Cryptography Reference
In-Depth Information
This is th e j -i nvariant of an elliptic curve that has complex multiplication
by Z 1+ 17 2 . The others are j ( τ 2 ) ,j ( τ 3 ) ,j ( τ 4 ), which are given below,
along with j 1+ 1 2 , which corresponds to an elliptic curve with a larger
endomorphism ring. We can evaluate x numerically using Proposition 9.12.
This yields
j 1+ 171
2
=
694282057876536664 . 01228868670830742604436745364124466 ....
This number is an algebraic integer by Theorem 10.9. Suppose we want a
polynomial that has x as its root. One way to do this is to find the Galois
conjugates of x , namely, the other roots of a polynomial satisfied by x . We'll
show how to proc eed fo r this particular x , then describe the general method.
Let τ 0 =(1+ 171) / 2. Then
K = Q ( τ 0 )= Q (
171) = Q (
19) .
Let
R = Z 1+ 171
2
Z 1+ 19
2
=
O K .
The endomorphism ring of the lattice R ⊂ C is R . Asweshowedinthe
proof of Proposition 10.4, the Galois conjugates of j ( R )are j -invariants of
lattices with the same endomorphism ring, namely R . These have the form
j ( I ), where I is an ideal of R . However, I cannot be an ideal for any order
larger than R since then I has an endomorphism ring larger than R .
If I is an ideal of R , it has the form
I = γ ( Z τ + Z )
C × and some τ
for some γ
∈H
. By an appropriate change of basis, we can
assume τ
, the fundamental domain for SL 2 ( Z ) acting on the upper half
plane. See Proposition 9.15. As we saw in Equation 10.1, τ
∈F
K .Let
2 + + c =0 ,
with a, b, c ∈ Z . We may assume that gcd( a, b, c )=1andthat a> 0. The
fact that I is an ideal for R but not for any larger order can be shown to
imply that the discriminant is exactly
171:
b 2
4 ac = 171 .
(On the other hand, the polynomial X 2 + X +5 has a root τ =(1+ 19) / 2,
which corresponds to the ideal 3 O K ⊂ R . This is an ideal not only of R , but
also of O K .) The fact that τ ∈F means that
 
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