Cryptography Reference
In-Depth Information
8.8 A Nontrivial Shafarevich-Tate Group
Let
E
be the elliptic curve over
Q
given by
y
2
=
x
(
x −
2
p
)(
x
+2
p
)
,
where
p
is a prime. If we do a 2-descent on
E
, we encounter the equations
x
=
u
2
x −
2
p
=
pv
2
x
+2
p
=
pw
2
.
These yield the curve defined by the intersection of two quadratic surfaces:
C
1
,p,p
:
u
2
pv
2
=2
p,
u
2
pw
2
=
−
−
−
2
p.
(8.14)
THEOREM 8.28
If
p
≡
9 (mod 16)
,then
C
1
,p,p
has
q
-adicpointsfor allprimes
q
≤∞
,but
has no rationalpoints.
PROOF
First, we'll show that there are no rational points. Suppose there
is a rational point (
u, v, w
). We may assume that
u, v, w >
0. If
p
divides
the denominator of
v
, then an odd power of
p
is in the denominator of
pv
2
andanevenpowerof
p
is in the denominator of
u
2
,so
u
2
pv
2
cannot be
an integer, contradiction. Therefore,
u, v
, and hence also
w
have no
p
in their
denominators. It follows easily that the denominators of
u, v, w
are equal.
Since
u
2
=2
p
+
pv
2
,wehave
u ≡
0(mod
p
). Write
−
u
=
pr
v
=
s
t
e
,
e
,
e
,
w
=
with positive integers
r, s, t, e
and with
gcd(
r, e
)=gcd(
s, e
)=gcd(
t, e
)=1
.
The equations for
C
1
,p,p
become
pr
2
− s
2
=2
e
2
,
2
− t
2
=
−
2
e
2
.
Subtracting yields
s
2
+4
e
2
=
t
2
.
If
s
is even, then
pr
2
=
s
2
+2
e
2
is even, so
r
is even. Then 2
e
2
=
pr
2
−
s
2
≡
0 (mod 4), which implies that
e
is even. This contradicts the fact that
gcd(
s, e
) = 1. Therefore,
s
is odd, so
gcd(
s,
2
e
)=1
.
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