Cryptography Reference
In-Depth Information
Suppose there are constants
c
α
1
,c
α
2
su ch that
R
α
1
(
x
)
S
α
1
(
x
)
R
α
2
(
x
)
S
α
2
(
x
)
=
c
α
1
,
=
c
α
2
.
Then
R
α
3
(
x
)
S
α
3
(
x
)
=
c
α
1
+
c
α
2
.
PROOF
Let (
x
1
,y
1
)and(
x
2
,y
2
)bevariablepointson
E
.Write
(
x
3
,y
3
)=(
x
1
,y
1
)+(
x
2
,y
2
)
,
where
(
x
1
,y
1
)=
α
1
(
x, y
)
,
(
x
2
,y
2
)=
α
2
(
x, y
)
.
Then
x
3
and
y
3
are rational functions of
x
1
,y
1
,x
2
,y
2
,whichinturnare
rational functions of
x, y
. By Lemma 2.24, with (
u, v
)=(
x
2
,y
2
),
∂x
3
∂x
1
+
∂x
3
dy
1
dx
1
=
y
3
y
1
.
∂y
1
Similarly,
∂x
3
∂x
2
+
∂x
3
dy
2
dx
2
=
y
3
y
2
.
∂y
2
By assumption,
dx
j
dx
=
c
α
j
y
j
y
for
j
=1
,
2. By the chain rule,
dx
3
dx
∂x
3
∂x
1
dx
1
dx
+
∂x
3
∂y
1
dy
1
dx
1
dx
1
dx
+
∂x
3
∂x
2
dx
2
dx
+
∂x
3
∂y
2
dy
2
dx
2
dx
2
dx
=
y
3
y
1
y
1
y
c
α
1
+
y
3
y
2
=
y
c
α
2
y
2
=(
c
α
1
+
c
α
2
)
y
3
y
.
Dividing by
y
3
/y
yields the result.
REMARK 2.27
In terms of differentials (see the previous Remark), we
have (
dx/y
)
◦α
is a translation-invariant differential on
E
. Therefore it must be
a scalar multiple
c
α
dx/y
of
dx/y
. It follows that every nonzero endomorphism
α
satisfies the hypotheses of Lemma 2.26.
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