PROJECTIVE SCHUR GROUPS OF HENSELIAN FIELDS

Eli Aljadeff, Jack Sonn and Adrian R. Wadsworth

Technion–Israel Institute of Technology, Haifa, Israel, University of California, San Diego

Abstract.

One of the open questions that has emerged in the study of the projective Schur group P S(F) of a field F is whether or not P S(F) is an algebraic relative Brauer group over F, i.e. does there exist an algebraic extension L/F such that P S(F) = Br(L/F)? We show that the same question for the Schur group of a number field has a negative answer. For the projective Schur group, no counterexample is known. In this paper we prove that P S(F) is an algebraic relative Brauer group for all Henselian valued fields F of equal characteristic whose residue field is a local or global field. For this, we first show how P S(F) is determined by P S(k) for an equicharacteristic Henselian field with arbitrary residue field k.

1. Introduction

Let F be a field, Br(F) its Brauer group. The Schur group S(F) of F is the subgroup of Br(F) consisting of classes represented by Schur algebras over F. A finite dimensional central simple F -algebra A is called Schur over F if it is a homomorphic image of a group algebra F G with G finite. Equivalently, A is Schur over F if it is spanned as a vector space over F by a finite subgroup G of the group A^{∗} of invertible elements of A. In 1978 [LO], Lorenz and Opolka introduced projective analogues to these notions. They defined the projective Schur group P S(F) of F to be the subgroup of Br(F) consisting of classes represented by projective Schur algebras over F. A finite dimensional central simple F - algebra A is projective Schur over F if it is spanned as a vector space over F by a subgroup G of the group A^{∗} of invertible elements of A which is finite modulo F^{∗}, i.e. GF^{∗}/F^{∗} is finite. In either case, when a subgroup G of A^{∗} spans A over F, we write A = F (G).

In view of the fact that these two subgroups of Br(F) are defined in the language of algebras, we can ask for a natural characterization of them in the language of Galois cohomology, just as Br(F) is a Galois cohomology group H^{2}(G_{F} , F_{s}^{∗}), where F_{s} denotes the separable closure of F. In the case of the Schur group, the Brauer-Witt theorem can be viewed as a positive answer to this question: Let F_{cyc} be the maximal cyclotomic extension

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 11R52, 11S25, 12F05, 12G05, 13A20.

The research was supported by the Fund for the Promotion of Research at the Technion.

of F and let μ denote the group of all roots of unity in F_{s}. Then S(F) is the image of the canonical map H^{2}(G(F_{cyc}/F), μ) ⟶ H^{2}(G_{F} , F_{s}^{∗}) ≅ Br(F) (cf. [Y, Cor. 3.11]).

In the case of P S(F), all known examples of projective Schur algebras are Brauer equivalent to radical abelian algebras, defined as follows: Let A = (L/F, G, f) be a crossed product algebra, where L is a finite Galois extension field of F, G is the Galois group G(L/F), and f ∈ H^{2}(G, L^{∗}). Then A is said to be a radical algebra if L = F (T) for some subgroup T of L^{∗} containing F^{∗} such that T/F^{∗} is finite and f is represented by a 2-cocycle with values in T. This A is called a radical abelian algebra if in addition G(L/F) is abelian. It is easy to see that every radical algebra over F lies in P S(F). The first two authors have conjectured that all projective Schur algebras are Brauer equivalent to radical algebras, and even to radical abelian algebras. The radical algebra conjecture is equivalent to the conjecture that P S(F) is the image in Br(F) = H^{2}(G(F_{s}/F), F_{s}^{∗}) of H^{2}(G(F (T_{s})/F), T_{s}), where T_{s} is the subgroup of F_{s}^{∗} consisting of elements of finite order modulo F^{∗}. This would provide an analogue for P S(F) of the Brauer-Witt theorem for S(F). The radical abelian algebra conjecture has an analogous homological interpretation. The radical abelian algebra conjecture has been proved for all fields of nonzero characteristic [AS_{4}, Cor. 1.5]. In characteristic 0 only partial results are known [AS_{1}], [AS_{3}], [AS_{4}], [AS_{6}].

Another way of describing some subgroups of Br(F) is as algebraic relative Brauer groups. Let M/F be a field extension. The relative Brauer group Br(M/F) is the kernel of the restriction map res_{M/F} : Br(F) → Br(M). A subgroup H of Br(F) is called an algebraic relative Brauer group if there exists an algebraic extension M/F such that Br(M/F) = H. It is known that every subgroup of Br(F) is a relative Brauer group, by taking M to be an iterated generic splitting field of the division algebras in H, cf. [FS, Th. 1]; but it is not true in general that every subgroup is an algebraic relative Brauer group (e.g., let H be any nontrivial finite subgroup of Br(F), if F is a global field [FKS, Cor. 4]). Of course Br(F) itself is an algebraic relative Brauer group by definition. We ask if S(F) and P S(F) are algebraic relative Brauer groups. The answer is negative for S(F) even for F a number field, as we will show in §6 below. For P S(F) this question has an obvious affirmative answer for local and global fields F since in that case P S(F) = Br(F) by [LO, Satz 3], or see [AS_{2}, p. 531]. There is no good reason to believe that P S(F) is an algebraic relative Brauer group for every field F, but so far no counterexample has been found.

This paper is concerned with the radical (abelian) algebra conjecture and the algebraic relative Brauer group question for P S(F) for fields F with Henselian valuation such that the residue field k has the same characteristic as F. We show in Cor. 4.6 that the radical (resp. radical abelian) conjecture holds for F if it holds for k. We prove in §5 that if k is a local or global field, then P S(F) is an algebraic relative Brauer group.

The proofs of our main results require detailed information about the Brauer group of a Henselian valued field F, which we give in §3. Beyond the known results, which we recall, we construct explicit splitting maps for the inertially split part of Br(F) and for the tame part of Br(F). The splitting maps are used in §4 to show exactly how P S(F) is built from P S(k), where k is the residue field of the Henselian valuation on F (assuming char(k) = char(F)). This generalizes to arbitrary equicharacteristic Henselian fields results in [AS_{6}] for iterated power series fields.

Whenever a projective Schur algebra F (G) has abelian finite group GF^{∗}/F^{∗}, there is an associated symplectic pairing on GF^{∗}/F^{∗} given by commutators. We will show in §2 how such pairings and their associated Lagrangians can elucidate the structure of an arbitrary reduced projective Schur algebra. This provides a unified approach to a number of previous results on projective Schur algebras, as well as being needed for the analysis of the Henselian situation in §4.

We point out in passing that the algebraic relative Brauer group question has been studied for the m-torsion subgroups _{m}Br(F) of Br(F). In general _{m}Br(F) is not an algebraic relative Brauer group. Counterexamples exist for F a power series field k((t)), with k a local field [AS_{5}, Sec. 4]. On the other hand, for F a global field _{m}Br(F) is an algebraic relative Brauer group for every m, see [AS_{5}], [KS_{1}], [Po], [KS_{2}].

We will use the following notation throughout the paper: If C is a torsion abelian group, we write exp(C) for the exponent of C; _{n}C for the n-torsion subgroup of C; and C(p) for the p-primary component of C. For c ∈ C, o(c) denotes the order of c. If C is associated to a field F, C′ denotes the prime-to-p part of C if char(F) = p ≠ 0, while C′ = C if char(F) = 0. We write μ(F) for the group of roots of unity in a field F; we write μ_{n} for the group of n n-th roots of unity. If S is a central simple algebra over F, deg(S) = \sqrt{dim_{F} (S)} is the degree of S, and exp(S) is the exponent of S, which is the order of the class [S] of S in the Brauer group Br(F). If μ_{n} ⊆ F and a, b ∈ F^{∗} we write (a, b; F)_{n} for the symbol algebra of dimension n^{2} over F with generators i, j and relations i^{2} = a, j^{2} = b, and ij = ωji, where ω is some primitive n-th root of unity in F.

2. Projective Schur algebras of abelian type and Langrangians

A projective Schur algebra A = F (A) is said to be of abelian type if the finite group AF^{∗}/F^{∗} is abelian. Associated to such an A is a nonsingular symplectic pairing on AF^{∗}/F^{∗}. We will first recall some properties of such algebras, which can be seen easily by using this pairing. The data about the abelian case are relevant for more general projective Schur algebras because we will see in Prop. 2.2 below that every reduced projective Schur algebra admits after scalar extension a decomposition into a Schur algebra and a projective Schur algebra of abelian type. Furthermore, we will show that Lagrangian subgroups of AF^{∗}/F^{∗} with respect to the symplectic pairing yield useful refinements of such tensor decompositions. The results in this section provide a unified approach to arguments in several papers by the first two authors.

Proposition 2.1. Let A = F (A) be a projective Schur algebra of abelian type, where F^{∗} ⊆ A ⊆ A^{∗} (so A/F^{∗} is a finite abelian group). Then,

(a) There is a well-defined pairing B_{A}: A/F^{∗} × A/F^{∗} → F^{∗} given by (aF^{∗}, bF^{∗}) ↦ aba^{−1}b^{−1}.

(b) The pairing B_{A} is nondegenerate, bimultiplicative and symplectic, and im(B_{A}) is a finite (cyclic) subgroup of μ(F).

(c) |A/F^{∗}| = dim_{F} (A) and exp(A/F^{∗}) = |im(B_{A})|.

(d) A ≅ S_{1} ⊗_{F} ⋯ ⊗_{F} S_{m}, where each S_{i} is a symbol algebra, with exp(A/F^{∗}) =

lcm_{1≤i≤m} deg(S_{i}).

Proof. This mostly known, cf. [LO, Hilfsatz 1] and [AS_{3}, Th. 1.1], but we will show how the use of the pairing B_{A} facilitates the proof. For a ∈ A, let ã = aF^{∗} ∈ A/F^{∗}. The pairing B_{A} is well-defined because F^{∗} is central in A^{∗}. That B_{A} is symplectic means that B_{A}(ã, ã) = 1 for all a ∈ A; this is evident here. Let [a, b] = aba^{−1}b^{−1}. Since B_{A} maps into a central subgroup of A^{∗}, the commutator identities [a, bc] = [a, b]b[a, c]b^{−1} and [ab, c] = a[b, c]a^{−1}[a, c] show that B_{A} is bimultiplicative. Let e = exp(A/F^{∗}). The bimultiplicativity of B_{A} shows that im(B_{A}) ⊆ μ_{e}. The nondegeneracy of B_{A} means that for any a ∈ A, if B_{A}(ã, b̃) = 1 for all b ∈ A, then a ∈ F^{∗}; this holds as Z(F (A)) = F. Because B_{A} is nondegenerate, for any a ∈ A the function fã: A/F^{∗} → F^{∗} given by fã(b̃) = B_{A}(ã, b̃) must have the same order in the group Hom(A/F^{∗}, F^{∗}) as the order o(ã) of ã in A/F^{∗}. Hence if we take ã ∈ A/F^{∗} with o(ã) = e we see that im(fã) = μ_{e}; thus im(B_{A}) = μ_{e}; so |im(B_{A})| = e = exp(A/F^{∗}). Since we have a nondegenerate bimultiplicative symplectic pairing defined on a finite abelian group, it is known (see, e.g., [Wa, Th. (3)]) that there is a symplectic base of A/F^{∗}, i.e., a generating set ã_{1}, …, ã_{m}, b̃_{1}, …, b̃_{m} of A/F^{∗}, such that B(ã_{i}, b̃_{i}) = ω_{i}, where o(ω_{i}) = o(ã_{i}) = o(b̃_{i}) for each i, and B_{A}(ã_{i}, ã_{j}) = B_{A}(b̃_{i}, b̃_{j}) = 1 for all i, j, and B_{A}(ã_{i}, b̃_{j}) = 1 whenever i ≠ j. Let n_{i} = o(ã_{i}) = o(b̃_{i}) = o(ω_{i}). Then we have a_{i}^{n_{i}}, b_{i}^{n_{i}} ∈ F^{∗} and a_{i}b_{i} = ω_{i}b_{i}a_{i} for all i, and a_{i}a_{j} = a_{j}a_{i}, b_{i}b_{j} = b_{j}b_{i} for all i, j, and a_{i}b_{j} = b_{j}a_{i} whenever j ≠ i. Let c_{i} = a_{i}^{n_{i}} and d_{i} = b_{i}^{n_{i}} ∈ F^{∗}. Then the relations just listed among all the a_{i} and b_{i} show that F (A) is a homomorphic image of S = ⊗^{m}_{i=1}(c_{i}, d_{i}; F)_{n_{i}}. Since S is simple, the map of S onto A must be an isomorphism. So dim_{F} (A) = dim_{F} (S) = n_{1}^{2} ⋯ n_{m}^{2} = |A/F^{∗}|, since A/F^{∗} is the direct product of its cyclic subgroups generated by the ã_{i} and the b̃_{i}. This direct product decomposition shows that exp(A/F^{∗}) = lcm(n_{1}, ..., n_{m}) = lcm_{1≤i≤m}(deg(S_{i})), where S_{i} = (c_{i}, d_{i}; F)_{n_{i}}. □

A radical abelian extension of a field F is an abelian Galois field extension K of F such that K = F (U), where U is a subgroup of K^{∗} with U ⊇ F^{∗} and U/F^{∗} finite.

A projective Schur algebra A = F (G) is said to be reduced if for every subgroup H of G with H ⊇ G′ (the derived group of G) and every subfield L, F ⊆ L ⊆ A such that G acts on L by conjugation, the subalgebra L(H) is simple. Recall from [AS_{1}, Th. 1.4] that every projective Schur algebra is Brauer equivalent to a reduced projective Schur algebra. We now collect basic properties of subalgebras F (H) of a reduced projective Schur algebra F (G), where G′ ⊆ H ⊆ G.

Let A = F (G) be a projective Schur algebra with G ⊇ F^{∗}. Assume A is reduced. Let H be any subgroup of G such that H ⊇ G′. Let

B = F (H), a simple algebra, as A is reduced;

L = Z(B), a field;

Ĥ = C_{G}(L), a normal subgroup of G;

E = F (Ĥ);

T = C_{ĤB^{∗}}(B);

T = L(T ).

Proposition 2.2. In the setting just described,

(a) L is abelian Galois over F with Galois group G(L/F) ≅ G/Ĥ, and L lies in a radical abelian extension of F.

(b) E is a simple algebra with Z(E) = L and E = C_{A}(L) = B ⊗_{L} T.

(c) L^{∗} ⊆ T and T /L^{∗} ≅ Ĥ/(Ĥ ∩ B^{∗}), a finite abelian group.

(d) T is a projective Schur algebra of abelian type over L.

(e) Let G = G/Ĥ, a finite abelian group. Then A = E ∗ G, a ring-theoretic crossed product.

(f) If e = exp(T), then μ_{e} ⊂ F^{∗}.

Proof. Since G′ ⊆ H, we have H ◁ G, so G acts by conjugation on B. Hence B is simple as A is reduced. So L is a field with [L : F] < ∞, and G acts by conjugation on L. Clearly, F ⊆ L^{G} ⊆ A^{G} = F, so L is Galois over F and G maps onto G(L/F) with kernel Ĥ. Therefore, G(L/F) ≅ G/Ĥ, which is abelian as Ĥ ⊇ H ⊇ G′. Since F (H) is a simple F -algebra with H/F^{∗} finite and L = Z(F (H)) is Galois over F, by [AS_{2}, Th. 1.3] L lies in a radical extension of F. Because L is also abelian Galois over F, [AS_{2}, Prop. 2.1] shows that L lies in a radical abelian extension of F, proving (a).

As for (e), note first that since Ĥ ⊇ G′ and A is reduced, E = F (Ĥ) is simple. Let n = |G/Ĥ| = [L : F], and let g_{1}, ..., g_{n} be a set of coset representatives of Ĥ in G. Then G ⊆ ∑^{n} ; i=1Eg_{i}, so ∑^{n} ; i=1Eg_{i} = A. Hence dim_{F} (E) ≥ dim_{F} (A)/n. Clearly E ⊆ C_{A}(L). So using the Double Centralizer Theorem,

dim_{F} (E) ≤ dim_{F} (C_{A}(L)) = dim_{F} (A)/[L : F] ≤ dim_{F} (E).

So, equality holds throughout, which shows that E = C_{A}(L); hence Z(E) = L. The dimension equality shows that the sum A = ∑^{n} ; i=1Eg_{i} = A is direct. Hence A = E ∗ (G/Ĥ), a

ring-theoretic crossed product, proving (e). We have also proved the first three assertions of (b).

Now consider T . Since Ĥ normalizes H (as H ◁ G), Ĥ also normalizes B^{∗}. So ĤB^{∗} is a subgroup of A^{∗}, and the definition of T as C_{ĤB^{∗}}(B) makes sense. Observe also that B = F (H) ⊆ F (Ĥ) = E; so, T ⊆ E. Since L^{∗} ⊆ T ∩ B^{∗} ⊆ Z(B^{∗}) = L^{∗}, we have T ∩ B^{∗} = L^{∗}. Also, for any ĥ ∈ Ĥ, conjugation by ĥ gives an L-linear automorphism of B. By Skolem-Noether, there is a b ∈ B^{∗} conjugation by which produces the same automorphism of B. Then ĥb^{−1} ∈ T . This shows that Ĥ ⊆ T B^{∗}; since T ⊆ ĤB^{∗} by definition, we have ĤB^{∗} = T B^{∗}. Then,

T /L^{∗} = T /(T ∩ B^{∗}) ≅ T B^{∗}/B^{∗} = ĤB^{∗}/B^{∗} ≅ Ĥ/(Ĥ ∩ B^{∗}).

Because F^{∗}H ⊆ Ĥ ∩ B^{∗}, Ĥ/(Ĥ ∩ B^{∗}) is finite (as Ĥ/F^{∗} is finite) and abelian (as Ĥ/H is abelian), proving (c).

Next, consider T = L(T ). We have E = F (Ĥ) ⊆ BT ⊆ E, since we just saw Ĥ ⊆ B^{∗}T ⊆ E. Since E = B ⊗_{L} C_{E}(B) by the Double Centralizer Theorem and T ⊆ C_{E}(B) = E, we have E = BT = B ⊗_{L} T ⊆ B ⊗_{L} C_{E}(B) = E. So T = C_{E}(B). This completes the proof of (b) and shows also that L = Z(T). Since T = L(T ) is a central simple L-algebra, part (c) shows that T is a projective Schur algebra of abelian type, proving (d).

Because T /L^{∗} is abelian, we have the pairing B_{T} : T /L^{∗} × T /L^{∗} → L^{∗} described in Prop. 2.1. The key to proving (f) is to see that im(B_{T}) ⊆ F^{∗}. For this, observe that G acts by conjugation on H, B, L, and Ĥ, so on ĤB^{∗}, so on T = C_{ĤB^{∗}}(B) and on T /L^{∗}. Take any g ∈ G and t ∈ T ; write t = ĥb with ĥ ∈ Ĥ and b ∈ B^{∗}. Let c = gtg^{−1}t^{−1}. Then c ∈ T , as gtg^{−1} ∈ T ; likewise, gbg^{−1}b^{−1} ∈ B^{∗}. So

c = [gĥg^{−1}ĥ^{−1}]ĥ[gbg^{−1}b^{−1}]ĥ^{−1} ∈ G′ĥB^{∗}ĥ^{−1} ⊆ B^{∗}.

Hence c ∈ T ∩ B^{∗} = L^{∗}. Since gtg^{−1} = ct, this shows that G acts trivially on T /L^{∗}. Because the pairing B_{T} is clearly compatible with the G-action, G must also act trivially on im(B_{T}), i.e. im(B_{T}) ⊆ L^{∗G} = F^{∗}, as claimed. We have im(B_{T}) = μ_{ℓ}, where ℓ = |im(B_{T})| (see Prop. 2.1(b)); so μ_{ℓ} ⊆ F^{∗}. Let e = exp(T ). Then by Prop. 2.1(d) and (c), e| exp(T /L^{∗}) = ℓ. Thus μ_{e} ⊆ μ_{ℓ} ⊆ F^{∗}, proving (f). □

We digress to show that the preceding results yield a considerably simplified proof of the exponent reduction theorem that was the main result of [AS_{4}]. For this, let F_{cyc} be the maximal cyclotomic extension of F.

Proposition 2.3 [AS_{4}, Th. 1.3]. Let A be a projective Schur algebra over F, and let e = exp(A ⊗_{F} F_{cyc}). Then μ_{e} ⊆ F^{∗}.

Proof. Let A = F (G) with G ⊇ F^{∗}. We may assume that A is reduced. We apply Prop. 2.2 with H = G′. Recall (see [Sc, p. 443]) that since G has a central subgroup F^{∗} with G/F^{∗} finite, |G′| < ∞. Hence B = F (G′) is a Schur algebra. By the Brauer splitting theorem [CR, pp. 385, 418], B is split by F_{cyc} and L = Z(B) ⊆ F_{cyc}. Now E = C_{A}(L) ∼ A ⊗_{F} L in Br(L). Since E = B ⊗_{L} T as in Prop. 2.2, we have A ⊗_{F} F_{cyc} ∼ E ⊗_{L} F_{cyc} ∼ T ⊗_{L} F_{cyc}. Let d = exp(T). Then

e = exp(A ⊗_{F} F_{cyc}) = exp(T ⊗_{L} F_{cyc}) | exp(T) = d.

Since μ_{d} ⊆ F^{∗} by Prop. 2.2(f), this shows μ_{e} ⊆ F^{∗}. □

We now return to the setup of Prop. 2.2, with A = F (G), a reduced projective Schur algebra (F^{∗} ⊆ G), and H a subgroup of G with G′ ⊆ H, and the objects associated to H described there, including the projective Schur algebra of abelian type T = L(T ). We will show how to use subgroups L/L^{∗} of T /L^{∗} to build new subgroups H_{1} of G containing H so that the objects associated to H_{1} by Prop. 2.2 have a nice description in terms of H and the corresponding objects for H. This will be needed in §4. It also provides a unified approach to constructions that were given in [AS_{1}],[AS_{2}],[AS_{4}],[AS_{6}]. In each of those papers an H_{1} is chosen after starting with some H, so that H_{1} ⊇ H and F (H_{1}) is maximal with respect to some property. In examining these constructions, one sees that what was needed was primarily an H_{1} so that its associated T as in Prop. 2.2 is trivial and F (H_{1}) = F (H) ⊗_{L} Z(F (H_{1})). We will see that this occurs whenever L/L^{∗} is a Lagrangian of T /L^{∗}.

For the projective Schur algebra of abelian type T = L(T ) of Prop. 2.2 (with L^{∗} ⊆ T ), we have the nondegenerate symplectic pairing B_{T} : T /L^{∗} × T /L^{∗} → L^{∗} described in Prop. 2.1. Take any subgroup Λ of T /L^{∗}, and let L be the inverse image of Λ in T . Then L(L) is an L-subalgebra of T with dim_{L}(L(L)) = |Λ| (see [TW, Ex. 2.4(c)]). Let Λ^{⊥} = {γ ∈ T /L^{∗} | B_{T} (γ, λ) = 1 for all λ ∈ Λ}, a subgroup of T /L^{∗}. Because B_{T} is nondegenerate, we have |Λ||Λ^{⊥}| = |T /L^{∗}| (cf. [TW, (2.2)]), so Λ^{⊥⊥} = Λ. Abusing notation slightly, let L^{⊥} denote the inverse image of Λ^{⊥} in T . It is easy to check that L(L^{⊥}) = C_{T} (L(L)) (see [TW, Lemma 2.5(i)]). So Z(L(L)) = L(L) ∩ L(L^{⊥}) = L(L ∩ L^{⊥}). In particular, L(L) is commutative iff Λ ⊆ Λ^{⊥}, iff B_{T} is trivial on Λ × Λ. The subgroup Λ is called a Lagrangian of T /L^{∗} with respect to B_{T} if Λ^{⊥} = Λ. It is easy to see that if Λ is any subgroup of T /L^{∗} with Λ ⊆ Λ^{⊥}, then Λ lies in some Lagrangian of T /L^{∗}. Thus the algebras L(L) for L/L^{∗} a Langrangian are the maximal commutative subalgebras of T of the form L(Λ) for L ⊆ T .

Now fix a subgroup Λ of T /L^{∗}, let L be its inverse image in T . Let Λ = {a_{1}L^{∗}, ..., a_{m}L^{∗}}, with a_{i} ∈ L ⊆ T ⊆ ĤB^{∗}. Write each a_{i} = ĥ_{i}b_{i} with ĥ_{i} ∈ Ĥ and b_{i} ∈ B^{∗}. Let H_{1} = ⟨H, ĥ_{1}, ..., ĥ_{m}⟩, a subgroup of G with G′ ⊆ H ⊆ H_{1} ⊆ Ĥ. Associated to H_{1} we have the objects of Prop. 2.2: B_{1} = F (H_{1}), L_{1} = Z(B_{1}), Ĥ_{1} = C_{G}(L_{1}), E_{1} = F (Ĥ_{1}) = B_{1} ⊗_{L_{1}} T_{1}, where T_{1} = L_{1}(T_{1}) with T_{1} = C_{Ĥ_{1}B_{1}^{∗}}(B_{1}).

Proposition 2.4. In the situation described in the preceding paragraph,

(a) B_{1} = B ⊗_{L} L(L); L_{1} = L(L ∩ L^{⊥}), which is a Kummer extension field of L, and lies in a radical abelian extension of F; H ⊆ H_{1} ⊆ Ĥ_{1} ⊆ Ĥ; E_{1} = C_{A}(L_{1}) = B ⊗_{L} L(LL^{⊥}) ⊆ E; and T_{1} = L(L^{⊥}) ⊆ T.

(b) [T_{1} : L_{1}] = |L^{⊥}/(L ∩ L^{⊥})|

(c) If Λ is a Lagrangian of T /L^{∗}, then L_{1} = L(L) = T_{1} and B_{1} = E_{1}.

Proof. For each generator a_{i} of L, we have a_{i} = ĥ_{i}b_{i}, with each b_{i} ∈ B = F (H) ⊆ F (H_{1}) = B_{1}, and ĥ_{i} ∈ F (H_{1}) = B_{1}; so a_{i} ∈ B_{1}. Also, L ⊆ B ⊆ B_{1}. Thus B · L(L) ⊆ B_{1}. On the other hand, each ĥ_{i} = a_{i}b_{i}^{−1} = b_{i}^{−1}a_{i} ∈ B · L(L), and H ⊆ B; so B_{1} = F (H_{1}) ⊆ B · L(L). Hence B_{1} = B · L(L). But, B · L(L) = B ⊗_{L} L(L) because L(L) ⊆ T and E = B ⊗_{L} T by Prop. 2.2(b); so, B_{1} = B ⊗_{L} L(L). Since C_{T} (L(L)) = L(L^{⊥}), as noted above, we have Z(L(L)) = L(L) ∩ L(L^{⊥}) = L(L ∩ L^{⊥}). Hence L_{1} = Z(B_{1}) = Z(B ⊗_{L} L(L)) = Z(B) ⊗_{L} Z(L(L)) = L ⊗_{L} L(L ∩ L^{⊥}) = L(L ∩ L^{⊥}). This L_{1} is a field as A is reduced and B_{1} = F (H_{1}) with G′ ⊆ H ⊆ H_{1}. Moreover, L_{1} is a Kummer extension of L since exp((L ∩ L^{⊥})/L^{∗}) | exp(T /L^{∗}) = |im(B_{T})| and im(B_{T}) ⊆ μ(L) by Prop. 2.1(b),(c). Prop. 2.2(a), applied with H_{1} in place of H, says that L_{1} lies in a radical abelian extension of F. Because L_{1} ⊇ L, we have Ĥ_{1} = C_{G}(L_{1}) ⊆ C_{G}(L) = Ĥ. The other inclusions in H ⊆ H_{1} ⊆ Ĥ_{1} ⊆ Ĥ are clear. Now by Prop. 2.2(b), E_{1} = C_{A}(L_{1}) ⊆ C_{A}(L) = E, as L ⊆ L_{1}. Therefore, T_{1} = C_{E_{1}}(B_{1}) ⊆ C_{E}(B_{1}) = C_{B⊗_{L}T} (B⊗_{L}L(L)) = C_{T} (L(L)) = L(L^{⊥}). But also, L(L^{⊥}) = C_{T} (L(L)) ⊆ C_{A}(L(L ∩ L^{⊥})) = E_{1}. Since L(L^{⊥}) centralizes B and L(L), it centralizes B ⊗_{L} L(L) = B_{1}. Thus, L(L^{⊥}) ⊆ T_{1}; so equality holds. Finally, E_{1} = B_{1} ⊗_{L_{1}} T_{1} = B ⊗_{L} L(L) ⊗_{L(L∩L^{⊥})} L(L^{⊥}) = B ⊗_{L} L(LL^{⊥}), completing the proof of (a).

For (b), note that

\begin{aligned}[T_{1} : L_{1}] & = [L(L^{⊥}) : L(L ∩ L^{⊥})] = [L(L^{⊥}) : L]/[L(L ∩ L^{⊥}) : L] \\ & = |Λ^{⊥}|/|Λ ∩ Λ^{⊥}| = |L^{⊥}/L ∩ L^{⊥}|. \end{aligned}

Part (c) follows immediately from (a) and (b), since Λ is a Lagrangian just when Λ = Λ^{⊥}, so L = L^{⊥}. □

3. Splitting maps for the Brauer group of a Henselian field

In this section we give the properties of division algebras over Henselian fields that will be needed for the analysis of the projective Schur groups of such fields. We first recall some known results, then give analogues for an arbitrary Henselian valuation to Witt’s direct sum decomposition theorem for the Brauer group of a complete discretely valued field.

Let F be a field with a valuation v: F^{∗} → Γ_{F}, where Γ_{F} is the value group of v, a totally ordered abelian group, written additively. Let V_{F} be the valuation ring of v and M_{F} the unique maximal ideal of V_{F}; let \overline{F} = V_{F} /M_{F}, the residue field of v; and let U_{F} = V_{F} − M_{F}, the group of valuation units. Let Δ be the divisible hull of Γ_{F}; so Δ ⊇ Γ_{F}, and the ordering on Γ_{F} extends uniquely to Δ making Δ an ordered abelian group. Note that Δ ≅ ℚ ⊗_{ℤ} Γ_{F}. If L is a field algebraic over F and w is any extension of V to L, then there are canonical injections which we view as inclusions \overline{F} ↪ \overline{L}, Γ_{F} ↪ Γ_{L}, and Γ_{L} ↪ Δ. Recall the Fundamental Inequality [E, (13.10)], which says that whenever [L : F] < ∞, we have

[\overline{L} : \overline{F}] |Γ_{L} : Γ_{F} | ≤ [L : F]  (3.1)

Assume now and throughout the rest of this section that the valuation v on F is Henselian. This means that Hensel’s Lemma holds for v, or equivalently (see, e.g., [E, Cor. 16.6]), that v has a unique extension to each field L ⊇ F with L algebraic over F. Thus the extension of v to any such L, which we again denote by v, is also Henselian.

Examples 3.1. (a) If v is a valuation on a field F with Γ_{F} embedding in ℝ (equivalently, if V_{F} has Krull dimension 1) and F is complete with respect to v, then v is Henselian [E, (16.7)].

(b) Suppose F_{0} is a field with Henselian valuation v_{0}. Let F be the Laurent series field F = F_{0}((x)). Then the canonical extension v of v_{0} to F (given by v( ∞ ; ∑_{i=k}c_{i}x^{i}) =

(v_{0}(c_{k}), k) ∈ Γ_{F_{0}} × ℤ if c_{k} ≠ 0) is Henselian, with V_{F} = V_{0} + xF_{0}[[x]], \overline{F} = \overline{F_{0}}, and Γ_{F} = Γ_{F_{0}} × ℤ. The ordering on Γ_{F} is the right-to-left lexicographical ordering, in which (γ, i) ≤ (δ, j) just when i < j or both i = j and γ ≤ δ. That v is Henselian is a special case of the fact that composites of Henselian valuations are Henselian [Rib, p. 211, Prop. 10].

Let L be an algebraic extension of the Henselian field F. If [L : F] < ∞, we say that L is unramified over F if [\overline{L} : \overline{F}] = [L : F] and \overline{L} is separable over \overline{F}. When this occurs, the Fundamental Inequality shows that Γ_{L} = Γ_{F}. At the other extreme, we say that L is totally ramified over F if |Γ_{L} : Γ_{F} | = [L : F]. Also, L is said to be tamely ramified over F if char(\overline{F}) ∤ |Γ_{L} : Γ_{F} |, \overline{L} is separable over \overline{F}, and [\overline{L} : \overline{F}] |Γ_{L} : Γ_{F} | = [L : F]. If L is algebraic over F but [L : F] = ∞, we say that L is unramified (resp. totally ramified, tamely ramified) over F if L is a union of finite degree extensions of F each of which is unramified (resp. totally ramified, tamely ramified) over F.

Let F_{s} be a fixed separable closure of our Henselian valued field F. Let F_{nr} denote the maximal unramified extension of F in F_{s}. That is, F_{nr} is the inertia field for the extension of v from F to F_{s}. It is known (see [E, (19.10), (19.11)]) that for fields L with F ⊆ L ⊆ F_{s}, L is unramified over F iff L ⊆ F_{nr}. Furthermore (see [E, (19.12), (19.8), (19.6)]), \overline{F_{nr}} ≅ \overline{F}_{s}, and F_{nr} is Galois over F, with Galois group G(F_{nr}/F) ≅ G(\overline{F}_{s}/\overline{F}), which is the absolute Galois group of \overline{F}, also denoted G\overline{F}. From the isomorphism of Galois groups, one can see that there is a one-to-one correspondence L ↦ \overline{L} between unramified field extensions L of F (with L ⊆ F_{s}) and separable algebraic field extensions of \overline{F}. If E is a separable extension of \overline{F}, we call the field L ⊇ F with \overline{L} = E the inertial lift of E over \overline{F}; we will often write F (E) for this inertial lift of E.

When the valuation on F is complete and discrete, Witt gave a description of the Brauer group Br(F). We now give generalizations of Witt’s theorem, which are valid for any Henselian valued field. The basic exact sequence (3.3) below was derived in [JW, pp. 154-156], but the splitting maps in Th. 3.2 and Prop. 3.3 were not given there. We will review the derivation of (3.3), since it is essential to this paper, and we need to know the actual maps in it, not just the existence of an exact sequence. Let Γ = Γ_{F}, Δ = ℚ ⊗_{ℤ} Γ, and let G = G(F_{nr}/F) ≅ G\overline{F}. We first describe

SBr(F) = Br(F_{nr}/F)

which is called the inertially split part of Br(F).

The short exact sequence of discrete G-modules

1 ⟶ U_{F_{nr}} ⟶ F_{nr}^{∗} ⟶^{v} Γ ⟶ 0

induces an exact sequence in cohomology

H^{1}(G, Γ) ⟶ H^{2}(G, U_{F_{nr}}) ⟶ H^{2}(G, F_{nr}^{∗}) ⟶ H^{2}(G, Γ).  (3.2)

We interpret the terms in the sequence. We have H^{1}(G, Γ) = Hom_{c}(G, Γ) = 0 (continuous homomorphisms; there are none, as G is profinite and Γ is torsion-free). It is known (see [JW, Th. 5.6(a)]) that the residue map U_{F_{nr}} ⟶ \overline{F_{nr}}^{∗} ≅ \overline{F} _{s}^{∗} induces an isomorphism H^{2}(G, U_{F_{nr}}) ≅ H^{2}(G, \overline{F} _{s}^{∗}) ≅ Br(\overline{F}). The next term in (3.2) is SBr(F). Since Δ is uniquely divisible, H^{1}(G, Δ) = H^{2}(G, Δ) = 0, so H^{2}(G, Γ) ≅ H^{1}(G, Δ/Γ) = Hom_{c}(G, Δ/Γ). We shall give a splitting map which shows that the last map in (3.2) is onto, so (3.2) becomes the short exact sequence

0 ⟶ Br(\overline{F}) ⟶ SBr(F) ⟶^{β} Hom_{c}(G, Δ/Γ) ⟶ 0  (3.3)

It is easy to see that (3.3) is functorial with respect to algebraic field extensions, i.e., for any field L ⊇ F with L algebraic over F, and for the unique Henselian extension of v to L, the following diagram is commutative with exact rows:

0 ⟶ Br(\overline{F}) ⟶ SBr(F) ⟶ Hom_{c}(G\overline{F} , Δ/Γ_{F}) ⟶ 0
↓  ↓  ↓
0 ⟶ Br(\overline{L}) ⟶ SBr(L) ⟶ Hom_{c}(G\overline{L}, Δ/Γ_{L}) ⟶ 0

(3.4)

Here the right vertical map arises from the canonical inclusion G\overline{L} ↪ G\overline{F} and the surjection Δ/Γ_{F} → Δ/Γ_{L}.

Theorem 3.2. For any Henselian valued field F, there is a homomorphism f: Hom_{c}(G, Δ/Γ) → SBr(F) splitting the β of (3.3). Hence

SBr(F) ≅ Br(\overline{F}) ⊕ Hom_{c}(G\overline{F} , Δ/Γ_{F})

Proof. The abelian torsion group Δ/Γ has its canonical primary decomposition Δ/Γ = ⨁_{p prime}(Δ/Γ)(p), with (Δ/Γ)(p) = ⋃^{∞} ; j=1p^{−j}Γ/Γ. Fix a prime number p. Choose {γ_{i}}_{i∈I_{p}} ⊆ Γ

so that the γ_{i} map to a ℤ/pℤ-vector space base of Γ/pΓ. Then for each n ∈ ℕ, the γ_{i} also map to a base of Γ/p^{n}Γ as a free ℤ/p^{n}ℤ-module. So, {p^{−n}γ_{i}}_{i∈I_{p}} maps to a base of the free ℤ/p^{n}ℤ-module p^{−n}Γ/Γ. Let p^{−∞}γ_{i}/Γ denote ⋃^{∞} ; j=1⟨p^{−j}γ_{i} + Γ⟩ ⊆ Δ/Γ. Then,

(Δ/Γ)(p) = ⨁_{i∈I_{p}}p^{−∞}γ_{i}/Γ.

Let γ be any of the γ_{i}, and take any t ∈ F^{∗} with v(t) = γ. We use t to define a homomorphism from Hom_{c}(G, p^{−∞}γ/Γ) to the Brauer group. For this, note that since ℤ[1/p]γ ≅ ℤ[1/p] and ℤ[1/p]γ ∩ Γ = ℤγ (as γ ∉ pΓ), we have p^{−∞}γ/Γ ≅ ℤ[1/p]γ/ℤγ ≅ ℤ[1/p]/ℤ. Thus there is a homomorphism

θ: H^{1}(G, p^{−∞}γ/Γ) ⟶ H^{1}(G, ℤ[1/p]/ℤ) ⟶^{δ} H^{2}(G, ℤ)  (3.5)

where δ is the connecting homomorphism arising from the short exact sequence 0 → ℤ → ℤ[1/p] → ℤ[1/p]/ℤ → 0 of trivial G-modules. We use the cup product pairing H^{0}(G, F_{nr}^{∗}) ∪ H^{2}(G, ℤ) → H^{2}(G, F_{nr}^{∗}) ≅ SBr(F) to define our map

f_{t}: Hom_{c}(G, p^{−∞}γ/Γ) ⟶ SBr(F) by χ ↦ (t) ∪ θ(χ).  (3.6)

Then f_{t} is a group homomorphism, as the cup product is ℤ-bilinear. To describe f_{t}(χ) as an algebra, let N = ker(χ), an open normal subgroup of G, and let K be the fixed field of N. So F ⊆ K ⊆ F_{nr} and G(K/F) ≅ im(χ), which is a finite subgroup of p^{−∞}γ/Γ; thus im(χ) is the cyclic group p^{−r}γ/Γ, where [K : F] = p^{r}. Choose σ ∈ G(F_{nr}/F) = G with χ(σ) = p^{−r}γ + Γ; then σ|_{K} is a generator of G(K/F). It is easy to check that θ(χ) is represented by the cocycle z ∈ Z^{2}(G, ℤ) given by: for τ, ρ ∈ G, if τ|_{K} = (σ|_{K})^{i} and ρ|_{K} = (σ|_{K})^{j} with 0 ≤ i, j ≤ p^{r} − 1,

z(τ, ρ) = \{ \begin{matrix}0 if i + j < p^{r}, \\ 1 if i + j ≥ p^{r}. \end{matrix}

Then f_{t} is represented by z′ ∈ Z^{2}(G, F_{nr}^{∗}) given by

z′(τ, ρ) = \{ \begin{matrix}1 if i + j < p^{r}, \\ t if i + j ≥ p^{r}, \end{matrix}

(cf. [Se, p. 176, Lemma 1]). From the formula for z′ it is easy to see that for the β of (3.3), β(f_{t}(χ)) = χ, as v(t) = γ. Also we can read off from the formula for z′ that the algebra in SBr(F) represented by f_{t}(χ) is the cyclic algebra (K/F, σ, t).

Our splitting map f is obtained by aggregating such f_{t}: For each prime p, we have {γ_{i}}_{i∈I_{p}} ⊆ Γ mapping to a ℤ/pℤ-base of Γ/pΓ. For each i ∈ I_{p} choose some t_{i} ∈ F^{∗} with v(t_{i}) = γ_{i}. We have Hom_{c}(G, Δ/Γ)(p) = ⨁_{i∈I_{p}}Hom_{c}(G, p^{−∞}γ_{i}/Γ). Define f_{p}: Hom_{c}(G, Δ/Γ)(p) ⟶

SBr(F) by f_{p} = ⨁_{i∈I_{p}} f_{t_{i}}. Then define f: Hom_{c}(G, Δ/Γ) → SBr(F) by f = ⨁_{p prime}f_{p}.

Since each β ∘ f_{t_{i}} is the identity on Hom_{c}(G, p^{−∞}γ_{i}/Γ), we have β ∘ f is the identity on Hom_{c}(G, Δ/Γ), so f is a splitting map for (3.3). □

Note that the splitting map f of Th. 3.2 depends on the choice of the t_{i}. Similarly, in Witt’s theorem for complete discrete valuations, the splitting map depends on the choice of a uniformizing parameter. The reason for using the primary decomposition of Δ/Γ in the proof of Th. 3.2 is that the index sets I_{p} could have different sizes for different primes p. (By definition, |I_{p}| = dim_{ℤ/pℤ}(Γ/pΓ).) Of course if Γ were a free ℤ-module, we could choose a family {t_{i}} ⊆ F^{∗} mapping to a ℤ-base of Γ. We could then use the same t_{i} for each prime p, which would simplify the description of the splitting map.

We now show that the splitting of SBr(F) is compatible with suitably chosen scalar extensions. In the proof of Th. 3.2, we chose {γ_{i}}_{i∈I_{p}} ⊆ Γ and {t_{i}}_{i∈I_{p}} ⊆ F^{∗} with v(t_{i}) = γ_{i}, for any fixed prime p. To allow for consideration of all primes at once, we write γ_{p,i} for the earlier γ_{i} and t_{p,i} for t_{i}.

Proposition 3.3. Let L be an algebraic extension of the Henselian valued field F.

(a) Suppose L is unramified over F. Then Γ_{L} = Γ_{F}, and we can use the same t_{p,i} for the splitting map of SBr(L) as for SBr(F). Then we have a commutative diagram

Br(\overline{F}) ⊕ Hom_{c}(G\overline{F} , Δ/Γ_{F}) ⟶^{≅} SBr(F)
res⊕res↓  ^{res}↓
Br(\overline{L}) ⊕ Hom_{c}(G\overline{L}, Δ/Γ_{L}) ⟶^{≅} SBr(L)

(b) With the t_{p,i} chosen for the splitting map f of Th. 3.2, take any subset {j_{p,i}| p prime, i ∈ I_{p}} ⊆ ℕ. For each such p and i, let s_{p,i} be some p^{j_{p,i}}th root of t_{p,i}, and suppose L = F (s_{p,i}| p prime, i ∈ I_{p}). Then L is totally ramified over F, {v(s_{p,i})| i ∈ I_{p}} maps to a ℤ/pℤ-base of Δ/Γ_{L}, and if we use the t_{p,i} for the splitting of SBr(F) and the s_{p,i} for the splitting of SBr(L), then there is a commutative diagram

Br(\overline{F}) ⊕ Hom_{c}(G\overline{F} , Δ/Γ_{F}) ⟶^{≅} SBr(F)
res⊕can↓  ^{res}↓
Br(\overline{L}) ⊕ Hom_{c}(G\overline{L}, Δ/Γ_{L}) ⟶^{≅} SBr(L)

where can is induced by the canonical surjection Δ/Γ_{F} → Δ/Γ_{L}.

Proof. (a) Assume L is unramified over F, so Γ_{L} = Γ_{F}. Take any prime p, any γ ∈ Γ_{F}, γ ∉ pΓ_{F}, and any t ∈ F^{∗} with v(t) = γ. We have an evidently commutative diagram

H^{1}(G\overline{F} , p^{−∞}γ/Γ_{F}) ⟶^{θ} H^{2}(G\overline{F} , ℤ) ⟶^{(t)∪·} SBr(F)
res↓  ^{res}↓  ^{res}↓
H^{1}(G\overline{L}, p^{−∞}γ/Γ_{L}) ⟶^{θ} H^{2}(G\overline{L}, ℤ) ⟶^{(t)∪·} SBr(L)

(3.7)

where the horizontal maps θ are as in (3.5), the right horizontal maps are cup product with (t), and the first two vertical restriction maps arise from the canonical inclusion G\overline{L} ↪ G\overline{F}. The composition in the top row is the map f_{t} of (3.6) for F, and in the bottom row is the f_{t} for L. Since the splitting map f: Hom_{c}(G\overline{F} , Δ/Γ_{F}) → SBr(F) is a direct sum of such f_{t} and likewise for L replacing F (using the same family of t’s) the commutativity of (3.7) for each t yields the commutativity of the diagram in (a).

(b) For each p, i let δ_{p,i} = v(s_{p,i}) = p^{1/j_{p,i}}γ_{p,i}. The Fundamental Inequality (3.1) shows that the field L_{p,i} = F (s_{p,i}) is totally ramified over F with Γ_{L_{p,i}} = ⟨δ_{p,i}⟩ + Γ_{F}. Since the relative value groups Γ_{L_{p,i}}/Γ_{F} are all independent in Δ/Γ_{F}, our field L, which is the compositum of all the L_{p,i}, is totally ramified over F with Γ_{L} = ∑_{p,i}⟨δ_{p,i}⟩ + Γ_{F}; so

\overline{L} = \overline{F}. We have Γ_{L}/Γ_{F} = ⨁_{p,i} (⟨δ_{p,i}⟩ + Γ_{F})/Γ_{F} with each summand (⟨δ_{p,i}⟩ + Γ_{F})/Γ_{F} ⊆

p^{−∞}γ_{p,i}/Γ_{F}. Since Δ/Γ_{F} = ⨁_{p,i} p^{−∞}γ_{p,i}/Γ_{F}, we have a compatible decomposition Δ/Γ_{L} ≅

⨁_{p,i} (p^{−∞}γ_{p,i}/Γ_{F})/(⟨δ_{p,i}⟩+Γ_{F}) ≅ ⨁_{p,i} p^{−∞}δ_{p,i}/Γ_{L}. Hence {δ_{p,i}| i ∈ I_{p}} maps to a ℤ/pℤ-base

of Γ_{L}/pΓ_{L}, so the s_{p,i}, with v(s_{p,i}) = δ_{p,i}, are a valid set to use for the splitting map of SBr(L). Because the canonical map Δ/Γ_{F} → Δ/Γ_{L} sends p^{−∞}γ_{p,i}/Γ_{F} onto p^{−∞}δ_{p,i}/Γ_{L}, it suffices for (b) to verify the commutativity of the diagrams

Hom_{c}(G\overline{F} , p^{−∞}γ_{p,i}/Γ_{F}) ⟶^{f_{p,i}} SBr(F)
↓  ^{res}↓
Hom_{c}(G\overline{L}, p^{−∞}δ_{p,i}/Γ_{L}) ⟶^{f′_{p,i}} SBr(L)

(3.8)

where the maps f_{p,i} and f′_{p,i} are as in (3.6). Of course G\overline{L} = G\overline{F} here, as \overline{L} = \overline{F}. To analyze (3.8), take any χ ∈ Hom_{c}(G\overline{F} , p^{−∞}γ_{p,i}/Γ_{F}), and let ψ be the image of χ in Hom_{c}(G\overline{L}, p^{−∞}δ_{p,i}/Γ_{L}). We have the diagram

\begin{matrix}H^{1}(G\overline{F} , p^{−∞}γ_{p,i}/Γ_{F}) & ⟶ & H^{2}(G\overline{F} , ℤγ_{p,i}) & ⟶ & H^{2}(G\overline{F} , ℤ) \\ ↓ & & ↓ & & _{p^{j_{p,i}}} ↓ \\ H^{1}(G\overline{L}, p^{−∞}δ_{p,i}/Γ_{L}) & ⟶ & H^{2}(G\overline{L}, ℤδ_{p,i}) & ⟶ & H^{2}(G\overline{L}, ℤ) \end{matrix}  (3.9)

where the middle vertical map arises from the inclusion ℤγ_{p,i} ↪ ℤδ_{p,i}, and the right vertical map is multiplication by p^{j_{p,i}}. In (3.9), the left rectangle is commutative since the horizontal maps are connecting homomorphisms arising from compatible short exact sequences of G\overline{F} -modules. The right rectangle in (3.9) is evidently commutative. The composition of the top (resp. bottom) maps in (3.9) is the θ of (3.5) for γ_{p,i} (resp. δ_{p,i}). So the commutativity of (3.9) shows that p^{j_{p,i}}θ(χ) = θ(ψ). Hence res_{L/F} (f_{p,i}(χ)) = res_{L/F} ((t_{p,i}) ∪ θ(χ)) = (s_{p,i}^{p^{j_{p,i}}}) ∪ θ(χ) = (s_{p,i}) ∪ p^{j_{p,i}}θ(χ) = (s_{p,i}) ∪ θ(ψ) = f′_{p,i}(ψ), proving the commutativity of (3.8), as desired. □

There is a further well-described part of Br(F) for F Henselian, which is what we get when we add in the tame totally ramified division algebras. A central division algebra T over our Henselian field F is said to be tame and totally ramified (TTR) if (with respect to the unique extension of the valuation v on F to T) |Γ_{T} : Γ_{F} | = [T : F] and char(\overline{F}) ∤ [T : F]. The theory of such division algebras is described in [TW]. In particular, it is known (see [Dr, Th. 1]) that every such T is isomorphic to a tensor product of symbol algebras (so lies in P S(F)), and that exp(T) = exp(Γ_{T} /Γ_{F}) (by [TW, Ex. 4.4(ii)]). The possible TTR algebras are thus constrained by the roots of unity in F. Here is a typical example of a TTR symbol algebra:

Example 3.4. Suppose μ_{n} ⊆ F^{∗}, char(\overline{F}) ∤ n, and s, t ∈ F^{∗} such that v(s) and v(t) generate a group of order n^{2} in Γ_{F} /nΓ_{F}. Then the symbol algebra T = (s, t; F)_{n} is TTR, with Γ_{T} = ⟨\frac{1}{n}v(s), \frac{1}{n}v(t)⟩ + Γ_{F} (see [TW, Prop. 3.5]).

A central division algebra D over a Henselian field F is said to be tame if D is split by the maximal tamely ramified extension F_{tr} of F. This field F_{tr} is the ramification field of the extension of v from F to F_{s}. It has the property that for any field L with F ⊆ L ⊆ F_{s}, L is tamely ramified over F iff L ⊆ F_{tr} (see [E, (20.7), (20.16)]). It is known [JW, Lemma 6.2] that if D is tame, then D ∼ S ⊗_{F} T in Br(F), where S is inertially split and T is TTR. (But the S and T are not uniquely determined, and it is not in general possible to express D ≅ S ⊗_{F} T with S inertially split and T TTR.) Let the tame part of Br(F) be denoted by

T Br(F) = {[D] ∈ Br(F)| D is tame} = Br(F_{tr}/F).

For the primary components of T Br(F) we have (see [HW, Prop. 4.3]) for every prime p,

T Br(F)(p) = \{ \begin{matrix}Br(F)(p) if p ≠ char(\overline{F}), \\ SBr(F)(p) if p = char(\overline{F}). \end{matrix}  (3.10)

There is a noncanonical splitting for the inclusion of SBr(F) in T Br(F), expressed in terms of the primary components as follows. Take any prime p ≠ char(\overline{F}), and as above take {t_{i} | i ∈ I_{p}} ⊆ F^{∗} with {v(t_{i})} mapping to a ℤ/pℤ-base of Γ_{F} /pΓ_{F}. Fix some total ordering on the index set I_{p}. For any n ∈ ℕ with μ_{p^{n}} ⊆ F^{∗}, let T_{p^{n}} be the subgroup of T Br(F) generated by the symbol algebras {(t_{i}, t_{j}; F)_{p^{n}}| i, j ∈ I_{p}, i < j}. As the proof of Prop. 3.5 below shows, T_{p^{n}} is a free ℤ/p^{n}ℤ-module, and these symbol algebras are a base.

If μ_{p^{n}} ⊆ F^{∗} for all n ∈ ℕ, then let T_{p^{∞}} = ⋃^{∞} ; n=1T_{p^{n}}.

Proposition 3.5. Fix any Henselian valued field F and any prime p ≠ char(\overline{F}). If r ∈ ℕ is maximal such that μ_{p^{r}} ⊆ F^{∗}, then Br(F)(p) = SBr(F)(p) ⊕ T_{p^{r}}. If μ_{p^{n}} ⊆ F^{∗} for every n, then Br(F)(p) = SBr(F)(p) ⊕ T_{p^{∞}}.

Proof. We have the exact sequence

0 ⟶ SBr(F)(p) ⟶ Br(F)(p) ⟶^{ρ} Br(F_{nr})(p),  (3.11)

where ρ is the restriction map. Assume first that r is maximal such that μ_{p^{r}} ⊆ F^{∗}. For D ∈ Br(F)(p), since p ≠ char(\overline{F}), D is tame (see (3.10)). So D ∼ S ⊗_{F} T in Br(F), where S ∈ SBr(F) and T is TTR, with S and T both p-primary. It follows that the division algebra T has degree a power of p. In a tensor decomposition of T into symbol algebras, each symbol algebra factor has degree p^{m} for some m with μ_{p^{m}} ⊆ F^{∗}, so m ≤ r. Hence,

exp(D ⊗_{F} F_{nr}) = exp(T ⊗_{F} F_{nr}) | exp(T) | p^{r}.

That is, im(ρ) ⊆ _{p^{r}}Br(F_{nr}).

We claim that _{p^{r}}Br(F_{nr}) is a free ℤ/p^{r}ℤ-module with base B = {(t_{i}, t_{j}; F_{nr})_{p^{r}} | i, j ∈ I_{p}, i < j}. For this, note first that every division algebra D ∈ _{p^{r}}Br(F_{nr}) is tame (see (3.10)), so TTR, since SBr(F_{nr}) = 0; so D is a tensor product of TTR symbol algebras. Since F_{nr}^{∗}/F_{nr}^{∗p^{r}} ≅ Γ/p^{r}Γ, {t_{i} | i ∈ I_{p}} maps to a base of the free ℤ/p^{r}ℤ-module F_{nr}^{∗}/F_{nr}^{∗p^{r}}. Hence, B is a generating set for _{p^{r}}Br(F_{nr}). To see ℤ/p^{r}ℤ-independence of B it suffices to check independence for {(t_{i}, t_{j}; F_{nr})_{p^{r}} | i, j ∈ J, i < j} for any finite subset J ⊆ I_{p}. Say J = {i_{1}, …, i_{ℓ}} with i_{1} < i_{2} < … < i_{ℓ}. Take any u = t_{i_{1}}^{m_{1}} … t_{i_{ℓ−1}}^{m_{ℓ−1}} with not all m_{j} ∈ p^{r}ℤ, and let A = (u, t_{i_{ℓ}}; F_{nr})_{p^{r}}. By induction on ℓ, it suffices to check that [A] is not in the subgroup S of _{p^{r}}Br(F) generated by {(t_{i}, t_{j}; F_{nr})_{p^{r}} | i, j ∈ {i_{1}, …, i_{i_{ℓ−1}}}, i < j}. But, if s is maximal such that p^{s} divides each of the m_{j}, then s < r and u = u_{0}^{p^{s}} with u_{0} ∉ F_{nr}^{∗p}. Then in Br(F_{nr}), A ∼ A_{0} = (u_{0}, t_{i_{ℓ}}; F_{nr})_{p^{r−s}}. Ex. 3.4 shows that A_{0} is a TTR division algebra with Γ_{A_{0}} = \frac{1}{p^{r−s}} ⟨v(u_{0}), v(t_{i_{ℓ}})⟩ + Γ_{F}. On the other hand, by [TW, Ex. 4.4(i)], every division algebra in S has value group lying in

∑_{j=1}^{ℓ−1} ⟨\frac{1}{p^{r}} v(t_{i_{j}})⟩ + Γ_{F}, and this group does not contain \frac{1}{p^{r−s}} v(t_{i_{ℓ}}). Thus, [A] ∉ S, and the

claim is proved.

Because _{p^{r}}Br(F_{nr}) is a free ℤ/p^{r}ℤ-module with base B, there is a well-defined group homomorphism h_{r}: _{p^{r}}Br(F_{nr}) → _{p^{r}}Br(F) given by (t_{i}, t_{j}; F_{nr})_{p^{r}} ↦ (t_{i}, t_{j}; F)_{p^{r}}. Clearly im(h_{r}) = T_{p^{r}} and ρ ∘ h_{r} = id on _{p^{r}}Br(F_{nr}); combined with what we have proved above, this shows that im(ρ) = _{p^{r}}Br(F_{nr}) and that h_{r} is a splitting map for (3.11). Thus Br(F)(p) = ker(ρ) ⊕ im(h_{r}) = SBr(F)(p) ⊕ T_{p^{r}}.

Now assume instead that μ_{p^{n}} ⊆ F^{∗} for every n ∈ ℕ. For each n, we have a homomorphism h_{n}: _{p^{n}}Br(F_{nr}) → _{p^{n}}Br(F) defined as above. Since h_{n} |_{_{p^{m}} Br(F_{nr})} = h_{m} for m < n, these maps are compatible, and we can take their union, obtaining h: Br(F_{nr})(p) → Br(F)(p) with ρ ∘ h = id. Thus h is a splitting map for (3.11), and Br(F)(p) = ker(ρ) ⊕ im(h) = SBr(F)(p) ⊕ T_{p^{∞}}, as desired. □

Remark 3.6. F. Chang tells us that for p odd, all the division algebras in T_{p^{r}} are TTR, whenever μ_{p^{r}} ⊆ F^{∗}. A proof of this is given in [C, Th. 2.3.2]. But this is not true in general for p = 2. For example if |I_{2}| ≥ 3, let T be the underlying division algebra of (t_{1}, t_{2}; F)_{2} ⊗ (t_{1}, t_{3}; F)_{2} ⊗ (t_{2}, t_{3}; F)_{2}. Then T ∼ (−1, t_{2}; F)_{2} ⊗ (t_{1}t_{2}, t_{2}t_{3}; F)_{2} in Br(F). If μ_{4} ⊈ F^{∗}, then this equivalence is an isomophism, and T is not TTR. But if μ_{4} ⊆ F^{∗}, then T ≅ (t_{1}t_{2}, t_{2}t_{3}; F)_{2} which is TTR.

4. Projective Schur groups of Henselian valued fields

Let F be an equicharacteristic Henselian valued field with residue field k. (Equicharacteristic means char(k) = char(F).) In this section we show how P S(F) is related to P S(k). We show that if every projective Schur algebra over k is a radical (resp. radical abelian) algebra, then the same is true for projective Schur algebras over F. The results here generalize those in [AS_{6}], which treated the case where F = k((t_{1}))...((t_{n})).

Let F be any field, and let K be a finite radical field extension of F, i.e., K = F (U) where U is a subgroup of K^{∗} with U ⊇ F^{∗} and U/F^{∗} is finite. If L is an intermediate field, F ⊆ L ⊆ K, and L is abelian Galois over F, then it is shown in [AS_{2}, Prop. 2.1] that L ⊆ M where M is a compositum of a finite cyclotomic extension of F and a finite Kummer extension of F. In particular, if L is a radical abelian extension of F, then L lies in such an M. Let F_{ra} denote the maximal radical abelian extension of F, i.e., the compositum F_{ra} = F_{cyc} · F_{kum}, where F_{cyc} is the maximal cyclotomic extension of F and F_{kum} is the maximal Kummer extension of F. (The notation F_{radab} was used for F_{ra} in [AS_{6}].)

Now, suppose F has a Henselian valuation, with residue field k. As in §3 we write F_{nr} for the maximal unramified extension of F; SBr(F) for Br(F_{nr}/F); and F (k_{ra}) for the unramified extension of F with residue field k_{ra}. We have,

F (k_{ra}) = F_{nr} ∩ F_{ra} .  (4.1)

To see this, note first that the maximal unramified extension of F in F_{kum} is F (k_{kum}). Therefore, F_{nr} ∩ F_{ra} = F_{nr} ∩ (F_{cyc} · F_{kum}) = F_{cyc} · (F_{nr} ∩ F_{kum}) = F (k_{cyc}) · F (k_{kum}) = F (k_{ra}), where the second equality is immediate from the corresponding equality of associated subgroups of the absolute Galois group of F, as F_{nr} is Galois over F.

Proposition 4.1. (cf. [AS_{6}, Th. 2.3]) Let F be a Henselian valued field with residue field k, and assume char(F) = char(k). Then P S(F) ∩ SBr(F) ⊆ Br(F (k_{ra})/F).

Proof. Let A ∈ P S(F) ∩ SBr(F), say A = F (G), where G is a subgroup of A^{∗} spanning A as an F -vector space, with F^{∗} ⊆ G and |G/F^{∗}| < ∞. Assume A is reduced. Let H = G′ (the derived group of G, a finite group), and as in Prop. 2.2 let B = F (H), L = Z(B), Ĥ = C_{G}(L), and E = F (Ĥ) = C_{A}(L) = B ⊗_{L} T, where T = C_{E}(B) = L(T ), where T = C_{ĤB^{∗}}(B). So, L is a field, and as B is a Schur algebra, by the Brauer splitting theorem ([CR, pp. 385, 418]) L ⊆ F_{cyc} and F_{cyc} splits B. Note that F_{cyc} = F (k_{cyc}) ⊆ F_{nr}, since F is Henselian and char(k) = char(F), so that F and k contain “the same” roots of unity. Because A ∈ SBr(F) and E = C_{A}(L) ∼ L ⊗_{F} A in Br(L), we have E ∈ SBr(L). Since E = B ⊗_{L} T and B ∈ SBr(L), we must also have T ∈ SBr(L). By Prop. 2.2(d), T = L(T ) is a projective Schur algebra of abelian type over L. Therefore, if we let Λ = T /L^{∗}, Prop. 2.1 shows that we have the nondegenerate symplectic pairing B_{T} : Λ × Λ → μ(L) induced by commutators of elements of T . Let n = exp(Λ). Since μ_{n} = im(B_{T}) ⊆ F^{∗}, we have char(F) ∤ n. Therefore, char(k) = char(F) ∤ |Λ| = dim_{L}(T), using Prop. 2.1(c).

Let v: F^{∗} ⟶ Γ be our Henselian valuation on F, where Γ is the value group of v, and let Δ = ℚ ⊗_{ℤ} Γ, the divisible hull of Γ. Because commutators of elements of T are roots of unity, the valuation v induces a well-defined group homomorphism w: Λ ⟶ Δ/Γ given by w(tL^{∗}) = \frac{1}{n}v(t^{n}) + Γ, where n = exp(Λ). Let Λ_{0} = ker(w) ⊆ Λ. Because T = L(T ) is a projective Schur algebra of abelian type, the same is true for L_{nr} ⊗_{L} T = L_{nr}(T̃), where T̃/L_{nr}^{∗} is the image of T /L^{∗} under the canonical injection T^{∗}/L^{∗} ↪ (L_{nr} ⊗_{L} T)^{∗}/L_{nr}^{∗}. Let Λ̃ = T̃/L_{nr}^{∗}, and let BT̃: Λ̃ × Λ̃ → μ(L_{nr}^{∗}) be the associated nondegenerate pairing. Since L_{nr} is unramified over L, and hence over F, the value group of L_{nr} is Γ. Let w̃: Λ̃ → Δ/Γ be the homomorphism defined the same way as w, and let Λ̃_{0} = ker(w̃). The isomorphism j: Λ → Λ̃ is clearly an isometry between the pairings B_{T} and BT̃; also, clearly w = w̃ ∘ j. By [TW, Th. 4.3] L_{nr} ⊗_{L}T is similar to a tame totally ramified division algebra D over L_{nr} with relative value group Γ_{D}/Γ ≅ Λ̃_{0}^{⊥}/(Λ̃_{0} ∩ Λ̃_{0}^{⊥}). But, we saw above that T ∈ SBr(L). Hence, D = L_{nr}, so Λ̃_{0}^{⊥}/(Λ̃_{0} ∩ Λ̃_{0}^{⊥}) must be trivial, i.e., Λ̃_{0}^{⊥} ⊆ Λ̃_{0}. From the isometry between B_{T} and BT̃ it follows that Λ_{0}^{⊥} ⊆ Λ_{0}. Therefore, by the comments preceding Prop. 2.4, there is a Lagrangian Λ_{1} of Λ with Λ_{0}^{⊥} ⊆ Λ_{1}, so Λ_{1} = Λ_{1}^{⊥} ⊆ Λ_{0}^{⊥⊥} = Λ_{0}. Let L_{1} = L(L_{1}), where Λ_{1} = L_{1}/L^{∗} ⊆ T /L^{∗}. By Prop. 2.4(a), L_{1} is a field which lies in a radical abelian extension of F, so L_{1} ⊆ F_{ra}. Note that for any t ∈ L_{1}, we have t^{n} ∈ L^{∗} as exp(Λ) = n, and v(t^{n}) ∈ nΓ as tL^{∗} ∈ Λ_{1} ⊆ Λ_{0} = ker(w). So, there is ℓ ∈ L^{∗} with v((tℓ)^{n}) = 0. Since L_{1} is generated over L by valuation units of n-th roots of elements of L and char(k) ∤ n, we have L_{1} is a compositum of unramified extensions of L, hence L_{1} is unramified over L. Because L is unramified over F, it follows that L_{1} ⊆ F_{nr}. Thus, L_{1} ⊆ F_{nr} ∩ F_{ra} = F (k_{ra}), with the last equality given by (4.1). Prop. 2.4(a) shows that for E_{1} = C_{A}(L_{1}) we have E_{1} = B ⊗_{L} L_{1}. So E_{1} is a Schur algebra over L_{1}, since B is a Schur algebra over L. Hence, there is a root of unity ω such that L_{1}(ω) splits E_{1}. Then L_{1}(ω) splits A as L_{1} ⊗_{F} A ∼ E_{1} in Br(L_{1}); also L_{1}(ω) ⊆ F (k_{ra}) since L_{1} ⊆ F (k_{ra}). So, A ∈ Br(F (k_{ra})/F), as desired. □

In order to relate unramified phenomena over a Henselian field to corresponding phenomena over the residue field, one often uses the valuation ring V as a bridge. To employ that bridge here for projective Schur algebras, we need to know about the structure of “tame” twisted group rings over V. The next proposition gives what is needed. It may be of some interest in its own right.

Proposition 4.2. Let V be a Henselian valuation ring, M its maximal ideal, F its quotient field, and k = V/M its residue field. Let V^{z}G be a twisted group ring over V, where z ∈ H^{2}(G, V^{∗}) and G is a finite group acting trivially on V. Suppose char(k) ∤ |G|. Then V^{z}G = ⨁_{i=1}^{m} S_{i}, where each S_{i} is an Azumaya algebra over Z(S_{i}), and Z(S_{i}) is a

valuation ring unramified over V. For the corresponding twisted group ring k\overline{z}G over k, we have k\overline{z}G ≅ ⨁_{i=1}^{m} S_{i}/MS_{i}, with each S_{i}/MS_{i} a simple algebra with center the residue

field of Z(S_{i}).

Proof. Let R = V^{z}G. We have R/MR ≅ k\overline{z}G, where \overline{z} is the image of z in H^{2}(G, k^{∗}). Because char(k) ∤ |G| it is known by [P, p. 30, Th. 4.2] that k\overline{z}G is a semisimple k-algebra. So, Z(k\overline{z}G) = ⨁_{i=1}^{m} L_{i}, where each L_{j} is a field with [L_{j} : k] < ∞. Moreover, we can see

that each L_{j} must be separable over k. For, L_{j} ⊗_{k} k\overline{z}G ≅ L\overline{z} ; jG, which is semisimple; so

its center ⨁_{i=1}^{m} L_{j} ⊗_{k} L_{i} is a direct sum of fields. Since L_{j} ⊗_{k} L_{j} is a direct sum of fields,

L_{j} must be separable over k. Consequently, k\overline{z}G is a separable algebra over k. Because R/MR is a separable V/M-algebra, it follows by [DI, p. 72, Th. 7.1] that R is a separable V -algebra. Let Z = Z(R). Then, R is a central separable algebra (= Azumaya algebra) over Z and Z is separable over V by [DI, p. 55, Th. 3.8]. Hence, Z is a direct summand of R as a Z-module, so as a V -module, by [DI, p. 51, Lemma 3.1]. Therefore, as R is a free V -module of rank |G| < ∞, Z is a finitely generated projective V -module, hence a free V -module as V is local. Now, R/MR ≅ R ⊗_{Z} (Z/MZ), which is a central separable Z/MZ-algebra (see [DI, p. 61, Lemma 5.1]); so Z/MZ ≅ Z(R/MR) ≅ ⨁_{i=1}^{m} L_{i}.

Suppose first that Z/MZ is a field, i.e., m = 1. Now, Z ⊗_{V} F is a commutative separable F -algebra by [DI, p. 44, Cor. 1.7], so Z ⊗_{V} F = ⨁_{j=1}^{n} K_{j}, where each K_{j} is a field separable

over F. Let W_{j} be the unique (as V is Henselian) valuation ring of K_{j} extending V; so, W_{j} is the integral closure of V in K_{j}. Let T = ⨁_{j=1}^{n} W_{j}, which is the integral closure of V

in ⨁_{j=1}^{n} K_{j}. When we view Z ⊆ Z ⊗_{V} F = ⨁_{j=1}^{n} K_{j}, we have Z ⊆ T, since Z is integral over

V. Because MZ was assumed to be a maximal ideal of Z, we have MT ∩ Z = MZ, so Z/MZ ⊆ T/MT. Thus,

dim_{k}(T/MT) ≥ dim_{k}(Z/MZ) = rk_{V} (Z) = dim_{F} (Z ⊗_{V} F)

= ∑_{j=1}^{n}[K_{j} : F] ≥ ∑_{j=1}^{n} dim_{k}(W_{j}/MW_{j}) = dim_{k}(T/MT).

See [Bou, Ch. VI, § 8, No. 5, Cor. to Prop. 5] for the last inequality here. Therefore, equality holds throughout. Hence, T/MT = Z/MZ ≅ L_{1}, which is a field separable over k. So, n = 1, T = W_{1}, and W_{1}/MW_{1} = T/MT ≅ L_{1}. So, MW_{1} is the maximal ideal of W_{1}, and dim_{k}(W_{1}/MW_{1}) = [K_{1} : F]. Hence, W_{1} is unramified over V. So, W_{1} is a finitely generated V -module (see [Bou, Ch. VI, § 8, No. 5, Th. 2]); since dim_{k}(Z/MZ) = dim_{k}(W_{1}/MW_{1}), Nakayama’s Lemma shows that Z = W_{1}, as desired.

Now drop the assumption that Z/MZ is a field. We still have Z/MZ ≅ ⨁_{i=1}^{m} L_{i}, where each L_{i} is a field separable over k. Let ẽ_{1}, …, ẽ_{m} be the primitive orthogonal idempotents of Z/MZ, with the ẽ_{i} numbered so that each ẽ_{i}(Z/MZ) ≅ L_{i}. Because Z is a finitely generated module over the Henselian local ring V, the ẽ_{i} lift to pairwise orthogonal idempotents e_{1}, …, e_{m} ∈ Z with each ẽ_{i} the image of e_{i} in Z/MZ, and e_{1}+…+e_{m} = 1, by [Az, Th. 24] or [MMU, p. 180, Th. A.18]. Then, Z = ⨁_{i=1}^{m} Z_{i}, where each Z_{i} = e_{i}Z. Since each Z_{i} is a direct summand of Z, Z_{i} is a finitely generated projective, hence free, V -module. Moreover, each Z_{i} is a separable V -algebra with Z_{i}/MZ_{i} = e_{i}Z/Me_{i}Z ≅ ẽ_{i}(Z/MZ) ≅ L_{i}, which is a field separable over k. Therefore, the argument of the preceding paragraph, applied to Z_{i}, shows that Z_{i} is a valuation ring unramified over V. The e_{i} are orthogonal central idempotents of R; so R = ⨁_{i=1}^{m} S_{i}, where S_{i} = e_{i}R. Since R is central separable over Z, each S_{i} is central separable over e_{i}Z = Z_{i}, i.e., S_{i} is an Azumaya algebra over Z_{i}, which is a valuation ring unramified over V. Also, k\overline{z}G ≅ R/MR = ⨁_{i=1}^{m} S_{i}/MS_{i}. Each

S_{i}/MS_{i} ≅ S_{i} ⊗_{Z_{i}} (S_{i}/MZ_{i}), so S_{i}/MS_{i} is an Azumaya algebra over the field Z_{i}/MZ_{i}, i.e., a simple algebra with center Z_{i}/MZ_{i}, which is the residue field of Z_{i}. □

For the rest of this section, we adopt the following standing hypotheses:

F is a field with Henselian valuation v with residue field k, with char(k) = char(F).

We write V for the valuation ring of v; Γ = Γ_{F} for the value group of v; and Δ = ℚ ⊗_{ℤ} Γ. We routinely identify Br(k) with its canonical image in Br(F) via the map of (3.3) above. Also, recall that C′ denotes the prime to p part of a torsion abelian group C associated to F if char(F) = p ≠ 0, while C′ = C if char(F) = 0.

Proposition 4.3. (cf. [AS_{6}, Prop. 2.6]) P S(F)′ ∩ Br(k) = P S(k)′.

Proof. Recall how the canonical inclusion Br(k) ↪ Br(F) can be obtained (cf. [JW, Th. 5.6(a), Th. 2.8]): The map θ: Br(V) → Br(k) given by [A] ↦ [A ⊗_{V} k] is an isomorphism, as V is Henselian. The map φ: Br(V) → Br(F) given by [A] ↦ [A ⊗_{V} F] is injective as V is a valuation ring. The inclusion Br(k) ↪ Br(F) is φ ∘ θ^{−1}. Of course if V contains a coefficient field, which is a subfield k_{0} ⊆ V, which maps isomorphically onto k under the composition k_{0} ↪ V → V/M = k (with M the maximal ideal of V), then we can identify k with k_{0}, and the map Br(k) ↪ Br(F) is given by scalar extension [A] ↦ [A ⊗_{k} F]. If char(k) = 0, then there always is a coefficient field, as V is Henselian. This is provable in the same way as for a complete discrete valuation ring (see e.g. [ZS, p. 280, Cor. 2]), since only Hensel’s Lemma is used. If char(k) = char(F) = p ≠ 0, then there may not always exist a coefficient field; if k is separably generated over its prime field, then there is a coefficient field.

The inclusion P S(k)′ ⊆ P S(F)′∩Br(k) is now clear if char(k) = 0, since then V contains a coefficient field. On the other hand, if char(k) = p ≠ 0, then we have the description in [AS_{4}, Th. 1.4] of algebras in P S(k)′ as tensor products of symbol algebras together with a cyclic algebra (ℓ/k, σ, a), with ℓ ⊆ k_{cyc}. Every such algebra has an obvious lift to an Azumaya algebra of the same type over V, which then extends by − ⊗_{V} F to a central simple algebra of the same type over F. So again it is clear that P S(k)′ ⊆ P S(F)′ ∩Br(k).

For the reverse inclusion, the argument is based on that in [AS_{6}, Prop. 2.6], but adapted to work even if V does not contain a coefficient field, and to apply for an arbitrary value group.

Let α ∈ Br(k)′ with image α̃ in Br(F)′ which lies in P S(F)′. Let Ã = F (G) be a projective Schur algebra over F representing α̃, where G is a subgroup of Ã^{∗} with F^{∗} ⊆ G, G spans Ã as an F -vector space, and |G/F^{∗}| < ∞. Let G = G/F^{∗}. By [AS_{6}, Lemma 2.5], we may assume that |G| is prime to p if char(F) = p ≠ 0. We have the central extension

1 ⟶ F^{∗} ⟶ G ⟶ G ⟶ 1.

Denote by z ∈ H^{2}(G, F^{∗}) the cohomology class of this extension (with G acting trivially on G^{∗}). There is a corresponding surjective F -algebra homomorphism,

η: F^{z}G ⟶ F (G),

where F^{z}G is the group algebra twisted by z.

Suppose for now that z lies in the image of the map H^{2}(G, V^{∗}) → H^{2}(G, F^{∗}). Let V^{z}G be the corresponding twisted group ring over V. We have V^{z}G = ⨁_{i=1}^{m} S_{i}, as in

Prop. 4.2. So, F^{z}G ≅ F ⊗_{V} V^{z}G ≅ ⨁_{i=1}^{m}(F ⊗_{V} S_{i}). Each F ⊗_{V} S_{i} is an Azumaya algebra

(= central simple algebra) over the field F ⊗_{V} Z(S_{i}), which is the quotient field of the valuation ring Z(S_{i}). The surjection η above shows that one of the simple summands F ⊗_{V} S_{j} of F^{z}G is isomorphic to F (G). By comparing centers, we find F ⊗_{V} Z(S_{j}) = F, so Z(S_{j}) = Z(S_{j}) ∩ F = V. Let A = k ⊗_{V} S_{j} ≅ S_{j}/MS_{j}. Then, A is a central simple k-algebra which by Prop. 4.2 is a direct summand of k^{z}G. Hence, A is a projective Schur algebra over k. Since [A] maps to [Ã] under the injective map Br(k) → Br(F), we have [A] = α ∈ Br(k), showing that α ∈ P S(k)′, as desired.

So far we have assumed that z was in the image of H^{2}(G, V^{∗}) ⟶ H^{2}(G, F^{∗}). Now drop this assumption. The short exact sequence of trivial G-modules 1 → V^{∗} → F^{∗} → Γ → 1 yields the exactness of H^{2}(G, V^{∗}) → H^{2}(G, F^{∗}) → H^{2}(G, Γ). Since Δ is a uniquely divisible group, we have H^{2}(G, Γ) ≅ H^{1}(G, Δ/Γ) = Hom(G, Δ/Γ) (cf. the comments preceding (3.3) above). Hence we have an exact sequence H^{2}(G, V^{∗}) → H^{2}(G, F^{∗}) → Hom(G, Δ/Γ). This is clearly functorial in F, i.e., for any field K algebraic over F, we have a commutative diagram with exact rows:

H^{2}(G, V^{∗}) ⟶ H^{2}(G, F^{∗}) ⟶ Hom(G, Δ/Γ)
↓  ↓  ↓
H^{2}(G, V_{K}^{∗}) ⟶ H^{2}(G, K^{∗}) ⟶ Hom(G, Δ/Γ_{K})

Let ψ ∈ Hom(G, Δ/Γ) be the image of z ∈ H^{2}(G, F^{∗}). Then, im(ψ) is a finite subgroup of Δ/Γ. Let K be a totally ramified finite degree extension of F such that im(ψ) ⊆ Γ_{K}/Γ. Then ψ maps to 0 in Hom(G, Δ/Γ_{K}), so the commutative diagram shows that the image z_{1} of z in H^{2}(G, K^{∗}) lies in the image of H^{2}(G, V_{K}^{∗}). Thus the preceding argument applies over K (we work with K^{z_{1}}G = K ⊗_{F} F^{z}G, which maps to the central simple K-algebra K(G_{1}) = K ⊗_{F} F (G), where G_{1} = K^{∗}G). The argument shows that α̃_{K} lies in the image of P S(\overline{K})′ in Br(K), where \overline{K} is the residue field of V_{K}. This α̃_{K} is the image in Br(K) of the image α\overline{K} of α in Br(\overline{K}), by the commutative diagram (3.4) with K replacing L; so α\overline{K} ∈ P S(\overline{K})′. But \overline{K} ≅ \overline{F} = k, as K is totally ramified over F. Hence α ∈ P S(k)′, as desired. □

Theorem 4.4. Assuming the standing hypotheses stated above, let SBr(F) denote the inertially split part of Br(F) and let SP S(F) = P S(F) ∩ SBr(F). Then there is a split exact sequence:

0 ⟶ P S(k)′ ⟶^{j} SP S(F)′ ⟶^{η} Hom_{c}(G(k_{ra}/k), Δ/Γ)′ ⟶ 0

If k is perfect, then the above sequence is split exact without (−)′.

Proof. Since the valuation v is Henselian, F (k_{ra}) ⊆ F_{ra}, and v extends uniquely to F (k_{ra}). Moreover since F (k_{ra}) is an unramified extension of F, the value groups Γ_{F (k_{ra})} and Γ_{F} are equal. Applying Th. 3.2 to F and F (k_{ra}) we obtain (using the functoriality noted in (3.4)) a commutative and exact diagram

0 ⟶ Br(k_{ra})′  ⟶^{i_{ra}} SBr(F (k_{ra}))′  ⟶^{π_{ra}} Hom_{c}(G_{k_{ra}}, Δ/Γ)′ ⟶ 0
res↑  ^{res}↑  ^{res}↑
0 ⟶ Br(k)′  ⟶^{i}  SBr(F)′  ⟶^{π}  Hom_{c}(G_{k}, Δ/Γ)′ ⟶ 0
↑  ↑  ↑
0 ⟶ Br(k_{ra}/k)′ ⟶^{res(i)} SBr(F (k_{ra})/F)′ ⟶^{res(π)} Hom_{c}(G(k_{ra}/k), Δ/Γ)′ ⟶ 0
↑  ↑  ↑
0  0  0

The third row is split exact because the first two rows are split exact with compatible splitting maps by the proof of Prop. 3.3(a). Now P S(k)′ ⊆ Br(k_{ra}/k)′ [AS_{2}, Cor. 2.3], and by Prop. 4.1, SP S(F)′ ⊆ Br(F (k_{ra})/F)′. Moreover by Prop. 4.3 we have P S(k)′ ⊆ P S(F)′ ∩ Br(k)′, hence P S(k)′ ⊆ P S(F)′ ∩ SBr(F)′ = SP S(F)′. We therefore obtain a commutative diagram

0 ⟶ Br(k_{ra}/k)′ ⟶^{res(i)} SBr(F (k_{ra})/F)′ ⟶^{res(π)} Hom_{c}(G(k_{ra}/k), Δ/Γ)′ ⟶ 0
↑  ↑  ↑
0 ⟶ P S(k)′  ⟶^{j}  SP S(F)′  ⟶^{η} Hom_{c}(G(k_{ra}/k), Δ/Γ)′ ⟶ 0

where the top row is split exact and the bottom row is exact at P S(k)′. Again by Prop. 4.3 we have P S(k)′ ⊇ P S(F)′ ∩ Br(k)′ = P S(F)′ ∩ Br(k_{ra}/k)′ so the bottom row is exact at SP S(F)′. To complete the proof of the first assertion of the theorem, we prove that the splitting map f: Hom(G(k_{ra}/k), Δ/Γ)′ ⟶ SBr(F (k_{ra})/F)′ takes values in P S(F)′ hence in SP S(F)′. Indeed, for χ ∈ Hom_{c}(G(k_{ra}/k), Δ/Γ)′, f(χ) is a tensor product of cyclic algebras of the form (F (E)/F, σ, t), where E is a cyclic extension of k lying in k_{ra} with [F (E) : F] = [E : k] dividing the order of χ. Since F (E) ⊆ F_{ra} each such cyclic algebra is a radical abelian algebra of degree not a multiple of char(F), so lies in SP S(F)′.

For the second assertion of the theorem, we may assume char(k) = p ≠ 0, and we need only prove the assertion for the p-primary components. By (3.3) we have an exact sequence

0 ⟶ Br(k) ⟶ SBr(F) ⟶ Hom_{c}(G_{k}, Δ/Γ) ⟶ 0.

Since the p-power map is an automorphism of k (k is perfect of characteristic p), Br(k)(p) = 0. Hence SBr(F)(p) = Hom_{c}(G_{k}, Δ/Γ)(p). Similarly, since k has no nontrivial Kummer pextensions, SBr(F (k_{ra}))(p) = SBr(F (k_{cyc}))(p) = Hom_{c}(G_{k_{cyc}}, Δ/Γ)(p), and we get a commutative diagram

SBr(F)(p) ⟶^{≅} Hom_{c}(G_{k}, Δ/Γ)(p)
res↓  ^{res}↓
SBr(F (k_{cyc}))(p) ⟶^{≅} Hom_{c}(G_{k_{cyc}}, Δ/Γ)(p)

It follows that the corresponding kernels of the restriction maps are isomorphic:

SBr(F (k_{cyc})/F)(p) ⟶^{≅} Hom_{c}(G_{k_{cyc}}, Δ/Γ)(p).

But by Prop. 4.1, SP S(F)(p) ⊆ SBr(F (k_{cyc})/F)(p). Furthermore, using the splitting map, we see that the map SP S(F)(p) ⟶ Hom(G_{k_{cyc}}, Δ/Γ)(p) is surjective, and the result follows. □

For an equicharacteristic Henselian field F, for any prime p ≠ char(F), Th. 3.2 and Prop. 3.5 yield a direct sum decomposition

Br(F)(p) = Br(k)(p) ⊕ f(Hom_{c}(G_{k}, Δ/Γ)(p)) ⊕ T ,  (4.2)

where f: Hom_{c}(G_{k}, Δ/Γ) ⟶ SBr(F) is the (injective) splitting map of Th. 3.2 above and T = T_{p^{r}} if r is maximal such that μ_{p^{r}} ⊆ k, and T = T^{p^{∞}} if there is no such r. We can now see that there is a compatible direct sum decomposition of P S(F)(p). For this, let k_{ra,p} be the maximal p-extension of k in the abelian Galois extension k_{ra}. Then, Proposition 4.5. For any prime p ≠ char(k),

P S(F)(p) = P S(k)(p) ⊕ f(Hom_{c}(G(k_{ra,p}/k), Δ/Γ)) ⊕ T ,

where f and T, are as in (4.2) above.

Proof. Since T is generated by symbol algebras, T ⊆ P S(F)(p). Therefore, it suffices to see that SP S(F)(p) = P S(k)(p)⊕f(Hom_{c}(G(k_{ra,p}/k, Δ/Γ)). But this is clear from Th. 4.4 since G(k_{ra,p}/k) is the p-part of the abelian profinite group G(k_{ra}/k). □

Corollary 4.6. Suppose char(k) = char(F) = 0. If every element of P S(k) is represented by a radical (resp. radical abelian) algebra, then the same holds for P S(F).

Proof. We recall first [AS_{3}, Lemma 2.4] that the tensor product of radical algebras is Brauer equivalent to a radical algebra, and the tensor product of radical abelian algebras is Brauer equivalent to a radical abelian algebra. Now as char(k) = 0 every element of Br(F) is tame, so is represented by a tensor product S ⊗_{F} T, where [S] ∈ SBr(F) and T is a tensor product of symbol algebras (which are clearly radical abelian algebras). We therefore need to prove the assertion for P S(F)∩SBr(F) = SP S(F). But this follows easily from Theorem 4.4 and the fact that every element of Hom_{c}(G(k_{ra}/k), Δ/Γ) is represented by a tensor product of cyclic radical abelian algebras. □

Corollary 4.7. Suppose the residue field k is a local or global field. Then every projective Schur algebra over F is Brauer equivalent to a radical abelian algebra.

Proof. If char(F) = 0, the result follows from the preceding corollary since it holds for k, by [O,(4.3)]. (To complete the argument in [O] for the case p = 2 one can observe that for any number field k, the field k(μ_{2^{∞}}) obtained by adjoining to k all 2^{n}-th roots of unity for all n contains nonreal cyclic extensions of arbitrary 2-power degree.) If char(F) ≠ 0, the result holds because it holds for any field of characteristic not zero [AS_{4}, Cor. 1.5]. □

5. Projective Schur groups of Henselian fields as algebraic relative Brauer groups

In this section we prove

Theorem 5.1. Let F be a field with Henselian valuation v, value group Γ = Γ_{F}, and residue field k. Assume k is a local or global field and that char(k) = char(F). Then P S(F) is an algebraic relative Brauer group Br(M/F) with M/F a radical abelian extension.

Proof. We consider the cases k = ℝ and k = ℂ separately. In the case k = ℂ, we have from Prop. 3.5 that for any prime p,

Br(F)(p) = SBr(F)(p) ⊕ T_{p^{∞}},

and since SBr(F)(p) = 0, Br(F)(p) = T_{p^{∞}} ⊆ P S(F), so Br(F) = P S(F) = Br(F_{kum}/F) and we are done.

We turn next to the case k = ℝ. Then the maximal unramified extension of F has residue field ℂ, so F_{nr} = F (\sqrt{−1}). Hence, SBr(F) = Br(F (\sqrt{−1})/F), a 2-torsion group. By Prop. 3.5, for p ≠ 2, Br(F)(p) = 0, so Br(F) = Br(F)(2) ≅ SBr(F)(2) ⊕ T_{2}. Each summand is generated by quaternion algebras, so P S(F) = Br(F) = Br(F_{kum}/F) and we are done.

We now assume that k is either a nonarchimedean local field or a global field.

Let p be a prime number. We will prove the theorem for p-primary components. More precisely, we will prove that P S(F)(p) = Br(M_{p}/F) with M_{p}/F a radical abelian pextension. It then follows that P S(F) = Br(M/F) with M equal to the composite of all the M_{p}.

For a given Galois extension E/k, we set X(E/k) := Hom_{c}(G(E/k), ℚ/ℤ), the character group of E/k, written additively. When G(E/k) is abelian there is an isomorphism between the lattice of intermediate fields K, k ⊆ K ⊆ E and the lattice of subgroups of X(E/k) given by K ↔ X(K/k). If m ∈ ℕ, let E^{(m)} denote the subfield of E corresponding to the subgroup mX(E/k). So, E^{(m)} is the smallest subfield of E containing k such that G(E/E^{(m)}) is m-torsion. Since k_{ra} = k_{cyc}k_{kum} (see §4), we have X(k_{ra}/k) = X(k_{cyc}/k) + X(k_{kum}/k). It follows that mX(k_{ra}/k) = mX(k_{cyc}/k) + mX(k_{kum}/k). Note that mX(k_{cyc}/k) = X(k_{cyc}/k) when char(k) ≠ 0 since X(k_{cyc}/k) is divisible in this case.

Let m = p^{r} be the number of p-power roots of unity in k (or equivalently in F). Then m is finite. Denote by k_{ra,p} the p-part of k_{ra}, i.e., the maximal p-extension of k contained in k_{ra}. Similarly let k_{cyc,p} and k_{kum,p} denote the p-parts of k_{cyc} and k_{kum}, respectively.

Proposition 5.2. Let F be a field with Henselian valuation v, value group Γ, and residue field k. Let p ≠ char(k) be a prime number. Set m = p^{r} with r maximal such that k contains the p^{r}th roots of unity. Set L := k_{ra,p}^{(m)}. Assume P S(k)(p) = Br(L/k). Then P S(F)(p) = Br(M_{p}/F), where M_{p} = F (L)( \sqrt[m]{t_{i}} | i ∈ I_{p}), where v(t_{i}) = γ_{i} (i ∈ I_{p}), and \{\overline{γ_{i}} | i ∈ I_{p}\} is a ℤ/pℤ-base of Γ/pΓ. (\overline{γ_{i}} = γ_{i} + pΓ.)

Proof. As usual, let Δ = ℚ ⊗_{ℤ} Γ. For fields E ⊇ K ⊇ k with E/K Galois, let Y (E/K) = Hom_{c}(G(E/K), Δ/Γ). Note that Y (E/K) has similar functorial properties to X(E/K). Indeed, since Δ/Γ(p) ≅ ⊕_{i∈I_{p}} ℚ/ℤ(p), we have

Y (E/K)(p) = Hom_{c}(G(E/K), Δ/Γ(p)) ≅ ⊕_{i∈I_{p}} Hom_{c}(G(E/K), ℚ/ℤ(p)) = ⊕_{i∈I_{p}} X(E/K)(p).

The isomorphism holds because every continuous homomorphism from a profinite group to a discrete group has finite image. Clearly, the direct sum decomposition is compatible with the canonical inclusion Y (N/K) ↪ Y (M/K) for fields K ⊆ N ⊆ M.

We compute Br(M_{p}/F) using the decomposition

Br(F)(p) = Br(k)(p) ⊕ f(Y (k_{s}/k)(p)) ⊕ T_{p^{r}}

in (4.2). Clearly T_{p^{r}} ⊆ Br(M_{p}/F) since each generator of T_{p^{r}} has a maximal subfield in M_{p}. Now, take any α ∈ Br(k(p)) and χ ∈ Y (k_{s}/k)(p). By Prop. 3.3(a) (for res_{F (L)/F}) and 3.3(b) (for res_{M_{p}/F (L)}), we have α+f(χ) ∈ Br(M_{p}/F) iff α and f(χ) each lie in Br(M_{p}/F), iff α ∈ Br(L/k) (as \overline{M_{p}} = L) and χ maps to 0 in Hom_{c}(G_{L}, Δ/Γ_{M_{p}}). Since Γ_{M_{p}}/Γ = \frac{1}{m}Γ/Γ is the m-torsion subgroup of Δ/Γ, the condition on χ is equivalent to: 0 = mres_{L/k}(χ) =

res_{L/k}(mχ), i.e., mχ ∈ Y (L/k). Now, Y (L/k) ≅ ⊕_{i∈I_{p}} X(L/k) = ⊕_{i∈I_{p}} mX(k_{ra,p}/k) ≅

mY (k_{ra,p}/k). Hence, mχ ∈ Y (L/k) iff mχ = mψ with ψ ∈ Y (k_{ra,p}/k). We claim that this last equality holds iff χ ∈ Y (k_{ra,p}/k). “If” is clear, taking ψ = χ. Conversely, suppose mχ = mψ. Since G_{k}/ ker(χ − ψ) ≅ im(χ − ψ) which is m-torsion, the fixed field of ker(χ − ψ) is an m-Kummer extension of k. So, χ − ψ ∈ Y (k_{kum,p}/k) ⊆ Y (k_{ra,p}/k). Then, χ = ψ + (χ − ψ) ∈ Y (k_{ra,p}/k), proving the claim. We have thus shown that Br(M_{p}/F) = Br(L/k) ⊕ f(Y (k_{ra,p}/k)) ⊕ T_{p^{r}}. Since Br(L/k) = P S(k)(p) by hypothesis, Prop. 4.5 shows that Br(M_{p}/F) = P S(F)(p). This completes the proof of Prop. 5.2. □

We now apply Prop. 5.2 to prove Th. 5.1. We will handle separately below an exceptional case when k is a number field which is not totally imaginary and p = 2. If k is a local field or a global field in the nonexceptional case and p ≠ char(k), we will verify that the hypothesis P S(k)(p) = Br(k_{ra,p}^{(m)}/k) is satisfied, proving Theorem 5.1. If k is a local or global field, then P S(k) = Br(k) [LO, Satz 3]. We need to check that L = k_{ra,p}^{(m)} splits every element of Br(k)(p). If k is local, then since X(k_{cyc,p}/k) contains a copy of ℚ/ℤ(p), so does mX(k_{cyc,p}/k), and mX(k_{cyc,p}/k) ⊆ mX(k_{ra,p}/k) = X(L/k). Hence, L contains extensions of k of all p-power degrees; so Br(L/k) = Br(k)(p). If k is global, then for any finite prime 𝔭 of k and any divisor 𝔓 of 𝔭 in L, we have L_{𝔓} contains k_{𝔭,cyc,p}^{(m)}, so Br(L_{𝔓}/k_{𝔭}) = Br(k_{𝔭})(p). Since p is odd or char(k) ≠ 0 or k is totally imaginary, the local global principle for Br(k) [R, p. 276, Th. 32.11] shows that Br(L/k) = Br(k)(p).

It remains to treat the exceptional case, for which we will need the following lemma:

Lemma 5.3. Let k be a number field which is not totally imaginary. Then there exists a totally imaginary quadratic extension ℓ = k(\sqrt{β}) of k such that ℓ/k does not embed into a cyclic degree 4 extension of k.

Proof. By the approximation theorem there is β ∈ k with β < 0 in each real completion of k. Then, ℓ = k(\sqrt{β}) is totally imaginary. Since there is a real place of k, this β cannot be a sum of two squares in k. Therefore, by Albert’s criterion [A, p. 207, Th. 11, p. 208, Ex. 1] ℓ cannot embed in a cyclic degree 4 extension of k. □

We now return to the proof of Th. 5.1 in the exceptional case. So, p = 2 and m = 2, and we replace the previous L = k_{ra,2}^{(2)} by L′ = k_{ra,2}^{(2)}(\sqrt{β}) = L(\sqrt{β}) with k(\sqrt{β}) as in Lemma 5.3, and replace M_{2} by M′_{2} = M_{2}(\sqrt{β}). The earlier argument shows that Br(L′_{𝔓}/k_{𝔭}) = Br(k_{𝔭})(2) for each finite prime 𝔭 of k. Since L′ has no real embeddings, it follows that Br(L′/k) = Br(k)(2) = P S(k)(2). We compute Br(M′_{2}) by a variant of the proof of Prop. 5.2. For χ ∈ Y (k_{s}/k)(2), the condition for res_{M′_{2}/F} (f(χ)) = 0 is now that 2χ ∈ Y (L′/k) = Y (L/k) + Y (k(\sqrt{β})/k) = 2Y (k_{ra,2}/k) + Y (k(\sqrt{β})/k). We claim that this holds iff χ ∈ Y (k_{ra,2}/k). Again, “if” is clear; for the converse, we suppose 2χ = 2ψ + φ with ψ ∈ Y (k_{ra,2}/k) and φ ∈ Y (k(\sqrt{β})/k). We have

Y (k(\sqrt{β})/k) ∩ 2Y (k_{s}/k)(2) ≅ ⊕_{i∈I_{2}} (X(k(\sqrt{β})/k) ∩ 2X(k_{s}/k)(2)) = 0,

since k(\sqrt{β}) lies in no cyclic extension of k of degree 4. But, φ = 2(χ − ψ) lies in this intersection, so φ = 0. Then, 2χ = 2ψ, and the rest of the argument to see that P S(F)(2) = Br(M′_{2}/F) is the same as for Prop. 5.2.

It remains only to prove that P S(F)(p) = Br(M_{p}/F) when p = char(k). For this, note first that for any field K with char(K) = p, we have Br(K_{cyc,p}/K) ⊆ P S(K)(p) ⊆ Br(K_{cyc,p}/K). The first inclusion holds as every finite subextension of K_{cyc,p}/K is cyclic, so Br(K_{cyc,p}/K) consists of cyclotomic cyclic algebras, which clearly lie in P S(K). The second inclusion holds because P S(K)(p) ⊆ Br(K_{ra,p}/K) by [AS_{2}, Cor. 2.3] (this is also deducible from Propositions 2.2(a) and 2.4(c) above) and K_{ra,p} = K_{cyc,p} as K contains no pth roots of unity. Thus, P S(K)(p) = Br(K_{cyc,p}/K). For the fields in the proof of Th. 5.1 with p = char(k), we have m = 1, so L = k_{ra,p} = k_{cyc,p} and M_{p} = F (L) = F (k_{cyc,p}) = F_{cyc,p}. Hence, Br(M_{p}/F) = Br(F_{cyc,p}/F) = P S(K)(p). This completes the proof of Theorem 5.1. □

6. The Schur group case

For F a number field, P S(F) = Br(F), so trivially P S(F) is the algebraic relative Brauer group Br(F_{s}/F). On the other hand, we now show that this need not be the case for the classical Schur group S(F). We are grateful to Allan Herman for suggesting that S(F) is not an algebraic relative Brauer group, because the local invariants of an element of the Schur group are uniformly distributed [BS, Th. 1]. Here is an explicit example: Let ζ_{n} be a primitive n-th root of unity in ℚ_{cyc}. Let F = ℚ(ζ_{12}), and let M = F (ζ_{13}). Let B be the cyclic algebra (M/F, σ, ζ_{12}). This B is a Schur (division) algebra, of exponent 12 over F as one can check by looking at it over the completion F_{𝔭}, where 𝔭 is a prime of F over (13). (Indeed, tensoring B with F_{𝔭}, we get an algebra B_{𝔭} and it suffices to show that it has order 12. This is equivalent to ζ_{12} having order 12 mod norms from M_{𝔭}. But this is the case by local class field theory because M_{𝔭}/F_{𝔭} is totally and tamely ramified: the norm group is generated by a local parameter and the one-units.) (13) splits completely in F into a product of four primes, and the local invariant of B at each of these is of order 12 (in fact these are the only nontrivial invariants of B, so they must be 1/12, 5/12, 7/12, 11/12 by [BS, Th. 1]). Any splitting field must have local degree divisible by 12 at these four places. Now take an algebra with local invariants say 1/2, 1/2, 0, 0 at these four places and 0 everywhere else. It is not in the Schur group because these local invariants are not uniformly distributed [BS, Th. 1] (in fact, since these four invariants are not of the same order, it is enough to invoke [B, Th. 1]) and is split by any field that splits B.

One can also prove that S(F) is not an algebraic relative Brauer group for F a formal power series field k((t)) over certain fields k. Here is a sketch of the proof. If A = F (G) is a Schur algebra over F with G finite, then A_{0} := k(G) is a Schur algebra over k and A = A_{0} ⊗_{k} F. It follows that S(F) = res_{F/k}(S(k)). Recall that S(k)(p) = 0 if char(k) = p ≠ 0; so S(k) ⊆ Br(k)′. Discretely valued fields have no TTR algebras (as |Γ/pΓ| < p^{2} for Γ = ℤ, see the definition of T_{p^{r}} in §3). Thus, Prop. 3.5 and Th. 3.2 above reduce to the classical Witt decomposition

Br(k((t)))′ = SBr(k((t)))′ ≅ Br(k)′ ⊕ Hom_{c}(G_{k}, ℚ/ℤ)′

Here, S(F) sits in the first component. For the sake of simplicity let k = ℝ (but the same type of argument works for any field k with S(k) nontrivial and G_{k} pronilpotent). Suppose S(F) were an algebraic relative Brauer group Br(L/F) with L/F algebraic. Let L_{0}/F be the maximal unramified subextension of L/F, and let ℓ = \overline{L_{0}}. Then, l is either ℝ or ℂ. If l = ℂ, then L will split the nontrivial quaternion algebra (−1, t). This algebra belongs to the second component in the Witt decomposition hence is not in S(F), contradiction. It follows that L/F is totally ramified, hence its residue field is k. So, (see (3.4)) Br(k) injects into Br(L), hence L does not split the nontrivial quaternion Schur algebra (−1, −1), another contradiction.

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Department of Mathematics, Technion, 32000 Haifa, Israel, Department of Mathematics, University of California, San Diego, 9500 Gilman Dr., La Jolla, California 92093-0112, USA

E-mail address: aljadeff@math.technion.ac.il, sonn@math.technion.ac.il, arwadsworth@ucsd.edu