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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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    $\begingroup$ I answered a similar question to your surjectivity question here: mathoverflow.net/questions/41187/a-coverage-question $\endgroup$ Commented Feb 10, 2012 at 0:27
  • $\begingroup$ @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! $\endgroup$ Commented Feb 10, 2012 at 0:38
  • $\begingroup$ @Frank: I concur entirely. Still, it seems very likely to be the true. $\endgroup$ Commented Feb 10, 2012 at 2:30

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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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    $\begingroup$ 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 $n$. $\endgroup$ Commented Feb 10, 2012 at 4:27
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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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  • $\begingroup$ 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. $\endgroup$ Commented Feb 10, 2012 at 0:47
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    $\begingroup$ 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. $\endgroup$ Commented Feb 10, 2012 at 0:54
  • $\begingroup$ @Frank: OK, after seeing the results by Soundararajan, mentioned in the comments above, I see your point. $\endgroup$ Commented Feb 10, 2012 at 0:58

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