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

Let $T=(0,0)$, the rational point of order two. Definebywhere the right sides denote square classes.

To prove multiplicativity, let a nonvertical line $y=mx+c$ meet $E$ in points with $x$-coordinates $x_1,x_2,x_3$. Substitution gives the monic cubicso Vieta formulas give $x_1{x}_2{x}_3=c^2$. If the first two intersection points are $P,Q$, the third is $-(P+Q)$ and has the same $x$-coordinate as $P+Q$. Thereforein the square-class group. Tangencies follow by taking a repeated root, and the vertical and exceptional cases give the stated values at $O$ and $T$. Thus $\alpha$ is a group homomorphism.

The curve has a two-isogeny descent. Its quotient by $⟨T⟩$ isand the dual two-isogeny gives a corresponding map ${\alpha }^{\prime }:E^{\prime }(\mathbb{Q})\to {\mathbb{Q}}^{\times }/({\mathbb{Q}}^{\times }{)}^2$. The standard kernel calculation yields

If $d$ represents a class in $\mathrm{im}\alpha$, writewith coprime integers $u,v$. Substitution into the equation and clearing squares shows that the class occurs exactly when the quartic covering in a two-isogeny descenthas a nontrivial rational, equivalently primitive integral, solution. Valuation parity shows that only the finitely many square classes represented by square-free divisors $d$ of $b$ need be considered. Repeating the construction for $E^{\prime }$ reduces the rank calculation to finitely many explicit quartic solubility tests.

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 $1$.