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§5. An Explicit 2-Isogeny 95

2
5
with discriminant = b (b − 11b − 1) = 0.
(c) 6P = 0 is equivalent to 3P =−3P which by (4.5) reduces to the relation
2
c + c = b. Hence f 6 (b, c) = 0 is a conic parametrized rationally by c.Alsothe
discriminant is
6 3
(b, c) = c (c + 1) (9c + 1) = 0.
(d) 7P = 0 is equivalent to 4P =−3P which by (4.5) reduces to the relations
2
2
2
c = d(d −1) = d −d and c = d (c−d +1). The second relation is a consequence
3
2
2
2
of the first since c − d + 1 = d − 2d + 1. Then b = d − d , c = d − d is the
rational parametrization of the cubic curve
3
c = b(b − c)
with a double point at the origin. In this case the discriminant is
7 7 3 2
(b, c) = d (d − 1) (d − 8d + 5d + 1) = 0.
(e) 8P = 0 is equivalent to 4P =−4P which by (4.5) reduces to the relations
2
2
d (c − d + 1) = d(d − 1) . Since b = 0 implies d = 0, we can divide by d to obtain
2
d(c − d + 1) = (d − 1) . Solving for c, we obtain
(d − 1)(2d − 1)
c = or b = cd = (d − 1)(2d − 1).
d
(f) 9P = 0 is equivalent to 5P =−4P which by (4.5) reduces to the relations
2
de(e − 1) = d(d − 1) or e(e − 1) = d − 1since d = b/c = 0. Thus d = e − e + 1
is the equation which for (b, c) becomes
2
3
c = de − e = e − e ,

4
2
5
3
2
3
b = cd = e − e 2 e − e + 1 = e − 2e + 2e − e .
3
2
5
4
2
3
Then b = e − 2e + 2e − e , c = e − e is the rational parametrization of the
fifth-order curve with singularity at the origin.
We return to this subject when we study elliptic curves over the complex numbers
and construct the curves Y 1 (n) and X 1 (n) along with related families.
§5. An Explicit 2-Isogeny
We consider an example of a 2-isogeny ϕ which is explicitly given by equations
and which we will use later to illustrate other ideas in the global theory. The dual
isogeny ˆϕ is also given explicitly, and it is shown that ˆϕϕ is multiplication by 2. These
formulas work over any field k of characteristic = 2. The curve for this isogeny ϕ is
2
2
3
E = E[a, b]: y = x + ax + bx,
and the kernel of ϕ is normalized to be {0,(0, 0)}.
Using the formulas of 6(2.1) for a 2 = a, a 4 = b,and a 1 = a 3 = a 6 = 0, we
2
have b 2 = 4a, b 4 = 2b, b 6 = 0, and b 8 =−b . This leads to the following relations.