2.9. Descent and the weak Mordell-Weil theorem 57
Thus, the image of δ is formed by the 16 elements that one obtains
by multiplying out δ(T1), δ(T2), δ(P ) and δ(Q), in all possible ways.
Thus, δ(E(Q)/2E(Q)) is the group:
{(1, 1,1), (−7, −14, 2), (14,161,46), (−2, −46, 23),
(−14, −7, 2), (2,2,1), (−1, −23, 23), (7,322,46),
(2,1,2), (−14, −14, 1), (7,161,23), (−1, −46, 46),
(−7, −7, 1), (1,2,2), (−2, −23, 46), (14,322,23)}.
(Exercise: Check that the elements listed above form a group under
multiplication.) We see that the only primes that appear in the fac-
torization of the coordinates of elements in the image of δ are: 2,7
and 23. Therefore, the coordinates of δ are not just in Q×/(Q×)2 but
in a much smaller subgroup of 16 elements:
Γ = {±1, ±2, ±7, ±23, ±14, ±46, ±161, ±322}
Q×/(Q×)2.
And the image of E(Q)/2E(Q) embeds into
ΓΔ = {(δ1,δ2,δ3) Γ × Γ × Γ : δ1 · δ2 · δ3 = 1
Q×/(Q×)2}
Γ × Γ × Γ .
Since Γ has 16 elements and E(Q)/2E(Q) embeds into )3, we
conclude that E(Q)/2E(Q) has at most (16)3 = 212 elements. In fact,
ΓΔ has only 162 elements, so E(Q)/2E(Q) has at most 28 elements.
Notice also the following interesting “coincidence”: the prime divisors
that appear in ΓΔ coincide with the prime divisors of the discriminant
of E, which is ΔE = 6795034624 =
218·72·232.
In the next proposition
we explain that, in fact, this is always the case.
Proposition 2.9.5. Let E :
y2
= (x−e1)(x−e2)(x−e3), with ei Z.
Let P = (x0,y0) E(Q) and write
(x0 e1) =
au2,
(x0 e2) =
bv2,
(x0 e3) =
cw2,
y0
2
=
abc(uvw)2,
where a, b, c, u, v, w Q, the numbers a, b, c Z are square-free, and
abc is a square (in Z). Then, if p divides a · b · c, then p also divides
the quantity Δ = (e1 e2)(e2 e3)(e1 e3).
Note: the discriminant of E equals ΔE = 16(e1
e2)2(e2

e3)2(e1

e3)2.
So a prime p divides Δ if and only if p divides ΔE. If
p 2, then this is clear (see Exercise 2.12.19 for p = 2).
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