# On the class number

If $K = \mathbb{Q}(\alpha)$ is a number field, where $\alpha$ is algebraic, and $\mathcal{O}_K$ the ring of integers in $K$, then the set of fractional ideals over $\mathcal{O}_K$ forms a group and if we mod out by the set of principal ideals, the resulting group is finite and we call its size the class number of $K$, which we denote $h(K)$.

I have two questions regarding the class number of imaginary quadratic fields:

If we consider the function $f(d) = h(\mathbb{Q}(\sqrt{d}))$ which maps the negative integers to the positive integers, do we know that this function is surjective? That is, can every positive integer be realized as the class number of an imaginary quadratic field extension of $\mathbb{Q}$?

We know that $h(\mathbb{Q}(\sqrt{d}))$ tends to infinity as $d$ tends to negative infinity, since there are at most finitely many imaginary quadratic fields with a given class number. However, do we have a rough estimate at how large class numbers can be relative to $|d|$? That is, if we consider the function $f(D) = \max_{|d| \leq D} h(\mathbb{Q}(\sqrt{d}))$ where $d < 0$, do we have any idea how large $f$ can be relative to $D$?

Thanks for any insights.

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I answered a similar question to your surjectivity question here: mathoverflow.net/questions/41187/a-coverage-question – Cam McLeman Feb 10 '12 at 0:27
@Cam: I saw your other answer there and after only briefly skimming Sound's paper, I would be very surprised if it could be turned into a proof of surjectivity. That said, I would be quite pleased to be proved wrong! – Frank Thorne Feb 10 '12 at 0:38
@Frank: I concur entirely. Still, it seems very likely to be the true. – Cam McLeman Feb 10 '12 at 2:30

This is from Buell, Binary Quadratic Forms. From page 84, the class number for a negative discriminant $\Delta$ is about $$\frac{\sqrt{|\Delta|}}{\pi},$$ which comes from an $L$-function calculation on page 83.

Let's see, on page 101, he points out that for negative field discriminants, class group and narrow class group are identical. Then on page 103, the group of classes of binary quadratic forms is isomorphic to the narrow class group. So that works out.

I don't know about surjectivity of class numbers. I imagine so. See OEIS

I wrote a little program up to 1000, here it is up to 111. The first number that achieves a given class number tends to be squarefree, an exception being h=104. EDIT: I've run this up to 4000 so far. To get a class number $h,$ there was always some $k$ with $k < 4 h^2.$ The largest $h$ where $h^2$ did not suffice was $h=677,$ with smallest $k = 601247.$ For $678 \leq h \leq 4000,$ there was always some $k \leq 0.751517... h^2,$ equality at $h=857, k=551951.$

    1       3 = 3
2      15 = 3 * 5
3      23 = 23
4      39 = 3 * 13
5      47 = 47
6      87 = 3 * 29
7      71 = 71
8      95 = 5 * 19
9     199 = 199
10     119 = 7 * 17
11     167 = 167
12     231 = 3 * 7 * 11
13     191 = 191
14     215 = 5 * 43
15     239 = 239
16     399 = 3 * 7 * 19
17     383 = 383
18     335 = 5 * 67
19     311 = 311
20     455 = 5 * 7 * 13
21     431 = 431
22     591 = 3 * 197
23     647 = 647
24     695 = 5 * 139
25     479 = 479
26     551 = 19 * 29
27     983 = 983
28     831 = 3 * 277
29     887 = 887
30     671 = 11 * 61
31     719 = 719
32     791 = 7 * 113
33     839 = 839
34    1079 = 13 * 83
35    1031 = 1031
36     959 = 7 * 137
37    1487 = 1487
38    1199 = 11 * 109
39    1439 = 1439
40    1271 = 31 * 41
41    1151 = 1151
42    1959 = 3 * 653
43    1847 = 1847
44    1391 = 13 * 107
45    1319 = 1319
46    2615 = 5 * 523
47    3023 = 3023
48    1751 = 17 * 103
49    1511 = 1511
50    1799 = 7 * 257
51    1559 = 1559
52    1679 = 23 * 73
53    2711 = 2711
54    2759 = 31 * 89
55    4463 = 4463
56    1991 = 11 * 181
57    2591 = 2591
58    2231 = 23 * 97
59    2399 = 2399
60    2159 = 17 * 127
61    3863 = 3863
62    2471 = 7 * 353
63    2351 = 2351
64    2519 = 11 * 229
65    3527 = 3527
66    3431 = 47 * 73
67    3719 = 3719
68    2831 = 19 * 149
69    3119 = 3119
70    3239 = 41 * 79
71    5471 = 5471
72    3311 = 7 * 11 * 43
73    2999 = 2999
74    4151 = 7 * 593
75    4703 = 4703
76    3071 = 37 * 83
77    6263 = 6263
78    5111 = 19 * 269
79    4391 = 4391
80    5183 = 71 * 73
81    3671 = 3671
82    3839 = 11 * 349
83    3911 = 3911
84    4031 = 29 * 139
85    4079 = 4079
86    6767 = 67 * 101
87    5279 = 5279
88    4199 = 13 * 17 * 19
89    6311 = 6311
90    5951 = 11 * 541
91    4679 = 4679
92    4991 = 7 * 23 * 31
93    5351 = 5351
94    7367 = 53 * 139
95    6959 = 6959
96    6071 = 13 * 467
97    5519 = 5519
98    6191 = 41 * 151
99    5591 = 5591
100    7991 = 61 * 131
101    5879 = 5879
102    9383 = 11 * 853
103   13799 = 13799
104    9359 = 7^2 * 191
105    6719 = 6719
106    7631 = 13 * 587
107    8231 = 8231
108    5759 = 13 * 443
109    5711 = 5711
110    7751 = 23 * 337
111   15359 = 15359

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Smallest number $k\equiv3\pmod4$ such that ${\bf Q}(\sqrt{-k})$ has class number $n$ is tabulated at oeis.org/A060649 out to 50 terms, and agrees with Will's table. Also of interest is oeis.org/A081319, Smallest squarefree integer $k$ such that ${\bf Q}(\sqrt{-k})$ has class number $n$. – Gerry Myerson Feb 10 '12 at 4:27

As Will Jagy explained, $h(-D)$ is roughly $\sqrt{D}$, given by Dirichlet's class number formula $$h(d) = \frac{w \sqrt{d}}{2 \pi} L(1, \chi_d)$$ where $L(1, \chi_d)$ is the L-function associated to the quadratic character of $\mathbb{Q}(\sqrt{-D})$. Upper bounds for $L(1, \chi_d)$ are easy to prove; you can get $\log(d)$ by partial summation, implying the bound $h(-d) \ll d^{1/2} \log(d)$. Effective lower bounds are notoriously more difficult.

I am fairly sure that the function $f(d)$ you describe is not known to be surjective, although it is widely expected to be; class numbers are fairly difficult to get a handle on, although a variety of divisibility results are known. However I am not entirely sure of this!

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Hi Frank! "although it is widely expected to be" really? it might be ignorance of my part, but I've never heard this before. In fact, I'm more inclined to guess that $f$ is non surjective. – Guillermo Mantilla Feb 10 '12 at 0:47
Since typically $h(d)$ is of size about $d^{1/2}$, a naïve probabilistic heuristic suggests that each $h$ should arise about $h$ times. There are confounding factors, most notably coming from genus theory which gives a lower bound on the 2-valuation of $h$; e.g. odd $h$ should be particularly rare because they arise only when $d$ is prime. But that's a $\log h$ effect so surjectivity should still hold once it's been checked out to say $h=10^4$ by exhaustive computation. As with the Goldbach conjecture, though, the expected existence of numerous preimages doesn't mean a proof is forthcoming. – Noam D. Elkies Feb 10 '12 at 0:54
@Frank: OK, after seeing the results by Soundararajan, mentioned in the comments above, I see your point. – Guillermo Mantilla Feb 10 '12 at 0:58
@Noam:I see, thanks! – Guillermo Mantilla Feb 11 '12 at 7:23