Can you give me a specific representation of q-Weyl algebra over field k,i.e $K<x,y>/<yx=qxy>$ for proving that {$x^{i}y^{j}$, i,j$\in$$Z$} is indeed a basis in $k<x,y>/<yx=qxy>$ as k vector space.
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$\begingroup$I believe you also need the condition that $$ xx^{-1} = x^{-1}x = yy^{-1} = y^{-1}y = 1 $$ Now note that any word in $x,y,x^{-1}, y^{-1}$ can be written in terms of $\{x^iy^j : i,j\in \mathbb{Z}\}$ (every time you flip $yx$ to $xy$ at the price of a $q$)
For linear independence, you can represent the algebra on the $k-$vector space $V = k[u,v,u^{-1},v^{-1}]$ with basis $\{u^i v^j : i,j \in \mathbb{Z}\}$. Let $x$ and $y$ act on $V$ by $$ x(u^i v^j) = u^{i+1}v^j $$ $$ y(u^i v^j) = q^i u^i v^{j+1} $$ Now note that $x$ and $y$ are invertible linear operators and $yx = qxy$. Now if $$ \sum c_{i,j}x^iy^j = 0 $$ Apply this to $1 \in V$ to get $$ \sum c_{i,j}u^iv^j = 0 $$ which implies $c_{i,j} = 0$
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