Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 24 / 4 / b / Solution

Choose an integral Minimal Weierstrass equation over ${\mathbb{Z}}_p$. Let $E_0({\mathbb{Q}}_p)$ be the points with nonsingular reduction, and let $E_1({\mathbb{Q}}_p)$ be those reducing to $O$. This distinction matters at bad reduction of an elliptic curve: the whole singular reduced cubic is not a group. The smooth locus is a group under the induced elliptic curve group law, and Hensel's lemma lifts each of its rational points, givingThe formal kernel of a minimal Weierstrass equation identifies $E_1({\mathbb{Q}}_p)$ with $p{\mathbb{Z}}_p$ equipped with $F$. Indeed, the integral series $w(t)$ converges there, and $x=t/w(t)$, $y=-1/w(t)$ give the unique point for each nonzero parameter. If $v_p(t)=r>0$, then $v_p(x)=-2r$ and $v_p(y)=-3r$; the parameter zero gives $O$. Conversely a point reducing to $O$ has these local coordinates of positive valuation, so is recovered by the series.

Define $E_m$ for $m\ge 1$ by $t\in p^m{\mathbb{Z}}_p$. The linear part of the formal group law is addition, and the higher terms have degree at least two. Hence the filtration of elliptic-curve points over a local field hasAt good reduction, $E_0=E$ and $E/E_1\cong \tilde{E}({\mathbb{F}}_p)$. At bad reduction, there is also the finite quotient $E/E_0$: $E({\mathbb{Q}}_p)$ is a compact group because the curve is projective, and $E_0$ is an open subgroup, so its index of a subgroup is finite. Together these quotients and the formal filtration describe the group, including the extra components at bad primes.

The formal logarithm makes the deepest part explicit. Write $L(T)=T+{\sum }_{n\ge 2}{c}_n{T}^n$. Since its derivative is the integral invariant differential of a formal group law, $n{c}_n\in {\mathbb{Z}}_p$. ConsequentlyThe inequality $v_p(n)\le (n-1)/(p-1)$ shows that for an integer $m>1/(p-1)$ every nonlinear term on $p^m{\mathbb{Z}}_p$ has strictly higher valuation than the linear term. After scaling, $L$ is the identity plus a strictly contracting series; solving $L(t)=u$ by successive correction gives its inverse formal group exponential. Thus the deep logarithm subgroup of a formal group givesOne may take $m=1$ for odd $p$ and $m=2$ for $p=2$. In particular, $E({\mathbb{Q}}_p)$ contains an open subgroup isomorphic to the additive p-adic integers, and the remaining information is in finite quotients and their group extensions.