Cryptography Reference
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(4) By (1) and (3),
1= e n ( S + T,S + T )= e n ( S, S ) e n ( S, T ) e n ( T,S ) e n ( T,T )
= e n ( S, T ) e n ( T,S ) .
Therefore e n ( T,S )= e n ( S, T ) 1 .
(5) Let σ be an automorphism of K such that σ is the identity on the
coe cients of E . Apply σ to everything in the construction of e n .Then
div( f σ )= n [ σT ]
n [
]
and similarly for g σ ,where f σ and g σ denote the functions obtained by ap-
plying σ to the coecients of the rational functions defining f and g (cf.
Section 8.9). Therefore,
σ ( e n ( S, T )) = σ g ( P + S )
g ( P )
= g σ ( σP + σS )
g σ ( σP )
= e n ( σS, σT ) .
(6) Let
{
Q 1 ,...,Q k }
=Ker( α ).
Since α is a separable morphism, k =
deg( α ) by Proposition 2.21. Let
div( f T )= n [ T ]
n [
] ,
div( f α ( T ) )= n [ α ( T )]
n [
]
and
g T = f T ◦ n, g α ( T ) = f α ( T ) ◦ n.
As in (3), let τ Q denote adding Q .Wehave
div( f T ◦ τ −Q i )= n [ T + Q i ] − n [ Q i ] .
Therefore,
[ T ] − n
α ( Q )=
div( f α ( T ) ◦ α )= n
[ Q ]
α ( T )= α ( T )
= n
i
([ T + Q i ]
[ Q i ])
=div(
i
( f T ◦ τ −Q i )) .
For each i , choose Q i with nQ i = Q i .Then
g T ( P − Q i ) n = f T ( nP − Q i ) .
Consequently,
div
i
( g T ◦ τ −Q i ) n =div(
i
f T ◦ τ −Q i ◦ n )
=div( f α ( T )
α
n )
=div( f α ( T )
α )
=div( g α ( T ) ◦ α ) n .
n
 
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