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
Tm and then note that iL(X)/F
(X)
iF
(X)/F
)) 2KmL(X). This means the
class σ iF
(X)/F
) 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
) 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]) = ] 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
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.
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