For $n \in \mathbb{Z}\_{\geq 0}$, let $[n]\_q := (1-q^n)/(1-q) = (1+q+...+q^{n-1})$ as is customary, with $[0]\_q=0$.


Let $R$ be the subring of $\mathbb{Q}(q)[x]$ consisting of all $f$ such that $f([n]\_q) \in \mathbb{Z}[q]$ for all $n \in \mathbb{Z}\_{\geq 0}$.


Define $f\_0(x) = 1$ and $f\_k(x) = f\_{k-1}(x)\cdot \frac{x-[k-1]\_q}{q^{k-1}[k]\_q}$ for $k \geq 1$. Note $f\_k([n]\_q) = \frac{[n]\_q!}{[n-k]\_q! [k]\_q!}=\binom{n}{k}\_q \in \mathbb{Z}[q]$, so indeed $f\_k \in R$ for all $k \geq 0$.


Is it the case that $R$ is spanned as a $\mathbb{Z}[q]$-module by the $f\_k$? If so, what is known about the structure constants $c\_{ij}^{k}(q)$?


EDIT: More information about this ring, such as the structure constants and a classification of all maps to fields, can be found in this preprint with Nate Harman: <http://arxiv.org/abs/1601.06110v1>.



q-Integer-valued polynomials
(Below is the proof that module $R$ is generated by $f\_i(x)$, without calculation of structure constants.)


Polynomial $f(x)$ of degree $n$ may be interpolated in points $[0],[1],\dots,[n]$ (I omit index $q$): $f(x)=\sum\_{i=0}^n f([i])\prod\_{j\ne i} \frac{x-[j]}{[j]-[i]}$. For any summand leading term $f([i])\prod\_{j\ne i} \frac1{[j]-[i]}$ is divisible in $\mathbb{Z}[q]$ by the leading term of $f\_n(x)$. Hence $f(x)-Af\_n(x)$ has degree less than $n$ for some $A\in \mathbb{Z}[q]$. This is induction step in the proof that $f$ is $\mathbb{Z}[q]$-linear combination of $f\_0,f\_1,\dots,f\_n$. 


