dL(
s
i=1
αi) =
s
i=1
(
(Qri ; {mi(
i
+

pri )}) + (T∞; {(mi
2( 2
i
+ pri
))−1})
)
χi
=
s
i=1
(Qri ; {mi(
i
+

pri )}) χi)
T∞;
s
i=1
(
{(mi
2( 2
i
+ pri ))} χi
)
The result follows from this.
Lemma 2.3 gives part (ii) of Theorem 2.1. For part (iii) the calculation of
NL(X)/F
(X)
) is somewhat more tricky, because we need to realize it as an element
of iF
(X)/F
(KmF ) up to some error terms. The next lemma gets this process started.
Lemma 2.4. Given the notation of Theorem 2.1 we have
iL(X)/F
(X)
(σ)
s
i=1
(
{ i(ri + 1 + β), mi
2( 2
i
+ pri )} χi
)
(mod (iL(X)/F
(X)
(Tm) + 2KmL(X))).
Proof. The calculation is given in Section 4.
The next lemma will enable us to rewrite σ as an image of an element from
KmF modulo 2KmF (X).
Lemma 2.5. Suppose that L is a separable quadratic extension of F and α
KnL. If NL/F (α) = for some γ KnF , then there exists KnL so that
α iL/F NL/F ( )) (mod 2KnL).
Proof. According to the Aravire-Baeza sequence, νF (n) νL(n) νF (n) is exact.
So the hypothesis shows we can express α = iL/F (δ)+2 for δ KnF and KnL.
As iL/F (δ) = α 2 we compute
NL/F (iL/F γ + NL/F ( )) = NL/F 2 iL/F (γ) + iL/F (NL/F ( )))
= NL/F (α) 2NL/F ( ) + 2NL/F ( )
= 0.
Since KnF KnL KnF is multiplication by 2, it is injective by Izboldin’s
Theorem, and therefore δ γ + NL/F ( ) = 0 KnF . Consequently, α = iL/F (δ) +
2 iL/F NL/F ( )) (mod 2KnL) as required.
This next lemma, where we apply Lemma 2.5, is where it becomes clear why a
computation in full K-theory is needed instead of only working with ν-groups.
Lemma 2.6. Suppose
∑s
i=1
(
{mi 2( 2
i
+ pri )} χi
)
= 2χ∞ KmF . Then
σ1 :=
s
i=1
(
{
i
(ri + 1 + β), mi
2( 2
i
+ pri )} χi
)
= iL/F (θ) + 2σ2 KmL
for some θ Tm and σ2 KmL.
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