αi,0 χi and χi Km−2F .
For part (ii) we compute the norm NL(X)/F
(X)
(αi,0). We express αi,0 := αi,1 +
αi,2 where
αi,1 :=
t2
+ auri
(ri + β)2(
2
i
+ pri )
, mi K2L(X), and
αi,2 :=
t2
+ auri
(ri + β)2(
2
i
+ pri )
,
i
+
t
ri + β
K2L(X).
By reciprocity, and as NL(X)/F
(X)
(t2
+ auri ) =
a
℘(ri)+b
(
y+ri
x
)2
we find that
NL(X)/F
(X)
(αi,1) = {a(℘(ri) + b), mi} + 2
y + ri
x(℘(ri) + b)3( 2
i
+ pri )
, mi
which lies in iL(X)/F (aNL/F (K2L)) + 2iL(X)/F (X)K2F (X).
The calculation of NL(X)/F (X)(αi,2) is more intricate. We recall Lri = L(

pri )
and Fri = F (

pri ). By definition, NL(X)/F (X)(t) = a and NL(X)/F (X)(uri ) = 1.
We set tri := t/(ri + β) and then NL(X)/F
(X)
(tri ) = a/(℘(ri) + b) = pri . Also
auri /(ri + β)2 = a(ri + β)2/(℘(ri) + b)(ri + β)2 = pri . We have
iLri
/L
(αi,2) =



t2
(ri+β)2
+
auri
(ri+β)2
2
i
+ pri
,
i
+
t
ri + β



=
t
2
ri
+ pri
2
i
+ pri
,
i
+ tri
= 2
t
ri
+

pri
i
+

pri
,
i
+ tri
Using the K2-identity that {z/w, z + w} = {z, w} over Lri = L(

pri ) gives
iLri
/L
(αi,2) = 2{tri +

pri ,
i
+

pri }.
Now NLri
(X)/Fri (X)
(tri +

pri ) = NLri
(X)/Fri (X)
(tri ) + NLri
(X)/Fri (X)
(

pri ) +
T rLri
(X)/Fri (X)
(tri

pri ) = pri + pri +

pri T rLri
(X)/Fri (X)
(tri ). One calculates
T rLri
(X)/Fri
(X)(tri ) = (y+ri)/x(℘(ri)+b) so NLri
(X)/Fri
(X)(tri +

pri ) =

pri (y+
ri)/x(℘(ri) + b) which gives that
NLri
(X)/Fri (X)
(αi,2 Lri ) = 2{

pri (y + ri)/x(℘(ri) + b),
ri
+

pri }.
From this calculation, (ii) follows. This proves Lemma 2.2.
To close we need to prove Lemma 2.4.
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))).
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