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 (Γ )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).