Let $k$ be an algebraically closed field.
Let $R := k\langle x,y \rangle/(yx-qxy) (q \in k^*)$.
We often call $R$ a quantum plane.
If $q$ is a primitive $n$-th root, then for any $(\zeta, \xi) \in k^* \times k^*$ we can construct a simple right $R$-module $V$ such that it has $k$-basis $v_i (i \in \mathbb{Z}_n)$ and $$ v_i.x = \zeta q^i v_i, \quad v_i.y = \xi v_{i+1} $$ as its module structure.
In Lemma 3.3 in page 203 of this note, it is shown that the annihilator of $V$ is $(x^n-\zeta^n,y^n-\xi^n)$ and the ideal is maximal.
It seems that we have $R/(x^n-\zeta^n,y^n-\xi^n) \simeq V$ if we assume the lemma because $(x^n-\zeta^n,y^n-\xi^n)$ is maximal. In detail, we have a surjection $R/(x^n-\zeta^n,y^n-\xi^n) \rightarrow V$. Then, the kernel seems to be the zero module.
Of course, this argument contradicts the above.
Question 1. Where is the error in the above argument ?
Moreover, there is another description of simple modules of quantum planes in Lemma 4.19 of this paper. There, it is shown that $V \simeq R/(x^n-\zeta^n,\xi x- \zeta y)$ implicitly. Note that isomorphisms between $V$ and $R/(x^n-\zeta^n,\xi x- \zeta y)$ are not constructed in the paper. (There, the fat points of quantum planes are classified.)
Question 2. How do we construct an explicit isomorphism between them ?
Any comments are welcome. Thank you.
Edit: We have the same question on MO. I would appreciate it if you could see it too.