On Hilbert class field tower for some quartic number fields

Abstract

We determine the Hilbert 2-class field tower for some quartic number fields k whose 2-class group Ck,2 is isomorphic to ℤ/2ℤ×ℤ/2ℤ.

Citation

Azizi, A., Talbi, M. and Talbi, M. (2021), "On Hilbert class field tower for some quartic number fields", Arab Journal of Mathematical Sciences, Vol. 27 No. 1, pp. 20-25. https://doi.org/10.1016/j.ajmsc.2019.05.005

1. Introduction

Let k be a number field, Ck,2 its 2-class group, it is the 2-Sylow subgroup of the class group (in the wide sense) of k,k(0)=k⊆k(1)⊆k(2)…k(i)…, 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≥1.

Lemma 1.

If G is a 2-group of order 2m , m≥2 , such that G/G′≃ℤ/2ℤ×ℤ/2ℤ , then G is isomorphic to Qm (respectively Dm,Sm,(2,2)) the quaternion (respectively dihedral, semidihedral, Klein) group of order 2m . In particular G′, the commutator subgroup of G , is cyclic.

Proof. See [3]. □

Let G=Gal(k(2)/k), be the Galois group of k(2)/k, then, with class field theory we have that G′=Gal(k(2)/k(1))≃Ck(1),2 and G/G′=Gal(k(1)/k)≃Ck,2. According to [3] and [8], if Ck,2 is an elementary group of rank 2, then Ck(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=jF/k (respectivelyNF/k) denote the conorm (respectively the norm) of F/k.

• F/k is of type (A) if and only if |ker(j)∩NF/k(CF)|>1,

• F/k is of type (B) if and only if |ker(j)∩NF/k(CF)|=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 Ck,2 is isomorphic to ℤ/2ℤ×ℤ/2ℤ,F1,F2,F3 the three unramified quadratic extensions of k within k(1) and G be the Galois group of k(2)/k, then

1. G is abelian if and only if the four classes of Ck,2 capitulate in each extension Fi/k ;

2. G≃Q3 if and only if the three extensions Fi/k are of type (A) and in each extension Fi/k only two classes of Ck,2 capitulate;

3. G≃Qm avec m>3 if and only if uniquely one extension Fi/k is of type (A) and in each extension Fi/k two classes of Ck,2 capitulate;

4. G≃Sm if and only if the three extensions Fi/k are of type (B) and in each extension Fi/k two classes of Ck,2 capitulate;

5. G≃Dm if and only if the four classes of Ck,2 capitulate uniquely in one extension Fi/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(ℓ) and L=ℚ(εℓ).

The extension L/ℚ is real cyclic of degree 4, of Galois group H=〈σ〉 and quadratic subfield ℚ(ℓ). Since L is of conductor ℓ, then L is a subfield of the ℓth cyclotomic field ℚ(ζℓ) and there is a character χ′ of Gal(ℚ(ζℓ)/ℚ)≃(ℤ/ℓℤ)∗, where its kernel is Gal(ℚ(ζℓ)/L), we will call the character of L.

Let χ=χ′+χ′−1, then χ is a rational character of ℚ(ζℓ) and L is fixed by the Common kernel of χ′ and χ−1. Let EL be the group of units of L, Eχ={ω∈EL|ω1+σ2=±1} the group of χ-relatives units of L (according to definition of H. W. Leopoldt in [10]), |EL| (respectively |Eχ|) the group of the absolute values of EL, (respectively of Eχ), |EL|=|EL|⊕|Eχ| and εχ a generator of Eχ, then:

Theorem 2.

Let L=ℚ(εℓ) where ℓ be a prime number congruent to 1 modulo 8 and ε the fundamental unit of ℚ(ℓ) , then there exists ξ in EL , such as ξ2=±εεχ1−σ and {ξ,ξσ,ξσ2} is a fundamental system of units of L

Proof. See [4]. □

Remark 1.

Since ξ2=±εεχ1−σ, then:

1. ξ1+σ=±εχ;

2. ξ1+σ2=±ε;

3. ξ1+σ+σ2+σ3=εχ1+σ2=ε1+σ;

4. Nℚ(ℓ)/ℚ(ε)=NL/ℚ(ℓ)(εχ)=NL/ℚ(ξ)=−1.

Lemma 2.

With the same notation of Theorem 2 , {ξ,ξσ,ξσ2} is also a fundamental system of units of F=L(−n) where n is a positive integer prime to ℓ and squarefree.

Proof. If L(−n)≠L(−1), then, according to [1, Proposition 3], to show that {ξ,ξσ,ξσ2} is a fundamental system of units of F it suffices to show that nμ is not a square in L, for μ=ξ1′j1ξ2′j2ξ3′ where {ξ1′,ξ2′,ξ3′}={ξ,ξσ,ξσ2} and j1,j2∈ {0,1}.

Indeed, if μ=ξ, then if nξ=x2 in L, by calculating the norm in L/k, we find that ξ1+σ2=±ε is a square in k, which is impossible.

If μ=ξσ, then if nξσ=±nεχξ=x2 in L, by calculating the norm in L/k, we find that ±ε1+σε=±1ε is a square in k, which is absurd.

If μ=ξσ2, then if nξσ2=±nεξ=x2 in L, by calculating the norm in L/k, we find that ±ε2ε=±ε is a square in k, which is not the case.

If μ=ξ1+σ=±εχ, then if ±nεχ=x2 in L, by calculating the norm in L/k, we find that εχ1+σ2=ε1+σ=−1 is a square in k, which is absurd.

If μ=ξ1+σ2=±ε, then ±nε cannot be a square in L.

If μ=ξσ+σ2, then if nξσ+σ2=x2 in L, by calculating the norm in L/k, we find that ξ1+σ+σ2+σ3=−1 is a square in k, which is not the case.

If μ=ξ1+σ+σ2, then if nξ1+σ+σ2=x2 in L, by calculating the norm in L/k, we find that ξ1+σ+σ2+σ3ξ1+σ2=±ε is a square in k, which is impossible.

If F=L(−1), then we have m=2 is the largest integer such that ζm∈F, so, from [1, Proposition 2], to show that {ξ,ξσ,ξσ2} is a fundamental system of units of F it suffices to show that 2μ is not a square in L, for μ=ξ1′j1ξ2′j2ξ3′ where {ξ1′,ξ2′,ξ3′}={ξ,ξσ,ξσ2} and j1,j2∈{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(−2εℓ)

Let k=ℚ(−2εℓ) where ℓ is a prime number such that ℓ≡1(mod  8) and (2ℓ)4=−1, according to [2], Ck,2 is isomorphic to ℤ/2ℤ×ℤ/2ℤ, thus k(1)/k has three intermediate subfields, F1,   F2,   F3 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(∗)=F3=k(−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 (2ℓ)=1, then there exist two prime ideals

B1=〈2,a+bℓ2〉,B2=〈2,a−bℓ2〉 of  ℚ(ℓ) such as (2)=B1B2 in ℚ(ℓ)

with a and b are integers, but B1h(ℓ) and B2h(ℓ) are principal ideals of ℚ(ℓ) where h(ℓ) is the class number of ℚ(ℓ), so we can choose c and d positive integers such as

B1h(ℓ)=(α) and B2h(ℓ)=(α¯) in ℚ(ℓ)

with α=c+dℓ2 and α¯=c−dℓ2 are positive. Let H1 and H2 be prime ideals of k above B1 and B2 respectively and H=H1H2.

Proposition 1.

Let k=ℚ(−2εℓ) where ℓ is a prime number such that ℓ≡1(mod  8) and (2ℓ)4=−1 . Then the class [H] is of order 2 in k . Moreover H capitulates in k(−2).

Proof. The class of H is of order 2, indeed, we have H2=(2) in k. Suppose that H=(λ) for some λ in k, which is equivalent to (λ2)=(2) in k, therefore, there exists a unit η of k such that 2η=λ2, although there exist x and y in ℚ(ℓ) such that λ=x+y−2εℓ, thus

2η=x2−2εℓy2+2xy−2εℓ.

Since ε is a fundamental unit of k and i=−1∉k, then 2η∈ℚ(ℓ), therefore x or y is equal to 0. If y=0, then 2η=x2, so 2=x′2 or 2ε=x″2 in ℚ(ℓ), which gives that 2∈ ℚ(ℓ) in the first case and −1∈  ℚ(ℓ) in the second case, which is impossible. Similarly, if x=0 we find that ±ℓ is a square in ℚ, consequently [H] is of order 2.

To show that H is capitulated in F3=k(−2), it suffices to remark that −2 is in F3 and (−22)=(2) in F3, thus H capitulates in F3□.

Theorem 3.

Let k=ℚ(−2εℓ) where ℓ is a prime number such that ℓ≡1(mod  8) and (2ℓ)4=−1 and F3=k(−2). Then CF3,2, the 2-part of the class group of F3, is cyclic of order h2(F3)=4.

Proof. Let α, α¯, B1, B2, H1, H2 and H=H1H2 defined as above and let L=ℚ(εℓ), then we have kL=F3, Nk/ℚ(ℓ)(H1)=B1 and B1 is unramified in L/ℚ(ℓ). So, to prove that H1 is inert in F3/k it suffices to show that B1 is inert in L/ℚ(ℓ) (Translation Theorem), and for this, we calculate the following norm residue symbol (α,εℓB1). According to [2], we have

(α,εℓB1)=(2ℓ)4=−1,

so B1 is inert in L/ℚ(ℓ), which gives H1 remains inert in F3/k and in the same way we have shown that H2 remains inert in F3/k and since H=H1H2 capitulates in F3, then by the Artin reciprocity law, we find that F3/K is of type (A), therefore, according to [8], we find that CF3,2 is cyclic.

Since F3/ℚ(ℓ) is a normal biquadratic extension with Galois group of type (2,2) and where k,L,k0=ℚ(ℓ,−2) are these intermediate subfields, then, by [9], we have

h2(F3)=22−1−2−0q(F3/ℚ(ℓ))h2(k)h2(L)h2(k0)h2(ℓ)2.

But h2(ℓ)=1, h2(k)=4, h2(L)=1 (see [12]), and by [7,11] we have h2(k0)=2, so h2(F3)=4q(F3/ℚ(ℓ)) and we have ε is a fundamental unit of k and since 2ε is not a square in ℚ(ℓ), then, by [1, Proposition 3], {ε} is also a fundamental system of units of k0 and according to Lemma 2, L and F3 have the same fundamental system of units which is {ξ,ξσ,ξσ2}, thus q(F3/ℚ(ℓ)), the unit index of F3/ℚ(ℓ), is equal to 1, which gives that h2(F3)=4. □

Remark 2.

Let k=ℚ(−2εℓ) where ℓ is a prime number such that ℓ≡1  (mod  8) and (2ℓ)4=−1 and let G be the Galois group of k(2)/k, then G is of order 8.

Indeed, we have k(1)/F3 is an unramified extension and the 2-class group of F3 is cyclic, thus F3 and k(1) have the same Hilbert 2-class field, namely k(2), so # G=2h2(F3)=8.

Theorem 4.

Let k=ℚ(−2εℓ) where ℓ is a prime number such that ℓ≡1(mod  8) and (2ℓ)4=−1 , F1, F2, F3 the quadratic subfields of k(1)/k . Then, in each extension Fi/k , i∈{1,2,3} , there exist exactly two classes of Ck,2 which capitulates and G is the quaternion group of order 8.

Proof. According to Lemma 2, we have {ξ,ξσ,ξσ2} is a fundamental system of units of F3. Since

NF3/k(ξ)=NF3/k(ξσ2)=±ε and NF3/k(ξσ)=±ε−1,

then NF3/k(EF3)=Ek, consequently, by [5], we have only two classes of Ck,2 which capitulates in F3, namely [H] and its square and since the extension F3/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≃Q3. □

Remark 3.

As G≃Q3, then, by [8],

CF1,2≃CF2,2≃CF3,2≃ℤ/4ℤ.

Example 1.

Let k=ℚ(−2(17+417)), we have 17≡1(mod  8) and (217)4=−1, then the Galois group of k(2)/k is the quaternion group of order 8, there exist exactly two classes of Ck,2 which capitulates in F3=k(−2), as well for F1 and for F2 and the Hilbert 2-class field tower of k stops at k(2).