Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
1  
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
add comment

2 Answers

up vote 5 down vote accepted

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 
share|improve this answer
1  
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
add comment

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!

share|improve this answer
    
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
6  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.