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# On ℤp-embeddability of cyclic p-class fields

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## Abstract

It is investigated when a cyclic p-class field of an imaginary quadratic number field can be embedded in an infinite pro-cyclic p-extension.
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
1
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 Kis 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 Kis 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 α1 (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/pe1×Z/pe1if p-,
Z/pe1×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/3e1×Z/3e1×Z/3.
Taking the inverse limit gives Gal(K/K0)
=Z3×Z3×Z/3. In particular,
T
=Z/3if K0=K.
3
Proof. There is a natural exact sequence
1→ O(O/pe)P/Pe1.
The exclusion of the case p= ∆ = 3 ensures that Ohas 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, . . . , pe1.
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)|=(p2e2if p-∆ ,
p2e1if 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 a1 (mod p) and pi||bfor some i=1. Write
(a+b∆)p=a0+b0∆. Then a01 (mod p) and pi+1||b0.
It follows from () that the cyclic subgroups U:= h1 + piand V:= h1 + pi
of (O/pe)both have order pe1. It follows from (∗∗) that they have trivial
intersection. So for p-∆,
(p-part of (O/pe)) = U×V
=Z/pe1×Z/pe1.
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/pe1×Z/pe
for V0:= h1 + i. In case p= 3 and ∆ 3 (mod 9), the 3-part of (O/9)is
h4i×h1+3i×h1 + i
=(Z/3)3, and therefore
(3-part of (O/3e)) = U×V×(group of order 3)
=Z/pe1×Z/pe1×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
Theorem 3. Assume K0/K is cyclic of degree p. Pick a prime q-pof Kof
4
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 αa+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
Kcontains 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
5
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/K0Kanti), i.e. Tis
isomorphic to a subgroup of Gal(K0/K) with cyclic quotient. If p= 3 and
3 (mod 9), then Khas degree 3 over KcyclKantiK0, and therefore Thas a
subgroup of index 3 which is isomorphic to Gal(K0/K0Kanti).
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 αα5and 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+7107)/2.
Modulo 9 we have αα3and 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+3237.
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+2129.
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 + 93387)/21 (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.
6
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:=
K0KcyclKanti =K0Kanti, and let pibe the degree of F0/K. It is the maximal
Zp-embeddable subextension of K0/K. Then T
=Z/pni. 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=LKcyclKanti.
Therefore [L:F0]=[Kcycl KantiL:Kcycl Kanti]5pniand 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.
7
... Chapter 1 is an introduction to the theory of procyclic Galois extensions. Chapters 2 and 3 are extended versions of my papers [3] and [4]. Chapters 4 and 5 are based on two papers still in preparation. ...
... Prime decomposition in the anti-cyclotomic extension 3 ...
... This notion was introduced in [19]. Some criteria for l-rationality are given there and in [3]. ...
Thesis
Full-text available
Denote by $Z_p$ the additive group of $p$-adic integers. The main theme of this thesis is the existence and properties of Galois extensions of algebraic number fields with Galois group $Z_p$, in short $Z_p$-extensions. We shall however also consider some non-abelian pro-$p$-groups as Galois groups.
... Let l = 3 and consider K = Q( √ −21). This field is not 3-rational, indeed it has (two linearly disjoint and hence) four Z/3-extensions which are unramified outside 3 and dihedral over Q (see Brink [2]). Using Lemma 4 and a computer, we easily find four polynomials f 1 , . . . ...
... We now aim at finding the first step of the anti-cyclotomic extension of K = Q( √ −107) for l = 3. Again, there are four Z/3-extensions of K which are unramified outside 3 and dihedral over Q (see Brink [2]), and we find four candidate polynomials: The class number of K is 3, and since f 1 generates a cubic field with discriminant −107, the splitting field of f 1 is the Hilbert class field of K. The anti-cyclotomic decomposition law depends on whether this class field is contained in K anti (and thus equals K (1) anti ) or not. ...
Article
Full-text available
For an imaginary quadratic number field K and an odd prime number l, the anti-cyclotomic Zl-extension of K is defined. For primes p of K, decomposition laws for p in the anti-cyclotomic extension are given. We show how these laws can be applied to determine if the Hilbert class field (or part of it) of K is Zl-embeddable. For some K and l, we find explicit polynomials whose roots generate the first step of the anti-cyclotomic extension and show how the prime decomposition laws give nice results on the splitting of these polyniomials modulo p. The article contains many numerical examples.
Article
For a CM-field K and an odd prime number p, let $$\widetilde{K}'$$ be a certain multiple $$\mathbb {Z}_p$$-extension of K. In this paper, we study several basic properties of the unramified Iwasawa module $$X_{\widetilde{K}'}$$ of $$\widetilde{K}'$$ as a $$\mathbb {Z}_p\llbracket \mathrm{Gal}(\widetilde{K}'/K)\rrbracket$$-module. Our first main result is a description of the order of a Galois coinvariant of $$X_{\widetilde{K}'}$$ in terms of the characteristic power series of the unramified Iwasawa module of the cyclotomic $$\mathbb {Z}_p$$-extension of K under a certain assumption on the splitting of primes above p. The second result is that if K is an imaginary quadratic field and if p does not split in K, then, under several assumptions on the Iwasawa $$\lambda$$-invariant and the ideal class group of K, we determine a necessary and sufficient condition such that $$X_{\widetilde{K}}$$ is $$\mathbb {Z}_p\llbracket \mathrm{Gal}(\widetilde{K}/K)\rrbracket$$-cyclic. Here, $$\widetilde{K}$$ is the $$\mathbb {Z}_p^2$$-extension of K.
Preprint
Let $p$ be an odd prime number and $k$ an imaginary quadratic field in which $p$ splits. In this paper, we consider a weak form of Greenberg's generalized conjecture for $p$ and $k$, which states that the non-trivial Iwasawa module of the maximal multiple $\mathbb{Z}_p$-extension field over $k$ has a non-trivial pseudo-null submodule. We prove this conjecture for $p$ and $k$ under the assumption that the Iwasawa $\lambda$-invariant for a certain $\mathbb{Z}_p$-extension over a finite abelian extension of $k$ vanishes and that the characteristic ideal of the Iwasawa module associated to the cyclotomic $\mathbb{Z}_p$-extension over $k$ has a square-free generator.
Only (I) is proved since the proof of (II) is very similar Assume F/K is not Z p -embeddable, i.e. that F is a proper extension of F
• Proof
Proof. Only (I) is proved since the proof of (II) is very similar. Put F := K 0 ∩ K cycl K anti = K 0 ∩ K anti, and let p i be the degree of F /K. It is the maximal Z p -embeddable subextension of K 0 /K. Then T ∼ = Z/p n−i. Assume F/K is not Z p -embeddable, i.e. that F is a proper extension of F. Assume that F/K is embedded in a cyclic extension L/K inside K ∞. Then F = L ∩ K cycl K anti.
• I R Shafarevich
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.