Journal of Number Theory 145 (2014) 210–229

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Journal of Number Theory

www.elsevier.com/locate/jnt

About the embedding of a number field in a Pólya

field

Amandine Leriche a,b,∗

a LAMFA CNRS UMR 7352, UFR des Sciences, 33, rue Saint-Leu,80039 Amiens Cedex 1, Franceb École Centrale de Lille, Cité Scientifique, 59651 Villeneuve d’Ascq, France

a r t i c l e i n f o a b s t r a c t

Article history:Received 22 March 2014Received in revised form 23 May 2014Accepted 23 May 2014Available online 2 July 2014Communicated by David Goss

Keywords:Integer-valued polynomialsHilbert class fieldGenus fieldPólya field

One knows the classical problem of the embedding of a number field K in a field with class number one. This problem has a negative answer. In this article, we consider a new embedding problem: Is every number field contained in a Pólya field? A Pólya field is a number field K such that all the characteristic ideals In(K) are principal. We give a positive answer to this problem: the Hilbert class field HK of K is a Pólya field. However, HK is not necessarily the smallest Pólya field containing K. Thus, we give upper bounds for the Pólya number of K, namely the minimal degree of a Pólya field containing K.

© 2014 Elsevier Inc. All rights reserved.

1. Introduction

Recall the classical embedding problem: Is every number field K contained in a field L with class number one? In 1964, Golod and Shafarevich [10] gave a negative answer to this question. Here, we are interested in an analogous question where the number field

* Correspondence to: LAMFA CNRS UMR 7352, UFR des Sciences, 33, rue Saint-Leu, 80039 Amiens Cedex 1, France.

E-mail address: amandine.leriche@ec-lille.fr.

http://dx.doi.org/10.1016/j.jnt.2014.05.0020022-314X/© 2014 Elsevier Inc. All rights reserved.

A. Leriche / Journal of Number Theory 145 (2014) 210–229 211

L has a weaker property: Is every number field K contained in a number field L which is a Pólya field? Let us recall the definition of a Pólya field.

For every number field K with ring of integers OK , Pólya [17] and Ostrowski [16]introduced the set formed by the integer-valued polynomials on OK , that is,

Int(OK) ={P ∈ K[X]

∣∣P (OK) ⊆ OK

}.

They try to characterize the fields that, following Zantema [20], we now call Pólya fields:

Definition 1.1. (See [20].) A number field K is said to be a Pólya field if the OK -module Int(OK) admits a regular basis, that is, a basis (fn)n∈N such that, for each n, the polynomial fn has degree n.

A useful tool to characterize Pólya fields is the following:

Definition 1.2. (See [5, Prop. I.3.1].) For each n ∈ N, let In(K) be the subset of Kformed by 0 and the leading coefficients of the polynomials in Int(OK) with degree n. This subset is a fractional ideal of OK called the characteristic ideal of index n of K.

It is well known that K is a Pólya field if and only if all the characteristic ideals In(K)are principal [5, II.1.4]. In particular, if OK is a principal ideal domain, K is a Pólya field but the converse is false. For instance, Q(

√−23 ) is a Pólya field (see Proposition 6.3)

and its class number is 3. Thus, the hypothesis “K is a Pólya field” is strictly weaker than “K has a class number one”.

Hence, it is natural to ask the following question:

Is every number field contained in a Pólya field?

The counterexample given by Golod and Schafarevich used for the classical embedding problem, that is, the quadratic field Q(

√−2 · 3 · 5 · 7 · 11 · 13 ), is not a counterexample

for this new embedding problem. Indeed, each quadratic field may be embedded in a cyclotomic field which is always a Pólya field (see Proposition 6.2).

In the next section, we recall the notion of Pólya extension which will be useful to study our embedding problem analogously to the Hilbert class field with respect to the classical embedding problem. In Section 3, we will give a positive answer to our “new embedding problem”: the Hilbert class field HK of every number field K is a Pólya field (Theorem 3.3). It is not, however, generally the smallest Pólya field containing K. For instance, the genus field ΓK of an Abelian number field K is also a Pólya field (Theorem 3.8). Although we have a positive answer to our new embedding problem, we construct in Section 4 an infinite tower of number fields which are not Pólya fields, but such that each subextension is a Pólya extension (Proposition 4.5). Since HK is not always the smallest Pólya field containing K, we try to give in Section 6 upper

212 A. Leriche / Journal of Number Theory 145 (2014) 210–229

bounds for the minimal degree poK of a Pólya field containing K, that we call the Pólya number of K (Proposition 6.6 and Corollary 6.10). In order to do this, we first study in Section 5 the behavior of Pólya groups in linearly disjoint extensions (Theo-rem 5.6).

2. Pólya fields and Pólya extensions

We already studied the question of the embedding in a Pólya field in [14] using the notion of a Pólya extension introduced in [6]. We first recall some of these results.

Definition 2.1. (See [5].) The Pólya group of K is the subgroup Po(K) of the class group Cl(K) of K generated by the classes of the characteristic ideals In(K) of K.

Notation. For each q ≥ 2, let Πq(K) be the product of all the maximal ideals of OK

with norm q:

Πq(K) =∏

m∈Max(OK)N(m)=q

m.

If q is not the norm of an ideal, then Πq(K) = OK .

In fact, we know [7, Proposition 3.1] that Po(K) is also the subgroup of Cl(K) gen-erated by the classes of the ideals Πq(K). Pólya fields are then characterized in several ways. The following proposition is essentially due to Pólya [17] and Ostrowski [16].

Proposition 2.2. The field K is a Pólya field if and only if one of the following assertions is satisfied:

(1) Int(OK) has a regular basis;(2) for each n ∈ N, the ideal In(K) is principal;(3) for each q ≥ 2, Πq(K) is principal;(4) Po(K) = {1}.

We may give an answer to the embedding problem in a Pólya field by introducing the notion of a Pólya extension. More precisely, the new embedding problem is equivalent to the following one:

Is there a field L containing K such that all the ideals In(L) are principal?

Consider the Hilbert class field HK of a number field K, that is the maximal Abelian extension of K which is unramified at all places. Since all the ideals of OK become

A. Leriche / Journal of Number Theory 145 (2014) 210–229 213

principal by extension to OHK(principal ideal theorem), in particular the ideals In(K)

become principal by extension to OHK. By analogy, we define the notion of Pólya exten-

sion:

Definition 2.3. An extension L/K is said to be a Pólya extension if all the characteristic ideals In(OK) extended to OL are principal.

Remark 2.4. Analogously to the fact that a Hilbert class field has not necessarily a class number equal to one, a Pólya extension is not necessarily a Pólya field. For instance, the extension Q[

√−5,

√2 ]/Q[

√−10 ] is a Pólya extension, but L = Q[

√−5,

√2 ] is not

a Pólya field (Π2(L) is not principal).

Example 2.5. For every number field K, HK/K is a Pólya extension.

This notion may be linked to the integer-valued polynomials in the following way:

Proposition 2.6. (See [14, Proposition 1.2].) An extension L/K is a Pólya extension if and only if the ring of integer-valued polynomials on OK with respect to OL, that is,

Int(OK ,OL) ={P ∈ L[X]

∣∣P (OK) ⊆ OL

}admits a regular basis as an OL-module.

3. An answer to the embedding problem in a Pólya field: the Hilbert class field and the genus field

When K is an Abelian field (that is, when the extension K/Q is Abelian), the answer to the embedding in a Pólya field is given by the Kronecker–Weber theorem: every Abelian number field is contained in a cyclotomic field, and we know with Zantema [20]that every cyclotomic field is a Pólya field.

When K is no more Abelian, what can we say?

3.1. The Hilbert class field

Recall first that, when the extension L/K is Galois, for every prime p of K, the gp = gp(L/K) primes of L lying over p have the same ramification index ep = ep(L/K)and the same residual degree fp = fp(L/K) in the extension L/K and we have

epfpgp = [L : K] and pOL =∏

M∈Max(OL)M|p

Mep . (1)

214 A. Leriche / Journal of Number Theory 145 (2014) 210–229

In the spirit of the capitulation theorem, we have the following result:

Proposition 3.1. Let L/K be a Galois extension of number fields such that every ideal of OK becomes principal by extension to OL. If all finite places are unramified in the extension L/K, then L is a Pólya field.

Proof. Let q be a power of a prime number and consider the ideal

Πq(L) =∏

m∈Max(OL)N(m)=q

m.

We have to prove that Πq(L) is principal. Denote by Mq the set formed by the primes of L with norm q and let {pj}1≤j≤s be the distinct primes of K lying under the elements of Mq, that is:

{p1, . . . , ps} = {m ∩K | m ∈ Mq}.

Since L/K is not ramified, for every j ∈ {1, . . . , s}, we have:

pjOL =∏

m∈Max(OL)m∩K=pj

m.

Since L/K is Galois, all the primes m ∈ Max(OL) such that m ∩K = pj have the same residual degree fpj

, and hence, the same norm N(pj)fpj which is q. Consequently,

(s∏

j=1pj

)OL =

∏m∈Mq

m = Πq(L).

Thus, Πq(L) is the extension to OL of an ideal of OK . It follows from the hypotheses that Πq(L) is principal. �Corollary 3.2. Let K be any number field. Then, the Hilbert class field HK of K is a Pólya field.

Consequently, the embedding problem in a Pólya field has a positive answer:

Theorem 3.3. Every number field is contained in a Pólya field, namely its Hilbert class field.

A. Leriche / Journal of Number Theory 145 (2014) 210–229 215

3.2. The genus field

It could exist Pólya fields containing K and strictly contained in HK . We will see that this may happen with the genus field of an Abelian field. If we do not assume that all ideals of OK become principal by extension to OL, but only the characteristic ideals In(K), in order to preserve the conclusion of Proposition 3.1, we have to strengthen the Galois hypothesis: both number fields K and L have to be Galois (that is, the extensions K/Q and L/Q are Galois).

Note that, when K is Galois, for every prime number p, formula (1) leads to:

pOK =∏p|p

pep = Πq(K)ep where q = pfp . (2)

Then, when considering Galois number fields, we have the following link between Pólya fields and Pólya extensions:

Proposition 3.4. Let L/K be a Pólya extension of Galois number fields. If all finite places are unramified in the extension L/K, then L is a Pólya field.

Proof. Let p be a prime number. Denote by f = fp(K/Q) (resp. F = fp(L/Q)) the residual degree of p in K/Q (resp. L/Q). Since the extension L/K is not ramified, we have the following equality:

Πpf (K)OL = ΠpF (L).

Moreover, since the extension L/K is a Pólya extension, Πpf (K)OL is a principal ideal of OL. Consequently, ΠpF (L) is principal. �

Recall that if K is Galois, an ambiguous ideal of K is an ideal which is invariant under the action of the Galois group Gal(K/Q), so that, Po(K) is the subgroup of Cl(K)generated by the classes of the ambiguous ideals of K. We are going to consider genus fields because we know that the ambiguous ideals of K become principal by extension to the genus field of K when K is Abelian. We first recall the definition.

Definition 3.5. Let K be a number field. The genus field (resp. the genus field in the narrow sense) of K is the maximal Abelian extension ΓK (resp. Γ ′

K) of Q which is the compositum of K with an Abelian number field and which is unramified over K at all places (resp. all finite places) of K.

Obviously, K ⊆ ΓK ⊆ Γ ′K and ΓK ⊆ HK .

Proposition 3.6. (See [9].) Let K be an Abelian number field. The extension of every ambiguous ideal of OK to the genus field ΓK of K is principal.

216 A. Leriche / Journal of Number Theory 145 (2014) 210–229

Corollary 3.7. If K is an Abelian number field, then the genus field ΓK of K is a Pólya extension of K.

As a corollary of Proposition 3.4, we obtain:

Theorem 3.8. If K is an Abelian number field, then the genus field ΓK (resp. Γ ′K , the

genus field in the narrow sense) of K is a Pólya field.

The previous proposition does not hold in general when K is not a Galois field as shown by the cubic field Q( 3

√20 ) (see Example 3.10 below). We first recall the form of

the genus field in the narrow sense of pure cubic fields.

Proposition 3.9. (See [11, IV, §4.2.6].) Let

L := Q(

3√

3n0pn11 · · · pnt

t

),

where t ≥ 0, n0 ∈ {0, 1}, ni ∈ {1, 2} for 1 ≤ i ≤ t, and the pi’s are distinct prime numbers different from 3. Then

[Γ ′L : L

]= 3γ

where γ is the number of pi congruent to 1 modulo 3.

Example 3.10. Let K = Q( 3√

20 ). The prime divisors of 20 are not congruent to 1 modulo 3. Thus, [Γ ′

K : K] = 1, that is, Γ ′K = ΓK = K. It follows from the decomposition

of the primes in pure cubic fields (see [8, Thm. 6.4.16]) that 2OK = P3 and P = Π2(K)is not a principal ideal. Consequently, K is not a Pólya field, and ΓK and Γ ′

K are neither Pólya fields, nor Pólya extensions of K.

The following example shows that Proposition 3.8 still does not always hold when K is Galois but not Abelian as shown by the sextic field Q( 3

√30, j) (see Example 3.13

below). But, once more, we first recall some results on the genus fields.

Proposition 3.11. (See [11, §4.2.9].) Let K = Q(√d ) be a quadratic field where d is a

squarefree integer. Write d in the form d := s2δ∏t

i=1 sipi where the pi’s are the odd

prime divisors of d, δ ∈ {0, 1}, si := (−1)pi−1

2 , s ∈ {−1, 1}. Then

Γ ′K = Q

(√s2δ,√s1p1, . . . ,

√stpt

)and ΓK is obtained by the following way:

(1) If d < 0, then ΓK = Γ ′K . If d > 0, ΓK = Γ ′

K if and only if s1 = s2 = . . . = st = 1.(2) In the other cases, ΓK is the maximal real subfield of Γ ′

K .

A. Leriche / Journal of Number Theory 145 (2014) 210–229 217

Proposition 3.12. (See [12, Chap. IV, Prop. 2].) If an algebraic number field K is the compositum of two subfields K1 and K2 such that ([K1 : Q], [K2 : Q]) = 1, then Γ ′

K =Γ ′K1

Γ ′K2

.

Example 3.13. Consider the Galois sextic field K = Q( 3√

20, j). Let K1 = Q[j] and K2 = Q( 3

√20 ). Following Proposition 3.11, Γ ′

K1= K1 and Γ ′

K2= K2 (see Example 3.10).

Thus, Γ ′K = K. In [13] (see also [15]), we studied Pólya fields which are the Galois closure

of pure cubic fields. Since Π2(K2) is not a principal ideal, it follows from [15, Lemma 6.4]that Π2(K) is not principal, and hence, Q( 3

√20, j) is not a Pólya field.

4. An infinite tower of Pólya extensions

We know that a positive answer to the classical embedding problem is equivalent to the finiteness of the Hilbert class field tower [18]. Since we defined the notion of Pólya extensions following the capitulation property of Hilbert class fields and since we answered positively the new embedding problem, we could think that a tower of Pólya extensions such that each one of the fields is not a Pólya field should be finite. In fact, we are going to construct such a tower which is infinite. In order to do this, we recall first two facts about Galois number fields:

Proposition 4.1. (See [14, Corollary 3.5].) Let K and L be two Galois number fields such that K ⊆ L. If L is a Pólya field, then L/K is a Pólya extension.

The example in Remark 2.4 shows that the converse does not hold in general.

Proposition 4.2. (See [20, Thm. 3.4].) Let K and K ′ be two Galois extensions of Q whose degrees are coprime and let L be the compositum of K and K ′. Then, L is a Pólya field if and only if K and K ′ are Pólya fields.

This last assertion will be proved in a more general setting in the next section (see Theorem 5.6).

Lemma 4.3. For every prime number p > 2, there is a cyclic field K with degree p which is not a Pólya field.

Proof. This is a consequence of [20, Proposition 3.2] which asserts that, if K/Q is an Abelian extension with degree pα (α ∈ N∗), the field K is Pólya field if and only if there is exactly one prime number which ramifies in K/Q. Indeed, it is well known that there exist cyclic fields of any prime degree having an arbitrary number of ramified primes (see [11, Chap. V, Cor. 2.4.5]). �Remark 4.4. Furuya’s Theorem 4.8 in [9] implies that, if K/Q is a cyclic extension with degree pα where α ≥ 1, g′K = [Γ ′

K : K] is necessarily of the form pβ where β ≥ 0 since it

218 A. Leriche / Journal of Number Theory 145 (2014) 210–229

is composite of cyclic extensions of Q whose degrees are powers of p. Moreover, if K is not a Pólya field then, by Theorem 3.8, β ≥ 1.

Proposition 4.5. One may construct an infinite tower of Galois number fields Q = L0 ⊂L1 ⊂ L2 ⊂ . . . ⊂ Ln ⊂ . . . such that for each integer i ≥ 1, Li/Li−1 is a Pólya extension and Li is never a Pólya field.

Proof. Let {pn}n≥1 be a sequence of distinct odd prime numbers. For each n ≥ 1, let Kn be a cyclic number field of degree pn which is not a Pólya field. By Remark 4.4, we have [Γ ′

Kn: Kn] = pβn

n where βn ≥ 1. The successive floors of our tower are

Q ⊂ K1 ⊂ Γ ′K1

K2 ⊂ Γ ′K1

Γ ′K2

K3 ⊂ Γ ′K1

Γ ′K2

Γ ′K3

K4 ⊂ . . .

Consider the following extensions:

�����

����������

�����

����������������������

Q

K1

p1K2

pβ11

Γ ′K1

K2

Γ ′K1

p2

The extensions Γ ′K1

/Q and K2/Q are Galois and their degrees are coprime. Since the field K2 is not a Pólya field, by Proposition 4.2, Γ ′

K1K2 is not a Pólya field. On the other

hand, Γ ′K1

K2/K1 is a Pólya extension because the subextension ΓK1/K1 is already a Pólya extension (Corollary 3.7).

Then, K3 is a cyclic field with degree p3 which is not a Pólya field. As Γ ′K1

and Γ ′K2

are Galois Pólya fields whose degree are coprime, Proposition 4.2 asserts that Γ ′K1

Γ ′K2

is a Galois Pólya field. The extensions Γ ′K1

Γ ′K2

/Q and K3/Q are Galois and their de-grees are coprime. Since K3 is not a Pólya field, it follows from Proposition 4.2 that Γ ′K1

Γ ′K2

K3 is not a Pólya field. Since the extensions Γ ′K1

Γ ′K2

/Q and Γ ′K1

K2/Q are Galois and Γ ′

K1Γ ′K2

is a Pólya field, Proposition 4.1 asserts that Γ ′K1

Γ ′K2

/Γ ′K1

K2 is a Pólya ex-tension. Hence, Γ ′

K1Γ ′K2

K3/Γ′K1

K2 is also a Pólya extension. We set L0 = Q, L1 = K1, L2 = Γ ′

K1K2, L3 = Γ ′

K1Γ ′K2

K3 and we pursue the construction so that, for i ≥ 2, Li = Γ ′

K1. . . Γ ′

Ki−1Ki. �

A. Leriche / Journal of Number Theory 145 (2014) 210–229 219

5. Compositum of linearly disjoint Galois extensions

In order to obtain bounds for the Pólya numbers in the next section, we need to study the behavior of Pólya groups in linearly disjoint Galois extensions. Such a study was undertaken first in [7], then in [14], but we need more general results, and specially Theorem 5.6 below.

Notation. Let L/K be a finite extension. Consider the norm morphism [19, Chap. I, §5]between I(L) and I(K), the groups of fractional ideals of L and K:

NKL : I(L) → I(K)

which is determined by its value on the maximal ideals N of OL

NKL (N ) = MfN (L/K)

where M = N ∩ OK and fN (L/K) = [OL/N : OK/M]. The morphism NKL induces a

morphism:

νKL : I ∈ Cl(L) → NKL (I) ∈ Cl(K).

On the other hand, the injective morphism defined by:

jLK : I ∈ I(K) → IOL ∈ I(L)

induces a morphism

εLK : I ∈ Cl(K) → IOL ∈ Cl(L).

Moreover, since the extension L/K is separable, for each ideal I of I(K), we have:

NKL ◦ jLK(I) = I [L:K].

The Pólya groups of Galois number fields behave nicely with respect to these mor-phisms:

Proposition 5.1. (See [7].) If K and L are two Galois extensions of Q such that K ⊆ L

then

εLK(Po(K)

)⊆ Po(L) and νKL

(Po(L)

)⊆ Po(K)

Now we study Pólya group in the following context:

Hypotheses for this section. Let K, K1 and K2 be Galois number fields such that K1 ∩K2 = K and denote by L = K1K2 the compositum of K1 and K2.

220 A. Leriche / Journal of Number Theory 145 (2014) 210–229

We obtained in [14] the following result:

Proposition 5.2. (See [14, Prop. 4.2].) If, for each prime ideal p of K, the ramification indices ep(K1/K) and ep(K1/K) are coprime, then:

εLK1

(Po(K1)

).εLK2

(Po(K2)

)= Po(L).

In particular, if K1 and K2 are Pólya fields, L is a Pólya field too.

We don’t know if, under the hypothesis of the previous proposition, the νLKi’s are onto

and if the εLKi’s are injective. However, one will obtain these properties under conditions

that are weaker than in [7], that is, ([K1 : Q], [K2 : Q]) = 1, but stronger than the previous one, that is, (ep(K1/Q), ep(K2/Q)) = 1 for each prime p of K. First, we recall a result from Ostrowski [16] about the Pólya group of a Galois number field.

Proposition 5.3. (See [16].) Let K be a finite Galois extension of Q. The group Po(K)is generated by the classes of the ideals Πq(K) where q = pf and the prime number p is ramified in K/Q.

Remark 5.4. Consequently, if K is a finite Galois extension of Q,

• the exponent of Po(K) divides [K : Q],• the order of Po(K) divides the product

∏p ep.

Indeed, pOK = Πpfp (K)ep , thus the order of the class of Πpfp (K) divides the ramification index ep which divides itself [K : Q].

Proposition 5.5. If, for each prime number p, (ep(K1/Q), [K2 : K]) = 1, then

(1) νK1L (Po(L)) = Po(K1),

(2) εLK1: Po(K1) → Po(L) is injective.

Proof. (1) Let ei = ep(Ki/Q), ni = [Ki : K], f1 = fp(K1/Q) and f = fp(L/Q). Let Π1 = Πpf1 (Ki) and Π = Πpf (L). We have:

pOK1 = Πe11 and Π1OL = Πe2

Let u and v be two integers such that ue1 + vn2 = 1. Then

NK1L (Π)ve2 = NK1

L

(Πe2

)v = NK1L (Π1OL)v

=(Πn2

1)v = Π1−e1u

1 = Π1 ×(Πe1

1)−u = p−uΠ1.

These equalities in Cl(K1) imply that Π1 ∈ νK1L (Po(L)).

A. Leriche / Journal of Number Theory 145 (2014) 210–229 221

(2) Since |Po(K1)| divides ∏

p ep(K1/Q) and, for each prime p, ep(K1/Q) and [K2 : K]are coprime, we have:

(∣∣Po(K1)∣∣, [K2 : K]

)= 1.

As a consequence, the order of any element I of Po(K1) is coprime with [K2 : K]. Thus, the morphism

νK1L ◦ εLK1

: I ∈ Po(K1) → I [K2:K] ∈ Po(K1)

is injective. A fortiori, the morphism εLK1is injective. �

Theorem 5.6. Let K, K1 and K2 be Galois extensions of Q such that K1 ∩K2 = K and let L = K1K2. If all the ramification indices in the extension K1/Q are coprime with [K2 : Q] and symmetric, then we have the isomorphism

Po(L) � Po(K1) × Po(K2).

Proof. Following the previous proposition (with K = Q), the morphisms εLK1and εLK2

are injective. The hypotheses of Proposition 5.2 are satisfied: for each prime p ∈ P, (ep(K1/Q), [K2 : Q]) = 1 and ep(K2/Q) divides n2, thus (ep(K1/Q), ep(K2/Q)) = 1. Consequently,

εLK1

(Po(K1)

)· εLK2

(Po(K2)

)= Po(L).

Moreover, consider an element I ∈ εLK1(Po(K1)) ∩ εLK2

(Po(K2)). Following Remark 5.4, its order l divides n2. Since l divides

∏p ep(K1/Q) and (ep(K1/Q), n2) = 1 for each

prime p, l is coprime with n2. The integer l must be equal to 1. Thus, εLK1(Po(K1)) ∩

εLK2(Po(K2)) = {1}. �

6. The Pólya number poK of a number field K

In Section 2, we proved that any algebraic number field may be embedded in a Pólya field, namely its Hilbert class field. If a field K is an Abelian extension of Q, we also proved that the genus field ΓK of K is a Pólya field. However, in this section, we prove that generally the genus field is not the smallest Pólya field containing K and we give upper bounds for the minimal degree of a Pólya field containing K.

Definition 6.1.

(1) A minimal Pólya field over K is a finite extension L of K which is a Pólya field and such that no intermediate extension K ⊆ M � L is Pólya field.

222 A. Leriche / Journal of Number Theory 145 (2014) 210–229

(2) The Pólya number of K is the following integer

poK = min{[L : K]

∣∣ K ⊆ L, L Pólya field}.

At the end of this section we prove that there is not necessarily a unique minimal Pólya field over a field K.

6.1. A bound for the Pólya number of an Abelian number field

In this section, we assume that K is an Abelian number field. Following Kronecker–Weber’s theorem, K is contained in a cyclotomic field Q[ζf ] where f denotes the conductor of K. We recall the following results due to Zantema:

Proposition 6.2. (See [20].)

(1) Every cyclotomic field is a Pólya field.(2) The maximal real subfield K = Q[ζm + ζ−1

m ] of the cyclotomic field Q[ζm] is a Pólya field.

Then, denoting Euler’s function by ϕ, we have:

poK ≤ ϕ(f)[K : Q] and poK ≤ ϕ(f)

2[K : Q] if K is real.

By Theorem 3.8, the genus field ΓK of K is a Pólya field, and hence,

poK ≤ gK = [ΓK : K].

However, Example 6.4 below shows that the genus field of an Abelian field K is not necessarily a minimal Pólya field over K. First recall the characterization of the quadratic Pólya fields given by Zantema.

Proposition 6.3. (See [20].) A quadratic field Q[√d ] is a Pólya field if and only if d is of

one of the following forms where p and q denote two distinct odd prime numbers:

(1) d = 2, or d = −1, or d = −2, or d = −p where p ≡ 3 (mod 4), or d = p,(2) d = 2p, or d = pq where pq ≡ 1 (mod 4) and, in both cases, the fundamental unit

has norm 1 if p ≡ 1 (mod 4).

Example 6.4. Consider the following extensions:

Q ⊂ K = Q[√

170 ] ⊂ L = Q[√

5,√

34 ] ⊂ M = Q[√

2,√

5,√

17 ]

A. Leriche / Journal of Number Theory 145 (2014) 210–229 223

Following Proposition 3.11, Γ ′K = M . In fact, Γ ′

K = ΓK because the si’s in Proposi-tion 3.11 are equal to 1 (see [11, §4.2.9]). Since Q[

√5 ] and Q[

√34 ] are Pólya fields, by

Proposition 5.2, the field L = Q[√

5, √

34 ] is a Pólya field. The genus fields Γ ′K and ΓK

are not minimal Pólya fields over K.

Proposition 6.5. The genus field ΓK of an Abelian extension K/Q is not necessarily a minimal Pólya field over K.

In fact, according to the number of ramified primes in K/Q, the following proposition shows that the genus field of K is quite far from being a minimal Pólya field over K.

Proposition 6.6. Let K = Q[√d ] (where d ∈ Z is a squarefree integer) be a quadratic num-

ber field. Denote by σ (resp. τ) the number of prime divisors of d congruent to 3 (mod 4)(resp. 1 (mod 4)). We have the following bounds:

(1) If d ≡ 1 (mod 4), poK ≤ 2σ2 +τ−1 if d > 0 and 2σ−1

2 +τ if d < 0.(2) If d ≡ 3 (mod 4), poK ≤ 2σ−1

2 +τ if d > 0 and 2σ2 +τ if d < 0.

(3) If d ≡ 2 (mod 8), poK ≤ 2σ2 +τ if d > 0 and 2σ−1

2 +τ+1 if d < 0.(4) If d ≡ 6 (mod 8), poK ≤ 2σ−1

2 +τ if d > 0 and 2σ2 +τ if d < 0.

Proof. Let d = (−1)e2γp1 . . . pσq1 . . . qτ where pi ≡ 3 (mod 4) and qj ≡ 1 (mod 4), e = 0 or 1, γ = 0 or 1. We will use the fact that, if p and p′ denote odd distinct prime numbers such that p, p′ ≡ 3 (mod 4), the following quadratic fields Q[

√δ ] are Pólya

fields: Q[i], Q[√

2 ], Q[√−2 ], Q[√p ], Q[

√−p ], Q[√

2p ], Q[√pp′ ]. In each of these fields,

the prime numbers which ramify in Q[√δ ]/Q are the prime divisors of δ except for

Q[i] where 2 is ramified. Now we consider the compositum of some of these quadratic Pólya fields in order to obtain a field containing K which is a Pólya field thanks to Proposition 5.2.

Let σ′ = 2�σ2 �.

By Proposition 5.2, L = Q[√p1p2, √p3p4, . . . ,

√pσ′−1pσ′ ,

√q1, . . . ,

√qτ ] is a Pólya

field where the only ramified primes are the pi’s (1 ≤ i ≤ σ′) and the qj ’s (1 ≤ j ≤ τ).Assume first that σ is even, that is, σ′ = σ. Assume also that d is odd, that is,

d = p1 . . . pσ′q1 . . . qτ . If d > 0 (and hence, d ≡ 1 (mod 4)), then K ⊂ L and poK ≤ [L :K] = 2σ

2 +τ−1. If d < 0 (d ≡ 3 (mod 4)), then K ⊂ L[i]. Once more by Proposition 5.2, L[i] is a Pólya field and poK ≤ [L[i] : K] = 2σ

2 +τ .Assume now that d is even. If d > 0 (d ≡ 2 (mod 8)), K ⊂ L[

√2 ] and, if d < 0

(d ≡ 6 (mod 8)), K ⊂ L[√−2 ]. Once more by Proposition 5.2, L[

√2 ] and L[

√−2 ] are

Pólya fields and poK ≤ 2σ2 +τ .

Now assume that σ is odd, that is, σ = σ′ + 1. Assume also that d is odd. If d > 0(d ≡ 3 (mod 4)), then K ⊂ L[√pσ ] and poK ≤ 2σ−1

2 +τ . If d < 0 (d ≡ 1 (mod 4)), then K ⊂ L[

√−pσ ] and poK ≤ 2σ−12 +τ .

224 A. Leriche / Journal of Number Theory 145 (2014) 210–229

Finally, assume that d is even. If d > 0 (d ≡ 6 (mod 8)), then K ⊂ L[√

2pσ ] and poK ≤2σ−1

2 +τ . If d < 0 (d ≡ 2 (mod 8)), then K ⊂ L[√pσ, √−2 ] and poK ≤ 2σ−1

2 +τ+1. �Now we compare these bounds with the genus number of quadratic fields.

Corollary 6.7. Let K = Q[√d ] (where d ∈ Z is a squarefree integer) be a quadratic num-

ber field. Denote by σ the number of primes p ≡ 3 (mod 4) which are ramified in K/Q, then:

gKpoK

≥ 2σ2 −2 if d is even and gK

poK≥ 2

σ+12 −2 if d is odd.

In particular, for σ ≥ 4, ΓK is not a minimal Pólya field over K.

Proof. Following Proposition 3.11, one has

Γ ′K = Q

(√(−1)e2γ ,

√−p1, . . . ,

√−pσ,

√q1, . . . ,

√qτ

).

The field M = Γ ′K∩R = Q(

√pe12γ ,

√p1p2, . . . ,

√p1pσ,

√q1, . . . ,

√qτ ) is the real maximal

subfield of Γ ′K = M(

√−p1 ). Thus, by Proposition 3.11, gK = [ΓK : K] ≥ g′K

2 = [Γ ′K :K]2 .

Then, gK ≥ 2σ+τ−2 if d is odd and ≥ 2σ+τ−1 if d is even. Moreover, the previous proposition proves that poK ≤ 2σ

2 +τ if d is odd and ≤ 2σ−12 +τ+1 if d is even. Thus, if d

is even,

gKpoK

≥ 2σ2 −2

and if d is odd,

gKpoK

≥ 2σ+12 −2.

The right hand sides of the previous inequalities are > 1 as soon as σ ≥ 4. �6.2. A bound for the Pólya number of a Galois number field

In Example 3.10, there is a non-Abelian Galois extension K/Q where the genus field is not a Pólya field. The upper bound of the previous section does not hold anymore. For any algebraic number field, the only upper bound that one may give a priori is the following one (see Proposition 3.3):

poK ≤ hK

where hK denotes the class number of K.

A. Leriche / Journal of Number Theory 145 (2014) 210–229 225

However, if K is a Galois extension, one may have a better upper bound.

Proposition 6.8. Let K ⊂ L be two Galois number fields such that the extension L/K is Abelian. Let n = [K : Q] and G = Gal(L/K). We have

G � Gn ×Gn̂

where Gn (resp. Gn̂) denotes the subgroup of G formed by the elements whose order divides a power of n (resp. whose order is coprime to n). Denote by Ln = LGn̂ (resp. Ln̂ = LGn) the subfield of L fixed by Gn (resp. Gn̂). If the fields Ln and Ln̂ are Galois extensions of Q, then Po(Ln) may be considered as a subgroup of Po(L).

��

���

��

���

��

���

��

���

Q

K1 = Ln

Gn̂

Gn̂

Gn

Gn

K

K2 = Ln̂G

n

L

Proof. We refer to Proposition 5.5 where K1 = Ln, K2 = Ln̂ and K1 ∩ K2 = K. The ramification indices of the primes in K1/Q are coprime to [K2 : K]. Indeed, the ramification indices in the Galois extension K1/Q divide [K1 : Q] = [K1 : K][K : Q] =n × |Gn| which is a divisor of a power of n, whereas [K2 : K] is coprime to n. By Proposition 5.5, the restriction to Po(K1) of the morphism εK1K2

K1is injective. �

Corollary 6.9. Let K be a Galois extension of Q. Let n = [K : Q] and HK,n be the subfield of HK fixed under the action of the elements of Gal(HK/K) � Cl(K) whose order is coprime to n. The field HK,n is a Pólya field.

Proof. Following Proposition 6.8, we denote by Cl(K)n (resp. Cl(K)n̂) the subgroup of Cl(K) formed by the elements whose order divides n (resp. whose order is coprime to n). Thus, identifying Gal(HK/K) with Cl(K), HK,n = H

Cl(K)n̂K (resp. HK,n̂ = H

Cl(K)nK ) is

the subfield of HK fixed by Cl(K)n̂ (resp. Cl(K)n).

226 A. Leriche / Journal of Number Theory 145 (2014) 210–229

��

���

��

���

��

���

��

���

Q

K1 = HK,n

Cl(K)n̂

Cl(K)n̂

Cl(K)n

Cl(K)n

K

K2 = HK,n̂Cl(K)

n

HK

The extension HK/Q is Galois. Indeed, let σ be a morphism from HK into C. Since K/Q is Galois, one has σ(K) = K. The extension σ(HK)/K is Abelian and unramified. By maximality of HK , σ(HK) ⊂ HK . Similarly, HK,n/Q and HK,n̂/Q are Galois. By Proposition 6.8, Po(HK,n) is contained in Po(HK). However, HK being a Pólya field, Po(HK) = {1}. Thus, HK,n is a Pólya field. �Corollary 6.10. Let K/Q be a Galois extension with degree n, then

poK ≤∣∣Cl(K)n

∣∣ =∏p|n

pvp(hK).

6.3. The non-unicity of the minimal Pólya fields

In this subsection, we prove that there could be several unramified minimal Pólya fields over an Abelian number field with same degree.

Notations. Let k = Q[√l ] where l is a prime number such that l = 2 or l ≡ 5 (mod 8).

Let p be a prime number such that p ≡ 1 (mod 4) and (pl )4 = −1 where ( .. )4 denotes the biquadratic symbol. Let ε be the fundamental unit of k. We consider the cyclic quartic

field K = k(√

−pε√l ).

The extension K/Q is quartic and Galois. Thus, by Corollary 3.2 in [14], Po(K) ⊆Cl2(K) where Cl2(K) is the 2-class group of K. Moreover, about Cl2(K), one has the following result from Brown and Parry ([3] for l ≡ 3 (mod 8) and [4] for l = 2):

A. Leriche / Journal of Number Theory 145 (2014) 210–229 227

Proposition 6.11. With the previous hypotheses, if H(2)K denotes the 2-Hilbert class field

of K, we have:

Gal(H

(2)K /K

)� Cl2(K) � Z/2Z× Z/2Z.

We have to specify the elements of Cl2(K). Since, (pl ) = 1, there exist two ideals p1and p2 of k such that

pOk = p1p2

and thus two prime ideals m1 and m2 of K such that

piOK = m2i (i = 1, 2).

Consequently, Po(K) is the subgroup of order 2 generated by the class of Π2(K) = m1m2. The other non-trivial elements of Cl2(K) are the classes of m1 and m2. Proposition 6.11proves that there exist three intermediate fields F1, F2, F3 between K and H(2)

K . A. Azizi and M. Talbi studied the capitulation of the 2-classes of ideals of K in the fields Fi. For this purpose, they introduce the conjugate elements π1 and π2 of k over Q such that:

phki = πiOk (i = 1, 2)

where hk is the class number of k. They obtain the following result:

Proposition 6.12. (See [2].) With the previous hypotheses and notation, one has:

(1) If ( lp )4 = 1, then F1 = K(

√−π1 ), F2 = K(√−π2 ) and F3 = K(√p ) = ΓK . The

four classes of Cl2(K) become principal by extension in each Fi for i ∈ 1, 2, 3.(2) If ( l

p )4 = −1, then F1 = K(√π1 ), F2 = K(√π2 ) and F3 = K(√p ) = ΓK . In each extension Fi, i ∈ 1, 2, 3, there are exactly two classes Cl2(K) which become principal.

Thus, in the second case, only the extension F3 = ΓK is a Pólya extension. But in the first case, when ( l

p )4 = 1, the three quadratic extensions of K contained in the

2-Hilbert class field H(2)K of K are minimal Pólya fields over K. Indeed, K is not a Pólya

field and the three subextensions are unramified (see Proposition 3.4). We pick up a counterexample given in [2]:

Example 6.13. Consider the field K = Q(√−89(2 +

√2 )). By Theorem 4.4 in [15], the

field K is a cyclic quartic field which is not a Pólya field. But 89 ≡ 9 mod 16 and

( 289 )4 = 1, thus, F1 = K(

√−(11 + 4

√2 )), F2 = K(

√−(11 − 4

√2 )) and F3 = K(

√89 ).

By the previous proposition, each one of the extensions Fi is a quadratic Pólya extension of K. Consequently, since K is not a Pólya field and since the extensions Fi/K are

228 A. Leriche / Journal of Number Theory 145 (2014) 210–229

unramified Galois Pólya extensions, by Proposition 3.4, the fields Fi are minimal Pólya fields over K.

Remark 6.14. The previous fields F2 and F3 are not isomorphic.

Thus, Theorem 3.3 may be precised as follows:

Proposition 6.15. A number field K is not necessarily contained in a unique minimal Pólya field (even if we consider Pólya fields contained in the Hilbert class field of K). Moreover, such distinct minimal Pólya fields are not necessarily isomorphic.

At last, a natural question arises:

Do all the minimal Pólya fields over a field K have the same degree?

There exists also a notion of Pólya field in the case of function fields. Some of these Pólya fields are characterized by Adam [1], so that the question of the embedding of a function field in a Pólya field could also be raised.

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