A BRUMER-STARK CONJECTURE FOR NON-ABELIAN GALOISEXTENSIONSGAELLE DEJOU AND XAVIER-FRANÇOIS ROBLOT

Abstract. Let K/k be an abelian extension of number fields. The Brumer-Stark conjecture
predicts that a group ring element constructed from special values of L-functions associated toK/k annihilates the ideal class group of K. Moreover it specifies that the generators obtained
have special properties. The aim of this article is to state and study a generalization ofthis conjecture to non-abelian Galois extensions that is, in spirit, very similar to the original
conjecture.

1. Introduction
The Brumer-Stark conjecture was first stated by Tate [21] and applies to abelian extensionsof number fields. It combines a conjecture of Brumer and ideas coming from conjectures ofStark. Let K/k be an abelian extension. The main ingredient of the conjecture is a certain
group-ring element in Z[Gal(K/k)], called the Brumer-Stickelberger element, constructed fromthe values at s = 0 of the L-functions associated to the extension K/k. The Brumer part of theconjecture states that the Brumer-Stickelberger element annihilates the class group of the field
K. The Stark part of the conjecture predicts that the principal ideals obtained in this way admit
generators satisfying special properties. A very nice reference for the Brumer-Stark conjecture,and Stark conjectures in general, is the book of Tate [22], see also [4] and [6]. The aim of thisarticle is to generalize the Brumer-Stark conjecture to Galois non-abelian extensions.
The plan of this paper is the following. In the second section, we state the Brumer-Stark
conjecture, some of its properties and say a few words about its current status. To avoid confusionin the setting of this paper, we will call this conjecture the abelian Brumer-Stark conjecture
and will call the conjecture that we propose the Galois Brumer-Stark conjecture. The third
section is devoted to the generalization of the Brumer-Stickelberger element to the Galois case.
There, we rely on an earlier work of Hayes [13] that constructs this generalization and studies
its properties. We show that it also satisfies additional properties very similar to the abeliancase. It is known that the Brumer-Stickelberger element is rational and a suitable denominator
is known in the abelian case. We make a first conjecture, called the Integrality Conjecture,
on a suitable denominator for this element in the general case. This conjecture is part of ourgeneralization of the abelian Brumer-Stark conjecture. The next section introduces the notion ofstrong central extensions. These extensions play a fundamental role in our generalization. The
Galois Brumer-Stark conjecture is stated in Section 5 and we study its properties in Section 6
with the generalization of the properties of the abelian Brumer-Stark conjecture in view. The
last section is devoted to the study of the conjecture in the special case where the Galois groupof the extension contains an abelian normal subgroup of prime index. In this setting, we provethat the abelian Brumer-Stark conjecture implies the Galois Brumer-Stark conjecture.
Different generalizations to the non-abelian case of the Brumer conjecture and Brumer-Stark
conjecture are stated by Nickel [14] (see also the work of Burns [2]). In an appendix at the endof the paper, we state the weak version of Nickel’s non-abelian Brumer-Stark conjecture andcompare it with our conjecture.
1

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GAELLE DEJOU AND XAVIER-FRANÇOIS ROBLOT

Note. Many of the results of this article are extracted from the PhD thesis [7] of the first named
author or generalizations of results contained in this thesis.
Convention. We denote the action of elements of Galois groups on elements, ideals, etc., using
the exponent notation with the convention that they act on the left, that is ασγ = (αγ )σ .
2. The abelian Brumer-Stark conjecture
In this section, we state the abelian Brumer-Stark conjecture and review some of its properties.
Let K/k be an abelian extension of number fields. Denote by G its Galois group. Fix S a finite
set of places of k containing the infinite places of k and the finite places of k that ramify in K/k.
To simplify the exposition, we assume from now on that the cardinality of S is at least two.1
The interested reader can refer to [22, IV§6] for the statement of the conjecture when |S| = 1.
To a character χ of G is associated the S-truncated Hecke L-function of χ defined for Re(s) > 1
byY
LK/k,S (s, χ) :=(1 − χ(σp )N (p)−s )−1p∈S/

where p runs through the prime ideals of k not in S, σp is the Frobenius automorphism of p in G,
and N (p) is the absolute norm of the ideal p. This function admits a meromorphic continuation
to C, which is in fact analytic if the character χ is non-trivial. A main object of the abelian
Brumer-Stark conjecture is the Brumer-Stickelberger element. It is a relative analogue of the
Stickelberger element of cyclotomic fields and is defined by the formulaX
θK/k,S :=LK/k,S (0, χ) eχ̄ ∈ C[G]χ∈Ĝwhere Ĝ denotes the group of characters of G and, for χ ∈ Ĝ, eχ is the associated idempotent.
Another characterization of this element is that it is the only element in C[G] such that
χ(θK/k,S ) = LK/k,S (0, χ̄)for all character χ ∈ Ĝ. A third characterization of this element is in term of partial zetafunctions. For σ ∈ G, the partial zeta function associated to g (and the extension K/k and theset S) is defined, for Re(s) > 1, byX
ζK/k,S (s, σ) :=N (a)−1(a,S)=1
σa =σwhere a runs through the integral ideals of k, not divisible by any prime ideal in S, and whose
Artin symbol σa in G is equal to σ. This function also admits meromorphic continuation to thecomplex plane and the partial zeta functions are related to Hecke L-functions by the formula
X
LK/k,S (s, χ) =ζK/k,S (s, σ)χ(σ).
(1)σ∈G

From this we deduce the third characterization of the Brumer-Stickelberger element
XθK/k,S =ζK/k,S (0, σ)σ −1 .

(2)

g∈G

It follows from works of Deligne and Ribet [8] (see also the works of Barsky [1] and Pi. CassouNoguès [5]) that, for any ξ ∈ AnnZ[G] (µK ), the annihilator in Z[G] of the group µK of roots ofunity in K, we have ξ θK/k,S ∈ Z[G]. In particular, if we let wK denote the cardinality of µK ,we havewK θK/k,S ∈ Z[G].
(3)1The only non-trivial case that we are excluding is when k is a complex quadratic field and K is a subfield of

the Hilbert class field of k.

A BRUMER-STARK CONJECTURE FOR NON-ABELIAN GALOIS EXTENSIONS

3

We need one last notation before stating the abelian Brumer-Stark conjecture. We say that a
non-zero element α in K is an anti-unit if all its conjugates have absolute value equal to 1. The
group of anti-units of K is denoted by K ◦ .
Conjecture (The abelian Brumer-Stark conjecture BS(K/k, S)).
For any fractional ideal A of K, the ideal AwK θK/k,S is principal and admits a generator α ∈ K ◦such that K(α1/wK )/k is abelian.
Remark. The last assertion that K(α1/wK )/k is abelian does not depend upon the choice of
the wK -th root of α since all these roots generate the same extension of K.
Remark. The Brumer conjecture states that the ideal AnnZ[G] (µ(K)) θK/k,S of Z[G] annihilates
the class group ClK of K. The Brumer-Stark conjecture implies the Brumer Conjecture.
P
Let v be a place in S and denote by Nv := σ∈Dv σ ∈ Z[G] the sum of all the elements inthe decomposition group Dv of v in G. Then, one can prove, see [22, Chap. IV], that
Nv θK/k,S = 0.

(4)

In particular, if the set S contains a place that is totally split in K/k, the Brumer-Stickelberger
element is equal to 0 and the abelian Brumer-Stark conjecture is trivially true. Therefore, theconjecture is only meaningful when both k is totally real and K is totally complex.2 In [21], Tateproves equivalent formulations of the conjecture that are very useful for its study. We will lateron generalize this result to the non-abelian Galois case. For α ∈ K × and A an integral ideal ofK, we write α ≡ 1 (mod∗ A) if vP (α − 1) ≥ vP (A) for all prime ideals P of K dividing A, wherevP is the valuation associated to P. This is equivalent to the usual notion α ≡ 1 (mod A) when
α is an algebraic integer.
Theorem 2.1 (Tate). Let A be a fractional ideal of K. Then the following statements areequivalent.
(i ). There exists an anti-unit α ∈ K ◦ such that AwK θK/k,S = αOK and K(α1/wK )/k is abelian.
(ii ). There exist an extension L/K such that L/k is abelian and an anti-unit γ ∈ L◦ such that
(AOL )θK/k,S = γOL .
(iii ). For almost all prime ideals p of k, there exists αp ∈ K ◦ such that αp ≡ 1 (mod∗ pOK )and A(σp −N (p))θK/k,S = αp OK where σp is the Frobenius automorphism of p in G.
(iv ). There exist a family (ai )i∈I of element of Z[G] generating AnnZ[G] (µK ) and a family(αi )i∈I of anti-units in K such that Aai θK/k,S = αi OK for all i ∈ I, and αi aj = αj ai for
all i, j ∈ I.
Remark. Here and in the rest of the paper, when we say “for almost all prime ideals”, we
implicitly exclude the ramified primes; therefore the Frobenius automorphism is uniquely defined.
1/nRemark. In part (ii), (AOL )θK/k,S is defined by the formula (AOL )nθK/k,S
where n ≥ 1
is any integer such that nθK/k,S ∈ Z[G]. This is well-defined, when it exists, since the group of
ideals of a number field is torsion-free.
Let A be a fractional ideal of K. We say that BS(K/k, S; A) holds if the ideal A satisfies
the equivalent conditions of Theorem 2.1. The conjecture BS(K/k, S) is thus the collection of
the conjectures BS(K/k, S; A) where A ranges through the fractional ideals of K. In [21], Tateproves that the set of fractional ideals A of K such that BS(K/k, S; A) holds is a subgroup of
the group of ideals of K, stable under the action of G, and that contains the principal ideals of
K. In particular, BS(K/k, S) holds if the field K is principal. Now, let p0 be a prime ideal ofk not in S, thenθK/k,S∪{p0 } = (1 − σp0 −1 )θK/k,S .
(5)
2Note that K ◦ = {±1} if K is not totally complex.

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GAELLE DEJOU AND XAVIER-FRANÇOIS ROBLOT

It follows from this formula that the validity of BS(K/k, S) implies that of BS(K/k, S ∪ {p0 }).
Therefore, the conjecture is true for any admissible set of places S if it is true for the minimal
set that contains exactly the infinite places of k and the finite places that ramify in K/k. The
validity of the abelian Brumer-Stark Conjecture is also preserved under change of extension asa consequence of part (ii) of Proposition 2.1. That is, if K/K 0 /k is a tower of number fields,
then the validity of BS(K/k, S) implies that of BS(K 0 /k, S). It also preserved under change ofbase, that is if BS(K/k, S) holds then so does BS(K/k 0 , S 0 ) where K/k 0 /k is a tower of number
fields and S 0 denotes the set of places of k 0 above the places in k, see [12]. The following casesof the conjecture are proved by Tate (see [21] and [22]).
Theorem 2.2 (Tate). The abelian Brumer-Stark conjecture BS(K/k, S) is true in the following
cases.
• The field k is the field Q of rational numbers.3• The extension K/k is quadratic.
• The extension K/k is of degree 4 and is contained in a non-abelian Galois extension
K/k0 of degree 8.
Sands proves the abelian Brumer-Stark conjecture when the group G is isomorphic to Z/2Z ×
Z/2Z and, more generally, when the group G has exponent 2 with some additional technicalhypothesis, see [19]. A local version of the conjecture is stated and is proved for some types ofextensions of degree 2p (with p odd) and numerically studied in some others by Greither et al.
in [11]. The local abelian Brumer-Stark conjecture at p holds for so-called “non-exceptional”
primes p provided some appropriate Iwasawa µ-invariant vanishes by results of Nickel [15], andwhen S contains all the prime ideals above p and, again, some appropriate Iwasawa µ-invariant
vanishes by results of Greither and Popescu [10]. Nickel shows in [14] that the local abelian
Brumer-Stark conjecture outside of 2 is implied by the relevant special case of the Equivariant
Tamagawa Number Conjecture (ETNC) plus some additional technical hypothesis. Since thisspecial case of the ETNC was proved by Burns and Greither [3], this implies, in particular, thepart outside of 2 of the abelian Brumer-Stark conjecture holds if K/k is a tame extension with
K an abelian extension of Q.
As mentioned in the introduction, generalizations to the non-abelian case of the Brumer-Stark
conjecture (and also the Brumer conjecture) due to Nickel are stated in [14] (see also [2] for much
more general conjectures due to Burns), we state these conjectures and study the links with ourconjecture in an appendix at the end of this article.
3. The Galois Brumer-Stickelberger element
We assume from now on that the extension K/k is Galois, but not necessarily abelian. The
set S still denotes a finite set of places of k containing the infinite places of k and the finite places
that ramify in K/k. As in the abelian case, we assume also that S is cardinality at least 2. Notethat the only non-trivial case we are excluding is when k is a complex quadratic field and K is
an unramified extension of k. The first step in the generalization of the abelian Brumer-Stark
conjecture is the construction of the Brumer-Stickelberger element associated to non-abelian
Galois extensions. Fortunately, such a construction is provided by the work of Hayes [13]. We
now review his construction and the first properties of the Brumer-Stickelberger element. Denoteby Ĝ the set of irreducible characters of G. For χ ∈ Ĝ, let LK/k,S (s, χ) denote the Artin Lfunction of χ with Euler factors at primes in S deleted. The Brumer-Stickelberger element isdefined byX
θK/k,S :=LK/k,S (0, χ) eχ̄
(6)χ∈Ĝ
3In this situation, it boils down to Stickelberger’s theorem on cyclotomic sums.

A BRUMER-STARK CONJECTURE FOR NON-ABELIAN GALOIS EXTENSIONS

where eχ :=5χ(1) X
χ(σ)σ −1 is the central idempotent associated to χ.
|G|σ∈G

The following results are extracted from [13].
Theorem 3.1 (Hayes). Denote by CG the set of conjugacy classes of G. The Brumer-Stickelberger
element lies in the center Z(C[G]) of C[G] and is the only element of Z(C[G]) such thatφχ (θK/k,S ) = LK/k,S (0, χ̄)(7)for all χ ∈ Ĝ, where φχ is the ring homomorphism from Z(C[G]) to C defined byφχ (C) :=

χ(C)
χ(1)

for all C ∈ CG .
Let B be a normal subgroup of G. Then we have
θK B /k,S = π(θK/k,S )B

where π : Gal(K/k) → Gal(K /k) is the canonical surjection induced by the restriction to K B .
Let H be a subgroup of G. Denote by SH the set of places of K H above the places in S. Let
INormG→H : Z(C[G]) −→ Z(C[H]) be the inhomogeneous norm defined byXYGa(χ)hχ,IndH φiG eφ
INormG→H (a) :=
φ∈Ĥ

χ∈Ĝ

P
for a := χ∈Ĝ a(χ)eχ ∈ Z(C[G]), where h·, ·iG is the inner product on the characters of G andeφ is the central idempotent of C[H] associated to φ. Then we have
θK/K H ,SH = INormG→H (θK/k,S ).
We are now interested in generalizing properties (4) and (5). We start with (4).
Proposition 3.2. For v a place of k, define
X
Nv :=
σ∈Dw

1
Cσ ∈ Q[G]|Cσ |where w is a fixed place of K above v, Dw is the decomposition group of w in G and Cσ ∈ CG
is the conjugacy class of σ in G. Then, for any place v in S, we have
Nv θK/k,S = 0.
Proof. Since Nv is in Z(C[G]), it is enough, with the notations of Theorem 3.1, to prove thatφχ (Nv θK/k,S ) = φχ (Nv )φχ (θK/k,S ) = 0 for all χ ∈ Ĝ. Let χ ∈ Ĝ be such that φχ (Nv ) 6= 0. By(7), we need to prove that the order r(χ̄) = r(χ) of vanishing at s = 0 of LK/k,S (s, χ) is at least
1. Let ρ : G → GL(V ) be an irreducible representation with character χ. By [22, Prop. I.3.4],
we haveXr(χ) =
dim V Dw0 − dim V G(8)v 0 ∈S

where w0 is a fixed place of K above v 0 and Dw0 denotes the decomposition group of w0 in G.
Assume first that χ is the trivial character. Then the above formula yields r(χ) = |S| − 1 andthe result follows from our hypothesis that S contains at least two places. Assume now that χ
is non-trivial. We compute
X 1
1 X
|Dw |
φχ (Nv ) =φχ (Cσ ) =χ(σ) =
1Dw , χ|Dw Dw
|Cσ |
χ(1)
χ(1)
σ∈Dw

σ∈Dw

where 1Dw is the trivial character of Dw . By the above hypothesis, φχ (Nv ) 6= 0 and thus
the trivial character 1Dw appears in the decomposition of χ|Dw . Therefore the space V Dw has

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GAELLE DEJOU AND XAVIER-FRANÇOIS ROBLOT

dimension at least 1. On the other hand, V G = {0} since χ is irreducible. It follows thatr(χ) ≥ 1 and the result is proved.

Assume that there exists v ∈ S that is totally split in K/k. Then Nv = 1 and the BrumerStickelberger element is trivial in this case. Therefore, as in the abelian case, we assume thatboth k is totally real and K is totally complex, otherwise the Brumer-Stickelberger element istrivial. In fact, we can say more than that. Recall that a number field E is CM if it is a totallycomplex quadratic extension of a totally real field. If furthermore E is Galois over some totallyreal subfield F , then Gal(E/F ) has a unique complex conjugation and we say that a characterχ of Gal(E/F ) is totally odd if the eigenvalues of an associated representation evaluated at thecomplex conjugation are all equal to −1. The following result is due to Tate, see [22, p. 71].
Proposition 3.3 (Tate). Let χ ∈ Ĝ be a character such that LK/k,S (0, χ) 6= 0. Then χ is the
inflation of a totally odd character of a Galois CM sub-extension F/k of K/k.
Corollary 3.4. If K/k does not contain a Galois CM sub-extension then θK/k,S = 0.
Proof. Assume that θK/k,S 6= 0. Then, by Theorem 3.1 and the fact that (φχ )χ∈Ĝ is a basisof the dual of Z(C[G]), we get that there exists an irreducible character χ ∈ Ĝ such thatφχ (θK/k,S ) = LK/k,S (0, χ) 6= 0. This character comes from a Galois CM sub-extension by the
proposition.

Corollary 3.5. Let τ be a complex conjugation of G. Then (τ + 1) · θK/k,S = 0.
Proof. By the proposition, it is enough to prove that (τ + 1) · eχ = 0 for any character χ ∈ Ĝthat is the inflation of a totally odd character χ̃ of a Galois CM sub-extension. Since χ̃ is totally
odd, we have χ(gτ ) = −χ(g) for all g ∈ G. Let R be a set of representatives of G/{1, τ }. We
now computeχ(1) X 
χ(ρ)ρ−1 + χ(ρτ )(ρτ )−1(τ + 1) · eχ = (τ + 1) ·
|G|
ρ∈R

χ(1) X 
χ(ρ)ρ−1 − χ(ρ)τ ρ−1= (τ + 1) ·|G|ρ∈R

= (τ + 1)(1 − τ ) ·χ(1) X
χ(ρ)ρ−1 = 0.
|G|



ρ∈R

The following result generalizes (5) to the Galois case.
Proposition 3.6. Let p0 be a prime ideal of k not in S. ThenXθK/k,S∪{p0 } = θK/k,Sdet(1 − ρχ (σP0 ))eχ̄χ∈Ĝ

where P0 is a fixed prime ideal of K above p0 , σP0 is the Frobenius automorphism of P0 in G,
and, for χ ∈ Ĝ, ρχ denotes a fixed irreducible representation of G with character χ.
Proof. With the notations of Theorem 3.1, it is enough to prove, for all ψ ∈ Ĝ, thatXφψ (θK/k,S∪{p0 } ) = φψ (θK/k,S ) φψdet(1 − ρχ (σP0 ))eχ̄χ∈Ĝ= LK/k,S (0, ψ̄)X

det(1 − ρχ (σP0 ))φψ (eχ̄ ).

χ∈ĜOn the other hand, from the definition of Artin L-functions, we see thatφψ (θK/k,S∪{p0 } ) = LK/k,S∪{p0 } (0, ψ̄) = LK/k,S (0, ψ̄) det(1 − ρψ̄ (σP0 )).

A BRUMER-STARK CONJECTURE FOR NON-ABELIAN GALOIS EXTENSIONS

The result follows from the fact that φψ (eχ̄ ) = 1 if ψ = χ̄ and zero otherwise.

7

We now turn to the question of the rationality of the Brumer-Stickelberger element θK/k,Swhen G is non-abelian. As noted on page 2584 of [14], it is a consequence of the principal rankzero Stark conjecture that was proved by Tate [22]. We recall the argument of the proof. For
any character χ of G, the principal rank zero Stark conjecture states that
LK/k,S (0, χα ) = LK/k,S (0, χ)α for all α ∈ AutQ (C)(9)

where χα := α ◦ χ. We writeθK/k,S =X

LK/k,S (0, χ)Xχ̄(1) X
χ(σ)σ =
xσ σ
|G|
σ∈G

χ∈Ĝ

σ∈G

where
xσ :=

1 X
χ̄(1)χ(σ)LK/k,S (0, χ).
|G|
χ∈Ĝ

Let α be an automorphism of C. We compute
1 X αχ̄ (1)χα (σ)LK/k,S (0, χ)αα(xσ ) =
|G|
χ∈Ĝ

1 X α
=
χ̄ (1)χα (σ)LK/k,S (0, χα ) = xσ
|G|
χ∈Ĝ

since the map χ 7→ χα is a bijection on the set Ĝ. It follows that xσ ∈ Q for all σ ∈ G, and thus
the Brumer-Stickelberger element θK/k,S lies in Q[G].
An interesting problem is to find a suitable denominator for the Brumer-Stickelberger elementin the non-abelian case. In the abelian case, as noted above, wK θK/k,S is always integral. In
the Galois case, however, one can see from examples that it is not true anymore. Let [G, G] be
the commutator subgroup of G, that is the subgroup generated by the commutators [g1 , g2 ] :=g1 g2 g1−1 g2−1 with g1 , g2 ∈ G. We make the following conjecture.
Conjecture (The Integrality Conjecture).
Define mG to be the lcm of the cardinalities of the conjugacy classes of G and let sG be the orderof the commutator subgroup [G, G] of G. Let dG be the lcm of mG and sG . Then, for almost allprime ideals P of K, we havedG (σP − N (p))θK/k,S ∈ Z[G](10)where p is the prime ideal of k below P and σP is the Frobenius automorphism of P in G.
One could weaken the Integrality Conjecture by just asking that there exists an integer dG ,depending only on the isomorphism class of G, such that (10) holds without specify its value.
However, heuristic arguments lead us to predict this specific value of dG . First, observe thatmG = 1 if and only if sG = 1 if and only if G is abelian. Therefore, when the extension K/k
is abelian, the Integrality Conjecture is equivalent to the statement before (3) using Lemma 3.7
below. We now explain why we conjecture that the factor sG is necessary. Let Gab := G/[G, G]
be the maximal abelian quotient of G and K ab := K [G,G] be the maximal sub-extension of K/kthat is abelian over k; we have Gal(K ab /k) = Gab . Denote by π ab : G → Gab the canonicalsurjection induced by the restriction to K ab . Let ν ab be the map from C[Gab ] to C[G] definedfor g̃ ∈ Gab by
X
1
g
(11)
ν ab (g̃) :=sG abπ(g)=g̃

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GAELLE DEJOU AND XAVIER-FRANÇOIS ROBLOT

where the sum is over elements g ∈ G whose image by π ab is equal to g̃. The map ν ab isextended to C[Gab ] by linearity.4 Let κ ∈ C[Gab ], we have (π ab ◦ ν ab )(κ) = κ and, if ξ ∈ C[G],then ξν ab (κ) = ν ab (π ab (ξ)κ). The 1-dimensional characters of G are exactly the ones that are
inflations of characters of Gab . For such a character χ, denote by χ̃ the character of Gab suchthat χ = χ̃ ◦ π ab . One checks readily that eχ = ν ab (eχ̃ ) where eχ̃ is the idempotent of C[Gab ]associated to χ̃. By the properties of Artin L-functions, we have
X
X
LK ab /k,S (0, χ̃)ν ab (eχ̃¯ )LK/k,S (0, χ)eχ̄ =χ∈Ĝχ(1)=1χ̃∈Ĝab= ν ab X

LK ab /k,S (0, χ̃)eχ̃¯ = ν ab (θK ab /k,S ).

χ̃∈ĜabWe define(>1)

θK/k,S :=X

LK/k,S (0, χ)eχ̄ .

χ∈Ĝχ(1)>1

By the above computation, we find that(>1)

θK/k,S = ν ab (θK ab /k,S ) + θK/k,S .

(12)

˜ K ab /k,S ) ∈ Z[G] by the remark
For all ξ ∈ AnnZ[G] (µK ), we have sG ξ ν ab (θK ab /k,S ) = sG ν ab (ξθab˜before (3) since ξ := π (ξ) ∈ AnnZ[Gab ] (µK ab ). Therefore the factor sG is there to ensurethat the part of the Brumer-Stickelberger element coming from the 1-dimensional characters is
integral.
The first open case for the Integrality Conjecture is when G ' SL2 (F3 ), see Theorem 5.2. In
fact, for relative Galois extensions K/k of degree ≤ 31 and with Galois group not isomorphic
to SL2 (F3 ), one can prove that sG is a suitable denominator for θK/k,S . However, numericalexperiments in the case G ' SL2 (F3 ) show that sG is not a suitable denominator in general and,in fact, it is necessary to use 3sG in some cases, see [7, Chap. 5]. It is therefore necessary to add
an extra factor. After Hayes, define, for s ∈ C, the meromorphic function
XΘK/k,S (s) :=LK/k,S (s, χ) eχ̄ .
χ∈ĜNote that ΘK/k,S (0) = θK/k,S . Using this function, Hayes defines in [13, §5] the partial zetafunction ζK/k,S (s, C) of a class C ∈ CG by the formulaX
1 −1
ΘK/k,S (s) =ζK/k,S (s, C)
C .
(13)
|C|
C∈CG

Note that this definition makes sense because the values of ΘK/k,S are in Z(C[G]). Applying φχ
on both sides, for χ ∈ Ĝ, he gets
1 X
LK/k,S (s, χ) =ζK/k,S (s, C) χ(σC )
(14)
χ(1)
C∈CG

where σC denotes a fixed element in C. Equations (13) and (14) should be thought as generalizations to the non-abelian case of equations (2) and (1) respectively. Assuming that thepartial zeta functions satisfy similar properties in the non-abelian case as in the abelian caseand comparing (2) and (13) evaluated at s = 0, it is therefore natural to assume that the factor
mG , the lcm of the cardinalities of the conjugacy classes of G, is needed to make the Galois
4Note that the image of ν ab is in fact contained in Z(C[G]).

A BRUMER-STARK CONJECTURE FOR NON-ABELIAN GALOIS EXTENSIONS

9

Brumer-Stickelberger element integral. This explains the value of dG given in the Integrality
Conjecture.5
The next result is proved in [22, Lemme IV.1.1] for abelian extensions. It is straightforward
to extend the proof to Galois extensions, also see [14, Lemma 2.2].
Lemma 3.7. Let T be a set of prime ideals containing all the unramified prime ideals of K
that do not divide wK except, possibly, a finite number. Then AnnZ[G] (µK ) is generated as a
Z-module by the elements σP − N (p) where P runs through the prime ideals in T and p denotes
the prime ideal of k below P. Furthermore, we have
wK = gcd (1 − N (p)).



P∈T
σP =1

From this, we deduce equivalent formulations of the Integrality Conjecture.
Proposition 3.8. The following assertions are equivalent(1). For almost all prime ideals P of K, dG (σP − N (p))θK/k,S ∈ Z[G].
(2). For all ξ ∈ AnnZ[G] (µK ), dG ξ θK/k,S ∈ Z[G].
(>1)(3). For almost all prime ideals P of K, dG (σP − N (p))θK/k,S ∈ Z[G].
(>1)(4). For all ξ ∈ AnnZ[G] (µK ), dG ξ θK/k,S ∈ Z[G].
Proof. The equivalences (1) ⇔ (3) and (2) ⇔ (4) are consequences of (12) and the discussionthat follows. The direction (2) ⇒ (1) is trivial. The other direction comes from Lemma 3.7. 
4. Strong central extensions
Before we generalize the abelian Brumer-Stark conjecture to Galois extensions, we introduce
the notion of strong central extensions that will play a crucial role. For that, we stop assuming
for a moment that G is the Galois group of the extension K/k and just consider G as a finitegroup. Let Γ and ∆ be two other finite groups with ∆ a normal subgroup of Γ such that thefollowing sequence is exact
/∆

1

/Γ

π

/G

/ 1,

(15)

that is, Γ is a group extension of G by ∆. Recall that the extension is said to be central if ∆is a subgroup of the center of Γ. This implies, in particular, that ∆ is an abelian group. If,furthermore, the extension is split, that is there exists an homomorphism s : G → Γ such thats ◦ π is the identity, then the extension is trivial, that is Γ ' ∆ × G.
We say that Γ is a strong central extension of G by ∆ if ∆ ∩ [Γ, Γ] = 1 where [Γ, Γ] is thecommutator subgroup of Γ. The choice of terminology is explained by the following lemma.
Lemma 4.1. Let Γ be a strong central extension of G by ∆. Then Γ is a central extension of
G by ∆.
Proof. Let γ ∈ Γ and δ ∈ ∆. We see that[γ, δ] = (γδγ −1 )δ −1 ∈ ∆since ∆ is normal in Γ. Thus, [γ, δ] = 1 and γ and δ commute. Therefore ∆ is in the center ofΓ and the extension is central.

The trivial extension ∆ × G is always a strong central extension. As noted above, a strong
central extension is trivial if and only if it is split. By the Schur-Zassenhaus theorem, this is
the case when the orders of ∆ and G are relatively prime. For strong central extensions, theextension is also trivial in an additional case. First, we have the following characterization ofstrong central extensions.
5Note that, for G ' SL (F ), we have s
2

3

G = 8 and mG = 12.

10GAELLE DEJOU AND XAVIER-FRANÇOIS ROBLOT

Lemma 4.2. Consider the group extension (15). This extension is strong central if and only if
the map π restricts to an isomorphism between [Γ, Γ] and [G, G].
Proof. It is straightforward to see that π restricts to a surjective map from [Γ, Γ] to [G, G]. Thismap is injective if and only if [Γ, Γ] ∩ Ker(π) = 1. The result follows since Ker(π) = ∆.
Lemma 4.3. Let Γ be a strong central extension of G by ∆. Assume that G = [G, G]. Then
Γ ' ∆ × G.
Proof. Indeed, by Lemma 4.2, the sequence is split.

It is not true however that all strong central extensions are split and give rise to a direct
product as we show in the following example.
Example. Let Γ be the dicyclic group of order 12. It is the group generated by the two elementsa and b with the following relations: a3 = b4 = 1 and bab−1 = a−1 . Let ∆ := hb2 i ' Z/2Z;
it is the center of Γ and one can verify that Γ/∆ ' S3 , the symmetric group on 3 letters. We
compute [Γ, Γ] = hai, thus ∆ ∩ [Γ, Γ] = {1} and we have the strong central extension1/ Z/2Z

/Γ

/ S3

/ 1.

However, the group Γ is not isomorphic to Z/2Z × S3 since the latter group does not have any
element of order 4.
The following lemma provides us with yet another characterization of strong central extensions.
Lemma 4.4. Consider the group extension (15). This extension is strong central if and only if,for every abelian subgroup H of G, the subgroup π −1 (H) of Γ is abelian.
Proof. Assume that the extension is strong central. Let H be an abelian subgroup of G. Let
γ1 , γ2 ∈ π −1 (H), say π(γ1 ) = h1 , π(γ2 ) = h2 with h1 , h2 ∈ H. We compute
π([γ1 , γ2 ]) = [h1 , h2 ] = 1.
By hypothesis, this implies that [γ1 , γ2 ] = 1 and therefore π −1 (H) is abelian.
Reciprocally, we assume that, for any abelian subgroup H of G, the group π −1 (H) is abelian.
Let γ1 , γ2 ∈ Γ be such that [γ1 , γ2 ] ∈ ∆. Then π([γ1 , γ2 ]) = 1 and π(γ1 ) and π(γ2 ) commute.
The subgroup of G that they generate is abelian and, by hypothesis, it follows that γ1 and γ2commute, that is [γ1 , γ2 ] = 1. Therefore the extension Γ of G by ∆ is strong central.

We note another property of strong central extensions that will be useful later on. For a finitegroup A, recall that mA denote the lcm of the cardinalities of the conjugacy classes of A, sA is
the order of the commutator subgroup [A, A] of A and dA is the lcm of mA and sA .
Lemma 4.5. Consider the group extension (15). Assume that the extension is strong central.
Then we have dΓ = dG .
Proof. It is enough to show that mΓ = mG and sΓ = sG . The fact that sΓ = sG is a directconsequence of Lemma 4.2. We now show that mΓ = mG . Let γ ∈ Γ. Denote by C and Z
respectively the conjugacy class of γ in Γ and the centralizer of γ in Γ. We have
|C| = (Γ : Z) = (π(Γ) : π(Z))(Ker(π) : Ker(π) ∩ Z) = (G : π(Z))(∆ : ∆ ∩ Z)
= (G : Z0 )(Z0 : π(Z))(∆ : ∆ ∩ Z) = |C0 |(Z0 : π(Z))(∆ : ∆ ∩ Z)
where C0 is the conjugacy class of π(γ) in G and Z0 is the centralizer of π(γ) in G. Since ∆is in the center of Γ by Lemma 4.1, we have ∆ ⊂ Z and (∆ : ∆ ∩ Z) = 1. Now, let ρ0 ∈ Z0
and let ρ ∈ π −1 (ρ0 ). We have π([ρ, γ]) = [ρ0 , π(γ)] = 1 since ρ0 commutes with π(γ). Therefore[ρ, γ] ∈ [Γ, Γ] ∩ ∆ = {1} and ρ ∈ Z. Thus, π(Z) = Z0 and we have finally |C| = |C0 |. As any
conjugacy class of G is the image by π of a conjugacy class of Γ, we see that mΓ = mG and theresult is proved.
A BRUMER-STARK CONJECTURE FOR NON-ABELIAN GALOIS EXTENSIONS

11

We now come back to our previous setting and assume that G is the Galois group of an
extension K/k. Let L be a finite extension of K. We say that L is a strong central extension of
K/k if L/k is Galois and the group extension
/∆

1

/Γ

/G

/1

is strong central where ∆ := Gal(L/K) and Γ := Gal(L/k). The following result is a directconsequence of Lemma 4.2 (see also Figure 1).
Lemma 4.6. Denote by Lab the maximal sub-extension of L/k that is abelian over k. Then
L is a strong central extension of K/k if and only if L = KLab . Furthermore, in that case,restriction to Lab yields an isomorphism between Gal(L/K) and Gal(Lab /K ab ) where K ab is
the maximal sub-extension of K/k that is abelian over k.

L
[Γ,Γ]

Lab

∆

Γ

K

[G,G]

∆

K ab

G

k

Gab

Figure 1. Some subfields of the strong central extension L/k of K/k
We conclude this section with a lemma that shows strong central extensions behave somewhat
nicely.
Lemma 4.7. Let L be a strong central extension of K/k.
(1) Let L0 /K be a sub-extension of L/K. Then L0 is a strong central extension of K/k.
(2) Let M be another strong central extension of K/k. Then LM is a strong central extensionof K/k.
Proof. We use repeatedly the characterization of strong central extensions given by Lemma 4.6.
We prove the first assertion. The group Gal(L/L0 ) is a subgroup of Gal(L/K) and thus it isabnormal in Gal(L/k). Therefore, L0 /k is a Galois extension. Let Lab∩ L0 be the maximal
0 =L
ab
ab
abelian sub-extension of L0 /k, then [L0 : K ] = [L0 : K] since Gal(L/K) ∼= Gal(Lab /K ab ).
ababFurthermore, since L0 ∩ K = K , we find that[KLab
0 : k] =[Lab
0 : k][K : k]ab
= [Lab0 : K ][K : k] = [L0 : k],[K ab : k]thus KLab
0 = L0 and L0 is a strong central extension of K/k.
We now prove the second assertion. The extension LM/k is Galois as the compositum of two
Galois extensions of k. Let F = L ∩ M . It is an extension of K. Then, a direct computationshows that [LM : K] = [Lab M ab : K ab ]. We find that
[KLab M ab : k] =[Lab M ab : k][K : k]
= [Lab M ab : K ab ][K : k] = [LM : k].
[K ab : k]Thus, KLab M ab = LM . Since the maximal abelian sub-extension (LM )ab of LM/k that isabelian over k contains Lab M ab , it follows that K(LM )ab = LM and LM is a strong central
extension of K/k.


12

GAELLE DEJOU AND XAVIER-FRANÇOIS ROBLOT

5. The Galois Brumer-Stark conjecture
We are now ready to state our generalization of the abelian Brumer-Stark conjecture to Galois
extensions.
Conjecture (The Galois Brumer-Stark conjecture BSGal (K/k, S)).
Let K/k be a Galois extension of number fields and let S be a finite set of places of k that
contains the infinite places and the finite places that ramify in K with |S| ≥ 2. The Integrality
Conjecture holds for the extension K/k and the set of places S, and, for any fractional ideal Aof K, the ideal AdG wK θK/k,S is principal and admits a generator α ∈ K ◦ such that K(α1/wK ) isa strong central extension of K/k.
Remark. As in the abelian case, the last assertion that K(α1/wK ) is a strong central extensionof K/k does not depend on the choice of the wK -th root of α since all of these generate the same
extension of K.
Before studying conjecture BSGal (K/k, S), we discuss briefly our evidence for it. Observe
first that it is in some ways a natural generalization of the abelian Brumer-Stark conjecture.
Indeed, we have the following result.
Proposition 5.1. Assume that K/k is abelian. Then the Galois Brumer-Stark conjecture
BSGal (K/k, S) is equivalent to the abelian Brumer-Stark conjecture BS(K/k, S).
Proof. This is clear since dG = 1 in that case and, by Lemma 4.4, we see that K(α1/wK )/k isabelian if and if only if K(α1/wK ) is a strong central extension of K/k.

Another piece of evidence is provided by the following result that sums up the cases where
the conjecture is proved or reduces to the abelian Brumer-Stark conjecture. Examples where
the conjecture is numerically proved are also given in [7, Chap. 5].
Theorem 5.2. The Galois Brumer-Stark conjecture is satisfied in the following cases(1) Gal(K/k) is a non-abelian simple group,(2) Gal(K/k) ' D2n where D2n is the dihedral group of order 2n with n odd,
(3) Gal(K/k) ' Sn where Sn is the symmetric group on n letters with n ≥ 1,(4) Gal(K/k) is non-abelian of order 8.
Assume that the abelian Brumer-Stark conjecture holds. Then the Galois Brumer-Stark conjecture is satisfied in the following cases(5) Gal(K/k) is abelian,(6) Gal(K/k) contains a normal abelian subgroup of prime index,(7) Gal(K/k) is of order < 32 and not isomorphic to SL2 (F3 ).
Proof. Cases 1, 2, 3, 4 and 5 follow respectively from Propositions 6.6, 6.7, 6.8, 7.7, and 5.1.
The results of Section 7, and in particular Theorem 7.4, imply case 6. Finally, case 7 follows froma direct inspection using the GAP system [9] and verifying that, in each case, one can reduce tothe abelian case, one of the other listed cases or an application of Proposition 6.5 below.

Remark. Using the GAP system [9], one can verify also by similar techniques that the Galois
Brumer-Stark conjecture holds or reduces to the abelian Brumer-Stark conjecture for 730 out of
the 1048 possible isomorphism types of Galois groups when [K : k] ≤ 100.
Remark. The Integrality Conjecture actually holds in all the cases listed in Theorem 5.2 without
having to assume the abelian Brumer-Stark conjecture for cases 5, 6, 7. It also holds for the 730
isomorphism types of Galois groups mentioned in the previous remark.
The following result is the generalization to the non-abelian case of Theorem 2.1. Recall that,for a prime ideal P of K, we denote by p the prime ideal of k below P and by σP the Frobenius
automorphism of P in G.

A BRUMER-STARK CONJECTURE FOR NON-ABELIAN GALOIS EXTENSIONS

13

Theorem 5.3. Assume that the Integrality Conjecture holds for the extension K/k and the setof places S. Let A be a fractional ideal of K. The following assertions are equivalent.
(i ). There exists an anti-unit α ∈ K ◦ such that AdG wK θK/k,S = αOK and K(α1/wK ) is a strong
central extension of K/k.
(ii ). There exists a strong central extension L of K/k and an anti-unit γ ∈ L◦ such that
(AOL )dG θK/k,S = γOL(iii ). For almost all prime ideals P of K, there exists an anti-unit αP ∈ K ◦ such that AdG (σP −N (p))θK/k,S =αP OK and αP ≡ 1 (mod∗ Q) for all prime ideals Q of K above p such that σQ = σP .
(iv ). For any abelian subgroup H of G, there exists a family (ai )i∈I of elements of Z[H] generating AnnZ[H] (µK ) as a Z-module and a family of anti-units (αi )i∈I of K such that
AdG ai θK/k,S = αi OK and αj ai = αi aj for all i, j ∈ I.
1/nRemark. In part (ii), (AOL )dG θK/k,S is defined by the formula (AOL )ndG θK/k,S
wheren ≥ 1 is any integer such that ndG θK/k,S ∈ Z[G]. This is well-defined since the group of ideals
of a number field is torsion-free.
Proof. We use repeatedly the fact that θK/k,S lies in the center of C[G].
(i) ⇒ (ii). Let γ := α1/wK and L := K(γ). Then, L is a strong central extension of K/k and γ
is an anti-unit in L. Furthermore, we have
wKγOL
= αOL = (AOL )dG wK θK/k,S
and the result follows since the group of ideals of a number field is torsion-free.
(ii) ⇒ (iii). Denote by Γ the Galois group of L/k and by ∆ the Galois group of L/K. Let T be
the set of prime ideals of K, unramified in L/K and K/Q, relatively prime with wK and with A
and all its conjugates over k. Note that T contains all but finitely many prime ideals of K. Let
P ∈ T and let P̃ be a prime ideal of L above P. Denote by σP̃ the Frobenius automorphismof P̃ in Γ. We set αP̃ := γ σP̃ −N (p) . Let Q̃ be another prime ideal of L above p such that
π(σP̃ ) = π(σQ̃ ) where π : Γ → G is the canonical surjection induced by the restriction to K
and σQ̃ is the Frobenius automorphism of Q̃ in Γ. There exists ρ ∈ Γ such that Q̃ = ρ(P̃),
and we have σQ̃ = ρσP̃ ρ−1 . Since π([ρ, σP̃ ]) = π(σQ̃ )π(σP̃ )−1 = 1, this commutator lies in ∆and is therefore trivial. Thus σQ̃ = σP̃ and αQ̃ = αP̃ . In particular, αP̃ does not depend on
the choice of the prime ideal P̃ of L above P, and we can just denote it by αP . Furthermore,
αP = γ σQ̃ −N (p) ≡ 1 (mod∗ Q̃) for all prime ideals Q̃ of L above p such that σQ = σP where Q
is the prime ideal of K below Q̃. We now prove that αP lies in K. Let δ ∈ ∆. We haveσ −N (p) δ−1  σ −N (p) δ−1
δ−1 wKαP= γ wK P̃
= α P
=1
δ−1
since α lies in K. Thus, there exists a root of unity ξ ∈ µK such that αP= ξ. We have∗∗δαP ≡ αP ≡ 1 (mod P̃) by the above remark, hence ξ ≡ 1 (mod P̃) and thus ξ = 1 by the
choice of P. Therefore, αP ∈ K as desired. Furthermore, it is clear from its construction that
it is an anti-unit and that we have αP ≡ 1 (mod∗ Q) for all prime ideals Q above p such that
σQ = σP by the above. Finally, we computeσ −N (p)σ −N (p)αP OL = (γOL ) P̃
= (AOL )dG θK/k,S P̃
= (AOL )dG (σP̃ −N (p))θK/k,S ,and, since A is an ideal of K and dG (σP − N (p))θK/k,S ∈ Z[G] by the Integrality Conjecture,
we getαP OK = AdG (σP −N (p))θK/k,S .
The implication is proved.
(iii) ⇒ (iv). Let H be an abelian subgroup of G. Denote by TH the set of prime ideals of K
for which (iii) applies and that are unramified in L/K and K/k, relatively prime with wK and14

GAELLE DEJOU AND XAVIER-FRANÇOIS ROBLOT

with A and all its conjugates over k, and whose Frobenius automorphism in G lies in H. Let I
be a set indexing TH , so that TH = {Pi : i ∈ I}. For i ∈ I, we set ai := σPi − N (pi ) ∈ Z[H]and αi := αPi ∈ K ◦ . It follows from Lemma 3.7 that the family (ai )i∈I generates AnnZ[H] (µK ).
By construction, we have also AdG ai θK/k,S = αi OK . It remains to prove that, for i, j ∈ I, weaσ −N (q)andhave αjai = αi j , that is, for two prime ideals P and Q in TH , the two elements αPQ
σ −N (p)αQPare equal. We have(αP OK )σQ −N (p) = AdG (σP −N (p))θK/k,S

σQ −N (p)= AdG (σQ −N (p))θK/k,S

σP −N (p)= (αQ OK )σP −N (p)where we used the fact that σP and σQ commute since they both belong to H. Since αP and
σ −N (p)σ −N (q).
αQ are both anti-units, there exists a root of unity ξ ∈ µK such that αPQ= ξαQP∗
Reasoning as above, we see that ξ ≡ 1 (mod P), thus ξ = 1 and the equality is proved.
(iv) ⇒ (i). Let H be an abelian subgroup of G. Let (ai )i∈I and (αi )i∈I be the corresponding
families. There exists a family (λi )i∈I of integers, with only finitely many non-zero terms, such
that
X
wK =
λi ai .
i∈I

We set αH :=λii∈I αi . It is clear that αH is an anti-unit of K and we haveP
αH OK = AdG ( i λi ai )θ = AdG wK θK/k,S .

Q

In particular, up to a root of unity in K, αH does not depend upon the choices made, and wewill therefore denote it simply by α. For any h ∈ H, there exists an integer nh ∈ N such thath − nh annihilates µK . Therefore, there exists a family (λh,i )i∈I of integers, with only finitely
many non-zero terms, such thatX
h − nh =λh,i ai .
i∈I

Furthermore, we have
αh−nh =

Y Y
i∈I

Q

a λ

αi j h,j

λi

=

j∈I

Y Y
i∈I

λ

αj h,jλi aiP= αh i∈I

λi ai= αhwKj∈I

λwhere αh := i∈I αi h,i . For g, another element of H, one can prove in the same way thatg−nαh g = αgh−nh . Let γ := α1/wK and L := K(γ). We now prove that L/K H is an abelian
extension. First, we prove that L/K H is a Galois extension. For h ∈ H, let h̃ be any lift of h to
L. We compute(γ h̃−nh )wK = (γ wK )h̃−nh = αh−nh = αhwK .
Thus, there exists ξh ∈ µK such that γ h̃−nh = ξh αh . Therefore, we haveγ h̃ = ξh αh γ nh ∈ L
and L/K H is a Galois extension. Observe, in passing, that since we can take H = hgi, whereg ∈ G is arbitrary, this implies that L/k is Galois. We now prove that Gal(L/K H ) is abelian.
Let h̃, g̃ be two elements of Gal(L/K H ); denote by h and g their restriction to K. We haveg−ng

γ (g̃−ng )(h̃−nh ) = (ξh αh )g−ng = αh

= αgh−nh = (ξg αg )h−nh = γ (h̃−nh )(g̃−ng )and therefore γ g̃h̃ = γ h̃g̃ . Thus Gal(L/K H ) is abelian as desired. Since this is true for any
abelian subgroup H of G, we get by Lemma 4.4 that L is a strong central extension of K/k.
This concludes the proof.
A BRUMER-STARK CONJECTURE FOR NON-ABELIAN GALOIS EXTENSIONS

15

For a fractional ideal A of K, we say that BSGal (K/k, S; A) is satisfied if the Integrality
Conjecture holds for the extension K/k and the set of places S, and the ideal A verifies theequivalent properties of Theorem 5.3. Conjecture BSGal (K/k, S) is thus equivalent to the collection of conjectures BSGal (K/k, S; A) where A ranges through the fractional ideals of K.
6. Some properties of the Galois Brumer-Stark conjecture
In this section, we look at some properties satisfied by the Galois Brumer-Stark conjecture
and, in particular, the generalizations of the properties of the abelian Brumer-Stark conjecturestated in Section 2.
Proposition 6.1. The set of fractional ideals A of K that satisfy BSGal (K/k, S; A) is a subgroupof the group of ideals of K, stable under the action of G and that contains the principal idealsof K.
Proof. We first prove that this set is a group. Let A and B be two fractional ideals of K suchthat BSGal (K/k, S; A) and BSGal (K/k, S; B) hold. Let α and β be anti-units satisfying part(i) of Theorem 5.3 for the ideals A and B respectively. Then αβ is an anti-unit such thatαβOK = (AB)dG wK θK/k,S . Furthermore, since K((αβ)1/wK ) ⊂ K(α1/wK , β 1/wK ), it is a strong
central extension of K/k by Lemma 4.7 and therefore BSGal (K/k, S; AB) is satisfied. Thus theset of ideals A such that BSGal (K/k, S; A) holds is a subgroup of the group of fractional idealsof K.
Let σ be an element of G. We now prove that BSGal (K/k, S; Aσ ) is satisfied assuming
BSGal (K/k, S; A) holds. Since θK/k,S is in the center of C[G], ασ is a generator of
AdG wK θK/k,S

σ

= Aσ

dG wK θK/k,S.

Furthermore, ασ is clearly an anti-unit. Let γ := α1/wK and δ := (ασ )1/wK . Denote by σ̃ a liftof σ to L := K(γ). Then there exists ξ ∈ µK such that δ = ξγ σ̃ . Since L/k is Galois, we getthat L0 := K(δ) ⊂ L. This proves that L0 is a strong central extension of K/k by Lemma 4.7
and thus concludes the proof that BSGal (K/k, S; Aσ ) is satisfied.
Finally, we prove that BSGal (K/k, S; A) is satisfied if A is a principal ideal, say A = ηOK .
For that, we use the equivalent formulation (iv) of Theorem 5.3. Let H be an abelian subgroupof G. For h ∈ H, let nh ∈ N be such that ξ h = ξ nh for all ξ ∈ µK with the convention thatn1 = wK + 1. Then the family ah := h − nh , for h ∈ H, generates AnnZ[H] (µK ). For h ∈ H, wedefine αh := η dG ah θK/k,S . Note that dG ah θK/k,S ∈ Z[G] by the Integrality Conjecture. For allh ∈ H, we have (ηOK )dG ah θK/k,S = αh OK by construction. Furthermore, let w be an infinite(complex) place of K. Denote by τw ∈ G the complex conjugation at w. By Corollary 3.5, wehave that (1 + τw )θK/k,S = 0 and thus αh1+τw = 1 for all complex places w of K. Therefore αhais an anti-unit for all h ∈ H. It remains to prove that αh g = αgah for all g, h ∈ H. But this is adirect consequence of the fact that (h − nh )(g − ng ) = (g − ng )(h − nh ) since H is abelian. Thisconcludes the proof.
Corollary 6.2. Assume that K is principal. Then BSGal (K/k, S) is satisfied.

Using the decomposition of the Brumer-Stickelberger element given by (12), we can prove thefollowing result that relates BS(K ab /k, S) and BSGal (K/k, S).
Theorem 6.3. Assume that the Integrality Conjecture is satisfied for the extension K/k andthe set of places S and that BS(K ab /k, S) holds. Then BSGal (K/k, S) is satisfied if, for any
d w θ

(>1)

fractional ideal A of K, the ideal A G K K/k,S is principal, and admits a generator β ∈ K ◦ suchthat K(β 1/wK ) is a strong central extension of K/k.

16

GAELLE DEJOU AND XAVIER-FRANÇOIS ROBLOT

Proof. Let A be a fractional ideal of K. Set a := NK/K ab (A). An direct computation shows that
AdG wK νab(θK ab /k,S )= a(dG /sG )wK θK ab /k,S OK .

By hypothesis, there exists α0 , an anti-unit in K ab , such thata(dG /sG )wK θK ab /k,S = α0 OK ab
1/wKand K ab (α0)/k is abelian. Let α := α0 β. Then α is an anti-unit of K and by (12), we have
αOK = AdG wK θK/k,S .

It remains to prove that K(α1/wK ) is a strong central extension of K/k. It is a sub-extension
1/w
of K(α0 K , β 1/wK )/K. But K(β 1/wK ) is a strong central extension of K/k by hypothesis and
1/wK
K(α0
) is a strong central extension of K/k by Lemma 4.6. Thus, K(α1/wK ) is a strong
central extension of K/k by Lemma 4.7 and the result is proved.
For χ ∈ Ĝ, recall that K χ denote the subfield of K fixed by the kernel of χ.
Corollary 6.4. Assume that BS(K ab /k, S) is satisfied and that, for all χ ∈ Ĝ such that χ(1) >1, K χ is not a CM extension. Then BSGal (K/k, S) holds.
(>1)

Proof. Indeed, in that case, θK/k,S = 0 by Proposition 3.3.



As an application of Corollary 6.4, we can prove that BS(K ab /k, S) implies BSGal (K/k, S)
for some isomorphic types of group Gal(K/k).
Proposition 6.5. Let G be a finite group such that, for all irreducible characters χ of G with
χ(1) > 1, the center of G/ ker(χ) does not contain an element of order 2. Then BSGal (K/k, S)
holds for any Galois extension K/k of number fields with Gal(K/k) ' G and such that BS(K ab /k, S)
is satisfied.
Proof. The result is trivial if k is not totally real or if K is not totally complex. Assume therefore
that k is totally real and K is totally complex. Let χ be an irreducible character of Gal(K/k)
with χ(1) > 1. It is enough to prove that K χ is not a CM extension. Assume it is a CM
extension. Then the complex conjugation is an element of order 2 in its Galois group, which
is isomorphic to G/ ker(χ), and it commutes with all the elements of the group since it is theunique complex conjugation. This is a contradiction, thus K χ is not CM and the result follows
from Corollary 6.4.

We give several applications of this result.
Proposition 6.6. Assume that Gal(K/k) is a non-abelian simple group. Then BSGal (K/k, S)
holds.
Proof. The commutator subgroup [G, G] is normal in G, thus it is equal to G and BS(K ab /k, S)
trivially holds since K ab = k. Now, let χ be an irreducible character of G with χ(1) > 1. Then χ
is faithful because ker(χ) is a normal subgroup of G. But the center of G is trivial and therefore
BSGal (K/k, S) holds by Proposition 6.5.
Proposition 6.7. Assume that Gal(K/k) is isomorphic to the dihedral group D2n of order 2nwhere n ≥ 3 is odd. Then BSGal (K/k, S) holds.
Proof. The group D2n is the group generated by two elements a and b with the following relations:a2 = bn = 1 and aba = b−1 . When n is odd, its maximal abelian quotient is the cyclic groupof order 2, thus K ab /k is quadratic and BS(K ab /k, S) holds. Furthermore, by [20, §I.5.3], its
non-linear irreducible representations are the representations ρh , for 1 ≤ h ≤ (n − 1)/2, defined
by
 kh



ω
0
0
ω −hk
k
ρh (bk ) =
and
ρ
(ab
)
=
h
0
ω −kh
ω hk
0

A BRUMER-STARK CONJECTURE FOR NON-ABELIAN GALOIS EXTENSIONS

17

for k ∈ Z, where ω is a fixed primitive n-th root of unity. In particular, the kernel of ρh is a
subgroup of hbi, distinct from hbi. It follows that D2n / ker(ρh ) is isomorphic to D2m for some
integer m ≥ 3 dividing n. But the center of D2m , for m ≥ 3 odd, is trivial. The result follows
from Proposition 6.5.
Proposition 6.8. Assume that Gal(K/k) is isomorphic to the symmetric group Sm on m letterswith m ≥ 2. Then BSGal (K/k, S) holds.
Proof. The result is clear if m = 2. Assume m ≥ 3. We use Proposition 6.5 again. The
commutator subgroup of Sm is the alternating group Am . Therefore K ab is a quadratic extensionof k and BS(K ab /k, S) holds. Assume first that m ≥ 5. Then Am is the only non-trivial normal
subgroup of G and therefore the non-trivial irreducible representations of Sm are either faithfulor have Am as kernel. In particular, the non-linear irreducible representations of Sm must be
faithful and the result follows since the center of Sm is trivial. For m = 3 and m = 4, theresult follows from direct inspection. Indeed, for m = 3, the unique non-linear irreducible
representation is faithful and the center of S3 is trivial. For m = 4, there is only one non-linear
irreducible representation ρ that is not faithful. Its kernel is isomorphic to the Klein group
Z/2Z × Z/2Z and the quotient is S4 / ker(ρ) ' S3 and thus has again trivial center.
Remark. Using Proposition 6.5 and similar techniques, one can prove that BSGal (K/k, S)
follows from BS(K ab /k, S) for some other families of groups, eg. the group of affine bijective
maps of a finite field Fq which is isomorphic to Fq o F×q .
We now turn to the question of the change of extension for the Galois Brumer-Stark conjecture.
We will prove that it is satisfied in many cases up to a factor.
Proposition 6.9. Let K 0 /k be a Galois sub-extension of K/k with G0 := Gal(K 0 /k). Denote
gGal (K 0 /k, S) the Galois Brumer-Stark conjecture for the extension K 0 /k and the set ofby BS
places S with the factor dG0 replaced by dG including in the statement of the Integrality Conjecture. Assume that wK is relatively prime with the degree of the extension K/K 0 K ab . Then
gGal (K 0 /k, S).
BSGal (K/k, S) implies BS
Remark. If G is abelian then K ab = K, thus K = K 0 K ab and the condition of the propositionis always satisfied. Furthermore, we have dG = dG0 = 1 and we recover the fact that BS(K/k, S)
implies BS(K 0 /k, S).
Remark. We prove actually a slighter stronger statement: if BSGal (K/k, S) holds then, for all
fractional ideal A0 of K 0 , there exists an anti-unit α ∈ K 0 such that
A0 dG wK 0 θK 0 /k,S = (α).
The extra hypothesis that wK is relatively prime with the degree of K/K 0 K ab is only used toprove the fact that K 0 (α1/wK 0 ) is a strong central extension of K 0 /k.
In order to see that the statement of Proposition 6.9 makes sense, we have the following
lemma.
Lemma 6.10. Let A be a finite group and let B be a quotient group of A. Then dB divides dA .
Proof. It is enough to prove that sB divides sA and mB divides mA . Let π : A → B be thecanonical surjection and denote by D its kernel. It is clear that sB divides sA since π([A, A]) =[B, B]. We now prove that mB divides mA . Let b ∈ B and let a ∈ A be such that π(a) = b.
Denote by Z the centralizer of a in A and by Z0 the centralizer of b in B. Note that Z := π −1 (Z0 )is a subgroup of A containing Z and that
|Z0 | =

(Z : Z) |Z|
|Z|
=
.
|D|
|D|

18

GAELLE DEJOU AND XAVIER-FRANÇOIS ROBLOT

Denote by C and C0 the conjugacy classes of a and b in A and B respectively. We find that
|A|
|A|(Z : Z)
|B|
|C| =
=
= (Z : Z)= (Z : Z) |C0 |.
|Z||D| |Z0 ||Z0 |Thus |C0 | divides |C| and therefore mB divides mA .



Proof of Proposition 6.9. To start, observe that, thanks to Theorem 3.1, the Integrality Conjecture for the extension K/k and the set of places S implies the Integrality Conjecture forthe extension K 0 /k and the set of places S with dG0 replaced by dG . We first prove theresult when K = K 0 K ab . In this situation, we shall actually prove that BSGal (K/k, S) implies BSGal (K 0 /k, S). Indeed, we have dG = dG0 by Lemma 4.5 since one can see, thanks to
Lemma 4.6, that K is a strong central extension of K 0 /k. Let A0 be a fractional ideal of K 0 .
By our assumption that BSGal (K/k, S) holds, taking A := A0 OK , we see that there exists ananti-unit α in K such thatαOK = (AOK )dG wK θK/k,S = A0 dG wK θK/k,S OK = A0 dG0 wK θK 0 /k,S OK .

(16)

1/wK

Furthermore, L := K(γ) is a strong central extension of K/k where γ := α
. Clearly, wehave
γOL = (A0 OL ) dG0 θK 0 /k,S .
We now use Theorem 5.3(ii) with the extension L/K 0 and the element γ. The only assertion
that needs to be checked is the fact that L is a strong central extension of K 0 /k. By Lemma 4.6,
this is equivalent to the fact that L = K 0 Lab where Lab is the maximal sub-extension of L/k thatis abelian over k. Clearly, K ab ⊂ Lab thus we have K 0 K ab = K ⊂ K 0 Lab . Since KLab = L, it
follows that L ⊂ K 0 Lab , thus K 0 Lab = L and L is a strong central extension of K 0 /k. Therefore
BSGal (K 0 /k, S; A0 ) holds for all fractional ideals A0 of K 0 and BSGal (K 0 /k, S) is satisfied.
We now prove the general case. By the first part, replacing K 0 by K 0 K ab if necessary, we
can assume that K 0 contains K ab and therefore, by hypothesis, wK is relatively prime with thedegree of K/K 0 . Let A0 be a fractional ideal of K 0 . Reasoning as above, we see that there exists
α ∈ K ◦ such thatαOK = A0 dG wK θK 0 /k,S OK
and L is a strong central extension of K/k where L := K(γ) and γ := α1/wK . Denote by Γ
the Galois group of L/k. For σ ∈ Γ, Lσ = L is a Kummer extension of K σ = K generated byγ σ . Thus there exist an integer nσ relatively prime to wK with 1 ≤ nσ ≤ d := [L : K], and anelement κσ ∈ K × such that γ σ = κσ γ nσ . Observe that, for δ ∈ ∆ := Gal(L/K), we have nδ = 1
and κδ is a root of unity in K. Furthermore, using the fact that σ and δ commute, we getγ δσ = (κσ γ nσ )δ = κσ κnδ σ γ nσ = γ σδ = (κδ γ)σ = κσδ κσ γ nσ
and thus κσδ = κnδ σ . As δ runs through the elements of ∆, κδ runs through the roots of unity oforder d, thus σ − nσ annihilates the group µd of d-th roots of unity. Assume now that σ lies in
A := Gal(L/K 0 ). Therefore, σ fixes the group of roots of unity µK 0 = µK and nσ = 1. Using
the fact that θK 0 /k,S is in the center of C[G], we get
ασ OK = (A0σ ) dG wK θK 0 /k,S OK = A0 dG wK θK 0 /k,S OK = αOK .
Since α is an anti-unit, there exists a root of unity ξσ in K × such that ασ = ξσ α. Combining
Kwith the above expression for γ σ , we find that κw= ξσ . Thus κσ is a root of unity in K andσ0gGal (K/k, S) holds
ξσ = 1. It follows that α ∈ K . Again we use Theorem 5.3(ii) to prove that BS00
for A . It remains to prove that there is a strong central extension of K /k containing γ. Let
L0 := K 0 Lab where Lab is the maximal sub-extension of L/k that is abelian over k. The Galois
group of the extension L/L0 is [Γ, Γ] ∩ A. Hence, by Lemma 4.6, L0 is the maximal sub-extensionof L/k that is strong central for K 0 /k. We now prove that γ ∈ L0 . Denote by π : Γ → G
the canonical surjection induced by the restriction to K. Its kernel is ∆, thus it restricts to an
isomorphism between [Γ, Γ] and [G, G] (see also Lemma 4.2). We have γ ∈ L0 if and only if

A BRUMER-STARK CONJECTURE FOR NON-ABELIAN GALOIS EXTENSIONS

19

π(Gal(L/L0 )) ⊂ π(Gal(L/K 0 (γ))), that is π([Γ, Γ] ∩ A) ⊂ Gal(K/N ) where N = K ∩ K 0 (γ). But
N/K 0 is a sub-extension of K/K 0 of degree dividing wK and therefore N = K 0 and the above
gGal (K 0 /k, S) holds and this concludes the proof.
condition is always satisfied. Hence BSWe conclude this section with a proof of when the validity of the conjecture is preserved
when one enlarges the set S. For χ ∈ Ĝ, denote by ρχ a fixed irreducible representation of G ofcharacter χ.
Lemma 6.11. Let P1 , . . . , Pt be prime ideals of K. We havet X
Ydet(1 − ρχ (σPi ))eχ̄ ∈i=1 χ∈Ĝ

1
Z(Z[G]).
|G|

Proof. Let α ∈ Gal(Q̄/Q). One can see that the above expression is invariant under the action of
α using the fact that the map χ 7→ χα is a bijection on Ĝ. Therefore, it lies in Q[G] ∩ Z(C[G]) =Z(Q[G]). Now, by the orthogonality of characters, we havet X
Ydet(1 − ρχ (σPi ))eχ̄ =i=1 χ∈Ĝt
XY

det(1 − ρχ (σPi ))eχ̄ .

χ∈Ĝ i=1

Finally, for all χ ∈ Ĝ, |G| eχ and det(1 − ρχ (σPi )), for i = 1, . . . , t, are algebraic integers andthus the result follows.
Proposition 6.12. Let p1 , . . . , pt be distinct prime ideals of k not belonging to S. Define
ω :=

t X
Y

det(1 − ρχ (σPi ))eχ̄ ∈i=1 χ∈Ĝ

1
Z(Z[G])|G|

where Pi is a prime ideal of K above pi for i = 1, . . . , t. Let d ≥ 1 be the smallest integer
such that dω ∈ Z[G]. Assume that BSGal (K/k, S) holds and let S 0 := S ∪ {p1 , . . . , pt }. Then
BSGal (K/k, S 0 ; A) is satisfied for any fractional ideal A of K whose class in ClK has orderrelatively prime to d.
Proof. Assume that BSGal (K/k, S) holds. Let A be an ideal of K whose class in ClK has orderrelatively prime to d. Thus there exists an ideal A0 of K and η ∈ K × such that A = ηAd0 . Let
α0 be an anti-unit of K such thatdG wK θK/k,S

α0 OK = A0
1/wK

and the extension K(α0) is a strong central extension of K/k. Define
α := α0dω η dG wK θK/k,S0 .

One checks directly thatαOK = AdG wK θK/k,S0 .
From the proof of Proposition 6.1, we see that δ := η dG wK θK/k,S0 is an anti-unit and that theextension K(δ 1/wK ) is a strong central extension of K/k. Therefore, α is an anti-unit and the
1/w
extension K(α1/wK ) ⊂ K(α0 K , δ 1/wK ) is a strong central extension of K/k by Lemma 4.7.
0Thus BSGal (K/k, S ; A) holds.
20

GAELLE DEJOU AND XAVIER-FRANÇOIS ROBLOT

7. Groups with a normal abelian subgroup of prime index
In this section, we consider the case where the Galois group G contains an abelian normal
subgroup H of prime index. We prove in this setting that the Integrality Conjecture is satisfied
and that the Galois Brumer-Stark conjecture follows from the abelian Brumer-Stark conjecture
for suitable abelian sub-extensions. The methods used in this section are similar in spirit to theones used by Nomura [17] to prove that the weak non-abelian Brumer-Stark conjecture of Nickel
for monomial groups follows from the abelian Brumer-Stark conjecture (and similar results forthe weak Brumer conjecture of Nickel). However, a big difference is that, in Nomura’s paper,
he can work (rational) character by character whereas this does not seem to possible with theGalois Brumer-Stark conjecture.
We assume from now on that the group G is not abelian and contains a normal abeliansubgroup H of index `, a prime number. Let m denote the order of H, thus |G| = m`. We
have [G, G] ⊂ H since G/H is cyclic of order ` and therefore K H is a subfield of K ab . Let SH
denote the set of places of K H that are above the places in S. The set SH contains the infinite
places of K H and the finite places that ramify in K/K H . The first result of this section gives adecomposition of the Brumer-Stickelberger element in this situation.
Theorem 7.1. With the notations and setting as above, we have


1
(>1)
N[G,G] θK/K H ,SHθK/k,S = 1 −sG
Pwhere N[G,G] :=c ∈ Z[G].
c∈[G,G]

Proof. Since the group G contains an abelian normal subgroup of index `, the dimensions of theirreducible characters of G divide `. Hence any character in Ĝ with χ(1) > 1 is of dimension `.
Denote by Ĝ` the set of irreducible characters of G of dimension `.
Lemma 7.2. Let Ĥ` be the set of irreducible characters of H whose kernel does not contain[G, G]. For χ ∈ Ĝ` , define Ĥ` (χ) to be the subset of those characters in Ĥ` whose induction to
G is χ. Then, we have[Ĥ` =Ĥ` (χ) (disjoint union)
χ∈Ĝ`and each Ĥ` (χ) has exactly ` elements. Furthermore, for all χ ∈ Ĝ` and g ∈ G, we have0
if g 6∈ H,
P
χ(g) =ϕ(g)ifg ∈ H.
ϕ∈Ĥ` (χ)

Proof of the lemma. Let ϕ be a character in Ĥ` and let χ := IndG
H (ϕ). Then χ is of dimension `.
Assume χ is not irreducible. Then it is a sum of 1-dimensional characters and all these characters
are trivial on [G, G]. By Frobenius reciprocity, the restriction of any of these characters to H isequal to ϕ. Thus ϕ is trivial on [G, G], a contradiction. Therefore χ is irreducible and lies inĜ` . The restriction of χ to H is the sum of ` characters of H, and using once again Frobenius
reciprocity, we see that these characters are exactly the characters of H whose induction to G
is χ and that they are all distinct. Therefore, we have proved that, if χ ∈ Ĝ` is the induction ofsome character in Ĥ` , then the set Ĥ` (χ) contains ` distinct characters, say ϕ1 , . . . , ϕ` , such that
χ|H = ϕ1 + · · · + ϕ` . Furthermore, if χ0 is a character of Ĝ` induced from a character in Ĥ` with
χ 6= χ0 , the sets Ĥ` (χ) and Ĥ` (χ0 ) are clearly disjoint. This implies that Ĥ` is the disjoint unionof the Ĥ` (χ)’s for χ ∈ Ĝ` . We now prove that Ĥ` (χ) is non-empty for all χ ∈ Ĝ` . This amounts
to proving that any character in Ĝ` is the induction of some character in Ĥ` . Characters of H
whose kernel contains [G, G] are in bijection with characters of H/[G, G]. Denote by t the index

A BRUMER-STARK CONJECTURE FOR NON-ABELIAN GALOIS EXTENSIONS

21

of [G, G] in H. The number of characters in Ĥ` is therefore m − t and, by the above discussion,
the inductions of these characters yield (m − t)/` characters in Ĝ` . On the other hand, we havethe formula m` = t` + a`2 , where a is the number of characters in Ĝ` , since the sum of thesquare of the dimensions of the irreducible characters of G is equal to |G| and using the fact that
(G : [G, G]) = t`. Therefore, we have a = (m − t)/` and all the characters of Ĝ` are inductionsof characters in Ĥ` . To conclude, it remains to prove the expression for χ ∈ Ĝ` . Let ϕ ∈ Ĥ` (χ).
For all g ∈ G, we have
1 X
χ(g) =
ϕ(rgr−1 ).
mr∈Grgr −1 ∈H

Since the group H is normal in G, rgr−1 ∈ H if and only
P g ∈ H. Thus χ(g) = 0 if g 6∈ H. If
g ∈ H, the expression follows from the fact that χ|H = ϕ∈Ĥ` (χ) ϕ.
As a consequence of Lemma 7.2, we have, for χ ∈ Ĝ` , that
X
eϕ
eχ =
ϕ∈Ĥ` (χ)

where eϕ is the idempotent of C[H] associated to the character ϕ. We now compute
X
X
X
(>1)
θK/k,s =LK/k,S (0, χ) eχ̄ =LK/k,S (0, χ)
eϕ̄
χ∈Ĝ`

=

X

χ∈Ĝ`

X

ϕ∈Ĥ` (χ)

LK/k,S (0, IndG
H ϕ) eϕ̄ =χ∈Ĝ` ϕ∈Ĥ` (χ)

=

X

X

X

LK/K H ,SH (0, ϕ) eϕ̄χ∈Ĝ` ϕ∈Ĥ` (χ)

LK/K H ,SH (0, ϕ) eϕ̄ =

ϕ∈Ĥ`

X
ϕ∈Ĥ

X

= θK/K H ,SH −X

LK/K H ,SH (0, ϕ) eϕ̄ −LK/K H ,SH (0, ϕ) eϕ̄ϕ∈Ĥ\Ĥ`LK/K H ,SH (0, ϕ) eϕ̄ .

ϕ∈Ĥ
[G,G]⊂Ker ϕ

Let ϕ be a character of H whose kernel contains [G, G]. Let ϕ̃ be the only character of J :=H/[G, G] such that the inflation of ϕ̃ to H is equal to ϕ. From the properties of Artin L-functions,ab(eϕ̃ )
we have LK/K H ,SH (0, ϕ) = LK ab /K H ,SH (0, ϕ̃) and a direct calculation shows that eϕ = νHab
: C[J] → C[H] is the map defined forwhere eϕ̃ is the idempotent of C[Gab ] associated to ϕ̃, νHg̃ ∈ J by
X
1
ab
g,
νH
(g̃) :=sG ab
πH (g)=g̃aband extended by linearity to C[J], and πH: H → J is the canonical surjection.

Therefore, we

have
X
ϕ∈Ĥ
[G,G]⊂Ker ϕ

LK/K H ,SH (0, ϕ) eϕ̄ =Xab
LK ab /K H ,SH (0, ϕ̃)νH(eϕ̃¯ )

ϕ̃∈Jˆ
ab
= νH

X

LK ab /K H ,SH (0, ϕ̃) eϕ̃¯



ϕ̃∈Jˆ
ab
= νH
(θK ab /K H ,SH ).
abab
ab
Now, for α ∈ C[H] and β ∈ C[J], one checks readily that ανH(β) = νH(α̃β) where α̃ := πH(α).
Thus, we find that1abab
N[G,G] θK/K H ,SH .
νH(θK ab /K H ,SH ) = θK/K H ,SH νH
(1) =
sG

22

GAELLE DEJOU AND XAVIER-FRANÇOIS ROBLOT

The result then follows by substituting in the above expression.

The main advantage of the decomposition given by Theorem 7.1 is the fact that the extensions
involved are all abelian. Therefore, in our study of BSGal (K/k, S) in that setting, we can reduceto the abelian case. In particular, it follows from (3) that(>1)

dG wK θK/k =dG(sG − N[G,G] )wK θK/K H ,SH ∈ Z[G].
sG

However, for p a prime ideal of k, unramified in K/k, and P a prime ideal of K above p, we do
not have necessarily that (σP − N (p))θK/K H ,SH lies in Z[G], since σP might not belong to H.
However, we can still prove the Integrality Conjecture is satisfied.
Proposition 7.3. We have(sG − N[G,G] )θK/K H ,SH ∈ Z[H].
In particular, the Integrality Conjecture holds for the extension K/k and the set S.
Proof. First note that, by Theorem 7.1 and the discussion after (12), the first assertion implies
the Integrality Conjecture in this case. Now, we haveX(sG − N[G,G] )θK/K H ,SH =(1 − c)θK/K H ,SH .
c∈[G,G]

But 1 − c ∈ AnnZ[H] (µK ) for all c ∈ [G, G] and thus, by the properties of the abelian BrumerStickelberger element, all the terms in the sum are in Z[H]. The first assertion and the proof ofthe proposition follow.

We now prove that, in this situation, the Galois Brumer-Stark conjecture is a consequence of
the abelian Brumer-Stark conjecture.
Theorem 7.4. Let K/k be a Galois extension of number fields whose Galois group G contains
an abelian normal subgroup H of prime index. Assume that BS(K ab /k, S) and BS(K/K H , SH )hold where SH denotes the set of places of K H above the places in S. Then BSGal (K/k, S) is
satisfied.
Proof. We will prove the result using Theorem 6.3. Let A be a fractional ideal of K. By our
hypothesis, there exists α1 ∈ K ◦ such thatw θ

A K K/K H ,SH = α1 OK
1/w

and the extension K(γ1 )/K H is abelian where γ1 := α1 K . Define Y
dG /sG(dG /sG )(sG −N[G,G] )β := α1=α11−c.
c∈[G,G]

By construction, β is an anti-unit of K and satisfies
d w θ

(>1)

βOK = A G K K/k,S .
It remains to prove that K(β 1/wK ) is a strong central extension of K/k. We will actually provethat K(β 1/wK ) = K. Let L1 be the Galois closure of K(γ1 )/k. Denote by Γ1 the Galois group
Gal(L1 /k). Let c1 ∈ [Γ1 , Γ1 ]. Note that c1 ∈ Gal(L1 /K H ) since K H /k is abelian. Thus,
by Theorem 2.1, there exists a prime ideal P1 of L1 , relatively prime to the order of µL1 ,whose Frobenius automorphism in Γ1 is equal to c1 , and an anti-unit α1,pH ∈ K ◦ such that
α1,pH ≡ 1 (mod∗ pH OK ) and(σpH −N (pH ))θK/K H ,S

α1,pH OK = A

HA BRUMER-STARK CONJECTURE FOR NON-ABELIAN GALOIS EXTENSIONS

23

where pH is the prime ideal of K H below P1 and σpH is the Frobenius automorphism of pH in
H. We have(c −1)θK/K H ,S

γ1c1 −1 OL1 = A 1
=A

H

OL1

(σpH −N (pH ))θK/K H ,S(N (pH )−1)/wK= α1,pH α1H(N (pH )−1)θK/K H ,S

A

H

OL1

OL1 .

Observe that γ1 , α1,pH and α1 are anti-units, thus there exists a root of unity ξ ∈ µL1 such that
(N (p )−1)/wK
ξγ1c1 −1 = α1,pH α1 H
. Furthermore, since c1 acts trivially on the group µK , wK divides(N (p )−1)/wKN (pH ) − 1 and α1,pH α1 H
belongs to K ◦ . Raising to the power wK , we get
σp −1

ξ wK α1 H
and therefore

N (pH )−1wK
= α1,pα1H

N (pH )−σpHξ wK ≡ α1≡ 1 (mod∗ pH OK ).

Therefore we find that ξ wK = 1, hence ξ ∈ µK and γ1c1 −1 ∈ K.
Now, for all c ∈ [G, G], fix an element c1 in [Γ1 , Γ1 ] whose restriction to K is equal to c, anddefine Y
dG /sGγ11−c1.
δ :=c∈[G,G]

By the above computation, we see that δ ∈ K and, by construction, that δ wK = β. ThereforeK(β 1/wK ) = K and the result follows.

Corollary 7.5. Assume that the order of H is odd. Then BS(K ab /k, S) implies BSGal (K/k, S).
Proof. Indeed, since the degree of K/K H is odd, BS(K/K H , S) is trivially true as we cannot
have both K totally complex and K H totally real.

We proved already that BSGal (K/k, S) holds when G is isomorphic to the dihedral group oforder 2m with m odd (see Proposition 6.7). We prove a similar statement when m is even.
Proposition 7.6. Assume that G is isomorphic to the dihedral group D2m of order 2m, withm even, and that BS(K/K H , SH ) holds where H is the unique cyclic subgroup of G of order m.
Then BSGal (K/k, S) is satisfied.
Proof. The cyclic subgroup H is normal and of index 2. Therefore, by Theorem 7.4 and the
hypothesis, it is enough to prove that BS(K ab /k, S) is satisfied. The maximal abelian quotientof D2m is isomorphic to Z/2Z × Z/2Z, thus BS(K ab /k, S) holds by the results of [19].

We conclude this section with another application of Theorem 7.4.
Proposition 7.7. Assume that G is non-abelian of order 8. Then BSGal (K/k, S) is satisfied.
Proof. The non-abelian groups of order 8 are, up to isomorphism, the dihedral group D8 of order
8 and the quaternion group Q8 . We start with Gal(K/k) ' D8 . Thanks to Proposition 7.6, it
remains to prove that BS(K/K H , S) holds where H is the cylic subgroup of order 4 containedin Gal(K/k). But K/K H is a degree 4 abelian extension contained in the degree 8 non-abelian
Galois extension K/k, so BS(K/K H , S) is true by results of Tate (see Theorem 2.2).
Assume now that Gal(K/k) ' Q8 . A presentation for the group Q8 is given by the three
generators a, b, c with the relations a2 = b2 = c2 = abc. It is customary to denote the element
abc by −1 since it is of order 2 and lies in the center of Q8 . (In fact, it generates the centerof Q8 .) The subgroup H generated by a is cyclic of order 4. Thus, we can apply Theorem 7.4
and BSGal (K/k, S) follows from BS(K ab /k, S) and BS(K/K H , SH ). Now, the commutator
subgroup of Q8 is the subgroup {±1} and Gal(K ab /k) ' Z/2Z × Z/2Z. Thus, BS(K ab /k, S)
is satisfied by the results of [19]. Finally, as above, the extension K/K H is a degree 4 abelian

24

GAELLE DEJOU AND XAVIER-FRANÇOIS ROBLOT

extension contained in the degree 8 non-abelian Galois extension K/k so BS(K/K H , SH ) follows
again from Theorem 2.2.
Appendix. The non-abelian Brumer-Stark conjecture of Nickel
In this appendix, we state the weak non-abelian Brumer-Stark conjecture of Nickel [14] andcompare it with our conjecture. Note that Nickel states also a strong version of the non-abelian
Brumer-Stark conjecture and similar generalizations of the Brumer conjecture. We changeslightly the notations used by Nickel to match the notations used in the previous sections.
Let K/k be a Galois CM-extension with group G. Fix a finite set S of places of k such that
S contains the infinite places of k and the finite places of k that ramify in K/k. Let Hyp(S) be
the set of finite set T of places of k such that
• S and T are disjoint,• the group EK (S, T ) is torsion-free.
Here, EK (S, T ) denotes the group of (S, T )-units of L, that is the group of elements u ∈ K ×such that vP (u) = 0 for all prime ideals P of K such that (P ∩ k) 6∈ S and u ≡ 1 (mod∗ Q) forall prime ideals Q of K such that (Q ∩ k) ∈ T . For T ∈ Hyp(S), defineY−1δT := nr
1 − σPN (p)p∈T

where P is a fixed prime ideal of K above p and nr : Q[G] → Z(Q[G]) is the reduced norm (see[18, §9]). Let Λ0 denote a fixed maximal order of Q[G] containing Z[G]. Denote by F(G) := {x ∈Z(Λ0 ) : xΛ0 ⊂ Z[G]} the central conductor of Λ0 over Z[G].
Conjecture (The weak non-abelian Brumer-Stark conjecture of Nickel).
Let wK := nr(wK ). Then wK θK/k,S ∈ Z(Λ0 ). Furthermore, for any fractional ideal A of K andfor each x ∈ F(G), there exists an anti-unit αx ∈ K ◦ such that
AxwK θK/k,S = αx OK
and, for any set of places T ∈ Hyp(S ∪ Sαx ), there exists αx,T ∈ EK (Sαx , T ) such that, for allz ∈ F(G)zwK
αxzδT = αx,T
where Sαx is the set of prime ideals p of k such that vp (NK/k (αx )) 6= 0.
Remark. The strong version of the conjecture is similar with the modules Z(Λ0 ) and F(G)replaced respectively by the modules I(G) and H(G) where I(G) is the module generated by
the reduced norms of matrices with coefficients in Z[G]; the definition of H(G) is more intricate,see [14, p. 2582]
The results of Greither-Popescu [10], mentioned at the end of Section 2, have been generalizedby Nickel in [16] where he proves that the p-part of his non-abelian Brumer conjecture andnon-abelian Brumer-Stark conjecture hold if S contains all the prime ideals above p and someappropriate µ-invariant vanishes. As mentioned in the previous section, Nomura [17] proves thatthe weak non-abelian Brumer-Stark conjecture of Nickel is implied by the abelian Brumer-Stark
conjecture when the Galois group of K/k is monomial (and also obtain additional results on thestrong version and local versions of the conjecture).
We now briefly compare our conjecture with the weak non-abelian Brumer-Stark conjecture of
Nickel. Both conjectures have two parts: an integrality statement for the Brumer-Stickelberger
element and an annihilation of the class group statement that also predicts special properties
for the generators obtained. We first prove that the Galois Brumer-Stark conjecture implies theannihilation statement in the weak non-abelian Brumer-Stark conjecture of Nickel up to a factor
dG , including the existence of generators that satisfy almost all the required properties.

A BRUMER-STARK CONJECTURE FOR NON-ABELIAN GALOIS EXTENSIONS

25

Proposition 7.8. Assume that BSGal (K/k, S) holds. Let A be a fractional ideal of K. Then,for all x ∈ F(G), there exists an anti-unit αx ∈ K ◦ such that
AdG xwK θK/k,S = αx OK
wK
and, for any set of places T ∈ Hyp(S ∪ Sαx ), there exists αx,T ∈ K × with αx,T
∈ EK (Sαx , T )such that, for all z ∈ F(G)zwK
αxzδT = αx,T.

Proof. Let α ∈ K ◦ be such that AdG wK θK/k,S = αOK and K(α1/wK ) is a strong central extension−1of K/k. For x ∈ F(G), let αx := αxmK where mK = wK
wK . Note that by (18) below, we have0mK ∈ Z(Λ ) and thus xmK ∈ Z[G]. We compute
AdG xwK θK/k,S = (AdG wK θK/k,S )xmK = αx OK .
Now, let T ∈ Hyp(S ∪ Sαx ). Let MT be the product of the prime ideals of K above the prime
ideals in T . Then, for any a ∈ F(G), the element aδT ∈ Z[G] kills (OK /MT )× . We give the proofof this result due to Nickel (personal communication). Let ` be a prime number. The module(OK /MT )× ⊗ Z` admits a quadratic presentation induced by the following exact sequencesZ` [DP ]/ Z` [DP ]/ (OK /P)× ⊗ Z`

/1

where p ranges through the prime ideals in T , P is a fixed prime ideal of K above p, DP is−1
the decomposition group of P in G, the first map is the multiplication by 1 − σP
N (p) and the×second map is induced by the action of DP on a fixed generator of (OK /P) . Thus, the Fitting
invariant of (OK /MT )× ⊗ Z` is generated by δT , see [14, p. 2580]. By Theorem 1.2, ibid., we
get that aδT annihilates (OK /MT )× ⊗ Z` . Since this is true for all primes ` and (OK /MT )× isa finite abelian group, the result follows. In particular, because EK (S, T ) is torsion-free, we getthat xδT lies in AnnZ[G] (µK ). Using Lemma 3.7, we can write
xδT =

t
X

λi (σPi − N (pi )),

i=1

where λi ’s are rational integers and the Pi ’s are prime ideals of K such that Theorem 5.3(iii)applies. Let γ := α1/wK , L := K(γ) and Γ := Gal(L/K). Definex̃T :=tXλi (σP̃i − N (pi )) ∈ Z[Γ]i=1where P̃i is a fixed prime ideal of L above Pi and σP̃i is the Frobenius automorphism of P̃iin Γ. We set αx,T := γ x̃T . From the proof of Theorem 5.3(iii), we see that γ σP̃i −N (pi ) is anelement of K × (it is equal to αPi with the notations of Theorem 5.3(iii)) and thus αx,T ∈ K × .
Furthermore, we have
αx,T OL = (AdG θK/k,S OL )x̃T = AdG xδT θK/k,S OL
and, using the fact that dG xδT θK/k,S ∈ Z[G] by the Integrality Conjecture, we get that αx,T OK =AdG xδT θK/k,S . Now, let N be a large enough integer such that N m−1
K δT ∈ Z[G]. Then, we find(compare with proof of [14, Lemma 2.12])
N m−1
N m−1 δNK δT
αx,TOK = AdG xmK θK/k,S= αx K T OK
wK
and αx,T is supported only by prime ideals above Sαx . Also, we have αx,T
= αxδT ≡ 1 (mod∗ MT )wK
and therefore αx,T ∈ EK (Sαx , T ). Finally, for z ∈ F(G), we computezwK= γ x̃T zwK = (γ wK )x̃T zmK = αxδT zmK = (αxmK )zδT = αxzδT .
αx,T



26

GAELLE DEJOU AND XAVIER-FRANÇOIS ROBLOT

It is an interesting question to ask if there is any result in the other direction: assuming
the non-abelian Brumer-Stark conjecture of Nickel, can we deduce some results on our Galois
Brumer-Stark conjecture? We were not able yet to obtain significant results on this question.
Similarly, it appears that the connections between the Integrality Conjecture and the integrality
statement in the conjectures of Nickel are quite thin and, indeed, the two statements appear
to be of quite different nature. To illustrate this point, we look at the proof of the integrality
statement of the weak non-abelian Brumer-Stark conjecture in the setting of the previous section.
(Of course, this case also follows from the results of [17].)
We start with some general results and facts. Let X be the set of irreducible Q-charactersof G or,Pequivalently, the set of orbits of Ĝ under the action of Gal(Q̄/Q). For X ∈ X, define
eX := φ∈X eφ ∈ Q[G]. Then, as a Q[G]-module, one has
M
eX Q[G].
Q[G] =X∈X

For each X ∈ X, fix a character φX ∈ X and set QX := Q(φX ) L
and nX := φX (1). Indeed
these do not depend on the choice of the character φX ∈ X. Then X∈X QX is isomorphic to
Z(Q[G]) by the mapXXσ(αX )X∈X 7→
αX
eφσX
(17)
X∈X σ∈Gal(QX /Q)
Lσwhere φX := σ ◦ φX . This map restricts to an isomorphism between X∈X OX and Z(Λ0 ) where

OX denotes the ring of integers of QX . A direct computation shows that
X
nX
wK =
wK
eX

(18)

X∈X

and thus
wK θK/k,S =

X

X

nX
wK
LK/k,S (0, φ̄X )σ eφσXX∈X σ∈Gal(QX /Q)

using (9). Thus, the assertion that wK θK/k,S lies in Z(Λ0 ) is equivalent to the fact thatnX
wK
LK/k,S (0, φX ) lies in OX for all X ∈ X. Assume that X ∈ X is such that nX = 1. Let φ̃X
be the unique character of Gab whose inflation to G is equal to φX . Then, we havenX
wK
LK/k,S (0, φX ) = wK LK ab /k,S (0, φ̃X ) ∈ OX(>1)

using (1) and (3). Thus wK θK/k,S ∈ Z(Λ0 ) if and only if wK θK/k,S ∈ Z(Λ0 ).
We now specialize to the setting of the last section and assume that G contains a normal
abelian subgroup H of prime index `. Thanks to the above remark and Theorem 7.1, we have`wK θK/k,S ∈ Z(Λ0 ) if and only if ewK
θK/K H ,SH ∈ Z(Λ0 ) whereX
1
N[G,G] =eX ∈ Z(Λ0 ).
(19)
e := 1 −
sG
X∈X
nX >1

Since wK θK/K H ,SH ∈ Z[H] ⊂ Z(Λ0 ) by (3), we recover the fact that wK θK/k,S ∈ Z(Λ0 ) in thatsetting.
The different roles played by the idempotent e in both proofs show, in our opinion, that the
integrality statements of the two conjectures are, somewhat, of different nature (at least in thissetting). Indeed, for the weak non-abelian Brumer-Stark conjecture of Nickel, this factor plays
no role at all whereas it plays an essential part in the proof of the Integrality Conjecture, see theproof of Proposition 7.3. This leads us to believe that there is no direct easy connection between
the two integrality conjectures.
Acknowledgements. We are grateful to A. Nickel for helpful discussions concerning his nonabelian Brumer-Stark conjectures. We are in addition very grateful to the anonymous referee

A BRUMER-STARK CONJECTURE FOR NON-ABELIAN GALOIS EXTENSIONS

27

for his or her very careful reading of the manuscript and may helpful corrections, suggestions
and remarks.
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GAELLE DEJOU AND XAVIER-FRANÇOIS ROBLOT

E-mail address: dejou@math.univ-lyon1.fr
E-mail address: roblot@math.univ-lyon1.fr
Université de Lyon, CNRS UMR 5208, Université Lyon 1, Institut Camille Jordan, 43 blvd. du 11
novembre 1918, F-69622 Villeurbanne cedex, France