Lifting identities of formal power series

I am looking for a possibly general class of algebraic structures (maybe special topological rings) in which one can deduce identities of concrete power series from formal ones. This class should especially contain $\mathbb{R}$ and $\mathbb{C}$.

If e.g. $f \circ g = h$ holds for formal power series $f, g, h \in R[[X]]$, I want to be able to conclude $f(g(r)) = h(r)$ for all $r \in R$ for which both sides of the equation converges in $R$.

For those structures $R$ it is obviously necessary that the Cauchy product of two convergent series (if it converges) equals the ordinary product. I know that this always is the case for series over the complex numbers, but does this also remain true in a more abstract setting?

• I can't think of any algebraic structure you would use here other than a topological ring. One case where this identity always holds is the completion of a local ring, where a power series that's not a polynomial converges only when $r$ lies in the maximal ideal, which allows one to truncate the power series when working in $R/m^n$ and then take the inverse limit. – Will Sawin Aug 31 '12 at 19:26
• Hi Will. You are talking about the universal property of formal power series, also mentioned at Wikipedia. This is indeed the solution for lifting those identities into certain rings with an I-adic topology, but unfortunately it does not generalize the theory of real or complex power series. I forgot mentioning that I am looking for such a generalization. – Dune Aug 31 '12 at 19:51
• In normed rings I think everything works out, and otherwise I'm not willing to speculate. It seems to me that without a norm you do not get very much control over what a convergent series looks like. – Qiaochu Yuan Aug 31 '12 at 23:42

Take $g(x)=x-x^2\in\mathbb{Z}[[x]]$. There is an $f(x)\in x\mathbb{Z}[[x]]$ which is an inverse to $g(x)$ under composition (this is true because $g(x)$ has constant term 0 and linear term a unit; we could also write down $f(x)$ explicitly). We have $g(1)=0$, and $f(0)=0$ (both expressions converge in any topological ring, as they have only finitely many non-zero terms), so the expression $f(g(1))$ converges, and equals 0. However, $h(x):=f(g(x))=x$, and $h(1)=1\neq 0$. This construction works in every topological ring, so the statement "$h=f\circ g\Rightarrow h(r)=f(g(r))$ for all $r\in R$ for which both sides converge" does not hold in ANY non-zero topological ring $R$.
• Can you give an explicit formula for the coefficients of $f$? Thank you very much for this great (although surprising) answer! – Dune Sep 2 '12 at 22:00
• I just figured it out. The n-th coefficient of $f$ is actually the (n-1)-th Catalan number! – Dune Sep 3 '12 at 0:02