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

+ 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