t2
+ auri = a
(y +
β)2
℘(y) + b
+
(ri + β)2
℘(ri) + b
= a
(y2
+
β2)(℘(ri)
+ b) +
(ri2
+
β2)(℘(y)
+ b)
(℘(y) + b)(℘(ri) + b)
= a
yri(y + ri) +
b(y2
+
ri2)
+
β2℘(y
+ ri)
(℘(y) + b)(℘(ri) + b)
= a
(y + ri)(yri + b(y + ri) + β2(y + ri + 1)
(℘(y) + b)(℘(ri) + b)
= a
(y + ri)((β + ri)y + (bri +
β2ri)
+
β2)
(℘(y) + b)(℘(ri) + b)
= a
(y + ri)(β + ri)(y + β)
(℘(y) + b)(℘(ri) + b)
= a
(y + ri)(β + ri)
(y + β + 1)(℘(ri) + b)
where the second to last equality uses bri + β2ri = riβ. We note for use in the
proof of (ii) below that this gives
NL(X)/F
(X)(t2
+ auri ) =
a2
y + ri
℘(ri) + b
2
·
℘(ri) + b
℘(y) + b
=
a
℘(ri) + b
y + ri
x
2
.
For each i we set
αi,0 :=
t2
+ auri
(ri + β)2(
2
i
+ pri )
, mi
i
+
t
ri + β
K2L(X).
Then αi = αi,0 ⊗χi is as defined in Section 2. We compute differentials over L, that
is we compute the image dL(αi,0) for dL : K2L(X) ⊕q∈XL K1L(q). Considering
the parameter t, it follows that the only non-zero differentials of αi,0 can occur
where t =

auri , t = i(ri + β) and t = ∞. From the calculation of
t2
+ auri in
terms of y just given, we see that t =

auri corresponds to Qri lifted to XL. Here
we have the L-differential
∂Qri (αi,0) = mi
i
+
y + β
x(ri + β)
= mi
i
+
1
x
= mi
(
i
+

pri
)
.
As T∞ is the infinite point on XL with respect to the parameter t we have vT∞ (t) =
−1. Applying the tame symbol it follows that
∂T∞ (αi,0) =





t2
(ri+β)2( 2+pri
i
)
mit
ri+β
2





= {(mi
−2( 2
i
+ pri
)−1}.
There is only one other possible non-zero differential, namely at Nri where t =
i
(ri + β). As ΣN(αi,0) = 0 we see ∂Nri (αi,0) = 0. (This can also be checked
directly, for when t = (ri + β)
ri
we find
t2 + auri
(ri + β)2( 2
i
+ pri )
=
(ri +
β)2 2
i
+ a(ri +
β)2/(℘(ri)
+ b)
(ri + β)2( 2
i
+ pri )
= 1
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