Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 125 / 5

An isogeny of elliptic curves is a nonconstant morphism preserving identity points; it is automatically a finite surjective group homomorphism. On the affine chart $v\ne 0$, putThe equation of $E$ is $t^3+d=s^3$, and the proposed map isIt lands on $E^{\prime }$ becauseThe rational formulas extend across $v=0$ to a morphism sending $O_E$ to $O_{E^{\prime }}$. It is nonconstant, hence an isogeny. On function fields, $t$ satisfies $t^3=y$, so the degree is at most three; generically the three cube roots give three distinct preimages. Equivalently, the points with $v=0$ form its three-element geometric kernel. Therefore

The principal divisor criterion on an elliptic curve says that $D=\sum n_P(P)$ is principal exactly when $\mathrm{deg}D=0$ and $\sum [n_P]P=O$.

On $E^{\prime }:y^2+dy=x^3$, takeThe line $y=0$ meets the cubic three times at $T$, while $y$ has a triple pole at the point at infinity. HenceWith $g=u/v\in \mathbb{Q}(E)$, part (a) givesTaking divisors and cancelling the factor three yieldsPullback on degree-zero divisor classes is the dual isogeny, so the pulled-back class is represented by $\hat{\varphi }(T)$. It is principal, and therefore

The three-isogeny descent connecting map, identified through the Weil pairing with $H^1(\mathbb{Q},{\mu }_3)={\mathbb{Q}}^{\times }/({\mathbb{Q}}^{\times }{)}^3$, isThe long exact sequence attached to $0\to E[\varphi ]\to E\stackrel{\varphi }{\to }{E}^{\prime }\to 0$ makes it a group homomorphism withThe function $f=y$ from part (b) gives the explicit formulaAt $T$, the value $d^{-1}$ is the leading coefficient of $y$ relative to the local parameter $x$, since $y(y+d)=x^3$ gives $y/x^3\to d^{-1}$. At $-T=(0,-d)$ the ordinary formula gives $-d$, whose class is the inverse of $d^{-1}$ because $-1$ is a cube.

Let $S$ be the primes dividing $d$. For a prime $\ell \notin S$, useIf $v_{\ell }(y)>0$, then $y+d$ is an $\ell$-adic unit, so $v_{\ell }(y)=3v_{\ell }(x)$. If $v_{\ell }(y)<0$, then $v_{\ell }(y+d)=v_{\ell }(y)$, so $2v_{\ell }(y)=3v_{\ell }(x)$. In either case $v_{\ell }(y)$ is divisible by three; the special values at $O$ and $T$ have the same property. ThereforeWhen $d=1$, this power-class group is trivial: a rational number whose valuation at every prime is divisible by three is a cube up to sign, and $-1=(-1{)}^3$. Thus $\alpha$ is trivial, its kernel is all of $E^{\prime }(\mathbb{Q})$, and