have shown that when S is split of degree d ≥ 2 how to rewrite (S; η) as a sum of

elements where the degrees of the points are less than d and in fact are less than

d − 1 in case d 2.

Combining the non-split and split results we can obtain the ﬁrst statement

of the lemma as follows. Let Cd,e := {N ∈ X | N is non-split with π-degree at

most d} ∪ {S ∈ X | S is split with π-degree at most e} and set Sd,e = {δ ∈

⊕p∈XKm−1F (p) | supp(δ) ∈ Cd,e}. For d ≥ 2, Step 1 shows that any element in

Sd,d can is equivalent modulo the image of dF (KmF (X)) to an element of Sd−1,d+1.

Now, if d 1 Step 2 shows that any element in Sd−1,d+1 can is equivalent modulo

the image of dF (KmF (X)) to an element of Sd−1,d−1. So by induction we can

reduce to see that every element of ⊕p∈XKm−1F (p) is equivalent modulo the image

of dF (KmF (X)) to an element of S1,1. This gives the ﬁrst statement of the lemma.

For the second statement, it suﬃces to show that the summands of any element

with support among the Ps and Q∞ can be represented by an equivalent sum with

support among the Qr and Q∞. As each F (Ps) is a separable quadratic extension

of F we know that Km−1F (Ps) = K1F (Ps)⊗Km−2F . Therefore to prove the result

it suﬃces to show that any (Ps; {u + vy}) ∈ ⊕pK1F (p) where u, v ∈ F is equivalent

modulo dF K2F (X) to a sum with support in the Qr and Q∞. We now observe

that if v = 0, dF {x + s, u + vy} = (Ps; {u + vy}) − (Qu/v; {x + s}) + (Q∞; {x/y})

and that d{x + s, u} = (Ps; {u}) + (Q∞;

{u−1})

when v = 0. These calculations

show that the support of γ can be arranged to include only the Qr and possibly

Q∞.

Finally we show that by a change of variables we assume that no Qri = Q∞. If

F is ﬁnite, then F

2

= F and the results of this paper are trivial, so we assume F is

inﬁnite. Since the support among the Qr is ﬁnite, by some translation of the form

y → y + s we can assume that the support does not include Q0. Then the change

of variables ˜ y = b/y and ˜ x = y/x gives ˜2 y + ˜ y + b = ˜˜2 a x where ˜ a = ab. Under this

change of variables we have

˜

Q

r

↔ Qb/r whenever r = 0,

˜

Q

0

↔ Q∞ and

˜

Q

∞

↔ Q0.

In this way, as there is no support at Q0 there will be no support at

˜

Q

∞

in the

˜-˜-coordinates. x y This concludes the proof of Lemma 1.3.

We next prove Lemma 2.2. We recall ui = (ri +

β)2/(℘(ri)

+ b) ∈ L.

Lemma 2.2. For αi ∈ KmL(X) as deﬁned in Theorem 2.1 and where T∞

denotes the inﬁnite point of XL with respect to the parameter t = (y + β)/x, we

have:

(i) dL(αi) =

(

(Qri ; {mi(

i

+

√

pri )}) + (T∞; {(mi

2( 2

i

+ pri

))−1})

)

⊗ χi.

(ii) NLri

(X)/Fri (X)

(iLri

(X)/L(X)

αi)) ≡ 2{

√

pri (y+r)/x(℘(ri)+b), mi( i+

√

pri )}⊗χi

(mod iLri

(X)/F

(Tm) + 2iLri

(X)/F (X)

KmF (X)).

elements where the degrees of the points are less than d and in fact are less than

d − 1 in case d 2.

Combining the non-split and split results we can obtain the ﬁrst statement

of the lemma as follows. Let Cd,e := {N ∈ X | N is non-split with π-degree at

most d} ∪ {S ∈ X | S is split with π-degree at most e} and set Sd,e = {δ ∈

⊕p∈XKm−1F (p) | supp(δ) ∈ Cd,e}. For d ≥ 2, Step 1 shows that any element in

Sd,d can is equivalent modulo the image of dF (KmF (X)) to an element of Sd−1,d+1.

Now, if d 1 Step 2 shows that any element in Sd−1,d+1 can is equivalent modulo

the image of dF (KmF (X)) to an element of Sd−1,d−1. So by induction we can

reduce to see that every element of ⊕p∈XKm−1F (p) is equivalent modulo the image

of dF (KmF (X)) to an element of S1,1. This gives the ﬁrst statement of the lemma.

For the second statement, it suﬃces to show that the summands of any element

with support among the Ps and Q∞ can be represented by an equivalent sum with

support among the Qr and Q∞. As each F (Ps) is a separable quadratic extension

of F we know that Km−1F (Ps) = K1F (Ps)⊗Km−2F . Therefore to prove the result

it suﬃces to show that any (Ps; {u + vy}) ∈ ⊕pK1F (p) where u, v ∈ F is equivalent

modulo dF K2F (X) to a sum with support in the Qr and Q∞. We now observe

that if v = 0, dF {x + s, u + vy} = (Ps; {u + vy}) − (Qu/v; {x + s}) + (Q∞; {x/y})

and that d{x + s, u} = (Ps; {u}) + (Q∞;

{u−1})

when v = 0. These calculations

show that the support of γ can be arranged to include only the Qr and possibly

Q∞.

Finally we show that by a change of variables we assume that no Qri = Q∞. If

F is ﬁnite, then F

2

= F and the results of this paper are trivial, so we assume F is

inﬁnite. Since the support among the Qr is ﬁnite, by some translation of the form

y → y + s we can assume that the support does not include Q0. Then the change

of variables ˜ y = b/y and ˜ x = y/x gives ˜2 y + ˜ y + b = ˜˜2 a x where ˜ a = ab. Under this

change of variables we have

˜

Q

r

↔ Qb/r whenever r = 0,

˜

Q

0

↔ Q∞ and

˜

Q

∞

↔ Q0.

In this way, as there is no support at Q0 there will be no support at

˜

Q

∞

in the

˜-˜-coordinates. x y This concludes the proof of Lemma 1.3.

We next prove Lemma 2.2. We recall ui = (ri +

β)2/(℘(ri)

+ b) ∈ L.

Lemma 2.2. For αi ∈ KmL(X) as deﬁned in Theorem 2.1 and where T∞

denotes the inﬁnite point of XL with respect to the parameter t = (y + β)/x, we

have:

(i) dL(αi) =

(

(Qri ; {mi(

i

+

√

pri )}) + (T∞; {(mi

2( 2

i

+ pri

))−1})

)

⊗ χi.

(ii) NLri

(X)/Fri (X)

(iLri

(X)/L(X)

αi)) ≡ 2{

√

pri (y+r)/x(℘(ri)+b), mi( i+

√

pri )}⊗χi

(mod iLri

(X)/F

(Tm) + 2iLri

(X)/F (X)

KmF (X)).