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)

(τ ) 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) = 2γ for some γ ∈ KnF , then there exists ∈ KnL so that

α ≡ iL/F (γ − 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 (δ − γ + NL/F ( )) = NL/F (α − 2 − iL/F (γ) + iL/F (NL/F ( )))

= NL/F (α) − 2NL/F ( ) − 2γ + 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 (γ − 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.

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)

(τ ) 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) = 2γ for some γ ∈ KnF , then there exists ∈ KnL so that

α ≡ iL/F (γ − 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 (δ − γ + NL/F ( )) = NL/F (α − 2 − iL/F (γ) + iL/F (NL/F ( )))

= NL/F (α) − 2NL/F ( ) − 2γ + 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 (γ − 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.