This was an old problem I had years ago, but never really solved. Maybe it can be cracked here?
The situation is as follows.
Denote by $\mathbb{Q}(q)[X,Y]$ the algebra of polynomials over $\mathbb{Q}(q)$, the field of rational functions. Moreover, $X$ and $Y$ are indeterminates but do not commute. Let $\mathbf{Q}[x,y]=\mathbb{Q}(q)[X,Y]/I$, with $I$ being the (two-sided) ideal generated by $YX-qXY$. So $\mathbf{Q}[X,Y]$ is the ring whose generators satisfy the relation $YX=qXY$.
Why then is \[ (X+Y)^n=\sum_i\binom{n}{i}_qX^iY^{n-i} \] an identity in $\mathbf{Q}[X,Y]$?
Thank you for the tips. I expand $ \begin{align*} (X+Y)^{n+1} &= (X+Y)^n(X+Y)\\ &= (X+Y)^nX+(X+Y)^nY\\ &= \sum_i\binom{n}{i}_qX^{i}Y^{n-i}X+\sum_i \binom{n}{i}_qX^iY^{n-i+1}\\ &= \sum_iq^{n-i}\binom{n}{i}_qX^{i+1}Y^{n-i}+\sum_i \binom{n}{i}_qX^iY^{n-i+1}. \end{align*} $
I also find $ \begin{align*} \sum_i\binom{n+1}{i}_qX^iY^{n+1-i} &= X^{n+1}+Y^{n+1}+\sum_{i=1}^n\binom{n+1}{i}_qX^iY^{n+1-i}\\ &= X^{n+1}+Y^{n+1}+\sum_{i=1}^n\left[q^i\binom{n}{i}_q+\binom{n}{i-1}_q\right]X^iY^{n+1-i}\\ &= \sum_{i=0}^nq^i\binom{n}{i}_qX^iY^{n+1-i}+\sum_{i=1}^{n+1}\binom{n}{i-1}_qX^iY^{n+1-i}. \end{align*} $ I also prove the commutation formula $ Y^{k+1}X=YY^kX=Yq^kXY^k=q^kYXY^k=q^kqXYY^k=q^{k+1}XY^{k+1}. $
How does one reorder the monomials now to get the equality?