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
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