About the formal exponential power series over an arbitrary field of characteristic 0

Question

Assume $K$is any field of characteristic $0$ which ensues that it makes sense to define in the ring $K[[[X]]$ of formal power series with coefficients in $K$ the formal exponential power series $exp(X)=\sum_{n=0}^{\infty }\frac{1}{n!}\cdot X^{n}$. Of course, rational numbers $\frac{r}{s}\ (r\in \mathbb{Z},s\in \mathbb{N})$ regarded as elements of$K$ have to be interpreted as $\frac{r\cdot 1}{s\cdot 1}$, where $1$is the neutral element of multiplication in $K$-. Let the operator $\left[ X^{k} \right]$ extract the k-th coefficient from a formal power series: $\left[ X^{k} \right](\sum_{n=0}^{\infty }c_{n}\cdot X^{n})=c_{k}$- Now it is easy to see that for every power series $q$ with $[X^{0}] (q)=0$ the composition $exp(q(X))$ is well-defined - not involving any convergence considerations - because for each fixed $k_{0}\in \mathbb{N}$ only finitely many powers $q(X)^{k}$ contribute a non-zero term to the coefficient $[X^{k_{0}}] (exp(q(X)))$. And it is clear that always $[X^{0}] (exp(q(X)))=1$ then. My question now is: What about the opposite direction? Can you show that for every $p\in K[[[X]]$ with $[X^{0}] (p(X))=1$ there is a $q\in K[[[X]]$ with $[X^{0}] (q(X))=0$ satisfying $exp(q(X))=p(X)$ ?

Answer 1

You can do this using the formal power series for $\log(1+x)$ which is $$L(x)=\sum_{n=1}^\infty(-1)^{n+1}x^n/n.$$ To make sense of this, note that more generally a composition of formal power series $p(q(x))$ is well-defined as long as $q$ has no constant term, and this composition is associative. In particular, the compositions $L(\exp(x)-1)$ and $\exp(L(x))-1$ both make sense, and both can be shown to be equal to just $x$. (You can either verify this by some explicit calculations with coefficients, or you can observe that the composition of formal power series will converge for $x$ near $0$ in $\mathbb{C}$ to the composition of the exponential and logarithm functions on $\mathbb{C}$; since the composition of those analytic functions is the identity function it follows that the composed formal power series must be the identity as well.)

So, given $p(x)\in K[[x]]$ with constant term $1$, the power series $$q(x)=L(p(x)-1)$$ will satisfy $$\exp(q(x))=\exp(L(p(x)-1))=(p(x)-1)+1=p(x)$$ by associativity of composition of power series.