σ1 = iL/F (θ) + 2σ2 ∈ KmL where θ ∈ Tm and σ2 ∈ KmL. This means that we can

express iL(X)/F

(X)

(σ) = iL(X)/F (σ6 +θ)+2(iL(X)/L(σ2)+σ7). We set θ = σ6 +θ ∈

Tm and then note that iL(X)/F

(X)

(σ − iF

(X)/F

(θ )) ∈ 2KmL(X). This means the

class σ − iF

(X)/F

(θ ) ∈ νF

(X)

(m) lies in the kernel ker(νF

(X)

(m) → νL(X)(m)).

However, νF

(X)

(m) → νL(X)(m) is injective as L/F is separable and we see that

σ − iF

(X)/F

(θ ) ∈ 2KmF (X). The lemma follows.

3. Proofs of the Main Theorems.

Most of this section involves recording a diagram chase in Diagram 1.2. For

this we need to deﬁne a collection of quotient groups. As

H0

1(X,

ν(m)) := ker

(

H1(X,

ν(m)) → H0

1(XL,

ν(m))

)

,

by Lemma 1.7 we see that H0

1(X,

ν(m)) is the subgroup of elements that map to

zero under ΣN : ⊕q∈XL νL(q)(m − 1) → νL(m − 1). We denote by

νL(X)(m) :=

νL(X)(m)

(iL(X)/F

(X)

νF

(X)

(m) + iL(X)/LνL(m))

and we denote by νL(X)(m)0 the subgroup of elements of νL(X)(m) that map to

zero under the composite

νL(X)(m)

NL(X)/F

(X)

−→ νF

(X)

(m)

dF

−→ ⊕p∈XνF

(p)

(m − 1)

in Diagram 1.2. One checks that elements in iL(X)/F

(X)

νF

(X)

(m) + iL(X)/LνL(m)

vanish under this composite so νL(X)(m)0 is well-deﬁned.

Lemma 3.1. Chasing in Diagram 1.2 gives a well-deﬁned homomorphism

ψ1 : H0

1(X,

ν(m)) → νL(X)(m)0

deﬁned by ψ1([γ]) = [τ ] where dL(τ ) = iL(X)/F (X)(γ).

Proof. We suppose that γ1, γ2 ∈ ⊕p∈XνF

(p)

(m − 1) represent the same class in

H0

1(X,

ν(m)). This means that γ1 − γ2 = dF (τ0) for τ0 ∈ νF

(X)

(m). Suppose

τ1, τ2 ∈ νL(X)(m) satisfy dL(τ1) = iL/F (γ1) and dL(τ2) = iL/F (γ2). Then dL(τ1 −

τ2) = iL/F (γ1) − iL/F (γ2) = iL/F (dF (τ0)) = dL(iL/F (τ0)). Taken together, dL(τ1 −

τ2 − iL/F (τ0)) = 0 so by Lemma 1.7 τ1 − τ2 − iL/F (τ0) = iL(X)/L(τ3) for some

τ3 ∈ νL(m). As τ1−τ2 = iL/F (τ0)+iL(X)/L(τ3) it follows that [τ1] = [τ2] ∈ νL(X)(m)

which shows that ψ1 is well-deﬁned. Finally we note that such [τ1] ∈ νL(X)(m)

actually lies in νL(X)(m)0 because of exactness of the middle column of Diagram

1.2. The lemma follows.

Lemma 3.2. The map ψ1 : H0 1(X, ν(m)) → νL(X)(m)0 in Lemma 3.1 is an

isomorphism.

Proof. Suppose that ψ1([γ]) = 0 for [γ] ∈ H0

1(X,

ν(m)). Then τ0 ∈ νL(m) and

τ1 ∈ νF

(X)

we know that iL/F (γ) = dL

(

iL(X)/L(τ0) + iL/F (τ1)

)for

∈ ⊕q∈XL νL(q)(m−

1). As dL(iL(X)/L(τ0)) = 0 we have iL/F (γ) − dL(iL/F (τ1)) = 0 so iL/F (γ −

dF (τ1)) = 0. However, as L is separable over F we have that ⊕p∈XνF (p)(m − 1) →

⊕q∈XL νL(q)(m − 1) is injective. It follows that γ = dF (τ1), that is, [γ] = 0 ∈

H1(X, ν(m)). This shows that ψ1 is injective.

express iL(X)/F

(X)

(σ) = iL(X)/F (σ6 +θ)+2(iL(X)/L(σ2)+σ7). We set θ = σ6 +θ ∈

Tm and then note that iL(X)/F

(X)

(σ − iF

(X)/F

(θ )) ∈ 2KmL(X). This means the

class σ − iF

(X)/F

(θ ) ∈ νF

(X)

(m) lies in the kernel ker(νF

(X)

(m) → νL(X)(m)).

However, νF

(X)

(m) → νL(X)(m) is injective as L/F is separable and we see that

σ − iF

(X)/F

(θ ) ∈ 2KmF (X). The lemma follows.

3. Proofs of the Main Theorems.

Most of this section involves recording a diagram chase in Diagram 1.2. For

this we need to deﬁne a collection of quotient groups. As

H0

1(X,

ν(m)) := ker

(

H1(X,

ν(m)) → H0

1(XL,

ν(m))

)

,

by Lemma 1.7 we see that H0

1(X,

ν(m)) is the subgroup of elements that map to

zero under ΣN : ⊕q∈XL νL(q)(m − 1) → νL(m − 1). We denote by

νL(X)(m) :=

νL(X)(m)

(iL(X)/F

(X)

νF

(X)

(m) + iL(X)/LνL(m))

and we denote by νL(X)(m)0 the subgroup of elements of νL(X)(m) that map to

zero under the composite

νL(X)(m)

NL(X)/F

(X)

−→ νF

(X)

(m)

dF

−→ ⊕p∈XνF

(p)

(m − 1)

in Diagram 1.2. One checks that elements in iL(X)/F

(X)

νF

(X)

(m) + iL(X)/LνL(m)

vanish under this composite so νL(X)(m)0 is well-deﬁned.

Lemma 3.1. Chasing in Diagram 1.2 gives a well-deﬁned homomorphism

ψ1 : H0

1(X,

ν(m)) → νL(X)(m)0

deﬁned by ψ1([γ]) = [τ ] where dL(τ ) = iL(X)/F (X)(γ).

Proof. We suppose that γ1, γ2 ∈ ⊕p∈XνF

(p)

(m − 1) represent the same class in

H0

1(X,

ν(m)). This means that γ1 − γ2 = dF (τ0) for τ0 ∈ νF

(X)

(m). Suppose

τ1, τ2 ∈ νL(X)(m) satisfy dL(τ1) = iL/F (γ1) and dL(τ2) = iL/F (γ2). Then dL(τ1 −

τ2) = iL/F (γ1) − iL/F (γ2) = iL/F (dF (τ0)) = dL(iL/F (τ0)). Taken together, dL(τ1 −

τ2 − iL/F (τ0)) = 0 so by Lemma 1.7 τ1 − τ2 − iL/F (τ0) = iL(X)/L(τ3) for some

τ3 ∈ νL(m). As τ1−τ2 = iL/F (τ0)+iL(X)/L(τ3) it follows that [τ1] = [τ2] ∈ νL(X)(m)

which shows that ψ1 is well-deﬁned. Finally we note that such [τ1] ∈ νL(X)(m)

actually lies in νL(X)(m)0 because of exactness of the middle column of Diagram

1.2. The lemma follows.

Lemma 3.2. The map ψ1 : H0 1(X, ν(m)) → νL(X)(m)0 in Lemma 3.1 is an

isomorphism.

Proof. Suppose that ψ1([γ]) = 0 for [γ] ∈ H0

1(X,

ν(m)). Then τ0 ∈ νL(m) and

τ1 ∈ νF

(X)

we know that iL/F (γ) = dL

(

iL(X)/L(τ0) + iL/F (τ1)

)for

∈ ⊕q∈XL νL(q)(m−

1). As dL(iL(X)/L(τ0)) = 0 we have iL/F (γ) − dL(iL/F (τ1)) = 0 so iL/F (γ −

dF (τ1)) = 0. However, as L is separable over F we have that ⊕p∈XνF (p)(m − 1) →

⊕q∈XL νL(q)(m − 1) is injective. It follows that γ = dF (τ1), that is, [γ] = 0 ∈

H1(X, ν(m)). This shows that ψ1 is injective.