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\begin{document}
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\title{Commutative Algebra}


\date{John Cremona\\
Notes by Florian Bouyer}

\maketitle
\begin{center}
Copyright (C) Bouyer 2012. 
\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{}

Books: Introduction to Commutative Algebra by Atiyah and Macdonald.
Commutative Algebra by Miles Reid.


\section{Rings and Ideals}

All rings $R$ in this course will be commutative with a $1=1_{R}$.
\\
We include the zero ring $0=\{0\}$ with $1=0$. (in all other rings
$1\neq0$)
\begin{example*}
Algebraic geometry: $k[x_{1},...,x_{n}]$ with $k$ a field. (The
polynomial ring)\\
Number Theory: $\mathbb{Z}$, + rings of algebraic integers e.g. $\mathbb{Z}[i]$\\
Plus other rings from these by taking quotients, homomorphic images,
localization,...
\end{example*}
Ring homomorphisms: $R\rightarrow S$ (maps $1_{R}\mapsto1_{S}$)

Subrings: $S\leq R$ ($\leq$ means subring) is a subset which is
also a ring with the same operations and the same $1_{S}=1_{R}$.

Ideals: $I\lhd R$: a subgroup such that $RI\subseteq I$

Quotient Ring: $R/I$ the set of cosets of $I$ in $R$ ($x+I$) with
a natural multiplication $(x+I)(y+I)=xy+I$

Associated surjective homomorphism: $\pi:R\rightarrow R/I$ defined
by $x\mapsto x+I$

1 to 1 correspondence: $\{\mbox{ideals }J\mbox{ of }R\mbox{ with}J\geq I\}\leftrightarrow\{\mbox{ideals }\tilde{J}\mbox{ of }R/I\}$
defined by $J\mapsto\tilde{J}=\pi(J)=\{x+I:x\in J\}$ and $\tilde{J}\mapsto J=\pi^{-1}(\tilde{J})$

More generally if $f:R\rightarrow S$ is a ring homomorphism then
$\ker(f)=f^{-1}(0)\lhd R$ and $\mbox{im}(f)=f(R)\leq S$ and $R/\ker(f)\cong\mbox{im}(f)$
defined by $x+\ker(f)\mapsto f(x)$ and we have a bijection $\{\mbox{ideals}J\mbox{ of }R,J\geq\ker(f)\}\leftrightarrow\{\mbox{ideals }\widetilde{J}\mbox{ of im}(f)\}$.
\begin{example*}
$f:\mathbb{Z}\rightarrow\mathbb{Z}/n\mathbb{Z}$. $\ker(f)=n\mathbb{Z},\mbox{im}(f)=\mathbb{Z}/n\mathbb{Z}$.
Ideal of $\mathbb{Z}/n\mathbb{Z}\leftrightarrow$ideals of $\mathbb{Z},$$\geq n\mathbb{Z}$
i.e. $m\mathbb{Z}/n\mathbb{Z},m|n$
\end{example*}

\subsection{Special elements, special rings}
\begin{defn}
$x\in R$ is a \emph{zero-divisor} if $xy=0$ for some $y\neq0$

$x\in R$ is \emph{nilpotent} if $x^{n}=0$ for some $n\geq1$ ($\Rightarrow$
$x$ is a zero divisor except in $0$ ring)

$x\in R$ is a \emph{unit} if $xy=1$ for some $y\in R$ (then $y$
is uniquely determined by $x$ and hence is denoted $x^{-1}$)

The set of all units in $R$ forms a group under multiplication and
is called the \emph{Unit Group. }Denoted $R^{\times}$ (or $R^{*}$)

$R$ is an \emph{integral domain} (or domain) if $R\neq0$ and $R$
has no zero divisors.

\emph{Principal ideals}: Every element $x\in R$ generates an ideal
$xR=(x)=\{xr:r\in R\}$. $(x)=R=(1)\iff x\in R^{\times}$. $(x)=\{0\}=(0)\iff x=0$

A \emph{field} is a ring in which every non-zero element is a unit.
In a field $k$ the only ideals are $(0)=\{0\}$ and $(1)=k$.\end{defn}
\begin{example*}
$\mathbb{Z},k[x_{1},...,x_{n}]$ are domains but not fields $(n\geq1)$. 

$\mathbb{Q},k(x_{1},...,x_{n})$ are fields.

$\mathbb{Z}/n\mathbb{Z}=\begin{cases}
0 & \mbox{if }n=1\\
\mbox{a field} & \mbox{if }n\mbox{ is prime}\\
\mbox{not a domain} & \mbox{if }n\mbox{ is not prime}
\end{cases}$\end{example*}
\begin{defn}
\emph{Prime ideal:}\textbf{\emph{ $P\lhd R$ }}is prime if $R/P$
is an integral domain. i.e. $P\neq R$ and $xy\in P\iff x\in P$ or
$y\in P$

\emph{Maximal ideal:} $M\vartriangleleft R$ is maximal if $R/M$
is a field. i.e. $R\geq I\geq M\Rightarrow I=R$ or $I=M$

An ideal $I\lhd R$ is \emph{proper }if $I\neq R$ ($\iff I$ does
not contain $1$$\iff$$I$ does not contain any units)
\end{defn}
Every maximal ideal is prime, but not conversely in general.
\begin{note*}
$0$ (the $0$ ideal) is prime $\iff R$ is a domain. $0$ is maximal
$\iff R$ is a field.\end{note*}
\begin{example*}
$R=\zz$. $0$ ideal is prime but not maximal. $p\zz$ ($p$ is prime)
is maximal.

If $R$ is a PID (Principal Ideal Domain) then every non-zero prime
is maximal:
\begin{proof}
$R\supseteq(y)\supseteq(x)=P\neq0$ $\Rightarrow x=yz$ for some $z\in R$.
$P$ prime $\Rightarrow y\in P$ or $z\in P$. If $y\in P$ then $(y)=(x)=P$.
On the other hand if $z\in P$ then $z=xt=ytz\Rightarrow z(1-yt)=0$,
but $z\neq0$ since $x\neq0$ but $R$ is a domain $\Rightarrow yt=1\Rightarrow(y)=R$
\end{proof}
\end{example*}
\begin{defn}
The set of all prime ideals of $R$ is called the \emph{spectrum}
of $R$, written $\mbox{Spec}(R)$

The set of all maximal ideals is $\mbox{Max}(R)$ and is less important.
\end{defn}
Let $f:R\rightarrow S$ be a ring homomorphism, and let $P$ be a
prime ideal of $S$ then $f^{-1}(P)$ is a prime ideal of $R$. $R\overset{f}{\rightarrow}S\overset{\pi}{\rightarrow}S/P$
has kernel $f^{-1}(P)$ and $S/P$ is a domain so $f^{-1}(P)$ is
prime.\\
Alternatively: If $x,y\notin f^{-1}(P)$ $\Rightarrow f(x),f(y)\notin P\Rightarrow f(xy)=f(x)f(y)\notin P\Rightarrow xy\notin f^{-1}(P)$.\\
Hence $f:R\rightarrow S$ induces a map $f^{*}:\Spec(S)\rightarrow\Spec(R)$
by $P\mapsto f^{-1}(P)$ 

e.g. If $f$ is surjective we have a bijection between $\{\mbox{ideals of }R\geq\ker(f)\}\leftrightarrow\{\mbox{ideals of }S\}$
which restricts to $\Spec(R)\supseteq\{\mbox{primes ideals of }R\geq\ker(f)\}\leftrightarrow\{\mbox{prime ideals }S\}=\Spec(S)$
with $P\mapsto f^{*}(P)$. So $f^{*}$ is injective
\begin{example*}
If $f:\zz\hookrightarrow\mathbb{Q}$ is the inclusion. $0\in\mbox{Max}(\mathbb{Q})$
but $f^{-1}(0)=0\notin\mbox{Max}(Z)$

$\Spec(\zz)=\{0\}\cup\{p\zz:p\mbox{ prime}\}$,

$\Spec(\mathbb{Q})=\{0\}=\Spec(k)$ for any field $k$

$\Spec\cc[x]\,'='\,\underset{0\mbox{\ensuremath{\mathrm{\, ideal}}}}{\{\infty\}}\cup\underset{a\in\cc\rightarrow(X-a)}{\cc}=\mathbb{P}^{1}(\cc)$\\


$\Spec\cc[x,y]\,'='\,\underset{0}{\{\infty\}}\cup\underset{\mathrm{e.g.\, lines}\, X+Y=0}{\{\mbox{irreducible curves in }\cc^{2}\}}\cup\underset{(a,b)\leftrightarrow(X-a,X-b)=\{f:f(a,b)=0\}}{\cc^{2}}$\end{example*}
\begin{thm}
Every non-zero ring has a maximal ideal\end{thm}
\begin{proof}
Uses Zorn's Lemma:
\begin{lem*}
Let $S,\leq$ be a partially ordered set (so $\leq$ is transitive
and antisymmetric $x\leq y$ and $y\leq x\iff x=y$)

If $S$ has the property that every totally ordered subset $T\subseteq S$
has an upper bound in $S$, then $S$ has a maximal element.
\end{lem*}
We apply this to the set of all proper ideals in $R$. Let $T$ be
a totally ordered set of proper ideals of $R$. Set $I=\bigcup_{J\in T}J$.
Claim: $I\lhd R,\, I\neq R$ then $I$ is an upper bound for the set
$T$ so Zorn $\Rightarrow\exists$ maximal proper ideal. 
\begin{enumerate}
\item Let $x\in I,\, r\in R\Rightarrow x\in J$ for some $J\in T\Rightarrow rx\in J\subseteq I\Rightarrow rx\in I$
\item Let $x,y\in I$ then $x\in J_{1}$ and $y\in J_{2}$. Either $J_{1}\subseteq J_{2}\Rightarrow x,y\in J_{2}\Rightarrow x+y\in J_{2}\subseteq I$
or similarly $J_{2}\subseteq J_{1}$.
\end{enumerate}
Notice that $1\notin J\,\forall J$ hence $1\notin\cup J$ so $I$
is a proper ideal of $R$
\end{proof}
The same proof can be used to show
\begin{cor}
Every proper ideal $I$ is contained in a maximal ideal (Apply theorem
to $R/I$)
\end{cor}

\begin{cor}
Every non-unit of $R$ is contained in a maximal ideal (can use corollary
1.5)\end{cor}
\begin{defn}
A \emph{local ring} is one with exactly one maximal ideal (it may
have other prime ideals!)\end{defn}
\begin{example*}
$p$ prime number $\zz_{(p)}=\{\frac{a}{b}\in\qq:p\nmid b\}\underset{\geq\zz}{\leq\qq}$
has unique maximal ideal $p\zz_{(p)}$ with $\zz_{(p)}/p\zz_{(p)}\equiv\zz/p\zz=\{\frac{a}{b}:p\mid a,p\nmid b\}$.
$\zz_{(p)}\setminus p\zz_{(p)}=\{\frac{a}{b}:p\nmid a,p\nmid b\}=$
set of units in $\zz_{(p)}$ In general in a local ring $R$ with
maximal ideal $M$ the set of units $R^{\times}=R\setminus M$. Note
that $(0)$ is a prime ideal of $\zz_{(p)}$

$k$ field, $R=k[[x]]=\{\mbox{power series in }X\mbox{ with coefficients in }k\}=\{f=\sum_{i=1}^{\infty}a_{i}x^{i}:a_{i}\in k\}$.
Can check $f$ is a unit $\iff a_{0}\neq0$. $f$ is not a unit $\iff a_{0}=0\iff f\in(x)\Rightarrow(x)=M$
is the unique maximal ideal.
\end{example*}

\subsection{Two radicals: The nilradical $N(R)$ and the Jacobson radical $J(R)$}
\begin{defn}
$N(R)=\{x\in R:x\mbox{ is nilpotent}\}$\end{defn}
\begin{prop}
~
\begin{enumerate}
\item $N(R)\lhd R$
\item $N(R/N(R))=0$
\end{enumerate}
\end{prop}
\begin{proof}
~
\begin{enumerate}
\item {}

\begin{enumerate}
\item Let $x\in N(R),r\in R$. So $x^{n}=0$ for some $n\geq1\Rightarrow(rx)^{n}=r^{n}x^{n}=0\Rightarrow rx\in N(R)$.
\item $x^{n}=0$, $y^{m}=0$ $(m,n\geq1$)$\Rightarrow(x+y)^{m+n+1}=0$,
$cx^{i}y^{j}=0$ since $i+j=m+n+1\Rightarrow$ either $i\geq n$ or
$j\geq m$
\end{enumerate}
\item Need to show that $R/N(R)$ has no non-zero nilpotents. \\
\[
x^{n}+N(R)=(x+N(R))^{n}=0=0+N(R)\mbox{ (in }R/N(R))
\]
\begin{eqnarray*}
 & \Rightarrow & x^{n}\in N(R)\\
 & \Rightarrow & (x^{n})^{m}=0\\
 & \Rightarrow & x^{mn}=0\\
 & \Rightarrow & x\in N(R)\\
 & \Rightarrow & x+N(R)=0\mbox{ in }R/N(R)
\end{eqnarray*}

\end{enumerate}
\end{proof}
\begin{prop}
$N(R)$ is the intersection of all the prime ideals of $R$\end{prop}
\begin{proof}
Let $x\in N(R)$ so $x^{n}=0$ but since $0\in P\,\forall P\in\Spec R$
hence $x^{n}\in P\,\forall P\in\Spec R\Rightarrow x\in P$ since $P$
is prime $\Rightarrow x\in\bigcap_{P\in\Spec R}P$

For the other way we use the contrapositive. Let $x\notin N(R)$.
So $x,x^{2},x^{3},\dots$ are all non-zero. Consider all ideals $I$
which contain no power of $x$ e.g.~$0$. In this collection there
is a maximal element say $P$. Then $P\lhd R$ and $x\notin P$. We
need to show that $P$ is prime. Let $y,z\notin P$, then $P+(y)\supsetneq P$
and $P+(z)\supsetneq P$. By maximality of $P$ each of $P+(y),P+(z)$
contains a power of $x$. Say ($p_{1},P_{2}\in P,y',z'\in R)$ 
\begin{eqnarray*}
x^{n} & = & p_{1}+yy'\\
x^{m} & = & p_{2}+zz'\\
 & \Rightarrow & x^{m+n}=\underset{\in P}{\underbrace{p_{1}p_{2}+p_{1}zz'+p_{2}yy'}}+yz(y'z')\\
 & \Rightarrow & x^{m+n}\in P+(yz)\\
 & \Rightarrow & P+(yz)\neq P\\
 & \Rightarrow & yz\notin P
\end{eqnarray*}
\end{proof}
\begin{defn}
$J(R)=$ intersection of all maximal ideals of $R$. $N(R)\subseteq J(R)$
(since maximals are primes)\end{defn}
\begin{prop}
\label{prop:J(R) iff unit}$x\in J(R)\iff1-xy\in R^{\times}\,\forall y\in R$.\end{prop}
\begin{proof}
$"\Rightarrow"$: If $1-xy\notin R^{\times}$ then $1-xy\in M$ for
some ideal maximal ideal $M\Rightarrow x\notin M$ (else $1\in M$
contradicting maximality of $M$) $\Rightarrow x\notin J(R)$

$"\Leftarrow"$: $ $ 
\begin{eqnarray*}
x\notin J(R) & \Rightarrow & x\notin M\mbox{ for some }M\\
 & \Rightarrow & M+(x)=R\\
 & \Rightarrow & 1=m+xy\,(m\in M,y\in R)\\
 & \Rightarrow & 1-xy=m\notin R^{\times}
\end{eqnarray*}
\end{proof}
\begin{example*}
$R=A\left[[x]\right]$ $(A$ is a ring). $R^{\times}=\{\sum_{i=0}^{\infty}a_{i}x^{i}:a_{0}\in A^{\times}\}$
(Exercise).\\
$\Rightarrow x\in J(R)$ since $1-xf\in R^{\times}\,\forall x\in R$.
\end{example*}

\subsection{New ideals from old}
\begin{description}
\item [{Sum}] If $I,J\lhd R$ then $I+J=\{x+y:x\in I,y\in J\}\lhd R$.
(The smallest ideal $\supseteq$ both $I$ and $J$)
\item [{Intersection}] $I\cap J\lhd R$ (The largest ideal $\subseteq$
both $I$ and $J$)
\item [{Product}] $IJ=$ ideal generated by all $xy$ with $x\in I,y\in J$
$=\{\sum_{i=1}^{n}x_{i}y_{i}:x_{i}\in I,y_{i}\in J\}$. $IJ\subseteq I\cap J$,
equality does not hold in general.
\item [{Powers:}] $I^{n}=$ideal generated by all product $x_{1}x_{2}\dots x_{n}$
($x_{i}\in I$)\end{description}
\begin{example*}
$R=\zz$. 
\begin{itemize}
\item $(m)+(n)=(d)$ where $d=\gcd(m,n)$
\item $(m)\cap(n)=(l)$ where $l=\lcm(m,n)$
\item $(m)(n)=(mn)$
\item $(m)^{k}=(m^{k})$
\end{itemize}
$R=k[x_{1},\dots,x_{n}]$. Let $M=(x_{1},x_{2},\dots,x_{n})=(x_{1})+(x_{2})+\dots+(x_{n})$.
($M=\ker(\phi:R\to k)$ where $\phi(f)=f(0,0,\dots,0)$) $R/M\cong k$\\
$M^{2}=(\dots,x_{i}x_{j},\dots)=\{$polynomials with $0$ constant
terms and $0$ linear terms$\}$
\end{example*}
These operation are commutative and associative, not all distributive.
\begin{itemize}
\item $I(J+K)=IJ+IK$

\begin{proof}
Each side is generated by $xy,xz$ for $x\in I,y\in J,z\in K$
\end{proof}
\item If $I+J=(1)$ then $I\cap J=IJ$

\begin{proof}
Take $(I+J)(I\cap J)=I(I\cap J)+J(I\cap J)\subseteq IJ+JI=IJ$ so
$I+J=(1)$ then $I\cap J\subseteq IJ$
\end{proof}
\end{itemize}
\begin{defn}
$I$ and $J$ are \emph{coprime/comaximal/relatively prime} if and
only if $I+J=(1)$$\iff x+y=1$ for some $x\in I,y\in J$.\end{defn}
\begin{example*}
For $R=\qq[x,y]$ we have $(x)+(y)=(x,y)=$ \{elements $f\in R$ such
that $f(0,0)=0\}\neq(1)$. So $(x)$ and $(y)$ are distinct prime
ideals but they are not coprime. \end{example*}
\begin{lem}
\label{lem:I,J coprime Im coprime}If $I$ and $J$ are coprime then
$I^{m}$ and $J^{n}$ are coprime for any $n,m\geq1$.\end{lem}
\begin{proof}
$x+y=1$ for certain $x\in I,y\in J$. Consider $1=(x+y)^{m+n-1}\in I^{m}+J^{n}$
hence $I^{m}$ and $J^{n}$ are coprime.\end{proof}
\begin{namedthm}[Chinese Remainder Theorem]
If $I_{1},\dots,I_{n}$ are pairwise coprime ideals of $R$ then
\[
\prod_{i=1}^{n}I_{i}=\bigcap_{i=1}^{n}I_{i}
\]
\[
R/\prod_{i=1}^{n}I_{i}=\prod_{i=1}^{n}\left(R/I_{i}\right)
\]
 \end{namedthm}
\begin{proof}
The first equation is true for $n=2$. We are going to use induction
so assume $n>2$ and the statement is true for $n-1$. Let $J=\prod_{i=1}^{n-1}I_{i}=\bigcap_{i=1}^{n-1}I_{i}$
by the induction hypothesis. We have $I_{i}+I_{n}=(1)$ for all $i=1,\dots,n-1$.
So take $x_{i}+y_{i}=1$ for some $x_{i}\in I_{i}$ and $y_{i}\in I_{n}$
then $\underset{\in J}{\underbrace{\prod_{i=1}^{n-1}x_{i}}}=\prod_{i=1}^{n-1}(1-y_{i})\equiv1\mod I_{n}$
so $J+I_{n}=(1)$. Hence $\prod_{i=1}^{n}I_{i}=JI_{n}=J\cap I_{n}=\bigcap_{i=1}^{n}I_{i}$

Define $\varphi:R\to\prod_{i=1}^{n}R/I_{i}$ by $x\mapsto(x+I_{1},x+I_{2},...,x+I_{n})$.
Kernel is $\bigcap_{i=1}^{n}I_{i}=\prod_{i=1}^{n}I_{i}$, now we just
need to show surjectivity. The element $\prod_{i=1}^{n-1}x_{i}$ maps
to $(0,\dots,0,1)$ (the $x_{i}$ are taken from the first paragraph).
By symmetry all ``unit vectors'' of $\prod(R/I_{i})$ are in the
image hence $\varphi$ is surjective. Then we use the first isomorphism
theorem to get $R/\prod I_{i}\to\prod(R/I_{i})$
\end{proof}
If ideals are not coprime, still get a ring homomorphism$R/\left(\bigcap_{i=1}^{n}I_{i}\right)\hookrightarrow\prod(R/I_{i})$
but not surjective.
\begin{prop}
\label{prop:Intersection of prime contains prime}{}
\begin{enumerate}
\item If $I\subseteq\bigcup_{i=1}^{n}P_{i}$ with $P_{i}$ prime, then $I\subseteq P_{i}$
for some $i$
\item If $P\supseteq\bigcap_{i=1}^{n}I_{i}$ and $P$ is prime, then $P\supseteq I_{i}$
for some $i$
\item 2. is also true with ``$=$''
\end{enumerate}
\end{prop}
\begin{proof}
{}
\begin{enumerate}
\item We prove by induction if $I\nsubseteq P_{i}$ for all $i$ then $I\nsubseteq\bigcup_{i=1}^{n}P_{i}$.
In the case $n=1$ it is obvious. So suppose $n>1$ and the statement
is true for $n-1$. Suppose $I\nsubseteq P_{i}\,\forall i$. Then
by induction $I\nsubseteq\bigcup_{j\neq i}P_{j}$ hence $\exists x_{i}\in I$
such that $x_{i}\notin\bigcup_{j\neq i}P_{j}$ so for all $j\neq i$
we have $x_{i}\notin P_{j}.$ If for some $i$ we have $x_{i}\notin P_{i}$
then $x_{i}\in I\setminus\bigcup_{j=1}^{n}P_{j}$ and we are done.
So assume $x_{i}\in P_{i}$ for all $i$. Let $y=\sum_{i=1}^{n}x_{1}x_{2}\dots x_{i-1}x_{i+1}\dots x_{n}\in I$.
The $i$th term is in $P_{j}$ for all $j\neq i$ but not in $P_{i}$.
Given $j$ we see that all but the $j$th term are in $P_{j}$ so
$y\notin P_{j}$, hence $y\notin\bigcup_{j=1}^{n}P_{j}$
\item Suppose $P\nsupseteq I_{i}\,\forall i$, then $\exists x_{i}\in I_{i}\setminus P$
for every $i$. Then $\prod x_{i}\in\left(\bigcap I_{i}\right)\setminus P$
\item If $P=\bigcap I_{i}$ then $P\supseteq I_{i}$ for some $i$ by part
2 and $P=\bigcap I_{i}\subseteq I_{i}$ hence $P=I_{i}$
\end{enumerate}
\end{proof}

\subsection{Quotients and radicals}
\begin{defn}
Let $I,J$ be ideals, define the \emph{quotient} $(I:J)=\{x\in R\mid xJ\subseteq I\}$
(This is an ideal, but not exactly the same as in algebraic number
theory)

Special case: $(0:J)=$\emph{annihilator of $J=\mbox{Ann}(J)$}\end{defn}
\begin{example*}
IF $R=\zz,$ $((15):(6))=(5)$. More generally if $m=\prod p_{i}^{ei}$
and $n=\prod p_{i}^{f_{i}}$ then $((m):(n))=(a)$ where $a=\prod p_{i}^{\max\{e_{i}-f_{i},0\}}$. \end{example*}
\begin{fact*}
\label{Fact:colon fact}{}
\begin{enumerate}
\item $I\subseteq(I:J)$ (since $IJ\subseteq I$)
\item $(I:J)J\subseteq I$ 
\item $((I:J):K)=(I:JK)=((I:K):J)$
\item $\left(\bigcap I_{i}:J\right)=\bigcap\left(I_{i}:J\right)$
\item $\left(I:\sum J_{i}\right)=\bigcap(I:J_{i})$
\end{enumerate}
\end{fact*}
\begin{defn}
Let $I$ be an ideal, define the \emph{radical }of $I$ to be $r(I):=\{x\in R|x^{n}\in I\mbox{ for some }n\geq1\}$

Special case: $r(0)=N(R)$
\end{defn}
Given $I$, let $\varphi:R\to R/I$. Then $\varphi^{-1}(N(R/I))=\{x\in R:\varphi(x)^{n}=0\mbox{ for some }n\}=r(I)$.
Hence $r(I)$ is an ideal. 
\begin{example*}
$R=\zz$. If $m=\prod p_{i}^{k_{i}},k_{i}\geq1$ then $r((m))=\left(\prod p_{i}\right)$\end{example*}
\begin{fact*}
{}
\begin{enumerate}
\item If $I\subseteq J$ then $r(I)\subseteq r(J)$.
\item $r(I)\supseteq I$ (take $n=1$ in the definition)
\item $r(r(I))=r(I)$ ($(x^{m})^{n}=x^{mn}$)
\item $r(IJ)=r(I\cap J)=r(I)\cap r(J)$ 
\item $r(I)=(1)\iff I=(1)$ (use $1\in r(I)$)
\item $r(I+J)=r(r(I)+r(J))$
\item $r(P^{n})=P$ where $P$ is a prime ideal and $n\geq1$
\item $r(I)=\bigcap_{\underset{P\mathrm{\, prime}}{P\supseteq I}}P$
\end{enumerate}
\end{fact*}
\begin{prop}
$I,J$ are coprime if and only if $r(I),r(J)$ are coprime if and
only if $I^{m},J^{n}$ are coprime for every/any $m,n\geq1$\end{prop}
\begin{proof}
$I$ and $J$ coprime then $I^{m},J^{n}$ coprime for all $m,n$ was
lemma \ref{lem:I,J coprime Im coprime} . If $\forall m,n$ $I^{m},J^{n}$
are coprime $\Rightarrow\exists m,n\, I^{m},J^{n}$ are coprime is
trivial.If $\exists m,n\geq1$ such that $I^{m},J^{n}$ are coprime
then $I+J\supseteq I^{m}+J^{n}=(1)$ hence $I+J=(1)$ (i.e~they are
coprime)

We now just need to prove $I,J$ coprime $\iff r(I),r(J)$ are coprime

``$\Rightarrow$'' obvious because $r(I)+r(J)\supseteq I+J=(1)$,
so $r(I)+r(J)=(1)$

``$\Leftarrow$'' $r(I+J)=r(r(I)+r(J))=r((1))=(1)$ hence by fact
5. we have $I+J=(1)$
\end{proof}

\subsection{Extension and Contractions}
\begin{defn}
Let $f:R\to S$ be a ring homomorphism. For $I\lhd R$, let the \emph{extension}
of $I$, $I^{e}$ be the ideal generated by $\{f(x)\in S\mid x\in I\}$.
So $I^{e}=\left\{ \sum_{\mathrm{finite}}s_{i}f(x_{i})\mid s_{i}\in S,x_{i}\in I\right\} $

For $J\lhd S$, let the \emph{contraction }of $J$, $J^{c}=f^{-1}(J)\subseteq R$
(this is an ideal)\end{defn}
\begin{example*}
If $R\hookrightarrow S$ then $J^{c}=J\cap R,I^{e}=\left\{ \sum s_{i}x_{i}\mid s_{i}\in S,x_{i}\in I\right\} =$
the $S$-ideal generated by $I$\end{example*}
\begin{fact*}
If $P$ is a prime ideal of $S$ then $P^{c}$ is a prime ideal of
$R$ (seen). This is not true for extensions:
\begin{example*}
$\zz\hookrightarrow\zz[i]$. If we take $(5)^{e}=5\zz[i]=(2+i)(2-i)\zz[i]$
is not a prime ideal.
\end{example*}
\end{fact*}
\begin{prop}
Let $I\lhd R$ and $J\lhd S$
\begin{enumerate}
\item $I\subseteq I^{ec}$ (since $x\in f^{-1}\left(f(x)\right)$ )
\item $J\supseteq J^{ce}$ (easy)
\item $I^{e}=I^{ece}$ and $J^{c}=J^{cec}$
\item Let $C=$ set of contracted ideals in $R$ and $E=$ set of extended
ideals in $S$. Then $C=\{I\lhd R|I=I^{ec}\}$, $E=\{J\lhd S|J=J^{ce}\}$
and there is a bijection $C\to E$ given by $e$ whose inverse is
$c$.
\end{enumerate}
\end{prop}
\begin{proof}
1 and 2 are easy. For 3 we have $I^{e}\supseteq I^{ece}$ by 2 applied
to $J=I^{e}$ but by 1 we have $I\subseteq I^{ec}$ and apply extension
hence $I^{e}\subseteq I^{ece}$. 4 is easy to prove using 3\end{proof}
\begin{example*}
Counter example to reverse inclusion of 1. $\zz\hookrightarrow\qq$,
$(2)^{ec}=\qq^{c}=\zz=(1)\neq(2)$\end{example*}
\begin{thm}
Let $f:R\to S$ be a ring homomorphism and $I\to I^{e}$ and $J\to J^{c}$
be the extension and contraction maps. Then
\begin{itemize}
\item Extension:

\begin{enumerate}
\item $(I_{1}+I_{2})^{e}=I_{1}^{e}+I_{2}^{e}$
\item $(I_{1}\cap I_{2})^{e}\subseteq I_{1}^{e}\cap I_{2}^{e}$
\item $(I_{1}I_{2})^{e}=I_{1}^{e}I_{2}^{e}$
\item $(I_{1}:I_{2})^{e}\subseteq I_{1}^{e}:I_{2}^{e}$
\item $r(I)^{e}\subseteq r(I^{e})$
\end{enumerate}
\item Contraction:

\begin{enumerate}
\item $(J_{1}+J_{2})^{c}\supseteq J_{1}^{c}+J_{2}^{c}$
\item $(J_{1}\cap J_{2})^{c}=J_{1}^{c}\cap J_{2}^{c}$
\item $(J_{1}J_{2})^{c}\supseteq J_{1}^{c}J_{2}^{c}$
\item $(J_{1}:J_{2})^{c}\subseteq J_{1}^{c}:J_{2}^{c}$
\item $r(J)^{c}=r(J^{c})$
\end{enumerate}
\end{itemize}
\end{thm}
\begin{proof}
None of these is too hard to show\end{proof}
\begin{example*}
Counter example to show cases where equality does not hold
\begin{itemize}
\item Contraction 1: Take $f:k\hookrightarrow k[x]$ (with $k$ any field),
$J_{1}=(x)$ and $J_{2}=(x+1)$. Then $J_{1}^{c}=J_{2}^{c}=(0)$ but
$J_{1}+J_{2}=(1)$ which contracts to $(1)$.
\item Extension 2: Take $f:\zz[x]\to\zz$ to be the ``evaluation homomorphism''
which maps $x\mapsto2$. Let $I_{1}=(x)$ and $I_{2}=(2)$ then $I_{1}\cap I_{2}=(2x)$
so $(I_{1}\cap I_{2})^{e}=(2x)^{e}=4\zz$ while $I_{1}^{e}=I_{2}^{e}=2\zz$
so $I_{1}^{e}\cap I_{2}^{e}=2\zz$
\item Contraction 3: Take $f:\zz\hookrightarrow\zz[i]$, $J_{1}=(2+i),J_{2}=(2-i)$.
Then $J_{1}^{c}=J_{2}^{c}=(J_{1}J_{2})^{c}=(5)$
\item Extension 4: Take $f:\zz[x]\to\zz$ to be the ``evaluation homomorphism''
which maps $x\mapsto2$. Let $I_{1}=(x)$ and $I_{2}=(2)$ then $(I_{1}:I_{2})=I_{1}$
(since $x|2f\iff x|f$) so $(I_{1}:I_{2})^{e}=(x)^{e}=2\zz$ while
$I_{1}^{e}=I_{2}^{e}=2\zz$ with quotient $\zz$
\item Contraction 4: Take $f:\zz\hookrightarrow\zz[i],J_{1}=(2+i),J_{2}=(2-i)$.
Then $J_{1}^{c}=J_{2}^{c}=(5)$ so $(J_{1}^{c}:J_{2}^{c})=\zz$ but
$(J_{1}:J_{2})=J_{1}$ which contracts to $(5)$.
\item Extension 5: Take $f:\zz\hookrightarrow\zz[i],I=2\zz$. Then $r(I)^{e}=(2\zz)^{e}=2\zz[i]$
while $r((2)^{e})=r(2\zz[i])=(1+i)\zz[i]$
\end{itemize}
\end{example*}
From the theorem we can see that the set of extended ideals of $S$
is closed under the sum and product, while the set of contracted ideals
of $R$ is closed under intersection and radical.

\newpage{}


\section{Modules}
\begin{defn}
An \emph{$R$-module} is an abelian group $M$ with a scalar multiplication
$\underset{(r,m)\mapsto rm}{R\times M\to M}$ satisfying
\begin{enumerate}
\item $(r_{1}+r_{2})m=r_{1}m+r_{2}m$
\item $r(m_{1}+m_{2})=rm_{1}+rm_{2}$
\item $r_{1}(r_{2}m)=(r_{1}r_{2})m$
\item $1_{R}m=m$
\end{enumerate}
\end{defn}
For each $r\in R$ the map $M\to M,m\mapsto rm$ is an endomorphism
of $M$ (by 2.) 1,3,4 says $R\to\mbox{End}(M)$ is a ring homomorphism
\begin{example*}
{}
\begin{enumerate}
\item $R$ itself is an $R$-module. So are all ideals of $R$
\item If $R$ is a field $k$ then an $R$-module is a $k$-vector space
\item Every abelian group $A$ is a $\zz$-module
\item A $k[x]$-module is $k$vector space $V$ together with a $k$-linear
map $V\to V$ given the scalar multiplication by $x$
\item Let $G$ be a finite group (abelian). Let $R=k[G]$ the group algebra.
Then a $k[G]$ module is a representation of $G$.
\end{enumerate}
\end{example*}
\begin{defn}
An $R$-module homomorphism $f:M\to N$ is a map $M\to N$ which satisfies
\begin{enumerate}
\item $f(m_{1}+m_{2})=f(m_{1})+f(m_{2})$
\item $f(rm)=rf(m)$
\end{enumerate}
Where $M,N$ are both $R$-module. $f$ is called $R$\emph{-linear}

$\Hom_{R}(M,N)=\{$all $R$-linear map $f:M\to N\}$ is another $R$-module
with point-wise operations\end{defn}
\begin{example*}
$\Hom_{R}(R,M)\cong M$ by $f\leftrightarrow f(1_{R})$ since $f(r)=f(r\cdot1)=rf(1)$ \end{example*}
\begin{defn}
$N\subseteq M$ is a \emph{submodule} if it is closed under addition
and scalar multiplication, (in particular $0\in N$). We will use
$N\leq M$ as notation.\end{defn}
\begin{example*}
$R$-submodules of $R$ are the ideals of $R$.\end{example*}
\begin{defn}
\emph{Quotient Modules: }If $N\leq M$ then $M/N$ is again an $R$-module
via $r(x+N)=rx+N$ (well-defined since $rN\subseteq N$)

\emph{Kernels and Cokernels: }If \emph{$f\in\Hom_{R}(M,N)$ }then
$\ker(f)\leq M$, $\im(f)\leq N$ and $\mbox{coker}(f)=N/\im(f)$
\end{defn}
So $f$ is injective $\iff\ker(f)=0$. $f$ is surjective $\iff\mbox{coker}(f)=0\iff\im(f)=N$
\begin{namedthm}[First Isomorphism Theorem]
 If $f\in\Hom_{R}(M,N)$ then $M/\ker(f)\cong\im(f)$ via $m+\ker(f)\mapsto f(m)$\end{namedthm}
\begin{defn}
\emph{Sums of Submodules:} Let $M_{i}\leq M$ for $i\in I$. Then
$\sum_{i\in I}M_{i}=\{$all finite sums $\sum_{i\in I}m_{i}$ with
$m_{i}\in M_{i}\}\leq M$

\emph{Intersection of Submodules: }Let $M_{i}\leq M$ for $i\in I$.
Then $\bigcap_{i\in I}M_{i}\leq M$\end{defn}
\begin{namedthm}[Second Isomorphism Theorem]
 Let $N\leq M\leq L$ be submodules of $R$. Then 
\[
\frac{L/N}{M/N}\cong\frac{L}{M}
\]
\end{namedthm}
\begin{proof}
The map $L/N\to L/M$ defined by $x+N\mapsto x+M$ ($x\in L$) is
surjective with kernel $M/N$, then use the first isomorphism theorem.\end{proof}
\begin{namedthm}[Third Isomorphism Theorem]
 Let $M_{1},M_{2}\leq M$ be $R$-modules. Then
\[
\frac{M_{1}+M_{2}}{M_{1}}\cong\frac{M_{2}}{M_{1}\bigcap M_{2}}
\]
\end{namedthm}
\begin{proof}
The map $M\to M_{1}+M_{2}\to(M_{1}+M_{2})/M_{1}$ defined by $y\mapsto0+y\mapsto y+M_{1}$is
surjective with kernel $M_{1}\bigcap M_{2}$. Then use the first isomorphism
theorem.\end{proof}
\begin{defn}
\emph{Product of Ideal and Modules: }Let $I\lhd R$ and $M$ a $R$-module.
Define the product of $I$ and $M$ to be $IM=\{\sum_{i=1}^{n}a_{i}m_{i}|a_{i}\in I,m_{i}\in M\}\leq M$.

A special case $I=(r)$ we write $rM=\{rm|m\in M\}\leq M$

\emph{Quotient:} Let $M,N$ be $R$-module such that they both are
submodules of $L$, we define the quotient to be $(M:N)=\{r\in R:rN\subseteq M\}\lhd R$

Special case: $M=0$, $(0:N)=\{r\in R:rN=0\}=\Ann_{R}(N)\lhd R$

$M$is a \emph{faithful} $R$-module if $\Ann_{R}M=0$
\end{defn}
If $I\subseteq\Ann_{R}M$ then $M$ may be regarded as an $R/I$-module
via $(r+I)m=rm$. In particular taking $I=\Ann_{R}M$ we may view
$M$ as a faithful $R/\Ann_{R}M$-module. 
\begin{example*}
If $A$ is an abelian group (hence a $\zz$-module) which is \emph{$p$-torsion
}(meaning $pA=0$ for some prime $p$) then $A$ is $\zz/p\zz$-module,
i.e., a vector space over $\ff_{p}$.\end{example*}
\begin{defn}
\emph{Cyclic Submodules: $x\in M$ }an $R$-module generates $(x)=Rx=\{rx|r\in R\}\leq M$
is \emph{the cyclic submodule} generated by $x$. In particular if
$M=Rx$ for some $x$ then $M$ is \emph{cyclic }and $M\cong R/\Ann_{R}x$
(as $R$-modules)

\emph{Finitely Generated Module:} We say $M$ is \emph{finitely generated
}(f.g.) if $M=\sum_{i=1}^{n}Rx_{i}$ for some finite collection $x_{1},\dots,x_{n}\in M$.
More generally $\{x_{i}\}_{i\in I}$ generates $M$ if every $x\in M$
is a finite $R$-linear collection of the $x_{i}\in M$.\end{defn}
\begin{example*}
$M=R[x]$ is generated by $1,x,x^{2},x^{3},\dots$ but $M$ is \uline{not
}finitely generated.\end{example*}
\begin{defn}
Let $M,N$ be $R$-modules. We define:

\emph{Direct Sum:} $M\oplus N=\{(m,n):m\in M,n\in N\}$ is an $R$-module
with coordinate operations.

\emph{Direct Product: }$M\times N=\{(m,n):m\in M,n\in N\}$ is an
$R$-module with coordinate operations.

Similarly if $M_{i}$ ($i=1,\dots,n$) are $R$-modules we can form\emph{
$\oplus_{i=1}^{n}M_{i}=\{(m_{1},\dots,m_{n})|m_{i}\in M_{i}\,\forall i\leq n\}=\prod_{i=1}^{n}M_{i}$}

\emph{Infinite Direct Sum:} If we start with $\{M_{i}\}_{i\in I}$
we define $\oplus_{i\in I}M_{i}=\{(m_{i})_{i\in I}:m_{i}\in M_{i}\,\forall i,$all
but finitely many $m_{i}=0\}$

\emph{Infinite Direct Product: }If we start with $\{M_{i}\}_{i\in I}$
we define $\prod_{i=I}M_{i}=\{(m_{i})_{i\in I}:m_{i}\in M_{i}\,\forall i\}$\end{defn}
\begin{example*}
As an $R$-module $R[x]\cong\oplus_{i=0}^{\infty}R$ where the isomorphism
is defined by $\sum_{i=0}^{d}r_{i}x^{i}\mapsto(r_{0},r_{1},r_{2},\dots,r_{d},0,0,\dots)$

$R\left[[x]\right]\cong\prod_{i=0}^{\infty}R$ (as $R$-modules)\end{example*}
\begin{defn}
\emph{Free Modules}: $M$ is \emph{free} if $M\cong\oplus_{i\in I}M_{i}$
where each $M_{i}\cong R$.

A \emph{finitely generated free module} $M\cong\underset{n}{\underbrace{R\oplus\dots\oplus R}}=R^{n}$\end{defn}
\begin{lem}
$M$ is finitely generated if and only if $M\cong$ a quotient of
$R^{n}$ for some $n$\end{lem}
\begin{proof}
``$\Rightarrow$'': If $x_{1},\dots,x_{n}$ generates $M$ then
map $R^{n}\to M$ by $(r_{1},\dots,r_{n})\mapsto\sum_{i=1}^{n}r_{i}x_{i}$
is surjective (since $M$ is finitely generated) so $R^{n}/\ker\cong M$

``$\Leftarrow$'': $R^{n}$ is finitely generated by $(1,0,\dots0),(0,1,0,\dots,0),\dots$
So $R^{n}/K$ is finitely generated by images of these in $R^{n}/K$\end{proof}
\begin{prop}
Let $M$ be a finitely generated $R$-module, $J\lhd R$ and $\varphi\in\End_{R}(M)=\Hom_{R}(M,M)$.
Suppose that $\varphi(M)\subseteq JM$. Then $\exists a_{1},a_{2},\dots,a_{n}\in J$
such that 
\[
\varphi^{n}+a_{1}\varphi^{n-1}+a_{2}\varphi^{n-2}+\dots+a_{n}I_{M}=0
\]
 in $\End_{R}(M)$ and $I_{M}$ is the identity map $M\to M$\end{prop}
\begin{proof}
Let $x_{1},\dots,x_{n}$ generate $M$. $\forall i\leq n,\varphi(x_{i})=\sum_{j=1}^{n}a_{j}x_{j}$
where $a_{j}\in J$. 
\[
\sum_{j=1}^{n}(\delta_{ij}\varphi-a_{ij}I)x_{i}=0
\]
for $i=1,\dots,n$ where $\delta_{ij}=\begin{cases}
0 & i\neq j\\
1 & i=j
\end{cases}$. We can rewrite this as $(I\varphi-A)X=0$ where $A=(a_{ij}),I=(\delta_{ij}),X=(x_{1},\dots,x_{n})^{T}$.
Multiply by $\mbox{adj}(I\varphi-A)$ whose entries are all in $\End_{R}(M)$
$\Rightarrow\det(I\varphi-A)x_{i}=0\,\forall i\Rightarrow\det(I\varphi-A)=0\in\End_{R}(M)$.
If we multiply out $\det(I\varphi-A)$ to get the equation above.
\end{proof}
\textbf{Applications}: 
\begin{enumerate}
\item $x\in\cc$. If $M$ is a non-zero finitely generated $\qq$-submodule
of $\cc$ such that $xM\subseteq M$ then $x$ is algebraic.

\begin{cor}
The set of all algebraic numbers in $\cc$ forms a field. 
\end{cor}
\item $x\in\cc,M\subseteq\cc$ a non-zero finitely generated $\zz$-submodule
such that $xM\subseteq M\Rightarrow x$ is an algebraic integer

\begin{cor}
The set of algebraic integers in $\cc$ is a ring.
\end{cor}
\end{enumerate}
\begin{proof}[Proof Of Applications and Corollary]
$\alpha\in\cc$ is algebraic $\iff\exists$monic $f\in\qq[x]$ such
that $\deg f=n\geq1$ and $f(\alpha)=0\iff\exists M\subseteq\cc$
a finitely generated $\qq$-submodule of $\cc$ with $\alpha M\subseteq M$.
(For $\Rightarrow:$ $M=\qq[\alpha]=\qq+\qq\alpha+\qq\alpha^{2}+\dots+\qq\alpha^{n-1}$)

$\alpha\in\cc$ is an algebraic integer $\iff\exists$ monic $f\in\zz[x],\mbox{such that}\deg f=n\geq1\mbox{ and }f(\alpha)=0\iff M\subset\cc$
a finitely generated $\zz$-module with $\alpha M\subseteq M$ (Again
for $\Rightarrow$: $M=\zz[\alpha]=\zz+\zz\alpha+\dots+\zz\alpha^{n-1}$)

Let $R=\qq$ or $\zz$ and let $\alpha,\beta$ be \{algebraic numbers
or algebraic integers respectively\}, then $\alpha\pm\beta,\alpha\beta$
are also \{algebraic numbers, algebraic integers\}. Let the polynomial
of $\alpha\mbox{ be }f(x),\deg f=n$ and of $\beta\mbox{ be }g(x),\deg g=m$
with $f,g\in R[x]$ monic. Let $M$ be the $R$-submodule of $\cc$
generated by $\alpha^{i}\beta^{j},0\leq i\leq n-1,0\leq j\leq m-1$,
i.e., $M=\sum_{i,j}R\alpha^{i}\beta^{j}$. Clearly $\alpha M\subseteq M$
and $\beta M\subseteq M$. Then $(\alpha\pm\beta)M\subseteq M$ and
$\alpha\beta M\subseteq M$ quite clearly hence $\alpha\pm\beta$
are \{algebraic numbers, algebraic integers\}. Hence both sets are
subrings of $\cc$. If $\alpha$ is an algebraic number $\alpha\neq0$
then $\alpha^{-1}$ is also algebraic (easy) so \{algebraic numbers\}
is a subfield of $\cc$.\end{proof}
\begin{cor}
\label{cor:1}If $M$ is an finitely generated $R$-module and $J\lhd R$
such that $JM=M$ then $\exists r\in R$ such that $rM=0$ and $r\equiv1\mod J$
(i.e., $r-1\in J$)\end{cor}
\begin{proof}
Apply the proposition with $\varphi=$ identity map. So the proposition
tells us $(1+a_{1}+\dots+a_{n-1})M=0$ with $a_{i}\in J$. So let
$r=1+a_{1}+\dots+a_{n-1}$.\end{proof}
\begin{cor}[Nakayama's Lemma]
\label{cor:Nakayama's Lemma} If $M$ is a finitely generated $R$-module
and $I\lhd R$ such that $I\subseteq J(R)$. If $IM=M$ then $M=0$\end{cor}
\begin{proof}
By Corollary \ref{cor:1} $\exists r\in R$ such that $rM=0$ and
$r-1\in I\Rightarrow r-1\in J(R)$ but this implies (by Proposition
\ref{prop:J(R) iff unit} )$r\in R^{*}$ so $M=r^{-1}rM=0$\end{proof}
\begin{cor}
\label{cor:3}Let $M$ be finitely generated and $I\lhd R$ such that
$I\subseteq J(R)$. Let $N\leq M$. If $M=IM+N$ then $M=N$.\end{cor}
\begin{proof}
Apply Corollary \ref{cor:Nakayama's Lemma} to $M/N$ (which is still
finitely generated), using $I(M/N)=(IM+N)/N\,(*)$, since $M=IM+N\Rightarrow I(M/N)=M/N\Rightarrow M/N=0\Rightarrow M=N$.
To check $(*)$ holds we use the map $\phi:IM+N\to I(M/N)$ defined
by $am+n\mapsto a(m+N)$. $\phi$ is clearly surjective and has kernel
$=N$ (hence use the first isomorphism theorem)\end{proof}
\begin{cor}
\label{cor:Naka cor}Let $M$ be a finitely generated $R$-module,
where $R$ is a local ring with (unique) maximal ideal $P$ and residue
field $k=R/P$. Then 
\begin{enumerate}
\item $M/PM$ is a finite dimensional vector space over $k$
\item $x_{1},\dots,x_{n}$ generates $M$ as an $R$-module $\iff\overline{x_{1}},\dots,\overline{x_{n}}$
generates $M/PM$ as a $k$-vector space. (Here $\overline{x}=x+PM\in M/PM$) 
\end{enumerate}
\end{cor}
\begin{proof}
{}
\begin{enumerate}
\item $M/PM$ is an $R$-module which is annihilated by $P$ hence is a
module over $R/P=k$. 
\item ``$\Rightarrow$'': Clear. $\overline{x}\in M/PM$ $\Rightarrow\exists x_{i}\in R$
such that $x=\sum_{i=1}^{n}r_{i}x_{i}\Rightarrow\overline{x}=\sum_{i=1}^{n}r_{i}\overline{x_{i}}$.
(Note that this also proves the finite dimensional claim of part 1)\\
``$\Leftarrow$'': Let $x_{1},\dots,x_{n}\in M$ be such that $\overline{x_{1}},\dots,\overline{x_{n}}$
generates $M/PM$. Set $M=\sum_{i=1}^{n}Rx_{i}\leq M$. We want to
show $M=N$. We are going to use Corollary \ref{cor:3}, noting that
$J(R)=P$, with $I=P$. Then we can apply the Corollary if $M=PM+N$.
Let $x\in M$, then $\overline{x}\in M/PM$ so $\exists r_{i}$ such
that $\overline{x}=\sum r_{i}\overline{x_{i}}$ in $M/PM\Rightarrow x-\sum r_{i}x_{i}\in PM\Rightarrow x\in N+PM$
\end{enumerate}
\end{proof}
\begin{example*}
$R=\zz_{(5)}=\{\frac{a}{b}\in\qq\mid5\nmid b\}$. This is a local
ring with maximal ideal $P=5R$. We can check that $R/P\cong\zz/5\zz$.
Let $M=\qq$, but $P\qq=\qq\Rightarrow\qq/P\qq$ is $0$ but $\qq$
is not finitely generated as an $R$-module. (see exercise)
\end{example*}

\subsection{Exact Sequences}
\begin{defn}
Let $L,M,N$ be $R$-module. A sequence $\xymatrix{L\ar^{\alpha}[r] & M\ar^{\beta}[r] & N}
$ of $R$-module homomorphism is \emph{exact} if $\im(\alpha)=\ker(\beta)$.

Note: This implies $\beta\cdot\alpha=0$ ($\iff\im(\alpha)\subseteq\ker(\beta)$)\end{defn}
\begin{example*}
Key Examples:
\begin{itemize}
\item $\xymatrix{L\ar^{\alpha}[r] & M\ar[r] & 0}
$ is exact $\iff$$\alpha$ is surjective
\item $\xymatrix{0\ar[r] & M\ar^{\alpha}[r] & N}
$ is exact $\iff\alpha$ is injective
\item A longer sequence $\xymatrix{\dots\ar[r] & M_{i-1}\ar^{\alpha_{i-1}}[r] & M_{i}\ar^{\alpha_{i}}[r] & M_{i+1}\ar^{\alpha_{i+1}}[r] & \dots}
$ is exact $\iff\ker(\alpha_{i})=\im(\alpha_{i-1})\,\forall i$
\item \emph{Short Exact Sequence $\xymatrix{0\ar[r] & L\ar^{\alpha}[r] & M\ar^{\beta}[r] & N\ar[r] & 0}
$ }is exact $\iff$

\begin{itemize}
\item $\alpha$ is injective $(L\hookrightarrow M$)
\item $\beta$ is surjective (so $N\cong M/\ker\beta$)
\item $\im(\alpha)=\ker(\beta)$
\item That is $L\cong\alpha(L)\leq M$ and $M/\alpha(L)\cong N$
\end{itemize}
\end{itemize}
\end{example*}

\subsection{Tensor products of modules}

Let $R$ be a ring. Given two $R$-modules, $A,B$ we will define/construct
an $R$-module $C=A\otimes_{R}B$ with the following properties
\begin{enumerate}
\item $C$ is an $R$-module and there is an $R$-bilinear map $g:A\times B\to C$
\item (Universal property) For any $R$-bilinear map $f:A\times B\to D$
(with $D$ any $R$-module) there is a \emph{unique} $R$-linear map
$h:C\to D$ such that $f=h\circ g$ 
\[
\xyR{1pc}\xymatrix{ & C=A\otimes_{R}B\ar^{h}[dd]\\
A\times B\ar^{g}[ur]\ar_{f}[dr]\\
 & D
}
\]

\end{enumerate}
These properties uniquely determine $A\otimes_{R}B$ up to unique
isomorphism. This is because:
\begin{itemize}
\item Taking $D=C$ shows that $\id_{C}:C\to C$ is the only map such that
$g=\id_{C}\circ g$
\item If $D$ also satisfies 1., 2. then $\exists h_{1}:C\to D$ such that
$f=h_{1}\circ g$ and $\exists h_{2}:D\to C$ such that $g=h_{2}\circ f$.
Then we see that $f=h_{1}\circ h_{2}\circ f\Rightarrow h_{1}\circ h_{2}=\id_{D}$
and $g=h_{2}\circ h_{1}\circ g\Rightarrow h_{2}\circ h_{1}=\id_{C}$
\end{itemize}
\textbf{Existence:}

We construct $C$ as follows
\begin{itemize}
\item Take the free $R$-module $F$ with $A\times B$ as generating set
i.e. generators $(a,b)\,\forall a\in A,b\in B$. $F=\{\sum_{i=1}^{n}r_{i}(a_{i},b_{i})|r_{i}\in R,a_{i}\in A,b_{i}\in B\}$
\item Factor out the submodule $L$ consisting of all elements of the form
$(r_{1}a_{1}+r_{2}a_{2},b)-r_{1}(a_{1},b)-r_{2}(a_{2},b)$ and $(a,r_{1}b_{1}-r_{2}b_{2})-r_{1}(a,b_{1})-r_{2}(a,b_{2})\,\forall r_{1},r_{2}\in R,a,a_{1},a_{2}\in A,b,b_{1},b_{2}\in B$
\item Set $C=F/L$. Denote the image in $F/L$ of $(a,b)$ by $a\otimes b$.
Then $F/L$ is generated by $\{a\otimes b|a\in A,b\in B\}$ with ``relations''
$(r_{1}a_{1}+r_{2}a_{2})\otimes b=r_{1}(a_{1}\otimes b)+r_{2}(a_{2}\otimes b)$
and $a\otimes(r_{1}b_{1}+r_{2}b_{2})=r_{1}(a\otimes b_{1})+r_{2}(a\otimes b_{2})$
$(*)$
\end{itemize}
So each elements of $A\otimes_{R}B$ has the form $\sum_{i=1}^{n}r_{i}(a_{i}\otimes b_{i})$.
But (by $(*)$) we have $r(a\otimes b)=(ra)\otimes b=a\otimes(rb)$.
Using this, every element of $A\otimes_{R}B$ is a finite sum of ``atomic
tensors'' $a\otimes b$. Can we simplify these sums further? Not
in general! e.g. $a_{1}\otimes b_{1}+a_{2}\otimes b_{2}$ can not,
in general, be rewritten as a single ``atom'' $a\otimes b$.
\begin{example*}
If $A,B$ are both cyclic $R$-modules, say $A=Rx,B=Ry$ then every
$a\in A$ has the form $a=rx$ for some $r\in R$ and similarly every
$b\in B$ has the form $b=sy$ for some $s\in R$. Then $a\otimes b=rx\otimes sy=rs(x\otimes y)$.
A general element of $A\otimes_{R}B$ is thus a finite sum of $\sum_{i=1}^{n}t_{i}(x\otimes y)=t(x\otimes y)$
where $t=\sum_{i=1}^{n}t_{i}\in R$. Hence $A\otimes_{R}B$ is cyclic,
generated by $x\otimes y$\end{example*}
\begin{fact*}
More generally if $A,B$ are finitely generated by $x_{1},\dots,x_{n}$
for $A$ and $y_{1},\dots,y_{m}$ for $B$. Then $(\sum r_{i}x_{i})\otimes(\sum s_{j}y_{j})=\sum_{i,j}(r_{i}s_{j})(x_{i}\otimes y_{j})$.
Hence $A\otimes_{R}B$ is also finitely generated by $x_{i}\otimes y_{j}$\end{fact*}
\begin{xca*}
$R=k$ a field. $x_{1},\dots,x_{n}$ a basis for $A$ and $y_{1},\dots,y_{n}$
a basis for $B$ then the $x_{i}\otimes y_{j}$ are a basis for $A\otimes_{k}B$
and hence $\dim_{k}A\otimes_{k}B=mn=(\dim_{k}A)(\dim_{k}B)$
\end{xca*}
Similarly we can define $A\otimes_{R}B\otimes_{R}C$ for any three
$R$-modules $A,B,C$ and $A_{1}\otimes_{R}A_{2}\otimes_{R}\dots\otimes_{R}A_{n}$
for any $n$ $R$-modules $A_{1},\dots,A_{n}$. We get nothing essentially
new since $A\otimes_{R}B\otimes_{R}C$ turns out to be isomorphic
to $(A\otimes_{R}B)\otimes_{R}C$ and to $A\otimes_{R}(B\otimes_{R}C)$
\begin{lem}
\label{lem:Properties of tensor products}{}
\begin{enumerate}
\item $A\otimes_{R}B\cong B\otimes_{R}A$
\item $A\otimes_{R}R\cong A$
\item $(A\oplus B)\otimes_{R}C\cong(A\otimes_{R}C)\oplus(B\otimes_{R}C)$
\end{enumerate}
\end{lem}
\begin{proof}
{}
\begin{enumerate}
\item We have an $R$-bilinear map $A\times B\to B\otimes_{R}A$ via $(a,b)\mapsto b\otimes a$.
(Since $(r_{1}a_{1}+r_{2}a_{2},b)\mapsto b\otimes(r_{1}a_{1}+r_{2}a_{2})=r_{1}(b\otimes a_{1})+r_{2}(b\otimes a_{2})\leftarrow r_{1}(a_{1},b)+r_{2}(a_{2},b)$).
Hence there is a unique $R$-linear map $h_{1}:A\otimes_{R}B\to B\otimes_{R}A$
with $a\otimes b\mapsto b\otimes a$. Similarly we get $h_{2}:B\otimes_{R}A\to A\otimes_{R}B$
with $b\otimes a\mapsto a\otimes b$, hence $h_{1}\circ h_{2}=\id$
and $h_{2}\circ h_{1}=\id$
\item Define a map $A\times R\to A$ by $(a,r)\mapsto ra$. It is surjective
(take $r=1$) and $R$-bilinear, hence induces a map $f:A\otimes_{R}R\to A$
with $a\otimes r\mapsto ra$ surjective. Define $g:A\to A\otimes_{R}R$
by $g(a)=a\otimes1\in A\otimes_{R}R$. We can easily check that $f\circ g=\id_{A}$
and $g\circ f=\id_{A\otimes_{R}R}$ . 
\item Exercise
\end{enumerate}
\end{proof}
\begin{defn}
\emph{Tensoring maps }(i.e., $R$-module homomorphism): Let $f:A_{1}\to A_{2},g:B_{1}\to B_{2}$
be $R$-linear maps where $A_{1},A_{2},B_{1},B_{2}$ are $R$-modules.
Then there is an $R$-linear map $f\otimes g:A_{1}\otimes_{R}B_{1}\to A_{2}\otimes_{R}B_{2}$
which sends $a\otimes b\mapsto f(a)\otimes g(b)$. This is induced
by the $R$-bilinear map $A_{1}\times B_{1}\to A_{2}\otimes_{R}B_{2}$
which sends $(a_{1},b_{1})\mapsto f(a_{1})\otimes g(b_{1})$
\end{defn}

\subsection{Restriction and Extension of Scalars}


\subparagraph{Or: How we usually think about tensor products}

Let $f:R\to S$ be a ring homomorphism. Then every $S$-module becomes
an $R$-module via $rx=f(r)x$.
\begin{example*}
Special Cases:
\begin{enumerate}
\item $S$ is an $R$-module ($rs=f(r)s$)
\item $R$ a subring of $S$ and $f$ the inclusion map $R\hookrightarrow S$.
Then every $S$-module \uline{is} an $R$-module too.

\begin{example*}
If $K,L$ are fields with $K\subset L$ (i.e., $L$ is an extension
of $K$) then $L$-vector space is a $K$-vector space. (Restriction
of scalars). In particular $L$ is a vector space over $K$. $\dim_{K}L$
is the \emph{degree }of the extension ($\leq\infty$). 

Standard Fact: If $L\supset K\supset F$ (fields) and $L$ is a finite
extension of $K$ and $K$ is finite over $F$ then $L$ is finite
over $F$.
\end{example*}
\end{enumerate}
\end{example*}
\begin{prop}
Let $f:R\to S$ be as above. If $M$ is a finitely generated $S$-module
and $S$ is a finitely generated $R$-module then $M$ is a finitely
generated $R$-module.\end{prop}
\begin{proof}
Straightforward
\end{proof}
We are now going to try to go the other way. Let $f:R\to S$ and $M$
be an $R$-module. Let $M_{S}=S\otimes_{R}M$, this is an $R$-module.
It can be made into an $S$-module via $s'(s\otimes m)=(s's)\otimes m$.
(The $R$-module structure of $M_{S}$ can be done in two ways $r(s\otimes m)=(f(r)s)\otimes m=s\otimes rm$).
If $R=S$ and $f=\id$ we just get $R\otimes_{R}M\cong M$ ($=M_{R}$)
\begin{defn}
We say that $M_{S}$ is obtained from $M$ by \emph{extension of scalars}\end{defn}
\begin{rem*}
If $\{x_{i}\}_{i\in I}$ generates $M$ as an $R$-module then $\{1\otimes x_{i}\}_{i\in I}$
generates $M_{S}$ as an $S$-module. i.e., $M=\sum_{i\in I}Rx_{i}\Rightarrow M_{S}=\sum_{i\in I}S(1\otimes x_{i})$.
By abuse of notation we often just write $M_{S}=\sum_{i\in I}Sx_{i}$
where $\sum s_{i}x_{i}$ is shorthand for $\sum s_{i}\otimes x_{i}$.\end{rem*}
\begin{example*}
{}
\begin{enumerate}
\item $\qq(i)\otimes_{\qq}\rr\cong\cc$. $\qq(i)$ is generated as $\qq$-module
by $1,i$ hence $\qq(i)\otimes_{\qq}\rr$ is generated as an $\rr$-module
by $1\otimes1,i\otimes1$. And we abbreviate $x(1\otimes1)+y(1\otimes i)$
as $x+yi$ where $x,y\in\rr$.
\item Let $R$ and $S$ be two ring with $f:R\to S$ is the ``structure
map'' giving $S$ the structure of an $R$-module. Then $R[x]\otimes_{R}S\cong S[x]$.
Strictly: elements of the left side are polynomials in $x\otimes1$
\item $R^{n}\otimes_{R}S\cong S^{n}$. If $e_{1},\dots,e_{n}$are the ``standard''
generators $(1,0,\dots,0),\dots,(0,\dots0,1)$ for $R^{n}$ then $R^{n}\otimes_{R}S$
is freely generated by $e_{i}\otimes1$.
\end{enumerate}
\end{example*}

\subsection{Algebras}
\begin{defn}
{}
\begin{enumerate}
\item Let $R$ be a ring. An \emph{$R$-algebra }is a ring $A$ with a ring
homomorphism $f:R\to A,$ which turns $A$ into an $R$-module. (via
$ra=f(r)a$)
\item Conversely if $A$ is both a ring and an $R$-module ($(r,a)\mapsto r\cdot a$)then
it is an $R$-algebra if the two structures of $A$ are compatible,
i.e.:

\begin{itemize}
\item $(r_{1}+r_{2})\cdot a=r_{1}\cdot a+r_{2}\cdot a$
\item $r_{1}(r_{2}\cdot a)=(r_{1}r_{2})\cdot a$
\item $1\cdot a=a$
\item $r\cdot(a_{1}a_{2})=(r\cdot a_{1})a_{2}=a_{1}\cdot(ra_{2})$
\end{itemize}
\end{enumerate}
We recover the structure map $f:R\to A$ by setting $f(r)=r\cdot1_{A}\in A$.
\end{defn}
To go from one definition to the other: $1\Rightarrow2$: Define $r\cdot a=f(r)a$
(show that this satisfy the axiom given).

$2\Rightarrow1$: Define $f(r)=r\cdot1_{a}\in A$ (Show that this
does give a ring homomorphism)
\begin{defn}
Let $A,B$ be $R$-algebra with structure maps $f:R\to A,g:R\to B$.
Then an \emph{$R$-algebra homomorphism} from $A\to B$ is a map $h:A\to B$
which is both a ring homomorphism and $R$-linear such that $g=h\circ f$
\end{defn}
\[
\xyR{0.8pc}\xymatrix{ & A\ar^{h}[dd]\\
R\ar^{f}[ur]\ar_{g}[dr]\\
 & B
}
\]
\begin{eqnarray*}
h(a_{1}+a_{"}) & = & h(a_{1})+h(a_{2})\\
h(ra) & = & rh(a)\,\forall a\in A,r\in R\\
 & \iff & h(f(r)a)=g(r)h(a)\\
 & \iff & h(f(r))h(a)=g(r)h(a)\\
 & \iff & h(f(r))=g(r)\\
 & \iff & h\circ f=g
\end{eqnarray*}
What we have proved: A ring homomorphism $h:A\to B$ is an $R$-module
homomorphism $\iff h\circ f=g$

\textbf{Special Cases:}
\begin{enumerate}
\item $R=k$ a field, $A\neq0$ then the structure map $f:k\to A$ must
be injective $(f(1_{k})=1_{A}$ so $f\neq0$). So $A$ is a ring with
$k$ as a subring. 

\begin{example*}
$A=k[X]$ is a $k$-algebra, $\cc$ is an $\rr$-algebra (and a $\qq$-algebra)
\end{example*}
\item $R=\zz$. Any ring $A$ is a $\zz$-algebra whose structure map is
the unique ring homomorphism $\zz\to A$, $n\mapsto n\cdot1_{A}=\underset{n>0}{\underbrace{1+1+\dots+1}}$
\item $k$ a field. Extension fields of $k$ are $k$-algebra. If $k\subset L_{1},k\subset L_{2}$
($L_{1},L_{2}$ are fields). Then a map $h:L_{1}\to L_{2}$ is a $k$-algebra
homomorphism if it is a ring homomorphism (necessarily injective)
such that $h(x)=x\,\forall x\in k$.
\[
\xyR{0.8pc}\xymatrix{ & L_{1}\ar^{h}[dd]\\
k\ar@{^{(}->}[ur]\ar@{^{(}->}[dr]\\
 & L_{2}
}
\]

\end{enumerate}

\subsection{Finite conditions}

Let $A$ be an $R$-algebra.
\begin{defn}
$A$ is a \emph{finite }$R$-algebra if it is finitely generated as
an $R$-module, i.e., $\exists a_{1},\dots,a_{2}\in A$ such that
$A=Ra_{1}+\dots+Ra_{n}$

$A$ is a \emph{finitely generated $R$-}algebra if there is a surjective
ring homomorphism $R[x_{1},\dots x_{n}]\to A$ for some $n$ defined
by $x_{i}\mapsto a_{i}$. Denote this by $A=R[a_{1},\dots,a_{n}]$.
Hence every element of $A$ is a \uline{polynomial} in the finite
set $a_{1},\dots,a_{n}$\end{defn}
\begin{example*}
$A=R[x]$ is a finitely generated $R$-algebra (generator $=x$),
but it is not a finite $R$-algebra since it is \uline{not} finitely
generated as an $R$-module. (it is generated by $1,x,x^{2},\dots$
but not by any finite set of polynomials)

If $\alpha\in\cc$ then $\qq[\alpha]$ is a finitely generated $\qq$-algebra,
and is a finite $\qq$-algebra $\iff\alpha$ is an algebraic number.

$A=\zz[\alpha]$ is finitely generated $\zz$-algebra, and is a finite
$\zz$-algebra $\iff\alpha$ is an algebraic integer.
\end{example*}

\subsection{Tensoring Algebras}

Let $A,B$ be $R$-algebras with structure maps $f:R\to A,g:R\to B$.
The $R$-module $C=A\otimes_{R}B$ may be turned into a ring and hence
an $R$-algebra by setting $(a_{1}\otimes b_{1})(a_{2}\otimes b_{2})=a_{1}a_{2}\otimes b_{1}b_{2}$.
(extended by linearity)
\begin{proof}[Proof that this is well defined and turns $A\otimes_{R}B$ into a
ring]
 Map $A\times B\times A\times B\to C$ by $(a_{1},b_{1},a_{2},b_{2})\mapsto a_{1}a_{2}\otimes b_{1}b_{2}$.
This is clearly $R$-multilinear and hence induces an $R$-linear
map from $(A\otimes_{R}B)\otimes_{R}(A\otimes_{R}B)\to C$, i.e, $C\otimes_{R}C\to C$
is a well defined map, which in turns gives our multiplication. $1_{C}=1_{A}\otimes1_{B}$
and $0_{C}=0_{A}\otimes0_{B}$. Checking $C$ is a ring is straightforward.
The structure map $R\to C$ is $r\mapsto r\cdot(1\otimes1)=1\otimes g(r)=f(r)\otimes1$

\selectlanguage{english}%
\[
\xyC{1.2pc}\xymatrix{ & A\ar^{\id\otimes1\mathrm{:}a\mapsto a\otimes1}[dr]\\
R\ar^{f}[ur]\ar_{g}[dr]\ar[rr] &  & C=A\otimes_{R}B\\
 & B\ar_{1\otimes\id:b\mapsto1\otimes b}[ur]
}
\]


\newpage{}
\end{proof}
\selectlanguage{english}%

\section{Localization}


\paragraph{Rings and Modules of Quotients}

Recall: If $R$ is an \uline{integral domain} then we construct
its field of fractions as follows: take the set of ordered pairs $(r,s),r\in R,s\in R\setminus\{0\}$
with equivalence relation $(r_{1},s_{1})\sim(r_{2},s_{2})\iff r_{1}s_{2}=r_{2}s_{1}$.
Denote the class of $(r,s)$ by $\frac{r}{s}$. Define ring operations
by via the usual formulas $\frac{r_{1}}{s_{1}}+\frac{r_{2}}{s_{2}}=\frac{r_{1}s_{2}+r_{2}s_{1}}{s_{1}s_{2}}$.
Lots of checking of well-defined-ness and axioms shows that this is
a field $K$. $0=\frac{0}{1},1=\frac{1}{1},\frac{r}{s}=0\iff r=0$
so we get $R\hookrightarrow K$ by $r\mapsto\frac{r}{1}$, so if $\frac{r}{s}\neq0\Rightarrow\frac{s}{r}\in K$
and $\frac{r}{s}\frac{s}{r}=\frac{1}{1}$
\selectlanguage{british}%
\begin{defn}
A \emph{multiplicatively closed set} (MCS) in a ring $R$ is a subset
$S$ of$R$ such that:
\begin{enumerate}
\item $1\in S$
\item $s_{1},s_{2}\in S\Rightarrow s_{1}s_{2}\in S$
\end{enumerate}
\end{defn}
We'll often assume $0\notin S$
\begin{example*}
If $R$ is an integral domain, $S=R\setminus\{0\}$.

$R$ any ring, $P$ prime ideal of $R$, $S=R\setminus P$
\end{example*}
Given a MCS $S$ take the set of pairs $R\times S$ with the relation:
$(r_{1},s_{1})\sim(r_{2},s_{2})\iff\exists s\in S$ such that $s(r_{1}s_{2}-r_{2}s_{1})=0$.
This is an equivalence relation: Reflexivity and Symmetry are trivial.
For Transitivity: $(r_{1},s_{2})\sim(r_{2},s_{2})$ and $(r_{2},s_{2})\sim(r_{3},s_{3})\Rightarrow\exists s,t\in S$
such that $s(r_{1}s_{2}-r_{2}s_{1})=0,t(r_{2}s_{3}-r_{3}s_{2})=0\Rightarrow s_{2}st(r_{1}s_{3}-r_{3}s_{1})=sts_{1}r_{2}s_{3}-sts_{3}r_{2}s_{1}=0$.

Let $S^{-1}R=\{\frac{r}{s}:r\in R,s\in S\}$ where $\frac{r}{s}$
is the equivalence class of $(r,s)$. So $\frac{r_{1}}{s_{1}}=\frac{r_{2}}{s_{2}}\iff s(r_{1}s_{2}-r_{2}s_{1})=0$
for some $s\in S$. This forms a ring under the usual addition and
multiplication of fractions. (Check ring axioms + well-defined-ness).
$0_{S^{-1}R}=\frac{0}{1}$, $1_{S^{-1}R}=\frac{1}{1}$ and we have
a ring homomorphism $f:R\to S^{-1}R$ defined by $r\mapsto\frac{r}{1}$
which is not injective in general. $\frac{r_{1}}{1}=\frac{r_{2}}{1}\iff\exists s\in S$
such that $s(r_{1}-r_{2})=0$, i.e., $r_{1}-r_{2}\in\{r\in R:rs=0$
for some $s\in S\}=\ker(f)\lhd R$.
\begin{note*}
$f(s)$ is a \uline{unit} in $S^{-1}R$: since $f(s)=\frac{s}{1}$
and $\frac{s}{1}\frac{1}{s}=\frac{1}{1}=1$. \end{note*}
\begin{prop}
Let $S$ be a MCS in $R$ and $f:R\to S^{-1}R$ as above. If $g:R\to R'$
is a ring homomorphism such that $g(s)$ is a unit in $R'$ for all
$s\in S$ then there is a unique map $h:S^{-1}R\to R'$ such that
$g=h\circ f$
\[
\xyR{1pc}\xymatrix{ & S^{-1}R\ar^{h}[dd]\\
R\ar^{f}[ur]\ar_{g}[dr]\\
 & R'
}
\]
 ``$g$ factors through $h$''\end{prop}
\begin{proof}
Uniqueness: Suppose such an $h$ exists. Let $\frac{r}{s}\in S^{-1}R$,
$\frac{s}{1}\frac{r}{s}=\frac{r}{1}\Rightarrow h(\frac{s}{1})h(\frac{r}{s})=h(\frac{r}{1})$
but $h(\frac{r}{1})=h(f(r))=g(r)\Rightarrow g(s)h(\frac{r}{s})=g(r)\Rightarrow h(\frac{r}{s})=g(r)g(s)^{-1}$

Existence: Define $h:S^{-1}R\to R'$ by $h(\frac{r}{s})=g(r)g(s)^{-1}$.
It it well-defined? $\frac{r_{1}}{s_{1}}=\frac{r_{2}}{s_{2}}\Rightarrow s(r_{1}s_{2}-r_{2}s_{1})=0$
for some $s\in S\Rightarrow g(s)(g(r_{1})g(s_{2})-g(r_{2})g(s_{1}))=0\Rightarrow g(r_{1})g(s_{2})=g(r_{2})g(s_{1})$
(Since $g(s)$ is a unit) $\Rightarrow g(r_{1})g(s_{1})^{-1}=g(r_{2})g(s_{2})^{-1}$
(again because $g(s_{1})$ and $g(s_{2})$ are units). It is easy
to check that $h$ is a ring homomorphism. $h(f(r))=h(\frac{r}{1})=g(r)g(1)^{-1}=g(r)$
$\forall r\in R\Rightarrow h\circ f=g$
\end{proof}
So the pair $(S^{-1}R,f)$ with $f:R\to S^{-1}R$ is determined up
to isomorphism by: 
\begin{enumerate}
\item $s\in S\Rightarrow f(s)$ is a unit
\item $f(r)=0\iff rs=0$ for some $s\in S$
\item $S^{-1}R=\{f(r)f(s)^{-1}|r\in R,s\in S\}$\end{enumerate}
\begin{example*}
{}
\begin{enumerate}
\item $P\lhd R$ prime ideal and $S=R\setminus P$. Set $R_{P}=S^{-1}R$
in this case. ``\emph{the localization of $R$ at $P$}''. $f:R\to R_{P}$,
$r\mapsto\frac{r}{1}$, the extension of $P$ to $R_{P}$ is $PR_{P}=\{\frac{r}{s}:r\in P,s\notin P\}$
which is the set of non-units in $R_{P}$. So this is the unique maximal
ideal in $R_{P}$, so $R_{P}$ is a local ring.\\
Special Case:

\begin{enumerate}
\item $R$ an integral domain, $P=0\mbox{ then }R_{P}$ is the field of
fractions of $R$. (e.g., $R=\zz$ then $R_{P}=\qq$)
\item $R=\zz$, $P=p\zz$ ($p$ a prime number) $\Rightarrow R_{P}=\zz_{(p)}=\{\frac{r}{s}\in\qq:r\in\zz,s\in\zz\setminus p\zz\}\subseteq\qq$\\
Let $f\in\zz$. Write $f(p)$ to be the image of $f$ in $\zz/p\zz=\ff_{p}$.
Then $p$ is a zero of $f\iff f(p)=0\iff f\in p\zz$. What about $f\in\qq$?
Write $f=\frac{r}{s}$, $f(p)=\begin{cases}
r(p)s(p)^{-1} & \mbox{if }p\nmid s\,(\iff s(p)\neq0)\\
\infty & \mbox{otherwise}
\end{cases}$. So $f$ gives a function on $\Spec\zz$ with $f(p)\in\begin{cases}
\ff_{p}\cup\{\infty\} & \mbox{if }p\,\mbox{is a prime}\\
\qq & \mbox{if\,}p=0
\end{cases}$
\item $R=k[x_{1},\dots,x_{n}]$ where $k$ is an algebraically closed field
(e.g., $k=\cc$). $M\lhd R,M=(x_{1}-a_{1},\dots,x_{n}-a_{n})$ where
$(a_{1},a_{2},\dots,a_{n}):=\underline{a}\in k^{n}$.

\begin{note*}
{}
\begin{enumerate}
\item $M$ is $\ker(\mbox{eval}_{\underline{a}}:R\to k\mbox{ defined by }f\mapsto f(\underline{a}))\Rightarrow M$
is maximal since $R/M\cong k$
\item Every maximal ideal of $R$ has this form (by the Hilbert's Nullstellensatz)
\end{enumerate}
\end{note*}

$R\subset R_{M}\subset k(x_{1},\dots,x_{n})$ and $R_{M}=\{\frac{f}{g}:f,g\in R,g(\underline{a})\neq0\}=$
subring of $k(x_{1},\dots,x_{n})$ consisting of rational functions
which are ``defined at $\underline{a}$''. The unique maximal ideal
in $R_{m}$ is $MR_{M}=\{\frac{f}{g}:f(\underline{a})=0,g(\underline{a})\neq0\}$.
Finally $R_{M}/MR_{M}\cong k=R/M$

\end{enumerate}
\item $0\in S\Rightarrow S^{-1}R=0$ (The zero ring)
\item If $S\subset R^{\times}$ then $f:R\to S^{-1}R$ is an isomorphism
(and conversely)
\item $f\in R,S=\{1,f,f^{2},\dots\}$ then $S^{-1}R$ is denoted $R_{f}=\{\frac{r}{f^{n}}|r\in R,n\geq0\}$

\begin{example*}
$R=\zz,f=2$, $R_{f}=\zz[\frac{1}{2}]$
\end{example*}
\end{enumerate}
\end{example*}

\subsection{Localization of Modules}

Given an $R$-module $M$ and a multiplicatively closed set $S\subset R$,
let $S^{-1}M=\{$equivalence classes: $\frac{m}{s}$ of pairs $(m,s)$
with $m\in M,s\in S$ modulo the relation $(m,s)\sim(m',s')\iff r(sm'-s'm)=0$
for some $t\in S\}$. Define $\frac{m_{1}}{s_{1}}+\frac{m_{2}}{s_{2}}=\frac{s_{2}m_{1}+s_{1}m_{2}}{s_{1}s_{2}}$
and $\frac{r}{s_{1}}\frac{m}{s_{2}}=\frac{rm}{s_{1}s_{2}}$. This
turns $S^{-1}M$ into an $S^{-1}R$-module.

Also if $\phi:M\to N$ is an $R$-linear map then we define $S^{-1}\phi:S^{-1}M\to S^{-1}N$
by $(S^{-1}\phi)(\frac{m}{s})=\frac{\phi(m)}{s}$. This is an $S^{-1}R$-linear
map. 

If we have $M_{1}\overset{\psi}{\to}M_{2}\overset{\phi}{\to}M_{3}$
is a sequence of $R$-linear map then $S^{-1}(\phi\psi)=(S^{-1}\phi)(S^{-1}\psi):S^{-1}M_{1}\to S^{-1}M_{3}$
since they both map $\frac{m}{s}\to\frac{\phi(\psi(m))}{s}\,\forall\frac{m}{s}\in S^{-1}M_{1}$
\begin{prop}
If $M_{1}\overset{\psi}{\to}M_{2}\overset{\phi}{\to}M_{3}$ is an
exact sequence of $R$-modules then $S^{-1}M_{1}\overset{S^{-1}\psi}{\to}S^{-1}M_{2}\overset{S^{-1}\phi}{\to}S^{-1}M_{3}$
is an exact sequence of $S^{-1}R$-modules.\end{prop}
\begin{proof}
We need to prove that: $\im\psi=\ker\phi\Rightarrow\im(S^{-1}\psi)=\ker(S^{-1}\phi)$

$\im\psi\subseteq\ker\phi\Rightarrow\phi\psi=0\Rightarrow(S^{-1}\phi)(S^{-1}\psi)=S^{-1}(\phi\psi)=S^{-1}0=0\Rightarrow\im(S^{-1}\psi)\subseteq\ker(S^{-1}\phi)$

Conversely: Let $\frac{m_{2}}{s}\in\ker(S^{-1}\phi)$. Then $0=\frac{\phi(m_{2})}{s}$
so $\exists t\in S$ such that $t\phi(m_{2})=0\Rightarrow\phi(tm_{2})=0$.
So $\exists m_{1}\in M_{1}$ such that $tm_{2}=\psi(m_{1})$. Now
$\frac{m_{1}}{ts}\overset{S^{-1}\psi}{\mapsto}\frac{\psi(tm_{1})}{ts}=\frac{tm_{2}}{ts}=\frac{m_{2}}{s}$.
So $\frac{m_{2}}{s}\in\im(S^{-1}\psi)$ as required.
\end{proof}
\uline{Special Case}: $M_{1}=0$, i.e., $\phi$ injective: If $M\leq N$
then $S^{-1}M\leq S^{-1}N$
\begin{cor}
Let $N,N_{1},N_{2}$ be $R$-modules of $M$. Then:
\begin{enumerate}
\item $S^{-1}(N_{1}+N_{2})=S^{-1}N_{1}+S^{-1}N_{2}$ (as submodules of $S^{-1}M$)
\item $S^{-1}(N_{1}\cap N_{2})=S^{-1}N_{1}\cap S^{-1}N_{2}$ (as submodules
of $S^{-1}M$)
\item $S^{-1}(M/N)\cong S^{-1}M/S^{-1}N$
\end{enumerate}
\end{cor}
\begin{proof}
{}
\begin{enumerate}
\item Trivial: Both sides consist of elements of $\frac{x_{1}+x_{2}}{s}=\frac{x_{1}}{s}+\frac{x_{2}}{s}$
($x_{i}\in N_{i},s\in S$), and $\frac{x_{1}}{s_{1}}+\frac{x_{2}}{s_{2}}=\frac{s_{2}x_{1}+s_{1}x_{2}}{s_{1}s_{2}}$,
the numerator is in $N_{1}+N_{2}$ and denominator in $S$, hence
the whole fraction is in $S^{-1}(N_{1}+N_{2})$
\item Exercise
\item Apply the proposition to the short exact sequence $0\to N\to M\to M/N\to0$
to get that $0\to S^{-1}N\to S^{-1}M\to S^{-1}(M/N)\to0$ is exact
then by first isomorphism theorem $S^{-1}(M/N)\cong S^{-1}M/S^{-1}N$
\end{enumerate}
\end{proof}
\begin{prop}
$S^{-1}M\cong S^{-1}R\otimes_{R}M$ via the map $\frac{r}{s}\otimes m\mapsto\frac{rm}{s}$.
That is $S^{-1}M$ is obtain via ``extension of scalars'' using
the standard map $f:R\to S^{-1}R$ as the structure map\end{prop}
\begin{proof}
Map $S^{-1}R\times M\to S^{-1}M$ by $(\frac{r}{s},m)\mapsto\frac{rm}{s}$.
This is bilinear so it induces a well defined map $g:S^{-1}R\otimes_{R}M\to S^{-1}M$
as in the theorem. We check $g$ is an isomorphism.

$g$ is surjective: $g(\frac{1}{s}\otimes m)=\frac{m}{s}$

Observe that every element of $S^{-1}R\otimes_{R}M$ has the form
$\frac{1}{s}\otimes m$ since $\sum_{i=1}^{n}\frac{r_{i}}{s_{i}}\otimes m_{i}=\sum_{i=1}^{n}\frac{r_{i}'}{s}\otimes m_{i}$
where $s=s_{1}s_{2}\dots s_{n}$. But $\sum_{i=1}^{n}\frac{r_{i}'}{s}\otimes m_{i}=\sum_{i=1}^{n}\frac{1}{s}\otimes r_{i}'m_{i}=\frac{1}{s}\otimes\left(\sum_{i=1}^{n}r_{i}'m_{i}\right)$.
Now we show $g$ is injective. Suppose $g(\frac{1}{s}\otimes m)=0\Rightarrow\frac{m}{s}=0\Rightarrow\exists t\in S$
such that $tm=0$. Now $\frac{1}{s}\otimes m=\frac{t}{ts}\otimes m=\frac{1}{ts}\otimes tm=\frac{1}{ts}\otimes0=0$.
Hence $g$ is injective.\end{proof}
\begin{prop}
Let $M,N$ be $R$-modules and $S$ a MCS. Then $S^{-1}M\otimes_{s^{-1}R}S^{-1}N\cong S^{-1}(M\otimes_{R}N)$
(as $S^{-1}R$-modules)\end{prop}
\begin{proof}
{}

\begin{eqnarray*}
S^{-1}M\otimes_{S^{-1}R}S^{-1}N & \cong & (M\otimes_{R}S^{-1}R)\otimes_{S^{-1}R}S^{-1}N\mbox{ by the preceding proposition}\\
 & \cong & M\otimes_{R}(S^{-1}R\otimes_{S^{-1}R}S^{-1}N)\mbox{ by associativity}\\
 & \cong & M\otimes_{R}S^{-1}N\mbox{ by Lemma \ref{lem:Properties of tensor products}}\\
 & \cong & M\otimes_{R}(S^{-1}R\otimes_{R}N)\mbox{ by preceding proposition}\\
 & \cong & S^{-1}R\otimes_{R}(M\otimes_{R}N)\mbox{ rearranging terms}\\
 & \cong & S^{-1}(M\otimes_{R}N)\mbox{ by preceding proposition}
\end{eqnarray*}
 
\end{proof}
\uline{Special Case:} Let $P\lhd R$ be a prime ideal. Let $S=R\setminus P$
and denote $S^{-1}M$ by $M_{P}$. (which is a module over the local
ring $R_{P}=S^{-1}R$). Then $M_{P}\otimes_{R_{P}}N_{P}\cong(M\otimes_{R}N)_{P}$


\subsection{Local Properties}
\begin{defn}
A property of $R$-modules is called \emph{local }if: $M$ has the
property if and only if $M_{P}$ has the property $\forall P\in\Spec R$\end{defn}
\begin{prop}[Being zero is a local property]
 Let $M$ be an $R$-module. Then the following are equivalent:
\begin{enumerate}
\item $M=0$
\item $M_{P}=0$ for all prime $P\lhd R$
\item $M_{P}=0$ for all maximals $P\lhd R$
\end{enumerate}
\end{prop}
\begin{proof}
1$\Rightarrow$2$\Rightarrow3$ is trivial. To show $3\Rightarrow1$,
suppose $M\neq0$. Let $x\in M,x\ne0$, set $I=\Ann_{R}x=\{r\in R:rx=0\}\lhd R$,
$\neq R$ (as $1\notin I$), so there exists a maximal ideal $P\supseteq I$.
Then $\frac{x}{1}\in M_{P}$ is non-zero: for $\frac{x}{1}=0\iff sx=0$
for some $s\in R\setminus P$, which is a contradiction. \end{proof}
\begin{prop}
Let $\phi:M\to N$ be a homomorphism of $R$-modules. The following
are equivalent:
\begin{enumerate}
\item $\phi$ is injective
\item $\phi_{P}:M_{P}\to N_{P}$ is injective for all primes $P$
\item $\phi_{P}:M_{P}\to N_{P}$ is injective for all maximals $P$
\end{enumerate}
Moreover the same holds with ``injective'' replaced by ``surjective''
throughout.\end{prop}
\begin{proof}
Surjective case: $1\Rightarrow M\overset{\phi}{\to}N\to0$ is exact
$\Rightarrow M_{P}\overset{\phi_{P}}{\to}N_{P}\to0$ is exact for
all primes $P$$\Rightarrow\phi_{P}$ is surjective for all $P$$\Rightarrow2$.

$2\Rightarrow3$ is trivial

$3\Rightarrow1$: Let $N'=\phi(M)\leq N$. Then $M\to N\to N/N'\to0$
is exact. $\Rightarrow M_{P}\overset{\phi_{P}}{\to}N_{P}\to(N/N')_{P}\to0$
is exact $\forall$ maximal $P$ $\Rightarrow(N/N')_{P}=0$ for all
maximal $P\Rightarrow N/N'=0$ (by previous proposition) $\Rightarrow N=N'$
hence $\phi$ is surjective.

(Injective case uses the same argument with the exact sequence $0\to M\to N$)
\end{proof}

\subsection{Localization of Ideals}

$R$ is a ring, $S$ a multiplicatively closed set $\subset R$, $f:R\to S^{-1}R$
defined by $r\mapsto\frac{r}{1}$. Recall that for $I\lhd R$ we have
$I^{e}=S^{-1}I=\{\frac{r}{s}:r\in I,s\in S\}\lhd S^{-1}R$. (We will
use $I\lhd R$ and $J\lhd S^{-1}R$)
\begin{note*}
Any \uline{finite} sum $\sum\frac{r_{i}}{s_{i}}$ can be put over
a common denominator\end{note*}
\begin{prop}
{}\label{prop:Properties of localisation}
\begin{enumerate}
\item Every ideal $J\lhd S^{-1}R$ is the extension of an ideal $I\lhd R$.
(Namely $J=J^{ce}$)
\item If $I\lhd R$ then $I^{ec}=\cup_{s\in S}(I:s)$; hence $I^{e}=(1)$
if and only if $I\cap S\neq\emptyset$.
\item If $I\lhd R$ then $I$ is the contraction of some ideal $J\lhd S^{-1}R$
if and only if no element of $S$ is a zero divisor in $R/I$.
\item The correspondence $P\leftrightarrow S^{-1}P$ gives an order-preserving
bijection between the prime ideals $P$ of $R$ which do not meet
$S$ and the prime ideals $S^{-1}P$ of $S^{-1}R$.
\item $S^{-1}$ commutes with sums, products, intersections and radicals:

\begin{enumerate}
\item $S^{-1}(I_{1}+I_{2})=S^{-1}I_{1}+S^{-1}I_{2}$
\item $S^{-1}(I_{1}I_{2})=(S^{-1}I_{1})(S^{-1}I_{2})$
\item $S^{-1}(I_{1}\cap I_{2})=S^{-1}I_{1}\cap S^{-1}I_{2}$
\item $S^{-1}(r(I))=r(S^{-1}I)$
\end{enumerate}
\end{enumerate}
\end{prop}
\begin{proof}
{}
\begin{enumerate}
\item We always have $J\supseteq J^{ce}$. We prove the containment the
other way, let $\frac{r}{s}\in J\lhd S^{-1}R\Rightarrow\frac{r}{1}\in J\Rightarrow r\in J^{c}\Rightarrow\frac{r}{s}=\frac{1}{s}\frac{r}{1}\in(J^{c})^{e}$.
Hence $J=J^{ec}$.
\item {}
\begin{eqnarray*}
r\in I^{ec}=(S^{-1}I)^{c} & \iff & \frac{r}{1}=\frac{a}{s}\mbox{ for some }a\in I,s\in S\\
 & \iff & t(sr-a)=0\mbox{ for some }a\in I,s,t\in S\\
 & \iff & rs_{1}\in I\mbox{ for some }s_{1}\in S\\
 & \iff & r\in(I:s_{1})\mbox{ for some }s_{1}\in S\\
 & \iff & r\in\cup_{s\in S}(I:S)
\end{eqnarray*}
So $\underset{I^{e}=I^{ece}}{\underbrace{I^{e}=(1)\iff I^{ec}=(1)}}\iff1\in\cup_{s\in S}(I:s)\iff I\cap S\neq\emptyset$
\item $I$ is a contraction $\iff I^{ec}\subseteq I\iff(sr\in I$ for some
$s\in S\Rightarrow r\in I$) $\iff(\bar{s}\bar{r}=0$in $R/I$ for
some $s\in S\Rightarrow\bar{r}=0$) $\iff\forall s\in S,\bar{s}$
is not a zero divisor in $R/I$
\item One way is clear: If $Q$ is a prime of $S^{-1}R$ then $Q^{c}$ is
a prime of $R$. Conversely: let $P$ be a prime of $R\Rightarrow R/P$
is a domain. Now $\bar{S}^{-1}(R/P)\cong S^{-1}R/S^{-1}P$ (where
$\bar{S}$ is the image of $S$ in $R/P$). But $\bar{S}^{-1}(R/P)$
is a subring of the field of fractions of $R/P$, so is either $0$
or an integral domain. If $0$ then $S^{-1}P=S^{-1}R=(1)$. If $\neq0$
then $S^{-}P$ is a prime ideal of $S^{-1}R$. The first case occurs
$\iff0\in\bar{S}\iff S\cap P\neq\emptyset$.
\item Easy Exercise
\end{enumerate}
\end{proof}
\begin{rem*}
Here's a quick proof that $f\in R$ not nilpotent $\Rightarrow\exists P$
with $f\notin P$ and $P$ prime.

Take $S=\{1,f,f^{2},\dots\}\not\ni0$$\Rightarrow S^{-1}R$ is a non-zero
ring, so it has a maximal ideal $Q\Rightarrow Q^{c}=P$ is a prime
of $R,P\cap S=\emptyset\Rightarrow f\notin P$.\end{rem*}
\begin{cor}
$N(S^{-1}R)=S^{-1}(N(R))$
\end{cor}

\begin{cor}[Special case when $S=R\setminus P,P$ prime]
 $I\cap S=\emptyset\iff I\subseteq P$. Hence the proper ideals of
$R_{P}$ are in bijection with the ideals of $R$ which are contained
in $P$. 
\[
\xyR{1pc}\xymatrix{R\ar@{-}[d] & S^{-1}R\ar@{-}[ddd]\\
P\ar@{-}[dd]\\
-I\ar@{<->}[r] & J-\\
0 & 0
}
\]

\end{cor}

\begin{cor}
The field of fractions of the domain $R/P$ ($P$ is prime) is isomorphic
to the residue field of $R_{P}$\end{cor}
\begin{proof}
$S=R\setminus P$. The residue field of $R_{P}$ is $R_{P}/S^{-1}P=S^{-1}R/S^{-1}P=\bar{S}^{-1}(R/P)=$
field of fraction of $R/P$ since $\bar{S}=(R/P)\setminus\{0\}$.\end{proof}
\begin{cor}
If $P_{1}\subset P_{2}$ are primes of $R$ then $(R/P_{1})_{P_{2}}=R_{P_{2}}/P_{1_{P_{2}}}$
- a ring whose prime correspond to primes of $R$ \uline{between
}$P_{1}$ and $P_{2}$

\newpage{}
\end{cor}

\section*{Geometrical Interlude I}

Let $k$ be an algebraically closed field (e.g. $k=\cc$). Let $k^{n}$
be affine $n$-space over $k$: $\{\underline{a}=(a_{1},\dots,a_{n}):a_{j}\in k\}$.
Algebraic geometry studies solutions to polynomial equations $S=\{f_{j}(x_{1},\dots,n_{n})\}\subseteq k[x_{1},\dots,x_{n}]$.
$V(S)=\{\underline{a}\in k^{n}:f_{j}(\underline{a})=0\,\forall f_{j}\in S\}$. 
\begin{defn*}
The set $V(S)$ is an \emph{affine algebraic set}
\end{defn*}
Clearly $V(S)=V(I)$ where $I$ is the \uline{ideal} of $k[x_{1},\dots,x_{n}]$
generated by $S$ and $V(I)=V(r(I))$, since $f\in r(I)\iff f^{n}\in I$
($n\geq1$)
\begin{namedthm}[Hilbert Basis Theorem]
 Every ideal $I\lhd k[x_{1},\dots,x_{n}]$ is finitely generated\end{namedthm}
\begin{proof}
Later
\end{proof}
If $I=(f_{1},\dots,f_{k})$ then $V(I)=V(\{f_{1},\dots,f_{k}\})$.
It is not hard to check that:
\begin{itemize}
\item $V(0)=k^{n}$
\item $V(1)=\emptyset$
\item $V(\cup_{j}S_{j})=\cap_{j}V(S_{j})$
\item $V(IJ)=V(I)\cup V(J)$
\end{itemize}
Hence the collection of all algebraic subsets of $k^{n}$ is closed
under intersections and finite unions, so they form the \uline{closed}
sets of a topology on $k^{n}$ called the \emph{Zariski} \emph{topology}
on $k^{n}$.

In the other direction: let $S\subset k^{n}$ and define $I(S)=\{f\in k[x_{1},\dots,x_{n}]:f(\underline{a})=0\,\forall\underline{a}\in S\}$,
which is an ideal of $k[x_{1},\dots,x_{n}]$, and in fact $r(I(S))=I(S)$.
\begin{fact*}
$V(I(S))=\overline{S}$ (for $S\subset k^{n},\overline{S}$ is the
closure of $S$ in $k^{n}$)
\end{fact*}

\begin{fact*}
$I(V(J))=r(J)$ for $J\lhd k[x_{1},\dots,x_{n}]$. This is called
``Hilbert's Nullstellensatz'', we will prove this later.
\end{fact*}
The conclusion is that $V$ and $I$ gives (inclusion order-reversing)
bijections between radical ideals of $k[x_{1},\dots,x_{n}]$ and closed
subsets of $k^{n}$. 
\begin{defn}
An algebraic set is \emph{irreducible} if it is not the union of two
proper closed subsets. ($\iff$ any two non-empty open subsets intersects
non-trivially). These are $V(P)$ for $P$ a prime ideal of $k[x_{1},\dots,x_{n}${]}.
Irreducible algebraic sets are often called \emph{algebraic varieties}\end{defn}
\begin{example*}
{}
\begin{description}
\item [{$n=1$:}] $k^{n}=k^{1}=k$. Now $k[x]$ is a UFD so the primes
are $(0)$ and $(x-a)$ with $a\in k$ (since $k$ is algebraically
closed). Note $(x-a)$ are maximals and correspond to points of $k$
while $(0)$ is not maximal and correspond to the whole of $k$. The
closed sets are $k$ itself and all the \uline{finite} subsets
of $k$. (So every infinite subset of $k$ is dense)
\item [{$n=2$:}] $k[x_{1},x_{2}]=k[x,y]$. Primes have $3$ types: 

\begin{itemize}
\item $(0)\leftrightarrow V(0)=k^{2}$
\item $P=(f(x,y))\leftrightarrow V(f)=$ irreducible curves in $k^{2}$
($f$ irreducible). e.g., $V(x^{2}+y^{2}-1)=$ circle in $k^{2}$
\item $M=(x-a,y-b)\leftrightarrow V(M)=\{(a,b)\}$ singleton in $k^{2}$
($a,b\in k$)
\end{itemize}
\end{description}
\end{example*}

\subsection*{Coordinate rings (of algebraic sets)}

Every element $f\in k[x_{1},\dots,x_{n}]$ defines a polynomial function
$k^{n}\to k$ (defined by $\underline{a}\mapsto f(\underline{a})$).
$f,g$ agree on $V(I)\iff f-g\in I(V(I))$. Without loss of generality
we can assume $I=r(I)$ so $f,g$ agree on $V(I)\iff f-g\in I$.
\begin{defn*}
Define $k[V]=k[x_{1},\dots,x_{n}]/I$.\emph{ }Then $k[V]$ is the
ring of polynomial function on $V$. This is called the \emph{coordinate
ring of $V$.}
\end{defn*}
Ideals of $k[V]\leftrightarrow$ ideals $J$ with $I\subseteq J\lhd k[x_{1},\dots,x_{n}]$.
$M_{\underline{a}}=$ Maximal ideals of $k[V]\leftrightarrow$ maximal
ideals $M\supseteq I$, i.e., $M=(x_{1}-a_{1},\dots,x_{n}-a_{n})$
with $\underline{a}=(a_{1},\dots,a_{n})\in V$. $M_{\underline{a}}=\{\overline{f}\in k[V]:f(\underline{a})=0\}=$kernel
of map $k[V]\to k$ defined by $\overline{f}\mapsto f(\underline{a})$.

If $V$ is a variety then $k[V]$ is an integral domain, (since $V=V(P)$
so $K[V]=k[x_{1},\dots,k_{n}]/P$ where $P$ is prime)

We have a correspondence between
\begin{itemize}
\item Algebraic sets (or varieties) in $k^{n}$
\item finitely generated $k$-algberas (or domains)
\end{itemize}
This correspondence extends to one which takes polynomial maps \uline{between}
algebraic sets to morphism of $k$-algebras. 

\newpage{}


\section{Integral Dependence}
\begin{defn}
Let $A$ be a subring of the ring $B$. An element $b\in B$ is \emph{integral
over }$A$ if it satisfies an equation
\begin{equation}
b^{n}+a_{n-1}b^{n-1}+\dots+a_{0}=0,a_{i}\in A\label{eq:Integrak}
\end{equation}

\end{defn}
Let $f(x)=x^{2}+a_{n-1}x^{n-1}+\dots+a_{0}\in A[x]$. If $a\in A$
then $a$ is a root of $x-a$, so $a$ is integral over $A$
\begin{example*}
$A=\zz,B=\cc$, $z\in\cc$ is integral over $\zz\iff z$ is an algebraic
integer

$A=\qq,B=\cc$ gives algebraic numbers

$A=\zz,B=\qq,z\in\qq$ integral over $\zz\iff z\in\zz$, i.e., let
$x=\frac{r}{s},r,s\in\zz$ coprime. If $\left(\frac{r}{s}\right)^{n}+\dots+a_{0}=0$
then $r^{n}+a_{n-1}r^{n}s+\dots+a_{0}s^{n}=0\Rightarrow s|r^{n}\Rightarrow s=\pm1,x\in\zz$.

$A$ is a UFD, $B$ its field of fraction gives similar result as
the previous example.\end{example*}
\begin{thm}
Let $A$ be a subring of $B$, $b\in B$. Then the following are equivalent:
\begin{enumerate}
\item $b$ is integral over $A$
\item $A[b]$ is a finitely generated $A$-module
\item $B$ contains a subring $C\supseteq A[b]$ which is finitely generated
as an $A$-module
\item There exists a faithful $A[b]$-module $M$ which is finitely generated
as an $A$-module
\end{enumerate}
\end{thm}
\begin{proof}
$1\Rightarrow2$: If $b$ satisfies \eqref{Integrak} then $A[b]$
is generated by $1,b,\dots,b^{n-1}$ since \eqref{Integrak} $\Rightarrow b^{n}=-(a_{n-1}b^{n-1}+\dots+a_{0})$

$2\Rightarrow3$: Take $C=A[b]$

$3\Rightarrow4$: $M=C$. This is a faithful $A[b]$-module as $A[b]$
is a subring $C$ and $1\in C$. So if $rx=0\,\forall r\in A[b],x\in M=C$
then $r1=0$, hence $r=0$.

$4\Rightarrow1$: Given $M$ as in 4. let $m_{1},\dots,m_{n}$be generators
of $M$ as an $A$-module. Let $\phi:M\to M$ be the map defined by
$x\mapsto bx$. This is $A$-linear so $\phi\in\End_{A}(M)$. Hence
there exists $a_{0},\dots,a_{n-1}\in A$ such that $\phi^{n}+a_{n-1}\phi^{n-1}+\dots+a_{0}=0$
(in $\End_{A}(M)$), i.e., $(\phi^{n}+\dots+a_{0})y=0\,\forall y\in M\Rightarrow(b^{n}+a_{n-1}b^{n-1}+\dots+a_{0})y=0\,\forall y\in M\underset{M\mathrm{\, faithful}}{\Rightarrow}b^{n}+a_{n-1}b^{n-2}+\dots+a_{0}=0$\end{proof}
\begin{cor}
For all $n\geq1$, if $b_{1},\dots,b_{n}\in B$ are all integral over
$A$ then $A[b_{1},\dots,b_{n}]$ is finitely generated as an $A$-module. \end{cor}
\begin{proof}
We prove this using induction on $n$.

$n=1$: Use the previous theorem.

In general: Let $A_{1}=A[b_{1},\dots,b_{n-1}]$. Then $A_{1}$ is
finitely generated as an $A$ module. $A[b_{1},\dots,b_{n}]=A_{1}[b_{n}]$,
but $b_{n}$ is integral over $A_{1}$, hence $A_{1}[b_{n}]$ is finitely
generated as an $A_{1}$-module, so $A_{1}[b_{n}]$ is finitely generated
as an $A$-module\end{proof}
\begin{cor}
\label{cor:Def of integral closure}Let $C=\{b\in B|b$ integral over
$A\}\subseteq B$. Then $C$ is a subring of $B$ containing $A$.\end{cor}
\begin{proof}
We need to show that for all $x,y\in C$ then $x\pm y,xy\in C$. Since
$x,y\in C$ by the previous corollary we know $A[x,y]$ is finitely
generate as an $A$-module and it contains $x\pm y,xy$. By the previous
theorem (3.$\Rightarrow$1.) all elements of $A[x,y]$ are integral
over $A$\end{proof}
\begin{defn}
Using the notation of Corollary \corref{Def of integral closure},
$C$ is the \emph{integral closure} of $A$ in $B$.

If $C=B$ we say $B$ is \emph{integral} over $A$

If $C=A$ we say $A$ is \emph{integrally closed} in $B$\end{defn}
\begin{example*}
$\zz$ is integrally closed over $\qq$

The integral closure of $\zz$ in $\cc$ is the ring of algebraic
integers.\end{example*}
\begin{defn}
If $A$ is an integral domain, we say that $A$ is \emph{integrally
closed} if $A$ is integrally closed in its field of fractions.\end{defn}
\begin{example*}
$\zz$ is integrally closed

Any UFD is integrally closed\end{example*}
\begin{cor}
If $A\subseteq B\subseteq C$ then $C$ is integral over $A\iff B$
is integral over $A$ and $C$ is integral over $B$\end{cor}
\begin{proof}
``$\Rightarrow$'': Obvious

``$\Leftarrow$'': Let $c\in C$. Then $c^{n}+b_{n-1}c^{n-1}+\dots+b_{0}=0$,
$b_{i}\in B$. Define $B_{0}:=A[b_{0},\dots,b_{n-1}]$. Then $c$
is integral over $B_{0}$ and $B_{0}$ is finitely generated as an
$A$-module. By the theorem $c$ is integral over $A$\end{proof}
\begin{cor}
The integral closure of $A$ in $B$ is integrally closed in $B$\end{cor}
\begin{proof}
Trivially follows from previous corollary\end{proof}
\begin{example*}
Let $K$ be a number field (that is a field containing $\qq$ with
finite degree). Then the integral closure of $\zz$ in $K$ is the
ring of algebraic integers of $K$, called the \emph{ring of integers.
}That is, the ring of integers is $K\cap\{$ring of all algebraic
integers$\}$. We will denote this $\mathcal{O}_{K}$ (or $\zz_{K}$).
e.g.:
\begin{itemize}
\item $K=\qq(i)$, $\mathcal{O}_{K}=\zz[i]$ (the Gaussian integers)
\item $K=\qq(\sqrt{-3})$, $\mathcal{O}_{K}$ contains $\zz[\sqrt{-3}]$.
In fact $\mathcal{O}_{K}=\zz[\frac{1+\sqrt{-3}}{2}]$
\item $K=\qq(\sqrt[3]{10})$. The integral closure of $\zz$ in $K$ is
$\mathcal{O}_{K}=\zz[\frac{1+\sqrt[3]{10}+\sqrt[3]{100}}{3}]$
\end{itemize}
\end{example*}
\begin{prop}
\label{prop:Proposiiton}Let $B$ be an integral extension of $A$.
Then:
\begin{enumerate}
\item For all $J\lhd B,$ $I=J^{c}=J\cap A$ we have $B/J$ is integral
over $A/I$
\item If $S$ is a multiplicatively closed set in $A$ then $S^{-1}B$ is
integral over $S^{-1}A$.\\
\emph{Special Case: }$P$ a prime of $A$, $S=A\setminus P\Rightarrow B_{P}$
is integral over $A_{P}$
\end{enumerate}
\end{prop}
\begin{proof}
Let $b\in B$ satisfy $b^{n}+a_{1}b^{n-1}+\dots+a_{n}=0$ (in $B$)
$\Rightarrow\bar{b}^{n}+\bar{a_{1}}\bar{b}^{n-1}+\dots+\bar{a_{n}}=0$
(in $B/J$)$\Rightarrow\bar{b}$ is integral over $A/I$

$\frac{b}{s}\in S^{-1}B\Rightarrow\left(\frac{b}{s}\right)^{n}+\frac{a_{1}}{s}(\frac{b}{s})^{n-1}+\dots+\frac{a_{n-1}}{s^{n-1}}(\frac{b}{s})+\frac{a_{n}}{s^{n}}=0\Rightarrow\frac{b}{s}$
is integral over $S^{-1}A$\end{proof}
\begin{lem}
\label{lem:Lemma-1}Let $B$ be an integral extension of $A$, with
$A$ and $B$ both domains. Then $B$ is a field if and only if $A$
is a field\end{lem}
\begin{proof}
Assume $A$ is a field. Let $b\in B,b\neq0$. Let $b^{n}+a_{1}b^{n-1}+\dots+a_{n-1}b+a_{n}=0$
be an integral equation of \uline{minimal} degree $n$. Then $a_{n}\neq0$
so $a_{n}^{-1}$ exists in $A$. Hence the equation can be rewritten
as $b(b^{n-1}+\dots+a_{n-1})=-a_{n}\Rightarrow b^{-1}=-a_{n}^{-1}(b^{n-1}+\dots+a_{n-1})\in B$.
Hence $b$ as an inverse, so $B$ is a field.

Conversely suppose $B$ is a field. Let $a\in A,a\neq0$. Then $a^{-1}$
exists in $B$. So there is an equation: $(a^{-1})^{n}+a_{1}(a^{-1})^{n-1}+\dots+a_{n}=0$
($a_{i}\in A$), which can be rearranged to give $a^{-1}=-(a_{1}+a_{2}a+\dots+a_{n}a^{n-1})\in A$.\end{proof}
\begin{lem}
\label{lem:Lemma-2}Let $B$ be an integral extension of $A$. Let
$Q\lhd B$ be prime and $P=Q\cap A$, a prime of $A$. Then $P$ is
maximal if and only if $Q$ is maximal\end{lem}
\begin{proof}
By Proposition \ref{prop:Proposiiton} $B/Q$ is integral over $A/P$
so by Lemma \ref{lem:Lemma-1} $Q$ is maximal $\iff B/Q$ is a field
$\iff A/P$ is a field $\iff P$ is maximal\end{proof}
\begin{thm}
Let $B$ be an integral extension of $A$ and $P$ a prime of $A$.
Then:
\begin{enumerate}
\item There exists a prime $Q$ of $B$ with $P=Q\cap A$
\item If $Q_{1},Q_{2}$ are primes of $B$ with $Q_{1}\cap A=P=Q_{2}\cap A$
and $Q_{1}\supseteq Q_{2}$ then $Q_{1}=Q_{2}$.
\end{enumerate}
\end{thm}
\begin{proof}
Consider the following commutative diagram:
\[
\xymatrix{A\ar^{\mathrm{int}}@{^{(}->}[r]\ar_{\alpha}[d] & B\ar^{\beta}[d]\\
A_{P}\ar^{\mathrm{int}}@{^{(}->}[r] & B_{P}
}
\]
Let $M$ be a maximal ideal in $B_{P}$. Let $Q=\beta^{-1}(M)$, a
prime in $B$. Now $Q\cap A=P$ since $M\cap A_{P}$ is maximal in
$A_{P}$ (Lemma \ref{lem:Lemma-2}) but $A_{P}$ has only one maximal
ideal namely $PA_{P}$, which contracts to $P$: $\alpha^{-1}(M\cap A_{P})=P=A\cap\beta^{-1}(M)=A\cap Q$.

Let $Q_{1}$ and $Q_{2}$ be as in the statement. Then let $N_{1}=Q_{1}B_{P}$
and $N_{2}=Q_{2}B_{P}$ their extension in $B_{P}$. These are primes
of $B_{P}$ (by Proposition \ref{prop:Properties of localisation},
and the fact that $Q_{j}\cap S=\emptyset$ where $S=A\setminus P$).

\uline{Claim:} $N_{1},N_{2}$ are maximal.

This follow from $N_{j}\cap A_{P}$ are maximal (using Lemma \ref{lem:Lemma-2}),
but $N_{1}\cap A_{P}=N_{2}\cap A_{P}=PA_{P}$ since both contract
to $P$. Hence each $N_{j}$ is maximal. But if $Q_{1}\supseteq Q_{2}\Rightarrow N_{1}\supseteq N_{2}\Rightarrow N_{1}=N_{2}\Rightarrow Q_{1}=\beta^{-1}(N_{1})=\beta^{-1}(N_{2})=Q_{2}$\end{proof}
\begin{example*}[Counter-Example showing the requirement of part 2]
 $A=\zz,B=\zz[i],P=5\zz$, then if we let $Q_{1}=(2+i),Q_{2}=(2-i)$
we find $Q_{1}\cap\zz=5\zz$ and $Q_{2}\cap\zz=5\zz$\end{example*}
\begin{namedthm}[The ``Going Up'' Theorem]
Consider the following set-up. 
\[
\xymatrix{B & Q_{1}\subseteq\dots\subseteq Q_{m}\mbox{(primes of }B)\\
A\ar^{\mathrm{int}}@{^{(}->}[u] & P_{1}\subseteq P_{2}\subseteq\dots\subseteq P_{m}\subseteq\dots\subseteq P_{n}\mbox{ (primes of }A)
}
\]
 with $Q_{i}\cap A=P_{i}$ (for all $1\leq i\leq m$). With that set-up
there exists $Q_{m+1},\dots,Q_{n}$ primes of $B$ with $Q_{m}\subseteq Q_{m+1}\subseteq\dots\subseteq Q_{n}$
and $Q_{i}\cap A=P_{i}$ (for $m+1\leq i\leq n$)\end{namedthm}
\begin{proof}
By induction we reduce to the case $m=1,n=2$. That is we must find
$Q_{2}$ such that $Q_{1}\subseteq Q_{2}$ and $Q_{2}\cap A=P_{2}$.
(where $P_{1}\subseteq P_{2}$ and $Q_{1}\cap A=P_{1})$

Let $\bar{A}=A/P_{1},\bar{B}=B/Q_{1}$. Then $\bar{B}$ is integral
over $\bar{A}$ (by Proposition \ref{prop:Proposiiton}) and $P_{2}/P_{1}$
is a prime of $\bar{A}$ so there exists a prime of $\bar{B}$ above
it. This prime has the form $Q_{2}/Q_{1}$ with $Q_{2}\supseteq Q_{1}$
and $Q_{2}$ a prime of $B$. Then $(Q_{2}/Q_{1})\cap\bar{A}=P_{2}/P_{1}\Rightarrow Q_{2}\cap A=P_{2}$
\end{proof}

\subsection{Valuation Rings}
\begin{defn}
A \emph{valuation ring} is an integral domain $R$ such that for every
$x\in K$ (the field of fractions of $R$) either $x\in R$ or $x^{-1}\in R$\end{defn}
\begin{example*}
$\zz$ is \uline{not} a valuation ring ($\frac{2}{3}\notin\zz,\frac{3}{2}\notin\zz$)

$\zz_{(p)}$ is a valuation ring

$R=K$: any field is a valuation ring.\end{example*}
\begin{prop}
\label{prop:valuation ring properties}Let $R$ be a valuation ring
with field $K$. Then:
\begin{enumerate}
\item $R$ is a local ring
\item $R\subseteq R'\subseteq K$ $\Rightarrow R'$ is a valuation ring
\item $R$ is integrally closed
\end{enumerate}
\end{prop}
\begin{proof}
2. trivial

1. The units of $R$ are the (non-zero) $x\in K$ with \uline{both}
$x,x^{-1}\in R$. Let $M=\{$non-units in $R\}=\{x\in R:x^{-1}\notin R\}\cup\{0\}$.
We'll show that $M\lhd R$, then it's the unique maximal ideal of
$R$. Let $x\in M,r\in R$. Then $rx$ is not a unit since otherwise
$x^{-1}=r(rx)^{-1}\in R$, contradiction, i.e., $rx\in M$. Let $x,y\in M$
be non-zero. Then either $\frac{x}{y}\in R$ or $\frac{y}{x}\in R$.
If $\frac{x}{y}\in R$ then $x+y=y(\frac{x}{y}+1)\in M$. Otherwise
if $\frac{y}{x}\in R$ then $x+y=x(1+\frac{y}{x})\in M$

3. Let $x\in K$ be integral over $R$. Then $x^{n}+r_{1}x^{n-1}+\dots+r_{n}=0$
($r_{i}\in R$). If $x\in R$ there is nothing to prove. If $x^{-1}\in R$
then $x+(r_{1}+r_{2}x^{-1}+\dots+r_{n}(x^{-1})^{n-1})=0\Rightarrow x\in R$\end{proof}
\begin{defn}
Let $K$ be a field. A \emph{discrete valuation }on $K$ is a function
$v:K^{*}\to\zz$ such that:
\begin{enumerate}
\item $v(xy)=v(x)+v(y)\,\forall x,y\in K^{*}$
\item $v(x+y)\geq\min\{v(x),v(y)\}\,\forall x,y\in K^{*}$ with $x+y\neq0$
\end{enumerate}
\end{defn}
We extend $v$ to a function $K\to\zz\cup\{\infty\}$ by setting $v(0)=\infty$.
Now 1., 2. holds for all $x,y\in K$ with the obvious conventions. 
\begin{example}
$K=\qq$, $p$ a prime number, $v=\ord_{p}$ defined as follows: for
$x\in\qq^{*}$ write $x=p^{n}\frac{a}{b}$ where $a,b\in\zz$ and
$p\nmid a,b$ and $n\in\zz$. Set $\ord_{p}(x)=n$.
\end{example}
Associated to every discrete valuation of $K$ there is a valuation
ring $R_{v}$. $R_{v}=\{x\in K:v(x)\geq0\}$. Clearly $R_{v}$ is
a ring (by 1. and 2.). Also $R_{v}$ is a valuation ring since $v(x^{-1})=-v(x)$
for all $x\in K^{*}$.
\begin{defn}
These $R_{v}$ are called \emph{discrete valuation ring} (DVR)\end{defn}
\begin{example*}
$K=\qq$ has a DVR for each prime $p$, namely $v=\ord_{p}$ then
$R_{v}=\zz_{(p)}$.\end{example*}
\begin{note*}
$\cap_{p}\zz_{(p)}=\zz$\end{note*}
\begin{xca*}
Every valuation ring of $\qq$ is $\qq$ itself or $\zz_{(p)}$ for
some prime $p$.\end{xca*}
\begin{example*}
Let $K=k(x)$ where $k$ is a field. $K$ is the field of fractions
of $k[x]$. Let $p(x)$ be a monic irreducible polynomial in $k[x]$.
Every element of $K^{*}$ can be written as $p^{n}\frac{a}{b}$ where
$a,b\in k[x]$ and $p\nmid a,b$ with $n\in\zz$. In this case $n$
is uniquely determined. Define $\ord_{p}(p^{n}\frac{a}{b})=n$, just
as for $K=\qq$ this is a discrete valuation. The associated valuation
ring is $\{\frac{f(x)}{g(x)}\in k[x]:p(x)\nmid g(x)\}$

e.g. $K=\cc(x)$. The monic irreducible monic polynomial are $p(x)=x-a$
($x\in\cc$). Then $\ord_{p}(h)=\begin{cases}
n>0 & \mbox{if }h\mbox{ has a zero of order }n\mbox{ at }a\\
n<0 & \mbox{if }h\mbox{ has a pole of order }n\mbox{ at }a\\
0 & \mbox{if neither}
\end{cases}$.

e.g. $K=k(x)$. Define $v(\frac{f}{g})=\deg(g)-\deg(f)$ then $v$
is a discrete valuation. Note $k(x)=k(\frac{1}{x})$. This $v$ is
just $\ord_{1/x}$
\end{example*}
Let $v$ be a discrete valuation on $K$ such that $v:K^{*}\to\zz$
is surjective. (This only involves rescaling $v$, unless $v$ is
identically $0$). Let $\pi\in K$ be such that $v(\pi)=1$. 

$R_{v}=\{x\in K:v(x)\geq0\}$ - $M_{v}\cup U_{v}$

$M_{v}=\{x\in K:v(x)>0\}$ - maximal ideal of $R_{v}$

$U_{v}=\{x\in K:v(x)=0\}$ - set of units in $R_{v}$

If $x,y\in R_{v}$ then $x|y\iff\frac{y}{x}\in R_{v}\iff v(\frac{y}{x})\geq0\iff v(y)\geq v(x)$.
So if $x_{n}$ is an element with $v(x_{n})=n$ (for all $n\in\zz$)
then $x_{n}|x_{n+1}$ $\forall n$ hence $R_{v}\supset(x_{1})\supset(x_{2})\supset\dots$

Every $x\in R\setminus\{0\}$ can be written uniquely as $x=\pi^{n}u$
where $n=v(x)\geq0$ and $u\in U_{v}$. (Since if $n=v(x)$ then $u=\pi^{-n}x\Rightarrow v(u)=-n+v(x)=0\Rightarrow u\in U_{v}$),
i.e., $R_{v}$ is a UFD with only one prime, namely $\pi$. 

\uline{Every} ideal in $R_{v}$ is principal: the only non-zero
ideals are $(\pi^{n})$, $n\geq0$. $M_{v}=(\pi)$ since $x\in M_{v}\iff v(x)\geq1=v(\pi)\iff\pi|x$.
If $I\lhd R_{v},I\neq0$ let $n=\min\{v(x):x\in I\}$. Then $I=(\pi^{n})$
since $\exists x\in I$ with $v(x)=n$ $\forall y\in I$, $v(y)\geq n\Rightarrow x|y$
so $I=(x)$, and $v(x)=v(\pi^{n})\Rightarrow x=\pi^{n}u\Rightarrow(x)=(\pi^{n})=(\pi)^{n}$.

\newpage{}


\section*{Geometrically Interlude II: Hilbert's Nullstellensatz.}
\begin{namedthm}[Algebraic form of Nullstellensatz]
 Let $k$ be a field and let $F$ be a field which is a finitely
generated $k$-algebra. Then $F$ is a finite algebraic extension
of $k$. In particular if $k$ is algebraically closed then $F=k$.
\end{namedthm}

\begin{namedthm}[Weak form of Nullstellensatz]
 Let $k$ be an algebraically closed field and $I\lhd k[x_{1},\dots,x_{n}]$.
If $I\neq(1)$ then $V(I)\neq\emptyset$ (i.e., $\exists\underline{a}\in k^{n}$
such that $f(\underline{a})=0\,\forall f\in I$)\end{namedthm}
\begin{cor}
The maximal ideals in $k[x_{1},\dots,x_{n}]$ ($k$ algebraically
closed) are precisely the ideals $M_{\underline{a}}=(x_{1}-a_{1},\dots,x_{n}-a_{n}),\,\underline{a}\in k^{n}$\end{cor}
\begin{namedthm}[Strong form of Nullstellensatz]
 Let $k$ be an algebraically closed field and $I\lhd k[x_{1},\dots,x_{n}]$.
Then $I(V(I))=r(I)$ (i.e., if $g(\underline{a})=0$ whenever $f(\underline{a})=0\,\forall f\in I$
then $g^{N}\in I$)\end{namedthm}
\begin{proof}[Proof that Algebraic form $\Rightarrow$ Weak form]
 Let $k$ be a algebraically closed field and $I\underset{\neq}{\lhd}k[x_{1},\dots,x_{n}]\Rightarrow I\subseteq M$
a maximal ideal. Consider $k\to k[x_{1},\dots,x_{n}]\to k[x_{1},\dots,x_{n}]/M$.
Now $k[x_{1},\dots,x_{n}]/M$ is a field which is a finitely generated
$k$-algebra. By the Algebraic form the composite of the previous
map is surjective $(k[x_{1},\dots,x_{n}]/M\cong k$ as $k$ is algebraically
closed), so for all $i$, $\exists a_{i}\in k$ such that $ $$x_{i}-a_{i}\in M$.
So $M\supseteq(x_{1}-a_{1},\dots,x_{n}-a_{n})=M_{\underline{a}}.$
But $M_{\underline{a}}$ is maximal so $M=M_{\underline{a}}.$ Now
for all $f\in I\Rightarrow f\in M\Rightarrow f(\underline{a})=0$
\end{proof}

\begin{proof}[Proof that Weak form $\Rightarrow$ Strong form]
 $I\lhd k[x_{1},\dots,x_{n}]$. We know that $I(V(I))\supseteq r(I)$
since $g\in r(I)\Rightarrow g^{N}\in I\Rightarrow g^{N}(\underline{a})=0\,\forall\underline{a}\in V(I)\Rightarrow g(\underline{a})=0\Rightarrow g\in I(V(I))$.

Conversely let $g\in I(V(I))$, then $(*)\,(f(\underline{a})=0\,\forall f\in I)\Rightarrow g(\underline{a})=0$.

Extend the ring $k[x_{1},\dots,x_{n}]$ by adding a new variable $y$
to get $k[x_{1},\dots,x_{n},y]$. In $k[x_{1},\dots,x_{n},y]$ form
the ideal $J$ generated by all $f\in I$ and $1-g(x_{1},\dots,x_{n})y$,
i.e., $J=(1-g(x)y)+I\cdot k[x_{1},\dots,x_{n},y]$. Now $V(J)=\emptyset$
(in $k^{n+1}$) since if $(a_{1},\dots,a_{n},b)\in V(J)$ then 
\begin{enumerate}
\item $f(a_{1},\dots,a_{n})=0\,\forall f\in I$
\item $1-g(a_{1},\dots,a_{n})b=0$
\end{enumerate}
This is clearly a contradiction to $(*)$. So by the Weak form, we
have $J=k[x_{1},\dots,x_{n},y]$, i.e., $1\in J$. So 
\[
1=h(x_{1},\dots,x_{n},y)(1-g(x_{1},\dots,x_{n})y)+\sum_{j}h_{j}(x_{1},\dots,x_{n},y)f_{j}(x_{1},\dots,x_{n})\, f_{j}\in I.
\]
Substitute $y=\frac{1}{g(x_{1},\dots x_{n})}$ to get an equation
in $k(x_{1},\dots,x_{n})$. 
\[
1=\sum_{j}h_{j}(x_{1},\dots,x_{n},\frac{1}{g(x_{1},\dots,x_{n}})f_{j}(x_{1},\dots,x_{n})
\]
The RHS is a rational function whose denominator is a power of $g$.
So for large enough $N\geq0$:
\[
g^{N}=\sum_{j}\tilde{h}_{j}(x_{1},\dots,x_{n})f_{j}(x_{1},\dots,x_{n})\in I
\]
for some $\tilde{h}_{j}\in k[x_{1},\dots,x_{n}]$. Hence $g\in r(I)$
\end{proof}

\begin{proof}[Proof of Algebraic Form of Nullstellensatz]
 $F=k[x_{1},x_{2},\dots,x_{n}]$ (where $x_{i}\in F$ are the generators
of $F$) is a field. We must show that each $x_{i}$ is algebraic
over $k$. We are going to use induction on $n$

$n=1$: $F=k[x_{1}]$. Write $x_{1}^{-1}$ as a polynomial in $x_{1}$,
then we can get an equation for $x_{1}$ over $k$. (Alternative:
if $x_{1}$ were not algebraic then $k[x_{1}]$ is a polynomial ring,
not a field)

Inductive Step: $F=k(x_{1})[x_{2},\dots,x_{n}]$ (since $F$ is a
field) is a finitely generated algebra over $k(x_{1})$ with only
$n-1$ generators. So each $x_{j}$ for $j\geq2$ is algebraic over
$k(x_{1})$. If we can show that $x_{1}$ is algebraic over $k$ then
we are done. For all $j\geq2$, we have a polynomial equation for
$x_{j}$ over $k(x_{1})$. Let $f\in A:=k[x_{1}]$ be a common denominator
for all coefficient for all these polynomials. Consider the ring $A_{f}=S^{-1}A$
where $S=\{1,f,f^{2},f^{3},\dots\}$. All the $n-1$ polynomials are
monic in with coefficients in $A_{f}$. Hence each $x_{j}$ ($j\geq2$)
is integral over $A_{f}$. It follows that $F$ is integral over $A_{f}$
since $F=A[x_{2},\dots,x_{n}]=A_{f}[x_{2},\dots,x_{n}].$ By Lemma
\ref{lem:Lemma-1}, since $F$ is a field, so is $A_{f}$. Let $K=k(x_{1})$,
a subfield of $F$, the field of fractions of both $A$ and $A_{f}$.
Now $A=k[x_{1}]\subseteq A_{f}\subseteq K=k(x_{1})$ and $A_{f}$
a field implies that $A_{f}=K$ (since $K$ is the smallest field
containing $A$, being its field of fractions)

If $x_{1}$ were not algebraic over $k$ then $A=k[x_{1}]$ would
be the polynomial ring in one variable over $k$ and $k(x_{1})=K$
its field of fractions. Take any irreducible $g\in k[x_{1}]$ with
$g\nmid f$, then $\frac{1}{g}\notin A_{f}$. (NB: $k[x_{1}]$ would
have infinitely many irreducible) This leads to a contradiction hence
$x_{1}$ is algebraic

\newpage{}
\end{proof}

\section{Noetherian and Artinian modules and rings}
\begin{prop}[Definition]
\label{prop:Definition of Noetherina} An $R$-module $M$ is \emph{Noetherian}
if it satisfies one of the following equivalent conditions:
\begin{enumerate}
\item \emph{ACC (Ascending Chain Condition)}: any ascending chain $M_{1}\subseteq M_{2}\subseteq M_{3}\subseteq\dots$
of submodules of $M$ terminates, i.e., for some $n$ we have $M_{n}=M_{n+1}=\dots$
\item Every non-empty collection of submodules of $M$ has a maximal element
\item Every submodule of $M$ is finitely generated
\end{enumerate}
\end{prop}
\begin{defn}
A ring $R$ is \emph{Noetherian} if it is so as an $R$-module, i.e.,
the ideals of $R$ satisfies ACC and every ideal if finitely generated.\end{defn}
\begin{prop}[Definition]
 \label{prop:defn Artinian}An $R$-module $M$ is \emph{Artinian}
if it satisfies the following equivalent conditions
\begin{enumerate}
\item \emph{DCC (Descending Chain Condition):} any descending chain $M_{1}\supseteq M_{2}\supseteq M_{3}\supseteq\dots$
of submodule of $M$ terminates, i.e., for some $n$ we have $M_{n}=M_{n+1}=\dots$
\item Every non-empty collection of submodules has a minimal element
\end{enumerate}
\end{prop}
\begin{proof}[Proof of Proposition\ref{prop:Definition of Noetherina}]
 $1)\iff2)$: If we had an infinite AC $M_{1}\subsetneqq M_{2}\subsetneqq M_{3}\subsetneqq\dots$
then $\{M_{n}:n\geq1\}$ has no maximal elements. Conversely if $S$
is a non-empty set of submodules of $M$ with no maximal elements,
then pick $M_{1}\in S$, $\exists M_{2}\supsetneqq M_{1},\exists M_{3}\supsetneqq M_{2},\dots$

$2)\Rightarrow3)$: Let $S$ be the set of finitely generated submodules
of $N$, where $N\leq M$. $0\in S$ so $S$ has a maximal elements,
say $N_{0}$. So $N_{0}\leq N$ and $N_{0}$ is finitely generated,
if $N_{0}\neq N$ take $x\in N\setminus N_{0}$, then $N_{0}+Rx\supsetneqq N_{0}$
and is finitely generated, contradiction.

$3)\Rightarrow1)$: Given $M_{1}\subseteq M_{2}\subseteq M_{3}\subseteq\dots$,
let $N=\cup_{n=1}^{\infty}M_{n}$. Then $N$ is a submodule of $M$.
Let $x_{1},\dots,x_{n}$ generate $N$. For large enough $k$, $M_{k}$
contains contain all of the $x_{i}$. Then $M_{k}=N=M_{k+1}=M_{k+1}=\dots$
\end{proof}
Note that the proof of $1)\iff2)$ can easily be adapted to prove
Proposition \ref{prop:defn Artinian}
\begin{example*}
{}
\begin{enumerate}
\item Every finite $\zz$-module is both Noetherian and Artinian
\item If $R$ is a field $k$ then $R$-modules are $k$-vector spaces and
they are Noetherian $\iff$ they are finite dimensional $\iff$ they
are Artinian.
\item $\zz$ is a Noetherian ring (every ideal is generated by 1 element)
but is not Artinian: $\zz\supset(2)\supset(4)\supset(8)\supset\dots\supset(2^{n})\supset\dots$.
\item $R=k[x_{1},x_{2},\dots]$ polynomials in a countable (non-finite)
number of variables. $R$ is neither Noetherian nor Artinian: $(x_{1})\subset(x_{1},x_{2})\subset(x_{1},x_{2},x_{3})\subset\dots$
and $(x_{1})\supset(x_{1}^{2})\supset(x_{1}^{3})\supset\dots$
\end{enumerate}
\end{example*}
\begin{prop}
\label{prop:s.e.s and Noetherian}If $0\to M_{1}\overset{\alpha}{\to}M_{2}\overset{\beta}{\to}M_{3}\to0$
is a short exact sequence of $R$-modules then $M_{2}$ is Noetherian
$\iff$ both $M_{1},M_{3}$ are. Similarly $M_{2}$ is Artinian $\iff$
both $M_{1},M_{3}$ are.\end{prop}
\begin{proof}
The proof for both cases are the similar, so we are just going to
prove the Artinian case.

``$\Rightarrow$'' : Suppose $M_{2}$ is Artinian. Any Descending
Chain in $M_{1}$maps isomorphically under $\alpha$ to a Descending
Chain in $M_{2}$ which terminates. Similarly any Descending Chain
in $M_{3}$ lifts to a Descending Chain in $M_{2}$ via $\beta^{-1}$
, hence terminates

``$\Leftarrow$'': Suppose $M_{1},M_{3}$ Artinian. Let $N_{1}\supseteq N_{2}\supseteq N_{3}\supseteq\dots$
be a Descending Chain in $M_{2}$. Then $\alpha^{-1}(N_{1})\supseteq\alpha^{-1}(N_{2})\supseteq\dots$
is a Descending Chain in $M_{1},$hence stops, and $\beta(N_{1})\subseteq\beta(N_{2})\subseteq\dots$
is a Descending Chain in $M_{3}$, hence stops. So there exists $n$
such that $\alpha^{-1}(N_{n})=\alpha^{-1}(N_{n+1})=\dots$ and $\beta(N_{n})=\beta(N_{n+1})=\dots$.
This implies $N_{n}=N_{n+1}$ since let $x\in N_{n}$, then $\beta(x)\in\beta(N_{n})=\beta(N_{n+1})\Rightarrow\exists y\in N_{n+1}$
with $\beta(x)=\beta(y)$. So $x-y\in\ker(\beta)=\im(\alpha)$, so
$x-y=\alpha(z)$ for some $z\in M_{1}$ and since $\alpha(z)=x-y\in N_{n},z\in\alpha^{-1}(N_{n})=\alpha^{-1}(N_{n+1})\Rightarrow\alpha(z)\in N_{n+1}$
so $x=y+\alpha(z)\in N_{n+1}$.\end{proof}
\begin{cor}
Any finite sum of Noetherian (respectively Artinian) modules is again
Noetherian (respectively Artinian)\end{cor}
\begin{proof}
The sequence $0\to M_{1}\to M_{1}\oplus M_{2}\to M_{2}\to0$ is exact. \end{proof}
\begin{note*}
A subring of a Noetherian ring is not necessarily Noetherian, e.g.
$R=k[x_{1},x_{2},\dots]\subset k(x_{1},x_{2},\dots)$.\end{note*}
\begin{cor}
\label{cor:Noetherian f.h}If $R$ is Noetherian and $M$ is a finitely
generated $R$-module then $M$ is Noetherian. Same for Artinian.\end{cor}
\begin{proof}
$R^{n}=R\oplus R\oplus\dots\oplus R$ is a Noetherian $R$-module,
since $R$ is. Every finitely generated $R$-module $M=Rx_{1}+\dots+Rx_{n}$
is the homomorphic image of some $R^{n}$, i.e., $0\to\ker\to R^{n}\to M\to0$
is exact. 
\end{proof}
Later we'll prove that $R$ Noetherian $\Rightarrow$ $R[x]$ is Noetherian
(Hilbert Basis Theorem). Hence $R[x_{1},\dots,x_{n}]$ is Noetherian,
e.g, $R=k$ a field. Hence any finitely generated $R$-algebra is
Noetherian. 


\subsection{Noetherian Rings}
\begin{lem}
If $R$ is a Noetherian ring and $f:R\to S$ a surjective ring homomorphism
then $S$ is Noetherian\end{lem}
\begin{proof}
$R/\ker(f)\cong S$$\Rightarrow S$ is Noetherian as an $R$-module$\Rightarrow S$
is Noetherian.\end{proof}
\begin{lem}
Let $R\leq S$ with $R$ Noetherian. If $S$ is finitely generated
as an $R$-module then $S$ is Noetherian.\end{lem}
\begin{proof}
$S$ is Noetherian as $R$-module by Corollary \corref{Noetherian f.h}
hence is also Noetherian as $S$-module.\end{proof}
\begin{example*}
$\zz$ is Noetherian $\Rightarrow$ any ring which is finitely generated
as $\zz$-module is Noetherian.

$\zz[\alpha]$ with $\alpha$ an algebraic integer is Noetherian\end{example*}
\begin{lem}
If $R$ is a Noetherian ring and $S$ a multiplicatively closed set
in $R$ then $S^{-1}R$ is Noetherian.\end{lem}
\begin{proof}
By Proposition \propref{Properties of localisation} there is a bijection,
preserving inclusion, between the set of ideals of $S^{-1}R$ and
a subset of the ideals of $R$. So Ascending Chain Condition for $R\Rightarrow$
Ascending Chain Condition for $S^{-1}R$\end{proof}
\begin{cor}
If $R$ is Noetherian and $P\lhd R$ prime then $R_{P}$ is a Noetherian
local ring\end{cor}
\begin{namedthm}[Hilbert Basis Theorem]
 If $R$ is a Noetherian ring then so is $R[x]$\end{namedthm}
\begin{proof}
Let $J\lhd R[x]$. For $n\geq0$ let $I_{n}$ be the ideal of $R$
consisting of all leading coefficients of $f\in J$ with $\deg(f)=n$
and $0$. It is easy to check that $I_{n}$ is an ideal. Then $I_{0}\subseteq I_{1}\subseteq I_{2}\subseteq\dots$
since $\deg(f)=n\Rightarrow\deg(xf)=n+1$ and they have the same leading
coefficients. By Ascending Chain Condition for $R$ there exists $n$
such that $I_{n}=I_{n+1}=\dots$. Let $f_{1,n},f_{2,n},\dots,f_{k_{n},n}\in J$
be polynomials of degree $n$ whose leading coefficients generates
$I_{n}$. For each $0\leq m<n$ let $f_{1,m},\dots,f_{k_{m},m}$ ($k_{m}\geq0$)
be polynomials in $J$ of degree $m$ whose leading coefficients generates
$I_{m}$. (Use $k_{m}=0$ if $I_{m}=0$)

\uline{Claim:} $J$ is generated by all $f_{i,m}$, with $m\leq n,i\leq k_{m}$.

Let $g\in J$. Proceed by induction on $\deg(g)$. Our base case is
the $0$ polynomial, since this is trivial.
\begin{casenv}
\item $\deg(g)\geq n$: Then the leading coefficient of $g$ are in $I_{n}\Rightarrow\exists r_{1},\dots r_{k_{n}}\in R$
such that $\mbox{lc}(g)=\mbox{lc}(\sum_{i=1}^{k_{n}}r_{i}f_{i,n})$
where $\mbox{lc}(f)=$ leading coefficient of $f$. $\Rightarrow$
leading term of $g$ $=$ leading term of $(g_{1}=\sum r_{i}x^{\deg(g)-n}f_{1,n})$,
$g_{1}\in(f_{i,j})$. So $g_{2}=g-g_{1}$ has $\deg(g_{2})<\deg(g_{1})$.
By induction $g_{2}\in(f_{i,j})$ so $g\in(f_{i,j})$
\item $\deg(g)=m<n$: Now an $R$-linear combination of $f_{i,m}$ ($1\leq i\leq k_{m}$)
has the same leading term as $g$. The rest is as in Case 1.
\end{casenv}
Hence $J$ is generated by the finite set $\{f_{i,j}:1\leq i\leq k_{m},0\leq j\leq n\}$.
Hence every ideal in $R[x]$ is finitely generated, so $R[x]$ is
a Noetherian ring\end{proof}
\begin{cor}
If $R$ is Noetherian so is $R[x_{1},x_{2},\dots,x_{n}]$ for all
$n\geq1$\end{cor}
\begin{proof}
Since $R[x_{1},\dots,x_{n-1}][x_{n}]=R[x_{1},x_{2},\dots,x_{n}]$
\end{proof}
In particular if $k$ is a field then $k[x_{1},\dots,x_{n}]$ is Noetherian.
Hence any system of polynomial equation has the same set of zeros
as a \uline{finite} system
\begin{cor}
If $R$ is Noetherian then so is any finitely generated $R$-algebra.\end{cor}
\begin{proof}
Any finitely generated $R$-algebra is of the form $R[\alpha_{1},\dots,\alpha_{n}]$
- a quotient of $R[x_{1},\dots,x_{n}]$\end{proof}
\begin{example*}
Any finitely generated $k$-algebra ($k$ a field) is Noetherian.

Any finitely generated $\zz$-algebra is Noetherian. (e.g., the ring
of integers in a number field is Noetherian: NB theses do not all
have the form $\zz[\alpha]$ with a \uline{single} generator)

\newpage{}
\end{example*}

\section{Primary Decomposition}

In general rings we don't have a factorization theory which expresses
\uline{elements} as \uline{products} of \uline{prime powers.}
Instead we make do with writing \uline{ideals} as \uline{intersections
}of \uline{primary ideals.} 
\begin{defn}
A \emph{primary} ideal $Q\lhd R$ is a proper ideal such that $xy\in Q\Rightarrow x\in Q$
or $y^{n}\in Q$ for some $n\geq1$, i.e., $xy\in Q\Rightarrow$ either
$x\in Q$ or $y\in r(Q)$.

Equivalently: $R/Q\neq0$ and every zero-divisor is nilpotent.\end{defn}
\begin{prop}
{}
\begin{enumerate}
\item Every prime ideal is primary.
\item The contraction of a primary is primary.
\item If $Q$ is primary then $r(Q)$ is prime. It is the smallest prime
containing $Q$.
\end{enumerate}
\end{prop}
\begin{proof}
{}
\begin{enumerate}
\item Clear from the definition ($n=1$)
\item Let $f:A\to B$ be a ring homomorphism, $Q\lhd B$ primary $\Rightarrow Q^{c}=f^{-1}(Q)\lhd A$
is primary. To see this: $1\notin Q^{c}$ since $f(1)=1\notin Q$,
hence $A/Q^{c}\neq0$. Also note that $f$ induces an injective map
$A/Q^{c}\hookrightarrow B/Q$ so $A/Q^{c}$ also has the property
that zero-divisors are nilpotent.
\item Let $P=r(Q)$. Suppose $xy\in P$. Then $x^{n}y^{n}\in Q$ (for some
$n\geq1$) so either $x^{n}\in Q$ or $(y^{n})^{m}\in Q$ (for some
$m\geq1$), so either $x\in P$ or $y\in P$. For the last sentence
use the fact that the radical of $I$ is the intersection of prime
ideals containing $I$
\end{enumerate}
\end{proof}
\begin{defn}
\label{def of P-primary}If $Q$ is primary and $r(Q)=P$ we say that
$Q$ is $P$\emph{-primary}\end{defn}
\begin{example*}
{}
\begin{enumerate}
\item In $\zz$ the primary ideals are $(0)$ and $(p^{n})$, $p$ prime,
$n\geq1$.
\item $R=k[x,y]$. Let $Q=(x,y^{2})\Rightarrow P=r(Q)=(x,y)$. $P^{2}=(x^{2},xy,y^{2})\subsetneqq Q\subsetneqq P$.
Now $R/Q\cong k[y]/(y^{2})$ in which we see that $\{$nilpotent$\}=\{$zero-divisors$\}=\{$multiples
of $y\}$. This is an example of a primary which is not a prime power.
\item An example of prime power needs not be primary. Let $R=k[X,Y,Z]/(XY-Z^{2})=k[x,y,z]$
where $x,y,z$ satisfies the relation $xy=z^{2}$. Let $P=(x,y)$,
then $R/P\cong k[X,Y,Z]/(X,Y)\cong k[Y]\Rightarrow P$ prime. Now
$xy=z^{2}\in P^{2}$ which is not primary, since $x\notin P^{2}$
and $y\notin P$.
\end{enumerate}
\end{example*}
\begin{prop}
{}
\begin{enumerate}
\item If $r(I)$ is maximal then $I$ is primary
\item If $M$ is maximal then $M^{n}$ is $M$-primary for all $n\geq1$
\end{enumerate}
\end{prop}
\begin{proof}
{}
\begin{enumerate}
\item Let $M=r(I)$. Then $M/I$ is the nilradical of $R/I$, and $M/I$
is prime so $R/I$ has a unique prime ideal, namely $M/I$. So every
non-nilpotent element of $R/I$ is a unit, so it is not a zero-divisor.
\item $r(M^{n})=M$ (since $r(M^{n})\supseteq M$ and $M$ is maximal)
\end{enumerate}
\end{proof}
\begin{lem}
Any finite intersection of $P$-primary ideals is again $P$-primary.\label{lem:Any-finite-intersection of primary}\end{lem}
\begin{proof}
Let $Q_{i}$ be $P$-primary for $i=1,\dots,n$. Set $Q=\cap_{i=1}^{n}Q_{i}$.
Then $r(Q)=r(\cap_{i=1}^{n}Q_{i})=\cap_{i=1}^{n}r(Q_{i})=\cap_{i=1}^{n}P=P$.
If $xy\in Q$ and $x\notin Q$ then $\exists i$ such that $xy\in Q_{i}$
but $x\notin Q_{i}$. Hence $y\in r(Q_{i})=P\Rightarrow y\in r(Q)$\end{proof}
\begin{lem}
\label{lem:(Q:x) primary}Let $Q$ be $P$-primary and let $x\in R$.
Then:
\begin{enumerate}
\item $x\in Q\Rightarrow(Q:x)=R$
\item $x\notin Q\Rightarrow(Q:x)$ is $P$-primary
\item $x\notin P\Rightarrow(Q:x)=Q$
\end{enumerate}

To make sense of the three cases remember that $Q\subseteq P\subseteq R$.

Recall: $(Q:x)=\{y\in R:xy\in Q\}\supseteq Q$\end{lem}
\begin{proof}
{}
\begin{enumerate}
\item If $x\in Q$ then $xy\in Q\,\forall y\in R$.
\item We have $Q\subseteq(Q:x)\subseteq P$, where the second containment
holds because $xy\in Q,x\notin Q\Rightarrow y\in P$. So $r(Q)=P\subseteq r(Q:x)\subseteq r(P)=P\Rightarrow r(Q:x)=P$.
Now suppose $yz\in(Q:x)$ with $y\notin P\Rightarrow yxz\in Q\Rightarrow y(xz)\in Q\Rightarrow xz\in Q\Rightarrow z\in(Q:x)$.
So $(Q:x)$ is indeed $P$-primary.
\item If $xy\in Q$ but $x\notin P\Rightarrow y\in Q$.
\end{enumerate}
\end{proof}
\begin{defn}
A \emph{primary decomposition }of an ideal $I\lhd R$ is an expression
$I=Q_{1}\cap Q_{2}\cap\dots\cap Q_{n}$ with each $Q_{i}$ primary.\end{defn}
\begin{rem*}
Such a decomposition may or may not exist. It does always exists when
$R$ is Noetherian.
\end{rem*}
Let $P_{i}=r(Q_{i})$ - the primes associated with the decomposition.
\begin{description}
\item [{Minimality~Condition~1}] If some $Q_{j}\supseteq\cap_{i\neq j}Q_{i}$
then $Q_{j}$ may be omitted.
\item [{Minimality~Condition~2}] If more than one $Q_{i}$ has the same
radical we may combine them (using Lemma \ref{lem:Any-finite-intersection of primary})
\end{description}
We call the decomposition \emph{minimal }if:
\begin{enumerate}
\item No $Q_{j}\supseteq\cap_{j\neq i}Q_{i}$.
\item The $P_{i}$ are distinct. 
\end{enumerate}
It will turn out that the primes $P_{i}$ \uline{are} uniquely
determined by $I$, but the $Q_{i}$ need not be. 
\begin{example*}
Let $I=(x^{2},xy)\lhd k[x,y]$ where $k$ is any field. Then $I=P_{1}\cap P_{2}^{2}$
where $P_{1}=(x)$ and $P_{2}=(x,y)$ (note $P_{1}$ is prime hence
primary, and $P_{2}$ is maximal hence $P_{2}^{2}$ is primary). This
is a minimal primary decomposition. Note that $P_{1}\subset P_{2}$
(this means $V(P_{2})\subset V(P_{1})$, we say $P_{2}$ is an \uline{embedded}
prime). Also $I=P_{1}\cap Q_{2}$ where $Q_{2}=(x^{2},y)$ with $r(Q_{2})=P_{2}$
again. \end{example*}
\begin{thm}
Let $I=Q_{1}\cap Q_{2}\cap\dots\cap Q_{n}$ be a minimal primary decomposition.
Let $P_{i}=r(Q_{i})$. Then $P_{i},\dots,P_{n}$ are all the prime
ideals in the set $\{r(I:x)|x\in R\}$. Hence the set of $P_{i}$
is uniquely determined by $I$, independent of the decomposition.\end{thm}
\begin{proof}
Consider $(I:x)=(\cap_{i=1}^{n}Q_{i}:x)=\cap_{i=1}^{n}(Q_{i}:x)$
by the Fact \vpageref{Fact:colon fact}. This means $r(I:x)=r(\cap_{i=1}^{n}(Q_{i}:x))=\cap_{i=1}^{n}r(Q_{i}:x)$.
But $r(Q_{i}:x)=\begin{cases}
R & x\in Q_{i}\\
P_{i} & x\notin Q_{i}
\end{cases}$ by Lemma \ref{lem:(Q:x) primary}. Hence $r(I:x)=\cap_{i:x\notin Q_{i}}P_{i}$.

If $r(I:x)$ is prime, $P$ say, then $P=\cap_{x\notin Q_{i}}P_{i}\Rightarrow P=P_{i}$
by Proposition \ref{prop:Intersection of prime contains prime}.

Conversely for each $i$ choose $x\in Q_{j}$ ($\forall j\neq i)$,
$x\notin Q_{i}$ (this is possible by minimality condition 1) then
$r(I:x)=P_{i}$.\end{proof}
\begin{notation}
To each $I$ with a primary decomposition we have a set of primes
$P_{i}$ called the \emph{associated primes} of $I$. Any minimal
elements of this set is called an \emph{isolated} or \emph{minimal}
prime of $I$. Any other primes associated to $I$ are called \emph{embedded
}primes.
\end{notation}
We'll prove later that $P_{i}$ isolated $\Rightarrow Q_{i}$ is uniquely
determined. 
\begin{cor}
Suppose that $0$ is decomposable. Then $D:=\{$zero-divisors in $R\}=$union
of all primes associated to $0$.

$N=\{$nilpotent in $R\}=N(R)=$ intersection of all minimal primes
associated to $0$\end{cor}
\begin{proof}
Note that $D$ is not an ideal (in general), but we can still define
$r(D)=\{x\in R:x^{n}\in D$ for some $n\geq1\}=D$ (exercise: if $x^{n}$
is a zero-divisor, so is $x$). Note that $D=\cup_{x\neq0}(0:x)$
so if we take radicals $D=r(D)=\cup_{x\neq0}r(0:x)$. Let $0=Q_{1}\cap Q_{2}\cap\dots\cap Q_{n}$
be minimal primary decomposition. Let $x\neq0$, $r(0:x)=\cap_{x\notin Q_{j}}P_{j}\subseteq P_{j_{0}}$
where $x\notin Q_{j_{0}}$. Note that $j_{0}$ exists since $x\neq0$.
Hence $D=\cup_{x\neq0}r(x:0)\subseteq\cup_{j=1}^{n}P_{j}$. But each
$P_{j}=r(0:x)$ for some $x\neq0$ so each $P_{j}\subseteq D$

$N(R)=r(0)=\cap r(Q_{i})=\cap P_{i}$.\end{proof}
\begin{cor}
Let $I=Q_{1}\cap Q_{2}\cap\dots\cap Q_{n}$ be a minimal primary decomposition
and $P_{i}=r(Q_{i})$. Then $\cup_{i=1}^{n}P_{i}=\{x\in R:(I:x)\neq I\}\,(*)$\end{cor}
\begin{proof}
Apply the previous corollary to $R/I$: Note that $I=Q_{1}\cap Q_{2}\cap\dots\cap Q_{n}\Rightarrow0=\overline{Q_{1}}\cap\overline{Q_{2}}\cap\dots\cap\overline{Q_{n}}$
where as usual $\overline{Q_{i}}\lhd R/I$. Each $\overline{Q_{i}}$
is primary in $R/I$ since $(R/I)/\overline{Q_{i}}\cong R/Q_{i}$.
So the zero-divisors in $R/I$ are the union of all $r(\overline{Q_{i})}=\overline{r(Q_{i})}=\overline{P_{i}}$
and $\overline{y}$ is a zero-divisors in $R/I\iff\exists x\notin I:yx\in I\iff y$
in RHS of $(*)$. While $\overline{y}\in\cup\overline{P_{i}}\iff y\in\cup P_{i}$
\end{proof}

\subsection{Primary Decomposition and Localization}
\begin{prop}
Let $Q$ be $P$-primary and $S$ a multiplicatively closed set in
$R$
\begin{enumerate}
\item $S\cap P\neq\emptyset\Rightarrow S\cap Q\neq\emptyset$ and $S^{-1}Q=S^{-1}R$
\item $S\cap P=\emptyset\Rightarrow S^{-1}Q$ is $S^{-1}P$-primary and
$(S^{-1}Q)^{c}=Q$
\end{enumerate}
\end{prop}
\begin{proof}
{}
\begin{enumerate}
\item $S\cap P\neq\emptyset$, then there exists $s\in S\cap P\Rightarrow s^{m}\in S\cap Q$
for some $m$. We can now use Proposition \ref{prop:Properties of localisation}
(part 2.) to show $S^{-1}Q=S^{-1}R$.
\item $Q^{ec}=\cup_{s\in S}(Q:s)$ by Proposition \ref{prop:Properties of localisation}
(part 2.) but $x\in(Q:s)\Rightarrow x\cdot s\in Q,s\neq P\supset Q\Rightarrow S^{n}\notin Q\,\forall n\Rightarrow x\in Q$$\Rightarrow Q^{ec}=Q$.
To show that $S^{-1}Q$ is $S^{-1}P$-primary, note $r(Q^{e})=r(S^{-1}Q)=S^{-1}r(Q)=S^{-1}P$,
also if $\frac{x}{s}\cdot\frac{y}{t}\in S^{-1}Q$ (so there exist
$u\in S$ such that $uxy\in Q$) and $\frac{x}{s}\notin S^{-1}Q\Rightarrow x\notin Q$
but $Q$ is still primary, hence $uy\in P,u\in S$ and $S\cap P=\emptyset\Rightarrow y\in P\Rightarrow\frac{y}{t}=\frac{uy}{ut}\in S^{-1}P\Rightarrow S^{-1}Q$
is $S^{-1}P$-primary. 
\end{enumerate}
\end{proof}
\begin{notation*}
We denote $S(I)=(S^{-1}I)^{c}=\cup_{s\in S}(I:s)$\end{notation*}
\begin{prop}
Let $S$ be a multiplicatively closed set in $R$ and $I=Q_{1}\cap\dots\cap Q_{n}$
be a minimal primary decomposition of $I$ numbered so that $\begin{cases}
S\cap P_{i}=\emptyset & 1\leq i\leq m\\
S\cap P_{i}\neq\emptyset & m+1\leq i\leq n
\end{cases}$. Then $S^{-1}I=\cap_{i=1}^{m}S^{-1}Q_{i}$ and $S(I)=\cap_{i=1}^{m}Q_{i}$.
Both of these decomposition are \uline{minimal} primary decompositions.\end{prop}
\begin{proof}
For $i\in\{1,\dots,m\}$ we have $S^{-1}Q_{i}$ is $S^{-1}P_{i}$-primary
by the previous proposition, furthermore $S^{-1}P_{i}$ are distinct
primes of $S^{-1}R$ (by Proposition \ref{prop:Properties of localisation}
part 4.) therefore $S^{-1}I=S^{-1}(\cap_{i=1}^{n}Q_{i})=\cap_{i=1}^{n}S^{-1}Q_{i}\underset{i>m\Rightarrow S^{-1}Q_{i}=S^{-1}R}{=}\cap_{i=1}^{m}S^{-1}Q_{i}$
is a minimal primary decomposition. From this it is clear that $S(I)=\cap_{i=1}^{m}Q_{i}$.
\end{proof}
Recall: A prime $P$ is minimal (or isolated) for an ideal $I$ if
it is minimal under inclusion in the set of associated primes of $I$.
More generally we define:
\begin{defn}
A set $\mathscr{P}$ of primes associated to $I$ to be \emph{isolated}
if $P\in\mathscr{P}$, $P'\subset P$ and $P'$ is associated to $I$
then we have $P'\in\mathscr{P}$.\end{defn}
\begin{thm}
Let $I$ be an ideal of the ring $R$. Let $\mathscr{P}=\{P_{1},\dots,P_{n}\}$
be an isolated set of primes associated to $I$. Then $Q_{1}\cap\dots\cap Q_{m}$
is independent of the minimal primary decomposition of $I$.\end{thm}
\begin{proof}
Let $S=R\setminus\cup_{i=1}^{m}P_{i}$ then $S$ is a multiplicatively
closed set and $P_{j}\cap S=\emptyset\iff P_{j}\in\mathscr{P}$. Indeed
$P_{j}\in\mathscr{P}$ means $P_{j}\cap S=\emptyset$ and conversely
$P_{j}\notin\mathscr{P}\Rightarrow P_{j}\nsubseteq P_{i}\,\forall P_{i}\in\mathscr{P}\Rightarrow P_{j}\nsubseteq\cup_{i=1}^{m}P_{i}\Rightarrow P_{j}\cap S\neq\emptyset$.
Therefore $S(I)=Q_{1}\cap Q_{2}\cap\dots\cap Q_{m}$. This ideal only
depends on the primes in $\mathscr{P}$. \end{proof}
\begin{cor}
The isolated primary component of $I$ are uniquely determined.\end{cor}
\begin{proof}
Choose $\mathscr{P}=\{P\}$ where $P$ is a minimal prime, let $S=R\setminus P$,
then $S(I)=Q$ with $Q$ is the unique $P$-primary factor of $I$. 
\end{proof}

\subsection{Primary Decomposition in a Noetherian Ring}

The main aim of this sub-section is to prove the existence of primary
decomposition in a Noetherian ring.
\begin{defn}
An ideal $I$ is \emph{irreducible}\emph{\uline{ }}if $I=J_{1}\cap J_{2}$
then $I=J_{1}$ or $I=J_{2}$.\end{defn}
\begin{lem}
\label{lem:Lemma A} In a Noetherian ring $R$, every ideal is a finite
intersection of irreducible ideals.\end{lem}
\begin{proof}
Let $S$ be the set of ideals which are \uline{not} finite intersections
of irreducible ideals. If $S\neq\emptyset$ then $S$ has a maximal
element, $I$ (since $R$ is Noetherian). Then $I$ is not irreducible,
therefore $I=J_{1}\cap J_{2}$ with $J_{1},J_{2}\supsetneqq I$. So
$J_{1},J_{2}\notin S$, hence they are finite intersection of irreducible
ideals. Since the intersection of two finite intersection of irreducible
ideals, $I$ is the intersection of irreducible ideals, i.e., $I\notin S$.
This is a contradiction. Hence $S=\emptyset$\end{proof}
\begin{lem}
\label{lem:Lemma B}In a Noetherian ring $R$, all irreducible ideals
are primary.\end{lem}
\begin{proof}
Let $I$ be irreducible. Let $x,y\in R$ with $xy\in I$. We must
show that either $x\in I$ or $y^{n}\in I$ for some $n\geq1$. 

Define $I_{n}=(I:y^{m})$ for $m=1,2,\dots$. Then $I\subseteq I_{1}\subseteq I_{2}\subseteq\dots$,
since $R$ is Noetherian there exists $N$ such that $I_{n}=I_{n+1}$

\uline{Claim:} $I=(I+(x))\cap(I+(y^{n}))$

It is clear that $I\subseteq(I+(x))\cap(I+(y^{n}))$. Let $z\in(I+(x))\cap(I+(y^{n}))$,
so $z=i_{1}+r_{1}x=i_{2}+r_{2}y^{n}$ for some $i_{1},i_{2}\in I$
and $r_{1},r_{2}\in R$. Then $yz=i_{1}y+r_{1}xy\in I$ (since $i_{1},xy\in I$).
So $r_{2}y^{n+1}=yz-i_{2}y\in I\Rightarrow r_{2}\in(I:y^{n+1})=I_{n+1}=I_{n}\Rightarrow r_{2}y^{n}\in I$,
hence $z\in I$. So $(I+(x))\cap(I+(y^{n}))\subseteq I$

Since $I$ is irreducible, either:
\begin{itemize}
\item $I+(x)=I$, in which case $x\in I$
\item or $I+(y^{n})=I$, in which case $y^{n}\in I$
\end{itemize}
\end{proof}
\begin{thm}
In a Noetherian ring $R$, every ideal $I$ has a primary decomposition. \end{thm}
\begin{proof}
This follows directly from the previous two lemma.\end{proof}
\begin{prop}
\label{prop:every ideal contains a power of its}Let $R$ be a Noetherian
ring, every ideal $I$ contains a power of its radical. In particular,
the nilradical is nilpotent.\end{prop}
\begin{proof}
Let $x_{1},\dots,x_{k}$ generate $r(I)$ ($R$ is Noetherian). For
large enough $n$ we have $x_{i}^{n}\in I\,\forall i$. Now $r(I)^{kn}\subseteq I$
since $r(I)^{kn}$ is generated by elements of the form $x_{1}^{m_{1}}x_{2}^{m_{2}}\dots x_{k}^{m_{k}}$
where $\sum m_{i}=nk$, so at least of one of the $m_{i}\geq n\Rightarrow$
the generators of $r(I)^{kn}$ is in $I$, hence $r(I)^{kn}\subseteq I$. 

For the in particular part, just apply the proposition to $I=0$.\end{proof}
\begin{cor}
Let $R$ be a Noetherian ring, $M$ a maximal ideal and $Q$ an ideal.
Then the following are equivalent:
\begin{enumerate}
\item $Q$ is $M$-primary
\item $r(Q)=M$
\item $M^{n}\subseteq Q\subseteq M$ for some $n\geq1$.
\end{enumerate}
\end{cor}
\begin{proof}
$1.\iff2.$ (by Definition \ref{def of P-primary})

$2.\Rightarrow3.$: By the previous Proposition

$3.\Rightarrow2.$: Take the radicals $M=r(M^{n})\subseteq r(Q)\subseteq r(M)=M\Rightarrow r(Q)=M$\end{proof}
\begin{namedthm}[Krull's Theorem]
 Let $I$ be an ideal in a Noetherian ring $R$. Then $\cap_{n=1}^{\infty}I^{n}=0$
if and only if $1+I$ contains no zero-divisors.\end{namedthm}
\begin{proof}
``$\Rightarrow$'': If $1+I$ contains a zero-divisor $1-x$, with
$x\in I$, such that $(1-x)y=0$ for some $y\neq0$, then $y=xy=x^{2}y=x^{3}y=\dots=x^{n}y\in I^{n}$.
So $y\in\cap_{n=1}^{\infty}I^{n}$, hence $\cap_{n=1}^{\infty}I^{n}\neq0$

``$\Leftarrow$'': Let $J=\cap_{n=1}^{\infty}I^{n}$

\uline{Claim}: $IJ=J$.

Certainly $IJ\subseteq J$. Let $IJ=Q_{1}\cap Q_{2}\cap\dots\cap Q_{n}$
be a minimal primary decomposition of $IJ$ with $r(Q_{i})=P_{i}$,
so we must show that $J\subseteq Q_{i}\,\forall i$. We have $IJ\subseteq Q_{i}$.
\begin{casenv}
\item If $I\subseteq P_{i}$ then $Q_{i}\supseteq P_{i}^{m}$ (by Proposition
\ref{prop:every ideal contains a power of its}) $\supseteq I^{m}\supseteq J\Rightarrow J\subseteq Q_{i}$
\item If $I\nsubseteq P_{i}$ then $J\subseteq Q_{i}$ since if $x\in I,x\notin P_{i}$
then $xJ\subseteq IJ\subseteq Q_{i}$ so for all $y\in J,xy\in Q_{i}$
but $x\notin r(Q_{i})=P_{i}\Rightarrow y\in Q_{i}$.
\end{casenv}
Hence $J\subseteq\cap Q_{i}=IJ$ so $J=IJ$.

By Nakayama's Lemma since $J$ is finitely generated, $xJ=0$ for
some $x\in1+I$. If $1+I$ has no zero-divisors then $x$ is not a
zero-divisor, so $xJ=0\Rightarrow J=0$.\end{proof}
\begin{cor}
In a Noetherian domain $R$, if $I\neq R$ then $\cap_{n=1}^{\infty}I^{n}=0$\end{cor}
\begin{proof}
Obvious\end{proof}
\begin{cor}
If $I\subset J(R)$ then $\cap_{n=1}^{\infty}I^{n}=0$\end{cor}
\begin{proof}
Obvious from Proposition \ref{prop:J(R) iff unit}\end{proof}
\begin{cor}
\label{cor:3 of last lecture}In a Noetherian local ring with maximal
idea $M$, $\cap_{n=1}^{\infty}M^{n}=0$\end{cor}
\begin{proof}
Obvious since $M=J(R)$.

\newpage{}
\end{proof}

\section{Rings of small dimension}
\begin{prop}
\label{prop:Artinian iff Noetherian}In the ring $R$, suppose $0=M_{1}M_{2}\dots M_{n}$
with $M_{i}$ maximal ideals. Then $R$ is Noetherian if and only
if $R$ is Artinian.\end{prop}
\begin{proof}
$R\supset M_{1}\supseteq M_{1}M_{2}\supseteq M_{1}M_{2}M_{3}\supseteq\dots\supseteq M_{1}M_{2}\dots M_{n}=0$.
Let $V_{i}:=M_{1}M_{2}\dots M_{i-1}/M_{1}M_{2}\dots M_{i}$, notice
that each $V_{i}$ is a module over the \uline{field} $R/M_{i}$,
i.e, is a vector space. So each $V_{i}$ is Noetherian $\iff$ Artinian
$\iff$ finite dimensional. We then use Proposition \ref{prop:s.e.s and Noetherian},
over and over again on the following set of short exact sequences.
\[
\xyC{1pc}\xyR{0.5pc}\xymatrix{0\ar[r] & M_{1}\ar[r] & R\ar[r] & V_{1}\ar[r] & 0 & R\mbox{ Noetherian} & \iff & M_{1},V_{1}\mbox{ are both Noetherian}\\
0\ar[r] & M_{1}M_{2}\ar[r] & M_{1}\ar[r] & V_{2}\ar[r] & 0 &  & \iff & M_{1}M_{2},V_{1},V_{2}\mbox{ are all Neotherian}\\
0\ar[r] & M_{1}M_{2}M_{3}\ar[r] & M_{1}M_{2}\ar[r] & V_{3}\ar[r] & 0 &  & \iff & M_{1}M_{2}M_{3},V_{1},V_{2},V_{3}\mbox{ are all Noetherian}\\
 &  & \vdots &  &  &  & \dots & \dots\\
0\ar[r] & \overset{M_{1}M_{2}\dots M_{n}}{=0}\ar[r] & M_{1}M_{2}\dots M_{n-1}\ar[r] & V_{n}\ar[r] & 0 &  & \iff & V_{1},V_{2},\dots V_{n}\mbox{ are all Noetherian}\\
0\ar[r] & \overset{M_{1}M_{2}\dots M_{n}}{=0}\ar[r] & M_{1}M_{2}\dots M_{n-1}\ar[r] & V_{n}\ar[r] & 0 &  & \iff & V_{1},V_{2},\dots,V_{n}\mbox{ are all Artinian}\\
 &  & \vdots &  &  &  & \dots & \dots\\
0\ar[r] & M_{1}\ar[r] & R\ar[r] & V_{1}\ar[r] & 0 &  & \iff & R\mbox{ is Artinian}
}
\]
\end{proof}
\begin{prop}
Let $R$ be a Noetherian local ring with maximal ideal $M$. Then
\uline{either} $M^{n}\neq M^{n+1}$ for all $n\geq1$. \uline{Or}
$M^{n}=0$ for some $n$ in which case $R$ is Artinian and $M$ is
its only prime ideal.\end{prop}
\begin{proof}
Suppose $M^{n}=M^{n+1}$ for some $n$. Then $M^{n}=M^{n+1}=M^{n+1}=\dots$.
So $\cap_{k=1}^{\infty}M^{k}=M^{n}$, but by Corollary \ref{cor:3 of last lecture}
we have $\cap_{k=1}^{\infty}M^{k}=0$, hence $M^{n}=0$. By previous
proposition, $R$ is Artinian. 

Let $P$ be a prime of $R$. Then $P\supseteq0=M^{n}$, taking radicals
$P=r(P)\supseteq r(M^{n})=M$, so $P=M$\end{proof}
\begin{defn}
A ring in which every prime is maximal is said to have \emph{dimension
$0$.}\end{defn}
\begin{example*}
Any field

$\zz/n\zz$ (since primes are $p\zz/n\zz$, $p\nmid n$)

Any finite ring (since every finite integral domain is a field)\end{example*}
\begin{prop}
Artinian rings have dimension $0$.\end{prop}
\begin{proof}
Let $P\lhd R$ be a prime. Let $\overline{R}=R/P$, a domain. Let
$\overline{x}\in\overline{R},$ $\overline{x}\neq0$ (so $x\in R\setminus P$).
Now in $\overline{R}$ we have $(\overline{x})\supseteq(\overline{x}^{2})\supseteq(\overline{x}^{3})\supseteq\dots$.
By Descending Chain Condition in $\overline{R}$ (which is also Artinian)
there exists $n$ such that $(\overline{x}^{n})=(\overline{x}^{n+1})$,
so $\overline{x}^{n}=\overline{x}^{n+1}\overline{y}$ for some $\overline{y}\in\overline{R}$.
Since $\overline{x}\neq0$ and $\overline{R}$ is a domain, cancel
$\overline{x}$ from both sides $n$ times to get $1=\overline{x}\overline{y}$.
Hence $\overline{R}$ is a field and $P$ is maximal.\end{proof}
\begin{prop}
\label{prop:Artinian has finite maximal}An Artinian ring $R$ has
only finitely many maximal ideals.\end{prop}
\begin{proof}
Consider the set of all finite intersections of maximal ideals $M_{1}\cap M_{2}\cap\dots\cap M_{n},n\geq1$.
Since $R$ Artinian, this set has a minimal element $M_{1}\cap M_{2}\cap\dots\cap M_{n}=I$.
Let $M$ be any maximal ideal in $R$. Then $M\cap I\subseteq I$,
so by minimality of $I$ we have $M\cap I=I\Rightarrow M\supseteq I=M_{1}\cap\dots\cap M_{n}\Rightarrow M\supseteq M_{i}$
for some $i$ by Proposition \ref{prop:Intersection of prime contains prime},
hence $M=M_{i}$ for some $i$.\end{proof}
\begin{prop}
Let $R$ be an Artinian ring, then $N(R)=J(R)$ is nilpotent, i.e.,
$(N(R))^{k}=0$ for some $k\geq1$.\end{prop}
\begin{proof}
Let $N:=N(R)$, and consider $N\supseteq N^{2}\supseteq N^{3}\supseteq\dots$
so by the Descending Chain Condition there exists $k$ such that $N^{k}=N^{k+1}=N^{k+2}=\dots=:I$.
We want to show that $I=0$. Suppose $I\neq0$. Let $S=\{\mbox{ideals }J\lhd R\mbox{ such that }IJ\neq0\}$.
Notice $S\neq\emptyset$ since $R\in S$ as $I\neq0$. So let $J\in S$
be minimal (which exists since $R$ is Artinian). Then $\exists x\in J$
such that $xI\neq0$, so $(x)\subseteq(J)$ and $(x)I\neq0$ so $(x)\in S$
and by minimality $J=(x)$. Now $((x)I)I=(x)I^{2}=(x)I\neq0$ since
$I^{2}=I$, so $(x)I\in S$ and $(x)I\subseteq(x)=J$ so by minimality
of $J$ we have $(x)I=(x)$. So there exist $y\in I$ such that $xy=x\Rightarrow xy=xy^{2}=xy^{3}=\dots=xy^{n}=\dots$,
but $y\in I\subseteq N$ so $y$ is nilpotent, so $y^{n}=0$ for some
$n\Rightarrow x=0$. This contradicts the fact $I\neq0=(x)$\end{proof}
\begin{prop}
Every Artinian ring $R$ is Noetherian\end{prop}
\begin{proof}
Let $M_{1},M_{2},\dots,M_{n}$ be the complete set of all maximal
ideals of $R$ (by Proposition \ref{prop:Artinian has finite maximal}).
So $N=N(R)=J(R)=\cap_{i=1}^{n}M_{i}$. Also $N^{k}=0$ for some $k\geq1$.
Consider $M_{1}^{k}M_{2}^{k}\dots M_{n}^{k}=(M_{1}M_{2}\dots M_{n})^{k}\subseteq(M_{1}\cap M_{2}\cap\dots\cap M_{n})^{k}=N^{k}=0$.
So $M_{1}^{k}M_{2}^{k}\dots M_{n}^{k}=0$ so by Proposition \ref{prop:Artinian iff Noetherian}
we have that $R$ Artinian $\Rightarrow$ $R$ Noetherian.\end{proof}
\begin{rem*}
Every Noetherian ring of dimension $0$ is Artinian. (c.f. Atiyah
and Macdonald pg.90)\end{rem*}
\begin{namedthm}[The Structure Theorem for Artinian Rings]
 Every Artinian ring is uniquely isomorphic to a \uline{finite}
direct product of Artinian \uline{local }rings.\end{namedthm}
\begin{proof}
Existence: Let $M_{1},\dots,M_{n}$ be the maximal ideals of $R$.
Then $\prod_{i=1}^{n}M_{i}^{k}=0$ for some $k$. The ideals $M_{i}^{k}$
are pairwise comaximal so by the Chinese Remainder Theorem we have
\begin{eqnarray*}
R & = & R/0\\
 & = & R/\prod_{i=1}^{n}M_{i}^{k}\\
 & = & R/\bigcap_{i=1}^{n}M_{i}^{k}\mbox{ by comaximality}\\
 & \cong & \bigoplus_{i=1}^{n}R/M_{i}^{k}\mbox{ by Chinese Remainder Theorem}
\end{eqnarray*}
Now each $R/M_{i}^{k}$ has only one maximal ideal, $M_{i}/M_{i}^{k}$
so is an Artinian local ring.

Uniqueness: c.f. Atiyah and Macdonald pg. 90
\end{proof}

\subsection{Noetherian integral domains of dimension 1}

\textbf{Including Dedekind domain and Discrete Valuation Rings}
\begin{defn}
The \emph{dimension} of a ring $R$ is the maximal length ($\geq0$)
of a chain of prime ideals $P_{0}\subsetneqq P_{1}\subsetneqq\dots\subsetneqq P_{n}$
in $R$. \end{defn}
\begin{description}
\item [{Dim$ $$0$:}] All primes are maximal
\item [{Dim$ $$1$:}] e.g., $R=\zz$ and any integral domain, not a field
in which all non-zero primes are maximal.\end{description}
\begin{example*}
$k[x_{1},\dots,x_{n}]$ has dimension $n$. \end{example*}
\begin{prop}
Let $R$ be a Noetherian domain of dimension $1$. Then every non-zero
ideal $I$ of $R$ has a unique expression as a product of primary
ideals with distinct radicals.\end{prop}
\begin{proof}
Let $I=Q_{1}\cap\dots\cap Q_{n}$ with each $Q_{i}$ primary and each
$P_{i}=r(Q_{i})$ maximal. ($P_{i}\supseteq Q_{i}\supseteq I\neq0$).
No $P_{i}\subseteq P_{j}$ ($i\neq j$) so no embedding primes, hence
the $Q_{i}$ are unique. The $P_{i}$ are pairwise comaximal ($P_{i}+P_{j}=R$
for all $i\neq j$) hence so are the $Q_{i}$. To see this $r(Q_{i}+Q_{j})=r(P_{i}+P_{j})=r(1)=(1)\Rightarrow Q_{i}+Q_{j}=(1)$.
Hence $Q_{1}\cap\dots\cap Q_{n}=Q_{1}\dots Q_{n}$.

Conversely if $I=Q_{1}'Q_{2}'\dots Q_{m}'$ where $Q_{i}'$ are primary
with distinct radicals $r(Q_{i}')$ . As before the $Q_{i}'$ are
comaximal, so $I=\prod Q_{i}'=\cap Q_{i}'$. By uniqueness of primary
decomposition, $m=n$ and $Q_{i}=Q_{i}'$ after permuting.
\end{proof}
Recall: A DVR (Discrete Valuation Ring) is the valuation ring $ $$R$
of a ($\zz$-valued) discrete valuation $\nu:R\to\zz\cup\{\infty\}$.
Such an $R$ has the properties:
\begin{itemize}
\item $R$ is local with maximal ideal $M=\{x:\nu(x)\geq1\}$
\item $M$ is principal: $M=(\pi)$ with $\nu(\pi)=1$
\item All non-zero ideals of $R$ are $M^{n}=(\pi^{n})$, $n\geq0$.
\item Hence $R$ is Noetherian (it's a PID) and a domain 
\item $R$ has dimension $1$ since the only primes are $0$ and $M$\end{itemize}
\begin{lem}
Let $R$ be a Noetherian integral domain of dimension $1$ which is
local, with maximal ideal $M$ and residue field $k=R/M$. Then 
\begin{enumerate}
\item Every ideal $I\neq(0),(1)$ is $M$-primary, so $I\supseteq M^{n}$
for some $n$.
\item $M^{n}\neq M^{n+1}\,\forall n\geq0$
\end{enumerate}
\end{lem}
\begin{proof}
$R$ has two prime ideal, $(0)$ and $M$. Let $I\lhd R$ with $I\neq(0),(1)$,
then $r(I)=$ intersections of the primes containing $I$. So $r(I)=M$,
and $M$ is maximal, hence $I$ is $M$-primary. Now $I\supseteq M^{n}$
for some $n\geq1$ since $R$ is Noetherian.

If $M^{n}=M^{n+1}$ then $M^{n}=0$ which implies $R$ has dimension
$0$.\end{proof}
\begin{prop}
\label{prop:DVR proposition}Let $R$ be a Noetherian integral domain
of dimension $1$ which is local, with maximal ideal $M$ and residue
field $k=R/M$. Then the following are equivalent:
\begin{enumerate}
\item $R$ is a DVR,
\item $R$ is integrally closed,
\item $M$ is principal,
\item $\dim_{k}(M/M^{2})=1$,
\item every non-zero ideal of $R$ is a power of $M$,
\item there exists $\pi\in R$ such that every non-zero ideal is principal,
of the form $(\pi^{n}),\, n\geq0$.
\end{enumerate}
\end{prop}
\begin{proof}
{}
\begin{description}
\item [{$1\Rightarrow2$:}] Every valuation ring is integrally closed (See
Proposition \ref{prop:valuation ring properties})
\item [{$2\Rightarrow3$:}] Let $a\in M$, $a\neq0$. If $(a)=M$ we are
done. Otherwise $(a)\subsetneqq M$. Choose $n\geq0$ such that $M^{n}\subseteq(a),M^{n-1}\nsubseteq(a)$.
Such an $n$ exists since (by the previous lemma) $r((a))$ is a power
of $M$ and $(a)\supseteq M^{n}$ for some $n$. Choose $b\in M^{n-1}\setminus(a)$
so $\frac{b}{a}\notin R$. Let $x=\frac{a}{b}\in K$, the field of
fractions of $R$. 


\uline{Claim} $M=(x)$.


Since $b\notin(a)$, $x^{-1}\notin R$. Since $R$ is integrally closed,
$x^{-1}$ is not integral over $R$. This means that $x^{-1}M\nsubseteq M$.
To see this suppose $x^{-1}M\subseteq M$, then $M$ is a module over
the ring $R[x^{-1}]$ which is a finitely generated $R$-module, since
$R$ is Noetherian, and faithful as an $R[x^{-1}]$-module (since
$K$ has no zero-divisors so if $y\in R[x^{-1}]$ satisfies $yM=0$
then $y=0$); and these would imply that $x^{-1}$ is integral over
$R$. But $x^{-1}M\subseteq R$, since $bM\subseteq M^{n-1}M=M^{n}\subseteq(a)$.
So $x^{-1}M$ is an ideal of $R$ not contained in its unique maximal
ideal. Hence $x^{-1}M=R$, and hence $M=(x)$ proving the claim.

\item [{$3\Rightarrow4$:}] Let $M=(x)$, i.e., $x$ generates $M$ (as
$R$-module), so $\overline{x}$ generates $M/M^{2}$ (as $k=R/M$-module),
i.e., $\dim_{k}M/M^{2}\leq1$. But $M\neq M^{2}\Rightarrow M/M^{2}\neq0$
hence $\dim_{k}M/M^{2}\geq1$.
\item [{$4\Rightarrow5$:}] For any $\overline{x}$ which generates $M/M^{2}$,
the element $x\in R$ generates $M$. (By Corollary \ref{cor:Naka cor}).
So $M=(x)$, so $M^{n}=(x^{n})$ $(\forall n\geq0$). Let $I$ be
a proper non-zero ideal of $R$. So $I\subseteq M$, since $\cap_{k=1}^{\infty}M^{k}=0$
there exists $n\geq1$ such that $I\subseteq M^{n}$ and $I\nsubseteq M^{n+1}$.
Let $y\in I\setminus M^{n+1}$, since $y\in I\subseteq M^{n}=(x^{n})$,
we have $y=cx^{n}$, with $c\notin M=(x)$. So $c$ is a unit of $R$,
so $M^{n}=(x^{n})=(y)\subseteq I\subseteq M^{n}$. Therefore $I=M^{n}$
\item [{$5\Rightarrow6$:}] Let $\pi\in M\setminus M^{2}$. Then $(\pi)=M$
by $5.$ so every non-zero ideal $I=M^{n}=(\pi^{n})$.
\item [{$6\Rightarrow1$:}] Note that $M=(\pi)$ where $\pi$ is given
as in 6. So $M^{n}=(\pi^{n})$ $\forall n\geq0$. Let $a\in R,a\neq0$,
then $(a)=M^{n}$ for some $n\geq0$. Define $\nu(a)=n$. Extend to
a function $\nu:K^{*}\to\zz$ by setting $\nu(\frac{a}{b})=\nu(a)-\nu(b)\in\zz$.
Easy check that: 

\begin{enumerate}
\item $\nu$ is well define
\item $\nu$ is a group homomorphism. ($\nu(xy)=\nu(y)+\nu(x)$)
\item $\nu(\pi)=1\Rightarrow\nu$ is surjective
\item $\nu(x+y)\geq\min\{\nu(x),\nu(y)\}$
\end{enumerate}

So $\nu$ is a discrete valuation and $R=\{x\in K:\nu(x)\geq0\}$

\end{description}
\end{proof}

\subsection{Dedekind Domains}

These are Noetherian integral domains $R$ of dimension $1$ such
that every localization $R_{p}$ (for all maximal $p$) is a DVR. 
\begin{lem}[Definition]
 A \emph{Dedekind Domain} $R$ is a Noetherian integral domain of
dimension $1$ satisfying any of the following equivalent conditions:
\begin{enumerate}
\item $R$ is integrally closed.
\item Every primary ideal of $R$ is a prime power.
\item Every localization $R_{p}$ (at non-zero primes $P$) is a DVR.
\end{enumerate}
\end{lem}
\begin{proof}
{}
\begin{description}
\item [{$1\iff3$:}] Since being integrally closed is a local property,
so we use the Proposition \ref{prop:DVR proposition}.
\item [{$2\Rightarrow3$:}] Let $P$ be a non-zero prime and let $M=P_{p}$
be the extension of $P$ to $R_{p}$, so $M$ is the unique maximal
ideal in $R_{p}$. Every ideal ($\neq(0),(1)$) in $R_{p}$ is $M$-primary.
Every $P$-primary ideal of $R$ is a power of $P$ (by condition
2.) so its extension to $R_{p}$ is $M$-primary and is a power of
$M$. So all non-zero ideals of $R_{p}$ are powers of $M$. So we
can use 5. from Proposition \ref{prop:DVR proposition} and hence
$R_{p}$ is a DVR.
\item [{$3\Rightarrow2$:}] Let $Q$ be $P$-primary in $R$ (where $P$
is a non-zero prime). Its extension to $R_{p}$ is $M$-primary so
is a power of $M$, hence $Q$ is a power of $P$. Since $Q=(M^{n})^{c}=(M^{c})^{n}=P^{n}$
\end{description}
\end{proof}
\begin{cor}
In a Dedekind domain, every non-zero ideal has a unique factorization
as a product of prime ideals.
\end{cor}
Let $I$ be an ideal of a Dedekind domain $R$. Then $I=P_{1}^{n_{1}}P_{2}^{n_{2}}\dots P_{k}^{n_{k}}$
with each $P_{i}$ distinct maximal and $n_{i}\geq1$. If $P$ is
any non-zero prime the extension of $I$ in $R_{P}$ is the product
of the extensions of the $P_{i}^{n_{i}}$in $R_{p}$. If $P_{i}\neq P$,
the extension is the whole ring $R_{p}$. If $P_{i}=P$ the extension
is the maximal ideal of $R_{p}$, $P_{p}$. So $I_{p}=P_{p}^{n}$
where $n$ is the exponent of $P$ in the factorization of $I,n\geq0$. 

Define $\nu_{p}$ to be the discrete valuation which has valuation
ring $R_{p}$, so $\nu_{p}$ is a discrete valuation of the field
of fractions $K$ of $R$. Hence 
\[
I=\prod_{P\mathrm{\, non-zero\, prime}}P^{\nu_{P}(I)}.
\]
Consequences:
\begin{itemize}
\item $\xyR{3pc}\xyC{2pc}\xymatrix{I\subseteq J\ar@{<=>}[d]\ar@{<=>}[r] & J|I\ar@{<=>}[d]\\
\forall P:I_{P}\subseteq J_{P}\ar@{<=>}[r] & \nu_{p}(J)\leq\nu_{p}(I)\,\forall P
}
$Note $I_{p}=P_{p}^{\nu_{p}(I)},J_{p}=P_{p}^{\nu_{p}(J)}$. Therefore
$I\subseteq J\iff J|I$ ``to contain is to divide''
\item $\nu_{p}(I+J)=\min\{\nu_{p}(I),\nu_{p}(J)\}$
\item $\nu_{p}(I\cap J)=\max\{\nu_{p}(I),\nu_{p}(J)\}$
\item $\nu_{p}(IJ)=\nu_{p}(I)+\nu_{p}(J)$
\end{itemize}

\subsection{Examples of Dedekind Domains}
\begin{enumerate}
\item Every PID is a Dedekind Domain.

\begin{itemize}
\item Noetherian (every ideal has 1 generator)
\item Integrally closed (since a UFD)
\item Dimension 1 (the non-zero primes are $(\pi)$ with $\pi$ irreducible
- these are maximal)
\end{itemize}
\item Let $K$ be a number field, i.e, a finite extension (field) of $\qq$,
of degree $n$. $n=[K:\qq]=\dim_{\qq}K$.


The ring of integers $\mathcal{O}_{K}$ is the integral closure of
$\zz$ in $K$, i.e., $\mathcal{O}_{K}$ is the set of all algebraic
\uline{integers} in $K$. 


\uline{Claim}: $\mathcal{O}_{K}$ is a Dedekind Domain
\begin{prop*}
$\mathcal{O}_{K}$ is a free $\zz$-module of rank $n$, i.e., there
exists $\alpha_{1},\dots,\alpha_{n}\in\mathcal{O}_{K}$ such that
$\mathcal{O}_{K}=\zz\alpha_{1}+\zz\alpha_{2}+\dots+\zz\alpha_{n}$
(\emph{``integral basis''}). This implies $K=\qq\alpha_{1}+\dots\qq\alpha_{n}$. \end{prop*}
\begin{proof}
Omitted (See Algebraic Number Theory Course)\end{proof}
\begin{cor}
$\mathcal{O}_{K}$ is Noetherian.
\end{cor}

$\mathcal{O}_{K}$ is integrally closed, being in the integral closure
of $\zz$ in $K$. We need to check that it has dimension 1. Let $P$
be a non-zero prime of $\mathcal{O}_{K}$. We want to show that $P$
is maximal.
\begin{description}
\item [{Method~1:}] Show $\mathcal{O}_{K}/P$ is finite. (In fact $P$
is also a free $\zz$-module of rank $n$). Now every finite integral
domain is a field so $P$ is maximal.
\item [{Method~2:}] Consider $P\cap\zz$, this is a prime ideal of $\zz$.
It is non-zero since $\mathcal{O}_{K}$ is an integral extension of
$\zz$ so we cannot have both $0$ and $P$ (prime of $\mathcal{O}_{K}$)
contracting to $0$, primes of $\zz$. So $P\cap\zz=p\zz$ for some
prime number $p$. Now $p\zz$ is maximal so $P$ is maximal.
\end{description}

All of this proves that $\mathcal{O}_{K}$ is a Dedekind Domain.


Two special properties of $\mathcal{O}_{K}$, not shared by Dedekind
Domains in general:
\begin{enumerate}
\item (Dirichlet) $\mathcal{O}_{K}^{\times}$ (the group of units) is finitely
generated. If $K=\qq(\alpha)$, where $\alpha$ has minimal polynomial
$f(x)\qq[x]$, irreducible of degree $n$ (the degree of the number
field). Let $m$ be the number of irreducible factors of $f$ in $\rr[x]$.
Then there exists units $\epsilon_{1},\dots,\epsilon_{m-1}\in\mathcal{O}_{K}^{\times}$
such that every unit is uniquely $\zeta\epsilon_{1}^{n_{1}}\epsilon_{2}^{n_{2}}\dots\epsilon_{m-1}^{n_{m-1}}$,
where $\zeta$ is a root of unity and $n_{j}\in\zz$. 
\item Let $I,J\lhd\mathcal{O}_{K}$ be non-zero ideals. Define an equivalence
relation: $I\sim J\iff\alpha I=\beta J$ with $\alpha,\beta\in\mathcal{O}_{K}$
and non-zero. In particular $I\sim\mathcal{O}_{K}$ if and only if
$I$ is principal. 

\begin{xca*}
$I\sim J\iff I\cong J$ as $\mathcal{O}_{K}$-module
\end{xca*}

The equivalence classes form a group (induced by ideal multiplication),
i.e., $\forall I$ there exists $J$ such that $IJ$ is principal.
This is called the ideal class group (attached to any Dedekind Domain).
For rings of integers $\mathcal{O}_{K}$ it is a \uline{finite}
group.

\end{enumerate}
\item The coordinate ring of a smooth irreducible plane curve $C$. Let
$f\in\cc[X,Y]$ be irreducible then $C=\{(a,b)\in\cc^{2}:f(a,b)=0\}$.
The coordinate ring of $C$ is $R=\cc[X,Y]/(f)=\cc[x,y]$ with $f(x,y)=0$.
This is an integral domain (since $f$ is irreducible) 


\uline{Claim }$R$ is a Dedekind Domain:
\begin{itemize}
\item $R$ is Noetherian (By the Hilbert Basis Theorem)
\item Every non-zero prime of $R$ is maximal. 

\begin{proof}
Let $P$ be a prime of $\cc[X,Y]$ with $P\supsetneqq(f)$. Let $g\in P\setminus(f)$,
so $\gcd(f,g)=1$. View $f,g\in\cc(X)[Y]$ (as this as Euclidean property),
then there exists $a,b\in\cc(X)[Y]$ such that $af+bg=1$. Write $a=\frac{a_{1}}{d},b=\frac{b_{1}}{d}$
where $a_{1},b_{1}\in\cc[X,Y]$ and $d\in\cc[X]$, $d\neq0$. So $a_{1}f+b_{1}g=d$
$\Rightarrow$ the set of common zero of $f,g$ has only finitely
many $x$-coordinate (roots of $d$). So $f,g$ have only finitely
many common zeroes. In fact there is only \uline{one} common zero,
$(x_{0},y_{0})$, (after some work) this implies $P=(X-x_{0},Y-y_{0})$
which is maximal. (Fill in the gaps yourself)
\end{proof}
\item We'll show that every localization $R_{P}$ is a DVR, where $P$ a
non-zero prime of $R$. Without loss of generality, $P=(x,y)$, i.e.,
$P$ is associated to the point of $(0,0)$. $P$ is smooth: $\frac{\partial f}{\partial Xx},\frac{\partial df}{\partial Y}$
do not vanish at $(0,0)$. So $f=aX+bY+$higher term, $a,b$ not both
zero. Without loss of generality, we can assume $a=0$ and $b=1$.
So $Y=0$ at the tangent to $C$ at $(0,0)$. Now $f(X,Y)=Y\cdot G(X,Y)+X^{2}H(X)$
with $G(0,0)=1$. Module $f$ we have $0=y\cdot g+x^{2}\cdot h$ where
$g=G(x,y),h=h(x)\in R$. The maximal ideal of $R_{P}$ is generated
by $x,y$. $R_{P}=\{\frac{r(x,y)}{s(x,y)}|r,s\in R,s(0,0)\neq0\}$.
The maximal ideal $PR_{P}$ is $\{\frac{r}{s}:r(0,0)=0,s(0,0)\neq0\}$,
i.e., $r\in P$. But $yg=-x^{2}h$ so $y=-x^{2}\frac{h}{g}$ where
$g(0,0)=1\neq0$, so $-x^{2}\frac{h}{g}\in R_{P}$. So $x$ alone
generates $P\cdot R_{P}$, hence $R_{P}$ is a DVR.\end{itemize}
\end{enumerate}

\end{document}
