Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 21 / 2 / d / Solution

Only the prime two can ramify. More precisely, every field $\mathbb{Q}(P_n)$ is unramified at every odd prime; actual ramification at two depends on $P$ and the chosen division point. Use $[2^n]P_n=P$, as printed in the original PDF.

Fix an odd prime $p$ and put $N=2^n$. The elliptic-curve discriminant $-2^8$ gives good reduction at $p$, and $N$ is a unit in ${\mathbb{Z}}_p$. The unramified division torsors at good primes imply that all division points of an elliptic curve above $P$ are defined over finite unramified extensions of ${\mathbb{Q}}_p$. Here is a direct lifting proof, which does not require $P$ to have integral affine coordinates.

Properness gives a reduction $\tilde{P}$. Choose a point $\overline{Q}$ with $[N]\overline{Q}=\tilde{P}$ over the algebraic closure of the residue field, and a finite residue extension containing it. Pass to the corresponding finite unramified extension $K/{\mathbb{Q}}_p$. The surjectivity of good reduction over a local field lifts $\overline{Q}$ to $Q_0\in E(K)$. Its error $[N]Q_0-P$ lies in the kernel of reduction of an elliptic curve. The prime-to-residue-characteristic multiplication on a formal group makes $[N]$ bijective on that kernel. Choose the unique kernel point $R$ with $[N]R=[N]Q_0-P$; then $Q=Q_0-R$ satisfies $[N]Q=P$ and has reduction $\overline{Q}$.

Every algebraic division point has some such reduction and is the unique lift in its division fibre: the difference of two lifts lies in the reduction kernel, which has no $N$-torsion. Thus every chosen $P_n$ lies in an unramified local extension. For every prime of its possibly non-Galois coordinate field above $p$, the ramification index is one. Globally,One cannot universally assert that two actually ramifies: $P=O$ and $P_n=O$ give the field $\mathbb{Q}$. It can ramify. For example, a half of $T=(-1,0)$ isThe elliptic curve group law verifies this by doubling. Its coordinate field contains $\mathbb{Q}(\sqrt{2})$, which has ramification in the quadratic extension generated by the square root of two, so the only possible bad prime does occur for this choice.