Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2007 / B / 7 / iii

Here $a=1$, $b=-2$ and the two-isogenous curve is $D:v^2=u(u^2-2u+9)$. Let ${\delta }_C$ and ${\delta }_D$ be the first-coordinate square-class maps, using the coefficient $b$ or $b_1$ at the point $(0,0)$. The same chord argument shows that both maps are homomorphisms.

For an equation $y^2=x(x^2+ax+b)$ with integer coefficients, its first-coordinate square class has a square-free representative dividing $b$, including a possible sign. Indeed, at a prime not dividing $b$, a negative valuation of $x$ must be even because $x^3$ dominates; a positive valuation must be even because $x^2+ax+b$ is a unit. Thus only primes dividing $b$ can occur to odd order. For $C$, the possibilities are $1,-1,2,-2$. The classes $1,-2$ occur at the identity and $(0,0)$. If class two occurred, write $x=2M^2/e^2$ with coprime integers $M,e$. Substituting in the equation would give a rational square, necessarily an integer square,Modulo three, the right side is two whenever $M,e$ are not both divisible by three: this holds whether just one or both are units. That is impossible. Since the image is a subgroup containing $-2$, exclusion of two also excludes minus one. Therefore $|{\delta }_C(C(\mathbb{Q}))|=2$.

For $D$, the quadratic $u^2-2u+9=(u-1{)}^2+8$ is positive, so every nonzero first coordinate is positive. Its square class can only be one or three. Both occur, with three supplied by $(3,6)$, so $|{\delta }_D(D(\mathbb{Q}))|=2$.

Let $\psi :D\to C$ be the dual two-isogeny, with $\psi \varphi =[2]$. The kernel identities of part (ii), applied to both curves and the usual coordinate scaling, give $[C:\psi D]=2$ and $[D:\varphi C]=2$. Moreover $\ker \psi =\{O,T_1\}$ lies in $\varphi C$, since $b_1=9$ is a square. Thus $[\psi D:2C]=[D:\varphi C]=2$, andThis is a rank-zero two-isogeny descent for roots zero one and minus two. The rational two-torsion has four points, since the cubic factors as $x(x-1)(x+2)$. By the Mordell-Weil theorem, the last index is $2^r|C(\mathbb{Q})[2]|=2^{r+2}$. Therefore