Consider the quantum group $U_q(\mathfrak{sl}_2)$, with generators $E,F,K$ such that $[E,F]=\frac{K-K^{-1}}{q-q^{-1}}$. Write $[n]=\frac{q^n-q^{-n}}{q-q^{-1}}$, and $[n]!=[n][n-1]\dotsm[1]$.
In Quantum deformations of certain simple modules over enveloping algebras, Lusztig defined the following elements inside his integral form for $U_q(\mathfrak{sl}_2)$ (I am replacing $q$ by its square root in his notation): $$ \newcommand\qbinom{\genfrac[]0{}}\qbinom{K;\ c}t=\frac{1}{[t]!}\prod_{s=0}^{t-1}\frac{(q^{c-s}K-q^{s-c}K^{-1})}{(q-q^{-1})}.$$ These are $q$-deformations of the element $\binom{H+c}{t}$ in Kostant's $\mathbb{Z}$-form for $U(\mathfrak{sl}_2)$.
In Kostant's $\mathbb{Z}$-form, we have identities such as $\binom{H}{1}^2=2\binom{H}{2}+\binom{H}{1}$. However, if one tries to quantize both sides, the result is $$ \qbinom{K;\ 0}1^2=q^2[2]\qbinom{K;\ 0}2+qK^{-1}\qbinom{K;\ 0}1.$$
For various reasons, I am interested in identities whose coefficients do not involve powers of $K$. For instance, it turns out $$ \qbinom{K;\ 0}1^2=\qbinom{K;\ 1}2+\qbinom{K;\ 0}2.$$
In general, I believe one may be able to apply iteratively a relation in Lusztig's paper, to prove that products between such elements are always $\mathbb{Z}(q)$-linear combinations of $\qbinom{K;c}t$'s. A possible caveat, is that these elements are not linearly independent over $\mathbb{Z}(q)$. For instance, one has the relation $$ \qbinom{K;\ 0}2-[3]\qbinom{K;\ 1}2+[3]\qbinom{K;\ 2}2-\qbinom{K;\ 3}2=0.$$
[EDIT: Assuming a certain PBW-like result, I think one may be able to prove that Lusztig’s elements for $c\le t$ form a basis.]
This is such a well studied object, so I was hoping this ring structure would be worked out somewhere. In short:
What is the product $\qbinom{K;\ a}t\qbinom{K;\ b}s$ in terms of Lusztig's elements?
\genfrac
is meant for situations like yours. For example, $\genfrac()0{}{H + c}t$\genfrac()0{}{H + c}t
produces the un-quantised version, $\genfrac[]0{}{K; c}t$\genfrac[]0{}{K; c}t
produces the quantised version, and, just to illustrate, $\genfrac(){}{}a q$ is a Legendre symbol\genfrac(){}{}a q
. I always have to Google the syntax, but the first two arguments are the delimiters, and the last two are the "numerator" and "denominator" of a generalised fraction. Anyway, I edited accordingly. $\endgroup$