\documentclass[12pt]{amsart}
\usepackage{amsmath,amsthm,amsfonts,latexsym,amssymb}
\pagestyle{headings}
\setlength{\textwidth}{36true pc}
\setlength{\headheight}{8true pt} % needed for 12pt documents
\setlength{\oddsidemargin}{15 truept}
\setlength{\evensidemargin}{15 truept}
\setlength{\textheight}{572true pt}
\newtheorem{thm}{Theorem}[section]
\newtheorem{cor}[thm]{Corollary}
\newtheorem{lm}[thm]{Lemma}
\newtheorem{prp}[thm]{Proposition}
\newtheorem{clm}[thm]{Claim}
%
\theoremstyle{definition}
\newtheorem{df}[thm]{Definition}
\newtheorem{exmp}[thm]{Example}
\newtheorem{prb}[thm]{Problem}
%
\theoremstyle{remark}
\newtheorem{rem}[thm]{Remark}
%
\numberwithin{equation}{section}
%
\newcommand{\thref}[1]{Theorem~\ref{#1}}
\newcommand{\secref}[1]{\S\ref{#1}}
\newcommand{\lmref}[1]{Lemma~\ref{#1}}
%
\DeclareMathOperator{\satf}{\models}
\DeclareMathOperator{\miff}{\Longleftrightarrow} 
                                           % math `if and only if'
\DeclareMathOperator{\mimplies}{\Longrightarrow} 
                                           % math `implies'
\DeclareMathOperator{\mimplied}{\Longleftarrow}  
                                           % math `implied by'
\newcommand{\set}[2]{\{#1\,:\,\text{#2}\}} % { x : ... }
\DeclareMathOperator{\id}{\textsl{id}}     % The identity map
\DeclareMathOperator{\miffdef}{\stackrel{def}{\miff}}
                                           % defining iff
\DeclareMathOperator{\eqdef}{\stackrel{def}{=}}
                                           % defining equation

\DeclareMathOperator{\neareq}{\stackrel{\ae}{=}}
\DeclareMathOperator{\aeqv}{\stackrel{ae}{\equiv}}
\DeclareMathOperator{\aeq}{\stackrel{ae}{=}}
%
\DeclareMathOperator{\BZ}{\mathbb{Z}} % integers
\DeclareMathOperator{\BQ}{\mathbb{Q}} % rationals
\DeclareMathOperator{\BR}{\mathbb{R}} % reals
\DeclareMathOperator{\BC}{\mathbb{C}} % complex numbers
\DeclareMathOperator{\BE}{\mathbb{E}} % 
\DeclareMathOperator{\BF}{\mathbb{F}} % 
\DeclareMathOperator{\BH}{\mathbb{H}} % 
\DeclareMathOperator{\BK}{\mathbb{K}} % 
\DeclareMathOperator{\BL}{\mathbb{L}} % 
\DeclareMathOperator{\BO}{\mathbb{O}} % 
\DeclareMathOperator{\re}{Re}         % real part
\DeclareMathOperator{\im}{Im}         % imaginary part
\DeclareMathOperator{\Ker}{Ker}
%
\DeclareMathOperator{\meet}{\wedge}
\DeclareMathOperator{\join}{\vee}
\DeclareMathOperator{\union}{\cup}
\DeclareMathOperator{\inter}{\cap}  % intersection
\DeclareMathOperator{\iso}{\cong}   % isomorphic
\DeclareMathOperator{\piso}{\simeq} % polynomially isomorphic
\newcommand{\Star}{^ *}
\DeclareMathOperator{\compose}{{\scriptstyle \circ}} % f o g
%
\newcommand{\wec}[1]{{\mathbf{#1}}}  % notation for vectors in algebras
%\newcommand{\wec}[1]{{\bar{#1}}} % which one do you prefer?
\newcommand{\wrel}[1]{\;#1\;}     % a\wrel{\alpha}b = a \alpha b
%              % this provides a flexible way to change the spacing
%
\newcommand{\m}[1]{{\mathbf{\uppercase{#1}}}}
\newcommand{\vr}[1]{{\mathcal{\uppercase {#1}}}}
%
\DeclareMathOperator{\Var}{\mathbf{V}}
\DeclareMathOperator{\Ho}{\mathbf{H}}
\DeclareMathOperator{\Su}{\mathbf{S}}
\DeclareMathOperator{\Pd}{\mathbf{P}}
\DeclareMathOperator{\Iso}{\mathbf{I}}
\DeclareMathOperator{\Ps}{\mathbf{P}_s}
\newcommand{\fin}[1]{{\mathbf{#1}}_{\text{\upshape fin}}}
%
\DeclareMathOperator{\Pol}{Pol}
\DeclareMathOperator{\Clo}{Clo}
\DeclareMathOperator{\Con}{Con}
\DeclareMathOperator{\Conb}{\mathbf{Con}} % Con(A) as a lattice
\DeclareMathOperator{\cg}{Cg}
\DeclareMathOperator{\sg}{Sg}
\newcommand{\Cg}[1]{{\cg}^{\m{#1}}}
\newcommand{\Sg}[1]{{\sg}^{\m{#1}}}
%
\newcommand{\M}[3]{\text{\upshape M}_{\m{#1}} (#2, #3)}
\newcommand{\minl}[1]{\text{\upshape Min}_{\m{#1}}}
%
\DeclareMathOperator{\typ}{typ}
\DeclareMathOperator{\ann}{ann}
\DeclareMathOperator{\Sep}{Sep}
\DeclareMathOperator{\ssb}{ssb}
\DeclareMathOperator{\sst}{sst}
%
\DeclareMathOperator{\utyp}{\mathbf{1}}% unary      
\DeclareMathOperator{\atyp}{\mathbf{2}}% affine     
\DeclareMathOperator{\btyp}{\mathbf{3}}% boolean    
\DeclareMathOperator{\ltyp}{\mathbf{4}}% lattice    
\DeclareMathOperator{\styp}{\mathbf{5}}% semilattice
\DeclareMathOperator{\ityp}{\mathbf{i}}
\DeclareMathOperator{\jtyp}{\mathbf{j}}
%
\DeclareMathOperator{\solv}{\stackrel{s}{\sim}}
\DeclareMathOperator{\ssolv}{\stackrel{ss}{\sim}}
%
\newcommand{\bd}{\begin{description}}
\newcommand{\ed}{\end{description}}
%
\newcommand{\lb}{\langle} % <...>
\newcommand{\rb}{\rangle}
\newcommand{\lv}{{[\![}}  % [[...]]
\newcommand{\rv}{{]\!]}}
%
\newcommand{\ul}{\underline}
\newcommand{\ol}{\overline}
%
\newcommand{\vep}{\varepsilon}
\newcommand{\di}{\diamond}
%


\begin{document}

\noindent
{\large \bf Additional notes \#7.}

\bigskip

\noindent
{\bf Chain Conditions.}

\bigskip

\noindent
{\bf Definition.}
An order set $\m p = \lb P; \leq \rb$ is {\it well-founded} if
every nonempty subset of $P$ has a minimal element.
($\m p$ is {\it well-ordered} if it is 
a well-founded linear order.)

$\m p$ {\it satisfies the descending chain condition (DCC)}
if it has no infinite strictly decreasing chain
$p_1 > p_2 > \cdots$.
Dually, $\m p$ {\it satisfies the ascending chain condition (ACC)}
if it has no infinite strictly increasing chain.

\medskip

\noindent
{\bf Exercise.} Show that $\m p$ is well-founded if and only
if it satisfies the DCC.

\medskip

Well-founded sets are important in mathematics because they are 
the ordered sets over which induction is possible: let 
$S_p, p\in P$ be a family of statements. Suppose that 
\begin{enumerate}
\item[(1)] $S_p$ is true for all minimal $p\in P$, and
\item[(2)] If $S_q$ is true for all $q<p$, then $S_p$ is true.
\end{enumerate}
Then all $S_p$ are true if $\m p$ is well-founded.
(To see why, let $F$ be the set of all $p$
such that $S_p$ is false. If $F\neq\emptyset$ then it has
a minimal element. Conditions (1) and (2) prevent this, so
$F$ is empty.)

\medskip

\noindent
{\bf Exercise.} 
Show that this type of induction is not valid over sets
that are not well-founded. (Hint: assume that $C$
is an infinite strictly decreasing chain in $\m p$. 
For each $p\in P$
let $S_p$ be the statement ``$p\not\in C$''. Show that 
(1) and (2) hold, but that not all $S_p$ are true.)

\medskip

\noindent
{\bf Definition.}
A ring $\m r$ is {\it (left, right) Artinian} if its lattice
of (left, right) ideals satisfies the DCC.
$\m r$ is {\it (left, right) Noetherian} if its lattice
of (left, right) ideals satisfies the ACC.

\medskip

\noindent
{\bf Exercise.} 
Show that $\m r$ is (left, right) Noetherian if and only
if each (left, right) ideal is finitely generated.

\medskip

\noindent
{\bf Examples.} 
\begin{enumerate}
\item[(1)] Any finite ring is 
is Artinian and Noetherian.
\item[(2)] Any matrix ring over a division ring
is left and right Artinian, as well as left and right 
Noetherian.
\item[(3)] If ${\BF}$ is a field, then any 
finite dimensional $\BF$-algebra 
is left and right Artinian, as well as left and right 
Noetherian (since any left or right ideal is 
a subspace).
\item[(4)] Any finitely generated commutative ring
is Noetherian.
\item[(5)] A finite product of (left, right) Artinian rings
is Artinian. The same is true for for
(left, right) Noetherian rings.
\item[(6)] A quotient of a (left, right) 
Artinian ring is (left, right) 
Artinian, The same is true for 
(left, right) Noetherian rings.
\end{enumerate}

\vfill\pagebreak

\noindent
{\bf Major Theorems About Rings Satisfying Chain Conditions.}

\bigskip

\noindent
{\bf Hopkins-Levitski Theorem.} A left Artinian ring is 
left Noetherian.

\bigskip

\noindent
{\bf Wedderburn-Artin Theorem.} If $\m r$ is left Artinian,
then $\m r$ has a largest {\it nilpotent}\footnote{An ideal
$I$ is nilpotent if $I^k = \{0\}$ for some $k$.} ideal $J$
(called {\it the radical} of $\m r$). The quotient
$\m r/J$ is isomorphic to a finite direct product
$M_{n_1}(\m d_1)\times \cdots \times M_{n_k}(\m d_k)$
of matrix rings over division rings.

\bigskip

Although $\m r/J$ has a direct product 
factorization into simple rings in the previous theorem, 
this decomposition 
usually cannot be pulled back to a direct product factorization
of $\m r$ unless $\m r$ is commutative. 
In the commutative case we can pull this factorization back
and get the following strengthening:

\bigskip

\noindent
{\bf Artin's Theorem.} If $\m r$ is an Artinian
commutative ring, then 
$\m r\cong \m r_1\times \cdots\times \m r_k$
where each $\m r_i$ has the property
that $\m r_i/J_i$ is a field.
($J_i$ is the radical of
$\m r_i$.)

\bigskip

The previous theorm shows that an arbitrary 
commutative ring satisfying the DCC behaves
like ${\BZ}_n$: if $n = p_1^{e_1}\cdots p_k^{e_k}$, then
\[{\BZ}_n\cong {\BZ}_{{p_1}^{e_1}}\times\cdots\times
{\BZ}_{{p_k}^{e_k}}\]
where $J_i = p_i{\BZ}_{{p_i}^{e_i}}$ satisfies
$\left(J_i\right)^{e_i} = \{0\}$ and
${\BZ}_{{p_i}^{e_i}}/J_i\cong {\BZ}_{p_i}$ is a field.

\bigskip

\noindent
{\bf Hilbert Basis Theorem.} If $\m r$ is a commutative
Noetherian ring, then $\m r[x]$ is Noetherian.
(Hence so is $\m r[x,y], \m r[x,y,z]$, etc.)

\bigskip






\end{document}

