Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 32 / 3 / ii / Solution

Near $O$, put $t=-x/y$ and $w=-1/y$, or projectively $t=-X/Y$, $w=-Z/Y$. The integral Weierstrass equation of an elliptic curve becomesThe linear coefficient of $w$ at the origin is a unit, so successive coefficient comparison gives a unique series $w(t)\in {\mathbb{Z}}_p[[t]]$ beginning $t^3$. Expressing the addition morphism in the local parameters of its two inputs givesIts identity, inverse, commutativity and associativity identities follow from the corresponding elliptic group identities. This one-dimensional commutative formal group law is the formal group of an elliptic curve $\hat{E}$ attached to the model.

A point reducing to $O$ lies in the chart with $Y$ a unit and has $t,w\in p{\mathbb{Z}}_p$. The implicit equation determines $w$ uniquely from $t$. Conversely every $t\in p{\mathbb{Z}}_p$ gives the convergent solution $w(t)$ and the point $[-t:1:-w(t)]$; $t=0$ gives $O$. Hence the parameter is a bijection from the reduction kernel to $p{\mathbb{Z}}_p$. The series $F$ converges there and describes its addition. ThereforeFor nonzero $t$ with $v_p(t)=r>0$, $w(t)=t^3$ times a unit, so $v_p(x)=-2r$ and $v_p(y)=-3r$. These are the formal kernel of a minimal Weierstrass equation valuations used earlier.