42 2. Elliptic curves
does not divide 13, it follows that E(Q) must be trivial. Similarly,
one can show that E(Q) is trivial, and we conclude that E(Q)torsion
However, notice that P = (1,2) ∈ E(Q) is a point on the curve.
Since we just proved that E does not have any points of finite order,
it follows that P must be a point of infinite order, and, hence, we have
shown that E has infinitely many rational points: ±P, ±2P, ±3P,....
In fact, E(Q)
= Z and (1,2) is a generator of its Mordell-Weil group.
In the previous example, the Nagell-Lutz theorem (Theorem 2.5.5)
would have yielded the same result, i.e., the torsion is trivial, in an
easier way. Indeed, for the curve E :
+ 3, the quantity
so the possibilities for y(P
where P is a tor-
sion point of order ≥ 3, are 1, 9 or 81 (it is easy to see that there are no
2-torsion points). Therefore, the possibilities for x(P
are −2, 6 or 78, respectively. Since x(P ) is an integer, we reach a con-
tradiction. In the following example, the Nagell-Lutz theorem would
be a lengthier and much more tedious alternative, and Proposition
2.6.15 is much more effective.
Example 2.6.17. Let E/Q :
+ 4249388. In this case
is divisible by 192 distinct positive squares,
which makes it very tedious to use the Nagell-Lutz theorem. The
(minimal) discriminant of E/Q is ΔE = −16(4A3 +27B2) and there-
fore E has good reduction at 5 and 7. Moreover, B = 4249388 ≡
3 mod 35 and therefore, by our calculations in Example 2.6.16, N5 = 6
and N7 = 13. Thus, Proposition 2.6.15 and the same argument we
used in Ex. 2.6.16 show that the torsion of E(Q) is trivial.
Incidentally, the curve E/Q : y2 = x3 + 4249388 has a rational
point P =
. Since the torsion of E(Q) is trivial, P
must be of infinite order. Another way to see this: since P has
rational coordinates that are not integral, the Nagell-Lutz theorem
implies that the order of P is infinite. In fact, E(Q) is isomorphic to
Z and it is generated by P .