# Polynomials mapping the twisted cubic power series ring to itself

If $f(X) \in \mathbb{F}_2((T))[X]$ and $f(\mathbb{F}_2[[T^2,T^3]]) \subseteq \mathbb{F}_2[[T^2,T^3]]$, then does it follow that $f'(0) \in \frac{1}{T}\mathbb{F}_2[[T]]$?

This might seem like an unmotivated question, but if it is true then I can show that the ring consisting of all such polynomials $f(X)$ is not flat over $\mathbb{F}_2[[T^2,T^3]]$, which will settle a more general problem concerning "integer-valued polynomials" on integral domains. (These are the subject of the book "Integer-Valued Polynomials," by Cahen and Chabert.) The latter problem is as follows.

If $D$ is an integral domain with quotient field $K$, then one denotes by $\operatorname{Int(D)}$ the subring of $K[X]$ consisting of all $f \in K[X]$ such that $f(D) \subseteq D$. There is no known integral domain $D$ such that $\operatorname{Int(D)}$ is not free as a $D$-module. I suspect that $\operatorname{Int(D)}$ is not flat over $D$ if $D = \mathbb{F}_2[[T^2,T^3]]$. Note that for this domain $D$ one has $T(X^2+X) \in \operatorname{Int}(D) \backslash D[X]$, and $\operatorname{Int}(D)$ is dense in the ring of continuous functions from $D$ to $D$ in the topology of uniform convergence by Theorem III.5.3 of the book mentioned above.

ADDED TWO YEARS LATER: It turns out that $\operatorname{Int}(D)$ is free over $D$ for $D = \mathbb{F}_2[[T^2,T^3]]$. I will be writing up the proof in a paper at some point. The real problem I am after a solution to is to find an integral domain $D$ such that $\operatorname{Int}(D)$ is not free as a $D$-module.

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Counterexample: $f(X)=T^{-2}X+T^{-5}X^2+T^{-9}X^4+(T^{-2}+T^{-5}+T^{-9})X^8$
Thanks! I was really hoping I could show that $X^2+X$ is not of the form $T^2f(X) + T^3g(X)$ for $f,g \in \operatorname{Int}(D)$. I'll have to take a different route. Any ideas? –  Jesse Elliott Sep 12 '11 at 17:38