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.