Cryptography Reference
In-Depth Information
1. if
n
is odd,
f
n
=
P
n
(
℘
)
,where
P
n
(
X
)
isapolynom ialofdegree
(
n
2
−
1)
/
2
with eading coe cient
n
,
2. if
n
is even,
f
n
=
℘
P
n
(
℘
)
,where
P
n
(
X
)
isapo ynom ial of degree
(
n
2
−
4)
/
2
with eading coe cient
n/
2
.
Theexpansion of
f
n
at
0
is
f
n
(
z
)=
(
−
1)
n
+1
n
z
n
2
−
1
+
···
.
T he zeros of
f
n
are at the points
0
=
u ∈
(
C
/L
)[
n
]
,andthese are simp e
zeros.
PROOF
The product is over the nonzero
n
-torsion in
C
/L
.Since
℘
(
u
)=
℘
(
−u
), the factors for
u
and
−u
are equal. Suppose
n
is odd. Then
u
is
never congruent to
−u
mod
L
, so every factor in the product occurs twice.
Therefore,
f
n
can be taken to be
n
(
℘
(
z
)
−
℘
(
u
)), where we use only one
member of each pair (
u,
−
u
). This is clearly a polynomial in
℘
(
z
)ofdegree
(
n
2
1)
/
2 and leading coe
cient
n
.When
n
is even, there are three values
of
u
that are congruent to their negatives mod
L
,namely,
ω
j
/
2for
j
=1
,
2
,
3.
Since
−
(
℘
)
2
=4
j
(
℘ − ℘
(
ω
j
/
2))
,
these factors contribute
℘
/
2to
f
n
. The remaining factors can be paired up,
as in the case when
n
is odd, to obtain a polynomial in
℘
of degree (
n
2
−
4)
/
2
and leading coecient
n
. Therefore,
f
n
has the desired form.
Since
℘
(
z
)=
z
−
2
+
···
and
℘
(
z
)=
−
2
z
−
3
+
···
, we immediately obtain
the expansion of
f
n
at 0.
Clearly
f
n
has a zero at each nonzero
u ∈
(
C
/L
)[
n
]. There are
n
2
−
1such
points. Since the only pole mod
L
of
f
n
is one of order
n
2
1at
z
=0,and
since the number of zeros equals the number of poles (counting multiplicities),
these zeros must all be simple.
−
LEMMA 9.28
Let
n ≥
2
.Then
f
n
−
1
(
z
)
f
n
+1
(
z
)
f
n
(
z
)
2
℘
(
nz
)=
℘
(
z
)
−
.
PROOF
Let
g
(
z
)=
℘
(
nz
)
− ℘
(
z
). We'll show that
g
and
f
n−
1
f
n
+1
/f
n
have the same divisors.
The function
g
(
z
) has a double pole at each
u ∈
(
C
/L
)[
n
]with
u
=0. At
z
= 0, it has the expansion
1
n
2
z
2
−
1
z
2
+
··· ,
g
(
z
)=
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