Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 125 / 5 / c / Solution

Here $a=p$ and $b=p^2$. For $E$, the four possible square classes arewith coveringsThe classes $-1$ and $-p$ have no real points, because their right sides are respectivelyfor every nonzero pair $(u,v)$. Hence

The two-isogenous curve isIts candidate classes and coverings areThe image of ${\alpha }^{\prime }$ is a subgroup and contains $-3$, the image of the rational two-torsion point $(0,0)$.

Suppose now that $p\equiv 7(\mathrm{mod}12)$. Then both $-1$ and $3$ are quadratic nonresidues modulo $p$. The covering $C_{-1}^{\prime }$ has no ${\mathbb{Q}}_p$-point. Indeed, after choosing primitive $p$-adic coordinates, if $p\nmid u$, reduction modulo $p$ would make $-1$ a square. If $p\mid u$, then $p\nmid v$; the right side has valuation two, and division by $p^2$ would make $3$ a square modulo $p$. Both alternatives are impossible.

Thus $-1\notin \mathrm{im}{\alpha }^{\prime }$. A subgroup of the three-dimensional square-class group generated by $-1,3,p$ that contains $-3$ but not $-1$ has order at most four. ThereforeThe rank is consequently $0$ or $1$.