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](https://doi.org/10.1016/0001-8708(88)90056-4), 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?
> 
> 
> 



What is the ring structure on Lusztig's integral form of quantum $\mathfrak{sl}(2)$?
I figured it out, but I would really appreciate a reference!


The formula is quite nice. For $c,d\ge0$ such that $c\le t $ and $d\le s$, the following holds:
$$ \newcommand\qbinom{\genfrac[]0{}}\qbinom{K;\ c}t\qbinom{K;\ d}s=\sum\_{i\ge 0}\qbinom{t-c+d}{i-c}\qbinom{s-d+c}{i-d}\qbinom{K;\ i}{t+s}.$$


(Here, the binomial $\qbinom{a}{b}$ is set to zero whenever $b<0$ or $b>a$)


