New York Journal of Mathematics New York J. Math.

New York Journal of Mathematics

New York J. Math. 11(2005)291–302.

When is Galois cohomology free or trivial?

Nicole Lemire, J´ an Min´ aˇ c and John Swallow

Abstract. Letpbe a prime,Fa field containing a primitivepth root of unity, andE/F a cyclic extension of degreep. Using the Bloch–Kato Conjecture we determine precise conditions for the cohomology groupHⁿ(E) :=Hⁿ(GE,Fp) to be free or trivial as anFp[Gal(E/F)]-module, and we examine when these properties forHⁿ(E) are inherited by H^k(E), k > n. By analogy with co- homological dimension, we introduce notions of cohomological freeness and cohomological triviality, and we give examples of Hⁿ(E) free or trivial for eachn∈Nwith prescribed cohomological dimension.

Contents

1. Free cohomology 292

2. Hereditary freeness 295

3. Examples of free cohomology 296

4. Trivial cohomology 298

5. Hereditary triviality 300

6. Examples of trivial cohomology 300

Acknowledgement 301

References 301

Letpbe a prime andF a field containing a primitivepth root of unityξ_p. Let E/F be a cyclic extension of degree p, G_E the absolute Galois group of E, and G= Gal(E/F). In [LMS] we determined the structure of Hⁿ(E) :=Hⁿ(G_E,Fp), n ∈N, as an Fp[G]-module. In this paper we study more closely the question of whenHⁿ(E) is free or trivial as anFp[G]-module.

Received October 26, 2004.

Mathematics Subject Classification. 12G05 (primary), 19D45 (secondary).

Key words and phrases. Galois cohomology, MilnorK-theory, Bloch–Kato conjecture, pro-p- group, free product.

The first author was supported in part by NSERC grant R3276A01.

The second author was supported in part by NSERC grant R0370A01, by the Institute for Advanced Study, Princeton, by a Distinguished Professorship during 2004–2005 at the University of Western Ontario, and by the Mathematical Sciences Research Institute, Berkeley.

The third author was supported in part by NSA grant MDA904-02-1-0061.

ISSN 1076-9803/05

291

1. Free cohomology

Let a ∈ F satisfy E = F(√^p

a). We write ann_nx for the annihilator of x ∈ H^m(F), m ∈ N, under the cup-product operation on Hⁿ(F). Let (f) ∈ H¹(F) denote the class off under the Kummer isomorphism ofH¹(F) with thepth-power classes ofF^× :=F\ {0}, and let (f, g)∈H²(F) denote the cup-product of (f) and (g)∈H¹(F).

Theorem 1. Let n∈N. Ifp >2, then the following are equivalent:

(1) Hⁿ(E)is a free Fp[G]-module.

(2) Hⁿ⁻¹(F) = ann_n−1(a).

(3) res :Hⁿ(F)→Hⁿ(E) is injective.

(4) cor :Hⁿ⁻¹(E)→Hⁿ⁻¹(F)is surjective.

If p= 2, then the following are equivalent:

(1) Hⁿ(E)is a free F2[G]-module.

(2) ann_n−1(a) = ann_n−1(a,−1) andHⁿ(F) = corHⁿ(E) + (a)∪Hⁿ⁻¹(F).

(3) ann_n−1(a) = ann_n−1(a,−1) andHⁿ(F) = ann_n(a) + (a)∪Hⁿ⁻¹(F).

(4) Hⁿ(F) = ann_n(a)⊕(a)∪Hⁿ⁻¹(F).

We prove this theorem using the sequence of results below.

Fori≥0, letK_iF denote theith MilnorK-group of the fieldF, with standard generators denoted by {f₁, . . . , f_i}, f₁, . . . , f_i ∈F^×. For α∈ K_iF, we denote by α the class ofα modulop, and we abbreviate K_nF/pK_nF by k_nF. Throughout the paper we use the Bloch–Kato Conjecture in identifying k_nF and Hⁿ(F), as well as Hilbert 90 for Milnor K-theory. (See [V1, Lemma 6.11 and §7] and [V2,

§6 and Theorem 7.1]. For further expositions of the work of Rost and Voevodsky on Bloch–Kato Conjecture, see [Ro] and [Su].) The image of an elementα∈K_iF in Hⁱ(F) we also denote byα. We omit the bars over elements {a}, {ξ_p},{a, a}, {a, ξ_p}, and {√^p

a}. We denote the norm map K_nE → K_nF by N_E/F, the map K_nF →K_nEinduced by the inclusion map byi_E, and multiplication by{a} from K_n−1F to K_nF by{a} · −. We use the same notation for the analogous maps of K-theory modulop.

The following theorem is a strengthening of [V1, Prop. 5.2].

Theorem 2([LMS, Thm. 5]). Let F be a field containing a primitive pth root of unity. Then for any cyclic extensionE/F of degreepandm≥1 the sequence

k_m−1E−−−−→^NE/F k_m−1F −−−−→^{a}·− k_mF −→^iE k_mE is exact.

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

(1) ann_n−1{a}= ann_n−1{a,−1} andk_nF =N_E/Fk_nE+{a} ·k_n−1F. (2) ann_n−1{a}= ann_n−1{a,−1} andk_nF = ann_n{a}+{a} ·k_n−1F. (3) k_nF = ann_n{a} ⊕ {a} ·k_n−1F.

Proof. (1) =⇒(2) follows from N_E/Fk_nE= ann_n{a}.

(2) =⇒ (3) Let α ∈ ({a} ·k_n−1F)∩ann_n{a}. Then α = {a} ·f for some f ∈K_n−1F, and 0 ={a, a} ·f ={a,−1} ·f. By the first hypothesis,{a} ·f = 0.

Thenα= 0 and the sum is direct.

(3) =⇒(1) The second claim follows from the fact that ann_n{a} =N_E/Fk_nE.

For the first, suppose {a,−1} ·f = 0 for f ∈K_n−1F. Because {a,−1} ={a, a}, we have {a} ·f ∈ (ann_n{a})∩({a} ·k_n−1F) = {0}. Hence f ∈ ann_n−1{a} and

ann_n−1{a}= ann_n−1{a,−1}.

For anyFp[G]-module W we denote by W^G the submodule of W consisting of elements fixed by G. For γ ∈ K_nE, let l(γ) denote the dimension of the cyclic Fp[G]-submoduleγofk_nE generated byγ. Then we have, forl(γ)≥1,

(σ−1)^l(γ)−1γ=γ^G= 0 and (σ−1)^l(γ)γ= 0.

We denote byN the map (σ−1)^p−1 onk_nE. Because (σ−1)^p−1= 1 +σ+· · ·+σ^p−1 inFp[G], we shall usei_EN_E/F andN interchangeably onk_nE.

Lemma 1. Let n∈N. Suppose that either

• p >2 andN_E/F:k_n−1E→k_n−1F is surjective, or

• p= 2 andk_nF =N_E/Fk_nE⊕ {a} ·k_n−1F.

Then we have:

(1) For eachγ∈K_nE, there existsα∈K_nE withN α=γ^G. (2) (k_nE)^G=i_EN_E/Fk_nE = (σ−1)^p−1k_nE=i_Ek_nF.

Proof. (1) Assume first that p > 2. By hypothesis, N_E/F: k_n−1E → k_n−1F is surjective, and then using the projection formula ([FW, p. 81]) we see that N_E/F:k_nE→k_nFis also surjective. Hence ifγ∈i_Ek_nFthen there existsα∈k_nE such thatN α=γand we are done. Otherwise, letl=l(γ) and supposeγ∈i_Ek_nF and 1≤l≤i≤p.

If l ≥ 2 we show by induction on i that there exists α_i ∈ K_nE such that (σ−1)ⁱ⁻¹α_i=γ^G. Then settingα:=α_p, the proof will be complete in the case when 2≤l. Assume then that l ≥2. If i=l then α_i =γ suffices. Assume now that 2≤l ≤i < p and that our statement is true for i. Set c =N_E/Fα_i. Since i_Ec=N α_i= (σ−1)^p−1α_i andi < p,i_Ec= 0. By our hypothesis and Theorem2, we havec= 0, that is,c=pffor somef ∈K_nF. HenceN_E/F(α_i−i_E(f)) = 0. By Hilbert 90, there existsω∈K_nEsuch that (σ−1)ω=α_i−i_E(f). Sincel(α_i)>1 we have (σ−1)²ω = (σ−1)α_i = 0. Therefore (σ−1)ⁱω = γ^G and we can set α_i+1=ω. We have proved that ifl ≥2 then there exists α∈K_nE such that N α= (σ−1)^l(γ)−1γ.

Now assume that l(γ) = 1 but γ /∈ i_Ek_nF. Then γ = α₁ and proceeding as above we have (σ−1)ω=α₁−i_E(f)= 0. Thusl(ω) = 2 and our argument above shows that there exists β ∈K_nE such that N β = (σ−1)ω =α₁−i_E(f). As we observed at the beginning of our proof there exists an elementδ∈K_nE such that N δ =i_E(f). Therefore we have N(β+δ) = α₁ = γ. Hence for each γ ∈ K_nE, there exists α∈K_nE such thatN α=γ^G.

Now consider the casep= 2. In this case from our hypothesisk_nF =N_E/Fk_nE+

{a} ·k_n−1F we again have i_EN_E/Fk_nE = i_Ek_nF. Therefore if γ ∈ i_Ek_nF our statement follows. Assume thatγ∈k_nE\i_Ek_nF. (We use this hypothesis only to exclude the trivial case γ= 0.) Sincep= 2 we see that l(γ)≤2, and ifl(γ) = 2 we may setα=γ and (1) follows again. Next we shall assume thatl(γ) = 1 and therefore γ ∈ (k_nE)^G. Set c = N_E/Fγ. Then i_Ec = 0. From Theorem 2, we

conclude that c = {a} ·g+ 2f for g ∈ K_n−1F and f ∈ K_nF. Hence from the projection formula,N_E/F(γ− {√

a} ·i_E(g)−i_E(f)) ={−1} ·g. Using Theorem2 again, we obtain that{a,−1} ·g= 0. Our hypothesis and theCorollary imply that {a} ·g = 0. Hence{a} ·g = 2hfor someh∈K_nF and N_E/Fγ = 2(h+f). Thus N_E/F(γ−i_E(h+f)) = 0. Then by Hilbert 90 there exists α ∈ K_nE such that (σ−1)α=γ−i_E(h+f). Henceγ= (σ−1)α+i_E(h+f)∈N k_nE.

(2) Suppose γ ∈ (k_nE)^G. Then by (1) we see that γ = N α for α ∈ K_nE. Hence (k_nE)^G ⊂ i_EN_E/Fk_nE. Then i_EN_E/Fk_nE ⊂i_Ek_nF ⊂ (k_nE)^G ⊂ i_EN_E/Fk_nE, and so all inclusions are equalities.

Proof of Theorem 1. First we show that for allp,k_nE free implies i_EN_E/Fk_nE=i_Ek_nF = (k_nE)^G.

If k_nE is free, then we have (σ−1)^p−1k_nE = (k_nE)^G. Hence i_EN_E/Fk_nE = (σ−1)^p−1k_nE = (k_nE)^G. Then i_EN_E/Fk_nE ⊂ i_Ek_nF ⊂ (k_nE)^G and we have established our claim.

Assume first that p > 2. First we show (1) =⇒ (2). Let f ∈ K_n−1F be arbitrary, and set α={√^p

a} ·i_E(f). Now because (k_nE)^G =i_EN_E/Fk_nE, there existsβ∈K_nE such thati_EN_E/Fβ =i_E({ξ_p} ·f)∈(k_nE)^G. Setγ= (σ−1)^p−2β.

Sinceγis in the image ofσ−1 we haveN_E/Fγ= 0. ThenN_E/F(α−γ) ={a} ·f. Also, (σ−1)(α−γ) = 0 and therefore α−γ∈(k_nE)^G =i_Ek_nF. But oni_Ek_nF the norm mapN_E/F is trivial. Therefore{a} ·f = 0 and ann_n−1{a}=k_n−1F. By Theorem 2, items (2), (3) and (4) are all equivalent. Now we show (4) =⇒ (1).

Assume thatN_E/F:k_n−1E→k_n−1F is surjective. By Lemma 1 we have (k_nE)^G = (σ−1)^p−1k_nE.

HenceH²(G, k_nE) ={0} and sok_nE is free. (See [La, p. 63].)

Now suppose thatp= 2. By the Corollary, we need only show that (1) and (2) are equivalent. We show first that (1) =⇒ (2). We established that (1) implies i_EN_E/Fk_nE = i_Ek_nF. Since keri_E = {a} ·k_n−1F, this equality is equivalent to k_nF =N_E/Fk_nE+{a} ·k_n−1F, so we have the second part of (2). We now show that ann_n−1{a,−1} ⊂ann_n−1{a}. Letf ∈ann_n−1{a,−1}. Set α={√

a} ·i_E(f).

Since {a} · {−1} ·f = 0, Theorem 2 tells us that there existsβ ∈K_nE such that N_E/Fβ ={−1} ·f. Now we calculate by the projection formula N_E/F(α−β) = {−a}·f−{−1}·f ={a}·f. On the other hand, (σ−1)(α−β) ={−1}·i_E(f)−{−1}·

i_E(f) = 0. Henceα−β∈(k_nE)^G. Sincek_nEis free, (k_nE)^G= (σ−1)k_nE. But on (σ−1)k_nE the norm mapN_E/F is trivial. Hence{a} ·f = 0, andf ∈ann_n−1{a}, so ann_n−1{a,−1} ⊂ ann_n−1{a}. Now we show that (2) =⇒ (1). Assume that ann_n−1{a,−1}= ann_n−1{a}and thatk_nF =N_E/Fk_nE+{a}·k_n−1F. By Lemma 1 and the Corollary, (k_nE)^G = (σ−1)k_nE and sok_nE is free.

Ifp >2 thenN_E/Fk₀E={0} =k₀F ∼=Fp. Therefore:

Corollary 1. Forp >2,k₁E is never free.

If p= 2 and−1 ∈F^×2, then ann₀{a,−1} =k₀F ∼=F₂ = ann₀{a} ={0} and hence:

Corollary 2. Forp= 2 and√

−1∈F,k₁E is never free.

2. Hereditary freeness

We say that a property of Milnor k-groups k_nE and k_nF is hereditary if the validity of the property for a given n implies the validity of the property for all integers greater thann. We consider the zeroFp[G]-module{0}to be a freeFp[G]- module.

Theorem 3. Suppose that either p >2 or p= 2 and a∈(F²+F²)\F². Then free cohomology is hereditary: ifn∈N, then for allm≥n,Hⁿ(E)is a freeFp[G]- module =⇒ H^m(E) is a free Fp[G]-module. Moreover, if H^m(E), m ∈N, is a freeFp[G]-module, then the sequence

0→H^m(F)−−→^res H^m(E)−−→^cor H^m(F)→0.

is exact at the first and third terms. Whenp= 2 this sequence is exact.

Lemma 2. Let n∈N. The following are hereditary properties:

(1) k_n−1F = ann_n−1{a}= ann_n−1{a, ξ_p}. (2) i_E: k_nF →k_nE is injective.

(3) N_E/F:k_n−1E→k_n−1F is surjective.

(4) for some fixedα₁, α₂∈K₁F,α₁·k_n−1F ⊂α₂·k_n−1F.

(5) for some fixedα∈K₁E,k_nE=i_Ek_nF+α·i_Ek_n−1F.

Proof. (1)k_nF =k_n−1F·k₁F, and since ann_n−1{a}=k_n−1F, we have ann_n{a}= k_nF as well. The other equality follows from ann_n{a} ⊂ ann_n{a, ξ_p}. The result follows by induction.

(2)–(3) By Theorem 2 the first three properties are equivalent, hence (2) and (3) are hereditary.

(4)K_nF is generated by elements of the form{f₁, f₂, . . . , f_n}={f₁, . . . , f_n−1} · {f_n},f_i∈F^×. For each such generator, we calculateα₁·{f₁, . . . , f_n}=α₂·g·{f_n} for someg∈K_n−1F, whenceα₁·k_nF ⊂α₂·k_nF. The result follows by induction.

(5) The condition on k_nE gives us that k_n+1E is generated by elements of the form γ₁ = {δ} ·i_E({f₁, . . . , f_n}), δ ∈ E^×, f_i ∈ F^× and γ₂ = {δ} ·α· i_E({f₁, . . . , f_n−1}),δ∈E^×,f_i∈F^×. Ifn−1≥1 then we see thatk_n+1E is gener- ated by the elements ink_nE·i_Ek₁F. By hypothesisk_nE=i_Ek_nF+α·i_Ek_n−1F and thereforek_n+1Eis generated by elements ini_Ek_n+1F+α·i_Ek_nF. If insteadn= 1, then using our hypothesisk₁E=i_Ek₁F+α·i_Ek₀F we may write the second type of generatorsγ₂ofk₂Eas γ₂= (i_E({f}) +cα)·α=−α·i_E({f})−α·i_E(c{−1}), f ∈F^×,c∈Z. Thus in this case both types of generators ofk₂Ehave the required form of elements in i_Ek₂F +α·i_Ek₁F. The result now follows by induction as

above.

Lemma 3. Suppose thatp= 2 and for some n∈N,ann_n−1{a} =k_n−1F. Then k_mE is a free F₂[G]-module for allm≥n.

Proof. We show that the two conditions of part (2) of thep= 2 portion of Theo- rem 1 hold forK-theory degree at leastn. From Lemma 2, part (1), we deduce that k_mF = ann_m{a}= ann_m{a,−1} for allm≥n−1. By Theorem 2 and Lemma 2, we havek_mF =N_E/Fk_mE for allm≥n−1 and therefore we see in particular that k_mF =N_E/Fk_mE+{a} ·k_m−1F for all m ≥n−1. We conclude that k_mE is a

freeF2[G]-module for allm≥n.

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

Assume then that p = 2 and a = x²+y² for some x, y ∈ F^×. It is well- known that then {a,−1} = 0 ∈ k₂F and hence ann_n−1{a,−1} = k_n−1F. Now observe that sincek_nEis a freeF₂[G]-module, by Theorem 1 we have ann_n−1{a}= ann_n−1{a,−1}, and so ann_n−1{a}=k_n−1F. We deduce from Lemma 3 thatk_mE is a freeF₂[G]-module for allm≥n.

For the exact sequence forp= 2, we have shown thatk_n−1F = ann_n−1{a}, and so by Theorem 2 and Lemma 2, we have k_mF = N_E/Fk_mE for all m ≥ n−1.

Furthermore, we conclude from Theorem 2 that i_E is injective fromk_mF to k_mE for allm≥n. The exactness of our sequence in the middle term follows from [Ar,

Satz 4.5].

Freeness is not generally hereditary whenp= 2. For example, letp= 2,F =Q₂, anda=−1, soE=Q₂(√

−1). Thenk₁F ∼=F^×/F^×2=[−1],[2],[5]. N_E/Fk₁E∼= N_E/F(E^×)F^×2/F^×2 =[2],[5]. Therefore k₁F =N_E/Fk₁E+{−1} ·k₀F. Since [−1]∈/ N_E/F(E^×)F^×2/F^×2, we have{−1,−1} = 0∈ k₂F and ann₀{−1,−1} = {0}= ann₀{−1}. Hence the conditions of part (2) of thep= 2 portion of Theorem 1 are satisfied, whence k₁E is a free F2-module. It is well-known, however, that k₂E ∼=F2.

3. Examples of free cohomology

Define cf(E/F)∈ {0} ∪N∪ {∞}by

cf(E/F) = sup{n∈N∪ {0} | Hⁿ(E) is not a freeFp[G]-module}. Further for any pro-p-groupT we denote by cd(T) the cohomological dimension of T.

Theorem 4. Given 1 ≤n ≤ m ∈N∪ {∞} and a prime p, there exists a cyclic extensionE/F of degreepwith ξ_p ∈F such that:

(1) G_F is a pro-p-group;

(2) cf(E/F) =n; and (3) cd(G_E) =m.

LetZ_(p):=_c

d ∈Qc, d∈Z, d= 0; ifc= 0 then (c, d) = 1, pd

,I be a well- ordered set of cardinalitym, and Γ be a direct sum ofmcopies ofZ_(p), indexed by I. We order Γ lexicographically. Thenm= dim_FpΓ/pΓ. It is well-known that 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. The absolute Galois group ofF_mis known to beZ^m_p , the topological product ofmcopies of Zp

[K, pp. 3–4].

Lemma 4([Wad, Thm. 3.6]). For m, n ∈ N∪ {ℵ₀}, Hⁿ(F_m) ∼= _n

H¹(Z^m_p) ∼= _n

⊕_mF_p, where the cup-product is sent to the wedge product.

Lemma 5. Suppose that m₁, m₂ are nonzero cardinal numbers, and let F_m1 and F_m2 be as above. There exists a field F_m1,m2 of characteristic 0, containing a primitive p²th root of unity ξ_p2, such that the absolute Galois group G_Fm

1,m2 ∼= G_Fm

1 pro-pG_Fm

2 ∼=Z^m_p¹ pro-pZ^m_p², where the free products are taken in the category of pro-p-groups, and the natural restriction maps res:Hⁿ(F_m1,m2)→Hⁿ(F_m1)⊕ Hⁿ(F_m2) are isomorphisms.

Proof. The existence of a fieldF_m1,m2 with

char(F_m1,m2) = char(F_m1) = char(F_m2) = 0

and the given absolute Galois group follows from [EH, Prop. 1.3]. Additionally using the construction of F_m1,m2 following [EH, proof of Prop. 1.3] we assume thatF_m1,m2 is the intersection of two henselian valued fields (L_i, V_i), i= 1,2, with residue fields isomorphic to some fieldsF_m1andF_m2closely related toF_m1andF_m2, respectively. HereV_i is a henselian valuation onL_i. Then by Hensel’s Lemma (see [Ri, pp. 12-13, condition (3)]) and by the fact thatF_m1andF_m2have characteristic 0 and both contain a primitivep²th root of unity, we see thatF_m1,m2 also contains a primitivep²th root of unity. The fact that the restriction maps are isomorphisms

follows from [N, S¨atze (4.1),(4.2)].

Proof of Theorem 4. The casem∈N:

(1) Let F := F_n,m be a field of characteristic 0 with G_F ∼=Zⁿ_p pro-pZ^m_p and ξ_p2 ∈F, given by Lemma 5. LetE =F(√^p

a) for any a∈F^× such that under the restriction map onH¹, res(a) = (a)₁⊕(a)₂, (a)₁= 0, (a)₂= 0. We use here the fact that res is an isomorphism, by Lemma 5. Observe that there exists anawith the required conditions because by Lemma 4,H¹(F_n)={0}.

(2) We first show Hⁿ(E) is not free. We claim that ann_n−1(a)=Hⁿ⁻¹(F). If n = 1 this statement is true since (a) = 0 ∈ H¹(F). Assume now that n > 1.

Let a₁ ∈ F_n^× satisfy (a₁) = (a)₁, and extend {(a₁)} to a basis {(a₁), (a₂), · · ·, (a_n)} of H¹(F_n). By Lemma 4, the element (a₁)∪(a₂)∪ · · · ∪(a_n) ∈ Hⁿ(F_n) is nontrivial, so that 0= (a₂)∪ · · · ∪(a_n)∈Hⁿ⁻¹(F_n). Let b∈Hⁿ⁻¹(F) satisfy b₁= (a₂)∪· · ·∪(a_n)∈Hⁿ⁻¹(F_n),b₂= 0∈Hⁿ⁻¹(F_m). Then since the cup-product commutes with res, ((a)∪b)₁= (a₁)∪b₁= 0∈Hⁿ(F_n), so that (a)∪b= 0∈Hⁿ(F) and hence ann_n−1(a)=Hⁿ⁻¹(F). Ifp >2, we conclude by Theorem 1 thatHⁿ(F) is not free. If p = 2, observe that since √

−1 ∈ F, we have ann_n−1(a,−1) = Hⁿ⁻¹(F), so that ann_n−1(a,−1) = ann_n−1(a). We deduce from Theorem 1 that Hⁿ(F) is not free.

We now showH^k(E) is free for allk≥n+ 1. We claim that ann_n(a) =Hⁿ(F).

Letc∈Hⁿ(F). Then sinceHⁿ⁺¹(F_n) = 0 by Lemma 4, res(a)∪c=

(a₁)∪c₁

⊕ 0∪c₂

= res0.

Hence (a)∪c= 0 and ann_n(a) =Hⁿ(F). Ifp >2 then we conclude by Theorem 1 that Hⁿ⁺¹(E) is free, and by Theorem 3, H^k(E) is free for all k ≥ n+ 1. If p= 2, observe that ann_n(a,−1) =Hⁿ(F). By Theorem 2, cor :Hⁿ(E)→Hⁿ(F) is surjective. Then by Lemma 3 and Theorem 1, we have thatH^k(E) is free for all k≥n+ 1.

(3) First we claim thatG_F does not contain an element of orderp. By the Artin–

Schreier theorem, finite subgroups of absolute Galois groups are either trivial or of order 2, and since√

−1∈F no element of order 2 exists inG_F. Then, by Serre’s

theorem [S], we obtain cd(G_E) = cd(G_F). From Lemmas 4 and 5 we find that cd(G_F) = max{cd(F_n),cd(F_m)}=m. Thus cd(G_E) =m. (One can also conclude thatG_F contains no element of orderpfrom the fact that cd(G_F) =m <∞.)

The case n < m= ∞: set F_∞ := C((Γ)), where m = ℵ0 and Γ is the direct sum of m copies of Z_(p). By Lemma 5, there exists a field F := F_n,∞ such that G_F ∼= G_{Fn pro-p}G_F∞ and ξ_p2 ∈ F. Then set E = F(√^p

a) for any a ∈ F^× such that under the restriction map res:H¹(F)→H¹(F_n)⊕H¹(F_∞), we have res(a) = (a)₁⊕0, (a)₁ = 0. Then cd(G_F) = cd(G_E) =∞, and as above we see that cf(E/F) =n.

The casen=∞=m: let Γ be a direct sum ofℵ₀ copies ofZ_(p). Then setF :=

F_∞=C((Γ)). Leta∈F^×such thatv(a)∈Γ\pΓ, wherevis a natural valuation on F. Then from the description of Galois cohomology ofp-henselian fields (see [Wad, Thm. 3.6]), we obtain ann_n(a) = (a)∪Hⁿ⁻¹(F) and (a)∪Hⁿ⁻¹(F)=Hⁿ(F) for alln∈N. SettingE=F(√^p

a), we obtain cf(E/F) =∞.

4. Trivial cohomology

By a trivialFp[G]-module we mean an Fp[G]-moduleW such thatW =W^G. Theorem 5. Let n∈N. Ifp >2, then the following are equivalent:

(1) Hⁿ(E)is a trivialF_p[G]-module.

(2) (ξ_p)∪Hⁿ⁻¹(F)⊂(a)∪Hⁿ⁻¹(F)andann_n(a) = (a)∪Hⁿ⁻¹(F).

(3) (ξ_p)∪Hⁿ⁻¹(F)⊂(a)∪Hⁿ⁻¹(F)and Hⁿ(E) = resHⁿ(F) + (√^p

a)∪resHⁿ⁻¹(F).

If p= 2, then the following are equivalent:

(1) Hⁿ(E)is a trivialF₂[G]-module.

(2) ann_n(a)⊂(a)∪ann_n−1(a,−1).

In thep= 2case, suppose additionally thata∈(F²+F²)\F². Then the conditions above are also equivalent to:

(3) Hⁿ(E) = resHⁿ(F) + (δ)∪resHⁿ⁻¹(F), where (δ) ∈ H¹(E)^G satisfies cor(δ) = (a).

Remark. Forp >2 andn= 1 the second condition in (3) was observed in [War, Lemma 3].

Lemma 6. Letn∈N. Suppose thati_E({ξ_p} ·k_n−1F) =i_EN_E/Fk_nE={0}. Then (k_nE)^G =k_nE.

Proof. Letγ∈K_nE. We show thatl(γ)>1 leads to a contradiction, whence we will have the result. Suppose thatl =l(γ)≥2 and 2 ≤l ≤i ≤p. We show by induction onithat there existsα_i∈K_nEsuch that(σ−1)ⁱ⁻¹α_i=(σ−1)^l−1γ. Ifi=l thenα_i=γsuffices. Assume now thatl≤i < pand that our statement is true fori. Setc=N_E/Fα_i. By our hypothesis,i_Ec= 0. By Theorem2,c={a} ·b for some b ∈ K_nF. Hence c = {a} ·b+pf for f ∈ K_nF. Since 2 ≤ i < p, N_E/F(α_i− {√^p

a} ·i_E(b)−i_E(f)) = 0. By Hilbert 90, there existsω ∈K_nE such that

(σ−1)ω=α_i− {√^p

a} ·i_E(b)−i_E(f).

Then

(σ−1)²ω= (σ−1)α_i−i_E({ξ_p} ·i_E(b)) = (σ−1)α_i = 0, and we can setα_i+1=ω. Observe that here we use our hypothesis

i_E({ξ_p} ·k_n−1F) ={0}.

Hence by induction there existsα_p∈K_nE such thatN α_p=(σ−1)^l−1γ. But i_EN_E/Fα_p= 0, whence (σ−1)^l−1γ= 0, a contradiction.

Lemma 7. Suppose thatp= 2. Then {a} ·ann_n−1{a,−1} ⊂N_E/Fk_nE.

Proof. Let β ∈ ann_n−1{a,−1}. Then {−1} ·β ∈ann_n{a} =N_E/Fk_nE by The- orem 2. Let γ ∈ K_nE such that {−1} ·β =N_E/F(γ). Then we have {a} ·β = N_E/F({√

a} ·i_E(β) +γ). Thus {a} ·ann_n−1{a,−1} ⊂N_E/Fk_nE.

Proof of Theorem 5. We consider first the casep >2.

(1) =⇒(3) Assume that k_nE is a trivial Fp[G]-module. Suppose f ∈K_n−1F, and set β = {√^p

a} ·i_E(f). Then (σ−1)β = 0 =⇒ i_E({ξ_p} ·f) = 0. But then by Theorem 2, {ξ_p} · f ∈ {a} ·k_n−1F. Now let γ ∈ K_nE be arbitrary.

Then i_EN_E/Fγ = (σ−1)^p−1γ = 0 and so by Theorem 2, N_E/Fγ = {a} ·f for f ∈ K_n−1F. By the projection formula, N_E/F({√^p

a} ·i_E(f)) = {a} ·f. Then N_E/F(γ− {√^p

a} ·i_E(f)) = 0, and henceN_E/F(γ− {√^p

a} ·i_E(f)) =pg, for some g∈K_n−1F. Setδ=γ− {√^p

a} ·i_E(f)−i_E(g). ThenN_E/F(δ) = 0. By Hilbert 90, there existsα∈K_nE such that (σ−1)α=δ. But since k_nE is fixed byG,δ= 0.

Hencek_nE=i_E(k_nF) +{√^p

a} ·i_E(k_n−1F).

(3) =⇒ (2) Since p > 2, N_E/F({√^p

a} ·i_E(f)) = {a} ·f for f ∈ K_n−1F, and N_E/F(i_E(g)) = 0 forg∈K_nF. HenceN_E/Fk_nE ={a} ·k_n−1F. Thus ann_n{a}= {a} ·k_n−1F.

(2) =⇒(1) Assume that{ξ_p} ·k_n−1F ⊂ {a} ·k_n−1F and ann_n{a}={a} ·k_n−1F. Hence{ξ_p}·k_n−1F ⊂N_E/Fk_nE={a}·k_n−1F. We then apply Lemma 6 to deduce thatk_nE= (k_nE)^G.

Now we consider the casep= 2:

(1) =⇒ (2) Assume that k_nE is a trivial F₂[G]-module. Let α ∈K_nE. Then i_EN_E/Fα = (σ−1)α = 0 implies that N_E/Fα = {a} ·b for some b ∈ K_n−1F. Now {a,−1} = {a, a} in k₂F, and then {a,−1} ·b = {a} ·N_E/Fα = 0. Hence ann_n{a} ⊂ {a} ·ann_n−1{a,−1}.

(2) =⇒ (1) Assume that ann_n{a} ⊂ {a} ·ann_n−1{a,−1}. Then N_E/Fk_nE ⊂ {a} ·ann_n−1{a,−1}. By Lemma 7, {a} ·ann_n−1{a,−1} ⊂ N_E/Fk_nE and hence N_E/Fk_nE={a}·ann_n−1{a,−1}. Letγ∈K_nEbe arbitrary. ThenN_E/Fγ={a}·b for someb∈ann_n−1{a,−1}. Hence (σ−1)γ=i_EN_E/Fγ=i_E({a} ·b) = 0. Hence (σ−1)γ= 0, and (k_nE)^G=k_nE.

Now assumep= 2 anda∈(F²+F²)\F². Then−1 =N_E/Fγfor someγ∈E^×. Thus forδ=γ√

awe haveN_E/Fδ=a. Then{δ} ∈(k₁E)^G andN_E/F{δ}={a}. (1) =⇒ (3) Let γ ∈ K_nE be arbitrary. Then, replacing √^p

a by δ and using N_E/F{δ}={a}in the proof of the (1) =⇒(3) portion of thep >2 case, we obtain γ∈i_E(k_nF) +{δ} ·i_E(k_n−1F).

(3) =⇒ (1). If β ∈ K_nE, then β = i_E(g) +{δ} ·i_E(f) for g ∈ K_nF and f ∈K_n−1F. Then (σ−1)β = 0, which implies that (k_nE)^G =k_nE.

Corollary 3. Suppose n ∈ N and (k_nE)^G = k_nE. Then we have the following exact sequence:

0→ann_n−1{a} →kn−1F −−−−−→^{a}·− knF −−→^iE knE−−−−→ {a} ·^NE/F ann_n−1{a, ξp} →0.

Proof. We first consider exactness at the fifth term. In thep= 2 case, Theorem5 tells us thatN_E/Fk_nE = ann_n{a} ⊂ {a} ·ann_n−1{a,−1}. By Lemma 7 we have the reverse inclusion, so thatN_E/Fk_nE ={a} ·ann_n−1{a,−1}. In thep >2 case, observe that{ξ_p} ·k_n−1F ⊂ {a} ·k_n−1F implies thatk_n−1F= ann_n−1{a, ξ_p}, since {a, a}= 0. Therefore, by part (2) of Theorem 5, we knowN_E/Fk_nE= ann_n{a}= {a} ·ann_n−1{a, ξ_p}. Hence the sequence is exact at the fifth term in thep >2 case as well.

For exactness at the fourth term, supposeγ∈K_nEandN_E/Fγ= 0. Then there existsf ∈K_nF such thatN_E/Fγ=pf, and thenN_E/F(γ−i_E(f)) = 0. By Hilbert 90, there existsα∈K_nEsuch that (σ−1)α=γ−i_E(f). But (σ−1)α= 0 because (k_nE)^G =k_nE. Henceγ=i_E(f). Since exactness in the first two terms is obvious and exactness at the third term follows from Theorem2, our proof is complete.

5. Hereditary triviality

Theorem 6. TrivialFp[G]-module cohomology is hereditary: ifn∈N, then for all m≥n,Hⁿ(E)^G=Hⁿ(E) =⇒ H^m(E)^G=H^m(E).

Proof. In thep >2 case, the result on heredity follows from Theorem5, part (3), together with two hereditary properties from Lemma 2: item (4), withα₁={ξ_p} andα₂={a}, and item (5).

In the p= 2 case, since m > n ≥ 1, by [BT, Cor. 5.3] K_mE is generated by symbols {u, f₁, . . . , f_m−1} where u ∈ E^× and f_i ∈ F^× for all i = 1, . . . , m−1.

Using the projection formula and a straightforward induction onm, we prove that ifHⁿ(E)^G =Hⁿ(E) then ann_m{a} ⊂ {a} ·ann_m−1{a,−1}. Hence by Theorem 5

we conclude thatH^m(E)^G =H^m(E) as well.

6. Examples of trivial cohomology

Define ct(E/F)∈ {0} ∪N∪ {∞}by

ct(E/F) = sup{n∈N∪ {0} |Hⁿ(E) is not a trivialF_p[G]-module}, where we set sup∅= 0.

Theorem 7. Given1≤n≤m∈N∪ {∞}, or 0 =n < m∈N∪ {∞}, and a prime p, there exists a cyclic extensionE/F of degreepwithξ_p∈F such that:

(1) G_F is a pro-p-group;

(2) ct(E/F) =n; and (3) cd(G_E) =m.

Proof. The casem∈N:

(1) Let F := F_n,m be a field of characteristic 0 with G_F ∼=Zⁿ_p pro-pZ^m_p and ξ_p2 ∈ F, given by Lemma 5. (Observe that if n = 0 then the first factor is {1}.) Let E =F(√^p

a) wherea∈F^× such that under the restriction map on H¹,