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§6. Examples of Noncyclic Subgroups of Torsion Points 101

§6. Examples of Noncyclic Subgroups of Torsion Points

Under the curve E[a, b] studied in the previous section, we look for examples of
subgroups of E(k) of the form Z/nZ ⊕ Z/2Z. We start with the cases n = 2, 4, and
8 always assuming that k is of characteristic different from 2.

(6.1) Example. For n = 2wehave Z/2Z ⊕ Z/2Z ⊂ E(k) for E = E[a, b]ifand
2 2 3 2
only if the quadratic polynomial x + ax + b factors, that is, y = x + ax + bx =
x(x −r 1 )(x −r 2 ) and {0,(0, 0), (r 1 , 0), (r 2 , 0)} is the subgroup of 2-division points.
2
Now assuming that x + ax + b factors,wegoonto n = 4 by applying 1(4.1) to
see when there exists a point P ∈ E(k) with 2P = (0, 0).

(6.2) Example. For n = 4wehave Z/4Z ⊕ Z/2Z ⊂ E(k) for E = E[a, b] =
2 3
E[−r 1 − r 2 ,r 1 r 2 ] with some P satisfying 2P = (0, 0) if and only if y = x +
2
2
2
ax + bx = x(x + r )(x + s ),thatis, −r 1 and −r 2 are squares in k. The four
solutions to the equation 2P = (0, 0) are (rs, ±rs(r − s)) and (−rs, ±rs(r + s)).
2
Assuming that 2P = (0, 0) has a solution as in (6.2), then 2P = (−r , 0) has a
2
2
2
2
solution by (4.1) if and only if the elements −r − 0 =−r , −r − (−r ) = 0, and
2
2
2
2
−r − (−s ) = s − r are all squares in k. This cannot happen in Q or even R,as
we know, but over Q(i), the Gaussian numbers, we need only choose s and r such
2
2
2
that s = r + t , and this takes us back to Pythagorean triples again.
Continuing with the curve given in (6.2), we ask when is it true that
(rs, ±rs(r − s)) ∈ 2E(k) or (−rs, ±rs(r + s)) ∈ 2E(k)?
Again we apply 1(4.1) to see that 2P = (rs,rs(r − s)) has a solution if and only if
2
2
rs,rs + r = r(r + s),and rs + s = s(r + s) are all three squares in k. This is the
2
2
2
2
2
2
case when r = u , s = v ,and r + s = t for t = u + v .
2 2
2
2
(6.3) Example. For n = 8wehave Z/8Z⊕Z/2Z ⊂ E(k) for E = E[r +s ,r s ]
2
2
2
2
2
when r = u , s = v , and u + v = t in k. This is equivalent to E is a curve
4
4
4 4
of the form E[u + v , u v ]. In particular, over the rational numbers such curves
correspond to Pythagorean triples (u,v, t).
Applying 1(4.1) to the equation 2P = (−rs,rs(r + s)), we obtain almost the
2
2
same result, that is, −rs, −rs + r ,and −rs + s must be squares. This can be
2
2
2
2
2
realized for r = u , −s = v ,and r − s = t = u + v .
For the question of a point of order 3 on E[a, b] we recall from (5.3) that if
2
2
2
(x 2 , y 2 ) = 2(x 1 , y 1 ), then x 2 = (x − b) /4y . The case 2(x 1 , y 1 ) = (x 1 , −y 1 )
1 1
2
2
2
2
2
occurs for x = x 1 = x 2 , that is, 4xy = (x − b) .For x = t , y = st we see that
3
2
2
3
b = t (t − 2s) and from ax = y − x − bx, we obtain
2 2 3
a = (s − t) − 3t and b = t (t − 2s).