HASSE PRINCIPLES FOR MULTINORM EQUATIONS

E. BAYER-FLUCKIGER, T.-Y. LEE, AND R. PARIMALA

Abstract. A classical result of Hasse states that the norm principle holds for finite cyclic extensions of global fields, in other words local norms are global norms. We investigate the norm principle for finite dimensional commutative étale algebras over global fields; since such an algebra is a product of separable extensions, this is often called the multinorm principle. Under the assumption that the étale algebra contains a cyclic factor, we give a necessary and sufficient condition for the Hasse principle to hold, in terms of an explicitly constructed element of a a finite abelian group. This can be seen as an explicit description of the Brauer-Manin obstruction to the Hasse principle. Keywords: multinorm, tori, Shafarevich groups, local-global principles, Brauer-Manin obstructions MSC 2000: 11G35, 12G05,14G05, 14G25.

0. Introduction

Let k be a global field, and L be a finite dimensional commutative étale algebra over k. We say that the Hasse norm principle holds for L if the localglobal principle holds for the equation

(0.1)  N_{L/k}(t) = c

for all c ∈ k^{×}; this terminology is inspired by Hasse’s result that the norm principle holds in the case of cyclic extensions ([Ha31], [Ha32] §I (3.11) and §II (15)). Over the years, the norm principle for separable field extensions attracted a lot of attention; it is known not to hold in general, and many positive results are also available, see for instance [PlR94], pages 308-309 for a survey; for more recent results, see [BN16], [FLN], and the references therein.

It is natural to ask for Hasse principles in the case when L is a finite dimensional commutative étale algebra, and not just a field extension. Since L is by definition a product of separable extensions, the equation (0.1) is often called a multinorm equation.

This more general problem was also studied extensively, in particular by Hürlimann ([Hu84]), Colliot-Thélène and Sansuc (unpublished), Platonov and Rapinchuk (see [PlR94], sections 6.3 and 9.3), Prasad and Rapinchuk ([PR10], Section 4), Pollio and Rapinchuk ([PoR13]), Demarche and Wei ([DW14]), Pollio ([Po14]). Multinorm equations also arise when dealing with classical groups of type A_{n} (see for instance [PR10] Prop. 4.2).

Date: October 20, 2018.

In spite of many interesting results, some quite simple cases were still open. We illustrate this, as well as our results, by the following example:

Example. Assume that L is a product of n non-isomorphic quadratic field extensions of k. If n = 1 or n = 2, then the Hasse principle holds for L - this is clear for n = 1, and easy for n = 2 (for instance, it is a consequence of [Hu84], Proposition 3.3). It is also well-known that it does not hold in general when n = 3 (see for instance [CT 14]). In the present paper, we show that the Hasse principle holds if n ≥ 4.

To obtain this result and others, let us assume that one of the factors of L is a cyclic field extension of k. Under this hypothesis, we construct a finite abelian group � (L) having the property that

� (L) = 0 ⟺ the Hasse principle holds for L

(cf. Section 5). Assume now that � (L) ≠ 0, and that c ∈ k^{×} is such that (0.1) has a solution locally everywhere. Then we construct a homomorphism

α_{c} : � (L) → ℚ/ℤ

such that

(0.1) has a solution over k ⟺ α_{c} = 0

(see Sections 6 and 7, in particular Theorem 7.1).

These results can be summarized as follows : let I_{L} be the idèle group of L. Then sending c ∈ k^{×} to α_{c} gives rise to an isomorphism

k^{×} ∩ N_{L/k}(I_{L})/N_{L/k}(L^{×}) → � (L)^{∗}

(where � (L)^{∗} is the dual of � (L), cf. Corollary 7.16).

We also give a necessary and sufficient condition for the Hasse principle to hold when one of the factors is metacyclic (see Proposition 7.17).

The results are easy to use. To illustrate this, we consider the case where L is a product of cyclic extensions; assume that L = ∏_{i∈J}K_{i}, where K_{i}/k is a

cyclic extension of degree d_{i}. Let P be the set of prime numbers dividing ∏_{i∈J}d_{i}.

For all p ∈ P and all i ∈ J, let K_{i}(p) be the largest subfield of K_{i} such that [K_{i}(p) : k] is a power of p, and set L(p) = ∏_{i∈J}K_{i}(p). Then we have

� (L) = ⊕_{p∈P}� (L(p)),

(see Proposition 8.6).

For any cyclic field extension K/k of prime power degree, we denote by K_{prim} the unique subfield of K of degree p over k. Set

L(p)_{prim} = ∏_{i∈J}K_{i}(p)_{prim}.

Then we have

� (L) = 0 ⟺ ⊕_{p∈P(L)}� (L(p)_{prim}) = 0,

(cf Theorem 8.1), and

� (L(p)_{prim}) ≃ (ℤ/pℤ)^{m_{p}(L)},

where P(L) is a set of prime numbers (subset of P), and m_{p}(L) is a positive integer; both are determined explicitly (see Theorem 8.2).

The paper is structured as follows. Sections 1-4 contain some preliminary results, including a new proof of a proposition of Hürlimann, [Hu84] Prop. 3.3. The group � (L) is defined in Section 5, and the homomorphism α_{c} in Section 6. In both sections, we start with the case where the étale algebra L has a cyclic factor of prime power degree, which is the essential case. We also show how one can reduce the exponent of the prime number, using the exact sequence of Proposition 5.9 - this is then used in inductive arguments. The main result is proved in Section 7 (see Theorem 7.1). Section 8 contains the application of the above results to the special case where all the factors of the étale algebra are cyclic.

Note that the results of this paper are related to the Brauer-Manin obstruction. Indeed, for c = 1, the equation (0.1) yields the so-called norm-one torus defined by L/k (see 1.2 for details); we denote this torus by T_{L/k}. When k is an algebraic number field, then one can deduce from [San81] that the only obstruction to the Hasse principle is the Brauer-Manin obstruction, and is an element of the group � ^{2}(k, T̂_{L/k})^{∗}. We show that � (L) ≃ � ^{2}(k, T̂_{L/k}) (see Proposition 5.10), hence our results provide an explicit description of the Brauer-Manin obstruction.

1. Notation, definitions and basic facts

1.1. Weil restriction

If f : R → R′ is a homomorphism of commutative rings such that R′ is a projective R-module of finite type, and if W is an affine R′-scheme, then we denote by R_{R′/R}W the Weil restriction (see for instance [O 84], Appendice 2).

1.2. Etale algebras, tori and characters

Let k be a field, let k_{s} be a separable closure of k and set Γ_{k} = Gal(k_{s}/k). We use standard notation in Galois cohomology; in particular, if M is a discrete Γ_{k}-module and i is an integer ≥ 0, we set H^{i}(k, M) = H^{i}(Γ_{k}, M).

If L is a commutative étale k-algebra of finite rank, we denote by N_{L/k} the norm map, and set T_{L/k} = R_{L/k}^{(1)}(𝔾_{m}); then T_{L/k} is the k-torus determined by the exact sequence

(1.1)  \xymatrix{1 \ar[r] & T_{L/k} \ar[r] & R_{L/k}(𝔾_{m}) ⟶^{N_{L/k}} 𝔾_{m} \ar[r] & 1} .

For a k-torus T, we denote by T̂ = Hom(T, 𝔾_{m}) its character group. If K/k is a finite separable extension, set Γ_{K} = Gal(k_{s}/K). If moreover M is a discrete Γ_{K}-module, set I_{K/k}(M) = Ind_{Γ_{K}}^{Γ_{k}}(M).

The following lemmas will be used several times in the sequel

Lemma 1.1. Let F/k be a separable extension of finite degree, and let L be the product of n copies of F. Then we have

(i) T_{L/k} ≃ R_{F/k}(𝔾_{m})^{n−1} × T_{F/k}.

(ii) H^{1}(k, T̂_{L/k}) ≃ H^{1}(k, T̂_{F/k}).

Proof. The isomorphism (R_{F/k}(𝔾_{m}))^{n} → (R_{F/k}(𝔾_{m}))^{n} sending (b_{1}, ..., b_{n}) to (b_{1}, ..., b_{n−1}, b_{1}...b_{n−1}b_{n}) induces an isomorphism T_{L/k} ≃ R_{F/k}(𝔾_{m})^{n−1} × T_{F/k}. This proves (i). By (i), we have T̂_{L/k} ≃ I_{F/k}(ℤ)^{n−1} ⊕ T̂_{F/k}; since H^{1}(k, I_{F/k}(ℤ)) = 0, this implies (ii).

Lemma 1.2. Let K/k be a cyclic extension of degree d. Then we have

H^{1}(k, T̂_{K/k}) ≃ ℤ/dℤ.

Proof. Let σ be a generator of Gal(K/k). Consider the exact sequence

1 → 𝔾_{m} → R_{K/k}(𝔾_{m}) → T_{K/k} → 1,

where the map from R_{K/k}(𝔾_{m}) to T_{K/k} sends x to x/σ(x), and its dual sequence

0 → T̂_{K/k} → I_{K/k}(ℤ) → ℤ → 0.

This exact sequence induces

\xymatrix{I_{K/k}(ℤ)^{Γ_{k}} ⟶^{ϵ} ℤ \ar[r] & H^{1}(k, T̂_{K/k}) \ar[r] & H^{1}(k, I_{K/k}(ℤ)) = 0} .

We have I_{K/k}(ℤ)^{Γ_{k}} ≃ ℤ, and the map ϵ is multiplication by d; hence H^{1}(k, T̂_{K/k}) ≃ ℤ/dℤ.

1.3. The multinorm problem

Let L be an étale k-algebra, and let c ∈ k^{∗}. Let X_{c} be the affine k-variety determined by the equation N_{L/k}(t) = c. Then X_{c} is a torsor over the torus T_{L/k} defined in 1.2, hence defines a class [X_{c}] ∈ H^{1}(k, T_{L/k}); the variety X_{c} has a k-point if and only if [X_{c}] = 0. Hence we have

c ∈ N_{L/k}(L^{×}) ⟺ X_{c}(k) ≠ ∅ ⟺ [X_{c}] = 0.

2. A construction

Let k be a field, and let L be a commutative étale k-algebra of finite rank; assume that L is not a field. We keep the notation of the previous section. The aim of this section is to introduce a k-torus that will play a basic role in the study of the cohomology of the torus T_{L/k}, and of the multinorm problem.

Let us write L = K × K′, where K and K′ are étale k-algebras, and set E = K ⊗_{k} K′.

The norm maps N_{K/k} : K → k and N_{K′/k} : K′ → k induce N_{E/K′} : E → K and N_{E/K} : E → K′. Let f : R_{E/k}(𝔾_{m}) → R_{L/k}(𝔾_{m}) be defined by f(x) = (N_{E/K}(x)^{−1}, N_{E/K′}(x)). The image of f is T_{L/k}, and we consider the torus S_{K,K′} defined by the exact sequence

\xymatrix{1 \ar[r] & S_{K,K′} \ar[r] & R_{E/k}(𝔾_{m}) ⟶^{f} T_{L/k} \ar[r] & 1} .

Note that S_{K,K′} also fits in the exact sequence

(2.1)  \xymatrix{1 \ar[r] & S_{K,K′} \ar[r] & R_{K′/k}(T_{E/K′}) ⟶^{N_{E/K}} T_{K/k} \ar[r] & 1} .

where T_{E/K′} is defined by the exact sequence

1 → T_{E/K′} → R_{E/K′}(𝔾_{m}) ⟶^{N_{E/K′}}𝔾_{m} → 1.

3. Tate-Shafarevich groups

We keep the notation of the previous sections, and assume that k is a global field. Let Ω_{k} be the set of all places of k; if v ∈ Ω_{k}, we denote by k_{v} the completion of k at v.

For any k-torus T, set � ^{i}(k, T) = Ker(H^{i}(k, T) → ∏_{v∈Ω_{k}}H^{i}(k_{v}, T)). If M is

a Γ_{k}-module, set � ^{i}(k, M) = Ker(H^{i}(k, M) → ∏_{v∈Ω_{k}}H^{i}(k_{v}, M)). Recall that

by Poitou-Tate duality, we have � ^{2}(k, T̂) ≃ � ^{1}(k, T)^{∗}.

3.1. Hasse principle for the multinorm problem

Let L be an étale k-algebra, and let c ∈ k^{×}. If X_{c}(k_{v}) ≠ ∅ for all v ∈ Ω_{k}, then we have [X_{c}] ∈ � ^{1}(k, T_{L/k}). In particular, the Hasse principle holds for all c ∈ k^{×} if and only if � ^{1}(k, T_{L/k}) = 0.

We have the following relationship between the Tate-Shafarewich groups of the torus T_{L/k}, and the torus S_{K,K′} defined in §2 :

Lemma 3.1. We have � ^{1}(k, T_{L/k}) ≃ � ^{2}(k, S_{K,K′}).

Proof. By the definition of the torus S_{K,K′}, we have the exact sequence

\xymatrix{1 \ar[r] & S_{K,K′} \ar[r] & R_{E/k}(𝔾_{m}) ⟶^{f} T_{L/k} \ar[r] & 1} ,

giving rise to the cohomology exact sequence

0 → H^{1}(k, T_{L/k}) → H^{2}(k, S_{K,K′}) → H^{2}(k, R_{E/k}(𝔾_{m})).

By the Brauer-Hasse-Noether Theorem, we have � ^{2}(k, R_{E/k}(𝔾_{m})) = 0, hence � ^{1}(k, T_{L/k}) ≃ � ^{2}(k, S_{K,K′}), as claimed.

We now compute the group � ^{2}(k, T̂_{K/k}) for a cyclic extension K/k - note that by Poitou-Tate duality, this is equivalent to Hasse’s cyclic norm principle, which is the following proposition :

Proposition 3.2. Let K/k be a cyclic extension. Then � ^{1}(k, T_{K/k}) = 0.

Proof. We give a proof for the convenience of the reader. Let σ be a generator of Gal(K/k). Consider the exact sequence

1 → 𝔾_{m} → R_{K/k}(𝔾_{m}) → T_{K/k} → 1,

where the map from R_{K/k}(𝔾_{m}) to T_{K/k} sends x to x/σ(x). This sequence gives rise to an injection H^{1}(k, T_{K/k}) → H^{2}(k, 𝔾_{m}). By the Brauer-Hasse-Noether theorem, we have � ^{2}(k, 𝔾_{m}) = 0, hence � ^{1}(k, T_{K/k}) = 0.

Corollary 3.3. Let K/k be a cyclic extension. Then � ^{2}(k, T̂_{K/k}) = 0.

This follows from the previous proposition, combined with Poitou-Tate duality.

4. A result of Hürlimann

Using the above lemmas, we generalize a result of Hürlimann ([Hu84] Prop. 3.3).

Proposition 4.1. Let K/k be a cyclic extension of k, and let K′/k be a separable extension of finite degree. Let c ∈ k^{×}. Then the local-global principle holds for the multinorm equation N_{K/k}(x)N_{K′/k}(y) = c.

Proof. Set L = K × K′; the assertion is equivalent to the vanishing of � ^{1}(k, T_{L/k}). By Lemma 3.1, we have � ^{1}(k, T_{L/k}) ≃ � ^{2}(k, S_{K,K′}). By Poitou-Tate duality we have � ^{2}(k, S_{K,K′}) ≃ � ^{1}(k, Ŝ_{K,K′})^{∗}, hence it suffices to prove that � ^{1}(k, Ŝ_{K,K′}) = 0. Since K/k is a cyclic extension, the algebra E = K ⊗_{k} K′ is isomorphic to a product of copies of F, where F/K′ is some cyclic field extension. Set d = [K : k] and f = [F : K′].

Consider the dual sequence of (2.1) :

(4.1)  \xymatrix{0 \ar[r] & T̂_{K/k} ⟶^{ι} I_{K′/k}(T̂_{E/K′}) ⟶^{ρ} Ŝ_{K,K′} \ar[r] & 0} ,

and the sequence induced by (4.1)

(4.2) H^{1}(k, T̂_{K/k}) ⟶^{ι^{1}} H^{1}(k, I_{K′/k}(T̂_{E/K′})) ⟶^{ρ^{1}} H^{1}(k, Ŝ_{K,K′}) ⟶^{δ} H^{2}(k, T̂_{K/k}).

We have H^{1}(k, I_{K′/k}(T̂_{E/K′})) ≃ H^{1}(K′, T̂_{E/K′}); lemmas 1.1 (ii) and 1.2 imply that H^{1}(K′, T̂_{E/K′}) ≃ H^{1}(K′, T̂_{F/K′}) ≃ ℤ/fℤ. The map ι^{1} is the natural projection from ℤ/dℤ → ℤ/fℤ, therefore ι^{1} is surjective. This implies that δ : H^{1}(k, Ŝ_{K,K′}) → H^{2}(k, T̂_{K/k}) is injective; moreover, δ induces an injection � ^{1}(k, Ŝ_{K,K′}) → � ^{2}(k, T̂_{K/k}). Since K/k is a cyclic extension, we have � ^{2}(k, T̂_{K/k}) = 0 by Lemma 3.3. The proposition then follows.

5. The group � (K, K′)

We keep the notation of the previous sections : in particular, L = K × K′, where K and K′ are étale k-algebras, and E = K ⊗ K′. In addition, we now assume that K/k is a cyclic extension. Under this hypothesis, we define a finite abelian group � (K, K′), and we show that � ^{1}(k, T_{L/k}) is isomorphic to the dual of � (K, K′).

Let K′ = ∏_{i∈I} K_{i}, where the K_{i}/k are field extensions. Then we have E = ∏_{i∈I} E_{i}, with E_{i} = K ⊗ K_{i}.

5.1. The prime power degree case

Suppose that K is a cyclic extension of degree p^{e}, where p is a prime number. We start with some notation and definitions. For each i ∈ I, let M_{i} be a cyclic extension of K_{i} such that E_{i} is isomorphic to a product of copies of M_{i}. Let p^{e_{i}} = [M_{i} : K_{i}]; without loss of generality, we assume that e_{i} ≥ e_{i+1} for 1 ≤ i ≤ m − 1.

Let s and t be positive integers. For s ≥ t, let π_{s,t} be the canonical projection ℤ/p^{s}ℤ → ℤ/p^{t}ℤ. For x ∈ ℤ/p^{s}ℤ and y ∈ ℤ/p^{t}ℤ, we say that x dominates y if s ≥ t and π_{s,t}(x) = y; if this is the case, we write x ⪰ y. For x ∈ ℤ/p^{s}ℤ and y ∈ ℤ/p^{t}ℤ, let δ(x, y) be the greatest nonnegative integer d ≤ min{s, t} such that π_{s,d}(x) = π_{t,d}(y). We have δ(x, y) = min{s, t} if and only if x ⪰ y or y ⪰ x.

Let I = {1, ..., m}. For a = (a_{1}, ..., a_{m}) ∈ ⊕_{i∈I}ℤ/p^{e_{i}}ℤ and n ∈ ℤ/p^{e_{1}}ℤ, let

I_{n}(a) be the set {i ∈ I| n ⪰ a_{i}} and let I(a) = (I_{0}(a), ..., I_{p^{e_{1}}−1}(a)).

Let E be the set of p^{e_{1}}-tuples (I_{0}, ..., I_{p^{e_{1}}−1}), where I_{0}, ..., I_{p^{e_{1}}−1} are subsets of I such that ⋃_{0≤n≤p^{e_{1}}−1}I_{n} = I. Now we characterize the image of the map

I : ⊕_{i∈I}ℤ/p^{e_{i}}ℤ → E.

An element (I_{0}, ..., I_{p^{e_{1}}−1}) ∈ E is said to be coherent if for all n_{1}, n_{2} ∈ ℤ/p^{e_{1}}ℤ we have:

(1) If i ∈ I_{n_{1}} ∩ I_{n_{2}}, then π_{e_{1},e_{i}}(n_{1}) = π_{e_{1},e_{i}}(n_{2}).

(2) If i ∈ I_{n_{1}} and π_{e_{1},e_{i}}(n_{1}) = π_{e_{1},e_{i}}(n_{2}), then i ∈ I_{n_{2}}.

Let E^{c} be the subset of all coherent elements in E. For a ∈ ⊕_{i∈I}ℤ/p^{e_{i}}ℤ, it

is clear that I(a) is a coherent element. Conversely for a coherent element (I_{0}, ..., I_{p^{e_{1}}−1}) ∈ E^{c}, we set a_{i} = π_{e_{1},e_{i}}(n) for i ∈ I_{n}. Note that condition (1) of the definition of a coherent element ensures that the a_{i}’s are well-defined. Hence a = (a_{1}, ..., a_{m}) is a well-defined element in ⊕_{i∈I}ℤ/p^{e_{i}}ℤ; condition (2)

implies that I(a) = (I_{0}, ..., I_{p^{e_{1}}−1}). This shows that I is a bijection between

⊕_{i∈I}ℤ/p^{e_{i}}ℤ and E^{c}.

Given a positive integer 0 ≤ d ≤ e and i ∈ I, let Σ_{i}^{d} be the set of all places v ∈ Ω_{k} such that at each place w of K_{i} above v, the algebra K ⊗ K_{i}^{w} is isomorphic to a product of isomorphic field extensions of degree at most p^{d} of K_{i}^{w}. Let Σ_{i} = Σ_{i}^{0}, in other words, Σ_{i} is the set of all places v ∈ Ω_{k} where E_{i}^{v} is isomorphic to a product of copies of K_{i}^{v}.

Let (I_{0}, ..., I_{p^{e_{1}}−1}) ∈ E^{c}. For n_{1} ∈ ℤ/p^{e_{1}}ℤ and i ∈ I, set δ(n_{1}, i) = δ(n_{1}, π_{e_{1},e_{i}}(n_{2})), where n_{2} is an element in ℤ/p^{e_{1}}ℤ such that i ∈ I_{n_{2}}. Since (I_{0}, ..., I_{p^{e_{1}}−1}) is coherent, δ(n_{1}, i) is independent of the choice of n_{2} and hence is well-defined. Note that if we let a = (a_{1}, ..., a_{m}) be the element in ⊕_{i∈I}ℤ/p^{e_{i}}ℤ

corresponding to (I_{0}, ..., I_{p^{e_{1}}−1}), then δ(n_{1}, i) = δ(n_{1}, a_{i}). For I_{n} ⫋ I, define

(5.1)  Ω(I_{n}) = ∩_{i∉I_{n}}Σ_{i}^{δ(n,i)}.

For I_{n} = I, we set Ω(I_{n}) = Ω_{k}.

Set

G = G_{k}(K, K′) = {(a_{1}, ..., a_{m}) ∈ ⊕_{i∈I}ℤ/p^{e_{i}}ℤ| ⋃_{n∈ℤ/p^{e_{1}}ℤ}Ω(I_{n}(a)) = Ω_{k}}.

Lemma 5.1. The set G is a subgroup of ⊕_{i∈I}ℤ/p^{e_{i}}ℤ.

Proof. Let a = (a_{1}, ..., a_{m}) and b = (b_{1}, ..., b_{m}) be elements of G. By the definition of G, for each v ∈ Ω_{k}, there exist some n, n′ ∈ ℤ/p^{e_{1}}ℤ such that v ∈ Ω(I_{n}(a)) and v ∈ Ω(I_{n′}(b)). We claim that

v ∈ Ω(I_{n+n′}(a + b)).

This is clear when I_{n+n′}(a + b) = I. Suppose that I_{n+n′}(a + b) ≠ I. First note that δ(n+n′, a_{i}+b_{i}) ≥ min{δ(n, a_{i}), δ(n′, b_{i})} and that min{δ(n, a_{i}), δ(n′, b_{i})} ≤ e_{i} for all i ∈ I. Pick an arbitrary i ∉ I_{n+n′}(a + b). Without loss of generality, we suppose that min{δ(n, a_{i}), δ(n′, b_{i})} = δ(n, a_{i}). If i ∉ I_{n}(a), we have v ∈ Σ_{i}^{δ(n,a_{i})} ⊆ Σ_{i}^{δ(n+n′,a_{i}+b_{i})}; hence we have v ∈ Ω(I_{n+n′}(a + b)).

If i ∈ I_{n}(a), then by definition δ(n, a_{i}) = e_{i}. We have δ(n, a_{i}) ≤ δ(n′, b_{i}) by assumption, hence δ(n′, b_{i}) ≥ e_{i}. But δ(n′, b_{i}) ≤ e_{i}, therefore we have δ(n′, b_{i}) = e_{i}, and hence i ∈ I_{n′}(b). This implies that i ∈ I_{n+n′}(a + b), and this is a contradiction. This completes the proof of the lemma.

Let D be the subgroup of ⊕_{i∈I}ℤ/p^{e_{i}}ℤ generated by the diagonal element

(1, ..., 1), and note that D is contained in G. Set

� _{k}(K, K′) = G/D.

Example 5.2. Assume that k = ℚ, and that L = ℚ(\sqrt{a}) × ℚ(\sqrt{b}) × ℚ(\sqrt{ab}), where a, b are distinct square-free integers. Set K = ℚ(\sqrt{a}), K_{1} = ℚ(\sqrt{b}) and K_{2} = ℚ\sqrt{ab}). Then with the above notation we have I = \{1, 2\}, and E_{1} = E_{2} = ℚ(\sqrt{a}, \sqrt{b}), hence e = e_{1} = e_{2} = 1. This implies that either � (K, K′) = 0, or � (K, K′) ≃ ℤ/2ℤ. Note that

there exists v ∈ Ω_{k} such that E_{1}^{v} is a field ⟺ Σ_{1} ∪ Σ_{2} ≠ Ω_{k},

hence

� (K, K′) = 0 ⟺ there exists v ∈ Ω_{k} such that E_{1}^{v} is a field.

Set now a = 13, b = 17 : then there exists no v ∈ Ω_{k} such that E_{1}^{v} is a field, therefore � (K, K′) = ℤ/2ℤ. Note that it is well-known that the multinorm principle fails in this case (see for instance [CT 14], Proposition 5.1).

Theorem 5.3. Suppose that K/k is a cyclic extension of degree p^{e}, where p is a prime number. Then � ^{1}(k, Ŝ_{K,K′}) ≃ � (K, K′).

Proof. Consider the dual sequence of (2.1),

(5.2)  \xymatrix{0 \ar[r] & T̂_{K/k} ⟶^{ι} I_{K′/k}(T̂_{E/K′}) ⟶^{ρ} Ŝ_{K,K′} \ar[r] & 0} ,

and the exact sequence induced by (5.2),

(5.3) H^{1}(k, T̂_{K/k}) ⟶^{ι^{1}} H^{1}(k, I_{K′/k}(T̂_{E/K′})) ⟶^{ρ^{1}} H^{1}(k, Ŝ_{K,K′}) → H^{2}(k, T̂_{K/k}).

We have � ^{2}(k, T̂_{K/k}) = 0 by Corollary 3.3, therefore � ^{1}(k, Ŝ_{K,K′}) is in the image of ρ^{1}.

Note that H^{1}(k, I_{K_{i}/k}(T̂_{E_{i}/K_{i}})) ≃ H^{1}(K_{i}, T̂_{E_{i}/K_{i}}), and that by Lemma 1.1 (ii), we have H^{1}(K_{i}, T̂_{E_{i}/K_{i}}) ≃ H^{1}(K_{i}, T̂_{M_{i}/K_{i}}). Moreover, by Lemma 1.2, we have H^{1}(K_{i}, T̂_{M_{i}/K_{i}}) ≃ ℤ/p^{e_{i}}ℤ.

In the following we identify H^{1}(k, T̂_{K/k}) to ℤ/p^{e}ℤ and H^{1}(k, I_{K_{i}/k}(T̂_{E_{i}/K_{i}})) to ℤ/p^{e_{i}}ℤ for 1 ≤ i ≤ m. Under this identification, the map

ι^{1} : H^{1}(k, T̂_{K/k}) → H^{1}(k, I_{K′/k}(T̂_{E/K′})) = ⊕_{i∈I}H^{1}(k, I_{K_{i}/k}(T̂_{E_{i}/K_{i}}))

sends ℤ/p^{e}ℤ to ⊕_{i∈I}ℤ/p^{e_{i}}ℤ by the natural projections. Therefore we can rewrite

the exact sequence (5.3) as follows :

(5.4)  ℤ/p^{e}ℤ ⟶^{ι^{1}} ⊕_{i∈I}ℤ/p^{e_{i}}ℤ ⟶^{ρ^{1}} H^{1}(k, Ŝ_{K,K′}) → H^{2}(k, T̂_{K/k}),

where ι^{1} is the natural projection from ℤ/p^{e}ℤ to ℤ/p^{e_{i}}ℤ for each i. Note that the image of ι^{1} is the subgroup D, and we have the exact sequence

(5.5)  0 → ( ⊕_{i∈I}ℤ/p^{e_{i}}ℤ)/D ⟶^{ρ^{1}} H^{1}(k, Ŝ_{K,K′}) → H^{2}(k, T̂_{K/k}).

Let a = (a_{1}, ..., a_{m}) ∈ ⊕_{i∈I}ℤ/p^{e_{i}}ℤ and [a] be its image in ( ⊕_{i∈I}ℤ/p^{e_{i}}ℤ)/D. We

claim that ρ^{1}([a]) is in � ^{1}(k, Ŝ_{K,K′}) if and only if a ∈ G.

We denote by a^{v} the image of a in ⊕^{m}_{i=1}H^{1}(k_{v}, I_{K_{i}^{v}/k_{v}}(T̂_{E_{i}^{v}/K_{i}^{v}})), and by D_{v} the

image of D in this sum.

By the exact sequence (5.5), we have ρ^{1}([a]) ∈ � ^{1}(k, Ŝ_{K,K′}) if and only if a^{v} ∈ D_{v} for all places v ∈ Ω_{k}. Therefore, it suffices to prove that a ∈ G if and only if a^{v} ∈ D_{v} for all places v ∈ Ω_{k}.

Suppose that a ∈ G, and let v ∈ Ω_{k}. Then there exists n ∈ ℤ/p^{e_{1}}ℤ such that v ∈ Ω(I_{n}(a)). If I_{n}(a) = I, then clearly a ∈ D ⊆ G. Suppose that I_{n}(a) ≠ I. This implies that for each i ∉ I_{n}(a) and for each place w of K_{i} above v, the étale algebra K_{i}^{w} ⊗ K is isomorphic to a product of field extensions of K_{i}^{w} of degree at most δ(n, i). Let δ_{i} = δ(n, i) = δ(n, a_{i}). Note that

H^{1}(k_{v}, I_{K_{i}^{v}/k}(T̂_{E_{i}^{v}/K_{i}^{v}})) = H^{1}(K_{i}^{v}, T̂_{E_{i}^{v}/K_{i}^{v}}).

We have

H^{1}(K_{i}^{v}, T̂_{E_{i}^{v}/K_{i}^{v}}) ≃ ⊕_{w|v}H^{1}(K_{i}^{w}, T̂_{K_{i}^{w}⊗K/K_{i}^{w}}) ≃ ⊕_{w|v}ℤ/p^{e_{i,w}}ℤ,

where e_{i,w} ≤ δ_{i}, and the localization map H^{1}(K_{i}, T̂_{E_{i}/K_{i}}) → H^{1}(K_{i}^{v}, T̂_{E_{i}^{v}/K_{i}^{v}}) is the canonical projection π_{e_{i},e_{i,w}} from ℤ/p^{e_{i}}ℤ to each component ℤ/p^{e_{i,w}}ℤ. Since for all i ∉ I_{n}(a) we have e_{i,w} ≤ δ_{i}, and π_{e_{i},δ_{i}}(a_{i}) = π_{e_{1},δ_{i}}(n), this implies that a^{v} = (n, ..., n)^{v}.

Suppose conversely that a^{v} ∈ D_{v} for all v ∈ Ω_{k} and a ∉ G. Then a ∉ D, and there exists a place v ∈ Ω_{k} such that v ∉ ∪_{n∈ℤ/p^{e_{1}}ℤ}Ω(I_{n}(a)). Since a^{v} ∈ D_{v},

there exists n′ ∈ ℤ/p^{e}ℤ such that a^{v} = (ι^{1}(n′))_{v}. Let n = π_{e,e_{1}}(n′). As v ∉ Ω(I_{n}(a)), there exists i ∉ I_{n}(a) and a place w of K_{i} above v such that K_{i}^{w} ⊗K is isomorphic to a product of field extensions of degree e_{i,w} > δ_{i} of K_{i}^{w}. Then by the definition of δ_{i} = δ(n, a_{i}),we have π_{e_{i},e_{i,w}}(a_{i}) ≠ π_{e_{1},e_{i,w}}(n). Hence the localization a_{i}^{v} of the i-th coordinate of a is not equal to the localization of the i-th coordinate of (n, ..., n), which is a contradiction. Our claim then follows. Therefore, we have � ^{1}(k, Ŝ_{K,K′}) ≃ � (K, K′).

5.2. The group � (K/K_{0}, K′)

Let K_{0} be the unique subfield of K such that [K_{0} : k] = p^{e−1}. The proof of the main theorem in the prime power case uses induction on e, and the comparison of the groups � (K, K′) and � (K_{0}, K′). We first define a homomorphism F : � (K_{0}, K′) → � (K, K′), and then determine the cokernel of F, denoted by � (K/K_{0}, K′).

Note that if e = 1, then K_{0} = k, and hence � (K_{0}, K′) is trivial; in this case, � (K/K_{0}, K′) is the group � (K, K′) itself.

The homomorphism F : � (K_{0}, K′) → � (K, K′).

Recall that we have K′ = ∏_{i∈I}K_{i}, that E_{i} = K ⊗ K_{i}, and that E_{i} is the

product of copies of a cyclic extension of degree p^{e_{i}} of K_{i}. Set E_{i}^{0} = K_{0} ⊗ K_{i}. Then E_{i}^{0} also splits as a product of copies of a cyclic extension of K_{i}; let us denote by p^{f_{i}} the degree of this extension.

Proposition 5.4. For all i ∈ I, we have f_{i} ≤ e_{i}. If moreover e_{i} ≠ 0, then e_{i} = f_{i} + 1.

This is an immediate consequence of the following proposition :

Proposition 5.5. Let F/k be a field extension, and let K ⊗_{k} F be a product of cyclic field extensions of F of degree p^{e_{F}}; let K_{0} ⊗_{k} F be a product of cyclic field extensions of F of degree f_{F}. Then we have

(i) f_{F} ≤ e_{F};

(ii) f_{F} ≥ e_{F} − 1;

(iii) If e_{F} ≠ 0, then e_{F} = f_{F} + 1.

Proof. If n is a positive integer, let us denote by C_{n} the cyclic group of order n. Let us consider the homomorphisms

Γ_{F} →^{ι}Γ_{k}⟶^{ϕ_{K}}C_{p^{e}} ⟶^{π}C_{p^{e−1}} → 1,

where ι is the inclusion of Γ_{F} into Γ_{k}, the homomorphism ϕ_{K} : Γ_{k} → C_{p^{e}} corresponds to the cyclic extension K/k, and π : C_{p^{e}} → C_{p^{e−1}} is the quotient of C_{p^{e}} by its unique subgroup of order p. Note that the image of ϕ_{K} ∘ ι is the Galois group of the cyclic factors of K ⊗_{k} F, and hence is of order p^{e_{F}}; similarly, the image of π ∘ ϕ_{K} ∘ ι is the Galois group of the cyclic factors of K_{0} ⊗_{k} F, and hence is of order p^{f_{F}}. Therefore we have f_{F} ≤ e_{F}. Moreover, if e_{F} ≠ 0, then the image of ϕ_{K} ∘ ι contains the unique subgroup of order p of C_{p^{e}}, and hence e_{F} = f_{F} + 1. This completes the proof of the proposition.

For all i ∈ I, let F_{i} : ℤ/p^{f_{i}}ℤ → ℤ/p^{e_{i}}ℤ be the inclusion of the subgroup of order p^{f_{i}} in the group ℤ/p^{e_{i}}ℤ, and set F_{K/K_{0}} = F = ⊕_{i∈I}F_{i}.

Proposition 5.6. The map F : ⊕_{i∈I}ℤ/p^{f_{i}}ℤ → ⊕_{i∈I}ℤ/p^{e_{i}}ℤ induces an injective

homomorphism F : � (K_{0}, K′) → � (K, K′).

Proof. Let us recall some notation from 5.1, for K and K_{0} : For all i ∈ I and for all positive integers d, we denote by Σ(K)_{i}^{d} (respectively Σ(K_{0})_{i}^{d}) the set of all places v ∈ Ω_{k} such that at each place w of K_{i} above v, the algebra K ⊗ K_{i}^{w} (respectively K_{0} ⊗ K_{i}^{w} ) is isomorphic to a product copies of a cyclic extension of degree at most p^{d} of K_{i}^{w}. Recall that

G = G(K, K′) = {a ∈ ⊕_{i∈I}ℤ/p^{e_{i}}ℤ | ⋃_{n∈ℤ/p^{e_{1}}ℤ}Ω(I_{n}(a)) = Ω_{k}},

and that D is the diagonal subgroup of G. Similarly, set

G_{0} = G(K_{0}, K′) = {b ∈ ⊕_{i∈I}ℤ/p^{f_{i}}ℤ | ⋃_{n∈ℤ/p^{f_{1}}ℤ}Ω(I_{n}(b)) = Ω_{k}},

and let D_{0}, be the diagonal subgroup of G_{0}. Then we have � (K, K′) = G/D and � (K_{0}, K′) = G_{0}/D_{0}.

Let b ∈ G_{0}, and let us show that F (b) ∈ G. Let v ∈ Ω_{k}. Then there exists r ∈ ℤ/p^{f_{1}}ℤ such that v ∈ Σ(K_{0})_{i}^{δ(r,i)} for all i ∈ I such that i ∉ I_{r}(b). Note that for all positive integers δ, we have Σ(K_{0})_{i}^{δ} ⊂ Σ(K)_{i}^{δ+1}. Set n = F_{1}(r) ∈ ℤ/p^{e_{1}}ℤ; then we have δ(n, F_{i}(b_{i})) = δ(r, b_{i}) + 1. Hence we have v ∈ Σ(K)_{i}^{δ(r,b_{i})+1}, and therefore F (b) ∈ G.

It is clear that F is injective.

Remark 5.7. For any subextension N/k of K/k, let F_{K/N} : � (N, K′) → � (K, K′) be the injective homomorphism obtained by successive applications of Proposition 5.6.

The group � (K/K_{0}, K′).

As we will see, the cokernel of F is isomorphic to the group � (K/K_{0}, K′), defined as follows :

For all i ∈ I, set r_{i} = min{1, e_{i}}. For all c ∈ ⊕_{i∈I}ℤ/p^{r_{i}}ℤ and n ∈ ℤ/pℤ, set

I_{n}^{1}(c) = {i ∈ I |n ⪰ c}. If I_{n}^{1}(c) ≠ I, set Ω(I_{n}^{1}(c)) = ∩_{i∉I_{n}^{1}(c)}Σ_{i}; if I^{n}(c) = I, set

Ω(I_{n}^{1}(c)) = Ω_{k}. Set

G(K/K_{0}, K′) = {c ∈ ⊕_{i∈I}ℤ/p^{r_{i}}ℤ | ⋃_{n∈ℤ/p^{r_{1}}ℤ}Ω(I_{n}^{1}(c)) = Ω_{k}},

let D(K/K_{0}, K′) be the diagonal subgroup of G(K/K_{0}, K′), and set

� (K/K_{0}, K′) = G(K/K_{0}, K′)/D(K/K_{0}, K′).

Lemma 5.8. The projection π : ⊕_{i∈I}ℤ/p^{e_{i}}ℤ → ⊕_{i∈I}ℤ/p^{r_{i}}ℤ induces a homomor-

phism π : � (K, K′) → � (K/K_{0}, K′).

Proof. Let a ∈ G, and set \overline{a} = π(a). Let us show that \overline{a} ∈ G(K/K_{0}, K′). Let v ∈ Ω_{k}; then there exists s ∈ ℤ/p^{e_{1}}ℤ such that v ∈ Ω(I_{s}(a)). Set n = π_{e_{1},1}(s), and let us prove that v ∈ Ω(I_{n}^{1}(\overline{a})). This is clear if I_{n}^{1}(\overline{a}) = I. Suppose that I_{n}^{1}(\overline{a}) ≠ I. If i ∈ I is such that i ∉ I_{n}^{1}(\overline{a}), then we have i ∉ I_{s}(a), and therefore v ∈ Σ_{i}^{δ(s,a_{i})}. Since n = π_{e_{1},1}(s) and i ∉ I_{n}^{1}(\overline{a}), we have δ(s, a_{i}) = 0, and hence v ∈ Σ_{i}. Therefore we have \overline{a} ∈ G(K/K_{0}, K′), as claimed, and this completes the proof of the lemma.

Proposition 5.9. The sequence

0 → � (K_{0}, K′) ⟶^{F} � (K, K′) ⟶^{π} � (K/K_{0}, K′) → 0

is exact.

Proof. It is clear that F is injective, and that π ∘ F = 0; it remains to check that π is surjective, and that Ker(π) ⊂ Im(F). Let us check the second assertion first. Let a ∈ � (K, K′) be such that π(a) = 0. Then there exists b ∈ ⊕_{i∈I}ℤ/p^{f_{i}}ℤ such that F (b) = a; let us check that b ∈ � (K_{0}, K′). Let

v ∈ Ω_{k}. Then there exists n ∈ ℤ/p^{e_{i}}ℤ such that v ∈ Ω(I_{n}(a)). If i ∈ I_{n}(a), then we have π_{e_{1},e_{i}}(n) = a_{i}. Since a_{i} = F_{i}(b_{i}), this implies that there exists r ∈ ℤ/p^{f_{1}}ℤ such that n = F_{1}(r). Let us show that v ∈ Ω(I_{r}(b)). For all i ∈ I such that i ∉ I_{n}(a), we have v ∈ Σ(K)_{i}^{δ(n,a_{i})}. Note that δ(n, a_{i}) = δ(r, b_{i}) + 1, hence this implies that v ∈ Σ(K_{0})_{i}^{δ(r,b_{i})}. Therefore we have v ∈ Ω(I_{r}(b)), as claimed, and this implies that b ∈ � (K_{0}, K′). Let us now prove that π is surjective. Let \overline{a} ∈ � (K/K_{0}, K′). For each n ∈ ℤ/pℤ, let us fix a lifting r(n) ∈ ℤ/p^{e_{1}}ℤ. If i ∈ I_{n}^{1}(\overline{a}), set a_{i} = π_{e_{1},e_{i}}(r(n)). Let us check that a_{i} ∈ ℤ/p^{e_{i}}ℤ is well-defined. Suppose that n_{1}, n_{2} ∈ ℤ/pℤ are such that i ∈ I_{n_{1}}^{1}(\overline{a}) ∩ I_{n_{2}}^{1}(\overline{a}); then we have π_{1,r_{i}}(n_{1}) = π_{1,r_{i}}(n_{2}). If n_{1} ≠ n_{2}, then this implies that r_{i} = 0, hence e_{i} = 0. We have π_{e_{1},e_{i}}(r(n_{1})) = π_{e_{1},e_{i}}(r(n_{2})) in this case, hence a_{i} is well-defined. Let us check that a ∈ � (K, K′). Since \overline{a} ∈ � (K/K_{0}, K′), we have ⋃_{n∈ℤ/p^{r_{1}}ℤ}Ω(I_{n}^{1}(\overline{a})) = Ω_{k}. Let v ∈ Ω_{k}; then there exists n ∈ ℤ/pℤ such that

v ∈ Ω(I_{n}^{1}(\overline{a})). Let r = r(n); we claim that v ∈ Ω(I_{r}(a)). If I_{n}^{1}(\overline{a}) = I, then we have I_{r}(a) = I, and the claim is clear. Suppose that I_{n}^{1}(\overline{a}) ≠ I. If i ∉ I_{r}(a), then we have i ∉ I_{n}^{1}(\overline{a}) by construction, hence v ∈ Σ_{i}. Since Σ_{i} ⊂ Σ_{i}(K)^{δ(r,a_{i})}, the claim follows. This completes the proof of the proposition.

5.3. The general case

Recall that K/k is a cyclic extension of degree d, and let P be the set of prime numbers dividing d. For all p ∈ P, let K(p) be the largest subfield of K such that [K(p) : k] is a power of p. Set

� (K, K′) = ⊕_{p∈P}� (K(p), K′).

Proposition 5.10. We have � ^{1}(k, Ŝ_{K,K′}) ≃ � (K, K′).

Proof. By 5.3 we have � ^{1}(k, Ŝ_{K(p),K′}) ≃ � (K(p), K′), hence it suffices

to show that � ^{1}(k, Ŝ_{K,K′}) ≃ ∏_{p∈P} � ^{1}(k, Ŝ_{K(p),K′}). For every p ∈ P, set

E(p) = K(p) ⊗_{k} K′ and L(p) = K(p) × K′. The inclusion K(p) → K induces maps ϵ_{p} : T_{K(p)/k}→T_{K/k}, ϵ_{p} : T_{E(p)/K′} → T_{E/K′} and ϵ_{p} : S_{K(p),K′} → S_{K,K′}. We have the commutative diagram, coming from cohomology exact sequences associated to the dual sequences of (2.1) :

(5.6)

\xymatrix{H^{1}(k, T̂_{K/k}) \ar[d]_{⊕ϵ̂_{p}^{1}} \ar[r]^{ι^{1}} & H^{1}(k, I_{K′/k}(T̂_{E/K′})) \ar[d]_{⊕ϵ̂_{p}^{1}} \ar[r]^{ρ^{1}} & H^{1}(k, Ŝ_{K,K′}) \ar[d]_{⊕ϵ̂_{p}^{1}} \ar[r] & ... \\ ⊕_{p∈P}H^{1}(k, T̂_{K(p)/k}) \ar[r]^{ι^{1}} & ⊕_{p∈P}H^{1}(k, I_{K′/k}(T̂_{E(p)/K′})) \ar[r]^{ρ^{1}} & ⊕_{p∈P}H^{1}(k, Ŝ_{K(p),K′}) \ar[r] & ...}

where the vertical maps are induced by the maps ϵ_{p}.

For all i ∈ I, let M_{i} be a cyclic extension of K_{i} such that E_{i} is isomorphic to a product of copies of M_{i}, and let d_{i} = [M_{i} : K_{i}]. If p is a prime divisor of [K : k], set E_{i}(p) = K(p) ⊗_{k} K_{i}, and let M_{i}(p) be a cyclic extension of K_{i} such that E_{i}(p) is isomorphic to a product of copies of M_{i}(p); set d_{i}(p) = [M_{i}(p) : K_{i}]. Note that d_{i}(p) is the highest power of p dividing d_{i}, and that d_{i} = ∏_{p∈P}d_{i}(p).

Note that H^{1}(k, I_{K_{i}/k}(T̂_{E_{i}/K_{i}})) ≃ H^{1}(K_{i}, T̂_{E_{i}/K_{i}}), and that by Lemma 1.1 (ii), we have H^{1}(K_{i}, T̂_{E_{i}/K_{i}}) ≃ H^{1}(K_{i}, T̂_{M_{i}/K_{i}}). Moreover, by Lemma 1.2, we have H^{1}(K_{i}, T̂_{M_{i}/K_{i}}) ≃ ℤ/d_{i}ℤ. Similarly, we have H^{1}(k, I_{K_{i}(p)/k}(T̂_{E_{i}(p)/K_{i}})) ≃ ℤ/d_{i}(p)ℤ. Note that the morphism

ϵ̂^{1}(p) : H^{1}(k, I_{K_{i}/k}(T̂_{E_{i}/K_{i}})) ≃ ℤ/d_{i}ℤ → H^{1}(k, I_{K_{i}/k}(T̂_{E_{i}(p)/K_{i}})) ≃ ℤ/d_{i}(p)ℤ

is the canonical projection ℤ/d_{i}ℤ → ℤ/d_{i}(p)ℤ. Hence the morphism

⊕_{p∈P} ϵ̂^{1}(p) : ⊕_{i∈I}H^{1}(k, I_{K_{i}/k}(T̂_{E_{i}/K_{i}})) → ⊕_{p∈P} ⊕_{i∈I} H^{1}(k, I_{K_{i}/k}(T̂_{E_{i}(p)/K_{i}}))

is an isomorphism. Similarly,

⊕_{p∈P} ϵ̂^{1}(p) : H^{1}(k, T̂_{K/k}) ≃ ℤ/dℤ → ⊕_{p∈P}H^{1}(k, T̂_{K(p)/k}) ≃ ⊕_{p∈P}ℤ/d_{i}ℤ

is also an isomorphism. By Corollary 3.3, we have � ^{2}(k, T̂_{K/k}) = 0 and � ^{2}(k, T̂_{K(p)/k}) = 0, hence � ^{1}(k, Ŝ_{K,K′}) and � ^{1}(k, Ŝ_{K(p),K′}) are in the image of the maps ρ^{1}. Since the localization map commutes with ⊕_{p∈P}ϵ̂^{1}(p), by diagram

chasing we see that ϵ̂^{1} : � ^{1}(k, Ŝ_{K,K′}) → ⊕_{p∈P}� ^{1}(k, Ŝ_{K(p),K′}) is an isomorphism.

This completes the proof of the proposition.

Note that the proposition, together with Lemma 3.1, implies that � (K, K′) does not depend on the decomposition of L as L = K × K′. We will also use the notation � (L) = � (K, K′), where L = K × K′ is any decomposition of L with K/k a cyclic extension.

In summary, we proved

Corollary 5.11. We have � (L)^{∗} ≃ � ^{1}(k, T_{L/k}).

Example 5.12. Let p and q be two distinct odd prime numbers, with p > q. For all positive integers n, let ζ_{n} be a primitive nth root of unity. Let

L = ℚ(ζ_{p^{2}}) × ℚ(ζ_{pq}) × ℚ(ζ_{q^{2}}).

Since ℚ(ζ_{p^{2}}) and ℚ(ζ_{q^{2}}) are both cyclic, we can determine � (L) in two ways; this shows that the order of � (L) divides p − 1, and that

� (L) = � (ℚ(ζ_{p}) × ℚ(ζ_{pq}) × ℚ(ζ_{q^{2}})).

But since ℚ(ζ_{p}) is a subfield of ℚ(ζ_{pq}), we have � (ℚ(ζ_{p}) × ℚ(ζ_{pq}) × ℚ(ζ_{q^{2}})) = � (ℚ(ζ_{p})×ℚ(ζ_{q^{2}})). Note that by Proposition 4.1 we have � (ℚ(ζ_{p})×ℚ(ζ_{q^{2}})) = 0, hence we have

� (L) = 0.

6. The Brauer-Manin map

We keep the notation of the previous section : in particular, L = K × K′, where K is a cyclic extension of k of degree d, K′ is an étale k-algebra, and E = K ⊗ K′. We write K′ = ∏_{i∈I} K_{i}, where the K_{i}/k are field extensions, and E = ∏_{i∈I} E_{i}, with E_{i} = K ⊗K_{i}. The group � (L) = � (K, K′) is defined in the previous section.

Let c ∈ k^{×} and recall that X_{c} is the affine k-variety defined by the equation

N_{L/k}(t) = c.

Assume that X_{c}(k_{v}) ≠ ∅ for all v ∈ Ω_{k}. In the following, we define a homomorphism α_{c} : � (L) → ℚ/ℤ such that X_{c}(k) ≠ ∅ if and only if α_{c} = 0; the map α_{c} will be called the Brauer-Manin map associated to c.

6.1. Local points

We start with some preliminary results. We are assuming that ∏X_{c}(k_{v}) ≠ ∅; as we will see, this set contains elements satisfying certain finiteness conditions.

We first recall the notion of cyclic algebra. Let us choose a generator g of the cyclic group Gal(K/k), and let ϕ : Γ_{k} → ℤ/dℤ be given by the composition of the isomorphism Gal(K/k) → ℤ/dℤ sending g to 1 with the surjection Γ_{k} → Gal(K/k). Let us consider the exact sequence 0 → ℤ⟶^{×d}ℤ → ℤ/dℤ → 0, and let δ : H^{1}(k, ℤ/dℤ) → H^{2}(k, ℤ) be the connecting homomorphism of the associated cohomology exact sequence. If c ∈ k^{×}, let us denote by (c) the corresponding element of H^{0}(k, 𝔾_{m}). The cup product δ(ϕ).(c) is an element of H^{2}(k, 𝔾_{m}), and via the identification H^{2}(k, 𝔾_{m}) ≃ Br(k) it is mapped to the class of the cyclic algebra defined by K and c (see for instance [GS06], Proposition 4.7.3). We denote this cyclic algebra by (K, c).

The first observation is the following:

Lemma 6.1. Suppose that E_{i} is isomorphic to a product of copies of a field M_{i}, and set [M_{i} : K_{i}] = d_{i}. Then for any x ∈ K_{i}^{×}, the order of the cyclic algebra (K, N_{K_{i}/k}(x)) divides d_{i}. In particular, if d_{i} = 1, then for any x ∈ k^{×}, the algebra (K, N_{K_{i}/k}(x)) splits.

Proof. Given x ∈ K_{i}^{×}, consider the cyclic algebra (M_{i}, x) = δ(ϕ|_{K_{i}}).(x), where ϕ|K_{i} : Γ_{K_{i}} → ℤ/dℤ is the restriction of ϕ to Γ_{K_{i}}. Let r be the order of (M_{i}, x) in Br(K_{i}). Since M_{i} is of degree d_{i} over K_{i}, we have r|d_{i}. By the projection formula ([GS06] Prop. 3.4.10), the corestriction of (M_{i}, x) is (K, N_{K_{i}/k}(x)). Therefore the order of (K, N_{K_{i}/k}(x)) divides d_{i}.

Let x = (x^{v}) ∈ ∏_{v∈Ω_{k}}X_{c}(k_{v}), and let us write x^{v} = (x_{0}^{v}, x_{1}^{v}, …, x_{m}^{v}), with

x_{0}^{v} ∈ K^{v} and x_{i}^{v} ∈ K_{i}^{v} for i ∈ I. Let us consider the invariant map inv : Br(k_{v}) → ℚ/ℤ and set b_{i}^{v}(x) = b_{i}^{v}(x_{i}^{v}) = inv(K^{v}, N_{K_{i}^{v}/k_{v}}(x_{i}^{v})) ∈ \frac{1}{d_{i}} ℤ/ℤ ⊂ ℚ/ℤ. Note that if d_{i} is odd, then b_{i}^{v}(x) = 0 for all infinite places v ∈ Ω_{k}.

We say that x = (x_{i}^{v}) ∈ ∏_{v∈Ω_{k}}X_{c}(k_{v}) is a local point of X_{c} if for each i ∈ I,

we have b_{i}^{v}(x) = 0 for almost all v ∈ Ω_{k}. The following lemma implies the existence of local points whenever ∏_{v∈Ω_{k}}X_{c}(k_{v}) ≠ ∅.

Lemma 6.2. Assume that ∏_{v∈Ω_{k}}X_{c}(k_{v}) ≠ ∅. Then there exists

x = (x_{i}^{v}) ∈ ∏_{v∈Ω_{k}}X_{c}(k_{v})

such that b_{i}^{v}(x) = 0 for almost all v ∈ Ω_{k}, and for all i ∈ I.

Proof. For each v ∈ Ω_{k} such that (K, c)^{v} is split, there exists x_{0}^{v} ∈ K^{v} such that N_{K^{v}/k_{v}}(x_{0}^{v}) = c. Then x = (x_{0}^{v}, 1, ..., 1) is a k_{v}-point of X_{c}, and b_{i}^{v}(x) = 0 for all i ∈ I. Since (K, c)^{v} is split for almost all places v ∈ Ω_{k}, the lemma follows.

We now prove some properties of local points which will be used later.

Lemma 6.3. Let x = (x_{i}^{v}) ∈ ∏_{v∈Ω_{k}}X_{c}(k_{v}) be a local point of X_{c}. Then we have

∑_{v∈Ω_{k}}∑_{i∈I}b_{i}^{v}(x) = 0.

Proof. Let us write x^{v} = (x_{0}^{v}, x_{1}^{v}, …, x_{m}^{v}), with x_{0}^{v} ∈ K^{v} and x_{i}^{v} ∈ K_{i}^{v} for i ∈ I. For all i ∈ I, set y_{i}^{v} = N_{L_{i}^{v}/k_{v}}(x_{i}^{v}). Set y_{0}^{v} = N_{K^{v}/k_{v}}(x_{0}^{v}), and note that

y_{0}^{v} ∏_{i∈I}y_{i}^{v} = c. We have

∑_{i∈I}b_{i}^{v}(x) = ∑_{i∈I}inv(K^{v}, y_{i}^{v}) = inv(K^{v}, ∏_{i∈I}y_{i}^{v}) = inv(K^{v}, c/y_{0}^{v}) = inv(K^{v}, c).

Since c ∈ k^{×}, the Brauer-Hasse-Noether Theorem implies that ∑_{v∈Ω_{k}}inv(K^{v}, c) =

0. Hence we have ∑_{v∈Ω_{k}}∑_{i∈I}b_{i}^{v}(x) = 0, as claimed.

Lemma 6.4. Let x = (x_{i}^{v}) be a local point of X_{c}, and set b_{i}^{v} = b_{i}^{v}(x) = inv(K^{v}, N_{K_{i}^{v}/k_{v}}(x_{i}^{v})). For all i ∈ I, let x̃_{i}^{v} ∈ K_{i}^{v} and set

b̃_{i}^{v} = inv(K^{v}, N_{K_{i}^{v}/k_{v}}(x̃_{i}^{v})).

Suppose that for all i ∈ I we have b̃_{i}^{v} = 0 for almost all v ∈ Ω_{k}, and that ∑_{i∈I}b_{i}^{v} = ∑_{i∈I}b̃_{i}^{v}. Then for all v ∈ Ω_{k}, there exists x̃_{0}^{v} ∈ K^{v} such that x̃ = (x̃_{i}^{v}) is

a local point of X_{c}.

Proof. Let ỹ_{i}^{v} = N_{K_{i}^{v}/k_{v}}(x̃_{i}^{v}) and y_{i}^{v} = N_{K_{i}^{v}/k_{v}}(x_{i}^{v}). Since ∑_{i∈I}b_{i}^{v} = ∑_{i∈I}b̃_{i}^{v}, the

algebras (K^{v}, ∏_{i∈I}y_{i}^{v}) and (K^{v}, ∏_{i∈I}ỹ_{i}^{v}) are isomorphic, hence there exists some

z ∈ K^{v} such that (∏_{i∈I}y_{i}^{v})(∏_{i∈I}ỹ_{i}^{v})^{−1} = N_{K^{v}/k_{v}}(z). Therefore (x_{0}^{v}z, x̃_{1}^{v}, ..., x̃_{m}^{v}) is a

k_{v}-point of X_{c}.

Lemma 6.5. Let x = (x_{i}^{v}) be a local point of X_{c}, and set b_{i}^{v} = b_{i}^{v}(x) = inv(K^{v}, N_{K_{i}^{v}/k_{v}}(x_{i}^{v})). Suppose that for all i ∈ I, we have ∑_{v∈Ω_{k}}b_{i}^{v} = 0. Then X_{c}

has a k-point.

Proof. By the Brauer-Hasse-Noether Theorem, for every i ∈ I there exists a central simple algebra A_{i} such that inv(A_{i}) = b_{i}^{v} for all v ∈ Ω_{k}. Set y_{i}^{v} = N_{K_{i}^{v}/k_{v}}(x_{i}^{v}). Since (K^{v}, y_{i}^{v}) splits over K^{v} for all v, the algebra A_{i} also splits over K. Hence there exists ỹ_{i} ∈ k such that A_{i} is Brauer equivalent to (K, ỹ_{i}) (see [GS06] Cor. 4.7.6). Since (K, ∏_{i∈I}ỹ_{i})_{v} ≃ (K^{v}, ∏_{i∈I}y_{i}^{v}) ≃ (K, c)_{v}, the

Brauer-Hasse-Noether Theorem implies that (K, ∏_{i∈I}ỹ_{i}) ≃ (K, c), and hence

∏_{i∈I}ỹ_{i} = cN_{K/k}(w) for some w ∈ K^{×}. Moreover, we claim that the element ỹ_{i}

belongs to the group N_{K/k}(K^{×})N_{K_{i}/k}(K_{i}^{×}). To see this, we note that

(K, ỹ_{i})_{v} = (K, y_{i}^{v}) = (K, N_{K_{i}^{v}/k_{v}}(x_{i}^{v})).

Hence we have ỹ_{i} ∈ N_{K/k}(J_{K})N_{K_{i}/k}(J_{i}) where J_{i} is the idèle group of K_{i}, for all i ∈ I, and J_{K} is the idèle group of K. By Proposition 4.1, we have ỹ_{i} = N_{K/k}(w_{i})N_{K_{i}/k}(z_{i}) for some w_{i} ∈ K^{×} and z_{i} ∈ K_{i}^{×}. Therefore

∏_{i∈I}ỹ_{i} = ∏_{i∈I}N_{K/k}(w_{i})N_{K_{i}/k}(z_{i}) = cN_{K/k}(w) and (w^{−1} ∏_{i∈I}w_{i}, z_{1}, ..., z_{m}) is a k-

point of X_{c}. This completes the proof of the lemma.

6.2. Brauer-Manin map - the prime power degree case

Now suppose that K is a cyclic extension of degree d = p^{e}, where p is a prime. Let x = (x_{i}^{v}) ∈ ∏_{v∈Ω_{k}}X_{c}(k_{v}) be a local point of X_{c}. Let M_{i} be a cyclic

extension of K_{i} such that the algebra E_{i} is isomorphic to a product of copies of M_{i}; then the degree of M_{i} is p^{e_{i}} for some 0 ≤ e_{i} ≤ e. Without loss of generality, we assume that (e_{1}, ..., e_{m}) is a decreasing sequence. Let us define

α_{c} : � (K, K′) → ℚ/ℤ

by α_{c}(a_{1}, ..., a_{m}) = ∑_{v∈Ω_{k}}∑_{i∈I}a_{i}b_{i}^{v}(x), where (a_{1}, ..., a_{m}) ∈ G ⊆ ⊕_{i∈I}ℤ/p^{e_{i}}ℤ. Note

that by Lemma 6.1, we have b_{i}^{v}(x) ∈ \frac{1}{p^{e_{i}}} ℤ/ℤ. Hence a_{i}b_{i}^{v}(x) is well-defined. Moreover, by Lemma 6.3, the map α_{c} vanishes on the subgroup D of G; hence, the map α_{c} : � (K, K′) → ℚ/ℤ is well-defined.

Proposition 6.6. The map α_{c} : � (K, K′) → ℚ/ℤ is independent of the choice of the local point x = (x_{i}^{v}).

Proof. We use the notation of section 5.1. Let a ∈ G and I(a) = (I_{0}, ...., I_{p^{e_{1}}−1}). If a ∈ D, then by Lemma 6.3, we have α_{c}(a) = 0. In the following, we assume that a ∉ D.

By the definition of G, we have Ω(I_{0}) ∪ ... ∪ Ω(I_{p^{e_{1}}−1}) = Ω_{k}. Given a place v ∈ Ω_{k}, there exists n(v) ∈ ℤ/p^{e_{1}}ℤ such that v ∈ Ω(I_{n(v)}). Set δ_{i} = δ(n(v), a_{i}) and let K_{i}^{v} = ∏_{w|v}K_{i}^{w}, where K_{i}^{w} are field extensions of k_{v}. Then for all i ∉ I_{n(v)},

the algebra E_{i}^{v} is isomorphic to a products of field extensions of K_{i}^{w} of degree at most p^{δ_{i}}. Set b_{i}^{v} = b_{i}^{v}(x); by Lemma 6.1, we have b_{i}^{v} ∈ \frac{1}{p^{δ_{i}}} ℤ/ℤ. By the definition of δ_{i}, we have π_{e_{i},δ_{i}}(a_{i}) = π_{e_{1},δ_{i}}(n(v)). Hence for i ∉ I_{n(v)}, we have

a_{i}b_{i}^{v} = π_{e_{i},δ_{i}}(a_{i})b_{i}^{v} = π_{e_{1},δ_{i}}(n(v))b_{i}^{v} = n(v)b_{i}^{v}.

Hence for all v ∈ Ω_{k}, we have

∑_{i∈I}a_{i}b_{i}^{v} = n(v)∑_{i∈I}b_{i}^{v} = n(v)inv(K, c)_{v},

which is again independent of the x_{i}^{v}’s. Therefore, the map α_{c} is independent of the choice of the local point, and the proposition is proved.

The map α_{c} : � (K, K′) → ℚ/ℤ will be called the Brauer-Manin map for X_{c}.

Let K_{0} be the unique subfield of K such that [K_{0} : k] = p^{e−1}, and set L_{0} = K_{0} × K′. If c ∈ k^{×}, let X_{c}^{0} be the affine k-variety determined by N_{L_{0}/k}(t) = c. If X_{c}^{0}(k_{v}) ≠ ∅ for all v ∈ Ω_{k}, we denote by α_{c}^{0} : � (K_{0}, K′) → ℚ/ℤ the corresponding Brauer-Manin map.

If t_{i} ∈ K_{i}^{v}, set

b_{i}^{v}(K, t_{i}) = inv(K^{v}, N_{K_{i}^{v}/k_{v}}(t_{i})), and b_{i}^{v}(K_{0}, t_{i}) = inv(K_{0}^{v}, N_{K_{i}^{v}/k_{v}}(t_{i})).

Recall that a local point of X_{c} is x = (x_{i}^{v}) ∈ ∏_{v∈Ω_{k}}X_{c}(k_{v}) such that for each

i ∈ I, we have b_{i}^{v}(K, x_{i}^{v}) = 0 for almost all v ∈ Ω_{k}.

Lemma 6.7. Assume that X_{c}(k_{v}) ≠ ∅ for all v ∈ Ω_{k}. Then we have

(i) X_{c}^{0}(k_{v}) ≠ ∅ for all v ∈ Ω_{k}.

(ii) α_{c} ∘ F = α_{c}^{0}.

Proof. If x^{v} ∈ X_{c}(k_{v}), then N_{L^{v}/L_{0}^{v}}(x^{v}) ∈ X_{c}^{0}(k_{v}). This proves (i). Let us check (ii). Let x = (x_{i}^{v}) be a local point of X_{c}. Note that b_{i}^{v}(K_{0}, x_{i}^{v}) = pb_{i}^{v}(K, x_{i}^{v}). Let a ∈ � (K_{0}, K′). Then we have

α_{c}(F (a)) = ∑_{v∈Ω_{k}}∑_{i∈I}a_{i}(pb_{i}^{v}(K, x_{i}^{v})) = ∑_{v∈Ω_{k}}∑_{i∈I}a_{i}b_{i}^{v}(K_{0}, x_{i}^{v}) = α_{c}^{0}(a).

This completes the proof of the lemma.

6.3. Brauer-Manin map - the general case

Recall that K/k is a cyclic extension of degree d, and that L = K × K′, where K′ is an étale k-algebra. We keep the notation of 5.3, in particular, P is the set of prime divisors of d. For all p ∈ P, we denote by K(p) the largest subfield of K of degree a power of p, and we set L(p) = K(p) × K′. For all c ∈ k^{×} and p ∈ P, we let X_{c}(p) be the T_{L(p)/k}-torsor defined by

N_{L(p)/k}(x) = c.

Let x = (x_{i}^{v}) ∈ ∏_{v∈Ω_{k}}X_{c}(k_{v}) be a local point of X_{c}, and let us write x = (x_{0}^{v}, x^{′v})

with x_{0}^{v} ∈ K^{v} and x^{′v} ∈ K^{′v}. Then (N_{K^{v}/K(p)^{v}}(x_{0}^{v}), x^{′v}) is a local point of X_{c}(p).

Let α(p) be the Brauer-Manin map of X_{c}(p), as defined above. By Proposition 6.6 the map α(p) is independent of the choice of the local point. Recall that � (K, K′) = ⊕_{p∈P}� (K(p), K′), and let us define α_{c} : � (K, K′) → ℚ/ℤ

by α_{c} = ⊕_{p∈P}α_{c}(p). Hence α_{c} is also independent of the choice of the local point.

We call α_{c} the Brauer-Manin map for X_{c}.

7. Necessary and sufficient condition

We keep the notation of the previous sections. The main theorem is the following:

Theorem 7.1. The affine k-variety X_{c} has a k-point if and only if X_{c} has a k_{v}-point at each place v ∈ Ω_{k} and α_{c} is the zero map.

7.1. The prime power degree case

We suppose that K is cyclic of degree p^{e}, where p is a prime number and e ≥ 1. The proof of Theorem 7.1 uses induction on e. We start with some preliminary results.

Recall that E_{i} = K ⊗_{k} K_{i}.

Lemma 7.2. Suppose that K is a cyclic extension of degree p^{e}, where p is a prime and e ≥ 1. Let v ∈ Ω_{k} and i ∈ I be such that E_{i}^{v} is not isomorphic to a product of copies of K_{i}^{v}. Then for all b ∈ \frac{1}{p}ℤ/ℤ ⊆ ℚ/ℤ, there exists x ∈ K_{i}^{v} such that inv(K^{v}, N_{K_{i}^{v}/k_{v}}(x)) = b.

Proof. Suppose that K^{v} is isomorphic to a product of copies of a field extension M of k_{v}, and set [M : k_{v}] = p^{f}. Since by hypothesis E_{i}^{v} is not isomorphic to a product of copies of K_{i}^{v}, we have f ≥ 1. Assume that K_{i}^{v} ≃ ∏_{j∈J}M_{i,j}, where M_{i,j} is a field extension of k_{v} for all j ∈ J.

It suffices to prove that \frac{1}{p}ℤ/ℤ ⊆ inv(M, N_{M_{i,j}/k_{v}}(M_{i,j}^{×})) for some j ∈ J; hence we may assume that K^{v} is a field extension of k_{v} of degree p^{e} with e ≥ 1, and that K_{i}^{v} is a field.

Let Br(K^{v}/k_{v}) be the subgroup of the Brauer group of k_{v} split by K^{v}; this group is isomorphic to ℤ/p^{e}ℤ ≃ k_{v}^{×}/N((K^{v})^{×}).

For all i ∈ I, let M_{i} be a field such that E_{i}^{v} = K ⊗_{k} K_{i}^{v} is a product of copies of M_{i}, and set [M_{i} : K_{i}^{v}] = p^{e_{i}}; the hypothesis implies that e_{i} ≥ 1. The corestriction map Br(K_{i}^{v}) → Br(k_{v}) is an injection and restricts to an injection of Br(M_{i}/K_{i}^{v}) into Br(K^{v}/k_{v}), the image being the unique subgroup of order p^{e_{i}} of the cyclic group of order p^{e}. By the projection formula ([GS06] Prop. 3.4.10), the image consists of cyclic algebras of the type (K^{v}, N_{K_{i}^{v}/k_{v}}(z)) with z an element of K_{i}^{v}. Hence \frac{1}{p}ℤ/ℤ ⊆ \frac{1}{p^{e_{i}}} ℤ/ℤ = inv(K^{v}, N_{K_{i}^{v}/k_{v}}(K_{i}^{v})^{×}). This completes the proof of the lemma.

Let K_{0} be the unique subfield of K such that [K_{0} : k] = p^{e−1}.

Recall that we have K′ = ∏_{i∈I}K_{i}, that E_{i} = K ⊗ K_{i}, and that E_{i} is the

product of copies of a cyclic extension of degree e_{i} of K_{i}. Set E_{i}^{0} = K_{0} ⊗ K_{i}. Then E_{i}^{0} also splits as a product of copies of a cyclic extension of K_{i}; let us denote by f_{i} the degree of this extension. Recall that for all i ∈ I, we have f_{i} ≤ e_{i}. If moreover e_{i} ≠ 0, then e_{i} = f_{i} + 1 (cf. lemma 5.4). For all i ∈ I, the map F_{i} : ℤ/p^{f_{i}}ℤ → ℤ/p^{e_{i}}ℤ is the inclusion of the unique subgroup of order p^{f_{i}} in the group ℤ/p^{e_{i}}ℤ, and we set F = ⊕_{i∈I}F_{i}.

The map F : ⊕_{i∈I}ℤ/p^{f_{i}}ℤ → ⊕_{i∈I}ℤ/p^{e_{i}}ℤ induces a homomorphism F :

� (K_{0}, K′) → � (K, K′). Recall that the cokernel of F is isomorphic to the group � (K/K_{0}, K′), defined in section 5.

For all i ∈ I, set r_{i} = min{1, e_{i}}. For all c ∈ ⊕_{i∈I}ℤ/p^{r_{i}}ℤ and n ∈ ℤ/pℤ, set

I_{n}^{1}(c) = {i ∈ I |n ⪰ c}. If I_{n}^{1}(c) ≠ I, set Ω(I_{n}^{1}(c)) = ∩_{i∉I_{n}^{1}(c)}Σ_{i}; if I^{n}(c) = I, set

Ω(I_{n}^{1}(c)) = Ω_{k}. Set

G(K/K_{0}, K′) = {c ∈ ⊕_{i∈I}ℤ/p^{r_{i}}ℤ | ⋃_{n∈ℤ/p^{r_{1}}ℤ}Ω(I_{n}(c)) = Ω_{k}},

let D(K/K_{0}, K′) be the diagonal subgroup of G(K/K_{0}, K′), and recall that

� (K/K_{0}, K′) = G(K/K_{0}, K′)/D(K/K_{0}, K′).

Recall that the projection π : ⊕_{i∈I}ℤ/p^{e_{i}}ℤ → ⊕_{i∈I}ℤ/p^{r_{i}}ℤ induces a homomor-

phism F : � (K, K′) → � (K/K_{0}, K′) (cf. lemma 5.8)

Set L_{0} = K_{0} × K′. If c ∈ k^{×}, let X_{c}^{0} be the affine k-variety determined by N_{L_{0}/k}(t) = c. If X_{c}^{0}(k_{v}) ≠ ∅ for all v ∈ Ω_{k}, we denote by α_{c}^{0} : � (K_{0}, K′) → ℚ/ℤ the corresponding Brauer-Manin map.

If t_{i} ∈ K_{i}^{v}, set b_{i}^{v}(K, t_{i}) = inv(K^{v}, N_{K_{i}^{v}/k_{v}}(t_{i})), and b_{i}^{v}(K_{0}, t_{i}) = inv(K_{0}^{v}, N_{K_{i}^{v}/k_{v}}(t_{i})).

Recall that a local point of X_{c} is x = (x_{i}^{v}) ∈ ∏_{v∈Ω_{k}}X_{c}(k_{v}) such that for each

i ∈ I, we have b_{i}^{v}(K, x_{i}^{v}) = 0 for almost all v ∈ Ω_{k}.

Lemma 7.3. Let x = (x_{i}^{v}) be a local point of X_{c}, and let z = (z_{i}) be a global point of X_{c}^{0}. Then for all v ∈ Ω_{k}, we have

p ∑_{i∈I}b_{i}^{v}(K, x_{i}^{v}) = p ∑_{i∈I}b_{i}^{v}(K, z_{i}).

Proof. Since x is a local point of X_{c}, we have ∑_{i∈I}b_{i}^{v}(K, x_{i}^{v}) = inv(K^{v}, c) for all

v ∈ Ω_{k}. Similarly, we have ∑_{i∈I}b_{i}^{v}(K_{0}, z_{i}) = inv(K_{0}^{v}, c) for all v ∈ Ω_{k}. Note that

inv(K_{0}^{v}, c) = p inv(K^{v}, c), and inv(K_{0}^{v}, z_{i}) = p inv(K^{v}, z_{i}) for all i ∈ I. Hence we have

p ∑_{i∈I}b_{i}^{v}(K, x_{i}^{v}) = p inv(K^{v}, c) = inv(K_{0}^{v}, c) = ∑_{i∈I}b_{i}^{v}(K_{0}, z_{i}) = p ∑_{i∈I}b_{i}^{v}(K, z_{i}),

as claimed.

Lemma 7.4. Let x = (x_{i}^{v}) be a local point of X_{c}, and assume that X_{c}^{0}(k) ≠ ∅. Then there exist x̃_{i}^{v} ∈ K_{i}^{v} such that

(i) For each i ∈ I, we have b_{i}^{v}(K, x̃_{i}^{v}) = 0 for almost all v ∈ Ω_{k}.

(ii) For all i ∈ I, we have ∑_{v∈Ω_{k}}b_{i}^{v}(K, x̃_{i}^{v}) ∈ \frac{1}{p}ℤ/ℤ.

(iii) For all v ∈ Ω_{k}, we have

∑_{i∈I}b_{i}^{v}(K, x_{i}^{v}) = ∑_{i∈I}b_{i}^{v}(K, x̃_{i}^{v}).

Proof. Let z = (z_{i}) be a global point of X_{c}^{0}. Set b_{i}^{v} = b_{i}^{v}(K, x_{i}^{v}) and h_{i}^{v} = b_{i}^{v}(K, z_{i}). By Lemma 7.3, we have p∑_{i∈I}b_{i}^{v} = p∑_{i∈I}h_{i}^{v}. Since b_{i}^{v} = 0 and h_{i}^{v} = 0 for

almost all v ∈ Ω_{k}, for almost places v ∈ Ω_{k} we have ∑_{i∈I}h_{i}^{v} = ∑_{i∈I}b_{i}^{v}. Suppose that

v ∈ Ω_{k} is such that ∑_{i∈I}h_{i}^{v} ≠ ∑_{i∈I}b_{i}^{v}. Then there exists i ∈ I such that v ∉ Σ_{i}.

Since p∑_{i∈I}b_{i}^{v} = p∑_{i∈I}h_{i}^{v} in ℚ/ℤ, we know that ∑_{i∈I}b_{i}^{v} − ∑_{i∈I}h_{i}^{v} ∈ \frac{1}{p}ℤ/ℤ. By Lemma

7.2, there exists x̃_{i}^{v} ∈ K_{i}^{v} such that inv(K^{v}, N_{K_{i}^{v}/k_{v}}(x̃_{i}^{v})) = h_{i}^{v} − ∑_{i∈I}(h_{i}^{v} − b_{i}^{v}).

Set h̃_{i}^{v} = inv(K^{v}, N_{K_{i}^{v}/k_{v}}(x̃_{i}^{v})); for all j ≠ i, let x̃_{j}^{v} = z_{j}, h̃_{j}^{v} = h_{j}^{v} = b_{j}^{v}(K, z_{j}). Then we have ∑_{i∈I}h̃_{i}^{v} = ∑_{i∈I}b_{i}^{v}; this proves (iii).

Since h̃_{i}^{v} = h_{i}^{v} for almost all v ∈ Ω_{k}, (i) holds. As z = (z_{i}) is a global point of X_{c}^{0}, we have ∑_{v∈Ω_{k}}b_{i}^{v}(K_{0}, z_{i}) = 0, hence ∑_{v∈Ω_{k}}h_{i}^{v} ∈ \frac{1}{p}ℤ/ℤ; moreover,

h_{i}^{v} −h̃_{i}^{v} ∈ \frac{1}{p}ℤ/ℤ for all i ∈ I and all v ∈ Ω_{k}. Therefore we have ∑_{v∈Ω_{k}}h̃_{i}^{v} ∈ \frac{1}{p}ℤ/ℤ,

and this proves (ii).

Lemma 7.5. Assume that X_{c}(k_{v}) ≠ ∅ for all v ∈ Ω_{k}, and that X_{c}^{0}(k) ≠ ∅. Then there exists a local point x̃ = (x̃_{i}^{v}) of X_{c} such that for all i ∈ I, we have

∑_{v∈Ω_{k}}b_{i}^{v}(K, x̃_{i}^{v}) ∈ \frac{1}{p}ℤ/ℤ.

Proof. Let x = (x_{i}^{v}) be a local point of X_{c}. By Lemma 7.4, there exist x̃_{i}^{v} ∈ K_{i}^{v} such that b_{i}^{v}(K, x̃_{i}^{v}) = 0 for almost all v ∈ Ω_{k}, that ∑_{i∈I}b_{i}^{v}(K, x_{i}^{v}) = ∑_{i∈I}b_{i}^{v}(K, x̃_{i}^{v}),

and that for all i ∈ I, we have ∑_{v∈Ω_{k}}b_{i}^{v}(K, x̃_{i}^{v}) ∈ \frac{1}{p}ℤ/ℤ. By Lemma 6.4, for

all v ∈ Ω_{k}, there exists x̃_{0}^{v} ∈ (K^{v})^{×} such that (x̃_{0}^{v}, x̃_{1}^{v}, ..., x̃_{m}^{v}) ∈ X_{c}(k_{v}). This completes the proof of the lemma.

Recall that if X_{c}(k_{v}) ≠ ∅ for all v ∈ Ω_{k}, and that for all c ∈ k^{×}, we have a homomorpism α_{c} : � (K, K′) → ℚ/ℤ. We now show that α_{c} induces a homomorphism \overline{α}_{c} : � (K/K_{0}, K′) → ℚ/ℤ such that \overline{α}_{c} ∘ π = α_{c}.

Lemma 7.6. Assume that X_{c}(k_{v}) ≠ ∅ for all v ∈ Ω_{k}, and let c ∈ k^{×}, and that X_{c}^{0}(k) ≠ ∅. Then there exists a homomorphism \overline{α}_{c} : � (K/K_{0}, K′) → ℚ/ℤ such that \overline{α}_{c} ∘ π = α_{c}.

Proof. Let x = (x_{i}^{v}) be a local point of X_{c}, and set b_{i}^{v} = inv(K^{v}, N_{K_{i}^{v}/k_{v}}(x_{i}^{v})) for all i ∈ I and all v ∈ Ω_{k}. By lemma 7.5, we may assume that ∑_{v∈Ω_{k}}b_{i}^{v} ∈ \frac{1}{p}ℤ/ℤ

for all i ∈ I. Let a = (a_{1}, ..., a_{m}) ∈ G and \overline{a} = π(a) = (\overline{a}_{1}, ..., \overline{a}_{m}) ∈ G(K/K_{0}, K′). We have

α_{c}(a_{1}, ..., a_{m}) = ∑_{i∈I}a_{i}( ∑_{v∈Ω_{k}}b_{i}^{v}) = ∑_{i∈I}\overline{a}_{i}( ∑_{v∈Ω_{k}}b_{i}^{v}).

Hence α_{c} induces a homomorphism \overline{α}_{c} : � (K/K_{0}, K′) → ℚ/ℤ, as claimed.

The group � (K/K_{0}, K′) and partitions

We now need more information about the group � (K/K_{0}, K′). This will be useful in the inductive step, as well as in dealing with the case e = 1, in other words, in the case where K/k is of prime degree.

Lemma 7.7. The set G(K/K_{0}, K′) is in bijective correspondence with the partitions (J_{0}, ..., J_{p−1}) of the set {i ∈ I | r_{i} = 1} such that ∪_{n∈ℤ/pℤ}Ω(J_{n}) = Ω_{k}.

Proof. Recall that we have r_{i} = 0 or 1. Set I′ = {i ∈ I | r_{i} = 1}. Let a ∈ G(K/K_{0}, K′), and set I^{1}(a) ∩ I′ = (I_{1}^{0}(a) ∩ I′, ..., I_{p−1}^{1}(a) ∩ I′); note that I^{1}(a) ∩ I′ is a partition of I′. Hence the set G(K/K_{0}, K′) is then in bijective correspondence with the partitions (J_{0}, ..., J_{p−1}) of I′ such that ∪_{n∈ℤ/pℤ}Ω(J_{n}) = Ω_{k}, as claimed.

In the sequel, we identify G(K/K_{0}, K′) with the set of these partitions. We also note a consequence for the case where K is of degree p, in other words, if e = 1. If K/k is of prime degree, then either E_{i} is a field extension of K_{i}, or E_{i} is a product of copies of K_{i}. Let J be the subset of I such that E_{i} is a field extension of K_{i} if i ∈ J, and that E_{i} is a product of copies of K_{i} if i ∉ J.

Lemma 7.8. Assume that K/k is a degree p extension. Then G(K, K′) is in bijective correspondence with the partitions (J_{0}, ..., J_{p−1}) of J such that

∪_{n∈ℤ/pℤ}Ω(J_{n}) = Ω_{k}.

Proof. Since e = 1, we have K_{0} = k and G(K/K_{0}, K′) = G(K, K′). Hence the lemma follows from Lemma 7.7.

Let K_{prim} be the unique subfield of K of degree p over k.

Proposition 7.9. The group � (K/K_{0}, K′) is a subgroup of � (K_{prim}, K′).

Proof. J be the subset of i ∈ I such that K_{prim} ⊗_{k} K_{i} is a field extension. Then G(K_{prim}, K′) is in bijection with the set of partitions (I_{0}, …, I_{p−1}) of J such that ∪_{n∈ℤ/pℤ}Ω_{K_{prim}}(I_{n}) = Ω_{k}, where Ω_{K_{prim}}(I_{n}) = ∩_{i∉I_{n}}Σ_{i}(K_{prim}) (see

Lemma 7.8).

Note that J = {i ∈ I | r_{i} = 1}. Hence by Lemma 7.7, the set G(K/K_{0}, K′) is in bijection with the set of partitions (I_{0}, …, I_{p−1}) of J such that ∪_{n∈ℤ/pℤ}Ω_{K}(I_{n}) =

Ω_{k}, where Ω_{K}(I_{n}) = ∩_{i∉I_{n}}Σ_{i}(K).

Note that Σ_{i}(K_{prim}) ⊂ Σ_{i}(K) for all i ∈ I. Hence G(K/K_{0}, K′) ⊂ G(K_{prim}, K′), and this implies that � (K/K_{0}, K′) is a subgroup of � (K_{prim}, K′).

Proposition 7.10. If � (K_{prim}, K′) = 0, then � (K, K′) = 0.

Proof. By Proposition 7.9, we have � (K/K_{0}, K′) = 0; hence Proposition 5.9 implies that � (K, K′) = � (K_{0}, K′). Repeating this argument, we see that � (K, K′) = � (K_{prim}, K′). But � (K_{prim}, K′) = 0 by hypothesis, hence � (K, K′) = 0, as claimed.

The following lemma will be useful in the sequel.

Lemma 7.11. Let (I_{0}, …, I_{p−1}) ∈ G(K/K_{0}, K′), and let r, r′ be two distinct elements of ℤ/pℤ. Set J_{r} = I_{r}, J_{r′} = ∪_{n≠r}I_{n}, and J_{n} = ∅ if n ≠ r, r′. Then

(J_{0}, …, J_{p−1}) ∈ G(K/K_{0}, K′). If moreover (I_{0}, …, I_{p−1}) ∉ D(K/K_{0}, K′) and I_{r} ≠ ∅, then (J_{0}, …, J_{p−1}) ∉ D(K/K_{0}, K′).

Proof. Let us show that Ω(J_{r}) ∪ Ω(J_{r′}) = Ω_{k}. Let v ∈ Ω_{k} be such that v ∉ Ω(J_{r}). Since we have ∪_{n∈ℤ/pℤ}Ω(I_{n}) = Ω_{k}, there exists n(v) ∈ ℤ/pℤ with

n(v) ≠ r such that v ∈ Ω(I_{n(v)}). Since n(v) ≠ r, we have Ω(I_{n(v)}) ⊂ ∩_{i∈I_{r}}Σ_{i} =

Ω(J_{n′}). Therefore we have Ω(J_{r}) ∪ Ω(J_{r′}) = Ω_{k}, and hence (J_{0}, …, J_{p−1}) ∈ G(K/K_{0}, K′).

Let us prove the second statement. If (J_{0}, …, J_{p−1}) ∈ D(K/K_{0}), then either J_{r} = I′ or J_{r′} = I′; we have I_{r} = I′ in the first case, hence (I_{0}, …, I_{p−1}) ∈ D(K/K_{0}, K′), and I_{r} = ∅ in the second case. This completes the proof of the lemma.

Lemma 7.12. Let x = (x_{i}^{v}) be a local point of X_{c}, and set b_{i}^{v} = b_{i}^{v}(K, x_{i}^{v}). Assume that α_{c} = 0, and that X_{c}^{0}(k) ≠ ∅. Let (I_{0}, …, I_{p−1}) ∈ G(K/K_{0}, K′). Then we have ∑_{i∈I_{n}} ∑_{v∈Ω_{k}}b_{i}^{v} = 0 for all n ∈ ℤ/pℤ.

Proof. Let n ∈ ℤ/pℤ. The statement is trivial if I_{n} is empty, and it follows from Lemma 6.3 if I_{n} = I′. Assume that I_{n} is not empty, and I_{n} ≠ I′. Let

n′ ∈ ℤ/pℤ such that n′ ≠ n, and set J_{n} = I_{n}, J_{n′} = ∪_{r≠n}I_{r}, and J_{r} = ∅ if

r ≠ n, n′. Then by Lemma 7.11, we have (J_{0}, …, J_{p−1}) ∈ G(K/K_{0}, K′). Since X_{c}^{0}(k) ≠ ∅, by Lemma 7.6 there exists a homomorphism \overline{α}_{c} : � (K/K_{0}, K′) → ℚ/ℤ such that \overline{α}_{c} ∘ π = α_{c}. By hypothesis α_{c} is the zero map, hence we have \overline{α}_{c} = 0. Therefore we have

∑_{r∈ℤ/pℤ} ∑_{i∈J_{r}} ∑_{v∈Ω_{k}}rb_{i}^{v} = ∑_{i∈J_{n}} ∑_{v∈Ω_{k}}nb_{i}^{v} + ∑_{i∈J_{n′}} ∑_{v∈Ω_{k}}n′b_{i}^{v} = 0.

By Lemma 6.3 we have ∑_{i∈I′} ∑_{v∈Ω_{k}}b_{i}^{v} = 0, hence (n − n′) ∑_{i∈J_{n}} ∑_{v∈Ω_{k}}b_{i}^{v} = 0. Recall

that n′ ≠ n by hypothesis, therefore we have ∑_{i∈J_{n}} ∑_{v∈Ω_{k}}b_{i}^{v} = 0; since J_{n} = I_{n}, we

have ∑_{i∈I_{n}} ∑_{v∈Ω_{k}}b_{i}^{v} = 0, as claimed.

Lemma 7.13. Let x = (x_{i}^{v}) be a local point of X_{c}. Assume that α_{c} = 0, and that X_{c}^{0}(k) ≠ ∅. Let (I_{0}, …, I_{p−1}) ∈ G(K/K_{0}, K′), and let n ∈ ℤ/pℤ. Then there exists a local point x̃ = (x̃_{i}^{v}) of X_{c} such that x̃_{i}^{v} = x_{i}^{v} if i ∉ I_{n}, and that

∑_{v∈Ω_{k}}b_{i}^{v}(K, x̃) = 0 for all i ∈ I_{n}.

Proof. We prove this by induction on the cardinality of I_{n}. If |I_{n}| = 0 then the claim is trivial; if |I_{n}| = 1, then it follows from lemma 7.12, since we have ∑_{v∈Ω_{k}}b_{i}^{v}(x) = 0 for all i ∈ I_{n}. Suppose that the claim is true for |I_{n}| < h. For

|I_{n}| = h, suppose that there are nonempty disjoint subsets I_{n}^{0} and I_{n}^{1} of I_{n} satisfying I_{n}^{0} ∪ I_{n}^{1} = I_{n} and ( ∩_{i∈I_{n}^{0}}Σ_{i}) ∪ ( ∩_{i∈I_{n}^{1}}Σ_{i}) = Ω_{k}. Then consider the element

(J_{0}, ...J_{p−1}) where J_{r} = I_{r} if r ≠ n, n + 1, J_{n} = I_{n}^{0} and J_{n+1} = I_{n}^{1} ∪ I_{n+1}. Note that Ω(I_{r}) = Ω(J_{r}) if r ≠ n, n + 1 and that Ω(I_{n+1}) ⊂ Ω(J_{n+1}). Let us prove that (J_{0}, ...J_{p−1}) represents an element of � (K, K′); for this, we have to check that Ω(J_{0}) ∪ ⋯ ∪ Ω(J_{p−1}) = Ω_{k}. Since Ω(I_{r}) ⊂ Ω(J_{r}) if r ≠ n and Ω(I_{0}) ∪ ⋯ ∪ Ω(I_{p−1}) = Ω_{k}, it suffices to check that if v ∈ Ω(I_{n}), then v ∈ Ω(J_{0}) ∪ ⋯ ∪ Ω(J_{p−1}). If v ∈ ∩_{i∈I_{n}^{1}}Σ_{i}, then we have v ∈ Ω(J_{n}).

Otherwise, we have v ∈ ∩_{i∈I_{n}^{0}}Σ_{i} because ( ∩_{i∈I_{n}^{0}}Σ_{i}) ∪ ( ∩_{i∈I_{n}^{1}}Σ_{i}) = Ω_{k}. Hence we have

v ∈ ( ∩_{i∉I_{n}∪I_{n+1}}Σ_{i}) ∩ ( ∩_{i∈I_{n}^{0}}Σ_{i}) = Ω(J_{n+1}). Therefore (J_{0}, ...J_{p−1}) represents an

element of � (K, K′). Since |J_{n}| < h, we can apply the induction hypothesis, and hence there exists a local point x̃ = (x̃_{i}^{v}) such that x̃_{i}^{v} = x_{i}^{v} if i ∉ J_{n} = I_{n}^{1}, and that ∑_{v∈Ω}b_{i}^{v}(K, x̃) = 0 for all i ∈ J_{n} = I_{n}^{1}. The same argument with I_{n}^{0}

instead of I_{n}^{1} gives the desired result.

Assume now that I_{n} does not have any non-trivial subpartitions, in other words, that there are no nonempty disjoint subsets I_{n}^{0} and I_{n}^{1} of I_{n} satisfying I_{n}^{0} ∪ I_{n}^{1} = I_{n} and ( ∩_{i∈I_{n}^{0}}Σ_{i}) ∪ ( ∩_{i∈I_{n}^{1}}Σ_{i}) = Ω_{k}. Let us consider the graph with

vertex set I_{n}, and edge set E = {(i, j)|Σ_{i} ∪ Σ_{j} ≠ Ω_{k}}; since I_{n} has no nontrivial subpartitions, this graph is connected. Set b_{i}^{v} = b(K, x_{i}^{v})_{i}^{v}, and for all i ∈ I_{n}, set d_{i} = ∑_{v∈Ω_{k}}b_{i}^{v}. Let us fix an ordering of I_{n}, say I_{n} = {i_{0}, ..., i_{t}}. Since

the graph is connected, there exists a loop-free path between i_{0} and i_{1}. Along this path, for any two adjacent vertices i, j, there exists v ∈ Ω_{k} such that v ∉ Σ_{i} ∪ Σ_{j}. By Lemma 7.5 we may assume that b_{i}^{v} ∈ \frac{1}{p}ℤ/ℤ for all i ∈ I_{n}. Applying Lemma 7.2, by modifying x_{i}^{v} and x_{j}^{v} we can modify b_{i}^{v} to b_{i}^{v} −d_{i_{0}} and b_{j}^{v} to b_{j}^{v} + d_{i_{0}}. Note that this modification does not change ∑_{i∈I}b_{i}^{v}. Therefore by

Lemma 6.4, after changing also x_{0}^{v} if necessary, the modified (x_{i}^{v}) is still a local point of X_{c}. After these modifications, we have ∑_{v∈Ω_{k}}b_{i_{0}}^{v} = 0, ∑_{v∈Ω_{k}}b_{i_{1}}^{v} = d_{i_{1}} + d_{i_{0}},

and all the other d_{i}’s remain unchanged. We repeat this process along a loopfree path from i_{1} to i_{2}, and we modify each adjacent pair along the path from i_{1} to i_{2} by d_{i_{0}} + d_{i_{1}} and so on. At the end, we modify each adjacent pair along

the path from i_{t−1} to i_{t} by ∑^{t−1} ; r=0d_{i_{r}}. After this process, we have ∑_{v∈Ω_{k}}b_{i_{r}}^{v} = 0 for

r = 0, ..., t − 1 and ∑_{v∈Ω_{k}}b_{i_{t}}^{v} = d_{i_{t}} + ∑^{t−1} ; r=0d_{i_{r}}. However, by Lemma 7.12, we know

that ∑^{t} ; r=0d_{i_{r}} = 0; hence, we have ∑_{v∈Ω_{k}}b_{i_{t}}^{v} = 0. Moreover, only finitely many b_{i}^{v}’s

are modified, so b_{i}^{v} = 0 for almost all v; the lemma then follows.

Proposition 7.14. Let x = (x_{i}^{v}) be a local point of X_{c}. Assume that α_{c} = 0, and that X_{c}^{0}(k) ≠ ∅. Then there exists a local point x̃ = (x̃_{i}^{v}) of X_{c} such that for all i ∈ I, we have

∑_{v∈Ω_{k}}b_{i}(K, x̃_{i}^{v}) = 0.

Proof. This follows from Lemma 7.13.

Proof of Theorem 7.1 for K of prime power degree.

It is clear that if X_{c} has a k-point, then X_{c} has a k_{v}-point for all v ∈ Ω_{k} and α_{c} = 0. Conversely, suppose that X_{c} has a k_{v}-point for all v ∈ Ω_{k} and that α_{c} = 0. Let us show that the variety X_{c} has a k-point. We show our claim by induction on the exponent e. Suppose that e = 1. Then K_{0} = k, and X_{c}^{0}(k) ≠ ∅. By Proposition 7.14, there exists a local point x = (x_{i}^{v}) of X_{c} such that for all i ∈ I, we have ∑_{v∈Ω_{k}}b_{i}(x_{i}^{v}) = 0. Lemma 6.5 implies that

X_{c}(k) ≠ ∅. Assume now that e > 1. Since X_{c} has a k_{v}-point for all v ∈ Ω_{k}, the variety X_{c}^{0} also has a k_{v}-point for all v ∈ Ω_{k}. As α_{c} is the zero map, by Lemma 6.7 the Brauer-Manin map α_{c}^{0} for X_{c}^{0} is also the zero map. Therefore X_{c}^{0} has a k-point by induction hypothesis. By Proposition 7.14, there exists a local point x = (x_{i}^{v}) of X_{c} such that for all i ∈ I, we have ∑_{v∈Ω_{k}}b_{i}(x_{i}^{v}) = 0.

Lemma 6.5 implies that X_{c}(k) ≠ ∅.

7.2. The general case

Recall that K/k is a cyclic extension of degree d, and that L = K × K′, where K′ is an arbitrary étale k-algebra. We keep the notation of 5.3, in particular, P is the set of prime divisors of d. For all p ∈ P, we denote by K(p) the largest subfield of K of order a power of p, and L(p) = K(p) × K′. For all c ∈ k^{×} and p ∈ P, the affine k-variety defined by

N_{L(p)/k}(x) = c.

is denoted by X_{c}(p). We denote by α_{c}(p) be the Brauer-Manin map of X_{c}(p). Recall that � (K, K′) = ⊕_{p∈P}� (K(p), K′), and that α_{c} : � (K, K′) → ℚ/ℤ is

given by α_{c} = ⊕_{p∈P}α_{c}(p).

Lemma 7.15. Let c ∈ k^{×}. Then X_{c} has a k-point if and only if X_{c}(p) has a k-point for all p ∈ P.

Proof. Let z ∈ X_{c}(k) be a k-point of X_{c}, and let us write z = (x, y) with x ∈ K and y ∈ K′. Then (N_{K/K(p)}(x), y) is a k-point of X_{c}(p) for all p ∈ P. Conversely, suppose that for all p ∈ P, the k-variety X_{c}(p) has a k-point (x_{p}, y_{p}) ∈ K(p) × K′. For all p ∈ P, set

r_{p} = ∏_{q∈P,q≠p}[K(q) : k],

and let s_{p} ∈ ℤ such that ∑_{p∈P}r_{p}s_{p} = 1. Set x = ∏_{p∈P}x_{p}^{s_{p}}, and y = ∏_{p∈P}y_{p}^{r_{p}s_{p}}. Then

(x, y) is a k-point of X_{c}.

Proof of Theorem 7.1. Suppose that X_{c} has a k-point. Then by lemma 7.15, X_{c}(p) has a k-point for all p ∈ P. This implies that α_{c}(p) = 0 for all p ∈ P, and hence α_{c} = 0. Conversely, suppose that X_{c} has a k_{v}-point for all v ∈ Ω_{k} and that α_{c} = 0. Then X_{c}(p) has a k_{v}-point for all v ∈ Ω_{k}. Since α_{c} = 0, we have α_{c}(p) = 0 for all p ∈ P. But K(p) is a cyclic extension of prime power degree, hence this implies that X_{c}(p) has a k-point for all p ∈ P. Therefore X_{c} has a k-point by Lemma 7.15.

Corollary 7.16. Let I_{L} be the idèle group of L. Then sending c ∈ k^{×} to α_{c} gives rise to an isomorphism

(k^{×} ∩ N_{L/k}(I_{L}))/N_{L/k}(L^{×}) → � (L)^{∗}.

Proof. It is clear from the definition of α_{c} that sending c ∈ k^{×} to α_{c} is a homomorphism; Theorem 7.1 implies that this homomorphism is injective. That it is an isomorphism follows from the fact that � (L)^{∗} ≃ � ^{1}(k, T_{L/k}) (see Corollary 5.11).

Metacyclic extensions

In the following we apply the main theorem to the case where K is a metacyclic extension of k (recall that a metacyclic extension is a Galois extension such that all the Sylow subgroups of its Galois group are cyclic). As before, let X_{c} be the k-variety defined by the equation (0.1). Assume that K/k is a

metacyclic extension of degree q = ∏^{s} ; j=1p_{j}^{e_{j}}, where p_{j}’s are distinct primes. Let

q_{j} = p_{j}^{e_{j}} and r_{j} = q/q_{j}. For 1 ≤ j ≤ s, let G_{j} be a p_{j}-Sylow subgroup of Gal(K/k) and let F_{j} be the subfield of K fixed by G_{j}. Note that [F_{j} : k] = r_{j}. Let X_{c}^{j} be X_{c} ⊗_{k} F_{j}. Then the injection k → F_{j} induces a natural injection of X_{c}(k) to X_{c}(F_{j}) = X_{c}^{j}(F_{j}).

Suppose that X_{c} has a k_{v}-point for all v ∈ Ω_{k}. Then X_{c}^{j} has a F_{j,w}-point for all w ∈ Ω_{F_{j}}. Since K is a cyclic extension of F_{j}, we can define the Brauer-Manin map α_{j} for X_{c}^{j}. The necessary and sufficient condition for the Hasse principle for X_{c} to hold is the following :

Proposition 7.17. Assume that K is a metacyclic extension. Then X_{c} has a k-point if and only if X_{c} has a k_{v}-point for all v ∈ Ω_{k} and α_{j} = 0 for 1 ≤ j ≤ s.

Proof. Assume that X_{c} has a k_{v}-point for all v ∈ Ω_{k}, and that α_{j} = 0 for 1 ≤ j ≤ s. Then the variety X_{c}^{j} has a F_{j,w}-point for all w ∈ Ω_{F_{j}}. Since α_{j} = 0 for all 1 ≤ j ≤ s, by Theorem 7.1 the variety X_{c}^{j} has a F_{j}-point. Let (x_{j,i}) be a

F_{j}-point of X_{c}^{j}, where x_{j,i} ∈ (F_{j} ⊗_{k} K_{i})^{×}. Let b_{j} ∈ ℤ such that ∑^{s} ; j=1b_{j}r_{j} = 1, and

set z_{i} = ∏^{s} ; j=1N_{F_{j}⊗K_{i}/K_{i}}(x_{j,i})^{b_{j}}; then (z_{i}) is a point of X_{c}. The other direction is

trivial.

8. Products of cyclic extensions

In this section, we suppose that L is a product of cyclic extensions, and we denote by � (L) the obstruction group. In the following, we give a simple criterion for the vanishing of � (L); in other words, an easy way to decide whether the Hasse principle holds for L.

Assume that L = ∏_{i∈J}K_{i}, where K_{i}/k is a cyclic extension of degree d_{i}. Let

P be the set of prime numbers dividing ∏_{i∈J}d_{i}. For all p ∈ P and all i ∈ J, let

K_{i}(p) be the largest subfield of K_{i} such that [K_{i}(p) : k] is a power of p, and

set L(p) = ∏_{i∈J}K_{i}(p).

For any cyclic field extension K/k of prime power degree, we denote by K_{prim} the unique subfield of K of degree p over k. Set L(p)_{prim} = ∏_{i∈J}K_{i}(p)_{prim}.

The aim of this section is to prove the following two results :

Theorem 8.1.

� (L) = 0 ⟺ ⊕_{p∈P(L)}� (L(p)_{prim}) = 0,

where P(L) is a set of prime numbers, subset of P.

The set P(L) is determined in Theorem 8.3, see below.

Theorem 8.2.

� (L(p)_{prim}) ≃ (ℤ/pℤ)^{m_{p}(L)},

where m_{p}(L) is a positive integer.

The value of m_{p}(L) is given in Theorem 8.3.

We start with the proof of Theorem 8.2, which amounts to treating the case where L is a product of cyclic extensions of prime degree.

Theorem 8.3. Let p be a prime number, and assume that L is a product of n non-isomorphic cyclic extensions of degree p. Then we have

(a) If n ≤ 2, then � (L) = 0.

(b) If 3 ≤ n ≤ p + 1, then either � (L) = 0, or � (L) ≃ (ℤ/pℤ)^{n−2}.

(c) If n ≥ p + 2, then � (L) = 0.

Note that Theorem 8.3 implies immediately Theorem 8.2, and gives the value of the integer m_{p}(L).

In order to prove Theorem 8.3, we need to come back to the definition of � (L) = � (K, K′) in the case where K is cyclic of prime degree, and give a description of this group in terms of partitions.

We keep the notation of 5.1, with e = 1. In particular, p is a prime number, and L = K × K′, where K is a cyclic extension of k of degree p. Recall that E_{i} = K ⊗ K_{i}, and note that E_{i} is either a cyclic field extension of K_{i} or a product of p copies of K_{i}. Let J be the subset of i ∈ I such that E_{i}/K_{i} is a field extension, and let r = |J|.

Recall that Σ_{i} is the set of v ∈ Ω_{k} such that E_{i}^{v} is the product of p copies of K_{i}^{v}. For all J′ ⊂ J with J′ ≠ J, set Ω(J′) = ∩_{i∉J′}Σ_{i}, and let Ω(J) = Ω_{k}.

By lemma 7.8, the group G(K, K′) is in bijection with the set of partitions (J_{0}, …, J_{p−1}) of J such that ∪_{n∈ℤ/pℤ}Ω(J_{n}) = Ω_{k}. We identify G(K, K′) with

the set of these partitions. Note that under this identification, D(K, K′) corresponds to the partitions where one of the subsets is J, and all the others are empty; these will be called the trivial partitions of J.

For all n ∈ ℤ/pℤ and all a ∈ (ℤ/pℤ)^{r}, set J_{n}(a) = {i ∈ J | a_{i} = n}. Then lemma 7.8 can be reformulated as follows :

Lemma 8.4. G(K, K′) is in bijection with the set

{a ∈ (ℤ/pℤ)^{r} | ∪_{n∈ℤ/pℤ} Ω(J_{n}(a)) = Ω_{k}}.

Proof of Theorem 8.3

Note first that (a) follows from Proposition 4.1. From now on, we assume that n ≥ 3. Theorem 8.3, as well as a precise condition for when � (L) = 0 in case (b), is a consequence of Proposition 8.5 below.

For any positive integer d, a finite separable extension F of k is said to have local degrees ≤ d if for all places v ∈ Ω_{k}, the étale algebra F ⊗_{k} k_{v} is a product of field extensions of k_{v} with degrees ≤ d.

Proposition 8.5. Let p be a prime number, and assume that L is a product of distinct field extensions of degree p of k, at least one of which is cyclic.

Then � (L) ≠ 0 ⟺ the factors of L are distinct subfields of a field extension F/k of degree p^{2}, and all the local degrees of F are ≤ p.

Moreover, if � (L) ≠ 0, and if L is a product of n distinct degree p field extensions of k, then � (L) ≃ (ℤ/pℤ)^{n−2}.

Proof. Let K be a cyclic factor of L, and let us write L = K ×K′, where K′ is a product of field extensions of degree p of k. Suppose that � (L) ≠ 0. Then there exists a partition (I_{0}, I_{1}) of J such that Ω(I_{0}) ∪ Ω(I_{1}) = Ω_{k}. Indeed, let (J_{0}, …, J_{p−1}) be a non-trivial partition of J such that ∪_{r∈ℤ/pℤ}Ω(J_{i}) = Ω_{k}.

Without loss of generality, we can assume that J_{0} is not empty. Set I_{0} = J_{0}, and let I_{1} = ∪_{i≠0}J_{i}; then we have Ω(I_{0}) = Ω(J_{0}), and Ω(J_{r}) ⊂ Ω(I_{1}) for all

r ≠ 0. Therefore Ω(I_{0}) ∪ Ω(I_{1}) = Ω_{k}, as claimed. Let K_{i} and K_{j} be two distinct factors of K′, and let K_{i}K_{j} be the composite of K_{i} and K_{j}. For all v ∈ Σ_{i}, we have

K ⊗_{k} (K_{i}K_{j})^{v} ≃ K ⊗_{k} K_{i}^{v} ⊗_{K_{i}^{v}} (K_{i}K_{j})^{v},

and, since v ∈ Σ_{i}, this is isomorphic to the product of p copies of (K_{i}K_{j})^{v}.

Let i ∈ I_{0} and j ∈ I_{1}. As we have Ω(I_{0}) ∪ Ω(I_{1}) = Ω_{k}, the tensor product K ⊗_{k} (K_{i}K_{j})^{v} is isomorphic to the product of p copies of (K_{i}K_{j})^{v} for all v ∈ Ω_{k}. This implies that K is a subfield of K_{i}K_{j}. Recall that K is cyclic, and that K_{i}, K_{j} are not isomorphic; hence we have K ⊗_{k} K_{i} ≃ KK_{i} ⊂ K_{i}K_{j}. The degree of K_{i}K_{j} is at most p^{2}, hence we have KK_{i} = K_{i}K_{j} = KK_{j}, and K_{i} ⊗_{k} K_{j} ≃ K_{i}K_{j} is of degree p^{2} over k.

Let i ∈ I_{0}, and set F = KK_{i}; we just saw that F is independent of the choice of i, and that F = K_{i}K_{j} for all j ∈ I_{1}. This shows that K_{i} is a subfield of F for all i ∈ J. Since (I_{0}, I_{1}) represents a non-trivial element of � (L), for all v ∈ Ω_{k} there exists i ∈ J such that F^{v} ≃ K ⊗_{k} K_{i}^{v} is isomorphic to a product of p copies of K_{i}^{v}. Therefore all the local degrees of F are ≤ p.

Conversely, let F be a separable extension of degree p^{2} of k such that all the factors of L are distinct subfields of F. It suffices to prove that all nontrivial partitions (J_{0}, …, J_{p−1}) of J satisfy ∪_{r∈ℤ/pℤ}Ω(J_{i}) = Ω_{k}. Suppose that

this is not the case. Let (J_{0}, …, J_{p−1}) be a non-trivial partition of J with ∪_{r∈ℤ/pℤ}Ω(J_{i}) ≠ Ω_{k}. Let v ∈ Ω_{k} with v ∉ ∪_{r∈ℤ/pℤ}Ω(J_{i}). Since v ∉ Ω(J_{0}), there

exists i ∉ J_{0} such that iv ∉ Σ_{i}. Let r ∈ ℤ/pℤ such that i ∈ J_{r}; since v ∉ Ω(J_{r}), there exists j ∉ J_{r} such that v ∉ Σ_{j}.

Since the degree p extensions K, K_{i} and K_{j} are distinct subfields of F, we have F ≃ K ⊗ K_{i} ≃ K ⊗ K_{j}. Note that [K^{v} : k_{v}] = p, because v ∉ Σ_{i}. Let us write K_{i}^{v} as a product of separable extensions of k_{v}. If one of the factors M_{s} of K_{i}^{v} is such that 1 < [M_{s} : k_{v}] < p, then M_{s} and K^{v} are linearly disjoint, and this contradicts the assumption that all the local degrees of F are ≤ p. Hence K_{i}^{v} is either a degree p field extension of k_{v}, or a product of p copies of k_{v}. However, if K_{i}^{v} and K^{v} are both fields, then E_{i}^{v} is a field extension of degree p^{2} of k_{v}. Since F^{v} ≃ E_{i}^{v}, this contradicts the hypothesis that all the local degrees of F are ≤ p. Therefore K_{i}^{v} is a product of p copies of k_{v}, and hence F^{v} ≃ E_{i}^{v} is a product of p copies of K^{v}.

Set d = [K_{i}K_{j} : k]. Since v ∉ Σ_{j}, the same argument shows that K_{j}^{v} is a product of p copies of k_{v}, hence (K_{i}K_{j})^{v} is a product of d copies of k_{v}. Note that (K_{i}K_{j})^{v} is a subalgebra of F^{v}, and that F^{v} is a product of p copies of K^{v}; hence we have d ≤ p. As K_{i} and K_{j} are distinct subfields of K_{i}K_{j}, we have d = rp for some integer r > 1, and this leads to a contradiction.

Hence for all non-trivial partitions (J_{0}, …, J_{p−1}) of J we have ∪_{r∈ℤ/pℤ}Ω(J_{i}) =

Ω_{k}. This shows that � (K, K′) = � (L) ≃ (ℤ/pℤ)^{n−2}.

Proof of Theorem 8.1

Assume now that L is a product of n cyclic extensions, L = K_{1} × ⋯ × K_{n}, where K_{i}/k is a cyclic extension of degree d_{i}, and let J = {1, …, n}. Note that � (L) = � (K_{i}, K′_{i}) for any i ∈ J, where L = K_{i} × K′_{i}. This will be used repeatedly in the sequel.

Let P be the set of prime numbers dividing d_{1} … d_{n}. For all p ∈ P and all i ∈ J, let K_{i}(p) be the largest subfield of K_{i} such that [K_{i}(p) : k] is a power of p, and set L(p) = K_{1}(p) × ⋯ × K_{n}(p).

Proposition 8.6. We have

� (L) = ⊕_{p∈P(L)}� (L(p)).

Proof. This follows from Proposition 5.10, and from the fact that L is a product of cyclic extensions.

Lemma 8.7. Let p be a prime number, and let K_{i}/k, i ∈ J, be cyclic extensions of degree a power of p of k. For all i ∈ J, let N_{i}/k be a subextension of K_{i}/k. Then � (∏_{i∈J}N_{i}) injects into � (∏_{i∈J}K_{i}).

Proof. This follows from Proposition 5.6, and Remark 5.7 following this proposition.

Proof of Theorem 8.1 Assume that � (L) = 0. By Lemma 8.7 and Proposition 8.6, the group � (L_{prim}) injects into � (L), hence this implies that � (L_{prim}) = 0. Conversely, suppose that � (L_{prim}) = 0. By Proposition 8.6, we may assume that L is a product of extensions of degree a power of a prime p. Let us write L = K × K′, for some cyclic field extension K/k; then L_{prim} = K_{prim} × K′_{prim}. Since � (L_{prim}) = 0, by Proposition 7.10 we have � (K, K′_{prim}) = 0. Permuting K with one of the other cyclic factors and repeating the same procedure, we obtain � (L) = 0.

Example 8.8. Let p be a prime number, and let F/k be an extension with Galois group C_{p} × C_{p}, where C_{p} denotes the cyclic group of order p. Let K_{1}, …, K_{p+1} be the distinct subfields of degree p of F. Set L = K_{1}×⋯×K_{p+1}. Then by Proposition 8.5, we have � (L) = 0 or � (L) = (ℤ/pℤ)^{p−1}. Moreover, we have

� (L) = 0 ⟺ there exists v ∈ Ω_{k} such that F^{v} is a field.

• Assume first that there exists v ∈ Ω_{k} such that F^{v} is a field. Then � (L) = 0, hence for all c ∈ k^{×}, we have X_{c}(k) ≠ ∅. In other words, we have

N_{L/k}(L^{×}) = k^{×}

in this case.

• Assume now that all the local degrees of F are ≤ p. Then by Proposition 8.5 we have � (L) = (ℤ/pℤ)^{p−1}.

Let Ω_{i} be the set of v ∈ Ω_{k} such that K_{i}^{v} is split. Note that we have Ω_{1} ∪ ⋯ ∪ Ω_{p+1} = Ω_{k}. This implies that X_{c}(k_{v}) ≠ ∅ for all v ∈ Ω_{k} and for all c ∈ k×.

Set K = K_{p+1}. For all c ∈ k^{×} and for all v ∈ Ω_{k}, let us denote by [K, c]_{v} ∈ ℤ/pℤ the image of inv(K, c)_{v} by the isomorphism \frac{1}{p}ℤ/ℤ ≃ pℤ/ℤ. Then the map

f : k^{×}/N_{L/k}(L^{×}) → (ℤ/pℤ)^{p−1}

given by

c ↦ (∑_{Ω_{1}}[K, c]_{v}, …, ∑_{Ω_{p−1}}[K, c]_{v}),

is an isomorphism.

When p = 2, we recover a well-known result of Serre and Tate, see [CF 67], Exercise 5.2, page 360; see also [CT 14], Proposition 5.1.

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Eva Bayer–Fluckiger and Ting-Yu Lee

EPFL-FSB-MATHGEOM-CSAG

Station 8

1015 Lausanne, Switzerland

eva.bayer@epfl.ch

ting-yu.lee@epfl.ch

Raman Parimala

Department of Mathematics & Computer Science Emory University Atlanta, GA 30322, USA.

parimala@mathcs.emory.edu