Cryptography Reference
In-Depth Information
left when we divide the cubic polynomial by the product of the two linear factors
corresponding to the first two roots; then the third root is the
x
-coordinate of the third
intersection point (if the line is vertical, the third intersection point is either
or
one of the two remaining points obtained by eliminating the variable
x
and solving
the resulting quadratic equation). Note that this point might be equal to either
P
or
Q
because when we say that the line intersects the curve at three points, we are
counting multiplicities and the third root of the cubic might be a multiple root; we
illustrate this phenomenon with some examples below. We also remark that if the
field is algebraically closed then every line intersects the curve at exactly three points
but this is not necessarily so if the field is not algebraically closed, as is the case with
Q
O
,
R
, or any finite field. However, the preceding argument shows that, for any field
K
, if the line intersects the curve at two points of
E
(
K
)
, then it also intersects it at a
third point.
Example 11.1
Let us consider the curve
E
defined by the equation:
y
2
x
3
=
−
4
x
+
1
.
The coefficients of the polynomial defining the curve belong to the rational field
Q
(
Q
)
and we may consider the set of points
E
corresponding to the elliptic curve over
(
K
)
K
Q
this field (we may also consider
E
for any extension
of
such as, for example,
K = R
or
K = C
). Since
Δ
=
0 we see that
E
is indeed an elliptic curve. Consider
the points
P
=
(
−
1
,
2
)
,
Q
=
(
1
/
4
,
1
/
8
)
. Then the line joining these two points is
the line of equation 2
y
+
3
x
−
1
=
0. We use Maple to compute the third intersection
point as follows:
> solve([yˆ2-xˆ3+4*x-1 = 0, 2*y+3*x-1 = 0], [x, y]);
[[x = 3, y = -4], [x = 1/4, y = 1/8], [x = -1, y = 2]]
. Note that, in contrast
with the line
L
PQ
,the
x
-axis does not intersect the rational curve
E
because the
polynomial
x
3
We see that the third intersection point is
R
=
(
3
,
−
4
)
−
4
x
+
1
∈ Q[
x
]
is irreducible as can be checked with Maple:
> irreduc(xˆ3-4*x+1);
true
and
hence the curve has no points on the
x
-axis despite its appearance when plotted as we
do in Example 11.2 below. In the graph it looks like the curve intersects the
x
-axis
at three points and this is indeed the case if we consider the
real curve
defined by
the same equation; these intersection points do not have rational coordinates and lie
in
E
As seen in Examples 2.4, this means that the polynomial has no zeros in
Q
(
R
)
but not in
E
(
Q
)
.
In the previous discussion we assumed that
P
=
Q
but we have also to consider
the case in which
P
P
. Then it is still true that
the elliptic curve determines a unique line passing through
P
, namely the tangent
=
Q
in order to be able to define
P
+
