_{-5}\mp 5} =\gpgen{3,1\pm\tau_{-5}}, \end{gather*} hence \begin{equation*} \begin{array}{c||c|c} I&(2,1+\tau_{-5})&(3,1\pm\tau_{-5})\\\hline N(I)&2&3 \end{array}. \end{equation*} So these ideals are maximal, hence prime. Ideals of the rings $\Z[\tau_d]$ were originally called \textbf{ideal numbers.} \section{Integral domains}\label{sect:int-dom} \begin{theorem} In an integral domain, if $a$ and $b$ are non-zero associates, and $a=bx$, then $x$ is a unit. \end{theorem} \begin{proof} We have also $b=ay=bxy$, $b(1-xy)=0$, $1=xy$ since $b\neq0$ and we are in an integral domain. \end{proof} \begin{corollary} In an integral domain, prime elements are irreducible. \end{corollary} \begin{proof} If $p$ is prime, and $p=ab$, then $p$ is an associate of $a$ or $b$, so the other is a unit. \end{proof} A \textbf{unique factorization domain}\index{unique factorization domain}\index{domain!unique factorization ---} or UFD is an integral domain whose every non-zero element is `unique\-ly' a product of irreducibles. This means that, if \begin{equation*} \prod_{iUFD} A PID is a UFD.\hfill\qedsymbol \end{theorem} Recall how the Euclidean algorithm for finding greatest common divisors works. To find $\gcd(201,27)$, compute: \begin{align*} 201&=87\cdot 2+27,\\ 87&=27\cdot3+6,\\ 27&=6\cdot 4+3,\\ 6&=3\cdot 2. \end{align*} So $\gcd(201,27)=3$. In general, if $a_0\geq a_1>0$, then $\gcd(a_0,a_1)=a_n$, where there is a descending sequence $(a_0,\dots a_n)$ of positive integers such that $a_{k+2}=a_{k+1}\cdot b_k+a_k$ for some $b_k$. A \textbf{Euclidean domain}\index{Euclidean domain}\index{domain!Euclidean ---} is then an integral domain in which the Euclidean algorithm works. More precisely, a Euclidean domain is a domain $R$ equipped with a map $\phi$ from $R\setminus\{0\}$ to $\vnn$ such that, and, for all $a$ and $b$ in $R\setminus\{0\}$, one of the following holds: \begin{itemize} \item there exist $q$ in $R$ and $r$ in $R\setminus\{0\}$ such that $a=qb+r$ and $\phi(r)<\phi(b)$, or \item $b\divides a$ and $\phi(b)\leq\phi(a)$. \end{itemize} For example: \begin{asparaenum} \item $\Z$ is Euclidean with respect to $x\mapsto\abs x$; \item a field, $x\mapsto 0$; \item a polynomial-ring $K[X]$ over a field $K$, $f\mapsto\deg f$ (see \S~\ref{sect:fact-pol}). \end{asparaenum} The \textbf{Gaussian integers}\index{Gaussian integer} are the elements of $\Z[\tau_{-1}]$, where $\tau_{-1}=\sqrt{-1}=\mi$ as in \S~\ref{sect:ant}. This domain is Euclidean with respect to the norm function, namely $z\mapsto\abs z^2$, where $\abs{x+y\mi}^2=x^2+y^2$. Indeed, if $a$ and $b$ are nonzero Gaussian integers, then there is a Gaussian integer $q$ such that $\abs{a/b-q}\leq\sqrt 2/2$. Let $r=a-bq$; then $\abs r^2=\abs b^2\cdot\abs{a/b-q}^2\leq\abs b^2/2$. \begin{theorem} Euclidean domains are \pid s. \end{theorem} \begin{proof} An ideal of a Euclidean domain is generated by any non-zero element $x$ such that $\phi(x)$ is minimal. \end{proof} \section{Localization} A subset of a ring is \textbf{multiplicative}\index{multiplicative} if it is closed under multiplication. For example, the complement of a prime ideal is multiplicative. \begin{lemma} If $S$ is a multiplicative subset of a ring $R$, then on $R\times S$ there is an equivalence-relation $\sim$ given by \begin{equation}\label{eqn:q} (a,b)\sim (c,d)\iff (ad-bc)\cdot e=0\text{ for some $e$ in }S. \end{equation} \end{lemma} \begin{proof} Reflexivity and symmetry are obvious. For transitivity, note that, if $(a,b)\sim(c,d)$ and $(c,d)\sim(e,f)$, so that, for some $g$ and $h$ in $S$, \begin{align*} 0&=(ad-bc)g=adg-bcg,&0&=(cf-de)h=cfh-deh, \end{align*} then \begin{equation*} (af-be)cdgh =afcdgh-becdgh =adgcfh-bcgdeh =bcgcfh-bcgcfh=0, \end{equation*} so $(a,b)\sim(e,f)$. \end{proof} In the notation of the lemma, the equivalence-class of $(a,b)$ is denoted by \begin{equation*} \frac ab, \end{equation*} and the quotient $R\times S\modsim$ is denoted by \begin{equation*} S\inv R. \end{equation*} If $R$ is an integral domain, and $0\notin S$, then~\eqref{eqn:q} can be simply \begin{equation*} (a,b)\sim (c,d)\iff ad-bc=0. \end{equation*} If $0\in S$, then $S\inv R$ has a unique element. An instance where $R$ is not an integral domain will be considered in the next section. \begin{theorem}\label{thm:loc} Suppose $R$ is a ring with multiplicative subset $S$. \begin{enumerate} \item In $S\inv R$, if $c\in S$, \begin{equation*} \frac ab=\frac{ac}{bc}. \end{equation*} \item $S\inv R$ is a ring in which the operations are given by \begin{align*} \frac ab\cdot\frac cd&=\frac{ac}{bd},& \frac ab\pm\frac cd&=\frac{ad\pm bc}{bd}. \end{align*} \item There is a ring-homomorphism $\phi$ from $R$ to $S\inv R$ where, for every $a$ in $S$, \begin{equation*} \phi(x)=\frac{xa}a. \end{equation*} \newcounter{local} \setcounter{local}{\value{enumi}} \end{enumerate} Suppose in particular $R$ is an integral domain and $0\notin S$. \begin{enumerate}\setcounter{enumi}{\value{local}} \item $S\inv R$ is an integral domain, and the homomorphism $\phi$ is an embedding. \item If $S=R\setminus\{0\}$, then $S\inv R$ is a field, and if If $\psi$ is an embedding of $R$ in a field $K$, then there is an embedding $\tilde{\psi}$ of $S\inv R$ in $K$ such that $\tilde{\psi}\circ\phi=\psi$. \end{enumerate} \end{theorem} In the most important case, $S$ is the complement of a prime ideal $\primei$, and then $S\inv R$ is called the \textbf{localization}\index{local!---ization} of $R$ at $\primei$, denoted by \begin{equation*} R_{\primei}. \end{equation*} If $R$ is an integral domain, so that $(0)$ is prime, then $R_{(0)}$ (which is a field by the theorem) is the \textbf{quotient-field}% \index{quotient!--- field}\index{field!quotient ---} of $R$. A \textbf{local ring}\index{ring!local ---}\index{local!--- ring} is a ring with a unique maximal ideal. The connection between localizations and local rings is made by the theorem below. \begin{lemma} An ideal $\maxi$ of a ring $R$ is a unique maximal ideal of $R$ if and only if $\unit R=R\setminus\maxi$. \end{lemma} \begin{theorem} The localization of a ring at a prime ideal is a local ring. \end{theorem} \begin{proof} The ideal generated by the image of $\primei$ in $R_{\primei}$ consists of those $a/b$ such that $a\in\primei$. In this case, if $c/d=a/b$, then $cb=da\in\primei$, so $c\in\primei$ since $\primei$ is prime. Hence the following are equivalent: \begin{enumerate} \item $x/y\notin R_{\primei}\primei$; \item $x\notin\primei$; \item $x/y$ has an inverse, namely $y/x$. \end{enumerate} By the lemma, we are done. \end{proof} \section{Ultraproducts of fields} Suppose $\family K$ is an indexed family $(K_i\colon i\in A)$ of fields. If $a\in\prod\family K$, there is an element $a^*$ of $\prod K$ given by \begin{equation*} \pi_i(a^*)=\begin{cases} \pi_i(a)\inv,&\text{ if }\pi_i(a)\neq0,\\ 0,&\text{ if }\pi_i(a)=0. \end{cases} \end{equation*} Then \begin{equation*} aa^*a=a. \end{equation*} Because of this, $\prod\family K$ is an example of a \textbf{regular ring} (in the sense of von Neumann).\footnote{In general, a regular ring need not be commutative; see \cite[IX.3, ex.~5, p.~442]{MR600654}.} \begin{theorem} In a regular ring, all prime ideals are maximal. \end{theorem} \begin{proof} Let $R$ be a regular integral domain. If $a\in R\setminus\{0\}$, then, since \begin{equation*} 0=aa^*a-a=a(a^*a-1), \end{equation*} we have $a^*a=1$. Thus $R$ is a field. \end{proof} \begin{theorem} If $\primei$ is a prime ideal of a regular ring $R$, then \begin{equation*} R/\primei\cong R_{\primei}, \end{equation*} the isomorphism being $x+\primei\mapsto x/1$. \end{theorem} \begin{proof} If $a\in R$ and $b\in R\setminus\primei$, then $a/b=ab^*/1$ since \begin{equation*} (a-bab^*)b=ab-abb^*b=ab-ab=0. \end{equation*} Thus the homomorphism $x\mapsto x/1$ guaranteed by Theorem~\ref{thm:loc} is surjective. We also have $a/1=0/1$ if and only if $ab=0$ for some $b$ in $R\setminus\primei$; but the latter implies $ab\in\primei$, so $a\in \primei$ since the ideal is prime. Conversely, if $a\in\primei$, then $a^*a\in\primei$, so $a^*a-1\notin\primei$ since the ideal is proper; but $a(a^*a-1)=0$, so $a/1=0/1$. Therefore the kernel of the homomorphism is $\primei$. \end{proof} With $\family K$ as above, there is a one-to-one correspondence between ideals of $\prod\family K$ and ideals of the Boolean ring $\pow A$. To define this correspondence, we first define the \textbf{support} of an element $a$ of $\prod\family K$ to be the set of those $i$ in $A$ such that $\pi_i(a)\neq0$. We may denote this set by $\supp a$. Then \begin{align*} \supp{ab}&=\supp a\cap\supp b,& \supp{a+b}&\included\supp a\cup\supp b. \end{align*} So $x\mapsto\supp x$ is not quite a ring-homomorphism from $\prod\family K$ to $\pow A$. However, if $I$ is an ideal of $\prod\family K$, then $\Supp I$ is an ideal of $\pow A$. Indeed, for every subset $B$ of $A$, there is an element $e_B$ of $\prod\family K$ given by \begin{equation*} \pi_i(e_B)=\begin{cases} 1,&\text{ if }i\in B,\\ 0,&\text{ if }i\notin B. \end{cases} \end{equation*} Then $\supp{e_B}=B$. If $a\in\prod\family K$, and $B=\supp a$, then $e_B=aa^*$. If, further, $a\in I$, and $C\included B$, then $e_C=e_Caa^*$, so this is in $I$ and therefore $C\in\Supp I$. Also, if $B$ and $C$ are in $\Supp I$, then $B\symdiff C=\supp{e_B-e_C}$, which is in $\Supp I$. So $\Supp I$ is indeed an ideal of $\pow A$. If $J$ is an ideal of $\pow A$, then $J=\Supp I$, where $I$ is the ideal of $\prod\family K$ generated by those $e_B$ such that $B\in J$. Since every ideal $I$ is generated by those $e_B$ such that $B\in\Supp I$, we conclude that $\phi$ is the claimed bijection. Let $\primei$ be a prime ideal of $\prod\family K$. Then the quotient $\prod\family K/\primei$ is a field, called an \textbf{ultraproduct} of $\family K$. Now, $\primei$ could be principal, in which case $\phi(\primei)$ would be principal; but since it is also maximal, it would have a set $A\setminus\{i\}$ as a generator. In this case $\prod\family K/\primei\cong K_i$. However, $\pow A$ has the ideal $I$ consisting of the the finite subsets of $A$. If $A$ itself is infinite, then $I$ is a proper ideal. In this case, if $I\included\Supp{\primei}$, then $\primei$ is not principal, and the field $\prod\family K/\primei$ is called a \textbf{nonprincipal ultraproduct} of $\family K$. This is a sort of `average' of the $K_i$. In particular, we have \begin{align*} a\equiv b\pmod{\primei} &\iff a-b\in\primei\\ &\iff \supp{a-b}\in\Supp{\primei}\\ &\iff \{i\in A\colon \pi_i(a)\neq\pi_i(b)\}\in\Supp{\primei}. \end{align*} We may think of the elements of $\Supp{\primei}$ as `small' sets; their complements are `large'. (Then every subset of $A$ is small or large.) So all finite subsets of $A$ are small, and all cofinite subsets of $A$ are large. Then elements of $\prod\family K$ represent the same element in the ultraproduct if they agree on a large set. Say for example $A$ is the set of prime numbers in $\vnn$, along with $0$, and each $K_p$ has characteristic $p$. Then $\prod\family K/\primei$ has characteristic $0$, since for each prime $p$, the element $p1$ of $\prod\family K$ disagrees with $0$ on a large set. \section{Factorization of polynomials}\label{sect:fact-pol} \begin{theorem} If $R$ is a ring, then $R[X_0,\dots,X_{n-1}]$ is the unique ring-extension $A$ of $R$ such that, for all rings $S$, and all homomorphisms $\phi$ from $R$ to $S$, and all $\vec a$ in $S^n$, there is a unique homomorphism $\tilde{\phi}$ from $A$ to $S$ such that $\tilde{\phi}|_R=\phi$ and $\tilde{\phi}(X^i)=a^i$ in each case. \end{theorem} An arbitrary element of $R[X]$ can be written \begin{equation*} \sum_{i\leq n}a_iX^i; \end{equation*} the \textbf{degree}\index{degree} of this is $n$, if $a_n\neq0$; then $a_n$ is the \textbf{leading coefficient}\index{leading coefficient} of the polynomial. We said in \S~\ref{sect:int-dom} that $K[X]$ is a Euclidean domain when equipped with $\deg$. More generally: \begin{lemma} If $f$ and $g$ are polynomials over $R$, then: \begin{itemize} \item $\deg (f+g)\leq\max(\deg f, \deg g)$; \item $\deg (f\cdot g)\leq \deg f+\deg g$, with equality if the product of the leading coefficients is not $0$. \end{itemize} In particular, if $R$ is an integral domain, then so is $R[X]$. \end{lemma} \begin{proof} The leading coefficient of a product is the product of the leading coefficients. \end{proof} \begin{lemma}[Division Algorithm]\index{Division Algorithm} \index{theorem!Division Algorithm} \index{algorithm!Division A---} If $f$ and $g$ are polynomials in $X$ over $R$, and the leading coefficient of $g$ is $1$, then \begin{equation*} f=qg+r \end{equation*} for some unique $q$ and $r$ in $R[X]$ such that $\deg r<\deg g$. \end{lemma} \begin{proof} If $\deg g\leq \deg f$, and $a$ is the leading coefficient of $f$, then \begin{equation*} f=aX^{\deg f-\deg g}\cdot g + (f-aX^{\deg f-\deg g}\cdot g), \end{equation*} the second term having degree less than $f$. Continue as necessary. \end{proof} \begin{lemma}[Remainder Theorem]\index{Remainder Theorem} \index{theorem!Remainder Th---} If $c\in R$, then any $f$ in $R[X]$ can be written uniquely as $q(X)\cdot (X-c)+f(c)$. \end{lemma} \begin{proof} By the Division Algorithm, $f=q(X)\cdot (X-c)+d$ for some $d$ in $R$; letting $X$ be $c$ yields the claim. \end{proof} \begin{theorem} A ring-element $c$ is a zero of a polynomial $f$ if and only if $(X-c)\divides f$. If $f$ is over an integral domain, then the number of its distinct zeros is at most $\deg f$. \end{theorem} \begin{proof} By the Remainder Theorem, $c$ is a zero of $f$ if and only if $f=q(X)\cdot(X-c)$ for some $q$. In this case, if the ring is an integral domain, and $d$ is another zero of $f$, then, since $d-c\neq0$, we must have that $d$ is a zero of $q$. Hence, if $\deg(f)=n$, and $f$ has the distinct zeros $r_0$, \dots, $r_{n-1}$, then repeated application of the Remainder Theorem yields \begin{equation*} f=(X-r_0)\dotsm(X-r_{m-1}). \end{equation*} Then every zero of $f$ is a zero of one of the $X-r_k$, so it must be $r_k$. \end{proof} Recall however from the proof of Theorem~\ref{thm:Boole} that every element of a Boolean ring is a zero of $X(1+X)$, that is, $X+X^2$; but some Boolean rings have more than two elemments. In $\Zmod 6$, the same polynomial has the zeros $0$, $2$, $3$, and $5$. \begin{theorem} If $K$ is a field, then $K[X]$ is a Euclidean domain whose units are precisely the elements of $K$. \end{theorem} \begin{proof} Over a field, the Division Algorithm does not require the leading coefficient of the divisor to be $1$. \end{proof} A zero $c$ of a polynomial over an integral domain has \textbf{multiplicity} $m$ if the polynomial can be written as $g(X)\cdot(X-c)^m$, where $c$ is not a zero of $g$. A zero with multiplicity greater than $1$ is \textbf{multiple.} Derivations were defined in \S~\ref{sect:nna-rings}; they will be useful for recognizing the existence of multiple roots. \begin{lemma} If $\delta$ is a derivation of a ring $R$, then for all $x$ in $R$ and $n$ in $\vnn$, \begin{equation*} \delta(x^n)=nx^{n-1}\delta(x). \end{equation*} \end{lemma} \begin{proof} Since $\delta(1)=\delta(1\cdot1)=\delta(1)\cdot 1+1\cdot\delta(1)=2\cdot\delta(1)$, we have $\delta(1)=0$, so the claim holds when $n=0$. If it holds when $n=k$, then \begin{equation*} \delta(x^{k+1})=\delta(x)x^k+x\delta(x^k)=\delta(x)x^k+kx^k\delta(x)=(k+1)x^k\delta(x), \end{equation*} so the claim holds when $n=k+1$. \end{proof} \begin{theorem} On a polynomial ring $R[X]$, there is a unique derivation $f\mapsto f'$ such that \begin{enumerate} \item $X'=1$, \item $c'=0$ for all $c$ in $R$. \end{enumerate} This derivation is given by \begin{equation}\label{eqn:der} \Bigl(\sum_{k=0}^na_kX^k\Bigr)'= \sum_{k=0}^{n-1}(k+1)a_{k+1}X^k. \end{equation} \end{theorem} \begin{proof} Uniqueness and~\eqref{eqn:der} follow from the lemma and the definition of a derivation. If $\delta$ is a derivation, then $\delta(x\cdot (y+z))=\delta(xy+xz)$. Also,~\eqref{eqn:der} does define an endomorphism of the underlying group of $R[X]$ that meets the given conditions. This endomorphism is a derivation, because \begin{equation*} (X^k)'(X^{\ell})+X^k(X^{\ell})' =kX^{k-1}X^{\ell}+\ell X^kX^{\ell-1} =(k+\ell)X^{k+\ell+1} =(X^{k+\ell})'.\qedhere \end{equation*} \end{proof} In the notation of the theorem, $f'$ is the \textbf{derivative} of $f$. \begin{lemma} Say $R$ is an integral domain, $f\in R[X]$ and $f(c)=0$. Then $c$ is a multiple zero of $f$ if and only if $f'(c)=0$. \end{lemma} \begin{proof} Write $f$ as $(X-c)^m\cdot g$, where $g(c)\neq0$. Then $m\geq1$, so \begin{equation*} f'=m(X-c)^{m-1}\cdot g+(X-c)^m\cdot g'. \end{equation*} If $m>1$, then $f'(c)=0$. If $f'(c)=0$, then $m\cdot 0^{m-1}\cdot g(c)=0$, so $m>1$. \end{proof} If $L$ is a field with subfield $K$, then a polynomial over $K$ may be irreducible over $K$, but not over $L$. For example, $X^2+1$ is irreducible over $\R$, but not over $\C$. Likewise, the polynomial may have zeros from $L$, but not $K$. Hence it makes sense to speak of zeros of an irreducible polynomial. \begin{theorem} Supppose $K$ is a field and $f\in K[X]$. \begin{enumerate} \item If $\gcd(f,f')=1$, then $f$ has no multiple zeros. \item If $f$ is irreducible, then $\gcd(f,f')$ is $1$ or $0$. \item If $\gcd(f,f')=0$, then $K$ has a positive characteristic $p$, and $f=g(X^p)$ for some polynomial $g$ over $K$. \end{enumerate} \end{theorem} \begin{proof} If $\gcd(f,f')=1$, then $1=g\cdot f+h\cdot f'$ for some polynomials $g$ and $h$, so $f$ and $f'$ can have no common zero. Since $\deg(f')<\deg(f)$ by~\eqref{eqn:der}, if $f$ is irreducible and $\gcd(f,f')\neq1$, then $\gcd(f,f')=0$. The rest also follows from~\eqref{eqn:der}. \end{proof} A polynomial over a UFD is \textbf{primitive} if $1$ is a greatest common divisor of its coefficients. \begin{lemma}[Gauss] The product of primitive polynomials is primitive. \end{lemma} \begin{proof} Let $f=\sum_{k=0}^ma_kX^k$ and $g=\sum_{k=0}^nb_kX^k$. Then $fg=\sum_{k=0}^{mn}c_kX^k$, where \begin{equation*} c_k=\sum_{i+j=k}a_ib_j=a_0b_k+a_1b_{k-1}+\dotsb+a_kb_0. \end{equation*} Suppose the $c_k$ have a common prime factor $\pi$, but $f$ is primitive. There is some $\ell$ such that $\pi\divides a_i$ when $i<\ell$, but $\pi\ndivides a_{\ell}$. Since $\pi\divides c_{\ell}$, we have $\pi\divides b_0$; then, since $\pi\divides e_{\ell+1}$, we have $\pi\divides b_1$, and so on. So $g$ is not primitive. \end{proof} Henceforth let $R$ be a UFD with quotient field $K$. \begin{lemma} Primitive polynomials over $R$ that are associated over $K$ are associated over~$R$. \end{lemma} \begin{proof} If $f$ and $g$ are polynomials defined over $R$, but associated over $K$, then they must have the same degree, and so we have $af=bg$ for some $a$ and $b$ in $R$. If $f$ and $g$ are primitive, then $a$ and $b$ must be associates, so $b=ua$ for some unit in $R$, and then $f=ug$, so $f$ and $g$ are associates. \end{proof} \begin{lemma} Primitive polynomials over $R$ are irreducible over $R$ if and only if irreducible over~$K$. \end{lemma} \begin{proof} Say $f$ and $g$ are defined over $K$, but $fg$ is over $R$ and primitive. Then $af$ and $bg$ are over $R$ and primitive for some $a$ and $b$ in $R$. By a previous lemmma, $abfg$ is primitive; but so is $fg$, so $ab$ must be a unit in $R$. Hence $a$ and $b$ are units in $R$, so $f$ and $g$ are over $R$. Since units of $R[X]$ are units of $K[X]$, it follows that a primitive polynomial irreducible polynomial over $R$ is still irreducible over $K$. Also, any non-unit factor of a \emph{primitive} polynomial over $R$ is still not a unit over $K$, so the polynomial is reducible over $K$. \end{proof} Note however that if $f$ is primitive and irreducible over $R$, and $a$ in $R$ is not a unit or $0$, then $af$ is still irreducible over $K$ (since $a$ is a unit in $K$) but not over $R$. \begin{theorem} $R[X]$ is a UFD. \end{theorem} \begin{proof} Every element of $R[X]$ can be written as $af$, where $a\in R$ and $f$ is primitive. Then $f$ has a prime factorization over $K$ (since $K[X]$ is a Euclidean domain): say $f=f_0\dotsm f_{n-1}$. There are $b_k$ in $R$ such that $a_kf_k$ is a primitive polynomial over $R$. The product of these is still primitive, so the product of the $a_k$ must be a unit in $R$, hence each $a_k$ is a unit in $R$. Thus $f$ has an irreducible factorization over $R$. Its uniqueness follows from its uniqueness over $K$ and the next-to-last lemma. \end{proof} \begin{theorem}[Eisenstein's Criterion] If $f$ is a polynomial $\sum_{k=0}^na_kX^k$ over $R$, and $\pi$ is an irreducible element of $R$ such that \begin{align*} \pi^2&\ndivides a_0,& \pi&\divides a_0,& \pi&\divides a_1,& &\dots,& \pi&\divides a_{n-1},& \pi\ndivides a_n, \end{align*} then $f$ is irreducible over $K$ and, if primitive, over $R$. \end{theorem} \begin{proof} Suppose $f=gh$, where $g=\sum_{k=0}^nb_kX^k$ and $h=\sum_{k=0}^nc_kX^k$, all coefficients from $R$ (and some being $0$). We may assume $f$ is primitive, so $g$ and $h$ must be primitive. We may assume $\pi$ divides $b_0$, but not $c_0$. Let $\ell$ be such that $\pi\divides b_k$ when $k<\ell$. If $\ell=n$, then (since $g$ is primitive) we must have $b_n\neq0$, so $\deg g=n$, and $h=c_0$ and is a unit. If $\ell