Cryptography Reference
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Ω 1
1
II
2 Ω 1
I
Ω 2
0
Figure 9.5
The Real Points on C/L
In the case of a skewed parallelogram, that is, when ω 2 is real and the
imaginary part of ω 1 is half of ω 2 , the real axis is mapped to the reals, but
the analogue of path (II) is not mapped to the reals. This corresponds to the
situation of Figure 2.1(b) on page 10.
9.4 Computing Periods
Suppose E is an elliptic curve over C . From Theorem 9.21, we know that
E corresponds to a lattice L = Z ω 1 + Z ω 2 via the doubly periodic functions
and , but how do we find the periods ω 1 and ω 2 ?
For simplicity, let's consider the case where E is defined over R and E [2]
E ( R ). Then the equation for E can be put in the form
y 2 =4 x 3
− g 2 x − g 3 =4( x − e 1 )( x − e 2 )( x − e 3 )wi h e 1 <e 2 <e 3 .
We may assume ω 2
( ω 1 ) > 0, as in
Figure 9.5. The graph of E is as in Figure 2.1(a) on page 10. The Weierstrass
-function and its derivative map C /L to E via
R with ω 2 > 0and ω 1
i R with
( x, y )=( ( z ) ,℘ ( z )) .
As z goes from 0 to ω 2 / 2, the function ( z ) takes on real values, starting with
x = . The first point of order two is encountered when z = ω 2 / 2. Which
point ( e i , 0) is it? The graph of the real points of E has two components. The
one connected to contains the point ( e 3 , 0) of order two, so x = ( z )must
run from to e 3 as z goes from 0 to ω 2 / 2. The expansion of ( z )starts
with the term 2 /z 3 , from which it follows that y = ( z ) < 0near z =0,
hence ( z ) < 0for0 <z<ω 2 / 2.
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