q-Integer-valued polynomials

Question

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.

Answer 1

(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 $f_0,f_1,\dots,f_n$.