Imaginary quadratic number fields K = Q(\sqrt{d}) with |d| < 1000, where d is the discriminant of K, can be classified simply as follows:

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

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

d = −856, −996

d = d_{E} { d_{E} = −p: prime

d_{E} : 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_{g})_{1}

Here, we denote by K_{g} the genus field of K. We mean by d_{E} | d that the discriminant d of K is divisible by the discriminant of some quartic number field E. In this classification, there are some possible exceptions. More precisely, for 18 fields K with d_{E} ∤ d ≠ −856, −996, we have not succeeded in showing K_{ur} = (K_{g})_{1}. (Among them, for 15 fields, we have l = 1 and in order to conclude K_{ur} = K_{1}, we must show that K_{1} does not have an unramified A_{5}-extension. For the other 3 fields we have K_{ur} = K_{l}, 2 ≦ l ≦ 3, and we expect l = 2.) For (12 fields) K with K_{ur} ⫌ (K_{g})_{1}, there exists an S_{4}-extension M of Q such that the compositum K_{g}M is an unramified A_{4}-extension of K_{g}.

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 1. 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.

It is reasonable to expect that the field Q(\sqrt{−1507}) gives the answer to the question above.

we may expect that [K_{ur} : (K_{g})_{1}] is small for fields we consider on the ground of the following.

Proposition 2. Let L be the Hilbert class field of the gunus field K_{g} 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_{g}) > h(K)/[K_{g} : K], which is equivalent to (K_{g})_{1} ⫌ K_{1}, we have K_{2} = (K_{g})_{1}. For h(K_{g}) > h(K)/[K_{g} : K], h(K) must necessarily be even. h(K) is even if and only if d has (at least) two distinct prime factors, however, for most K, this inequality holds. In fact, if a quadratic subfield≠ K of K_{g} has class number divisible by an odd prime p, then we have h(K_{g}) ≧ ph(K)/[K_{g} : K]. Thus, the following problem arises naturally:

Typeset by AMS-TEX

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

This is very difficult. (But we can easily find K with K_{2} ⫌ (K_{g})_{1} ⫌ K_{1}. In fact, if the p-rank of the p-class group of K is larger than one for some odd prime p, then K_{2} ⫆ (K_{g})_{1}K_{2}^{(p)} ⫌ (K_{g})_{1}, and therefore for such K with (K_{g})_{1} ≠ K_{1}, we have K_{2} ≠ (K_{g})_{1} ≠ K_{1}. For example, let K = Q(\sqrt{−3896}). Then we have Cl(K) ≅ C_{12} × C_{3}, K_{g} = Q(\sqrt{2}, \sqrt{−487}), and Cl(K_{g}) ≅ C_{42} × C_{3}. Therefore K_{2} ⫌ K_{1}^{(3)}(K_{g})_{1} = (K_{g})_{1} ⫌ K_{1}.) Thus, it is reasonable to consider the following before treating Problem 2.

Problem 2. Characterize the imaginary quadratic number fields K with

K_{2}^{(2)} = (K_{g})_{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_{g})_{1}^{(2)} ≠ K_{1}^{(2)} holds. No example of K with K_{2}^{(2)} ⫌ (K_{g})_{1}^{(2)} ⫌ K_{1}^{(2)} is known.

Note that h(K_{g}) = h(K)/[K_{g} : K] is equivalent to (K_{g})_{1} = K_{1} and that all fields with odd class number satisfy this. For some K with (K_{g})_{1} = K_{1}, we can estimate h(K_{1}) from below under some assumptions by considering the action of Galois groups on class groups.

Proposition 3. 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 4. 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_{g}) = 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_{g}) = 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 K_{g} is the unique unramified quadratic extension of K.)