Abdelmalek Azizi , Idriss Jerrari , Abdelkader Zekhnini and Mohammed Talbi

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

Published online: October 12, 2018

# Abstract

Let p 3 ( mod 4 ) and l 5 ( mod 8 ) be different primes such that p l = 1 and 2 p = p l 4 . Put k = ( l ) , and denote by ϵ its fundamental unit. Set K = k ( - 2 p ϵ l ) , and let K 2 ( 1 ) be its Hilbert 2-class field, and let K 2 ( 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 ( K 2 ( 2 ) / K ) , and thus to study the capitulation of the 2-ideal classes of K in all its unramified abelian extensions within K 2 ( 1 ) . Additionally, these extensions are constructed, and their abelian-type invariants are given.

## 1 Introduction

Let K be an algebraic number field. For a prime number p, let K p ( 0 ) = K and K p ( i ) denote the Hilbert p-class field of K p ( i - 1 ) for i 1 . Then the sequence ( K p ( i ) ) i is called the p-class field tower of K. Put K ~ p = i = 0 K p ( 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 ( mod 4 ) and l 5 ( mod 8 ) be different primes. Put k = ( l ) , and denote by ϵ its fundamental unit. Let K = k ( - 2 p ϵ l ) and K 2 ( 1 ) be its Hilbert 2-class field; denote by K 2 ( 2 ) the Hilbert 2-class field of K 2 ( 1 ) and by K ( * ) the genus field of K that is the maximal absolute abelian subfield of K 2 ( 1 ) / K . If p l = - 1 , then 𝐂 K , 2 , the 2-class group of K, is isomorphic to / 2 × / 2 (see [9]) and K ( * ) = K 2 ( 1 ) ; this case is studied in [3]. But if p l = 1 , then K ( * ) K 2 ( 1 ) and the rank of 𝐂 K , 2 is greater than 3, and the situation becomes more interesting. Assuming that 𝐂 K , 2 is isomorphic to / 2 × / 2 × / 2 , we aim, in this paper, the following four objectives.

• Construct the fourteen unramified subextensions of K 2 ( 1 ) / K .

• Prove that the class field tower of K stops at the second stage, and thus G = Gal ( K ~ 2 / K ) = Gal ( K 2 ( 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 K 2 ( 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 ( mod 8 ) , only one maximal normal abelian subgroup of index 2, and its order is equal to 2 m 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 = ( 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 ( m ) (resp. k)

• h 2 ( m ) , h 2 ( k ) : the 2-class number of ( m ) (resp. k)

• 𝐂 k , 2 : the 2-part of the class group of k

• k + : the maximal real subfield of k if it is a CM-field

• E k : the unit group of k

• W k : the group of roots of unity contained in k

• w k : the order of W k

• Q k = [ E k : W k E k + ] : Hasse’s unit index if k is a CM-field

• N L / k : the relative norm for an extension L / k

• κ L / k : the capitulation kernel of L / k

• 𝔭 k : a prime ideal of k

• x , y 𝔭 k , x 𝔭 k : the Hilbert symbol (resp. the quadratic residue symbol) for the prime 𝔭 k applied to ( x , y ) (resp. x)

• ξ m : a primitive m-th root of unity

• ( ξ m ) : 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 ( mod 4 ) , l 5 ( mod 8 ) , p l = 1 and 2 p = p l 4 .

Let k = ( l ) , and denote by ϵ its fundamental unit and by h 0 its class number. Then K = k ( - 2 p ϵ l ) is a complex cyclic quartic field [9].

Let K 2 ( 1 ) be the Hilbert 2-class field of K, K 2 ( 2 ) the Hilbert 2-class field of K 2 ( 1 ) , and let K ( * ) be the absolute genus field of K. Since p l = 1 , so K ( * ) K 2 ( 1 ) (as mentioned in the introduction), and there exist two prime ideals 1 , 2 of k such that 1 2 = ( p ) . Thus 1 h 0 and 2 h 0 are principal in k, say 1 h 0 = ( π 1 ) and 2 h 0 = ( π 2 ) , where

π 1 = a + b l , π 2 = a - b l and p h 0 = a 2 - l b 2 .

One can easily show that four prime ideals of k ramify in K, which are the prime ideal of k above 2, ( l ) , 1 and 2 . Let 𝒫 1 , 𝒫 2 be the prime ideals in K above 1 , 2 , respectively, let 𝒱 be the prime ideal in K above l , and let be the prime ideal in K above 2. Therefore, 𝒫 i 2 h 0 = i h 0 𝒪 K = ( π i ) , 𝒱 2 = ( l ) and 2 = ( 2 ) . So, according to [9], the rank of the 2-class group of K is 4 - 1 = 3 , and the 2-class number h 2 of K is strictly greater than 4. Furthermore, by [9, Theorem 4, p. 68], h 2 = 8 , i.e., 𝐂 K , 2 is isomorphic to / 2 × / 2 × / 2 if and only if 2 p = p l 4 .

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

### The unramified extensions of K

As 𝐂 K , 2 ( 2 , 2 , 2 ) , by class field theory, K admits, within its first Hilbert 2-class field K 2 ( 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 P 1 , P 2 are the prime ideals in K above p, and H is the prime ideal in K above 2. Moreover, P i 2 h 0 = ( π i ) and H 2 = ( 2 ) .

1. (1)

The 2 -class group of K is given by 𝐂 K , 2 = [ 𝒫 1 h 0 ] , [ 𝒫 2 h 0 ] , [ ] ( 2 , 2 , 2 ) .

2. (2)

The seven unramified quadratic extensions of K are as follows:

1. (a)

If 2 a b p = 1 , then

F 1 = K ( - ϵ π 1 ) , F 2 = K ( - 2 ϵ π 2 ) , F 3 = K ( 2 p ) , F 4 = K ( - 2 ) , F 5 = K ( - p ) , F 6 = K ( 2 ϵ π 1 ) 𝑎𝑛𝑑 F 7 = K ( ϵ π 2 ) .

2. (b)

If 2 a b p = - 1 , then

F 1 = K ( - 2 ϵ π 1 ) , F 2 = K ( - ϵ π 2 ) , F 3 = K ( 2 p ) , F 4 = K ( - 2 ) , F 5 = K ( - p ) , F 6 = K ( ϵ π 1 ) 𝑎𝑛𝑑 F 7 = K ( 2 ϵ π 2 ) .

The fields F 3 , F 4 and F 5 are intermediate between K and K ( * ) . The fields F 1 F 7 and F 2 F 6 are pairwise conjugate and isomorphic. Consequently, F 3 , F 4 and F 5 are absolutely abelian, whereas F 1 , F 2 , F 6 and F 7 are not.

3. (3)

The seven unramified biquadratic extensions of K are

L 2 = F 3 F 4 F 5 = K ( * ) = K ( - p , - 2 ) , the absolute genus field of K ,
L 1 = F 1 F 2 F 3 , L 3 = F 3 F 6 F 7 , L 4 = F 2 F 5 F 6 , L 5 = F 1 F 5 F 7 , L 6 = F 1 F 4 F 6 , L 7 = F 2 F 4 F 7 .

Moreover, L 1 L 3 and L 6 L 7 are pairwise conjugate and thus isomorphic and absolutely non-normal, whereas L 4 and L 5 are absolutely Galois over Q .

### Structure of G = Gal ⁡ ( K 2 ( 2 ) / K )

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

### Theorem 2.2.

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

1. (1)

𝐂 F 5 , 2 ( 2 , 2 n + 1 ) , where n 2 or n = 1 , according as p 3 ( mod 8 ) or p 7 ( mod 8 ) .

2. (2)

The length of the 2 -class field tower of K is 2.

3. (3)

The second 2 -class group G = Gal ( K 2 ( 2 ) / K ) of K is given by

G = { ρ , τ , σ : ρ 4 = σ 4 = τ 2 n + 1 = 1 , σ 2 = ρ 2 = τ 2 n , n 2 , [ τ , σ ] = 1 , [ σ , ρ ] = σ 2 , [ ρ , τ ] = τ 2 𝑖𝑓 p 3 ( mod 8 ) , ρ , τ , σ : ρ 4 = σ 4 = τ 2 = 1 , σ 2 = ρ 2 , [ τ , σ ] = [ τ , ρ ] = 1 , [ σ , ρ ] = σ 2 𝑖𝑓 p 7 ( mod 8 ) .

4. (4)

The derived subgroup of G is G = τ 2 𝐂 K 2 ( 1 ) , 2 or G = σ 2 𝐂 K 2 ( 1 ) , 2 , according as p 3 ( mod 8 ) or p 7 ( mod 8 ) . It is cyclic of order 2 n or 2 , respectively.

5. (5)

The coclass of G is equal to 2 , and its nilpotency class is n + 1 .

6. (6)

The center Z ( G ) of G is of type ( 2 , 2 ) .

### Abelian-type invariants and capitulation kernels

Let κ F / K denote the capitulation kernel of the natural class extension homomorphism J F / K : 𝐂 K , 2 𝐂 F , 2 , where F is an unramified extension of K within K 2 ( 1 ) .

### Theorem 2.3.

Keep the preceding assumptions and notations, then:

1. 1

The abelian-type invariants of the 2 -class groups 𝐂 F i , 2 are given by:

1. (i)

𝐂 F 3 , 2 and 𝐂 F 4 , 2 are of type ( 2 , 2 ) , and 𝐂 F 5 , 2 is of type ( 2 , 2 n + 1 ) with n 1 .

2. (ii)

If 2 a b p = 1 , then:

1. (a)

𝐂 F 1 , 2 𝐂 F 7 , 2 ( 2 , 4 ) and 𝐂 F 2 , 2 𝐂 F 6 , 2 ( 2 , 2 ) if 2 p = 1 .

2. (b)

𝐂 F 1 , 2 𝐂 F 7 , 2 ( 2 , 2 , 2 ) and 𝐂 F 2 , 2 𝐂 F 6 , 2 ( 2 , 2 ) if 2 p = - 1 .

3. (iii)

If 2 a b p = - 1 , then:

1. (a)

𝐂 F 1 , 2 𝐂 F 7 , 2 ( 2 , 2 ) and 𝐂 F 2 , 2 𝐂 F 6 , 2 ( 2 , 4 ) if 2 p = 1 .

2. (b)

𝐂 F 1 , 2 𝐂 F 7 , 2 ( 2 , 2 ) and 𝐂 F 2 , 2 𝐂 F 6 , 2 ( 2 , 2 , 2 ) if 2 p = - 1 .

2. 2

The abelian-type invariants of the 2 -class groups 𝐂 L i , 2 are given by:

1. (i)

If 2 a b p = 1 , then:

1. (a)

𝐂 L 1 , 2 , 𝐂 L 2 , 2 , 𝐂 L 3 , 2 , 𝐂 L 4 , 2 , 𝐂 L 6 , 2 and 𝐂 L 7 , 2 are cyclic of order 4 , and 𝐂 L 5 , 2 is of type ( 2 , 2 ) if 2 p = 1 .

2. (b)

𝐂 L 1 , 2 𝐂 L 3 , 2 𝐂 L 6 , 2 𝐂 L 7 , 2 ( 2 , 2 ) , 𝐂 L 5 , 2 ( 2 , 2 n ) , and 𝐂 L 2 , 2 and 𝐂 L 4 , 2 are cyclic of order 2 n + 1 if 2 p = - 1 .

2. (ii)

If 2 a b p = - 1 , then:

1. (a)

𝐂 L 1 , 2 , 𝐂 L 2 , 2 , 𝐂 L 3 , 2 , 𝐂 L 5 , 2 , 𝐂 L 6 , 2 and 𝐂 L 7 , 2 are cyclic of order 4 , and 𝐂 L 4 , 2 is of type ( 2 , 2 ) if 2 p = 1 .

2. (b)

𝐂 L 1 , 2 𝐂 L 3 , 2 𝐂 L 6 , 2 𝐂 L 7 , 2 ( 2 , 2 ) , 𝐂 L 4 , 2 ( 2 , 2 n ) , and 𝐂 L 2 , 2 and 𝐂 L 5 , 2 are cyclic of order 2 n + 1 if 2 p = - 1 .

3. 3

The capitulation is given by:

1. (i)

For all i, # κ F i / K = 4 or 8 (for more details, see Tables 9 and 10), and all the extensions F i satisfy Taussky’s condition ( A ) , i.e., # ( κ F i / K N i ) > 1 ; see [16].

2. (ii)

For all i, κ L i / K = 𝐂 K , 2 (total 2 -capitulation), and each L i 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:

1. (1)

The equations - ϵ π 1 X 2 ( mod 4 ) and ϵ π 2 X 2 ( mod 4 ) have solutions in k if and only if 2 a b p = 1 .

2. (2)

The equations ϵ π 1 X 2 ( mod 4 ) and - ϵ π 2 X 2 ( mod 4 ) have solutions in k if and only if 2 a b p = - 1 .

## Proof.

First, we put ϵ = x + y l , where x and y are two integers. Then x 2 ( mod 4 ) and y 1 ( mod 4 ) since l 5 ( mod 8 ) .

As p h 0 = a 2 - l b 2 , then:

• If p 3 ( mod 8 ) , we have a 0 ( mod 4 ) and b ± 1 ( mod 4 ) .

• If p 7 ( mod 8 ) , we have a 2 ( mod 4 ) and b ± 1 ( mod 4 ) .

Thus we have four cases to discuss:

1. (1)

If a 2 ( mod 4 ) and b 1 ( mod 4 ) , then we have ( mod  4 )

ϵ π 1 = ( x + y l ) ( a + b l ) ( 2 + l ) ( 2 + l ) 4 + 4 l + l 1 ,
- ϵ π 2 = - ( x + y l ) ( a - b l ) - ( 2 + l ) ( 2 - l ) - 4 + l 1 .

2. (2)

If a 2 ( mod 4 ) and b - 1 ( mod 4 ) , then we have ( mod  4 )

- ϵ π 1 = - ( x + y l ) ( a + b l ) - ( 2 + l ) ( 2 - l ) - 4 + l 1 ,
ϵ π 2 = ( x + y l ) ( a - b l ) ( 2 + l ) ( 2 + l ) 4 + 4 l + l 1 .

3. (3)

If a 0 ( mod 4 ) and b 1 ( mod 4 ) , then

- ϵ π 1 = - ϵ ( a + b l ) - ϵ l ( mod 4 ) ,
ϵ π 2 = ϵ ( a - b l ) = - ϵ l ( mod 4 ) ,

which is a square in k (see [9, Lemma A]).

4. (4)

If a 0 ( mod 4 ) and b - 1 ( mod 4 ) , then

ϵ π 1 = ϵ ( a + b l ) = - ϵ l ( mod 4 ) ,
- ϵ π 2 = - ϵ ( a - b l ) = - ϵ l ( mod 4 ) ,

which is a square in k (see [9, Lemma A]).

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

1. (1)

If - ϵ π 1 X 2 ( mod 4 ) and ϵ π 2 X 2 ( mod 4 ) have solutions in k, then ( a 2 and b - 1 ( mod 4 ) ) or ( a 0 and b 1 ( mod 4 ) ) .

2. (2)

If ϵ π 1 X 2 ( mod 4 ) and - ϵ π 2 X 2 ( mod 4 ) have solutions in k, then ( a 2 and b 1 ( mod 4 ) ) or ( a 0 and b - 1 ( mod 4 ) ) .

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

a b p = ( - 1 ) b - 1 2 ( - 1 ) j 2 p j p l 4 ,

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

a b p = ( - 1 ) b - 1 2 ( - 1 ) j 2 p j + 1 ,

and hence

2 a b p = ( - 1 ) b - 1 2 ( - 1 ) j 2 p j .

Finally, we have:

1. (1)

- ϵ π 1 X 2 ( mod 4 ) and ϵ π 2 X 2 ( mod 4 ) have solutions in k if and only if ( a 2 and b - 1 ( mod 4 ) ) or ( a 0 and b 1 ( mod 4 ) ) , i.e., 2 a b p = 1 .

2. (2)

ϵ π 1 X 2 ( mod 4 ) and - ϵ π 2 X 2 ( mod 4 ) have solutions in k if and only if ( a 2 and b 1 ( mod 4 ) ) or ( a 0 and b - 1 ( mod 4 ) ) , i.e., 2 a b p = - 1 . ∎

## Lemma 3.2.

Let p 3 ( mod 4 ) and l 5 ( mod 8 ) 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 h 2 ( - p l ) , the 2-class number of k 1 = ( - p l ) , is 2 n + 1 , where n 1 (see [13]). Put k 2 = ( - p ) and R = k k 1 k 2 . Then we get the following lemma:

## Lemma 3.3.

The 2-class group C R , 2 of R is cyclic of order 2 n . It is generated by the class of an ideal I R such that I 2 n - 1 p i h 0 V i v , where p i (resp. V i ) is a prime ideal of R above p (resp. l) and v is the odd part of the class number of k 2 .

## Proof.

First, prove that P, the prime ideal of k 1 above p, is not principal. If not, there exists α k 1 such that P = ( α ) . This yields P 2 = ( p ) = ( α 2 ) . Hence, by putting α = x + y - p l with x , y , we get

x 2 - p l y 2 + 2 x y - p l = ± p .

This is equivalent to

{ x = 0 , - l y 2 = ± 1    or    { y = 0 , x 2 = ± p ,

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

Let 𝔭 i and V i be the prime ideals of R above p and l, respectively. Then 𝔭 i h 0 and V i v are not principal in R since their norms are not principal in k 1 . Therefore, ( 𝔭 i h 0 ) 2 ( V i v ) 2 1 .

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

Finally, the ideal I generates a subgroup of 𝐂 R , 2 of order 2 n since I 2 n P (in 𝐂 R , 2 ) and P capitulates in R. Then 𝐂 R , 2 = [ ] and 2 n - 1 𝔭 i h 0 V i v . ∎

## Remark 3.4.

Note the following:

1. (1)

If p 3 ( mod 8 ) , then, by genus theory, 2 k 1 , a prime ideal of k 1 above 2, is not a square in k 1 since

2 , - p l p = 2 p = - 1 and 2 , - p l l = 2 l = - 1 .

Hence I 2 k 1 u , where u is the odd part of the class number of k 1 . Thus 2 R u , where 2 R is a prime ideal of R above 2. Moreover, n 2 (see [13, Théorème 8]).

2. (2)

If p 7 ( mod 8 ) , then 2 R v 1 since 2 k 2 v 1 and 2 k 2 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 = σ L and of quadratic subfield k, and let ϵ be the fundamental unit of k. Let f L (resp. m k ) be the conductor of L (resp. k). Then f L = m k g , and the discriminant of L is equal to m k f L 2 . It is known that L ( ξ f L ) .

Denote by χ L a rational character of ( ξ f L ) , and put E χ L = { ω E L : ω 1 + σ L 2 = ± 1 } . Let | E L | (resp. | E k | , | E χ L | ) be the group of the absolute values of E L (resp. E k , E χ L ), | E L | = | E k | | E χ L | , Q L = [ | E L | : | E L | ] , and let ϵ χ L be a generator of E χ L . From [12], we have Q L = 1 or 2, which gives two possible structures for the unit group of L:

1. (i)

If Q L = 1 , we have E L = - 1 , ϵ , ϵ χ L , ϵ χ L σ L ;

2. (ii)

If Q L = 2 , we have E L = - 1 , ξ L , ξ L σ L , ξ L σ L 2 with ξ L 2 = ± ϵ ϵ χ L 1 - σ L , ξ L 1 + σ L = ± ϵ χ L and ξ L 1 + σ L 2 = ± ϵ .

Note that ξ L is called a Minkowski unit.

## Proposition 3.5.

Let l 5 ( mod 8 ) be a prime number. Put L = k ( 2 ϵ l ) ; then Q L = 1 .

## Proof.

It is known that Q L = 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 v 𝔭 ( 2 ϵ l ) = 0 , so

ϵ , 2 ϵ l 𝔭 = ϵ 𝔭 v 𝔭 ( 2 ϵ l ) = 1 .

• If 𝔭 lies above l, then v 𝔭 ( 2 ϵ l ) = 1 . So, by [2, Proof of Proposition 4.1], we get

ϵ , 2 ϵ l 𝔭 = ϵ 𝔭 v 𝔭 ( 2 ϵ l ) = ϵ 𝔭 = 2 l = - 1 .

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

## Lemma 3.6.

Let p 3 ( mod 4 ) and l 5 ( mod 8 ) be different primes. Then the Hasse unit index of

F 5 = k ( 2 ϵ l , - p )

is Q F 5 = 1 .

## Proof.

Let λ be a unit of F 5 + . Prove that the index Q F 5 equals 1, which is equivalent, by [1, Proposition 3], to proving that p λ is not a square in F 5 + . Suppose that p λ is a square in F 5 + . This is equivalent to ( p ) F 5 + = H 2 , where H is an ideal of F 5 + , which contradicts the fact that p cannot ramify in F 5 + . ∎

## Theorem 3.7.

Keep the assumptions and notations of Lemma 3.6. Then

E F 5 = { - 1 , ϵ , ϵ χ F 5 + , ϵ χ F 5 + σ F 5 + 𝑖𝑓 p 3 , ξ 6 , ϵ , ϵ χ F 5 + , ϵ χ F 5 + σ F 5 + 𝑖𝑓 p = 3 .

## Proof.

Since the Hasse unit index of F 5 is Q F 5 = 1 , then E F 5 = W F 5 E F 5 + . 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, Q F = 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 𝐂 F + , 2 of F + is given by rank ( 𝐂 F + , 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 2 e = [ E k : E k N F + / k ( F + × ) ] . We easily check that the primes ( 2 ) k and l ramify in F + . Hence t = 2 , and therefore, rank ( 𝐂 F + , 2 ) = 1 - e . We know, from the proof of Proposition 3.5, that ϵ is not a norm in F + , whereas -1 is. Then E k / E k N F + / k ( F + × ) = { 1 ¯ , ϵ ¯ } . Therefore, e = 1 , so 𝐂 F + , 2 is trivial.

Let us prove that Q F = 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 F 5 + . In other words, it suffices to prove that ( 2 ) F + = H 2 , where H is a principal ideal in F + . As 2 ramifies in F + , there exists H in F + such that ( 2 ) F + = H 2 . If H is not principal, then the class of H is of order 2, which is absurd since 𝐂 F + , 2 is trivial.

Next, we will compute the 2-class number h 2 ( F ) of F. For this, notice that F / k is a V 4 -extension of CM-type fields (Figure 1), of quadratic subextensions F + , K = k ( - 2 ) and L = k ( - ϵ l ) . Then, from [14], we have

h 2 ( F ) = Q F Q K Q L w F w L w K h 2 ( F + ) h 2 ( K ) h 2 ( L ) h 2 ( l ) 2 .

Note that w K = w F = w L = 2 . Furthermore, from [17, Theorem 10.4], we have h 2 ( L ) = 1 since f L = l and Q K = Q L = 1 . From this, we deduce that h 2 ( F ) = h 2 ( K ) . As K is an unramified extension of ( - 2 l ) and the 2-class group of ( - 2 l ) is cyclic of order 2, then h 2 ( K ) = h 2 ( - 2 l ) 2 = 1 . Thus h 2 ( 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,

E L = - 1 , ϵ , ϵ χ L , ϵ χ L σ L .

Then, from [11] again, the rank of 2-class group 𝐂 F , 2 of F is given by rank ( 𝐂 F , 2 ) = t - 1 - e , where t is the number of primes (finite and infinite) of L that ramify in F and 2 e = [ E L : E L N F / L ( F × ) ] . We easily check that t = 2 , and therefore, rank ( 𝐂 F , 2 ) = 1 - e . In the following, we show that e = 1 , which is equivalent to showing that E L N F / L ( F × ) = - 1 , ϵ , ϵ χ L ϵ χ L σ L .

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:

• If 𝔭 is not above p and 2, then v 𝔭 ( 2 p ) = 0 , so

α , 2 p 𝔭 = α 𝔭 v 𝔭 ( 2 p ) = 1 .

• If 𝔭 lies above p, then v 𝔭 ( 2 p ) = 1 , and thus

α , 2 p 𝔭 = α 𝔭 v 𝔭 ( 2 p ) = α 𝔭 = N L / k ( α ) 𝔭 k = α 2 𝔭 k = 1 .

• If 𝔭 lies above 2, then

α , 2 p 𝔭 = N L / k ( α ) , 2 p 𝔭 k = α 2 , 2 p 𝔭 = 1 .

• ϵ χ L is not a norm in F . In fact, we have:

• If 𝔭 is not above p and 2, then v 𝔭 ( 2 p ) = 0 , and thus

ϵ χ L , 2 p 𝔭 = ϵ χ L 𝔭 v 𝔭 ( 2 p ) = 1 .

• If 𝔭 lies above p, then v 𝔭 ( 2 p ) = 1 , and thus

ϵ χ L , 2 p 𝔭 = ϵ χ L 𝔭 v 𝔭 ( 2 p ) = ϵ χ L 𝔭 = N L / k ( ϵ χ L ) 𝔭 k = - 1 𝔭 k = - 1 p = - 1 .

• Similarly, we prove that ϵ χ L σ L is not a norm in F .

Consequently, E L / E L N F / L ( F × ) = { 1 ¯ , ϵ χ L ¯ } . Therefore, e = 1 , and thus 𝐂 F , 2 is trivial. ∎

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

## Theorem 4.3.

Put F 5 = K ( - p ) . Then h 2 ( K ( * ) ) = h 2 ( F 5 ) 2 = 2 n + 1 , where n 1 .

## Proof.

First, we compute the 2-class number h 2 ( F 5 ) of F 5 . For this, consider Figure 2.

### Figure 2

Subfields of F 5 / k .

It is easy to see that F 5 / k is a V 4 -extension of CM-type fields of quadratic subextensions L, K and R = k ( - p ) . Then, from [14], we have

h 2 ( F 5 ) = Q F 5 Q K Q R w F 5 w K w R h 2 ( R ) h 2 ( K ) h 2 ( L ) h 2 ( l ) 2 .

Note that w R = w F 5 and w K = 2 . Moreover, we have Q F 5 = 1 (see Lemma 3.6), and we easily see that Q K = Q R = 1 . Then,

h 2 ( F 5 ) = 1 2 8 h 2 ( R ) = 4 h 2 ( R )

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

h 2 ( F 5 ) = 2 n + 2 .

### Figure 3

Subfields of K ( * ) / L .

Now, we prove that h 2 ( K ( * ) ) = h 2 ( F 5 ) 2 = 2 n + 1 with n 1 . For this, note that K ( * ) / L is a V 4 -extension of CM-type fields (Figure 3), of quadratic subextensions F 5 , F = L ( 2 p ) and F = L ( - 2 ) . Then, from [14] again, we have

h 2 ( K ( * ) ) = Q K ( * ) Q F 5 Q F w K ( * ) w F 5 w F h 2 ( F ) h 2 ( F ) h 2 ( F 5 ) h 2 ( L ) 2

Note that

W F 5 = W K ( * ) = { { + 1 , - 1 } if p 3 , ξ 6 if p = 3 , and W F = { + 1 , - 1 } .

Then w F 5 = w K ( * ) and w F = 2 . So Lemma 3.6 and Propositions 4.1 and 4.2 yield that Q F 5 = 1 , Q F = 2 and h 2 ( F ) = h 2 ( F ) = h 2 ( L ) = 1 , and therefore,

h 2 ( K ( * ) ) = Q K ( * ) 2 1 2 h 2 ( F 5 ) .

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

h 2 ( K ( * ) ) = h 2 ( F 5 ) 2 = 2 n + 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 h 0 the fundamental unit and the class number of k = ( l ) , respectively. Put K = k ( - 2 p ϵ l ) . Recall that 𝒫 1 , 𝒫 2 are the prime ideals in K above p and is the prime ideal in K above 2. Moreover, 𝒫 i 2 h 0 = ( π i ) and 2 = ( 2 ) .

### 5.1 Generators of the 2-class group of K

Let us show that 𝐂 K , 2 = [ 𝒫 1 h 0 ] , [ 𝒫 2 h 0 ] , [ ] . For this, it suffices to show that [ 𝒫 1 h 0 ] , [ 𝒫 2 h 0 ] and [ ] are of order 2 and pairwise different.

• [ 𝒫 1 h 0 𝒫 2 h 0 ] is of order 2. In fact, if 𝒫 1 h 0 𝒫 2 h 0 ( α ) for some α K , this yields that ( p h 0 ) = ( α 2 ) . Then there exists ϵ , a unit of K, such that p h 0 ϵ = α 2 , and thus

p h 0 ϵ = ( c + d - 2 p ϵ l ) 2 = c 2 - 2 p ϵ l d 2 + 2 c d - 2 p ϵ l

with c and d in k. As { ϵ } is an FSU of K and - 1 K , then p h 0 ϵ k , and therefore, c = 0 or d = 0 . If d = 0 , then p h 0 ϵ = c 2 , and thus ± p h 0 = c 2 or p h 0 ϵ = c 2 in k, which gives that ± p k in the first case and - 1 in the second case, but this is impossible. Similarly, if c = 0 , we find that ± l is a square in , which is absurd.

• [ 𝒫 i h 0 ] is of order 2. In fact, assume that 𝒫 i h 0 = ( α ) for α K . Then 𝒫 i 2 h 0 = ( α ) 2 , which is equivalent to ( π i ) = ( α 2 ) . Then there exists ϵ , a unit of K, such that

π i ϵ = α 2 = ( c + d - 2 p ϵ l ) 2 = c 2 - 2 p ϵ l d 2 + 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 π j for j i , we get p h 0 ϵ = π j c 2 . So, by applying the norm in k / , we find that ± p h 0 = N k / ( c ) 2 . This means that ± p is a square in , which is impossible. In a similar way, if c = 0 , we get ± l p is a square in , which is impossible too.

Similarly, we can show that [ ] , [ 𝒫 i h 0 ] and [ 𝒫 1 h 0 𝒫 2 h 0 ] are of order 2. Thus [ 𝒫 1 h 0 ] , [ 𝒫 2 h 0 ] , [ ] is of type ( 2 , 2 , 2 ) , and then 𝐂 K , 2 = [ 𝒫 1 h 0 ] , [ 𝒫 2 h 0 ] , [ ] .

### Remark 5.1.

Let 𝒱 be a prime ideal of K above l. Then 𝒱 𝒫 1 h 0 𝒫 2 h 0 since 𝒫 1 h 0 𝒫 2 h 0 𝒱 = ( - 2 p ϵ l ) , where - 2 p ϵ l K .

### 5.2 The unramified extensions of K within K 2 ( 1 )

We know from Lemma 3.2 that the genus field of K is K ( * ) = K ( - p , - 2 ) , and therefore, its subextensions F 3 = K ( 2 p ) , F 4 = K ( - 2 ) and F 5 = K ( - p ) are unramified over K and absolutely abelian.

From Lemma 3.1, if 2 a b p = 1 , then the equation - ϵ π 1 X 2 ( mod 4 ) admits a solution in k. Moreover, since ( - ϵ π 1 ) = ( 𝒫 1 h 0 ) 2 , where 𝒫 1 is an ideal of K, so K ( - ϵ π 1 ) is an unramified quadratic extension of K. This implies that F 1 = K ( - ϵ π 1 ) is non-Galois over .

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

F 2 = K ( - 2 ϵ π 2 ) , F 6 = K ( 2 ϵ π 1 ) and F 7 = K ( ϵ π 2 )

are pairwise different and non-Galois over .

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

F 1 = K ( - ϵ π 1 ) = K ( ϵ π 1 ) and F 2 = K ( - 2 ϵ π 2 ) = K ( 2 ϵ π 2 ) .

Thus F 1 F 7 and F 2 F 6 .

Regarding the biquadratic extensions L i , note that each extension is gotten by composing two different unramified quadratic extensions of K. Then L i , 1 i 7 , are the unramified biquadratic extensions of K within K 2 ( 1 ) . By using the fact that F 1 F 7 and F 2 F 6 , we deduce that L 1 L 3 and L 6 L 7 .

If 2 a b p = - 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 F 5

From Figure 2, we see that F 5 / R , F 5 / L are ramified quadratic extensions, whereas F 5 / K is not. Hence, according to class field theory,

[ 𝐂 K , 2 : N F 5 / K ( 𝐂 F 5 , 2 ) ] = 2 , 𝐂 R , 2 = N F 5 / R ( 𝐂 F 5 , 2 ) and 𝐂 L , 2 = N F 5 / L ( 𝐂 F 5 , 2 ) .

Since F 5 = K ( - p ) = K ( 2 ϵ l ) and

N F 5 / K ( 𝐂 F 5 , 2 ) = { [ Q ] 𝐂 K , 2 | 2 ϵ l Q = 1 } = { [ Q ] 𝐂 K , 2 | - p Q = 1 } ,

then N F 5 / K ( 𝐂 F 5 , 2 ) = [ ] , [ 𝒫 1 h 0 𝒫 2 h 0 ] .

In fact, from [9], we have

2 ϵ l 𝒫 i h 0 = - 2 p p l 4 = - 2 p 2 = - 1 ,

and then

2 ϵ l 𝒫 1 h 0 𝒫 2 h 0 = 2 ϵ l 𝒫 1 h 0 2 ϵ l 𝒫 2 h 0 = 1 .

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

- p = - p 2 k = - p , 2 2 k = N k / ( - p ) , 2 2 = p 2 , 2 2 = 1 .

Thus [ 𝒫 1 h 0 𝒫 2 h 0 ] and [ ] are norms in F 5 / K .

Let us now compute κ F 5 / K . From Theorem 3.7, we get

E F 5 = - 1 , ϵ , ϵ χ F 5 + , ϵ χ F 5 + σ F 5 + and E K = - 1 , ϵ , so N F 5 / K ( E F 5 ) = - 1 , ϵ 2 .

Therefore, [ E K : N F 5 / K ( E F 5 ) ] = 2 , which implies that # κ F 5 / K = 4 . Moreover, we showed in the proof of Proposition 4.1 that ( 2 ) L = H 2 , H is principal in L = F + and H 𝒪 F 5 = 𝒪 F 5 . So, necessarily, capitulates in F 5 . In addition, it is known that 𝒫 1 2 h 0 𝒫 2 2 h 0 = ( p h 0 ) , which means that ( 𝒫 1 h 0 𝒫 2 h 0 ) = ( p h 0 ) , and hence 𝒫 1 h 0 𝒫 2 h 0 capitulates in F 5 . Thus κ F 5 / K = [ ] , [ 𝒫 1 h 0 𝒫 2 h 0 ] .

On the other hand, we know from Proposition 4.1 and Lemma 3.3 that 𝐂 L , 2 is trivial and 𝐂 R , 2 = [ ] . Let 𝔍 be a prime ideal of F 5 above l. Then we claim ( 𝔍 v ) 4 1 .

To this end, let s and t be the elements of order 2 in Gal ( F 5 / 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 [ Gal ( F 5 / k ) ] and taking into account that the 2-class number of k is 1, we get

( 𝔍 v ) 2 ( 𝔍 v ) 1 + s ( 𝔍 v ) 1 + t ( 𝔍 v ) 1 + s t V i v 𝒱 v V i v 𝒫 1 h 0 𝒫 2 h 0 V i v

since 𝒱 v 𝒱 𝒫 1 h 0 𝒫 2 h 0 (by Remark 5.1) and 𝒫 1 h 0 𝒫 2 h 0 κ F 5 / K ( V i is the prime ideal of R = k 1 k 2 above l, see Lemma 3.3). Therefore, ( 𝔍 v ) 4 ( V i v ) 2 1 .

Let 0 be a prime ideal of F 5 above 2. So, from Remark 3.4, we have:

1. (1)

If p 3 ( mod 8 ) , then ( 0 u ) 2 ( 0 u ) 1 + s ( 0 u ) 1 + t ( 0 u ) 1 + s t 2 R u u 2 R u , and therefore, ( 0 u ) 2 n + 1 ( 2 R u ) 2 n 1 .

2. (2)

If p 7 ( mod 8 ) , then ( 0 v ) 2 ( 0 v ) 1 + s ( 0 v ) 1 + t ( 0 v ) 1 + s t 2 R v v 1 .

Moreover, we have:

1. (1)

If p 3 ( mod 8 ) , then ( 𝔍 v ) 2 V i v 2 n - 1 ( 0 u ) 2 n . Furthermore, 𝔍 v ( 0 u ) 2 n - 1 . If not, by applying the norm of F 5 / K , we get 𝒫 1 h 0 𝒫 2 h 0 ( u ) 2 n - 1 1 (in K), which is absurd.

2. (2)

If p 7 ( mod 8 ) , then ( 𝔍 v ) 2 0 v . If not, by applying the norm of F 5 / K , we get 2 𝒫 1 2 h 0 𝒫 2 2 h 0 1 (in K), which is absurd.

Consequently,

𝐂 F 5 , 2 = { [ 0 u ] , [ 𝔍 v ] ( 2 , 2 n + 1 ) if p 3 ( mod 8 ) , [ 0 v ] , [ 𝔍 v ] ( 2 , 4 ) if p 7 ( mod 8 ) .

Note that 𝒫 i remains inert and does not capitulate in F 5 . Then 𝒫 1 2 h 0 1 (in F 5 ), and therefore,

𝒫 1 h 0 { ( 𝔍 v ) 2 ( 0 u ) 2 n in the first case , 0 v or ( 𝔍 v ) 2 in the second case .

If 𝒫 1 h 0 0 v , by applying the norm of F 5 / K , we get 𝒫 1 2 h 0 1 , which is absurd. Thus 𝒫 1 h 0 ( 𝔍 v ) 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 K 2 ( 2 ) . From Theorem 4.3, we know that h 2 ( K ( * ) ) = h 2 ( F 5 ) 2 and the 2-rank of 𝐂 F 5 , 2 equals 2. So, by [7, Proposition 7], the Hilbert 2-class field tower of F 5 stops at the first stage F 5 , 2 ( 1 ) . Therefore, the Hilbert 2-class field tower of K stops at K 2 ( 2 ) since K F 5 .

### 6.3 The generators of G = Gal ⁡ ( K 2 ( 2 ) / K )

Let us compute the generators of G = Gal ( K 2 ( 2 ) / K ) . Let K 2 ( 2 ) / F 5 P denote the Artin symbol for K 2 ( 2 ) / F 5 , and put

σ = K 2 ( 2 ) / F 5 𝔍 v    and    τ = { K 2 ( 2 ) / F 5 0 u if p 3 ( mod 8 ) , K 2 ( 2 ) / F 5 0 v if p 7 ( mod 8 ) .

Then σ and τ generate the abelian subgroup of G,

Gal ( K 2 ( 2 ) / F 5 ) { ( 2 , 2 n + 1 ) if p 3 ( mod 8 ) , ( 2 , 4 ) if p 7 ( mod 8 ) ,

with

{ τ 2 n + 1 = σ 4 = 1 and τ 2 n = σ 2 if p 3 ( mod 8 ) , τ 2 = σ 4 = 1 if p 7 ( mod 8 ) .

Also put

ρ = K 2 ( 2 ) / K 𝒫 1 h 0 ,

so ρ restricts to the non-trivial automorphism of F 5 / K since 𝒫 1 h 0 is not a norm from F 5 / K , and hence G = σ , τ , ρ .

Furthermore, the following relations hold:

• Since 𝒫 1 h 0 ( 𝔍 v ) 2 ,

ρ 2 = K 2 ( 2 ) / K 𝒫 1 2 h 0 = K 2 ( 2 ) / K N F 5 / K ( 𝒫 1 h 0 ) = K 2 ( 2 ) / F 5 𝒫 1 h 0 ,

and thus ρ 2 = σ 2 .

• Since ( 𝔍 v ) 1 + ρ = N F 5 / K ( 𝔍 v ) 𝒫 1 h 0 𝒫 2 h 0 1 in F 5 ,

σ ρ - 1 σ ρ = K 2 ( 2 ) / F 5 ( 𝔍 v ) 1 + ρ = 1 ,

and thus [ σ , ρ ] = [ ρ , σ ] = σ 2 .

• Since ( 0 u ) 1 + ρ = N F 5 / K ( 0 u ) 1 and ( 0 v ) 1 + ρ = N F 5 / K ( 0 v ) 1 in F 5 , according as p 3 ( mod 8 ) and p 7 ( mod 8 ) ,

τ ρ - 1 τ ρ = { K 2 ( 2 ) / F 5 ( 0 u ) 1 + ρ = 1 if p 3 ( mod 8 ) , K 2 ( 2 ) / F 5 ( 0 v ) 1 + ρ = 1 if p 7 ( mod 8 ) ,

and thus [ τ , ρ ] = τ - 2 .

Hence

G = { ρ , τ , σ : ρ 4 = σ 4 = τ 2 n + 1 = 1 , σ 2 = ρ 2 = τ 2 n , n 2 , [ τ , σ ] = 1 , [ σ , ρ ] = σ 2 , [ ρ , τ ] = τ 2 if p 3 ( mod 8 ) , ρ , τ , σ : ρ 4 = σ 4 = τ 2 = 1 , σ 2 = ρ 2 , [ τ , σ ] = [ τ , ρ ] = 1 , [ σ , ρ ] = σ 2 if p 7 ( mod 8 ) .

1. (1)

If p 3 ( mod 8 ) , then G = τ 2 since [ τ , σ ] = 1 , [ ρ , τ ] = τ 2 , [ ρ , σ ] = σ 2 and σ 2 = ( τ 2 ) 2 n - 1 , and therefore, 𝐂 K 2 ( 1 ) , 2 is cyclic of order 2 n .

2. (2)

If p 7 ( mod 8 ) , we easily check that G = σ 2 , and thus 𝐂 K 2 ( 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 γ 1 ( G ) = G and γ i + 1 ( G ) = [ γ i ( G ) , G ] , which is the subgroup of G generated by the set { [ a , b ] = a - 1 b - 1 a b a γ i ( G ) , b G } . So the coclass of G is defined to be c c ( G ) = h - c , where # G = 2 h and c = c ( G ) is the nilpotency class of G, which is the smallest positive integer satisfying γ c ( G ) { 1 } and γ c + 1 ( G ) = { 1 } .

1. (1)

If p 3 ( mod 8 ) , then γ 1 ( G ) = G , γ 2 ( G ) = G = τ 2 and

γ n + 1 ( G ) = [ γ n ( G ) , G ] = τ 2 n , γ n + 2 ( G ) = [ γ n + 1 ( G ) , G ] = τ 2 n + 1 = 1 .

So c ( G ) = n + 1 , and c c ( G ) = 2 .

2. (2)

If p 7 ( mod 8 ) , then γ 1 ( G ) = G , γ 2 ( G ) = G = σ 2 and γ 3 ( G ) = [ G , G ] = 1 . So c c ( G ) = c ( G ) = 2 .

Finally, prove assertion (6) of Theorem 2.2.

1. (1)

If p 3 ( mod 8 ) , then [ ρ , σ 2 ] = σ 4 = 1 , [ ρ , σ τ 2 n - 1 ] = 1 and σ 2 σ τ 2 n - 1 . So σ 2 , σ τ 2 n - 1 = Z ( G ) ( 2 , 2 ) .

2. (2)

If p 7 ( mod 8 ) , then [ ρ , σ 2 ] = σ 4 = 1 , [ ρ , τ ] = 1 and σ 2 τ . So σ 2 , τ = Z ( G ) ( 2 , 2 ) .

## 7 Proof of Theorem 2.3

To prove this theorem, we need the following norm class groups N F i / K ( 𝐂 F i , 2 ) . The results are summarized in Tables 1 and 2.

Table 1

Norm class groups for the case 2 a b p = 1 .

 N F i / K ⁢ ( 𝐂 F i , 2 ) F i 2 p = - 1 2 p = 1 F 1 = K ⁢ ( - ϵ ⁢ π 1 ) 〈 [ 𝒫 2 h 0 ] , [ ℋ ⁢ 𝒫 1 h 0 ] 〉 〈 [ 𝒫 1 h 0 ] , [ ℋ ] 〉 F 2 = K ⁢ ( - 2 ⁢ ϵ ⁢ π 2 ) 〈 [ 𝒫 1 h 0 ] , [ ℋ ] 〉 〈 [ 𝒫 1 h 0 ] , [ ℋ ⁢ 𝒫 2 h 0 ] 〉 F 6 = K ⁢ ( 2 ⁢ ϵ ⁢ π 1 ) 〈 [ 𝒫 2 h 0 ] , [ ℋ ] 〉 〈 [ 𝒫 2 h 0 ] , [ ℋ ⁢ 𝒫 1 h 0 ] 〉 F 7 = K ⁢ ( ϵ ⁢ π 2 ) 〈 [ 𝒫 1 h 0 ] , [ ℋ ⁢ 𝒫 2 h 0 ] 〉 〈 [ 𝒫 2 h 0 ] , [ ℋ ] 〉 F 3 = K ⁢ ( 2 ⁢ p ) 〈 [ ℋ ⁢ 𝒫 i h 0 ] , [ 𝒫 1 h 0 ⁢ 𝒫 2 h 0 ] 〉 〈 [ 𝒫 1 h 0 ] , [ 𝒫 2 h 0 ] 〉 F 4 = K ⁢ ( - 2 ) 〈 [ 𝒫 1 h 0 ] , [ 𝒫 2 h 0 ] 〉 〈 [ ℋ ⁢ 𝒫 i h 0 ] , [ 𝒫 1 h 0 ⁢ 𝒫 2 h 0 ] 〉 F 5 = K ⁢ ( - p ) 〈 [ ℋ ] , [ 𝒫 1 h 0 ⁢ 𝒫 2 h 0 ] 〉
Table 2

Norm class groups for the case 2 a b p = - 1 .

 N F i / K ⁢ ( 𝐂 F i , 2 ) F i 2 p = - 1 2 p = 1 F 1 = K ⁢ ( - 2 ⁢ ϵ ⁢ π 1 ) 〈 [ 𝒫 2 h 0 ] , [ ℋ ] 〉 〈 [ 𝒫 2 h 0 ] , [ ℋ ⁢ 𝒫 1 h 0 ] 〉 F 2 = K ⁢ ( - ϵ ⁢ π 2 ) 〈 [ 𝒫 1 h 0 ] , [ ℋ ⁢ 𝒫 2 h 0 ] 〉 〈 [ 𝒫 2 h 0 ] , [ ℋ ] 〉 F 6 = K ⁢ ( ϵ ⁢ π 1 ) 〈 [ 𝒫 2 h 0 ] , [ ℋ ⁢ 𝒫 1 h 0 ] 〉 〈 [ 𝒫 1 h 0 ] , [ ℋ ] 〉 F 7 = K ⁢ ( 2 ⁢ ϵ ⁢ π 2 ) 〈 [ 𝒫 1 h 0 ] , [ ℋ ] 〉 〈 [ 𝒫 1 h 0 ] , [ ℋ ⁢ 𝒫 2 h 0 ] 〉 F 3 = K ⁢ ( 2 ⁢ p ) 〈 [ ℋ ⁢ 𝒫 i h 0 ] , [ 𝒫 1 h 0 ⁢ 𝒫 2 h 0 ] 〉 〈 [ 𝒫 1 h 0 ] , [ 𝒫 2 h 0 ] 〉 F 4 = K ⁢ ( - 2 ) 〈 [ 𝒫 1 h 0 ] , [ 𝒫 2 h 0 ] 〉 〈 [ ℋ ⁢ 𝒫 i h 0 ] , [ 𝒫 1 h 0 ⁢ 𝒫 2 h 0 ] 〉 F 5 = K ⁢ ( - p ) 〈 [ ℋ ] , [ 𝒫 1 h 0 ⁢ 𝒫 2 h 0 ] 〉

Let us check some examples. Recall that, for 1 i 7 , [ 𝐂 K , 2 : N F i / K ( 𝐂 F i , 2 ) ] = 2 .

## Example.

For F 5 , we already proved that N F 5 / K ( 𝐂 F 5 , 2 ) = [ ] , [ 𝒫 1 h 0 𝒫 2 h 0 ] .

## Example.

For F 3 = K ( 2 p ) = K ( - ϵ l ) , we have

- ϵ l 𝒫 i h 0 = - 1 𝒫 i h 0 ϵ l 𝒫 i h 0 = - - 1 p h 0 p l 4 = 2 p ,