of PF 1 corresponding to the irreducible polynomial x2 (℘(r) + b)/a, and therefore
are split points.
We study summands of δ one at a time. There are two steps.
Step 1. For the first step we assume N is a non-split point with π(N) = N cor-
responding to p(x) where the degree of p(x) is d 2. By Lemma 4.2, we know
that when d is even, K1F (N) = K1F (N)[y] is generated by elements of the form
{f0(x) + yf1(x)} where deg(f0(x)) d
2
and deg(f1(x)) d
2
1, and when d
is odd we have such generators with deg(f0(x))
d−1
2
and deg(f1(x))
d−1
2
.
We call such generators low-degree generators. Because F (N)/F (N) is quadratic,
Km−1F (N) = K1F (N) Km−2F (N). Let η = {f0(x) + yf1(x)} η0 where
{f0(x) + yf1(x)} is a low-degree generator and η0 Km−2F (N) is a sum symbols
{g1,j(x), g2,j (x), . . . gm−2,j (x)} with each gi,j (x) F [x] having degree less than d.
Such η generate Km−1F (N) and for such an (N; η) we can express
dF ({p(x)} {f0(x) + yf1(x)} η0) = (N; η) +
i
(Si; γSi ) +
j
(Ti, γTi )
where the Si X are points with π(Si) corresponding to irreducible factors of
f0(x)2
+ f0(x)f1(x) +
(abx2
+
b)f1(x)2
and the Ti X are points with π(Ti) corre-
sponding to irreducible factors of the gi,j (x). This means the π-degrees of each Ti
are less than d.
Now,
deg(f0(x)2 +f0(x)f1(x)+(abx2 +b)f1(x)2)
d when d is even and d+1
when d is odd. Moreover, if qi(x) is one of its irreducible factors of degree d or d +1
then we have that h = f0(x) · f1(x)
−1
F [x]/(qi(x)) satisfies ℘(h) = b + abx2.
So Si corresponding to such a qi(x) is a split point. This observation shows that
when N is a non-split point of degree d 2, we can rewrite any (N, η) modulo the
image im(dF : KmF (x)
p∈X
Km−1F (p)) as a sum of elements with support
over split points of degree at most d when d is even or at most degree d + 1 when
d is odd, together with other points of π-degree strictly less than d.
Step 2. Now suppose that S is a split point on X of π-degree d 2. This means
that π(S) = S corresponds to an irreducible polynomial p(x) and
℘−1(b
+
abx2)

F [x]/(p(x)) = F (S). Since a, b]] = 0 WqF (S), by Springer’s Theorem it must
happen that the degree of p is even, say d = 2d0. We suppose r(x) F [x] and
℘(r(x)) = b +
abx2
F (S). Let Ve F [x] be the (e + 1)-dimensional subspace
of polynomials of degree at most e and view Ve F (S). Consider the linear map
T : Vd0 F (S) defined by T (f(x)) = r(x) · f(x). By dimension count there exists
nonzero f(x) Vd0 with T (f(x)) = g(x) Vd0−1. Then as p(x) |
f(x)2
+f(x)g(x)+
g(x)2(b
+
ax2)
and
deg(f(x)2
+ f(x)g(x) +
g(x)2(b
+
ax2))
2d0 = d we see that
p(x) =
λ(f(x)2
+ f(x)g(x) +
g(x)2(b
+
ax2))
for some scalar λ F .
As F (S) = F (S), for η Km−1F (S) we can express η as a sum of a product of
symbols {hi(x)} where by Lemma 4.1 we can assume hi(x) F [x] and deg(hi(x))
d/2. If we let η Km−1F (x) denote the same sum of symbols but where the hi(x)
are lifted to the polynomials hi(x), we then find that
d({f(x) + g(x)y} η) = (S; η) +
j
(Uj ; θj ))
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