Cryptography Reference
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ignoring the ambiguities (i.e., up to n th powers) in the definition of the terms
on the right side, since they cancel out.
Therefore, we see that computing the Weil pairing and computing the Tate-
Lichtenbaum pairing both reduce to finding a function f with
div( f )= n [ P + R ] − n [ R ]
for points P ∈ E [ n ]and R ∈ E and evaluating f ( Q 1 ) /f ( Q 2 )forpoints Q 1 ,Q 2 .
The following algorithm due to Victor Miller [83] shows how to do this e -
ciently.
The idea is to use successive doubling (see page 18) to get to n . But the
divisors j [ P + R ]
j [ R ]for j<n are not divisors of functions, so we introduce
the divisors
D j = j [ P + R ] − j [ R ] [ jP ]+[ ] .
(11.7)
Then D j is the divisor of a function, by Theorem 11.2:
div( f j )= D j .
(11.8)
Suppose we have computed f j ( Q 1 ) /f j ( Q 2 )and f k ( Q 1 ) /f k ( Q 2 ). We show
how to compute f j + k ( Q 1 ) /f j + k ( Q 2 ). Let
ax + by + c =0
be the line through jP and kP (the tangent line if jP = kP ), and let x + d =0
be the vertical line through ( j + k ) P . Then (see the proof of Theorem 11.2),
div ax + by + c
x + d
=[ jP ]+[ kP ]
[( j + k ) P ]
[
] .
Therefore,
div( f j + k )= D j + k = D j + D k +div ax + by + c
x + d
=div f j f k ax + by + c
.
x + d
This means that there exists a constant γ such that
f j + k = γf j f k ax + by + c
x + d
.
Therefore,
( ax + by + c ) / ( x + d )
| ( x,y )= Q 1
( ax + by + c ) / ( x + d ) | ( x,y )= Q 2
f j + k ( Q 1 )
f j + k ( Q 2 ) = f j ( Q 1 )
f k ( Q 1 )
f k ( Q 2 )
.
(11.9)
f j ( Q 2 )
We conclude that passing from f j and f k to f j + k can be done quite quickly.
 
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