# Finding an explicit isomorphism between $\mathbb{C}_q[x^{\pm 1},y^{\pm 1}]/\langle x^k-\alpha, y^k-\beta \rangle$ and $M_k(\mathbb{C})$

Consider the quantum torus \begin{align*} \mathbb{C}_q[x^{\pm 1},y^{\pm 1}] = \frac{\mathbb{C}\langle x,y \rangle}{\langle yx-qxy \rangle }[x^{-1}, y^{-1}] \end{align*} where $q \in \mathbb{C}^\times$ is a primitive $k$th root of unity. This is an Azumaya algebra, and as a consequence of this we have \begin{align*} \frac{\mathbb{C}_q[x^{\pm 1},y^{\pm 1}]}{\langle x^k - \alpha, y^k-\beta \rangle} \cong M_k(\mathbb{C}), \end{align*} provided $\alpha \neq 0 \neq \beta$. The only proof I know of this is non-constructive. Is it easy to write down an explicit isomorphism? I can do it for $k=2$, but as soon as I hit $k=3$ I can't even find three orthogonal idempotents in $\mathbb{C}_q[x^{\pm 1},y^{\pm 1}]/\langle x^k - \alpha, y^k-\beta \rangle$ to play the roles of the matrix units $e_{ii}$.

## 1 Answer

We may as well assume that $\alpha=\beta=1$. Consider the matrices $$X=\pmatrix{1&0&0&\cdots&0\\0&q&0&\cdots&0\\ 0&0&q^2&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&q^{k-1}}$$ and $$Y=\pmatrix{0&1&0&\cdots&0\\ 0&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&1\\ 1&0&0&\cdots&0}.$$

Then $X^k=Y^k=I$ and $YX=qXY$. Mapping $x\to X$ and $y\to Y$ gives an isomorphism between your ring and $\text{GL}_k(\Bbb C)$.