2.9. Descent and the weak Mordell-Weil theorem 53

If we evaluate p(x) at x = ei, we obtain

p(ei) = (aei +

b)2

= −(ei − x0)(ei − x1)(ei − x2)

or, equivalently, (x0 − ei)(x1 − ei)(x2 − ei) = (aei +

b)2.

Thus, the

product δ(P ) · δ(Q) · δ(R) equals

δ(P ) · δ(Q) · δ(R) = (x0 − e1,x0 − e2,x0 − e3)

·(x1 − e1,x1 − e2,x1 − e3)

·(x2 − e1,x2 − e2,x2 − e3)

= ((x0 − e1)(x1 − e1)(x2 − e1),

(x0 − e2)(x1 − e2)(x2 − e2),

(x0 − e3)(x1 − e3)(x2 − e3))

= ((ae1 +

b)2,

(ae2 +

b)2,

(ae3 +

b)2)

= (1,1,1) ∈

(Q×/(Q×)2)3.

Hence, δ(P ) · δ(Q) · δ(R) = 1. If we multiply both sides by δ(R) and

notice that

a2

= 1 for any a ∈

Q×/(Q×)2,

we conclude that

δ(P ) · δ(Q) = δ(R) = δ(−(P + Q)) = δ(P + Q),

as desired. In order to completely prove that δ is a homomorphism,

we would need to check the cases when P , Q or R is one of the points

(ei,0) or O, but we leave those special cases for the reader to check

(Exercise 2.12.15).

If δ(P ) = (δ1,δ2,δ3), then it follows directly from the definition

of δ that δ1 · δ2 · δ3 = 1 in

Q×/(Q×)2.

Indeed, this is clear for

P = O or P = (ei,0), and if P = (x0,y0) with y0 = 0, then (x0 −

e1)(x0 − e2)(x0 − e3) = y0,

2

which is a square, and is therefore trivial

in

Q×/(Q×)2.

Next, let us show that the kernel of δ is 2E(Q). Clearly, 2E(Q)

is in the kernel of δ, because δ is a homomorphism with image in

(Q×/(Q×)2)3,

as we just proved. Indeed, if P ∈ E(Q), then

δ(2P ) = δ(P ) · δ(P ) = δ(P

)2

= (δ1,δ2,δ3)

2 2 2

= (1,1,1),

because squares are trivial in

Q×/(Q×)2.

Now let us show the reverse inclusion, i.e., that the kernel of δ

is contained in 2E(Q). Let Q = (x1,y1) ∈ E(Q) such that δ(Q) =

(1,1,1). We want to find P = (x0,y0) such that 2P = Q. Notice that