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rational function that gives the same function as
R
(
x, y
)onpointsin
E
(
K
).
Therefore, we may assume that
R
(
x, y
)=
p
1
(
x
)+
p
2
(
x
)
y
p
3
(
x
)+
p
4
(
x
)
y
.
Moreover, we can rationalize the denominator by multiplying the numerator
and denominator by
p
3
− p
4
y
and then replacing
y
2
by
x
3
+
Ax
+
B
.This
yields
R
(
x, y
)=
q
1
(
x
)+
q
2
(
x
)
y
q
3
(
x
)
.
(2.10)
Consider an endomorphism given by
α
(
x, y
)=(
R
1
(
x, y
)
,R
2
(
x, y
))
,
as above. Since
α
is a homomorphism,
α
(
x, −y
)=
α
(
−
(
x, y
)) =
−α
(
x, y
)
.
This means that
R
1
(
x, −y
)=
R
1
(
x, y
) d
R
2
(
x, −y
)=
−R
2
(
x, y
)
.
Therefore, if
R
1
is written in the form (2.10), then
q
2
(
x
) = 0, and if
R
2
is
written in the form (2.10), then the corresponding
q
1
(
x
) = 0. Therefore, we
may assume that
α
(
x, y
)=(
r
1
(
x
)
,r
2
(
x
)
y
)
with rational functions
r
1
(
x
)
,r
2
(
x
).
We can now say what happens when one of the rational functions is not
defined at a point. Write
r
1
(
x
)=
p
(
x
)
/q
(
x
)
with polynomials
p
(
x
)and
q
(
x
) that do not have a common factor. If
q
(
x
)=0
for some point (
x, y
), then we assume that
α
(
x, y
)=
=0,then
Exercise 2.19 shows that
r
2
(
x
) is defined; hence the rational functions defining
α
are defined.
We define the
degree
of
α
to be
∞
.If
q
(
x
)
deg(
α
)=Max
{
deg
p
(
x
)
,
deg
q
(
x
)
}
if
α
is nontrivial. When
α
= 0, let deg(0) = 0. Define
α
=0tobea
separable
endomorphism if the derivative
r
1
(
x
) is not identically zero. This
is equivalent to saying that at least one of
p
(
x
)and
q
(
x
) is not identically
zero. See Exercise 2.22. (In characteristic 0, a nonconstant polynomial will
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