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 =
⊕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∞;
when v = 0. These calculations
show that the support of γ can be arranged to include only the Qr and possibly
Finally we show that by a change of variables we assume that no Qri = Q∞. If
F is finite, then F
= 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
Qb/r whenever r = 0,
Q∞ and

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

in the
˜-˜-coordinates. x y This concludes the proof of Lemma 1.3.
We next prove Lemma 2.2. We recall ui = (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
(i) dL(αi) =
(Qri ; {mi(

pri )}) + (T∞; {(mi
2( 2
+ pri
(ii) NLri
(X)/Fri (X)
αi)) 2{

pri (y+r)/x(℘(ri)+b), mi( i+

pri )}⊗χi
(mod iLri
(Tm) + 2iLri
(X)/F (X)
KmF (X)).
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