Compositional inverse of a formal power series ($f(T)=a_{1}T+O(T^2),a_1\in R^{\times }$)

A formal power series with zero constant coefficient has a compositional inverse over $R$ exactly when its linear coefficient is a unit. Starting with $b_1=a_1^{-1}$, the degree-$k$ coefficient of $f(b_{1}T+\cdots +b_k{T}^k)$ is $a_1{b}_k$ plus an expression already determined by lower coefficients, so recursion gives a unique inverse. For a homomorphism of formal group laws, cancellation in the homomorphism identity proves that this inverse is also a formal-group homomorphism.

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 $f\circ g=g\circ f=T$.