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all) divisors
D
in the divisor class. These correspond to the points on the
Jacobian variety that are defined over
F
q
. We denote this set by
J
(
F
q
).
Suppose
D
is a divisor of degree 0 such that
φ
(
D
) is in the same divisor class
as
D
. The divisor class of
D
contains a unique reduced divisor
R
, and the
divisor class of
φ
(
D
)contains
φ
(
R
) (proof:
D
φ
(
R
)=
div(
φ
(
F
))), and
φ
(
R
) is also reduced. The uniqueness implies that
R
=
φ
(
R
).
Therefore, the divisor class contains a divisor fixed by
φ
. The reduced divisor
R
corresponds to a unique pair (
U, V
) of polynomials. The divisor
φ
(
R
)
corresponds to the pair (
U
φ
,V
φ
), where
U
φ
denotes the polynomial obtained
by applying
φ
to the coecients of
U
.Since
φ
(
R
)=
R
,wehave
U
φ
=
U
and
V
φ
=
V
, because the pair corresponding to a divisor is unique. It follows
that a divisor class that is mapped to itself by
φ
corresponds to a unique
pair (
U, V
) of polynomials, with
U, V ∈
F
q
[
x
] (= the set of polynomials with
coe
cients in
F
q
). Conversely, if
U, V ∈
F
q
[
x
], then gcd (div(
U
)
,
div(
y − V
))
is fixed by
φ
, hence yields a divisor class fixed by
φ
. Wehaveprovedthe
following.
−
R
=div(
F
), so
φ
(
D
)
−
PROPOSITION 13.11
Thereisaone-to-one correspondence between
J
(
F
q
)
and pairs
(
U, V
)
ofpoly-
nom ialswith coe cientsin
F
q
satisfying
1.
U
ismonic.
2.
deg
V<
deg
U
≤
g
.
3.
V
2
− f
(
x
)
isamultipleof
U
.
Since there are only finitely many polynomials
U ∈
F
q
[
x
]ofdegreeatmost
g
, there are only finitely many divisor classes of degree 0 that are defined over
F
q
. In other words,
J
(
F
q
) is finite. It is easy to see that it is closed under
addition, so it forms a group. Alternatively, Cantor's algorithm clearly takes
pairs of polynomials defined over
F
q
to pairs defined over
F
q
.Infact,we
could ignore the geometry completely and consider a group whose elements
are suitable pairs of polynomials and whose law of composition is given by
Cantor's algorithm. This defines a group (although the associativity might
be di
cult to prove without the geometric interpretation).
Example 13.2
Let's consider the case where deg
U
=1. Then
U
=
x − a
for some
a
,and
V
=
b
for some
b ∈
F
q
.Also,
f
(
x
)
≡ f
(
a
)(mod
x − a
), so
b
2
=
V
2
≡ f
(
a
),
hence
b
2
=
f
(
a
). This means that (
a, b
) is a point on the curve. The divisor
class for (
U, V
) is defined over
F
q
if and only if the polynomials
x − a
and
b
have coe
cients in
F
q
, which happens if and only if the point (
a, b
)has
coordinates in
F
q
.
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