WHEN IS GALOIS COHOMOLOGY FREE OR TRIVIAL?

NICOLE LEMIRE^{⋆}, JÁN MINÁČ^{∗†}, AND JOHN SWALLOW^{‡}

Abstract. Let p be a prime and F a field, perfect if p > 2, containing a primitive pth root of unity. Let E/F be a cyclic extension of degree p and G_{E} ◁ G_{F} the associated absolute Galois groups. We determine precise conditions for the cohomology group H^{n}(E) = H^{n}(G_{E}, 𝔽_{p}) to be free or trivial as an 𝔽_{p}[Gal(E/F)]- module. We examine when these properties for H^{n}(E) are inherited by H^{k}(E), k > n, and, by analogy with cohomological dimension, we introduce notions of cohomological freeness and cohomological triviality. We give examples of H^{n}(E) free or trivial for each n ∈ ℕ with prescribed cohomological dimension.

Let p be a prime and F a field containing a primitive pth root of unity ξ_{p}, and if p > 2 suppose additionally that F is perfect. Let E/F be a cyclic extension of degree p and G_{E} the absolute Galois group of E. In our previous paper [LMS] we determined the structure of H^{n}(G_{E}, 𝔽_{p}), n ∈ ℕ, as an 𝔽_{p}[G]-module. In this paper we study more closely the question of when H^{n}(G_{E}, 𝔽_{p}) is free or trivial as an 𝔽_{p}[G]-module.

Let a ∈ F satisfy E = F (\sqrt[p]{a}). We write H^{n}(F) for H^{n}(G_{F} , 𝔽_{p}) and ann_{n} x for the annihilator of x under the cup-product operation on H^{n}(F). (Thus ann_{n} x ⊂ H^{n}(F).) Let (f) ∈ H^{1}(F) denote the class of f under the Kummer isomorphism of H^{1}(F) with the pth-power classes of F^{×} := F \ {0}, and let (f, g) ∈ H^{2}(F) denote the cup-product of (f) and (g) ∈ H^{1}(F).

We first give precise conditions for free 𝔽_{p}[G]-module cohomology.

Theorem 1. Let n ∈ ℕ.

Suppose p > 2 and F is perfect. Then the following are equivalent:

Date: October 30, 2004. ⋆Research supported in part by NSERC grant R3276A01. ∗Research supported in part by NSERC grant R0370A01 and by a Distinguished

Professorship during 2004–2005 at the University of Western Ontario.

†Supported by the Mathematical Sciences Research Institute, Berkeley. ‡Research supported in part by NSA grant MDA904-02-1-0061.

(1) H^{n}(E) is a free 𝔽_{p}[G]-module

(2) H^{n−1}(F) = ann_{n−1}(a)

(3) res: H^{n}(F) → H^{n}(E) is injective

(4) cor: H^{n−1}(E) → H^{n−1}(F) is surjective.

Suppose p = 2. Then the following are equivalent:

(1) H^{n}(E) is a free 𝔽_{2}[G]-module

(2) ann_{n−1}(a) = ann_{n−1}(a, −1) and H^{n}(F) = cor H^{n}(E) + (a) ∪ H^{n−1}(F)

(3) ann_{n−1}(a) = ann_{n−1}(a, −1) and H^{n}(F) = ann_{n}(a) + (a) ∪ H^{n−1}(F)

(4) H^{n}(F) = ann_{n}(a) ⊕ (a) ∪ H^{n−1}(F).

In the following theorem we examine to what extent free cohomology is hereditary.

Theorem 2. Suppose that either

• p > 2 or

• p = 2 and a ∈ (F^{×2} + F^{×2}) \ F^{2}.

Then free cohomology is hereditary: if n ∈ ℕ, then for all m ≥ n,

H^{n}(E) is a free 𝔽_{p}[G]-module ⟹ H^{m}(E) is a free 𝔽_{p}[G]-module.

Moreover, if H^{m}(E), m ∈ ℕ, is a free 𝔽_{p}[G]-module, then the sequence

0 → H^{m}(F) ⟶^{res} H^{m}(E) ⟶^{cor} H^{m}(F) → 0.

is exact in the first and third terms.

We consider Theorems 1 and 2 in section 3. We moreover show that when p > 2, H^{1}(E) is never free. When p = 2 we show that free cohomology is not generally hereditary and establish a condition for hereditary freeness that is more general than the one given above.

We next give precise conditions for trivial 𝔽_{p}[G]-module cohomology.

Theorem 3. Let n ∈ ℕ.

Suppose p > 2 and F is perfect. Then the following are equivalent:

(1) H^{n}(E) is a trivial 𝔽_{p}[G]-module (2) (ξ_{p}) ∪ H^{n−1}(F) ⊂ (a) ∪ H^{n−1}(F) and

ann_{n}(a) = (a) ∪ H^{n−1}(F)

(3) (ξ_{p}) ∪ H^{n−1}(F) ⊂ (a) ∪ H^{n−1}(F) and H^{n}(E) = res H^{n}(F) + (\sqrt[p]{a}) ∪ res H^{n−1}(F).

Suppose p = 2. Then the following are equivalent:

(1) H^{n}(E) is a trivial 𝔽_{2}[G]-module

(2) ann_{n}(a) ⊂ (a) ∪ ann_{n−1}(a, −1).

In the p = 2 case, suppose additionally that a ∈ (F^{×2} + F^{×2}) \ F^{2}. Then the conditions above are also equivalent to

(3) H^{n}(E) = res H^{n}(F) + (δ) ∪ res H^{n−1}(F) where (δ) ∈ H^{1}(E)^{G} satisfies N_{E/F} (δ) = (a).

For p > 2 and n = 1 the second condition in (3) was observed in [War, Lemma 3].

We deduce that trivial 𝔽_{p}[G]-module cohomology is a hereditary property.

Theorem 4. Trivial 𝔽_{p}[G]-module cohomology is hereditary: if n ∈ ℕ, then for all m ≥ n,

H^{n}(E)^{G} = H^{n}(E) ⟹ H^{m}(E)^{G} = H^{m}(E).

Moreover, if H^{m}(E)^{G} = H^{m}(E), m ∈ ℕ, then the following sequence is exact:

0 → ann_{m−1}(a) → H^{m−1}(F) ⟶^{(a)∪−} H^{m}(F) ⟶^{res}

H^{m}(E) ⟶^{cor} (a) ∪ ann_{m−1} ((a) ∪ (ξ_{p})) → 0,

where the map ann_{m−1}(a) → H^{m−1}(F) is the natural inclusion.

We consider Theorems 3 and 4 in section 5.

In section 4 we introduce cf(E/F), the largest degree n ∈ ℕ for which H^{n}(E) is not free or ∞ if H^{n}(E) is never free, and we give examples, for each m ≥ n ≥ 1, of extensions E/F with cf(E/F) = n and G_{E} a pro-p-group of cohomological dimension m.

In section 6 we introduce ct(E/F), the largest degree n ∈ ℕ for which H^{n}(E) is not a trivial 𝔽_{p}[G]-module or ∞ if H^{n}(E) is never trivial, and we give examples, for each m ≥ n ≥ 1, of extensions E/F with ct(E/F) = n and G_{E} a pro-p-group of cohomological dimension m.

Our proof relies on two recent results of Voevodsky in his proof of the Bloch-Kato Conjecture. (Before Voevodsky’s proof these results were standard conjectures in Galois cohomology and they were proved in important special cases.) In section 1 we recall these results and present two corollaries deducing collections of equivalent statements in Milnor K-theory. In section 2 we introduce various lemmas that give sufficient conditions for our 𝔽_{p}[G]-modules to be free or trivial, demonstrate that some properties in Milnor K-theory are hereditary, and establish some basic facts about certain p-henselian fields we will use to construct our examples in sections 4 and 6. For the convenience of the reader we have made our paper quite independent of [LMS].

1. Bloch-Kato and Milnor K-theory

The main ingredient for our determination of the G-module structure of H^{n}(E) is Milnor K-theory. (See [M] and [FV, Chap. IX].) For i ≥ 0, let K_{i}F denote the ith Milnor K-group of the field F, with standard generators denoted by {f_{1}, …, f_{i}}, f_{1}, …, f_{i} ∈ F \ {0}. For α ∈ K_{i}F, we denote by ᾱ the class of α modulo p, and we use the usual abbreviation k_{n}F for K_{n}F/pK_{n}F. The image of an element α ∈ K_{i}F in H^{i}(F) we also denote by α. Because we will often use the elements \overline{\{a\}}, \overline{\{ξ_{p}\}}, \overline{\{a, a\}}, and \overline{\{a, ξ_{p}\}}, we omit the bars for these elements. We also omit the bar in the element \overline{\{\sqrt[p]{a}\}}.

We write N_{E/F} for the norm map K_{n}E → K_{n}F, and we use the same notation for the induced map modulo p. We write cor = cor_{E/F} for the corresponding map of cohomology H^{n}(E) → H^{n}(F). We denote by i_{E} the natural homomorphism from K_{n}F to K_{n}E, and we use the same notation for the induced map modulo p. We denote by res = res_{E/F} the corresponding map of cohomology H^{n}(F) → H^{n}(E). We use a well-known projection formula in Milnor K-theory several times. (See [FW, page 81].)

Our proof relies on the following two results in Voevodsky’s proof of the Bloch-Kato Conjecture. The first is the Bloch-Kato Conjecture itself:

Theorem 5 ([V1, Definition 5.1] and [V2, Theorem 7.1]).

(1) Let F be a field of characteristic not p and m ∈ ℕ. Then the norm residue homomorphism

k_{m}F → H^{m}(G_{F} , μ_{p}^{⊗m})

is an isomorphism.

(2) For any cyclic extension E/F of degree p, the sequence

K_{m}E ⟶^{1−σ} K_{m}E ⟶^{N_{E/F}} K_{m}F

is exact.

The second result establishes an exact sequence connecting k_{m}F and k_{m}E for consecutive m. (We translate the statement of the original result to K-theory using the previous theorem.) In the following result a is chosen to satisfy E = F (\sqrt[p]{a}).

Theorem 6 ([V1, Proposition 5.2]). Let F be a field of characteristic not p with no extensions of degree prime to p. Then for any cyclic extension E/F of degree p and m ≥ 1, the sequence

k_{m−1}E ⟶^{N_{E/F}} k_{m−1}F ⟶^{{a}·−} k_{m}F ⟶^{i_{E}} k_{m}E

is exact.

Now if F is a perfect field containing a primitive pth root of unity, we observe that we may remove the hypothesis that the field F has no extensions of degree prime to p.

Theorem 7 (Modification of Theorem 6: [LMS, Theorem 5]). Let F be a field containing a primitive pth root of unity, and if p > 2 assume that F is perfect. Then for any cyclic extension E/F of degree p and m ≥ 1 the sequence

k_{m−1}E ⟶^{N_{E/F}} k_{m−1}F ⟶^{{a}·−} k_{m}F ⟶^{i_{E}} k_{m}E

is exact.

We have the following corollaries of Theorem 7. For an element ᾱ of k_{i}F, let

ann_{n−1} ᾱ = ann_{k_{n−1}F} ᾱ = ann (k_{n−1}F ⟶^{ᾱ·−} k_{n−1+i}F )

denote the annihilator of the product with ᾱ.

Corollary 1. Assume the same hypotheses. The following are equivalent for n ∈ ℕ:

(1) k_{n−1}F = ann_{n−1}{a}

(2) k_{n−1}F = ann_{n−1}{a} = ann_{n−1}{a, ξ_{p}}

(3) i_{E}: k_{n}F → k_{n}E is injective

(4) N_{E/F} : k_{n−1}E → k_{n−1}F is surjective.

Proof. The equivalence of the items (1), (3), and (4) follows directly from the exact sequence. Assuming (1) we see that

k_{n−1}F = ann_{n−1}{a} ⊂ ann_{n−1}{a, ξ_{p}} ⊂ k_{n−1}F,

whence (2) follows, and (2) implies (1) trivially. □

In Lemma 4 we show that all of the properties in Corollary 1 are hereditary.

Corollary 2. Assume the same hypotheses. The following are equivalent for n ∈ ℕ:

(1) ann_{n−1}{a} = ann_{n−1}{a, −1} and k_{n}F = N_{E/F} k_{n}E + {a} · k_{n−1}F

(2) ann_{n−1}{a} = ann_{n−1}{a, −1} and k_{n}F = ann_{n}{a} + {a} · k_{n−1}F

(3) k_{n}F = ann_{n}{a} ⊕ {a} · k_{n−1}F.

Proof. (1) ⟹ (2). This implication follows directly from N_{E/F} k_{n}E = ann_{n}{a}.

(2) ⟹ (3). Let ᾱ ∈ ({a} · k_{n−1}F) ∩ ann_{n}{a}. Then ᾱ = {a} · f̄ for some f ∈ K_{n−1}F. Since {a} · ᾱ = 0, {a, a} · f̄ = 0. Because {a, a} = {a, −1}, we have {a, −1} · f̄ = 0, and by the first hypothesis, {a} · f̄ = 0. Then ᾱ = 0 and the sum is direct.

(3) ⟹ (1). The second claim follows from the fact that ann_{n}{a} = N_{E/F} k_{n}E. For the first, suppose {a, −1} · f̄ = 0 for f ∈ K_{n−1}F. Because {a, −1} = {a, a}, we have

{a} · f̄ ∈ (ann_{n}{a}) ∩ ({a} · k_{n−1}F) = {0}.

Hence f̄ ∈ ann_{n−1}{a} and ann_{n−1}{a} = ann_{n−1}{a, −1} as required. □

2. Notation and Lemmas

For a field F, we let F^{×} denote its multiplicative group F \ {0}. We assume throughout the entire paper that F is perfect if p = [E : F] > 2 and F is arbitrary of characteristic not 2 if p = [E : F] = 2.

For the remainder of the paper n ∈ ℕ denotes an arbitrary natural number, E/F a cyclic extension of fields of degree p with a primitive pth root of unity ξ_{p} ∈ F, and a ∈ F^{×} an element such that E = F (\sqrt[p]{a}). Let G = Gal(E/F), and choose σ ∈ G to satisfy \sqrt[p]{a}^{σ−1} = ξ_{p}. For f, g ∈ F^{×}, we write (f) for the class of f in H^{1}(F) ≅ F^{×}/F^{×p} and (f, g) for (f) ∪ (g) ∈ H^{2}(F).

2.1. Module Structure.

For γ ∈ K_{n}E, let l(γ) denote the dimension of the cyclic 𝔽_{p}[G]- submodule ⟨γ̄⟩ of k_{n}E generated by γ̄. Then we have, for l(γ) ≥ 1,

(σ − 1)^{l(γ)−1}⟨γ̄⟩ = ⟨γ̄⟩^{G} ≠ 0 and (σ − 1)^{l(γ)}⟨γ̄⟩ = 0.

We denote by N the map (σ − 1)^{p−1} on k_{n}E. Because (σ − 1)^{p−1} = 1 + σ + ⋯ + σ^{p−1} in 𝔽_{p}[G], we may use i_{E}N_{E/F} and N interchangeably on k_{n}E.

Our first lemma establishes that in certain situations, all elements in (k_{n}E)^{G} are norm classes.

Lemma 1. Let n ∈ ℕ. Suppose that either

• p > 2 and N_{E/F} : k_{n−1}E → k_{n−1}F is surjective, or
• p = 2, ann_{n−1}{a} = ann_{n−1}{a, −1}, and
k_{n}F = N_{E/F} k_{n}E + {a} · k_{n−1}F.

Then we have

(1) For each γ ∈ K_{n}E, there exists α ∈ K_{n}E such that

⟨Nᾱ⟩ = ⟨γ̄⟩^{G}.

(2) (k_{n}E)^{G} = i_{E}N_{E/F} k_{n}E = (σ − 1)^{p−1}k_{n}E = i_{E}k_{n}F.

Proof. (1). Assume first that p > 2. By hypothesis, N_{E/F} : k_{n−1}E → k_{n−1}F is surjective, and then using the projection formula ([FW, p. 81]) we see that N_{E/F} : k_{n}E → k_{n}F is also surjective. Hence if γ̄ ∈ i_{E}k_{n}F then there exists ᾱ ∈ k_{n}E such that Nᾱ = γ̄ and we are done. Otherwise, let l = l(γ) and suppose γ̄ ∉ i_{E}k_{n}F and 1 ≤ l ≤ i ≤ p.

If l ≥ 2 we show by induction on i that there exists α_{i} ∈ K_{n}E such that ⟨(σ − 1)^{i−1}ᾱ_{i}⟩ = ⟨γ̄⟩^{G}. Then setting α := α_{p}, the proof will be complete in the case when 2 ≤ l. The case l = 1 we then handle at the end of the proof, using the case 2 ≤ l.

Assume then that l ≥ 2. If i = l then α_{i} = γ suffices. Assume now that 1 ≤ l ≤ i < p and that our statement is true for i. Set c = N_{E/F} α_{i}. Since i_{E}c̄ = Nᾱ_{i} = (σ − 1)^{p−1}ᾱ_{i} and i < p, i_{E}c̄ = 0.

By Corollary 1, we have c̄ = 0, that is, c = pf for some f ∈ K_{n}F. Hence

N_{E/F} (α_{i} − i_{E}(f)) = 0.

By Theorem 5, there exists ω ∈ K_{n}E such that

(σ − 1)ω = α_{i} − i_{E}(f).

If i > 1, l(α_{i}) > 1 and so l(α_{i} − i_{E}(f)) > 1. Hence (σ − 1)^{2}ω̄ = (σ − 1)ᾱ_{i} ≠ 0. Therefore ⟨(σ − 1)^{i}ω̄⟩ = ⟨γ̄⟩^{G} and we can set α_{i+1} = ω.

Therefore we have proved that if l(γ) ≥ 2 then there exists α ∈ K_{n}E such that Nᾱ = (σ − 1)^{l(γ)−1}γ̄.

Now assume that l(γ) = 1 but γ̄ ∉ i_{E}k_{n}F. Then γ̄ = ᾱ_{1} and (σ − 1)ω̄ = ᾱ_{1} − i_{E}(f̄) ≠ 0. Thus l(ω) = 2 and our argument above shows that there exists β ∈ K_{n}E such that Nβ̄ = (σ−1)ω̄ = ᾱ_{1}−i_{E}(f̄). As we observed at the beginning of our proof there exists an element δ ∈ K_{n}E such that Nδ̄ = i_{E}(f̄). Therefore we have:

N(β̄ + δ̄) = ᾱ_{1} = γ̄.

Thus we have established in all cases that for each γ ∈ K_{n}E there exists α ∈ K_{n}E such that ⟨Nᾱ⟩ = ⟨γ̄⟩^{G}.

Now consider the case p = 2. In this case from our hypothesis k_{n}F = N_{E/F} k_{n}E +{a}·k_{n−1}F we again have i_{E}N_{E/F} k_{n}E = i_{E}k_{n}F. Therefore if γ̄ ∈ i_{E}k_{n}F our statement follows. Assume that γ̄ ∈ k_{n}E \ i_{E}k_{n}F. Then l(γ) ≤ 2, and if l(γ) = 2 we may set α = γ and (1) follows again. Next we shall assume that l(γ) = 1 and therefore γ̄ ∈ (k_{n}E)^{G}. Set c = N_{E/F} γ. Then i_{E}c̄ = 0.

From Theorem 7, we conclude that c = {a} · g + 2f for g ∈ K_{n−1}F and f ∈ K_{n}F. Hence from the projection formula,

N_{E/F} (γ − \{\sqrt{a}\} · i_{E}(g) − i_{E}(f)) = (\{a\} · g + 2f) − \{−a\} · g − 2f
= {−1} · g.

Using Theorem 7 again, we obtain that {a, −1}·ḡ = 0. Our hypothesis ann_{n−1}{a} = ann_{n−1}{a, −1} gives us that {a}·ḡ = 0. Hence {a}·g = 2h for some h ∈ K_{n}F and N_{E/F} γ = 2(h + f). Thus

N_{E/F} (γ − i_{E}(h + f)) = 0.

Then by Theorem 5 there exists α ∈ K_{n}E such that

(σ − 1)α = γ − i_{E}(h + f).

Observe that since γ̄ ∉ i_{E}k_{n}F we have γ̄ − i_{E}(\overline{h + f}) ≠ 0. Hence

γ̄ = (σ − 1)ᾱ + i_{E}(\overline{h + f}) ∈ Nk_{n}E,

as required.

(2). Suppose γ̄ ∈ (k_{n}E)^{G}. Then l(γ) = 1 and the preceding part of our proof shows that ⟨γ̄⟩ = ⟨Nᾱ⟩ for α ∈ K_{n}E. Hence (k_{n}E)^{G} ⊂ i_{E}N_{E/F} k_{n}E. Then

i_{E}N_{E/F} k_{n}E ⊂ i_{E}k_{n}F ⊂ (k_{n}E)^{G} ⊂ i_{E}N_{E/F} k_{n}E,

and so all inclusions are equalities. □

Our second lemma establishes a situation in which all elements in k_{n}E are fixed by G.

Lemma 2. Let n ∈ ℕ. Suppose that

i_{E}({ξ_{p}} · k_{n−1}F) = i_{E}N_{E/F} k_{n}E = {0}.

Then (k_{n}E)^{G} = k_{n}E.

Remark. The hypothesis i_{E}({ξ_{p}} · k_{n−1}F) = {0} can be omitted in the case p = 2.

Proof. Let γ ∈ K_{n}E. We show that l(γ) > 1 leads to a contradiction, whence we will have the result.

Suppose that l = l(γ) ≥ 2 and 1 ≤ l ≤ i ≤ p. We show by induction on i that there exists α_{i} ∈ K_{n}E such that ⟨(σ−1)^{i−1}ᾱ_{i}⟩ = ⟨(σ−1)^{l−1}γ̄⟩. If i = l then α_{i} = γ suffices. Assume now that l ≤ i < p and that our statement is true for i. Set c = N_{E/F} α_{i}. Since i_{E}c̄ = Nᾱ_{i} = (σ−1)^{p−1}ᾱ_{i} and i < p, i_{E}c̄ = 0.

By Theorem 7, c̄ = {a} · b̄ for some b ∈ K_{n}F. Hence c = {a} · b + pf for f ∈ K_{n}F. Then since 2 ≤ i < p in this case,

N_{E/F} (α_{i} − \{\sqrt[p]{a}\} · i_{E}(b) − i_{E}(f)) = 0.

By Theorem 5, there exists ω ∈ K_{n}E such that

(σ − 1)ω = α_{i} − \{\sqrt[p]{a}\} · i_{E}(b) − i_{E}(f).

Then (σ − 1)^{2}ω = (σ − 1)α_{i} − i_{E}({ξ_{p}} · i_{E}(b)) = (σ − 1)α_{i} ≠ 0, and we can set α_{i+1} = ω. Observe that here we use our hypothesis

i_{E}({ξ_{p}} · k_{n−1}F) = {0}.

Hence by induction there exists α_{p} ∈ K_{n}E such that

⟨Nᾱ_{p}⟩ = ⟨(σ − 1)^{l−1}γ̄⟩.

But i_{E}N_{E/F} ᾱ_{p} = 0, whence (σ − 1)^{l−1}γ̄ = 0, a contradiction. □

Finally, we record a necessary and sufficient condition for an 𝔽_{p}[G]- module to be free.

Lemma 3. Let M be an 𝔽_{p}[G]-module. Then the following are equivalent:

(1) M is a free 𝔽_{p}[G]-module

(2) M^{G} = (σ − 1)^{p−1}M.

Proof. Condition (2) is equivalent to

H^{2}(G, M) = {0}.

But this condition is known to be equivalent with (1) (for any p-group G!). (See for example [Lg2, p. 63].) □

2.2. Hereditary Properties.

We say that a property of Milnor k-groups k_{n}E and k_{n}F is hereditary if the validity of the property for a given n implies the validity of the property for all integers greater than n.

The next lemma establishes various hereditary properties, including the properties in Corollary 1.

Lemma 4. Let n ∈ ℕ.

The following are hereditary properties:

(1) k_{n−1}F = ann_{n−1}{a} = ann_{n−1}{a, ξ_{p}}.

(2) i_{E}: k_{n}F → k_{n}E is injective

(3) N_{E/F} : k_{n−1}E → k_{n−1}F is surjective

(4) for some fixed α_{1}, α_{2} ∈ K_{1}F, ᾱ_{1} · k_{n−1}F ⊂ ᾱ_{2} · k_{n−1}F

(5) for some fixed α ∈ K_{1}E, k_{n}E = i_{E}k_{n}F + ᾱ · i_{E}k_{n−1}F

Proof. (1). k_{n}F = k_{n−1}F · k_{1}F, and since ann_{n−1}{a} = k_{n−1}F, we have ann_{n}{a} = k_{n}F as well. The other equality follows from ann_{n}{a} ⊂ ann_{n}{a, ξ_{p}}. The result follows by induction.

(2-3). By Corollary 1, the first three properties are equivalent, hence (2) and (3) are hereditary.

(4). K_{n}F = K_{n−1}F · K_{1}F, so K_{n}F is generated by elements of the form

{f_{1}, f_{2}, …, f_{n}} = {f_{1}, …, f_{n−1}} · {f_{n}}, f_{i} ∈ F^{×}.

For each such generator, we calculate

ᾱ_{1} · \overline{\{f_{1}, …, f_{n}\}} = ᾱ_{2} · ḡ · \overline{\{f_{n}\}}

for some g ∈ K_{n−1}F, whence ᾱ_{1} · k_{n}F ⊂ ᾱ_{2} · k_{n}F. The result follows by induction.

(5). k_{n+1}E = k_{1}E ·k_{n}E, so the condition on k_{n}E gives us that k_{n+1}E is generated by elements of the form

γ̄_{1} = \overline{\{δ\}} · i_{E}(\overline{\{f_{1}, …, f_{n}\}}), δ ∈ E^{×}, f_{i} ∈ F^{×}

and

γ̄_{2} = \overline{\{δ\}} · ᾱ · i_{E}(\overline{\{f_{1}, …, f_{n−1}\}}), δ ∈ E^{×}, f_{i} ∈ F^{×}.

If n − 1 ≥ 1 then we see that k_{n+1}E is generated by the elements in k_{n}E · i_{E}k_{1}F. By hypothesis k_{n}E = i_{E}k_{n}F + ᾱ · i_{E}k_{n−1}F and therefore k_{n+1}E is generated by elements in i_{E}k_{n+1}F + ᾱ · i_{E}k_{n}F.

If n = 1 then using our hypothesis k_{1}E = i_{E}k_{1}F + ᾱ · i_{E}k_{0}F we may write the generators γ̄_{2} of k_{2}E as

γ̄_{2} = (i_{E}(\overline{\{f\}}) + cᾱ) · ᾱ, f ∈ F^{×}, c ∈ ℤ.

Since ᾱ · ᾱ = {−1} · ᾱ,

γ̄_{2} = i_{E}(\overline{\{f\}}) · ᾱ + c i_{E}(\{−1\}) · ᾱ = −ᾱ · i_{E}(\overline{\{f\}}) − ᾱ · i_{E}(c\{−1\}).

Thus in this case both types of generators of k_{2}E have the required form of elements in i_{E}k_{2}F + ᾱ · i_{E}k_{1}F.

The result now follows by induction. □

2.3. Fields of the Form ℂ((⊕_{I}ℤ_{(p)})).

For our examples in sections 4 and 6 we introduce the following notation and results.

Let

ℤ_{(p)} := \{\frac{c}{d} ∈ ℚ | c, d ∈ ℤ, d ≠ 0; if c ≠ 0 then (c, d) = 1, p ∤ d\} .

Observe that ℤ_{(p)} carries a natural ordering induced from ℚ. Let I be a well-ordered set of cardinality m, and let Γ be a direct sum of m copies of ℤ_{(p)}, indexed by I. Then m = dim_{𝔽_{p}} Γ/pΓ. Order Γ lexicographically.

Then Γ is a linearly ordered abelian group. (Recall that each nonempty set can be well-ordered (see [Lg1, Appendix 2, Theorem 4.1]).) Now it is well-known that since Γ is a totally ordered abelian group, the field

F_{m} := ℂ((Γ)) := {f: Γ → ℂ | supp(f) is well-ordered}

is a henselian valued field with value group Γ and residue field ℂ. (See [R1, Chapitre D, Théorèmes 2 et 3, page 103, et Chapitre F, Théorème 4, page 198].) Thus a typical element f ∈ F_{m} may be written as a formal sum

f = ∑_{g∈Γ} a_{g}t^{g}

such that the set supp(f) := {g ∈ Γ | a_{g} ≠ 0} is a well-ordered subset of Γ. The absolute Galois group of F_{m} is known to be ℤ_{p}^{m}, the topological product of m copies of ℤ_{p} [K, pages 3 and 4]. We record one property of F_{m} in the following lemma.

Lemma 5. For m, n ∈ ℕ ∪ {ℵ_{0}},

H^{n}(F_{m}) ≅ ⋀^{n} H^{1}(ℤ_{p}^{m}) ≅ ⋀^{n} ⊕_{m}𝔽_{p}

where the cup-product is sent to the wedge product.

Proof. Since F_{m} is a henselian valued field, the second result follows from [Wad, Theorem 3.6], observing that under the Kummer isomorphism F_{m}^{×}/F_{m}^{×p} ≅ H^{1}(F_{m}), 0 = (−1) ∈ H^{1}(F_{m}), and H^{j}(ℂ) = {0} for all j ∈ ℕ. □

Of particular interest to us will be certain fields with absolute Galois groups which are pro-p free products of groups of the form ℤ_{p}^{m}.

Lemma 6. Suppose that m_{1}, m_{2} are non-zero cardinal numbers, and let F_{m_{1}} and F_{m_{2}} be as above. There exists a field F_{m_{1},m_{2}} of characteristic 0, containing a primitive p^{2}th root of unity ξ_{p^{2}}, such that the absolute Galois group

G_{F_{m_{1},m_{2}}}  ≅ G_{F_{m_{1}}} ⋆_{pro-p} G_{F_{m_{2}}}  ≅ ℤ_{p}^{m_{1}} ⋆_{pro-p} ℤ_{p}^{m_{2}},

where the free products are taken in the category of pro-p-groups, and the natural restriction maps

res_{⋆}: H^{n}(F_{m_{1},m_{2}}) → H^{n}(F_{m_{1}}) ⊕ H^{n}(F_{m_{2}})

are isomorphisms.

Note that we use the notation res_{⋆} to distinguish this restriction map from restriction maps H^{n}(F) → H^{n}(E). For h ∈ H^{n}(F_{m_{1},m_{2}}), we will write res_{⋆} h = h_{1} ⊕ h_{2}.

Proof. The existence of a field F_{m_{1},m_{2}} with char(F_{m_{1},m_{2}}) = char(F_{m_{1}}) = char(F_{m_{2}}) = 0 and the given absolute Galois group follows from [EH, Proposition 1.3].

Additionally using the construction of F_{m_{1},m_{2}} following [EH, proof of Proposition 1.3] we assume that F_{m_{1},m_{2}} is the intersection of two henselian valued fields (L_{i}, V_{i}), i = 1, 2, with residue fields isomorphic to F_{m_{1}} and F_{m_{2}} respectively. Here V_{i} is a henselian valuation on L_{i}. Then by Hensel’s Lemma (see [R2, pages 12 and 13, condition (3)]) and by the fact that F_{m_{1}} and F_{m_{2}} have characteristic 0 and both contain a primitive p^{2}th root of unity, we see that F_{m_{1},m_{2}} also contains a primitive p^{2}th root of unity. The fact that the restriction maps are isomorphisms follows from [N, Sätze (4.1) und (4.2)]. □

Remark. From the proof above it follows that F_{m_{1},m_{2}} contains all p^{k}th primitive roots, k ∈ ℕ. However we shall not need this observation.

3. When is Galois Cohomology Free?

Proof of Theorem 1. Here and elsewhere we use Theorem 5 to translate between Galois cohomology H^{n} and K-theory k_{n}.

First we show that for all p, k_{n}E free implies that

i_{E}N_{E/F} k_{n}E = i_{E}k_{n}F = (k_{n}E)^{G}.

If k_{n}E is free, then by Lemma 3, (σ − 1)^{p−1}k_{n}E = (k_{n}E)^{G}. Observing that (σ − 1)^{p−1} and i_{E}N_{E/F} are equivalent,

i_{E}N_{E/F} k_{n}E = (σ − 1)^{p−1}k_{n}E = (k_{n}E)^{G}.

Then since i_{E}k_{n}F ⊂ (k_{n}E)^{G} and i_{E}N_{E/F} k_{n}E ⊂ i_{E}k_{n}F, we have established our claim.

Assume first that p > 2. First we show (1) ⟹ (2). Let f ∈ K_{n−1}F be arbitrary, and set α = \{\sqrt[p]{a}\}·f. Now because (k_{n}E)^{G} = i_{E}N_{E/F} k_{n}E, there exists β ∈ K_{n}E such that i_{E}N_{E/F} β̄ = i_{E}({ξ_{p}} · f̄) ∈ (k_{n}E)^{G}. Set γ = (σ − 1)^{p−2}β. Since p > 2, γ is in the image of σ − 1 and hence has trivial norm. We calculate

N_{E/F} (ᾱ − γ̄) = {a} · f̄.

On the other hand, observing that i_{E}N_{E/F} = (σ − 1)^{p−1} on k_{n}E,

(σ − 1)(ᾱ − γ̄) = i_{E}({ξ_{p}} · f̄) − i_{E}({ξ_{p}} · f̄) = 0.

Hence ᾱ − γ̄ ∈ (k_{n}E)^{G} = i_{E}k_{n}F. But on i_{E}k_{n}F the norm map N_{E/F} is trivial. Hence {a} · f̄ = 0 and ann_{n−1}{a} = k_{n−1}F.

By Corollary 1, (2), (3), and (4) are all equivalent. Now we show (4) ⟹ (1). Assume that N_{E/F} : k_{n−1}E → k_{n−1}F is surjective. By Lemma 1 we have (k_{n}E)^{G} = (σ − 1)^{p−1}k_{n}E. Hence by Lemma 3, k_{n}E is free.

Now suppose that p = 2. By Corollary 2, we need only show that (1) and (2) are equivalent. We show first that (1) ⟹ (2). We established that (1) implies i_{E}N_{E/F} k_{n}E = i_{E}k_{n}F. Since ker i_{E} = {a} · k_{n−1}F, this equality is equivalent to k_{n}F = N_{E/F} k_{n}E + {a} · k_{n−1}F, so we have the second part of (2). Clearly ann_{n−1}{a} ⊂ ann_{n−1}{a, −1}, so we show that ann_{n−1}{a, −1} ⊂ ann_{n−1}{a}.

We adapt the argument above. Let f̄ ∈ ann_{n−1}{a, −1}. Set ᾱ = \{\sqrt{a}\} · f̄. Since \{a\} · \{−1\} · f̄ = 0, Theorem 7 tells us that there exists β ∈ K_{n}E such that N_{E/F} β̄ = {−1} · f̄. Now we calculate by the projection formula

N_{E/F} (ᾱ − β̄) = {−a} · f̄− {−1} · f̄ = {a} · f̄.

On the other hand, using the fact that σ − 1 = σ + 1 when p = 2,

(σ − 1)(ᾱ − β̄) = {−1} · f̄− {−1} · f̄ = 0.

Hence ᾱ − β̄ ∈ (k_{n}E)^{G}. By Lemma 3, (k_{n}E)^{G} = (σ − 1)k_{n}E = i_{E}N_{E/F} k_{n}E ⊂ i_{E}k_{n}F. Therefore ᾱ − β̄ ∈ i_{E}k_{n}F. But on i_{E}k_{n}F, the norm map N_{E/F} is trivial. Hence {a} · f̄ = 0, and f̄ ∈ ann_{n−1}{a}, so ann_{n−1}{a, −1} ⊂ ann_{n−1}{a}, as required.

Now we show that (2) ⟹ (1). Assume that ann_{n−1}{a, −1} = ann_{n−1}{a} and that k_{n}F = N_{E/F} k_{n}E + {a} · k_{n−1}F. We use Lemmas 1 and 3 to deduce that k_{n}E is free. □

It follows easily that Corollary 3. For p > 2, k_{1}E is never free.

Proof. Since G acts trivially on k_{0}E ≅ 𝔽_{p},

N_{E/F} k_{0}E = 0 ≠ k_{0}F ≅ 𝔽_{p}.

Alternatively, i_{E}: k_{1}F → k_{1}E is not injective, since {a} ∈ k_{1}F is a nontrivial element of the kernel. □

With Theorem 1 in hand, Lemma 4 is enough to establish hereditary freeness in the p > 2 case, and for the p = 2 case we show that an additional condition, analogous to the p > 2 case, is sufficient:

Corollary 4. Suppose that p = 2 and for some n ∈ ℕ,

ann_{n−1}{a} = k_{n−1}F.

Then k_{m}E is a free 𝔽_{2}[G]-module for all m ≥ n.

Proof. We show that the two conditions of part (2) of the p = 2 portion of Theorem 1 hold for K-theory degree at least n. From Lemma 4, part (1), we deduce that k_{m}F = ann_{m}{a} = ann_{m}{a, −1} for all m ≥ n − 1.

By Theorem 7 and Lemma 4, we have k_{m}F = N_{E/F} k_{m}E for all m ≥ n − 1 and therefore we see that

k_{m}F = N_{E/F} k_{m}E + {a} · k_{m−1}F for all m ≥ n.

We conclude that k_{m}E is a free 𝔽_{2}[G]-module for all m ≥ n. □

Just as before it follows easily that

Corollary 5. For p = 2 and \sqrt{−1} ∈ F, k_{1}E is never free.

Proof. Since −1 ∈ F^{×2}, we have {−1} = 0 ∈ k_{1}F and {a, −1} = 0 ∈ k_{2}F, so that ann_{0}{a, −1} = k_{0}F ≅ 𝔽_{2} ≠ ann_{0}{a} = {0}. □

We are now ready to prove Theorem 2.

Proof of Theorem 2. For p > 2, the fact that free cohomology is hereditary follows from Lemma 4 and condition (2) in Theorem 1. The exactness of the first term of the sequence follows from Theorem 1, part (3), while the exactness at the third term follows from Theorem 1, part (4) and Lemma 4, part (3).

Assume then that p = 2 and a = x^{2} + y^{2} for some x, y ∈ F^{×}. If −1 ∈ F^{×2} then {a, −1} ∈ 2K_{2}F and so {a, −1} = 0 ∈ k_{2}F. Otherwise let K = F (\sqrt{−1}), and observe that a = N_{K/F} (x + y\sqrt{−1}). Then {a, −1} = 0 ∈ k_{2}F. Hence ann_{n−1}{a, −1} = k_{n−1}F.

Now observe that since k_{n}E is a free 𝔽_{2}[G]-module, by Theorem 1 we have ann_{n−1}{a} = ann_{n−1}{a, −1}, and so ann_{n−1}{a} = ann_{n−1}{a, −1} = k_{n−1}F. We deduce from Corollary 4 that k_{m}E is a free 𝔽_{2}[G]-module for all m ≥ n.

For the exact sequence in the case p = 2, we have shown that k_{m−1}F = ann_{m−1}{a}, and so by Theorem 7 and Lemma 4, we have k_{n}F = N_{E/F} k_{n}E for all n ≥ m − 1. Hence we have exactness at the third term. Furthermore, we conclude from Corollary 1 that i_{E} is injective from k_{m}F to k_{m}E. Hence we have exactness at the first term as well. □

We now provide an example of k_{1}E free but k_{2}E nonfree, showing that freeness is not generally hereditary when p = 2.

Example. Let p = 2, F = ℚ_{2}, and a = −1, so E = ℚ_{2}(\sqrt{−1}). Then

k_{1}F ≅ F^{×}/F^{×2} = ⟨[−1], [2], [5]⟩.

and

N_{E/F} k_{1}E ≅ N_{E/F} (E^{×})F^{×2}/F^{×2} = ⟨[2], [5]⟩.

(See [Lam, page 162, Corollary 2.24].)

Therefore k_{1}F = N_{E/F} k_{1}E + {−1} · k_{0}F. Moreover, since [−1] ∉ N_{E/F} (E^{×}) F^{×2}/F^{×2}, we have {−1, −1} ≠ 0 ∈ k_{2}F. (Again see [Lam, page 162, Corollary 2.24].)

Hence ann_{0}{−1, −1} = {0}. Since {−1} ≠ 0 in k_{1}F we see that ann_{0}{−1} = {0}. Hence the conditions of part (2) of the p = 2 portion of Theorem 1 are satisfied, whence k_{1}E is a free 𝔽_{2}-module.

Observe, however, that since ann_{0}{−1, −1} = {0} ≠ k_{0}F, the first hypothesis in Corollary 4 does not hold. Therefore we cannot conclude that k_{2}E is a free 𝔽_{2}[G]-module—and of course it is not, as it is well known that k_{2}E ≅ 𝔽_{2}. (See [Lam, page 158, Corollary 2.15] and [M, Theorem 4.1].)

4. Examples of H^{k}(E) Free for all n < k and H^{n}(E) Nonfree, with Given Cohomological Dimension

We have shown in Theorem 2 that if p > 2 then the property H^{n}(E) is a free 𝔽_{p}[G]-module is hereditary. Moreover, the same property is hereditary in the case p = 2 as well if i = \sqrt{−1} ∈ F, since then a = ((a + 1)/2)^{2} + ((a − 1)i/2)^{2}.

These results lead naturally to the definition of an interesting invariant cf(E/F) ∈ {0} ∪ ℕ ∪ {∞}:

cf(E/F) = sup {n ∈ ℕ ∪ {0} | H^{n}(E) is not a free 𝔽_{p}[G]-module} .

We have chosen cf to indicate that after degree cf(E/F), Galois cohomology is cohomologically free. Of course, if H^{n}(E) is never free then cf(E/F) = ∞, and otherwise cf(E/F) ∈ ℕ ∪ {0}.

Assume for the moment that either p > 2 or \sqrt{−1} ∈ F. If cf(E/F) = n, then by definition H^{m}(E) is a free 𝔽_{p}[G]-module for all m > n. On the other hand, by the hereditary property we also have that H^{k}(E) is not free for all k ≤ cf(E/F). Finally, Corollaries 3 and 5 tell us that H^{1}(E) is never free and hence cf(E/F) ≥ 1. A natural question arises: can we choose a suitable field extension E/F so that cf(E/F) is a given natural number or ∞? We show that the answer is affirmative.

Before formulating our result precisely, let us recall that for any prop-group T we may define cd(T), the cohomological dimension of T, as

cd(T) = sup {k ∈ ℕ | H^{k}(T, 𝔽_{p}) ≠ {0}} ∈ ℕ ∪ {∞}.

(See [RZ, Chapter 7].) Suppose that the absolute Galois group G_{E} is a pro-p-group. Then, adopting the convention that {0} is considered a free 𝔽_{p}[G]-module, we have:

cf(E/F) ≤ cd(G_{E}).

From Corollaries 3 and 5 above it follows that if p > 2 or p = 2 and \sqrt{−1} ∈ F then cf(E/F) ≥ 1.

Our result is then the following.

Given 1 ≤ n ≤ m ∈ ℕ ∪ {∞} and a prime p, there exists a cyclic extension E/F of degree p with ξ_{p} ∈ F such that (1) G_{E} is a pro-p-group;

(2) cf(E/F) = n; and

(3) cd(G_{E}) = m.

Observe that if we choose n ∈ ℕ and m > n, then we have obtained examples as promised in the title of this section.

4.1. The case m ∈ ℕ.

(1). Let F := F_{n,m} be a field of characteristic 0 with G_{F} ≅ ℤ_{p}^{n} ⋆_{pro-p} ℤ_{p}^{m} and ξ_{p^{2}} ∈ F, given by Lemma 6. Observe particularly that \sqrt{−1} ∈ F in the case p = 2. Since char(F) = 0, F is perfect. Let

E = F (\sqrt[p]{a})

for any a ∈ F^{×} such that under the restriction map on H^{1},

res_{⋆}(a) = (a)_{1} ⊕ (a)_{2}, (a)_{1} ≠ 0, (a)_{2} = 0.

We use here, and later without mention, the fact that res_{⋆} is an isomorphism, by Lemma 6. Observe that there exists an a with the required conditions because by Lemma 5, H^{1}(F_{n}) ≠ {0}.

(2a). H^{n}(E) is not free. We claim that

ann_{n−1}(a) ≠ H^{n−1}(F).

If n = 1 this inequality is true as (a) ≠ 0 ∈ H^{1}(F). Assume now that n > 1. We shall use Lemma 5 together with the fact that the restriction map in the cohomology ring of a profinite group to the cohomology ring of a closed subgroup is a ring homomorphism. (See for example [RZ, Proposition 7.9.4].)

Let a_{1} ∈ F_{n}^{×} satisfy (a_{1}) = (a)_{1}, and extend {(a_{1})} to a basis {(a_{1}), (a_{2}), ⋯, (a_{n})} of H^{1}(F_{n}). By Lemma 5, the element

(a_{1}) ∪ (a_{2}) ∪ ⋯ ∪ (a_{n}) ∈ H^{n}(F_{n})

is nontrivial, so that 0 ≠ (a_{2})∪⋯∪(a_{n}) ∈ H^{n−1}(F_{n}). Let b ∈ H^{n−1}(F) satisfy

b_{1} = (a_{2}) ∪ ⋯ ∪ (a_{n}) ∈ H^{n−1}(F_{n}), b_{2} = 0 ∈ H^{n−1}(F_{m}).

Then since the cup-product commutes with res_{⋆},

((a) ∪ b)_{1} = (a_{1}) ∪ b_{1} ≠ 0 ∈ H^{n}(F_{n}),

so that (a) ∪ b ≠ 0 ∈ H^{n}(F) and hence ann_{n−1}(a) ≠ H^{n−1}(F).

If p > 2, we conclude by Theorem 1 that H^{n}(F) is not free. If p = 2, observe that since \sqrt{−1} ∈ F, we have ann_{n−1}(a, −1) = ann_{n−1} 0 = H^{n−1}(F), so that ann_{n−1}(a, −1) ≠ ann_{n−1}(a). We deduce from Theorem 1 that H^{n}(F) is not free.

(2b). H^{k}(E) is free for all k ≥ n + 1. We claim that

ann_{n}(a) = H^{n}(F).

Let c ∈ H^{n}(F). Then since H^{n+1}(F_{n}) = 0 by Lemma 5,

res_{⋆}(a) ∪ c = ((a_{1}) ∪ c_{1}) ⊕ (0 ∪ c_{2}) = 0 ⊕ 0 = res_{⋆} 0

Hence (a) ∪ c = 0 and ann_{n}(a) = H^{n}(F).

If p > 2 then we conclude by Theorem 1 that H^{n+1}(E) is free, and by Theorem 2, H^{k}(E) is free for all k ≥ n + 1. If p = 2, observe that ann_{n}(a, −1) = ann_{n} 0 = H^{n}(F). Furthermore, we use Corollary 1 to obtain that cor: H^{n}(E) → H^{n}(F) is surjective. Then by Corollary 4, we have that H^{k}(E) is free for all k ≥ n + 1.

(3). cd(G_{E}) = m. First we claim that G_{E} does not contain an element of order p. By Artin-Schreier’s theorem (see for instance [J, Chapter VI, Theorem 17]), finite subgroups of absolute Galois groups are either trivial or of order 2, and since \sqrt{−1} ∈ E no element of order 2 exists in G_{E}.

Then, by Serre’s well-known theorem [S], we obtain

cd(G_{E}) = cd(G_{F}).

From Lemmas 5 and 6 we find that

cd(G_{F}) = max{cd(F_{n}), cd(F_{m})} = m.

Thus cd(G_{E}) = m as required.

4.2. The case n < m = ∞. Set

F_{∞} := ℂ ((ℤ_{(p)}^{m})) , where m = ℵ_{0}.

With the same argument as in the proof of Lemma 6, there exists a field F := F_{n,∞} such that G_{F} ≅ G_{F_{n}} ⋆_{pro-p} G_{F_{∞}} and ξ_{p^{2}} ∈ F. Then set

E = F (\sqrt[p]{a})

for any a ∈ F^{×} such that under the restriction map

res_{⋆}: H^{1}(F) → H^{1} (F_{n}) ⊕ H^{1} (F_{∞}) ,

we have

res_{⋆}(a) = (a)_{1} ⊕ 0, (a)_{1} ≠ 0.

Then cd(G_{F}) = cd(G_{E}) = ∞, and with the same argument as above we see that cf(E/F) = n.

4.3. The case n = ∞ = m. As above we let Γ be a direct sum of ℵ_{0} copies of ℤ_{(p)}. Then we set F := F_{∞} = ℂ((Γ)). Let a ∈ F^{×} such that v(a) ∈ Γ \ pΓ, where v is a natural valuation on F. Then from the description of Galois cohomology of p-henselian fields (see [Wad, Theorem 3.6]), we obtain

ann_{n}(a) = (a) ∪ H^{n−1}(F)

and

(a) ∪ H^{n−1}(F) ≠ H^{n}(F) for all n ∈ ℕ.

(Observe that when p = 2 we use the fact that \sqrt{−1} ∈ F in the cited result.) Setting E = F (\sqrt[p]{a}), just as before we have that

cf(E/F) = ∞,

as required.

5. When is Galois Cohomology Trivial?

First we need a lemma.

Lemma 7. Suppose that p = 2. Then

{a} · ann_{n−1}{a, −1} ⊂ N_{E/F} k_{n}E.

Proof. Let β̄ ∈ ann_{n−1}{a, −1}. Then {−1} · β̄ ∈ ann_{n}{a} = N_{E/F} k_{n}E by Theorem 7. Let γ ∈ K_{n}E such that {−1} · β̄ = N_{E/F} (γ̄). Then we have

\{a\} · β̄ = N_{E/F} (\{\sqrt{a}\} · i_{E}(β̄) + γ̄).

Thus {a} · ann_{n−1}{a, −1} ⊂ N_{E/F} k_{n}E as asserted. □

It is worth observing that if n = 1, Lemma 7 is equivalent to

{a, −1} = 0 if and only if {a} ∈ N_{E/F} k_{1}E,

and therefore Lemma 7 can be viewed as a generalization of this statement.

Now we are ready to prove Theorem 3.

Proof of Theorem 3. As before, we translate to K-theory using Theorem 5. We first consider the case p > 2.

(1) ⟹ (3). Assume that k_{n}E is a trivial 𝔽_{p}[G]-module. Suppose f ∈ K_{n−1}F, and set β = \{\sqrt[p]{a}\} · i_{E}(f). Then

(σ − 1)β̄ = 0 ⟹ {ξ_{p}} · i_{E}(f̄) = i_{E}({ξ_{p}} · f̄) = 0.

But then by Theorem 7, {ξ_{p}} · f̄ ∈ {a} · k_{n−1}F.

Now let γ ∈ K_{n}E be arbitrary. Again, (σ − 1)γ̄ = 0. Then

i_{E}N_{E/F} γ̄ = (σ − 1)^{p−1}γ̄ = 0

and so by Theorem 7, N_{E/F} γ̄ = {a} · f̄ for f ∈ K_{n−1}F. By the projection formula, N_{E/F} (\{\sqrt[p]{a}\} · i_{E}(f̄)) = \{a\} · f̄. Then

N_{E/F} (γ̄ − \{\sqrt[p]{a}\} · i_{E}(f̄)) = 0,

and hence

N_{E/F} (γ − \{\sqrt[p]{a}\} · i_{E}(f)) = pg,

for some g ∈ K_{n−1}F. Set

β = γ − \{\sqrt[p]{a}\} · i_{E}(f) − i_{E}(g).

Then N_{E/F} (β) = 0. By Theorem 5, there exists α ∈ K_{n}E such that (σ − 1)α = β. But since k_{n}E is fixed by G, β̄ = 0. Hence k_{n}E = i_{E}(k_{n}F) + \{\sqrt[p]{a}\} · i_{E}(k_{n−1}F).

(3) ⟹ (2). Since p > 2, N_{E/F} (\{\sqrt[p]{a}\}·i_{E}(f̄)) = \{a\}·f̄for f ∈ K_{n−1}F, and N_{E/F} (i_{E}(ḡ)) = 0 for g ∈ K_{n}F. Hence N_{E/F} k_{n}E = {a} · k_{n−1}F. Since by Theorem 7, ann_{n}{a} = N_{E/F} k_{n}E, we are done.

(2) ⟹ (1). Assume that {ξ_{p}}·k_{n−1}F ⊂ {a}·k_{n−1}F and {a}·k_{n−1}F = ann_{n}{a}. By Theorem 7, ann_{n}{a} = N_{E/F} k_{n}E. Hence {ξ_{p}} · k_{n−1}F ⊂ N_{E/F} k_{n}E = {a} · k_{n−1}F. But by Theorem 7, {a} · k_{n−1}F = ker i_{E}. We then apply Lemma 2 to deduce that k_{n}E = (k_{n}E)^{G}.

Now we consider the case p = 2.

(1) ⟹ (2). Assume that k_{n}E is a trivial 𝔽_{2}[G]-module. Let α ∈ K_{n}E. Then i_{E}N_{E/F} ᾱ = (σ − 1)ᾱ = 0 implies that N_{E/F} ᾱ = {a} · b̄ for some b ∈ K_{n−1}F, by Theorem 7. Now {a, −1} = {a, a} in k_{2}F, and then

{a, −1} · b̄ = {a} · N_{E/F} ᾱ = 0,

again by Theorem 7. Hence N_{E/F} k_{n}E ⊂ {a} · ann_{n−1}{a, −1}. By Theorem 7, N_{E/F} k_{n}E = ann_{n}{a}, whence ann_{n}{a} ⊂ {a} · ann_{n−1}{a, −1}.

(2) ⟹ (1). Assume ann_{n}{a} ⊂ {a} · ann_{n−1}{a, −1}. By Theorem 7, we have ann_{n}{a} = N_{E/F} k_{n}E and therefore N_{E/F} k_{n}E ⊂ {a} · ann_{n−1}{a, −1}. By Lemma 7, {a} · ann_{n−1}{a, −1} ⊂ N_{E/F} k_{n}E and hence N_{E/F} k_{n}E = {a} · ann_{n−1}{a, −1}. Let γ ∈ K_{n}E be arbitrary. Then N_{E/F} γ̄ = {a} · b̄ for some b̄ ∈ ann_{n−1}{a, −1}. Hence

(σ − 1)γ̄ = i_{E}N_{E/F} γ̄ = i_{E}({a} · b̄).

But by Theorem 7, i_{E}({a} · b̄) = 0. Hence (σ − 1)γ̄ = 0, and (k_{n}E)^{G} = k_{n}E as required.

Now assume p = 2 and a ∈ (F^{×2} + F^{×2}) \ F^{2}. As in the proof of Theorem 2, we have that {a, −1} = 0 ∈ k_{2}F. Therefore {−1} ∈ ann_{1}{a}. Since ann_{1}{a} = N_{E/F} k_{1}E by Theorem 7, we obtain {−1} ∈ N_{E/F} k_{1}E. Equivalently, −1 ∈ N_{E/F} (E^{×}). By [A, Theorem 3], E/F embeds in an extension E′/F cyclic of degree 4 with E′ = E(\sqrt{δ}) for δ ∈ E^{×}. Kummer theory tells us that \overline{\{δ\}} ∈ (k_{1}E)^{G}, so (σ − 1)(\overline{\{δ\}}) = (σ + 1)(\overline{\{δ\}}) = 0. Therefore (σ + 1)(\{δ\}) ∈ 2K_{1}E, whence N_{E/F} (δ) ∈ F^{×} ∩ E^{×2}. On the other hand, Kummer theory gives that (F^{×} ∩ E^{×2})/F^{×2} = {F^{×2}, aF^{×2}}. If N_{E/F} (δ) = f^{2} for f ∈ F^{×} then we have:

(\sqrt{δ})^{σ^{2}−1} = (\sqrt{δ}\sqrt{δ}^{σ})^{σ−1} = (\sqrt{δ}\sqrt{δ^{σ}})^{σ−1} = (\sqrt{N_{E/F} δ})^{σ−1}
= (±f)^{σ−1} = 1.

(The choice of sign in the square roots above is irrelevant as we apply σ − 1 afterwards.) Hence σ extends to an order 2 automorphism of E′/F, a contradiction. We may conclude that \overline{\{δ\}} ∈ (k_{1}E)^{G} satisfies N_{E/F} \overline{\{δ\}} = \{a\}, as required.

We now follow the proof of the p > 2 case to show that (1) and (3) are equivalent in the p = 2 case as well.

(1) ⟹ (3). Assume that k_{n}E is a trivial 𝔽_{2}[G]-module. Let γ ∈ K_{n}E be arbitrary. Then (σ − 1)γ̄ = 0. Hence

i_{E}N_{E/F} γ̄ = (σ − 1)γ̄ = 0

and so by Theorem 7, N_{E/F} γ̄ = {a} · f̄ for f ∈ K_{n−1}F. By the projection formula, N_{E/F} ({δ} · i_{E}(f̄)) = {a} · f̄. Then

N_{E/F} (γ̄ − \overline{\{δ\}} · i_{E}(f̄)) = 0,

and hence

N_{E/F} (γ − {δ} · i_{E}(f)) = 2g,

for some g ∈ K_{n−1}F. Set

β = γ − {δ} · i_{E}(f) − i_{E}(g).

Then N_{E/F} (β) = 0. By Theorem 5, there exists α ∈ K_{n}E such that (σ − 1)α = β. But since k_{n}E is fixed by G, β̄ = 0. Hence k_{n}E = i_{E}(k_{n}F) + \overline{\{δ\}} · i_{E}(k_{n−1}F).

(3) ⟹ (1). If β ∈ K_{n}E, then β̄ = i_{E}(ḡ) + \overline{\{δ\}} · i_{E}(f̄) for g ∈ K_{n}F and f ∈ K_{n−1}F. Clearly (σ − 1)β̄ = 0, which implies that (k_{n}E)^{G} = k_{n}E. □

It is worth looking more closely at the case n = 1 of Theorem 3.

Observe that the condition

(ξ_{p}) ∪ H^{0}(F) ⊂ (a) ∪ H^{0}(F),

for n = 1 and p > 2, is equivalent with the condition ξ_{p^{2}} ∈ E^{×}. (See [MS, Corollary 1].)

In the case n = 1 and p = 2 the condition

ann_{1}(a) ⊂ (a) ∪ ann_{0}(a, −1)

can be reformulated as follows:

If (a, −1) = 0 then N_{E/F} (k_{1}E) ⊂ ⟨{a}⟩

and if (a, −1) ≠ 0 then ann_{1}(a) = {0}.

Since {−a, a} = 0 we see that ann_{1}(a) = 0 implies that {−a} = 0 or equivalently {a} = {−1}. Recall that a field F is called Pythagorean if F^{2} + F^{2} ⊂ F^{2}. In the first case when (a, −1) = 0 the equality \{−a, a\} = 0 and N_{E/F} (k_{1}E) ⊂ ⟨\{a\}⟩ implies \sqrt{−1} ∈ E^{×}.

Summarizing our discussion for p = 2 we have:

H^{1}(E) is a trivial 𝔽_{2}[G]-module if and only if either (a, −1) = 0 ∈ H^{2}(F) and N_{E/F} (k_{1}E) ⊂ ⟨{a}⟩ or a = −1 and F is a Pythagorean field. In both cases \sqrt{−1} ∈ E^{×}. (See [MS, Corollary 1].)

Corollary 6. Suppose n ∈ ℕ and (k_{n}E)^{G} = k_{n}E. Then we have the following exact sequence:

0 → ann_{n−1}{a} → k_{n−1}F ⟶^{{a}·−} k_{n}F ⟶^{i_{E}}

k_{n}E ⟶^{N_{E/F}} {a} · ann_{n−1}{a, ξ_{p}} → 0.

Here the map ann_{n−1}{a} → k_{n−1}F is the natural inclusion.

Proof. Exactness at the first and second terms is obvious, and exactness at the third term follows from Theorem 7.

We consider exactness at the fifth term. In the p = 2 case, Theorem 3 tells us that ann_{n}{a} ⊂ {a} · ann_{n−1}{a, −1}. By Theorem 7, we have ann_{n}{a} = N_{E/F} k_{n}E, hence N_{E/F} k_{n}E ⊂ {a} · ann_{n−1}{a, −1}. By Lemma 7 we have the reverse inclusion, so that N_{E/F} k_{n}E = {a} · ann_{n−1}{a, −1} and the sequence is exact at the fifth term.

In the p > 2 case, observe that {ξ_{p}} · k_{n−1}F ⊂ {a} · k_{n−1}F implies that k_{n−1}F = ann_{n−1}{a, ξ_{p}}, since {a, a} = 0. Therefore, by part (2) of Theorem 3, we know ann_{n}{a} = {a} · ann_{n−1}{a, ξ_{p}}. By Theorem 7, we have ann_{n}{a} = N_{E/F} k_{n}E and hence the sequence is exact at the fifth term in the p > 2 case as well.

Hence it remains to show exactness at the fourth term. Suppose γ ∈ K_{n}E and N_{E/F} γ̄ = 0. Then there exists f ∈ K_{n}F such that N_{E/F} γ = pf, and then N_{E/F} (γ − i_{E}(f)) = 0. By Theorem 5, there exists α ∈ K_{n}E such that (σ − 1)ᾱ = γ̄ − i_{E}(f̄). But (σ − 1)ᾱ = 0 because (k_{n}E)^{G} = k_{n}E. Hence γ̄ = i_{E}(f̄) and we are done. □

We are now ready to prove Theorem 4.

Proof of Theorem 4. In the p > 2 case, the result on heredity follows from Theorem 3, part (3), together with two hereditary properties from Lemma 4: item (4), with α_{1} = {ξ_{p}} and α_{2} = {a}, and item (5). The exact sequence, in turn, follows from Corollary 6.

In the case p = 2, by Theorem 3, it is sufficient to prove that condition (2) in the p = 2 case is also hereditary. Assume (2) holds for n and m > n. By a well-known fact in Milnor K-theory, the group K_{m}E is generated by the symbols

α = {u, f_{1}, …, f_{n−1}, …, f_{m−1}},
if n > 1 and by α = {u, f_{1}, …, f_{m−1}} if n = 1,

where u ∈ E^{∗} and f_{i} ∈ F^{∗} for all i = 1, …, m − 1. (See [FV, page 291, Corollary 2].)

Assume now that n > 1. By the projection formula, we obtain

N_{E/F} ᾱ = \overline{\{N_{E/F} u, f_{1}, …, f_{n−1}\}} · \overline{\{f_{n}, …, f_{m−1}\}}.

Since ann_{n}{a} = N_{E/F} k_{n}E by Theorem 7, condition (2) gives us that

\overline{\{N_{E/F} u, f_{1}, …, f_{n−1}\}} = N_{E/F} \overline{\{u, f_{1}, …, f_{n−1}\}} ∈ \{a\} · ann_{n−1}\{a, −1\}.

Hence we may write

\overline{\{N_{E/F} u, f_{1}, …, f_{n−1}\}} = \{a\} · c̄,

where c̄ ∈ ann_{n−1}{a, −1}. Observe that this last equality holds also in the case when n = 1, provided that we interpret the left-hand side as \overline{\{N_{E/F} u\}}. Thus

N_{E/F} ᾱ = \{a\} · c̄ · \overline{\{f_{n}, …, f_{m−1}\}}

and

c̄ · \overline{\{f_{n}, …, f_{m−1}\}} ∈ ann_{m−1}\{a, −1\}.

Therefore N_{E/F} k_{m}E ⊂ {a} · ann_{m−1}{a, −1}, and we see that condition (2) is indeed hereditary. □

6. Examples of H^{k}(E) Trivial for all n < k and H^{n}(E) Nontrivial, with Given Cohomological Dimension

We have shown in Theorem 4 that the property H^{n}(E) is a trivial 𝔽_{p}[G]-module is hereditary. This result leads naturally to the definition of an interesting invariant ct(E/F) ∈ {0} ∪ ℕ ∪ {∞}:

ct(E/F) = sup {n ∈ ℕ ∪ {0} | H^{n}(E) is not a trivial 𝔽_{p}[G]-module} .

As with cf, we have chosen ct to indicate that after degree ct(E/F), Galois cohomology consists of trivial 𝔽_{p}[G]-modules. Of course, if H^{n}(E) is never trivial for n ≥ 1 then ct(E/F) = ∞, and otherwise ct(E/F) ∈ ℕ ∪ {0}. (Observe that since H^{0}(E) ≅ 𝔽_{p} and there are no nontrivial G-actions on 𝔽_{p}, we always have that H^{0}(E) is a trivial 𝔽_{p}[G]-module. However, Theorem 4 establishes the hereditary property only when n > 0.)

If ct(E/F) = n ∈ ℕ, then by definition H^{m}(E) is a trivial 𝔽_{p}[G]- module for all m > n. On the other hand, by the hereditary property we also have that H^{k}(E) is not trivial for all 1 ≤ k ≤ ct(E/F). A natural question arises: can we choose a suitable field extension E/F so that ct(E/F) is a given natural number or ∞? We show that the answer is affirmative. In fact, we can arrange that both values ct(E/F) and cd(G_{E}) are any natural numbers or ∞ and the absolute Galois group G_{F} is a pro-p-group modulo an obvious restriction, the inequality described below.

Suppose that the absolute Galois group G_{E} is a pro-p-group. Then, observing that {0}^{G} = {0}, we have:

ct(E/F) ≤ cd(G_{E}).

Our result is then the following.

Given 1 ≤ n ≤ m ∈ ℕ ∪ {∞} and a prime p, there exists a cyclic extension E/F of degree p with ξ_{p} ∈ F such that

(1) G_{E} is a pro-p-group;

(2) ct(E/F) = n; and

(3) cd(G_{E}) = m.

It is quite an interesting feature of our construction that it parallels the construction made in the rather opposite free case dealt with before. The only difference is the choice of a in our field extension of the form E = F (\sqrt[p]{a}).

6.1. The case m ∈ ℕ.

(1). Let F := F_{n,m} be a field of characteristic 0 with G_{F} ≅ ℤ_{p}^{n} ⋆_{pro-p} ℤ_{p}^{m} and ξ_{p^{2}} ∈ F, given by Lemma 6. Let

E = F (\sqrt[p]{a})

where a ∈ F^{×} such that under the restriction map on H^{1},

res_{⋆}(a) = (a)_{1} ⊕ (a)_{2}, (a)_{1} = 0, (a)_{2} ≠ 0.

Observe that there exists an a with the required conditions because by Lemma 5, H^{1}(F_{m}) ≠ {0}.

(2a). H^{n}(E) is not trivial. We claim that

ann_{n}(a) ⊄ (a) ∪ H^{n−1}(F).

By Lemma 5, H^{n}(F_{n}) contains a nontrivial element c. Let b ∈ H^{n}(F) such that

b_{1} = c ∈ H^{n}(F_{n}) and b_{2} = 0 ∈ H^{n}(F_{m}).

Then b ≠ 0 and since the cup-product commutes with res_{⋆},

res_{⋆}(a) ∪ b = (0 ∪ b_{1}) ⊕ ((a_{1}) ∪ 0) = 0 ∈ H^{n+1}(F).

Therefore b ∈ ann_{n}(a).

Not let f ∈ H^{n−1}(F) be arbitrary. Then

((a) ∪ f)_{1} = 0 ∪ f_{1} = 0

and therefore b ∉ (a) ∪ H^{n−1}(F). Thus ann_{n}(a) ⊄ (a) ∪ H^{n−1}(F).

For the case p > 2, Theorem 3, part (2) implies that H^{n}(E) is not trivial.

In the case p = 2 we have (−1) = 0 since \sqrt{−1} ∈ F^{×}. Therefore 0 = (a, −1) ∈ H^{2}(F). Thus ann_{n−1}(a, −1) = H^{n−1}(F) and (a) ∪ ann_{n−1}(a, −1) = (a) ∪ H^{n−1}(F). Hence by our claim above ann_{n}(a) ⊄ (a) ∪ ann_{n−1}(a, −1), and we can again apply Theorem 3 to conclude that H^{n}(E) is not trivial.

(2b). H^{k}(E)^{G} = H^{k}(E) for all k ≥ n + 1.

Let a_{1} ∈ F_{m}^{×} satisfy (a_{1}) = (a)_{2} and extend {(a_{1})} to a basis {(a_{1}), …, (a_{m})} of H^{1}(F_{m}). Recall that by Lemma 5, H^{k}(F_{m}) is just the kth homogenous summand of the exterior algebra over 𝔽_{p} generated by H^{1}(F_{m}). Using this fact and writing each element in H^{k}(F_{m}) as a sum of elements of the form

(a_{i_{1}}) ∪ ⋯ ∪ (a_{i_{k}}) , 1 ≤ i_{1} < i_{2} < ⋯ < i_{k} ≤ m,

and also the fact that H^{k}(F_{n}) = {0} we see that

ann_{k}(a) = (a) ∪ H^{k−1}(F).

Now again using Theorem 3 as in the case (2a), we conclude that H^{k}(E)^{G} = H^{k}(E).

(3). cd(G_{E}) = m. Indeed cd(G_{E}) = cd(G_{F}) by Serre’s theorem. (See [S] and the discussion in section 4.1 which guarantees that the hypothesis of Serre’s theorem is valid.)

But from Lemma 5 and Lemma 6 we see that

cd(G_{F}) = max{cd(G_{F_{n}}), cd(G_{F_{m}})} = m.

Thus we see that in the case when m < ∞ we constructed a cyclic field extension E/F of degree p with required properties (1), (2) and (3).

6.2. The case m = ∞. We first consider the subcase of this case when n < ∞. As in section 4.2 set

F_{∞} := ℂ((ℤ_{(p)}^{m})), where m = ℵ_{0}.

By Lemma 6 we see that there exists a field F := F_{n,∞} such that G_{F} ≅ G_{F_{n}} ⋆_{pro-p} G_{F_{∞}} and ξ_{p^{2}} ∈ F. Let a ∈ F^{×} such that under the restriction map

res_{⋆} : H^{1}(F) → H^{1}(F_{n}) ⊕ H^{1}(F_{∞})

we have

res_{⋆}(a) = 0 ⊕ (a)_{2}, (a)_{2} ≠ 0.

Then cd(F) = ∞ and with the same argument as above we see that ct(E/F) = n.

Finally we consider the case n = ∞ = m. Set again F_{∞} := ℂ((ℤ_{(p)}^{m})), where m = ℵ_{0} and F = F_{∞,∞}. Also let a ∈ F^{×} such that

res_{⋆}(a) = 0 ⊕ (a)_{2}, (a)_{2} ≠ 0.

Then using the same argument as in (2b) we see that ct(F) = ∞.

Our construction is now completed.

References

[A] A. Albert. On cyclic fields. Trans. Amer. Math. Soc. 37 (1935), 454–462.

[EH] I. Efrat and D. Haran. On Galois groups over pythagorean and semi-real closed fields. Israel J. Math. 85 (1994), no. 1-3, 57–78.

[FV] I. Fesenko and S. Vostokov. Local fields and their extensions, 2nd ed. Translations of Mathematical Monographs 121. Providence, RI: American Mathematical Society, 2002.

[FW] E. Friedlander and C. Weibel. An overview of algebraic K-theory. Algebraic K-theory and its applications (Trieste, 1997), edited by H. Bass, A. O. Kuku, and C. Pedrini, 1–119. River Edge, NJ: World Scientific Publishing, 1999.

[J] N. Jacobson. Lectures in abstract algebra. Vol. III: Theory of fields and Galois theory. Princeton: Van Nostrand Company, 1964.

[K] J. Koenigsmann. Encoding valuations in absolute Galois groups. Proceedings of the international conference and workshop on Valuation Theory (Saskatoon, 1999), edited by F.-V. Kuhlmann, S. Kuhlmann, and M. Marshall, 107–132. Fields Institute Communications 33. Providence, RI: American Mathematical Society, 2003.

[Lam] T.-Y. Lam. The algebraic theory of quadratic forms, revised 2nd printing. Mathematics Lecture Note Series. Reading, MA: Benjamin/Cummings Publishing Company Advanced Book Program, 1986.

[Lg1] S. Lang. Algebra, revised 3rd ed. Graduate Texts in Mathematics 211. New York: Springer-Verlag, 2002.

[Lg2] S. Lang. Topics in cohomology of groups. Lecture Notes in Mathematics 1625. New York: Springer-Verlag, 1996.

[LMS] N. Lemire, J. Mináč, and J. Swallow. Galois module structure of Galois cohomology. Preprint, ArXiv: Math.NT/0409484v1 (2004).

[M] J. Milnor. Algebraic K-theory and quadratic forms. Invent. Math. 9 (1970), 318–344.

[MS] J. Mináč and J. Swallow. Galois module structure of pth-power classes of extensions of degree p. Israel J. Math. 138 (2003), 29–42.

[N] J. Neukirch. Freie Produkte pro-endlicher Gruppen und ihre Kohomologie. Arch. Math. (Basel) 22 (1971), 337–357.

[R1] P. Ribenboim. Théorie des valuations. Deuxième édition multigraphiée. Séminaire de Mathématiques Supérieures 9 (Été, 1964). Montréal: Les Presses de l’Université de Montréal, 1968.

[R2] P. Ribenboim. Equivalent forms of Hensel’s Lemma. Expo. Math. 3 (1985), 3–24.

[RZ] L. Ribes and P. Zalesskii. Profinite groups. Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. A Series of Modern Surveys in Mathematics 40. Berlin: Springer-Verlag, 2000.

[S] J.-P. Serre. Sur la dimension cohomologique des groupes finis. Topology 3 (1965), 413–420.

[V1] V. Voevodsky. Motivic cohomology with ℤ/2-coefficients. Publ. Inst. Hautes Études Sci., No. 98 (2003), 59–104.

[V2] V. Voevodsky. On motivic cohomology with ℤ/l coefficients. Ktheory preprint archive 639 (2003). http://www.math.uiuc.edu/K-theory/ 0639/.

[Wad] A. Wadsworth. p-Henselian fields: K-theory, Galois cohomology, and graded Witt rings. Pacific J. Math. 105 (1983), 473–496.

[War] R. Ware. Galois groups of maximal p-extensions. Trans. Amer. Math. Soc. 333 (1992), no. 2, 721–728.

Department of Mathematics, Middlesex College, University of Western Ontario, London, Ontario N6A 5B7 CANADA

  E-mail address: nlemire@uwo.ca

Department of Mathematics, Middlesex College, University of Western Ontario, London, Ontario N6A 5B7 CANADA

E-mail address: minac@uwo.ca

Department of Mathematics, Davidson College, Box 7046, Davidson, North Carolina 28035-7046 USA

  E-mail address: joswallow@davidson.edu