GALOIS COHOMOLOGY OF FIELDS WITHOUT ROOTS OF UNITY

UZI VISHNE

Abstract. There is a standard correspondence between elements of the cohomology group H^{1}(F, μ_{n}) (with the trivial action of Γ_{F} = Gal(F_{s}/F) on μ_{n}) and cyclic extensions of dimension n over F. We extend this to a correspondence between the cohomology groups H^{1}(F, μ_{n}) where the action of Γ_{F} on μ_{n} varies, and the extensions of dimension n of K which are Galois over F, where K = F [μ_{n}] and [K:F] is prime to n. The cohomology groups are also related to eigenspaces of H^{1}(K, ℤ/n) with respect to the natural action of Gal(K/F).

As a result, we extend Albert’s cyclicity criterion, stated in the 1930s for division algebras of prime degree, to algebras of primepower degree over F, under the assumption stated above. We also extend the Rosset-Tate result on the corestriction of cyclic algebras in the presence of roots of unity, to extensions in which roots of unity live in an extension of dimension ≤ 3 over the base field. In particular if roots of unity live in a quadratic extension of the base field, then corestriction of a cyclic algebra along a quadratic extension is similar to a product of two cyclic algebras. Another application is that F -central algebras which are split by a certain semidirect product extension of F, are cyclic. In particular, if [K:F] = 2 then algebras over F which are split by an odd order dihedral extension, are cyclic.

We also construct generic examples of algebras which become cyclic after extending scalars by roots of unity, and show the existence of elements for which most powers have reduced trace zero.

1. Introduction

The purpose of this paper is to generalize some standard constructions of Galois cohomology to fields without roots of unity. Let n be a prime power, F a field of characteristic prime to n, and K = F [ρ],

Date: August 1, 2002. 2000 Mathematics Subject Classification. 16K20, 12G05. I thank J.-P. Tignol and L.H. Rowen for many stimulating discussions. I grate-

fully acknowledge a post-doctoral grant from the Fulbright program, United States Department of State.

where ρ is a primitive unity root of order n. Let F_{s} denote the separable closure of F. In [11] and [12] A. Merkurjev use decomposition into eigenspaces of the cohomology groups of K to study the relation between the Brauer group of F to the Brauer group of K, which is better understood. We apply his technique first to the first cohomology group H^{1}(K, ℤ/n), and use the cup product to study the Brauer groups. This somewhat different approach provides numerous new results on central simple algebras over F.

We start with a decomposition of the cohomology group H^{1}(K, ℤ/n) to eigenspaces under the natural action of Gal(K/F). It turns out that the components are naturally isomorphic to the cohomology groups H^{1}(F, μ_{n}(χ)), where this notation stands for the cohomology group of Γ_{F} with an action on μ_{n} which is twisted by a character χ of Gal(K/F) (so in particular the restricted action of Γ_{K} is trivial).

It is well known that the cocycles of order n in H^{1}(F, μ_{n}), with the trivial action, correspond to cyclic Galois extensions of dimension n over F. In a similar manner we interpret the other groups H^{1}(F, μ_{n}(χ)), in terms of field extensions of dimension n over K which are Galois over F.

The theorem of Merkurjev-Suslin is often quoted as saying that if K contains n roots of unity, then_{n}Br(K) ≅ K_{2}(K)/n (where_{n}Br(K) is the exponent n part of the Brauer group and K_{2}(K) is the second Milnor K-group of K). In fact, the theorem holds more generally, giving the isomorphism H^{2}(F, μ_{n}) ≅ K_{2}(F)/n whenever F has characteristic prime to n (see [20, Sec. 8]). Since _{n}Br(F) = H^{2}(F, μ_{n}^{⊗2}), this is useful (in terms of Brauer group theory) mainly when the two actions coincide, i.e. when μ_{n} ⊆ F.

Our description of the various cohomology groups H^{1}(F, μ_{n}) can be used, taking cup products, to provide information on the groups H^{2}(F, μ_{n}(χ)), including H^{2}(F, μ_{n}) and H^{2}(F, μ_{n}^{⊗2}) which are both isomorphic to eigenspaces of H^{2}(K, μ_{n}). This approach is similar to that of [9], where the authors study cohomology groups twisted by quadratic extensions of F. They handle arbitrary quadratic extensions, while here the focus is on the extension of F by the n-roots of unity; however, the main technique here, decomposition into eigenspaces of characters (from the Galois group to U_{n}), should work for arbitrary Abelian Galois extensions with dimension prime to n. The price paid in [9] for the generality is that the characteristic of F is assumed to be different than 2, while here we do not make this assumption. More recently, decomposition into eigenspaces with respect to the action of Gal(K/F) is used in [22] to study cohomology groups. The main focus of this paper is in the Galois groups of maximal p-extensions, and it also develops methods to avoid the assumption that [K:F] is prime to n. Our Theorem 11.4 is Theorem 3.6 in [22].

Using a projection formula for the corestriction of cyclic algebras from K to F, we show that if D is a division algebra of degree n over F which restricts to a cyclic algebra (a, β)_{K} with a ∈ K and β ∈ F, then D is cyclic over F. One application is a generalization of Albert’s cyclicity criterion, that a central simple algebra of degree n with an n-central elements is cyclic, from prime degree to any prime power degree, assuming [K:F] is prime to n.

An algorithm due to Rosset and Tate shows that the corestriction of a cyclic algebra from K_{2} to K_{1}, when K_{1} has enough roots of unity, is similar to at most [K_{2} :K_{1}] cyclic algebras over K_{1}. A similar result follows for separable extensions L/F when we assume [K:F] ≤ 3, namely that in this case the corestriction of a cyclic algebra over L is similar to at most [K:F] · ([L:F] − 1) + 1 cyclic algebras over F. If [K:F] = 2 and L/F is a quadratic extension, then the corestriction is a product of two (and not three) cyclic algebras.

If roots of unity are present in the base field, then it is known [16] that algebras with certain semidirect product splitting fields are cyclic. It is easy to show that _{n}Br(F) is generated by algebras which we call quasi-symbols, which are algebras of degree n over F, which when restricted to K are cyclic of some special form. We show that a cyclicity result for algebras with semi-direct product splitting field is equivalent to the cyclicity of quasi-symbols, which is not known to hold. We find splitting fields of smaller dimension for certain quasi-symbols, and in particular show that algebras with splitting field containing K whose Galois group is G = ⟨σ, τ| σ^{n} = τ^{d} = 1, τστ^{−1} = σ^{t}⟩ (where τ is a generator of Gal(K/F) such that τ(ρ) = ρ^{t}), are cyclic. In a related direction, we extend the theorem of Rowen and Saltman [17] that dihedral algebras are cyclic, where they assume the base field contains n-roots of unity, to dihedral algebras over F in which the roots of unity live in a quadratic extension of the base field. Moreover, F -algebras with dihedral splitting field which intersects K non-trivially are quasisymbols.

The paper is organized as follows. After setting the notation of decomposition into eigenspaces in Section 2, we decompose the isomorphic Gal(K/F)-modules K^{×}/K^{×}^{n} and H^{1}(K, ℤ/n) in Sections 3 and 4, respectively. We derive some technical results on representatives of classes in K^{×}/K^{×}^{n}, which are frequently used later on.

In Section 5 we explicitly describe an isomorphism from an eigenspace of K^{×}/K^{×}^{n} to a group H^{1}(F, μ_{n}(χ)), for a suitable character χ. Also, the eigenspaces of H^{1}(K, ℤ/n) are shown to be isomorphic to those H^{1}(F, μ_{n}(χ)). The isomorphisms are natural in the sense that they commute with the restriction map. The correspondence between the first cohomology groups of F with coefficients in μ_{n} to Galois extensions of F which are cyclic over K is presented in Section 6, where we also give explicit formulas for the inverse of the isomorphisms presented in previous sections. We always assume that [K:F] is prime to n. An example in which this is not the case is given in Section 7, where we explain how the constructions fail without this assumption.

The cup products enables to lift the information we gather on the first cohomology groups, to the groups H^{2}(F, μ_{n}) (again for various actions of Γ_{F} on μ_{n}), and in Section 8 we compute the corestriction of cyclic algebras from K to F, and show that the various restriction maps take the groups H^{2}(F, μ_{n}) to distinct eigenspaces of H^{2}(K, ℤ/n), which is the Brauer group of K. In Section 10 we extend the action of Gal(K/F) to certain cyclic algebras over K.

The action of Gal(K/F) on cyclic algebras is applied in Sections 11–13 to derive the results described above.

Finally in Section 14 we present a candidate for being a non-cyclic algebra of prime degree over fields without roots of unity. We also show that some generic algebras of this type have elements w for which the reduced trace Tr(w^{ℓ}) is zero for most values of ℓ < n.

We use the notation suggested by J.-P. Tignol and S.A. Amitsur [21], that a central simple algebra A is split by a group G is A is similar (in the Brauer sense) to a crossed product with respect to a subgroup of G. In particular A is split by the cyclic group C_{n} iff it is similar to a cyclic algebra of degree dividing n.

Frequent use is made of the fact that if the index [G:H] is prime to n and M is a G-module of exponent n, then the restriction of cohomology groups (in particular Brauer groups) from G to H is injective, and the corestriction (from H to G) is onto.

2. Decomposition into Eigenspaces

Let G be a finite Abelian group, and M a faithful G-module of finite torsion n (i.e. n · M = 0). Since we are interested in decomposing M, we assume that n = p^{m} is a prime power.

Let U_{n} = (ℤ/n)^{×} denote the Euler group of n. For a multiplicative character ϕ: G→U_{n}, we define a homomorphism of G-modules tr_{ϕ}: M→M, by

(1)  tr_{ϕ}(a) = ∑_{τ∈G} ϕ(τ)^{−1}τ(a).

Let

M^{(ϕ)} = {a ∈ M : τ(a) = ϕ(τ) · a}

be the eigenspace of ϕ, and observe that Im(tr_{ϕ}) ⊆ M^{(ϕ)}. Note that for a ∈ M^{(ϕ)} we have tr_{ϕ}(a) = |G| · a, so in particular if |G| is prime to n, tr_{ϕ}: M→M^{(ϕ)} is onto.

The following easy observation will be frequently used.

Remark 2.1. If 1 ≠ g ∈ U_{n} has order prime to p, then g ≢ 1 (mod p).

Proof. The kernel of the projection U_{n}→U_{p} is of order n/p which is a p-power, and by assumption g is not in this kernel. □

Corollary 2.2. If H ≤ U_{n} is a non-trivial subgroup and d = |H| is prime to p, then s = ∑_{h∈H} h ≡ 0 (mod n).

Proof. Choose some 1 ≠ g ∈ H, then gs = ∑ gh = s, so (g − 1)s ≡ 0 (mod n), but g − 1 is prime to p by the last remark. □

Proposition 2.3. If d = |G| is prime to n, then

M = ⨁_{ϕ} M^{(ϕ)}.

Proof. Applying the above corollary to the subgroups {ϕ(τ)}_{ϕ} (for every τ ∈ G), we have for arbitrary a ∈ M that

d^{−1} ∑_{ϕ} tr_{ϕ}(a) = d^{−1} ∑_{ϕ} ∑_{τ} ϕ(τ)τ(a)
= d^{−1} ∑_{τ}(∑_{ϕ} ϕ(τ))τ(a)
= ∑_{τ}(δ_{τ,1})τ(a) = a,

where tr_{ϕ}(a) ∈ M^{(ϕ)} as we noted above. □

3. The module K^{×}/K^{×}^{n}

For the rest of this paper, let F be a field of characteristic prime to n, where n = p^{m} is a prime power, and let K = F [ρ] where ρ = ρ_{n} is a primitive root of unity of order n.

The dimension d = [K:F] obviously divides |U_{n}| = (p − 1)p^{m−1}. We assume throughout that d is prime to n. In particular, p is odd, and G = Gal(K/F) is cyclic. For every n = p^{m}, setting K = F [ρ_{n}], we have that [K:F [ρ_{p}]] is a power of p, so the following holds:

Remark 3.1. [K:F] is prime to n iff K = F [ρ_{n}] = F [ρ_{p}].

For a given field F, there is a maximal α ≤ ∞ such that the dimension of K = F [ρ_{p^{α}}] over F is prime to p.

The decomposition given in the last section applies to M = K^{×}/K^{×}^{n}, which is an n-torsion module over G = Gal(K/F). For a character φ: G→U_{n}, the lift to K^{×} of the eigenspace (K^{×}/K^{×}^{n})^{(φ)} defined in Section 2, is

K^{(φ)} = {a ∈ K^{×}: ∀τ, τ(a) ≡ a^{φ(τ)} (mod K^{×}^{n})},

and the twisted norm

N^{(φ)} : K^{×}→K^{(φ)}/K^{×}^{n}

is defined by

(2)  N^{(φ)}a = ∏_{τ∈G} τ(a)^{φ(τ^{−1})}.

For every a ∈ K^{×}, Proposition 2.3 gives a decomposition

(3)  a = Πa_{φ}

where φ runs over the characters of Gal(K/F), and each a_{φ} ∈ K^{(φ)} is unique modulo K^{×}^{n}.

It is obvious that F^{×} ⊆ K^{(1)}. On the other hand,

Proposition 3.2. We have that K^{(1)} = F^{×}K^{×}^{n}.

Proof. The inclusion F^{×}K^{×}^{n} ⊆ K^{(1)} is obvious. Fix a generator τ of Gal(K/F) and let a ∈ K^{(1)}. then by assumption there is some μ ∈ K such that τ(a) = μ^{n}a. Taking norms, we see that N_{K/F} (μ)^{n} = 1, but since the norm is in F which does not contain roots of unity even of order p (Remark 3.1), we have that N_{K/F} (μ) = 1. Thus, there is some g ∈ K such that μ = τ(g)g^{−1}. Let α = g^{−n}a. Then τ(α) = τ(g)^{−n}τ(a) = μ^{−n}g^{−n}μ^{n}a = α, so that α ∈ F and a ∈ F^{×}K^{×}^{n}, as asserted. □

Of similar nature is the following observation:

Remark 3.3. We have that F^{×} ∩ K^{×}^{n} = F^{×}^{n}.

Proof. Let a ∈ K^{×} be an element such that a^{n} ∈ F. Let τ be a generator of Gal(K/F). Since τ(a^{n}) = a^{n}, we have that τ(a) = ρ^{i}a for some i. Write τ(ρ) = ρ^{t}. Since ⟨t⟩ is a subgroup of order d of U_{n}, t − 1 is prime to p (Remark 2.1). Let b = ρ^{i/(1−t)}a (where the inverse is taken mod n), and compute that τ(b) = ρ^{it/(1−t)+i}a = ρ^{i(t/(1−t)+1)}a = b, showing that b ∈ F, hence a^{n} = b^{n} ∈ F^{×}^{n}. □

Since K has n roots of unity, the cyclic extensions of dimension n over K are classified, by Kummer correspondence, by elements of K^{×}/K^{×}^{n}. In a similar manner, the subspaces K^{(ψ)} classify cyclic extensions of dimension n over K, which are Galois over F. The following is in [4, p. 211] for p prime. The condition that [K:F] is prime to n is not needed here.

Proposition 3.4. Let a ∈ K, α = \sqrt[n]{a}, and assume K_{1} = K[α] is a field. Then K_{1} is Galois over F iff a ∈ K^{(φ)} for some character φ.

Proof. Let ϖ denote the generator of Gal(K_{1}/K), defined by ϖ(α) = ρα. Assume K_{1}/F is Galois; then every τ ∈ Gal(K/F) extends to K_{1}. Fix τ. Since τ(α)^{n} = τ(α^{n}) = τ(a) ∈ K, we have that ϖ(τ(α))^{n} = τ(α)^{n}, and ϖ(τ(α)) = ρ^{φ(τ)}τ(α) for some number φ(τ) ∈ ℤ. Now ϖ(\frac{τ(α)}{α^{φ(τ)}} ) = \frac{ϖτ(α)}{ϖ(α)^{φ(τ)}} = \frac{ρ^{φ(τ)}τ(α)}{(ρα)^{φ(τ)}} = \frac{τ(α)}{α^{φ(τ)}} , so that τ(α)α^{−φ(τ)} ∈ K^{×} and τ(a) ∈ K^{×}^{n}a^{φ(τ)}. It is then immediate that φ is multiplicative mod n, so that φ is a character and a ∈ K^{(φ)}.

Now assume that a ∈ K^{(φ)}. Then for every τ ∈ Gal(K/F), we have τ(a) = k^{n}a^{φ(τ)} for some k ∈ K^{×}. For every i ∈ ℤ/n, α ↦ ρ^{i}kα^{φ(τ)} is a well defined extension of τ to K_{1} (since it preserves the defining relation α^{n} = a), so there are n|Gal(K/F)| = [K_{1} :F] distinct automorphisms, showing that K_{1}/F is Galois. □

We can omit the assumption that K_{1} is a field. If the order n′ of a in K^{×}/K^{×}^{n} is a proper divisor of n, then K_{1} = K[\sqrt[n]{a}] is a direct product of n/n′ copies of the field K[ \sqrt[n′]{a}]. Letting φ_{1} : Gal(K/F)→U_{n′} denote the composition of φ with the projection U_{n}→U_{n′}, we see that K_{1} is a Galois ring over F iff a ∈ K^{(φ_{1})} for some character φ.

The following remark allows to choose a ∈ K^{(φ)} with the convenient property N_{K/F} (a) = 1. In the case [K:F] = 2 this is [9, Prop. 25].

Remark 3.5. Let φ ≠ 1. Every class in K^{(φ)}/K^{×}^{n} has a representative a such that N_{K/F} (a) = 1.

Proof. Let N = N_{K/F} denote the usual norm. Let τ be a generator of Gal(K/F), and r ≡ φ(τ) (mod n). By Remark 2.1 there are i, j ∈ ℤ such that i(r − 1) + jn = 1. Let a ∈ K^{(φ)}, then by assumption τ(a) = μ^{n}a^{r} for some μ ∈ K^{×}. The equivalent element a_{1} = (μ^{i}a^{−j})^{n}a satisfies N(a_{1}) = N(μ)^{ni}N(a)^{1−jn} = (N(μ)^{n}N(a)^{r−1})^{i} = (N(τ(a))/N(a))^{i} = 1.

□

If d = [K:F] is even and K_{0}/F the quadratic intermediate field, let μ denote the character of order 2 of Gal(K/F). A very similar argument proves the following

Remark 3.6. Let φ ≠ 1, μ. Every class in K^{(φ)}/K^{×}^{n} has a representative a such that N_{K/K_{0}}(a) = 1.

Proof. As in the previous remark let τ be a generator of Gal(K/F), r ≡ φ(τ) (mod n), and denote N′(a) = N_{K/K_{0}}(a) · τN_{K/K_{0}}(a)^{−1}. Since r ≢ −1 by assumption, there are i′, j′ ∈ ℤ such that i′(r + 1) + j′n = 1. Given an element a ∈ K^{(φ)}, write τ(a) = μ^{n}a^{r}, so that τ(N′(a)) = N′(μ)^{n}N′(a)^{r}. Now the element a_{1} = (μ^{i′}a^{−j′})^{n}a satisfies N′(a_{1}) = N′(μ)^{ni′}N′(a)^{1−j′n} = (N′(μ)^{n}N′(a)^{r+1})^{i′} = (τN′(a)N′(a))^{i′} = 1, so that N_{K/K_{0}}(a_{1}) ∈ F.

Since φ ≠ 1 we can by the previous remark assume N(a) = 1 to begin with, then N(μ)^{n} = N(a)^{1−r} = 1 so that N(μ) = 1, and N_{K/K_{0}}(a_{1})^{2} = N_{K_{0}/F} N_{K/K_{0}}(a_{1}) = N(a_{1}) = 1. Finally if N_{K/K_{0}}(a_{1}) = −1 we can replace a_{1} by a_{1}^{n+1} which will have norm 1. □

4. Decomposition of H^{1}(K, μ_{n})

Let F_{s} denote the separable closure of F, and let

Γ_{F} = Gal(F_{s}/F)

denote the absolute Galois group of F. As usual, we denote the cohomology groups of Γ_{F} by H^{i}(F, M) = H^{i}(Γ_{F} , M).

The long exact sequence of cohomology groups applied to the Kummer sequence

(4)  1 ⟶ μ_{n} ⟶ F_{s}^{×} ⟶ F_{s}^{×} ⟶ 1,

gives an isomorphism

(5)  j_{F} : F^{×}/F^{×}^{n} ⟶ H^{1}(F, μ_{n})

sending aF^{×}^{n} to (a), which is defined by (a): σ ↦ σ(α)α^{−1} where α = \sqrt[n]{a} is a fixed root.

As before, let K = F [ρ] where ρ is a primitive n-root of unity. The action of Γ_{K} on μ_{n} is always taken to be the trivial action; this is often expressed by writing H^{1}(K, ℤ/n) instead of H^{1}(K, μ_{n}). Considering (4) as a sequence of Γ_{K}-modules (rather than Γ_{F} -modules), gives an isomorphism

j_{K}: K^{×}/K^{×}^{n} ⟶ H^{1}(K, ℤ/n)

defined similarly to j_{F}.

Let a ∈ K^{×} and α = \sqrt[n]{a}. Since K has enough roots of unity, K[α] is Galois over K, and the value (a)_{σ} is determined by the restriction of σ to K[α]. In other words, (a) is an element of H^{1}(K, ℤ/n) induced from H^{1}(Gal(K[α]/K), ℤ/n).

The Galois group G = Gal(K/F) acts on H^{1}(K, ℤ/n) by

τ(a): σ ↦ τ((a)_{τ^{−1}στ}).

Since the Kummer sequence (4) is a sequence of G-modules, j_{K} is an isomorphism of G-modules. Indeed, let a ∈ K^{×} with α = \sqrt[n]{a}, τ ∈ Gal(K/F) and σ ∈ Γ_{K}, then by definition

τ(a): σ ↦ τ((a)_{τ^{−1}στ})
= τ(τ^{−1}στ(α)α^{−1})
= στ(α)τ(α)^{−1}
= (τa)_{σ}.

In particular, j_{K} is compatible with the decomposition of K^{×}/K^{×}^{n} and H^{1}(K, ℤ/n) into eigenspaces which is given in Proposition 2.3, and restricts to an isomorphism

j_{K}: K^{(φ)}/K^{×}^{n} ⟶ H^{1}(K, ℤ/n)^{(φ)}

for every character φ: Gal(K/F)→U_{n}. Since N^{(φ)} of Equation (2) is the twisted trace function defined on K^{×}/K^{×}^{n}, the following diagram commutes:

(6)  \xymatrix{K^{(φ)}/K^{×}^{n} \ar[d]^{j_{K}} \ar@{^{(}->}[r] & K^{×}/K^{×}^{n} \ar[d]^{j_{K}} \ar[r]^{^{N^{(φ)}}} & K^{(φ)}/K^{×}^{n} \ar[d]^{j_{K}} \\ _{H^{1}(K, ℤ/n)_{(φ)}} \ar@{^{(}->}[r] & H^{1}(K, ℤ/n) \ar[r]^{_{tr_{φ}}} & _{H^{1}(K, ℤ/n)_{(φ)}}}

5. First Cohomologies of F

Let F be a field of characteristic prime to n, where n = p^{m}. Let μ_{n} denote the group of roots of unity of order n in F_{s}. Fix a primitive root ρ ∈ μ_{n}.

In this section we study the Galois groups H^{1}(F, μ_{n}) for various actions of Γ_{F} on μ_{n}, which restrict to the trivial action of Γ_{K}. One case is well understood: if we let Γ_{F} act naturally on μ_{n} (that is, by the usual action of automorphisms on μ_{n} ⊆ F_{s}^{×}), then H^{1}(F, μ_{n}) ≅ F^{×}/F^{×}^{n} by the Kummer sequence mentioned earlier.

Let K = F [ρ]. A character χ: Gal(K/F)→U_{n} determines an action of Γ_{F} on μ_{n} by σ_{∗}(ρ) = ρ^{χ(σ)} (the star reminds us that this is not the usual action); and of course every extension of the trivial action of Γ_{K} to Γ_{F} is determined in this way. We use H^{1}(F, μ_{n}(χ)) to denote the cohomology group with respect to the action determined by χ. Since the action of Γ_{K} is trivial, we use H^{1}(K, ℤ/n) for the cohomology group with respect to Γ_{K}, and we have the restriction map res : H^{1}(F, μ_{n}(χ))→H^{1}(K, ℤ/n).

Denote G = Gal(K/F), and view μ_{n} as a Γ_{F} -module with the action via χ. As a subgroup, Γ_{K} ≤ Γ_{F} acts trivially on μ_{n}, and the short exact sequence

1 ⟶ Γ_{K} ⟶ Γ_{F} ⟶ G ⟶ 1,

gives rise to the Serre-Hochshild spectral sequence

H^{1}(G, μ_{n}) ⟶ H^{1}(F, μ_{n}(χ)) ⟶ H^{1}(K, μ_{n})^{G} ⟶ H^{2}(G, μ_{n}).

Since |Γ_{F} /Γ_{K}| = [K:F] is by assumption prime to n, we have that

H^{1}(G, A) = H^{2}(G, A) = 0,

so we obtain an isomorphism

H^{1}(F, μ_{n}(χ)) ≅ H^{1}(K, μ_{n})^{G},

which we will explicitly describe below. In particular we will see that for the action of G on H^{1}(K, μ_{n}) via χ, the invariant subgroup H^{1}(K, μ_{n})^{G} is the component H^{1}(K, ℤ/n)^{(νχ^{−1})} of the previous section.

From Proposition 3.2 and Remark 3.3 we see that

K^{(1)}/K^{×}^{n} = K^{×}^{n}F^{×}/K^{×}^{n} ≅ F^{×}/F^{×}^{n},

so that K^{(1)}/K^{×}^{n} ≅ H^{1}(F, μ_{n}) with the usual action. This motivates an attempt to express other cohomology groups in terms of the components K^{(φ)}/K^{×}^{n}.

Identify U_{n} with Aut(μ_{n}) is the usual way (where the action is by exponentiation). Fix a character

φ: Gal(K/F)→U_{n},

and let a ∈ K^{(φ)} be an element such that K_{1} = K[α] is a field, where α = \sqrt[n]{a} is a fixed root (the assumption that K_{1} is a field will eventually be removed). By Proposition 3.4, K_{1}/F is a Galois extension, and so every σ ∈ Γ_{F} restricts to an automorphism of K_{1}/F. The idea is to use this restriction to define the value of the cocycle at σ, making the cocycle induced from H^{1}(Gal(K_{1}/F), μ_{n}) (with an action which is yet to be determined), just as (c) ∈ H^{1}(K, ℤ/n) is induced from H^{1}(Gal(K[\sqrt[n]{c}]/F), ℤ/n) for c ∈ K^{×}. The analogy to the special case φ = 1 still holds: if a ∈ K^{(1)} then by Proposition 3.2 we can assume that a ∈ F^{×}, and then the value of (a) ∈ H^{1}(F, μ_{n}) (with the usual action) at σ ∈ Γ_{F} is determined by the restriction of σ to K_{1} = K⊗F [α], where α = \sqrt[n]{a} is a fixed root.

Let ϖ denote a generator of Gal(K_{1}/K) defined by

(7)  ϖ(α) = ρα.

Let ν: Gal(K/F)→U_{n} be the distinguished character defined by

(8)  τ(ρ) = ρ^{ν(τ)}.

Note that ν is uniquely determined by F, and that ν extends to a character ν: Γ_{F} →U_{n} in the obvious way. The natural action of Γ_{F} on μ_{n} is the action via ν.

Consider the short exact sequence

(9)  1 ⟶ Gal(K_{1}/K) ⟶ Gal(K_{1}/F) ⟶ Gal(K/F) ⟶ 1.

Since Gal(K/F) is Abelian, the action of

H = Gal(K_{1}/F)

on Gal(K_{1}/K) by conjugation reduces to an action of Gal(K/F). Let

χ: Gal(K/F)→U_{n}

be the character associated to this action, namely

(10)  τϖτ^{−1} = ϖ^{χ(τ)}.

The three characters φ, ν, χ are related by the following equation:

Remark 5.1. We have that χφ = ν.

Proof. Writing τ^{−1}(α) = kα^{φ(τ^{−1})} for some k ∈ K^{×}, we have that

ρ^{χ(τ)}α = ϖ^{χ(τ)}(α)
= τϖ(τ^{−1}(α))
= τϖ(kα^{φ(τ^{−1})})
= τ(kρ^{φ(τ^{−1})}α^{φ(τ^{−1})})
= τ(ρ)^{φ(τ^{−1})}τ(kα^{φ(τ^{−1})})
= ρ^{ν(τ)φ(τ^{−1})}α.

□

Let N = Gal(K_{1}/K) = ⟨ϖ⟩. Repeating the spectral sequence argument given above, we obtain an isomorphism

H^{1}(F, μ_{n}(χ)) ≅ H^{1}(K, μ_{n})^{H/N},

which we will use to construct the co-cycle in H^{1}(F, μ_{n}(χ)) associated to the element a. The group

H^{1}(N, μ_{n}) = Hom(N, μ_{n})

is cyclic of order n, generated by an element c which is defined by c(ϖ^{s}) = ρ^{s}. Moreover, the action of

H/N = Gal(K/F)

on this group is trivial: for every τ ∈ Gal(K/F), we have (τc)(ϖ^{s}) = τ_{∗}(c(τ^{−1}ϖ^{s}τ)) = τ_{∗}(c(ϖ^{sχ(τ)^{−1}})) = τ_{∗}(ρ^{sχ(τ)^{−1}}) = ρ^{s} = c(ϖ^{s}), where we used the definition τ_{∗}(ρ) = ρ^{χ(τ)}.

We will now define a map

J^{φ}: K^{(φ)}/K^{×}^{n}→H^{1}(F, μ_{n}(χ))

where φ and χ = νφ^{−1} are the characters fixed above.

Definition 5.2. Let F, n, K be as above, G = Gal(K/F), φ: G→U_{n} a character, and a ∈ K^{(φ)}. Let ν be the character defined by (8), and χ = νφ^{−1}.

First assume that a is of order n in K^{×}/K^{×}^{n}, and let α = \sqrt[n]{a} ∈ F_{s}, K_{1} = K[α] and H = Gal(K_{1}/K). Choose a splitting of (9) which maps τ ↦ τ′ ∈ Gal(K_{1}/F) (this is possible since H^{2}(G, H(χ)) = 0).

Let σ ∈ Γ be given, and let τ ∈ Gal(K/F) be the restriction of σ to K. Then for a unique s ∈ ℤ/n, the restriction of σ to K_{1} is σ = ϖ^{s}τ′. Now, define c_{a} ∈ H^{1}(F, μ_{n}(χ)) by

(11)  c_{a}(σ) = ρ^{s}.

If the order n′ of a in K^{×}/K^{×}^{n} strictly divides n, then repeating the argument for b = \sqrt[n′]{a} ∈ K and n/n′ (noting that b ∈ K^{(φ)} when φ is viewed as a character modulo n/n′), we define

c_{a}(ϖ^{s}τ′) = ρ^{n′s}.

Finally, we define J^{φ}: K^{(φ)}/K^{×}^{n}→H^{1}(F, μ_{n}(νφ^{−1})) by

J^{φ}(a) = c_{a}.

We need to show that the definition does not depend on the splitting of (9). Indeed, since H^{1}(G, H(χ)) = 0, every other splitting τ ↦ τ′′ is of the form

τ′′ = τ_{∗}(ϖ^{i})ϖ^{−i}τ′ = ϖ^{i(χ(τ)−1)}τ′

for some i. Let c′_{a} be the cocycle defined using this splitting. If σ = ϖ^{s}τ′ = ϖ^{s−i(χ(τ)−1)}τ′′, then c_{a}(σ) = ρ^{s} and c′_{a}(σ) = ρ^{s−i(χ(τ)−1)}, and the quotient is

g(σ) = ρ^{i(χ(τ)−1)} = τ_{∗}(ρ^{i})ρ^{−i} = σ_{∗}(ρ^{i})ρ^{−i},

which is cohomologous to 1.

Theorem 5.3. The following diagram commutes:

(12)  \xymatrix{K^{(φ)}/K^{×}^{n} \ar[d]^{J^{φ}} \ar@{^{(}->}[r] & K^{×}/K^{×}^{n} \ar[d]^{j_{K}} \ar[r]^{^{N^{(φ)}}} & K^{(φ)}/K^{×}^{n} \ar[d]^{J^{φ}} \\ H^{1}(F, μ_{n}(χ)) \ar[r]^{res} & H^{1}(K, ℤ/n) \ar[r]^{cor} & H^{1}(F, μ_{n}(χ))}

Proof. Let a ∈ K^{(φ)}, and let α, K_{1}, ϖ, χ be as above. Then j_{K}: a ↦ (a), where (a)_{σ} = σ(α)α^{−1} for every σ ∈ Γ_{K}. On the other hand, if σ = ϖ^{s} on K_{1}, then

J^{φ}(a) = c_{a} ∈ H^{1}(F, μ_{n}(χ))

has value ρ^{s} at σ by definition, and σ(α)α^{−1} = ρ^{s} = c_{a}(σ).

Now consider the right hand square. Since K^{×} is generated by the subgroups K^{(φ′)}, it is enough to check that J^{φ} ∘ N^{(φ)} = cor ∘ j_{K} on an element a ∈ K^{(φ′)}. The corresponding element in H^{1}(K, ℤ/n) is by definition σ ↦ \frac{σ(α)}{α} where α is a fixed nth root of a. Let K_{1} = K[α], and τ ↦ τ′ a splitting of (9) for this K_{1}. Define ϖ by (7), and set

χ′ = νφ^{′−1},

so that G = Gal(K/F) acts on ⟨ϖ⟩ via χ′.

We need to show that J^{φ}(N^{(φ)}(a)) and cor(a) agree on every σ ∈ Γ_{F}. Let ϖ^{s}τ′_{0} be the restriction of σ to K_{1}. Since Γ_{F} = ∪_{τ∈G}Γ_{K}τ, we have by definition of the corestriction that

cor(a): σ ↦ ∏_{τ} τ_{∗}^{−1}((a)_{τσ(ττ_{0})^{−1}})
= ∏_{τ} τ_{∗}^{−1}((a)_{τϖ^{s}τ^{−1}})
= ∏_{τ} τ_{∗}((a)_{ϖ^{sχ′(τ)^{−1}}})
= ∏_{τ} τ_{∗}(ρ^{sχ′(τ)^{−1}})
= ∏_{τ} ρ^{sχ(τ)χ′(τ)^{−1}},

where τ_{∗}(ρ) = ρ^{χ(τ)} since the computation is in H^{1}(F, μ_{n}(χ)).

If χ ≠ χ′, then

∑_{τ} χ(τ)χ′(τ)^{−1} = 0

by Remark 2.2, so that cor(a) = 1. This is what we need since N^{(φ′)}(a) ≡ 1 modulo K^{×}^{n}, again by the same remark.

Now assume χ′ = χ (i.e. φ′ = φ), then cor(a): σ ↦ ρ^{sd} (where d = [K:F]), and likewise N^{(φ)}(a) ≡ a^{d} so that

J^{φ}(N^{(φ)}(a)) = J^{φ}(a)^{d}

while

J^{φ}(a) = c_{a}: σ ↦ ρ^{s},

and the maps coincide. □

Corollary 5.4. For φ: Gal(K/F)→U_{n} and χ = νφ^{−1},

J^{φ}: K^{(φ)}/K^{×}^{n} ⟶ H^{1}(F, μ_{n}(χ))

is an isomorphism.

Proof. Since

cor ∘ res : H^{1}(F, μ_{n}(χ))→H^{1}(F, μ_{n}(χ))

is multiplication by d = [K:F] which is prime to n, res is injective and cor surjective. Our claim then follows from the commutativity of (12). □

Corollary 5.5. If χ ≠ χ′, the composition

H^{1}(F, μ_{n}(χ)) ⟶^{res} H^{1}(K, ℤ/n) ⟶^{cor} H^{1}(F, μ_{n}(χ′))

is the zero map.

Proof. Use the left hand square of (12) for φ = νχ^{−1} and χ, and the right hand square for φ′ = νχ^{−1} and χ′, and the fact (following from Proposition 2.3) that N^{(φ′)} is zero on K^{(φ)}/K^{×}^{n}. □

Notice that Equation (12) provides an expression for J^{φ} in terms of the restriction and corestriction, namely J^{φ} = d^{−1} · cor ∘ j_{K}, where the corestriction here is cor : H^{1}(K, ℤ/n)→H^{1}(F, μ_{n}(χ)).

For φ = 1 we have χ = ν, so that H^{1}(F, μ_{n}(χ)) = H^{1}(F, μ_{n}) is the group with usual action, and the isomorphism J^{1} is the composition of K^{(1)}/K^{×}^{n} ≅ F^{×}/F^{×}^{n} with j_{F}, which was discussed at the beginning of this section. In particular, if K = F, then J^{1} = j_{K}.

Corollary 5.6. For every character φ: Gal(K/F)→U_{n},

H^{1}(K, ℤ/n)^{(φ)} ≅ H^{1}(F, μ_{n}(χ)),

where χ = νφ^{−1}. Moreover, the following diagram commutes:

(13)  [figure: H^{1}(K, ℤ/n)^{(φ)}; H^{1}(K, ℤ/n); ^{tr^{φ}}; H^{1}(K, ℤ/n)^{(φ)}; H^{1}(F, μ_{n}(χ)); res; H^{1}(K, ℤ/n); cor; H^{1}(F, μ_{n}(χ))]

where the left and right arrows are J^{φ} ∘ j_{K}^{−1}.

Proof. Apply the commutativity of (6) and (12) to the following diagram:

[figure: K^{(φ)}/K^{×}^{n}; J^{φ}; j_{K}; K^{×}/K^{×}^{n}; j_{K}; ^{N^{(φ)}}; j_{K}; K^{(φ)}/K^{×}^{n}; J^{φ}; j_{K}; _{H^{1}(K, ℤ/n)_{(φ)}}; H^{1}(K, ℤ/n); _{tr_{φ}}; _{H^{1}(K, ℤ/n)_{(φ)}}; H^{1}(F, μ_{n}(χ)); res; H^{1}(K, ℤ/n); cor; H^{1}(F, μ_{n}(χ))]

□

In general, if M is a G-module and H ⊴ G, then the image of the restriction H^{1}(G, M)→H^{1}(H, M) is invariant under the action of G/H; one might thus expect that the image of

res : H^{1}(F, μ_{n}(χ))→H^{1}(K, μ_{n})

would be contained in H^{1}(K, μ_{n})^{(1)}, and not H^{1}(K, μ_{n})^{(φ)} as the corollary indicates. In fact, every character χ defines a different action of Gal(K/F) on H^{1}(K, μ_{n}), compatible with μ_{n} being a Γ_{F} -module with the action induced by χ (and indeed Im(res) is invariant under this action). However, notice that we decompose H^{1}(K, μ_{n}) to eigenspaces with respect to a fixed action of Γ_{F} on μ_{n}, namely the natural one, and Im(res) is invariant under this action iff χ = ν.

We end this section by showing that the maps J^{φ} commute with restriction when F is being changed. For that, we need to connect eigenspaces of K^{×}/K^{×}^{n} with respect to the action of Gal(K/F) and a subgroup.

Lemma 5.7. Let F ⊆ L ⊆ K be an intermediate field, and φ_{0} : Gal(K/L)→U_{n} a character. Let K^{(φ_{0})} be the φ_{0}-component of K^{×} with respect to the action of Gal(K/L).

Then K^{(φ_{0})} is the product of the components K^{(φ′)} for all the extensions of φ_{0} to characters φ′: Gal(K/F)→U_{n}.

Proof. Let τ be a generators of Gal(K/F), so that L = K^{τ^{e}} and Gal(K/L) = ⟨τ^{e}⟩. When viewed as a Gal(K/L)-module, the eigenspace of K^{×} with respect to φ_{0} is

K^{(φ_{0})} = {a ∈ K^{×}: τ^{e}(a) ≡ a^{φ_{0}(τ^{e})} (mod K^{×}^{n})}.

Now let φ′: Gal(K/F)→U_{n} be an arbitrary character, and compare K^{(φ′)} and K^{(φ_{0})}: if a belongs to the intersection, then τ^{e}(a) ≡ a^{φ_{0}(τ^{e})} ≡ a^{φ′(τ^{e})}, so if φ′ restricts to φ_{0} we have that K^{(φ′)} ⊆ K^{(φ_{0})}, while otherwise the intersection is K^{×}^{n}. □

Let L be a separable extension of F (viewed as a subfield of F_{s}), and let L_{0} = L ∩ K. Let

G_{0} = Gal(K/L_{0})

be the corresponding subgroup of G = Gal(K/F), and φ_{0} the restriction of φ: G→U_{n} to G_{0}. By the lemma, K^{(φ)} ⊆ K^{(φ_{0})}. Since T = L[ρ] = L⊗_{L_{0}}K, there is a natural identification

Gal(T/L) ⟶^{≅} Gal(K/L_{0}),

which makes φ_{0} a character on Gal(T/L). Likewise let χ_{0} be the restriction of χ = νφ^{−1} to Gal(T/L).

Proposition 5.8. Let L/F be a separable extension, T = L[ρ], and let φ_{0}, χ_{0} : Gal(T/L)→U_{n} be defined as above. Then the following diagram commutes.

(14)  \xymatrix{T^{(φ_{0})}/T^{×}^{n} \ar[r]^{J^{φ_{0}}} & H^{1}(L, μ_{n}(χ_{0})) \\ K^{(φ)}/K^{×}^{n} \ar[r]^{J^{φ}} \ar@{^{(}->}[u] & H^{1}(F, μ_{n}(χ)) \ar[u]^{res}}

Proof. It is enough to treat the extensions L/L_{0} and L_{0}/F, that is, to show that the following diagram commutes.

\xymatrix{T^{(φ_{0})}/T^{×}^{n} \ar[r]^{J^{φ_{0}}} & H^{1}(L, μ_{n}(χ_{0})) \\ K^{(φ_{0})}/K^{×}^{n} \ar[r]^{J^{φ_{0}}} \ar@{^{(}->}[u] & H^{1}(L_{0}, μ_{n}(χ_{0})) \ar[u]^{res} \\ K^{(φ)}/K^{×}^{n} \ar[r]^{J^{φ}} \ar@{^{(}->}[u] & H^{1}(F, μ_{n}(χ)) \ar[u]^{res}}

For the lower square, let a ∈ K^{(φ)}, K_{1} = K[\sqrt[n]{a}], and assume K_{1} is a field (the general case will then easily follows). We need to prove that the restriction of J^{φ}(a) to Γ_{L_{0}} coincides with J^{φ_{1}}(a). Recall that c_{a} = J^{φ}(a) is not well defined as a function on Γ_{F}, as it depends on the splitting of (9). Choose a splitting G→ Gal(K_{1}/F) which maps τ ↦ τ′. Let σ ∈ Γ_{L_{0}}, and write the restriction of σ to K_{1} as ϖ^{s}τ′ where τ ∈ Gal(K/L_{0}) ⊆ Gal(K/F); then c_{a}(σ) = s either as a cocycle on Γ_{F} or on Γ_{L_{0}}.

Next, for the upper square, let a ∈ K^{(φ_{0})}. Again, it is enough to assume that T_{1} = T [\sqrt[n]{a}] is a field. Let τ ↦ τ′ be a splitting of (9) (for K/L_{0} instead of K/F), and extend each τ′ to T = L⊗_{L_{0}}K as 1⊗τ′. For σ ∈ Γ_{L}, write the restriction to T_{1} as ϖ^{s}τ′ where ϖ is defined as usual. Then σ restricts to ϖ^{s}τ′ on K_{1}, so J^{φ_{0}}(σ) = ρ^{s} either over the extension T/L, or over K/L_{0}. □

Note that for the special case L = K, (14) is the left hand side of (12).

6. Subfields of K_{1}

As before, let F be a field of characteristic prime to n = p^{m}, K = F [ρ] where ρ is a primitive root of unity of order n. We assume that [K:F] is prime to n.

Let φ: Gal(K/F)→U_{n} be a character, χ = νφ^{−1} (where ν is defined by (8)) and a ∈ K^{(φ)} an element such that K_{1} = K[α] is a field for α = \sqrt[n]{a}. We discuss subfields of K_{1}, and give an explicit computation of the inverse of the map J^{φ} of Definition 5.2.

Let ϖ be the generator of Gal(K_{1}/K) defined by ϖ(α) = ρα. By Proposition 3.4, K_{1} is Galois over F, with Gal(K_{1}/F) the semidirect product of

Gal(K/F) = ℤ/d

acting (via χ, by Remark 5.1) on

Gal(K_{1}/K) = ⟨ϖ⟩ ≅ ℤ/n.

Let τ ↦ τ′ be a splitting of (9), and G′ the copy of G = Gal(K/F) in Gal(K_{1}/F) under this map. Let F_{1} denote the invariant subfield of K_{1} under G′, so that [F_{1} :F] = n. Notice that K_{1} = F_{1}⊗_{F} K as the dimensions are co-prime. The following is essentially in [19, Sec. 2].

Proposition 6.1. If χ = 1 then F_{1} = K_{1}^{G′} is cyclic over F. Every cyclic extension of dimension n over F has this form.

Proof. If χ = 1 then

Gal(K_{1}/F) = ℤ/n × Gal(K/F)

by Equation (10), so that G′ is normal in Gal(K_{1}/F), with cyclic quotient.

Now let F_{1}/F be cyclic extension of dimension n. Let K_{1} = F_{1}⊗_{F} K, then K_{1} is Galois over F (since F_{1} ∩ K = F as the dimensions are coprime) and thus of the form K_{1} = K[α] where α^{n} = a and a ∈ K^{(φ)} for some character φ: G→U_{n}. Let χ = νφ^{−1} (where ν is defined by Equation (8)). Let ϖ be the automorphism defined by ϖ(α) = ρα. Let 1 ≠ τ ∈ Gal(K/F), and let τ′ be the extension of τ to K_{1}. By the normality of G′ in Gal(K_{1}/F), ϖτ′ϖ^{−1} = ϖ^{1−χ(τ)}τ′ is in G′, so that χ(τ) = 1 (otherwise 1 − χ(τ) is prime to n by Remark 2.1). □

Example 6.2. When [K:F] = 2 we can, by Remark 3.5, choose a representative of [a] ∈ K^{(ν)}/K^{×}^{n} such that τ(a) = a^{−1}. Let α = \sqrt[n]{a}, then K_{1} = K[α] with τ(α) = α^{−1} and ϖ(α) = ρα. Then

F_{1} = F [α + α^{−1}].

Writing a = a_{0} + a_{1}ρ for a_{0}, a_{1} ∈ F, the condition τ(a) = a^{−1} is equivalent to

a_{0}^{2} + (ρ + ρ^{−1})a_{0}a_{1} + a_{1}^{2} = 1.

Of course the general solution to this quadratic equation is obtained by taking a = bτ(b)^{−1} for arbitrary b = b_{0} + b_{1}ρ ∈ K, namely

a_{0} = (b_{0}^{2} − b_{1}^{2})/N and a_{1} = (2b_{0}b_{1} + (ρ + ρ^{−1})b_{1}^{2})/N

where b_{0}, b_{1} ∈ F and N = b_{0}^{2} + (ρ + ρ^{−1})b_{0}b_{1} + b_{1}^{2} = N_{K/F} (b_{0} + b_{1}ρ).

Since H^{1}(G, ℤ/n(χ)) = 0, the splitting of (9), and hence G′, is unique up to conjugation by elements of ⟨ϖ⟩.

Remark 6.3. If χ ≠ 1, then there are d = [K:F] distinct conjugates of G′ in Gal(K/F).

Proof. Let τ be a generator of Gal(K/F). Applying Remark 2.1 to χ(τ), we see from the equality ϖ^{i}τ′ϖ^{−i} = ϖ^{i(χ(τ′)−1)}τ′ that ⟨τ′⟩ is its own normalizer in Gal(K_{1}/K), and the result follows. □

The various copies G′ of G correspond to d isomorphic subfields of K_{1}, all having dimension n over F. This can be used to invert the map

J^{φ}: K^{(φ)}/K^{×}^{n}→H^{1}(F, μ_{n}(χ))

in the following way. Suppose we are given a cocycle c ∈ H^{1}(F, μ_{n}(χ)). By assumption c(σ_{1}σ_{2}) = σ_{1}_{∗}(c(σ_{2}))c(σ_{1}), so that

H = {σ ∈ Γ_{F} : c(σ) = 1}

is a subgroup of Γ_{F} (though in general not a normal subgroup). It also follows that c is well defined on right cosets of H, so that [Γ_{F} :H] divides n. Adjusting n if necessary, we may assume H is of index n. Now, F_{1} = F_{s}^{H} has dimension n over F, and

K_{1} = F_{1}⊗K

is a cyclic extension, generated by some α ∈ K such that a = α^{n} ∈ K. Choosing α so that ϖ defined by (7) will satisfy c(ϖ) = 1, it is immediate that a ∈ K^{(φ)} and c = J^{φ}(a).

We now describe explicitly how the action of Gal(K/F) on K extends to K_{1}. Let τ be a generator of Gal(K/F), and fix a ∈ K^{(φ)}. Let

r ≡ φ(τ) (mod n)

be an integer. By the assumption τ(a) ≡ a^{r} (mod K^{×}^{n}), so for some μ ∈ K^{×} we have τ(a) = μ^{n}a^{r}. Let α = \sqrt[n]{a} be a generator of K_{1} over K. For every j, τ_{1}(α) = ρ^{j}μα^{r} is a well defined automorphism of K_{1}. The fact that (9) splits ensures that j can be chosen so that τ_{1}^{d}(α) = α. Changing μ accordingly, we may assume that

(15)  τ′(α) = μα^{r},

where τ ↦ τ′ splits (9). In particular, choosing F_{1} = K_{1}^{τ′}, we can decompose K_{1} = K⊗_{F} F_{1}. The action of Gal(K/F) on K_{1} will be used in Section 10 to define an action of Gal(K/F) on cyclic K-algebras.

The action of Gal(K/F) on K_{1} which we just described can be refined in some cases, namely when φ does not generate the full character group Gal(K/F)^{#}. This refinement will be used in the construction of generic examples in Section 14.

Lemma 6.4. Let F ⊆ L ⊆ K be an intermediate subfield and e = [L:F], a divisor of d = [K:F]. Then

L^{×}K^{×}^{n} = K^{(1)}K^{(ν^{d/e})} … K^{(ν^{(e−1)d/e})}.

Proof. Using Proposition 3.2, this is a special case of Lemma 5.7 with φ_{0} = 1. □

Let a ∈ K^{(φ)} as before, and let e denote the order of φ in the character group Gal(K/F)^{#}; equivalently, that Ker(φ) = ⟨τ^{e}⟩; or that φ ∈ ⟨τ^{d/e}⟩. From the lemma it then follows that a can be taken to be an element of L = K^{τ^{e}}.

Proposition 6.5. Let L = K^{τ^{e}}, and assume a ∈ K^{(φ)} ∩ L^{×} for a character φ of order e, a proper divisor of d = [K:F]. Let α = \sqrt[n]{a} and K_{1} = K[α]. Then Gal(K/F) can be extended to K_{1} in a way that satisfies τ^{e}(α) = α. Moreover, if τ(α) = μα^{c} for μ ∈ K and c ≡ φ(τ) (mod n), then μ ∈ L.

If e = 2 we can choose c = −1, and then μ ∈ F.

Proof. Choose c = φ(τ), and let μ ∈ K^{×} be an element such that τ(a) = μ^{n}a^{c}. Extend τ to K_{1} by setting τ(α) = μα^{c}. Since τ^{e}(a) = a, (τ^{e}(α)α^{−1})^{n} = 1. Replacing α by ρ^{j}α, we still have α^{n} = a, but τ^{e}(α)α^{−1} is now multiplied by τ^{e}(ρ)ρ^{−1} = ρ^{j(φ(τ^{e})−1)}. By Remark 2.1, φ(τ)^{e} −1 is invertible mod n, so there is a (unique) j for which τ^{e}(α) = α.

Now compute that

τ^{e−1}(μ)τ^{e−2}(μ)^{c} … μ^{c^{e−1}}α^{c^{e}−1} = τ^{e}(α)α^{−1} = 1,

so applying τ and dividing by the c-power of this equality, we find that τ^{e}(μ) = μ.

Finally, assume c = −1, so that τ(α) = μα^{−1}. Then μ = ατ(α) which is a norm from K^{τ^{2}} to F. □

This proposition can be viewed (in the case e = n) as a direct proof that Equation (9) is a split extension.

7. An example

Many of the statements in this paper hold only if d = [K:F] is prime to n, and providing counterexamples in this direction is an interesting and worthy task. In this section we give one such example, and hint on some of the problem it arises.

Let F = ℚ[\sqrt{−2} ], and n = 4. Adding the forth root of unity ρ = \sqrt{−1}, we obtain the field K = F [ρ], where of course d = [K:F] = 2. Let τ denote the non-trivial automorphism of K over F. Notice that K contains an eighth root of unity, namely ρ_{8} = \frac{\sqrt{−2}}{2}(1 − ρ) (and τ(ρ_{8}) = −ρ_{8}^{−1}).

Let a = 1 + \sqrt{2}, then N_{K/F} (a) = −1 and τ(a) = −a^{−1} = ρ^{2}a^{−1}, so that a ∈ K^{(ν)} (where ν is the non-trivial character from Gal(K/F) = ⟨τ⟩ to U_{4} = {±1}). Let α be the forth root of a. Then by Proposition 3.4, K_{1} = K[α] is Galois over F. Indeed, τ(α) = ρ_{8}α^{−1} preserves the defining relation α^{4} = a, and so extends from K to an automorphism of K_{1} over F. In fact, τ^{2}(α) = τ(ρ_{8}α^{−1}) = −ρ_{8}^{−1}(ρ_{8}α^{−1})^{−1} = ρα, τ^{4}(α) = τ^{2}(ρα) = −α, and τ^{8}(α) = α; thus, Gal(K_{1}/F) = ⟨τ⟩ is a cyclic group of order 8.

In particular, the short exact sequence (9) does not split in this case, and there is no subfield F_{1} of K_{1} such that K = K⊗_{F} F_{1}. Also, The character χ, which comes from the action of Gal(K/F) on Gal(K_{1}/K), is trivial.

Finally, suppose c ∈ H^{1}(F, μ_{4}(χ)) = Hom(Γ_{F} , ℤ/4) is defined with the basic properties of Section 5, namely that c(σ) only depends on σ|_{K_{1}}, and that c(σ) = 1 for σ ∈ Γ_{K}. Since τ^{2} is the identity on K we have that c(τ^{2}) = 1, so c has order at most 2, and cannot be of order n = 4.

For the rest of the paper we again assume that d = [K:F] is prime to n.

8. Central Simple Algebras

Denote by _{n}Br(F) the exponent n part of the Brauer group of F. Applying the long exact sequence of cohomology groups to the Kummer sequence (4), and using the identification H^{2}(F, F_{s}^{×}) = Br(F) of the crossed products construction [10], we obtain the isomorphism H^{2}(F, μ_{n}(ν)) = _{n}Br(F), where ν is the natural action of Γ_{F} on μ_{n}, defined by (8). Since in general the action of G on tensor products M⊗N is diagonal, the cup product ∪: H^{i}(G, M)⊗H^{j}(G, N)→H^{i+j}(G, M⊗N) is in our notation a map

H^{1}(F, μ_{n}(φ))⊗H^{1}(F, μ_{n}(φ′)) ⟶^{∪} H^{2}(F, μ_{n}(φφ′)).

In particular, for every character φ, letting χ = νφ^{−1}, we have a map

H^{1}(F, μ_{n}(φ))⊗H^{1}(F, μ_{n}(χ)) ⟶^{∪} _{n}Br(F),

which we explicitly describe below. The case φ = 1 is well known (e.g. [18, p. 555]): an element γ ∈ H^{1}(F, μ_{n}(1)) = Hom(Γ_{F} , ℤ/n) has kernel Γ_{1} = Ker(γ), of index equal to the order of γ. Assuming γ has order n, let F_{1} = F_{s}^{Γ^{1}} be the invariant subfield, then F_{1} is a cyclic extension of dimension n over F. Now, given (b) ∈ H^{1}(F, μ_{n}(ν)) (which corresponds to an element b ∈ F^{×} by (5)), γ ∪ (b) is the Brauer class of the cyclic algebra

(F_{1}/F, ϖ, b) = F_{1}[z| zfz^{−1} = ϖ(f), z^{n} = b],

where ϖ is the generator of Gal(F_{1}/F) specified by γ(ϖ) = 1. This is always an F -central simple algebra of degree n. When the construction is applied to K = F [ρ_{n}] instead of F, we get what is called a symbol algebra,

(a) ∪ (b) = (a, b)_{K} = K[x, y| x^{n} = a, y^{n} = b, yxy^{−1} = ρx],

which is another notation for (K[\sqrt[n]{a}]/K, ϖ, b).

We call a pair of generators x, y a standard pair if x^{n} = a, y^{n} = b and yxy^{−1} = ρx. The symbol (a, b)_{K} is multiplicative in both variables, that is

(a, b_{1}b_{2})_{K} ∼ (a, b_{1})_{K}⊗(a, b_{2})_{K},
(a_{1}a_{2}, b)_{K} ∼ (a_{1}, b)_{K}⊗(a_{2}, b)_{K}.

By Merkurjev-Suslin theorem the symbol algebras of degree n generate nBr(K). Strictly speaking, not every algebra of the form (a, b)_{K} is a cyclic algebra, since K[x] = K[\sqrt[n]{a}] need not be a field. However it is always split by the cyclic group C_{n} (in the sense of [21], given at the end of the introduction).

Let φ, φ′: Gal(K/F)→U_{n} be characters such that φφ′ = ν. To describe the cup product H^{1}(F, μ_{n}(φ)) ∪ H^{1}(F, μ_{n}(φ′)), recall that the maps

J^{φ}: K^{(φ)}/K^{×}^{n}→H^{1}(F, μ_{n}(φ′)),
J^{φ′}: K^{(φ′)}/K^{×}^{n}→H^{1}(F, μ_{n}(φ))

are onto (Corollary 5.4).

Proposition 8.1. Let φ, φ′: Gal(K/F)→U_{n} be characters such that φφ′ = ν. Then for every a ∈ K^{(φ)} and b ∈ K^{(φ′)}, we have that

J^{φ}(a) ∪ J^{φ′}(b) ∼ cor_{K/F} (a, b)_{K}^{⊗d′},

where d′ is the inverse of d modulo n.

Proof. From Theorem 5.3 we have that (a) = res_{K/F} J^{φ}(a) and (b) = res_{K/F} J^{φ′}(b), where (a), (b) ∈ H^{1}(K, ℤ/n). Thus, the projection formula gives

cor(a, b)_{K} = cor((a) ∪ (b))
= cor(resJ^{φ}(a) ∪ resJ^{φ′}(b))
= J^{φ}(a) ∪ cor(resJ^{φ′}(b))
= d · J^{φ}(a) ∪ J^{φ′}(b).
□

Notice that if D has degree n over K, then cor_{K/F} D has degree n^{[K:F]} over F. From time to time we will use degree argument to write that the corestriction is equal (and not just similar) to a given algebra.

Corollary 8.2. Under the assumptions of Proposition 8.1,

res_{K/F} cor_{K/F} (a, b) = (a, b)^{⊗d}.

Proof. By the commutativity of restriction with the cup product,

res cor(a, b)_{K} = d · res(J^{φ}(a) ∪ J^{φ′}(b))
= d · (a) ∪ (b)
= [(a, b)_{K}^{⊗d}].

□

We now want to compute the algebra cor_{K/F} (a, b)_{K} for a ∈ K^{(φ)} and b ∈ K^{(φ′)}, without the assumption φφ′ = ν. Just as above, the general projection formula yields

cor((a) ∪ (b)) = d · J^{φ}(a) ∪ J^{φ′}(b),

but the corestriction here is from_{n}Br(K) = H^{2}(K, ℤ/n) to H^{2}(F, μ_{n}(φφ′)), which is in general not the corestriction of central simple algebras. To compute the algebra cor_{K/F} (a, b)_{K}, we need to know what are the conjugate algebras τ(a, b) for τ ∈ Gal(K/F) (see [14, p. 220]).

If σ ∈ Γ_{K}, then σ(a, b)_{n;K} = (σa, σb)_{n;K}. More generally, let σ ∈ Γ_{F} be an automorphism. Letting

D = K[x, y| x^{n} = σa, y^{n} = σb, yxy^{−1} = σρ = ρ^{t}]

where t = ν(σ) and tt′ ≡ 1 (mod n), we see that x^{t′}, y generate D, with yx^{t′}y^{−1} = ρ, so by our definition σ(a, b)_{K} = D is the symbol algebra (σa^{t′}, σb)_{K}.

Proposition 8.3. Suppose a ∈ K^{(φ)} and b ∈ K^{(φ′)}, where φφ′ ≠ ν. Then the corestriction of algebras is

cor_{K/F} (a, b)_{K} ∼ F.

Proof. Let ϕ = φφ′ν^{−1} (ϕ ≠ 1 by the assumption). For a ∈ K^{(φ)}, b ∈ K^{(φ′)} we have σ(a, b)_{K} ∼ (σa, σb)_{K}^{⊗ν(σ)^{−1}} ∼ (a, b)_{K}^{⊗ϕ(σ)}.

It is known (e.g. [20]) that

res_{K/F} cor_{K/F} (a, b) = ∏_{τ∈Gal(K/F)} τ(a, b),

so we have

res_{K/F} cor_{K/F} (a, b) = ∏ (a, b)^{⊗ϕ(τ)}
= (a, b)^{⊗ ∑ ϕ(τ)}

and the result follows from Corollary 2.2, and the injectivity of res_{K/F}.

□

Let R_{n,K}: K_{2}(K)/n→_{n}Br(K) denote the norm residue map

R_{n,K}: {a, b} ↦ (a, b)_{K}.

Corollary 8.4. There is no map making the diagram (16) commutative, unless K_{2}(F)/n = 0.

(16)  \xymatrix{K_{2}(F)/n \ar@{.>}[d] \ar[r]^{res} & K_{2}(K)/n \ar[d]^{_{R_{n,K}}} \\ _{n}Br(F) \ar[r]^{res} & _{n}Br(K)}

Proof. Let R̃_{n,F} denote a map making the diagram commute. Then for α, β ∈ F we have that

res_{K/F} R̃_{n,F} {α, β}_{F} = R_{n,K}res_{K/F} {α, β}_{F} = (α, β)_{K}.

Taking corestriction, we get

(R̃_{n,F} {α, β}_{F} )^{⊗d} = cor_{K/F} (α, β)_{K} ∼ F,

so that R̃_{n,F} = 0. □

By the above computation, the Gal(K/F)-eigenspace of_{n}Br(K) with respect to a character ψ is the subgroup

⟨[(a, b)] : a ∈ K^{(φ)}, b ∈ K^{(φ′)}, φφ′ = ψ⟩.

Combining Corollary 8.2 and the proposition, we obtain the projection formula for K/F: given a, b ∈ K^{×}, decompose a = ∏ a_{ψ} and b = ∏ b_{ψ} mod K^{×}^{n} with a_{ψ}, b_{ψ} ∈ K^{(ψ)}; then

(17)  res_{K/F} cor_{K/F} (a, b) ∼ ⊗_{φφ′=ν}(a_{φ}, b_{φ′})^{⊗d}.

Corollary 8.5. Let φ, φ′ be characters with φφ′ = ν, and let a ∈ K^{(φ)}, b ∈ K^{(φ′)}, and D = (a, b)_{n,K}.

Let D_{0} be a central simple algebra of degree n over F. The following conditions are equivalent.

a. res_{K/F} D_{0} = D.

b. D_{0} ∼ cor_{K/F} D^{⊗d′} where d′ is an inverse of d modulo n.

Corollary 8.6. Let D be a central simple algebra of degree n over F. If D⊗_{F} K = (a, b)_{K} and a ∈ K^{(φ)}, then we can assume b ∈ K^{(φ′)} where φφ′ = ν.

Proof. Let b_{φ′} be the component of b in K^{(φ′)}, then cor_{K/F} (a, b)_{K} = cor_{K/F} (a, b_{φ′})_{K} by Proposition 8.3. By the previous corollary D ∼ cor_{K/F} (a, b)_{K}^{⊗d′} = cor_{K/F} (a, b_{φ′})_{K}^{⊗d′}, and then also res_{K/F} D = (a, b_{φ′})_{K}.

□

9. Second cohomologies of F

In this section we consider eigenspaces of the second cohomology group H^{2}(K, μ_{n}) for K = F [ρ]. Applying the cup product to the decomposition H^{1}(K, μ_{n}) = ⊕_{φ}H^{1}(K, μ_{n})^{(φ)}, we obtain the subspaces

H^{1}(K, μ_{n})^{(φ)} ∪ H^{1}(K, μ_{n})^{(φ′)} ⊆ H^{2}(K, μ_{n})^{(φφ′)}.

The ’onto’ part of the Merkurjev-Suslin Theorem [20, Thm 8.5], applied to K, shows that H^{1}(K, μ_{n}) ∪ H^{1}(K, μ_{n}) = H^{2}(K, μ_{n}), so in fact

(18)  H^{2}(K, μ_{n}) = ⊕_{ψ}( ∑_{φφ′=ψ} H^{1}(K, μ_{n})^{(φ)} ∪ H^{1}(K, μ_{n})^{(φ′)}).

and in particular for every ψ,

(19)  H^{2}(K, μ_{n})^{(ψ)} = ∑_{φφ′=ψ} H^{1}(K, μ_{n})^{(φ)} ∪ H^{1}(K, μ_{n})^{(φ′)}.

Similarly to Corollary 5.6, we have the following isomorphism:

Theorem 9.1. For every character ψ: Gal(K/F)→U_{n},

H^{2}(K, μ_{n})^{(ψ)} ≅ H^{2}(F, μ_{n}(θ)),

where θ = ν^{2}ψ^{−1}.

Proof. We will show that the restriction res : H^{2}(F, μ_{n}(θ))→H^{2}(K, μ_{n}) (which is obviously injective) covers H^{2}(K, μ_{n})^{(ψ)}.

Recall the general projection formula for cohomology groups: let M, N be Γ_{F} -modules, a ∈ H^{1}(F, M) and b ∈ H^{1}(K, N). Then

cor(res(a) ∪ b) = a ∪ cor(b),

where the restriction is H^{1}(K, M)→H^{1}(F, M), the first corestriction is H^{2}(F, M⊗N)→H^{2}(K, M⊗N) and the second is H^{1}(F, N)→H^{1}(K, N).

Let φ, φ′ be arbitrary characters. We choose M = N = μ_{n} with the actions of Γ_{F} on M, N via χ = νφ^{−1} and θχ^{−1}, respectively, so our restriction and corestriction are the maps

res_{χ} : H^{1}(F, μ_{n}(χ))→H^{1}(K, μ_{n}),
cor_{θ} : H^{2}(K, μ_{n})→H^{2}(F, μ_{n}(θ)),
cor_{θχ^{−1}} : H^{1}(K, μ_{n})→H^{1}(F, μ_{n}(θχ^{−1})).

We apply the projection formula to an element

u = c ∪ c′ ∈ H^{1}(K, μ_{n})^{(φ)} ∪ H^{1}(K, μ_{n})^{(φ′)}

in the decomposition (18). From Corollary 5.6 we have that

(20)  res_{χ}H^{1}(F, μ_{n}(χ)) = H^{1}(K, μ_{n})^{(φ)}

and similarly

(21)  res_{χ′}H^{1}(F, μ_{n}(χ′)) = H^{1}(K, μ_{n})^{(φ′)}

for χ′ = νφ^{′−1}, so we can write c = res_{χ}a and c′ = res_{χ′}a′ where a ∈ H^{1}(F, μ_{n}(χ)) and a′ ∈ H^{1}(F, μ_{n}(χ′)). It follows that

cor_{θ}(c ∪ c′) = cor_{θ}(res_{χ}(a) ∪ res_{χ′}(a′)) = a ∪ cor_{θχ^{−1}}res_{χ′}(a′).

If χχ′ = θ, then we find that cor_{θ}(c ∪ c′) = d · (a ∪ a′) since cor_{χ′}res_{χ′} is multiplication by d. On the other hand if χχ′ ≠ θ then by Corollary 5.5 we have that cor_{θχ^{−1}}res_{χ′}(a′) = 0, so that cor_{θ}(c ∪ c′) = 0.

Applying cor_{θ} to (18) and recalling that cor_{θ} is onto H^{2}(F, μ_{n}(θ)), we find that

(22)  H^{2}(F, μ_{n}(θ)) = ∑_{χχ′=θ} H^{1}(F, μ_{n}(χ)) ∪ H^{1}(F, μ_{n}(χ′)).

Finally applying res_{θ} = res_{χ} ∪ res_{χ′} (for every χχ′ = θ), and using Equations (19), (20) and (21), we obtain

res_{θ}H^{2}(F, μ_{n}(θ)) = ∑_{φφ′=ψ} H^{1}(K, μ_{n})^{(φ)} ∪ H^{1}(K, μ_{n})^{(φ′)} = H^{2}(K, μ_{n})^{(ψ)}.

□

The onto part of Merkurjev-Suslin for the field F gives the equality H^{2}(F, μ_{n}(ν^{2})) = H^{1}(F, μ_{n}(ν)) ∪ H^{1}(F, μ_{n}(ν)), which is of course much better than Equation (22) for ψ = 1 (and θ = ν^{2}). It is not known if H^{2}(F, μ_{n}(ν)) = H^{1}(F, μ_{n}(ν)) ∪ H^{1}(F, μ_{n}(1)) in general (see Question 10.6 below). A related negative result is given in the remark after Lemma 29 in [9].

10. Action of Gal(K/F) on (a, b)_{K}

In Section 8 we have seen that if a ∈ K^{(φ)} and b ∈ K^{(φ′)} where φφ′ = ν, then τ(a, b)_{K} = (a, b)_{K} for every τ ∈ Gal(K/F).

Proposition 10.1. Let φ, φ′ be characters such that φφ′ = ν, a ∈ K^{(φ)} and b ∈ K^{(φ′)}.

There is a natural extension of the action of Gal(K/F) on K to an action on D = (a, b)_{K}.

Proof. As in Section 6, fix a generator τ of Gal(K/F), let r ≡ φ(τ), r′ ≡ φ′(τ), and t = rr′ ≡ ν(τ). By the assumption on a, b, there exist μ, η ∈ K^{×} such that

τ(a) = μ^{n}a^{r} and τ(b) = η^{n}b^{r′}.

Define τ on

D = K[x, y| x^{n} = a, y^{n} = b, yxy^{−1} = ρx]

by

τ(x) = μx^{r},
τ(y) = ηy^{r′},

where μ, η are chosen (i.e. multiplied by appropriate powers of ρ) so that τ^{d}(x) = x and τ^{d}(y) = y. Since τ(yxy^{−1}) = μy^{r′}x^{r}y^{−r′} = μρ^{rr′}x^{r} = τ(ρx), this action respects the defining relations, and is thus well defined on D. Finally, τ^{d} is the identity on K[x, y] = D. □

Lemma 10.2. Let φ, φ′ be characters with φφ′ = ν, and let a ∈ K^{(φ)}, b ∈ K^{(φ′)}. Let D = (a, b)_{n,K} be a cyclic algebra over K.

Let D_{0} be a central simple algebra of degree n over F. The following conditions are equivalent.

a. res_{K/F} D_{0} = D.

b. D_{0} ∼ cor_{K/F} D^{⊗d′} where d′ is an inverse of d modulo n.

c. D_{0} is the invariant subalgebra D^{Gal(K/F)} under the action defined in Proposition 10.1.

Proof. The equivalence of a. and b. is Corollary 8.5.

Let G = Gal(K/F). In general the invariant subalgebra D^{G} is central simple of degree n over F = K^{G} [14, Cor. 7.2.15]. Since res_{K/F} D^{G} = D^{G}⊗_{F} K = D and the restriction _{n}Br(F)→_{n}Br(K) is injective, a. is equivalent to c. □

The following theorem is proved in [11] for n prime.

Theorem 10.3. Let F be a field of characteristic prime to n = p^{m}, K = F [ρ_{n}], and assume that d = [K:F] is prime to n.

Let D_{0} be a central simple algebra of degree n over F. D_{0} is cyclic if and only if there are a ∈ K^{(ν)} and β ∈ F^{×} where K_{1} = K[\sqrt[n]{a}] is a field, such that res_{K/F} D_{0} = (a, β)_{n,K}.

Proof. If a, β are as assumed, then G acts on D = (a, β)_{K} (by Proposition 10.1), K_{1} = K[\sqrt[n]{a}] is a subfield of D, and F_{1} = K_{1}^{G} is a cyclic subfield of D^{G} (Proposition 6.1).

On the other hand, if D_{0} is cyclic of degree n over F, it has a cyclic subfield F_{1} of dimension n over F, which again by Proposition 6.1 has the form F_{1} = K_{1}^{G} where K_{1} = K[\sqrt[n]{a}] is a field, for some a ∈ K^{(ν)}. As D_{0} is cyclic, we can write D_{0} = (F_{1}/F, ϖ, β) for some β ∈ F, and then D = D_{0}⊗K = (K_{1}/K, ϖ, β) = (a, β)_{K}. □

Definition 10.4. An algebra D_{0} of degree n over F is called a quasisymbol (of type (φ, φ′)) iff it satisfies the conditions of Lemma 10.2.

We give a description of quasi-symbols in terms of their splitting fields in Proposition 13.1 below.

Under the assumption that [K:F] is prime to n, the corestriction from _{n}Br(K) to _{n}Br(F) is onto, so by the projection formula (17), nBr(F) is generated by quasi-symbols.

Question 10.5. Are quasi-symbols cyclic over F?

By the Theorem 10.3, quasi-symbols of type (1, ν) (or (ν, 1)) are cyclic, but nothing is known in general about the cyclicity of other quasi-symbols.

Merkurjev proved in [11] that _{n}Br(F) is generated by cyclic algebras (of degree dividing n) if [K:F] ≤ 3. This is not known if [K:F] ≥ 4.

Question 10.6. Is _{n}Br(F) generated by (the classes of) algebras which are split by C_{n}?

Of course, a positive answer to the first question will answer the second. One approach is to compute the corestriction of cyclic algebras from K to F; we return to this in Section 12.

We conclude this section with a remark concerning involutions of the second kind. We touch this subject again in Proposition 11.6. In Proposition 8.3 we saw that algebras of the form A = (a, b)_{K} where a ∈ K^{(φ)} and b ∈ K^{(φ′)} and φφ′ ≠ ν, have trivial corestriction. If [K:F] = 2, this is known to be equivalent to the existence of an involution of the second kind on A for which the invariant subfield of the center is F. Since we assume [K:F] = 2, there are only two cases to consider: either a, b ∈ K^{(1)} = F^{×}K^{×}^{n}, or a, b ∈ K^{(ν)}. A similar configuration is studied in [7, Thm. 3.5]; it is shown there that if [K:F] = 2 and a cyclic algebra A = (α, b)_{K} has an involution of the second kind over F where α ∈ F, then it can be assumed that b ∈ F. This theorem covers also the case of even n (where it is assumed that τ(ρ) = ρ^{−1}).

Similarly to Proposition 10.1, we now use the action of Gal(K/F) on K[x] and K[y] (where x, y are a standard pair of generators of A) to define an involution of the second kind on A.

Assume that [K:F] = 2, and let τ ∈ Gal(K/F) be the non-trivial automorphism. Note that τ(ρ) = ρ^{−1}. Let φ(τ) = ϵ define a character, where ϵ = ±1 is fixed. Let a, b ∈ K^{(φ)} and

A = (a, b)_{K} = K[x, y| x^{n} = a, y^{n} = b, yx = ρxy].

By assumption there are μ, η ∈ K^{×} such that τ(a) = μ^{n}a^{ϵ} and τ(b) = η^{n}b^{ϵ}. Define a linear map (∗) on A by

(∑ k_{ij}x^{i}y^{j})^{∗} = ∑ τ(k_{ij})(y^{∗})^{j}(x^{∗})^{i}

where x^{∗} = μx^{ϵ} and y^{∗} = ηy^{ϵ}. The relations x^{n} = a and y^{n} = b are automatically preserved, and for both possible values of ϵ, (yx)^{∗} = (ρxy)^{∗} = ρ^{−1}μy^{ϵ}ηx^{ϵ} = μηx^{ϵ}y^{ϵ} = x^{∗}y^{∗}. Finally, x^{∗∗} = x and y^{∗∗} = y by the choice of μ, η, so that u ↦ u^{∗} is an involution.

11. Albert’s cyclicity criterion

Every cyclic algebra D = (F_{1}/F, ϖ, b) has by definition an element z ∈ D such that z^{n} = b is central. In the ”special results” chapter (Chapter XI) of his seminal book [5], Albert proves the converse, which is known as Albert’s cyclicity criterion: a central simple algebra D of prime degree p over a field F of characteristic prime to p is cyclic, if and only if u^{p} ∈ F for some non-central u ∈ D. Of course the assertion follows from Kummer theory if F has p-roots of unity. Albert remarks [5, p. 175] that the criterion fails if the degree is not a prime, and indeed in [6] he provides an example of a non-cyclic algebra of degree (and exponent) 4, with an element u such that u^{4} is central but u^{2} is not. In Theorem 11.4 below we generalize Albert’s criterion, and show that it holds for algebras of degree n = p^{m} over a field F, assuming that d = [K:F] is prime to n, where as always K = F [ρ_{n}].

The following is a slightly more general version of Theorem 10.3.

Proposition 11.1. Let n be a prime power, K = F [ρ] where ρ is a primitive n-root, and assume d = [K:F] is prime to n.

Let a ∈ K^{×} and β ∈ F^{×}. Then A = cor_{K/F} (a, β)_{n,K} is split by C_{n}.

Proof. Since β ∈ F^{×} ⊆ K^{(1)}, we can by Corollary 8.6 replace a by its ν-component, so we assume a ∈ K^{(ν)}. Let D = (a, β^{d})_{K}. The projection formula for K/F (Equation (17)) gives res_{K/F} A ∼ D, which is an algebra of degree n. Since d is prime to n, the index of A divides n.

Let D_{0} be a the algebra of degree n over F which is similar to A. Then res_{K/F} D_{0} = D, so if K[\sqrt[n]{a}] is a field, D_{0} is cyclic by Theorem 10.3. For the general case, let n′ denote the order of a in K^{×}/K^{×}^{n}; equivalently, n′ is the maximal divisor of n such that a ∈ K^{×}^{n/n′}. Write a = c^{n/n′} for some c ∈ K^{×}; then

D = (a, β^{d})_{n,K} = (c^{n/n′}, β^{d})_{n,K} ∼ (c, β^{d})_{n′,K},

and K[ \sqrt[n′]{c}] is a field. In particular,

A ∼ cor_{K/F} D^{⊗d′} ∼ cor_{K/F} (c, β)_{n′,K},

where d′ is the inverse of d modulo n. Let D_{1} be the algebra of degree n′ similar to A. By Lemma 10.2 we have that res_{K/F} D_{1} = cor_{K/F} (c, β)_{n′,K}. In order to show that D_{1} is cyclic, we need to know that c is in the ν-component of the decomposition of K^{×}/K^{×}^{n′}. Indeed, Let τ be a generator of Gal(K/F), and t = ν(τ). For some μ ∈ K^{×} we have τ(a) = μ^{n}a^{t}, so that τ(c) = (ρ^{i}μ)^{n′}c^{t} for some i. We are now done by Theorem 10.3. □

Example 11.2. Assume that the p-part of the group of units in K is of order n, and take a = ρ^{n/n′} in the above corollary. Then a is of order n′ in K^{×}/K^{×}^{n}, and

cor_{K/F} (ρ^{n/n′}, β)_{K} ∼ (F_{1}/F, ϖ, β^{d}),

a cyclic algebra of degree n′, with F_{1} ⊆ K_{1} = F [ \sqrt[n′]{ρ}].

Corollary 11.3. With F, n, K as in the proposition, let D_{0} be a division algebra of degree n over F. If res_{K/F} D_{0} = (a, β) for some a ∈ K^{×} and β ∈ F^{×}, then D_{0} is cyclic.

Proof. Let d′ be an inverse of d modulo n. By Lemma 8.5 D_{0} ∼ cor_{K/F} (a, β^{d′}), so by the proposition D_{0} is similar to a cyclic algebra of degree dividing n, and being a division algebra it is cyclic. □

This leads directly to the general form of Albert’s theorem.

Theorem 11.4. Let F be a field as above. Let D be a central division algebra of degree n over F, with an element u ∈ D such that [F [u]:F] = n and β = u^{n} ∈ F^{×}. Then D is cyclic.

Proof. Since F [u] splits D, we have that K[u] = K⊗F [u] splits K⊗D. Thus, there exists an element a ∈ K^{×} such that res_{K/F} D = (a, β)_{K}, and we are done by the corollary. □

A standard argument using idempotents gives the following more general result.

Corollary 11.5. Let F be a field as above. Let D be a central simple algebra of degree n over F, with an element u ∈ D such that [F [u]:F] = n and β = u^{n} ∈ F^{×}. Then D is split by C_{n}.

Proof. If F [u] is a field then D is cyclic by the proof given above. Otherwise, let n′ be the order of u in F^{×}/F^{×}^{n}. Let D_{1} = D⊗_{F} K. The subring K[u^{n′}] of D_{1} is isomorphic to a direct product of n/n′ copies of K. Let C = Cent_{D_{1}}(K[u^{n′}]). Taking a minimal idempotent e_{1} of K[u^{n′}], Ce_{1} is a central simple algebra of degree n′ over Ke_{1} ≅ K and Ke_{1}[ue_{1}] is a subfield of dimension n′ of Ce_{1}, so Ce_{1} is cyclic. We are done since D_{1} ∼ Ce_{1}. □

It would be interesting to have examples of non-cyclic algebras of odd degree n = p^{2} with an n-central element, when [K:F] is divisible by p. As mentioned above, Albert gave such an example for p = 2.

We prove another generalization of Albert’s theorem, due to N. Elkayam [7, Thm. 3.9]. We give only the version of prime degree, as the generalization to prime-power degree along the lines drawn above is easy.

Proposition 11.6. Let D be a central division algebra of degree p over F, with an involution of the second kind over F/F_{0}, and assume F_{0}[ρ] = F [ρ].

If D has a symmetric element u ∉ F such that u^{p} ∈ F, then D contains a subfield F_{1} Galois over F_{0}, such that Gal(F_{1}/F_{0}) is dihedral.

Proof. Let d = [K:F], and let μ be the element of order 2 in the character group of Gal(K/F_{0}) ≅ ℤ/2d. Let ν_{0} : Gal(K/F_{0})→U_{p} be the character defined by Equation (8) for the extension K/F_{0}, and ν the restriction to Gal(K/F); there are two characters of Gal(K/F_{0}) whose restriction is ν, namely ν_{0} and ν_{0}μ. By Lemma 5.7 we have that K^{(ν)} = K^{(ν_{0})}K^{(ν_{0}μ)}, where K^{(ν)} is the component with respect to the action of Gal(K/F) on K^{×}, and K^{(ν_{0})}, K^{(ν_{0}μ)} are components with respect to the action of Gal(K/F_{0}).

By the assumption, α = u^{p} ∈ F_{0}. Moreover, K[u] ⊆ D⊗_{F} K is a cyclic subfield, and we can write D_{1} = D⊗_{F} K = (α, b)_{K} for some b ∈ K^{×}. By Corollary 8.6, we may assume b ∈ K^{(ν)}. Let b = b_{ν_{0}}b_{ν_{0}μ} be the decomposition to eigenvectors with respect to the action of Gal(K/F_{0}). Since cor_{F/F_{0}}D ∼ F_{0}, we have that F_{0} ∼ cor_{F/F_{0}}D^{⊗d} = cor_{F/F_{0}}cor_{K/F} D_{1} = cor_{K/F_{0}}(α, b)_{K} = cor_{K/F_{0}}(α, b_{ν_{0}})_{K}, so that in fact D^{⊗d} ∼ cor_{K/F} res_{K/F} D ∼ cor_{K/F} (α, b_{νμ})_{K}. Letting D_{2} = (a, b_{νμ})_{K} we have that D = D_{2}^{Gal(K/F)} by Lemma 10.2.

But now, K_{1} = K[\sqrt[p]{b_{νμ}}] is Galois over F_{0} with Galois group ℤ/d ⋊ ℤ/p, where the action is by the character μ. In particular, letting τ be a generator of Gal(K/F_{0}), τ^{2} is central in Gal(K/F_{0}), and K_{1}^{τ^{2}} ⊆ D_{2}^{τ^{2}} = D is Galois over F_{0}, with Gal(K_{1}^{τ^{2}}/F_{0}) ≅ D_{p}. □

12. Corestriction of cyclic algebras

Rosset and Tate proved in [13] that if K_{2}/K_{1} is an extension of fields in which K_{1} has n-roots of unity, then the corestriction of a cyclic algebra of degree n over K_{2} to K_{1} is similar to a product of at most [K_{2} :K_{1}] cyclic algebras of the same degree. No similar result is known if K_{1} does not have roots of unity (see [14], the remark after Corollary 7.2.39). A general result in this direction should not be easy to obtain, as it would imply a positive answer to Question 10.6.

In [11, Lemma 2], Merkurjev proved that if [K:F] ≤ 3 (for an arbitrary fields extension K/F), then K_{2}(K) is generated by symbols of the form {K, F } (i.e. with one entry in F; this is how Question 10.6 is solved in this case). We give an explicit version of this useful lemma.

Lemma 12.1. Let K/F be any extension of fields, and let a, b ∈ K, a, b ∉ F. Then the following identities hold in K_{2}(K).

a. Assume [K:F] = 2, and write b = α + βa, α, β ∈ F. Then

(23)  \{a, b\} = \begin{cases}\{\frac{b}{β} , −\frac{β}{α}\} + \{a, α\} & if α ≠ 0 \\ \{a, −β\} & if α = 0 \end{cases}

b. Assume [K:F] = 3, and write b = g(a), g ∈ F [λ] a polynomial of degree ≤ 2. Since the minimal polynomial of a is of degree 3, there exist α + βλ, α′ + β′λ ∈ F [λ] such that (α + βa) · g(a) = α′ + β′a. Then we have (where in each case we only indicate the non-zero coefficients):

\{a, b\} = \begin{cases}\{a, −\frac{β′}{α} \} & α′ = β = 0 \\ \{a, −\frac{α′}{β} \} & α = β′ = 0 \\ \{a, −β′\} − \{b, −\frac{β}{α}\} − \{βa, α\} & α′ = 0 \\ \{a, α′\} − \{b, −\frac{β}{α}\} − \{βa, α\} & β′ = 0 \\ \{b, −\frac{β′}{α′} \} + \{β′a, α′\} − \{a, −β\} & α = 0 \\ \{b, −\frac{β′}{α′} \} + \{β′a, α′\} − \{a, α\} & β = 0 \\ \{a, \frac{α′}{α} \} + \{\frac{α′+β′a}{α′}, \frac{−β′}{α′} \} + \{\frac{α+βa}{α}, \frac{−α}{β}\} \end{cases}

Proof. For the [K:F] = 2 case, compute that

\{a, b\} − \{a, α\} = \{a, 1 + \frac{βa}{α}\}
= \{a, 1 + \frac{βa}{α}\} − \{−\frac{βa}{α}, 1 + \frac{βa}{α}\}
= \{−\frac{α}{β}, \frac{b}{α}\} = \{\frac{b}{α}, −\frac{β}{α}\} = \{\frac{b}{β} , −\frac{β}{α}\},

where the last equality follows from \{\frac{β}{α}, −\frac{β}{α}\} = 0. The case [K:F] = 3 follows from {a, b} = {a, α′ + β′b} − {a, α + βb}, applying the case of dimension 2 for each of these symbols. □

We can now prove a version of the Rosset-Tate theorem for the corestriction from K to F, if [K:F] ≤ 3 (under the usual assumptions that [K:F] is prime to n).

Proposition 12.2. Let D = (a, b)_{n,K} be a cyclic algebra over K. If [K:F] ≤ 3, then cor_{K/F} D is similar to a product of at most [K:F] algebras which are split by C_{n}.

Proof. The corestriction of an algebra of the form (c, γ), c ∈ K and γ ∈ F, is split by C_{n} by Proposition 11.1. Expressing (a, b)_{K} as a product of at most [K:F] symbols of this form (using the previous lemma), we are done by taking corestriction.

We remark that the case [K:F] = 2 also follows from decomposing a = a_{1}a_{ν} and b = b_{1}b_{ν} as in Equation (17), for then cor(a, b)_{K} ∼ cor(a_{1}, b_{ν})_{K}⊗cor(a_{ν}, b_{1})_{K}. □

Corollary 12.3. Let L/F be a separable extension of fields of characteristic prime to n, such that d = [K:F] ≤ 3 where K = F [ρ]. If D is cyclic of degree n over L, then cor_{L/F} D is similar to a product of at most

d · ([L:F] − 1) + 1

algebras which are split by C_{n}.

Proof. Let m = [L:F]. If K ⊆ L then in fact we only need [L:F] cyclic algebras, since cor_{L/K}D is similar to a product of [L:K] cyclics of degree n over K, and the corestriction down to F requires [K:F] cyclics to each of these by the proposition.

So suppose L does not contain K, and let T = L[ρ] = L⊗_{F} K. Since D is cyclic over L, we can (by Proposition 10.3) write res_{T/L}D = (a, β)_{n} for a ∈ T and β ∈ L. Let D_{2} = (a^{d′}, β)_{n}, where dd′ ≡ 1 (mod n), and note that

cor_{L/F} D ∼ cor_{L/F} cor_{T/L}D_{2} = cor_{K/F} cor_{T/K}D_{2}.

By the Rosset-Tate algorithm, cor_{T/K}D_{2} is a product of [T:K] = m symbols, the first of which is (c, N_{T/K}(β))_{K} for some c ∈ K [14, Cor. 7.2.38]. Since N_{T/K}(β) = N_{L/F} (β) ∈ F, the corestriction from K to F of this first symbol is split by C_{n} (Proposition 11.1). The corestriction of each of the other m − 1 symbols is similar to a product of d cyclic algebras over F by the previous proposition. □

The bound d · ([L:F] − 1) + 1 of the corollary is higher than the [L:F] value suggested by the Rosset-Tate result. In the proof we write res_{T/L}D = (a, β) for a ∈ T and β ∈ L, and ignores the fact that a ∈ T^{(ν)}. This can be used to improve the bound if [L:F] = [K:F] = 2.

Theorem 12.4. Suppose that K = F [ρ] is quadratic over F, and let L/F be any separable quadratic extension. Let D_{1} be a cyclic algebra of degree n (an odd prime power) over L.

Then cor_{L/F} D_{1} is similar to a product of at most two algebras over F which are split by C_{n}.

Proof. If L = K then the result follows from Proposition 12.2, and we assume this is not the case. Let

T = L[ρ] = L⊗_{F} K,

a quadratic extension of K. By Theorem 10.3,

D_{2} = res_{T/L}D_{1} = (a, β)_{T}

where a ∈ T^{(ν)} and β ∈ L; in fact we can assume N_{T/L}(a) = 1 (see Remark 3.5). We compute cor_{T/K}D_{2} using Equation (23). If β ∈ F ⊆ K then cor_{T/K}D_{2} = (N_{T/K}(a), β)_{K} is cyclic, and the corestriction from K to F is split by C_{n} by Proposition 11.1. We thus assume β ∉ F, so that K[β] = T.

Choose c, d ∈ K such that a = d + cβ. By the above argument we may assume a ∉ K, so that c ≠ 0. Similarly if d = 0 then (a, β) = (cβ, β) = (−c, β), and we are back in the case a ∈ K. We thus assume d ≠ 0.

By Equation (23) (after taking opposite of all the symbols involved),

D_{2} = (a, β)_{T} ∼ (−\frac{c}{d}, \frac{a}{c})_{T} ⊗_{T} (d, β)_{T} ,

so that

(24)  cor_{T/K}D_{2} = (−\frac{c}{d}, N_{T/K}(\frac{a}{c}))_{K}⊗_{K}(d, N_{L/F} (β))_{K}.

The right-hand symbol has at least one entry, namely N_{L/F} (β), in F, and so its corestriction from K to F is split by C_{n}. Let σ denote the non-trivial automorphism of L/F (which extends to T/K as the identity on K). We will show that N_{T/K}(a/c) ∈ F. Consider the polynomial

g(λ) = N_{K/F} (λ + \frac{d}{c}) − N_{K/F} (\frac{1}{c}),

which is quadratic over F. We have that

g(β) = N_{T/L}(a/c) − N_{K/F} (1/c) = 0,

so that g is the minimal polynomial of β, and σ(β) is its other root. It follows that g(λ) = λ^{2}−(β +σ(β))λ+β ·σ(β), so comparing coefficients we obtain

β + σ(β) = − (\frac{τ(d)}{τ(c)} + \frac{d}{c}) ,

β · σ(β) = \frac{τ(d)d − 1}{τ(c)c},

where τ generates Gal(K/F). We can now compute that

(25)  N_{T/K}(a/c) = N_{T/K}(d/c + β)
= (d/c + β)(d/c + σ(β))
= \frac{d^{2}}{c^{2}} − \frac{d}{c} (\frac{τ(d)}{τ(c)} + \frac{d}{c}) + \frac{τ(d)d − 1}{τ(c)c}
= −\frac{1}{τ(c)c}.

It follows that N_{T/K}(a/c) ∈ F, and the left-hand symbol in Equation (24) also has one entry in F. The corestriction from K to F of cor_{T/K}D_{2} is thus similar to the product of two algebras split by C_{n}. □

By induction, we obtain

Corollary 12.5. Let L/F be a separable extension and assume there is a chain of quadratic extensions from F to L. Also assume that [K:F] = 2 where K = F [ρ].

For every cyclic algebra D_{1} of degree n over L, cor_{L/F} D_{1} is similar to a product of at most [L:F] algebras split by C_{n}.

In the situation described in Theorem 12.4, it is obvious that D_{2} = res_{T/L}D_{1} is invariant under the action of Gal(K/F), and that the corestriction cor_{T/K}D_{2} is a product of two symbol algebras over K. This can be improved:

Remark 12.6. The algebra D_{2} of Theorem 12.4 satisfies

(26)  cor_{T/K}D_{2} = (τ(c), c)_{K}⊗_{K}(d, 1 − τ(d)d)_{K},

where both symbols in this expression are invariant under Gal(K/F), and their corestriction to F is split by C_{n}.

Proof. We continue the computation from Equation (24), using the equality N_{T/K}(a/c) = −1/τ(c)c proved in Equation (25) and the value N_{L/F} (β) = (τ(d)d − 1)/τ(c)c.

cor_{T/K}D_{2} = (−\frac{c}{d}, −\frac{1}{τ(c)c})_{K}⊗_{K}(d, N_{L/F} (β))_{K}
= (−c, −\frac{1}{τ(c)c})_{K}⊗_{K}(d, −τ(c)cN_{L/F} (β))_{K}
= (τ(c), c)_{K}⊗_{K}(d, 1 − τ(d)d)_{K},

where we use the fact that (c, −c) ∼ K and that −1 is an n-power in K (since n is odd). Writing c = c_{1}c_{ν} for c_{1} ∈ F^{×} and c_{1} ∈ K^{(ν)}, one can check that (τ(c), c) = (c_{1}^{2}, c_{ν}); likewise 1 − τ(d)d ∈ F, so the corestriction of both symbols is split by C_{n} by Proposition 11.1.

Recall that τ(ρ) = ρ^{−1} (since [K:F] = 2), so in general τ(u, v)_{K} = (τ(u), τ(v))^{op}. Now

τ(τ(c), c)_{K} = (c, τ(c))_{K}^{op} = (τ(c), c)_{K},

and

τ(d, 1 − τ(d)d)_{K} = (τ(d), 1 − τ(d)d)_{K}^{op}
∼ (τ(d)d, 1 − τ(d)d)_{K}^{op}⊗(d, 1 − τ(d)d)_{K}
∼ (d, 1 − τ(d)d)_{K}.
□

13. Semidirect crossed products

Rowen and Saltman [16] proved that if D is a division algebra of prime degree p over a field L with p roots of unity, and D is split by a field T such that Gal(T/L) = ℤ/e ⋊ ℤ/p, where e = 2, 3, 4 or 6 and divides p−1, then D is cyclic. Our version (Corollary 13.4 below) holds for any prime-power degree n = p^{k} and arbitrary e (dividing p−1), and rather than assuming the base field contains roots of unity, we assume that the splitting field contains them. On the other hand, the cyclicity result only holds for specific semidirect products. To make the notation clear, if τ is a generator of ℤ/d and φ: ℤ/d→U_{n} is a character defined by φ(τ) = t, then

ℤ/d ⋊_{φ} ℤ/n = ⟨ϖ, τ| ϖ^{n} = τ^{d} = 1, τϖτ^{−1} = ϖ^{t}⟩.

The cyclicity of semidirect crossed products is strongly related to Question 10.5, as the following observation on quasi-symbols (see Definition 10.4) shows.

Proposition 13.1. A central simple algebra D of degree n over F is a quasi-symbol, iff it has a splitting field S ⊃ K which is Galois over F with Galois group a semidirect product ℤ/d ⋊ ℤ/n.

The splitting field S has Gal(S/F) = ℤ/d ⋊_{φ} ℤ/n iff D is of type (φ, νφ^{−1}).

Proof. If res_{K/F} D = (a, b) for a ∈ K^{(φ)} and b ∈ K^{(φ′)}, then S = K[\sqrt[n]{a}] is a splitting field satisfying the assumptions (by Proposition 3.4).

On the other hand, let S be a splitting field satisfying the assumptions. Then S is Galois over K, and Gal(S/K) ≅ ℤ/n since this is the unique normal subgroup of order n of ℤ/d ⋊ ℤ/n. We can thus write S = K[α] where α^{n} = a ∈ K, and again by Proposition 3.4, a ∈ K^{(φ)} for some character φ. By Corollary 8.6 there is an element b ∈ K^{(φ′)} such that D⊗_{F} K = (a, b)_{K}, where φφ′ = ν. Finally, in this case Gal(S/F) = ℤ/d ⋊_{φ′} ℤ/n by Remark 5.1. □

It follows that semidirect product splitting fields come in pairs:

Corollary 13.2. Let φ, φ′ be characters with φφ′ = ν. A central simple algebra of degree n over F has a splitting field containing K with Galois group ℤ/d ⋊_{φ} ℤ/n iff it has a splitting field containing K with Galois group ℤ/d ⋊_{φ′} ℤ/n.

Proposition 13.1 can be extended to characterize quasi-symbols with splitting fields of smaller dimension.

Proposition 13.3. Let F ⊆ L ⊆ K be an intermediate field of dimension e = [L:F], and let D be a central simple algebra of degree n over F.

Then D has a splitting field S ⊃ L such that Gal(S/F) = ℤ/e⋊ℤ/n, iff D is a quasi-symbol of type (φ, φ′) for φ of order e in Gal(K/F)^{#}.

Proof. Let S ⊃ L be a splitting field which is Galois over F. We want to check that S_{1} = S⊗_{L}K is also Galois over F, and for that we need to count automorphisms of S_{1} over F.

Let ϖ denote a generator for Gal(S/L), and let τ be a generator of Gal(K/L). Since S/F is Galois, τ extends to an automorphism (of order e) of S. But τ is also defined on K, and thus on S_{1}. Likewise ϖ extends to S_{1} as the identity on K. Finally, we have that τϖτ^{−1} = ϖ^{r} on S for some r, and the same relation trivially holds on K (since ϖ is the identity on K) and thus on S_{1}. It follows that |⟨ϖ, τ⟩| = nd and S_{1}/F is Galois. In particular S = S_{1}^{τ^{e}}.

Now let φ denote the character defining the action of Gal(K/F) on Gal(S_{1}/K). Since S/F is Galois, ⟨τ^{e}⟩ is normal in Gal(S_{1}/F); as easily seen, this is equivalent to τ^{e} being central; which is equivalent to φ^{e} = 1.

Writing S_{1} = K[α] for α^{n} = a ∈ K^{×}, we have by Proposition 3.4 that a ∈ K^{(νφ^{−1})}, and when we write D⊗_{F} K = (a, b)_{K} by Corollary 8.6 we have b ∈ K^{(φ)}, so that D is a quasi-symbol of type (φ, νφ^{−1}). □

In particular, for the case e = 1, we have

Corollary 13.4. Let D be a central simple algebra of degree n over F. If D is split by a Galois extension S ⊃ K such that Gal(S/F) = ℤ/d ⋊_{ν} ℤ/n, then D is split by C_{n}.

Proof. By the above proposition res_{K/F} D = (a, b) for a ∈ K^{(ν)} and b ∈ K^{(1)}, so we are done by Proposition 11.1. □

Though essentially the same, we can formulate the result more generally for [K:F] = 2.

Proposition 13.5. Let m be an arbitrary odd number, F a field of characteristic prime to m, ρ ∈ F_{s} a primitive m-root of unity and K = F [ρ], and assume that [K:F] is prime to m.

Let D be a division algebra of degree m over F. If D is split by a field S ⊃ K which is dihedral of dimension 2m over F, then D is cyclic.

Proof. The only normal subgroups of the dihedral group

D_{m} = ⟨σ, τ| σ^{m} = τ^{2} = 1, τστ^{−1} = σ^{−1}⟩

are contained in ⟨σ⟩, so from S being dihedral over F it follows that [K:F] = 2, K = S^{σ} and S is cyclic over K.

Let m = n_{1} … n_{t} be the factorization of m into prime power factors. Decompose S = K_{1}⊗_{K} … ⊗_{K}K_{t} where [K_{i}:K] = n_{i}, and D = D_{1}⊗_{F} … ⊗_{F} D_{t} where deg D_{i} = n_{i}. The fields K_{i} are Galois over F, and since [K:F] = 2, there are only two possibilities for every i = 1, …, t: Gal(K_{i}/F) is either the dihedral group of order 2n_{i}, or the direct product ℤ/2×ℤ/n_{i}. Applying the proposition in the first case and standard arguments in the second, we see that every D_{i} is cyclic, and (tensoring the cyclic splitting fields over F) D is cyclic as well. □

Our final result on dihedral algebras is to remove the assumption that the splitting field contains roots of unity, made in the last proposition. For that we need to assume that [K:F] = 2. This theorem is proved in [9, Cor. 30] under the assumption char F ≠ 2.

Theorem 13.6. Let n be an odd prime power, K = F [ρ] where ρ is an n-root of unity, and assume [K:F] ≤ 2. Let D be a central simple algebra of degree n over F.

If D is split by a dihedral extension of dimension 2n over F, then D is cyclic.

Proof. The result was proved under the assumption K = F in [17], so we assume [K:F] = 2. Let S be the dihedral splitting field. The case K ⊆ S is covered by Proposition 13.5, so we assume S does not contain K.

Let ϖ, σ be generators of the Galois group Gal(S/F), such that ϖ^{n} = σ^{2} = 1 and σϖσ^{−1} = ϖ^{−1}. Let L = S^{ϖ} be the quadratic subfield, over which S is cyclic. By our assumption K is linearly independent with L, and we let T = L⊗_{F} K = L[ρ].

Since D_{1} = D⊗_{F} L is split by S and thus cyclic, Theorem 10.3 shows that

D_{2} = res_{T/L}D_{1} = (a, β)_{T}

for some a ∈ T and β ∈ L, which is the situation analyzed in Theorem 12.4. In fact (Section 6), letting α = \sqrt[n]{a} we have that S_{1} = T [α] decomposes as S_{1} = S⊗_{L}T. Moreover, letting τ be the non-trivial automorphism of T/L (so that τ(ρ) = ρ^{−1}), we may by Remark 3.5 assume that τ(a) = a^{−1}, and then τ extends to S_{1} by τ(α) = α^{−1}. We can also extend ϖ to S_{1} by ϖ(α) = ρα, and then T = S_{1}^{ϖ}.

The main feature of the proof is that to understand D, we first take the restriction to L and from there to T, and then the corestriction to K and down back to F. Indeed, we will apply the computations of Theorem 12.4 to res_{T/L}D_{1}; from this it will follow that D^{⊗4} ∼ cor_{T/F} res_{T/L}D_{1} is split by C_{n}, which proves the claim.

For that we need to know more about a. Since S_{1} = K⊗_{F} S, it is Galois over F (with Gal(S_{1}/F) = D_{n} × ℤ/2), and σ is extended to S_{1} as the identity on K. Since the relations σ^{2} = 1 and ϖσ = σϖ^{−1} hold both in K and S, they also hold in S_{1}. Now let α′ = σ(α) and note that σ(α′) = α. We have that

ϖ(α′) = ϖσ(α) = σϖ^{−1}(α) = σ(ρ^{−1}α) = ρ^{−1}α′.

It follows that g = αα′ ∈ S_{1}^{σ,ϖ} = K, and then aσ(a) = g^{n}. Checking that τ(α′) = τσ(α) = στ(α) = σ(α^{−1}) = α′^{−1}, we see that τ(g) = g^{−1}. Now, consider the element a_{1} = (g^{(n−1)/2}a^{−1})^{n}a: since τ(a_{1}) = a_{1}^{−1} and K[\sqrt[n]{a_{1}} ] = K[\sqrt[n]{a}] we can replace a by a_{1}; but σ(a_{1}) = a_{1}^{−1}, so from now on we may assume σ(a) = a^{−1} and g = 1.

Recall that D_{2} = res_{T/F} D = (a, β)_{T}. If β ∈ F then

cor_{T/K}D_{2} ∼ (N_{T/K}a, β)_{K} = (1, β)_{K} ∼ K,

so D is already split. We thus assume β ∉ F, and in particular T = K[β]. As in Theorem 12.4, let c, d ∈ K be such that a = cβ + d. We proved in Equation (25) that N_{T/K}(a/c) = −1/N_{K/F} (c), and adding the fact that N_{T/K}(a) = 1, we obtain the equality τ(c) = −c. Plugging this in Equation (26), we immediately obtain

cor_{T/K}D_{2} ∼ (d, 1 − τ(d)d)_{K},

and since 1 − τ(d)d ∈ F, we have that D^{⊗4} ∼ cor_{T/F} res_{T/F} D = cor_{T/F} D_{2} = cor_{K/F} cor_{T/K}D_{2} = cor_{K/F} (d, 1 − τ(d)d)_{K} is split by C_{n}.

□

We end this section with a curious property of_{n}Br(F) when [K:F] = 6, which is related to the fact that in the equation

3 + 4 ≡ 1 (mod 6),

the common divisor with 6 with each of the summands is greater than 1. Let F ⊆ L_{2}, L_{3} ⊆ K be the intermediate subfields (with [L_{t}:F] = t). Since [K:L_{t}] ≤ 3 (t = 2, 3), we have from [11] that_{n}Br(L_{t}) is generated by (classes of) cyclic algebras of degree n. Since cor_{L_{t}/F} is onto, we immediately get that _{n}Br(F) is generated by algebras which become cyclic after extending scalars to L_{2}, and also by algebras which become cyclic after extending scalars to L_{3}. But in fact _{n}Br(F) is generated by algebras with a better set of splitting fields:

Proposition 13.7. Assume that [K:F] = 6 and let L_{2}, L_{3} be as above. Then _{n}Br(F) is generated by (the classes of) algebras which are split by both ℤ/2 ⋊ ℤ/n and ℤ/3 ⋊ ℤ/n.

Proof. By Lemma 12.1, _{n}Br(K) is generated by cyclic algebras of the form (a, b) where a ∈ K^{×} and b ∈ L_{2}^{×}. We can decompose a = a_{0}a_{1} … a_{5} and b = b_{0}b_{3} where a_{i}, b_{i} ∈ K^{(ν^{i})} (see Lemma 6.4). By Proposition 8.3,

cor_{K/F} (a, b) ∼ cor_{K/F} (a_{1}, b_{0})⊗cor_{K/F} (a_{4}, b_{3}).

The algebras cor_{K/F} (a_{1}, b_{0}) are split by C_{n}, by Proposition 11.1.

By Lemma 6.4 we may assume that a_{4} ∈ L_{3} and b_{3} ∈ L_{2}, so the same argument works for res_{L_{t}/F} cor_{K/F} (a_{4}, b_{3}), and the result follows from Corollary 13.2. □

A similar result holds by the same proof whenever d = [K:F] is divisible by 6: _{n}Br(F) is generated by algebras which are split by both ℤ/(d/2) ⋊ ℤ/n and ℤ/(d/3) ⋊ ℤ/n.

14. Generic constructions

In Section 8 we have seen that_{n}Br(F) is generated by quasi-symbols, which are algebras D_{0} such that D_{0}⊗_{F} K = (a, b)_{K} for a ∈ K^{(φ)} and b ∈ K^{(φ′)}, where φ, φ′ range over the pairs of characters with product ν. If either φ = 1 or φ′ = 1, then D_{0} is cyclic, and we will assume this is not the case. We will describe a generic construction for quasisymbols, and present elements w with the property that Tr(w^{i}) = 0 for most values of i.

According to Theorem 10.3, D_{0} is split by C_{n} iff there are α ∈ F^{×} and c ∈ K^{(ν)} of order n, such that J^{φ}(a) ∪ J^{φ′}(b) = (α) ∪ J^{ν}(c). It is generally believed that there exist non-cyclic algebras of prime degree p ≥ 5, though no one was yet able to construct such an algebra. Based on the experience with other crossed products, it seems even more plausible that non-cyclic algebras exist over fields without roots of unity of order p, and the equation given above suggests that a generic quasi-symbol is a reasonable candidate to be non-cyclic.

Let n = p^{m} be a prime power, and k_{0} a field of characteristic prime to n. Let ρ be an n-root of unity in the separable closure (k_{0})_{s}, and k = k_{0}[ρ]. We assume that d = [k:k_{0}] is prime to n. We then define

K = k(a, b, μ_{0}, …, μ_{d−2}, η_{0}, …, η_{d−2}),

a transcendental extension of degree at most 2d (in some cases there are equations on the μ_{i}, η_{i}, see below). Let τ be a generator for Gal(k/k_{0}). The extension of τ to K will depend on our choice of characters. Define ν: Gal(k/k_{0})→U_{n} by Equation (8), and let φ, φ′ be two characters such that φφ′ = ν. Let

r = φ(τ), r′ = φ′(τ).

We extend τ to K by setting

τ(a) = μ_{0}^{n}a^{r}
τ(b) = η_{0}^{n}b^{r′}
τ(μ_{i}) = μ_{i+1 (mod d)}
τ(η_{i}) = η_{i+1 (mod d)},

where μ_{d−1} and η_{d−1} are defined in K by the following formulas:

μ_{d−1} = μ_{d−2}^{−r}μ_{d−3}^{−r^{2}} … μ_{1}^{−r^{d−2}}μ_{0}^{−r^{d−1}}a^{− \frac{r^{d}−1}{n}} ,
η_{d−1} = η_{d−2}^{−r′}η_{d−3}^{−r′^{2}} … η_{1}^{−r′^{d−2}}η_{0}^{−r′^{d−1}}b^{− \frac{r′^{d}−1}{n}} .

This definition makes τ an automorphism of order d of K, and F = K^{τ} satisfies F [ρ] = K.

If the order e of φ is strictly less than n, we can apply Lemma 6.4 and Proposition 6.5 to simplify K without losing the generality of the construction. Specifically, if L = K^{Ker(φ)}, we can have a ∈ L and μ_{0} ∈ L, so that μ_{e+i} = μ_{i}. In particular if φ^{2} = 1 we choose r = −1 and μ_{i} = 1, and then τ(a) = a^{−1}. More generally we can alter the definition so that N_{K/F} (a) = 1 (Remark 3.5), which can be used to express μ_{d−2} in terms of μ_{0}, …, μ_{d−3}, a, and similarly η_{d−2} in terms of η_{0}, …, η_{d−3}, b. If d = [K:F] is even (and F ⊂ K_{0} ⊆ K the intermediate quadratic extension), we can have N_{K/K_{0}}(a) = aτ^{2}(a) … τ^{d−2}(a) = 1 by Remark 3.6, and this can be used to express μ_{d−3} in terms of μ_{0}, …, μ_{d−4}, a.

After τ is defined on K, we let

(27)  D = (a, b)_{K} = K[x, y| x^{n} = a, y^{n} = b, yxy^{−1} = ρx],

and extend τ to D by

τ(x) = μ_{0}x^{r}, τ(y) = η_{0}y^{r′}.

The generic quasi-symbol D_{0} = D^{G} is, according to Lemma 10.2, similar to cor_{K/F} (D).

Example 14.1. Consider the case d = [K:F] = 4 (so in particular n ≡ 1 (mod 4)). There are only two options for the pair φ, φ′: either they are (1, ν) (in which case the quasi-symbol is surely cyclic), or φ = ν^{2}, φ′ = ν^{−1}.

We construct a quasi-symbol of type (ν^{2}, ν^{−1}). Let k, k_{0} be as above, Gal(k/k_{0}) = ⟨τ⟩, and t ≡ ν(τ) (mod n). We then choose r = −1 and r′ = −t; if τ(b) = η_{0}^{n}b^{−t} and τ(η_{0}) = η_{1} then the condition bτ^{2}(b) = 1 gives η_{1} = η_{0}^{t}b^{−(t^{2}+1)/n}. We thus set

K = k(a, b, η)

and define τ on K by

τ(a) = a^{−1},
τ(b) = η^{n}b^{−t},
τ(η) = η^{t}b^{−(t^{2}+1)/n}.

Now define D = (a, b)_{K} by Equation (27), and extend τ to D by τ(x) = x^{−1}, τ(y) = ηy^{−t}. Notice that τ^{2}(η) = η^{−1} and τ^{2}(y) = y^{−1}.

The problem we already stated (Question 10.5) is whether or not D^{τ} is cyclic. By Theorem 11.4, this is equivalent to the existence of u ∈ D^{τ} such that [F [u]:F] = n and u^{n} ∈ F. Naturally, we view the candidates u as elements of D⊗_{F} K = (a, b)_{K}.

We now change perspective, and study some special elements in quasi-symbols. Let w be an element in a central simple algebra. If the characteristic of the field is zero, the coefficients of the minimal polynomial of w are expressed by Newton formulas in terms of the reduced traces Tr(w^{i}). In particular, assuming [F [w]:F] = n, w^{n} ∈ F iff Tr(w) = ⋯ = Tr(w^{n−1}) = 0. One traditional approach to cyclicity problems is to find elements for which the trace of powers is often zero. The same technique works in our case: by Theorem 11.4 D_{0} = D^{G} is cyclic over F iff there is u ∈ D such that τ(u) = u and u^{n} ∈ K (in which case u^{n} ∈ F). Rowen has shown [15] that in an algebra of odd degree, there always exist an element u such that Tr(u) = Tr(u^{2}) = 0. A similar result was obtained by Haile [8]: in any division algebra there exists an element u with Tr(u) = Tr(u^{−1}) = 0. The technique of power traces was effectively used by Rowen and Saltman [17] to prove their above mentioned result on cyclicity of dihedral algebras. Our computations can be viewed as a partial analog to fields without roots of unity.

From now on, assume d = [K:F] is even, and let μ = ν^{d/2} denote the character of order 2 in Gal(K/F)^{#}. Consider a quasi-symbol D_{0} of type (μ, νμ). There exist a ∈ K^{(μ)} and b ∈ K^{(νμ)} such that res_{K/F} D_{0} = (a, b); letting τ be a generator of Gal(K/F), we set t = ν(τ), so that τ(a) ≡ a^{−1} and τ(b) ≡ b^{−t}.

Let W = {tr_{τ}(kxy) : k ∈ K}. This is a subspace of dimension d of D_{0} (its elements are all different since the coefficient of xy in tr_{τ}(kxy) equals k). We will show that powers of elements w ∈ W, often have zero reduced trace. To this end, note that the reduced trace in D_{0} is the same as in D, which is easily computed: writing D = ∑_{i,j=0}^{n−1} Kx^{i}y^{j}, the reduced trace of an element is n times the free coefficient.

Proposition 14.2. Let W be as above. Then for every w ∈ W,

Tr(w) = Tr(w^{3}) = Tr(w^{5}) = Tr(w^{7}) = ⋯ = Tr(w^{n−2}) = 0,

and

Tr(w^{2}) = 0.

If, moreover, char F = 2, then Tr(w^{ℓ}) = 0 for every ℓ = 0, …, n − 1.

Proof. Write D = K[x, y] for a standard pair of generators. Let w = tr_{τ}(kxy) ∈ ∑_{i=0}^{d−1} Kx^{(−1)^{i}}y^{s^{i}} where s = −ν(τ) and k ∈ K is arbitrary. For odd ℓ < n, w^{ℓ} ∈ ∑_{i=1,3,...,n−2}(K[y]x^{i}+K[y]x^{−i}), so that Tr(w^{ℓ}) = 0.

To see that Tr(w^{2}) = 0, notice that

w^{2} ∈ K[y]x^{−2} + K[y] + K[y]x^{2},

and the component in K[y] is in ∑ Ky^{s^{i}+s^{j}} where the sum is over i = 0, 2, …, d − 2 and j = 1, 3, …, d − 1. To have non-zero reduced trace there must be a pair i, j such that s^{i} + s^{j} ≡ 0 (mod n), but then (−s)^{i} ≡ (−s)^{j}, which is impossible since (−s) is of order d in U_{n}.

Finally, if char F = 2, the claim follows by induction on ℓ from Lemma 14.3 below, whose easy proof was pointed out to me by A. Wadsworth. □

Lemma 14.3. Let D be a central simple algebra over a field of characteristic 2. The reduced trace satisfies Tr(u^{2}) = Tr(u)^{2}.

Proof. Extending scalars to the separable closure of F, every element of D is conjugate to an upper triangular matrix, where the reduced trace is the sum of diagonal entries. But if u_{11}, …, u_{nn} is the diagonal of u, the diagonal of u^{2} is u_{11}^{2}, …, u_{nn}^{2} and Tr(u^{2}) = ∑ u_{ii}^{2} = (∑ u_{ii})^{2} = Tr(u)^{2}. □

The result Tr(w^{2}) = 0 of the proposition can be significantly improved if [K:F] = 4. Note that if [K:F] = 4, then _{n}Br(F) is generated by algebras which are split by C_{n}, and the quasi-symbols of type (μ, νμ).

Proposition 14.4. Let W be the space defined above, where we assume d = 4. Let ℓ ≥ 1 be odd. If n > ℓ^{2}, then for every w ∈ W,

Tr(w^{2ℓ}) = 0.

Proof. Let τ be a generator of Gal(K/F) and let t be a number such that τ(ρ) = ρ^{t}. Since

w = tr_{τ}(kxy) ∈ Kxy + Kx^{−1}y^{−t} + Kxy^{−1} + Kx^{−1}y^{t},

we have that

w^{2} ∈ (Ky^{±2t} + K)x^{−2} + Ky^{±1±t} + (Ky^{±2} + K)x^{2}.

The trace Tr(w^{2ℓ}) can be non-zero only if there is a product of ℓ monomials of w^{2} which is in K. Let g_{i,j} denote the number of monomials coming from Kx^{i}y^{j} in such a product, so that ∑ g_{ij} = ℓ. Let h_{i} denote the sum of g_{i,j} over all possible values of j. Since the exponent of x in the product is 2(h_{2} − h_{−2}), we have that h_{−2} = h_{2}, and in particular h_{0} = ℓ − 2h_{2} is odd.

It then follows that the exponent of y in the product is of the form d_{1}t + d_{2}, where the only condition on the d_{i} is that they are odd and |d_{i}| ≤ ℓ. We thus have that d_{1}t ≡ −d_{2} (mod n), and since t^{2} ≡ −1, we obtain n | d_{1}^{2}+d_{2}^{2}, which is even. But n is odd, so n | (d_{1}^{2}+d_{2}^{2})/2 ≤ ℓ^{2}. □

Example 14.5. Let D be the algebra of Example 14.1, and let

w = tr_{τ}(kxy) = kxy + τ(k)ηx^{−1}y^{−t} + τ^{2}(k)xy^{−1} + τ^{3}(k)η^{−1}x^{−1}y^{t}.

We proved that

Tr(w) = Tr(w^{3}) = Tr(w^{5}) = ⋯ = Tr(w^{n−2}) = 0

and

Tr(w^{2}) = Tr(w^{6}) = Tr(w^{10}) = ⋯ = Tr(w^{2ℓ}) = 0

for ℓ ≤ \sqrt{n} odd.

The only way to combine four monomials of w so that the product is in K, is if they are different. Summing all the possible products, we obtain

Tr(w^{4}) = 4n((ρ + ρ^{−1})(ρ^{t} + ρ^{−t}) + 2) · N_{K/F} (k).

The coefficient is never zero by the final proposition below, so that Tr(w^{4}) ≠ 0 whenever w ≠ 0.

Remark 14.6. A similar computation shows that if [K:F] = 6 and n > 7, and D_{0} is a quasi-symbol of type (ν^{3}, ν^{4}), then for every w ∈ W,

Tr(w^{4}) = 0.

We end this paper with a bit of trigonometry (which is needed for the Example 14.5).

Proposition 14.7. Let n be an odd prime power and t an integer such that t^{2} ≡ −1 (mod n), and let ρ be an n-root of unity. Then

(ρ + ρ^{−1})(ρ^{t} + ρ^{−t}) + 2 ≠ 0.

Proof. Let θ_{1} = \frac{2(t−1)π}{n} and θ_{2} = \frac{2(t+1)π}{n}. Since (ρ + ρ^{−1})(ρ^{t} + ρ^{−t}) = ρ^{t+1} + ρ^{t−1} + ρ^{1−t} + ρ^{−1−t} = 2 cos θ_{1} + 2 cos θ_{2}, we need to show that cos θ_{1} +cos θ_{2} ≠ −1. Otherwise, the average of cos θ_{1}, cos θ_{2} is −1/2, so for some {i, j} = {1, 2} we have −1/2 ≤ cos θ_{i} ≤ 0 and cos θ_{j} ≤ −1/2; in other words

θ_{i} ∈ [π/2, 2π/3] ∪ [4π/3, 3π/2],
θ_{j} ∈ [2π/3, 4π/3].

First assume j < i. Then since θ_{1} < θ_{2} we have 4π/3 ≤ θ_{2}, and (by changing t to n − t if necessary) θ_{1} < π. It follows that 4π/n = θ_{2} − θ_{1} > 4π/3 − π = π/3, so n < 12 and n ∈ {5, 9}.

Now assume i < j: again since θ_{1} < θ_{2}, we have that θ_{1} ≤ 2π/3 < θ_{2}, so substituting we obtain n − 3 < 3t ≤ n + 3. Writing n = 3p + u for u = 0, 1, 2, we get t ∈ {p, p + 1}. Recall that by assumption n | t^{2} + 1. If t = p, then n | 9(t^{2} + 1) − n(3p − u) = 9 + u^{2}; otherwise t = p + 1, and n | 9(t^{2} + 1) − (3p + 6 − u)n = 9 + (3 − u)^{2}. In both cases n ∈ {5, 9, 13}.

A direct computation now excludes the few possible values of (n, t). Note that for n = 5, (ρ + ρ^{−1})(ρ^{2} + ρ^{−2}) = −1. □

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Department of Mathematics, Yale University, 10 Hillhouse Avenue, New-Haven CT 06520, USA

  E-mail address: uv2@math.yale.edu