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Over an algebra over a field whose scalar field is $\mathbb{Q}$, the series $\exp (X)-1$ defines an isomorphism from the formal additive group to the formal multiplicative group, with inverse $\log (1+X)$.

For $K={\mathbb{Q}}_p$ and $\pi =p$, choose the Lubin–Tate series $f(T)=(1+T{)}^p-1$. The associated formal group law is the formal multiplicative group. Its torsion points are $\zeta -1$ for ${\zeta }^{p^n}=1$, so its Lubin–Tate extension is the cyclotomic extension of a p-adic field ${\mathbb{Q}}_p({\mu }_{p^n})$. The Galois action $\zeta \mapsto {\zeta }^a$ gives $(\mathbb{Z}/p^n\mathbb{Z}{)}^{\times }$, including the trivial first layer when $p=2$.

For a uniformizer $\pi$ of a non-Archimedean local field $K$ with residue field of cardinality $q$, a Lubin–Tate series is a series $f(T)\in {\mathcal{O}}_K[[T]]$ satisfyingIt determines the associated two-variable Lubin–Tate formal group. A useful example is $f(T)=\pi T+T^q$. For $K={\mathbb{Q}}_p$ and $\pi =p$, the choice $(1+T{)}^p-1$ gives the formal multiplicative group and the Lubin–Tate cyclotomic example.

The multiplicative algebraic group has ${\mathbb{G}}_m(K)=K^{\times }$, with multiplication as its group operation. Its coordinate ring is $K[z,z^{-1}]$. The formal multiplicative group describes its law near the identity $z=1$.

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.

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.

The formal additive group and formal multiplicative group have lawsIf $R$ is an algebra over a field whose scalar field is $\mathbb{Q}$, the serieshas linear coefficient one and satisfiesIt is therefore the exponential isomorphism between the additive and multiplicative formal groups, with inverse $\log (1+X)$.

For a one-dimensional commutative integral formal group law, the formal logarithm identifies $F(4{\mathbb{Z}}_2)$ with $4{\mathbb{Z}}_2$. This is an index-two subgroup of $F(2{\mathbb{Z}}_2)$. Choose a topological generator $h$ of the subgroup and an element $v$ outside it, and write $2v=ch$. If $c$ is odd, $v$ generates the whole group. If $c$ is even, $v-(c/2)h$ has order two and splits off a cyclic factor. The formal additive group and formal multiplicative group realize the two cases.