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<record version="3" id="410">
 <title>maximal ideal</title>
 <name>MaximalIdeal</name>
 <created>2001-10-20 01:43:56</created>
 <modified>2002-04-20 00:02:21</modified>
 <type>Definition</type>
 <creator id="24" name="djao"/>
 <author id="24" name="djao"/>
 <classification>
	<category scheme="msc" code="13A15"/>
	<category scheme="msc" code="16D25"/>
 </classification>
 <related>
	<object name="ProperIdeal"/>
	<object name="Module"/>
	<object name="Comaximal"/>
	<object name="PrimeIdeal"/>
	<object name="EveryRingHasAMaximalIdeal"/>
 </related>
 <preamble>\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{graphicx}
\usepackage{xypic}</preamble>
 <content>Let $R$ be a ring with identity. A proper left (right, two-sided) ideal $\mathfrak{m} \subsetneq R$ is said to be {\em maximal} if $\mathfrak{m}$ is not a proper subset of any other proper left (right, two-sided) ideal of $R$.

One can prove:
\begin{itemize}
\item A left ideal $\mathfrak{m}$ is maximal if and only if $R/\mathfrak{m}$ is a simple left $R$-module.
\item A right ideal $\mathfrak{m}$ is maximal if and only if $R/\mathfrak{m}$ is a simple right $R$-module.

\item A two-sided ideal $\mathfrak{m}$ is maximal if and only if $R/\mathfrak{m}$ is a simple ring.
\end{itemize}

All maximal ideals are prime ideals. If $R$ is commutative, an ideal $\mathfrak{m} \subset R$ is maximal if and only if the quotient ring $R/\mathfrak{m}$ is a field.</content>
</record>
