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§2. Illustrations of the Elliptic Curve Group Law 29

2
(2.2) Remark. The point (0, 0) is on the curve with y = f (x) if and only if the
3
2
equation has the form y 2 = x + ax + bx.If r is a root of f (x), then y 2 =
f (x +r) has this form, and we will use the equation of the elliptic curve in this form
frequently. If 3 = 0 in the field, i.e., the characteristic of k is different from 3, then in
2
the special normal form y = f (x) we can complete the cube in the right-hand side
and after translation of x by a constant we have the Weierstrass form of the cubic
2 3
y = x + ax + b.
Now we consider some examples of subgroups of points on E(Q) for elliptic
curves which arise frequently.
2
3
2
(2.3) Example. For the curve E defined by y + y = x − x we have four obvious
points, (1, 0), (0, 0), (0, −1) =−(0, 0) and (1, −1) =−(1, 0), on the curve. The
tangent line through (1, 0) intersects E at (0, −1) from which we deduce 2(1, 1) =
(0, 0) and thus also 2(1, −1) = (0, −1). The tangent line to (0, 0) intersects E at
(1, 0) from which we deduce 2(0, 0) = (1, −1).

From 2(1, 1) = (0, 0) and 2(0, 0) = (1, −1) =−(1, 1),weobtain4(1, 1) =
(1, −1) =−(1, 1) or 5(1, 1) = O. Hence the set

{0,(1, 0), (0, 0), (0, −1), (1, −1)}
is a cyclic subgroup of E(Q) of order 5. In fact we will see later that E(Q) is exactly
this cyclic group of order 5.
The next example differs from the previous one by a single change in sign of the
2
coefficient of x . The group of rational points is now infinite instead of finite as in
(2.3).
3
2
2
(2.4) Example. For the curve E defined by y + y = x + x we have four obvious
points (0, 0), (−1, 0)(0, −1) =−(0, 0) and (−1, −1) =−(−1, 0), on the curve.