Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 125 / 4

The Kummer map of an elliptic curve gives an injectionBecause $E[n]\subseteq E(K)$, the Galois action on $E[n]$ is trivial, so a cocycle in the image is a continuous homomorphism $G_K\to E[n]$. For $P\in E(K)$ and $Q$ with $nQ=P$, its kernel fixes the Kummer extension $K(Q)/K$, which is Galois of degree at most $n^2$ and exponent dividing $n$.

The local theory of reduction of an elliptic curve shows that these extensions are unramified outside the finite set consisting of primes dividing $n$, primes of bad reduction, and archimedean places. Local fields have only finitely many extensions of any bounded degree. Together with the Hermite-Minkowski finiteness theorem, this implies that only finitely many global extensions of degree at most $n^2$ with these ramification restrictions occur. Each has only finitely many homomorphisms to the finite group $E[n]$. Hence the image of $\delta$, and therefore $E(K)/nE(K)$, is finite.

Solved by gpt-5.6-sol high.

Foruse two-isogeny descent throughThe square-class maps send a nonexceptional point to the class of its $x$-coordinate. On $E$, possible classes are $\pm 1,\pm 7$; the defining quartics and positivity exclude the negative classes, while $(1,3)$ and $(7,21)$ realize $1$ and $7$. Thus the image has order two. On $E^{\prime }$, the possible classes are $\pm 1,\pm 3$; the classes $1$ and $-3$ occur, while the quartics for $-1$ and $3$ have no primitive solution modulo $16$. Hence this image also has order two. The two-isogeny descent formulatherefore gives $r=0$.

The displayed curve has good reduction at $5$ and $11$, where direct counting givesReduction injects rational torsion of order prime to these characteristics, so its order divides $\gcd (6,18)=6$. We already have the six distinct pointsSince the rank is zero, these are all the rational points and $E(\mathbb{Q})\cong \mathbb{Z}/6\mathbb{Z}$.

Solved by gpt-5.6-sol high.