46 2. Elliptic curves
Example 2.7.6. Let E/Q be the elliptic curve y2 = x(x + 1)(x +
2), which already appeared in Example 1.1.1. Since the Weierstrass
equation of E is
y2
= x(x + 1)(x + 2) =
x3
+
3x2
+ 2x,
it follows from Theorem 2.7.4 that the rank RE satisfies
RE
ν(A2
4B) + ν(B) 1 = ν(1) + ν(2) 1 = 0 + 1 1 = 0,
and therefore the rank is 0. The reader can check that
E(Q)torsion = {O, (0,0),(−1,0),(−2,0)}.
Since the rank is zero, the four torsion points on E/Q are the only
rational points on E.
Example 2.7.7. Let E :
y2
=
x3 +2308x2
+665858x. The primes 2
and 577 are the only prime divisors of (both) B and
A2
4B. Thus,
RE
ν(A2
4B) + ν(B) 1 = 2 + 2 1 = 3.
The points P1 = (−1681,25543), P2 = (−338,26), and P3 = (577/16,
332929/64) are of infinite order and the subgroup of E(Q) generated
by P1, P2 and P3 is isomorphic to
Z3.
Therefore, the rank of E is
equal to 3.
2.8. Linear independence of rational points
Let E/Q be the curve defined in Example 2.7.7. We claimed that
the subgroup generated by the points P1 = (−1681,25543), P2 =
(−338,26), and P3 = (577/16,332929/64) is isomorphic to
Z3.
But
how can we show that? In particular, why is P3 not a linear combi-
nation of P1 and P2? In other words, are there integers n1 and n2
such that P3 = n1P1 + n2P2? In fact, E/Q has a rational torsion
point T = (0,0) of order 2, so could some combination of P1, P2 and
P3 equal T ? This example motivates the need for a notion of linear
dependence and independence of points over Z.
Definition 2.8.1. Let E/Q be an elliptic curve. We say that the
rational points P1,...,Pm E(Q) are linearly dependent over Z if
there are integers n1,...,nm Z such that
n1P1 + n2P2 + · · · + nmPm = T,
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