Cryptography Reference
In-Depth Information
we can evaluate the right-hand side at
W
=
jU
for 0
≤
j<n
and multiply
theresultstoobtain
=
d
(
U, V
)
d
(
V,U
)
n
n
−
1
c
(
nU, nV
)
c
(
nV, nU
)
d
(
V,jU
)
d
(
U
+
jU,V
)
d
(
V,U
+
jU
)
d
(
jU,V
)
.
=
j
=0
All the factors in this product cancel except some of those for
j
=0and
j
=
n
−
1. We obtain
c
(
S, T
)
c
(
T,S
)
=
)
d
(
nU, V
)
d
(
V,nU
)
d
(
∞,V
)
.
d
(
V,
∞
In the first equation of Lemma 11.22, set
W
=
∞
to obtain
d
(
V,nU
)=
d
(
V,
∞
). In the second equation of Lemma 11.22, set
V
=
∞
and then set
W
=
V
to obtain
d
(
nU, V
)=
d
(
∞
,V
)
e
(
nU, nV
). This yields the result.
We now proceed with the proof of the theorem. The definition of
c
shows
that
e
n
(
S, T
)=
c
(
S, T
)
c
(
T,S
)
=
f
T
(
X
)
f
S
(
X
T
)
f
S
(
X
)
f
T
(
X − S
)
,
−
(11.13)
which is independent of
X
.Let
D
S
=[
S
]
D
T
−
[
∞
]
,
=[
X
0
]
−
[
X
0
−
T
]
,
where
X
0
is chosen so that
D
S
and
D
T
are disjoint divisors. Let
F
S
(
X
)=
f
S
(
X
)and
F
T
(
X
)=1
/f
T
(
X
0
−
X
). Then
div(
F
S
)=
n
[
S
]
]=
nD
S
,
div(
F
T
)=
n
[
X
0
]
T
]=
nD
T
.
−
n
[
∞
−
n
[
X
0
−
Therefore, (11.13) yields
e
n
(
S, T
)=
F
T
(
D
S
)
F
S
(
D
T
)
.
This shows that the theorem is true for the choice of divisors
D
S
and
D
T
.
We now need to consider arbitrary choices. Let
D
S
be any divisor of degree
0 such that sum(
D
S
)=
S
and let
D
T
be any divisor of degree 0 such that
sum(
D
T
)=
T
.Then
D
S
=div(
h
1
)+
D
S
and
D
T
=div(
h
2
)+
D
T
for some
functions
h
1
,h
2
.Let
F
S
=
h
1
F
S
and
F
T
=
h
2
F
T
.Then
nD
S
=div(
F
S
)and
nD
T
=div(
F
T
). First, assume that the divisors
D
S
and
D
S
are disjoint from
D
T
and
D
T
.Then
h
1
(
D
T
)
n
F
S
(
D
T
)
=
h
2
(div(
h
1
))
n
h
2
(
D
S
)
n
F
T
(div(
h
1
))
F
T
(
D
S
)
h
2
(
D
S
)
n
F
T
(
D
S
)
F
T
(
D
S
)
F
S
(
D
T
)
=
h
1
(div(
h
2
))
n
h
1
(
D
T
)
n
F
S
(div(
h
2
))
F
S
(
D
T
)
.
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