2.9. Descent and the weak Mordell-Weil theorem 53
If we evaluate p(x) at x = ei, we obtain
p(ei) = (aei +
= −(ei x0)(ei x1)(ei x2)
or, equivalently, (x0 ei)(x1 ei)(x2 ei) = (aei +
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 +
(ae2 +
(ae3 +
= (1,1,1)
Hence, δ(P ) · δ(Q) · δ(R) = 1. If we multiply both sides by δ(R) and
notice that
= 1 for any a
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
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,
which is a square, and is therefore trivial
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
as we just proved. Indeed, if P E(Q), then
δ(2P ) = δ(P ) · δ(P ) = δ(P
= (δ1,δ2,δ3)
2 2 2
= (1,1,1),
because squares are trivial in
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
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