Prime decomposition in the anti-cyclotomic extension

Abstract

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.

Full text

Prime decomposition in the anti-cyclotomic
extension
David Brink
July 2006
Abstract: For an imaginary quadratic number field K and an odd
prime number l, the anti-cyclotomic Z
l
-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 Z
l
-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 polynomials modulo p. The
article contains many numerical examples.
I. Introduction
Let l be an odd prime number, and denote by Z
l
the infinite pro-cyclic l-group
lim
←−
Z/l
n
. Consider an imaginary quadratic number field K. As is well known, K has
a unique Z
l
-extension which is pro-dihedral over Q. We call it the anti-cyclotomic
Z
l
-extension of K (for reasons later to be clear).
The purpose of this paper is to study the decomposition of primes p of K
in the anti-cyclotomic extension. Since this extension is pro-cyclic over K, the
decomposition type of p is completely determined by the number of steps of the
anti-cyclotomic extension in which p is unramified, and the number of steps in which
p splits totally. By the n
th
step of a Z
l
-extension we understand the subextension
of degree l
n
over the ground field.
Such decomposition laws are given in Section III (Theorem 1 and 2). The laws
involve representations of primes p or prime powers p
h
by certain quadratic forms.
As we shall see, the decomposition laws also depend on how many steps of the
anti-cyclotomic extension are unramified. This dependence may be turned around,
meaning that if we know how certain primes decompose, then we can compute the
number of unramified steps. In particular, we can answer whether the Hilbert class
field of K is contained in the anti-cyclotomic extension and thus is Z
l
-embeddable.
In Section IV we show how to find explicit polynomials whose roots generate
the first step of the anti-cyclotomic extension. When K is not l-rational (to be
1

defined in Section II), this involves using the decomposition laws to identify the right
polynomial f among a finite number of candidates. When this is done, one obtains
nice laws for the splitting of f modulo p. For instance we show that X
5
+ 20X + 32
splits into linear factor modulo a prime number p 6= 2, 5 iff p is of the form x
2
+125y
2
or 2x
2
+ 2xy + 63y
2
.
Throughout the article we use the following notation:
l : an odd prime number
∆ : a square–free natural number
K : the imaginary quadratic number field Q(
√
−∆)
d
K
: the discriminant of K
h, µ, u : we write the class number of K as h = l
µ
u with l - u
O : the ring of integral elements in K
p : a prime of K, i.e. a prime ideal in O
p : the rational prime divisible by p
K
H
: the Hilbert class field of K
K
max
: the maximal abelian extension of K unramified outside l
K
(n)
anti
: the n
th
step of the anti-cyclotomic extension K
anti
ν : the non-negative integer defined by K
anti
∩ K
H
= K
(ν)
anti
II. The cyclotomic and the anti-cyclotomic extension
In Iwasawa [4] it is shown that any Z
l
-extension of K is unramified outside l.
This result motivates the study of the maximal abelian extension K
max
of K which
is unramified outside l. If K
f
denotes the ray class field over K of conductor f,
then K
max
is the union of the tower
K ⊆ K
1
⊆ K
l
⊆ K
l
2
⊆ . . .
Here, K
1
is the Hilbert class field of K which we also denote K
H
.
Let τ denote complex conjugation. Clearly, K
max
is normal over Q, so τ operates
on Gal(K
max
/K) by conjugation.
Main Lemma 1. We may write Gal(K
max
/K) = U × W × T × T
0
such that
(i) U is isomorphic to Z
l
, and τ operates trivially on U,
(ii) W is isomorphic to Z
l
, and τ operates by inversion on W ,
(iii) T is a finite l-group, and τ operates by inversion on T ,
(iv) T
0
is finite of order prime to l.
Further, we may write Gal(K
max
/K
H
) = U ×V × T
∗
× S
0
where
(v) V is isomorphic to Z
l
, contained in W × T , and has |V : W ∩ V | ≤ |T |,
(vi) T
∗
is trivial unless l = 3, ∆ ≡ 3 mod 9, and ∆ 6= 3; in this exceptional case,
T
∗
has order 3 and is contained in T ,
(vii) S
0
is contained in T
0
(and thus finite of order prime to l).
2

Consider a conductor f = l
e
with e ≥ 1. Then Gal(K
max
/K
f
) = U
f
× V
f
where
(iix) U
f
is contained in U and has index |U : U
f
| = l
e−1
,
(ix) V
f
is contained in V . If l - ∆, then |V : V
f
| = l
e−1
. If l | ∆, then
|V : V
f
| = l
e
unless l = 3 and ∆ ≡ 3 mod 9; in this case, |V : V
f
| = 3
e−1
.
The subgroups U, W × T , T , T
0
, V , T
∗
, S
0
, U
f
, and V
f
are unique with these
properties.
A similar representation of Gal(K
max
/K) appears in Carroll and Kisilevsky [3],
proved using id`eles rather than ideals as here.
A proof will be given at the end of the section. At this point, we only note
that the uniqueness statement is seen as follows: U is the maximal subgroup of
the l-part of Gal(K
max
/K) on which τ operates trivially, W × T is the maximal
subgroup of the l-part of Gal(K
max
/K) on which τ operates by inversion, T is
the l-torsion and T
0
is the non-l-part
1
(and the non-l-torsion) of Gal(K
max
/K), V
equals Gal(K
max
/K
H
) ∩ (W × T ), T
∗
is the l-torsion, and S
0
is the non-l-part of
Gal(K
max
/K
H
). Note that W is not unique if T is non-trivial.
Proposition 1. (a) K has a unique Z
l
-extension which is pro-cyclic over Q. It
is called the cyclotomic extension and is denoted K
cycl
. Adjoin to Q all roots of
unity of l-power-order, and let Q
cycl
be the l-part of this extension. Then K
cycl
is
the composite of K and Q
cycl
.
(b) K has a unique Z
l
-extension which is pro-dihedral over Q. It is called the
anti-cyclotomic extension and is denoted K
anti
.
(c) K
cycl
and K
anti
are the only absolutely normal Z
l
-extensions of K. They are
linearly disjoint over K, and any Z
l
-extension of K is contained in the composite
K
cycl
K
anti
. The l-part of the Hilbert class field K
H
(or any other part of it) is
embeddable in a Z
l
-extension of K iff it is contained in K
anti
.
(d) The Galois group of the maximal abelian l-extension of K which is unrami-
fied outside l is isomorphic to Z
l
×Z
l
×T where T is a finite l-group. If T is trivial,
the l-part of K
H
is cyclic and Z
l
-embeddable.
Proof. Everything follows from the theorem: K
cycl
is the fixed field of W ×T ×T
0
,
and K
anti
is the fixed field of U × T × T
0
. Any Z
l
-extension of K is contained in
the fixed field of the torsion T × T
0
, i.e. in K
cycl
K
anti
. K
cycl
and K
anti
are the
only absolutely normal Z
l
-extensions of K since U and W are the only τ-invariant
subgroups of U ×W with quotient Z
l
. Since K
H
is generalised dihedral over Q, the
maximal Z
l
-embeddable subfield of it is K
H
∩K
anti
. If T is trivial, the l-part of K
H
is contained in K
anti
. It is clear that Q
cycl
is a Z
l
-extension of Q. Hence KQ
cycl
is a
Z
l
-extension of K and a (Z
l
×Z/2)-extension of Q. The uniqueness of K
cycl
implies
1
The l-part of an abelian pro-finite group is its Sylow-l-subgroup, the “non-l-part” is the
product of the l
0
-parts for l
0
6= l. The l-part of an abelian field extension is the fixed field of the
non-l-part of the Galois group, and vice versa.
3

K
cycl
= KQ
cycl
. 
The situation is particularly simple when the torsion T is trivial. If this is the
case, K is called l-rational. This notion was introduced in Jaulent and Nguyen
Quang Do [5]. Some criteria for l-rationality are given there and in Brink [2].
Lemma 2. (a) Let X be an infinite abelian pro-l-group, and assume V and T
∗
are subgroups of X of which V is pro-cyclic with finite index, and T
∗
is finite.
Then we may write X = W × T with W pro-cyclic, T finite containing T
∗
, and
|V : V ∩ W | ≤ |T |.
(b) Let X be an abelian pro-l-group with a subgroup V . Assume τ is an au-
tomorphism of order 2 on X that operates by inversion both on V and on X/V .
Then τ operates by inversion on X.
(c) Let X be an abelian pro-l-group with a subgroup U. Assume τ is an au-
tomorphism on X that operates trivially on U and by inversion on X/U. Then
X = U ×V where V = {x ∈ X | x
τ
= x
−1
}.
Proof. (a) Assume V × T
∗
to have index l in X; the general case will then follow
by induction. Pick an x ∈ X\(V × T
∗
) and write x
l
= vt with v ∈ V and t ∈ T
∗
.
If v is an l
th
power in V , then X = V × T with a T containing T
∗
. If v
l
is not an
l
th
power in V , then X = W ×T
∗
where W is the pro-cyclic group generated by x;
from x
l·|T
∗
|
= v
|T
∗
|
follows |V : W ∩ V | ≤ |T
∗
|.
(b) Let x ∈ X. Then x
τ
= x
−1
v for some v ∈ V . Hence x = x
ττ
= x
−τ
v
τ
= xv
−2
and therefore v
2
= e, v = e (since X has no elements of order 2), and x
τ
= x
−1
.
(c) Let x ∈ X. Then x
τ
= x
−1
u for a u ∈ U. Every element in X is a square,
so there is a u
0
∈ U with u
2
0
= u
−1
. Put v = xu
0
. Then v
τ
= x
τ
u
0
= x
−1
uu
0
= v
−1
,
i.e. v ∈ V . Hence x = u
−1
0
v ∈ U × V . 
Lemma 3. Let e ≥ 1. The group of units in the ring O/l
e
may be written
(O/l
e
)
∗
= U × V × S
0
such that the following hold:
(a) Complex conjugation τ operates trivially on U which is isomorphic to Z/l
e−1
.
(b) Complex conjugation τ operates by inversion on V , and
V
∼
=







Z/l
e−1
if l - ∆,
Z/l
e
if l | ∆, unless l = 3 and ∆ ≡ 3 mod 9,
Z/3
e−1
× Z/3 if l = 3 and ∆ ≡ 3 mod 9.
4

(c) S
0
is the non-l-part of (O/l
e
)
∗
and has order
|S
0
| =







(l − 1)
2
if (−∆/l) = 1,
l
2
− 1 if (−∆/l) = −1,
l − 1 if (−∆/l) = 0.
There is a subgroup S
00
of S
0
of order l − 1 such that (Z/l
e
)
∗
= U × S
00
.
Proof. To begin with, note that each coset of O/l
e
has a unique representative of
the form a + b
√
−∆ with a, b ∈ {0, 1, . . . , l
e
− 1}.
The order of (O/l
e
)
∗
depends on the decomposition of l in K as follows:
|(O/l
e
)
∗
| =







(l − 1)
2
l
2e−2
if l splits,
(l
2
− 1)l
2e−2
if l is inert,
(l − 1)l
2e−1
if l ramifies.
This gives the order of S
0
.
The subgroups U := h1 + li and V
0
:= h1 + l
√
−∆i of (O/l
e
)
∗
are both
∼
=
Z/l
e−1
and have trivial intersection. Clearly, τ operates trivially on U and by inversion on
(U ×V
0
)/U. So by Lemma 3 (c), U ×V
0
= U ×V for a group V
∼
=
Z/l
e−1
on which
τ operates by inversion. This shows (a) and (b) when l - ∆.
When l | ∆ the same arguments work for V
0
:= h1 +
√
−∆i unless l = 3 and
∆ ≡ 3 mod 9. In the exceptional case l = 3 and ∆ ≡ 3 mod 9, however, the 3-part
of (O/9)
∗
is h1+3i×h1+3
√
−∆i×h1+
√
−∆i
∼
=
(Z/3)
3
, showing V
∼
=
Z/3
e−1
×Z/3.
This finishes the proof of (b).
To see the last part of (c), note U = {u ∈ (Z/l
e
)
∗
| u ≡ 1 mod l}. 
Proof of Main Lemma 1. Consider conductors f = l
e
with e ≥ 1. Let J
f
K
be the
group of fractional ideals prime to f (i.e. prime to l) and let P
f
K
be the subgroup
generated by the principal ideals (α) with integral α ≡ 1 mod f. By class field
theory, the Artin symbol is a surjective homomorphism

K
f
/K

: J
f
K
→ Gal(K
f
/K)
with kernel P
f
K
. It maps the group P
K
of principal ideals prime to l onto Gal(K
f
/K
H
)
and behaves nicely with respect to restriction when e varies. Moreover, since
τ = τ
−1
, the Artin symbol satifies

K
f
/K
τ(p)

= τ

K
f
/K
p

τ .
Assume for simplicity ∆ 6= 1, 3. We then have the natural exact sequence
1 → {±1} → (O/l
e
)
∗
→ P
K
/P
f
K
→ 1
5

where an α ∈ (O/l
e
)
∗
is sent to the principal ideal (α). The Artin symbol thus
induces an isomorphism
lim
←−
(O/l
e
)
∗
/{±1}
∼
=
−→ Gal(K
max
/K
H
) .
Conclude from Lemma 3 that Gal(K
max
/K
H
) = U × V × T
∗
× S
0
with U, V , and
T
∗
as in the theorem, and S
0
finite of order
|S
0
| =







(l − 1)
2
/2 if (−∆/l) = 1,
(l
2
− 1)/2 if (−∆/l) = −1,
(l − 1)/2 if (−∆/l) = 0.
From Lemma 3 also follows Gal(K
max
/K
f
) = U
f
× V
f
with U
f
and V
f
as in the
theorem.
The rest is group theory: Write Gal(K
max
/K) = X ×T
0
with l-part X and non-
l-part T
0
. Then X contains U ×V ×T
∗
, and T
0
contains S
0
with index |T
0
: S
0
| = u.
It is well known that K
H
is a generalised dihedral extension of Q, so that τ operates
by inversion on X/(U × V × T
∗
). It follows from Lemma 2 (b) that τ operates by
inversion on X/U. By Lemma 2 (c), X = U × Y where Y = {x ∈ X | x
τ
= x
−1
}.
Clearly, Y contains V × T
∗
with finite index |Y : V × T
∗
| = l
µ
. By Lemma 2 (a),
Y = W × T with W
∼
=
Z
l
, T finite containing T
∗
, and |V : W ∩ V | ≤ |V |.
In the case ∆ = l = 3, the occurence of a factor of order 3 in O
∗
causes T
∗
to
vanish. So in this situation, we are not in the “exceptional case”. 
III. Prime decomposition laws
Consider a prime ideal p of K, and let p the rational prime it divides. Our main
objective is to give a law for the decomposition or factorisation of p in K
anti
. For
the sake of completeness, we start with the cyclotomic extension in which the law
has the simplest form possible.
Proposition 2. If p = l, then p is totally ramified in K
cycl
. If p 6= l, then p is
unramified in K
cycl
, and p splits totally in the n
th
step of K
cycl
iff p ≡ ±1 mod l
n+1
.
Proof. This is an immediate consequence of Proposition 1 (a) and the law on de-
composition of prime numbers in cyclotomic fields. 
Now we turn to the anti-cyclotomic extension. Recall that K
anti
/K is unramified
outside l by Iwasawa’s result. Define ν ≥ 0 such that K
anti
∩ K
H
= K
(ν)
anti
. Then
any prime P of K
(ν)
anti
dividing l ramifies totally in K
anti
.
The ring class field N
f
over K of conductor f is the maximal subfield of the ray
class field K
f
being dihedral over Q. So K
anti
is contained in the union of the tower
K ⊆ N
1
⊆ N
l
⊆ N
l
2
⊆ . . .
6

If p is inert (resp. ramified) in K/Q, then the ideal class of p is trivial (resp. of
order 2) in the ideal class group of K, and hence class field theory ([6], Theorem
7.3) gives that p splits totally in any ring class field N
f
(resp. in a subfield L of
N
f
with |N
f
: L| = 2) of conductor f prime to l. In particular, p splits totally in
K
anti
if p is different from l and non-split in K (another proof of this is given in
[3], Section III). So the remaining problem is the case where p 6= l splits in K. We
treat first the easier situation where K is l-rational (as defined in Section II).
Theorem 1. Assume K is l-rational, and consider a prime p - d
K
l and an integer
n ≥ 0. Write the class number of K as h = l
µ
u with l - u. For n ≤ µ, p splits in
K
(n)
anti
iff p is representable by a quadratic form of discriminant d
K
whose order in
the form class group is not divisible by l
µ−n+1
. For n > µ, p splits in K
(n)
anti
iff p is
representable by a quadratic form of discriminant
(
d
K
· l
2(n−µ+1)
if l - ∆ or ∆ = l = 3
d
K
· l
2(n−µ)
otherwise
whose order in the form class group is prime to l.
Proof. First some general observations. Consider a ring class field N
f
of K with
arbitrary conductor f. The Galois group Gal(N
f
/K) is isomorphic to the ring
class group of conductor f via the Artin isomorphism. This ring class group is
again isomorphic to the form class group C of discriminant −d
K
f
2
. Now let L be
any field with K j L j N
f
. By the main theorem of Galois theory and the above
isomorphisms, there corresponds to L some subgroup H of C . For a prime number
p dividing neither d
K
nor f, class field theory gives that p splits totally in L iff p is
representable by a quadratic form f whose equivalence class k belongs to H.
Assume n ≤ µ and let N be the ring class field of K with conductor f = 1 (which
equals the Hilbert class field). The l-part of N/K is K
(µ)
anti
since K is l-rational.
The subgroup H of the form class group C of discriminant d
K
corresponding to
L := K
(n)
anti
consists of the classes of forms of order not divisible by l
µ−n+1
. This
proves the first claim.
Now assume n > µ. We only prove the case l - ∆. Let N be the ring class field
of K with conductor f = l
n−µ+1
. By the Main Lemma, the l-part of N/K is K
(n)
anti
since K is l-rational. The subgroup H of the form class group C of discriminant
d
K
f
2
corresponding to L := K
(n)
anti
consists of the classes of forms of order prime to
l. This proves the second claim. 
Antoniadis [1] gives a prime decomposition law for ring class fields and their
subfields involving coefficients of L-series.
Example 1. (i) Let l = 3 and ∆ = 3. We seek the primes p 6= 3 that split in K
(1)
anti
.
The form class group of discriminant −3 · 3
4
= −243 has order 3. So p splits iff it
is representable by the principal form x
2
+ xy + 61y
2
.
7

(ii) Let l = 5 and ∆ = 5. The form class group of discriminant −20 ·5
2
= −500
has order 10. So a prime p 6= 2, 5 splits in K
(1)
anti
iff it is representable by either the
principal form x
2
+ 125y
2
or the form 2x
2
+ 2xy + 63y
2
of order 2.
(iii) Let l = 7 and ∆ = 1. The form class group of discriminant −4 ·7
4
= −9604
is cyclic of order 28. So a prime p 6= 2, 7 splits in K
(1)
anti
iff it is representable by either
the principal form x
2
+ 2401y
2
, or the form 2x
2
+ 2xy + 1201y
2
of order 2, or the
form 41x
2
+ 20xy + 61y
2
of order 4 (the other form of order 4 is 41x
2
−20xy + 61y
2
which represents the same numbers).
When K is not l-rational, the l-part of the the ring class fields of l-power conduc-
tor are not contained in K
anti
, and the problem lies in identifying their intersection.
Theorem 2. Assume that p is different from l and splits in K. We may then
write
p
h
=
(
a
2
+ ∆b
2
if ∆ 6≡ 3 mod 4,
a
2
+ ab + ((∆ + 1)/4)b
2
if ∆ ≡ 3 mod 4,
(1)
with relatively prime a, b ∈ Z. Put ω :=
√
−∆ if ∆ 6≡ 3 mod 4, otherwise ω :=
(1 +
√
−∆)/2. Let n ≥ 0 be an integer.
(a) Suppose l splits in K. Write (a + bω)
l−1
= a
∗
+ b
∗
ω. Then p splits totally in
K
(n)
anti
iff b
∗
≡ 0 mod l
n+1+µ−ν
.
(b) Suppose l is inert in K. Write (a + bω)
l+1
= a
∗
+ b
∗
ω. Then the conclusion
of (a) holds.
(c) Suppose l is ramified in K and we are not in the exceptional case (see below).
Then p splits totally in K
(n)
anti
iff b ≡ 0 mod l
n+µ−ν
.
(d) Suppose l = 3 and ∆ ≡ 3 mod 9 (the exceptional case). Write (a + bω)
3
=
a
∗
+ b
∗
ω. Then p splits totally in K
(n)
anti
iff b
∗
≡ 0 mod 3
n+2+µ−ν
.
In all cases, p only splits in a finite number of steps of K
anti
.
2
Proof. Write (p) = pq with conjugate prime ideals p, q of K. By definition of h, p
h
and q
h
are principal, i.e. p
h
= (a + bω) and q
h
= (a + b¯ω) for some a, b ∈ Z. When
∆ 6≡ 3 mod 4, we have (p
h
) = p
h
q
h
= (a + b
√
−∆)(a − b
√
−∆) = (a
2
+ ∆b
2
) and
consequently p
h
= a
2
+ ∆b
2
. The representation of p
h
in case ∆ ≡ 3 mod 4 is seen
similarly. If a and b were not relatively prime, then p
h
= (a +bω) and q
h
= (a +b¯ω)
would not be relatively prime either, a contradiction.
Now assume a representation
p
h
= (u + vω)(u + v ¯ω)
is given with relatively prime u, v ∈ Z. Then p
h
q
h
= (p
h
) = (u + vω)(u + v¯ω). If
(u + vω) and (u + v ¯ω) were not relatively prime, then one of these ideals would be
2
I wish to thank the referee for calling my attention to this.
8

divisible by pq = (p) which is not the case since u and v are relatively prime. Hence
the ideal (u + vω) equals either p
h
or q
h
, say (u + vω) = p
h
= (a + bω).
The remainder of the proof relies on Main Lemma 1 whose notation we adopt.
The different cases are now treated separately.
(a) Assume l splits in K. It follows immediately from the definition of ν that
Gal(K
max
/K
(ν)
anti
) = U × V × T × T
0
. Hence l
ν
= |W × T : V × T | and |T | = l
µ−ν
since |W × T : V | = l
µ
.
Consider the conductor f = l
e
with e = n + 1 + µ − ν. By Main Lemma 1 (v)
and (ix), V
f
is contained in W and hence
Gal(K
f
/K) =
¯
U ×
¯
W × T × T
0
where
¯
U = U/U
f
is cyclic of order l
e−1
= l
n+µ−ν
,
¯
W = W/V
f
is cyclic of order
l
e−1+ν
= l
n+µ
, and T
0
has order prime to l. The fixed field of
¯
U × T ×T
0
is K
(n+µ)
anti
It follows from Lemma 3 that the image of (Z/l
e
)
∗
under the Artin symbol

K
f
/K

: (O/l
e
)
∗
→ Gal(K
f
/K)
is
¯
U × S
00
where S
00
is a subgroup of T
0
with index u(l − 1).
Let W
0
be the subgroup of
¯
W of order l
µ
. Then K
(n)
anti
is the fixed field of
¯
U × W
0
× T × T
0
. Now class field theory yields (see Neukirch [6]),
p splits in K
(n)
anti
⇔

K
f
/K
p

∈
¯
U × W
0
× T × T
0
⇔

K
f
/K
p
h(l−1)

∈
¯
U × S
00
⇔ b
∗
≡ 0 mod l
e
if we write p
h(l−1)
= (a
∗
+ b
∗
ω).
To show that p only splits in a finite number of steps of K
anti
, we must show
b
∗
6= 0. But this follows from (a
∗
, b
∗
) = 1 which is seen the same way as we saw
(a, b) = 1 above.
(b) If l is inert in K, everything goes the same way except that T
0
has now order
u(l
2
− 1)/2.
(c) Suppose l ramifies in K and we are not in the exceptional case. Then T
0
has order u(l −1)/2, and everything goes as above using the conductor f = l
e
with
e = n + µ − ν.
(d) Suppose we are in the exceptional case. Then |T | = 3
µ−ν+1
and |T
0
| =
u(l −1)/2. Using the conductor f = l
e
with e = n + 2 + µ −ν, the same arguments
hold if we write p
3h
= (a
∗
+ b
∗
ω). 
Remark. Everything goes the same way if one uses the exponent of K’s class
group instead of h.
9

Corollary 1. No rational prime p splits completely in K
anti
.
Proof. The prime l is (infinitely) ramified in K
anti
, since otherwise K
anti
would be an
infinite unramified extension of K, contradicting the finiteness of the class number.
If p is different from l, the claim follows from the last statement of Theorem 2. 
When the l-Hilbert class field of K is non-trivial, the decomposition law depends
on how much of it is contained in K
anti
, expressed by the number ν. Since all primes
trivially split in K
(0)
anti
= K, but not all primes split in K
(1)
anti
, we can give the following
description of ν (here stated in the case where l splits in K, the other cases are
similar): Let p run through all primes 6= l that split in K, and compute b
∗
as in
Theorem 2 (a). Then ν is the minimal integer such that l
1+µ−ν
divides all the b
∗
.
We illustrate this principle by three examples.
Example 2. Let l = 5 and K = Q(
√
−599). The class group of K is cyclic
of order 25. Thus µ = 2 and ν = 0, 1, or 2. The prime p = 2 splits in K since
(−599/2) = 1. We therefore write 2
25
= a
2
+ ab + 150b
2
with a = 5737 and b = 49
and find b
∗
= 37079430566955 (Theorem 2 (a)). Since b
∗
is divisible by 5, but not
by 25, we conclude ν = 2. In other words: the entire Hilbert class field K
H
of K is
contained in K
anti
.
Example 3. Let l = 5 and K = Q(
√
−479). Again, the class group of K is cyclic
of order 25, so µ = 2 and ν = 0, 1, or 2. Further, p = 2 again splits in K. Writing
2
25
= a
2
+ ab + 120b
2
with a = −56 and b = 529 gives b
∗
= −14765386940175 which
is divisible by 25, but not by 125. This shows ν ≥ 1, so K
H
contains at least K
(1)
anti
.
Now class field theory gives a simple decomposition law for K
(1)
anti
: a prime ideal p
of K splits in K
(1)
anti
iff it has order 1 or 5 in the ideal class group. Since 2
5
is not of
the form a
2
+ ab + 120b
2
, a prime p of K dividing 2 has order 25 in the class group,
so it does not split in K
(1)
anti
. If ν were equal to 2, Theorem 2 (a) would contradict
this. Hence ν equals 1, and we conclude: K
anti
contains the subfield of K
H
of degree
5 over K, but not the entire K
H
.
Example 4. Let l = 5 and K = Q(
√
−2887). The class group of K is cyclic
of order 25. Writing 2
25
= a
2
+ ab + 722b
2
with a = 4771 and b = 119 gives
b
∗
= −503658527236874547125 which is divisible by 125. The same arguments as
in Example 2 show that p = 2 is inert in H
K
. This implies ν = 0 and therefore:
K
H
and K
anti
are linearly disjoint over K.
IV. The first step of the anti-cyclotomic extension
In this section we address the problem of finding the first step K
(1)
anti
of the anti-
cyclotomic extension K
anti
/K. By “finding” we understand displaying explicitly a
polynomial f over Q of degree l having K
(1)
anti
as its splitting field. The decomposition
10

laws from section II then dictate the factorisation of f modulo p. In some cases we
will actually use this knowledge of the factorisation to identify f among a number
of candidates.
To begin with, recall that K
(1)
anti
is a dihedral extension of Q of degree 2l hav-
ing K as its quadratic subfield, and that K
(1)
anti
/K is unramified outside l. If K is
l-rational, K
(1)
anti
is unique with these properties. We state without proof a lemma
that allows us easily to determine if a given dihedral extension is unramified, or
unramified outside l, over its quadratic subfield.
Lemma 4. Consider a dihedral extension M/Q of degree 2l having K as its
quadratic subfield. Let L be one of the l subfields of M of absolute degree l.
Then the cyclic extension M/K is unramified iff the field discriminants satisfy
d
L
= d
(l−1)/2
K
. Further, M/K is unramified outside l iff d
L
= (power of l) ·d
(l−1)/2
K
.
So when K is l-rational, we can find K
(1)
anti
by guessing a D
l
-polynomial f whose
splitting field contains K, and such that the discriminant condition of the lemma
is satisfied. Some examples are given in the following table.
∆ h f (for l = 3) f (for l = 5)
1 1 X
3
− 3X − 4 X
5
+ 2500X + 120000
2 1 X
3
− 3X − 10 X
5
+ 6875X + 17500
3 1 X
3
− 3 X
5
+ 10X
3
− 15X
2
+ 10X − 12
5 2 X
3
− 3X − 8 X
5
+ 20X + 32
6 2 X
3
+ 3X − 2 X
5
+ 15X
3
− 70X
2
+ 60X − 24
7 1 X
3
− 3X − 5 X
5
+ 15X
3
− 5X
2
+ 35X − 91
10 2 X
3
− 3X − 22 X
5
− 5X + 12
11 1 X
3
+ 6X − 1 X
5
− 15X
3
− 15X
2
+ 110X + 143
13 2 X
3
+ 9X − 36 X
5
+ 25772500X − 395460000
14 4 X
3
− 3X − 26 X
5
+ 10X
3
− 140X
2
+ 585X − 532
15 2 X
3
+ 3X − 1 X
5
+ 5X
2
+ 3
17 4 X
3
+ 6X − 28 X
5
− 35X
3
− 30X
2
+ 1060X − 2616
19 1 X
3
+ 6X − 5 X
5
+ 35X
3
− 40X
2
+ 160X − 232
Consider one of the polynomials f from the table, and let p be a prime not
dividing the discriminant of f. If p is inert in K, then it splits in K
(1)
anti
. It follows
that f is the product of one linear and (l − 1)/2 irreducible quadratic polynomials
modulo p. If, on the other hand, p splits as pq in K, then f is either irreducible
modulo p, or f is the product of linear factors modulo p – and this happens according
to whether p is inert or splits in K
(1)
anti
.
For example, the result mentioned in the introduction about the factorisation of
the polynomial X
5
+ 20X + 32 modulo p follows immediately from the above table
and Example 1 in section III.
When K is not l–rational, finding K
(1)
anti
is harder since it is no longer unique with
the property of being dihedral over Q and unramified outside l over K. But this case
11

can be dealt with by first finding all fields with that property, and then identifying
K
(1)
anti
using our knowledge of which primes split in that field. This method always
leads to a conclusive answer, for different Galois extensions have different sets of
splitting primes by a theorem of Bauer (see Neukirch [6], page 572). We illustrate
by two examples.
Example 1. 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
, . . . , f
4
whose splitting fields are the
above-mentioned four dihedral fields. The polynomials are shown in the below table
together with all primes < 200 modulo which they split into linear factors. These
prime lists are the “finger prints” of the polynomials, and we shall use them to
uncover the culprit among our four suspects.
i f
i
primes < 200 modulo which f
i
splits
1 X
3
− 3X + 16 17, 101, 107, 139, 179, 193
2 X
3
+ 9X + 12 11, 19, 89, 103, 191
3 X
3
+ 9X + 30 5, 71, 109, 199
4 X
3
+ 18X + 12 23, 31, 37, 41, 173
Now consider a prime p that splits in K, i.e. with (−21/p) = 1. The class group of
K has exponent 2, so we may write
p
2
= a
2
+ 21b
2
with relatively prime a, b ∈ N. This is shown in the table below for all p < 200. We
have
(a + b
√
−21)
3
= (a
3
− 63ab
2
) + (3a
2
b − 21b
3
)
√
−21 .
Therefore, by Theorem 2 (d), p splits in K
(1)
anti
iff b
∗
= 3a
2
b −21b
3
is divisible by 27.
The primes for which this is the case are typed with bold in the table.
p a b b
∗
5 2 1 −9
11 10 1 279
17 10 3 333
19 5 4 −1044
23 2 5 −2565
31 25 4 6156
37 5 8 −10152
41 34 5 14715
71 50 11 54549
89 86 5 108315
p a b b
∗
101 74 15 175545
103 47 20 −35460
107 82 15 231705
109 59 20 40860
139 85 24 229896
173 170 7 599697
179 10 39 −1233999
191 170 19 1503261
193 185 12 1195812
199 185 16 1556784
12

Comparing the bold primes with the ones in the previous table reveals f
4
as the
wanted polynomial.
Let us note additionally that p splits in the 3-part of K’s ray class field of
conductor 3 iff b is divisible by 3. The table shows that this is the case for the
primes 17, 101, 107 etc., i.e. the primes modulo which the polynomial f
1
splits. So
this ray class field is the splitting field of f
1
. Finally, all four polynomials f
i
split
modulo p iff b is divisible by 9.
Example 2. 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 (2005)), and we find four
candidate polynomials:
i f
i
primes < 200 modulo which f
i
splits
1 X
3
− X + 4 29, 47, 83, 137
2 X
3
+ 6X − 17 23, 37, 47, 61, 79, 101, 149
3 X
3
+ 15X − 28 11, 19, 47, 151, 163, 197
4 X
3
+ 18X − 45 13, 41, 47, 53, 89, 193, 199
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.
Let p 6= 3 be a prime that splits in K. Since K has class number 3, we write
p
3
= a
2
+ ab + 27b
2
with relatively prime a, b ∈ Z. This representation is shown in the table below for
all p < 200. We are in case (a) of Theorem 2 and must compute
(a + bω)
2
= (a
2
− 27b
2
) + (2ab + b
2
)ω .
Thus, p splits in K
(1)
anti
iff b
∗
= 2ab + b
2
is divisible by 3
3−ν
.
p a b b
∗
11 1 7 63
13 1 9 99
19 64 9 1233
23 89 11 2079
29 107 20 4680
37 163 27 9531
41 118 43 11997
47 253 34 18360
53 341 29 20619
61 442 27 24597
79 523 81 91287
p a b b
∗
83 109 142 51120
89 694 79 115893
101 962 47 92637
137 163 304 191520
149 953 281 614547
151 1412 207 627417
163 1360 279 836721
193 1189 441 1243179
197 2690 83 453429
199 316 531 617553
13

Now if the Hilbert class field were contained in K
anti
, that is if ν = 1, then all the
primes in the table would split in K
(1)
anti
since all the b
∗
are divisible by 9. Not only
does this seem unlikely, it is also demonstrably false since none of the polynomials
f
i
splits modulo all these primes. Hence ν = 0, and the Hilbert class field is not
contained in K
anti
. So the p that split in K
(1)
anti
are the ones for which 27 divides b
∗
.
These primes (typed bold in the table) are the ones in the second line of the previous
table, thereby identifying f
2
as the polynomial whose splitting field is K
(1)
anti
.
References
[1] J. A. Antoniadis, Diedergruppe und Reziprozit¨atsgesetz, J. Reine Angew. Math.
377 (1987), 197–209.
[2] D. Brink, On Z
p
-embeddability of cyclic p-class fields, C. R. Math. Acad. Sci.
Soc. R. Can. 27 (2005), 48–53.
[3] J. E. Carroll, H. Kisilevsky, Initial layers of Z
l
-extensions of complex quadratic
fields, Compositio Math. 32 (1976), no. 2, 157–168.
[4] K. Iwasawa, On Z
l
-extensions of algebraic number fields, Ann. of Math. (2) 98
(1973), 246–326.
[5] 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.
[6] J. Neukirch, Algebraische Zahlentheorie, Springer, Berlin, 1992.
14