On Hilbert class field tower for some quartic number fields

Azizi, Abdelmalek; Talbi, Mohamed; Talbi, Mohammed

doi:10.1016/j.ajmsc.2019.05.005

We determine the Hilbert 2-class field tower for some quartic number fields $k$ whose 2-class group $C_{k, 2}$ is isomorphic to $ℤ / 2 ℤ \times ℤ / 2 ℤ$ ⁠.

1. Introduction

Let $k$ be a number field, $C_{k, 2}$ its $2$ -class group, it is the $2$ -Sylow subgroup of the class group (in the wide sense) of $k, k^{(0)} = k \subseteq k^{(1)} \subseteq k^{(2)} \dots k^{(i)} \dots$ ⁠, the tower of Hilbert $2$ -class field of $k$ which means that $k^{(1)}$ is the Hilbert $2$ -class field of $k$ (this is the maximal abelian unramified extension of $k$ of degree a power of $2$ ⁠) and $k^{(i + 1)}$ is the Hilbert $2$ -class field of $k^{(i)}$ for $i \geq 1$ ⁠.

Lemma 1.

If $G$ is a $2$ -group of order $2^{m}$ , $m \geq 2$ , such that $G / G^{'} ≃ ℤ / 2 ℤ \times ℤ / 2 ℤ$ , then $G$ is isomorphic to $Q_{m}$ $(r e s p e c t i v e l y D_{m}, S_{m}, (2, 2))$ the quaternion (respectively dihedral, semidihedral, Klein) group of order $2^{m}$ . In particular $G^{'}$ , the commutator subgroup of $G$ , is cyclic.

Proof. See [3]. □

Let $G = G a l (k^{(2)} / k)$ ⁠, be the Galois group of $k^{(2)} / k$ ⁠, then, with class field theory we have that $G^{'} = G a l (k^{(2)} / k^{(1)}) ≃ C_{k^{(1)}, 2}$ and $G / G^{'} = G a l (k^{(1)} / k) ≃ C_{k, 2}$ ⁠. According to [3] and [8], if $C_{k, 2}$ is an elementary group of rank $2$ ⁠, then $C_{k^{(1)}, 2}$ is cyclic, which implies that the Hilbert $2$ -class field tower stops at $k^{(2)}$ ⁠.

Definition 1

(Taussky’s Conditions). Let $F / k$ be a cyclic unramified extension and $j = j_{F / k}$ $(respectively N_{F / k})$ denote the conorm (respectively the norm) of $F / k$ ⁠.

$F / k$ is of type $(A)$ if and only if $| \ker (j) \cap N_{F / k} (C_{F}) | > 1$ ⁠,
$F / k$ is of type $(B)$ if and only if $| \ker (j) \cap N_{F / k} (C_{F}) | = 1$ ⁠.

Note that $\ker (j)$ is the set of all the class ideals of $k$ which capitulate in $F$ ⁠.

Theorem 1.

Let $k$ be a number field such that $C_{k, 2}$ is isomorphic to $ℤ / 2 ℤ \times ℤ / 2 ℤ, F_{1}, F_{2}, F_{3}$ the three unramified quadratic extensions of $k$ within $k^{(1)}$ and $G$ be the Galois group of $k^{(2)} /k$ , then

$G$ is abelian if and only if the four classes of $C_{k, 2}$ capitulate in each extension $F_{i} / k$ ;
$G ≃ Q_{3}$ if and only if the three extensions $F_{i} / k$ are of type $(A)$ and in each extension $F_{i} / k$ only two classes of $C_{k, 2}$ capitulate;
$G ≃ Q_{m}$ avec $m > 3$ if and only if uniquely one extension $F_{i} / k$ is of type $(A)$ and in each extension $F_{i} / k$ two classes of $C_{k, 2}$ capitulate;
$G ≃ S_{m}$ if and only if the three extensions $F_{i} / k$ are of type $(B)$ and in each extension $F_{i} / k$ two classes of $C_{k, 2}$ capitulate;
$G ≃ D_{m}$ if and only if the four classes of $C_{k, 2}$ capitulate uniquely in one extension $F_{i} / k$ .

Proof. See [8]. □

2. Units of some number fieldss

In the remainder of this paper, let $ℓ$ be a prime number congruent to $1$ modulo 8, $ε$ the fundamental unit of $ℚ (\sqrt{ℓ})$ and $L = ℚ (\sqrt{ε \sqrt{ℓ}})$ ⁠.

The extension $L / ℚ$ is real cyclic of degree $4$ ⁠, of Galois group $H = 〈 σ 〉$ and quadratic subfield $ℚ (\sqrt{ℓ})$ ⁠. Since $L$ is of conductor $ℓ$ ⁠, then $L$ is a subfield of the $ℓ$ th cyclotomic field $ℚ (ζ_{ℓ})$ and there is a character $χ^{'}$ of $G a l (ℚ (ζ_{ℓ}) / ℚ) ≃ {(ℤ / ℓ ℤ)}^{*}$ ⁠, where its kernel is $G a l (ℚ (ζ_{ℓ}) / L)$ ⁠, we will call the character of $L$ ⁠.

Let $χ = χ^{'} + {χ^{'}}^{- 1}$ ⁠, then $χ$ is a rational character of $ℚ (ζ_{ℓ})$ and $L$ is fixed by the Common kernel of $χ^{'}$ and $χ^{- 1}$ ⁠. Let $E_{L}$ be the group of units of $L$ ⁠, $E_{χ} = {ω \in E_{L} | ω^{1 + σ^{2}} = \pm 1}$ the group of $χ$ -relatives units of $L$ (according to definition of H. W. Leopoldt in [10]), $| E_{L} |$ (respectively $| E_{χ} |$ ⁠) the group of the absolute values of $E_{L}$ ⁠, (respectively of $E_{χ}$ ⁠), $| E^{L} | = | E_{L} | \oplus | E_{χ} |$ and $ε_{χ}$ a generator of $E_{χ}$ ⁠, then:

Theorem 2.

Let $L = ℚ (\sqrt{ε \sqrt{ℓ}})$ where $ℓ$ be a prime number congruent to $1$ modulo $8$ and $ε$ the fundamental unit of $ℚ (\sqrt{ℓ})$ , then there exists $ξ$ in $E_{L}$ , such as $ξ^{2} = \pm ε ε_{χ}^{1 - σ}$ and ${ξ, ξ^{σ}, ξ^{σ^{2}}}$ is a fundamental system of units of $L$

Proof. See [4]. □

Remark 1.

Since $ξ^{2} = \pm ε ε_{χ}^{1 - σ}$ ⁠, then:

$ξ^{1 + σ} = \pm ε_{χ}$ ⁠;
$ξ^{1 + σ^{2}} = \pm ε$ ⁠;
$ξ^{1 + σ + σ^{2} + σ^{3}} = ε_{χ}^{1 + σ^{2}} = ε^{1 + σ}$ ⁠;
$N_{ℚ (\sqrt{ℓ}) / ℚ} (ε) = N_{L / ℚ (\sqrt{ℓ})} (ε_{χ}) = N_{L / ℚ} (ξ) = - 1$ ⁠.

Lemma 2.

With the same notation of Theorem 2 , ${ξ, ξ^{σ}, ξ^{σ^{2}}}$ is also a fundamental system of units of $F = L (\sqrt{- n})$ where $n$ is a positive integer prime to $ℓ$ and squarefree.

Proof. If $L (\sqrt{- n}) \neq L (\sqrt{- 1})$ ⁠, then, according to [1, Proposition 3], to show that ${ξ, ξ^{σ}, ξ^{σ^{2}}}$ is a fundamental system of units of $F$ it suffices to show that $n μ$ is not a square in $L$ ⁠, for $μ = {\begin{array}{l} ξ_{1}^{'} \end{array}}^{j_{1}} {\begin{array}{l} ξ_{2}^{'} \end{array}}^{j_{2}} ξ_{3}^{'}$ where ${ξ_{1}^{'}, ξ_{2}^{'}, ξ_{3}^{'}} = {ξ, ξ^{σ}, ξ^{σ^{2}}}$ and $j_{1}, j_{2} \in {0, 1}$ ⁠.

Indeed, if $μ = ξ$ ⁠, then if $n ξ = x^{2}$ in $L$ ⁠, by calculating the norm in $L / k$ ⁠, we find that $ξ^{1 + σ^{2}} = \pm ε$ is a square in $k$ ⁠, which is impossible.

If $μ = ξ^{σ}$ ⁠, then if $n ξ^{σ} = \pm n \frac{ε_{χ}}{ξ} = x^{2}$ in $L$ ⁠, by calculating the norm in $L / k$ ⁠, we find that $\pm \frac{ε^{1 + σ}}{ε} = \pm \frac{1}{ε}$ is a square in $k$ ⁠, which is absurd.

If $μ = ξ^{σ^{2}}$ ⁠, then if $n ξ^{σ^{2}} = \pm n \frac{ε}{ξ} = x^{2}$ in $L$ ⁠, by calculating the norm in $L / k$ ⁠, we find that $\pm \frac{ε^{2}}{ε} = \pm ε$ is a square in $k$ ⁠, which is not the case.

If $μ = ξ^{1 + σ} = \pm ε_{χ}$ ⁠, then if $\pm n ε_{χ} = x^{2}$ in $L$ ⁠, by calculating the norm in $L / k$ ⁠, we find that $ε_{χ}^{1 + σ^{2}} = ε^{1 + σ} = - 1$ is a square in $k$ ⁠, which is absurd.

If $μ = ξ^{1 + σ^{2}} = \pm ε$ ⁠, then $\pm n ε$ cannot be a square in $L$ ⁠.

If $μ = ξ^{σ + σ^{2}}$ ⁠, then if $n ξ^{σ + σ^{2}} = x^{2}$ in $L$ ⁠, by calculating the norm in $L / k$ ⁠, we find that $ξ^{1 + σ + σ^{2} + σ^{3}} = - 1$ is a square in $k$ ⁠, which is not the case.

If $μ = ξ^{1 + σ + σ^{2}}$ ⁠, then if $n ξ^{1 + σ + σ^{2}} = x^{2}$ in $L$ ⁠, by calculating the norm in $L / k$ ⁠, we find that $ξ^{1 + σ + σ^{2} + σ^{3}} ξ^{1 + σ^{2}} = \pm ε$ is a square in $k$ ⁠, which is impossible.

If $F = L (\sqrt{- 1})$ ⁠, then we have $m = 2$ is the largest integer such that $ζ_{m} \in F$ ⁠, so, from [1, Proposition 2], to show that ${ξ, ξ^{σ}, ξ^{σ^{2}}}$ is a fundamental system of units of $F$ it suffices to show that $2 μ$ is not a square in $L$ ⁠, for $μ = {\begin{array}{l} ξ_{1}^{'} \end{array}}^{j_{1}} {\begin{array}{l} ξ_{2}^{'} \end{array}}^{j_{2}} ξ_{3}^{'}$ where ${ξ_{1}^{'}, ξ_{2}^{'}, ξ_{3}^{'}} = {ξ, ξ^{σ}, ξ^{σ^{2}}}$ and $j_{1}, j_{2} \in {0,1}$ ⁠. Using the same reasoning as above we find the result, which completes the proof of the lemma.□

3. Hilbert 2-class field tower of $ℚ (\sqrt{−2ε \sqrt{ℓ}})$

Let $k = ℚ (\sqrt{- 2 ε \sqrt{ℓ}})$ where $ℓ$ is a prime number such that $ℓ \equiv 1 (\mod 8)$ and ${(\frac{2}{ℓ})}_{4} = - 1$ ⁠, according to [2], $C_{k, 2}$ is isomorphic to $ℤ / 2 ℤ \times ℤ / 2 ℤ$ ⁠, thus $k^{(1)} / k$ has three intermediate subfields, $F_{1}, F_{2}, F_{3}$ and let $G$ be the Galois group of $k^{(2)} / k$ ⁠. Using [6, Theorem 4, p. 48–49], it is easy to show that the genus field of $k$ is $k^{(*)} = F_{3} = k (\sqrt{- 2})$ ⁠, this is the maximal extension of $k$ of the form $kM$ which is unramified for all prime ideals of $k$ ⁠, finite and infinite, and such that $M$ is abelian over $ℚ$ ⁠.

As $(\frac{2}{ℓ}) = 1$ ⁠, then there exist two prime ideals

B_{1} = 〈 2, \frac{a + b \sqrt{ℓ}}{2} 〉, B_{2} = 〈 2, \frac{a - b \sqrt{ℓ}}{2} 〉 of ℚ (\sqrt{ℓ}) such as (2) = B_{1} B_{2} in ℚ (\sqrt{ℓ})

with $a$ and $b$ are integers, but $B_{1}^{h (ℓ)}$ and $B_{2}^{h (ℓ)}$ are principal ideals of $ℚ (\sqrt{ℓ})$ where $h (ℓ)$ is the class number of $ℚ (\sqrt{ℓ})$ ⁠, so we can choose $c$ and $d$ positive integers such as

B_{1}^{h (ℓ)} = (α) and B_{2}^{h (ℓ)} = (\bar{α}) in ℚ (\sqrt{ℓ})

with $α = \frac{c + d \sqrt{ℓ}}{2}$ and $\bar{α} = \frac{c - d \sqrt{ℓ}}{2}$ are positive. Let $H_{1}$ and $H_{2}$ be prime ideals of $k$ above $B_{1}$ and $B_{2}$ respectively and $H = H_{1} H_{2}$ ⁠.

Proposition 1.

Let $k = ℚ (\sqrt{- 2 ε \sqrt{ℓ}})$ where $ℓ$ is a prime number such that $ℓ \equiv 1 (\mod 8)$ and ${(\frac{2}{ℓ})}_{4} = - 1$ . Then the class $[H]$ is of order $2$ in $k$ . Moreover $H$ capitulates in $k (\sqrt{- 2})$ ⁠.

Proof. The class of $H$ is of order $2$ ⁠, indeed, we have $H^{2} = (2)$ in $k$ ⁠. Suppose that $H = (λ)$ for some $λ$ in $k$ ⁠, which is equivalent to $(λ^{2}) = (2)$ in $k$ ⁠, therefore, there exists a unit $η$ of $k$ such that $2 η = λ^{2}$ ⁠, although there exist $x$ and $y$ in $ℚ (\sqrt{ℓ})$ such that $λ = x + y \sqrt{- 2 ε \sqrt{ℓ}}$ ⁠, thus

2 η = x^{2} - 2 ε \sqrt{ℓ} y^{2} + 2 x y \sqrt{- 2 ε \sqrt{ℓ}} .

Since $ε$ is a fundamental unit of $k$ and $i = \sqrt{- 1} \notin k$ ⁠, then $2 η \in ℚ (\sqrt{ℓ})$ ⁠, therefore $x$ or $y$ is equal to $0$ ⁠. If $y = 0$ ⁠, then $2 η = x^{2}$ ⁠, so $2 = x^{' 2}$ or $2 ε = x^{″ 2}$ in $ℚ (\sqrt{ℓ})$ ⁠, which gives that $\sqrt{2} \in ℚ (\sqrt{ℓ})$ in the first case and $\sqrt{- 1} \in ℚ (\sqrt{ℓ})$ in the second case, which is impossible. Similarly, if $x = 0$ we find that $\pm ℓ$ is a square in $ℚ$ ⁠, consequently $[H]$ is of order $2$ ⁠.

To show that $H$ is capitulated in $F_{3} = k (\sqrt{- 2})$ ⁠, it suffices to remark that $\sqrt{- 2}$ is in $F_{3}$ and $({\sqrt{- 2}}^{2}) = (2)$ in $F_{3}$ ⁠, thus $H$ capitulates in $F_{3}$ □.

Theorem 3.

Let $k = ℚ (\sqrt{- 2 ε \sqrt{ℓ}})$ where $ℓ$ is a prime number such that $ℓ \equiv 1 (\mod 8)$ and ${(\frac{2}{ℓ})}_{4} = - 1$ and $F_{3} = k (\sqrt{- 2})$ . Then $C_{F_{3}, 2}$ , the $2$ -part of the class group of $F_{3}$ , is cyclic of order $h_{2} (F_{3}) = 4$ ⁠.

Proof. Let $α$ ⁠, $\bar{α}$ ⁠, $B_{1}$ ⁠, $B_{2}$ ⁠, $H_{1}$ ⁠, $H_{2}$ and $H = H_{1} H_{2}$ defined as above and let $L = ℚ (\sqrt{ε \sqrt{ℓ}})$ ⁠, then we have $k L = F_{3}$ ⁠, $N_{k / ℚ (\sqrt{ℓ})} (H_{1}) = B_{1}$ and $B_{1}$ is unramified in $L / ℚ (\sqrt{ℓ})$ ⁠. So, to prove that $H_{1}$ is inert in $F_{3} / k$ it suffices to show that $B_{1}$ is inert in $L / ℚ (\sqrt{ℓ})$ (Translation Theorem), and for this, we calculate the following norm residue symbol $(\frac{α, ε \sqrt{ℓ}}{B_{1}})$ ⁠. According to [2], we have

(\frac{α, ε \sqrt{ℓ}}{B_{1}}) = {(\frac{2}{ℓ})}_{4} = - 1,

so $B_{1}$ is inert in $L / ℚ (\sqrt{ℓ})$ ⁠, which gives $H_{1}$ remains inert in $F_{3} / k$ and in the same way we have shown that $H_{2}$ remains inert in $F_{3} / k$ and since $H = H_{1} H_{2}$ capitulates in $F_{3}$ ⁠, then by the Artin reciprocity law, we find that $F_{3} / K$ is of type $(A)$ ⁠, therefore, according to [8], we find that $C_{F_{3}, 2}$ is cyclic.

Since $F_{3} / ℚ (\sqrt{ℓ})$ is a normal biquadratic extension with Galois group of type $(2,2)$ and where $k, L, k_{0} = ℚ (\sqrt{ℓ}, \sqrt{- 2})$ are these intermediate subfields, then, by [9], we have

h_{2} (F_{3}) = \frac{2^{2 - 1 - 2 - 0} q (F_{3} / ℚ (\sqrt{ℓ})) h_{2} (k) h_{2} (L) h_{2} (k_{0})}{h_{2} {(ℓ)}^{2}} .

But $h_{2} (ℓ) = 1, h_{2} (k) = 4, h_{2} (L) = 1$ (see [12]), and by [7,11] we have $h_{2} (k_{0}) = 2$ ⁠, so $h_{2} (F_{3}) = 4 q (F_{3} / ℚ (\sqrt{ℓ}))$ and we have $ε$ is a fundamental unit of $k$ and since $2 ε$ is not a square in $ℚ (\sqrt{ℓ})$ ⁠, then, by [1, Proposition 3], ${ε}$ is also a fundamental system of units of $k_{0}$ and according to Lemma 2, $L$ and $F_{3}$ have the same fundamental system of units which is ${ξ, ξ^{σ}, ξ^{σ^{2}}}$ ⁠, thus $q (F_{3} / ℚ (\sqrt{ℓ}))$ ⁠, the unit index of $F_{3} / ℚ (\sqrt{ℓ})$ ⁠, is equal to $1$ ⁠, which gives that $h_{2} (F_{3}) = 4$ ⁠. □

Remark 2.

Let $k = ℚ (\sqrt{- 2 ε \sqrt{ℓ}})$ where $ℓ$ is a prime number such that $ℓ \equiv 1 (\mod 8)$ and ${(\frac{2}{ℓ})}_{4} = - 1$ and let $G$ be the Galois group of $k^{(2)} /k$ ⁠, then $G$ is of order $8$ ⁠.

Indeed, we have $k^{(1)} / F_{3}$ is an unramified extension and the $2$ -class group of $F_{3}$ is cyclic, thus $F_{3}$ and $k^{(1)}$ have the same Hilbert $2$ -class field, namely $k^{(2)}$ ⁠, so # $G = 2 h_{2} (F_{3}) = 8$ ⁠.

Theorem 4.

Let $k = ℚ (\sqrt{- 2 ε \sqrt{ℓ}})$ where $ℓ$ is a prime number such that $ℓ \equiv 1 (\mod 8)$ and ${(\frac{2}{ℓ})}_{4} = - 1$ , $F_{1}, F_{2}, F_{3}$ the quadratic subfields of $k^{(1)} / k$ . Then, in each extension $F_{i} / k$ , $i \in {1,2,3}$ , there exist exactly two classes of $C_{k, 2}$ which capitulates and $G$ is the quaternion group of order $8$ ⁠.

Proof. According to Lemma 2, we have ${ξ, ξ^{σ}, ξ^{σ^{2}}}$ is a fundamental system of units of $F_{3}$ ⁠. Since

N_{F_{3} / k} (ξ) = N_{F_{3} / k} (ξ^{σ^{2}}) = \pm ε and N_{F_{3} / k} (ξ^{σ}) = \pm ε^{- 1},

then $N_{F_{3} / k} (E_{F_{3}}) = E_{k}$ ⁠, consequently, by [5], we have only two classes of $C_{k, 2}$ which capitulates in $F_{3}$ ⁠, namely $[H]$ and its square and since the extension $F_{3} / k$ is of type $(A)$ ⁠, then, by Theorem 1, we find that $G$ is a quaternion group, and by Remark 2, the group $G$ is of order $8$ ⁠, therefore $G ≃ Q_{3}$ ⁠. □

Remark 3.

As $G ≃ Q_{3}$ ⁠, then, by [8],

C_{F_{1}, 2} ≃ C_{F_{2}, 2} ≃ C_{F_{3}, 2} ≃ ℤ / 4 ℤ .

Example 1.

Let $k = ℚ (\sqrt{- 2 (17 + 4 \sqrt{17})})$ ⁠, we have $17 \equiv 1 (\mod 8)$ and ${(\frac{2}{17})}_{4} = - 1$ ⁠, then the Galois group of $k^{(2)} / k$ is the quaternion group of order $8$ ⁠, there exist exactly two classes of $C_{k, 2}$ which capitulates in $F_{3} = k (\sqrt{- 2})$ ⁠, as well for $F_{1}$ and for $F_{2}$ and the Hilbert $2$ -class field tower of $k$ stops at $k^{(2)}$ ⁠.

Declaration of Competing Interest: Authors do not have any conflict of interest.The publisher wishes to inform readers that the article “On Hilbert class field tower for some quartic number fields” was originally published by the previous publisher of the Arab Journal of Mathematical Sciences and the pagination of this article has been subsequently changed. There has been no change to the content of the article. This change was necessary for the journal to transition from the previous publisher to the new one. The publisher sincerely apologises for any inconvenience caused. To access and cite this article, please use Azizi, A., Talbi, M., Talbi, M. (2019), “On Hilbert class field tower for some quartic number fields”, Arab Journal of Mathematical Sciences, Vol. 27 No. 1, pp. 20-25. The original publication date for this paper was 22/05/2019.

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2019

Abdelmalek Azizi, Mohamed Talbi and Mohammed Talbi

Published in the Arab Journal of Mathematical Sciences. Published by Emerald Publishing Limited. This article is published under the Creative Commons Attribution (CC BY 4.0) license. Anyone may reproduce, distribute, translate and create derivative works of this article (for both commercial and non-commercial purposes), subject to full attribution to the original publication and authors. The full terms of this license may be seen at http://creativecommons.org/licences/by/4.0/legalcode

On Hilbert class field tower for some quartic number fields

1. Introduction

2. Units of some number fieldss

3. Hilbert 2-class field tower of $ℚ (\sqrt{−2ε \sqrt{ℓ}})$

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On Hilbert class field tower for some quartic number fields Open Access

1. Introduction

2. Units of some number fieldss

3. Hilbert 2-class field tower of ℚ(−2εℓ)

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On Hilbert class field tower for some quartic number fields

3. Hilbert 2-class field tower of $ℚ (\sqrt{−2ε \sqrt{ℓ}})$