2.9. Descent and the weak Mordell-Weil theorem 51
Notice that abc is a square and y0 2 = (
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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