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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