Let K = Q(\sqrt{d}) be an imaginary quadratic number field with discriminant d and K_{ur} its maximal unramified extension. We denote by G the Galois group Gal(K_{ur}/K) and l be the length of the class field tower of K: K = K_{0} ⫅ K_{1} ⫅ K_{2} ⫅ ⋯ (K_{i+1} is the Hilbert class field of K_{i}), that is, l is the smallest number with K_{l+1} = K_{l}.

Under GRH we have determined the structure of G for |d| < 1000 except for d = −883, −907, −947, and −996. For |d| ≦ 467, d ≠ −427, the results are unconditional. We give a table of G for K with K_{ur} ≠ K_{1} below. For this table we omit the fields Q(\sqrt{−883}), Q(\sqrt{−907}), and Q(\sqrt{−947}), for which we expect K_{ur} = K_{1} because, if not, there wuold exist a PGL(2, 7)-extension of Q that is an unramified PSL(2, 7)-extenison of K, however it is unlikely from experimental numerical data.

Table of imaginary quadratic number fields K = Q(\sqrt{d}), |d| < 1000 with K_{ur} ≠ K_{1}

−d Cl(K) l  G  −d Cl(K) l  G
115  C_{2}  2  D_{3}  472  C_{6} 2  D_{3} × C_{3}
120  V_{4}  2  Q_{8}  483  V_{4}  2  D_{6}
155  C_{4}  2 Q_{12}  491  C_{9} 3  Ã_{4} ⋏ C_{9}
184  C_{4}  2 Q_{12}  515  C_{6} 2  D_{5} × C_{3}
195  V_{4}  2 Q_{16}  520  V_{4}  2  Q_{24}
235  C_{2}  2  D_{5}  527  C_{18} 2  D_{3} × C_{9}
248  C_{8}  2  I_{3}^{8}  535  C_{14} 2  D_{3} × C_{7}
255 C_{6} × C_{2} 2 Q_{8} × C_{3} 552 C_{4} × C_{2} 2 Q_{12} × C_{2}
260 C_{4} × C_{2} 2 M_{16}  555  V_{4}  2  Q_{32}
276 C_{4} × C_{2} 2 Q_{12} × C_{2} 563  C_{9} 3  Ã_{4} ⋏ C_{9}
280  V_{4}  2 Q_{16}  564 C_{4} × C_{2} 2 Q_{20} × C_{2}
283  C_{3}  3  Ã_{4}  568  C_{4} 2  Q_{28}
295  C_{8}  2  I_{3}^{8}  580 C_{4} × C_{2} 2 32Γ_{3}f ⋉ C_{3}
299  C_{8}  2  I_{3}^{8}  595  V_{4}  2  Q_{40}
312  V_{4}  2 Q_{16}  611  C_{10} 2  D_{5} × C_{5}
331  C_{3}  3  Ã_{4}  632  C_{8} 2  I_{5}^{8}
340  V_{4}  2 SD_{16}  635  C_{10} 2  D_{5} × C_{5}
355  C_{4}  2 Q_{28}  643  C_{3} 3  Ã_{4}
372  V_{4}  2  D_{6}  644 C_{8} × C_{2} 2  D_{3} × C_{8}
376  C_{8}  2  I_{5}^{8}  651 C_{4} × C_{2} 2  D_{3} × C_{4}
391  C_{14} 2 D_{3} × C_{7} 655  C_{12} 2 Q_{20} × C_{3}
395  C_{8}  2  I_{5}^{8}  660  C_{2}^{3} 2  64Γ_{15}f_{2}
403  C_{2}  2  D_{3}  663 C_{8} × C_{2} 2  64Γ_{3}p
408  V_{4}  2  D_{4}  664  C_{10} 2  D_{3} × C_{5}
415  C_{10} 2 D_{3} × C_{5} 667  C_{4} 2  Q_{12}
420  C_{2}^{3}  2 32Γ_{4}c_{3} 680 C_{6} × C_{2} 2 Q_{16} × C_{3}
435  V_{4}  2 Q_{16} ⋉ C_{3} 687  C_{12} 3 (S_{4} ⋏ C_{4}) × C_{3}
440 C_{6} × C_{2} 2 Q_{16} × C_{3} 695  C_{24} 2  I_{3}^{8} × C_{3}
455 C_{10} × C_{2} 2 Q_{8} × C_{5} 696 C_{6} × C_{2} 2 Q_{24} × C_{3}

Continued (under GRH)

−d Cl(K) l  G  −d Cl(K)  l  G
708  V_{4}  2  D_{6}  856  C_{6}  4  S̃_{4} × C_{3}
715  V_{4}  2 Q_{16} ⋉ C_{5} 868 C_{4} × C_{2}  2  Q_{12} × C_{2}
728 C_{6} × C_{2} 2 Q_{16} × C_{3} 884 C_{8} × C_{2}  2  64Γ_{3}n_{2}
731  C_{12} 3 Ã_{4} ⋎ C_{8} 888 C_{6} × C_{2}  2  Q_{32} × C_{3}
740 C_{8} × C_{2} 2 M_{32}  895  C_{16}  2  I_{5}^{16}
744 C_{6} × C_{2} 2 D_{3} × C_{6} 899  C_{14}  2  D_{3} × C_{7}
751  C_{15} 3 Ã_{4} × C_{5} 915 C_{4} × C_{2}  2  64Γ_{8}e
755  C_{12} 2 Q_{28} × C_{3} 916  C_{10}  3  S_{4} × C_{5}
759 C_{12} × C_{2} 2 D_{3} × C_{12} 920 C_{10} × C_{2}  2 (Q_{16} ⋉ C_{3}) × C_{5}
760  V_{4}  2 Q_{32} ⋉ C_{3} 923  C_{10}  2  D_{7} × C_{5}
767  C_{22} 2 D_{3} × C_{11} 935 C_{14} × C_{2}  2  Q_{16} × C_{7}
771  C_{6}  3 S_{4} × C_{3} 943  C_{16}  2  I_{3}^{16}
795  V_{4}  2 Q_{40}  948 C_{6} × C_{2}  2  D_{5} × C_{6}
799  C_{16} 2  I_{5}^{16}  952 C_{4} × C_{2}  2 (C_{4} ≀ C_{2}) ⋉ C_{5}
820 C_{4} × C_{2} 2 64Γ_{8}c_{2}  955  C_{4}  2  Q_{52}
824  C_{20} 2 Q_{20} × C_{5} 984 C_{6} × C_{2}  3 (Q_{16} ⋉ C_{3}^{2}) × C_{3}
835  C_{6}  2 D_{11} × C_{3} 987 C_{4} × C_{2}  2  Q_{20} × C_{2}
840  C_{2}^{3}  2 32Γ_{4}d  995  C_{8}  2  I_{9}^{8}
851  C_{10} 2 D_{3} × C_{5} 996 C_{6} × C_{2} 3 or 4  ?
852 C_{4} × C_{2} 2 Q_{28} × C_{2}

The fields with |d| < 1000, d ≠ −883, −907, −947 can be classified simply as follows:

{ \begin{matrix}d_{E} ∤ d\{ \\ d_{E} | d\{ \end{matrix}

d ≠ −856, −984, −996 \{ h(K_{gen}) = \frac{h(K)}{[K_{gen}:K]} \{ \begin{matrix}h(K) = 1 \\ h(K) > 1 \end{matrix} h(K_{gen}) > \frac{h(K)}{[K_{gen}:K]} \begin{matrix}⋯ & K_{ur} = K \\ ⋯ & K_{ur} = K_{1} \\ ⋯ & K_{ur} = K_{2} \end{matrix} \}K_{ur} = (K_{gen})_{1}

d = −856, −984, −996

d = d_{E} { − prime ; composite

d = d′d_{E} (d′ fundamental quadratic discriminant)

\begin{matrix}⋯ & l ≧ 3 \\ ⋯ & K_{ur} = K_{3} \\ ⋯ & l ≧ 2 \\ ⋯ & l ≧ 3 \end{matrix} \} K_{ur} ⫌ (K_{gen})_{1}

Here, we denote by K_{gen} the genus field of K and by (K_{gen})_{1} its Hilbert class field. We mean by ‘d_{E} | d’ that the discriminant d of K is divisible by the discriminant of some quartic number field E. As a remarkable observation, we note that for most K with |d| < 1000 we have K_{ur} = (K_{gen})_{1}. In fact, among 305 such fields, at most 16 are exceptional. Why is the degree [K_{ur} : (K_{gen})_{1}] small ? We may expect that [K_{ur} : (K_{gen})_{1}] is small for fields we consider on the ground of the following.

Proposition 1. Let L be the Hilbert class field of the gunus field K_{gen} of an imaginary abelian number field K. Then for any prime number p with p ∤ [L : Q], the p-class group Cl^{(p)}(L) of L is trivial or noncyclic.

For all K with |d| < 1000 such that h(K_{gen}) > h(K)/[K_{gen} : K], which is equivalent to (K_{gen})_{1} ⫌ K_{1}, we have K_{2} = (K_{gen})_{1}. Thus, the following problem arises naturally:

Problem 1. Characterize the imaginary quadratic number fields K with K_{2} = (K_{gen})_{1} ≠ K_{1}.

For h(K_{gen}) > h(K)/[K_{gen} : K], h(K) must necessarily be even and h(K) is even if and only if d has (at least) two distinct prime factors. Probably, for most K, this inequality holds. In fact, if a quadratic subfield ≠ K of K_{gen} has class number divisible by an odd prime p, then we have h(K_{gen}) ≧ ph(K)/[K_{gen} : K]. However, the characterization of such K is very difficult. For example, let consider K = Q(\sqrt{−q}) with q a prime ≡ 1 (mod 4). Then we have h(K_{gen}) = h(Q(\sqrt{q}))h(K)/[K_{gen} : K] and therefore the inequality holds if and only if h(Q(\sqrt{q})) > 1. Moreover the characterization of K with K_{2} = (K_{gen})_{1} is also difficult. But we can easily find K with K_{2} ⫌ (K_{gen})_{1} ⫌ K_{1}. The inclusion K_{2} ⫌ (K_{gen})_{1} holds for K whose p-class group has rank ≧ 2. For such K, the length of the p-class field tower of K: K = K_{0}^{(p)} ⫅ K_{1}^{(p)} ⫅ K_{2}^{(p)} ⫅ ⋯ (K_{i+1}^{(p)} is the Hilbert class field of K_{i}^{(p)}) is larger than one, that is, K_{2}^{(p)} ≠ K_{1}^{(p)}. Since K_{2}^{(p)} and K_{gen} is linearly independent over K, we have K_{2} ⫆ (K_{gen})_{1}K_{2}^{(p)} ⫌ (K_{gen})_{1}. Hence if moreover (K_{gen})_{1} ≠ K_{1}, we have K_{2} ⫌ (K_{gen})_{1} ⫌ K_{1}. For example, let K = Q(\sqrt{−3896}). Then we have Cl(K) ≅ C_{12} × C_{3}, K_{gen} = Q(\sqrt{2}, \sqrt{−487}), and Cl(K_{gen}) ≅ C_{42} × C_{3}. Therefore K_{2} ⫌ K_{1}^{(3)}(K_{gen})_{1} = (K_{gen})_{1} ⫌ K_{1}. Consequently, for K_{2} = (K_{gen})_{1} it is necessary that the odd part of the class group of K is cyclic. Thus, it is reasonable to consider the following before treating Problem 1.

Problem 2. Characterize the imaginary quadratic number fields K with

K_{2}^{(2)} = (K_{gen})_{1}^{(2)} ≠ K_{1}^{(2)}.

By known results on the 2-class field tower, if the 2-class group of K_{1}^{(2)} is nontrivial and cyclic, K_{2}^{(2)} = (K_{gen})_{1}^{(2)} ≠ K_{1}^{(2)} holds. No example of K with K_{2}^{(2)} ⫌ (K_{gen})_{1}^{(2)} ⫌ K_{1}^{(2)} is known.

For K satisfying the stronger condition E_{1} = K_{1} for all intermediate fields E of K_{1}/K than the condition (K_{gen})_{1} = K_{1}, we can estimate h(K_{1}) from below in the case h(K_{1}) > 1 by considering the action of Galois groups on class groups.

Proposition 2. Let K be an imaginary quadratic number field with cyclic class group of odd order n ≧ 7. Assume that for all intermediate fields E of K_{1}/K, we have E_{1} = K_{1}. Then if h(K_{1}) > 1 (this is equivalent to K_{2} ≠ K_{1}), we have h(K_{1}) ≧ 2^{6} = 64. (This is valid also for real quadratic number fields. In the case where K is imaginary and h(K) is a prime (≧ 7), h(K_{1}) ≧ 64 can be improved to h(K_{1}) ≧ 13^{2} = 169. Moreover, in the case where h(K) ≧ 11 and h(K) ≠ 21, 63, h(K_{1}) ≧ 64 can be improved to h(K_{1}) ≧ 2^{8} = 256 (without assuming the imaginarity of K.) )

The first known imaginary quadratic number field with h(K) = q ≧ 7 (q an odd prime) and l ≧ 2 is Q(\sqrt{−14947}). (For K = Q(\sqrt{−14947}), we have h(K) = 17 and 256 | h(K_{1}).) It is thought likely that this field is really the first such field.

Proposition 3. Let K be an imaginary quadratic number field with nontrivial cyclic 2-class group.

(i) Assume Cl(K) ≅ C_{2^{m}}(m ≧ 3) and that the unique unramified cyclic quartic extension E of K has class number 2^{m−2}, which implies h(K_{gen}) = h(K)/2. If h(K_{1}) > 1, then we have h(K_{1}) ≧ 7^{2} = 49. (ii) Assume h(K) = 2q (q an odd prime ≧ 7) and h(K_{gen}) = q (= h(K)/2). If h(K_{1}) > 1, then we have h(K_{1}) ≧ 2^{6} = 64. (These are valid also for real quadratic number fields if the wide genus field K_{gen} is the unique unramified quadratic extension of K.)

Question. What is the first imaginary quadratic number field having an unramified nonsolvable Galois extension ? (What is the first K with K_{ur} ≠ K_{l} ?)

Recent data for quintic number fields enable us to give a partial answer:

Proposition 4. The field Q(\sqrt{−1507}) is the first imaginary quadratic number field having an unramified A_{5}-extension which is normal over Q in the sense that none of Q(\sqrt{d}) of discriminant d with 0 > d > −1507 has such an extension. Moreover, such an extension of Q(\sqrt{−1507}) is given by the composite field of it with the splitting field of the quintic polynomial X^{5} − 5X^{3} + 5X^{2} + 24X + 4, which is an A_{5}-extension of Q.