2
$\begingroup$

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.

$\endgroup$

1 Answer 1

3
$\begingroup$

Counterexample: $f(X)=T^{-2}X+T^{-5}X^2+T^{-9}X^4+(T^{-2}+T^{-5}+T^{-9})X^8$

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$ Sep 12, 2011 at 17:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.