Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 125 / 3

The curve has good reduction when it admits a Minimal Weierstrass equation over ${\mathcal{O}}_K$ whose discriminant is a unit, so reduction modulo $\pi$ is a nonsingular elliptic curve $\tilde{E}/k$. Write a point of $E(K)$ in primitive projective coordinates $(X:Y:Z)$ over ${\mathcal{O}}_K$. Reducing all three coordinates definesPrimitivity ensures that the reduced triple is not zero. The addition morphism on the smooth Weierstrass model reduces to the addition morphism on $\tilde{E}$; equivalently, this follows from the valuative criterion for properness for the smooth proper group scheme. Therefore $\mathrm{red}(P+Q)=\mathrm{red}(P)+\mathrm{red}(Q)$.

The needed Hensel lemma says that if $f\in {\mathcal{O}}_K[X]$ and $a\in {\mathcal{O}}_K$ satisfy $f(a)\equiv 0(\mathrm{mod}\pi )$ and $f^{\prime }(a)\not\equiv 0(\mathrm{mod}\pi )$, then $a$ lifts uniquely to a root of $f$ in ${\mathcal{O}}_K$ with the prescribed residue. At every affine point of the smooth curve $\tilde{E}$, one partial derivative of its Weierstrass equation is nonzero. Fixing the other coordinate and applying Hensel's lemma lifts that point to $E(K)$; $O_E$ lifts itself. Thus reduction is surjective.

Use the local parametersso $x=t/z$ and $y=-1/z$. Substitution in a general integral Weierstrass equation givesFor fixed $0\ne t\in \pi {\mathcal{O}}_K$, the difference between the two sides, viewed as a polynomial in $z$, is congruent to $z$ modulo $\pi$ and has derivative congruent to one. Hensel's lemma gives a unique $z\in \pi {\mathcal{O}}_K$, and therefore a unique point $\theta (t)=(t/z,-1/z)$ in the kernel of reduction of an elliptic curve. Together with $\theta (0)=O_E$, this identifies that kernel with the parameters $t\in \pi {\mathcal{O}}_K$.

Let $K^{\mathrm{nr}}$ be the maximal unramified extension of $K$. Because $p\nmid n$, multiplication by $n$ on the special fibre is a separable isogeny and is surjective on $\tilde{E}(\overline{k})$. Choose a point $\tilde{Q}$ with $[n]\tilde{Q}=\tilde{P}$ and lift it, after a finite unramified extension, to $Q_0$. Then $R=P-[n]Q_0$ lies in the kernel of reduction.

On the formal group of an elliptic curve, multiplication by $n$ has the formSince $n$ is a unit in ${\mathcal{O}}_K$, the invertible morphism criterion for formal group laws makes $[n{]}_F$ an automorphism of $\pi {\mathcal{O}}_{K^{\mathrm{nr}}}$. Hence $R=[n]S$ for a unique point $S$ in the kernel of reduction, and $Q=Q_0+S$ satisfies $[n]Q=P$. Moreover, good reduction makes the finite group scheme $E[n]$ étale over ${\mathcal{O}}_K$, so all its points are defined over an unramified extension. Every point of $[n{]}^{-1}P=Q+E[n]$ is therefore unramified, proving that $K([n{]}^{-1}P)/K$ is unramified.

Multiplication by $n$ is already an automorphism of the formal kernel, so the exact reduction sequence shows that $E(K)/nE(K)$ is finite. Choose representatives $P_1,\dots ,P_s$. Each becomes $n$-divisible over a finite unramified extension; their compositum $L/K$ is still finite and unramified. Every class from $E(K)/nE(K)$ then maps to zero in $E(L)/nE(L)$.