Is it known whether every finite abelian group is isomorphic to the ideal class group of the ring of integers in some number field? If so, is it still true if we consider only imaginary quadratic fields?

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    $\begingroup$ The Cohen-Lenstra heuristics give predictions about the answers to these sorts of questions. $\endgroup$ Nov 19, 2010 at 15:25
  • $\begingroup$ every abelian group is the class group of some dedekind domain (not necessarily the ring of integers of a number field) - projecteuclid.org/euclid.pjm/1102994263 $\endgroup$
    – yoyo
    May 13, 2014 at 2:55
  • $\begingroup$ (The result cited by yoyo above is Claborn's theorem). $\endgroup$
    – Watson
    Aug 16, 2018 at 16:04
  • $\begingroup$ Related: mathoverflow.net/questions/252313 $\endgroup$
    – Watson
    Aug 16, 2018 at 16:06

3 Answers 3


The smallest abelian group which is not the class group of an imaginary quadratic field is $(\mathbf{Z}/3 \mathbf{Z})^3$. There are six other groups of order $< 100$ which do not occur in this way, of orders $32$, $27$, $64$, $64$, $81$, and $81$ respectively. The groups $(\mathbf{Z}/3 \mathbf{Z})^2$ and $(\mathbf{Z}/2 \mathbf{Z})^4$ occur as class groups of imaginary quadratic fields exactly once, for $D = -4027$ and $-5460$ respectively. These results follow from the "class number $100$" problem, solved by Mark Watkins.

If you restrict to the $p$-part of the class group, then the answer (for general number fields) is positive. That is, for any abelian $p$-group $A$, there exists a number field $K$ with class group $C$ such that $C \otimes \mathbf{Z}_p = A$. There is even a non-abelian analog of this. Namely, for any finite $p$-group $G$, there exists a number field $K$ such that the maximal Galois $p$-extension $L/K$ unramified everywhere has Galois group $G$. This is a very recent result of Ozaki.

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    $\begingroup$ Wow, Ozaki's result sounds quite impressive (there are a lot of $p$-groups). Could you provide a reference? $\endgroup$ Nov 20, 2010 at 3:20
  • $\begingroup$ springerlink.com/content/c887p5226753075v $\endgroup$ Nov 20, 2010 at 5:37
  • $\begingroup$ @Pete, I can't comment on your answer, but I think you meant to say that the order of the group $2 H_D$ goes to infinity. The order of $H_D/2 H_D$ (as you know) corresponds to the genus field, which has order $2^{d-1}$ where $d$ is the number of prime factors of the discriminant. $\endgroup$
    – ABC
    Nov 20, 2010 at 6:39
  • $\begingroup$ Yes, you're absolutely right: thank you. (This is not the first time I've been a little dyslexic on this issue.) $\endgroup$ Nov 20, 2010 at 12:27
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    $\begingroup$ That $({\mathbf Z}/3{\mathbf Z})^2$ is the ideal class group of a unique imag. quad. field, ${\mathbf Q}(\sqrt{-4027})$, is originally due not to Watkins (2004), but to Arno, Robinson, and Wheeler (1998). They found all imag. quadratic fields with class number 9, and a check of the list shows all have cyclic class group except for one. A list with reference is at numbertheory.org/classnos. $\endgroup$
    – KCd
    Jun 30, 2012 at 14:24

Virtually nothing is known about the question of which abelian groups can be the ideal class group of (the full ring of integers of) some number field. So far as I know, it is a plausible conjecture that all finite abelian groups (up to isomorphism, of course) occur in this way. Conjectures and heuristics in this vein have been made, but unfortunately for me I'm not so familiar with them.

The situation for imaginary quadratic fields is different. Here there is an absolute bound on the size of an integer $k$ such that the class group of an imaginary quadratic field can be isomorphic to $(\mathbb{Z}/2\mathbb{Z})^k$. Conditionally on the Generalized Riemann Hypothesis, the largest such $k$ is $4$. This has do to with idoneal numbers, of which the following paper provides a very fine survey:


Actually the truth is slightly stronger: let $H_D$ be the class group of the imaginary quadratic field $\mathbb{Q}(\sqrt{-D})$. Then, as $D$ tends to negative infinity through squarefree numbers, the size of $2H_D$ (the image of multiplication by $2$) tends to infinity. See for instance


for some recent explicit bounds on this.

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    $\begingroup$ Maybe I'm missing something... why is $H_D/2H_D \rightarrow \infty$ a stronger statement than the boundedness of $k$? If anything, it seems to give evidence to the contrary. $\endgroup$ Nov 20, 2010 at 6:00
  • $\begingroup$ @Kevin: again, absolutely correct. I've fixed it. $\endgroup$ Nov 20, 2010 at 12:30
  • $\begingroup$ Dear @PeteL.Clark, I mentioned your answer to report the "error" in the last paragraph of this page, which claims that « Every finite Abelian group occurs as the ideal class group for some suitable choice of number field. » ! $\endgroup$
    – Watson
    Aug 16, 2018 at 16:09

At least as of 1999, this was still an open question, according to the MathReview of a paper by Marc Perret (On the ideal class group problem for global fields, J. Number Theory 77 (1999), no. 1, pages 27-35; MR1695698 (2000d:11135), review by Bruno Anglès).

Luther Claborn proved in 1966 that every abelian group is the ideal class group of some Dedekind domain (Every abelian group is a class group, Pacific J. Math. 18 (1966), pages 219-222, MR0195889 (33 #4085)).

Gary Cornell (Abhyankar's lemma and the class group. Number theory, Carbondale 1979 (Proc. Southern Illinois Conf., Southern Illinois Univ., Carbondale, Ill., 1979), pp. 82–88, Lecture Notes in Math., 751, Springer, Berlin, 1979, MR0564924 (82c:12007)) proved that every finite abelian group is a subgroup of the ideal class group of a cyclotomic extension of $\mathbb{Q}$. There have been some refinements, but I did not find anything in MathSciNet that would suggest the problem has been settled in the interim.

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    $\begingroup$ Yeah, I was going to say something about the Claborn result. In fact, there's a nice result due to Mr. Clark above which says you can actually take the Dedekind domain to be a subring of the function field of some elliptic curve. $\endgroup$ Nov 20, 2010 at 5:48

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