\hfill \thepage} %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \section*{Problems and Comments for Section 17, 18, and 21} \vspace{1pt}$\sim $ \textbf{Problems: }17.6, 17.7, 18.1 (a), (b), (c), 21.11, 21.12 \textbf{Comments (and synopsis for these sections): }You should read 17 and 18 simultaneously. You may stop reading section 18 after the examples for Theorem 18.5. Add in the definition of a ring homomorphism the condition iii) $\varphi (1_{R})=1_{S}$ because all rings should have a unit. The kernel of a ring homomorphism $\varphi :R\rightarrow S$ is the set of all elements of $R$ which are mapped to the zero of $S.$By what we have learned about group homomorphisms, $\ker (\varphi )$ must be a subgroup $I$ of $(R,+,-,0)$. Moreover, if $\varphi (a)=0$ and if $b$ is any element in $R$ then $\varphi (ba)=\varphi (ab)=0.$That is, if $a\in I$ and if $b\in R$ then $ab\in I$ and $ba\in I$. This is how ideals are defined. If $I$ is an ideal then the group $(R/I,+,-,0=I)$ is also a ring under "representative wise" multiplications (see Theorem 17.3). The multiplicative unit \ is the class of $1,$ that is $1+I.$ If $I$ is the ideal (that is the kernel) for a homomorphism $\varphi $ then the ring $R/I\cong im(\varphi )$. That is the homomorphism theorem for rings, Theorem 18.5 If an ideal $I$ contains an element $a\ $which has an inverse $a^{-1}$ then $% a^{-1}a=1\in I$, hence $I=R$ If $\mathbf{F}$ is a field and $I\neq 0$ an ideal of $\mathbf{F}$ then $I=% \mathbf{F.}$ Assume that $R$ is commutative and $R/I$ is a domain. That is, whenever ($% a+I)(b+I)=ab+I=I,$one has that $(a+I)=I$ or $(b+I)=I.$ Thus $ab\in I$ iff $% a\in I$ or $b\in I.$ Such ideals are called \textit{prime }ideals. The converse is also easy to see, that is $R/I$ is a domain if $I$ is prime. Let $I$ be any ideal of the commutative ring $R.$Let $a\in R.$ Then $% J=I+(a)=\{i+ab|i\in I,b\in R\}$ is an ideal, actually the smallest ideal that contains $I$ and $a.$ An ideal $M\ $is called \textit{maximal }if $M\neq R$ and if for any ideal $% I\supseteq M$ one has that $I=M$ or $I=R.$\newline If $M$ is maximal and $a\notin M$ then $M+(a)=R$. Hence $m+ab=1$ for some $% m\in M$ and $b\in R.$ Now, $(a+M)$ is not the zero in $R/M$ is equivalent to $a\notin M.$ By what we just said, one has some $b$ and $m\ $such that $m+ab=1.$ But this is: $% (a+M)(b+M)=(ab+M)=1+M.$ Hence every element $(a+M)\neq 0\ $of $R/M$ has an inverse $(b+M).$ We proved: \textit{\ If }$M$\textit{\ is a maximal ideal of the commutative ring }$R$% \textit{\ then }$R/M$\textit{\ is a field. } Now, if $R/I$ is a field then every class $(a+I)\neq I$ has an inverse $% (b+I).$Thus $(a+I)(b+I)=1+I.$ This is $ab-1=i$ for some $i\in I.$We conclude that $I+(a)$ contains $1$ if $a\notin I.$ Hence $I$ has to be maximal. A (commutative) domain $D\ $is called a principal ideal domain (PID) if every ideal is principal. $% %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion $ and polynomial rings, like $% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion \lbrack x]$ are PId's. For domains the divisibility relation is all important: \begin{equation*} a|b\text{ iff }a\cdot q=b\text{ for some }q\in D\text{ iff }(a)\supseteq (b) \end{equation*} Every element $a\in D$ has trivial divisors: $a$ and $1.$ We have that $a|b$ and $b|a$ iff $b=ea$ and $a=fb$. Hence $a=fea$ This is $% fe=1$ because $D$ is a domain. Hence $a$ and $b$ differ only by an invertible element. In this case we say that $a$ and $b$ are associates and write $a\sim b.$ For example, in $% %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion $ one has that $a\sim \pm a$ because $1$ and $-1$ are the only elements which have an inverse. One always has $a|0$, that is with respect to divisibility, $0$ is the largest element and because $1|a$, $1$ is the smallest element. An element $q\in D$ is called \textit{irreducible} if $q$ has only tivial divisors. Trivial divisors of an element $a$ are all $e\sim 1,$that is the invertible elements, and $a^{\prime }\sim a.$ An element $p\in D$ is called\textit{\ prime }if whenever $p|ab$ one has that $p|a$ or $p|b.$ \begin{remark} A prime element is irreducible. \end{remark} \begin{proof} Assume that $p=a\cdot b$. Because $p\cdot 1=a\cdot b$ we have that $p|a\cdot b.$ Hence $p|a$ or $p|b.$On the other hand, $p=a\cdot b$ tells us that $a|p$ and $b|p.$ Thus $a\sim p$ or $b\sim p.$ \end{proof} \begin{theorem} In a PID, every irreducible element is prime. \end{theorem} \begin{proof} That $q$ is irreducible means that $(q)$ is a maximal ideal. Hence $D/(q)$ is a field, thus a domain. So $(q)$ is a prime ideal and (easy to see), $q$ has to be prime. \end{proof} \begin{theorem} In a PID, every ascending chain $I_{1}\subseteq I_{2}\subseteq \ldots $ of ideals is finite. That is for some $k$ one has that $I_{k}=I_{k+1}=\ldots $ \end{theorem} \begin{proof} It is quite obvious that the union of an ascending chain of ideals is an ideal. Thus $\bigcup I_{n}=I$ $=(d).$ If $d\in I_{k}$ then all ideals are equal from $k$ on. \end{proof} \begin{theorem} Let $a$ be a non invertible element of the PID $D.$Then there is some irreducble $p$ which divides $a$. \end{theorem} \begin{theorem} If $a$ is not irreducible then it has a proper divisor $a_{1}.$Thus $% (a)\subset (a_{1}).$If $a_{1}$is irreducible, we are done. Otherwise, $a_{1}$ has a proper divisor $a_{2}$ and we have $(a_{1})\subset (a_{2})$. If If $% a_{2\ }$is irreducible, we are done. Otherwise, $a_{2}$ has a proper divisor $a_{3}$ and we have $(a_{2})\subset (a_{3})$. By the previous theorem, this has to stop at some point. Thus $a$ has an irreducible divisor $q=a_{k}.$ \end{theorem} \begin{theorem} In a PID, any non invertible element $a\ $different from zero is a product of irreducible elements. The factorization is essentially unique. \end{theorem} \begin{proof} The element $a\neq 0$ has an irreducible divisor $p_{1}.$If $q_{1}=a/p_{1}$ is invertible, we are done. Otherwise $q_{1}$ has an irreducible divisor $% p_{2}$. If $q_{2}=q_{1}/p_{2}=a/p_{1}p_{2}$ is invertible, we are done. Otherwise $q_{2}$ has an irreducible divisor $p_{3}$. If $% q_{3}=q_{2}/p_{3}=a/p_{1}p_{2}p_{3}$ is invertible, we are done.....Notice that $\ldots q_{3}|q_{2}|q_{1}$or $(q_{1})\subset (q_{2})\subset (q_{3})\subset \ldots $ Hence for some $k$ we must have that $% q_{k}=a/p_{1}p_{2}p_{3}\ldots p_{k}=\epsilon $ is an invertible element, hence $a=(\epsilon p_{1})p_{2}p_{3}\ldots p_{k}$ where $\epsilon p_{1}$ as an associate of $p_{1}$ is also irreducible. Assume that \begin{equation*} a=p_{1}p_{2}p_{3}\ldots p_{k}=q_{1}q_{2}q_{3}\ldots q_{l} \end{equation*} then $k=l$ and after some re-enumeration one has that $p_{i}\sim q_{i}.$ This follows from the fact that irreducible elements are prime. Thus, because $p_{1}|q_{1}(q_{2}q_{3}\ldots q_{l})$ we have that $p_{1}|q_{1}$ or $% p_{1}|q_{2}(q_{3}\ldots q_{l}).$If$p_{1}|q_{1}$ then becasue $q_{1}$ is irreducible one has that $p_{1}\sim q_{1}.$Otherwise $p_{1}|q_{2}$ which leads to $p_{1}\sim q_{2}$ or $p_{1}|q_{3}(\ldots q_{l}).$If $p_{1}|q_{3}$ then because $q_{3}$ is irreducible one has that $p_{1}\sim q_{3}.$ hence, we must get $p_{1}\sim q_{j}$ for some $j\leq l.$ After some re-arrangement of the $q^{\prime }s$ we can assume that $j=1.$We cancel on both sides $% p_{1} $ and continue or finish by induction. \ \end{proof} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vspace{1pt} \textit{\vspace{1pt}} \vspace{1pt} \textbf{\ } \end{document}