Steinitz classes of some abelian and nonabelian extensions of even degree

0.22 MB·English

by  Alessandro Cobbe

Steinitz classes of some abelian and nonabelian
extensions of even degree
0
Alessandro Cobbe
1
0
2
January 15, 2010

n
a
J

5
1

] Abstract
T
N
The Steinitz class of a number field extension K/k is anideal class in the ring
.
h
of integers of k, which, together with the degree [K : k] of the extension
t k
a O
determines the -module structure of . We call R (k,G) the classes
m k K t
O O
which are Steinitz classes of a tamely ramified G-extension of k. We will say
[
that those classes are realizable for the group G; it is conjectured that the

1
set of realizable classes is always a group.
v
1 In this paper we will develop some of the ideas contained in [8] to obtain
2
some results in the case of groups of even order. In particular we show that
7
2 to study the realizable Steinitz classes for abelian groups, it is enough to
.
1 consider the case of cyclic groups of 2-power degree.
0
0
1
Introduction
:
v
i
X
Let K/k be an extension of number fields and let and be their rings
r OK Ok
a of integers. By Theorem 1.13 in [18] we know that
[K:k]−1
= I
OK ∼ Ok ⊕
where I is an ideal of . By Theorem 1.14 in [18] the -module structure
k k
O O
of is determined by [K : k] and the ideal class of I. This class is called
K
O
the Steinitz class of K/k and we will indicate it by st(K/k). Let k be a
number field and G a finite group, then we define:
R (k,G) = x Cl(k) : K/k tame, Gal(K/k) = G,st(K/k) = x .
t ∼
{ ∈ ∃ }
1
1 Preliminary results
In this paper we will use the notations and some results from [8] to study
the realizable classes for some groups of even order. In particular we will
obtain some explicit results for abelian groups. In this case it follows by
[17] that the realizable Steinitz classes form a group, but there is no known
explicit characterization of them in the most general situation.
This paper is a slightly shortened version of parts of the author’s PhD
thesis [7]. For earlier results see [1], [2], [3], [4], [5], [6], [9], [10], [11], [13],
[14], [15], [16], [21], [22] and [23].
Acknowledgements
I am very grateful to Professor Cornelius Greither and to Professor Roberto
Dvornicich for their advice and for the patience they showed, assisting me
in the writing of my PhD thesis with a lot of suggestions. I also wish to
thank the Scuola Normale Superiore of Pisa, for its role in my mathematical
education and for its support during the time I was working on my PhD
thesis.
1 Preliminary results
We start recalling the following two fundamental results.
Theorem 1.1. If K/k is a finite tame Galois extension then
d(K/k) = p(ep−1)[Kep:k],
p
Y
where e is the ramification index of p.
p
Proof. This follows by Propositions 8 and 14 of chapter III of [12].
Theorem 1.2. Assume K is a finite Galois extension of a number field k.
(a) If its Galois group either has odd order or has a noncyclic 2-Sylow
subgroup then d(K/k) is the square of an ideal and this ideal represents
the Steinitz class of the extension.
(b) If its Galois group is of even order with a cyclic 2-Sylow subgroup and
α is any element of k whose square root generates the quadratic subex-
tension of K/k then d(K/k)/α is the square of a fractional ideal and
this ideal represents the Steinitz class of the extension.
2
Proof. This is a corollary of Theorem I.1.1 in [9]. In particular it is shown in
[9] that in case (b) K/k does have exactly one quadratic subextension.
Further, considering Steinitz classes in towers of extensions, we will need
the following proposition.
Proposition 1.3. Suppose K/E and E/k are number fields extensions.
Then
st(K/k) = st(E/k)[K:E]N (st(K/E)).
E/k
Proof. This is Proposition I.1.2 in [9].
We will use a lot of results proved in [8]. In this section we prove a few
facts we will use later.
Lemma 1.4. Foranye mthe greatestcommondivisor, forl e, of the integers
| |
(l 1) m divides (e 1)m.
− e(l) − e
Proof. First of all it is clear that we can assume that m = e.
Let I be the Z-ideal generated by the l 1, for all primes l e. Then e 1
− | ≡
(mod I), since it is the product of prime factors, each one congruent to 1
modulo I. It follows that for any prime l ∤ e, there exists an l e, such that
1
|
the l-component of l 1, which coincides with that of (l 1) e , divides
1 − 1 − e(l1)
that of e 1. Finally, for any l e, l does not divide (l 1) e .
− | − e(l)
Lemma 1.5. Let m,n,x be integers. If x 1 (mod m) and any prime q
≡
dividing n divides also m then
xn 1 (mod mn).
≡
Proof. Let n = q ...q be the prime decomposition of n (q and q with i = j
1 r i j
6
are allowed to be equal). We prove by induction on r that xn 1 (mod mn).
≡
If r = 1, then mn = mq must divide mq1 and there exists b N such that
1
∈
q1−1 q
xn = (1+bm)q1 = 1+ 1 (bm)i +(bm)q1 1 (mod mn).
i ≡
i=1 (cid:18) (cid:19)
X
Let us assume that the lemma is true for r 1 and prove it for r. Since q m,
r
− |
as above, for some c N we have
∈
qr q
xn = (1+cmq ...q )qr = 1+ r (cmq ...q )i 1 (mod mn).
1 r−1 1 r−1
i ≡
i=1 (cid:18) (cid:19)
X
3
2 Some general results
Definition 1.6. Let K/k be a finite abelian extension of number fields. Then
we define W(k,K) in the following equivalent ways (the equivalence is shown
in [8], Proposition 1.10):
W(k,K) = x J /P : x contains infinitely many primes of absolute
k k
{ ∈
degree 1 splitting completely in K
}
W(k,K) = x J /P : x contains a prime splitting completely in K
k k
{ ∈ }
W(k,K) = N (J ) P /P .
K/k K k k
·
In the case of cyclotomic extensions we will also use the shorter notation
W(k,m) = W(k,k(ζ )).
m
Lemma 1.7. If q n q m then W(k,m)n W(k,mn).
| ⇒ | ⊆
Proof. Let x W(k,m). By the definition and by Lemma 1.11 of [8], x
∈
contains a prime ideal p, prime to mn and such that N (p) Pm, where
k/Q ∈ Q
m = m p . Then by Lemma 1.5, N (pn) Pn, with n = mn p , and it
· ∞ k/Q ∈ Q · ∞
follows from Lemma 1.12 of [8] that xn W(k,mn).
∈
2 Some general results
We recall the following definition, from [8].
Definition 2.1. We will call a finite group G good if the following properties
are verified:
1. For any number field k, R (k,G) is a group.
t
2. For any tame G-extension K/k of number fields there exists an element
α k such that:
K/k
∈
(a) If G is of even order with a cyclic 2-Sylow subgroup, then a square
root of α generates the quadratic subextension of K/k; if G
K/k
either has odd order or has a noncyclic 2-Sylow subgroup, then
α = 1.
K/k
(b) For any prime p, with ramification index e in K/k, the ideal class
p
of
1
p(ep−1)emp−vp(αK/k) 2
(cid:16) (cid:17)
is in R (k,G).
t
4
3. For any tame G-extension K/k of number fields, for any prime ideal
p of k and any rational prime l dividing its ramification index e , the
p
class of the ideal
(l−1) m
p ep(l)
is in R (k,G) and, if 2 divides (l 1) m , the class of
t − ep(l)
l−1 m
p 2 ep(l)
is in R (k,G).
t
4. G is such that for any number field k, for any class x R (k,G) and
t
∈
any integer n, there exists a tame G-extension K with Steinitz class x
and such that every non trivial subextension of K/k is ramified at some
primes which are unramified in k(ζ )/k.
n
In [8] we prove that abelian groups of odd order are good and, more
generally, we construct nonabelian good groups by an iteration of direct and
semidirect products.
Let be a finite group of order m, let H = C(n ) C(n ) be an
1 r
G × ···×
abelian group of order n, with generators τ ,...,τ and with n n . Let
1 r i+1 i
|
µ : Aut(H)
G →
be an action of on H and let
G
ϕ ψ
0 H G 0
→ −→ −→ G →
be an exact sequence of groups such that the induced action of on H is µ.
G
We assume that the group G is determined, up to isomorphism, by the above
exact sequence and by the action µ. The following well-known proposition
shows a class of situations in which our assumption is true.
Proposition 2.2 (Schur-Zassenhaus, 1937). If the order of H is prime to
the order of then G is a semidirect product:
G
G = H ⋊ .
∼ µ
G
Proof. This is Theorem 7.41 in [20].
WearegoingtostudyR (k,G), considering inparticularthecaseinwhich
t
the order of H is even.
5
2 Some general results
We also define
1 if 2 ∤ n or the 2-Sylow subgroups of G are not cyclic
η =
G
(2 if 2 n and the 2-Sylow subgroups of G are cyclic
|
and in a similar way we define η and η . We will always use the letter l
H G
only for prime numbers, even if not explicitly indicated.
We say that (K,k ,k) is of type µ if k /k, K/k and K/k are Galois
1 1 1
extensions with Galois groups isomorphic to , H and G respectively and
G
such that the action of Gal(k /k) = on Gal(K/k ) = H is given by µ. For
1 ∼ 1 ∼
G
any -extension k of k we define R (k ,k,µ) as the set of those ideal classes
1 t 1
G
of k which are Steinitz classes of a tamely ramified extension K/k for which
1 1
(K,k ,k) is of type µ.
1
We will repeatedly use the following generalization of the Multiplication
Lemma on page 22 in [9] by Lawrence P. Endo.
Lemma 2.3. Let (K ,k ,k) and (K ,k ,k) be extensions of type µ, such
1 1 2 1
that (d(K /k ),d(K /k )) = 1 and K /k and K /k have no non-trivial
1 1 2 1 1 1 2 1
unramified subextensions. Then there exists an extension (K,k ,k) of type
1
µ, such that K K K and for which
1 2
⊆
st(K/k ) = st(K /k )st(K /k ).
1 1 1 2 1
Proof. This is Lemma 2.5 in [8].
Now we recall some further notations introduced in [8].
For any τ H we define
∈
G˜ = (g ,g ) Gal(k(ζ )/k) : µ(g )(τ) = τνk,τ(g2) ,
k,µ,τ 1 2 o(τ) 1
∈ G ×
where g (ζ ) =(cid:8) ζνk,τ(g2) for any g Gal(k(ζ )/k), (cid:9)
2 o(τ) o(τ) 2 ∈ o(τ)
G = g Gal(k(ζ )/k) : g , (g ,g) G˜
k,µ,τ o(τ) 1 1 k,µ,τ
∈ ∃ ∈ G ∈
n o
and E as the fixed field of G in k(ζ ).
k,µ,τ k,µ,τ o(τ)
Given a -extension k of k, there is an injection of Gal(k (ζ )/k) into
1 1 o(τ)
G
Gal(k(ζ )/k) (defined in the obvious way). We will always identify
o(τ)
G ×
Gal(k (ζ )/k) with its image in Gal(k(ζ )/k). So we may consider
1 o(τ) o(τ)
G ×
the subgroup
G˜ = G˜ Gal(k (ζ )/k)
k1/k,µ,τ k,µ,τ ∩ 1 o(τ)
of G˜ . Let Z be its fixed field in k (ζ ).
k,µ,τ k1/k,µ,τ 1 o(τ)
If k k(ζ ) = k then Gal(k (ζ )/k) = Gal(k(ζ )/k) and hence
1 o(τ) 1 o(τ) ∼ o(τ)
∩ G×
G˜ = G˜ .
k1/k,µ,τ k,µ,τ
Now we can state one of the principal theorems proved in [8].
6
Theorem 2.4. Let k be a number field and let be a good group of order
G
m. Let H = C(n ) C(n ) be an abelian group of odd order prime to
1 r
×···×
m and let µ be an action of on H. Then
G
R (k,H ⋊ ) = R (k, )n W (k,E )l−21om(τn) .
t µ t k,µ,τ
G G
l|n τ∈H(l)\{1}
Y Y
Furthermore G = H ⋊ is good.
µ
G
Proof. This is Theorem 2.19 of [8].
In this paper we will obtain some results also for abelian groups H of
even order, if all the other hypotheses of the above theorem continue to hold.
Lemma 2.5. Let k be a tame -extension of k and let x W(k,k (ζ )).
1 G ∈ 1 n1
Then there exist tame extensions of k of type µ, whose Steinitz classes (over
1
k ) are ι(x)ηHα, where
1
r
n 1 n n 1 n
i 1
α = − + − .
2 n 2 n
i 1
i=1
X
In particular there exist tame extensions of k of type µ with trivial Steinitz
1
class.
We can choose these extensions so that they are unramified at all infinite
primes, that the discriminants are prime to a given ideal I of and that
k
O
all their proper subextensions are ramified.
Proof. As in the proof of Lemma 2.10 in [8] we can construct an extension
of type µ with discriminant
r
d = q(ni−1)nni q(n1−1)nn1 ,
i r+1 Ok1
!
i=1
Y
where q ,...,q are prime ideals of k in the class of x.
1 r+1
If H is of odd order or the 2-Sylow subgroup of H is cyclic then the
result follows immediately by Theorem 1.2 (a). If this is not the case then
by Theorem 1.2 (b) we obtain extensions whose Steinitz classes have x2α
as their square. We may construct infinitely many such µ-extensions whose
discriminants over k are relatively prime andso, by the pigeonhole principle,
1
there are two of them, which we call K and K , with the same Steinitz class.
1 2
Thentheconclusionfollowsconsidering theextensionK givenbyLemma2.3.
As in the proof of Lemma 2.10 in[8] we can assume that all the additional
conditions are also verified.
7
2 Some general results
Lemma 2.6. Let k be a -extension of k, let H be a group of even order
1
G
n, let τ H(2) 1 and let x be any class in W(k,Z ). Then there
∈ \ { } k1/k,µ,τ
exist extensions of k of type µ, whose Steinitz classes (over k ) are ι(x)ηHαj,
1 1
where:
n
(a) α = ,
1
2
n
(b) α = (o(τ) 1) ,
2
− o(τ)
Further there exist extensions whose Steinitz classes have ι(x)2αj as their
square. We can choose these extensions so that they satisfy the additional
conditions of Lemma 2.5.
Proof. (a) As in the proof of Lemma 2.11 (a) in [8] we can construct an
extension of type µ with discriminant
n
d(K/k1) (q1q2)2Ok1 ,
(cid:0) (cid:1)
where q and q are prime ideals of k in the class of x and (K,k ,k)
1 2 1
is a µ-extension of k with trivial Steinitz class, obtained by Lemma
1
2.5. Its Steinitz class has ι(x)2α1 as its square and we conclude as in
Lemma 2.5.
(b) In this case, as in the proof of Lemma 2.11 (b) of [8], we obtain an
extension of type µ with discriminant
d(K/k ) (q q )(o(τ)−1)o(nτ) .
1 1 2 Ok1
(cid:16) (cid:17)
We conclude as in (a).
Lemma 2.7. Let k /k be a -extension of number fields, let H(2) be the
1
2-Sylow subgroup of H and H˜Gsuch that H = H(2) H˜. Let µ and µ
× H˜ H(2)
be the actions of induced by µ on H˜ and H(2) respectively. Then
G
R (k ,k,µ )n(2) R (k ,k,µ).
t 1 H˜ ⊆ t 1
˜
Proof. Let x R (k ,k,µ ) and let (K,k ,k) be a µ -extension of k with
∈ t 1 H˜ 1 H˜ 1
Steinitz class x, which is the class of
d(K˜/k1)21.
8
Let (K,k ,k) be a µ -extension of k with trivial Steinitz class and such
1 H(2) 1
that K/k and K˜/k are arithmetically disjoint (such an extension exists
1 1
because of Lemma 2.5). The Steinitz class of K/k is the class of
1
d(K/k ) 2
1
α
(cid:18) (cid:19)
˜
for a certain α k . Then the extension (KK,k ,k) is a µ-extension and its
1 1
∈
Steinitz class is the class of
1
d(KK˜/k) 2 = d(K˜/k)n(22) d(K/k) 2nn(2)
αn(n2) ! α
(cid:18) (cid:19)
which is xn(2).
At this point we can prove the following proposition.
Proposition 2.8. Let l = 2 be a prime dividing n and let τ H(l), then
6 ∈
l−1 n
ι W k,Z 2 o(τ) R (k ,k,µ)
k1/k,µ,τ ⊆ t 1
If 2 n then, for any(cid:0)τ (cid:0)H(2), (cid:1)(cid:1)
| ∈
ι W k,Z ηHo(nτ) R (k ,k,µ)
k1/k,µ,τ ⊆ t 1
and (cid:0) (cid:0) (cid:1)(cid:1)
ι W k,Z 2o(nτ) R (k ,k,µ)2.
k1/k,µ,τ ⊆ t 1
We can choose the correspondingextensions so that they satisfy the additional
(cid:0) (cid:0) (cid:1)(cid:1)
conditions of Lemma 2.5.
Proof. The first inclusion follows immediately by Lemma 2.12 of [8] and by
Lemma 2.7.
Now let us assume that 2 n, let τ H(2) and let x W(k,Z ). It
| ∈ ∈ k1/k,µ,τ
follows from Lemma 2.3 and Lemma 2.6 that ι(x)ηHβ2 is in R (k ,k,µ) and
t 1
ι(x)2β2 is in R (k ,k,µ)2, where
t 1
n n
β = gcd ,(o(τ) 1) .
2
2 − o(τ)
(cid:18) (cid:19)
So we obtain
ι(x)ηHo(nτ) R (k ,k,µ)
t 1
∈
and
ι(x)2o(nτ) R (k ,k,µ)2.
t 1
∈
To conclude we observe that applying Lemma 2.3 to extensions (K ,k ,k)
1 1
and(K ,k ,k)oftypeµwhichsatisfytheadditionalconditionsofLemma2.5,
2 1
we obtainanextension (K,k ,k)which still satisfies the sameconditions.
1
9
2 Some general results
Proposition 2.9. Let a be a multiple of a positive integer n . Let k be a
1
number field and let be a finite group such that for any class x R (k, )
t
G ∈ G
there exists a tame -extension k with Steinitz class x and such that every
1
G
subextension of k /k is ramified at some primes which are unramified in
1
k(ζ )/k.
a
Let H = C(n ) C(n ) be an abelian group of order n and let µ be
1 r
×···×
an action of on H. We assume that the exact sequence
G
ϕ ψ
0 H G 0,
→ −→ −→ G →
in which the induced action of on H is µ, determines the group G, up to
G
isomorphism. Further we assume that H is of odd order or with noncyclic
2-Sylow subgroup, or that is of odd order. Then
G
R (k,H⋊ ) R (k, )n W (k,E )l−21om(τn) W (k,E )ηoG(mτ)n .
t µ t k,µ,τ k,µ,τ
G ⊇ G
l|n τ∈H(l) τ∈H(2)
Y Y Y
l6=2
Further we can choose tame G-extensions K/k with a given Steinitz class
(of the ones considered above), such that every nontrivial subextension of
K/k is ramified at some primes which are unramified in k(ζ )/k.
a
Proof. Let x R (k, ) and let k be a tame -extension of k, with Steinitz
t 1
∈ G G
class x, and such that every subextension of k /k is ramified at some primes
1
which are unramified in k(ζ )/k. Thus, since a is a multiple of n , it follows
a 1
also that k k(ζ ) = k.
1 ∩ n1
By Lemma 2.3, Lemma 2.4 in [8], Proposition 2.8 and Proposition 1.3 we
obtain
R (k,H ⋊ ) xn W (k,E )l−21om(τn) W (k,E )ηHo(τm)n ,
t µ k,µ,τ k,µ,τ
G ⊇
l|n τ∈H(l) τ∈H(2)
Y Y Y
l6=2
from which we obtain the result we wanted to prove, if η = η .
H G
With our hypotheses η = η implies that the order of H is odd, i.e. that
H G
6
there does not exist any nontrivial τ H(2). Hence we obtain the desired
∈
result also in this case.
Proposition 2.10. Let τ,τ˜ H(2) be elements such that τ,τ˜,ττ˜ are all of
∈
the same order. Let k be a -extension of k. Then
1
G
n
ι(W(k,Z Z Z ))2o(τ) R (k ,k,µ).
k1/k,µ,τ k1/k,µ,τ˜ k1/k,µ,ττ˜ ⊆ t 1
10