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For a Weierstrass equation of an elliptic curve, use the local parameter $t=-x/y$ and put $w=-1/y$. The formal kernel of a minimal Weierstrass equation construction gives $w=t^3+a_1{t}^4+(a_1^2+a_2)t^5+(a_1^3+2a_1{a}_2+a_3)t^6+O(t^7)$. Substitution into the elliptic curve group law yields the displayed initial coefficients and the formal inverse $-t/(1-a_{1}t-a_{3}w(t))$. In characteristic zero, the invariant differential on an elliptic curve becomes $(1+a_{1}t+(a_1^2+a_2)t^2+O(t^3))dt$, giving the formal logarithm $t+a_1{t}^2/2+(a_1^2+a_2)t^3/3+O(t^4)$. The addition coefficients are integral polynomial expressions and hence also apply in positive characteristic of a field.

Work near the identity $O$ of a generalized Weierstrass equation of an elliptic curve. On the projective chart where $Y\ne 0$, use the local parameter $t=-x/y$ and the auxiliary coordinate $w=-1/y$. Substitution givesThe coefficient of $w$ in the linearized equation at $(0,0)$ is one. Consequently, successive coefficient comparison determines a unique formal power series $w(t)$, without dividing by any coefficient of the equation. Its initial terms areRecover the affine coordinates as Laurent series $x=t/w(t)$ and $y=-1/w(t)$; in particular $x=t^{-2}+O(t^{-1})$ and $y=-t^{-3}+O(t^{-2})$.

The elliptic curve group law is a regular morphism near $(O,O)$. Applying the local parameter to the sum and passing to completed local rings producesOne can construct the series without any nonunit division. PutThe divided difference is an integral formal power series, including its tangent line specialization at $T_1=T_2$. The line $w=\lambda t+\nu$ meets the cubic in the two chosen points and a third point $R$. Substitution in the equation shows that its cubic and quadratic coefficients in $t$ are $-A,-B$, whereSince $\lambda$ has total degree at least two and $\nu$ at least three, $A$ is a unit. The sum of the roots therefore givesThe last expression reflects the third intersection according to the chord-and-tangent group law. Both denominators have constant term one. This proves integrality of the universal construction and also provides an explicit coefficient algorithm. Expanding it, the resulting formal group of an elliptic curve beginsThe integral coefficient recursion and the unit-denominator addition calculation work universally over $\mathbb{Z}[a_1,a_2,a_3,a_4,a_6]$; thus these coefficients specialize in every characteristic of a field. This is the initial coefficients of the elliptic formal group at infinity construction, rather than a definition requiring division by the elliptic-curve discriminant.

The identity, commutativity and associativity of the elliptic curve group law give the defining identities of a one-dimensional commutative formal group law:Reflection in the generalized Weierstrass equation of an elliptic curve gives its formal inverse explicitly:Successive addition gives the multiplication series $[n{]}_F(T)=nT+O(T^2)$.

In characteristic zero, an especially useful description comes from the invariant differential on an elliptic curve:Integrating with zero constant term constructs the formal logarithmTranslation invariance of the invariant differential on an elliptic curve implies $L(F(X,Y))=L(X)+L(Y)$. Thus the formal logarithm converts the formal group law into the formal additive group in characteristic zero. Expanding this identity to degree three also recovers the displayed coefficients of $F$; in positive characteristic the integral addition construction remains valid although the logarithmic denominators may not exist.

Here $F$ is the usual one-dimensional commutative formal group law. Normalize the valuation $v$ by $v(\pi )=1$, and put $e=v(p)$. We show explicitly that for any integer $m>e/(p-1)$,This subgroup will have finite index in $F(\pi {\mathcal{O}}_K)$.

Set $a(T)={\partial }_{Y}F(T,0)$. Since $F(X,Y)=X+Y+$ terms of total degree at least two, $a(T)\in 1+T{\mathcal{O}}_K[[T]]$. Its inverse has integral coefficients, say $a(T{)}^{-1}={\sum }_{j\ge 0}{c}_j{T}^j$ with $c_0=1$. Define the formal logarithm over $K$ byDifferentiate associativity $F(F(X,Y),Z)=F(X,F(Y,Z))$ with respect to $Z$ at zero. This gives $a(F(X,Y))={\partial }_{Y}F(X,Y)a(Y)$, soEvaluating at $Y=0$ proves the formal identity $\ell (F(X,Y))=\ell (X)+\ell (Y)$.

The coefficients satisfy $v(c_{n-1}/n)\ge -e{v}_p(n)$. Thus the series converges on $\pi {\mathcal{O}}_K$, since for $v(T)\ge 1$ the valuations of its terms are at least $n-e{v}_p(n)\to \infty$. More quantitatively,Indeed, if $k=v_p(n)$ then $n\ge p^k\ge 1+k(p-1)$. If $v(T)\ge m>e/(p-1)$, every nonlinear term has valuation strictly greater than $v(T)$. Write $h(T)=\ell (T)-T$. For $S,T\in {\pi }^m{\mathcal{O}}_K$, use $T^n-S^n=(T-S){\sum }_{j=0}^{n-1}{T}^{n-1-j}{S}^j$ to obtainThe extra valuation is integral and positive because $(n-1)m-e{v}_p(n)>0$ for all $n\ge 2$.

For any $b\in {\pi }^m{\mathcal{O}}_K$, solve $\ell (T)=b$ by iterating $T\mapsto b-h(T)$. This map preserves the complete ball ${\pi }^m{\mathcal{O}}_K$ and improves differences by at least one valuation step. The contraction mapping theorem gives a unique solution. Therefore $\ell$ is a bijection of this ball onto itself, and the formal identity, now evaluated in convergent series, makes it a group isomorphism. Multiplication by ${\pi }^{-m}$ then identifies the additive target with ${\mathcal{O}}_K$.

Integral coefficients in $F$ and its formal inverse ensure that every $F({\pi }^j{\mathcal{O}}_K)$ is a subgroup. Reduction modulo ${\pi }^m$ respects the group law; two parameters represent the same coset of $F({\pi }^m{\mathcal{O}}_K)$ exactly when they are congruent modulo ${\pi }^m$. If the residue field has $q_K$ elements, the quotient consequently has $q_K^{m-1}$ elements. This proves both the finite-index assertion and the claimed additive structure, with an explicit depth that also works in ramified extensions and when $p=2$.

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.

Here a formal group law means a one-dimensional commutative law $F(X,Y)\in R[[X,Y]]$ with $F(X,0)=X$, $F(0,Y)=Y$, $F(X,Y)=F(Y,X)$ and $F(F(X,Y),Z)=F(X,F(Y,Z))$. It has leading part $X+Y$ and a recursively determined formal inverse $i_F(T)=-T+O(T^2)$. A homomorphism $h(T)\in TR[[T]]$ satisfies $h(F(X,Y))=G(h(X),h(Y))$. It is an isomorphism precisely when its linear coefficient is a unit of $R$: necessity follows by composing with an inverse, while sufficiency follows by recursive inversion of the series and then the homomorphism identity. Over a field the condition is $\boxed{h^{\prime }(0)\ne 0}$.

The characteristic-zero classification is that every such law over $K$ has a unique isomorphism to the formal additive group with linear coefficient one. This is its formal logarithm. To prove it, put $f(T)={\partial }_{Y}F(T,0)$, whose constant term is one, and defineIntegration is coefficientwise, which is possible over the characteristic-zero field. Differentiate associativity with respect to $Z$ at zero to obtainIt follows that ${\partial }_Y{\ell }_F(F(X,Y))=1/f(Y)={\ell }_F^{\prime }(Y)$. The difference ${\ell }_F(F(X,Y))-{\ell }_F(Y)$ is therefore independent of $Y$; evaluating at $Y=0$ givesIts linear coefficient one makes it invertible as a formal series. Conversely, a homomorphism from $F$ to the additive law with derivative $c$ satisfies $h^{\prime }(T)f(T)=c$, by differentiating its defining identity at the second variable zero. Thus $h=c{\ell }_F$. In particular the normalized logarithm is unique, and every homomorphism between two laws over $K$ is ${\ell }_G^{-1}(c{\ell }_F(T))$, with an isomorphism exactly when $c\ne 0$. This is a formal classification; analytic convergence is a separate issue.

Now suppose $F$ is defined over ${\mathcal{O}}_K$. Write $v_K(\pi )=1$, $e=v_K(p)$ and let the residue field have size $q$. Since $f(T)$ and its reciprocal have integral coefficients, writing ${\ell }_F(T)=T+{\sum }_{n\ge 2}{c}_n{T}^n$ gives the valuation bound for coefficients of an integral formal logarithmTherefore the logarithm converges on $\pi {\mathcal{O}}_K$, as $n{v}_K(t)-e{v}_p(n)$ tends to infinity. Choose an integer $s>e/(p-1)$ and put $g(t)={\ell }_F(t)-t$. For $t,u\in {\pi }^s{\mathcal{O}}_K$, factoring $t^n-u^n$ givesThe last inequality uses $v_p(n)\le (n-1)/(p-1)$, since $n\ge p^{v_p(n)}\ge 1+(p-1)v_p(n)$. It follows that $g$ maps this ball into ${\pi }^{s+1}{\mathcal{O}}_K$ and is strictly contracting, gaining at least one unit of valuation in differences.

For any $z\in {\pi }^s{\mathcal{O}}_K$, iterate $t_{j+1}=z-g(t_j)$ from $t_0=z$. The iterates remain in the ball, their differences gain valuation at each step, and completeness gives a limit solving ${\ell }_F(t)=z$. The same difference estimate makes this solution unique. Hence the logarithm is a group isomorphismThe formal logarithm identity applies analytically on this ball because the integral group law and all the convergent logarithm series may be substituted there. Finally reduction of the formal parameter modulo ${\pi }^s$ is a group homomorphism from $F(\pi {\mathcal{O}}_K)$ onto the finite group on $\pi {\mathcal{O}}_K/{\pi }^s{\mathcal{O}}_K$, with kernel $F({\pi }^s{\mathcal{O}}_K)$. Its index is $q^{s-1}$. Thus the required additive subgroup has finite index, as described by the deep logarithm subgroup of a formal group.

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 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.

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 $L/K$.

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 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 $i(X)\equiv -X(\mathrm{mod}{X}^2)$.

Repeated formal addition givesfor positive $n$, and the same formulas extend to all $n\in \mathbb{Z}$ using the formal inverse.

Now let $R$ be a field of characteristic $p>0$ and let $h:{\hat{\mathbb{G}}}_a\to {\hat{\mathbb{G}}}_m$ be a homomorphism. Compatibility with multiplication by $p$ givesThe left side is $h(0)=0$, while the Frobenius identity gives the right side asThe formal power series ring over a field is an integral domain, so $h(X{)}^p=0$ implies $h=0$. Hence there are no nonzero homomorphisms from the formal additive group to the formal multiplicative group.