Cryptography Reference
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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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