An explicit candidate for the set of Steinitz classes of tame Galois extensions with fixed Galois group of odd order
aa r X i v : . [ m a t h . N T ] M a r An explicit candidate for the set of Steinitz classesof tame Galois extensions with fixed Galois groupof odd order
Luca Caputo and Alessandro CobbeSeptember 30, 2018
Abstract -
Given a finite group G and a number field k , a well-known conjecture assertsthat the set R t ( k, G ) of Steinitz classes of tame G -Galois extensions of k is a subgroup of theideal class group of k . In this paper we investigate an explicit candidate for R t ( k, G ), when G is of odd order. More precisely, we define a subgroup W ( k, G ) of the class group of k andwe prove that R t ( k, G ) ⊆ W ( k, G ). We show that equality holds for all groups of odd orderfor which a description of R t ( k, G ) is known so far. Furthermore, by refining techniquesintroduced in [8], we use the Shafarevich-Weil Theorem in cohomological class field theory,to construct some tame Galois extensions with given Steinitz class. In particular, thisallows us to prove the equality R t ( k, G ) = W ( k, G ) when G is a group of order dividing ℓ ,where ℓ is an odd prime. Let
K/k be an extension of number fields and let O K and O k be the rings of integersof K and k respectively. Then O K ∼ = O [ K : k ] − k ⊕ I as O k -modules, where I is an ideal in O k which is determined by the extension K/k up to principal ideals. The ideal class of I is called the Steinitz class st( K/k ) of theextension.For a number field k , denote by Cl( k ) the class group of k . If G is a finitegroup, one can consider the set of all Steinitz classes which arise from tame G -Galoisextensions of k :R t ( k, G ) = { x ∈ Cl( k ) : ∃ K/k tame Galois, Gal(
K/k ) ∼ = G, st( K/k ) = x } . There is no known description of R t ( k, G ) in general, but in all cases for which itis known it turns out to be a subgroup of the ideal class group of k . This leads toformulate the following conjecture. Conjecture . For any number field k and any finite group G , R t ( k, G ) is a subgroupof Cl( k ).The conjecture is known to hold for abelian groups as a consequence of the resultsof the paper [14] by Leon McCulloh. An explicit description of R t ( k, G ) for abelian1roups of odd order was first found by Lawrence Paul Endo in his unpublished PhDthesis [10], in which he also obtained some explicit results in the even case.Concerning nonabelian groups, there are a lot of results for groups of some parti-cular shape, however sometimes with restrictions on the base field k . A list of recentresults includes [1], [2], [4], [5], [6], [7], [8], [9], [11], [12], [13], [18], [20], [21] and [22].The aim of this paper is to develop some new ideas to improve the results of [8]and [9]. In Section 2 we generalize [8, Lemma 3.14], obtaining some limitations onthe prime ideals which can ramify in tame Galois extensions of a number field withGalois group G . In particular, this gives an inclusion of R t ( k, G ) in an explicitlydescribed subgroup of Cl( k ), which we baptize W ( k, G ) (see Definition 2.11). Itseems natural to ask whether this inclusion is always an equality for groups of oddorder (see Question 2.18), while some results contained in [10] give counterexamplesalready for abelian groups of even order. Note, in particular, that a positive answerto Question 2.18 would have Conjecture 1.1 as a consequence (for groups of oddorder).In Section 3 we extend the result of [8, Proposition 3.13], using a theorem byShafarevich and Weil as a tool to construct Galois extensions of a given number field.The main idea is to use some cohomological information encoded in the reciprocitymap of an abelian extension of a number field, which is itself a Galois extension of asmaller field.In the last two sections we answer affirmatively to Question 2.18 for A ′ -groupsof odd order ( A ′ -groups have been introduced in [8], see also Definition 4.1) and for ℓ -groups of order up to ℓ , where ℓ is an odd prime.The case of groups of prime power order already shows some interesting com-plications, since for instance a two-step approach corresponding to a given splittingof the group as a semidirect product is not always enough to obtain all the neededSteinitz classes. The difficulty here comes from those extensions with prime idealsramifying in both steps which, contrary to the case of A ′ -groups, cannot be left out.More precisely, given an arbitrary element τ ∈ G , we need to exhibit tame G -Galoisextensions such that τ is a generator of the inertia group of some prime ideals be-longing to some prescribed ideal classes. To this aim, in Section 5, we develop atechnique involving the construction of a larger Galois extension of the ground field,whose Galois group has G as a quotient (see the proof of Lemma 5.8). This methodmay be used to prove the equality R t ( k, G ) = W ( k, G ) for a lot of other groups G of some particular kind. Anyway, we do not believe that it is worth to include moreexamples and we feel that one probably needs significantly new ideas to get somemore general results. For the rest of the paper we will use the letter G to denote a finite group. For any τ ∈ G , let N G ( τ ) (resp. C G ( τ )) be the normalizer (resp. the centralizer) of thesubgroup of G generated by τ and let o ( τ ) be the order of τ . For a positive integer s , ζ s will denote a primitive s -th root of unity. Finally we will always use the letter ℓ only for rational prime numbers, even if not explicitly indicated.2or any τ ∈ G , we define a homomorphism ϕ τ : N G ( τ ) → ( Z /o ( τ ) Z ) × by ϕ τ ( σ ) = α , where στ σ − = τ α . Furthermore, if k is a number field, we need thecyclotomic character ν k,τ : Gal( k ( ζ o ( τ ) ) /k ) → ( Z /o ( τ ) Z ) × , which is defined by ν k,τ ( g ) = β , where g ( ζ o ( τ ) ) = ζ βo ( τ ) . Note that ν k,τ is injective.We also set H k,G,τ = ν − k,τ ( ϕ τ ( N G ( τ ))) ⊆ Gal( k ( ζ o ( τ ) ) /k )and we call E k,G,τ the subfield of k ( ζ o ( τ ) ) which is fixed by H k,G,τ . Remark . We have the following commutative diagram ϕ − τ ( ν k,τ (Gal( k ( ζ o ( τ ) ) /k ))) (cid:31) (cid:127) ι / / ν − k,τ ◦ ϕ τ (cid:15) (cid:15) N G ( τ ) ϕ τ (cid:15) (cid:15) Gal( k ( ζ o ( τ ) ) /k ) ν k,τ / / ( Z /o ( τ ) Z ) × and it is easy to verify that ϕ − τ ( ν k,τ (Gal( k ( ζ o ( τ ) ) /k ))) is the pullback of ϕ τ and ν k,τ .Its image in Gal( k ( ζ o ( τ ) ) /k ) is exactly H k,G,τ .We start proving some elementary properties of the number fields E k,G,τ , whichwill be useful throughout the paper. Lemma 2.2.
Suppose that G = H ⋊ µ G is the (internal) semidirect product of twosubgroups, H and G . Then, for all σ ∈ G , we have E k,G,σ = E k, G ,σ . If H is abelian, then, for all τ ∈ H , E k,G,τ coincides with the E k,µ,τ defined in [8,Section 3.1].Proof. Let σ ∈ G and let σ ∈ G , τ ∈ H be such that σ τ ∈ N G ( σ ). Then σ ϕ σ ( σ τ ) = σ τ στ − σ − = ( σ τ σ − )( σ στ − σ − σ − ) σ σσ − . Since σ τ σ − , σ στ − σ − σ − ∈ H and each element of G can be written uniquelyas a product of an element of H and one of G , we deduce that( σ τ σ − )( σ στ − σ − σ − ) = 1and σ ϕ σ ( σ τ ) = σ σσ − . Hence σ ∈ N G ( σ ) and ϕ σ ( σ τ ) = ϕ σ ( σ ). Therefore we obtain ϕ σ ( N G ( σ )) = ϕ σ ( N G ( σ )) , i.e. E k,G,σ = E k, G ,σ . The second part of the lemma is trivial. 3 emma 2.3.
Let σ ∈ G and n ∈ N . Then E k,G,σ n ⊆ E k,G,σ . In particular if τ ∈ G generates the same cyclic subgroup as σ , then E k,G,σ = E k,G,τ . Proof.
We consider the restrictionres : Gal( k ( ζ o ( σ ) ) /k ) → Gal( k ( ζ o ( σ n ) ) /k )and the projection π : ( Z /o ( σ ) Z ) × → ( Z /o ( σ n ) Z ) × . Using that N G ( σ ) ⊆ N G ( σ n ), we get commutative diagrams N G ( σ ) ϕ σ / / (cid:127) _ ι (cid:15) (cid:15) ( Z /o ( σ ) Z ) × π (cid:15) (cid:15) N G ( σ n ) ϕ σn / / ( Z /o ( σ n ) Z ) × Gal( k ( ζ o ( σ ) ) /k ) ν k,σ / / res (cid:15) (cid:15) ( Z /o ( σ ) Z ) × π (cid:15) (cid:15) Gal( k ( ζ o ( σ n ) ) /k ) ν k,σn / / ( Z /o ( σ n ) Z ) × . Therefore H k,G,σ ⊆ res − ( H k,G,σ n ), since H k,G,σ = ν − k,σ ( ϕ σ ( N G ( σ ))) ⊆ res − ( ν − k,σ n ( ν k,σ n (res( ν − k,σ ( ϕ σ ( N G ( σ )))))))= res − ( ν − k,σ n ( π ( ν k,σ ( ν − k,σ ( ϕ σ ( N G ( σ ))))))) ⊆ res − ( ν − k,σ n ( π ( ϕ σ ( N G ( σ )))))= res − ( ν − k,σ n ( ϕ σ n ( ι ( N G ( σ ))))) ⊆ res − ( ν − k,σ n ( ϕ σ n ( N G ( σ n )))) = res − ( H k,G,σ n ) . Now the fixed field of res − ( H k,G,σ n ) in k ( ζ o ( σ ) ) coincides with the fixed field of H k,G,σ n in k ( ζ o ( σ n ) ), i.e. with E k,G,σ n . Therefore E k,G,σ n ⊆ E k,G,σ . Lemma 2.4.
Let σ, τ ∈ G , then E k,G,σ = E k,G,τστ − . Proof.
We observe that N G ( τ στ − ) = τ N G ( σ ) τ − . Further, for any ρ ∈ N G ( σ ), wehave ( τ στ − ) ϕ τστ − ( τρτ − ) = ( τ ρτ − )( τ στ − )( τ ρτ − ) − = τ ρσρ − τ − = τ σ ϕ σ ( ρ ) τ − = ( τ στ − ) ϕ σ ( ρ ) , i.e. ϕ τστ − ( τ ρτ − ) = ϕ σ ( ρ ) . Therefore ϕ τστ − ( N G ( τ στ − )) = ϕ τστ − ( τ N G ( σ ) τ − ) = ϕ σ ( N G ( σ ))and, since clearly ν k,σ = ν k,τστ − , we conclude that H k,G,σ = H k,G,τστ − , and therefore E k,G,σ = E k,G,τστ − . 4e recall the following important definition from [10, Section I.2] (see also [8,Definition 2.8 and Proposition 2.10]). Definition 2.5.
For any finite abelian extension
K/k of number fields, we define W ( k, K ) = N K/k J K · P k /P k ⊆ Cl( k ) , where J K is the group of fractional ideals of K , P k is the group of principal ideals in k and the map N K/k is induced by the norm from K to k . For any positive integer m , we use the notation W ( k, m ) = W ( k, k ( ζ m )).The following lemma is an immediate consequence of the definition. Lemma 2.6.
Let L ⊇ K ⊇ k be number fields. Then W ( k, L ) ⊆ W ( k, K ) . Proof.
Actually: N L/k J L = N K/k ( N L/K J L ) ⊆ N K/k J K .For an integer n and a prime number ℓ , we will always indicate by n ( ℓ ) thegreatest power of ℓ dividing n . For an element σ ∈ G of order o ( σ ), we will denoteby σ ( ℓ ) the element σ o ( σ ) /o ( σ )( ℓ ) , which is of order o ( σ )( ℓ ). Further we will denote by G { ℓ } the set of all the elements σ ∈ G such that σ ( ℓ ) = σ , i.e. of all the elements oforder a power of ℓ .Let A be a subgroup of an abelian group B (written multiplicatively). For t ∈ N ⊂ Q , we set A tB = ( { x ∈ A : ∃ a ∈ A, x = a t } if t ∈ N { x ∈ B : ∃ a ∈ A, x = a t } if t N . . In the above notation, we shall always omit the index B , since it will be clearfrom the context (actually B will be an ideal class group).We explicitly remark that in general A = ( A ) / . For example, if A is the trivialsubgroup of the group B with 2 elements, then ( A ) / = B = A . Lemma 2.7.
Let m , n be positive integers. If every prime q dividing n also divides m , then W ( k, m ) n ⊆ W ( k, mn ) .Proof. By the assumptions on m and n , we have that[ Q ( ζ mn ) : Q ( ζ m )] = φ ( mn ) φ ( m ) = n, and this is a multiple of [ k ( ζ mn ) : k ( ζ m )]. Now, for any x ∈ W ( k, m ), there exists anideal I in k ( ζ m ) such that N k ( ζ m ) /k I is in the class of x . Hence we can conclude that x n ∈ W ( k, mn ), since( N k ( ζ m ) /k I ) n = N k ( ζ mn ) /k ( I n/ [ k ( ζ mn ): k ( ζ m )] O k ( ζ mn ) ) . The inclusion W ( k, m ) n ⊆ W ( k, mn ) follows. Lemma 2.8.
Let A be a subgroup of an abelian group B . If d | m , then A m/ ⊆ A d/ . roof. Let x ∈ A m/ . If d is even, then also m must be even, and it is clear that x ∈ A d/ . If d is odd, then x ∈ A m ⊆ A d , i.e. x ∈ A d/ . Lemma 2.9.
Let A be a subgroup of an abelian group B , let m , . . . , m n be positiveintegers and let d be their greatest common divisor. Then n Y i =1 A m i / = A d/ . Proof. ⊆ By Lemma 2.8, we have A m i / ⊆ A d/ for i = 1 , . . . , n . ⊇ Let x ∈ A d/ and let c , . . . , c n ∈ Z be such that P ni =1 c i m i = d . If d is even,then also every m i is even, and there exists a ∈ A such that x = a d/ = a P ni =1 c i m i / = n Y i =1 ( a c i ) m i / ∈ n Y i =1 A m i / . If d is odd, then there exists a ∈ A such that x = a d . Now let i ∈ { , . . . , n } .If m i is odd, then ( x c i m i /d ) = a d ( c i m i /d ) = a c i m i , i.e. x c i m i /d ∈ A m i / . If m i is even, then x c i m i /d = x c i m i / (2 d ) = a dc i m i / (2 d ) = a c i m i / ∈ A m i / . Hence x = x P ni =1 c i m i /d = n Y i =1 x c i m i /d ∈ n Y i =1 A m i / . If G is a group and S is a subset of G containing 1, we use the notation S ∗ forthe subset S \ { } . Proposition 2.10.
Let k be a number field and let G be a finite group. Then, for i = 0 or , we have Y τ ∈ G ∗ W ( k, E k,G,τ ) o ( τ ) − i Go ( τ ) = Y ℓ | G Y σ ∈ G { ℓ } ∗ W ( k, E k,G,σ ) ℓ − i Go ( σ ) . Proof. ⊆ Using [8, Lemma 3.16], Lemma 2.8 and Lemma 2.9 (for the first in-clusion in the following formula), Lemma 2.3 and Lemma 2.6 (for the secondinclusion in the following formula), we see that W ( k, E k,G,τ ) o ( τ ) − i Go ( τ ) ⊆ Y ℓ | o ( τ ) W ( k, E k,G,τ ) ℓ − i Go ( τ )( ℓ ) ⊆ Y ℓ | o ( τ ) W ( k, E k,G,τ ( ℓ ) ) ℓ − i Go ( τ ( ℓ )) ⊆ Y ℓ | G Y σ ∈ G { ℓ } ∗ W ( k, E k,G,σ ) ℓ − i Go ( σ ) . Let σ ∈ G { ℓ } ∗ , where ℓ is a prime number dividing G , and set ˜ σ = σ o ( σ ) /ℓ .Clearly gcd (cid:18) ( ℓ −
1)
Gℓ , ( o ( σ ) −
1) Go ( σ ) (cid:19) = ( ℓ −
1) Go ( σ ) . (1)Hence, using Lemma 2.3, Lemma 2.6 and Lemma 2.9, W ( k, E k,G,σ ) ℓ − i Go ( σ ) = W ( k, E k,G,σ ) ℓ − i Gℓ W ( k, E k,G,σ ) o ( σ ) − i Go ( σ ) ⊆ W ( k, E k,G, ˜ σ ) o (˜ σ ) − i Go (˜ σ ) W ( k, E k,G,σ ) o ( σ ) − i Go ( σ ) ⊆ Y τ ∈ G ∗ W ( k, E k,G,τ ) o ( τ ) − i Go ( τ ) . Definition 2.11.
Let k be a number field and G be a finite group. Then we set W ( k, G ) = Y τ ∈ G ∗ W ( k, E k,G,τ ) o ( τ ) −
12 Go ( τ ) = Y ℓ | G Y σ ∈ G { ℓ } ∗ W ( k, E k,G,σ ) ℓ −
12 Go ( σ ) ⊆ Cl( k ) . We now come to the principal results of this section. We shall show that theclasses of primes ramifying in a tame G -Galois extension of k have to satisfy someconditions. As a consequence, we will get the inclusion R t ( k, G ) ⊆ W ( k, G ).If K/k is a Galois extension of number fields and P is a prime ideal in K , wedenote by D ( K/k, P ) (resp. I ( K/k, P )) the decomposition group (resp. the inertiagroup) of P in K/k . Further we denote by d ( K/k ) the discriminant of K over k . Proposition 2.12.
Let
K/k be a tame G -Galois extension of number fields, let p bea prime of k ramifying in K/k and let τ be a generator of I ( K/k, P ) , where P is aprime of K dividing p . Then x ∈ W ( k, E k,G,τ ) , where x is the class of p in Cl( k ) .Proof. Let ˜ P be a prime above P in K ( ζ o ( τ ) ). In order to show that x ∈ W ( k, E k,G,τ ),we will prove that D ( K ( ζ o ( τ ) ) /k, ˜ P ) is contained in Gal( K ( ζ o ( τ ) ) /E k,G,τ ). It willfollow that p has inertia degree 1 in E k,G,τ /k , and hence that it is the norm of anideal in E k,G,τ /k .Recall that there is a canonical isomorphism between I ( K/k, P ) and a subgroupof the units of the residue field κ P of K at P , which is easily seen to be invariant withrespect to the action of the decomposition group of P in K/k (see [19, PropositionIV.2.7]). Hence κ × P contains an element of order o ( τ ) and therefore κ P = κ ˜ P , wherethe latter denotes the residue field of K ( ζ o ( τ ) ) at ˜ P .Let now δ ∈ D ( K ( ζ o ( τ ) ) /k, ˜ P ). Then δ | K ∈ N G ( τ ), since δ | K ∈ D ( K/k, P ) and I ( K/k, P ) ⊳ D ( K/k, P ). By definition, τ ϕ τ ( δ | K ) = ( δ | K ) τ ( δ | K ) − and therefore, using the invariance of the above mentioned isomorphism, δ | K (andhence also δ ) acts as raising to the power ϕ τ ( δ | K ) on the subgroup of order o ( τ ) of7 × P = κ × ˜ P . Recalling that the powers of ζ o ( τ ) are distinct modulo ˜ P (since ˜ P ∤ o ( τ )),we get that ζ ν k,τ ( δ | k ( ζo ( τ )) ) o ( τ ) = δ ( ζ o ( τ ) ) = ζ ϕ τ ( δ | K ) o ( τ ) . In particular, ν k,τ ( δ | k ( ζ o ( τ ) ) ) = ϕ τ ( δ | K ), i.e. δ | k ( ζ o ( τ ) ) ∈ H k,G,τ = Gal( k ( ζ o ( τ ) ) /E k,G,τ ) . Thus δ ∈ Gal( K ( ζ o ( τ ) ) /E k,G,τ )and, since the choice of δ ∈ D ( K ( ζ o ( τ ) ) /k, ˜ P ) was arbitrary, we have D ( K ( ζ o ( τ ) ) /k, ˜ P ) ⊆ Gal( K ( ζ o ( τ ) ) /E k,G,τ ) . This shows that x ∈ W ( k, E k,G,τ ). Theorem 2.13.
Let k be a number field and G a finite group. Then R t ( k, G ) ⊆ W ( k, G ) . Proof.
Let
K/k be a tame Galois extension with Galois group G . Its discriminant is d ( K/k ) = Y p p ( e p − Ge p , where the product runs over the primes p of k which ramify in K/k and whoseramification index in
K/k is denoted by e p . Note that e p equals the order of anelement τ generating the inertia group of a prime ideal P of K dividing p .If G has odd order or noncyclic 2-Sylow subgroups, then d ( K/k ) is the squareof an ideal whose class is st(
K/k ), by [8, Theorem 2.1]. Hence, in this case, usingProposition 2.12 we obtainst(
K/k ) ∈ Y τ ∈ G ∗ W ( k, E k,G,τ ) o ( τ ) −
12 Go ( τ ) = W ( k, G ) . If G has nontrivial cyclic 2-Sylow subgroups, then by [8, Theorem 2.1] and Propo-sition 2.12 we only obtainst( K/k ) ∈ Y τ ∈ G ∗ W ( k, E k,G,τ ) ( o ( τ ) − Go ( τ ) ! / i.e., using Proposition 2.10,st( K/k ) ∈ Y ℓ | G Y σ ∈ G { ℓ } ∗ W ( k, E k,G,σ ) ( ℓ − Go ( σ ) . This means that, for every ℓ | G and every σ ∈ G { ℓ } ∗ , there exists x σ ∈ W ( k, E k,G,σ )such that st( K/k ) = Y ℓ | G Y σ ∈ G { ℓ } ∗ x ( ℓ − Go ( σ ) σ .
8n the above formula, the only odd exponents in the right-hand side are those corre-sponding to the elements σ ∈ G of order G (2) (see the notation after Lemma 2.6).For all the other terms we have Y ℓ | G Y σ ∈ G { ℓ } ∗ o ( σ ) = G (2) x ℓ −
12 Go ( σ ) σ ∈ Y ℓ | G Y σ ∈ G { ℓ } ∗ W ( k, E k,G,σ ) ℓ −
12 Go ( σ ) . Further st(
K/k ) Y ℓ | G Y σ ∈ G { ℓ } ∗ o ( σ ) = G (2) x − ℓ −
12 Go ( σ ) σ ∈ Y σ ∈ G : o ( σ )= G (2) W ( k, E k,G,σ ) Go ( σ ) . By the Sylow Theorems, all the 2-Sylow subgroups of G are conjugate to a singlecyclic group, generated, say, by an element ˜ σ . Hence, for every σ ∈ G of order G (2), there exists τ ∈ G such that h σ i = h τ ˜ στ − i . Therefore, by Lemma 2.4 andLemma 2.3, for all such σ , E k,G,σ = E k,G, ˜ σ . So we obtainst( K/k ) Y ℓ | G Y σ ∈ G { ℓ } ∗ o ( σ ) = G (2) x − ℓ −
12 Go ( σ ) σ ∈ W ( k, E k,G, ˜ σ )
12 Go (˜ σ ) . i.e. st( K/k ) ∈ Y ℓ | G Y σ ∈ G { ℓ } ∗ W ( k, E k,G,σ ) ℓ −
12 Go ( σ ) = W ( k, G ) . This completes the proof.
Remark . Using the results of L. McCulloh on realizable classes, one can obtaina different proof of Theorem 2.13. More precisely for a tame G -Galois extension K/k , let cl O k [ G ] ( O K ) denote the class defined by O K in the projective class groupCl( O k [ G ]). Then McCulloh proved that, for every number field k and every group G , the set R ( O k [ G ]) = { ξ ∈ Cl( O k [ G ]) : ∃ K/k tame Galois, Gal(
K/k ) ∼ = G, cl O k [ G ] ( O K ) = ξ } is contained in the kernel of a homomorphism defined on Cl( O k [ G ]) (see [15]). Hethen used this inclusion to derive an equivalent version of Theorem 2.13 (see [16]).One may wonder for which groups the inclusion of the above theorem is actuallyan equality (for every number field k ). It is easy to find groups of even order forwhich the inclusion may be strict for some number fields. Example . Let k = Q ( i, √
10) and G = C (8) be a cyclic group of order 8. Thefield k ( ζ ) is obtained extending k with a root of the polynomial x − i , whose rootsare ζ and ζ . Hence Gal( k ( ζ ) /k ) ⊆ Gal( Q ( ζ ) / Q ) = ( Z / Z ) × is of order 2 and is generated by the class of 5. In particular it is different from thesubgroups generated by the class of − −
25. We obtain by [10,II.2.6] that R t ( k, C (8)) = W ( k, .
9n the other hand, it is easy to verify, using Lemma 2.7, that W ( k, C (8)) = W ( k, .We now show that R t ( k, C (8)) = W ( k, C (8)). Actually, an easy calculation withPARI/GP shows that Cl( k ( ζ )) is trivial, while Cl( k ) is of order 2. It follows thatR t ( k, C (8)) = W ( k,
8) is the trivial group, while W ( k, C (8)) = W ( k, is equal toCl( k ), which is of order 2.On the other hand, for a lot of groups of odd order, including those for whichConjecture 1.1 is known to hold so far, we will show in Sections 4 and 5 that theinclusion of Theorem 2.13 is indeed an equality (for every number field k ). To simplifystatements in the next sections, we will adopt the following working definitions. Definition 2.16.
Given a number field F , we will say that an extension K/k ofnumber fields satisfies the property P ( F ) if it is tame, with no nontrivial unramifiedsubextensions and such that d ( K/k ) is prime to d ( F/ Q ). To simplify notations wewill say that K/k satisfies P ( s ) if it satisfies P ( k ( ζ s )). Definition 2.17.
We call a finite group G very good if the following conditions hold:1. For every number field k , we haveR t ( k, G ) = W ( k, G ) .
2. For every number field k , for every class x ∈ R t ( k, G ) and every positive integer s , there exists a G -Galois extension K/k with Steinitz class x and satisfying P ( s ).It is clear that a very good group of odd order is also good, in the sense of [8,Definition 3.15]. Further it seems reasonable to expect that the second condition ofthe above definition is satisfied for every group G , including those not satisfying thefirst condition.With these definitions and in view of the results that will be proved in Sections4 and 5, it seems natural to ask the following question. Question . Is every group G of odd order very good?Of course a positive answer would also imply Conjecture 1.1 for all groups of oddorder.We do not know the answer to Question 2.18, but we suspect that it might beaffirmative. The main point here is to construct tame G -Galois extensions with somegiven Steinitz classes; of course this could be very difficult in general, but in the nextsection we will give some results in this direction. In this section we develop some techniques, which we will need in the last two sectionsto construct some tame Galois extensions of number fields with given Galois groupsand Steinitz classes.Let G be a finite group and let H be an abelian group. Let µ : G →
Aut( H ) bean action of G on H and let G be a group such that there exists an exact sequence1 → H → G → G → G on H induced by the conjugation coincides with µ . Let k be a G -Galois extension of a number field k and K/k an H -Galois extension such that K/k is Galois with Galois group G . Let C k be the id`ele class group of k . Theorem 3.1 (Shafarevich-Weil) . With notation and hypotheses as above, the reci-procity map ( , K/k ) : C k → H induces a map H ( G , C k ) → H ( G , H ) , which sends the fundamental class u ∈ H ( G , C k ) to the class of the group extension(2). In particular, the image of u is trivial precisely when G is isomorphic to thesemidirect product H ⋊ µ G .Proof. See [17, Theorem 2.110].Using Theorem 3.1 we can improve the result of [8, Proposition 3.13]. In that pa-per, the action µ was assumed to be such that H ( G , H ) is trivial. This assumptionis satisfied in some particular cases (for example when the orders of H and G arecoprime, by a theorem of Schur-Zassenhaus), but of course not in general. We aregoing to use Theorem 3.1 to study groups G of the form H ⋊ µ G , where H ( G , H ) isno more assumed to be trivial. The proof will work exactly as in [8], except for somedetails, which are pointed out in what follows; for the unchanged parts we will referthe reader to [8].From now on, in this section, G will be of the form H ⋊ µ G , where H is an abeliangroup of odd order. Lemma 3.2.
Let s be a positive integer and let k be a tame G -Galois extension of k satisfying P ( s ) . Then there exists a tame H -Galois extension K of k , such that K/k is G -Galois and satisfying the following conditions.1. The prime ideals ramifying in K/k are principal.2. st( K/k ) = 1 .3. K/k satisfies P ( s ) .Proof. Choose a set { P , . . . , P t } of prime ideals in k which are unramified over Q and whose classes generate Cl( k ). Now recall the definition of the so-called contentmap: π : I k → J k , α Y p ∤ ∞ p v p ( α p ) , where I k and J k are the id`ele group and the ideal group of k , respectively. Forevery i ∈ { , . . . , t } , let π P i be a prime element in the completion ( k ) P i and let y i be the id`ele in I k whose component at P i (resp. at any prime different from P i ) is π P i (resp. 1).Let ˜ u : G × G → C k be a cocycle, whose class in H ( G , C k ) is the fundamentalclass u . Recall that we have an exact sequence1 → Y P U P /U k → C k π → Cl( k ) → , U k is the group of units of k and Q P U P ⊆ I k is the group of unit id`eles.Therefore, for all couples ( ρ , ρ ) ∈ G × G , we can find v ρ ,ρ ∈ Q P U P and λ ρ ,ρ ,j ∈ Z such that ˜ u ( ρ , ρ ) is represented by v ρ ,ρ Q tj =1 y λ ρ ,ρ ,j j .We add all the elements v ρ ,ρ to the set { u , . . . , u T } defined in [8], before Lemma3.9. The arguments of [8, Lemma 3.10], with x = 1, give a tame H -Galois extension K of k , which is also Galois over k , and such that the action of G on H inducedby the conjugation is µ . Further, by construction, the kernel of the reciprocity map( , K/k ) : C k → Gal(
K/k ) = H contains all the elements ˜ u ( ρ , ρ ) with ρ , ρ ∈ G .Therefore the image of the fundamental class by the reciprocity map correspondingto the extension K/k must be trivial and so, by Theorem 3.1, the Galois group of K/k must be isomorphic to H ⋊ µ G .Finally, the construction of the proof of [8, Lemma 3.10] implies that the primes q , . . . , q r +1 of k ramifying in K/k are in the class of x = 1, hence principal. Since H is of odd order, we deduce that st( K/k ) = 1, i.e. we have proved conditions 1and 2. As for condition 3, it follows, once more by the proof of [8, Lemma 3.10], that K/k has no nontrivial unramified subextensions and that the q i can be chosen sothat their restrictions to Q are unramified in k ( ζ s ) / Q . Lemma 3.3.
Let s be a multiple of H and let k be a tame G -Galois extension of k satisfying P ( s ) . For every prime number ℓ dividing H and every τ ∈ H { ℓ } ∗ , let x τ be any class in W ( k, E k,G,τ ) and let A τ , B τ be nonnegative integers, with A τ = 1 .Then there exists a tame H -Galois extension K of k , such that K/k is G -Galoisand satisfying the following conditions.1. The only nonprincipal prime ideals of k ramifying in K/k are (for every ℓ dividing H and every τ ∈ H { ℓ } ∗ ):(a) q τ, , q τ, , . . . , q τ,A τ , in the class of x τ and such that, for any i = 1 , . . . , A τ ,the inertia group of every prime of K dividing q τ,i is a conjugate of h τ o ( τ ) ℓ i ;(b) q τ,A τ +1 , q τ,A τ +2 , in the class of x B τ τ and such that the inertia group of everyprime of K dividing q τ,A τ +1 and q τ,A τ +2 is a conjugate of h τ i .2. We have st( K/k ) = Y ℓ | H Y τ ∈ H { ℓ } ∗ ι ( x τ ) A τ ℓ −
12 Hℓ + B τ ( o ( τ ) − Ho ( τ ) , where ι : Cl( k ) → Cl( k ) is induced by the inclusion k ⊆ k .3. K/k satisfies P ( s ) .Proof. By Lemma 3.2 there exists a tame H -Galois extension ˜ K of k , such that˜ K/k is G -Galois, ˜ K/k is ramified only at principal ideals and it satisfies P ( s ).Since k /k satisfies P ( s ), by [8, Lemma 3.4] and Lemma 2.2, we have for every τ ∈ H ∗ that W ( k, E k,G,τ ) = W ( k, Z k /k,µ,τ ), where Z k /k,µ,τ has been defined in[8]. This observation will allow us to argue as in [8, Lemma 3.11], but replacing W ( k, Z k /k,µ,τ ) with W ( k, E k,G,τ ).As in the proof of Lemma 3.2, we add all the v ρ ,ρ (defined in the proof of Lemma3.2) to the set { u , . . . , u T } defined in [8], before Lemma 3.9. Now, for every primenumber ℓ dividing H , for every τ ∈ H { ℓ } ∗ and for every i = 1 , . . . , A τ (resp.12 = A τ +1 , A τ +2 ) we choose prime ideals q τ,i in the class of x τ (resp. in the class of x B τ τ ). For any i = 1 , . . . , A τ + 2, we also choose prime ideals Q τ,i in k dividing the q τ,i and generators g Q τ,i of the units of the residue fields κ Q τ,i . As in the proof of [8,Lemma 3.11], we can assume that the q τ,i are unramified in k ( ζ s ) / Q and in ˜ K/k .For every prime number ℓ dividing H , for every τ ∈ H { ℓ } ∗ and for every i = 1 , . . . , A τ + 2, we define ϕ τ,i : κ × Q τ,i → H as follows. If A τ is odd, we set ϕ τ,i ( g Q τ,i ) = τ o ( τ ) ℓ if i = 1 , τ − o ( τ ) ℓ if i = 3 τ ( − i o ( τ ) ℓ if 4 ≤ i ≤ A τ τ ( − i if i = A τ + 1 , A τ + 2 . If A τ is even, we set ϕ τ,i ( g Q τ,i ) = ( τ ( − i o ( τ ) ℓ if 1 ≤ i ≤ A τ τ ( − i if i = A τ + 1 , A τ + 2 . As in the proof of [8, Lemma 3.11], using that x τ ∈ W ( k, E k,G,τ ), we can assumethat the Q τ,i and the g Q τ,i are such that the ϕ τ,i are D ( k /k, Q τ,i )-invariant. Hence,for i = 1 , . . . , A τ + 2, we extend ϕ τ,i to a G -invariant homomorphism˜ ϕ τ,i : ind G D ( k /k, Q τ,i ) κ × Q τ,i ∼ = Y δ ∈G /D ( k /k, Q τ,i ) κ × δ ( Q τ,i ) → H, using the Frobenius reciprocity. We define ϕ : Q P κ × P → H , setting ϕ | κ × δ ( Q τ,i ) = ˜ ϕ τ,i for ℓ | H , τ ∈ H { ℓ } ∗ , i = 1 , . . . , A τ + 2 and δ ∈ G ϕ | κ × P = 1 for all the other primes . As in the proof of [8, Lemma 3.10], we can assume that all the elements u j are in thekernel of ϕ , so that we can extend ϕ to a G -invariant homomorphism ϕ : C k → H ,whose kernel contains a congruence subgroup of C k . Therefore taking the product ϕ · ( , ˜ K/k ) : C k → H, we get a G -invariant homomorphism whose kernel contains a congruence subgroupand which is surjective, by the assumptions on the ideals q τ,i .By class field theory (see for example [8, Theorem 2.3 and Proposition 2,4]), weobtain a tame H -Galois extension K of k , which is Galois over k . The extension K/k satisfies P ( s ), by the assumptions on the prime ideals q τ,i and the fact that˜ K/k satisfies P ( s ). As in the proof of Lemma 3.2, actually Gal( K/k ) ∼ = G . All therequested properties hold, by construction, once more as in the proof of [8, Lemma3.11].For any G -Galois extension k of k , we define R t ( k , k, G ) ⊆ Cl( k ) as the setof those ideal classes of k which are Steinitz classes of a tamely ramified H -Galoisextension K/k , such that K/k is G -Galois.13 emma 3.4. Let s be a multiple of H and let k be a tame G -Galois extension of k satisfying P ( s ) . Then Y ℓ | H Y τ ∈ H { ℓ } ∗ ι ( W ( k, E k,G,τ )) ℓ −
12 Ho ( τ ) ⊆ R t ( k , k, G ) , where ι : Cl( k ) → Cl( k ) is induced by the inclusion k ⊆ k . Further any classin the left-hand side of the above inclusion is the Steinitz class of a tame H -Galoisextension K/k which satisfies P ( s ) and such that K/k is G -Galois.Proof. Let x = Y ℓ | H Y τ ∈ H { ℓ } ∗ x ℓ −
12 Ho ( τ ) τ ∈ Y ℓ | H Y τ ∈ H { ℓ } ∗ W ( k, E k,G,τ ) ℓ −
12 Ho ( τ ) , with x τ ∈ W ( k, E k,G,τ ). Let o ( x τ ) be the order of x τ ∈ Cl( k ). Since any prime ℓ dividing H is odd, for any τ ∈ H { ℓ } ∗ , using (1), we can find integers A τ , B τ > A τ ℓ −
12 Hℓ + B τ ( o ( τ ) −
1) Ho ( τ ) ≡ ℓ −
12 Ho ( τ ) (mod o ( x τ ))and, in particular, we have x A τ ℓ −
12 Hℓ + B τ ( o ( τ ) − Ho ( τ ) τ = x ℓ −
12 Ho ( τ ) τ . Hence we conclude by Lemma 3.3, with A τ and B τ as above. Theorem 3.5.
Let k be a number field and let G be a finite group of odd ordersatisfying the second condition of Definition 2.17. Let H be an abelian group of oddorder, let µ : G →
Aut( H ) be an action of G on H and let G = H ⋊ µ G . Then R t ( k, G ) ⊇ R t ( k, G ) H Y ℓ | H Y τ ∈ H { ℓ } ∗ W ( k, E k,G,τ ) ℓ −
12 Go ( τ ) . More precisely, for any class x in the right-hand side of the above inclusion and anypositive integer s , there exists a tame G -Galois extension K/k with Steinitz class x and satisfying P ( s ) .Proof. We will prove directly the last assertion of the theorem. Note that it sufficesto prove it for s big enough. So let s be a positive integer which is a multiple of theorder of H . Let x ∈ R t ( k, G ) H Y ℓ | H Y τ ∈ H { ℓ } ∗ W ( k, E k,G,τ ) ℓ −
12 Go ( τ ) and write x = y H z G with y ∈ R t ( k, G ) and z ∈ Q ℓ | H Q τ ∈ H { ℓ } ∗ W ( k, E k,G,τ ) ℓ −
12 Ho ( τ ) .By the hypothesis on G , we can find a tame G -Galois extension k /k with Steinitzclass y and which satisfies P ( s ).Further, by Lemma 3.4, there exists an H -Galois extension K/k , which satisfies P ( s ), which is Galois over k , with Galois group G , and with st( K/k ) = ι ( z ), where14 : Cl( k ) → Cl( k ) is induced by the inclusion k ⊆ k . Then, using a well-knownformula relating the Steinitz classes in towers of extensions (see for example [10,Proposition I.1.2]), we getst( K/k ) = st( k /k ) H N k /k (st( K/k )) = y H N k /k ( ι ( z )) = y H z G = x, where N k /k : Cl( k ) → Cl( k ) is induced by the norm from k to k . Remark . The techniques described in this section can be extended to groups ofeven order, as in [7]. We have restricted our attention to groups of odd order onlyto have cleaner proofs and statements (as already noticed, there are groups of evenorder which are not very good). However the complications arising from extensionsof even degree vanish if we assume that the class number of the base field k is odd. A ′ -groups of odd order In this section we find some slightly more explicit descriptions for the results of [8].In particular we prove that the answer to Question 2.18 is affirmative for the so-called A ′ -groups of odd order, whose definition we now recall. Definition 4.1.
We define A ′ -groups inductively:1. Finite abelian groups are A ′ -groups.2. If G is an A ′ -group and H is a finite abelian group of order prime to that of G ,then H ⋊ µ G is an A ′ -group, for any action µ : G →
Aut( H ) of G on H .3. If G and G are A ′ -groups, then G × G is an A ′ -group. Proposition 4.2.
Let G be a very good group of odd order, let H be an abelian groupof odd order prime to that of G and let µ : G →
Aut( H ) be an action of G on H .Then G = H ⋊ µ G is very good. In particular abelian groups are very good.Proof. Let π : G → G/H ∼ = G be the projection. For any ℓ | G and any σ ∈ G { ℓ } ,there exists ˜ σ ∈ G{ ℓ } such that π ( σ ) = π (˜ σ ). Further, since σ and π ( σ ) have thesame order, it is straightforward to verify that H k,G,σ ⊆ H k,G/H,π ( σ ) = H k, G , ˜ σ . Hence E k,G,σ ⊇ E k, G , ˜ σ and, by Lemma 2.6, W ( k, E k,G,σ ) ⊆ W ( k, E k, G , ˜ σ ) . We also observe that for ℓ | H , H { ℓ } = G { ℓ } . Using Theorem 3.5 (but actually also[8, Theorem 3.19] would work in this case) and the hypothesis that G is very good,we then have:R t ( k, G ) ⊇ R t ( k, G ) H Y ℓ | H Y τ ∈ H { ℓ } ∗ W ( k, E k,G,τ ) ℓ −
12 Go ( τ ) = Y ℓ | G Y σ ∈G{ ℓ } ∗ W ( k, E k, G ,σ ) ℓ −
12 Go ( σ ) Y ℓ | H Y τ ∈ H { ℓ } ∗ W ( k, E k,G,τ ) ℓ −
12 Go ( τ ) ⊇ Y ℓ | G Y σ ∈ G { ℓ } ∗ W ( k, E k,G,σ ) ℓ −
12 Go ( σ ) = W ( k, G ) . G is very good, also the second condition of Definition 2.17 is satisfied by G . Proposition 4.3.
Let G , G be very good groups of odd order. Then G = G × G isvery good.Proof. Let ℓ be a prime number, let σ ∈ G { ℓ } , σ ∈ G { ℓ } be not both trivial.Then o ( σ σ ) = max { o ( σ ) , o ( σ ) } > o ( σ σ ) = o ( σ ). Clearly N G ( σ σ ) ⊆ N G ( σ ) and, for every τ ∈ N G ( σ σ ), we have ϕ σ σ ( τ ) = ϕ σ ( τ ) . Therefore H k,G,σ σ ⊆ H k,G,σ , i.e. E k,G,σ σ ⊇ E k,G,σ . Using Lemma 2.6 and Lemma 2.2, we conclude that W ( k, E k,G,σ σ ) ℓ −
12 Go ( σ σ ⊆ W ( k, E k,G,σ ) ℓ −
12 G · G o ( σ = W ( k, E k, G ,σ ) ℓ −
12 G o ( σ G and an analogous result holds when o ( σ σ ) = o ( σ ). Hence, using the hypothesisthat G and G are very good groups of odd order, W ( k, G ) = Y ℓ | G Y σ ∈ G { ℓ } ∗ W ( k, E k,G,σ ) ℓ −
12 Go ( σ ) ⊆ Y ℓ | G Y σ ∈G { ℓ } ∗ W ( k, E k, G ,σ ) ℓ −
12 G o ( σ G Y ℓ | G Y σ ∈G { ℓ } ∗ W ( k, E k, G ,σ ) ℓ −
12 G o ( σ G = R t ( k, G ) G R t ( k, G ) G . In particular every class in W ( k, G ) is of the form x G x G , where x i ∈ R t ( k, G i ) for i = 1 ,
2. Now let s be a positive integer: then, since G is very good, there exists atame G -Galois extension k /k with Steinitz class x and satisfying P ( s ). Further let˜ s be a multiple of s such that all the primes ramifying in k /k ramify also in k ( ζ ˜ s ) /k .Now, since also G is very good, there exists a tame G -Galois extension k /k withSteinitz class x and satisfying P (˜ s ). The compositum K = k k is a tame G -Galoisextension of k , satisfying P ( s ). Since k and k are linearly disjoint over k , we have d ( K/k ) = d ( k /k ) G d ( k /k ) G and hence, by [8, Theorem 2.1],st( K/k ) = x G x G , since G and G are both odd.Therefore W ( k, G ) ⊆ R t ( k, G ) and the opposite inclusion is given by Theorem2.13. Actually the above construction also proves the second property characterizingvery good groups.Putting things together, we obtain a refinement of [8, Theorem 3.23]. Theorem 4.4.
Every A ′ -group of odd order is very good.Proof. Obvious, by induction, using the above results.16
Groups of order dividing ℓ ℓ is an odd prime. We want to show that the answer to Question2.18 is affirmative for all nonabelian groups of order dividing ℓ (the abelian case iscovered by Theorem 4.4). Further the same is true for all groups of order ℓ n andexponent ℓ n − , for any positive integer n (these are actually the groups studied in[9]). Lemma 5.1.
Let G be an ℓ -group and let τ ∈ G ∗ . Set e τ = o ( τ ) N G ( τ ) /C G ( τ )) . Then ℓ | e τ and E k,G,τ = k ( ζ e τ ) ; in particular ζ ℓ ∈ E k,G,τ .Proof. Since the kernel of ϕ τ is the centralizer C G ( τ ), ϕ τ ( N G ( τ )) is the subgroupof order N G ( τ ) /C G ( τ )) of the ℓ -Sylow subgroup of the cyclic group ( Z /o ( τ ) Z ) × .This already gives ℓ | e τ and we also have ϕ τ ( N G ( τ )) = 1 + o ( τ ) N G ( τ ) /C G ( τ )) Z (mod o ( τ )) = 1 + e τ Z (mod o ( τ )) . Therefore H k,G,τ = Gal( k ( ζ o ( τ ) ) /k ( ζ e τ ))and E k,G,τ = k ( ζ e τ ) ⊇ k ( ζ ℓ ) . Proposition 5.2 (Burnside) . A group of order ℓ n , whose center has order ℓ c , con-tains a normal abelian subgroup of order ℓ ν where ν ≥ −
12 + r n + c − c + 14 . In particular, for n = 3 , this becomes ν ≥ n − . Proof.
This result is shown in [3, Section 4].We shall use the above proposition for n = 3 ,
4. For these values of n the resultactually follows quite easily using standard arguments of the theory of ℓ -groups.From now on we will use the notation C ( n ) to denote a cyclic group of order n ,where n is a positive integer. Lemma 5.3.
Let G be a group of exponent ℓ and order ℓ n , where n = 3 , . Then G is very good and in particular R t ( k, G ) = W ( k, ℓ ) ℓ − ℓ n − . roof. By Proposition 5.2, there is an exact sequence1 → C ( ℓ ) n − → G → C ( ℓ ) → , which splits, since the exponent of G is ℓ .Using Lemma 5.1, we get W ( k, G ) = W ( k, ℓ ) ℓ − ℓ n − and the proof of the equalityR t ( k, G ) = W ( k, G ) is straightforward, using Theorem 2.13 and Theorem 3.5 (whichwe can apply with G = C ( ℓ ) and H = C ( ℓ ) n − , since C ( ℓ ) is very good by Theorem4.4).Since in particular every x ∈ R t ( k, G ) belongs to the right-hand side of theinclusion of Theorem 3.5, it follows by Theorem 3.5 that G satisfies the secondcondition of Definition 2.17.The above result for n = 3 is actually proved in [1, Corollaire 1.2]. The followingresult slightly improves [9, Theorem 3.8]. Proposition 5.4.
Let G be a nonabelian group of order ℓ n and exponent ℓ n − , where n > is an integer. Then G is very good and R t ( k, G ) = W ( k, ℓ n − ) ℓ − ℓ . Proof.
As it is well known (and easily verified), G can be written as G = h σ, τ : σ ℓ = τ ℓ n − = 1 , στ σ − = τ ℓ n − i . Clearly h τ ℓ i equals the center of G , which has therefore order ℓ n − . Let ρ = τ a σ b ∈ G ,with σ b = 1 (resp. σ b = 1), then τ a σ b (resp. τ ) is an element of the centralizer of τ a σ b which is not in the center of G . Thus, for any element ρ ∈ G , the order of C G ( ρ ) is strictly larger than that of the center of G and hence C G ( τ ) is a multipleof ℓ n − . Therefore N G ( τ ) /C G ( τ )) divides ℓ and o ( ρ ) ℓ | e ρ , i.e. k ( ζ o ( ρ ) /ℓ ) ⊆ E k,G,ρ , byLemma 5.1.Hence, using Lemma 2.6 and Lemma 2.7, if ℓ | o ( ρ ), then W ( k, E k,G,ρ ) ℓ − ℓno ( ρ ) ⊆ W ( k, o ( ρ ) /ℓ ) ℓ − ℓno ( ρ ) ⊆ W ( k, ℓ n − ) ℓ − ℓ . If ρ = 1 and ℓ ∤ o ( ρ ), i.e. o ( ρ ) = ℓ , then by Lemma 5.1 we must have E k,G,ρ = k ( ζ ℓ ).Hence, again using Lemma 2.6 and Lemma 2.7, W ( k, E k,G,ρ ) ℓ − ℓno ( ρ ) = W ( k, ℓ ) ℓ − ℓ n − ⊆ W ( k, ℓ n − ) ℓ − ℓ ⊆ W ( k, ℓ n − ) ℓ − ℓ . Further, we have N G ( τ ) = G and C G ( τ ) = h τ i and therefore, by Lemma 5.1, W ( k, E k,G,τ ) ℓ − ℓno ( τ ) = W ( k, ℓ n − ) ℓ − ℓ . Therefore W ( k, G ) = Y ρ ∈ G ∗ W ( k, E k,G,ρ ) ℓ − ℓno ( ρ ) = W ( k, E k,G,τ ) ℓ − ℓ . We easily conclude by Theorems 2.13, 3.5 and 4.4, as in the proof of Lemma 5.3.
Lemma 5.5.
Let G be a nonabelian group of order ℓ and exponent ℓ . Let τ ∈ G be of order ℓ and such that N G ( τ ) /C G ( τ )) is maximal. Then W ( k, G ) = W ( k, E k,G,τ ) ℓ − ℓ . roof. For any σ ∈ G of order ℓ , using Lemma 5.1, Lemma 2.7 and Lemma 2.6, wehave W ( k, E k,G,σ ) ℓ − ℓ o ( σ ) = W ( k, ℓ ) ℓ − ℓ ⊆ W ( k, ℓ ) ℓ − ℓ ⊆ W ( k, E k,G,τ ) ℓ − ℓ . Further for any σ ∈ G of order ℓ we have N G ( σ ) /C G ( σ )) ≤ N G ( τ ) /C G ( τ )),i.e., by Lemma 5.1, E k,G,σ ⊇ E k,G,τ , and hence, by Lemma 2.6, W ( k, E k,G,σ ) ⊆ W ( k, E k,G,τ ) . Therefore W ( k, G ) = Y ρ ∈ G ∗ W ( k, E k,G,ρ ) ℓ − ℓ o ( ρ ) = W ( k, E k,G,τ ) ℓ − ℓ . Lemma 5.6.
Let G be a (nonabelian) group of order ℓ and exponent ℓ . Then thereexists a normal abelian subgroup H of order ℓ and an element τ ∈ G of order ℓ maximizing N G ( τ ) /C G ( τ )) , such that we have one of the following possibilities:1. τ ∈ H and G ∼ = H ⋊ C ( ℓ ) .2. G ∼ = h τ i ⋊ C ( ℓ ) .3. τ H and there exists σ ∈ N G ( τ ) \ h τ i of order ℓ , which acts trivially (resp.nontrivially) on τ if N G ( τ ) = C G ( τ ) (resp. if N G ( τ ) = C G ( τ ) ).Proof. Let σ ∈ G be an element of order ℓ , which maximizes N G ( σ ) /C G ( σ ));since, by Lemma 5.1, ℓ divides e σ = o ( σ ) N G ( σ ) /C G ( σ )) = ℓ N G ( σ ) /C G ( σ )) , we have that N G ( σ ) /C G ( σ )) divides ℓ . We will study separately the cases C G ( σ ) ≥ ℓ and C G ( σ ) = ℓ . • If C G ( σ ) ≥ ℓ , then we set H to be a subgroup of order ℓ of C G ( σ ) containing σ . Clearly H is abelian and normal, being of index ℓ , and we have an exactsequence 1 → H → G → C ( ℓ ) → . (3)There are three cases, corresponding to the three possibilities in the statementof the lemma.1. Sequence (3) splits. Then we set τ = σ ∈ H and we have G ∼ = H ⋊ C ( ℓ ).2. Sequence (3) does not split and N G ( σ ) /C G ( σ )) = ℓ . In this case N G ( σ ) = G , C G ( σ ) = H and we consider an element ρ ∈ G \ H . If h σ, ρ i = ℓ , then h σ, ρ i ∼ = C ( ℓ ) ⋊ C ( ℓ ) and there would exist an ele-ment of order ℓ in N G ( σ ) \ C G ( σ ) = G \ H , contradicting the assumptionthat (3) does not split. Therefore h σ, ρ i = G and, setting τ = σ , we get G = h τ i ⋊ h ρ i . 19. Sequence (3) does not split and N G ( σ ) /C G ( σ )) = 1. In this case, by theassumption of maximality, an analogous condition holds for every elementof order ℓ . So we can take τ to be any element not in H , since it mustbe of order ℓ because of the nonsplitting of (3). Further let ˜ H be asubgroup of order ℓ of N G ( τ ) = C G ( τ ) (which has order at least ℓ , since τ is of order ℓ ) and containing τ . Clearly ˜ H is abelian and we can findan element σ of order ℓ in ˜ H \ h τ i ⊆ N G ( τ ) \ h τ i . In particular σ actstrivially on τ . • If C G ( σ ) = ℓ , i.e. C G ( σ ) = h σ i , then N G ( σ ) = ℓ , since it must beat least ℓ , being strictly larger than h σ i , and N G ( σ ) cannot be G , since N G ( σ ) /C G ( σ )) divides ℓ . By Proposition 5.2 there exists an abelian sub-group H < G of order ℓ . Clearly σ H , since C G ( σ ) = ℓ . Further N G ( σ )is nonabelian of order ℓ and exponent ℓ , so there exists σ ∈ N G ( σ ) \ h σ i oforder ℓ , acting nontrivially on σ . Hence we are again in case 3, taking τ = σ . Lemma 5.7.
Let G be of the first or second type described in Lemma 5.6. Then G is very good and R t ( k, G ) = W ( k, E k,G,τ ) ℓ − ℓ , where τ ∈ G is as in Lemma 5.6.Proof. By Lemma 5.5, we have W ( k, G ) = W ( k, E k,G,τ ) ℓ − ℓ . By Theorem 3.5, with G = C ( ℓ ) or C ( ℓ ) (note that both are very good groups by Theorem 4.4), we obtain W ( k, G ) = W ( k, E k,G,τ ) ℓ − ℓ ⊆ R t ( k, G ) . The opposite inclusion is given by Theorem 2.13. The second condition of the defi-nition of very good groups is immediate, as in the proof of Lemma 5.3.
Lemma 5.8.
Let G be of the third type described in Lemma 5.6. Then G is verygood and R t ( k, G ) = W ( k, E k,G,τ ) ℓ − ℓ , where τ ∈ G is as in Lemma 5.6.Proof. By hypothesis there exists a normal abelian subgroup H of G of order ℓ and an element τ ∈ G \ H of order ℓ maximizing N G ( τ ) /C G ( τ )). We can alsofind an element σ ∈ N G ( τ ) \ h τ i of order ℓ and acting trivially on τ if and only if N G ( τ ) = C G ( τ ). Let G be the subgroup of G generated by σ and τ , which must beof the form C ( ℓ ) ⋊ C ( ℓ ), possibly with trivial action. Clearly there exists an integer a such that σ τ σ − = τ aℓ (since the order of the conjugation divides ℓ ). Since τ H , we have that τ H generates G/H ; in particular, σ = στ b , for some σ ∈ H ∗ (not necessarily of order ℓ ) and b ∈ N . We also have στ σ − = τ aℓ . For later use, we note that, since G has order ℓ and exponent ℓ , there must exist c ∈ { , , . . . , ℓ − } such that σ ℓ = τ cℓ (an easy calculation shows that c = − b , butwe won’t need it), so that we have G = h σ, τ : τ ℓ = σ ℓ τ − cℓ = 1 , στ σ − = τ aℓ i .
20e now consider the (abstract) group ˜ G , generated by ˜ σ and ˜ τ , satisfying the samerelations as σ and τ in G . We also consider the element ˜ σ ∈ ˜ G , corresponding to σ ∈ G .We define an action µ : ˜ G →
Aut( H ) by setting, for any h ∈ H , ( µ (˜ τ )( h ) = τ hτ − µ (˜ σ )( h ) = σhσ − = h. We consider the semidirect product ˜ G = H ⋊ µ ˜ G . For every couple h , h of elements of H , we want to define a surjective homomor-phism π : ˜ G → G , such that π ( h ˜ τ ) and π ( h ˜ τ ) are both of order ℓ and ker( π ) ∩ H is trivial. There are two cases:1. For every h ∈ H , hτ is of order ℓ .In this case we define π : ˜ G → G by removing tildes. Then π is clearly asurjective homomorphism, π ( h ˜ τ ) = hτ has order ℓ for any h ∈ H and ker( π )has trivial intersection with H .2. There exists an element h ∈ H such that h τ is of order ℓ .Note that in this case, if for example h = h , we cannot define π simply byremoving tildes because π ( h ˜ τ ) = h τ would have order ℓ . Then we proceed asfollows: by assumption,( h τ ) ℓ = ℓ − Y i =0 τ i h τ − i ! τ ℓ = 1 , i.e. ℓ − Y i =0 ( τ i h τ − i ) = τ − ℓ . It follows that, for any j ∈ Z ,( h j τ ) ℓ = ℓ − Y i =1 τ i h j τ − i ! τ ℓ = ℓ − Y i =1 τ i h τ − i ! j τ ℓ = τ ℓ (1 − j ) . Now, for any j ∈ Z , we want to define a homomorphism π j : ˜ G → G by setting π j (˜ τ ) = h j τπ j (˜ σ ) = σ − j π j ( h ) = h, ∀ h ∈ H.
21e have to verify that the definition is compatible with the relations of ˜ G : π j (˜ σ ) π j (˜ τ ) π j (˜ σ − ) = σ − j h j τ σ − (1 − j ) = h j σ − j τ σ − (1 − j ) = h j τ aℓ (1 − j ) = h j τ ( h j τ ) aℓ = ( h j τ ) aℓ = π j (˜ τ aℓ ); π j (˜ τ ) π j ( h ) π j (˜ τ − ) = h j τ hτ − h − j = τ hτ − = π j ( τ hτ − ); π j (˜ σ ) π j ( h ) π j (˜ σ − ) = σ − j hσ − (1 − j ) = h = π j ( h ); π j (˜ σ ) ℓ = σ (1 − j ) ℓ = τ cℓ (1 − j ) = ( h j τ ) cℓ = π j (˜ τ ) cℓ ; π j (˜ τ ) ℓ = ( h j τ ) ℓ = τ ℓ (1 − j ) = 1= π j (1) . So, for any j ∈ Z , π j : ˜ G → G is indeed a homomorphism, which is clearlysurjective and whose kernel has trivial intersection with H . Now let h , h betwo elements in H . We want to choose j such that π j ( h ˜ τ ) and π j ( h ˜ τ ) areboth of order ℓ . For i = 1 ,
2, we can find h ′ i ∈ H such that ( h i ˜ τ ) ℓ = h ′ i ˜ τ ℓ .Then, for i = 1 , π j ( h ′ i ˜ τ ℓ ) = h ′ i ( h j τ ) ℓ = h ′ i τ ℓ (1 − j ) . Since h ′ i τ ℓ (1 − j ) = h ′ i τ ℓ (1 − j ) if j j (mod ℓ ), we have at least ℓ − ≥ j such that π j ( h ′ ˜ τ ℓ ) = 1. Among those choices, at least ℓ − ≥ π j ( h ′ ˜ τ ℓ ) = 1.Hence, in both cases, for every couple h , h of elements of H , we have defined aprojection π : ˜ G → G , such that π ( h ˜ τ ) and π ( h ˜ τ ) are of order ℓ and ker( π ) hastrivial intersection with H . With this in mind, we start constructing the extensionsneeded to prove the lemma.Let k be a number field, let x ∈ W ( k, E k,G,τ ) and let s be a positive multiple of ℓ (this hypothesis will be used later to apply Lemma 3.3). Since ℓ is an odd prime,there exist integers A, B > ℓA + 2( ℓ + 1) B ≡ o ( x ))and, in particular, we have x ℓA +2( ℓ +1) B = x. By Lemma 3.2 there exists a tame C ( ℓ )-Galois extension ˜ k /k ramified only atprincipal ideals and satisfying P ( s ). Since ˜ G = h ˜ τ i ⋊ h ˜ σ i , possibly with trivial action,and thanks to our assumption on σ , we have x ∈ W ( k, E k,G,τ ) = W ( k, E k, ˜ G , ˜ τ ). Sowe can apply Lemma 3.3 (with x ˜ τ = x , x ˜ ρ = 1 for every ˜ ρ ∈ h ˜ τ i ∗ \ { ˜ τ } , A ˜ τ = 0and B ˜ τ = B ) and construct a tame C ( ℓ )-Galois extension ˜ k / ˜ k , such that ˜ k /k is˜ G -Galois and it satisfies the following conditions. • The only nonprincipal prime ideals of k ramifying in ˜ k / ˜ k are p , p in theclass of x B and the inertia group of every prime of ˜ k dividing these p i is h ˜ τ i in ˜ G = Gal(˜ k /k ) ( h ˜ τ i is invariant by conjugation, since it is normal in ˜ G );22 ˜ k / ˜ k satisfies P ( s ).Now we want to construct an H -Galois extension ˜ K/ ˜ k . By Lemma 2.6 and Lemma5.1, we have x ∈ W ( k, E k,G,τ ) ⊆ W ( k, ℓ ) = W ( k, E k, ˜ G,ρ ) for every ρ ∈ H of order ℓ .Hence again we can use Lemma 3.3 (with x ρ = x , for a certain ρ ∈ H ∗ of order ℓ , x ρ = 1 for every ρ ∈ H ∗ \ { ρ } , A ρ = A and B ρ = 0) to find a tame H -Galoisextension ˜ K/ ˜ k , such that ˜ K/k is ˜ G -Galois and satisfies the following conditions. • The only nonprincipal prime ideals of k ramifying in ˜ K/ ˜ k are p , . . . , p A +2 inthe class of x and the inertia group of every prime of ˜ K dividing these p i isgenerated by a conjugate of ρ in ˜ G = Gal( ˜ K/k ). • ˜ K/ ˜ k satisfies P ( s ) (in particular it follows that { p , p }∩{ p , . . . , p A +2 } = ∅ ).Let us consider two prime ideals ˜ P and ˜ P of ˜ K dividing p and p respectively.Their inertia groups in the extension ˜ K/k must be generated by elements of the form h i ˜ τ for some h i ∈ H , i = 1 ,
2. We consider a projection π : ˜ G → G for which π ( h ˜ τ )and π ( h ˜ τ ) are both of order ℓ and such that ker( π ) has trivial intersection with H .We call K the fixed field of ker( π ). By construction the ramification index of p and p in K/k is ℓ . Further the inertia group of the primes of ˜ K dividing p , . . . , p A +2 lies in H , which has trivial intersection with ker( π ). Therefore p , . . . , p A +2 ramifyin K/k , with ramification index ℓ and all the other ramifying primes are principal.Hence the discriminant of K/k is d ( K/k ) = ( p p ) ( ℓ − ℓ A +2 Y i =3 p ( ℓ − ℓ i I , where I is a principal ideal. Hence, by [8, Theorem 2.1], the Steinitz class isst( K/k ) = x B ( ℓ − ℓ + A ℓ − ℓ = x ℓ − ℓ (2( ℓ +1) B + ℓA ) = x ℓ − ℓ . Hence, recalling also Lemma 5.5, we have W ( k, G ) = W ( k, E k,G,τ ) ℓ − ℓ ⊆ R t ( k, G ) , We conclude, by Theorem 2.13, thatR t ( k, G ) = W ( k, G ) . As in the proof of Lemma 5.3, the second condition of Definition 2.17 holds, sincethe extensions we have constructed all satisfy P ( s ′ ) for every s ′ dividing s . Theorem 5.9.
Let ℓ be an odd prime. Then every group of order dividing ℓ is verygood. In particular R t ( k, G ) = W ( k, G ) . Proof. If G is abelian, the theorem is a consequence of Theorem 4.4. So supposethat G is nonabelian. If G is of exponent ℓ , then the theorem follows from Lemma5.3. If G is of order ℓ n and exponent ℓ n − , for n = 3 ,
4, then the result is givenby Proposition 5.4. We are therefore left with the case where G has order ℓ andexponent ℓ and we conclude by Lemma 5.6, Lemma 5.7 and Lemma 5.8.23 cknowledgements We wish to thank Maurizio Monge for sharing with us some of his insights in thetheory of ℓ -groups. Further we wish to thank the institutions which are supportingour research, i.e. the Universit´e de Limoges and the Scuola Normale Superiore ofPisa. References [1]
C. Bruche , Classes de Steinitz d’extensions non ab´eliennes de degr´e p , ActaArith., 137 (2009), pp. 177–191.[2] C. Bruche and B. Soda¨ıgui , On realizable Galois module classes and Steinitzclasses of nonabelian extensions , J. Number Theory, 128 (2008), pp. 954–978.[3]
W. Burnside , On some properties of groups whose orders are powers of primes ,Lond. M. S. Proc. (2), 11 (1913), pp. 225–245.[4]
N. P. Byott, C. Greither, and B. Soda¨ıgui , Classes r´ealisablesd’extensions non ab´eliennes , J. Reine Angew. Math., 601 (2006), pp. 1–27.[5]
J. E. Carter , Steinitz classes of a nonabelian extension of degree p , Colloq.Math., 71 (1996), pp. 297–303.[6] J. E. Carter and B. Soda¨ıgui , Classes de Steinitz d’extensions quaternioni-ennes g´en´eralis´ees de degr´e p r , J. Lond. Math. Soc. (2), 76 (2007), pp. 331–344.[7] A. Cobbe , Steinitz classes of some abelian and nonabelian extensions of evendegree , J. Th´eor. Nombres Bordeaux, 22 (2010), pp. 607–628.[8] ,
Steinitz classes of tamely ramified galois extensions of algebraic numberfields , Journal of Number Theory, 130 (2010), pp. 1129 – 1154.[9] ,
Steinitz classes of tamely ramified nonabelian extensions of odd primepower order , Acta Arith., 149 (2011), pp. 347–359.[10]
L. P. Endo , Steinitz classes of tamely ramified Galois extensions of algebraicnumber fields , PhD thesis, University of Illinois at Urbana-Champaign, 1975.[11]
M. Godin and B. Soda¨ıgui , Classes de Steinitz d’extensions `a groupe deGalois A , J. Th´eor. Nombres Bordeaux, 14 (2002), pp. 241–248.[12] , Module structure of rings of integers in octahedral extensions , Acta Arith.,109 (2003), pp. 321–327.[13]
R. Massy and B. Soda¨ıgui , Classes de Steinitz et extensions quaternioni-ennes , Proyecciones, 16 (1997), pp. 1–13.[14]
L. R. McCulloh , Galois module structure of abelian extensions , J. ReineAngew. Math., 375/376 (1987), pp. 259–306.[15] ,
On realizable classes for nonabelian extensions (unpublished)
Informal report: Galois module to Steinitz classes (unpublished) [17]
A. N. Parshin and I. R. Shafarevich , eds.,
Number theory. II , vol. 62 ofEncyclopaedia of Mathematical Sciences, Springer-Verlag, Berlin, 1992.[18]
F. Sbeity and B. Soda¨ıgui , Classes de Steinitz d’extensions non ab´eliennes`a groupe de Galois d’ordre 16 ou extrasp´ecial d’ordre 32 et probl`eme de plonge-ment , Int. J. Number Theory, 6 (2010), pp. 1769–1783.[19]
J.-P. Serre , Local fields , vol. 67 of Graduate Texts in Mathematics, Springer-Verlag, New York, 1979.[20]
B. Soda¨ıgui , Classes de Steinitz d’extensions galoisiennes relatives de degr´eune puissance de 2 et probl`eme de plongement , Illinois J. Math., 43 (1999),pp. 47–60.[21] ,
Relative Galois module structure and Steinitz classes of dihedral extensionsof degree
8, J. Algebra, 223 (2000), pp. 367–378.[22]
E. Soverchia , Steinitz classes of metacyclic extensions , J. London Math. Soc.(2), 66 (2002), pp. 61–72.[23]
The PARI Group , PARI/GP, version , Bordeaux, 2011, available from http://pari.math.u-bordeaux.fr/http://pari.math.u-bordeaux.fr/