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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 $D$ ?

Since the finite abelian group case has been answered/linked-to satisfactorily, comments on the infinite case would be appreciated!

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$G$ must be abelian... – lhf Jan 24 '12 at 2:04
I know this is a repeat... Yes: here it is for finite abelian groups. – Arturo Magidin Jan 24 '12 at 4:27
@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! – Zhen Lin Jan 24 '12 at 7:40
@ZhenLin: My apologies for the notation abuse. I have corrected it. – Ravi Jan 24 '12 at 16:16
I 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. – Zhen Lin Jan 24 '12 at 16:24
up vote 7 down vote accepted

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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Our dear Pete Clark has a recent result on this subject. You should find it in his webpage. – Mariano Suárez-Alvarez Jan 24 '12 at 2:11
Is the proof constructive? – Ravi Jan 24 '12 at 2:15
Have you looked? – Mariano Suárez-Alvarez Jan 24 '12 at 2:19
Yes, I tried to read the proof but it seems to involve many details that I do not understand. – Ravi Jan 24 '12 at 2:24

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