2.9. Descent and the weak Mordell-Weil theorem 51 Notice that abc is a square and y2 0 = ( 66469980 4913 )2 = abc(uvw)2. Example 2.9.2. Let E : y2 = x3 − 556x + 3120 as before, with e1 = 6, e2 = 20 and e3 = −26. Let P = (−8,84), Q = (24,60) and S = P +Q = (− 247 16 , − 5733 64 ). The points P , Q and S are in E(Q). We would like to calculate the aforementioned numbers a, b, c for each of the points P, Q and S. For instance x(P ) − e1 = −8 − 6 = −14 = −14 · 12, x(P ) − e2 = −7 · 42, and x(P ) − e3 = 2 · 32. Thus, aP = −14, bP = −7 and cP = 2. Similarly, we calculate x(Q) − 6 = 2 · 32, x(Q) − 20 = 22, x(Q) + 26 = 2 · 52, x(S) − 6 = −7 · 7 4 2 , x(S) − 20 = −7 · 9 4 2 , x(S) + 26 = 13 4 2 . Thus aQ = 2, bQ = 1, cQ = 2, and aS = −7, bS = −7, cS = 1. Notice the following interesting fact: aP · aQ = −28 = −7 · 22, bP · bQ = −7, cP · cQ = 4. Therefore, the square-free part of aP · aQ equals aS = aP +Q = −7. And similarly, the square-free parts of bP ·bQ and cP ·cQ equal bS = −7 and cS = 1, respectively. Also, the reader can check that a2P = b2P = c2P = 1 and a2Q = b2Q = c2Q = 1. The previous example points to the fact that there may be a ho- momorphism between points on E(Q) and triples (a, b, c) of rational numbers modulo squares, or square-free parts of rational numbers formally, we are talking about Q×/(Q×)2 × Q×/(Q×)2 × Q×/(Q×)2. Here, the group Q×/(Q×)2 is the multiplicative group of non-zero rational numbers, with the extra relation that two non-zero rational numbers are equivalent if their square-free parts are equal (or, equiv- alently, if their quotient is a perfect square). For instance, 3 and 12 25 represent the same element of Q×/(Q×)2 because 12 25 = 3 · ( 2 5 )2 . The following theorem constructs such a homomorphism. Here we have adapted the proof that appears in [Was08], Theorem 8.14.

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