Capitulation of the 2-ideal classes of type (2, 2, 2) of some quartic cyclic number fields

Abdelmalek Azizi; Idriss Jerrari; Abdelkader Zekhnini; Mohammed Talbi

doi:10.1515/jmc-2017-0037

Article Open Access

Capitulation of the 2-ideal classes of type (2, 2, 2) of some quartic cyclic number fields

Abdelmalek Azizi , Idriss Jerrari , Abdelkader Zekhnini and Mohammed Talbi

Published/Copyright: October 12, 2018

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From the journal Journal of Mathematical Cryptology Volume 13 Issue 1

Abstract

Let p≡3(mod4) and l≡5(mod8) be different primes such that pl=1 and 2p=pl4. Put k=ℚ⁢(l), and denote by ϵ its fundamental unit. Set K=k⁢(-2⁢p⁢ϵ⁢l), and let K2(1) be its Hilbert 2-class field, and let K2(2) be its second Hilbert 2-class field. The field K is a cyclic quartic number field, and its 2-class group is of type (2,2,2). Our goal is to prove that the length of the 2-class field tower of K is 2, to determine the structure of the 2-group G=Gal⁡(K2(2)/K), and thus to study the capitulation of the 2-ideal classes of K in all its unramified abelian extensions within K2(1). Additionally, these extensions are constructed, and their abelian-type invariants are given.

Keywords: capitulation; Hilbert class field

MSC 2010: 11R11; 11R16; 11R29; 11R32; 11R37

1 Introduction

Let K be an algebraic number field. For a prime number p, let Kp(0)=K and Kp(i) denote the Hilbert p-class field of Kp(i-1) for i≥1. Then the sequence (Kp(i))i∈ℕ is called the p-class field tower of K. Put K~p=⋃i=0∞Kp(i). It is well known that the p-group G=Gal⁡(K~p/K) can be finite as it can be infinite, thanks to Golod and Shafarevich [10], but we do not know enough criteria to decide whether G is finite or not. In general, this group is far from understood. In addition, the structure of G incorporates a lot of arithmetic information about the field K and its unramified p-extensions. It is quite difficult to determine this structure, and we do not have efficient methods for describing it. An important method of describing a p-group is via a presentation by generators and relations. In this paper, we are interested in giving the structure of G=Gal⁡(K~2/K), via generators and relations, for some quartic cyclic number field K, such that G/G′ is of type (2,2,2), where G′ is the derived subgroup of G.

Let p≡3(mod4) and l≡5(mod8) be different primes. Put k=ℚ⁢(l), and denote by ϵ its fundamental unit. Let K=k⁢(-2⁢p⁢ϵ⁢l) and K2(1) be its Hilbert 2-class field; denote by K2(2) the Hilbert 2-class field of K2(1) and by K(*) the genus field of K that is the maximal absolute abelian subfield of K2(1)/K. If pl=-1, then 𝐂K,2, the 2-class group of K, is isomorphic to ℤ/2⁢ℤ×ℤ/2⁢ℤ (see [9]) and K(*)=K2(1); this case is studied in [3]. But if pl=1, then K(*)⊊K2(1) and the rank of 𝐂K,2 is greater than 3, and the situation becomes more interesting. Assuming that 𝐂K,2 is isomorphic to ℤ/2⁢ℤ×ℤ/2⁢ℤ×ℤ/2⁢ℤ, we aim, in this paper, the following four objectives.

Construct the fourteen unramified subextensions of K2(1)/K.
Prove that the class field tower of K stops at the second stage, and thus G=Gal⁡(K~2/K)=Gal⁡(K2(2)/K).
Give the structure of the 2-group G via generators and relations.
Concretely answer, via the transfer kernels for the finite 2-group G and their relation to capitulation kernels, the capitulation problem of the 2-ideal classes of K in the fourteen unramified subextensions of K2(1)/K.

Note that, in [2], we treated the same topics for the field K=k⁢(-p⁢ϵ⁢l), whose 2-class group is isomorphic to G/G′≃(2,2,2), but G, in this case, admits three maximal normal abelian subgroups of index 2, and its order is exactly 16. Whereas, in the present case, the 2-group G admits, whenever p≡3(mod8), only one maximal normal abelian subgroup of index 2, and its order is equal to 2m for some integer m>4. On the other hand, the same subjects are treated in the papers [5, 4, 6] by considering base fields of the form k=ℚ⁢(d,-1) for some square-free integer d, and in [15] by considering some imaginary quadratic number field as a base field.

Notations

Let m be a square free integer, and let k be a number field. Throughout this paper, we adopt the following notations:

FSU: a fundamental system of units of k
h⁢(m), h⁢(k): the class number of ℚ⁢(m) (resp. k)
h2⁢(m), h2⁢(k): the 2-class number of ℚ⁢(m) (resp. k)
𝐂k,2: the 2-part of the class group of k
k+: the maximal real subfield of k if it is a CM-field
Ek: the unit group of k
Wk: the group of roots of unity contained in k
wk: the order of Wk
Qk=[Ek:WkEk+]: Hasse’s unit index if k is a CM-field
NL/k: the relative norm for an extension L/k
κL/k: the capitulation kernel of L/k
𝔭k: a prime ideal of k
x,y𝔭k, x𝔭k: the Hilbert symbol (resp. the quadratic residue symbol) for the prime 𝔭k applied to (x,y) (resp. x)
ξm: a primitive m-th root of unity
ℚ⁢(ξm): the m-th cyclotomic number field

2 Main results

In all this paper, we consider the following conditions and notations: p and l are different primes satisfying

(2.1)p≡3(mod4),l≡5(mod8),pl=1 and 2p=pl4.

Let k=ℚ⁢(l), and denote by ϵ its fundamental unit and by h0 its class number. Then K=k⁢(-2⁢p⁢ϵ⁢l) is a complex cyclic quartic field [9].

Let K2(1) be the Hilbert 2-class field of K, K2(2) the Hilbert 2-class field of K2(1), and let K(*) be the absolute genus field of K. Since pl=1, so K(*)⊊K2(1) (as mentioned in the introduction), and there exist two prime ideals ℬ1, ℬ2 of k such that ℬ1⁢ℬ2=(p). Thus ℬ1h0 and ℬ2h0 are principal in k, say ℬ1h0=(π1) and ℬ2h0=(π2), where

π1=a+b⁢l,π2=a-b⁢l and ph0=a2-l⁢b2.

One can easily show that four prime ideals of k ramify in K, which are the prime ideal of k above 2, (l), ℬ1 and ℬ2. Let 𝒫1, 𝒫2 be the prime ideals in K above ℬ1, ℬ2, respectively, let 𝒱 be the prime ideal in K above l, and let ℋ be the prime ideal in K above 2. Therefore, 𝒫i2⁢h0=ℬih0⁢𝒪K=(πi), 𝒱2=(l) and ℋ2=(2). So, according to [9], the rank of the 2-class group of K is 4-1=3, and the 2-class number h2 of K is strictly greater than 4. Furthermore, by [9, Theorem 4, p. 68], h2=8, i.e., 𝐂K,2 is isomorphic to ℤ/2⁢ℤ×ℤ/2⁢ℤ×ℤ/2⁢ℤ if and only if 2p=pl4.

The main results of this paper are Theorems 2.1, 2.2 and 2.3.

The unramified extensions of K

As 𝐂K,2≃(2,2,2), by class field theory, K admits, within its first Hilbert 2-class field K2(1), seven unramified quadratic extensions and seven unramified biquadratic extensions.

Theorem 2.1.

Let p and l be different primes as specified by equation (2.1). Recall that P1, P2 are the prime ideals in K above p, and H is the prime ideal in K above 2. Moreover, Pi2⁢h0=(πi) and H2=(2).

The 2 -class group of K is given by 𝐂K,2=〈[𝒫1h0],[𝒫2h0],[ℋ]〉≃(2,2,2).
The seven unramified quadratic extensions of K are as follows:
1. If 2⁢a⁢bp=1, then
  F1=K⁢(-ϵ⁢π1),F2=K⁢(-2⁢ϵ⁢π2),F3=K⁢(2⁢p),F4=K⁢(-2),F5=K⁢(-p),F6=K⁢(2⁢ϵ⁢π1) 𝑎𝑛𝑑 F7=K⁢(ϵ⁢π2).
2. If 2⁢a⁢bp=-1, then
  F1=K⁢(-2⁢ϵ⁢π1),F2=K⁢(-ϵ⁢π2),F3=K⁢(2⁢p),F4=K⁢(-2),F5=K⁢(-p),F6=K⁢(ϵ⁢π1) 𝑎𝑛𝑑 F7=K⁢(2⁢ϵ⁢π2).
The fields F3, F4 and F5 are intermediate between K and K(*). The fields F1≃F7 and F2≃F6 are pairwise conjugate and isomorphic. Consequently, F3, F4 and F5 are absolutely abelian, whereas F1, F2, F6 and F7 are not.
The seven unramified biquadratic extensions of K are
L2=F3⁢F4⁢F5=K(*)=K⁢(-p,-2),the absolute genus field of⁢K,
L1=F1⁢F2⁢F3,L3=F3⁢F6⁢F7,L4=F2⁢F5⁢F6,L5=F1⁢F5⁢F7,L6=F1⁢F4⁢F6,L7=F2⁢F4⁢F7.

Moreover, L1≃L3 and L6≃L7 are pairwise conjugate and thus isomorphic and absolutely non-normal, whereas L4 and L5 are absolutely Galois over Q.

Structure of G=Gal⁡(K2(2)/K)

Recall that K2(2) is the second Hilbert 2-class field of K.

Theorem 2.2.

Keep the preceding assumptions and notations, and put h2⁢(-p⁢l)=2n+1, where n≥1. Then:

𝐂F5,2≃(2,2n+1), where n≥2 or n=1, according as p≡3(mod8) or p≡7(mod8).
The length of the 2 -class field tower of K is 2.
The second 2 -class group G=Gal⁡(K2(2)/K) of K is given by
G={〈ρ,τ,σ:ρ4=σ4=τ2n+1=1,σ2=ρ2=τ2n,n≥2,[τ,σ]=1,[σ,ρ]=σ2,[ρ,τ]=τ2〉𝑖𝑓⁢p≡3(mod8),〈ρ,τ,σ:ρ4=σ4=τ2=1,σ2=ρ2,[τ,σ]=[τ,ρ]=1,[σ,ρ]=σ2〉𝑖𝑓⁢p≡7(mod8).
The derived subgroup of G is G′=〈τ2〉≃𝐂K2(1),2 or G′=〈σ2〉≃𝐂K2(1),2, according as p≡3(mod8) or p≡7(mod8). It is cyclic of order 2n or 2 , respectively.
The coclass of G is equal to 2 , and its nilpotency class is n+1.
The center Z⁢(G) of G is of type (2,2).

Abelian-type invariants and capitulation kernels

Let κF/K denote the capitulation kernel of the natural class extension homomorphism JF/K:𝐂K,2→𝐂F,2, where F is an unramified extension of K within K2(1).

Theorem 2.3.

Keep the preceding assumptions and notations, then:

The abelian-type invariants of the 2 -class groups 𝐂Fi,2 are given by:
1. 𝐂F3,2 and 𝐂F4,2 are of type (2,2), and 𝐂F5,2 is of type (2,2n+1) with n≥1.
2. If 2⁢a⁢bp=1, then:
  1. 𝐂F1,2≃𝐂F7,2≃(2,4) and 𝐂F2,2≃𝐂F6,2≃(2,2) if 2p=1.
  2. 𝐂F1,2≃𝐂F7,2≃(2,2,2) and 𝐂F2,2≃𝐂F6,2≃(2,2) if 2p=-1.
3. If 2⁢a⁢bp=-1, then:
  1. 𝐂F1,2≃𝐂F7,2≃(2,2) and 𝐂F2,2≃𝐂F6,2≃(2,4) if 2p=1.
  2. 𝐂F1,2≃𝐂F7,2≃(2,2) and 𝐂F2,2≃𝐂F6,2≃(2,2,2) if 2p=-1.
The abelian-type invariants of the 2 -class groups 𝐂Li,2 are given by:
1. If 2⁢a⁢bp=1, then:
  1. 𝐂L1,2, 𝐂L2,2, 𝐂L3,2, 𝐂L4,2, 𝐂L6,2 and 𝐂L7,2 are cyclic of order 4 , and 𝐂L5,2 is of type (2,2) if 2p=1.
  2. 𝐂L1,2≃𝐂L3,2≃𝐂L6,2≃𝐂L7,2≃(2,2), 𝐂L5,2≃(2,2n), and 𝐂L2,2 and 𝐂L4,2 are cyclic of order 2n+1 if 2p=-1.
2. If 2⁢a⁢bp=-1, then:
  1. 𝐂L1,2, 𝐂L2,2, 𝐂L3,2, 𝐂L5,2, 𝐂L6,2 and 𝐂L7,2 are cyclic of order 4 , and 𝐂L4,2 is of type (2,2) if 2p=1.
  2. 𝐂L1,2≃𝐂L3,2≃𝐂L6,2≃𝐂L7,2≃(2,2), 𝐂L4,2≃(2,2n), and 𝐂L2,2 and 𝐂L5,2 are cyclic of order 2n+1 if 2p=-1.
The capitulation is given by:
1. For all i, #⁢κFi/K=4 or 8 (for more details, see Tables 9 and 10), and all the extensions Fi satisfy Taussky’s condition (A), i.e., #⁢(κFi/K∩Ni)>1; see [16].
2. For all i, κLi/K=𝐂K,2 (total 2 -capitulation), and each Li is of type (A).

3 Preliminary results

Let us first collect some results that will be useful in the sequel.

Lemma 3.1.

Let p and l be different primes as specified by equation (2.1). Keep the previous notations, then:

The equations -ϵ⁢π1≡X2(mod4) and ϵ⁢π2≡X2(mod4) have solutions in k if and only if 2⁢a⁢bp=1.
The equations ϵ⁢π1≡X2(mod4) and -ϵ⁢π2≡X2(mod4) have solutions in k if and only if 2⁢a⁢bp=-1.

Proof.

First, we put ϵ=x+y⁢l, where x and y are two integers. Then x≡2(mod4) and y≡1(mod4) since l≡5(mod8).

As ph0=a2-l⁢b2, then:

If p≡3(mod8), we have a≡0(mod4) and b≡±1(mod4).
If p≡7(mod8), we have a≡2(mod4) and b≡±1(mod4).

Thus we have four cases to discuss:

If a≡2(mod4) and b≡1(mod4), then we have (mod⁢ 4)
ϵ⁢π1=(x+y⁢l)⁢(a+b⁢l)≡(2+l)⁢(2+l)≡4+4⁢l+l≡1,
-ϵ⁢π2=-(x+y⁢l)⁢(a-b⁢l)≡-(2+l)⁢(2-l)≡-4+l≡1.
If a≡2(mod4) and b≡-1(mod4), then we have (mod⁢ 4)
-ϵ⁢π1=-(x+y⁢l)⁢(a+b⁢l)≡-(2+l)⁢(2-l)≡-4+l≡1,
ϵ⁢π2=(x+y⁢l)⁢(a-b⁢l)≡(2+l)⁢(2+l)≡4+4⁢l+l≡1.
If a≡0(mod4) and b≡1(mod4), then
-ϵ⁢π1=-ϵ⁢(a+b⁢l)≡-ϵ⁢l(mod4),
ϵ⁢π2=ϵ⁢(a-b⁢l)=-ϵ⁢l(mod4),
which is a square in k (see [9, Lemma A]).
If a≡0(mod4) and b≡-1(mod4), then
ϵ⁢π1=ϵ⁢(a+b⁢l)=-ϵ⁢l(mod4),
-ϵ⁢π2=-ϵ⁢(a-b⁢l)=-ϵ⁢l(mod4),
which is a square in k (see [9, Lemma A]).

By the disjunction of cases, one can easily show the following:

If -ϵ⁢π1≡X2(mod4) and ϵ⁢π2≡X2(mod4) have solutions in k, then (a≡2 and b≡-1(mod4)) or (a≡0 and b≡1(mod4)).
If ϵ⁢π1≡X2(mod4) and -ϵ⁢π2≡X2(mod4) have solutions in k, then (a≡2 and b≡1(mod4)) or (a≡0 and b≡-1(mod4)).

On the other hand, we know from [2, Proof of Lemma 8.2] that

a⁢bp=(-1)b-12⁢(-1)j⁢2pj⁢pl4,

where j=v2⁢(a)≥1 is the 2-adic valuation of a. So, under our conditions, we have

a⁢bp=(-1)b-12⁢(-1)j⁢2pj+1,

and hence

2⁢a⁢bp=(-1)b-12⁢(-1)j⁢2pj.

Finally, we have:

-ϵ⁢π1≡X2(mod4) and ϵ⁢π2≡X2(mod4) have solutions in k if and only if (a≡2 and b≡-1(mod4)) or (a≡0 and b≡1(mod4)), i.e., 2⁢a⁢bp=1.
ϵ⁢π1≡X2(mod4) and -ϵ⁢π2≡X2(mod4) have solutions in k if and only if (a≡2 and b≡1(mod4)) or (a≡0 and b≡-1(mod4)), i.e., 2⁢a⁢bp=-1. ∎

Lemma 3.2.

Let p≡3(mod4) and l≡5(mod8) be different primes. Then the genus field of K=k⁢(-2⁢p⁢ϵ⁢l) is K(*)=K⁢(-p,-2).

Proof.

The proof is the same as the one for [2, Lemma 3.4]. ∎

Let p and l be different primes as specified by equation (2.1). Then h2⁢(-p⁢l), the 2-class number of k1=ℚ⁢(-p⁢l), is 2n+1, where n≥1 (see [13]). Put k2=ℚ⁢(-p) and R=k⁢k1⁢k2. Then we get the following lemma:

Lemma 3.3.

The 2-class group CR,2 of R is cyclic of order 2n. It is generated by the class of an ideal I∈R such that I2n-1∼pih0∼Viv, where pi (resp. Vi) is a prime ideal of R above p (resp. l) and v is the odd part of the class number of k2.

Proof.

First, prove that P, the prime ideal of k1 above p, is not principal. If not, there exists α∈k1 such that P=(α). This yields P2=(p)=(α2). Hence, by putting α=x+y⁢-p⁢l with x,y∈ℚ, we get

x2-p⁢l⁢y2+2⁢x⁢y⁢-p⁢l=±p.

This is equivalent to

{x=0,-l⁢y2=±1 or {y=0,x2=±p,

which is absurd. Similarly, one can prove that V, the prime ideal of k1 above l, is not principal. Hence P2∼V2∼1. Moreover, according to [13], 𝐂k1,2 is cyclic of order 2n+1, so generated by some class [I] such that P∼V∼I2n.

Let 𝔭i and Vi be the prime ideals of R above p and l, respectively. Then 𝔭ih0 and Viv are not principal in R since their norms are not principal in k1. Therefore, (𝔭ih0)2∼(Viv)2∼1.

Since the class number of k2 is odd, then, from [11], the rank of the 2-class group 𝐂R,2 is t-1-e, where t is the number of primes (finite and infinite) of k2 which ramify in R and 2e=[Ek2:Ek2∩NR/k2(R×)]. Then we can easily prove that 𝐂R,2 is of rank 1, i.e., it is cyclic of order h2⁢(-p⁢l)2=2n.

Finally, the ideal I generates a subgroup of 𝐂R,2 of order 2n since I2n∼P (in 𝐂R,2) and P capitulates in R. Then 𝐂R,2=〈[ℐ]〉 and ℐ2n-1∼𝔭ih0∼Viv. ∎

Remark 3.4.

Note the following:

If p≡3(mod8), then, by genus theory, 2k1, a prime ideal of k1 above 2, is not a square in k1 since
2,-p⁢lp=2p=-1 and 2,-p⁢ll=2l=-1.
Hence I∼2k1u, where u is the odd part of the class number of k1. Thus ℐ∼2Ru, where 2R is a prime ideal of R above 2. Moreover, n≥2 (see [13, Théorème 8]).
If p≡7(mod8), then 2Rv∼1 since 2k2v∼1 and 2k2 remains inert in R. In this case, n=1; see [13, Théorème 8].

Now, using the results of M. N. Gras [12], we will determine an FSU of some real cyclic quartic number field. Let L be a real cyclic quartic number field of Galois group H=〈σL〉 and of quadratic subfield k, and let ϵ be the fundamental unit of k. Let fL (resp. mk) be the conductor of L (resp. k). Then fL=mk⋅g, and the discriminant of L is equal to mk⋅fL2. It is known that L⊂ℚ⁢(ξfL).

Denote by χL a rational character of ℚ⁢(ξfL), and put EχL={ω∈EL:ω1+σL2=±1}. Let |EL| (resp. |Ek|,|EχL|) be the group of the absolute values of EL (resp. Ek,EχL), |EL|=|Ek|⊕|EχL|, QL=[|EL|:|EL|], and let ϵχL be a generator of EχL. From [12], we have QL=1 or 2, which gives two possible structures for the unit group of L:

If QL=1, we have EL=〈-1,ϵ,ϵχL,ϵχLσL〉;
If QL=2, we have EL=〈-1,ξL,ξLσL,ξLσL2〉 with ξL2=±ϵ⁢ϵχL1-σL, ξL1+σL=±ϵχL and ξL1+σL2=±ϵ.

Note that ξL is called a Minkowski unit.

Proposition 3.5.

Let l≡5(mod8) be a prime number. Put L=k⁢(2⁢ϵ⁢l); then QL=1.

Proof.

It is known that QL=2 if and only if L admits a Minkowski unit, and, from [8, Théorème II.2], the field L admits a Minkowski unit if and only if the unit ϵ is norm in L.

Let 𝔭 be a prime ideal of k, so we have:

If 𝔭 is not above l and 2, then v𝔭⁢(2⁢ϵ⁢l)=0, so
ϵ,2⁢ϵ⁢l𝔭=ϵ𝔭v𝔭⁢(2⁢ϵ⁢l)=1.
If 𝔭 lies above l, then v𝔭⁢(2⁢ϵ⁢l)=1. So, by [2, Proof of Proposition 4.1], we get
ϵ,2⁢ϵ⁢l𝔭=ϵ𝔭v𝔭⁢(2⁢ϵ⁢l)=ϵ𝔭=2l=-1.

This means that the unit ϵ is not norm in L. Thus the result follows. ∎

Lemma 3.6.

Let p≡3(mod4) and l≡5(mod8) be different primes. Then the Hasse unit index of

F5=k⁢(2⁢ϵ⁢l,-p)

is QF5=1.

Proof.

Let λ be a unit of F5+. Prove that the index QF5 equals 1, which is equivalent, by [1, Proposition 3], to proving that p⁢λ is not a square in F5+. Suppose that p⁢λ is a square in F5+. This is equivalent to (p)F5+=H2, where H is an ideal of F5+, which contradicts the fact that p cannot ramify in F5+. ∎

Theorem 3.7.

Keep the assumptions and notations of Lemma 3.6. Then

EF5={〈-1,ϵ,ϵχF5+,ϵχF5+σF5+〉𝑖𝑓⁢p≠3,〈ξ6,ϵ,ϵχF5+,ϵχF5+σF5+〉𝑖𝑓⁢p=3.

Proof.

Since the Hasse unit index of F5 is QF5=1, then EF5=WF5⁢EF5+. Thus the result follows. ∎

4 The 2-class number of the genus field K(*)

Let p and l be different primes as specified by equation (2.1). In this paragraph, we will compute the 2-class number of K(*), the genus field of K.

Proposition 4.1.

Put F=k⁢(2⁢ϵ⁢l,-2). Then the class number h⁢(F+) is odd. Moreover, QF=2, and h⁢(F) is odd too.

Proof.

Since the class number of k is odd, then, by [11], the rank of 2-class group 𝐂F+,2 of F+ is given by rank⁡(𝐂F+,2)=t-1-e, where t is the number of primes (finite and infinite) of k that ramify in F+ and the integer e is defined by 2e=[Ek:Ek∩NF+/k(F+×)]. We easily check that the primes (2)k and l ramify in F+. Hence t=2, and therefore, rank⁡(𝐂F+,2)=1-e. We know, from the proof of Proposition 3.5, that ϵ is not a norm in F+, whereas -1 is. Then Ek/Ek∩NF+/k⁢(F+×)={1¯,ϵ¯}. Therefore, e=1, so 𝐂F+,2 is trivial.

Let us prove that QF=2. For this, it suffices to prove that 2⁢λ is a square in F+ (see [1, Proposition 3]), where λ is a unit of F5+. In other words, it suffices to prove that (2)F+=H2, where H is a principal ideal in F+. As 2 ramifies in F+, there exists H in F+ such that (2)F+=H2. If H is not principal, then the class of H is of order 2, which is absurd since 𝐂F+,2 is trivial.

Next, we will compute the 2-class number h2⁢(F) of F. For this, notice that F/k is a V4-extension of CM-type fields (Figure 1), of quadratic subextensions F+, K′=k⁢(-2) and L′=k⁢(-ϵ⁢l). Then, from [14], we have

h2⁢(F)=QFQK′⁢QL′⋅wFwL′⁢wK′⋅h2⁢(F+)⁢h2⁢(K′)⁢h2⁢(L′)h2⁢(l)2.

Note that wK′=wF=wL′=2. Furthermore, from [17, Theorem 10.4], we have h2⁢(L′)=1 since fL′=l and QK′=QL′=1. From this, we deduce that h2⁢(F)=h2⁢(K′). As K′ is an unramified extension of ℚ⁢(-2⁢l) and the 2-class group of ℚ⁢(-2⁢l) is cyclic of order 2, then h2⁢(K′)=h2⁢(-2⁢l)2=1. Thus h2⁢(F)=1. ∎

Figure 1

Subfields of F/k.

Proposition 4.2.

Put F′=k⁢(2⁢ϵ⁢l,2⁢p). Then the class number of F′ is odd.

Proof.

By Proposition 4.1, the class number of L=F+=k⁢(2⁢ϵ⁢l) is odd, and, by Proposition 3.5,

EL=〈-1,ϵ,ϵχL,ϵχLσL〉.

Then, from [11] again, the rank of 2-class group 𝐂F′,2 of F′ is given by rank⁡(𝐂F′,2)=t-1-e, where t is the number of primes (finite and infinite) of L that ramify in F′ and 2e=[EL:EL∩NF′/L(F′⁣×)]. We easily check that t=2, and therefore, rank⁡(𝐂F′,2)=1-e. In the following, we show that e=1, which is equivalent to showing that EL∩NF′/L⁢(F′⁣×)=〈-1,ϵ,ϵχL⋅ϵχLσL〉.

Let 𝔭 be a prime ideal of L. So we have:

For all α∈{-1,ϵ}, α is a norm in F′. In fact, we have:
1. If 𝔭 is not above p and 2, then v𝔭⁢(2⁢p)=0, so
  α,2⁢p𝔭=α𝔭v𝔭⁢(2⁢p)=1.
2. If 𝔭 lies above p, then v𝔭⁢(2⁢p)=1, and thus
  α,2⁢p𝔭=α𝔭v𝔭⁢(2⁢p)=α𝔭=NL/k⁢(α)𝔭k=α2𝔭k=1.
3. If 𝔭 lies above 2, then
  α,2⁢p𝔭=NL/k⁢(α),2⁢p𝔭k=α2,2⁢p𝔭=1.
ϵχL is not a norm in F′. In fact, we have:
1. If 𝔭 is not above p and 2, then v𝔭⁢(2⁢p)=0, and thus
  ϵχL,2⁢p𝔭=ϵχL𝔭v𝔭⁢(2⁢p)=1.
2. If 𝔭 lies above p, then v𝔭⁢(2⁢p)=1, and thus
  ϵχL,2⁢p𝔭=ϵχL𝔭v𝔭⁢(2⁢p)=ϵχL𝔭=NL/k⁢(ϵχL)𝔭k=-1𝔭k=-1p=-1.
Similarly, we prove that ϵχLσL is not a norm in F′.

Consequently, EL/EL∩NF′/L⁢(F′⁣×)={1¯,ϵχL¯}. Therefore, e=1, and thus 𝐂F′,2 is trivial. ∎

We are now able to compute the 2-class number of K(*).

Theorem 4.3.

Put F5=K⁢(-p). Then h2⁢(K(*))=h2⁢(F5)2=2n+1, where n≥1.

Proof.

First, we compute the 2-class number h2⁢(F5) of F5. For this, consider Figure 2.

$Figure 2 Subfields of F5/k{F_{5}/k}.$

Figure 2

Subfields of F5/k.

It is easy to see that F5/k is a V4-extension of CM-type fields of quadratic subextensions L, K and R=k⁢(-p). Then, from [14], we have

h2⁢(F5)=QF5QK⁢QR⋅wF5wK⁢wR⋅h2⁢(R)⁢h2⁢(K)⁢h2⁢(L)h2⁢(l)2.

Note that wR=wF5 and wK=2. Moreover, we have QF5=1 (see Lemma 3.6), and we easily see that QK=QR=1. Then,

h2(F5)=12⋅8⋅h2(R)=4⋅h2(R)⋅

On the other hand, we easily check that h2⁢(R)=h2⁢(-p⁢l)2, and from [13], we have h2⁢(-p⁢l)=2n+1, where n≥1 (under our conditions). Hence

h2⁢(F5)=2n+2.

$Figure 3 Subfields of K(*)/L{K^{(*)}/L}.$

Figure 3

Subfields of K(*)/L.

Now, we prove that h2⁢(K(*))=h2⁢(F5)2=2n+1 with n≥1. For this, note that K(*)/L is a V4-extension of CM-type fields (Figure 3), of quadratic subextensions F5, F′=L⁢(2⁢p) and F=L⁢(-2). Then, from [14] again, we have

h2(K(*))=QK(*)QF5⁢QF⋅wK(*)wF5⁢wF⋅h2⁢(F′)⁢h2⁢(F)⁢h2⁢(F5)h2⁢(L)2⋅

Note that

WF5=WK(*)={{+1,-1}if⁢p≠3,〈ξ6〉if⁢p=3, and WF={+1,-1}.

Then wF5=wK(*) and wF=2. So Lemma 3.6 and Propositions 4.1 and 4.2 yield that QF5=1, QF=2 and h2⁢(F′)=h2⁢(F)=h2⁢(L)=1, and therefore,

h2⁢(K(*))=QK(*)2⋅12⋅h2⁢(F5).

Finally, from the proof of Proposition 4.1, we have (2)L=HL2, where HL is a principal ideal in L. On the other hand, HL remains inert in K(*)+, so HK(*)+, the prime ideal of K(*)+ above 2, is principal. Therefore, (2)K(*)+=HK(*)+2, and thus [1, Proposition 3] implies that QK(*)=2. From this, we deduce that

h2⁢(K(*))=h2⁢(F5)2=2n+1.∎

5 Proof of Theorem 2.1

Let p and l be different primes as specified by equation (2.1). Denote by ϵ and by h0 the fundamental unit and the class number of k=ℚ⁢(l), respectively. Put K=k⁢(-2⁢p⁢ϵ⁢l). Recall that 𝒫1, 𝒫2 are the prime ideals in K above p and ℋ is the prime ideal in K above 2. Moreover, 𝒫i2⁢h0=(πi) and ℋ2=(2).

5.1 Generators of the 2-class group of K

Let us show that 𝐂K,2=〈[𝒫1h0],[𝒫2h0],[ℋ]〉. For this, it suffices to show that [𝒫1h0], [𝒫2h0] and [ℋ] are of order 2 and pairwise different.

[𝒫1h0⁢𝒫2h0] is of order 2. In fact, if 𝒫1h0⁢𝒫2h0≠(α) for some α∈K, this yields that (ph0)=(α2). Then there exists ϵ′, a unit of K, such that ph0⁢ϵ′=α2, and thus
ph0⁢ϵ′=(c+d⁢-2⁢p⁢ϵ⁢l)2=c2-2⁢p⁢ϵ⁢l⁢d2+2⁢c⁢d⁢-2⁢p⁢ϵ⁢l
with c and d in k. As {ϵ} is an FSU of K and -1∉K, then ph0⁢ϵ′∈k, and therefore, c=0 or d=0. If d=0, then ph0⁢ϵ′=c2, and thus ±ph0=c2 or ph0⁢ϵ=c2 in k, which gives that ±p∈k in the first case and -1∈ℚ in the second case, but this is impossible. Similarly, if c=0, we find that ±l is a square in ℚ, which is absurd.
[𝒫ih0] is of order 2. In fact, assume that 𝒫ih0=(α) for α∈K. Then 𝒫i2⁢h0=(α)2, which is equivalent to (πi)=(α2). Then there exists ϵ′, a unit of K, such that
πi⁢ϵ′=α2=(c+d⁢-2⁢p⁢ϵ⁢l)2=c2-2⁢p⁢ϵ⁢l⁢d2+2⁢c⁢d⁢-2⁢p⁢ϵ⁢l
with c and d in k, and then c=0 or d=0. If d=0, multiplying by πj for j≠i, we get ph0⁢ϵ′=πj⁢c2. So, by applying the norm in k/ℚ, we find that ±ph0=Nk/ℚ⁢(c)2. This means that ±p is a square in ℚ, which is impossible. In a similar way, if c=0, we get ±l⁢p is a square in ℚ, which is impossible too.

Similarly, we can show that [ℋ], [ℋ⁢𝒫ih0] and [ℋ⁢𝒫1h0⁢𝒫2h0] are of order 2. Thus 〈[𝒫1h0],[𝒫2h0],[ℋ]〉 is of type (2,2,2), and then 𝐂K,2=〈[𝒫1h0],[𝒫2h0],[ℋ]〉.

Remark 5.1.

Let 𝒱 be a prime ideal of K above l. Then 𝒱∼ℋ⁢𝒫1h0⁢𝒫2h0 since ℋ⁢𝒫1h0⁢𝒫2h0⁢𝒱=(-2⁢p⁢ϵ⁢l), where -2⁢p⁢ϵ⁢l∈K.

5.2 The unramified extensions of K within K2(1)

We know from Lemma 3.2 that the genus field of K is K(*)=K⁢(-p,-2), and therefore, its subextensions F3=K⁢(2⁢p), F4=K⁢(-2) and F5=K⁢(-p) are unramified over K and absolutely abelian.

From Lemma 3.1, if 2⁢a⁢bp=1, then the equation -ϵ⁢π1≡X2(mod4) admits a solution in k. Moreover, since (-ϵ⁢π1)=(𝒫1h0)2, where 𝒫1 is an ideal of K, so K⁢(-ϵ⁢π1) is an unramified quadratic extension of K. This implies that F1=K⁢(-ϵ⁢π1) is non-Galois over ℚ.

Since the extensions Fi are pairwise different for i∈{1,3,4,5}, then one can easily deduce that

F2=K⁢(-2⁢ϵ⁢π2),F6=K⁢(2⁢ϵ⁢π1) and F7=K⁢(ϵ⁢π2)

are pairwise different and non-Galois over ℚ.

Finally, as -ϵ=ϵ′⁣-1=ϵ′⁣-2⁢ϵ′, where ϵ′ is the conjugate of ϵ, hence

F1=K⁢(-ϵ⁢π1)=K⁢(ϵ′⁢π1) and F2=K⁢(-2⁢ϵ⁢π2)=K⁢(2⁢ϵ′⁢π2).

Thus F1≃F7 and F2≃F6.

Regarding the biquadratic extensions Li, note that each extension is gotten by composing two different unramified quadratic extensions of K. Then Li, 1≤i≤7, are the unramified biquadratic extensions of K within K2(1). By using the fact that F1≃F7 and F2≃F6, we deduce that L1≃L3 and L6≃L7.

If 2⁢a⁢bp=-1, the results are proved similarly.

6 Proof of Theorem 2.2

Let p and l be different primes as specified by equation (2.1). Recall that R=k⁢(-p) and L=k⁢(2⁢ϵ⁢l).

6.1 The 2-class group of F5

From Figure 2, we see that F5/R, F5/L are ramified quadratic extensions, whereas F5/K is not. Hence, according to class field theory,

[𝐂K,2:NF5/K(𝐂F5,2)]=2,𝐂R,2=NF5/R(𝐂F5,2) and 𝐂L,2=NF5/L(𝐂F5,2).

Since F5=K⁢(-p)=K⁢(2⁢ϵ⁢l) and

NF5/K⁢(𝐂F5,2)={[Q]∈𝐂K,2|2⁢ϵ⁢lQ=1}={[Q]∈𝐂K,2|-pQ=1},

then NF5/K⁢(𝐂F5,2)=〈[ℋ],[𝒫1h0⁢𝒫2h0]〉.

In fact, from [9], we have

2⁢ϵ⁢l𝒫ih0=-2p⁢pl4=-2p2=-1,

and then

2⁢ϵ⁢l𝒫1h0⁢𝒫2h0=2⁢ϵ⁢l𝒫1h0⁢2⁢ϵ⁢l𝒫2h0=1.

Moreover, let 2k be a prime ideal of k above 2. So

-pℋ=-p2k=-p,22k=Nk/ℚ⁢(-p),22=p2,22=1.

Thus [𝒫1h0⁢𝒫2h0] and [ℋ] are norms in F5/K.

Let us now compute κF5/K. From Theorem 3.7, we get

EF5=〈-1,ϵ,ϵχF5+,ϵχF5+σF5+〉 and EK=〈-1,ϵ〉,so NF5/K⁢(EF5)=〈-1,ϵ2〉.

Therefore, [EK:NF5/K(EF5)]=2, which implies that #⁢κF5/K=4. Moreover, we showed in the proof of Proposition 4.1 that (2)L=H2, H is principal in L=F+ and H⁢𝒪F5=ℋ⁢𝒪F5. So, necessarily, ℋ capitulates in F5. In addition, it is known that 𝒫12⁢h0⁢𝒫22⁢h0=(ph0), which means that (𝒫1h0⁢𝒫2h0)=(ph0), and hence 𝒫1h0⁢𝒫2h0 capitulates in F5. Thus κF5/K=〈[ℋ],[𝒫1h0⁢𝒫2h0]〉.

On the other hand, we know from Proposition 4.1 and Lemma 3.3 that 𝐂L,2 is trivial and 𝐂R,2=〈[ℐ]〉. Let 𝔍 be a prime ideal of F5 above l. Then we claim (𝔍v)4∼1.

To this end, let s and t be the elements of order 2 in Gal⁡(F5/k) which fix R and K, respectively, so st fixes L. Using the relation 2+(1+s+t+s⁢t)=(1+s)+(1+t)+(1+s⁢t) of the group ring ℤ⁢[Gal⁡(F5/k)] and taking into account that the 2-class number of k is 1, we get

(𝔍v)2∼(𝔍v)1+s⁢(𝔍v)1+t⁢(𝔍v)1+s⁢t∼Viv⁢𝒱v∼Viv⁢ℋ⁢𝒫1h0⁢𝒫2h0∼Viv

since 𝒱v∼𝒱∼ℋ⁢𝒫1h0⁢𝒫2h0 (by Remark 5.1) and ℋ⁢𝒫1h0⁢𝒫2h0∈κF5/K (Vi is the prime ideal of R=k1⁢k2 above l, see Lemma 3.3). Therefore, (𝔍v)4∼(Viv)2∼1.

Let ℋ0 be a prime ideal of F5 above 2. So, from Remark 3.4, we have:

If p≡3(mod8), then (ℋ0u)2∼(ℋ0u)1+s⁢(ℋ0u)1+t⁢(ℋ0u)1+s⁢t∼2Ru⁢ℋu∼2Ru, and therefore, (ℋ0u)2n+1∼(2Ru)2n∼1.
If p≡7(mod8), then (ℋ0v)2∼(ℋ0v)1+s⁢(ℋ0v)1+t⁢(ℋ0v)1+s⁢t∼2Rv⁢ℋv∼1.

Moreover, we have:

If p≡3(mod8), then (𝔍v)2∼Viv∼ℐ2n-1∼(ℋ0u)2n. Furthermore, 𝔍v≁(ℋ0u)2n-1. If not, by applying the norm of F5/K, we get ℋ⁢𝒫1h0⁢𝒫2h0∼(ℋu)2n-1∼1 (in K), which is absurd.
If p≡7(mod8), then (𝔍v)2≁ℋ0v. If not, by applying the norm of F5/K, we get ℋ2⁢𝒫12⁢h0⁢𝒫22⁢h0∼ℋ∼1 (in K), which is absurd.

Consequently,

𝐂F5,2={〈[ℋ0u],[𝔍v]〉≃(2,2n+1)if⁢p≡3(mod8),〈[ℋ0v],[𝔍v]〉≃(2,4)if⁢p≡7(mod8).

Note that 𝒫i remains inert and does not capitulate in F5. Then 𝒫12⁢h0∼1 (in F5), and therefore,

𝒫1h0∼{(𝔍v)2∼(ℋ0u)2nin the first case,ℋ0v⁢or⁢(𝔍v)2in the second case.

If 𝒫1h0∼ℋ0v, by applying the norm of F5/K, we get 𝒫12⁢h0∼ℋ∼1, which is absurd. Thus 𝒫1h0∼(𝔍v)2.

6.2 The Hilbert 2-class field tower of K

Let us prove that the Hilbert 2-class field tower of K stops at K2(2). From Theorem 4.3, we know that h2⁢(K(*))=h2⁢(F5)2 and the 2-rank of 𝐂F5,2 equals 2. So, by [7, Proposition 7], the Hilbert 2-class field tower of F5 stops at the first stage F5,2(1). Therefore, the Hilbert 2-class field tower of K stops at K2(2) since K⊂F5.

6.3 The generators of G=Gal⁡(K2(2)/K)

Let us compute the generators of G=Gal⁡(K2(2)/K). Let K2(2)/F5P denote the Artin symbol for K2(2)/F5, and put

σ=K2(2)/F5𝔍v and τ={K2(2)/F5ℋ0uif⁢p≡3(mod8),K2(2)/F5ℋ0vif⁢p≡7(mod8).

Then σ and τ generate the abelian subgroup of G,

Gal(K2(2)/F5)≃{(2,2n+1)if⁢p≡3(mod8),(2,4)if⁢p≡7(mod8),

with

{τ2n+1=σ4=1⁢and⁢τ2n=σ2if⁢p≡3(mod8),τ2=σ4=1if⁢p≡7(mod8).

Also put

ρ=K2(2)/K𝒫1h0,

so ρ restricts to the non-trivial automorphism of F5/K since 𝒫1h0 is not a norm from F5/K, and hence G=〈σ,τ,ρ〉.

Furthermore, the following relations hold:

Since 𝒫1h0∼(𝔍v)2,
ρ2=K2(2)/K𝒫12⁢h0=K2(2)/KNF5/K⁢(𝒫1h0)=K2(2)/F5𝒫1h0,
and thus ρ2=σ2.
Since (𝔍v)1+ρ=NF5/K⁢(𝔍v)∼ℋ⁢𝒫1h0⁢𝒫2h0∼1 in F5,
σ⁢ρ-1⁢σ⁢ρ=K2(2)/F5(𝔍v)1+ρ=1,
and thus [σ,ρ]=[ρ,σ]=σ2.
Since (ℋ0u)1+ρ=NF5/K⁢(ℋ0u)∼ℋ∼1 and (ℋ0v)1+ρ=NF5/K⁢(ℋ0v)∼ℋ∼1 in F5, according as p≡3(mod8) and p≡7(mod8),
τ⁢ρ-1⁢τ⁢ρ={K2(2)/F5(ℋ0u)1+ρ=1if⁢p≡3(mod8),K2(2)/F5(ℋ0v)1+ρ=1if⁢p≡7(mod8),
and thus [τ,ρ]=τ-2.

Hence

G={〈ρ,τ,σ:ρ4=σ4=τ2n+1=1,σ2=ρ2=τ2n,n≥2,[τ,σ]=1,[σ,ρ]=σ2,[ρ,τ]=τ2〉if⁢p≡3(mod8),〈ρ,τ,σ:ρ4=σ4=τ2=1,σ2=ρ2,[τ,σ]=[τ,ρ]=1,[σ,ρ]=σ2〉if⁢p≡7(mod8).

If p≡3(mod8), then G′=〈τ2〉 since [τ,σ]=1, [ρ,τ]=τ2, [ρ,σ]=σ2 and σ2=(τ2)2n-1, and therefore, 𝐂K2(1),2 is cyclic of order 2n.
If p≡7(mod8), we easily check that G′=〈σ2〉, and thus 𝐂K2(1),2 is cyclic of order 2.

Now, compute the coclass of G. Recall that the lower central series of G is inductively defined by γ1⁢(G)=G and γi+1⁢(G)=[γi⁢(G),G], which is the subgroup of G generated by the set {[a,b]=a-1⁢b-1⁢a⁢b∣a∈γi⁢(G),b∈G}. So the coclass of G is defined to be c⁢c⁢(G)=h-c, where #⁢G=2h and c=c⁢(G) is the nilpotency class of G, which is the smallest positive integer satisfying γc⁢(G)≠{1} and γc+1⁢(G)={1}.

If p≡3(mod8), then γ1⁢(G)=G, γ2⁢(G)=G′=〈τ2〉 and
γn+1⁢(G)=[γn⁢(G),G]=〈τ2n〉,γn+2⁢(G)=[γn+1⁢(G),G]=〈τ2n+1〉=〈1〉.
So c⁢(G)=n+1, and c⁢c⁢(G)=2.
If p≡7(mod8), then γ1⁢(G)=G, γ2⁢(G)=G′=〈σ2〉 and γ3⁢(G)=[G′,G]=〈1〉. So c⁢c⁢(G)=c⁢(G)=2.

Finally, prove assertion (6) of Theorem 2.2.

If p≡3(mod8), then [ρ,σ2]=σ4=1, [ρ,σ⁢τ2n-1]=1 and σ2≠σ⁢τ2n-1. So 〈σ2,σ⁢τ2n-1〉=Z⁢(G)≃(2,2).
If p≡7(mod8), then [ρ,σ2]=σ4=1, [ρ,τ]=1 and σ2≠τ. So 〈σ2,τ〉=Z⁢(G)≃(2,2).

7 Proof of Theorem 2.3

To prove this theorem, we need the following norm class groups NFi/K⁢(𝐂Fi,2). The results are summarized in Tables 1 and 2.

Table 1

Norm class groups for the case 2⁢a⁢bp=1.

	NFi/K⁢(𝐂Fi,2)
Fi	2p=-1	2p=1
F1=K⁢(-ϵ⁢π1)	〈[𝒫2h0],[ℋ⁢𝒫1h0]〉	〈[𝒫1h0],[ℋ]〉
F2=K⁢(-2⁢ϵ⁢π2)	〈[𝒫1h0],[ℋ]〉	〈[𝒫1h0],[ℋ⁢𝒫2h0]〉
F6=K⁢(2⁢ϵ⁢π1)	〈[𝒫2h0],[ℋ]〉	〈[𝒫2h0],[ℋ⁢𝒫1h0]〉
F7=K⁢(ϵ⁢π2)	〈[𝒫1h0],[ℋ⁢𝒫2h0]〉	〈[𝒫2h0],[ℋ]〉
F3=K⁢(2⁢p)	〈[ℋ⁢𝒫ih0],[𝒫1h0⁢𝒫2h0]〉	〈[𝒫1h0],[𝒫2h0]〉
F4=K⁢(-2)	〈[𝒫1h0],[𝒫2h0]〉	〈[ℋ⁢𝒫ih0],[𝒫1h0⁢𝒫2h0]〉
F5=K⁢(-p)	〈[ℋ],[𝒫1h0⁢𝒫2h0]〉

Table 2

Norm class groups for the case 2⁢a⁢bp=-1.

	NFi/K⁢(𝐂Fi,2)
Fi	2p=-1	2p=1
F1=K⁢(-2⁢ϵ⁢π1)	〈[𝒫2h0],[ℋ]〉	〈[𝒫2h0],[ℋ⁢𝒫1h0]〉
F2=K⁢(-ϵ⁢π2)	〈[𝒫1h0],[ℋ⁢𝒫2h0]〉	〈[𝒫2h0],[ℋ]〉
F6=K⁢(ϵ⁢π1)	〈[𝒫2h0],[ℋ⁢𝒫1h0]〉	〈[𝒫1h0],[ℋ]〉
F7=K⁢(2⁢ϵ⁢π2)	〈[𝒫1h0],[ℋ]〉	〈[𝒫1h0],[ℋ⁢𝒫2h0]〉
F3=K⁢(2⁢p)	〈[ℋ⁢𝒫ih0],[𝒫1h0⁢𝒫2h0]〉	〈[𝒫1h0],[𝒫2h0]〉
F4=K⁢(-2)	〈[𝒫1h0],[𝒫2h0]〉	〈[ℋ⁢𝒫ih0],[𝒫1h0⁢𝒫2h0]〉
F5=K⁢(-p)	〈[ℋ],[𝒫1h0⁢𝒫2h0]〉

Let us check some examples. Recall that, for 1≤i≤7, [𝐂K,2:NFi/K(𝐂Fi,2)]=2.

Example.

For F5, we already proved that NF5/K⁢(𝐂F5,2)=〈[ℋ],[𝒫1h0⁢𝒫2h0]〉.

Example.

For F3=K⁢(2⁢p)=K⁢(-ϵ⁢l), we have

-ϵ⁢l𝒫ih0=-1𝒫ih0⁢ϵ⁢l𝒫ih0=--1ph0⁢pl4=2p,

-ϵ⁢lℋ=-ϵ⁢l2k=-ϵ⁢l,22k=Nk/ℚ⁢(-ϵ⁢l),22=l,22=2l=-1.

Then NF3/K⁢(𝐂F3,2)=〈[𝒫1h0],[𝒫2h0]〉 or 〈[ℋ⁢𝒫ih0],[𝒫1h0⁢𝒫2h0]〉, according as 2p=1 or -1.

Example.

For F4=K⁢(-2)=K⁢(p⁢ϵ⁢l), we have

-2𝒫ih0=-2p,

p⁢ϵ⁢lℋ=p⁢ϵ⁢l2k=p⁢ϵ⁢l,22k=Nk/ℚ⁢(p⁢ϵ⁢l),22=p2⁢l,22=l,22=2l=-1.

Then NF4/K⁢(𝐂F4,2)=〈[𝒫1h0],[𝒫2h0]〉 or 〈[ℋ⁢𝒫ih0],[𝒫1h0⁢𝒫2h0]〉, according as 2p=-1 or 1.

Example.

Consider as a last example F6=K⁢(2⁢ϵ⁢π1)=K⁢(-π2⁢l), i.e., the case

2⁢a⁢bp=1⇔a⁢bp=2p.

The quadratic residue symbol is -π2⁢l𝒫1h0=-1. In fact,
-π2⁢l𝒫1h0=--1ph0⁢pl4⁢l𝒫1h0=2p⁢l𝒫1h0
since π1=a+b⁢l. Then
π1-a𝒫1h0=-a𝒫1h0=b⁢l𝒫1h0,
which implies
l𝒫1h0=-a⁢b𝒫1h0=-2p,
and thus
-π2⁢l𝒫1h0=-2p2=-1.
The quadratic residue symbol is 2⁢ϵ⁢π1𝒫2h0=1. In fact,
2⁢ϵ⁢π1𝒫2h0=2⁢ϵ𝒫2h0⁢π1𝒫2h0=-pl4⁢2p⁢ϵ𝒫2h0=-2p2⁢ϵ𝒫2h0=-ϵ𝒫2h0.
Similarly, as we just did for l𝒫1h0, we get
l𝒫2h0=a⁢bp=2p,
so
ϵ⁢l𝒫2h0=ϵ𝒫2h0⁢l𝒫2h0=ϵ𝒫2h0⁢2p,
Thus
ϵ𝒫2h0=-pl4⁢2p=-2p2=-1,
and finally,
2⁢ϵ⁢π1𝒫2h0=-ϵ𝒫2h0=1.
The quadratic residue symbol is -π2⁢lℋ=-2p. In fact,
-π2⁢lℋ=-π2⁢l2k=-π2⁢l,22k=Nk/ℚ⁢(-π2⁢l),22=-p⁢l,22=2p⁢2l=-2p.

Thus

if 2p=1, then NF6/K⁢(𝐂F6,2)=〈[𝒫2h0],[ℋ⁢𝒫1h0]〉;
if 2p=-1, then NF6/K⁢(𝐂F6,2)=〈[𝒫2h0],[ℋ]〉.

Proceeding similarly, we check the other tables entries.

7.1 The abelian-type invariants of 𝐂Fi,2 and capitulation kernels κFi/K

We now compute the Galois groups Gi=Gal⁡(K2(2)/Fi), their derived groups, the abelian-type invariants of 𝐂Fi,2 and the transfer kernels ker⁡VG→Gi, and we deduce κFi/K the capitulation kernels of Fi/K. The results are summarized in Tables 3–10.

Tables 3 and 4 are filled using Tables 1 and 2, the relations in Remark 7.1 and Proposition 7.2.

Remark 7.1.

Note the following relations:

σ=K2(2)/F5𝔍v=K2(2)/KNF5/K⁢(𝔍v)=K2(2)/Kℋ⁢𝒫1h0⁢𝒫2h0,

τ=K2(2)/F5ℋ0=K2(2)/KNF5/K⁢(ℋ0)=K2(2)/Kℋ

and

ρ-1⁢τ-1⁢σ=K2(2)/K𝒫2h0

since

σ=K2(2)/Kℋ⁢𝒫1h0⁢𝒫2h0,τ=K2(2)/Kℋ and ρ=K2(2)/K𝒫1h0.

Proposition 7.2.

Let G=〈ρ,τ,σ〉 denote the group defined in Section 6.3. Then the following relations hold:

ρ-1⁢σ⁢ρ=σ-1,ρ-1⁢τ⁢ρ=τ-1,

(τ⁢σ)2={τ2⁢σ2=τ2n+2𝑖𝑓⁢p≡3(mod8),σ2𝑖𝑓⁢p≡7(mod8),

(ρ⁢τ⁢σ)2=(τ⁢ρ)2=(ρ⁢σ)2=ρ2,

ρ⁢σ⁢ρ=σ

and

(τ⁢σ)2n+2=τ2n⁢τ2⁢σ2n⁢σ2=τ2n+1⁢τ2=τ2 𝑖𝑓⁢p≡3(mod8)

since σ2=τ2n and σ2n=1.

Table 3

Gi=Gal⁡(K2(2)/Fi) for the case 2⁢a⁢bp=1.

	Gi		Gi′
Fi	2p=-1	2p=1	2p=-1	2p=1
F1=K⁢(-ϵ⁢π1)	〈σ,τ⁢ρ,τ2〉	〈ρ,τ〉	〈τ4〉	〈1〉
F2=K⁢(-2⁢ϵ⁢π2)	〈ρ,τ〉	〈ρ,σ〉	〈τ2〉	〈σ2〉
F6=K⁢(2⁢ϵ⁢π1)	〈σ⁢ρ,τ〉	〈σ,τ⁢ρ〉	〈τ2〉	〈σ2〉
F7=K⁢(ϵ⁢π2)	〈ρ,σ,τ2〉	〈σ⁢ρ,τ〉	〈τ4〉	〈1〉
F3=K⁢(2⁢p)	〈τ⁢ρ,σ⁢ρ,τ2〉	〈ρ,τ⁢σ〉	〈τ2〉	〈σ2〉
F4=K⁢(-2)	〈ρ,τ⁢σ,τ2〉	〈τ⁢ρ,σ⁢ρ〉	〈τ2〉	〈σ2〉
F5=K⁢(-p)	〈τ,σ〉		〈1〉

Table 4

Gi=Gal⁡(K2(2)/Fi) for the case 2⁢a⁢bp=-1.

	Gi		Gi′
Fi	2p=-1	2p=1	2p=-1	2p=1
F1=K⁢(-2⁢ϵ⁢π1)	〈σ⁢ρ,τ〉	〈σ,τ⁢ρ〉	〈τ2〉	〈σ2〉
F2=K⁢(-ϵ⁢π2)	〈ρ,σ,τ2〉	〈σ⁢ρ,τ〉	〈τ4〉	〈1〉
F6=K⁢(ϵ⁢π1)	〈σ,τ⁢ρ,τ2〉	〈ρ,τ〉	〈τ4〉	〈1〉
F7=K⁢(2⁢ϵ⁢π2)	〈ρ,τ〉	〈ρ,σ〉	〈τ2〉	〈σ2〉
F3=K⁢(2⁢p)	〈τ⁢ρ,σ⁢ρ,τ2〉	〈ρ,τ⁢σ〉	〈τ2〉	〈σ2〉
F4=K⁢(-2)	〈ρ,τ⁢σ,τ2〉	〈τ⁢ρ,σ⁢ρ〉	〈τ2〉	〈σ2〉
F5=K⁢(-p)	〈τ,σ〉		〈1〉

Tables 5 and 6 are easily deduced by calculating 𝐂Fi,2≃Gi/Gi′. Take, for example, G7=〈ρ,σ,τ2〉 for the case

2⁢a⁢bp=1 and 2p=-1.

Then G7′=〈τ4〉 since [ρ,σ]=σ2, [σ,τ2]=1, [ρ,τ2]=τ4 and σ2=(τ4)2n-2. Thus

𝐂F7,2≃G7/G7′=〈ρ¯,σ¯,τ2¯〉≃(2,2,2).

Table 5

𝐂Fi,2 for the case 2⁢a⁢bp=1.

	𝐂Fi,2
Fi	2p=-1	2p=1
F1=K⁢(-ϵ⁢π1)	(2,2,2)	(2,4)
F2=K⁢(-2⁢ϵ⁢π2)	(2,2)	(2,2)
F6=K⁢(2⁢ϵ⁢π1)	(2,2)	(2,2)
F7=K⁢(ϵ⁢π2)	(2,2,2)	(2,4)
F3=K⁢(2⁢p)	(2,2)	(2,2)
F4=K⁢(-2)	(2,2)	(2,2)
F5=K⁢(-p)	(2,2n+1)	(2,4)

Table 6

𝐂Fi,2 for the case 2⁢a⁢bp=-1.

	𝐂Fi,2
Fi	2p=-1	2p=1
F1=K⁢(-2⁢ϵ⁢π1)	(2,2)	(2,2)
F2=K⁢(-ϵ⁢π2)	(2,2,2)	(2,4)
F6=K⁢(ϵ⁢π1)	(2,2,2)	(2,4)
F7=K⁢(2⁢ϵ⁢π2)	(2,2)	(2,2)
F3=K⁢(2⁢p)	(2,2)	(2,2)
F4=K⁢(-2)	(2,2)	(2,2)
F5=K⁢(-p)	(2,2n+1)	(2,4)

Compute now the transfer kernels ker⁡VG→Gi (Tables 7 and 8). First, recall that if G/Gi={1¯,z¯}, then

VG→Gi⁢(g⁢G′)={g⁢z-1⁢g⁢z⁢Gi′=g2⁢[g,z]⁢Gi′if⁢g∈Gi,g2⁢Gi′if⁢g∉Gi.

Table 7

ker⁡VG→Gi for the case 2⁢a⁢bp=1.

	ker⁡VG→Gi
Fi	2p=-1	2p=1
F1=K⁢(-ϵ⁢π1)	〈σ⁢G′,ρ⁢G′〉	〈ρ⁢G′,τ⁢G′〉
F2=K⁢(-2⁢ϵ⁢π2)	〈ρ⁢G′,σ⁢G′,τ⁢G′〉	〈ρ⁢G′,σ⁢G′,τ⁢G′〉
F6=K⁢(2⁢ϵ⁢π1)	〈ρ⁢G′,σ⁢G′,τ⁢G′〉	〈ρ⁢G′,σ⁢G′,τ⁢G′〉
F7=K⁢(ϵ⁢π2)	〈σ⁢G′,ρ⁢τ⁢G′〉	〈σ⁢ρ⁢G′,τ⁢G′〉
F3=K⁢(2⁢p)	〈ρ⁢G′,σ⁢G′,τ⁢G′〉
F4=K⁢(-2)	〈ρ⁢G′,σ⁢G′,τ⁢G′〉
F5=K⁢(-p)	〈τ⁢G′,σ⁢G′〉

Table 8

ker⁡VG→Gi for the case 2⁢a⁢bp=-1.

	ker⁡VG→Gi
Fi	2p=-1	2p=1
F1=K⁢(-2⁢ϵ⁢π1)	〈ρ⁢G′,σ⁢G′,τ⁢G′〉	〈ρ⁢G′,σ⁢G′,τ⁢G′〉
F2=K⁢(-ϵ⁢π2)	〈σ⁢G′,ρ⁢τ⁢G′〉	〈σ⁢ρ⁢G′,τ⁢G′〉
F6=K⁢(ϵ⁢π1)	〈σ⁢G′,ρ⁢G′〉	〈ρ⁢G′,τ⁢G′〉
F7=K⁢(2⁢ϵ⁢π2)	〈ρ⁢G′,σ⁢G′,τ⁢G′〉	〈ρ⁢G′,σ⁢G′,τ⁢G′〉
F3=K⁢(2⁢p)	〈ρ⁢G′,σ⁢G′,τ⁢G′〉
F4=K⁢(-2)	〈ρ⁢G′,σ⁢G′,τ⁢G′〉
F5=K⁢(-p)	〈τ⁢G′,σ⁢G′〉

Table 9

κFi/K for the case 2⁢a⁢bp=1.

	κFi/K
Fi	2p=-1	2p=1
F1=K⁢(-ϵ⁢π1)	〈[𝒫1h0],[ℋ⁢𝒫2h0]〉	〈[ℋ],[𝒫1h0]〉
F2=K⁢(-2⁢ϵ⁢π2)	〈[ℋ],[𝒫1h0],[𝒫2h0]〉	〈[ℋ],[𝒫1h0],[𝒫2h0]〉
F6=K⁢(2⁢ϵ⁢π1)	〈[ℋ],[𝒫1h0],[𝒫2h0]〉	〈[ℋ],[𝒫1h0],[𝒫2h0]〉
F7=K⁢(ϵ⁢π2)	〈[𝒫2h0],[ℋ⁢𝒫1h0]〉	〈[ℋ],[𝒫2h0]〉
F3=K⁢(2⁢p)	〈[ℋ],[𝒫1h0],[𝒫2h0]〉
F4=K⁢(-2)	〈[ℋ],[𝒫1h0],[𝒫2h0]〉
F5=K⁢(-p)	〈[ℋ],[𝒫1h0⁢𝒫2h0]〉

Table 10

κFi/K for the case 2⁢a⁢bp=-1.

	κFi/K
Fi	2p=-1	2p=1
F1=K⁢(-2⁢ϵ⁢π1)	〈[ℋ],[𝒫1h0],[𝒫2h0]〉	〈[ℋ],[𝒫1h0],[𝒫2h0]〉
F2=K⁢(-ϵ⁢π2)	〈[𝒫2h0],[ℋ⁢𝒫1h0]〉	〈[ℋ],[𝒫2h0]〉
F6=K⁢(ϵ⁢π1)	〈[𝒫1h0],[ℋ⁢𝒫2h0]〉	〈[ℋ],[𝒫1h0]〉
F7=K⁢(2⁢ϵ⁢π2)	〈[ℋ],[𝒫1h0],[𝒫2h0]〉	〈[ℋ],[𝒫1h0],[𝒫2h0]〉
F3=K⁢(2⁢p)	〈[ℋ],[𝒫1h0],[𝒫2h0]〉
F4=K⁢(-2)	〈[ℋ],[𝒫1h0],[𝒫2h0]〉
F5=K⁢(-p)	〈[ℋ],[𝒫1h0⁢𝒫2h0]〉

Tables 9 and 10 are deduced from Tables 7 and 8 and Remark 7.1.

Take some examples:

Example.

Consider F4. Then we know from Table 1 or 2 that

NF4/K⁢(𝐂F4,2)={〈[𝒫1h0],[𝒫2h0]〉if⁢(2p)=-1,〈[ℋ⁢𝒫1h0],[𝒫1h0⁢𝒫2h0]〉if⁢(2p)=1.

Then G4=〈ρ,τ-1⁢σ,τ2〉=〈ρ,τ⁢σ,τ2〉 or G4=〈τ⁢ρ,τ-1⁢σ〉=〈τ⁢ρ,σ⁢ρ〉 and G4′=〈τ2〉 or G4′=〈σ2〉, respectively, and hence G/G4={1¯,τ¯=σ¯} or G/G4={1¯,τ¯=σ¯=ρ¯}. Thus

VG→G4⁢(ρ⁢G′)=ρ2⁢[ρ,σ]⁢G4′=ρ2⁢σ2⁢G4′=G4′ since ρ2=σ2 and σ4=1,
VG→G4⁢(σ⁢G′)=σ2⁢G4′=G4′ since σ2=(τ2)2n-1∈G4′,
VG→G4⁢(τ⁢G′)=τ2⁢G4′=G4′;

VG→G4⁢(ρ⁢G′)=ρ2⁢G4′=σ2⁢G4′=G4′ since ρ2=σ2,
VG→G4⁢(σ⁢G′)=σ2⁢G4′=G4′,
VG→G4⁢(τ⁢G′)=τ2⁢G4′=G4′ since τ2=1.

So ker⁡VG→G4=〈τ⁢G′,ρ⁢G′,σ⁢G′〉, which gives

κF4/K=〈[ℋ],[𝒫1h0],[𝒫2h0]〉=𝐂K,2.

Example.

Now consider F1.

If 2⁢a⁢bp=1, from Table 1, we have
NF1/K⁢(𝐂F1,2)={〈[𝒫2h0],[ℋ⁢𝒫1h0]〉if⁢(2p)=-1,〈[𝒫1h0],[ℋ]〉if⁢(2p)=1.
Then G1=〈ρ-1⁢τ-1⁢σ,τ⁢ρ,τ2〉=〈σ,τ⁢ρ,τ2〉 or G1=〈ρ,τ〉, and G1′=〈τ4〉 or G1′=〈1〉, respectively, and hence G/G1={1¯,τ¯=ρ¯} or G/G1={1¯,σ¯}. Thus
1. VG→G1⁢(σ⁢G′)=σ2⁢[σ,τ]⁢G1′=σ2⁢G1′=τ2n⁢G1′=G1′ since n≥2,
2. VG→G1⁢(τ⁢G′)=τ2⁢G1′≠G1′,
3. VG→G1⁢(ρ⁢G′)=ρ2⁢G1′=τ2n⁢G1′=G1′,
so ker⁡VG→G1=〈σ⁢G′,ρ⁢G′〉, which implies κF1/K=〈[𝒫1h0],[ℋ⁢𝒫2h0]〉;
or
1. VG→G1⁢(ρ⁢G′)=ρ2⁢[ρ,σ]=ρ2⁢σ2=1,
2. VG→G1⁢(τ⁢G′)=τ2⁢[τ,σ]=τ2=1,
3. VG→G1⁢(σ⁢G′)=σ2≠1,
so ker⁡VG→G1=〈τ⁢G′,ρ⁢G′〉, which implies κF1/K=〈[ℋ],[𝒫1h0]〉.
If 2⁢a⁢bp=-1, from Table 2, we have
NF1/K⁢(𝐂F1,2)={〈[𝒫2h0],[ℋ]〉if⁢(2p)=-1,〈[𝒫2h0],[ℋ⁢𝒫1h0]〉if⁢(2p)=1.
Then G1=〈σ⁢ρ,τ〉 or G1=〈σ,ρ⁢τ〉, and G1′=〈τ2〉 or G1′=〈σ2〉, respectively, and hence G/G1={1¯,σ¯=ρ¯} or G/G1={1¯,τ¯=ρ¯}. Thus
1. VG→G1⁢(τ⁢G′)=τ2⁢[τ,σ]⁢G1′=τ2⁢G1′=G1′,
2. VG→G1⁢(σ⁢G′)=σ2⁢G1′=τ2n⁢G1′=G1′,
3. VG→G1⁢(ρ⁢G′)=ρ2⁢G1′=τ2n⁢G1′=G1′;
or
1. VG→G1⁢(σ⁢G′)=σ2⁢[σ,τ]⁢G1′=σ2⁢G1′=G1′,
2. VG→G1⁢(τ⁢G′)=τ2⁢G1′=G1′ since τ2=1,
3. VG→G1⁢(ρ⁢G′)=ρ2⁢G1′=G1′ since ρ2=σ2,
so ker⁡VG→G1=〈τ⁢G′,ρ⁢G′,σ⁢G′〉, which implies κF1/K=〈[ℋ],[𝒫1h0],[𝒫2h0]〉=𝐂K,2.

We similarly check the remaining tables entries.

7.2 The abelian-type invariants of 𝐂Li,2 and capitulation kernels κLi/K

Finally, we easily compute the Galois groups 𝒢i=Gal⁡(K2(2)/Li), their derived groups, the abelian-type invariants of 𝐂Li,2, the transfer kernels ker⁡VG→𝒢i and κLi/K, the capitulation kernels of Li/K. The results are summarized in Tables 11, 12, 13 and 14.

Let us take just one example:

Example.

Consider L1=F1⁢F2⁢F3 in the case where 2⁢a⁢bp=1. Then Gal⁡(K2(2)/L1)=𝒢1=G1∩G2∩G3, and hence

𝒢1={〈ρ,τ〉∩〈ρ,σ〉∩〈ρ,τ⁢σ〉=〈ρ〉if⁢(2p)=1.〈σ,τ⁢ρ,τ2〉∩〈ρ,τ〉∩〈τ⁢ρ,σ⁢ρ,τ2〉=〈τ⁢ρ,τ2〉if⁢(2p)=-1.

Thus 𝒢1′=〈1〉 or 𝒢1′=〈τ4〉 according as 2p=1 or -1, which implies that 𝐂L1,2≃ℤ/4⁢ℤ or 𝐂L1,2≃(2,2), respectively.

On the other hand, κL1/K=𝐂K,2 since κF3/K=𝐂K,2 and κF3/K⊂κL1/K.

Table 11

𝒢i=Gal⁡(K2(2)/Li) and 𝒢i′ for the case 2⁢a⁢bp=1.

	𝒢i		𝒢i′
Li	2p=-1	2p=1	2p=-1	2p=1
L1=K⁢(-ϵ⁢π1,-2⁢ϵ⁢π2)	〈τ⁢ρ,τ2〉	〈ρ〉	〈τ4〉	〈1〉
L2=K⁢(-p,-2)	〈τ⁢σ〉	〈τ⁢σ〉	〈1〉	〈1〉
L3=K⁢(2⁢ϵ⁢π1,ϵ⁢π2)	〈σ⁢ρ,τ2〉	〈τ⁢ρ⁢σ〉	〈τ4〉	〈1〉
L4=K⁢(-2⁢ϵ⁢π2,2⁢ϵ⁢π1)	〈τ〉	〈σ〉	〈1〉	〈1〉
L5=K⁢(-ϵ⁢π1,ϵ⁢π2)	〈σ,τ2〉	〈τ,σ2〉	〈1〉	〈1〉
L6=K⁢(-ϵ⁢π1,2⁢ϵ⁢π1)	〈τ⁢σ⁢ρ,τ2〉	〈τ⁢ρ〉	〈τ4〉	〈1〉
L7=K⁢(-2⁢ϵ⁢π2,ϵ⁢π2)	〈ρ,τ2〉	〈σ⁢ρ〉	〈τ4〉	〈1〉

Table 12

𝐂Li,2, ker⁡VG→𝒢i and κLi/K for the case 2⁢a⁢bp=1.

	𝐂Li,2
Li	2p=-1	2p=1	ker⁡VG→𝒢i	κLi/K
L1=K⁢(-ϵ⁢π1,-2⁢ϵ⁢π2)	(2,2)	ℤ/4⁢ℤ	〈σ⁢G′,τ⁢G′,ρ⁢G′〉	𝐂K,2
L2=K⁢(-p,-2)	ℤ/2n+1⁢ℤ	ℤ/4⁢ℤ	〈σ⁢G′,τ⁢G′,ρ⁢G′〉	𝐂K,2
L3=K⁢(2⁢ϵ⁢π1,ϵ⁢π2)	(2,2)	ℤ/4⁢ℤ	〈σ⁢G′,τ⁢G′,ρ⁢G′〉	𝐂K,2
L4=K⁢(-2⁢ϵ⁢π2,2⁢ϵ⁢π1)	ℤ/2n+1⁢ℤ	ℤ/4⁢ℤ	〈σ⁢G′,τ⁢G′,ρ⁢G′〉	𝐂K,2
L5=K⁢(-ϵ⁢π1,ϵ⁢π2)	(2,2n)	(2,2)	〈σ⁢G′,τ⁢G′,ρ⁢G′〉	𝐂K,2
L6=K⁢(-ϵ⁢π1,2⁢ϵ⁢π1)	(2,2)	ℤ/4⁢ℤ	〈σ⁢G′,τ⁢G′,ρ⁢G′〉	𝐂K,2
L7=K⁢(-2⁢ϵ⁢π2,ϵ⁢π2)	(2,2)	ℤ/4⁢ℤ	〈σ⁢G′,τ⁢G′,ρ⁢G′〉	𝐂K,2

Table 13

𝒢i=Gal⁡(K2(2)/Li) and 𝒢i′ for the case 2⁢a⁢bp=-1.

	𝒢i		𝒢i′
Li	2p=-1	2p=1	2p=-1	2p=1
L1=K⁢(-2⁢ϵ⁢π1,-ϵ⁢π2)	〈σ⁢ρ,τ2〉	〈τ⁢σ⁢ρ〉	〈τ4〉	〈1〉
L2=K⁢(-p,-2)	〈τ⁢σ〉	〈τ⁢σ〉	〈1〉	〈1〉
L3=K⁢(ϵ⁢π1,2⁢ϵ⁢π2)	〈τ⁢ρ,τ2〉	〈ρ〉	〈τ4〉	〈1〉
L4=K⁢(-ϵ⁢π2,ϵ⁢π1)	〈σ,τ2〉	〈τ,σ2〉	〈1〉	〈1〉
L5=K⁢(-2⁢ϵ⁢π1,2⁢ϵ⁢π2)	〈τ〉	〈σ〉	〈1〉	〈1〉
L6=K⁢(-2⁢ϵ⁢π1,ϵ⁢π1)	〈τ⁢σ⁢ρ,τ2〉	〈τ⁢ρ〉	〈τ4〉	〈1〉
L7=K⁢(-ϵ⁢π2,2⁢ϵ⁢π2)	〈ρ,τ2〉	〈σ⁢ρ〉	〈τ4〉	〈1〉

Table 14

𝐂Li,2, ker⁡VG→𝒢i and κLi/K for the case 2⁢a⁢bp=-1.

	𝐂Li,2
Li	2p=-1	2p=1	ker⁡VG→𝒢i	κLi/K
L1=K⁢(-2⁢ϵ⁢π1,-ϵ⁢π2)	(2,2)	ℤ/4⁢ℤ	〈σ⁢G′,τ⁢G′,ρ⁢G′〉	𝐂K,2
L2=K⁢(-p,-2)	ℤ/2n+1⁢ℤ	ℤ/4⁢ℤ	〈σ⁢G′,τ⁢G′,ρ⁢G′〉	𝐂K,2
L3=K⁢(ϵ⁢π1,2⁢ϵ⁢π2)	(2,2)	ℤ/4⁢ℤ	〈σ⁢G′,τ⁢G′,ρ⁢G′〉	𝐂K,2
L4=K⁢(-ϵ⁢π2,ϵ⁢π1)	(2,2n)	(2,2)	〈σ⁢G′,τ⁢G′,ρ⁢G′〉	𝐂K,2
L5=K⁢(-2⁢ϵ⁢π1,2⁢ϵ⁢π2)	ℤ/2n+1⁢ℤ	ℤ/4⁢ℤ	〈σ⁢G′,τ⁢G′,ρ⁢G′〉	𝐂K,2
L6=K⁢(-2⁢ϵ⁢π1,ϵ⁢π1)	(2,2)	ℤ/4⁢ℤ	〈σ⁢G′,τ⁢G′,ρ⁢G′〉	𝐂K,2
L7=K⁢(-ϵ⁢π2,2⁢ϵ⁢π2)	(2,2)	ℤ/4⁢ℤ	〈σ⁢G′,τ⁢G′,ρ⁢G′〉	𝐂K,2

Numerical examples

The examples in Tables 15 and 16 are computed by using PARI/GP.

Table 15

Invariants of Fi.

l.p	2⁢a⁢bp	2p	𝐂F1,2	𝐂F2,2	𝐂F3,2	𝐂F4,2	𝐂F5,2	𝐂F6,2	𝐂F7,2
5.19	-1	-1	(2,2)	(2,2,2)	(2,2)	(2,2)	(2,8)	(2,2,2)	(2,2)
5.19	1	-1	(2,2,2)	(2,2)	(2,2)	(2,2)	(2,8)	(2,2)	(2,2,2)
5.31	-1	1	(2,2)	(2,4)	(2,2)	(2,2)	(2,4)	(2,4)	(2,2)
5.31	1	1	(2,4)	(2,2)	(2,2)	(2,2)	(2,4)	(2,2)	(2,4)
101.43	1	-1	(2,2,2)	(2,2)	(2,2)	(2,2)	(2,64)	(2,2)	(2,2,2)
5.59	1	-1	(2,2,2)	(2,2)	(2,2)	(2,2)	(2,8)	(2,2)	(2,2,2)
5.59	-1	-1	(2,2)	(2,2,2)	(2,2)	(2,2)	(2,8)	(2,2,2)	(2,2)
37.67	-1	-1	(2,2)	(2,2,2)	(2,2)	(2,2)	(2,8)	(2,2,2)	(2,2)

Table 16

Invariants of Li.

l.p	2⁢a⁢bp	2p	𝐂L1,2	𝐂L2,2	𝐂L3,2	𝐂L4,2	𝐂L5,2	𝐂L6,2	𝐂L7,2
5.19	-1	-1	(2,2)	(8)	(2,2)	(2,4)	(8)	(2,2)	(2,2)
5.19	1	-1	(2,2)	(8)	(2,2)	(8)	(2,4)	(2,2)	(2,2)
5.31	-1	1	(4)	(4)	(4)	(2,2)	(4)	(4)	(4)
5.31	1	1	(4)	(4)	(4)	(4)	(2,2)	(4)	(4)
101.43	1	-1	(2,2)	(64)	(2,2)	(64)	(2,32)	(2,2)	(2,2)
5.59	1	-1	(2,2)	(8)	(2,2)	(8)	(2,4)	(2,2)	(2,2)
5.59	-1	-1	(2,2)	(8)	(2,2)	(2,4)	(8)	(2,2)	(2,2)
37.67	-1	-1	(2,2)	(8)	(2,2)	(2,4)	(8)	(2,2)	(2,2)

Communicated by Hugh Williams

Funding statement: This work is partially supported by CNRST (PBER), Hassan II Academy of Sciences and Technology (Morocco), URAC6 and ACSA laboratory (FSO-UMPO).

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Received: 2017-07-16

Accepted: 2018-09-15

Published Online: 2018-10-12

Published in Print: 2019-03-01

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https://doi.org/10.1515/jmc-2017-0037

Keywords for this article

capitulation; Hilbert class field

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Capitulation of the 2-ideal classes of type (2, 2, 2) of some quartic cyclic number fields

Article

Abstract

1 Introduction

Notations

2 Main results

The unramified extensions of K

Theorem 2.1.

Structure of G=Gal⁡(K2(2)/K)

Theorem 2.2.

Abelian-type invariants and capitulation kernels

Theorem 2.3.

3 Preliminary results

Lemma 3.1.

Proof.

Lemma 3.2.

Proof.

Lemma 3.3.

Proof.

Remark 3.4.

Proposition 3.5.

Proof.

Lemma 3.6.

Proof.

Theorem 3.7.

Proof.

4 The 2-class number of the genus field K(*)

Proposition 4.1.

Proof.

Proposition 4.2.

Proof.

Theorem 4.3.

Proof.

5 Proof of Theorem 2.1

5.1 Generators of the 2-class group of K

Remark 5.1.

5.2 The unramified extensions of K within K2(1)

6 Proof of Theorem 2.2

6.1 The 2-class group of F5

6.2 The Hilbert 2-class field tower of K

6.3 The generators of G=Gal⁡(K2(2)/K)

7 Proof of Theorem 2.3

Example.

Example.

Example.

Example.

7.1 The abelian-type invariants of 𝐂Fi,2 and capitulation kernels κFi/K

Remark 7.1.

Proposition 7.2.

Example.

Example.

7.2 The abelian-type invariants of 𝐂Li,2 and capitulation kernels κLi/K

Example.

Numerical examples

References

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Articles in the same Issue

Articles in the same Issue

Capitulation of the 2-ideal classes of type (2, 2, 2) of some quartic cyclic number fields