ON 2-CLASS FIELD TOWERS

OF IMAGINARY QUADRATIC NUMBER FIELDS

Franz Lemmermeyer

Institut für Mathematik Universität Heidelberg

Abstract. For a number field k, let k^{1} denote its Hilbert 2-class field, and put k^{2} = (k^{1})^{1}. We will determine all imaginary quadratic number fields k such that G = Gal (k^{2}/k) is abelian or metacyclic, and we will give G in terms of generators and relations. Moreover, we will correct a formula of Hasse on the 2-rank of the class group of relative quadratic extensions of number fields.

1. Introduction

Let k = ℚ(\sqrt{d}) be an imaginary quadratic number field with discriminant d < 0. It is well known that the structure of the 2-class group Cl_{2}(k) depends on the factorization of d into prime discriminants: these are discriminants which are prime powers, i.e. −4, ±8, −q (q ≡ 3 mod 4), and p (p ≡ 1 mod 4). We say that d has t factors if d is the product of exactly t prime discriminants. Here we will study the question how far these factorizations determine the 2-class field tower of k.

To this end let k^{1} denote the Hilbert 2-class field of k, i.e. the maximal unramified normal extension of k whose Galois group is an abelian 2-group. Moreover, let k^{2} = (k^{1})^{1}. We will classify the discriminants d of imaginary quadratic number fields according to the structure of G = Gal (k^{2}/k), and we will determine all d such that k = ℚ(\sqrt{d}) has abelian or metacyclic Gal (k^{2}/k). Partial classifications have been obtained in [1] and [10]; whereas Benjamin and Snyder used Koch’s Satz 1 of [9], we will employ his Satz 2 instead. The formulation of this theorem contains some errors; its correction reads (see [10]; a G-extension of k is an extension K/k with Gal (K/k) ≃ G, the notation of the groups is the one used in [2]):

Theorem 1. Let k be a quadratic number field; there exists a G-extension K/k which is unramified at the finite places and such that K/ℚ is normal if and only if there is a factorization d = disc k = d_{1}d_{2}d_{3} into relatively prime discriminants such that the Kronecker symbols (d_{i}/p_{j}) in (∗) below equal +1 (here p_{j} runs through all primes dividing d_{j}):

Typeset by AMS-TEX

1991 Mathematics Subject Classification. Primary 11 R 37; Secondary 11 R 20.

Key words and phrases. Class field tower.

G  (∗)  Gal (K/ℚ)
D_{4}  (d_{1}/p_{2}), (d_{2}/p_{1})  16.06
H_{8}  (d_{1}d_{2}/p_{3}), (d_{2}d_{3}/p_{1}), (d_{3}d_{1}/p_{2})  16.08
16.09  (d_{1}/p_{2}), (d_{1}/p_{3}), (d_{2}/p_{1}), (d_{3}/p_{1})  32.33
16.10 (d_{1}/p_{2}), (d_{2}/p_{1}), (d_{1}d_{2}/p_{3}), (d_{3}/p_{1}), (d_{3}/p_{2})  32.36
(4, 4)  all (d_{i}/p_{j}) with i ≠ j  32.34

If (d_{i}/p_{j}) = 1 for all i ≠ j, there also exists an unramified extension L/k such that Gal (L/k) ≃ 32.18 and Gal (L/ℚ) ≃ 64.144.

The following two propositions will help us in deciding if the p-class field tower of a quadratic number field terminates at some stage; the basic observation is due to Iwasawa [5]:

Proposition 1. Let k be a number field and suppose that its p-class field tower terminates with K. Then 𝔐(Gal (K/k)) ≃ E_{k}/N_{K/k}E_{K}, where 𝔐(G) denotes the Schur multiplier of a group G and E_{k} denotes the unit group of the ring of integers 𝔒_{k} of k.

If, for example, K/k is an unramified and normal 2-extension of an imaginary quadratic number field k with | 𝔐(Gal (K/k)) | ≥ 4, then K^{1} ≠ K. For showing that a certain class field tower terminates we use the following result (cf. [10]):

Proposition 2. Let k be a number field and let K^{1} be a normal unramified extension containing the p-class field k^{1} of k; if 𝔐(Gal (K/k)) = 1, then the p-class field tower of k terminates with K.

Proof. Suppose otherwise; then there exists an unramified extension L/K, which is central with respect to K/k, i.e. which satisfies Gal (L/K) ⊆ Z(Gal (L/k)). Moreover, Gal (L/K) ⊆ Gal (L/k)′ because K contains k^{1}. Recalling the most elementary properties of Schur multipliers (cf. [13] or [7]), this gives the contradiction 𝔐(Gal (K/k)) ≠ 1.

In particular, the p-class field tower of a field with cyclic p-class group terminates with k^{1}, because cyclic groups have trivial multiplier.

2. Abelian groups as Gal (k^{2}/k)

The fields ℚ(\sqrt{d}), d < 0, such that G = Gal (k^{2}/k) is abelian, have been determined in [10]; for fields with G/G′ ≃ (2, 2^{m}) this result has been obtained independently by Benjamin and Snyder [1].

Theorem 2. Let ℚ(\sqrt{d}) be an imaginary quadratic number field with discriminant d. Then Gal (k^{2}/k) is abelian if and only if d has at most three factors and if at most one of them is positive. Actually, we have (p and q, q′ etc. denote primes ≡ 1 and ≡ 3 mod 4, respectively):

G  disc k  |G|
1  −4, −8, −q  1
2^{m}, m ≥ 1  −4p, −8p, −pq  h_{2}(k)
(2, 2)  −4qq′, q ≡ 3 mod 8, (q/q′) = −1  4
−8qq′, q ≡ 3 mod 8, (q/q′) = −1  4
−qq′q′′, (q/q′) = (q′/q′′) = (q′′/q)  4
(2, 2^{m}), m ≥ 2 −4qq′, q ≡ 7 mod 8, (q/q′) = −1  h_{2}(k)
−8qq′, q ≡ 7 mod 8, (q/q′) = −1  h_{2}(k)
−qq′q′′, (q/q′) = (q′/q′′) = (q/q′′)  h_{2}(k)

The ”only-if”-part of the Theorem can be proved quite simply: suppose that G = Gal (k^{2}/k) is abelian. Then the 2-class field tower of k terminates with K = k^{1}. Since k is imaginary quadratic and disc k ≠ −3, −4 (recall that we have assumed d to have three prime factors), we find E_{k} = {−1, +1}, so E_{k}/N_{K/k}E_{K} has order ≤ 2. This implies that the abelian 2-groups possibly occuring as Gal (k^{2}/k) must have Schur multiplier of order ≤ 2. The only such 2-groups are the cyclic groups and those of type (2, 2^{m}), m ≥ 1 (see [7]).

If Gal (k^{2}/k) is cyclic, then so is Cl_{2}(k), and by genus theory this is equivalent to disc k being the product of exactly two prime discriminants. The theory of Rédei, Reichardt and Scholz allows us to find all d such that Cl_{2}(k) ≃ (2, 2^{m}), m ≥ 1; we will outline the method used to compute Gal (k^{2}/k) in case d = −4p · q is the corresponding factorization of the second kind. Here we have (−q/p) = 1 and q ≡ 7 mod 8; the ideal classes of 2 = (2, 1 + \sqrt{−pq}) and 𝔭 = (p, \sqrt{−pq}) have order 2 in Cl(k) and differ from each other. Genus theory shows that the ideal class of 𝔭 is no square in Cl(k); class field theory implies that the quadratic unramified extension K_{1} = k(\sqrt{−q}) belongs to a subgroup of index 2 in Cl_{2}(k), and since 𝔭 splits in K_{1}/k, this subgroup is not cyclic. Therefore, the fields K = k(\sqrt{−1} ) and k(\sqrt{q}) belong to the cyclic subgroups of index 2 in Cl_{2}(k), i.e. N_{K/k}Cl_{2}(K) is a cyclic group of order 2^{m}.

Now we notice that the quadratic field ℚ(\sqrt{pq}) has odd class number, and that K = k(\sqrt{−1} ) has unit group E_{K} = ⟨i, ε_{pq}⟩, where ε_{pq} is the fundamental unit of ℚ(\sqrt{pq}) (for a proof, see [8]). The class number formula now gives h_{2}(K) = 2^{m}, and since the norm of Cl_{2}(K) to k is cyclic of order 2^{m}, so is Cl_{2}(K). In particular, the 2-class field tower of k terminates with K^{1}, and the equality (K^{1} : k) = 2^{m+1} = (k^{1} : k) shows that in fact K^{1} = k.

Remark. The knowledge of k^{2}/k allows us to compute the structure of Cl_{2}(L) for the subfields L of k^{1}/k; we are also able to compute the subgroups κ_{L/K} of those ideal classes in Cl_{2}(L) which capitulate in any given extension L/K contained in k^{1}/k. For fields with k^{1} = k^{2}, however, this is hardly interesting: in this case, exactly the ideal classes of order ≤ (L : K) capitulate in any extension L/K such that k ⊂ K ⊂ L ⊂ k^{1} (this fact was already known to Scholz and can be proved immediately by computing the kernel of the corresponding transfer maps). If, on the other hand, k^{1} ≠ k^{2} = k^{3}, then we can compute the structure of Cl_{2}(L) (even as a Gal (L/k)-module) and κ_{L/K} for any extension L/K contained in the 2-class field tower of k.

3. Metacyclic groups as Gal (k^{2}/k)

Our primary aim is to show

Theorem 3. Let k = ℚ(\sqrt{d}) be an imaginary quadratic number field with discriminant d. Then Gal (k^{2}/k) is metacyclic if and only if

i) G is abelian and k is one of the fields described in Section 2, or ii) G/G′ ≃ (2, 2); then G is dihedral, semidihedral or quaternionic, and k is

one of the fields listed in [8] and [10], or iii) d = −4pq, where p and q are primes ≡ 5 mod 8.

For groups G = Gal (k^{2}/k) such that G/G′ ≃ (2, 2^{m}) this has been obtained by Benjamin and Snyder [1]; the proof of Theorem 3 that we will offer depends partly on their results. We begin our proof with the well known observation that factor groups of metacyclic groups are metacyclic. If, therefore, G = Gal (k^{2}/k) is metacyclic, then so is G/G′ ≃ Cl_{2}(k). Genus theory now implies that d = d_{1}d_{2}d_{3} for prime discriminants d_{j}. If Cl_{2}(k) has a subgroup of type (4, 4), then Theorem 1 shows that G has a factor group ≃ 32.18. Since 32.18 is not metacyclic, neither is G.

So far we have seen: if G = Gal (k^{2}/k) is metacyclic, then G/G′ ≃ (2, 2^{m}) for some m ≥ 1. If m = 1, G is dihedral, semidihedral or quaternionic, and this case has already been settled by Kisilevsky [8] (see also [10]). The case m > 1 has been studied by Benjamin and Snyder [1]: they have shown that metacyclic groups G with G/G′ ≃ (2, 2^{m}) occur as G = Gal (k^{2}/k) if and only if d = −4pq, where p and q are primes ≡ 5 mod 8. Using Theorem 1, we can prove this as follows: since Cl(k) is assumed to have a cyclic subgroup of order 4, we must have d = d_{1}d_{2}d_{3} for prime discriminants d_{j}, such that d_{1} · d_{2}d_{3} is a factorization of the second kind (in the terminology of Rédei and Reichardt): this is to say that (d_{2}d_{3}/p_{1}) = (d_{1}/p_{2}) = (d_{1}/p_{3}) = +1, where p_{j} is the unique prime dividing d_{j}.

Now we observe the following: if d_{1} and d_{2} are prime discriminants such that (d_{1}/p_{2}) = +1, then we also have (d_{2}/p_{1}) = +1 except when both d_{1} and d_{2} are negative or when d_{1} = −4, d_{2} = p ≡ 5 mod 8 (or vice versa).

Returning to the situation discussed above we see that there are three cases to consider:

1. All three d_{j} are negative: then G = Gal (k^{2}/k) is abelian, as we have seen in Section 2.

2. Exactly one d_{j} is negative, and d is not of the form −4pq, where p and q are primes ≡ 5 mod 8: then (d_{1}/p_{2}) = (d_{1}/p_{3}) = +1 implies, from what we have seen, that (d_{2}/p_{1}) = (d_{3}/p_{1}) = 1. By Theorem 1, these relations imply the existence of an unramified 16.09−extensions of k. Since 16.09 is not metacyclic, neither is G = Gal (k^{2}/k).

3. disc k = −4pq, p ≡ q ≡ 5 mod 8. We will now use the techniques sketched in [10] to compute the structure of

Gal (k^{2}/k) for the fields k in 3. To this end, suppose that p ≡ q ≡ 5 mod 8 are primes such that (p/q) = 1; suppose moreover that the fundamental unit ε_{pq} of F = ℚ(\sqrt{pq}) has norm −1. The theory of Rédei, Reichardt and Scholz now gives (p/q)_{4} = (q/p)_{4} and Cl_{2}(F) ≃ ℤ/2^{n}ℤ for some n ≥ 2. The prime ideal 𝔭 in F above p is not principal: for if 𝔭 = (π) for some π ∈ 𝔒_{F}, then π^{2}/p would be a unit with positive norm; this would imply that ±π^{2}/p and therefore ±p are squares in F. This contradiction shows that the ideal class [𝔭] has order 2; since Cl_{2}(F) is cyclic, it is generated by an ideal class [𝔞] such that 𝔞^{2^{n−1}} ∼ 𝔭. In fact we may choose 𝔞 as one of the two prime ideals above 2ℤ, because genus theory shows that their ideal classes are not squares in Cl(F).

Similarly, the prime ideals 2 = (2, 1 + \sqrt{−pq}) and 𝔭 = (p, \sqrt{−pq}) in k are not principal (the prime ideals 𝔭 in F and in k coincide in kF), and from genus theory we infer that the ideal class [𝔭] is a square in Cl(k). Kaplan ([6]) has shown that the condition (p/q)_{4} = (q/p)_{4} deduced above implies that Cl_{2}(k) ≃ (2, 4). Therefore, Cl_{2}(k) = ⟨2, 𝔟⟩ (we will write ⟨2, 𝔟⟩ for the rather clumsy ⟨[2], [𝔟]⟩), where 𝔟^{2} ∼ 𝔭.

Now let K = ℚ(\sqrt{−1}, \sqrt{pq}); this quadratic extension of k and F is contained in the 2-class field k^{1} of k. It is an easy exercise to show that the unit group E_{K} = ⟨i, ε_{pq}⟩ and that h_{2}(K) = 2^{n+2}. Since Nε_{pq} = −1, there exists exactly one non-trivial ideal class which capitulates in K/k; in fact, κ_{K/k} = ⟨2⟩ (here κ_{K/k} denotes the subgroup of ideal classes in Cl(k) becoming principal in K), because 2𝔒_{K} = (1 + i). From this we deduce that 𝔭𝔒_{K} is not principal, and that the ideal class of 𝔟𝔒_{K} generates a subgroup of order 4 in Cl(K). The same argument yields that the class of 𝔞O_{K} generates a subgroup of order 2^{n} in Cl(K). We claim that the subgroup ⟨𝔞, 𝔟⟩ of Cl(K) has index 2 in Cl_{2}(K). This index is certainly ≥ 2 because 𝔞^{2^{n−1}} ∼ 𝔭 ∼ 𝔟^{2}. Now suppose that 𝔞^{s} ∼ 𝔟 in K; taking the norm to k gives 1 ∼ 𝔟^{2} as a relation in Cl(k), which is clearly a contradiction.

The prime ideal 2 splits in K/k. Let 2𝔒_{K} = 𝔄𝔄′ for ideals 𝔄, 𝔄′ in 𝔒_{K}; then N_{K/k}⟨𝔞, 𝔟⟩ = ⟨𝔭⟩, because N_{K/k}𝔞 and 𝔟^{2} ∼ 𝔭, and therefore the subgroup of Cl_{2}(K) generated by the classes of 𝔞 and 𝔟 does not contain the ideal class of 𝔄, because N_{K/k}⟨𝔄⟩ = ⟨2⟩ is not contained in ⟨𝔭⟩. Since ⟨𝔞, 𝔟⟩ has index 2 in Cl_{2}(K), we must have Cl_{2}(K) = ⟨𝔄, 𝔞, 𝔟⟩.

Let ρ denote the automorphism of K/ℚ which fixes F; then 𝔄^{1+ρ} is a prime ideal in F above 2ℤ, and without loss of generality we may assume that it equals 𝔞. Similarly, let σ denote the automorphism fixing k; then 𝔄^{1+σ} = 2. We therefore get the following relations between ideal classes in K, keeping in mind that 1 + σ acts on Cl(K), whereas N_{K/k} maps Cl(K) to Cl(k):

𝔄^{1+ρ} ∼ 𝔞, 𝔄^{1+σ} ∼ 2 ∼ 1, 𝔄^{1+ρσ} ∼ 1;

the least relation holds because ρσ fixes ℚ(i) which has class number one. Now we see that

𝔄^{2} ∼ 𝔄^{1+ρ}𝔄^{1+σ}𝔄^{1+ρσ} ∼ 𝔞,

which shows that the ideal class of 𝔄 has order 2^{n+1} in Cl(K), and that Cl_{2}(K) = ⟨𝔄, 𝔟⟩. The relation 𝔄^{2^{n}} ∼ 𝔟^{2} = 𝔭 reveals that Cl_{2}(K) ≃ (2, 2^{n+1}).

The fact that we know the structure of Cl_{2}(K) as a Gal (K/k)-module allows us to compute Gal (K^{1}/k). To this end, let (L/k, 𝔞) denote the Artin symbol of a normal extension L/K (for properties of Artin symbols see Hasse’s Zahlbericht [3]), and let σ = (K^{1}/k, 𝔟) be an extension of the automorphism (k^{1}/k, 𝔟); then σ^{2} = (K^{1}/K, 𝔟) has order 4, and σ^{2} and τ = (K^{1}/K, 𝔄) generate Gal (K^{1}/K).

Therefore, Gal (K^{1}/k) = ⟨σ, τ⟩, and we have the relations

⟨σ^{2}, τ⟩ ≃ Cl_{2}(K), i.e., σ^{4} = τ^{2^{n+1}}, and
σ^{−1}τσ = (K^{1}/K, 𝔄^{σ}) = (K^{1}/K, 𝔄^{−1}) = τ^{−1}.

Now it is easy to see that 𝔐(G) = 1 (there is actually a formula for the order of 𝔐(G) for metacyclic groups G; cf. [7]); this implies that K^{1} = k^{2} (we also could have deduced this from [1, prop. 2]). Since G′ is cyclic, the 2-class field tower of k terminates with k^{2}. We note that the group G = Gal (k^{2}/k) in Theorem 4 is the group of type 2 in [1, prop. 2], with α = n − 1. We have proved

Theorem 4. Let p ≡ q ≡ 5 mod 8 be primes such that (p/q) = 1, and suppose that the fundamental unit ε_{pq} of ℚ(\sqrt{pq}) has norm −1. Then

1. Cl_{2}(k) ≃ (2, 4) for k = ℚ(\sqrt{−pq});

2. Cl_{2}(F) ≃ (2^{n}), n ≥ 2, for F = ℚ(\sqrt{pq});

3. Cl_{2}(K) ≃ (2, 2^{n+1}) for K = ℚ(i, \sqrt{pq});

4. G = Gal (K^{1}/k) = ⟨σ, τ | τ^{2^{n+1}} = 1, σ^{4} = τ^{2^{n}}, σ^{−1}τσ = τ^{−1}⟩;

5. K^{1} = k^{2} = k^{3}, and 𝔐(G) = 1.

Examples:

\begin{matrix}p & q & h(k) & h(F) \\ 5 & 29 & 8 & 4 \\ 5 & 229 & 24 & 4 \\ 13 & 101 & 24 & 4 \\ 5 & 349 & 40 & 4 \\ 13 & 173 & 40 & 4 \\ 5 & 461 & 24 & 16 \\ 5 & 509 & 24 & 4 \\ 5 & 541 & 72 & 8 \end{matrix}

Now we will examine the case where Nε_{pq} = +1. In F = ℚ(\sqrt{pq}), the prime ideals 2_{1} and 2_{2} above 2ℤ are not principal because the equation x^{2} − pqy^{2} = ±8 has no solutions mod p. Moreover, genus theory shows that [2_{1}] is no square in Cl(F); since Cl_{2}(F) is cyclic, the ideal class [2_{1}] generates Cl_{2}(F). We will need the fact that there is no non-trivial capitulation in K/F, where K = ℚ(i, \sqrt{pq}). This will follow from the slightly more general

Proposition 3. Let F be a real quadratic number field and K = F (i). There is non-trivial capitulation in K/F if and only if 2𝔒_{F} = 2^{2} for some non-principal ideal 2 (in particular, 2 is ramified in F/ℚ); in this case, the ideal class of 2 capitulates.

Proof. Let 𝔞 be a non-principal ideal in 𝔒_{F} which capitulates in K. Then there is an α ∈ 𝔒_{K} such that 𝔞𝔒_{K} = (α). Let σ denote the non-trivial automorphism of K/F; then α^{σ−1} = ε is a unit in 𝔒_{K}, and since K/ℚ is abelian, we have | ε | = 1. Now ε^{2} is a root of unity times a unit in 𝔒_{F} (this comes from the identity ε^{2} = ε^{1+σ}ε^{1+ρ}ε^{1+ρσ}, keeping in mind that the units of the two complex quadratic fields are roots of unity); the only units in 𝔒_{F} with absolute value equal to 1 are ±1. Therefore, α^{σ−1} must be a root of unity. There are the following possibilities:

1. α^{σ−1} = +1: then α ∈ F, and this contradicts the assumption that 𝔞 be non-principal in 𝔒_{F};

2. α^{σ−1} = −1: then iα ∈ F, and again 𝔞 = (iα) would be principal in 𝔒_{F};

3. α^{σ−1} = +i: then (1 − i)α ∈ F; since 𝔞 and α𝔒_{F} are ideals in 𝔒_{F}, it follows that 2 = (1 − i)𝔒_{K} ∩ 𝔒_{F} must be an ideal in 𝔒_{F}: but this implies 2^{2} = 2𝔒_{F};

4. α^{σ−1} = −i: then (1 + i)α ∈ F, and again we must have 2^{2} = 2𝔒_{F}.

Now assume that 2^{2} = 2𝔒_{F} and that 2 is a non-principal ideal in 𝔒_{F}; then 2𝔒_{K} = (1 + i) is principal in K, i.e. 2 capitulates.

Let 2^{n} be the order of [2_{1}] in Cl_{2}(F); since there is no capitulation in K/F, the prime ideal 𝔄 in K above 2_{1} generates an ideal class of order 2^{n+1}, because 𝔄^{2} = 2_{1}𝔒_{K}. In k = ℚ(\sqrt{−pq}) we have Cl_{2}(k) = ⟨2, 𝔟⟩, where 𝔟^{2^{m−1}} ∼ 𝔭. Since Nε_{pq} = +1, the ideal classes [2] and [𝔭] both capitulate in K/k, and so 𝔟 generates a subgroup of order 2^{m−1} in Cl(K). The computation of the 2-class number of K yields h_{2}(K) = \frac{1}{2}q(K)h_{2}(F)h_{2}(k) = 2^{m+n}; we claim that the ideal classes of 𝔟 and 𝔄 generate Cl_{2}(K). For suppose that 𝔟^{s} ∼ 𝔄^{t}: then 𝔟^{2s} = N_{K/k}𝔟^{s} ∼ N_{K/k}𝔄^{t} = 2^{t} shows that t is even and that s ≡ 0 mod 2^{m−1} (recall that 2 and 𝔟 generate different subgroups of Cl_{2}(k)). But the order of [𝔟] in Cl_{2}(K) equals 2^{m−1}, and therefore the relation 𝔟^{s} ∼ 𝔄^{t} is necessarily trivial. This proves that indeed Cl_{2}(K) = ⟨𝔟, 𝔄⟩.

Putting ρ = (K^{1}/K, 𝔄) and τ = (K^{1}/K, 𝔟), ⟨ρ, τ | ρ^{2^{n+1}} = τ^{2^{m−1}} = 1⟩ is an abelian subgroup of index 2 in G = Gal (K^{1}/k). Let σ ∈ (K^{1}/k, 𝔟) be an extension of (k^{1}/k, 𝔟) ∈ Gal (k^{1}/k); then we have G = ⟨ρ, σ, τ⟩, and we find the relations σ^{2} = (K^{1}/K, 𝔟) = τ, σ^{−1}ρσ = (K^{1}/K, 𝔄^{σ}) = ρ^{−1}; to prove the last relation we use the fact that 𝔄^{σ+1} = 2 = (1+i) ∼ 1 lies in the kernel of the Artin symbol of K^{1}/K. We finally remark that it follows from [6] that we either have n = 1 or m = 2, and that 𝔐(G) ≃ ℤ/2ℤ. This is in agreement with prop. 1, because −1 ∈ E_{k}\N_{K/k}E_{K}. The fact that k^{2} = K^{1} follows from [1]: they proved that Gal (k^{2}/k) has an abelian subgroup of order 2, which necessarily must be Gal (k^{2}/K). We also observe that G is their metacyclic group of type 1. We have shown

Theorem 5. Let p ≡ q ≡ 5 mod 8 be primes such that (p/q) = 1, and assume that the fundamental unit ε_{pq} of ℚ(\sqrt{pq}) has norm +1. Then

1. Cl_{2}(k) ≃ (2, 2^{m}), m ≥ 2, for k = ℚ(\sqrt{−pq});

2. Cl_{2}(F) ≃ (2^{n}), n ≥ 1, for F = ℚ(\sqrt{pq});

3. Cl_{2}(K) ≃ (2^{n+1}, 2^{m−1}) for K = ℚ(i, \sqrt{pq}), and either n = 1 or m = 2;

4. G = Gal (K^{1}/k) = ⟨ρ, σ | ρ^{2^{n+1}} = σ^{2^{m}} = 1, σ^{−1}ρσ = ρ^{−1}⟩;

5. K^{1} = k^{2} = k^{3}, and 𝔐(G) ≃ ℤ/2ℤ.

The only case left to consider is d = −4pq, p ≡ q ≡ 5 mod 8, (p/q) = −1. Then we have Nε_{pq} = −1 and h_{2}(F) = 2 for F = ℚ(\sqrt{pq}); we leave it as an exercise to the reader to prove the following Theorem, using the methods described in this paper. Theorem 6 has already been announced in [10], and a proof using Koch’s Satz 1 can be found in [1]:

Examples:

\begin{matrix}p & q & h(k) & h(F) \\ 5 & 61 & 16 & 2 \\ 13 & 29 & 16 & 2 \\ 5 & 101 & 8 & 4 \\ 5 & 109 & 32 & 2 \\ 13 & 53 & 40 & 4 \\ 5 & 149 & 16 & 2 \\ 13 & 61 & 8 & 4 \\ 5 & 181 & 24 & 4 \\ 5 & 269 & 16 & 6 \\ 29 & 53 & 16 & 2 \end{matrix}

Theorem 6. Let p ≡ q ≡ 5 mod 8 be primes such that (p/q) = −1. Then 1. Cl_{2}(k) ≃ (2, 2^{m}), m ≥ 2, for k = ℚ(\sqrt{−pq});

2. Cl_{2}(K) ≃ (2, 2^{m}) for K = ℚ(i, \sqrt{pq});

3. G = Gal (K^{1}/k) ≃ ⟨ρ, σ | ρ^{2} = σ^{2^{m}}, ρ^{4} = 1, σ^{−1}ρσ = ρ^{−1}⟩;

4. K^{1} = k^{2} = k^{3}, and 𝔐(G) = 1.

Obviously, Theorems 4, 5 and 6 prove the part of Theorem 3.iii) that was still open.

4. Hasse’s rank formula

On p. 52 of his book ”Über die Klassenzahl abelscher Zahlkörper” [4], Hasse gave a formula for computing the 2-rank of class groups in CM-fields K/K_{0} which reduces to the well known ambiguous class number formula in case K_{0} has odd class number. If K_{0} has even class number, however, Hasse’s formula does not always give the correct rank: the cyclic quartic field ℚ(\sqrt{−10 + 3\sqrt{10}} ) has cyclic 2-class group of order 4, and it is easily seen that this contradicts Hasse’s formula. In fact, the fields listed in Theorem 5, for example, provide infinitely many counterexamples. The formula stated in [4] reads rank Cl_{2}(K) = γ + r_{0}^{∗} + s_{0}^{∗} + κ, where

γ = t − 1 + rank E_{K_{0}} ∩ N_{K/K_{0}}K/E_{K_{0}}^{2}, where t is the number of (finite) prime

ideals ramifying in K/K_{0}, r_{0}^{∗} denotes the 2-rank of Cl(K_{0})^{j}, j : Cl(K_{0}) → Cl(K) is the transfer of ideal classes, s_{0}^{∗} = rank Cl(K_{0})^{j}/Cl(K_{0})^{j} ∩ Cl(K)^{2} denotes the rank of the group of ideal classes becoming squares in Cl(K). It follows from our proof below, that we also have s_{0}^{∗} = rank H/H ∩ Cl(K)^{2}, where H = Cl(K)^{1−σ};

κ = | ker j | is the order of the capitulation kernel. To see that the formula is incorrect, we let K_{0} = ℚ(\sqrt{pq}) as in Theorem 5, i.e.

we assume that p ≡ q ≡ 5 mod 8 are primes such that (p/q) = 1 and Nε_{pq} = −1; if we assume moreover that m = 2, then we find

E_{K_{0}} ∩ N_{K/K_{0}}K = ⟨1, ε_{pq}⟩: to verify this claim, it is sufficient to show that ε_{pq} is norm from K. This can be done as follows: since ε_{pq} has norm +1, the prime ideal 𝔭 above p is principal, i.e. 𝔭 = (π). Obviously, η = π^{σ−1} is a unit in 𝔒_{K}; if η were a square, so were p = π^{σ+1}. Now it is easily seen that we can choose π in such a way that π^{σ−1} = ε_{pq}. This implies ε_{pq} = π^{σ+1}/p, and since p = a^{2} + b^{2} = N_{K/k}(a + bi), we find that ε_{pq} = N_{K/k}(\frac{π}{a+bi}) is indeed a norm.

γ = t − 1 + 1 = 2, because only the two prime ideals above 2 ramify; r_{0}^{∗} = 1; H = Cl(K)^{1−σ} = 1: we have 𝔟^{1−σ} ∼ 𝔭 ∼ 1 and 𝔄^{1−σ} ∼ 1. In particular, we

have s_{0}^{∗} = 0; κ = 0, because there is only trivial capitulation in K/F.

Hasse’s formula gives r = 3, although rank Cl_{2}(K) = 2.

The correct rank formula can be deduced as follows: let σ denote complex conjugation (hence σ is the non-trivial automorphism of K/K_{0}), and put C = Cl_{2}(K), r = rank C, C_{0} = Cl_{2}(K_{0}), and H = Cl_{2}(K)^{1−σ}. Then the formula 2 = 1+σ+1−σ shows that C^{2}C^{1+σ} = C^{1−σ}C^{1+σ} = C_{0}^{j}H (the fact that also C^{2}C^{1−σ} = C_{0}^{j}H proves our previous claim that s_{0}^{∗} = rank H/H ∩ Cl(K)^{2}), and now we find

2^{r} = (C : C^{2}) = (C : C^{2}C^{1+σ})(C^{2}C^{1+σ} : C^{2})

= (C : C_{0}^{j}H)(C^{2}C_{0}^{j} : C^{2})

= \frac{(C : H)}{(C_{0}^{j}H : H)} (C_{0}^{j} : C_{0}^{j} ∩ C^{2})

Now, (C : H) equals the number of ideal classes in C fixed by σ, because

1 → C^{G} → C → C^{1−σ} → 1

is a short exact sequence (C^{G} is the subgroup of C fixed by G = ⟨σ⟩). Moreover,

(C_{0}^{j}H : H) = (C_{0}^{j} : C_{0}^{j} ∩ H)
= (C_{0}^{j} : _{2}C_{0}^{j})(_{2}C_{0}^{j} : C_{0}^{j} ∩ H),

where _{2}G denotes the subgroup of elements of order ≤ 2 of an abelian group G. Let c ∈ C_{0}^{j} ∩ H; then σ fixes c, and applying σ to the equation c = d^{1−σ} yields c = c^{−1}, i.e. c ∈ _{2}C_{0}^{j}. Hasse claimed that in fact C_{0}^{j} ∩ H = _{2}C_{0}^{j}: but an ideal class of order 2 is not necessarily of the form d^{1−σ} for some d ∈ C. Actually, in the counterexamples presented above, 1 − σ annihilates the whole 2-class group of K.

Using the ambiguous class number formula (C : H) = 2^{γ}h_{2}(K_{0}), where γ was defined above, we find that

rank Cl_{2}(K) = γ + r_{0}^{∗} + s_{0}^{∗} + κ − λ,

with 2^{λ} = (_{2}C_{0}^{j} : C_{0}^{j} ∩ H).

The main problem with this rank formula is the fact that λ usually is quite hard to compute. The trivial bounds 0 ≤ λ ≤ r_{0}^{∗} yield

γ + s_{0}^{∗} + κ ≤ rank Cl_{2}(K) ≤ γ + r_{0}^{∗} + s_{0}^{∗} + κ.

It should be noted that Hasse applied his formula only to fields K_{0} with odd class number; we have already observed that in this case it coincides with the ambiguous class number formula, because then obviously r_{0}^{∗} = s_{0}^{∗} = κ = 0, and the formula simplifies to rank Cl_{2}(K) = γ.

Acknowledgments

I would like to thank E. Benjamin and C. Snyder for sending me their preprint, and Prof. Y. Kida for providing UBasic; the examples in this paper (and many more) have been computed with version 8.65. Prof. M. Olivier has kindly verified that the 2-class group of the quartic cyclic field ℚ(\sqrt{−10 + 3\sqrt{10}} ) is cyclic of order 4 using PARI, and Prof. J. Martinet has suggested the possibility of bicyclic counterexamples to Hasse’s rank formula.

References

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