Cfrgtkky

A formal power series $f in R[[X]]$ is said to be invertible wrt. composition, iff there exists $g in R[[X]]$ s.t. $f circ g = g circ f = X$ holds.

It is easy to see, that for such $f = sum_{n=0}^infty f_n X^n$ necessarily $f_0 = 0$ and $f_1 in R^times$ holds. I am looking for a proof, that this condition is also sufficient. Is there an elegant proof which avoids using an explicit formula for the coefficients of compositions $f circ g$?

Thank you in advance!

You can use Hensel's lemma in the following variant:

Hensel's lemma: Let $A$ be an integral domain which is complete w.r.t. an ideal $I$. Suppose we are given $F in A[[Y]]$ and $a in I$ such that $F(a) equiv 0 bmod I^s$ for some $s geq 1$, and $F'(a) in A^times$. Then there exists a unique $b in I$ such that $F(b) = 0$ and $b equiv a bmod I^s$.

Given $f(X) = f_1X + ldots in R[[X]]$ with $f_1 in R^times$ (the dots mean terms of higher order), take $A = R[[X]]$, $F(Y) = f(Y)-X in R[[X]][[Y]]$, $a = f_1^{-1}X$ and $s = 2$ in the lemma. We have
$$F(f_1^{-1}X) = f_1(f_1^{-1}X) + ldots - X equiv 0 mod {X^2},$$
$$F'(f_1^{-1}X) = f_1 + ldots in R[[X]]^times,$$
so Hensel's lemma gives a unique $g in R[[X]]$ such that $f(g(X)) = X$ and $g(X) = f_1^{-1}X +ldots$. The same argument with $g$ instead of $f$ gives a unique $h(X) = f_1X + ldots in R[[X]]$ such that $g(h(X)) = X$. Now
$$f(X) = f(g(h(X))) = h(X),$$
so $h=f$ and thus $f(g(X)) = g(f(X)) = X$.

Edit 11/2018: I think you can prove the lemma as follows. Replacing $F(Y)$ with $F(Y+a)F'(a)^{-1}$ we can assume $a = 0$ and $F'(0) = 1$. Then we put $x_0 = 0$ and $x_{n+1} := x_n - F(x_n)$. Then $x_n equiv 0 bmod I^s$ for all $n geq 0$ by induction. We claim:
$$x_{n+1} equiv x_n bmod I^{n+s} quad text{for all $n geq 0$}.$$
The case $n=0$ is clear. Assume $x_n equiv x_{n-1} bmod I^{n+s-1}$ for the induction. We can write
$$F(Y) - F(Z) = (Y-Z)(1+YG(Y,Z)+ZH(Y,Z))$$
for certain $G, H in A[[Y,Z]]$. We find
$$F(x_n) - F(x_{n-1}) equiv x_n - x_{n-1} bmod I^{n+s},$$
hence $x_n - F(x_n) equiv x_{n-1} - F(x_{n-1}) bmod I^{n+s}$, or also $x_{n+1} equiv x_n bmod I^{n+s}$.
The claim shows that $(x_n)$ is a Cauchy sequence. By completeness, it converges to some $b in I^s$. It satisfies $b = b-F(b)$, hence $F(b) = 0$.
For uniqueness, assume $F(b) = F(c) = 0$ with $b,c in I^s$. We find
$$0 = (b-c)(1+bG(b,c)+cH(b,c)),$$
which implies $b = c$ since $1+I subseteq A^times$.
It seems that the assumption of $A$ to be integral is not necessary.

I'm going to basically copy a proof given by Cartan.

It is sufficient for there to exist a unique $g$ that $f(0) = 0$ and $f'(0) ne 0$, as you have said. We also know that $f_1g_1$ must equal $1$.

We want $fcirc g(X) = X$, So we try to find the conditions for which the coefficient of $X^n$ in $fcirc g$ is $0$. We know that this is equal to the coefficient of $X^n$ in
$$f_1g+f_2g^2+cdots + f_ng^ntag{1}$$
So we have a condition
$$f_1g_n + P_n(f_2,dots,f_n,g_1,dots,g_{n-1})= 0tag{2}$$
Where $P_n$ is a polynomial that may be calculated from $(1)$. Because $f_1 ne 0$, $f_1g_1 = 1$ determines $g_1$. We then take $(2)$ with $n = 2$ to find $g_2$, and by induction we can find all $g_n$.

We therefore have the existence and uniqueness of such an inverse compositional series.

We can also show that $f$ is the only series for which $g$ is an inverse. Construct $f'$ as we constructed $g$ such that $f'(0) = 0$ and $gcirc f' = X$. We have that
$$f' = Xcirc f' =fcirc g circ f' = fcirc X = f$$

This doesn't explicitly construct the coefficients, but proves that they may be constructed, which is a little more elegant.