13A99-ArithmeticalRing.tex

\documentclass[12pt]{article}
\usepackage{pmmeta}
\pmcanonicalname{ArithmeticalRing}
\pmcreated{2013-03-22 15:23:58}
\pmmodified{2013-03-22 15:23:58}
\pmowner{PrimeFan}{13766}
\pmmodifier{PrimeFan}{13766}
\pmtitle{arithmetical ring}
\pmrecord{8}{37237}
\pmprivacy{1}
\pmauthor{PrimeFan}{13766}
\pmtype{Theorem}
\pmcomment{trigger rebuild}
\pmclassification{msc}{13A99}
\pmrelated{QuotientOfIdeals}

\endmetadata

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\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}

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%\usepackage{psfrag}
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%\usepackage{graphicx}
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 \usepackage{amsthm}
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\theoremstyle{definition}
\newtheorem*{thmplain}{Theorem}
\begin{document}
\begin{thmplain}
If $R$ is a commutative ring, then the following three conditions are equivalent:
\begin{itemize}
 \item For all ideals $\mathfrak{a}$, $\mathfrak{b}$ and $\mathfrak{c}$ of $R$, one has\, $\mathfrak{a\cap(b+c) = (a\cap b)+(a\cap c)}$.
 \item For all ideals $\mathfrak{a}$, $\mathfrak{b}$ and $\mathfrak{c}$ of $R$, one has\, $\mathfrak{a+(b\cap c) = (a+b)\cap(a+c)}$.
 \item For each maximal ideal $\mathfrak{p}$ of $R$ the set of all ideals of $R_{\mathfrak{p}}$, the \PMlinkname{localisation}{Localization} of $R$ at\, $R\!\setminus\!\mathfrak{p}$,\, is totally ordered by set inclusion.
\end{itemize}
\end{thmplain}

The ring $R$ satisfying the conditions of the theorem is called an {\em arithmetical ring}.
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\end{document}