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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.

In this one-dimensional setting, a commutative formal group law over a commutative ring $R$ is a formal power series $F(X,Y)\in R[[X,Y]]$ withIn particular $F(X,Y)=X+Y+$ terms of total degree at least two. There is a unique formal inverse $i_F(T)=-T+O(T^2)$; coefficient recursion solves $F(T,i_F(T))=0$.

A morphism from formal group law $F$ to $G$ is $f(T)\in TR[[T]]$ satisfyingThe invertible morphism criterion for formal group laws isNecessity follows by differentiating $g\circ f=T$ at zero for an inverse $g$. Conversely, write $f(T)=a_{1}T+a_2{T}^2+\cdots$ with $a_1$ a unit. In constructing $g(T)=b_{1}T+b_2{T}^2+\cdots$, the coefficient of $T$ fixes $b_1=a_1^{-1}$; at degree $n$, the equation $f(g(T))=T$ has the form $a_1{b}_n+$ an already known expression $=0$. This determines every $b_n$ over $R$. The same construction gives an inverse on the other side, and uniqueness makes the two inverses agree. Finally apply $g$ to the morphism identity with $X=g(U),Y=g(V)$ to obtainThus the inverse is itself a morphism of formal group laws, not merely an inverse formal power series. Over a general ring, nonzero derivative is insufficient: it must be a unit.

Classical Kummer theory relates extraction of $n$th roots to Galois cohomology. In characteristic zero the exact sequenceand Hilbert theorem 90 identify $H^1(K,{\mu }_n)$ with $K^{\times }/K^{\times n}$. For an elliptic curve, the corresponding Kummer exact sequence of an elliptic curve isMultiplication by $n$ is surjective over the algebraic closure. If $nQ=P\in E(K)$, the cocycle $\sigma \mapsto \sigma Q-Q$ takes values in $E[n]$. Changing the choice of $Q$ changes it by a coboundary, and changing $P$ by an element of $nE(K)$ does not change its class. Conversely, a trivial cocycle class allows $Q$ to be adjusted by an $n$-torsion point to become $K$-rational. Thus the Kummer map of an elliptic curve is an injectionThis is the elliptic form of Kummer theory. The full cohomology group need not be finite; the arithmetic restriction on these classes is essential.

Choose a finite set $S$ of places containing the archimedean places, the primes over $n$, and all primes of bad reduction of an elliptic curve. At a finite place outside $S$, $E$ has good reduction and $n$ is a unit. A division point of the reduction of $P$ exists over the algebraic closure of the residue field. Smooth lifting gives a point over a finite unramified extension whose multiple differs from $P$ by an element of the kernel of reduction of an elliptic curve. In the formal group of an elliptic curve, $[n](T)=nT+O(T^2)$ is an isomorphism by the invertible morphism criterion for formal group laws, and its integral inverse converges on the maximal ideal. Correcting that difference produces a division point in the maximal unramified extension. Hence the Kummer map of an elliptic curve class is unramified outside $S$.

To prove finiteness explicitly, choose a finite Galois extension $L/K$ containing all $E[n]$ and all $n$th roots of unity, and enlarge $S$ to include its ramified places. A basis of $E[n]\cong (\mathbb{Z}/n\mathbb{Z}{)}^2$ identifies it over $L$ with ${\mu }_n^2$. Classical Kummer theory then identifiesThe restriction of every class in the image of $\delta$ belongs to $L(S_L,n{)}^2$, where the S-unramified power class group isIndeed an unramified local Kummer extension at residue characteristic prime to $n$ has valuation divisible by $n$: in an unramified field containing a root, $nv(a^{1/n})=v(a)$ with integral valuations.

The finiteness of S-unramified Kummer classes follows from the exact sequenceTo see the final map, write the ideal of $a$ away from $S_L$ as ${\mathfrak{a}}^n$ and take the ideal class of $\mathfrak{a}$. Its kernel is represented by an S-unit, after division by an $n$th power; conversely an $n$-torsion ideal class yields such an $a$. The S-unit group is finitely generated by the Dirichlet unit theorem together with the finitely many inverted primes. The ideal class group of the localized ring is a quotient of the finite ordinary ideal class group. Both outer groups are therefore finite.

Finally restriction has finite kernel: inflation-restriction puts it in the finite group $H^1(\mathrm{Gal}(L/K),E[n])$. Thus the image of $\delta$ has finite restriction image and finite kernel, andThis proves the weak Mordell-Weil theorem. Combining it with the height descent lemma proves the full Mordell-Weil theorem. Local restrictions at every place refine the finite group used here to the Selmer group of an elliptic curve, which is useful for explicit descent calculations.

In this setting a formal group law means a one-dimensional commutative series $F(X,Y)\in {\mathcal{O}}_K[[X,Y]]$ withThese identities force $F(X,Y)=X+Y+$ terms of total degree at least two. There is a unique formal inverse $i_F(T)=-T+O(T^2)$ with $F(T,i_F(T))=0$.

A morphism to another formal group law $G$ is a series $h(T)=cT+O(T^2)$ satisfying $h(F(X,Y))=G(h(X),h(Y))$. The invertible morphism criterion for formal group laws isNecessity follows by comparing linear coefficients in $h^{-1}\circ h=T$. For sufficiency, solve recursively for a compositional inverse $j(T)=c^{-1}T+\cdots$: at every stage the new coefficient is obtained by dividing by the unit $c$, so all coefficients stay in ${\mathcal{O}}_K$. Applying $j$ to the morphism identity shows that $j$ is itself a group homomorphism of formal group laws.

In particular, the multiplication series satisfies $[n{]}_F(T)=nT+O(T^2)$. Since $p\nmid n$, the multiplication isomorphism of a formal group law makes $[n{]}_F$ invertible over ${\mathcal{O}}_K$. Its series and inverse converge on ${\mathfrak{m}}_K$, so multiplication by $n$ is bijective on $E_1(K)$, and on $E_1(L)$ for every finite extension $L/K$.

A one-dimensional commutative formal group law over a ring $R$ is a series $F(X,Y)\in R[[X,Y]]$ satisfyingAn isomorphism $u:F\to G$ is a series $u(T)\in TR[[T]]$ with a compositional inverse and

The multiplication series is defined recursively by $[0{]}_F=0$, $[1{]}_F=T$ and $[n+1{]}_F=F([n{]}_F,T)$, with the formal inverse handling negative $n$. Its linear term isBy the invertible morphism criterion for formal group laws, it is an isomorphism exactly when its linear coefficient $n$ is a unit of $R$. Indeed, when $n\in R^{\times }$, recursive coefficient comparison constructs a unique compositional inverse; applying the morphism identity for $[n{]}_F$ shows that the inverse also respects $F$. Conversely, an invertible series must have a unit linear coefficient. Therefore

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 $E(L)/nE(L)$.

A one-dimensional commutative formal group law over ${\mathcal{O}}_K$ is a power series $F(X,Y)\in {\mathcal{O}}_K[[X,Y]]$ satisfyingA morphism $h:\mathcal{F}\to \mathcal{G}$ is a series $h(T)\in T{\mathcal{O}}_K[[T]]$ satisfyingIf $h(T)=uT+O(T^2)$, the invertible morphism criterion for formal group laws says that $h$ is an isomorphism whenever $u\in {\mathcal{O}}_K^{\times }$. Indeed, recursive coefficient comparison constructs a unique compositional inverse $j(T)$; applying $j$ to the morphism identity shows that $j$ is a morphism in the opposite direction.

The multiplication series hasSince $p\nmid n$, its linear coefficient is a unit, so $[n{]}_{\mathcal{F}}$ is an automorphism of the group $\mathcal{F}(\pi {\mathcal{O}}_K)$. Its kernel is therefore zero, and

Let $R$ be a complete discrete valuation ring of residue characteristic of a field $p$, with maximal ideal $\mathfrak{m}$, and let $F$ be a one-dimensional formal group law over $R$. Its multiplication series is $[n{]}_F(T)=nT+O(T^2)$. If $p\nmid n$, the invertible morphism criterion for formal group laws constructs its inverse with coefficients in $R$. Both formal power series converge on $\mathfrak{m}$, since their coefficients are integral and the powers of an element of $\mathfrak{m}$ tend to zero. Their compositions are the identity there. Thus multiplication by $n$ is bijective, including when $p=2$ and $n$ is odd. Every finite-order element of $F(\mathfrak{m})$ consequently has order of a group element equal to a prime power with prime $p$.