I have only recently encountered algebraic number theory and was wondering if this is the case. If the answer to the question is yes, then can we explicitly construct the domain $D$ ?
Given a group $G$, does there exist a domain $D$ with $G$ as its ideal class group?
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algebraic-number-theory
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2$G$ must be abelian... – 2012-01-24
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1I know this is a repeat... Yes: [here it is](http://math.stackexchange.com/questions/10949/finite-abelian-groups-as-class-groups) for finite abelian groups. – 2012-01-24
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0@Fortuon: It should be obvious from the definition of ideal class group that it is abelian. If this is not obvious, perhaps you should first learn what the ideal class group is! – 2012-01-24
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0@ZhenLin: My apologies for the notation abuse. I have corrected it. – 2012-01-24
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1I don't understand your edit. The ideal class group is commutative because multiplication is commutative in $\mathcal{O}_K$, and any reasonable definition of fractional ideal classes of an integral domain $D$ will make use of multiplication in $D$. Your question doesn't make sense for general non-commutative domains because there isn't even a reasonable definition of fraction field in that case. – 2012-01-24
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0The paper by Claborn cited in @Mariano's answer does not seem to be restricted to *finite* groups. – 2012-01-27
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0If you feel that @Mariano has answered your question fully then just accept his answer. – 2012-01-27
1 Answers
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See Claborn, Luther (1966), "Every abelian group is a class group", Pacific J. Math. 18: 219–222. You can get it here
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0Our dear Pete Clark has a recent result on this subject. You should find it in his webpage. – 2012-01-24
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1Have you looked? – 2012-01-24