σLs ≡

s

i=1

(

{(y + ri), mi

2

(

2

i

+ pri )} ⊗ χi

)

+

s

i=1

(

{ i, mi

2( 2

i

+ pri )} ⊗ χi

)

≡

s

i=1

(

{ i(y + ri), mi

2( 2

i

+ pri )} ⊗ χi

)

(mod (iLs(X)/F (Tm) + 2iLs(X)/F

(X)

KmF (X))).

Using the injectivity of KmL(X) → KmLs ([Ls : L(X)] is a power of 2 so this is a

consequence of Izboldin’s Theorem) we ﬁnd we ﬁnd that this congruence is in fact

valid (mod (iL(X)/F (Tm) + 2iL(X)/F

(X)

KmF (X))).

We next recall that t = (y+β)/x and that

t2

= ab+t/x so we have x =

t/(t2

+a)

and y = β + tx = β +

t2/(t2

+ a). Therefore for each i = 0

y + ri = ri + β +

t2

t2 + a

=

t2 + (t2 + a)(ri + β)

t2 + a

=

(ri + β +

1)t2

+ a(ri + β)

t2 + a

=

(ri + β +

1)(t2

+ a

(ri+β)2

℘(ri)+b

)

t2 + a

As

t2

+ a

(ri+β)2

℘(ri)+b

∈

(L(X))2(pri

) we ﬁnd for i ≥ 1 that each

{t2

+ a

(ri +

β)2

℘(ri) + b

, mi

2( 2

i

+ pri )} ∈ 2K2L(X),

so in particular we ﬁnd

∑s

i=1

{t2

+ a

(ri+β)2

℘(ri)+b

, mi

2( 2

i

+ pri )}⊗ χi ∈ 2KmL(X). Again

as

∑s

i=1

(

{mi

2( 2

i

+ pri )} ⊗ χi

)

= 2χ∞ ∈ 2Km−1L we see

{t2

+ a} ⊗

s

i=1

(

{mi

2( 2

i

+ pri )} ⊗ χi

)

∈ 2KmL(X).

Altogether, substituting

(ri+β+1)(t2+a

(ri+β)2

℘(ri)+b

)

t2+a

in place of y + ri in the previous

expression for σLs and again using the injectivity of KmL(X) → KmLs(X) gives

σL(X) ≡

s

i=1

(

{ i(ri + 1 + β), mi

2( 2

i

+ pri )} ⊗ χi

)

(mod (iL(X)/F

(X)

(Tm) + 2KmL(X))).

This proves the lemma.

References

[A] Arason, J., Cohomologische Invarianten quadratischer Formen, J. Algebra, 36 (1975), 448-

491

s

i=1

(

{(y + ri), mi

2

(

2

i

+ pri )} ⊗ χi

)

+

s

i=1

(

{ i, mi

2( 2

i

+ pri )} ⊗ χi

)

≡

s

i=1

(

{ i(y + ri), mi

2( 2

i

+ pri )} ⊗ χi

)

(mod (iLs(X)/F (Tm) + 2iLs(X)/F

(X)

KmF (X))).

Using the injectivity of KmL(X) → KmLs ([Ls : L(X)] is a power of 2 so this is a

consequence of Izboldin’s Theorem) we ﬁnd we ﬁnd that this congruence is in fact

valid (mod (iL(X)/F (Tm) + 2iL(X)/F

(X)

KmF (X))).

We next recall that t = (y+β)/x and that

t2

= ab+t/x so we have x =

t/(t2

+a)

and y = β + tx = β +

t2/(t2

+ a). Therefore for each i = 0

y + ri = ri + β +

t2

t2 + a

=

t2 + (t2 + a)(ri + β)

t2 + a

=

(ri + β +

1)t2

+ a(ri + β)

t2 + a

=

(ri + β +

1)(t2

+ a

(ri+β)2

℘(ri)+b

)

t2 + a

As

t2

+ a

(ri+β)2

℘(ri)+b

∈

(L(X))2(pri

) we ﬁnd for i ≥ 1 that each

{t2

+ a

(ri +

β)2

℘(ri) + b

, mi

2( 2

i

+ pri )} ∈ 2K2L(X),

so in particular we ﬁnd

∑s

i=1

{t2

+ a

(ri+β)2

℘(ri)+b

, mi

2( 2

i

+ pri )}⊗ χi ∈ 2KmL(X). Again

as

∑s

i=1

(

{mi

2( 2

i

+ pri )} ⊗ χi

)

= 2χ∞ ∈ 2Km−1L we see

{t2

+ a} ⊗

s

i=1

(

{mi

2( 2

i

+ pri )} ⊗ χi

)

∈ 2KmL(X).

Altogether, substituting

(ri+β+1)(t2+a

(ri+β)2

℘(ri)+b

)

t2+a

in place of y + ri in the previous

expression for σLs and again using the injectivity of KmL(X) → KmLs(X) gives

σL(X) ≡

s

i=1

(

{ i(ri + 1 + β), mi

2( 2

i

+ pri )} ⊗ χi

)

(mod (iL(X)/F

(X)

(Tm) + 2KmL(X))).

This proves the lemma.

References

[A] Arason, J., Cohomologische Invarianten quadratischer Formen, J. Algebra, 36 (1975), 448-

491