Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 28 / 3

A one-dimensional commutative formal group law over the p-adic integers is a power series $\mathcal{F}(X,Y)\in {\mathbb{Z}}_p[[X,Y]]$ withIts linear part is $X+Y$, and it has a unique formal inverse $i(T)=-T+O(T^2)$ satisfying $\mathcal{F}(T,i(T))=0$. A formal-group homomorphism is a series $h(T)\in T{\mathbb{Z}}_p[[T]]$ satisfying $h(\mathcal{F}(X,Y))=\mathcal{F}(h(X),h(Y))$.

Define $[n](T)$ by adding $T$ to itself $n$ times with the formal group law. Induction on the linear part givesAssociativity and commutativity make $[n]$ a formal-group homomorphism. Since $p\nmid n$, its linear coefficient is a unit in ${\mathbb{Z}}_p$. We now construct its compositional inverse of a formal power series, rather than merely assert that the nonzero coefficient is sufficient.

Write $[n](T)=nT+a_2{T}^2+\cdots$ and seek $g(T)=b_{1}T+b_2{T}^2+\cdots$ with $[n](g(T))=T$. The degree-one coefficient gives $b_1=n^{-1}$. At degree $k\ge 2$, the coefficient is $n{b}_k$ plus a polynomial in the already chosen $b_1,\dots ,b_{k-1}$ and $a_2,\dots ,a_k$. Division by the unit $n$ uniquely chooses $b_k\in {\mathbb{Z}}_p$. This recursively constructs $g$ over the same ring. A right inverse is constructed in the same way; associativity of composition identifies it with $g$, so $g\circ [n]=[n]\circ g=T$.

To prove that $g$ respects the group law, apply $[n]$ to the two candidate series:Cancel by composing with $g$. The result is $\mathcal{F}(g(X),g(Y))=g(\mathcal{F}(X,Y))$. ThusThis is the multiplication isomorphism of a formal group law, a special case of the invertible morphism criterion for formal group laws. If $p\mid n$, its linear coefficient is not a unit, so the same isomorphism assertion would fail.

Reduce $P$ to $\overline{P}\in \tilde{E}({\mathbb{F}}_p)$. Because $p\nmid n$, the multiplication-by-n morphism on the reduced elliptic curve is a separable, surjective morphism of degree $n^2$. Choose a point $\overline{Q}\in \tilde{E}({\overline{\mathbb{F}}}_p)$ with $n\overline{Q}=\overline{P}$. Its coordinates lie in some finite field ${\mathbb{F}}_{p^d}$.

Let $L/{\mathbb{Q}}_p$ be the finite unramified extension with residue field ${\mathbb{F}}_{p^d}$. The curve retains good reduction over $L$. Its smooth integral model and the Hensel lemma give surjectivity of good reduction over a local field, so lift $\overline{Q}$ to some $Q_0\in E(L)$. This lift need not satisfy $n{Q}_0=P$ exactly, but its errorhas zero reduction and belongs to $E_1(L)$, the kernel of reduction of an elliptic curve.

The parameter $t=-x/y$ at infinity identifies $E_1(L)$ with the maximal ideal ${\mathfrak{m}}_L$ equipped with the formal group of an elliptic curve. The integral power series for $[n]$ and its inverse from part (i) remain integral over ${\mathcal{O}}_L$. They converge on ${\mathfrak{m}}_L$: terms of degree $k$ tend to zero in valuation because their coefficients are integral and the argument has positive valuation. Hence multiplication by $n$ is bijective on $E_1(L)$. Choose its unique solution $T\in E_1(L)$ to $nT=R$, and put $Q=Q_0+T$. ThenThe residue-field extension supplies a division point modulo $p$, and the formal-group inverse corrects the error without introducing ramification. This is the local construction underlying the uniform unramified division field over a local field and unramified division torsors at good primes.