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On Zp-embeddability of cyclic p-class ﬁelds

David Brink

December 2005

Abstract. It is investigated when a cyclic p-class ﬁeld of an imag-

inary quadratic number ﬁeld can be embedded in an inﬁnite pro-cyclic

p-extension.

R´

esum´

e. On donne des conditions pour qu’un p-corps de classes

cyclique d’un corps de nombres quadratique imaginaire soit plongeable

dans une p-extension pro-cyclique inﬁnie.

Consider an imaginary quadratic number ﬁeld K. Let pbe an odd prime number,

and denote by Zpthe pro-cyclic pro-p-group lim

←−n(Z/pn). As shown by Iwasawa,

any Zp-extension of Kis unramiﬁed outside p. The lower steps of a such exten-

sion might well be unramiﬁed also at p. In this article the following question is

investigated: if the p-class group of Kis non-trivial and cyclic, is the p-Hilbert

class ﬁeld of K(or part of it) then embeddable in a Zp-extension of K? In doing

so, we are led to study the torsion subgroup of the Galois group over Kof the

maximal abelian p-extension of Kwhich is unramiﬁed outside p. First ﬁx some

notation:

p: an odd prime number

ζ: a primitive p-th root of unity

∆ : a square-free natural number

K: the imaginary quadratic number ﬁeld Q(√−∆)

O: the ring of integral elements in K

K0: the p-Hilbert class ﬁeld of K

Ke: the p-part of K’s ray class ﬁeld with conductor pe,e=0

K∞: the union S∞

e=0 Ke

T: the torsion subgroup of Gal(K∞/K)

Kcycl : the cyclotomic Zp-extension of K

Kanti : the anti-cyclotomic Zp-extension of K

I: the group of fractional ideals of Kprime to p

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P: the group of principal fractional ideals of Kprime to p

Pe: the ray modulo pe,e=0

Note that the ray class ﬁeld with conductor 1 is exactly the Hilbert class ﬁeld so

that the notation is consistent. We have the tower

KjK0jK1jK2j··· ⊂ K∞

and note that the union K∞is the maximal abelian p-extension of Kwhich

is unramiﬁed outside p. Thus, by Iwasawa’s result, any Zp-extension of Kis

contained in K∞. It is well known that K∞is the composite of three ﬁelds

Kcycl,Kanti, and KTwhich are linearly disjoint over K(see [1]). The cyclotomic

extension Kcycl is the unique Zp-extension of Kwhich is abelian over Q. The

anti-cyclotomic extension Kanti is the unique Zp-extension of Kwhich is pro-

dihedral over Q. Finally, KTis a ﬁnite extension of Kwith Gal(KT/K)∼

=Tand

dihedral over Q(but not unique with these properties). As we shall see, we may

usually for KTtake K0or a subﬁeld of K0. From the above discussion follows

the isomorphism

Gal(K∞/K)∼

=Zp×Zp×T

which will be important in the following. It may also be noted that the composite

KantiKTis the maximal abelian p-extension of Kwhich is unramiﬁed outside p

and dihedral over Q, and hence equals the union of all p-ring class ﬁelds over K

with conductor a power of p.

Theorem 1. (I) Assume p > 3and that K(ζ)has the same p-class number as

K. Then Tis trivial, and K0/K is cyclic (possibly trivial) and Zp-embeddable.

(II) Assume p= 3,∆6≡ 3 (mod 9), and that the class number of Q(√3∆) is

not divisible by 3. Then Tis trivial, and K0/K is cyclic (possibly trivial) and

Zp-embeddable.

Proof. Since the p-Hilbert class ﬁeld K0is dihedral over Q, it is contained in

KantiKT. So if we show T= 0, it will follow that K0is contained in Kanti.

Let rbe the p-rank of Gal(K∞/K). Shafarevich gives in [4] a formula for r

which implies r= 2 if the completion Qp(√−∆) does not contain ζ, and Khas

no element xsuch that the radical extension K(ζ, p

√x)/K(ζ) is non-trivial and

unramiﬁed (such elements are called hyperprimary, see [2] for more details on

this).

(I) For p > 3, Qp(√−∆) never contains ζ. The assumption on the class num-

ber of K(ζ) implies that K0(ζ) is the p-Hilbert class ﬁeld of K(ζ). In particular,

K0(ζ)/K is abelian. Assume for a contradiction that Khas a non-trivial hyperpri-

mary element x, i.e. that K(ζ, p

√x)/K(ζ) is an unramiﬁed Z/p-extension. Then

2

p

√xis contained in K0(ζ). But K(p

√x) is not normal over K, a contradiction. So

we have shown r= 2 and hence T= 0.

(II) The assumption ∆ 6≡ 3 (mod 9) implies ζ6∈ Q3(√−∆). The 3-class

number of K(ζ) = K(√−3) is the product of the 3-class numbers of K,Q(√−3),

and Q(√3∆). So the assumption on the class number of Q(√3∆) implies that

K0(ζ) is the 3-Hilbert class ﬁeld of K(ζ). The rest of the proof is the same as in

(I).

In [3], Jaulent and Nguyen Quang Do show T= 0 under the assumptions of (II).

For p > 3, they show T= 0 if Khas trivial p-class group.

For p= 3, this theorem applies for many values of ∆, for instance 23, 26, 29,

31, and 38. In all these cases, K0/K is cyclic of degree 3 and Z3-embeddable.

The smallest values of ∆ for which the theorem does not apply are 107 (because

Q(√3·107) has class number 3) and 129 (which is ≡3 (mod 9)).

For p= 5, the theorem likewise applies for many ∆, for instance 47, 74, 79, and

86. In these cases, K0/K is cyclic of degree 5 and Z5-embeddable. The smallest

values of ∆ for which the theorem does not apply are 127 and 166 (in both cases,

Khas 5-class number 5, whereas that of K(ζ) is 25 and 125, respectively).

It may be noted that the theorem never applies when pis an irregular prime,

i.e. when the class number of Q(ζ) is divisible by p.

We shall now see how the remaining cases can be dealt with. Recall that the

ray group modulo peis the subgroup Peof Pgenerated by the principal ideals

(α) with integral α≡1 (mod pe). The ray class group modulo peis the quotient

I/Pe. It is a central result in class ﬁeld theory that there is an isomorphism, the

Artin symbol, from the p-part of I/Peto Gal(Ke/K). It maps the p-part of

P/Peonto Gal(Ke/K0).

Lemma 2. (I) With notation as above, we have for p > 3,

Gal(Ke/K0)∼

=(Z/pe−1×Z/pe−1if p-∆,

Z/pe−1×Z/peif p|∆.

Taking the inverse limit gives Gal(K∞/K0)∼

=Zp×Zp. In particular, T= 0 if

K0=K.

(II) For p= 3, the above remains valid when ∆6≡ 3 (mod 9). Assume

∆≡3 (mod 9) and ∆6= 3. Then

Gal(Ke/K0)∼

=Z/3e−1×Z/3e−1×Z/3.

Taking the inverse limit gives Gal(K∞/K0)∼

=Z3×Z3×Z/3. In particular,

T∼

=Z/3if K0=K.

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Proof. There is a natural exact sequence

1−→ O∗−→ (O/pe)∗−→ P/Pe−→ 1.

The exclusion of the case p= ∆ = 3 ensures that O∗has trivial p-part. Hence

Gal(Ke/K0) is isomorphic to the p-part of (O/pe)∗by the Artin symbol. So we

compute the structure of (O/pe)∗.

To begin with, note that each coset of O/pehas a unique representative of

the form a+b√−∆ with a, b = 0,1, . . . , pe−1.

The order of (O/pe)∗, i.e. the norm of the ideal peO, depends on the prime

ideal decomposition of pin K. More precisely, the order of the p-part is

|p-part of (O/pe)∗|=(p2e−2if p-∆ ,

p2e−1if p|∆ .

We note the following two facts:

(∗) Let x∈ O and write (1 + x)p= 1 + x0. If pi||xfor some i=1 (meaning that

pi|x, but pi+1 -x), then pi+1||x0.

(∗∗) Let aand bbe integers with a≡1 (mod p) and pi||bfor some i=1. Write

(a+b√−∆)p=a0+b0√−∆. Then a0≡1 (mod p) and pi+1||b0.

It follows from (∗) that the cyclic subgroups U:= h1 + piand V:= h1 + p√−∆i

of (O/pe)∗both have order pe−1. It follows from (∗∗) that they have trivial

intersection. So for p-∆,

(p-part of (O/pe)∗) = U×V∼

=Z/pe−1×Z/pe−1.

Assume p|∆. Then U×Vhas index pin the p-part of (O/pe)∗. If p > 3, or

p= 3 and ∆ 6≡ 3 (mod 9), the same argument shows

(p-part of (O/pe)∗) = U×V0∼

=Z/pe−1×Z/pe

for V0:= h1 + √−∆i. In case p= 3 and ∆ ≡3 (mod 9), the 3-part of (O/9)∗is

h4i×h1+3√−∆i×h1 + √−∆i∼

=(Z/3)3, and therefore

(3-part of (O/3e)∗) = U×V×(group of order 3) ∼

=Z/pe−1×Z/pe−1×Z/3.

This ﬁnishes the proof of the lemma.

The question of Zp-embeddability in the case where the p-class ﬁeld is cyclic of

degree pcan now be answered.

Theorem 3. Assume K0/K is cyclic of degree p. Pick a prime q-pof Kof

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order pin the class group, and write qp= (α)with α∈ O.

I. Suppose p > 3.

(a) If αis not a p-th power in (O/p2)∗, then K0/K is Zp-embeddable (in fact K0

is contained in Kanti), and T= 0.

(b) If αis a p-th power in (O/p2)∗, then K0/K is not embeddable in Z/p2-

extension unramiﬁed outside p, and T∼

=Z/p.

II. Now suppose p= 3. If ∆6≡ 3 (mod 9), all the above remains valid. Assume

∆≡3 (mod 9), and write α≡a+b√−∆ (mod 9) with a, b ∈Z.

(c) If (a, b)≡(±1,0) modulo 3, but not modulo 9, then K0/K is Z3-embeddable

(in fact K0is contained in Kanti), and T∼

=Z/3.

(d) If (a, b)6≡ (±1,0) modulo 3, then K0/K is embeddable in a Z/9-extension

unramiﬁed outside 3, but not in a Z/27-extension unramiﬁed outside 3, and

T∼

=Z/9.

(e) If (a, b)≡(±1,0) modulo 9, then K0/K is not embeddable in a Z/9-extension

unramiﬁed outside 3, and T∼

=Z/3×Z/3.

Proof. I. Assume p > 3. By Lemma 2, Gal(K∞/K0)∼

=Zp×Zp. Since

Gal(K0/K)∼

=Z/p, there are two possibilities for T: 0 or Z/p.

(a) If T= 0, i.e. Gal(K∞/K)∼

=Zp×Zp, then K0is contained in KcyclKanti.

Since K0is dihedral over Q, it is in fact contained in Kanti. Both I /P2and P /P2

have p-rank 2. Therefore, qp= (α) is not ap-th power in P/P2. So αis not a

p-th power in (O/p2O)∗.

(b) If T∼

=Z/p, i.e. Gal(K∞/K)∼

=Zp×Zp×Z/p, then K0is linearly disjoint

from KcyclKanti. So we may for KTtake K0. Hence no Z/p2-extension of Kinside

K∞contains K0. Now the p-part of P/P2is a direct summand in the p-part of

I/P2. Therefore, qp= (α)is ap-th power in P/P2. So αis a p-th power in

(O/p2O)∗.

II. Now assume p= 3. If ∆ 6≡ 3 (mod 9), everything goes like above. Henceforth

assume ∆ ≡3 (mod 9). Then (O/9)∗∼

=Z/2×Z/3×Z/3×Z/3 (see the proof of

Lemma 2), so that α=a+b√−∆ is a cube in (O/9)∗iﬀ (a, b)≡(±1,0) modulo

9. Further, (O/3)∗∼

=Z/2×Z/3, so that αis a cube in (O/3)∗iﬀ (a, b)≡(±1,0)

modulo 3. By Lemma 2, Gal(K∞/K0)∼

=Z3×Z3×Z/3. Since Gal(K0/K)∼

=Z/3,

there are three possibilities for T:Z/3, Z/9, or Z/3×Z/3.

(c) If T∼

=Z/3, i.e. Gal(K∞/K)∼

=Z3×Z3×Z/3, then K0is contained

in KcyclKanti and therefore also in Kanti. Both I/P2and P/P2have 3-rank 3.

Therefore, q3= (α) is not a cube in P/P2. So αis not a cube in (O/9)∗. On

the other hand, the 3-part of P/P1is a direct summand in the 3-part of I/P1.

Therefore, q3= (α)is a cube in P/P1. So αis a cube in (O/3)∗. This shows the

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claims about aand b.

The cases (d) where T∼

=Z/9 and (e) where T∼

=Z/3×Z/3 are treated in a

similar manner, so their proofs are omitted.

The same arguments give a description of the torsion subgroup Tin the general

case where K0/K is not necessarily cyclic: If p > 3, or p= 3 and ∆ 6≡ 3 (mod 9),

then K∞=KcyclKantiK0, and therefore T∼

=Gal(K0/K0∩Kanti), i.e. Tis

isomorphic to a subgroup of Gal(K0/K) with cyclic quotient. If p= 3 and

∆≡3 (mod 9), then K∞has degree 3 over KcyclKantiK0, and therefore Thas a

subgroup of index 3 which is isomorphic to Gal(K0/K0∩Kanti).

The following examples answer the questions regarding Zp-embeddability from

the beginning of the article (i.e. for p= 3, ∆ = 107,129 and p= 5, ∆ = 127,166)

and show that all cases of Theorem 4 occur.

Examples. (i) Let p= 5 and ∆ = 127. The class number of Kis 5. The prime

number 2 is divisible by a non-principal prime ideal qof K. Further, q5= (α)

with α= (1 + √−127)/2 since 25=α¯α. Since αis not a ﬁfth power in (O/25)∗,

we are in case (a).

(ii) Let p= 5 and ∆ = 166. The class number of Kis 10. Here 7 is

divisible by a non-principal prime ideal qsuch that q5= (α) is principal, α=

(129 + √−166)/2. Modulo 25 we have α≡α5and conclude that we are in case

(b).

(iii) Let p= 3 and ∆ = 107. The class number of Kis 3. Here 11 is divisible by

a non-principal prime ideal qsuch that q3= (α) is principal, α= (9+7√−107)/2.

Modulo 9 we have α≡α3and conclude that we are in case (b).

(iv) Let p= 3 and ∆ = 237. The class number of Kis 12. Here 13 is divisible

by a non-principal prime ideal qsuch that q3= (α) is principal, α= 8+3√−237.

We are in case (c).

(v) Let p= 3 and ∆ = 129. The class number of Kis 12. Here 13 is divisible

by a non-principal prime ideal qsuch that q3= (α) is principal, α= 41+2√−129.

We are in case (d).

(vi) Let p= 3 and ∆ = 3387. The class number of Kis 12. Here the prime

43 is divisible by a non-principal prime ideal qsuch that q3= (α) is principal,

α= (209 + 9√−3387)/2≡1 (mod 9). We are in case (e).

As is the case for the ideal class group, there is numerical evidence that the torsion

subgroup T“prefers” having small rank, so that case (e) occurs quite rarely.

Finally a criterion for Zp-embeddability of a cyclic p-class ﬁeld (or part of it)

of arbitrary degree is given.

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Theorem 4. Assume K0/K is cyclic of degree pn>1, and let F/K be some

subextension

(I) Suppose p > 3. Then F/K is Zp-embeddable if it is embeddable in a

Z/pn+1-extension unramiﬁed outside p.

(II) Suppose p= 3. If ∆6≡ 3 (mod 9), the above holds. Assume ∆≡

3 (mod 9). Then F /K is Z3-embeddable if it is embeddable in a Z/3n+2-extension

unramiﬁed outside 3.

Proof. Only (I) is proved since the proof of (II) is very similar. Put F0:=

K0∩KcyclKanti =K0∩Kanti, and let pibe the degree of F0/K. It is the maximal

Zp-embeddable subextension of K0/K. Then T∼

=Z/pn−i. Assume F/K is not

Zp-embeddable, i.e. that Fis a proper extension of F0. Assume that F /K is

embedded in a cyclic extension L/K inside K∞. Then F0=L∩KcyclKanti.

Therefore [L:F0]=[Kcycl KantiL:Kcycl Kanti]5pn−iand we conclude that

[L:K]5pn.

References

[1] J. E. Carroll, H. Kisilevsky: Initial layers of Zl-extensions of complex

quadratic ﬁelds, Compositio Math. 32 (1976), no. 2, 157–168.

[2] K. Iwasawa: On Zl-extensions of algebraic number ﬁelds, Ann. of Math. (2)

98 (1973), 246–326.

[3] J.-P. Jaulent, T. Nguyen Quang Do: Corps p-rationnels, corps p-r´eguliers,

et ramiﬁcation restreinte, J. Th´eor. Nombres Bordeaux 5(1993), 343–363.

[4] A. Scholz: ¨

Uber die Beziehung der Klassenzahlen quadratischer K¨orper zu-

einander, J. Reine Angew. Math. 166 (1932), 201–203.

[5] I. R. Shafarevich: Extensions with given ramiﬁcation points (Russian), Publ.

Math., Inst. Hautes Etud. Sci. 18 (1964), 295–319. Translation: II. Ser., Am.

Math. Soc. 59 (1966), 128–149.

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