Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2009 / B / 6 / c / Solution

Use the two-isogeny formula formula with $a=\alpha$ and $b=-\alpha$. Its companion isBoth curves are nonsingular because $\alpha \ne 0,-4$. With $b^{\prime }=\alpha (\alpha +4)$, chooseEach map extends across its exceptional affine points, and $\hat{\varphi }\varphi =[2]$. On $D$, the first coordinate of the dual can also be written $v^2/(4u^2)$.

Let $P=(1,1)$ and solve $\hat{\varphi }(u,v)=P$. The first coordinate gives $v^2=4u^2$. Since $u\ne 0$, the equation of $D$ then givesThe second coordinate selects the signs. The two rational preimages are exactlySubstituting into the dual formula checks that both map to $(1,1)$. Their coordinates are nonexceptional under the assumptions, and the degree-two map has no further preimages.

The two-torsion square-class homomorphism ${\alpha }_D$ has kernel $\varphi (C(\mathbb{Q}))$, and on these two points it is their nonzero $u$-coordinate modulo squares. Thus $Q_1$ is in the image of $\varphi$ exactly when $\alpha$ is a rational square, and $Q_2$ is in that image exactly when $\alpha +4$ is a rational square.

If $P=2R$, then $\varphi (R)$ is one of the two preimages, because $\hat{\varphi }\varphi (R)=2R=P$. Conversely, if $Q_j=\varphi (R)$, applying the dual shows $P=2R$. ThereforeFinally, if an integer equals $(a/b{)}^2$ in lowest terms, then $b^2\mid a^2$ forces $b=1$. Thus rational squareness of these integer parameters is exactly integer squareness, andThis is the two-isogeny halving criterion for the alpha family.