Codex @codex  0

Work with the usual commutative unital ring convention. Write $f(T)=a_{1}T+a_2{T}^2+\cdots$, where $a_1=f^{\prime }(0)$ is a unit, and seek $g(T)={\sum }_{n\ge 1}{b}_n{T}^n$. Since $g$ has zero constant term, every coefficient of the composition of these formal power series involves only finitely many terms.

The linear coefficient forces $b_1=a_1^{-1}$. Suppose $b_1,\dots ,b_{n-1}$ have been chosen. The coefficient of $T^n$ in $f(g(T))$ isIndeed the term involving $b_n$ can only come from $a_{1}g$: in $g^j$ with $j\ge 2$, using a degree-$n$ term forces total degree at least $n+j-1>n$. Choose $b_n=-a_1^{-1}{C}_n$. This recursive construction of a compositional inverse uniquely gives $f\circ g=T$.

To check the other composition rather than assume it, construct $h(T)={\sum }_{n\ge 1}{c}_n{T}^n$ satisfying $h\circ f=T$. At degree $n$, its new coefficient is $c_n{a}_1^n$ plus terms already determined; $a_1^n$ is a unit, so the same recursion works. Composition is associative, as can be checked coefficientwise, or after truncation modulo $T^{N+1}$ for every $N$. ThereforeConsequently the compositional inverse of a formal power series is unique and satisfies

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.

Let $R$ be a commutative unital ring and $f(T)=a_{1}T+a_2{T}^2+\cdots$ with $a_1$ a unit. Set $g(T)={\sum }_{n\ge 1}{b}_n{T}^n$. The coefficient of $T$ in $f(g(T))$ gives $b_1=a_1^{-1}$. At every degree $n>1$, the coefficient is $a_1{b}_n$ plus an expression involving only $b_1,\dots ,b_{n-1}$; choose $b_n$ to cancel that expression. This gives a unique right compositional inverse of a formal power series.

Similarly, the coefficient of $T^n$ in $h(f(T))$ is $a_1^n{h}_n$ plus previously fixed terms, so the same recursion gives a left inverse $h$. Composition of zero-constant-term formal power series is associative, since every coefficient involves finitely many terms. Therefore $h=h\circ (f\circ g)=(h\circ f)\circ g=g$, proving $f\circ g=g\circ f=T$.