Cryptography Reference
In-Depth Information
Taking the ratio of the expressions for
t
−
2
and
t
−
3
gives
3
3
t
3
=
t
1
+
t
2
,
as desired.
If (
x
1
,y
1
)=(
x
2
,y
2
), the proof is similar. Finally, the cases where one or
more of the points (
x
i
,y
i
)=
∞
are easily checked.
Figure 2.8
y
2
=
x
3
+
x
2
We now consider the case where
x
3
+
Ax
+
B
has a double root. By trans-
lating
x
, we may assume that this root is 0 and the curve
E
has the equation
y
2
=
x
2
(
x
+
a
)
for some
a
=0. Thepoint(0
,
0) is the only singularity (see Figure 2.8). Let
E
ns
(
K
) be the nonsingular points on
E
with coordinates in
K
, including the
point
∞
.Let
α
2
=
a
(so
α
might lie in an extension of
K
). The equation for
E
may be rewritten as
y
x
2
=
a
+
x.
When
x
is near 0, the right side of this equation is approximately
a
. Therefore,
E
is approximated by (
y/x
)
2
=
a
,or
y/x
=
±α
near
x
= 0. This means that
the two “tangents” to
E
at (0
,
0) are
y
=
αx
and
y
=
−
αx
(for a different way to obtain these tangents, see Exercise 2.20).
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