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<record version="13" id="2854">
 <title>Dedekind domain</title>
 <name>DedekindDomain</name>
 <created>2002-04-19 22:33:03</created>
 <modified>2006-04-12 00:20:22</modified>
 <type>Definition</type>
 <creator id="2727" name="mathcam"/>
 <author id="2727" name="mathcam"/>
 <author id="225" name="saforres"/>
 <classification>
	<category scheme="msc" code="11R04"/>
	<category scheme="msc" code="11R37"/>
 </classification>
 <related>
	<object name="IntegralClosure"/>
	<object name="PruferDomain"/>
	<object name="MultiplicationRing"/>
	<object name="PrimeIdealFactorizationIsUnique"/>
	<object name="EquivalentCharacterizationsOfDedekindDomains"/>
	<object name="ProofThatADomainIsDedekindIfItsIdealsAreInvertible"/>
	<object name="ProofThatADomainIsDedekindIfItsIdealsAreProductsOfPrimes"/>
	<object name="ProofThatADomainIsDedekindIfItsIdealsAreProductsOfMaximals"/>
 </related>
 <keywords>
	<term>Noetherian</term>
	<term>finitely generated</term>
 </keywords>
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 <content>A {\em Dedekind domain} is a commutative integral domain $R$ for which:
  \begin{itemize}
  \item Every ideal in $R$ is finitely generated.
  \item Every nonzero prime ideal is a maximal ideal.
  \item The domain $R$ is integrally closed in its field of fractions.
  \end{itemize}

It is worth noting that the second clause above implies that the maximal length of a strictly increasing chain of prime ideals is 1, so the Krull dimension of any Dedekind domain is at most 1.  In particular, the affine ring of an algebraic set is a Dedekind domain if and only if the set is normal, irreducible, and 1-dimensional.

Every Dedekind domain is a noetherian ring.

If $K$ is a number field, then $\mathcal{O}_K$, the ring of algebraic integers of $K$, is a Dedekind domain.</content>
</record>
