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All geometric 2-torsion is rational: its points are $O,(\alpha ,0),(\beta ,0),(\gamma ,0)$. We will construct the required finite quotient directly, without assuming any theorem about finite generation or two-descent on an elliptic curve.

Choose $Q\in E(\overline{K})$ with $2Q=P$ and define the halving cocycle with rational two-torsionSuch a half exists because the nonconstant multiplication morphism $[2]$ on a smooth projective elliptic curve is surjective over an algebraic closure. Since $E[2]$ is rational, $c_P(\sigma \tau )=c_P(\sigma )+c_P(\tau )$. A different half is $Q+T$ with rational $T\in E[2]$ and gives the same cocycle. Replacing $P$ by $P+2R$, with $R\in E(K)$, allows the half $Q+R$ and again changes nothing. Adding chosen halves shows additivity in $P$. If the cocycle is zero, the half is fixed by the absolute Galois group and lies in $E(K)$, so $P\in 2E(K)$. This proves, rather than assumes, an injectionThe fixed field of $\ker c_P$ is exactly $K(Q)$. Its Galois group is the image of $c_P$, so its degree is at most four. This also follows from the permitted biquadratic description.

Let $S$ be the finite set of primes dividing $2(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$. At a place $v\notin S$ the displayed integral cubic has unit elliptic-curve discriminant, hence good reduction of an elliptic curve, and the residue characteristic is odd. Choose a half $\overline{Q}$ of the reduction of $P$ over the algebraic closure of the residue field. Its coordinates belong to a finite residue extension; let $L_v/K_v$ be the corresponding finite unramified extension. Smoothness and the Hensel lemma lift $\overline{Q}$ to some $Q_0\in E(L_v)$. The difference $P-2Q_0$ reduces to zero and belongs to the formal kernel of a minimal Weierstrass equation. In its integral formal group law the doubling series is $2t+O(t^2)$, with unit linear coefficient. The prime-to-residue-characteristic multiplication on a formal group gives a unique $R$ in this kernel such that $2R=P-2Q_0$. Thus $Q_0+R$ is a half of $P$ defined over an unramified local extension. Every other half differs by rational 2-torsion, so it too is unramified. We have proved the unramified halving fields for a split cubic assertion at every $v\notin S$.

There are only finitely many bounded-degree extensions with restricted ramification of $K$ of degree at most four. This is the permitted number-field finiteness consequence of the Hermite–Minkowski theorem: bounded local degrees bound the discriminant exponents at the finitely many allowed ramified primes, so absolute discriminants are bounded. Each possible halving field $L$ is Galois, and there are only finitely many homomorphisms $\mathrm{Gal}(L/K)\to E[2]$. The injection constructed above therefore has finite image. Consequently $E(K)/2E(K)$ is finite. No elliptic-curve finiteness or descent theorem was used in this argument.

For $K=\mathbb{Q}$, write $s=\#S$. Each coordinate of $c_P$ in a basis of $E[2]\cong (\mathbb{Z}/2\mathbb{Z}{)}^2$ is a quadratic Galois character unramified outside $S$. Its quadratic extension has a signed square-free integer representative $d=\pm {\prod }_{p\in S}{p}^{e_p}$, $e_p\in \{0,1\}$: an odd valuation outside $S$ would ramify there. There are at most $2^{s+1}$ choices, including the trivial character. Hence the pair of characters gives at most $2^{2s+2}$ cocycles. Each prime in $S$ is at most $2m$, because the three root differences have absolute values at most $2m$ and $2\le 2m$. Among the integers up to $2m$ there are at most $m$ primes: two, together with at most $m-1$ odd candidates. The rational halving quotient bound from root size is therefore

For an integral Minimal Weierstrass equation, let $E_0({\mathbb{Q}}_p)$ be the subgroup of points whose reduction is a smooth point of the reduced cubic. Its reduction is a homomorphism to the smooth-locus group of a singular Weierstrass cubic, or to the whole reduced elliptic curve in the good-reduction case. Define the formal kernel of a minimal Weierstrass equation byIt consists of the identity and the points reducing to the identity at infinity. For a nonidentity affine point this is equivalent to $v_p(x)=-2r$ and $v_p(y)=-3r$ for some positive integer $r$.

The local parameter $t=-x/y$, with $t(O)=0$, gives the group isomorphismwhere $\hat{E}$ is the formal group of an elliptic curve. To see the local coordinates, put $w=-1/y$ in the Weierstrass equation of an elliptic curve. The resulting relation is $w=t^3+a_{1}tw+a_2{t}^{2}w+a_3{w}^2+a_{4}t{w}^2+a_6{w}^3$, and its unit linear coefficient gives a unique integral series $w(t)=t^3+O(t^4)$. Thus $x=t/w$ and $y=-1/w$ have valuations $-2v_p(t)$ and $-3v_p(t)$. The chord-and-tangent group law expressed in this parameter is precisely the integral formal group law $\hat{E}$. This description includes bad reduction; only the smooth identity neighborhood is involved.

In the one-dimensional commutative convention, a formal group law over $R$ is a formal power series $F(X,Y)=X+Y+\text{terms of degree at least two}$ satisfying $F(X,0)=X$, $F(0,Y)=Y$, $F(X,Y)=F(Y,X)$ and $F(F(X,Y),Z)=F(X,F(Y,Z))$. The formal inverse is a series $I(T)=-T+O(T^2)$ with $F(T,I(T))=0$. For $R={\mathbb{Z}}_p$ all these series converge on $p{\mathbb{Z}}_p$, which becomes a group with operation $F$.

Repeated formal addition gives the multiplication isomorphism of a formal group law $[n{]}_F(T)=nT+O(T^2)$. If $p\nmid n$, the coefficient $n$ is a unit in the p-adic integers. We can construct an inverse $H(T)=n^{-1}T+{\sum }_{j\ge 2}{h}_j{T}^j$ by coefficient recursion. At degree $j$, the equation $[n{]}_F(H(T))=T$ has the form $n{h}_j+\text{known integral terms}=0$; dividing by the unit $n$ keeps $h_j$ integral. The inverse is two-sided by uniqueness of formal composition inverses. Since integral coefficients multiplied by successive powers of $t\in p{\mathbb{Z}}_p$ tend to zero, both series converge on that set and their formal identities hold there. This proves prime-to-residue-characteristic multiplication on a formal group:The argument also applies at $p=2$ for odd $n$; it does not require a logarithm or torsion-freeness.

In this setting a formal group law means a one-dimensional commutative series $F(X,Y)\in R[[X,Y]]$ with identity zero andIt starts with $X+Y$; there is a unique inverse series $\iota (T)=-T+\cdots$ satisfying $F(T,\iota (T))=0$. For coefficients in $\mathbb{Z}$, the series and its inverse converge on $p{\mathbb{Z}}_p$, making this ideal a topological abelian group under $F$.

Set $a(T)={\partial }_{Y}F(T,0)\in 1+T\mathbb{Z}[[T]]$, and write $a(T{)}^{-1}={\sum }_{j\ge 0}{b}_j{T}^j$. The formal logarithm isDifferentiate associativity in the third variable at zero. It gives $a(F(X,Y))={\partial }_{Y}F(X,Y)a(Y)$, so ${\partial }_{Y}L(F(X,Y))=L^{\prime }(Y)$. Subtracting $L(Y)$ leaves a series independent of $Y$, whose value at zero is $L(X)$. Thus $L(F(X,Y))=L(X)+L(Y)$.

For odd $p$, the logarithm converges on $I=p{\mathbb{Z}}_p$: the valuation of its degree-$n$ term is at least $n-v_p(n)$, which tends to infinity. Put $u(T)=L(T)-T$. For $x,y\in I$, factor $x^n-y^n$ to obtainfor every $n\ge 2$ and odd $p$. Hence $u$ is a contraction with constant at most $1/p$ and maps $I$ into $pI$. The equation $L(x)=z$ is equivalent to $x=z-u(x)$; the contraction mapping theorem gives a unique solution in $I$ for every $z\in I$. The same estimate shows $|L(x)-L(y){|}_p=|x-y{|}_p$, so the logarithm is a topological group isomorphism. Dividing its values by $p$ givesThis is the deep logarithm subgroup of a formal group argument, here with depth one.

For $p=2$, the degree-two estimate at depth one need not be strict. At depth two, however,so the same proof gives $F(4{\mathbb{Z}}_2)\cong (4{\mathbb{Z}}_2,+)\cong {\mathbb{Z}}_2$. This subgroup has index two in $F(2{\mathbb{Z}}_2)$, because $F(x,y)\equiv x+y(\mathrm{mod}4)$ when $x,y\in 2{\mathbb{Z}}_2$.

Indeed the topological structure of a formal group on twice the 2-adic integers has just two possibilities. Let $G=F(2{\mathbb{Z}}_2)$, let its index-two subgroup $H$ have topological generator $h$, and choose $v\notin H$. Write $2v=ch$, with $c\in {\mathbb{Z}}_2$, using group notation. If $c$ is odd, $2v$ generates $H$ and $v$ generates $G$, giving $G\cong {\mathbb{Z}}_2$. If $c$ is even, $v-(c/2)h$ has order two and lies outside $H$, giving $G\cong H\times \mathbb{Z}/2\mathbb{Z}$. The ${\mathbb{Z}}_2$ multiples are defined by continuity in this compact pro-two group.

For explicit contrasting examples, the formal additive group $F_a(X,Y)=X+Y$ gives $F_a(2{\mathbb{Z}}_2)\cong {\mathbb{Z}}_2$, which is a torsion-free group. The formal multiplicative group $F_m(X,Y)=X+Y+XY$ is identified by $x\mapsto 1+x$ with $1+2{\mathbb{Z}}_2$. Its element $x=-2$ corresponds to $-1$ and has order two. Moreoverand the logarithm identifies the second factor with $4{\mathbb{Z}}_2$. The two resulting groups are ${\mathbb{Z}}_2$ and $\mathbb{Z}/2\mathbb{Z}\times {\mathbb{Z}}_2$, so they are not isomorphic.

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.

Here a formal group means a one-dimensional commutative formal group law. Over a ring $R$, it is a formal power series $F(X,Y)\in R[[X,Y]]$ satisfyingIn particular, $F(X,Y)=X+Y+$ terms of total degree at least $2$. A formal inverse $\iota (T)=-T+\cdots$ with $F(T,\iota (T))=0$ is obtained recursively. An isomorphism of formal group laws from $F$ to $G$ is a series $h(T)\in TR[[T]]$ with invertible linear coefficient andIts compositional inverse exists over $R$ by coefficient recursion.

For $R={\mathcal{O}}_K$, all these series converge on $\pi {\mathcal{O}}_K$; the operations preserve that set and make it the group $F(\pi {\mathcal{O}}_K)$. Use the uniformizer $\pi$ and normalize the valuation by $v_K(\pi )=1$, and write $e=v_K(p)$.

We explicitly construct its formal logarithm. DefineThen putThis is a formal series over $K$ with linear coefficient $1$. Differentiating associativity in its last variable at $Z=0$ givesIt follows that the $Y$-derivative of ${\log }_F(F(X,Y))$ is $1/f(Y)$, the same as that of ${\log }_F(Y)$. Evaluating at $Y=0$ therefore provesThe differential $dT/f(T)$ is the invariant differential of a formal group law.

Choose an integer $m>e/(p-1)$. In degree $n\ge 2$, the coefficient of ${\log }_F(T)$ has valuation at least $-e{v}_p(n)$. Thus the scaled serieshasHere $v_p(n)\le (n-1)/(p-1)$, and the coefficient valuations tend to infinity. Consequently $H(U)=U+\pi R(U)$ with $R$ a convergent integral power series on ${\mathcal{O}}_K$. Such a series is $1$-Lipschitz. For any $a\in {\mathcal{O}}_K$, the equation $H(U)=a$ is equivalent to $U=a-\pi R(U)$, a strict contraction, to which the contraction mapping theorem applies on the complete ring ${\mathcal{O}}_K$. It has exactly one solution.

The deep logarithm subgroup of a formal group is obtained as follows. The formal group exponential is the formal compositional inverse ${\exp }_F(T)$ of ${\log }_F(T)$, obtained recursively; equivalently it solves ${\exp }_F^{\prime }(T)=f({\exp }_F(T))$, with ${\exp }_F(0)=0$. The preceding contraction, applied to the scaled series, constructs its convergent inverse on ${\pi }^m{\mathcal{O}}_K$ and agrees with that formal inverse. This supplies the required convergence sketch for ${\exp }_F$ as well as a bijectivity proof for the logarithm.

The integral formal law and inverse preserve ${\pi }^m{\mathcal{O}}_K$, so this is a subgroup. The logarithm identity and bijectivity giveReduction of the formal law modulo ${\pi }^m$ has kernel $F({\pi }^m{\mathcal{O}}_K)$. Its underlying image set is $\pi {\mathcal{O}}_K/{\pi }^m{\mathcal{O}}_K$, which has size $q^{m-1}$, where $q=\#({\mathcal{O}}_K/\pi )$. Thus this subgroup has finite index, explicitly $q^{m-1}$.

An elliptic curve has good reduction over $K$ if it admits an integral Weierstrass equation of an elliptic curve with unit elliptic-curve discriminant. Equivalently, the reduction of its Minimal Weierstrass equation is a nonsingular cubic over the residue field $k$. Its smooth integral projective model is a group scheme with the point $O$ as identity.

Choose homogeneous coordinates in the valuation ring ${\mathcal{O}}_K$ with at least one coordinate a unit, and reduce them modulo its maximal ideal ${\mathfrak{m}}_K$. This defines $\mathrm{red}:E(K)\to \tilde{E}(k)$. The chord-and-tangent group law extends over the smooth integral model, so reduction is a group homomorphism. For any point of $\tilde{E}(k)$, choose an affine projective chart containing it. Smoothness means some partial derivative of its equation is nonzero in $k$. Lift the other coordinate arbitrarily and apply the Hensel lemma to the coordinate with unit derivative. This constructs a point of $E(K)$ above the prescribed point, proving the surjectivity of good reduction over a local field.

The kernel of reduction of an elliptic curve consists precisely of points reducing to $O$. The parameter $t=-x/y$ at $O$ gives a bijection between this kernel and ${\mathfrak{m}}_K$: the integral formal-coordinate series solve the equation uniquely near $O$. Addition in these coordinates is the formal group of an elliptic curve, not ordinary addition in ${\mathfrak{m}}_K$. ThusHere the group operation on the last expression is its formal group law.

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 $L/K$.

For $p\nmid 2k$, the displayed integral equation has nonzero elliptic-curve discriminant modulo $p$, so it has good reduction. Let $\chi$ be the Legendre symbol, with $\chi (0)=0$. Counting the two, one, or zero possible ordinates over each $x$ givesIf $p\equiv 3(\mathrm{mod}4)$, then $\chi (-1)=-1$ and the terms for $x$ and $-x$ cancel. The term at zero vanishes. Thus the count is $p+1$.

For the converse, suppose $p\equiv 1(\mathrm{mod}4)$, and put $h=(p-1)/2$. The Euler criterion and the sum of powers over a finite field show that the Trace of Frobenius satisfiesIndeed, among the positive exponents in this polynomial, of degree $3(p-1)/2<2(p-1)$, only $p-1$ has a nonzero sum over ${\mathbb{F}}_p$, and that sum is $-1$. Since $(x^3+kx{)}^h=x^h(x^2+k{)}^h$, the coefficient formula for trace of Frobenius modulo p yieldsNeither $k$ nor the binomial coefficient vanishes modulo $p$. Hence $a_p\ne 0$, proving the point-count criterion for y squared equals x cubed plus k x:

We use the following formal logarithm fact, also useful for the next part. For a formal group law over ${\mathbb{Z}}_p$, its logarithm has coefficients $c_j/j$ with $c_j\in {\mathbb{Z}}_p$, obtained by integrating its integral invariant differential. If $p$ is odd and $t\in p{\mathbb{Z}}_p\setminus \{0\}$, every term of degree $j\ge 2$ has valuation strictly greater than $v_p(t)$, since $j{v}_p(t)-v_p(j)>v_p(t)$. Thus the logarithm converges and is nonzero at $t$. Its homomorphism into the additive characteristic-zero group proves torsion-freeness of the formal group over Qp for odd p. Consequently reduction injects the entire rational torsion subgroup at any odd prime of good reduction, including its $p$-primary torsion. In particular the torsion subgroup is finite, and its order divides $p+1$ at every good $p\equiv 3(\mathrm{mod}4)$.

By the Dirichlet theorem on primes in arithmetic progressions, choose a good prime $p\equiv 3(\mathrm{mod}8)$. Then $v_2(p+1)=2$, bounding the two-primary part by four. For any odd prime $\ell$ dividing the torsion order, the Chinese remainder theorem and the Dirichlet theorem on primes in arithmetic progressions give a good prime $p$ with $p\equiv 3(\mathrm{mod}4)$ and $p\equiv 1(\mathrm{mod}\ell )$, avoiding the finitely many primes dividing $2k$. But then $\ell \nmid p+1$, a contradiction. Therefore

The order four does occur. For $k=4$, the elliptic-curve addition formula gives $2(2,4)=(0,0)$, so $(2,4)$ has order four. More precisely, the only nonzero rational 2-torsion point when $k>0$ is $(0,0)$, so order four means a cyclic group with a rational half of this point. The duplication formula isSuch a half must have $x(P{)}^2=k$. Writing $k=s^2$, positivity of $y(P{)}^2=2kx(P)$ forces $x(P)=s>0$. Then $y(P{)}^2=2s^3$, so $2s$ is a rational square, equivalently $s=2t^2$ for an integer $t\ge 1$. Conversely $(2t^2,4t^3)$ is a half of $(0,0)$ when $k=4t^4$. The rational torsion on y squared equals x cubed plus positive k x is therefore cyclic of order four exactly for these $k$, and otherwise cyclic of order two.

Fix a non-Archimedean local field $K$, its valuation ring $R={\mathcal{O}}_K$, a uniformizer $\pi$, and a residue field with cardinality $q$. The construction below applies both in mixed and equal characteristic. A one-dimensional commutative formal group law is $F(X,Y)\in R[[X,Y]]$ with identity, associativity and commutativity, and $F(X,Y)=X+Y+$ higher-degree terms. It has a unique formal inverse. Its morphisms are series $h(T)\in TR[[T]]$ with $h(F(X,Y))=G(h(X),h(Y))$; a unit linear coefficient makes such a morphism invertible by recursive coefficient comparison. Evaluating its integral series on the maximal ideal of a finite extension of $K$ is legitimate because the series converge there.

Choose a Lubin–Tate series $f(T)$ satisfyingA useful choice is $f(T)=\pi T+T^q$. The Lubin–Tate functional equation lemma constructs a unique series with any prescribed linear form $\ell$ and satisfyingHere is its coefficient argument. If terms below degree $d$ have been found, the unknown homogeneous degree-$d$ term appears with multiplier $\pi -{\pi }^d$. The already known error has every coefficient divisible by $\pi$: after reduction, the equation becomes $H(X{)}^q=H(X_1^q,\dots ,X_r^q)$, valid because the residue coefficients lie in ${\mathbb{F}}_q$. Since $\pi -{\pi }^d$ is $\pi$ times a unit for $d\ge 2$, the next term is uniquely determined and integral. Recursion proves existence and uniqueness.

Use $\ell =X+Y$ to construct $F_f$, and $\ell =aT$ to construct $[a{]}_f$ for every $a\in R$. Uniqueness gives $F_f(X,0)=X$ and $F_f(0,Y)=Y$. Comparing $[a{]}_f(F_f(X,Y))$ with $F_f([a{]}_f(X),[a{]}_f(Y))$, which satisfy the same functional equation and have the same linear form, proves that $[a{]}_f$ is an endomorphism of the law. The same uniqueness applied to three variables proves associativity: $F_f(F_f(X,Y),Z)$ and $F_f(X,F_f(Y,Z))$ have the same linear form and functional equation. It also proves commutativity and the identitiesThus this Lubin–Tate formal group is equipped with an action of the whole ring $R$, not only its integers. For a unit $a$, the inverse of $[a{]}_f$ is $[a^{-1}{]}_f$. The same recursion, with two series for the fixed $\pi$, gives a unique strict Lubin–Tate change of series. Its integral series and inverse converge on torsion points, so it identifies their torsion fields. We may compute with $f(T)=\pi T+T^q$ without losing the general construction for that uniformizer.

Write $f_0(T)=T$ and $f_n=f^{\circ n}$. The Lubin–Tate torsion of level $n$ isFor $n\ge 1$, its primitive points are the roots ofThis is a monic polynomial of degree $D_n=(q-1)q^{n-1}$. Modulo $\pi$ it is $T^{D_n}$, and its constant coefficient is exactly $\pi$. Thus the Eisenstein layers of Lubin–Tate torsion are Eisenstein polynomials. A root ${\lambda }_n$ generates a totally ramified extension of degree $D_n$, and has $v_K({\lambda }_n)=1/D_n$ in the uniquely extended valuation. Also $f_{n-1}({\lambda }_n{)}^{q-1}=-\pi \ne 0$, so it has exact level $n$.

The mapis well-defined and injective. If $a={\pi }^{r}u$ with $r<n$ and $u$ a unit, then $[a]{\lambda }_n=[u]f_r({\lambda }_n)\ne 0$, because $[u]$ is invertible and ${\lambda }_n$ is primitive. Conversely every multiple of ${\pi }^n$ kills it. The image therefore contains $q^n$ distinct roots of the degree-$q^n$ polynomial $f_n$, so it is all of ${\Lambda }_n$. This proves, without merely assuming the torsion cardinality,Its primitive generators are exactly $[a]{\lambda }_n$ with $a\in (R/{\pi }^{n}R{)}^{\times }$.

Define the Lubin–Tate extension $K_{\pi ,n}=K({\Lambda }_n)$. Every integral endomorphism series evaluated at ${\lambda }_n$ converges in the complete field $K({\lambda }_n)$, so all torsion points already lie there. HenceIt is the splitting field of $f_n$, which has $q^n$ distinct roots by the module calculation. Thus it is Galois. Every K-automorphism preserves the uniquely extended absolute value on a field, hence is continuous and commutes with limits of the integral endomorphism series. Its automorphisms therefore commute with the endomorphism action and send a primitive generator to a primitive generator. There is a unique unit class $a_{\sigma }$ such that $\sigma ({\lambda }_n)=[a_{\sigma }]{\lambda }_n$. The Lubin–Tate Galois action on primitive torsion gives an injective homomorphismBoth sides have order $D_n=(q-1)q^{n-1}$, so it is an isomorphism. In particular every finite layer is abelian and totally ramified.

For an example, take $K={\mathbb{Q}}_p$, $\pi =p$, and the equally admissible series $f(T)=(1+T{)}^p-1$. Its formal group law is the formal multiplicative group $F(X,Y)=X+Y+XY$, with $[a](T)=(1+T{)}^a-1$ for $a\in {\mathbb{Z}}_p$. The binomial polynomials make its coefficients integral. Its level-$n$ torsion points are $\zeta -1$ with ${\zeta }^{p^n}=1$. Thus the Lubin–Tate cyclotomic example gives $K_{p,n}={\mathbb{Q}}_p({\mu }_{p^n})$ and Galois action $\zeta \mapsto {\zeta }^a$. The order formula includes $p=2,n=1$, when the first field and its Galois group are trivial over ${\mathbb{Q}}_2$.

Finally choose primitive points compatibly, $[\pi ]{\lambda }_{n+1}={\lambda }_n$. They exist by taking a root of $f(T)-{\lambda }_n$ in the algebraic closure; its roots have positive valuation and exact level $n+1$. Then $K_{\pi ,n}\subseteq K_{\pi ,n+1}$, and restriction of Galois automorphisms corresponds to reduction of unit classes. The Lubin–Tate tower $K_{\pi }={\bigcup }_n{K}_{\pi ,n}$ consequently has the topological Galois groupThe final isomorphism uses completeness of the discrete valuation ring $R$. This computes the finite-layer and infinite-tower Galois groups directly from the formal action and the primitive Eisenstein polynomials.

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

A one-dimensional commutative formal group law over ${\mathcal{O}}_K$ is a power series $F(X,Y)\in {\mathcal{O}}_K[[X,Y]]$ satisfyingFor $r\ge 1$, the ideal ${\pi }^r{\mathcal{O}}_K$ becomes a group, denoted $F({\pi }^r{\mathcal{O}}_K)$, under $x{+}_{F}y=F(x,y)$; convergence follows because both inputs lie in the maximal ideal.

Over the characteristic-zero field $K$, there is a unique formal logarithmsatisfying ${\log }_F(F(X,Y))={\log }_F(X)+{\log }_F(Y)$. Its coefficients have bounded denominator growth, so for sufficiently large $r$ both ${\log }_F$ and its inverse formal group exponential converge on ${\pi }^r{\mathcal{O}}_K$ and preserve that ideal. They givewhere the final isomorphism is multiplication by ${\pi }^{-r}$.

A one-dimensional commutative formal group law over a ring $R$ is a series $F(X,Y)\in R[[X,Y]]$ satisfyingThe identity axiom gives $F(X,Y)=X+Y+$ terms of total degree at least two. SeekAfter $a_2,\dots ,a_{n-1}$ have been chosen, the coefficient of $X^n$ in $F(X,i(X))$ is $a_n$ plus a known expression in the earlier coefficients. There is therefore a unique choice of $a_n$ making it zero. Recursion constructs the formal inverse $i(X)$ with $F(X,i(X))=0$ and $i(X)\equiv -X(\mathrm{mod}{X}^2)$.

A one-dimensional commutative formal group law over a ring $R$ is a power series $F(X,Y)\in R[[X,Y]]$ satisfying $F(X,0)=X$, commutativity, and associativity. A morphism $f:F\to G$ is a series $f(T)\in TR[[T]]$ satisfying

Express the isogeny of part b in the formal coordinates of part d and setPart c shows that points approaching $O$ map to points approaching $O^{\prime }$, so $f(t)\in tk[[t]]$. Since $\varphi$ is a group homomorphism, applying the parameter $t^{\prime }$ to $\varphi (P+Q)=\varphi (P)+\varphi (Q)$ givesThus $f$ is the formal-group morphism induced by an isogeny $\hat{E}\to \widehat{E^{\prime }}$.

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