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where
b
j
=
V
(
a
j
)
.If
b
j
=0
,then
d
j
=1
.
PROOF
Let
a ∈ K
.
Consider the function given by the polyn
omia
l
H
(
x, y
)=
x−a
.If
f
(
a
)
= 0
, this function has simple zeros at
P
=(
a,
f
(
a
))
and at
w
(
P
)=(
a, −
f
(
a
)). Theonlypossiblepoleof
x − a
is at
∞
.Since
the number of zeros equals the number of poles (Proposition 11.1), there is a
double pole at
∞
. Therefore,
div(
x − a
)=[
P
]+[
w
(
P
)]
−
2[
∞
]
.
If
f
(
a
)=0,then
a
f
(
x
)
.
Since
f
(
x
) has no multiple roots, (
x−a
)
/f
(
x
) does not have a zero or a pole at
(
a,
0). Therefore,
x−a
has a double zero at (
a,
0). Note that
w
(
a,
0) = (
a,
0),
so we can also write div(
x
x − a
=
y
2
x
−
−
a
) in the form [
P
]+[
w
(
P
)]
−
2[
∞
] in this case.
If
A
(
x
)=
j
(
x
a
j
)
c
j
, then div(
A
(
x
)) =
j
c
j
([
P
j
]+[
w
(
P
j
)]
−
−
2[
∞
]),
where
P
j
=
a
j
,
f
(
a
j
)
(for either choice of sign for the square root). We
will use this for polynomials
A
(
x
), but it also applies when
c
j
<
0 is allowed,
hence when
A
(
x
) is a rational function.
Consider now a function of the form
y
V
(
x
), where
V
(
x
) is a polynomial.
Let
P
=(
a, b
)beapointon
C
with
b
= 0. Assume that
y − V
(
x
) has a zero
at
P
,so
V
(
a
)=
b
.Since
b
=0,wehave
V
(
a
)+
b
= 0, so the function
y
+
V
(
x
)
does not have a zero at
P
. Therefore, the order of vanishing of
y − V
(
x
)at
P
is the same as the order of vanishing of
y
+
V
(
x
)
y − V
(
x
)
=
y
2
−
− V
(
x
)
2
=
f
(
x
)
− V
(
x
)
2
.
We conclude that, when
b
=0and
b
=
V
(
a
), the coecient of (
a, b
)in
div(
y − V
) equals the multiplicity of
x − a
in the factorization of
f − V
2
.
Now suppose (
a,
0) is a point on
C
at which
y − V
(
x
) has a zero. This
means that
f
(
a
)=0and
V
(
a
) = 0. Since the function
x−a
has a double zero
at (
a,
0), the function
V
(
x
) has at least a double zero at (
a,
0). But
y
has a
simple zero at (
a,
0), so the function
y −V
(
x
) has only a simple zero at (
a,
0).
Suppose (
x
a
)
2
is
afactorof
V
(
x
)
2
,itisalsoafactorof
f
(
x
), which is not possible since
f
(
x
)
has no multiple roots. Therefore, the polynomial
f
(
x
)
a
)
2
is a factor of the polynomial
f
(
x
)
V
(
x
)
2
.Since(
x
−
−
−
V
(
x
)
2
a
as
a simple factor. In other words, if
V
(
a
)=0and(
a,
0) is on
C
, the divisor of
y
−
has
x
−
V
(
x
)
2
−
V
(
x
)contains[(
a,
0)] with coe
cient 1, and the polynomial
f
(
x
)
−
has
x − a
as a simple factor.
So far, we have proved that every zero of
y − V
gives a root of
f − V
2
.We
need to show that
f − V
2
has no other roots. Write
V
)=
j
div(
y
−
d
j
([(
a
j
,b
j
)]
−
[
∞
])
.
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