Cryptography Reference
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An attractive feature of supersingular curves is that computations involving
an integer times a point can sometimes be done faster than might be expected.
Suppose E is a supersingular elliptic curve defined over F q and let P =( x, y )
be a point in E ( F q n )forsome n
1. Usually n is large. Let k be a positive
integer. We want to compute kP . This can be done quickly by successive
doubling, but it is possible to do even better. Let's assume that a =0. Then
φ q + q =0
by Theorem 4.10. Therefore
q ( x, y )= −φ q ( x, y )= x q 2 , −y q 2 .
The calculations of x q 2 and y q 2 involve finite field arithmetic, which is gener-
ally faster than elliptic curve calculations. Moreover, if x and y are expressed
in terms of a normal basis of F q n over F q ,then x q 2 and y q 2 are computed by
shift operations (see Appendix C). The procedure is now as follows:
1. Expand k in base q :
k = k 0 + k 1 q + k 2 q 2 + ··· + k r q r ,
with 0 ≤ k i <q .
2. Compute k i P =( x i ,y i )foreach i .
3. Compute q i k i P =( x q 2 i
1) i y q 2 i
, (
).
i
i
4. Sum the points q i k i P for 0 ≤ i ≤ r .
The main savings is in step (3), where elliptic curve calculations are replaced
by finite field computations.
We now show how to obtain all supersingular curves over F q . Note that
supersingularity means that there are no points of order p with coordinates
in the algebraic closure; hence, it is really a property of an elliptic curve over
an algebraically closed field. If we have two elliptic curves E 1 and E 2 defined
over a field such that E 1 can be transformed into E 2 by a change of variables
defined over some extension field, then E 1 is supersingular if and only if E 2
is supersingular.
For example, in Proposition 4.33, the curve y 1 = x 1 + B can be changed
into y 2 = x 2 + 1 via x 2 = x 1 /B 1 / 3 ,y 2 = y 1 /B 1 / 2 . Therefore, it would have
suced to prove the proposition for the curve y 2 = x 3 +1.
Recall (Section 2.5.1) that an elliptic curve E over an algebraically closed
field of characteristic not 2 can be put into the Legendre form y 2
= x ( x −
1)( x − λ )with λ =0 , 1.
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