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In this question, user begins shows that, for each $k\in \mathbb{N}$, there is a unique polynomial $P_k(x)$ of degree $k$ whose coefficients are in $\mathbb{Q}(q)$, the field of rational functions, such that $P_k(q^n)=\binom{n}{k}_q$ for all $n$.

I've read a bit more about this subject and want to ask this follow up. Suppose $f\in\mathbb{Q}(q)[x]$. Why is it true that $f(q^n)\in\mathbb{Z}[q,q^{-1}]$ for all $n$ if and only if the coefficients of $f$ w.r.t. the basis $\{P_k:k\in\mathbb{N}\}$ belong to $\mathbb{Z}[q,q^{-1}]$? Thanks.

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Do you know the proof that a polynomial in $\mathbb Q[x]$ takes integers to integers iff it is a linear combination withinteger coefficients of the polynomials $\binom{X}{k}$, with $k\geq0$? This should be similar... – Mariano Suárez-Alvarez Feb 17 '12 at 20:40
@MarianoSuárez-Alvarez I'm afraid not, I'll try to look into it. I was hoping to some explanation or even a sketch here. – Vika Feb 17 '12 at 20:44
The statement I mentioned is proved in Hartshorne's book on algebraic geometry, in Prasolov's book on Polynomials, and in many other places. I don't have thge book here to check but your question should probably be answered in Prasolov's book. – Mariano Suárez-Alvarez Feb 17 '12 at 22:51
Thanks, I think I found what I wanted in there. – Vika Feb 19 '12 at 3:27

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