%% LyX 2.0.6 created this file.  For more info, see http://www.lyx.org/.
%% Do not edit unless you really know what you are doing.
\documentclass[english,british]{article}
\usepackage[T1]{fontenc}
\usepackage[latin9]{inputenc}
\usepackage{geometry}
\geometry{verbose,tmargin=2cm,bmargin=2cm,lmargin=3cm,rmargin=3cm}
\usepackage{babel}
\usepackage{varioref}
\usepackage{mathrsfs}
\usepackage{amsthm}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage[all]{xy}
\PassOptionsToPackage{normalem}{ulem}
\usepackage{ulem}
\usepackage[unicode=true]
 {hyperref}
\usepackage{breakurl}

\makeatletter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Textclass specific LaTeX commands.
\numberwithin{equation}{section}
\numberwithin{figure}{section}
\numberwithin{table}{section}
\usepackage{enumitem}		% customizable list environments
\newlength{\lyxlabelwidth}      % auxiliary length 
  \theoremstyle{plain}
  \newtheorem*{thm*}{\protect\theoremname}
 \theoremstyle{definition}
 \newtheorem*{defn*}{\protect\definitionname}
\theoremstyle{plain}
\newtheorem{thm}{\protect\theoremname}[section]
  \theoremstyle{definition}
  \newtheorem{defn}[thm]{\protect\definitionname}
  \theoremstyle{remark}
  \newtheorem*{notation*}{\protect\notationname}
  \theoremstyle{plain}
  \newtheorem{prop}[thm]{\protect\propositionname}
  \theoremstyle{plain}
  \newtheorem{lem}[thm]{\protect\lemmaname}
  \theoremstyle{definition}
  \newtheorem*{example*}{\protect\examplename}
 \newcommand\thmsname{\protect\theoremname}
 \newcommand\nm@thmtype{theorem}
 \theoremstyle{plain}
 \newtheorem*{namedtheorem}{\thmsname}
 \newenvironment{namedthm}[1][Undefined Theorem Name]{
   \ifx{#1}{Undefined Theorem Name}\renewcommand\nm@thmtype{theorem*}
   \else\renewcommand\thmsname{#1}\renewcommand\nm@thmtype{namedtheorem}
   \fi
   \begin{\nm@thmtype}}
   {\end{\nm@thmtype}}
  \theoremstyle{plain}
  \newtheorem*{ax*}{\protect\axiomname}
  \theoremstyle{plain}
  \newtheorem{cor}[thm]{\protect\corollaryname}
  \theoremstyle{remark}
  \newtheorem*{rem*}{\protect\remarkname}
  \theoremstyle{remark}
  \newtheorem*{note*}{\protect\notename}
% labeling-like list based on enumitem's description list with
% mandatory second argument (label-pattern):
\newenvironment{elabeling}[2][]%
{\settowidth{\lyxlabelwidth}{#2}
\begin{description}[font=\normalfont,style=sameline,
leftmargin=\lyxlabelwidth,#1]}
{\end{description}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands.
\date{}
\newcommand{\xyR}[1]{
 \xydef@\xymatrixrowsep@{#1}}
\newcommand{\xyC}[1]{
  \xydef@\xymatrixcolsep@{#1}}

\makeatother

  \addto\captionsbritish{\renewcommand{\axiomname}{Axiom}}
  \addto\captionsbritish{\renewcommand{\corollaryname}{Corollary}}
  \addto\captionsbritish{\renewcommand{\definitionname}{Definition}}
  \addto\captionsbritish{\renewcommand{\examplename}{Example}}
  \addto\captionsbritish{\renewcommand{\lemmaname}{Lemma}}
  \addto\captionsbritish{\renewcommand{\notationname}{Notation}}
  \addto\captionsbritish{\renewcommand{\notename}{Note}}
  \addto\captionsbritish{\renewcommand{\propositionname}{Proposition}}
  \addto\captionsbritish{\renewcommand{\remarkname}{Remark}}
  \addto\captionsbritish{\renewcommand{\theoremname}{Theorem}}
  \addto\captionsenglish{\renewcommand{\axiomname}{Axiom}}
  \addto\captionsenglish{\renewcommand{\corollaryname}{Corollary}}
  \addto\captionsenglish{\renewcommand{\definitionname}{Definition}}
  \addto\captionsenglish{\renewcommand{\examplename}{Example}}
  \addto\captionsenglish{\renewcommand{\lemmaname}{Lemma}}
  \addto\captionsenglish{\renewcommand{\notationname}{Notation}}
  \addto\captionsenglish{\renewcommand{\notename}{Note}}
  \addto\captionsenglish{\renewcommand{\propositionname}{Proposition}}
  \addto\captionsenglish{\renewcommand{\remarkname}{Remark}}
  \addto\captionsenglish{\renewcommand{\theoremname}{Theorem}}
  \providecommand{\axiomname}{Axiom}
  \providecommand{\corollaryname}{Corollary}
  \providecommand{\definitionname}{Definition}
  \providecommand{\examplename}{Example}
  \providecommand{\lemmaname}{Lemma}
  \providecommand{\notationname}{Notation}
  \providecommand{\notename}{Note}
  \providecommand{\propositionname}{Proposition}
  \providecommand{\remarkname}{Remark}
  \providecommand{\theoremname}{Theorem}
\providecommand{\theoremname}{Theorem}

\begin{document}
\selectlanguage{english}%
\global\long\def\zz{\mathbb{Z}}
\global\long\def\im{\operatorname{im}}
\global\long\def\re{\operatorname{re}}
\global\long\def\rr{\mathbb{R}}
\global\long\def\cc{\mathbb{C}}
\global\long\def\vv{\mathbb{V}}
\global\long\def\SL{\operatorname{SL}}
\global\long\def\ord{\operatorname{ord}}
\global\long\def\Spec{\operatorname{Spec}}
\global\long\def\qq{\mathbb{Q}}
\global\long\def\pp{\mathbb{P}}
\global\long\def\Hom{\operatorname{Hom}}
\global\long\def\id{\operatorname{id}}
\global\long\def\gcd{\operatorname{gcd}}
\global\long\def\lcm{\operatorname{lcm}}
\global\long\def\Sym{\operatorname{Sym}}
\global\long\def\ff{\mathbb{F}}
\global\long\def\nn{\mathbb{N}}
\global\long\def\hh{\mathbb{H}}
\global\long\def\Ann{\operatorname{Ann}}
\global\long\def\End{\operatorname{End}}
\global\long\def\sgn{\operatorname{sgn}}
 \global\long\def\rk{\operatorname{rk}}
\global\long\def\Br{\operatorname{Br}}
 \global\long\def\Hasse{\operatorname{Hasse}}
\global\long\def\GL{\operatorname{GL}}
\global\long\def\Coll{\operatorname{Col}}
 \global\long\def\Tr{\operatorname{Tr}}
\global\long\def\aa{\mathbb{A}}
\global\long\def\ld{\operatorname{in}_{<}}
\global\long\def\ob{\operatorname{Ob}}
\global\long\def\mor{\operatorname{mor}}
\global\long\def\disc{\operatorname{disc}}
\global\long\def\ann{\operatorname{ann}}
\global\long\def\pd{\operatorname{pd}}


\selectlanguage{british}%

\title{Ring Theory (MA4H8)}


\author{Charudatta Hajarnavis\\
Notes by Florian Bouyer}

\maketitle
\begin{center}
Copyright (C) Bouyer 2011. 
\par\end{center}

\begin{center}
Permission is granted to copy, distribute and/or modify this document 
\par\end{center}

\begin{center}
under the terms of the GNU Free Documentation License, Version 1.3 
\par\end{center}

\begin{center}
or any later version published by the Free Software Foundation; 
\par\end{center}

\begin{center}
with no Invariant Sections, no Front-Cover Texts, and no Back-Cover
Texts. 
\par\end{center}

\begin{center}
A copy of the license can be found at \href{http://www.gnu.org/licenses/fdl.html}{http://www.gnu.org/licenses/fdl.html}
\par\end{center}

\tableofcontents{}

\newpage{}

Any reference to Commutative Algebra refer to the 2011-2012 Commutative
Algebra Lecture notes.

Rings studied will be mostly commutative. We aim to prove:
\begin{thm*}[Auslander - Buschsbaum 1959]
 A regular local ring is a unique factorization domain. 
\end{thm*}
Reason for selecting this theorem as our destination: 
\begin{enumerate}
\item It requires sophisticated results from the theory of commutative Noetherian
rings.
\item It requires methods from homological algebra. All known proofs require
this.
\item At a crucial stage it helps to think in terms of non-commutative rings.
\end{enumerate}
Prerequisite: MA3G6 Commutative Algebra

Topics assumed: 
\begin{enumerate}
\item Basic properties of Noetherian rings and modules.
\item Primary decomposition
\item Technicality of localization\end{enumerate}
\begin{defn*}
Let $R$ be a commutative Noetherian local ring with 1 and unique
maximal ideal $M$. Let $M=a_{1}R+\dots+a_{n}R$ ($a_{i}\in M$) be
chosen such that $n$ is as minimal as possible. Construct a chain
of prime ideals $M\supsetneq P_{1}\supsetneq\dots\supsetneq P_{r}$
($P_{i}$ prime) such that $r$ is greatest possible. Then $R$ is
\emph{regular} if $r=n$ (note that $r\leq n$ always in a Noetherian
ring)
\end{defn*}
Local rings arise naturally in geometry. In algebraic geometry points
correspond to local rings. 

Existence of an identity is not part of our definition of a ring.
For us a right, left or (two sided) ideal is a subring (Note that
in a non-commutative ring, by ideal we will mean a two sided ideal).
So for a right $R$-module $M$, $m\cdot1=m\,\forall m\in M$ is not
a part of our definition. \textbf{But} whenever $R$ has $1$, we
shall assume this. 

\newpage{}


\section{Chapter 1: Rings}


\subsection{Rings}
\begin{defn}
Let $R$ be a non-empty set which has tow law of composition defined
on it. (we call these law ``addition'' and ``multiplication''
respectively and use the familiar notation). We say that $R$ is a
\emph{ring} if the following hold:
\begin{enumerate}
\item $a+b\in R$ and $ab\in R$ $\forall a,b\in R$
\item $a+b=b+a\,\forall a,b\in R$ (Commutativity of addition)
\item $a+(b+c)=(a+b)+c\,\forall a,b,c\in R$ (Associativity of addition)
\item There exists an element $0\in R$ such that $a+0=a$ for all $a\in R$
\item Given $a\in R$ there exists an element $-a\in R$ such that $a+(-a)=0$
\item $a(bc)=(ab)c$ for all $a,b,c\in R$ (Associativity of multiplication)
\item $a(b+c)=ab+ac$ and $(a+b)c=ac+bc$ (Distributive Laws)
\end{enumerate}
Thus a ring is an additive Abelian group on which an operation of
multiplication is defined; this operation being associative and distributive
with respect to the addition.

$R$ is called a \emph{commutative ring} if it satisfies in addition
$ab=ba$ for all $a,b\in R$ . The term \emph{non-commutative ring
}usually stands for ``a not necessarily commutative ring''
\end{defn}

\subsection{Properties of Addition and Multiplication}

The following can be deduced from the axioms for a ring:
\begin{enumerate}
\item The element $0$ is unique
\item Given $a\in R$, $-a$ is uniquely
\item $-(-a)=a$ for all $a\in R$
\item $a+b=a+c$ if and only if $b=c$ for $a,b,c\in R$
\item Given $a,b\in R$, the equation $x+a=b$ has a unique solution $x=b+(-a)$

\begin{notation*}
We write $a-b$ to mean $a+(-b)$
\end{notation*}
\item $-(a+b)=-a-b$ for all $a,b\in R$
\item $-(a-b)=-a+b$ for all $a,b\in R$
\item $a\cdot0=0\cdot a=0$ for all $a\in R$
\item $a(-b)=(-a)b=-ab$ for all $a,b\in R$
\item $(-a)(-b)=ab$ for all $a,b\in R$
\item $a(b-c)=sb-ac$ for all $a,b,c\in R$\end{enumerate}
\begin{notation*}
$\zz$, the integers. $\qq$, the rational numbers. $\rr$, the real
numbers. $\cc$, the complex numbers. $M_{n}(R)$, the ring of $n\times n$
matrices whose entries are from the ring $R$.
\end{notation*}

\subsection{Subrings and Ideals}
\begin{defn}
A subset $S$ of a ring $R$ is called a \emph{subring} of $R$ if
$S$ itself is a ring with respect to the laws of composition of $R$\end{defn}
\begin{prop}
A non-empty subset $S$ of a ring $R$ is a subring of $R$ if and
only if $a-b\in S$ and $ab\in S$ whenever $a,b\in S$\end{prop}
\begin{proof}
If $S$ is a subring then obviously the given condition is satisfied.
Conversely, suppose that the condition holds. Take any $a\in S$.
We have $a-a\in S$ hence $0\in S$. Hence for any $x\in S$ we have
$0-x\in S$ so $-x\in S$. Finally, if $a,b\in S$ then by the above
$-b\in S$. Therefore $a-(-b)\in S$, i.e., $a+b\in S$. So $S$ is
closed with respect to both addition and multiplication. Thus $S$
is a subring since all the other axioms are automatically satisfied.\end{proof}
\begin{defn}
A subset $I$ of a ring $R$ is called an \emph{ideal} if 
\begin{enumerate}
\item $I$ is a subring of $R$
\item For all $a\in I,r\in R$ $ar\in I$ and $ra\in I$ 
\end{enumerate}
\end{defn}
If $I$ is an ideal of $R$ we denote this fact by $I\lhd R$.
\begin{prop}
A non-empty subset $I$ of a ring $R$ is an ideal of $R$ if and
only if $a-b\in I,ar\in I$ and $ra\in I$ whenever $a,b\in I$ and
$r\in R$ \end{prop}
\begin{proof}
Exercise
\end{proof}

\subsection{Cosets and Homomorphism}
\begin{defn}
Let $I$ be an ideal of a ring $R$ and $x\in R$ . Then the set of
elements $\{x+i:i\in I\}$ is called the \emph{coset} of $x$ in $R$
with respect to $I$. It is denoted by $x+I$
\end{defn}
When dealing with cosets, it is more important to realise that, in
general, a given coset can be represented in more than one way. The
next lemma shows how the coset representatives are related.
\begin{lem}
Let $R$ be a ring with an ideal $I$ and $x,y\in R$. Then $x+I=y+I\iff x-y\in I$\end{lem}
\begin{proof}
Exercise
\end{proof}
We denote the set of all cosets of $R$ with respect to $I$ by $R/I$.
We can give $R/I$ the structure of a ring as follows: Define $(x+I)+(y+I)=(x+y)+I$
and $(x+I)(y+I)=xy+I$ for $x,y\in R$.

The key point here is that the sum and the product of $R/I$ are well-defined,
that is, they are independent of the coset representatives chosen.
Check this and make sure that you understand why the fact that $I$
is an ideal is crucial to the proof.
\begin{defn}
$R/I$ is called the \emph{residue class ring} of $R$ with respect
to $I$
\end{defn}
The zero element of $R/I$ is $0+I=i+I$ for any $i\in I$ . If $S$
is a subset of $R$ with $S\supseteq I$ we denote by $S/I$ the subset
$\{s+I:s\in S\}$ of $R/I$.
\begin{prop}
Let $I$ be an ideal of a ring $R$. Then
\begin{enumerate}
\item Every ideal of the ring $R/I$ is of the form $K/I$ where $K\lhd R$
and $K\supseteq I$. Also conversely, $K\lhd R,K\supseteq I\Rightarrow K/I\lhd R/I$
\item There is a one to one correspondence between ideals of the ring $R/I$
and the ideals of $R$ containing $I$
\end{enumerate}
\end{prop}
\begin{proof}
{}
\begin{enumerate}
\item If $K^{*}\lhd R/I$, define $K]\{x\in R:x+I\in K^{*}\}$ . Then $K\lhd R,K\supseteq I$
and $K/I=K^{*}$
\item The correspondence is given by $K\leftrightarrow K/I$ where $K\lhd R$,
$K\supseteq I$
\end{enumerate}
\end{proof}
\begin{defn}
A mapping $\theta$ of a ring $R$ into a ring $S$ is said to be
a (ring) \emph{homomorphism} if $\theta(x+y)=\theta(x)+\theta(y)$
and $\theta(xy)=\theta(x)\theta(y)$ for all $x,y\in R$.

$\theta$ defined by $\theta(r)=0$ for all $r\in R$ is a homomorphism.
It is called the \emph{zero homomorphism.} 

$\phi$ defined by $\phi(r)=r$ for all $r\in R$ is also a homomorphism.
It is called the \emph{identity homomorphism}

Let $I\lhd R$. Then $\sigma:R\to R/I$ defined by $\sigma(x)=x+I$
for all $x\in R$ is a homomorphism of $R$ onto $R/I$. This is called
the \emph{natural }(or \emph{canonical}) \emph{homomorphism.}\end{defn}
\begin{prop}
Let $R,S$ be rings and $\theta:R\to S$ a homomorphism. Then :
\begin{enumerate}
\item $\theta(0_{R})=0_{S}$
\item $\theta(-r)=-q(r)$ for all $r\in R$
\item $K=\{x\in R:q\theta(x)=0_{S}\}$ is an ideal of $R$
\item $\theta R=\{\theta(r):r\in R\}$ is subring of $S$
\end{enumerate}
\end{prop}
\begin{proof}
Exercise
\end{proof}
$K$ is called the \emph{kernel} of $\theta$ and $\theta R$ is called
the (homomorphic) \emph{image} of $R$. The ideal $K$ is sometimes
denoted by $\ker\theta$.
\begin{defn}
Let $\theta$ be a homomorphism of a ring $R$ into a ring $S$. Then
$\theta$ is called an \emph{isomorphism} if $\theta$ is a one to
one and onto map. We say that $R$ and $S$ are isomorphic rings and
denote this by $R\cong S$.
\end{defn}

\subsection{The Isomorphism Theorems}

Question: Given a ring $R$, what rings can occur as its homomorphic
images?

The importance of the first isomorphism theorem lies in the fact that
it shows the answer to lie with $R$ itself. It tells us that if we
know all the ideals of $R$ then we know all the homomorphic images
of $R$.Only the first isomorphism theorem contains new information.
The other two are simply its application.
\begin{thm}
Let $\theta$ be a homomorphism of a ring $R$ into a ring $S$. Then
$\theta R\cong R/I$ where $I=\ker\theta$\end{thm}
\begin{proof}
Defined $\sigma:R/I\to R$ by $\sigma(x+I)=\theta(x)$ for all $x\in R$.
The map $\sigma$ is well defined since for $x,y\in R$, $x+I=y+I\Rightarrow x-y\in I=\ker\theta\Rightarrow\theta(x-y)=0\Rightarrow\theta(x)=\theta(y)$.
$\theta$ is easily seen to be the required isomorphism.\end{proof}
\begin{thm}
Let $I$ be an ideal and $L$ a subring of a ring $R$. Then $L/(L\cap I)\cong(L+I)/I$\end{thm}
\begin{proof}
Let $\sigma$ be the natural homomorphism $R\to R/I$. Restrict $\sigma$
to the ring $L$. We have $\sigma L=(L+I)/I$. The kernel of $\sigma$
restricted to $L$ is $L\cap I$. Now apply previous theorem.\end{proof}
\begin{thm}
Let $I,K$ be ideals of a ring $R$ such that $I\subseteq K$. Then
$(R/I)/(K/I)\cong R/K$\end{thm}
\begin{proof}
$K/I\lhd R/I$ and so $(R/I)/(K/I)$ is defined. Define a map $\gamma:R/I\to R/K$
by $\gamma(x+I)=x+K$ for all $x\in R$. The map $\gamma$ is easily
seen to be well defined and a homomorphism onto $R/K$. Further, 
\begin{eqnarray*}
\gamma(x+I)=K & \iff & x+K=K\\
 & \iff & x\in K\\
 & \iff & x+I\in K/I
\end{eqnarray*}
Therefore $\ker\gamma=K/I$. Now apply the first isomorphism theorem.
\end{proof}

\subsection{Direct Sums}
\begin{defn}
\emph{The internal direct sum: }Let $\{I_{\lambda}\}_{\lambda\in\Lambda}$
be a collection of ideals of a ring $R$. We define their \emph{sum}
to be $\sum_{\lambda\in\Lambda}I_{\lambda}=\{x\in R:x=x_{1}+\dots+x_{k},x_{i}\in I_{\lambda_{i}},k=1,2,3,\dots\}$.
That is the sum is the collection of \uline{finite }sums of elements
of the $I_{\lambda}$'s.

We say that the sum of the $I_{\lambda}$'s is \emph{direct} if each
element of $\sum_{\lambda\in\Lambda}I_{\lambda}$ is uniquely expressible
as $x_{1}+\dots+x_{k}$ with $x_{i}\in I_{\lambda_{i}}$. In this
case we denote this sum as $\sum_{\lambda\in\Lambda}\oplus I_{\lambda}$
or $I_{1}\oplus\dots\oplus I_{n}$ if $\Lambda$ is finite.\end{defn}
\begin{prop}
The sum $\sum_{\lambda\in\Lambda}I_{\lambda}$ is direct if and only
if $I\mu\cap(\sum_{\lambda\in\Lambda,\lambda\neq\mu}I_{\lambda})=0$
for all $\mu\in\Lambda$\end{prop}
\begin{proof}
Exercise\end{proof}
\begin{defn}
\emph{The external direct sum}: Let $R_{1},\dots,R_{n}$ be rings.
We define the \emph{external direct sum} $S$ to be the set of all
$n$-tuples $\{(r_{1},\dots,r_{n}):r_{i}\in R_{i}\}$. On $S$ we
define addition and multiplication component wise. This makes $S$
a ring. We write $S=R_{1}\oplus\dots\oplus R_{n}$.

The set $(0,\dots,0,R_{j},0,\dots,0)$ is an ideal of $S$. Clearly
$S$ is the internal direct sum of these ideals. But $(0,\dots,R_{j},\dots0)\cong R_{j}$.
Because of this $S$ can be considered as a ring in which the $R_{j}$
are ideals and $S$ is their internal direct sum. Also in internal
direct sum we can consider $I_{1}\oplus\dots\oplus I_{n}$ to be the
external direct sum of the rings $I_{j}$. Hence, in practice, we
do not need to distinguish between external and internal direct sums.
\end{defn}

\subsection{Division Rings}
\begin{defn}
Let $R$ be a ring with $1$. An element $u\in R$ is said to be a
\emph{unit} if there exists an element $v\in R$ such that $uv=vu=1$.
The element $v$ is called the \emph{inverse }of $u$ and is denoted
by $u^{-1}$

A ring $D$ with at least two elements is called a \emph{division
ring }(or a \emph{skew field}) if $D$ has an identity and every non-zero
element of $D$ has an inverse in $D$

A division ring in which the multiplication is commutative is called
a \emph{field-discriminant}\end{defn}
\begin{example*}
\uline{The Quaternions:} Let $D$ be the set of all symbols $a_{0}+a_{1}i+a_{2}j+a_{3}k$
where $a_{i}\in\rr$. Two such symbols are considered to be equal
if and only if $a_{i}=b_{i}$ for $i=0,1,2,3$. We make the ring as
follows: Addition is component-wise. Two such symbols are multiplied
term by term using the relations $i^{2}=j^{2}=k^{2}=-1$ and $ij=-jk=k,jk=-kj=i,ki=-ik=j$.
Then $D$ is a non-commutative ring with zero and identity. Let $a_{0}+a_{1}i+a_{2}j+a_{3}k$
be a non-zero element of $D$. Then not all the $a_{i}$ are zero.
We have
\[
(a_{0}+a_{1}i+a_{2}j+a_{3}k)(a_{0}-a_{1}i-a_{2}j-a_{3}k)=a_{0}^{2}+a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\neq0
\]
. So letting $n=a_{0}^{2}+a_{1}^{2}+a_{2}^{2}+a_{3}^{2}$, the element
$(a_{0}/n)+(a_{1}/n)i+(a_{2}/n)j+(a_{3}/n)k$ is the inverse of $a_{0}+a_{1}i+a_{2}j+a_{3}k$.
Thus $D$ is a division ring. It is called the division ring of \emph{real
quaternions.} \emph{Rational quaternions} can be defined similarly
where the coefficients are from $\qq.$
\end{example*}

\subsection{Modules}
\begin{defn}
Let $R$ be a ring. A set $M$ is called a \emph{right $R$-module
if:}
\begin{enumerate}
\item $M$ is an additive abelian group
\item A law of composition $M\times R\to M$ is defined, which satisfies
for $x,y\in M$ and $r_{1},r_{2}\in R$
\item $(x+y)r_{1}=xr_{1}+yr_{1}$
\item $x(r_{1}+r_{2})=xr_{1}+xr_{2}$
\item $x(r_{1}r_{2})=(xr_{1})r_{2}$
\end{enumerate}
A \emph{left} $R$-module is defined analogously. Here the product
of $m\in M$ and $r\in R$ is denoted by $rm$.\end{defn}
\begin{example*}
{}
\begin{enumerate}
\item $R$ and $\{0\}$ are left $R$-modules. They are also right $R$-modules.
\item Let $V$ be a vector space over a field $F$. Then $V$ is a left
$F$-module. The module axioms are part of the vector space axioms
\item Any abelian group can be considered a left $\zz$-module: 


Let $g\in A$ and $k\in\zz$. We defined $kg=\underset{k\mathrm{\, times}}{\underbrace{g+\dots+g}}$
if $k>0$, $0_{\zz}g=0_{A}$ and $kg=-[(-k)g]$ if $k<0$.

\item Let $R$ be a ring. Then $M_{n}(R)$ becomes a left $R$-module if
we define for $r\in R$ and $X\in M_{n}(R)$
\[
rX=\begin{pmatrix}r & 0 & 0 & \cdots & 0\\
0 & r & 0 & \cdots & 0\\
0 & 0 & r & \cdots & 0\\
\vdots & \vdots & \vdots & \ddots\\
0 & 0 & 0 &  & r
\end{pmatrix}X
\]
Clearly, we can also make $M_{n}(R)$ a right $R$-module.
\end{enumerate}
\end{example*}
The symbol $M_{R}$ will denote $M$ is a right $R$-module, while
the symbol $_{R}M$ will denote $M$ is a left $R$-module. For technical
reason it is sometimes easier to work with right $R$-modules while
dealing with non-commutative rings (when we choose to write maps on
the left). We say simply say that $M$ is a module if the other details
are clear from the context.
\begin{prop}
Let $M$ be a right $R$-module. Then:
\begin{enumerate}
\item $0_{M}r=0_{M}$ for all $r\in R$
\item $m0_{R}=0_{M}$ for all $m\in M$.
\item $(-m)r=m(-r)=-mr$ for all $m\in M$ and $r\in R$
\end{enumerate}
\end{prop}
\begin{proof}
Exercise\end{proof}
\begin{defn}
Let $K$ be a subset of a right $R$-module $M$. Then $K$ is called
a \emph{right $R$-submodule }(or just submodule) if $K$ is also
a right $R$-module under the laws of composition defined on $M$.\end{defn}
\begin{prop}
Let $K$ be a non-empty subset of $M_{R}.$ Then $K$ is a submodule
of $M\iff x-y\in K$ and $xr\in K$ for all $x,y\in K$ and $r\in R$\end{prop}
\begin{proof}
Exercise\end{proof}
\begin{defn}
Submodules of $R_{R}$ are called \emph{right ideals }of $R$ and
submodules of $_{R}R$ are called \emph{left ideals }of $R$.
\end{defn}

\subsection{Factor Modules and Homomorphisms}

Let $K$ be a submodule of a right $R$-module $M$. Consider the
facto group $M/K$. Elements of $M/K$ are cosets of the form $m+K$
with $m\in M$. We can make $M/K$ a right $R$-module by defining
$[m+K]r=mr+K$ for all $m\in M$ and $r\in R$. Check that this action
is well defined and the module axioms are satisfied to make $M/K$
a right $R$-module.
\begin{prop}
Let $K$ be a submodule of $M_{R}$. Then
\begin{enumerate}
\item every submodule of $M/K$ has the form $A/K$ where $A$ is a submodule
of $M$ and $A\supseteq K$.
\item There is a one to one correspondence between the submodules of $M/K$
and the submodules of $M$ containing $K$
\end{enumerate}
\end{prop}
\begin{defn}
Let $M$ and $M'$ be right $R$-modules. A mapping $\theta:M\to M'$
is called an $R$-\emph{homomorphism} if:
\begin{enumerate}
\item $\theta(x+y)=\theta(x)+\theta(y)$ for all $x,y\in M$
\item $\theta(xr)=\theta(x)r$ for all $x\in M$ and $r\in R$
\end{enumerate}
\end{defn}
If $K$ is a submodule of $M_{R}$ then the map $\sigma:M\to M/K$
defined by $\sigma(m)=m+K$ for all $m\in M$ is an $R$-homomorphism
of $M$ onto $M/K$. It is called the \emph{canonical $R$-homomorphism}
\begin{prop}
Let $\theta:M_{R}\to M'_{R}$ be an $R$-homomorphism. Then:
\begin{enumerate}
\item $\theta(0_{M})=0_{M'}$
\item $K=\{x\in M:\theta(x)=0_{M'}\}$ is a submodule of $M$
\item $\theta M=\{\theta(m):m\in M\}$ is a submodule of $M'$
\end{enumerate}
\end{prop}
\begin{proof}
Exercise
\end{proof}
$K$ is called the \emph{kernel} of $\theta$ and $\theta M$ is called
the \emph{image} of $\theta$. $\theta$ is a one to one correspondence
map if and only if $\ker\theta=0$
\begin{defn}
Let $\theta:M_{R}\to M'_{R}$ be an $R$-homomorphism. Then $\theta$
is called an $R$-\emph{isomorphism }if it is in addition a one to
one correspondence and onto map. In this case we write $M\cong M'$
\end{defn}

\subsection{The Isomorphism Theorem}

There are similar to those for rings
\begin{thm}
Let $M$ and $M'$ be right $R$-modules and $\theta:M\to M'$ and
$R$-homomorphism. Then $\theta M\cong M/K$ where $K=\ker\theta$
\end{thm}

\begin{thm}
Let $L,K$ be submodules of $M_{R}$. Then $(L+K)/K\cong L/(L\cap K)$
\end{thm}

\begin{thm}
If $K,L$ are submodules of $M_{R}$ and $K\subseteq L$ then $L/K$
is a submodule of $M/K$ and $(M/K)/(L/K)\cong M/L$.
\end{thm}
The proofs of these theorems are similar to those for rings


\subsection{Direct Sums of Modules\label{sub:1.13}}

Let $M_{1},\dots,M_{n}$ be right $R$-modules. The set of $n$-tuples
$\{(m_{1},\dots,m_{n}):m_{i}\in M_{i}\}$ becomes a right $R$-modules
if we define $(m_{1},\dots,m_{n})+(m_{1}',\dots,m_{n}')=(m_{1}+m_{1}',\dots,m_{n}+m_{n}')$
and $(m_{1},\dots,m_{n})r=(m_{1}r,\dots,m_{n}r)$. This is the \emph{external
direct sum }of the $M_{i}$ and is denoted $\sum_{i=1}^{n}\oplus M_{i}$
or $M_{1}\oplus\dots\oplus M_{n}$.

Let $\{M_{\lambda}\}_{\lambda\in\Lambda}$ be a collection of submodules
of a right $R$-modules $M$. We define their \emph{sum} $\sum_{\lambda\in\Lambda}M_{\lambda}$
to be $\{m_{\lambda_{1}}+\dots+m_{\lambda_{k}}:m_{\lambda_{i}}\in M_{\Lambda_{i}}$
for all possible subsets $\{\lambda_{1},\dots,\lambda_{k}\}$ of $\Lambda$\}.
Thus $\sum_{\lambda\in\Lambda}M_{\lambda}$ is the set of all \uline{finite}
sums of elements of the $M_{\lambda}$'s. It is easy to see that this
is a submodule of $M$.

$\sum_{\lambda\in\Lambda}M_{\lambda}$ is said to be \emph{direct}
if each $\sum_{\lambda\in\Lambda}M_{\lambda}$ has a unique expression
as $m_{\lambda_{1}}+\dots+m_{\lambda_{k}}$ for some $m_{\lambda_{i}}\in M_{\lambda_{i}}$.
As in 1.6 we can show that $\sum_{\lambda\in\Lambda}M_{\lambda}$
is direct $\iff M_{\mu}\cap\{\sum_{\lambda\in\Lambda,\lambda\neq\mu}M_{\lambda}\}=\{0\}$
for all $\mu\in\Lambda$. If $\sum_{\lambda\in\Lambda}M_{\Lambda}$
is direct, we denote it by $\sum_{\lambda\in\Lambda}\oplus M_{\lambda}$
or $M_{1}\oplus\dots\oplus M_{n}$ if $\Lambda$ is a finite set.
As explained for rings in 1.6, there is no real difference between
(finite) external and internal direct sums of modules.
\begin{defn}
Let $R$ be a ring with 1. A module $M_{R}$ is said to be \emph{unital}
if $m1=m$ for all $m\in M$
\end{defn}
We shall assume that all modules considered are unital whenever $R$
is a ring with identity.


\subsection{Products of subsets}

Let $M$ be a right $R$-module. Let $K,S$ be non-empty subsets of
$M$ and $R$ respectively. We defined their \emph{products} $KS$
to be $\{\sum_{i=1}^{n}k_{i}s_{i}|k_{i}\in K,s_{i}\in S;i=1,2,\dots\}$.
Thus $KS$ consists of all possible finite sums of elements of the
type $ks$ with $k\in K$ and $s\in S$. If $K$ is a non-empty subset
of $M$ and $S$ is a right ideal of $R$ then $KS$ is a submodule
of $M$. (Check that we require finite sums in our definition to make
this work)

The above definition applies, in particular, when $M=R$. Thus if
$S$ is a non-empty subset of $R$ then $S^{2}=\{\sum_{i=1}^{n}s_{i}t_{i}:s_{i},t_{i}\in S;n=1,2,\dots\}$.
Extending inductively, $S^{n}$consist of all finite sums of elements
of the type $x_{1}x_{2}\dots x_{n}$ with $x_{i}\in S$.

Note that if $S$ is a right ideal of $R$ then so is $S^{n}$


\subsection{A construction}

Let $R$ be a ring with an ideal $I$ and $M$ a right $R$-module.
In general, $M$ need not be a right $R/I$-module. However, we can
give $M$ a right $R/I$-module structure if $MI=0$. In this case
we define $mr=m[r+I]$ for all $m\in R$ and $r\in R$. It can be
checked that this is well-defined right $R/I$-module action. Further,
under this action the $R$ and $R/I$ submodules of $M$ coincide.

In particular, $I/I^{2}$ is naturally a right (and left) $R$-module.
This fact will be used repeatedly. In general same for $I^{n}/I^{n+1}$.


\subsection{Zorn's Lemma, Well-ordering Principle, The Axiom of Choice}
\begin{defn}
{}
\begin{enumerate}
\item A non-empty set $\mathscr{S}$ is said to be \emph{partially ordered
}if there exists a binary relation $\leq$ in $\mathscr{S}$ which
is defined for certain pairs of elements in $\mathscr{S}$ and satisfies:

\begin{enumerate}
\item $a\leq a$
\item $a\leq b$, $b\leq c\Rightarrow a\leq c$
\item $a\leq b$, $b\leq a$$\Rightarrow a=b$
\end{enumerate}
\item Let $\mathscr{S}$ be a partially ordered set. A non-empty subset
$\tau$ is said to be \emph{totally ordered} if for every pair $a,b\in\tau$
we have either $a\leq b$ or $b\leq a$
\item Let $\mathscr{S}$ be a partially ordered set. An elements $x\in\mathscr{S}$
is called a \emph{maximal element} if $x\leq y$ with $y\in\mathscr{S}\Rightarrow x=y$.
Similarly, for a \emph{minimal }element
\item Let $\tau$ be a totally ordered subset of a partially ordered set
$\mathscr{S}$. We say that $\tau$ has an \emph{upper bound} in $\mathscr{S}$
if there exists $c\in\mathscr{S}$ such that $x\leq c$ for all $x\in\tau.$
\end{enumerate}
\end{defn}
\begin{namedthm}[Zorn's Lemma (Axiom)]
 If a partially ordered set $\mathscr{S}$ has the property that
every totally ordered subset of $\mathscr{S}$ has an upper bound
in $\mathscr{S}$, then $\mathscr{S}$ contains a maximal element. 
\end{namedthm}
A non-empty set $\mathscr{S}$ is said to be \emph{well-ordered} if
it is totally ordered and every non-empty subset of $\mathscr{S}$
has a minimal element.
\begin{namedthm}[The Well ordering Principle]
 Any non-empty set can be well-ordered.\end{namedthm}
\begin{ax*}[The Axiom of Choice]
 Given a class of sets, there exists a ``choice function'', i.e.,
a function which assigns to each of these sets one of its elements. 
\end{ax*}
It can be shown that Axiom of Choice is logically equivalent to Zorn's
Lemma which is logically equivalent to the Well-ordering Principle. 

\newpage{}


\section{Chapter 2: The Jacobson Radical}

All rings considered in this chapter are assumed to have an identity.


\subsection{Quasi-regularity}
\begin{defn}
Let $M$ be a right ideal of $R$. $M$ is said to be a \emph{maximal
right ideal} if $M\neq R$ and $M'\supsetneq M$ with $M'\lhd_{r}R\Rightarrow M'=R$.

Similar definition is applied for a maximal two-sided ideal, and maximal
left ideal.\end{defn}
\begin{prop}
\label{prop:existence of maximal ideal}Let $I\neq R$ be a right
ideal of a ring $R$. Then there exists a maximal right ideal $M$
of $R$ such that $M\supseteq I$.\end{prop}
\begin{proof}[c.f. Commutative Algebra, Theorem 1.4]
 By Zorn's Lemma. Let $\mathscr{S}$ be the set of all proper right
ideals of $R$ containing $I$. Partially order $\mathscr{S}$ by
inclusion. Let $\{T_{\alpha}\}_{\alpha\in\Lambda}$ be a totally ordered
subset of $\mathscr{S}$. Let $T=\cup_{\alpha\in\Lambda}T_{\alpha}$.
Then $T\lhd_{r}R$ and $T\supseteq I$. Moreover $T$ is proper since
$T=R\Rightarrow1\in T\Rightarrow1\in T_{\alpha}$ for some $\alpha\in\Lambda\Rightarrow T_{\alpha}=R$.
Thus $T\neq R$ and so $T\in\mathscr{S}$. Thus $T\neq R$ and so
$T\in\mathscr{S}.$ Now $T\supseteq T_{\alpha}$ for all $\alpha\in\Lambda$.
Hence Zorn's Lemma applies and $\mathscr{S}$ contains a maximal element,
say $M$. Clearly $M$ is a maximal right ideal and $M\supseteq I$.\end{proof}
\begin{cor}
A ring with identity contains a maximal right ideal.\end{cor}
\begin{proof}
Take $I=0$ in the above theorem.\end{proof}
\begin{rem*}
This is not true for rings without 1\end{rem*}
\begin{defn}
The intersection of all maximal right ideals of a ring $R$ is called
its \emph{Jacobson radical}. It is usually denoted by $J(R)$ (or
simply $J$)\end{defn}
\begin{rem*}
Strictly speaking the above definition was for the \uline{right}
Jacobson radical. However we shall show that this is the same as the
left Jacobson radical. \end{rem*}
\begin{thm}[Crucial Lemma]
 Let $M$ be a maximal right ideal of a ring $R$ and let $a\in R$.
Define $K=\{r\in R:ar\in M\}$. Then $K\lhd_{r}R$ and:
\begin{enumerate}
\item if $a\in M$ then $K=R$
\item if $a\notin M$ then $K$ is also a maximal right ideal. 
\end{enumerate}
\end{thm}
\begin{proof}
Clear that $K\lhd_{r}R$, Now assume that $a\notin M$ so that $M+aR=R$
$(*)$. Define an $R$-module homomorphism $\theta:R\to R/M$ by $r\mapsto ar+M\,\forall r\in R$.
Check that this is a homomorphism of right $R$-modules. By $(*)$,
$\theta$ is an onto map. So by the isomorphism theorem for modules:
$R/M\cong R/\ker\theta=R/K$. It follows that $K$ is a maximal right
ideal.\end{proof}
\begin{thm}
$J\lhd R$\end{thm}
\begin{proof}
Clearly $J\lhd_{r}R$. Now let $j\in J$ and $a\in R$ and suppose
$aj\notin J$. Then by definition there exists a right ideal $M$
such that $aj\notin M$. Define $K=\{r\in R:ar\in M\}$. By the previous
theorem $K$ is a maximal right ideal. But $j\notin K$ since $aj\notin M$
hence $j\notin J$. This is a contradiction. Hence $aj\in J$ for
all $j\in J$ and $r\in R$. Thus $J\lhd R$.\end{proof}
\begin{defn}
Let $x$ be an element of a ring $R$. We say that $x$ is a \emph{right
quasi-regular }(rqr) if $1-x$ has a right inverse, i.e., if $\exists y\in R$
such that $(1-x)y=1$

A subset $S$ of $R$ is called \emph{right quasi-regular} if every
elements of $S$ is rqr

\emph{Left quasi-regular} (lqr) is defined analogously

We call an element or set \emph{quasi-regular} if it is both lqr and
rqr.\end{defn}
\begin{lem}
Let $I$ be a rqr right ideal of $R$. Then $I\subseteq J$\end{lem}
\begin{proof}
Let $M$ be a maximal right ideal of $R$. If $I\nsubseteq M$ then
$I+M=R$, so $1=x+m$ for some $x\in I$ and $m\in M$. Hence $1-x\in M$,
now there exits $y\in R$ such that $(1-x)y=1$, so $1\in M$ hence
$M=R$. A contradiction, thus $I\subseteq J$ as required.\end{proof}
\begin{lem}
Let $R$ be a ring, $J(R)$ is a right quasi-regular ideal.\end{lem}
\begin{proof}
Let $j\in J$. Suppose that $1-j$ has no right inverse. Then $(1-j)R\neq R$
so by Theorem \ref{prop:existence of maximal ideal} there exists
a maximal right ideal $M$ such that $(1-j)R\subseteq M$. But $j\in M$
by definition of $J(R)$ so $1=1-j+j\in M$, hence $M=R$. This is
a contradiction, hence $1-j$ has a right inverse for all $j\in J$.
So $J$ is a rqr.\end{proof}
\begin{lem}
Let $I$ be an ideal of a ring $R$. Then $I$ rqr if and only if
$I$ lqr.\end{lem}
\begin{proof}
Suppose that $I$ is rqr. Let $x\in I$, then there exists $a\in R$
such that $(1-x)(1-a)=1$. So $a=xa-x\in I$ since $I\lhd_{r}R$.
Hence there exists $t\in R$ such that $(1-a)(1-t)=1$, so $1-x=(1-x)1=(1-x)(1-a)(1-t)=1(1-t)=1-t$.
Hence $(1-a)(1-x)=1$, thus $x$ is lqr. By symmetry we can run the
converse argument.\end{proof}
\begin{thm}
The (right) Jacobson radical is a qr ideal and contains all the rqr
right ideals.\end{thm}
\begin{proof}
This is what we have proved above.\end{proof}
\begin{cor}
The Jacobson radical of a ring $R$ is left right symmetric, i.e.,
left Jacobson radical $J_{l}$ is equal to the right Jacobson radical
$J_{r}$\end{cor}
\begin{proof}
$J_{l}$ is a qr ideal by the left hand version of the theorem, so
$J_{l}\subseteq J_{r}$. Similarly $J_{r}\subseteq J_{l}$, hence
$J_{r}=J_{l}$.\end{proof}
\begin{thm}
Let $R$ be a ring with Jacobson radical $J$. Then $J(R/J)=0$\end{thm}
\begin{proof}
The maximal right ideals of the right $R/J$ are precisely the right
ideals of the form $M/J$ where $M$ is a maximal right ideal of $R$\end{proof}
\begin{rem*}
The theory can be adjusted to deal with rings without an identity.
\end{rem*}

\subsection{Commutative Local Rings}
\begin{defn}
Let $R$ be a commutative ring, $R$ is said to be a \emph{local ring}
if $R$ has a unique maximal ideal\end{defn}
\begin{note*}
This terminology is slightly different from Kaplansky's
\end{note*}
Let $R$ be a commutative local ring with 1. Let $M$ be the maximal
ideal of $R$, then:
\begin{enumerate}
\item $M$ is the Jacobson radical of $R$
\item $R/M$ is a field
\item If $x\in R$, $x\notin M$ then $x$ is a unit of $R$.\end{enumerate}
\begin{example*}
Let $R=\left\{ \frac{a}{b}|a,b\in\zz,b\mathrm{odd}\right\} $

Check that $R$ is a local ring. Find its unique maximal ideal. In
fact $R=\zz_{(2)}$, i.e., the ring $\zz$ localised at the prime
ideal $2\zz$\end{example*}
\begin{rem*}
There exists a non-commutative ring with unique maximal ideal (in
fact the only proper non-zero ideal) which is \uline{not} its Jacobson
radical.

\newpage{}
\end{rem*}

\section{Chapter 3: Chain conditions}

Rings need not have $1$ in this chapter


\subsection{Finitely Generated Modules}
\begin{defn}
Let $T$be a subset of $M_{R}$. The ``smallest'' submodule of $M$
containing $T$ is called \emph{the submodule of $M$ generated by
$T$,} i.e., it is the intersection of all submodules of $M$ containing
$T$.

By convention we take $\{0\}$ to be the submodule generated by the
empty set $\emptyset$.
\end{defn}
Of particular importance is the case when $T$ consists of a singles
element $a\in M$. In general the submodule generated by $a$ is $\{ar+\lambda a|r\in R,\lambda\in\zz\}$.
This equals $aR$ when $R$ has $1$ and $M$ is unital. 
\begin{defn}
If $M_{R}$ is generated by a single element then we say that $M$
is a \emph{cyclic module}

A right $R$-module $M$ is said to be \emph{finitely generated }(f.g.)
if it is the module generated by a finite subset. If $R$ has $1$
and $M$ is a finitely generated module then $\exists a_{1},\dots,a_{n}\in M$
such that $M=a_{1}R+\dots+a_{n}R$.

Cyclic submodules of $R_{R}$ {[}$_{R}R${]} are called \emph{principle
right }(\emph{left}) \emph{ideals}.
\end{defn}

\subsection{Finiteness Assumption}
\begin{defn}
Let $\mathscr{S}$ be a non-empty collection of submodules of a right
$R$-module $M$.
\begin{enumerate}
\item An element $K\in\mathscr{S}$is said to be \emph{maximal} in $\mathscr{S}$
if $\nexists K'\in\mathscr{S}$ such that $K'\supsetneq K$.


Similarly for a \emph{minimal} element of $\mathscr{S}$

\item $A$ is said to have the \emph{ascending chain condition }(ACC) for
submodules in $\mathscr{S}$ if every chain of submodules $A_{1}\subseteq A_{2}\subseteq\dots$
with $A_{i}\in\mathscr{S}$ has equal terms after a finite number
of terms.
\item $M$ is said to have the \emph{maximum condition} on submodules in
$\mathscr{S}$ if every non-empty collection of submodules in $\mathscr{S}$
has a submodules maximal in this collection.


The \emph{descending chain condition }(DCC) and \emph{minimum condition
}are defined analogously.

\end{enumerate}
\end{defn}
\begin{prop}
Let $\mathscr{S}$ be a non-empty collection of submodules of $M_{R}$
then the following are equivalent:
\begin{enumerate}
\item $M$ has ACC {[}DCC{]} on submodules in $\mathscr{S}$
\item $M$ has the maximum {[}minimum{]} condition on submodules in $\mathscr{S}$
\end{enumerate}
\end{prop}
\begin{proof}
Exercise
\end{proof}
Particularly important is the case when $\mathscr{S}$ consists of
all submodules in $M_{R}$. The abbreviation ``$M$ has ACC'' will
mean that $M$ has ACC on the set of all submodules of $M$. Similarly
for the other conditions.
\begin{prop}
The following are equivalent for a right $R$-module $M$.
\begin{enumerate}
\item $M$ has ACC
\item $M$ has the maximal condition
\item \uline{Every} submodule of $M$ is finitely generated.
\end{enumerate}
\end{prop}
\begin{proof}
This is Commutative Algebra Proposition 5.1\end{proof}
\begin{example*}
$\zz_{\zz}$ has ACC since every ideal is principle (this follows
from the Euclidean Algorithm)\end{example*}
\begin{rem*}
{}
\begin{enumerate}
\item ACC does not imply the existence of an integer $n$ such that all
chains stop after $n$ steps. This is easily checked on $\zz$
\item Similarly with DCC. Examples are harder but they do exists.
\item However if $M_{R}$ has both ACC and DCC then such an integer does
exists. This follows from the theory of composition series.
\end{enumerate}
\end{rem*}
\begin{lem}[Dedekind Modular Law]
 Let $A,B,C$ be submodules of $M_{R}$ such that $A\supseteq B$.
Then $A\cap(B+C)=B+(A\cap C)$.\end{lem}
\begin{proof}
Elementary\end{proof}
\begin{prop}[Commutative Algebra 5.4]
\label{prop:M ACC iff quotient ACC}Suppose that $K$ is a submodule
of $M_{R}$. Then $M$ has ACC {[}DCC{]} if and only if both $K$
and $M/K$ have ACC {[}DCC{]}\end{prop}
\begin{proof}
$\Rightarrow$: Straightforward

$\Leftarrow$: Let $M_{1}\subseteq M_{2}\subseteq\dots$ be an ascending
chain of submodules of $M$. Consider the chains $M_{1}\cap K\subseteq M_{2}\cap K\subseteq\dots$
and $M_{1}+K\subseteq M_{2}+K\subseteq\dots$. The first chain stops
since it consists of submodules of $K$. So there exists $k\geq1$
such that $M_{k}\cap K=M_{k+i}\cap K$ for all $i\geq1$. The second
chain stops since it consists of submodules of $M$ which are in 1
to 1 correspondence with those of $M/K$. So there exists an $l$
such that $M_{l}+K=M_{l+i}+K$ for all $i\geq1$. Let $n=\max\{k,l\}$.
Then $M_{n+i}=M_{n+i}\cap(M_{n+i}+K)=M_{n+i}\cap(M_{n}+K)=M_{n}+(M_{n+i}\cap K)$
by the Modular Law (since $M_{n+i}\supseteq M_{n}$). And $M_{n}+(M_{n+i}+K)=M_{n}+M_{n}\cap K=M_{n}$,
and this is true $\forall i\geq1$. So $M_{R}$ has ACC

Similarly for DCC
\end{proof}
This important proposition has many consequences
\begin{cor}[Commutative Algebra 5.5 ]
 \label{cor:M_1+dots+M_n has ACC}Let $M_{1},\dots,M_{n}$ be submodules
of a right $R$-modules $M$. If each $M_{i}$ has ACC {[}DDC{]} then
so does their sum $M_{1}+\dots+M_{n}=K$.\end{cor}
\begin{proof}
Take $K_{1}=M_{1}+M_{2}$. We have $K_{1}/M_{1}=\frac{M_{1}+M_{2}}{M_{1}}\cong\frac{M_{2}}{M_{1}\cap M_{2}}$.
So $\frac{K_{1}}{M_{1}}$ has ACC {[}DCC{]} since $\frac{M_{2}}{M_{1}\cap M_{2}}$
is a factor modules of $M_{2}$ and $M_{2}$ has ACC. Also $M_{1}$
has by assumption ACC {[}DCC{]}. So by the proposition \ref{prop:M ACC iff quotient ACC},
$K_{1}$ has ACC {[}DCC{]}.

This can easily be extended by induction.\end{proof}
\begin{cor}
\label{cor:3.2.8}Let $R$ be a ring with $1$. Suppose that $R$
has ACC {[}DCC{]} on right ideals. Let $M_{R}$ be a finitely generated
unital right $R$-module. Then $M_{R}$ has ACC {[}DCC{]} on submodules.\end{cor}
\begin{proof}
Since $M_{R}$ is finitely generated and unital, there exists $m_{1},\dots,m_{k}$
such that $M=m_{1}R+\dots m_{k}R$. So by Corollary \ref{cor:M_1+dots+M_n has ACC}
it is enough to show that each $m_{i}R$ has ACC {[}DCC{]}. The map
$r\to m_{i}r$ for all $r\in R$ is an $R$-homomorphism of $R_{R}$
onto $m_{i}R$. So $m_{i}R$ is isomorphic to a factor of $R_{R}$.
So $m_{i}R$ has ACC {[}DCC{]} on submodules.\end{proof}
\begin{rem*}
If $R$ does not have $1$, the ACC part of the corollary still holds
but the DCC part is false! This is because $(m_{i})=\{m_{i}r+\lambda m_{i}|r\in R,\lambda\in\zz\}$
and $\zz$ has ACC but not DCC\end{rem*}
\begin{defn}
A modules with ACC on submodules is called a \emph{Noetherian module.
}A modules with DCC on submodules is called an \emph{Artinian module}

A ring with ACC on right ideals is called a \emph{right Noetherian
ring}. A ring with ACC on left ideals is called a \emph{left Noetherian
ring.}

A ring with $1$ and DCC on right ideals is called a \emph{right Artinian
ring. }A ring with $1$ and DCC on left ideals is called a \emph{left
Artinian ring.}
\end{defn}

\subsection{Nil and Nilpotent Ideals}
\begin{defn}
Let $S$ be non-empty subset of a ring $R$. $S$ is said to be \emph{nil}
if given any $s\in S$ there exists an integer $k\geq1$ (which depends
on $s$) such that $s^{k}=0$. $S$ is said to be \emph{nilpotent}
if there exists an integer $k\geq1$ such that $S^{k}=0$
\end{defn}
If $S$ consists of a single element, there is no difference between
nil and nilpotent and we normally say that the element is nilpotent.
\begin{prop}
Let $R$ be a ring with $1$. Every nil one sided ideal of $R$ is
inside $J(R)$.\end{prop}
\begin{proof}
Let $I$ be a nil right ideal and $x\in I$. Then $x^{k}=0$ for some
$k\geq1$. We have $(1-x)(1+x+\dots+x^{k-1})=1$ so $x$ is r.q.r.
so $x\in J(R)$. Thus $I\subseteq J(R)$.\end{proof}
\begin{rem*}
This is also true without $1$. \end{rem*}
\begin{lem}
\label{lem:I and K nilpotent}Let $R$ be a ring:
\begin{enumerate}
\item If $I$ and $K$ are nilpotent right ideals then so are $I+K$ and
$RI$
\item Every nilpotent right ideal lies inside a nilpotent ideal.
\end{enumerate}
\end{lem}
\begin{proof}
Suppose that $I^{k}=0$ and $K^{l}=0$, $k,l\geq1$. Then $(I+K)^{k+l-1}=0$
since every term in the expansion lies in either $I^{k}$ or $K^{l}$
and hence is zero. So $I+K$ is nilpotent. $(RI)^{k}=(RI)(RI)\dots(RI)\subseteq R(IR)^{k-1}I\subseteq RI^{k}=0$.
So $RI$ is nilpotent.

Suppose that $I$ is a nilpotent right ideal. Then $I\subseteq I+RI$.
Now $I+RI\lhd R$ and is nilpotent by the first part.\end{proof}
\begin{defn}
The sum of all nilpotent ideals of $R$ is called the \emph{Nilpotent
radical} (or the Wedderburn radical). It is usually denoted by $N(R)$.\end{defn}
\begin{note*}
$N(R)\subseteq J(R)$ always.
\end{note*}
It follows from Lemma \ref{lem:I and K nilpotent} that $N(R)=\sum$
nilpotent right ideals $=\sum$ nilpotent left ideals. Clearly $N(R)$
is a nil ideal. It is in general not itself nilpotent.
\begin{example*}[Zassenhaus's Example]
 Let $F$ be a field, $I$ the open interval $(0,1)$ and $R$ a
vector space over $F$ with basis $\{x_{i}|i\in I\}$. Define a multiplication
on $F$ by extending the following product of basis elements $x_{i}x_{j}=\begin{cases}
x_{i+j} & \mathrm{if\,}i+j<1\\
0 & \mathrm{if\, i+j\geq1}
\end{cases}$. Thus every element of $R$ can be written uniquely in the form $\sum_{i\in I}a_{i}x_{i}$
where $a_{i}\in F$ and $a_{i}=0$ for all except a finite number
of $i$. Check that $N(R)=R$ but $R$ is not nilpotent.\end{example*}
\begin{prop}
\label{prop:section 3-6}Let $R$ be a commutative ring. Then $N(R)$
equals the set of all nilpotent elements of $R$.\end{prop}
\begin{proof}
Let $n$ be a nilpotent element. This implies that the principle ideal
generated by $n$ is nilpotent. (Prove!)\end{proof}
\begin{example*}
The above is false for non-commutative rings. e.g, let $R$ be the
ring of $2\times2$ matrices over $\qq$. Then $R$ has only two ideals
$0$ and $R$. So $N(R)=0$ but $\begin{pmatrix}0 & 1\\
0 & 0
\end{pmatrix}^{2}=0$.\end{example*}
\begin{defn}
An ideal $P$ of a ring $R$ is said to be a \emph{prime ideal} if
$AB\subseteq P$, $A,B\lhd R$ implies $A\subseteq P$ or $B\subseteq P$.
We exclude $R$ itself from the set of prime ideals.\end{defn}
\begin{prop}
Let $R$ be a commutative ring and $P\lhd R$. Then $P$ is a prime
ideal if and only if $(a,b\in R)$ we have $ab\in P\Rightarrow a\in P$
or $b\in P$.\end{prop}
\begin{proof}
Trivial if $R$ has $1$. Not so trivial but still true if $R$ does
not have $1$. \end{proof}
\begin{prop}[Commutative Algebra 1.10 ]
Let $R$ be a ring. The intersection of all prime ideals of $R$
is a nil ideal.\end{prop}
\begin{proof}
We shall show that if $x\in R$ is not nilpotent then there exists
a prime ideal excluding it. Suppose that $x\in R$ is not nilpotent.
Let $\mathscr{S}$ be the set of ideals which contains no power of
$x$. $\mathscr{S}\neq0$ since $\{0\}\in\mathscr{S}$ . Check that
Zorn's lemma applies. So $\mathscr{S}$ contains a maximal element,
say $P$. Claim: $P$ is a prime ideal. If not then $\exists$ ideals
$A$ and $B$ of $R$ such that $AB\subseteq P$ but $A\nsubseteq P$
and $B\subseteq P$. Then $A+P\supsetneq P$ and $B+P\supsetneq P$.
So $x^{k}\in A+P$ and $x^{l}\in B+P$ for some integers $k,l$. But
then $x^{k+l}\in(A+P)(B+P)\subseteq P$ which is a contradiction.
Thus $P$ is a prime ideal proving the proposition.\end{proof}
\begin{cor}
In a commutative ring $N(R)$ equals the intersection of all prime
ideals of $R$.\end{cor}
\begin{proof}
This follows from Theorem \ref{prop:section 3-6} and the previous
theorem.\end{proof}
\begin{cor}
\label{cor: 3.3.12??}In a commutative ring with $1$ a finitely generated
nil ideal is nilpotent. In particular when $R$ is Noetherian $N(R)$
is nilpotent.\end{cor}
\begin{proof}
Let $K$ be a finitely generated ideal of $R$. Let $K=k_{1}R+\dots+k_{s}R$
with $k_{i}\in K$. Each $k_{i}$ is nilpotent hence so is the ideal.
The result follows by \ref{lem:I and K nilpotent}. When $R$ is Noetherian
$N(R)$ is finitely generated and so nilpotent by above.
\end{proof}

\subsection{Nakayama\textquoteright{}s Lemma and an Application}
\begin{defn}
Let $I\lhd_{r}R$. We say that $a_{1},\dots,a_{n}$ is \emph{minimal
generated set} for $I$ if:
\begin{enumerate}
\item $a_{1},\dots,a_{n}$ generate $I$
\item No proper subset of $\{a_{1},\dots,a_{n}\}$ generates $I$.
\end{enumerate}
\end{defn}
\begin{namedthm}[Nakayama\textquoteright{}s Lemma]
 Let $R$ be a ring with $1$ and $M_{R}$ a finitely generated module.
Let $I$ be a subset of $J(R)$ Then $MI=M\Rightarrow M=0$. \end{namedthm}
\begin{proof}
Let $MI=M$. Then we have $MJ=M$. Suppose that $M\neq0$. Let $a_{1},\dots,a_{n}$
be a minimal generated set for $M$. We have $M=a_{1}R+\dots+a_{n}R$
so that $MJ=a_{1}J+\dots+a_{n}J$. Now $a_{1}\in M=MJ$ so $a_{1}=a_{1}x+\dots+a_{n}x_{n}$
for some $x_{i}\in J$. Now $a_{1}(1-x)=a_{2}x_{2}+\dots+a_{n}x_{n}$
($a_{1}(1-x_{1})=0$ if $n=1$). So $a_{1}=a_{2}x_{2}(1-x_{1})^{-1}+\dots+a_{n}x_{n}(1-x_{1})^{-1}$
($a_{1}=0$ if $n=1$). This contradicts the minimality of $n$. Hence
$M=0$\end{proof}
\begin{rem*}
This is also valid for rings without $1$.
\end{rem*}
Let $R$ be a commutative local ring with $1$ with unique maximal
ideal $J$. Then $R/J$ is a field. So $J/J^{2}$ is an $R/J$-module,
i.e., $J/J^{2}$ is a vector space over the field $R/J$. If $x\in R$
let $\overline{x}$ denote the coset $x+J^{2}$. So $\overline{x}\in R/J^{2}$. 
\begin{lem}[Commutative Algebra 2.17]
 Let $R$ be a commutative local ring with $1$. Let $J$ be the
maximal ideal of $R$. Suppose that $J$ is finitely generated and
$x_{1},\dots,x_{k}\in J$. Then $x_{1},\dots,x_{k}$ generate $J$
(as an $R$-module) $\iff\overline{x_{1}},\dots,\overline{x_{k}}$
is a set which spans the vector space $J/J^{2}$ (over the field $R/J$)\end{lem}
\begin{proof}
$\Rightarrow$) $\overline{x_{1}},\dots,\overline{x_{k}}$ generate
$J/J^{2}$ as an $R$-module so $\overline{x_{1}},\dots,\overline{x_{k}}$
generate $J/J^{2}$ as an $R/J$-module, i.e., they span the vector
space $J/J^{2}$.

$\Leftarrow$) Let $I=x_{1}R+\dots+x_{k}R$. Then $I\subseteq J$,
$\overline{x_{1}},\dots,\overline{x_{k}}$ generates $J/J^{2}$ as
an $R$-module, hence $I+J^{2}=J$. This implies that $(J/I)J=J/I$
where $J/I$ is considered as an $R$-module. So $J/I=0$ by Nakayama's
lemma, so $J\subseteq I$. Hence $J=x_{1}R+\dots+x_{k}R$. \end{proof}
\begin{cor}
\label{cor:3.4.4}In the above ring $x_{1},\dots,,x_{k}$ is a minimal
generated set for $J$ $\iff$ $\overline{x_{1}},\dots,\overline{x_{k}}$
is a basis for the vector space $J/J^{2}$ over $R/J$.\end{cor}
\begin{proof}
Follows from above\end{proof}
\begin{thm}
Let $R$ be a commutative Noetherian local ring with $1$. Let $J$
be the maximal ideal of $R$. Then any two minimal generating set
of $J$ contain the same number of elements.\end{thm}
\begin{proof}
This is a direct consequence of the corollary\end{proof}
\begin{notation*}
We shall denote this common number by $V(R)$. Thus $V(R)=\dim J/J^{2}$
as a vector space over the field $R/J$.

\newpage{}
\end{notation*}

\section{Commutative Noetherian Rings}

All rings considered in this chapter are assumed to be commutative
rings $1$.


\subsection{Primary Decomposition}
\begin{defn}
An ideal $Q$ is said to be \emph{primary} if $ab\in Q$ ($a,b\in R$)
implies that $a\in Q$ or $b^{n}\in Q$ for some integer $n$.
\end{defn}
Clearly a prime ideal is primary.
\begin{defn}
$R$ is called a \emph{primary ring} if $0$ is a primary ideal.
\end{defn}
Clearly an ideal $Q$ is primary if and only if $R/Q$ is a primary
ring.
\begin{defn}
We say that $R$ has \emph{primary decomposition} if every ideal of
$R$ is expressible as a finite intersection of primary ideals.
\end{defn}

\begin{defn}
An ideal is said to be \emph{meet-irreducible} if $I=A\cap B$, $A,B\lhd R$
implies $I=A$ or $I=B$.\end{defn}
\begin{note*}
The two different definitions: $M_{R}$ is \emph{irreducible} if $\{0\}$
and $M$ are the only submodules. $I\lhd R$ is \emph{meet-irreducible
}if $I=A\cap B$ implies $I=A$ or $I=B$\end{note*}
\begin{lem}[Commutative Algebra 6.18]
 Let $R$ be a Noetherian ring. Then every ideal of $R$ is expressible
as a finite intersection of meet-irreducible ideals.\end{lem}
\begin{proof}
Suppose not. Let $A\lhd R$ be a maximal counterexample. Then $A$
is not meet-irreducible. So $A=B\cap C$, $B,C\lhd R$, $B\supsetneq A,C\supsetneq A$.
By maximality of $A$, both $B$ and $C$ are finite intersection
of meet-irreducible ideals. Hence so is $A$. Contradiction hence
the result holds.\end{proof}
\begin{notation*}
Let $M$ be a subset of $M_{R}$. The \emph{annihilator }of $S$ in
$R$ is $\ann(S)=\{r\in R|Sr=0\}$. For $R$ is non-commutative $\ann(S)\lhd_{r}R$.
If $S$ is a submodule then typically $S$ is a subset of $R$.\end{notation*}
\begin{thm}[(Noether) Commutative Algebra 6.20]
\label{thm:has primary decomposition} A Noetherian ring has primary
decomposition\end{thm}
\begin{proof}
By the previous lemma it is enough to show that a meet-irreducible
ideal is primary. Without loss of generality assume $0$ to be meet-irreducible.
Suppose that $ab=0$, $a,b\in R$. 

\uline{Claim:} There exists an integer $n\geq1$ such that $b^{n}R\cap\ann(b^{n})=0$.

Since the chain $\ann(b)\subseteq\ann(b^{2})\subseteq\dots$stops
there is an integer $n\geq1$ such that $\ann(b^{n})=\ann(b^{2n})$.
Now $z\in b^{n}R\cap\ann(b^{n})\Rightarrow x=b^{n}t$ for some $t\in R$
and $b^{z}=0$. So $b^{2n}t=0\Rightarrow b^{n}t=0\Rightarrow z=0$.
Since $0$ is meet-irreducible either $b^{n}R=0$ or $\ann(b^{n})=0$.
Thus $b^{n}=0$ or $a=0$ and $0$ is a primary ideal\end{proof}
\begin{defn}
Let $Q$ be a primary ideal. Let $P/Q$ be the nilpotent radical of
the ring $R/Q$. $P$ is called the \emph{radical} of $Q$ and we
say that $Q$ is $P$-\emph{primary}. \end{defn}
\begin{notation*}
We denote the radical of $Q$ by $\sqrt{Q}$. 
\end{notation*}
Recall that for a commutative ring $R$, $N(R)=$set of all nilpotent
elements of $R$.
\begin{prop}
\label{prop:4.1.6}Let $Q$ be a primary ideal and let $P=\sqrt{Q}$.
Then:
\begin{enumerate}
\item $P$ is a prime ideal
\item If further $R$ is Noetherian, then $P^{k}\subseteq Q$ for some $k\geq1$.
\end{enumerate}
\end{prop}
\begin{proof}
{}
\begin{enumerate}
\item Let $ab\in P$ with $a,b\in R$. Then $(ab)^{n}\in Q$ for some $n\geq1$
so $a^{n}b^{n}\in Q$. If $a\notin P$ then $a^{n}\notin Q$ so $(b^{n})^{s}\in Q$
for some $s\geq1$ by definition of primary. Hence $b\in P$. Thus
$P$ is a prime ideal/
\item $P/Q$ is a nil ideal of $R/Q$. If $R/Q$ is Noetherian, $P/Q$ is
nilpotent (by Proposition \ref{lem:I and K nilpotent} ?(check reference
maybe)). Hence $P^{k}\subseteq Q$ for some $k\geq1$.
\end{enumerate}
\end{proof}
\begin{thm}[Commutative Algebra 6.24]
 \label{thm:4.1.7}Let $R$ be a commutative Noetherian ring. Then
$\cap_{n=1}^{\infty}J^{n}=0$ where $J=J(R)$.\end{thm}
\begin{proof}
Let $X=\cap_{n=1}^{\infty}J^{n}$. Let $XJ=Q_{1}\cap\dots\cap Q_{n}$
be a primary decomposition for $X$. Fix $i$ and let $P_{i}=\sqrt{Q_{i}}$,
if $X\nsubseteq Q_{i}$ then $J\subseteq P_{i}$. So $J^{k_{i}}\subseteq Q_{i}$
for some $k_{i}\geq1$ by the previous proposition. Thus $X\subseteq Q_{i}$
or $J^{k_{i}}\subseteq Q_{i}$. So $X\subseteq Q_{i}$ for all $i=1\dots,n$
in any case. Hence $X\subseteq XJ$. So $X=XJ$ hence by Nakayama's
lemma $X=0$. 
\end{proof}
This is a surprisingly useful result.
\begin{rem*}
For a right Noetherian ring this is false (Proven by Herstein in 1965).
While for left and right Noetherian rings the result is still an open
problem.\end{rem*}
\begin{defn}
A ring is called an \emph{integral domain} if the product of any two
non-zero elements of the ring is non-zero. \end{defn}
\begin{thm}
\label{thm:4.1.9}Let $R$ be a commutative, local, Noetherian ring.
Suppose that $J=J(R)$ is a principle ideal. Then every non-zero ideal
of $R$ is a power of $J$. In particular, $R$ is a principal ideal
ring.\end{thm}
\begin{proof}
Let $0\neq I\lhd R$, $I\neq R$. Then $I\subseteq J$. Since $\cap_{n=1}^{\infty}J^{n}=0$
there exists an integer $k\geq1$ such that $I\subseteq J^{k}$ but
$I\nsubseteq J^{k+1}$. Let $J=aR$ ($a\in J$), then $J^{m}=a^{m}R\,\forall m\geq1$.
Now there exists an element $x$ such that $x\in I$ but $x\notin a^{k+1}R$
$(*)$. Since $x\in a^{k}R$ we have $x=a^{k}t$ for some $t\in R$.
Now $t\notin J=aR$ by $(*)$. So $t$ must be a unit of $R$. So
$a^{k}=xt^{-1}\in I$. Hence $J^{k}=a^{k}R\subseteq I$. It follows
that $I=J^{k}$ proving the theorem.\end{proof}
\begin{cor}
\label{thm:4.1.10??}Let $R$ be a commutative, local, Noetherian
ring. 
\begin{enumerate}
\item If $J$ is not nilpotent then $R$ is an integral domain and $0$
and $J$ are the only prime ideals of $R$.
\item If $J$ is nilpotent then $R$ is Artinian and $J$ is the only prime
ideal of $R$.
\end{enumerate}
\end{cor}
\begin{proof}
Exercise. (Note that in 2. $J^{s}=0$ for some $s\geq1$ so $R,J,J^{2},\dots,J^{s}=0$
are the only ideals.
\end{proof}

\subsection{Decomposition of $0$}
\begin{defn}
Let $I=Q_{1}\cap\dots\cap Q_{n}$ be a primary decomposition for an
ideal $I$. Suppose that $Q_{i}$ are $P_{i}$-primary. We say the
decomposition is \emph{normal} {[}Commutative Algebra: minimal{]}
if
\begin{enumerate}
\item No $Q_{i}$ is superfluous
\item $P_{i}\neq P_{j}$ for all $i\neq j$
\end{enumerate}
\end{defn}
Given that $I$ has a primary decomposition, we can arrange a normal
decomposition for $I$ by:
\begin{enumerate}
\item Removing any superfluous primary ideals and
\item By applying the following:\end{enumerate}
\begin{lem}
If $Q_{1}$ and $Q_{2}$ are $P$-primary ideals then so is $Q_{1}\cap Q_{2}$\end{lem}
\begin{proof}
Let $ab\in Q_{1}\cap Q_{2}$, $a,b\in R$. If $a\notin Q_{1}\cap Q_{2}$
then $a\notin Q_{1}$ say. Then $b^{n}\in Q_{1}$ for some $n\geq1$.
So $b\in P$. Hence $b^{s}\in Q_{2}$ for some $s\geq1$ since $Q_{2}$
is $P$-primary. Let $k=\max(n,s)$ then $b^{k}\in Q_{1}\cap Q_{2}$.
Now $p\in P$ implies $p^{t}\in Q_{1}\cap Q_{2}$ for sufficiently
large $t\geq1$. Hence $P\subseteq\sqrt{Q_{1}\cap Q_{2}}$. But $Q_{1}\cap Q_{2}\subseteq Q_{1}$
so $\sqrt{Q_{1}\cap Q_{2}}\subseteq\sqrt{Q_{1}}=P$, thus $P=\sqrt{Q_{1}\cap Q_{2}}$. 
\end{proof}
Thus whenever necessary we shall assume that the primary decomposition
being considered is normal.
\begin{rem*}
We may still have $\sqrt{Q_{i}}\supsetneq\sqrt{Q_{j}}$ with a normal
primary decomposition {[}Commutative Algebra, example before 6.8{]}\end{rem*}
\begin{defn}
Let $R$ be a ring. We say that a prime ideal $P$ is a \emph{minimal}
prime ideal of $R$ if $Q\subseteq P$ with $Q$ prime implies $Q=P$.\end{defn}
\begin{lem}
Let $R$ be a commutative Noetherian ring. Suppose that $0=Q_{1}\cap\dots\cap Q_{n}$
be a primary decomposition of $0$. Let $P_{i}=\sqrt{Q_{i}}$ and
suppose (after possible renumbering) that $P_{1},\dots,P_{k}$ are
minimal in the set $\{P_{1},\dots,P_{n}\}$. Then $P_{1},\dots,P_{k}$
are precisely the minimal primes of $R$.\end{lem}
\begin{proof}
It is enough to show that if $P$ is a prime ideal of $R$ then $P\supseteq P_{j}$
for some $1\leq j\leq k$. By Theorem \ref{thm:has primary decomposition}
(? check reference) there exists integers $k_{i}\geq1$ such that
$P_{i}^{k_{i}}\subseteq Q_{i}$ for $i=1,\dots,n$. Then $P_{1}^{k_{1}}P_{2}^{k_{2}}\dots P_{n}^{k_{n}}\subseteq Q_{1}\cap\dots\cap Q_{n}=0$.
In particular, $P_{1}^{k_{1}}\dots P_{n}^{k_{n}}\subseteq P$ hence
$P_{m}\subseteq P$ for some $m$ with $1\leq m\leq n$. But since
$P_{1},\dots,P_{k}$ are minimal in the set $\{P_{1},\dots,P_{n}\}$
we have $P_{j}\subseteq P_{m}$ for some $j$, $1\leq j\leq m$. Thus
$P\supseteq P_{j}$ with $1\leq j\leq m$ as required. \end{proof}
\begin{defn}
Let $c\in R$, we say that $c$ is \emph{regular }if $ $$cx=0,x\in R\Rightarrow x=0$

An element which is not regular is called a \emph{zero-divisor}.\end{defn}
\begin{notation*}
Let $I\lhd R$. Write $\mathscr{C}(I)=\{x\in R|x+I$ is regular in
the ring $R/I$\}
\end{notation*}
Clearly $\mathscr{C}(0)=\{$regular elements of $R$\}. If $P$ is
a prime ideal, in a commutative ring then $\mathscr{C}(P)=R\setminus P$. 
\begin{prop}
\label{prop:4.2.6}Let $R$ be a Noetherian ring and $0=Q_{1}\cap\dots\cap Q_{n}$
a normal primary decomposition. Let $P_{i}=\sqrt{Q_{i}}$ and suppose
that $P_{1},\dots,P_{k}$ are the minimal primes of $R$. Then:
\begin{enumerate}
\item $N(R)=P_{1}\cap\dots\cap P_{k}$.
\item $\mathscr{C}(0)=R\setminus\cup_{i=1}^{n}P_{i}$ 
\item $\mathscr{C}(N)=R\setminus\cup_{i=1}^{k}P_{i}$
\end{enumerate}
\end{prop}
\begin{proof}
{}
\begin{enumerate}
\item Clearly $N\subseteq P_{1}\cap\dots\cap P_{k}$. Now $P_{1}\cap\dots\cap P_{k}\subseteq P_{j}$
for all $1\leq j\leq n$. By Proposition \ref{prop:4.1.6} there exists
an integer $t_{i}$ such that $(P_{1}\cap\dots\cap P_{k})^{t_{i}}\subseteq Q_{i}$.
Let $t=\max\{t_{i}\}$, then $(P_{1}\cap\dots\cap P_{k})^{t}\subseteq Q_{1}\cap\dots\cap Q_{n}=0$.
Thus $P_{1}\cap\dots\cap P_{k}\subseteq N$ and so $P_{!}\cap\dots\cap P_{k}=N$.
\item Let $c\in R\setminus\cup_{i=1}^{n}P_{i}$. Then $cx=0,x\in R\Rightarrow x\in Q_{i}$
for all $i$ $1\leq i\leq n$, since $c$ belong to no $P_{i}$. Hence
$x\in Q_{1}\cap\dots\cap Q_{n}=0$, so $c\in\mathscr{C}(0)$. 


Now $P_{i}^{n_{i}}\subseteq Q_{i}$ for some $n_{i}$. So $P_{i}^{n_{i}}[Q_{1}\cap\dots\cap Q_{i-1}\cap Q_{i+1}\cap\dots\cap Q_{n}]\subseteq Q_{1}\cap\dots\cap Q_{n}=0$.
Now $Q_{1}\cap\dots\cap Q_{i-1}\cap Q_{i+1}\cap\dots\cap Q_{n}\neq0$
since our decomposition is normal. So $P_{i}$ is does not contain
a regular elements and hence $\cup_{i=1}^{n}P_{i}$ does not contain
a regular element. Hence $\mathscr{C}(0)=R\setminus\cup_{i=1}^{n}P_{i}$

\item Exercise
\end{enumerate}
\end{proof}
\begin{lem}
\label{lem:4.2.7}Let $R$ be a commutative ring. Let $P_{1},\dots,P_{n}$
be ideals of $R$, at least $n-2$ of which are prime. Let $S$ be
a subring of $R$. Suppose that $S\subseteq\cup_{i=1}^{n}P_{i}$,
then $S\subseteq P_{k}$ for some $k$, $1\leq k\leq n$.\end{lem}
\begin{rem*}
Note that $S$ does not (necessarily) contain $1$, since our definition
of rings did not include $1$\end{rem*}
\begin{proof}
Proof by induction on $n$. For $n=1$, result is trivial.

For $n=2$ if $S\nsubseteq P_{1}$ and $S\nsubseteq P_{2}$ then choose
$x_{1},x_{2}\in S$ such that $x_{1}\notin P_{2}$ and $x_{2}\notin P_{1}$.
Then $x_{1}+x_{2}\in S$ but $x_{1}+x_{2}\notin P_{i},i=1,2$.

Now assume $n>2$ and that the result holds for values $<n$.

Clearly any selection of $n-1$ of the $P_{i}$ at most $2$ will
be non-prime. Suppose that $S\subseteq\cup_{i=1}^{n}P_{i}$ but $S\notin P_{i}$
for any $i$ ($i=1,2,\dots,n$). Then $S\nsubseteq P_{1}\cup\dots\cup P_{k-1}\cup P_{k+1}\cup\dots\cup P_{n}$
by induction hypothesis (as $k$ varies). Now choose $x_{k}\in S$
such that $x_{k}\notin P_{1}\cup\dots\cup P_{k-1}\cup P_{k+1}\cup\dots\cup P_{n}$.
Thus $x_{k}\in P_{k}$. Since $n>2$ at least of the $P_{i}$ must
be prime, say $P_{1}$. Let $y=x_{1}+x_{2}\dots x_{n}$, then $y\notin P_{i}$
for any $i=1,\dots,n$. This is a contradiction. This completes the
induction.\end{proof}
\begin{prop}
\label{prop:4.2.8}Let $R$ be a commutative Noetherian ring. Let
$I\lhd R$, then $I$ contains a regular element if and only if $\ann I=0$.\end{prop}
\begin{proof}
$\Rightarrow$: Trivial

$\Leftarrow$: Suppose that every element of $I$ is a zero divisor.
Then by the Proposition \ref{prop:4.2.6} part 2) $I\subseteq\cup_{i=1}^{n}P_{i}$
(where the $P_{i}$ are as in Proposition \ref{prop:4.2.6}. So $I\subseteq P_{j}$,
for some $j$, $1\leq j\leq n$. We have $\ann I\supseteq\ann P_{j}\neq0$.
This completes the proof.\end{proof}
\begin{prop}
\label{prop:4.2.9}Let $R$ be a commutative Noetherian ring and $I\lhd R$.
Suppose that $I$ contains a regular element. Then $I=c_{1}R+\dots+c_{n}R$
where each $c_{i}$ is regular.\end{prop}
\begin{proof}
Let $K$ be the right ideal generated by the regular elements in $I$.
So $I\setminus K$ is either empty or consists of zero divisors. Let
$P_{1},\dots,P_{n}$ be the primes associated with a primary decomposition
of $0$ (as in Proposition \ref{prop:4.2.6}). So $I\setminus K\subseteq P_{1}\cup\dots\cup P_{n}$
by Proposition \ref{prop:4.2.6} part 2, so $I\subseteq K\cup P_{1}\cup\dots\cup P_{n}$.
Hence $I\subseteq K$ or $I\subseteq P_{i}$ for some $i$ (by Lemma
\ref{lem:4.2.7}). But $I\nsubseteq P_{i}$ for any $i$ since $I$
contains a regular element but all $P_{i}$ contains zero-divisors.
Hence $I\subseteq K$ and so $I=K$. Since $R$ is Noetherian it follows
that we can find a finite generating set consisting of regular elements.
\end{proof}

\subsection{Localisation {[}Commutative Algebra Section 3{]}}
\begin{defn}
Let $S$ be a non-empty subset of a ring $R$. We say that $S$ is
\emph{multiplicatively closed} if $s_{1},s_{2}\in S\Rightarrow s_{1}s_{2}\in S$.
\end{defn}
Typical example: $\mathscr{C}(P)=R\setminus P$ where $P$ is a prime
ideal in a \uline{commutative} ring. We shall always assume $0\notin S$
and $1\in S$.

Define an equivalence relation $\sim$ on $R\times S$ as follows:
$(a,s)\sim(b,t)$ if there exists $s'\in S$ such that $(at-bs)s'=0$
(where $(a,s),(b,t)\in R\times S$)

Let $\frac{a}{c}$ be the equivalence class of $(a,b)$ and let $R_{S}$
denote the set of all such equivalence classes. For $\frac{a}{s},\frac{b}{t}\in R_{S}$
define $\frac{a}{s}+\frac{b}{t}=\frac{at+bs}{st}$ and $\frac{a}{s}\times\frac{b}{t}=\frac{ab}{st}$.

Check that this is well-defined and that $R_{S}$ is a ring. We have
a natural ring homomorphism $\phi:R\to R_{S}$ given by $\phi(r)=\frac{r}{1}$
for all $r\in R$
\begin{defn}
$R_{S}$ constructed above is called a \emph{localizations of $R$
at $S$}

Let $A,B$ be rings with $1$ and $\phi:A\to B$ a homomorphism of
rings. In this context we shall always assume $\phi(1_{A})=1_{B}$\end{defn}
\begin{namedthm}[The Universal Mapping Property]
\[
\xymatrix{A\ar[d]_{\phi}\ar[dr]^{\theta}\\
B & A_{S}\ar@{-->}[l]^{\psi}
}
\]
Let $A,B$ be rings and $S$ a multiplicatively closed subset of $A$.
Suppose that $\phi:A\to B$ is a ring homomorphism such that $\phi(s)$
is a unit in $B$ for all $s\in S$. Then there exists a unique ring
homomorphism $\psi:A_{S}\to B$ such that $\phi=\psi\theta$\end{namedthm}
\begin{proof}
See Commutative Algebra 3.2-point
\end{proof}
The ring homomorphism $\theta:R\to R_{S}$ has the following properties:
\begin{enumerate}
\item $s\in S$ implies $\theta(s)$ is a unit in $R_{S}$
\item Given $a\in R,\theta(a)=0$ if and only if $as=0$ for some $s\in S$
\item Every element of $R_{S}$ is expressible as $\theta(a)[\theta(s)]^{-1}$
for some $a\in R$, $s\in S$.
\end{enumerate}
These three properties determine $R_{S}$ to within isomorphism. 
\begin{thm}
\label{thm:Uniquness Thm}Let $A,B$ be rings and $S$ a multiplicatively
closed subset of $A$. Suppose that $\alpha:A\to B$ is a ring homomorphism
such that:
\begin{enumerate}
\item $s\in S$ implies $\alpha(s)$ is a unit of $B$
\item $\alpha(a)=0$ implies $as=0$ for some $s\in S$
\item Every element of $B$ is expressible as $\alpha(a)[\alpha(s)]^{-1}$
for some $a\in A,s\in S$.
\end{enumerate}
Then there exists a unique isomorphism $\psi:A_{S}\to B$ such that
$\alpha=\psi\theta$, where $\theta$ is the natural map $A\to A_{S}$.
\[
\xymatrix{A\ar[d]_{\alpha}\ar[dr]^{\theta}\\
B & A_{S}\ar@{-->}[l]^{\exists!\,\psi}
}
\]
\end{thm}
\begin{proof}
By the universal mapping property there is a unique homomorphism $\psi:A_{S}\to B$
such that $\alpha=\psi\theta$, where $\psi$ is given by $\psi(as^{-1})=\alpha(a)[\alpha(s)]^{-1}$
(used property 1.) Then use property 2 and 3 to check that $\psi$
is an isomorphism.
\end{proof}
In view of this we speak of \uline{the }localization of $R$ at
$S$. Also since $\frac{a}{s}=\frac{a}{1}\cdot\frac{1}{s}$ we usually
write $as^{-1}$ rather than $\frac{a}{s}$ for elements of $R_{S}$. 

Particularly important is the case when elements of $S$ are regular,
in this case the natural map $R\to R_{S}$ is a monomorphism. We identity
$R$ with its image in $R_{S}$. Thus we may assume that $R$ is a
subring of $R_{S}$, we write $r$ instead of $\frac{r}{1}$ for elements
of $R$. In particular when $R$ is an integral domain and $S=R\setminus\{0\}$
then $R_{S}$ is just the field of fractions of $R$. 
\begin{lem}
\label{lem:5i}Let $R$ be a ring and $S$ a multiplicatively closed
subset such that $S\subseteq\mathscr{C}(0)$. Then:
\begin{enumerate}
\item if $I\lhd R\Rightarrow IR_{S}\lhd R_{S}$ and every element of $IR_{S}$
is expressible as $xd^{-1}$ for some $x\in I$ and $d\in S$.
\item $K\lhd R_{S}\Rightarrow K\cap R\lhd R$ and $(K\cap R)R_{S}=K$.
\end{enumerate}
\end{lem}
\begin{proof}
We are assuming that $R$ is a subring of $R_{S}$. So a typical element
of $IR_{S}$ is $x_{1}r_{1}c_{1}^{-1}+\dots+x_{n}r_{n}c_{n}^{-1}$
for some $x_{i}\in I,r_{i}\in R$ and $c_{i}\in S$. Let $d=c_{1}c_{2}\dots c_{n}$
and $d_{i}=c_{1}c_{2}\dots c_{i-1}c_{i+1}\dots c_{n}$ then $x_{1}r_{1}c_{1}^{-1}+\dots+x_{n}r_{n}c_{n}^{-1}=(x_{1}r_{1}d_{1}+\dots+x_{n}r_{n}d_{n})d^{-1}=xd^{-1}$
where $x=x_{1}r_{1}d_{1}+\dots+x_{n}r_{n}d_{n}\in I$. 

The rest is an exercise.\end{proof}
\begin{rem*}
If $I\lhd R$ we have $IR_{S}\cap R\supseteq I$ but we do not have
equality in general. E.g. $R=\zz$ and $R_{S}=\qq$.

However, see Lemma \ref{lem:7ii} part 2 below.\end{rem*}
\begin{cor}
If $R$ is a Noetherian ring then so is the ring $R_{S}$. \end{cor}
\begin{proof}
Clear from the previous lemma (part 2)\end{proof}
\begin{lem}
\label{lem:7ii}Let $R$ be a ring and $S$ a multiplicatively closes
subset. Suppose that the elements of $S$ are regular. Then
\begin{enumerate}
\item If $\Pi$ is a prime ideal of $R_{S}$ then $\Pi\cap R$ is a prime
ideal of $R$
\item If $P$ is a prime ideal of $R$ and $P\cap S=\emptyset$ then $PR_{S}$
is a prime ideal of $R_{S}$ and $PR_{S}\cap R=P$
\end{enumerate}
\end{lem}
\begin{proof}
{}
\begin{enumerate}
\item Easy
\item We shall first need to show that $PR_{S}\cap R=P$. Clearly $PR_{S}\cap R\supseteq P$.
Let $z\in PR_{S}\cap R$, then $z=ps^{-1}$ for some $p\in P$ and
$s\in S$ Lemma \ref{lem:5i} part 1. So $zs=p\in P$ with $z,s\in R$.
Now $z\in P$ since $s\notin P$ and $P$ is prime. Thus $PR_{S}\cap R=P$.
Now let $\alpha\beta\in PR_{S}$ with $\alpha,\beta\in R_{S}$. Then
$\alpha=ac^{-1}$ and $\beta=bd^{-1}$ where $a,b\in R$, $c,d\in S$.
So $abc^{-1}d^{-1}\in PR_{S}$ hence $ab\in PR_{S}\cap R=P$. So $\alpha\in PR_{S}$
or $\beta\in PR_{S}$, hence $PR_{S}$ is a prime ideal of $R_{S}$.
(Note: $PR_{S}\neq R_{S}$ since $P\neq R$)
\end{enumerate}
\end{proof}
\begin{thm}
\label{thm:4.3.8}Let $R$, $S$ be as above. Then there is a one
to one order preserving correspondence between the prime ideals of
$R$ which do not intersect $S$ and the prime ideals of $R_{S}$\end{thm}
\begin{proof}
This follows from the previous lemma. The correspondence is $P\leftrightarrow PR_{S}$.\end{proof}
\begin{rem*}
Theorems analogous to the above hold even when the elements of $S$
are not assumed to be regular.\end{rem*}
\begin{notation*}
Of special importance is the case when $P$ is a prime ideal and $S=R\setminus P=\mathscr{C}(P)$.
In this case it is customary to write $R_{P}$ instead of $R_{\mathscr{C}(P)}$
or $R_{R\setminus P}$. \end{notation*}
\begin{prop}
Let $P$ be a prime ideal of a ring $R$ and suppose that the elements
of $\mathscr{C}(P)$ are regular. Then $PR_{P}$ is the unique maximal
ideal of $R_{P}$ and thus $R_{P}$ is a local ring. \end{prop}
\begin{proof}
Let $I\lhd R_{P},$ $I\neq R_{P}$. Then $I$ does not contain a unit
of $R_{P}$. $[I\cap R]\cap\mathscr{C}(P)=\emptyset$, i.e., $I\cap R\subseteq P$.
So $I=(I\cap R)R_{P}\subseteq PR_{P}$, since $P\cap\mathscr{C}(P)=\emptyset$,
$PR_{P}\neq R_{P}$. It follows that $PR_{P}$ is the unique maximal
ideal of $R_{P}$.\end{proof}
\begin{rem*}
Hence the name ``localization''\end{rem*}
\begin{example*}
$R=\zz,P=2\zz$, then $Z_{(2)}=\left\{ \frac{a}{b}|a,b\in\zz,b\mathrm{\, odd}\right\} $
\end{example*}

\subsection{Localisation of a Module {[}Commutative Algebra 3.1{]}}

Let $M$ be an $R$-module and $S$ a multiplicatively closed subset
of $R$ such that $0\notin S$, $1\in S$. Define an equivalence relation
on $M\times S$ as follows: $(m,s)\sim(m',s')$ if there exists $t\in S$
such that $(ms'-m's)t=0$. Check that $\sim$ is an equivalence relation.
Denote equivalence class of $(m,s)$ by $m/s$. Let $M_{S}$ be the
collection of all such equivalence classes. Define 
\[
\frac{m}{s}+\frac{m'}{s}=\frac{ms'+m's}{ss'},\frac{m}{s}\cdot\frac{r}{t}=\frac{mr}{st},m,m'\in M,s,s',t\in S,r\in R
\]
Check that this turns $M_{S}$ into an $R_{S}$-module. Uniqueness
corresponding to Theorem \ref{thm:Uniquness Thm} can also be proved.
We call $M_{S}$ the \emph{localization of $M$ at $S$.}

Note that if $A$ is an $R_{S}$-module then $A$ can be considered
an $R$-module via the action $a\cdot r=a\cdot\frac{r}{1}\,\forall a\in A,r\in R$.
In this case $A\cong A_{S}$ as $R_{S}$-module {[}Check that $\frac{a}{c}\to a\cdot\frac{1}{c}$
is an isomorphism $A_{S}\to S${]}


\subsection{Symbolic Powers}

Let $P$ be a prime ideal. Then the powers of $P$ need not be $P$-primary
{[}Commutative Algebra Example after 6.3{]}

$P^{(n)}=\{x\in R|xc\in P^{n}$ for some $c\in\mathscr{C}(P\}$. Check
that $P^{(n)}\lhd R$.
\begin{defn}
$P^{(n)}$ is called the \emph{$n^{\mathrm{nt}}$ symbolic power of
$P$}
\end{defn}
Clearly $P^{(1)}=P$ and $P^{(n)}\subseteq P$ for all $n$.
\begin{lem}
$P^{(n)}$ is $P$-primary\end{lem}
\begin{proof}
Let $ab\in P^{(n)}$, $a,b\in R$. Then $abc\in P^{n}$ for some $c\in\mathscr{C}(P)$.
If no power of $b$ lies in $P^{(n)}$ then $b\notin P$, i.e., $b\in\mathscr{C}(P)$,
We have $a(bc)\in P^{n}$ with $bc\in\mathscr{C}(P)$. Hence $a\in P^{(n)}$
and $P^{(n)}$ is primary. It is easy to see that $\sqrt{P^{(n)}}=P$\end{proof}
\begin{lem}
\label{lem:4.5.2}Let $P$ be a prime ideal and suppose that elements
of $\mathscr{C}(P)$ are regular. Then fro every $n\geq1$:
\begin{enumerate}
\item $(PR_{P})^{n}=P^{n}R_{P}$
\item $P^{n}R_{P}\cap R=P^{(n)}$
\item $P^{(n)}R_{P}=P^{n}R_{P}$
\end{enumerate}
\end{lem}
\begin{proof}
{}
\begin{enumerate}
\item $(PR_{P})^{n}=P^{n}R_{P}^{n}=P^{n}R_{P}$
\item $x\in P^{(n)}$$\Rightarrow xc\in P^{n}$ for some $c\in\mathscr{C}(P)$.
So $xcR_{P}\subseteq P^{n}R_{P}\Rightarrow xR_{P}\subseteq P^{n}R_{P}$
since $c$ is a unit of $R_{P}$. Hence $x\in P^{n}R_{P}\cap R$.


Conversely: $q\in P^{n}R_{P}\cap R\Rightarrow q=pc^{-1}$ with $p\in P^{n}$
and $c\in\mathscr{C}(P)$. Hence $qc=p\in P^{n}$, so $q\in P^{(n)}$
and noting that $q\in R$, we have$P^{(n)}=P^{n}R_{P}\cap R$

\item Exercise
\end{enumerate}
\end{proof}

\subsection{The Rank of a Prime Ideal}
\begin{defn}
A prime ideal $P$ is said to have \emph{rank $r$} (or \emph{height}
$r$) if there exists a chain of prime ideals $P_{1}\subsetneq P_{2}\subsetneq\dots\subsetneq P_{r}\subsetneq P$
but none longer. If there does not exists a maximal finite chain of
primes then we say $\rk P=\infty$. If $P$ contains no other primes,
we define $\rk P=0$
\end{defn}
Note that $\rk P=0$ if and only if $P$ is a minimal prime.
\begin{defn}
Let $a_{1},\dots,a_{n}\in R$, we say that prime $P$ is \emph{minimal
over $a_{1},\dots,a_{n}$ }if $P/(a_{1}R+\dots+a_{n}R)$ is a minimal
prime of the ring $R/(a_{1}R+\dots+a_{n}R)$.\end{defn}
\begin{lem}
\label{lem:4.6.3}Let $R$ be a Noetherian ring, $A\lhd R$. Suppose
that $R/A$ is an Artinian ring. Then $R/A^{n}$ is Artinian for all
$n\geq1$. \end{lem}
\begin{proof}
$R/A\cong\frac{R/A^{2}}{A/A^{2}}$ (by the third isomorphism theorem).
Note $A/A^{2}$ is finitely generated as an $R/A$-module, so by Corollary
\ref{cor:3.2.8} $A/A^{2}$ is Artinian. Since $R/A$ and $A/A^{2}$
are Artinian, it follows from Proposition \ref{prop:M ACC iff quotient ACC}
that $R/A^{2}$ is Artinian. The proof then extends by induction.\end{proof}
\begin{namedthm}[Krull's Principal Ideal Theorem]
 Let $R$ be a Noetherian Ring. Let $a\in R$ be a non-unit, suppose
that $P$ is a prime ideal minimal over $a$. Then $\rk P\leq1$.\end{namedthm}
\begin{proof}
We shall first deal with the case when $P$ is the unique maximal
ideal of $R$, i.e., when $R$ is a local ring with Jacobson radical
$P$. Suppose we have $Q_{1}\subseteq Q\subsetneq P$. Factoring out
by $Q_{1}$ we may without loss of generality assume that $R$ is
an integral domain. In the ring $R/aR$, $P/aR$ is both the unique
maximal ideal and a minimal prime. Hence by Proposition \ref{prop:4.2.6}
we have $P/aR=N(R/aR)$. By Proposition \ref{cor: 3.3.12??}(Check
this reference) there exists an integer $n\geq1n$ such that $P^{n}\subseteq aR$. 

Now $R/P$ is a field so by Lemma \ref{lem:4.6.3} $R/P^{n}$ is Artinian.
Hence $R/aR$ is an Artinian ring. Hence there exists $k\geq1$ such
that $Q^{(k)}+aR=Q^{(k+1)}+aR$. So $Q^{(k)}\subseteq Q^{(k+1)}+aR$.
Let $x\in Q^{(k)}$, then $x=y+at$ for some $y\in Q^{(k+1)}$, $t\in R$.
Hence $at=x-y\in Q^{(k)}$. Now $a\notin Q$ since $P$ is minimal
over $a.$ So $t\in Q^{(k)}$, thus $Q^{(k)}\subseteq Q^{(k+1)}+aQ^{(k)}$.
Hence $Q^{(k)}=Q^{(k+1)}+aQ^{(k)}$ (since the other containment is
true trivially). Hence $\left[\frac{Q^{(k)}}{Q^{(k+1)}}\right]=\left[\frac{Q^{(k)}}{Q^{(k+1)}}\right]aR$
where $\left[\,\right]$ is viewed as an $R$-module. 

So $\frac{Q^{(k)}}{Q^{(k+1)}}=0$ by Nakayama's Lemma since $aR\subseteq J(R)$,
so $Q^{(k)}=Q^{(k+1)}$. Now localize at $Q$. So $Q^{(k)}R_{Q}=Q^{(k+1)}R_{Q}$
and $Q^{k}R_{Q}=Q^{k+1}R_{Q}$ by Lemma \ref{lem:4.5.2} part 3. So
$(QR_{Q})^{k}=(QR_{Q})^{k+1}$ by Lemma \ref{lem:4.5.2} part 1. So
$(QR_{Q})^{k}=0$ by Nakayama's Lemma since $QR_{Q}=J(R_{Q})$. Hence
$Q^{k}=0$ and hence $Q=0$ since $R$ is a domain.

Now in the general case again suppose that $Q_{1}\subseteq Q\subsetneq P$.
Factor out $Q_{1}$ and assume that $R$ is an integral domain. Now
localize at $P$. Factor out $Q_{1}$ and assume that $R$ is an integral
domain. Now localise at $P$, by Theorem \ref{thm:4.3.8}, there exists
an inclusion preserving one to one correspondence between primes of
$R$ lying inside $P$ and primes of the ring $R_{P}$. Use this and
the first part of the proof applied to the ring $R_{P}$ to finish
the proof.\end{proof}
\begin{namedthm}[The Generalised Principal Ideal Theorem]
 Let $R$ be a commutative Noetherian ring. Suppose that $P$ is
a prime ideal minimal over the elements $x_{1},\dots,x_{r}\in R$.
Then $\rk P\leq r$. \end{namedthm}
\begin{proof}
We prove this by induction

For $r=1$ we use Krull's Principal Ideal Theorem.

Now assume the result is true for primes minimal over $\leq r-1$
elements. Suppose that $P$ is minimal over $x_{1},\dots,x_{r}$ and
suppose that we can construct a chain of primes $P=P_{0}\supsetneq P_{1}\supsetneq\dots\supsetneq P_{r+1}$.
If $x_{1}\in P_{r}$ then in the ring $R/x_{1}R$ we have a chain
of primes $P_{0}/x_{1}R\supsetneq P_{1}/x_{1}R\supsetneq\dots\supsetneq P_{r}/x_{1}R$
$(*)$ But $P_{0}/x_{1}R$ is minimal over the images of $x_{2},\dots,x_{r}$
in the ring $R/x_{1}R$. So $(*)$ contradicts the induction. So $x_{1}\notin P_{r}$.

Let $k$ be such that $x_{1}\in P_{k}$ but $x_{1}\notin P_{k+1}$.
So we have $P_{k}/P_{k+2}\supseteq\frac{P_{k+2}+x_{1}R}{P_{k+2}}\supsetneq P_{k+2}/P_{k+2}$.
By Krull's Principal Ideal Theorem $P_{k}/P_{k+2}$ can not be minimal
over $[x_{1}+P_{k+2}]$ (since otherwise we have $ $$P_{k}/P_{k+2}\supsetneq P_{k+1}/P_{k+2}\supsetneq P_{k+2}/P_{k+2}$).
So there exists a prime ideal $P_{k+1}'$ such that $P_{k}\supsetneq P_{k+1}'\supseteq P_{k+2}+x_{1}R\supsetneq P_{k+2}$.
Proceeding this way we can build a new chain $P=P_{0}\supsetneq P_{1}\supsetneq\dots\supsetneq P_{k}\supsetneq P'_{k+1}\supsetneq\dots\supsetneq P_{r}'\supsetneq P_{r+1}$.
Now we have $x_{1}\in P_{r}'$ and this leads to a contradiction as
in $(*)$.\end{proof}
\begin{defn}
Let $R$ be a commutative ring. We define the \emph{Krull} \emph{dimension
}of $R$ by $K\dim(R)=\sup_{P\mathrm{\, prime}}\rk P$.\end{defn}
\begin{note*}
$K\dim$ can be infinite in a Noetherian ring even thought the rank
of each prime ideal is finite.\end{note*}
\begin{prop}
Let $R$ be a commutative Noetherian local ring with Jacobson radical
$J$. Then $K\dim(R)=\rk J<\infty$. \end{prop}
\begin{proof}
Since $R$ is local, $K\dim(R)=\rk J$, and $\rk J<\infty$ by the
Generalised Principal Ideal Theorem as it is minimal over its generators. \end{proof}
\begin{lem}
\label{lem:4.6.8}Let $R$ be a commutative Noetherian local ring
with $K\dim(R)=n$. Then $K\dim(R/cR)\geq n-1$. Further, if $c$
is regular then equality holds.\end{lem}
\begin{proof}
Let $J$ be the maximal ideal of $R$. Then $\rk J=n$, so there exists
a chain of primes $J=P_{0}\supsetneq P_{1}\supsetneq\dots\supsetneq P_{n}$.
As in the Generalised Principal Ideal Theorem we can construct a new
chain of primes, $J=Q_{0}\supsetneq Q_{1}\supsetneq\dots\supsetneq Q_{n-1}$
with $c\in Q_{n-1}$. Hence $\rk(J/cR)\geq n-1$ $(*)$. 

Now assume that $c$ is regular. If $J/cR=T_{0}/cR\supsetneq\dots\supsetneq T_{k}/cR$
is a chain of primes in $R/cR$ then $J=T_{0}\supsetneq T_{1}\supsetneq\dots\supsetneq T_{k}$
is a chain of primes in $R$. Since $c$ is regular by Proposition
\ref{prop:4.2.6} $T_{k}$ can not be a minimal prime of $R$ since
$c\in T_{k}$. So $n=\rk J\geq\rk J/cr+1$. Hence $\rk J/cR=n-1$
from $(*)$ when $c$ is regular.
\end{proof}

\subsection{Regular Local Ring}

Let $R$ be a Noetherian local ring with Jacobson radical $J$. We
have $V(R)=\dim J/J^{2}$ as a vector space over the field $R/J$.
So $V(R)=$the number of elements in a minimal generator set for $J$
by Corollary \ref{cor:3.4.4}. By The Generalised Principal Ideal
Theorem we have $\rk J\leq V(R)$
\begin{defn}
A Noetherian local ring is called a \emph{regular local ring} if $\rk(J)=V(R)$.
\end{defn}
A local principal ideal domain is regular by Theorem \ref{thm:4.1.10??}
\begin{lem}
Let $R$ be a Noetherian local ring with Jacobson radical $J$ ($R$
not a field). Suppose that $x\in J\setminus J^{2}$, let $R^{*}=R/xR$.
Then $V(R^{*})=V(R)-1$. \end{lem}
\begin{proof}
Note that $R^{*}$ is a Noetherian local ring with Jacobson radical
$J^{*}=J/xR$. Let $y_{1}^{*},\dots,y_{k}^{*}$ be a minimal generating
set for $J^{*}$. Choose $y_{1},\dots,y_{k}\in J$ such that $y_{i}\mapsto y_{i}^{*}$
under the natural homomorphism $R\to R/xR$. Claim $x,y_{1},\dots,y_{k}$
is a minimal generating set for $J$. We shall now show that the homomorphic
images of $x,y_{1},\dots,y_{k}$ in the vector space $J/J^{2}$ are
linearly independent. Suppose that $xr+y_{1}r_{1}+\dots+y_{k}r_{k}\in J^{2}$
$(*)$. So $y_{1}^{*}r_{1}^{*}+\dots+y_{k}^{*}r_{k}^{*}\in(J^{*})^{2}$
where $r_{i}^{*}$ are the homomorphic images of $r_{i}$ under $R\to R/xR$.
It follows that $r_{i}^{*}\in J^{*}$ since $y_{1}^{*},\dots,y_{k}^{*}$
is a minimal generating set for $J^{*}$ and $\dim J^{*}/(J^{*})^{2}=k$.
So $r_{i}\in J$ for all $i$. It follows from $(*)$ that $xr\in J^{2}$
since $r_{i},y_{i}\in J$. So $r\in J$ since $x\notin J^{2}$. (Note
that $J^{2}$ is $J$-primary check!) This completes the proof.\end{proof}
\begin{thm}
\label{thm:4.7.3}Let $R$ be a regular local ring with Jacobson radical
$J$. Suppose that $x\in J\setminus J^{2}$. Then the ring $R^{*}=R/xR$
is also regular local.\end{thm}
\begin{proof}
{}

\begin{alignat*}{2}
V(R)-1 & =V(R^{*}) & \quad & \mbox{by the previous lemma}\\
 & \geq\rk J^{*} &  & \mbox{where }J^{*}=J/xR\mbox{ by the General Principal Ideal Theorem}\\
 & \geq\rk J-1 &  & \mbox{by Theorem \ref{lem:4.6.8}}\\
 & =V(R)-1
\end{alignat*}
So $V(R^{*})=\rk J^{*}$. Thus $R^{*}$ is a regular local ring\end{proof}
\begin{rem*}
We have also shown that $\rk J^{*}=\rk J-1$. \end{rem*}
\begin{lem}
\label{lem:4.7.4}Let $R$ be a Noetherian local ring which is not
an integral domain. Let $P=pR$ ($p\in P$) be a prime ideal. Then
$\rk P=0$. \end{lem}
\begin{proof}
Suppose that $Q\subsetneq P$ where $Q$ is a prime ideal. Then $p\notin Q$.
Now $q\in Q$ implies $q=pt$ for some $t\in R$. Hence $pt\in Q\Rightarrow t\in Q$
since $p\notin Q$. So $q\in pQ\subseteq P^{2}\subseteq p^{2}R$.
Preceding this way we have $Q\subseteq P^{n}$ for all $n\geq1$,
so $Q\subsetneq\cap_{n=1}^{\infty}P^{n}\subseteq\cap_{n=1}^{\infty}J$
where $J=J(R)$. But by Theorem \ref{thm:4.1.7} $\cap_{n=1}^{\infty}J^{n}=0$,
so $Q=0$ which is a contradiction since $R$ is not a domain. Hence
$\rk P=0$\end{proof}
\begin{thm}
\label{thm:4.7.5}A regular local ring is an integral domain.\end{thm}
\begin{proof}
By induction on $K\dim R=\rk J$. If $\rk J=0$ then $R$ must be
a field.

Suppose now that $\rk J=n>0$ and assume result for rings of $K\dim<n$.
Since $J\neq J^{2}$ by Nakayama's lemma choose $x\in J\setminus J^{2}$.
By Theorem \ref{thm:4.7.3}, $R^{*}=R/xR$ is regular local. Also
$K\dim R^{*}=K\dim R-1$. By induction hypothesis $R^{*}$ is an integral
domain, i.e., $xR$ is a prime ideal. Suppose that $R$ is not an
integral domain, then by Lemma \ref{lem:4.7.4} $xR$ is a minimal
prime. Let $P_{1},\dots,P_{k}$ be the minimal primes of $R$. We
have show that $J\setminus J^{2}\subseteq P_{1}\cup\dots\cup P_{k}$.
So $J\subseteq J^{2}\cup P_{1}\cup\dots\cup P_{2}$. So $J\subseteq P_{j}$
for some $j$ by Lemma \ref{lem:4.2.7} hence $J=P_{j}$. So $\rk J=0$,
which is a contradiction. So $R$ is an integral domain. 
\end{proof}

\section{Projective Modules}

All rings in this chapter are assumed to have $1$ but need not be
commutative. 

Suppose $R$ is regular local and $P$ prime. How about the ring $R_{P}$? 


\subsection{Free Modules}
\begin{defn}
A right $R$-module $M$ is said to be \emph{free} if:
\begin{enumerate}
\item $M$ is generated by a subset $S\subseteq M$ 
\item $\sum_{\mathrm{finite}}a_{i}r_{i}=0$ if and only if $r_{i}=0\forall r_{i}\in R,a_{i}\in S$.
\end{enumerate}
Then $S$ is called a \emph{free basis for }$M$.\end{defn}
\begin{rem*}
{}
\begin{enumerate}
\item $R_{R}$ is free with free basis $1$
\item In a free module not every minimal generating set is a free basis.
e.g: in the ring of $2\times2$ matrices over $\qq$, $\begin{pmatrix}1 & 0\\
0 & 0
\end{pmatrix}$ and $\begin{pmatrix}0 & 0\\
0 & 1
\end{pmatrix}$ is a minimal generating set but not a free basis.
\item By convention, $0$ is considered to be a free module on the empty
free basis.
\end{enumerate}
\end{rem*}
\begin{lem}
Let $R$ be a commutative ring, then any two free basis of a free
$R$-module have the same cardinality.\end{lem}
\begin{proof}
By Theorem \ref{prop:existence of maximal ideal}, $R$ contains a
maximal ideal, $M$ say. Then $R/M$ is a field. Let $A$ be a free
$R$-module with a free basis $\left\{ x_{\lambda}\right\} _{\lambda\in\Lambda}$.
We claim: $\frac{x_{\lambda}R}{x_{\lambda}M}\cong\frac{R}{M}$ (as
$R$ and hence as $R/M$-modules). To see this, define $\theta:R\to\frac{x_{\lambda}R}{x_{\lambda}M}$
by $\theta(r)=x_{\lambda}r+x_{\lambda}M$. Then $\theta$ is an $R$-homomorphism
and $\ker\theta\supseteq M$. But $M$ is maximal, so $\ker(\theta)=M$,
proving our claim. 

Write $B_{\lambda}=\frac{x_{\lambda}R}{x_{\lambda}M}$, since $B_{\lambda}\cong R/M$
each $B_{\lambda}$ is a 1-dimensional vector space over the field
$R/M$. From the external direct sum $\sum_{\lambda\in\Lambda}\oplus B_{\lambda}$.
Now $A/AM$ is an $R/M$-module. (see Section \ref{sub:1.13}). We
have $A/AM\cong\sum_{\lambda\in\Lambda}\oplus B_{\lambda}$ (as $R$-modules
and hence also as $R/M$-modules). Hence dimension of $A/AM$ as a
vector space is $|\Lambda|$. The dimension of $A/AM$ is invariant
by vector space theory, hence the result.\end{proof}
\begin{rem*}
Over a non-commutative ring it is possible to have $R\cong R\oplus R$
as right $R$-modules.
\end{rem*}
\textbf{The Free Module \label{5.1.3}$F_{A}$}. Let $A$ be a set
indexed by $\Lambda$. We define $F_{A}$ to be the set of all symbols
$\sum a_{\lambda}r_{\lambda}$ with $a_{\lambda}\in A,r_{\lambda}\in R,\lambda\in\Lambda$,
where all but a finite number of $r_{\lambda}$ are zero. We further
require these expression to satisfy $\sum a_{\lambda}r_{\lambda}=\sum a_{\lambda}s_{\lambda}\iff r_{\lambda}=s_{\lambda}\,\forall\lambda\in\Lambda$.
We can make $F_{A}$ a right $R$-module by defining $\sum a_{\lambda}r_{\lambda}+\sum a_{\lambda}s_{\lambda}=\sum a_{\lambda}(r_{\lambda}+s_{\lambda})$
and $\left(\sum a_{\lambda}r_{\lambda}\right)r=\sum a_{\lambda}(r_{\lambda}r)$
(for all $r_{\lambda},s_{\lambda},r\in R$) 

A is a \uline{free basis} for $F_{A}$ (identifying $a\in A$ with
$a\cdot1\in F_{A}$)
\begin{prop}
\label{prop:5.1.3}Every right $R$-module is a homomorphism image
of a free right $R$-module\end{prop}
\begin{proof}
Let $M$ be a right $R$-module. Index the elements of $M$ and form
the free right $R$-module $F_{M}$. Elements of $F_{M}$ are formal
sums of the form $\sum(m_{i})r_{i}$, $m_{i}\in M,r_{i}\in R$. Define
$F_{M}\to M$ by $\sum(m_{i})r\mapsto\sum m_{i}r_{i}\in M$. This
map is well-defined and is an $R$-homomorphism by the definition
of $F_{M}$. 
\end{proof}

\subsection{Exact Sequences}

Let $M_{i}$ be right $R$-modules and $f_{i}$ $R$-homomorphism
of $M_{i}$ into $M_{i-1}$. The sequence (which maybe finite or infinite)
$\xymatrix{\cdots\ar[r]^{f_{i+2}} & M_{i+1}\ar[r]^{f_{i+1}} & M_{i}\ar[r]^{f_{i}} & M_{i-1}\ar[r]^{f_{i-1}} & \cdots}
$ is said to be \emph{exact} if $\im f_{i+1}=\ker f_{i}$ for all $i$. 

A \emph{short exact sequence} (s.e.s.) is an exact sequence of the
form $\xyC{1.5pc}\xymatrix{0\ar[r] & M'\ar[r]^{f} & M\ar[r]^{g} & M''\ar[r] & 0}
$ . Note that since $\xymatrix{0\ar[r] & M'\ar[r]^{f} & M}
$ is exact we have $\ker(f)=0$, i.e., $f$ is a monomorphism. Similarly
we have $\xymatrix{M\ar[r]^{g} & M''\ar[r] & 0}
$ is exact so $M''=\im(g)$, i,e, $g$ is an epimorphism. We have $M'\cong f(M')$,
i.e., $M'$ is isomorphic to a submodule of $M$. Also $M''\cong M/\ker(g)=M/f(M')$.

Given modules $B\subseteq A$, we can construct the short exact sequence
$\xyC{1.5pc}\xymatrix{0\ar[r] & B\ar[r]^{i} & A\ar[r]^{\pi} & A/B\ar[r] & 0}
$ where $i$ is the inclusion map and $\pi$ the canonical homomorphism.
\begin{prop}[c.f. Graduate Algebra Theorem 5.3]
Given a short exact sequence $\xymatrix{0\ar[r] & A\ar@<1ex>[r]^{\alpha} & B\ar@<1ex>[r]^{\beta}\ar@{-->}@<1ex>[l]^{\delta} & C\ar[r]\ar@{-->}@<1ex>[l]^{\gamma} & 0}
$, the following conditions are equivalent.
\begin{enumerate}
\item $\im\alpha$ is a direct summand of $B$
\item There exists a homomorphism $\gamma:C\to B$ such that $\beta\gamma=1_{C}$
\item There exists a homomorphism $\delta:B\to A$ such that $\delta\alpha=1_{A}$
\end{enumerate}
\end{prop}
\begin{proof}
1. $\Rightarrow$ 2.) Let $B=\im(\alpha)+B_{1}=\ker\beta+B_{1}$.
Let $\beta_{1}$ be the restriction of $\beta$ to $B_{1}$. We have
$\beta B=\beta_{1}B_{1}=C$, so $\beta_{1}$ is an epimorphism. Also
$\ker\beta_{1}\subseteq\im\alpha\cap B_{1}=0$. Hence $\beta_{1}$
is an isomorphism and $C\cong B_{1}$. Define $\gamma:C\to B$ to
be the inverse of $\beta_{1}$. It follows that $\gamma$

2. $\Rightarrow$ 1.) We shall show that $B=\alpha(A)+\gamma\beta(B)=\ker\beta+\gamma\beta(B)$.
Let $b\in B$, then $b=(b-\gamma\beta b)+\gamma\beta b$. Now $b-\gamma\beta b\in\ker\beta$
since $\beta(b-\gamma\beta b)=\beta b-\beta\gamma\beta b=\beta b-1_{C}\beta b=\beta b-\beta b=0$.
If $z\in\ker\beta\cap\gamma\beta B$ means $z=\gamma\beta b$ for
some $b\in B$ and $\beta(z)=0$. This means $0=\beta(x)=\beta\gamma\beta b=\beta b\Rightarrow x=0$.
Thus $B=\ker(\beta)\oplus\gamma\beta(B)$ 

Similarly we can show $1\iff3$.\end{proof}
\begin{defn}
We say that the short exact sequence \emph{split} if any (and hence
all) of the above condition holds.
\end{defn}
Note that if the above short exact sequence split then we have $B=\im\alpha\oplus B_{1}\cong A\oplus C$
(external direct sum)
\begin{defn}
A right $R$-module $P$ is said to be \emph{projective} if every
diagram of the from 
\[
\xymatrix{ & P\ar[d]^{\mu}\\
A\ar[r]^{\pi} & B\ar[r] & 0\mbox{ exact}
}
\]
 can be embedded in he diagram 
\[
\xymatrix{ & P\ar[d]^{\mu}\ar@{-->}[ld]_{\overline{\mu}}\\
A\ar[r]^{\pi} & B\ar[r] & 0
}
\]
in such a way that $\pi\overline{\mu}=\mu$. (``the diagram commutes'')\end{defn}
\begin{lem}
\label{lem:5.2.4}A free module is projective.\end{lem}
\begin{proof}
Let $F$ be a free right module with a free basis $\{e_{\alpha}\}$.
Consider 
\[
\xymatrix{ & F\ar[d]^{\mu}\ar@{-->}[ld]_{\overline{\mu}}\\
A\ar[r]^{\pi} & B\ar[r] & 0\mbox{ exact}
}
\]
Let $b_{\alpha}=\mu e_{\alpha}$. As $\pi$ is an epimorphism, we
can choose $a_{\alpha}\in A$ such that $\pi a_{\alpha}=b_{\alpha}$.
Now define $\overline{\mu}:F\to A$ by $\overline{\mu}(\sum e_{\alpha}r_{\alpha})=\sum a_{\alpha}r_{\alpha}$,
$r_{\alpha}\in R$. Then $\overline{\mu}$ is an $R$-homomorphism
$F\to A$ and $\pi\overline{\mu}(\sum e_{\alpha}r_{\alpha})=\pi(\sum a_{\alpha}r_{\alpha})=\sum\pi(a_{\alpha})r_{\alpha}=\sum b_{\alpha}r_{\alpha}=\sum\mu(e_{\alpha})r_{\alpha}=\mu(\sum e_{\alpha}r_{\alpha})$.
Therefore $\pi\overline{\mu}=\mu$. 
\end{proof}
A projective module need not be free. To be shown later.
\begin{lem}
\label{lem:5.2.5}Let $P_{\alpha}$ ($\alpha\in\Lambda$) be right
$R$-modules. Then $\sum_{\alpha\in\Lambda}\oplus P_{\alpha}$ is
projective if and only if all $P_{\alpha}$ are projective\end{lem}
\begin{proof}
Let $i_{\alpha}$ be the injection map $P_{\alpha}\to\sum_{\alpha\in\Lambda}\oplus P_{\alpha}$
and let $p_{\alpha}$ be the projection map $\sum_{\alpha\in\Lambda}\oplus P_{\alpha}\to P_{\alpha}$
\begin{elabeling}{00.00.0000}
\item [{$\Leftarrow$}] Consider the diagram
\[
\xymatrix{ & \sum\oplus P_{\alpha}\ar[d]^{f}\\
A\ar[r]^{\pi} & B\ar[r] & 0
}
\]
Restrict $f$ to $P_{\alpha}$, $f|_{P_{\alpha}}=f_{\alpha}$ say.
Then $f_{\alpha}=fi_{\alpha}$. Since each $P_{\alpha}$ is projective,
there exists maps $\overline{f_{\alpha}}:P_{\alpha}\to A$ such that
$\pi\overline{f_{\alpha}}=f_{\alpha}$. Define $\overline{f}=\sum_{\alpha\in\Lambda}\overline{f_{\alpha}}p_{\alpha}$.
Then $\pi\overline{f}=\sum_{\alpha\in\Lambda}\pi\overline{f_{\alpha}}p_{\alpha}=\sum_{\alpha\in\Lambda}f_{\alpha}p_{\alpha}=\sum_{\alpha\in\Lambda}fi_{\alpha}p_{\alpha}=f$.
So $\sum_{\alpha\in\Lambda}\oplus P_{\alpha}$ is projective.
\item [{$\Rightarrow$}] For any $\beta\in\Lambda$ consider 
\[
\xymatrix{ & P_{\beta}\ar[d]^{f_{\beta}}\\
A\ar[r]^{\pi} & B\ar[r] & 0
}
\]
This gives rise to 
\[
\xyC{1.5pc}\xymatrix{ & \sum\oplus P_{\alpha}\ar[d]^{f_{\beta}p_{\beta}}\ar[dl]_{\overline{f}}\\
A\ar[r]^{\pi} & B\ar[r] & 0
}
\]
So there exists $\overline{f}:\sum_{\alpha\in\Lambda}\oplus P_{\alpha}\to A$
such that $ $$\pi\overline{f}=f_{\beta}p_{\beta}$. Hence $\pi\overline{f}i_{\beta}=f_{\beta}p_{\beta}i_{\beta}=f_{\beta}$
and $\overline{f}i_{\beta}$ maps $p_{\beta}\to A$.
\end{elabeling}
\end{proof}
\begin{prop}
The following conditions are equivalent:
\begin{enumerate}
\item $P$ is a projective right $R$-module
\item $P$ is a direct summand of a free module
\item Every short exact sequence $\xyC{1.5pc}\xymatrix{0\ar[r] & M'\ar[r] & M\ar[r] & P\ar[r] & 0}
$ splits.
\end{enumerate}
\end{prop}
\begin{proof}
{}
\begin{elabeling}{00.00.0000}
\item [{$3\Rightarrow2$}] Consider the short exact sequence $\xyC{1.5pc}\xymatrix{0\ar[r] & K_{P}\ar[r] & F_{p}\ar[r] & P\ar[r] & 0}
$ where $K_{P}$ is the kernel of the map $F_{P}\to P$. Since this
sequence splits, $F_{P}\cong P\oplus K_{P}$ 
\item [{$2\Rightarrow1$}] Follows from Lemma \ref{lem:5.2.4} and Lemma
\ref{lem:5.2.5}
\item [{$1\Rightarrow3$}] Consider 
\[
\xymatrix{ &  &  & P\ar[d]^{1_{P}}\ar@{-->}[ld]_{\overline{\mu}}\\
0\ar[r] & M'\ar[r]^{f} & M\ar[r]^{g} & P\ar[r] & 0
}
\]
Since $P$ is projective, there exists $\overline{\mu}:P\to M$ such
that $g\overline{\mu}=1_{P}$. Thus the short exact sequence splits.
\end{elabeling}
\end{proof}
\begin{example*}
Projective does not imply Free. Let $R=\zz/6\zz$, $A=2\zz/6\zz$
and $B=3\zz/6\zz$, then $A,B\lhd R$ and $R=A\oplus B$. $A$ being
a direct summand of $R$ is projective, but is not free since it has
fewer elements than $R$\end{example*}
\begin{thm}
\label{thm:5.2.8}Over a commutative local ring, finitely generated
projective modules are free.\end{thm}
\begin{proof}
Let $R$ be a commutative local ring with unique maximal ideal $J$.
Let $M$ be a finitely generated $R$-module. Let $\left\{ a_{1},\dots,a_{n}\right\} $
be a minimal set of generators for $M$. Then there exists a free
module with a free basis $\{x_{1},\dots,x_{n}\}$ and an $R$-homomorphism
$\phi:F\overset{\mathrm{onto}}{\to}M$ such that $\phi(x_{i})=a_{i}$
(See note\vpageref{5.1.3}, Question 1 on Exercise sheet 6 or Commutative
Algebra). Thus we have $\xyC{1.5pc}\xymatrix{0\ar[r] & K\ar[r] & F\ar[r]^{\phi} & M\ar[r] & 0}
$ where $K=\ker(\phi)$. 

Claim: $K\subseteq FJ$. If not there exists an element $k=x_{1}r_{1}+\dots+x_{n}r_{n}$
($r_{i}\in R$) of $F$ such that $k\in K$ but $r_{i}\notin J$ for
some $i$. Say $r_{1}\notin J$. Since $R$ is local, $r_{1}$ must
be a unit. Since $k\in\ker\phi$, $a_{1}r_{1}+\dots+a_{n}r_{n}=0$.
So $a_{1}=-r_{1}^{-1}(a_{2}r_{2}+\dots+a_{n}r_{n})$ contradiction
the fact that $\{a_{1},\dots,a_{n}\}$ was a minimal generating set.
Thus $K\subseteq FJ$. 

Now since $M$ is projective, the above short exact sequence split.
So $F=K\oplus M'$ where $M'\cong M$. Hence $FJ=KJ\oplus M'J$. So
$K=FJ\cap K=K\cap(KJ\oplus M'J)=KJ\oplus(K\cap M'J)$ by the modular
law. But $K\cap M'J\subseteq K\cap M'=0$, so $K=KJ$. Now $K$ is
finitely generated (check this!). By Nakayama's Lemma $K=0$, thus
$M'$ and hence $M$ is free. \end{proof}
\begin{rem*}
Kaplansky has shown that the result is true even without the finitely
generated assumption.
\end{rem*}

\subsection*{The Dual Basis Lemma}

Let $R$ be a commutative integral domain with a field of fraction
$K$. Let $0\neq A\lhd R$ and define $A^{*}=\{k\in K:kA\subseteq R\}$.
Then $A^{*}$ is an $R$-module.
\begin{lem}
Let $R,$$K,A$ be as above. Let $\theta:A\to R$ be an $R$-homomorphism.
Then there exists $q\in A^{*}$ such that $\theta(x)=qx$ for all
$x\in A$. \end{lem}
\begin{proof}
$AK=K$. So a typical element of $K$ is expressible as $ac^{-1}$
with $a,c\in R$, $c\neq0$. Now $\theta$ can be extended to a $K$-homomorphism,
$\theta^{*}:K\to K$ by $\theta^{*}(ac^{-1})=\theta(a)c^{-1}$. Check
that $\theta^{*}$ is well defined and $K$-homomorphism. Let $\theta^{*}(1)=q\in K$.
Then for $x\in A$, $\theta(x)=\theta^{*}(x)=\theta^{*}(1x)=\theta^{*}(1)x=qx$.
Clearly $q\in A^{*}$. \end{proof}
\begin{prop}[The Dual Basis Lemma - Special Case]
 With the notation as above: $A_{R}$ is projective if and only if
$1=x_{1}q_{1}+\dots+x_{n}q_{n}$ for some $x_{i}\in A$, $q_{i}\in A^{*}$.
(Or equivalently $A^{*}A=R$)\end{prop}
\begin{proof}
{}
\begin{elabeling}{00.00.0000}
\item [{$\Rightarrow$)}] Let $F$ be a free module with an $R$-homomorphism
$\phi:F\twoheadrightarrow A$. Since $A$ is projective, there exists
an $R$-homomorphism $\psi:A\to F$ such that $\phi\psi=1_{A}$
\[
\xymatrix{F\ar@<1ex>[r]^{\phi} & A\ar@{-->}@<1ex>[l]^{\psi}}
.
\]
Let $\{f_{\alpha}\}$ be a free basis for $F$. Then for each $y\in A$,
we have $\psi(y)=f_{1}r_{1}+\dots+f_{n}r_{n}$ uniquely for some $f_{i}\in\{f_{\alpha}\}$
and $r_{i}\in R$. So for each $i$, $y\to r_{i}$ is an $R$-homomorphism
$A\to R$. So by the previous lemma, there exists $q_{i}\in A^{*}$
such that $\psi(y)=f_{1}q_{1}y+\dots+f_{n}q_{n}y$. So 
\begin{alignat*}{2}
y & =\phi\psi(y)\\
 & =\phi(f_{1}q_{1}y+\dots+f_{n}q_{n}y)\\
 & =\phi(f_{1})q_{1}y+\dots+\phi(f_{n})q_{n}y &  & \mbox{since }q_{i}y\in R
\end{alignat*}
So $1=\phi(f_{1})q_{1}+\dots+\phi(f_{n})q_{n}=x_{1}q_{1}+\dots+x_{n}q_{n}$,
where $x_{i}=\phi(f_{i})\in A$.
\item [{$\Leftarrow$)}] Define $\psi:A\to\underset{n\mathrm{-times}}{\underbrace{R\oplus\dots\oplus R}}$
by $\psi(x)=(q_{1}x,\dots,q_{n}x)$ for all $x\in A$. 
\[
\xymatrix{A\ar@<1ex>[r]^{\psi} & R\oplus\dots\oplus R\ar@{-->}@<1ex>[l]^{\phi}}
\]
Note that $q_{i}x\in R$ since $q_{i}\in A^{*}$. Define $\phi:\underset{n\mathrm{-times}}{\underbrace{R\oplus\dots\oplus R}}\to A$
by $\phi(r_{1},\dots,r_{n})=x_{1}r_{1}+\dots+x_{n}r_{n}$, $r_{i}\in R$Then
$\phi$ is an $R$-homomorphism and for any $y\in A$ 
\begin{eqnarray*}
\phi\psi(y) & = & \phi(q_{1}y,\dots,q_{n}y)\\
 & = & x_{1}q_{1}y+\dots+x_{n}q_{n}y\\
 & = & y
\end{eqnarray*}
 So $\phi\psi=1_{A}$, hence $A_{R}$ is projective. 
\end{elabeling}
\end{proof}
\begin{prop}
\label{prop:5.2a.3}Let $R$ be a commutative Noetherian integral
domain and $I\lhd R$. Suppose that $IR_{M}$ is a projective $R_{M}$-module
for each maximal ideal $M$ of $R$. Then $I_{R}$ is projective. \end{prop}
\begin{proof}
$I=0$ is trivial so assume $I\neq0$.
\begin{proof}
Let $F$ be the field of fractions of $R$. Then $F$ is also the
field of fractions of each $R_{M}$ (check!). Consider a maximal ideal
$M$. Since $IR_{M}$ is $R_{M}$-projective by the Dual Basis Lemma,
there exists some $x_{i}'\in IR_{M}$ and $q_{i}\in F$ such that
$1=x_{1}'q_{1}+\dots+x_{n}'q_{n}$ and $q_{i}I\subseteq R_{M}$. Now
$q_{i}I$ is a finitely generated $R$-module. So $q_{i}I=z_{1}R+\dots+z_{k}R$
with $z_{i}\in R_{M}$. Let $a\in R$ be a common denominator of the
$x_{i}'$, let $b\in R$ be a common denominator of the $z_{j}$.
Let $d=ab$, then $d\in\mathscr{C}(M)$, $d=x_{1}(q_{1}b)+\dots+x_{n}(q_{n}b)$
where $x_{i}=x_{i}'a\in I$ and $q_{i}bI\subseteq R\,(\dagger)$.

Now $I^{*}I\lhd R$, by $(\dagger)$ $I^{*}I\cap\mathscr{C}(M)\neq\emptyset$.
This is true for all maximal ideal $M$. Hence $I^{*}I=R$. Thus $1\in I^{*}I$
and so $I_{R}$ is projective by the dual basis lemma.
\end{proof}
\end{proof}
\begin{rem*}
This is a special case of a standard result. If $A$ is a finitely
generated module over a commutative Noetherian ring $R$ then $A_{R}$
is projective if and only if $A_{M}$ is a projective $R_{M}$-module
for all maximal ideal $M$. See:
\begin{itemize}
\item Marsumura: Commutative ring Theory Theorem 7.12
\item Rotman: Intro to homological algebra Exercise 9.22 p258
\end{itemize}
\end{rem*}

\subsection{Projective Resolutions and Projective Dimension}
\begin{defn}
If $A$ is a right $R$-module, and exact sequence \foreignlanguage{english}{
\[
\xyC{1.5pc}\xymatrix{\dots\ar[r] & P_{n}\ar[r]^{\partial_{n}} & P_{n-1}\ar[r]^{\partial_{n-1}} & \dots\ar[r]^{\partial_{1}} & P_{0}\ar[r]^{\epsilon} & A\ar[r] & 0}
\]
 where each $P_{i}$ is projective is called a \emph{projective resolution
}for $A$. (This sequence may be finite or infinite)}
\end{defn}
\selectlanguage{english}%
\textbf{Construction of a Projective Resolution}

Let $A$ be a right $R$-module. A is a homomorphic image of a free
module, say $F_{0}$ (by Proposition \ref{prop:5.1.3}). So we have
the exact sequence $\xyC{1.5pc}\xymatrix{0\ar[r] & K_{0}\ar[r]^{i} & F_{0}\ar[r]^{\alpha} & A\ar[r] & 0}
$, where $\alpha$ is the homomorphism $F_{0}\twoheadrightarrow A$
and $K_{0}=\ker\alpha$ and $i=$inclusion map. If $K_{0}$ is projective
the above is a projective resolution.

Even if $K_{0}$ is not projective it is still a homomorphic image
of a free module, say $F_{1}$. So we have the exact sequence $\xyC{1.5pc}\xymatrix{0\ar[r] & K_{1}\ar[r] & F_{1}\ar[r]^{\beta} & K_{0}\ar[r] & 0}
$ where $K_{1}=\ker\beta$. Let $i\beta=\gamma$. Thus $\gamma$ maps
$F_{1}\to F_{0}$ and we have $\ker\alpha=K_{0}=\im\beta=\im\gamma$.
So we have the exact sequence 
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & K_{1}\ar[r] & F_{1}\ar[r] & F_{0}\ar[r] & A\ar[r] & 0}
\]
Here $F_{1}$ and $F_{0}$ are free and hence projective. If $K_{1}$
is not projective the procedure can be repeated. It may happen that
after a finite number of steps we get an exact sequence 
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & K_{n}\ar[r] & F_{n}\ar[r] & F_{n-1}\ar[r] & \dots\ar[r] & F_{1}\ar[r] & F_{0}\ar[r] & A\ar[r] & 0}
\]
where the $K_{n}$ are projective and all the $F_{i}$ are free.
\begin{defn}
A right $R$-module $A$ is said to have \emph{finite projective dimension}
if there exists an exact sequence
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & P_{k}\ar[r] & P_{k-1}\ar[r] & \dots\ar[r] & P_{1}\ar[r] & P_{0}\ar[r] & A\ar[r] & 0}
\]
where each $P_{i}$ is projective. $k$ is called the \emph{length}
of this sequence.

Further, we say that $A$ has \emph{projective dimension $n$ }if
$n$ is the least integer for which there exists a projective resolution
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & P_{n}\ar[r] & P_{n-1}\ar[r] & \dots\ar[r] & P_{1}\ar[r] & P_{0}\ar[r] & A\ar[r] & 0}
\]
We denote the projective dimension of $A$ by $\pd_{R}(A)$ (or simply
$\pd(A)$) If $A$ does not have finite projective dimension we write
$\pd A=\infty$. If $A=0$ we take $\pd A=-1$ conventionally.
\end{defn}
It is clear that $\pd A=0$ if and only if $A$ is projective.
\begin{namedthm}[\selectlanguage{english}%
Schanuel's Lemma\selectlanguage{british}%
]
 Let $M$ be a right $R$-module and let 
\[
\xymatrix{0\ar[r] & K\ar[r]^{\overline{f}} & A\ar[r]^{f} & M\ar[r] & 0 & 0\ar[r] & K'\ar[r]^{\overline{g}} & Y\ar[r]^{g} & M\ar[r] & 0}
\]
be two short exact sequence. If $X$ and $Y$ are projective then
$X\oplus K'\cong Y\oplus K$.\end{namedthm}
\begin{proof}
Define $L=\{(x,y)|x\in X,y\in Y\mbox{ such that }f(x)=g(y)\}$. Then
$L$ is a submodule of $X\oplus Y$. \foreignlanguage{british}{
\[
\xymatrix{ & X\ar[d]^{f}\ar@{-->}[ld]_{\alpha}\\
Y\ar[r]^{g} & M\ar[r] & 0
}
\]
Since $X$ is projective there exists an $R$ homomorphism $\alpha:X\to Y$
such that $f=g\alpha$. Define $\theta:X\oplus K'\to X\oplus Y$ by
$\theta(x,k')=(x,\alpha(x)+\overline{g}(k')$ with $x\in X,k'\in K'$.
$\theta$ is clearly an $R$-homomorphism, also $g(\alpha(x)+\overline{g}(k))=g\alpha(x)+g\overline{g}(k')=f(x)+0$.
Thus $\theta$ is an $R$-homomorphism $X\oplus K'\to L$. Now $\theta(x,k')=0\Rightarrow x=0$
and $\overline{g}(k')=0\Rightarrow x=0$ and $k'=0$. Thus $\theta$
is a monomorphism. }

\selectlanguage{british}%
Finally if $(x,y)\in L$ then $f(x)=g(y)$, so $g\alpha(x)=g(y)$.
So $g[-\alpha(x)+y]=0$. Hence $-\alpha(x)+y\in\ker g=\im(\overline{g})=\overline{g}(K')$.
Hence there exists $k_{1}'\in K'$ such that $g(k_{1}')=-\alpha(x)+y$.
Thus $\theta(x,k')=(x,y)$ and $\theta$ is an epimorphism.

So we have $X\oplus K'\cong L$ and $Y\oplus K\cong L$ and we are
done. \end{proof}
\selectlanguage{british}%
\begin{cor}
In the above situation $K$ is projective if and only if $K'$ is
projective.\end{cor}
\begin{rem*}
For free modules the result corresponding to Schanuel's Lemma does
not work.\end{rem*}
\begin{namedthm}[Generalised Schanuel's Lemma]
 Suppose that $A$ is a right $R$-module and we have two exact sequences
of $R$-modules\foreignlanguage{english}{
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & K_{n}\ar[r] & P_{n}\ar[r] & P_{n-1}\ar[r] & \dots\ar[r] & P_{1}\ar[r] & P_{0}\ar[r] & A\ar[r] & 0}
\]
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & K_{n}'\ar[r] & P_{n}'\ar[r] & P_{n-1}'\ar[r] & \dots\ar[r] & P_{1}'\ar[r] & P_{0}'\ar[r] & A'\ar[r] & 0}
\]
with $P_{j},P_{j}'$ projective for $j=1,2,\dots,n$. Then $K_{n}\oplus P_{n}'\oplus P_{n-1}\oplus\dots\oplus\begin{cases}
P_{0} & n\mbox{ odd}\\
P_{0}' & n\mbox{ even}
\end{cases}\cong K_{n}'\oplus P_{n}\oplus P_{n-1}'\oplus\dots\oplus\begin{cases}
P_{0}' & n\mbox{ odd}\\
P_{0} & n\mbox{ even}
\end{cases}$.}\end{namedthm}
\begin{proof}
By induction on $n$. If $n=0$ this is just Schanuel's lemma. 

So assume the result for $n=j-1$, i.e., $K_{j-1}\oplus P_{j-1}'\oplus\dots\cong K_{j-1}'\oplus P_{j-1}\oplus\dots$
where $K_{t}=\ker$ of map $P_{t}\to P_{t-1}$ and $K_{t}'=\ker$
of map $P_{t}'\to P_{t-1}'$. So we have the exact sequences 
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & K_{j}\ar[r] & P_{j}\ar[r] & K_{j-1}\ar[r] & 0}
\]
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & K_{j}'\ar[r] & P_{j}'\ar[r] & K_{j-1}'\ar[r] & 0}
\]
we obtain
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & K_{j}\ar[r] & P_{j}\oplus P_{j-1}'\oplus P_{j-2}\oplus\dots\ar[r] & K_{j-1}\oplus P_{j-1}'\oplus P_{j-2}\oplus\dots\ar[r] & 0}
\]
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & K_{j}'\ar[r] & P_{j}'\oplus P_{j-1}\oplus P_{j-2}'\oplus\dots\ar[r] & K_{j-1}'\oplus P_{j-1}\oplus P_{j-2}'\oplus\dots\ar[r] & 0}
\]
In both these sequences the middle terms are projective and the right
hand side terms are isomorphic by induction assumption. So by Schanuel's
lemma \foreignlanguage{english}{$K_{j}\oplus P_{j}'\oplus P_{j-1}\oplus\dots\cong K_{j}'\oplus P_{j}\oplus P_{j-1}'\oplus\dots$.
This completes the proof.}\end{proof}
\selectlanguage{english}%
\begin{cor}
\label{5.3.8}With the above notation we have $K_{n}$ projective
if and only if $K_{n}'$ is projective.
\end{cor}
\selectlanguage{british}%

\begin{cor}
If $\pd A_{R}=m$ and \foreignlanguage{english}{
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & K\ar[r] & P_{m}\ar[r] & P_{m-1}\ar[r] & \dots\ar[r] & P_{1}\ar[r] & P_{0}\ar[r] & A\ar[r] & 0}
\]
is an exact short sequence with $P_{j}$'s projective. Then $K$ is
projective.}\end{cor}
\begin{example*}
A module with infinite projective dimension. 

Consider $\zz/2\zz$ as a module over the ring $\zz/4\zz$ defined
by $[x+2\zz][a+4\zz]=[xa+2\zz]$, $x,a\in\zz$. Look at 
\[
\xyC{0.75pc}\xyR{0.75pc}\xymatrix{\dots\ar[rr] &  & \zz/4\zz\ar[rr]^{d_{2}} &  & \zz/4\zz\ar[rr]^{d_{1}}\ar[dr] &  & \zz/4\zz\ar[rr]^{\epsilon} &  & \zz/2\zz\ar[rr] &  & 0\\
 &  &  & 2\zz/4\zz\ar[ru] &  & 2\zz/4\zz\ar[dr]\ar[ru]\\
 &  & 0\ar[ur] &  & 0\ar[ru] &  & 0
}
\]
where $\epsilon:[a+4\zz]\to[a+2\zz]$ and $d_{i}:[a+4\zz]\to[2a+4\zz]$
for all $i$. The kernel at each stage is $2\zz/4\zz$ and thus cannot
be projective (why?).\end{example*}
\begin{prop}
\label{prop:5.3.11}Let $\left\{ A_{\lambda}\right\} _{\lambda\in\Lambda}$
be a family of right $R$-modules. Then $\pd\left(\sum_{\lambda\in\Lambda}\oplus A_{\lambda}\right)=\sup_{\lambda\in\Lambda}\pd A_{\lambda}$\end{prop}
\begin{proof}
We shall do this for the direct sum of two modules, the general case
just involves more notation. Let 
\[
\xyC{1.5pc}\xymatrix{\dots\ar[r] & P_{n}\ar[r]^{\alpha_{n}} & P_{n-1}\ar[r]^{\alpha_{n-1}} & \dots\ar[r]^{\alpha_{2}} & P_{1}\ar[r]^{\alpha_{1}} & P_{0}\ar[r]^{\alpha_{0}} & A\ar[r] & 0}
\]
\[
\xyC{1.5pc}\xymatrix{\dots\ar[r] & Q_{n}\ar[r]^{\beta_{n}} & Q_{n-1}\ar[r]^{\beta_{n-1}} & \dots\ar[r]^{\beta_{2}} & Q_{1}\ar[r]^{\beta_{1}} & Q_{0}\ar[r]^{\beta_{0}} & B\ar[r] & 0}
\]
be projective resolution for $A$ and $B$. Consider
\[
\xyC{1.5pc}\xymatrix{\dots\ar[r] & P_{n}\oplus Q_{n}\ar[r]^{\theta_{n}} & P_{n-1}\oplus Q_{n-1}\ar[r] & \dots\ar[r] & P_{1}\oplus Q_{1}\ar[r]^{\theta_{1}} & P_{0}\oplus Q_{0}\ar[r]^{\theta_{0}} & A\oplus B\ar[r] & 0}
\]
where $\theta_{n}(p_{n},q_{n})=(\alpha_{n}p_{n},\beta_{n}q_{n})$,
$p_{n}\in P_{n},q_{n}\in Q_{n}$. This is an exact sequence and each
$P_{i}\oplus Q_{i}$ is projective. It follows $\pd(A\oplus B)\leq\sup(\pd A,\pd B)$

Suppose that $\pd(A\oplus B)=m<\infty$. Consider 
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & T_{m}\ar[r] & P_{m-1}\oplus Q_{m-1}\ar[r]^{\theta_{m-1}} & \dots\ar[r] & P_{0}\oplus Q_{0}\ar[r]^{\theta_{0}} & A\oplus B\ar & 0}
\]
where $\theta_{1}$ are the maps defined above, since $\pd(A\oplus B)\cong m$.
But $T_{m}=\ker\theta_{m-1}\cong\ker\alpha_{m-1}\oplus\ker\beta_{m-1}$.
This implies $\pd A\leq\pd(A\oplus B)$ and $\pd(B)\leq\pd(A\oplus B)$. 

The above argument shows that if either $\pd A$ or $\pd B=\infty$
then $\pd(A\oplus B)=\infty$ and conversely. This completes the proof.\end{proof}
\begin{lem}
\label{lem:5.3.12} Suppose that 
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & K\ar[r] & P\ar[r] & A\ar[r] & 0}
\]
 is an exact sequence with $P$ projective and $A$ \uline{not}
projective. Then $\pd K<\infty$ if and only if $\pd A<\infty$ and
we have in this case $1+\pd K=\pd A$.\end{lem}
\begin{proof}
Follows from definition of projective dimension and generalised Schanuel's
Lemma. 
\end{proof}
Recall how build our projective resolution for $M_{k}$
\[
\xyC{1pc}\xyR{1pc}\xymatrix{ &  &  &  &  &  & 0\ar[rd] &  & 0\\
 &  &  &  &  &  &  & K_{1}\ar[ru]\ar[rd]\\
0\ar[rr] &  & P_{n}\ar[rr] &  & \dots\ar[rr] &  & P_{2}\ar[ru]\ar[rr] &  & P_{1}\ar[rd]\ar[rr] &  & P_{0}\ar[rr] &  & M\ar[rr] &  & 0\\
 &  &  &  &  & K_{2}\ar[ru] &  &  &  & K_{0}\ar[rd]\ar[ru]\\
 &  &  &  & 0\ar[ru] &  &  &  & 0\ar[ru] &  & 0
}
\]

\begin{thm}
\label{thm:5.3.13}Let $0\to B\to A\to C\to0$ be a short exact sequence.
If the projective dimension of any two module is the short exact sequence
is finite then so is the third. Furthermore we have
\begin{enumerate}
\item if $\pd A>\pd B$ then $\pd C=\pd A$
\item if $\pd A<\pd B$ then $\pd C=\pd B+1$
\item if $\pd A=\pd B$ then $\pd C\leq\pd A+1$. 
\end{enumerate}
\end{thm}
\begin{proof}
To prove the first part we induct on $n$ the sum of the finite projective
dimension. If $n=0$ then both modules must be projective. If one
of these is $C$ then the short exact sequence splits. So by Lemma
\ref{lem:5.2.5} if one of $A$ or $B$ is projective then so is the
other. On the other hand if $A$ and $B$ are projective then $\pd C\leq1$.

Now suppose that $n>0$ and the result is true when the sum of the
two projective dimension is $<n$. We may also assume that neither
$A$ nor $C$ is projective. Now there exists a projective $P$ such
that $0\to D\to P\to A\to0$ is exact $(*)$. So $A\cong P/D$. Hence
there exists a submodule $E$ with $P\supseteq E\supseteq D$ such
that $B\cong E/D$, moreover $C\cong A/B\cong(P/D)/(E/D)\cong P/D$
(by the third isomorphism theorem). Thus we have short exact sequences
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & E\ar[r] & P\ar[r] & C\ar[r] & 0}
\qquad(\dagger)
\]
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & D\ar[r] & E\ar[r] & B\ar[r] & 0}
\qquad(\ddagger)
\]
Now $(*)$ and $(\dagger)$ give $\pd D=\pd A-1$ if $\pd A<\infty$
and $\pd E=\pd C-1$ if $\pd C<\infty$ (by the previous lemma). So
by induction hypothesis $(\ddagger)$ gives that if any two of $D,E,B$
have finite projective dimension then so does the third. Hence the
same is true for $A,B$ and $C$. 

Now assume that all the projective dimension are finite. We prove
the second part by induction on the sum of all three projective dimension.
If $n=0$, nothing to prove (see the base case of the first part of
the proof)

Let $n>0$. If either $A$ or $C$ is projective, then the result
holds. So assume that neither is projective. Induction hypothesis
applied to $(\ddagger)$ gives:
\begin{enumerate}[label=\roman{enumi}]
\item If $\pd E>\pd D$ then $\pd B=\pd E$
\item if $\pd E<\pd D$ then $\pd B=\pd D+1$
\item if $\pd E=\pd D$ then $\pd B\leq\pd D+1$
\end{enumerate}
In terms of $A,B$ and $C$ these gives
\begin{enumerate}[label=\alph{enumi}]
\item If $\pd C>\pd A$ then $\pd B=\pd C-1$
\item If $\pd C<\pd A$ then $\pd B=\pd A$
\item If $\pd C=\pd A$ then $\pd B\leq A$. 
\end{enumerate}
It can be seen (check!) that  a. b. and c. are logically equivalent
to 1. 2. and 3. of the theorem.\end{proof}
\begin{thm}[Auslander]
\label{thm:5.3.14} Let $M$ be a right $R$-module, $I$ a non-empty
well-ordered set and $\left\{ M_{i}\right\} _{i\in I}$ a family of
submodules such that:
\begin{enumerate}
\item $M_{i}\subseteq M_{j}$ if $i\leq j$
\item $M=\cup_{i\in I}M_{i}$
\item $\pd(M_{i}/M_{i}')\leq n$ where $M_{i}'=\cup_{j<i}M_{j}$
\end{enumerate}
then $\pd M\leq n$\end{thm}
\begin{proof}
By induction on $n$. If $n=0$ then for all $i\in I$, $\pd(M_{i}/M_{i}')\leq0$
so $M_{i}/M_{i}'$ is projective. So each short exact sequence $0\to M_{i}'\to M_{i}\to M_{i}/M_{i}'\to0$
splits. So there exists submodules $C_{i}$ of $M_{i}$ such that
$M_{i}=M_{i}'\oplus C_{i}$ where $C_{i}\cong M_{i}/M_{i}'$. So each
$C_{i}$ is projective. 

We claim that $M=\sum_{i\in I}\oplus C$. The sum is direct for suppose
$c_{i_{1}}+c_{i_{2}}+\dots+c_{i_{m}}=0$ where $c_{i_{j}}\in C_{i_{j}}$
and $i_{1}<i_{2}<\dots<i_{m}$, then $-c_{i_{m}}=c_{i_{1}}+\dots+c_{i_{m-1}}\in M_{i_{m}}'\cap C_{m}=0$.
So $c_{i_{m}}=0$ and similarly $c_{i_{1}}=c_{i_{2}}=\dots=c_{i_{m-1}}=0$.
Suppose now that $M\neq\sum_{i\in I}\oplus C_{i}$, so there exists
$i\in I$ such that $M_{i}\nsubseteq\sum_{i\in I}C_{i}$. Suppose
that $j$ is the least index such that $M_{j}\nsubseteq\sum_{i\in I}\oplus c_{i}$.
So there exists $m\in M_{j}$ such that $m\notin\sum_{i\in I}\oplus C_{i}$.
Now $M_{j}=M_{j}'\oplus C_{j}$, so $m=b+c$ for some $b\in M_{j}'$,
$c\in C_{j}$. But $b\in\sum_{i\in I}\oplus C_{i}$ by the minimality
of $j$ ($b\in M_{k}$some $k<j$). So $m\in\sum_{i\in I}\oplus C_{i}$
a contradiction. Thus $M=\sum_{i\in I}\oplus C_{i}$ as required.
Hence $\pd M\leq0$ since $M$ is a direct sum of projective modules.

Now assume the result for $n-1$. We are given that $\pd(M_{i}/M_{i}')\leq n$
for all $i\in I$. Let $F$ ($=F_{M}$) be the free module with free
basis $M$, let $F_{i}$ be the free module with free basis $M_{i}$
and let $F_{i}'$ be the free module with free basis $M_{i}'$. We
have $F\supseteq F_{i}\supseteq F_{i}'$ so we have the short exact
sequence $0\to K\to F\to M\to0$. Define $K_{i}=F_{i}\cap K$ and
$K_{i}'=F_{i}'\cap K$. From the relations $M_{i}\supseteq M_{i}'$,
$F_{i}\supseteq F_{i}'$ and the short exact sequences $0\to K_{i}\to F_{i}\to M_{i}\to0$,
it follows that the sequences 
\[
\xymatrix{0\ar[r] & K_{i}/K_{i}'\ar[r] & F_{i}/F_{i}'\ar[r] & M_{i}/M_{i}'\ar[r] & 0}
\]
are exact. {[}Note that $(K_{i}+F_{i})/F_{i}'\cong K_{i}/(K_{i}\cap F_{i}')$
by the third isomorphism theorem. But this is $K_{i}/(K_{i}\cap F_{i}\cap F_{i}')=K_{i}/K_{i}'$.
{]} Each $F_{i}/F_{i}'$ is free since $F_{i}$ has a set of generators,
a subset of which generates $F_{i}'$. Hence $F_{i}/F_{i}'$ is projective
so by Lemma \ref{lem:5.3.12} $\pd K_{i}/K_{i}'\leq n-1$. It can
be checked that:
\begin{enumerate}[label=\roman{enumi}]
\item $i<j$, $i,j\in I$ implies $K_{i}\subseteq K_{j}$
\item $K=\cup_{i\in I}K_{i}$ and $K_{i}'=\cup_{j<i}K_{j}$.
\end{enumerate}
So by Lemma \ref{lem:5.3.12}, we have $\pd M\leq1+\pd K\leq n$.
This completes our proof.\end{proof}
\begin{defn}
Let $R$ be a ring. We define $D(R)=\sup_{\{M\}}\pd M$ where $M$
ranges over all right modules of $R$. $D(R)$ is called the \emph{right
global dimension of $R$.}\end{defn}
\begin{lem}
Let $M$ be a cyclic module over a ring $R$. Then $M\cong R/I$ where
$I$ is a right ideal of $R$.\end{lem}
\begin{proof}
Exercise sheet 2. Q4 i)\end{proof}
\begin{thm}
\label{thm:5.3.17}Let $R$ be a ring. We have
\begin{enumerate}
\item $D(R)=\sup_{\{B\}}\pd B$ where $B$ runs over all cyclic right $R$-modules
\item $D(R)=\sup_{\{I\}}\pd R/I$ where $I$ runs over all right ideals
of $R$
\item Further if $D(R)\neq0$ then $D(R)=1+\sup_{\{I\}}\pd I$ where $I$
runs over all right ideals of $R$.
\end{enumerate}
\end{thm}
\begin{proof}
The equivalence of 1 and 2 follows from the previous lemma. The equivalence
of 2 and 3 is clear from Lemma \ref{lem:5.3.12} using the short exact
sequence $0\to I\to R\to R/I\to0$. So we prove 1.

Let $M$ be a right $R$-module. Well order the elements $x_{i}$
of $M$ ($i\in I$) and denote by $M_{i}$ {[}respectively by $M_{i}'${]}
the submodule of $M$ generated by all $x_{j}$, $j\leq i$ {[}respectively
$j<i${]}. Then $M_{i}/M_{i}'$ is either $0$ or generated by a single
element $x_{i}$. So $\pd(M_{i}/M_{i}')\leq n$ where $n=\sup_{\{B\}}\pd B$
where $B$ ranges over all cyclic right $R$-modules. Since the family
$\{M_{i}\}_{i\in I}$ satisfies the hypothesis of Theorem \ref{thm:5.3.14},
we have $\pd M\leq n$, hence $D(R)\leq n$. But by definition $D(R)\geq n$,
hence $D(R)=n=\sup_{\{B\}}\pd B$.\end{proof}
\begin{rem*}
Auslander has shown that for a (left and right) Noetherian ring $R$,
left global dimension of $R$ is the same as the right global dimension
of $R$
\end{rem*}

\subsection{Localization and Global Dimension}

All rings are commutative in this section.

$S$ multiplicative subset of $R$, $0\notin S$, $1\in S$. Let $M,K$
be $R$-modules and $\phi:M\to K$ and $R$-homomorphism. Then we
can define a corresponding $R_{S}$-homomorphism $\phi^{*}:M_{S}\to K_{S}$
by $\phi^{*}\left(\frac{m}{s}\right)=\frac{\phi(m)}{s}$ with $m\in M,s\in S$.
(Check details, c.f. Commutative Algebra). If $\phi$ is an epimorphism,
so is $\phi^{*}$.
\begin{lem}
\label{lem:5.4.1}If $\xyC{1.5pc}\xymatrix{0\ar[r] & A\ar[r]^{\theta} & B\ar[r]^{\phi} & C\ar[r] & 0}
$ is an exact sequence of $R$-modules then\\
 $\xyC{1.5pc}\xymatrix{0\ar[r] & A_{S}\ar[r]^{\theta^{*}} & B_{S}\ar[r]^{\phi^{*}} & C_{S}\ar[r] & 0}
$ is an exact sequence of $R^{*}$-modules.\end{lem}
\begin{proof}
See Commutative Algebra 3.3\end{proof}
\begin{lem}
\label{lem:5.4.2}If $P$ is a projective $R$-module, then $P_{S}$
is a projective $R_{S}$-module.\end{lem}
\begin{proof}
Routine from first principle\end{proof}
\begin{lem}
\label{lem:5.4.3}$D(R_{S})\leq D(R)$\end{lem}
\begin{proof}
If $D(R)=\infty$ there is nothing to prove.

So assume$D(R)<\infty$. Let $A$ be an $R_{S}$-module. View $A$
as an $R$-module. Since $A_{S}\cong A$ (see section 4.4) using Lemma
\ref{lem:5.4.1} and \ref{lem:5.4.2} we get $\pd_{R_{S}}A\leq\pd_{R}A$.
It follows that $D(R_{S})\leq D(R)$\end{proof}
\begin{example*}
$D(\zz)=1$, $D(\zz/4\zz)=\infty$. $D(\zz_{(2)})=1$, $D(\zz_{(2)}/4\zz_{(2)})=\infty$
\end{example*}
\newpage{}


\section{Global Dimension of Regular Local Rings}


\subsection{Change of Rings Theorems}
\begin{thm}
\label{thm:6.1.1}Let $R$ be a commutative ring and suppose that
$x$ is a regular element of $R$. Denote the ring $R/xR$ by $R^{*}$.
Let $M$ be a non-zero $R^{*}$-module with $\pd_{R^{*}}M=n<\infty$.
Then $\pd_{R}M=n+1$\end{thm}
\begin{proof}
By induction on $n$.

Suppose that $n=0$, i.e., $M$ is $R^{*}$-projective, so there exists
a free module $F$ such that $F=M\oplus M'$ (for some submodule $M'$
of $F$). Now $0\to xR\to R\to R^{*}\to0$ is exact as $R$-modules.
$xR\cong R_{R}$, so $xR$ is $R$-projective. Hence $\pd_{R}(R^{*})\leq1$.
By Proposition \ref{prop:5.3.11}, it follows that 
\[
\pd_{R}F\leq1\,(*)
\]
So $\pd_{R}M\leq1$. Now $x$ does not annihilate any non-zero elements
of $R$. So $x$ does not annihilate any non-zero elements of a free
$R$-module and hence of a projective $R$-module. But $Mx=0$, so
it follows that $M_{R}$ cannot be projective. Thus $\pd M=1$.

So now let $n>0$ and assume the result for integers less than $n$.
Now there exists a free $R^{*}$-module $G$ such that $0\to K\to G\to M\to0$
is exact. Since $M$ is not $R^{*}$-projective, $\pd_{R^{*}}(K)=n-1$.
Hence $\pd_{R}(K)=n$ by induction hypothesis. Also $\pd_{R}(G)\leq1$
as in $(*)$. So by Theorem\ref{thm:5.3.13}$\pd_{R}M=n+1$ if $n\neq1$,
and $\pd_{R}M\leq2$ if $n=1$. 

In the first case we are done, so now we deal with the case $n=1$
and we must rule out the possibility that $\pd_{R}M\leq1$ when $\pd_{R^{*}}M=1$.
So assume that $\pd_{R}M\leq1$ and $\pd_{R^{*}}M=1$. So there exists
a free $R$-module $H$ such that 
\[
0\to T\to H\to M\to0\,(**)
\]
 is exact. So $T$ is projective since $\pd_{R}M\leq1$. Also $Hx\subseteq T$
since $Mx=0$. Therefore $(**)$ induces the exact sequence 
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & T/Hx\ar[r] & H/Hx\ar[r] & M\ar[r] & 0}
\]
Now $H/Hx$ is $R^{*}$-free (check!) and $\pd_{R^{*}}M=1$. Thus
$T/Hx$ is $R^{*}$-projective. But by the third isomorphism theorem
$\frac{T/Tx}{Hx/Tx}\cong T/Hx$ as $R^{*}$-modules. Hence $Hx/Tx$
is a direct summand of $T/Tx$. Since $T$ is $R$-projective, $T/Tx$
is $R^{*}$-projective. {[}If $\underset{R\mathrm{-free}}{F}=T\oplus K$
then $\underset{R^{*}\mathrm{-free}}{F/Fx}=T/Tx\oplus K/Kx${]}. Hence
$Hx/Tx$ is $R^{*}$-projective. But $Hx/Tx\cong H/T$ since $x$
is regular. But $H/T\cong M$, so $M$ is $R$-projective, contradiction.
So we have proved that $\pd_{R^{*}}M=1$ implies $\pd_{R}M=2$\end{proof}
\begin{cor}
\label{cor:6.1.2}In the above situation if $D(R^{*})=n<\infty$,
then $D(R)\geq n+1$\end{cor}
\begin{thm}
\label{thm:6.1.3}Let $R$ be a commutative ring. Let $M$ be a right
$R$-module. Suppose that $x$ is a regular element of $R$ such that
$x$ annihilates no non-zero elements of $M$. Write $R^{*}=R/xR$.
Then $\pd_{R^{*}}\left(M/Mx\right)\leq\pd_{R}M$. \end{thm}
\begin{proof}
If $\pd M_{R}=\infty$ then nothing to prove. So assume $\pd_{R}M=n<\infty$.
We prove the result by induction on $n$.

Suppose $n=0$. If $F$ is $R$-free then $F/Fx$ is $R^{*}$-free.
Hence if $M$ is a direct summand of an $R$-free module, then $M/Mx$
is a direct summand of $R^{*}$-free module. (This argument was used
before). Thus $M/Mx$ is $R^{*}$-projective, as required.

Now suppose that $n>0$ and the result holds for integers smaller
than $n$. There exists a $R$-module $F$ such that 
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & K\ar[r] & F\ar[r] & M\ar[r] & 0}
\quad(*)
\]
is exact, so $\pd_{R}(K)=n-1$. Hence $\pd_{R^{*}}(K/Kx)\leq n-1$
by induction hypothesis. From $(*)$ we get the exact sequence: 
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & \frac{K+Fx}{Fx}\ar[r] & F/Fx\ar[r] & M/Mx\ar[r] & 0}
\]
so we have 
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & \frac{F}{K\cap Fx}\ar[r] & F/Fx\ar[r] & M/Mx\ar[r] & 0}
\]
 is exact. We claim $K\cap Fx=Kx$, clearly $Kx\subseteq K\cap Fx$.
Suppose that $k=fx\in K\cap Fx$, where $k\in K$, $f\in F$. But
$x$ is not a zero divisor on $F/K\cong M$. Thus we have the exact
sequence of $R^{*}$-modules
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & K/Kx\ar[r] & F/Fx\ar[r] & M/Mx\ar[r] & 0}
\]
Since $\pd_{R^{*}}(K/Kx)\leq n-1$, it follows that $\pd_{R^{*}}(M/Mx)\leq n$.
This completes the proof
\end{proof}
We get equality if $R$ is Noetherian and $x$ lies in the Jacobson
Radical of $R$. 
\begin{lem}
Let $R$ be a commutative Noetherian ring. Let $M$ be a finitely
generated module and suppose that $x$ is a regular element lying
in $J(R)$. Suppose that $x$ does not annihilate any non-zero elements
of $M$. Write $R^{*}=R/xR$.

Then $M/Mx$ is $R^{*}$-projective implies that $M$ is $R$-projective.\end{lem}
\begin{proof}
First suppose that $M/Mx$ is $R^{*}$-free. Let $v_{1},\dots,v_{n}$
be a free basis of $M/Mx$. Let $u_{1},\dots,u_{n}$ be elements of
$M$ mapping onto $v_{1},\dots,v_{n}$ under the natural homomorphism
$M\to M/Mx$. 

Claim: $M$ is $R$-free with basis $u_{1},\dots,u_{n}$.

Let $C$ be the submodule of $M$ generated by $u_{1},\dots,u_{n}$.
Then clearly, $C+Mx=M$. This gives $[M/C]Rx=[M/C]$, so $M/C=0$
by Nakayama's lemma. Thus $M=C$ and $u_{1},\dots,u_{n}$ generate
$M$.

Suppose that $u_{1},\dots,u_{n}$ is not a free basis for $M$. Then
(after possible renumbering) there exists non-zero $r_{1},\dots,r_{k}\in R$
such that $u_{1}r_{1}+\dots+u_{k}r_{k}=0$, $k\leq n$ $(*)$. Thus
$v_{1}r_{1}+\dots+v_{k}r_{k}\in Mx$. Hence $r_{i}\in xR$ for all
$i$ since $v_{1},\dots,v_{k}$ is part of a free basis of an $R^{*}$-module.
Say $r_{i}=xs_{i}$ for $s_{i}\in R$. We claim $r_{k}R\subsetneq s_{k}R$.
Clearly $r_{k}R\subseteq s_{k}R$ and $r_{k}R=s_{k}R$ would imply
$s_{k}=r_{k}t_{k}$ for some $t_{k}\in R$, i.e., $s_{k}=xs_{k}t_{k}$
and so $s_{k}(1-xt_{k})=0$. Hence $x_{k}=0$ since $1-xt_{k}$ is
a unit since $x\in J(R)$. But is $s_{k}=0$ then $r_{k}=0$ contrary
to our assumption. Now cancelling out $x$, $(*)$ gives $u_{1}s_{1}+\dots+u_{k}s_{k}=0$
with $s_{k}\neq0$ since $r_{k}\neq0$. We can write this symbolically
as $u_{1}\left(\frac{r_{1}}{x}\right)+\dots u_{n}\left(\frac{r_{k}}{x}\right)=0$.
Repeating the above process we get an ascending chain of ideals
\[
r_{k}R\subsetneq\left(\frac{r_{k}}{x}\right)R\subsetneq\left(\frac{r_{k}}{x^{2}}\right)R\subsetneq\dots
\]
This is a contradiction since $R$ is a Noetherian ring. Hence $u_{1},\dots,u_{n}$
is a free basis for $M$ as claimed. So $M$ is $R$-free.

Next suppose that $M/Mx$ is $R^{*}$-projective. Then there exists
a free module $F$ such that 
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & K\ar[r] & F\ar[r] & M\ar[r] & 0}
\]
is exact. As before this induces the exact sequence of $R^{*}$-modules
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & K/Kx\ar[r] & F/Fx\ar[r] & M/Mx\ar[r] & 0}
\qquad(**)
\]
Now write $B=M\oplus K$ $(***)$(external direct sum). Then $Bx=Mx\oplus Kx$.
This gives $B/Bx=M/Mx\oplus K/Kx$. Since $M/Mx$ is $R^{*}$-projective,
$(**)$ splits so $F/Fx\cong M/Mx\oplus K/Kx\cong B/Bx$. Therefore
$B/Bx$ is $R^{*}$-free and by earlier part of the proof $B$ is
$R$-free. Hence from $(***)$ we have that $M$ is $R$-projective. \end{proof}
\begin{thm}
Let $R$ be a commutative Noetherian ring, $M_{R}$ a finitely generated
module. Suppose that $x\in R$ is a regular element such that $x\in J(R)$.
Suppose also that $x$ does not annihilate any non-zero elements of
$M$. Write $R^{*}=R/xR$. Then $\pd_{R^{*}}(M/Mx)=\pd_{R}(M)$\end{thm}
\begin{proof}
Let $\pd_{R^{*}}(M/Mx)=n$.

If $\pd_{R^{*}}(M/Mx)=\infty$ then $\pd_{R}(M)=\infty$ by Theorem
\ref{thm:6.1.3}

So assume that $n<\infty$. We induct on $n$. For $n=0$ the result
is proved by previous Lemma.

Assume that $n>0$ and the result for values smaller than $n$. There
exists a free module $F$ such that the sequence 
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & K\ar[r] & F\ar[r] & M\ar[r] & 0}
\]
is exact. As before this induces the short exact sequence
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & K/Kx\ar[r] & F/Fx\ar[r] & M/Mx\ar[r] & 0}
\qquad(*)
\]
Since $F/Fx$ is $R^{*}$-free we have that $\pd_{R^{*}}(K/Kx)=n-1$.
Since $R$ is Noetherian and $M$ is finitely generated we have $K$
is finitely generated. Clearly $x$ annihilates no non-zero elements
of $K$. Now $\pd_{R}(K)=n-1$ by induction hypothesis. So $(*)$
gives $\pd_{R}M=n$ (unless $\pd_{R}(M)=0$ but in this case $\pd_{R^{*}}(M/Mx)=0$
by Theorem \ref{thm:6.1.3}) This completes the proof.\end{proof}
\begin{cor}
\label{cor:6.1.6}Let $R$ be a commutative Noetherian ring. Let $x\in J(R)$
be regular and let $R^{*}/xR$. If $D(R^{*})=n<\infty$ then $D(R)=n+1$.\end{cor}
\begin{proof}
We have $D(R)\geq n+1$ by Corollary \ref{cor:6.1.2}. Now let $M$
be a finitely generated $R$-module. Let $\pd_{R}M=k$. We shall not
show that $k\leq n+1$. This is clear if $k=0$ , so assume that $M$
is not $R$-projective. So there exists a free $R$-module $F$ such
that 
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & K\ar[r] & F\ar[r] & M\ar[r] & 0}
\]
is exact. We have $\pd_{R}K=k=1$. Since $R$ is Noetherian and $F$
finitely generated, we have $K$ is finitely generated. Also since
$K\subseteq F$, $x$ does not annihilate any non-zero elements of
$K$. So by the previous theorem $\pd_{R}K=\pd_{R^{*}}(K/Kx)\leq n$.
So $\pd_{R}M=1+\pd_{R}K\leq n+1$ . But by Theorem \ref{thm:5.3.17}
$D(R)=\sup_{\{M_{R}\mathrm{\, f.g}\}}\pd M$. Hence $D(R)\leq n+1$.
Thus $D(R)=n+1$.
\end{proof}

\subsection{Regular Local Ring}
\begin{lem}
Let $R$ be a regular local ring of Krull dimension $n$. Then $D(R)=n$.\end{lem}
\begin{proof}
By induction on $n$. Let $J$ be the Jacobson radical of $R$. If
$n=0$ we have $J=0$, i.e., $R$ is a field and the result is true.

Let $n>0$ and assume the result holds for regular local ring of $K\dim\leq n-1$.
Since $n>0$, $J\neq0$ and so $J\neq J^{2}$ by Nakayama's lemma.
Let $x_{1},\dots,x_{n}$ be a minimal generating set for $J$. Then
there exists $x_{i}$ such that $x_{i}\notin J^{2}$. Write $x_{i}=x$.
Since $R$ is an integral domain, $x$ is regular. Let $R^{*}=R/xR$.
By Lemma \ref{lem:4.6.8} $K\dim R^{*}=n-1$. Clearly the images of
$x_{1},x_{2},\dots,x_{i-1},x_{i+1},\dots,x_{n}$ are a minimal generating
set for $J/xR$. Thus $R^{*}$ is a regular local ring, hence $D(R^{*})=n-1$
by induction hypothesis. Therefore $D(R)=n$ by Corollary \ref{cor:6.1.6}.
This completes the proof.\end{proof}
\begin{lem}
\label{lem:6.2.2}Let $R$ be a Noetherian commutative local ring.
Suppose that $\Ann J\neq0$ (where $J=J(R)$). Then $\pd M=0$ or
$\infty$. \end{lem}
\begin{proof}
If $\pd M\neq0$ or $\infty$ then there exists a module $B$ such
that $\pd B=1$. Now consider 
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & K\ar[r] & F\ar[r] & B\ar[r] & 0}
\]
 where $F$ is free and $K\subseteq FJ$ (as in Theorem\ref{thm:5.2.8}).
So $\Ann K\neq0$. But since $\pd B=1$, $K$ is projective and hence
free. This is a contradiction since a free module cannot have a non-zero
annihilator.\end{proof}
\begin{lem}
\label{lem:6.2.3}Let $R$ be a regular local ring with Jacobson radical
$J$. Let $x\in R$ be regular such that $x\in J$ but $x\notin J^{2}$.
Then $J/xR$ is isomorphic to a direct summand of $J/xJ$. \end{lem}
\begin{proof}
Since $x\notin J^{2}$ we can choose a minimal generating set $x,y_{1},\dots,y_{r}$
of $J$. Write $S=xJ+y_{1}R+\dots+y_{r}R$. Then clearly $S+xR=J$.
We claim that $S\cap xR=xJ$, clearly $xJ\subseteq S\cap xR$. Let
$z\in S\cap xR$. Then $z=x_{j}+u_{1}s_{1}+\dots+y_{r}s_{r}=xt$ for
some $h\in J,s_{i}\in R,t\in R$. So $xt-y_{1}s_{1}-\dots-y_{r}s_{r}\in J^{2}$,
since $x,y_{1},\dots,y_{r}$ is a minimal generating set for $J$,
we have $t\in J$, proving the claim.

Hence we have $J/xJ\cong S/xJ\oplus xR/xJ$ (check!). Now $J/xR\cong\frac{J/xJ}{xR/xJ}\cong S/xJ$
which is a direct summand of $J/xJ$.\end{proof}
\begin{prop}
Let $R$ be a Noetherian local ring with Jacobian radical $J$. If
$\pd J=m<\infty$ then $R$ is a regular local ring of Krull dimension
$m+1$\end{prop}
\begin{proof}
If $J=0$ then $R$ is a field, $\pd J=-1$ and $K\dim R=0$, so the
result is true.

We now deal with the case $m=0$. We can assume $J\neq0$. Since $J$
is projective it is free (Theorem \ref{thm:5.2.8}). So $J$ is a
principal ideal generated by a regular element, so by Theorem \ref{thm:4.1.10??},
$\rk J=K\dim R=1$ and the result holds.

We now prove the result by induction on $k$, the Krull dimension
of $R$.

If $k=0$ then $J$ is the unique minimal prime of $R$. Hence $\ann J\neq0$
(see Proposition \ref{prop:4.2.6}). Then by Lemma \ref{lem:6.2.2}
$\pd J=0$ and this is dealt with above (we get $J=0$)

So suppose that $k>0$ and that the result holds for rings of smaller
Krull dimension. Clearly we may also assume $m>0$. We have $0<m<\infty$.
So by \ref{lem:6.2.2} $\ann J=0$. So by Proposition \ref{prop:4.2.8},
$J$ contains a regular element, say $x$. By Proposition \ref{prop:4.2.9},
we may choose $x$ such that $x\notin J^{2}$. Write $R^{*}=R/xR$,
$J^{*}=R/xR$. Since $x$ is regular by Lemma \ref{lem:4.6.8} we
have $K\dim R^{*}=k-1$.

Claim: $\pd_{R^{*}}J^{*}=m-1$. We have $\pd_{R^{*}}(J/xJ)\leq\pd_{R}J$
by Theorem \ref{thm:6.1.3}, but by Lemma \ref{lem:6.2.3} $J^{*}$
is a direct summand of $J/xJ$, so $\pd J^{*}<\infty$. Since $m\geq1$,
applying Theorem \ref{thm:5.3.13} to 
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & xR\ar[r] & J\ar[r] & J^{*}\ar[r] & 0}
\]
we have $\pd_{R}J^{*}=\pd_{R}J=m$, so by Theorem \ref{thm:6.1.1}
$\pd_{R^{*}}J^{r}=m-1$.

So by induction hypothesis $R^{*}$ is a regular local ring of Krull
dimension $ $$m$. Hence $K\dim R=m+1$ and $R$ is regular local.
($J^{*}$ is generating by $m$ elements so $J$ is generated by $m+1$
elements. But $\rk J=m+1$)
\end{proof}
Collecting these results together we have
\begin{thm}[Serre]
 Let $R$ be a commutative Noetherian local ring. Then $R$ is regular
local ring of Krull dimension of $n$ if and only if $D(R)=n$.\end{thm}
\begin{cor}
If $P$ is a prime ideal of a regular local ring $R$ then the ring
$R_{P}$ is also regular local\end{cor}
\begin{proof}
$R_{P}$ is a Noetherian local ring, by the previous theorem $D(R)<\infty$.
Hence $D(R_{P})<\infty$ by Lemma \ref{lem:5.4.3}. $R$ is regular
local by the previous Theorem
\end{proof}
In fact, if $S$ is a multiplicatively closed subset of $R$ and $D(R)<\infty$
then $D(R_{S})\leq D(R)<\infty$.

\newpage{}


\section{Unique Factorization}

All rings are commutative with $1$


\subsection{Unique Factorization Domain}
\begin{defn}
An element $0\neq p\in R$ is said to be a \emph{prime} element if
$pR$ is a prime ideal.\end{defn}
\begin{note*}
If $p$ is a prime element, then so is $up$ where $u$ is a unit.\end{note*}
\begin{defn}
The ring $R$ is called a \emph{unique factorisation domain }(UFD)
if $R$ is an integral domain and every non-zero element $a\in R$
is expressible as $a=up_{1}\dots p_{n}$ where $u$ is a unit and
the $p_{i}$ are prime elements.\end{defn}
\begin{prop}
If an element of an integral domain is expressible as $p_{1}\dots p_{n}$
where the $p_{i}$ are primes, then this expression is unique up to
a permutation of the $p_{i}$'s and multiplication by a unit.\end{prop}
\begin{proof}
Algebra II course. (Or Hartley and Hawkes: Rings, Modules and Linear
Algebra; Theorem 4.10)\end{proof}
\begin{defn}
Let $R$ be an integral domain and $a,b\in R$. We say that $a$ \emph{divides}
$b$ and write $a|b$ if there exists $c\in R$ such that $b=ac$. \end{defn}
\begin{prop}
\label{prop:7.1.5}Let $R$ be a commutative Noetherian integral domain.
Then $R$ is a UFD if and only if every rank $1$ prime ideal of $R$
is principal.\end{prop}
\begin{proof}
{}
\begin{elabeling}{00.00.0000}
\item [{$\Rightarrow$:}] Let $P$ be a rank $1$ prime ideal of $R$.
Let $a\in P$. Then $a$ must be a non-unit, so $a=up_{1}\dots p_{n}$
where $u$ is a unit and the $p_{j}$ are primes. Hence $p_{i}\in P$
for some $i$ and so $P=p_{i}R$ since $P$ is a rank $1$ prime ideal
and $p_{i}R$ is a non-zero prime ideal.
\item [{$\Leftarrow$:}] Let $S$ be the set of all elements of $R$ which
are expressible in the form $up_{1}\dots p_{n}$ with $u$ a unit
and each $p_{i}$ is prime.


We shall first show that if $a\notin S$ then $aR\cap S=\emptyset$.
Suppose not. Let $b\in R$ such that $ab=up_{1}\dots p_{n}$ and $n$
is the least possible, where $u$ is a unit and the $p_{j}$ are primes.
(Note: $ab$ cannot be a unit since $a$ is not a unit). Now $p_{i}\nmid b$
for any $i$ since if $p_{i}|b\Rightarrow b=p_{i}t_{i}$ for some
$t_{i}\in R$. Hence $at_{i}p_{i}=up_{1}\dots p_{n}\Rightarrow at_{i}=up_{1}\dots p_{i-1}p_{i+1}\dots p_{n}$
which contradicts the choice of $n$. Now $p_{1}|ab$ so $p_{1}|a$.
Let $a=p_{1}a_{1}$ where $a_{1}\in R$. Then $p_{1}a_{1}b=up_{1}\dots p_{n}$
and so $a_{1}b=up_{2}\dots p_{n}$. Again $p_{2}|a_{1}$ since $p_{2}\nmid b$.
Proceeding this way we obtain that $b$ is a unit of $R$. Therefore
$a=b^{-1}up_{1}\dots p_{n}$, a contradiction since $a\notin S$. 


Now suppose that $R$ is not a UFD. Then there exists a non-zero element
$a\in R$ such that $a\notin S$. By the above $aR\cap S=\emptyset$.
Choose $P\supseteq aR$ to be an ideal maximal with respect to $P\cap S=\emptyset$.
Then $P$ is a prime ideal (check!). However, $P$ will contain a
rank $1$ prime ideal and hence, by assumption, a prime element. This
is a contradiction since $P\cap S=\emptyset$. Thus $R$ must be a
UFD. 

\end{elabeling}
\end{proof}
\begin{lem}
\label{lem:7.1.6}Let $s$ be a non-zero prime element of a Noetherian
local domain $R$. Let $A$ be a prime ideal with $s\notin A$. Let
$S=\left\{ s^{n}\right\} $. If $AR_{S}$ is a principal ideal of
$R_{S}$ then $A$ is a principal ideal of $R$\end{lem}
\begin{proof}
Let $AR_{S}=bR_{S}$. We may assume that $b\in A$ (why?). By Lemma
\ref{thm:4.1.7} $\cap_{n=1}^{\infty}s^{n}R=0$. So there exists $k\geq0$
such that $b\in s^{k}R$ but $b\notin s^{k+1}R$. Let $b=s^{k}a$
where $a\in R$. Then $a\notin sR$. We have $AR_{S}=bR_{S}=as^{k}R_{S}=aR_{S}$.
Also $as^{k}\in A$ gives $a\in A$ since $s\notin A$ and $A$ is
prime

Claim: $A=aR$

Let $x\in A$. Then $x\in aR_{S}$. So $x=ars^{-m}$ for some $m$,
suppose $m\geq1$. Hence $xs^{m}=ar$. Since $a\notin sR$, $r\in sR$
since $sR$ is prime. So $r=sr_{1}$ for some $r_{1}\in R$. Hence
$xs^{m}=asr_{1}$ and so $xs^{m-1}=ar_{1}\in sR$ if $m-1>0$. Proceeding
as above we finally obtain $x\in aR$. Thus $A=aR$ as required.
\end{proof}

\subsection{Stably Free Modules}

Let $A,B$ be $n\times n$ matrices over a commutative integral domain.
Then $|AB|=|A|\cdot|B|$ where $|\quad|$ denotes the determinant
of the matrix
\begin{notation*}
Let $R$ be a ring. We write $R^{n}$ (or sometimes $R^{(n)}$) for
$\underset{n\mathrm{\, times}}{\underbrace{R\oplus\dots\oplus R}}$\end{notation*}
\begin{thm}[Kaplansky ]
\label{thm:7.2.1} Let $R$ be a commutative integral domain and
$A$ a (non-zero) ideal of $R$ such that $A\oplus R^{n-1}\cong R^{n}$
as $R$-modules. Then $A$ is a principal ideal of $R$.\end{thm}
\begin{proof}
The isomorphism shows that $A\oplus R^{n-1}$ has a free basis consisting
of $n$ elements, say $\lambda_{1},\dots,\lambda_{n}$. Each $\lambda_{j}$
is an $n$-tuple, so let $\lambda_{j}=(\alpha_{1j},\beta_{2j},\dots,\beta_{nj})$
where $\alpha_{1j}\in A$ and $\beta_{ij}\in R$. Let 
\[
\Lambda=\begin{pmatrix}\alpha_{11} & \alpha_{12} & \dots & \alpha_{1n}\\
\beta_{21} & \beta_{22} &  & \beta_{2n}\\
\vdots &  & \ddots\\
\beta_{n1} & \beta_{n2} &  & \beta_{nn}
\end{pmatrix}
\]
Then $\Lambda\in M_{n}(R)$, note that $|\Lambda|\in A$. Now consider
\[
X=\begin{pmatrix}I & I & \dots & I\\
R & R &  & R\\
\vdots &  & \ddots\\
R & R &  & R
\end{pmatrix}
\]
Then $X\lhd_{r}M_{N}(R)$. Let 
\[
\begin{pmatrix}a_{11} & a_{12} & \dots & a_{1n}\\
b_{21} & b_{22} &  & b_{2n}\\
\vdots &  & \ddots\\
b_{n1} & b_{n2} &  & b_{nn}
\end{pmatrix}\in X
\]
where $a_{1j}\in A$ and $b_{ij}\in R$ for $2\leq i\leq n$. Writing
the elements of $A\oplus R\oplus\dots\oplus R$ as columns we have
\[
\begin{pmatrix}a_{1j}\\
b_{ij}\\
\vdots\\
b_{nj}
\end{pmatrix}=\underset{=\lambda_{1}}{\begin{pmatrix}\alpha_{11}\\
\beta_{21}\\
\vdots\\
\beta_{n1}
\end{pmatrix}}s_{1j}+\underset{=\lambda_{2}}{\begin{pmatrix}\alpha_{12}\\
\beta_{22}\\
\vdots\\
\beta_{n2}
\end{pmatrix}}s_{2j}+\dots+\underset{=\lambda_{n}}{\begin{pmatrix}\alpha_{1n}\\
\beta_{2n}\\
\vdots\\
\beta_{nn}
\end{pmatrix}}s_{nj}
\]
with $s_{ij}\in R$ since $\lambda_{1},\dots,\lambda_{n}$ is a free
basis for $A\oplus R^{n}$. In the matrix from these can be written
\[
\begin{pmatrix}a_{11} & a_{12} & \dots & a_{1n}\\
b_{21} & b_{22} &  & b_{2n}\\
\vdots &  & \ddots\\
b_{n1} & b_{n2} &  & b_{nn}
\end{pmatrix}=\begin{pmatrix}\alpha_{11} & \alpha_{12} & \dots & \alpha_{1n}\\
\beta_{21} & \beta_{22} &  & \beta_{2n}\\
\vdots &  & \ddots\\
\beta_{n1} & \beta_{n2} &  & \beta_{nn}
\end{pmatrix}\begin{pmatrix}s_{11} & s_{12} & \dots & s_{1n}\\
s_{21} & s_{22} &  & s_{2n}\\
\vdots &  & \ddots\\
s_{n1} & s_{n2} &  & s_{nn}
\end{pmatrix}
\]
Thus $X\subseteq\Lambda M_{n}(R)$, but $\Lambda M_{n}(R)\subseteq X$
since $X\lhd R$. Hence $X=\Lambda M_{n}(R)$. Now let $x\in A$ and
consider 
\[
\begin{pmatrix}\begin{matrix}x\\
 & 1\\
 &  & 1
\end{matrix} &  & 0\\
 & \ddots\\
0 &  & 1
\end{pmatrix}\in X
\]
 so by above there exists $B\in M_{n}(R)$ such that 
\[
\begin{pmatrix}\begin{matrix}x\\
 & 1\\
 &  & 1
\end{matrix} &  & 0\\
 & \ddots\\
0 &  & 1
\end{pmatrix}=\Lambda B
\]
Take determinants, we have $x=|\Lambda|\cdot|B|$. Thus $A\subseteq|\Lambda|R$,
but $|\Lambda|R\subseteq A$ since $A\lhd R$. Thus $A=|\Lambda|R$
and $A$ is principal. \end{proof}
\begin{defn}
$M_{R}$ is said to have a \emph{finite free resolution} if there
exists an exact sequence $0\to F_{n}\to F_{n-1}\to\dots\to F_{0}\to M\to0$
with each $F_{i}$ is free.
\end{defn}
Clearly, over a regular local ring each finitely generated module
has a finite free resolution
\begin{lem}
Let $S$ be a multiplicatively closed subset of a commutative ring
$R$. If $M_{R}$ has finite free resolution then so does the $R_{S}$-module
$M_{S}$\end{lem}
\begin{proof}
Exercise\end{proof}
\begin{defn}
An $R$-module $M$ is called \emph{stably free} if there exists finitely
generated free modules $F$ and $G$ such that $G\oplus M\cong F$.
\end{defn}
Clearly a stably free module is projective. A stably free module is
a finitely generated projective module with a finitely generated free
complement
\begin{lem}
Let $R$ be a commutative ring. A projective $R$-module with finite
free resolution is stably free\end{lem}
\begin{proof}
We prove this by induction on the length of the finite free resolution.
Let $M$ be a finite free resolution module.

For $n=1$ we have $0\to F_{1}\to F_{0}\to M\to0$. $M$ is projective.
So this splits, so $F_{0}\cong F_{1}\oplus M$ and $M$ is stably
free.

Now suppose we have 
\[
\xyR{1pc}\xymatrix{0\ar[r] & F_{n}\ar[r] & \dots\ar[r] & F_{1}\ar[rr]\ar[dr] &  & F_{0}\ar[r] & M\ar[r] & 0\\
 &  &  &  & K_{0}\ar[ur]\ar[dr]\\
 &  &  & 0\ar[ur] &  & 0
}
\]
We have $F_{0}\cong K_{0}\oplus M$ since $M$ is projective. $K_{0}$
has finite free resolution of length $n-1$ . By induction hypothesis
there exists a finitely free module $G$ such that $K_{0}\oplus G$
is free. Hence $F_{0}\oplus G\cong K_{0}\oplus G\oplus M$ with both
$F_{0}\oplus G$ and $K_{0}\oplus G$ free.
\end{proof}
If $R$ is a Noetherian domain and $0\neq A\lhd R$ such that $A$
is stably free then $A\oplus R^{m}\cong R^{n}$. In this case $m=n-1$
(Q4 on exercise sheet 7)
\begin{thm}[Auslander - Buchsbaum 1959]
\label{thm:7.2.6}  A regular local ring is a UFD.\end{thm}
\begin{proof}
Let $R$ be a regular local ring of dimension $n$. We prove the theorem
by induction on the (Krull) dimension $n$. 

If $n=0$ then $R$ is a field and there is nothing to prove.

Assume result for regular local rings of dimension less than $n$.
Let $J=J(R)$, choose $p\in J\setminus J^{2}$. By Theorem \ref{thm:4.7.3}
$R/pR$ is regular local. By Theorem \ref{thm:4.7.5} $pR$ is a prime
ideal and $p$ is a prime element. Let $S=\left\{ p^{n}\right\} $,
then clearly $K\dim R_{S}<K\dim R$. 

Now let $T$ be a rank $1$ prime of $R_{S}$. Let $M$ be a maximal
ideal of $R_{S}$ . Then either $T(R_{S})_{M}=TR_{S}$ or $T(R_{S})_{M}$
is a rank $1$ prime ideal of $(R_{S})_{M}$. By induction hypothesis
$(R_{S})_{M}$ is a UFD. So by Proposition \ref{prop:7.1.5} $T(R_{S})$
is principal and hence a projective (free) $(R_{S})_{M}$-module.
So by Proposition \ref{prop:5.2a.3} $T$ is a projective $R_{S}$-module.
Now let $A$ be a rank $1$ prime of $R$. By above $AR_{S}$ is a
projective $R_{S}$-module. Since every finitely generated module
over $R_{S}$ has finite free resolution by the previous lemma, $AR_{S}$
is stably free. So by Theorem \ref{thm:7.2.1} $AR_{S}$ is free.
Thus $AR_{S}$ is a principal ideal. So by Lemma \ref{lem:7.1.6}
$A$ is a principal ideal if $p\notin A$. However if $p\in A$ then
$pR=A$ since rank $A$ is $1$. So by Proposition \ref{prop:7.1.5}
$R$ is a UFD
\end{proof}
Key point. $R_{S}$ is \uline{not }local. 


\section*{Beyond the Course}
\begin{thm}
Let $R$ be a commutative Noetherian integral domain. The following
are equivalent:
\begin{enumerate}
\item Every ideal of $R$ is a product of prime ideals
\item $R_{M}$ is a PID for each maximal ideal $M$
\item $R$ is integrally closed and $K\dim R=1$
\end{enumerate}
(There are various other characterisation) Such a ring is called \emph{Dedekind
Domain.}
\end{thm}
Recall that if $R$ is a commutative integral domain, $I\lhd R$,
$F$ the field of fraction, then $I^{*}=\{q\in F|qI\subseteq R$\}
. Then $I^{*}I\subseteq R$, $I^{*}I\lhd R$.

$I$ is said to be \emph{invertible} if $I^{*}I=R$. By the dual basis
lemma $I$ invertible is the same as $I_{R}$ projective. So we can
add:
\begin{enumerate}[resume]
\item Every non-zero ideal of $R$ is invertible
\item Every ideal of $R$ is projective.\end{enumerate}
\begin{proof}
5) $\Rightarrow$ 2), $M_{R}$ projective implies $MR_{M}$ projective.
So $MR_{M}$ is free by Theorem \ref{thm:5.2.8}. Thus $MR_{M}$ is
principal, hence by Theorem \ref{thm:4.1.9} $R_{M}$ is a PID.

$2)\Rightarrow5)$. Let $I\lhd R$, then $IR_{M}$ is principal. So
for each maximal ideal $M$ of $R$. So each $IR_{M}$ is $R_{M}$-projective.
Hence by Proposition \ref{prop:5.2a.3} $I_{R}$ is projective.
\end{proof}
Thus a Dedekind domain is a Noetherian domain $R$ with $D(R)=1$. 
\end{document}
