SAID ESSAHEL,ALI MOUHIB
Sciences and Engineering Laboratory
Sidi Mohammed Ben Abdellah University, Fes
Polydisciplinary Faculty of Taza
MOROCCO
Abstract: Let dbe a square-free integer <0. In this paper, we will determine all imaginary quadratic
number fields Q(√d)that have a metacyclic Hilbert 2-class field tower. Finally, we will numerically
validate our theoretical results.
Key-Words: Keywords: class field tower; class group; imaginary quadratic number field; metacyclic
group.
Received: June 22, 2021. Revised: March 25, 2022. Accepted: April 23, 2022. Published: May 26, 2022.
1 Introduction
Let Kbe a number field. The maximal unramified
abelian 2-extension K(1)
2of K, is called the Hilbert
2-class field of K. We recall that by the class field
theory we have Gal(K(1)
2/K) = Cl2(K), the 2-
Sylow subgroup of the class group of Kdenoted
Cl(K).Cl2(K)is called the 2-class group of K.
For a nonnegative integer n, let K(n)
2be defined
inductively as K(0)
2=Kand K(n+1)
2=K(n)
2(1)
2;
then
K⊂K(1)
2⊂K(2)
2⊂... ⊂K(n)
2⊂...
is called the Hilbert 2-class field tower of K. If n
is the minimal integer such that K(n)
2=K(n+1)
2,
then this tower is called to be finite of length n.
If there is no such n, then the tower is called to
be infinite. We denote K(∞)
2=∪
i∈NK(i)
2. We recall
that K(∞)
2/Kis a Galois extension and the tower
of Kis finite iff K(∞)
2/Kis of finite degree.
The finiteness of the Hilbert 2-class field tower
of an imaginary quadratic number field Kis still
a problem of uncontrollable behavior for some
values of rank(Cl2(K)). It’s well known, that if
rank(Cl2(K)) ≥5, then, the tower is infinite [3].
For the case where rank(Cl2(K)) = 4, there is
no known imaginary quadratic field with finite
tower, and according to Martinet’s conjecture
the tower is infinite [6]. If rank(Cl2(K)) = 2
or 3, the tower may be finite or infinite ([5],
[6]), and there is no known procedure for decid-
ing if the tower is finite or not. Let p1= 73,
p2= 373. The class number of Q(√p1.p2)is 16.
Then, according to [7, Proposition 3.3], the field
K=Q(√−p1.p2.p)has infinite Hilbert 2- class
field tower for all prime psatisfying the conditions
p≡ −1mod (4) and 73.373
p=−1. This give
an infinite family of imaginary quadratic fields
Kwith rank(Cl2(K)) = 2 and infinite tower.
In this paper we give, in theorems 1, 2 and 3,
infinite family of imaginary quadratic number
fields Khaving finite Hilbert 2-class field tower
and satisfying rank(Cl2(K)) = 2. More precisely,
we give the list of all imaginary quadratic number
fields that have a metacyclic Hilbert 2-class field
tower.
Note that a group Gis said to be metacyclic if
there is a normal subgroup Nof Gsuch that
Nand G/Nare cyclic. For such a group, if
we denote N=< a > and G/N=< bN >,
then G=< a, b > and, thus, Gis generated
by 2 elements. Let Kbe a number field and
denote G2=Gal(K(∞)
2/K). The Hilbert 2-class
field tower of Kis said to be metacyclic if G2
is metacyclic. Note that in this case, the tower
terminate at most at the second steep.
2 Notations and useful results
2.1 Notations
• Let kbe a number field.
–k(1) denote the Hilbert class field of k
which is the maximal unramified abelian
extension of k.
–Okis the ring of integers of k.
–Ekis the unit group of Ok.
• Let K/kbe an extension of number fields.
–ram(K/k)is the number of all places of
kthat ramify in K.
On the Metacyclicity of the Hilbert 2-Class Field Tower of Imaginary
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– If K/kis a quadratic extension, then
e(K/k)is the rank of the elementary
2-group Ek/Ek∩NK/k(K×)where
K×={α∈K/α= 0}.
Note that an elementary 2-group Gcan
be seen as a F2-vector space and its di-
mension is said the 2-rank of Gwhich we
denote rank2(G).
– Let β∈kand pis a place of k.β,K/k
p
is the norm residue symbol of βin krela-
tively to p, we may denote it also β,K
p
or β,α
pif K=k(√α)is a quadratic
extension of k.
• If mis a square-free positive integer, then εm
denotes the fundamental unit of Q(√m).
2.2 Preliminary results
Lemma 1. Let K/kbe a quadratic extension of
number fields. We have
rank2(Cl(K)) ≥ram(K/k)−1−e(K/k).
If the 2-class group of kis trivial, then the preced-
ing inequality becomes an equality
Proof. For the inequality, see [4] and for the equal-
ity, see [1].
Lemma 2. Let Kbe a number field. If G2=
Gal(K(∞)
2/K)is metacyclic nonabelian, then
rank2(Cl(K)) = 2.
Proof. see [2]
Remarks. Let Kbe a quadratic number field.
• If G2=Gal(K(∞)
2/K)is metacyclic, then
rank2(Cl(K)) ≤2. If rank2(Cl(K)) =
1, then K(1)
2=K(2)
2, and the Hilbert 2-
class field tower of Kis cyclic. We deduce
that the important case to study is when
rank2(Cl(K)) = 2.
• If G2=Gal(K(∞)
2/K)is metacyclic non-
abelian, then rank2(Cl(K)) = 2, and Khas
three quadratic extensions L1, L2and L3con-
tained in K(1).
Lemma 3. Let Kbe a number field such that
rank2(Cl(K)) = 2 and denote L1, L2and L3the
three quadratic extensions of Kcontained in K(1).
If we denote G=Gal(K(∞)
2/K)and Ci=Cl2(Li)
for i= 1,2,3, Then Gis metacyclic if and only if
rank(Ci)≤2for i= 1,2,3.
Proof. see [2]
Lemma 4. Let m≤ −2and d≥2be two integers,
k=Q(√m)and K=k(√d). Suppose that 2
splits in kand ramifies in Q(√d). Then if Pis a
prime ideal of kthat divides 2, then
−1,K/k
P=−1
dif dis odd and
−1,K/k
P=−1
cif d= 2c.
Proof. Let Gal(K/k) =< σ >. We have 2Ok=
PP with P=σ(P). We have
−1,K/k
P=−1,d
P
=d,−1
P
The conductor of the extension Q(i)/Qis 4Zp∞
where p∞is the infinite prime of Q, then 4Ok=
P2P2is an admissible modulus for the extension
k(i)/k.
• Suppose that d≡ −1mod(4). Thus
d, −1
P=−1,−1
P.
Let b∈ksuch that b≡ −1mod(P2)and
b≡1mod(P2). Then
−1,−1
P=k(i)/k
bOk
=Q(i)/Q
Nk/Q(b)
=−1
Nk/Q(b)
We have b≡1mod(P2)then σ(b)≡
1mod(P2). We deduce that Nk/Q(b)≡
−1mod(4). Then
d,−1
P=−1,−1
P
=−1
Nk/Q(b)
=−1
=−1
d
If d≡1mod(4), Then d≡1mod(P2), and,
then
d,−1
P=1,−1
P= 1 = −1
d
• Suppose that d= 2cwhere c∈N∗. We have
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d,−1
P=2c,−1
P
=2,−1
Pc,−1
P
=−1,2
Pc,−1
P
=c,−1
P
=−1
c
3 Imaginary quadratic fields with
metacyclic Hilbert 2-class field
tower
Let d∈N∗be a positive square-free integer and
K=Q(√−d)such that rank2(Cl(K)) = 2. Ac-
cording to the genus theory, dwill be of one of the
following forms:
p1p2,2p1p2,2p1q1,p1p2q1,2q1q2,q1q2,q1q2q3,
where p1and p2are positive prime integers ≡1
mod(4) and q1, q2and q3are positive prime inte-
gers ≡3mod(4).
In what follows, we study all these cases in or-
der to see when the Hilbert 2-class field tower of
Q(√−d)is metacyclic.
Theorem 1. Let K=Q(√−d)with d= 2q1q2,
q1q2or q1q2q3. Then the Hilbert 2-class field tower
of Kis metacyclic.
Proof. The three quadratic extensions of Kcon-
tained in K(1) are L1=K(√−q1),L2=
K(√−q2)and L3=K(√q1q2).
Suppose that d=q1q2. Then L1=k1(√q2),
L2=k2(√q1)and L3=k3(√q1q2)with k1=
Q(√−q1),k2=Q(√−q2)and k3=Q(i). Since
Cl2(k1)is trivial, then by lemma 1, we have
rank2(Cl(L1)) = ram(L1/k1)−1−e(L1/k1).
We have rank2(Cl(L1)) ≤3and
rank2(Cl(L1)) = 3 iff 2and q2splits in
k1and e(L/k1)=0. But these conditions
cannot be satisfied simultaneously. In fact,
We have Ek1=<−1>if q1= 3 and
Ek1=< ζ6>=<−1, ζ6>if not. Then if
q2splits in k1and Qis a prime ideal of k1
dividing q2, then
−1,L1/k1
Q=−1,q2
Q
=q2,−1
Q
=−1
q2
=−1
Thus e(L1/k1)=1. We conclude that
rank2(Cl(L1)) ≤2. In the same way,
rank2(Cl(L2)) ≤2. We have ram(L3/k3)=2
then rank2(Cl(L3)) ≤2. Then Khas a meta-
cyclic Hilbert 2-class field tower.
Suppose that d=q1q2q3. Then L1=
k1(√q2q3),L2=k2(√q1q3)and L3=k3(√q1q2)
with k1=Q(√−q1),k2=Q(√−q2)and
k3=Q(√−q3). We have rank2(Cl(L1)) =
ram(L1/k1)−1−e(L1/k1). If q2is inert in k1,
then ram(L1/k1)≤3and thus
rank2(Cl(L1)) ≤2.
If q2splits in k1and Qis a prime ideal of k1di-
viding q2, then
−1,L1/k1
Q=−1,q2q3
Q
=q2,−1
Q
=−1
q2
=−1
Thus e(L1/k1)=1, and rank2(Cl(L1)) ≤
2. We conclude that in all cases we have
rank2(Cl(L1)) ≤2. In the same way, we have
rank2(Cl(Lj)) ≤2for j= 2,3. Then Khas a
metacyclic Hilbert 2-class field tower.
Suppose that d= 2q1q2. Then L1=k1(√2q2),
L2=k2(√2q1)and L3=k3(√−2) with k1=
Q(√−q1),k2=Q(√−q2)and k3=Q(√q1q2).
We have rank2(Cl(L1)) = ram(L1/k1)−1−
e(L1/k1). If 2is inert in k1, then ram(L1/k1)≤3
and thus rank2(Cl(L1)) ≤2.If 2splits in k1
and Qis a prime ideal of k1dividing 2, then, by
lemma 4,
−1,L1/k1
Q=−1
q2=−1.
Thus e(L1/k1)=1, and rank2(Cl(L1)) ≤2. We
conclude that in all cases, we have
rank2(Cl(L1)) ≤2.
In the same way, rank2(Cl(L2)) ≤2.
We have ram(L3/k3)≤4and if Pis an in-
finite place of k3then −1,L3/k3
P=−1, thus
e(L3/k3)≥1. We conclude that
rank2(Cl(L3)) ≤2.
Thus, using lemma 3, Khas a metacyclic Hilbert
2-class field tower.
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Theorem 2. Let K=Q(√−d)with d=p1p2or
d= 2p1p2. Then the Hilbert 2-class field tower
of Kis metacyclic iff, after a permutation of the
pi’s, we have one of the two following conditions:
(C1) p1≡p2≡5mod(8)
(C2) p1≡1mod(8),p2≡5mod(8) and p2
p1=−1
Proof. The three unramified quadratic extensions
of Kare L1=K(√p1),L2=K(√p2)and L3=
K(√p1p2).
• Suppose that d=p1p2. Then L1=
k1(√−p2),L2=k2(√−p1)and L3=
k3(√p1p2)with k1=Q(√p1),k2=Q(√p2)
and k3=Q(i). We have
rank2(Cl(L1)) = ram(L1/k1)−1−e(L1/k1)
and
Ek1=<−1, εp1>
Let P∞be one of the two infinite places of k1.
We have
−1,L1/k1
P∞=−εp1,L1/k1
P∞=−1
On the other hand, εp1can not be a norm
in L1/k1. In fact, if there is an element
v∈L1such that NL1/k1(v) = εp1, then
we will have NL1/Q(v) = Nk1/Q(εp1) = −1.
This is impossible since −1,L1/Q
P∞=−1.
We deduce that e(L1/k1)=2, and then
rank2(Cl(L1)) = ram(L1/k1)−3. The places
of k1that ramify in L1are exactly the 2 in-
finite places, the place(s) above 2 and the
place(s) above p2. Thus ram(L1/k1)=4,5
or 6and rank2(Cl(L1)) = 1,2or 3. In ad-
dition, rank2(Cl(L1)) = 3 iff p1≡1mod(8)
and p2
p1= 1.
In the same way, we have rank2(Cl(L2)) ≤3
and rank2(Cl(L2)) = 3 iff p2≡1mod(8) and
p2
p1= 1.
For the 2-rank of L3, we have ram(L3/k3) =
4. Then rank2(Cl(L3)) = 3 −e(L3/K3). Let
Pbe a prime ideal of k3dividing p1, for ex-
ample. We have
i,L3/k3
P=i,p1p2
P
=p1,i
P
=k3(ζ8)/k3
P
=Q(ζ8)/Q
p1
Thus e(L1/k1) = 0 iff p1≡p2≡1mod(8).
We conclude that rank2(Cl(L3)) ≤3and
rank2(Cl(L3)) = 3 iff p1≡p2≡1mod(8).
• Suppose that d= 2p1p2. We have L1=
k1(√−2p2),L2=k2(√−2p1)and L3=
k3(√p1p2)with k1=Q(√p1),k2=Q(√p2)
and k3=Q(√−2).
As in the previous case, We have, for j= 1,2,
e(Lj/kj)=2, and then rank2(Cl(Lj)) =
ram(Lj/kj)−3≤3. In addition,
rank2(Cl(Lj)) = 3 iff pj≡1mod(8) and
p2
p1= 1.
For the 2-rank of L3, we have ram(L3/k3)≤
4. Then rank2(Cl(L3)) ≤3with equality iff
p1≡p2≡1mod(8) and e(L1/k1)=0. If
p1≡p2≡1mod(8) and Pis a prime ideal of
k3dividing p1, for example, then We have
−1,L3/k3
P=−1,p1p2
P
=p1,−1
P
=−1
p1
= 1
Thus e(L1/k1) = 0. We conclude that
rank2(Cl(L3)) ≤3and rank2(Cl(L3)) = 3
iff p1≡p2≡1mod(8).
The above two cases can be summarized by:
The tower of kis not metacyclic iff p1≡
1mod(8) and p1
p1= 1or p2≡1mod(8) and
p2
p1= 1or p1≡p2≡1mod(8). The theorem
will be proved by discussing according first to (p1
mod (8), p2mod (8)) and then p2
p1.
Theorem 3. Let K=Q(√−d)with d=r1p2q1
and r1= 2 or r1=p1is a positive prime integer
≡1mod(4). Then the Hilbert 2-class field tower
of Kis metacyclic except in the following cases:
(C1) r1
p2=r1
q1= 1
(C2) r1
p2=q1
p2= 1
(C3) r1
q1=p2
q1= 1
Proof. The three quadratic extensions of Kcon-
tained in K(1) are L1=K(√r1),L2=K(√p2)
and L3=K(√r1p2).
• Suppose that d=p1p2q1. Then L1=
k1(√−p2q1),L2=k2(√−p1q1)and L3=
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k3(√p1p2)with k1=Q(√p1),k2=
Q(√p2)and k3=Q(√−q1). We have
rank2(Cl(Lj)) = ram(Lj/kj)−1−e(Lj/kj),
for j= 1,2and 3. As in the cases in the
previous theorem, we have e(L1/k1) = 2, and
then rank2(Cl(L1)) = ram(L1/k1)−3. Since
ram(L1/k1)≤6, we have rank2(Cl(L1)) ≤3
and rank2(Cl(L1)) = 3 iff p1
p2=p1
q1= 1.
In the same way, we have rank2(Cl(L2)) ≤3
and rank2(Cl(L2)) = 3 iff p2
p1=p2
q1= 1.
If q1
p1,q1
p2= (1,1), then ram(L3/k3)≤
3and thus rank2(Cl(L3)) ≤2.
Suppose that q1
p1=q1
p2= 1. Then
ram(L3/k3) = 4. let Pbe a prime ideal of
k3lying over pjwith j= 1 or 2. We have
−1,L3/k3
P=−1,p1p2
P
=pj,−1
P
=−1
pj
= 1
If q1= 3, then e(L3/k3)=0. If q= 3, then
Ek3=< ζ6>and
ζ6,L3/k3
P=ζ3
6,L3/k3
P
=−1,L3/k3
P
=−1,p1p2
P
=pj,−1
P
=−1
pj
= 1
Then e(L3/k3) = 0. In the two cases, we
have rank2(Cl(L3)) = 3. We conclude that
rank2(Cl(L3)) = 3 iff q1
p1=q1
p2= 1.
• Suppose that d= 2p2q1. Then L1=
k1(√−p2q1),L2=k2(√−2q1)and L3=
k3(√2p2)with k1=Q(√2),k2=Q(√p2)and
k3=Q(√−q1). We have rank2(Cl(L2)) =
ram(L2/k2)−1−e(L2/k2). Since e(L2/k2) =
2,rank2(Cl(L2)) = ram(L2/k2)−3≤
3. In addition rank2(Cl(L2)) = 3 iff
2
p2=q1
p2= 1. Similarly, we have
rank2(Cl(L1)) ≤3and rank2(Cl(L1)) = 3
iff 2
p2=2
q1= 1.
We have rank2(Cl(L3)) ≤3with equality iff
ram(L3/k3) = 4 and e(L3/k3) = 0.
Suppose that ram(L3/k3)=4(i.e p2
q1=
2
q1= 1). Let Pbe a prime ideal of k3lying
over p2. We have
−1,L3/k3
P=−1,2p2
P
=p2,−1
P
= 1
Let Qbe a prime ideal of k3lying over 2. By
lemma 4, we have
−1,L3/k3
Q=−1
p2= 1
Then e(L3/k3) = 0. We deduce that
rank2(Cl(L3)) ≤3and rank2(Cl(L2)) = 3
iff p2
q1=2
q1= 1.
4 Numerical verification
Our results can be checked. Indeed let kbe a
quadratic number field verifying rank(Cl2(k)) =
2. Using a computer algebra system, we can
compute the rank of the 2-class group of its
three quadratic unramified extensions. Then, by
lemma 3, we can decide if the tower of the Hilbert
2-class field of kis metacyclic or not. We will
do that, as an example, for the case where k=
Q(√−p1p2).
Let p1and p2be two positive prime integers
≡1mod (4) and k=Q(√−p1p2). The three
quadratic unramified extensions of kare L1=
Q(√p1,√−p2),L2=Q(√p2,√−p1)and L3=
Q(i, √p1p2). We denote rj=rank(Cl2(Lj)) for
j= 1,2,3.
Let B1and B2be respectively the truth values
(0 if false and 1 if true) of the proposition ” The
Hilbert 2-class field tower of kis metacyclic” and
the statement ” After a permutation of p1and
p2, we have (C1)or (C2), the two conditions in
theorem 2”. In the following table, We compute
B1and B2. By comparing them, we can see that
they are equivalent. This is in agreement with
Theorem 2.
The numerical results in this table were obtained
using Pari [8].
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p1p2r1r2r3B1p1mod (8) p2mod (8) p1
p2B2
5 13 1 1 1 1 5 5 -1 1
5 17 1 2 2 1 5 1 -1 1
5 29 2 2 2 1 5 5 1 1
5 241 2 3 2 0 5 1 1 0
13 17 2 3 2 0 5 1 1 0
13 29 2 2 2 1 5 5 1 1
13 149 1 1 2 1 5 5 -1 1
13 233 2 3 2 0 5 1 1 0
13 241 1 2 2 1 5 1 -1 1
17 41 2 2 3 0 1 1 -1 0
17 53 3 2 2 0 1 5 1 0
17 89 3 3 3 0 1 1 1 0
17 109 2 1 2 1 1 5 -1 1
17 149 3 2 2 0 1 5 1 0
29 37 1 1 2 1 5 5 -1 1
29 89 1 2 2 1 5 1 -1 1
29 149 2 2 2 1 5 5 1 1
29 233 2 3 2 0 5 1 1 0
37 41 2 3 2 0 5 1 1 0
37 53 2 2 2 1 5 5 1 1
37 149 2 2 2 1 5 5 1 1
37 193 1 2 2 1 5 1 -1 1
41 53 2 1 2 1 1 5 -1 1
41 61 3 2 2 0 1 5 1 0
41 89 2 2 3 0 1 1 -1 0
41 113 3 3 3 0 1 1 1 0
53 61 1 1 2 1 5 5 -1 1
53 73 1 2 2 1 5 1 -1 1
53 113 2 3 2 0 5 1 1 0
61 73 2 3 2 0 5 1 1 0
61 89 1 2 2 1 5 1 -1 1
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