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
aτ
2
+
bτ
+
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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