# Why is $(\mathbb{Z}/3\mathbb{Z})^3$ not a class group of an imaginary quadratic number field ?

In general, it seems not known which finite abelian groups are class groups of quadratic number fields. For imaginary quadratic number fileds, I read that $(\mathbb{Z}/3\mathbb{Z})^3$ is the smallest abelian group which does not occur. It does occur as the class group of $\mathbb{Q}(\sqrt{188184253})$. What other groups are known not to be the class groups of imaginary quadratic number fields ? How does one prove this ?

What is the situation for class groups of real quadratic number fields ?

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The only proof that I know that $(\mathbb{Z}/3\mathbb{Z})^3$ does not appear as the class group of a quadratic imaginary number field is by brute force search. Roughly, the idea is that since class numbers of quadratic fields increase with the discriminant, there are only finitely many such fields of any given class number. We can then enumerate all class groups of a given size, and check to see which class groups appear. Most recently, Watkins's computation of all quadratic imaginary number fields with class number up to 100 provides in particular a complete list of all class groups of order 27, and so by simple enumeration one can observer that $(\mathbb{Z}/3\mathbb{Z})^3$ does not appear. I should also mention in this regard the tremendous amount of data available in a recent article of Jacobson, Ramachandran, and Williams.

[Edit: As to another of your questions, the same technique shows 6 other class groups of order less than 100 do not appear as quadratic imaginary number fields, of orders 27,32,54,64,64,81,81 respectively, a fact I have stolen from http://math.stackexchange.com/a/11025/20762 (and now updated based on ABC's answer below).]

Lacking any particularly compelling arithmetic reasons which proscribe certain finite abelian groups from appearing as class groups of quadratic imaginary number fields, it's worth thinking about some heuristics. An argument of Soundararajan gives that, on average, the number of quadratic imaginary number fields of class number $p^n$ is roughly $p^n$ (or better, asymptotically between $p^n/n$ and $np^n$). On the other hand, the number of finite abelian groups of order $p^n$ is governed by the partition theory of $n$, and hence a standard asymptotic suggests there are $\frac{1}{4\sqrt{3}n}e^{\pi\sqrt{2n/3}}$ such groups. So roughly, there are increasingly more number fields of order $p^n$ than there are possible class groups of that size, so one might perhaps expect that (again, without any obvious arithemtical obstruction) that there are only finitely many examples like $(\mathbb{Z}/3\mathbb{Z})^3$ that do not appear, at least among $p$-groups (and especially for large $p$).

For real quadratic fields, there is no such bound pushing class numbers higher, and I can't see any reason not to expect every finite abelian group to appear as a class group of such a field. Of course, this result is nowhere near known.

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Thank you, Cam. Which groups are these, of orders $32$, $64$ and $81$ ? –  Dietrich Burde Jun 5 '13 at 15:30
The other ingredient is the Cohen-Lenstra heuristics, which suggest that among the groups of given order those with many automorphisms should be rarer. Since $({\bf Z}/3{\bf Z})^3$ has $26 \cdot 24 \cdot 18 = 11232$ automorphisms, and ${\bf Z}/27{\bf Z}$ only $18$, we expect the cyclic group to arise about $624$ times more often in the real quadratic case, and are not surprised that it does not arise at all among the few imaginary quadratic fields (IQFs) with $h=27$. It may be reasonable to conjecture that no odd elementary abelian group of rank $3$ or more is a class group of an IQF. –  Noam D. Elkies Jun 5 '13 at 16:50
P.S. The 2-part of the class group must be treated specially because of genus theory. –  Noam D. Elkies Jun 5 '13 at 16:52
Thanks, Noam. I had meant to mention Cohen-Lenstra at some point, but hadn't thought to do the actual automorphism count to make it explicit. That's pretty enlightening. –  Cam McLeman Jun 5 '13 at 21:42
The work of Watkins (or Arno et.al.) can be minutely simplified for $(Z/3)^3$. This is by the "structure of minima" they use. You cannot have a form of order $>3$, and forms of odd order pair conjugately, so every nonprincipal form $(a,b,c)$ must have $a^{(3+1)/2}\ge \sqrt{d/4}$ rather than $a^{(27+1)/2}\ge\sqrt{d/4}$ as with general order 27. I don't know how much this eases the situation, but the papers use similar facts to reduce the sieving problems in their cases. I would not be surprised if handling $(Z/3)^k$ for $k=5$ or even more is feasible. Even order case is much more difficult. –  v08ltu Jun 6 '13 at 4:26

I just gave an extended answer to this question which was eaten by captcha, but here is a short recap:

Real quadratic fields: Any odd abelian group $A$ should occur as the odd part of the class group for a positive density of discriminants (this is explicitly in Cohen-Lenstra). One thus expects that: any abelian group $A$ should occur as the class group with positive relative density amongst class groups which, by genus theory, have the same $2$-rank as $A$. These claims are all conjectural and nothing is known.

Imaginary quadratic fields: Conjecturally, the $p$-part of the class group is $O(|\Delta_F|^{\epsilon})$. On GRH, the $p$-part of the class group is $O(|\Delta_F|^{\delta +\epsilon})$ for some explicit constant $\delta < 1/2$ depending on $p$. For $p = 2$ this is unconditionally true by genus theory. For $p =3$ this is also unconditionally true, by Pierce, Helfgott-Venkatesh (independently), and later Ellenberg-Venkatesh.

By Brauer-Siegel, the class group has order at least $O(|\Delta_F|^{1/2 - \epsilon})$. Hence:

Unconditionally: For any abelian fixed group $A$, the groups $A \oplus (\mathbf{Z}/2 \mathbf{Z})^n$ and $A \oplus (\mathbf{Z}/3 \mathbf{Z})^n$ occur as class groups of imaginary quadratic fields for only finitely many $n$.

The result above is not effective, because Brauer-Siegel is not effective. (The effective lower bounds on class groups of Goldfeld-Gross-Zagier are not strong enough to prove these results either.) However, one should be able to produce the complete list (for any $A$) using GRH, and then prove unconditionally that there are at most an explicit bounded number of exceptions. (I think this has been done in this case $(\mathbf{Z}/2 \mathbf{Z})^n$, for example: look up idoneal numbers, e.g.: The missing Euler Idoneal numbers.)

Watkins: Finally: Watkins' computation is impressive, in part because he smashed the previous cases of the class number $\le N$ problem, which were only known for $N$ up to about $10$ or so.

Extra: Here's an email from Mark from 2008:

a search shows that the following groups do not occur:

(Z/3)^3

(Z/2)^5

(Z/2)x(Z/3)^3

(Z/2)^6

(Z/4)x(Z/2)^4

(Z/3)^4

(Z/9)x(Z/3)^2

Only occurring once:

(Z/3)^2 d=-4027

(Z/2)^4 d=-5460

(Z/5)^2 appears twice d=-12451,-37363

(Z/7)^2 appears twice d=-63499,-118843

(Z/9)^2 appears thrice d=-134059,-298483,-430411

(Z/3)x(Z/2)^5 appears thrice d=-87780,-145860,-106260

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Actually the unconditional results are effective thanks to the Goldfeld/Gross-Zagier bound, though the effective upper bounds are exponential in the size of the class group, which is what makes it nontrivial to provably list all cases of $h \leq 100$. –  Noam D. Elkies Jun 5 '13 at 21:26
Dear Noam, when I said "these unconditional results are ineffective", the word "these" was referring to the claim concerning class groups of the form $(\mathbf{Z}/3 \mathbf{Z})^n$ (i.e., finding all $n$ for which these groups occur). I've edited the post for clarity. –  Socky Jun 5 '13 at 22:44
OK, I get it now. Yes, even the idoneal-number conjecture (the case of $A \oplus ({\bf Z} / 2{\bf Z})^n$ where $A$ is the trivial group) is still unproved; the Goldfeld/Gross-Zagier bound is too weak. –  Noam D. Elkies Jun 6 '13 at 2:13
This is wrong, the hard part of Watkins's thesis wasnot the range near $e^{5077h}$, but in the reduction of the necessary sieving. The paper saves a log factor over Montgomery/Weinberger, so the sievelimit is up to $2^{52}$ typically, better by 1000x from previous approaches (Arno et.al. and Wagner). The range from $2^{162}$ to $e^{260000000}$ or something (the paper has two different numbers at places) is mechanical (5 hours) from low-lying zeros of auxiliary $L$-functions; the range from $2^{52}$ to $2^{162}$ also goes this way, though not easily. He says 100000x harder to do $h\le 1000$. –  v08ltu Jun 6 '13 at 2:18
@ABC: thank you for your nice answer. The list has now $7$ groups (at stackexchange it has $6$ groups). –  Dietrich Burde Jun 7 '13 at 9:58

Hello MO, the system will not let me make a comment only an answer, but I just wanted to point out that, contrary to what "v08ltu" commented, the case of $(Z/3)^k$ is not clearly easier for the methods of Arno and Watkins, but in fact in one aspect is close to the worst case (aside from genera at 2).

The point is, you can have heaps of $(a,b,c)$ forms with $a\sim d^{1/4}$, the multiplicative structure is all gone. In the notation of these papers, the $A(s)=\sum_a 1/a^s$ is only bounded by $|A(1/2+it)|\ge 1-(h-1)/d^{1/8}$, and you want to keep this away from zero (the trick of Watkins is to consider smaller $a$ multiplicatively, and larger ones additively in a variant of $A$, but here all is additive). It seems that Watkins has just enough to get $h=81$ to work, as $d\sim 80^8\sim 2^{52}$, which was his sieveing limit. I sincerely doubt that $(Z/3)^5$ is feasible the same way, as the bound jumps by $3^8$. Watkins barely mentions this at the end of his work, saying that workload is at least fourth-power growth in sieving limit from $h/d^{1/4}$ "but also the lower bounds on $|A(s)|$ become worse", and I think they particularly jam you up here, being 8th power not 4th.

On the other hand, once you actually get to the sieving problem, even with a higher limit, it is then easier as you can demand that all small primes (up to $d^{1/4}$ which is at least almost 10000) are inert, whereas Arno and Watkins split up into a myriad of cases to handle alternative behavior vis-a-vis accounting of smallish split primes and their effect. Sieving up to $728^8\sim 10^{23}$ could be feasible, as Sorenson has done $10^{25}$ for pseudosquares (using doubly-focussed methods).

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