Let $k$ be a field and $q\in k^{\*}$. The quantum plane $k\_{q}[x,y]$ is the algebra $k\langle x,y\rangle/\langle xy=qyx \rangle$ (i.e. the quotient of the free non-commutative $k$-algebra on two variables $x$ and $y$ modulo the ideal given). 



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> Question: For $q,r\in k^{\*}$ and $q\neq r$, when is $k\_{q}[x,y]$ isomorphic (as an algebra) to $k\_{r}[x',y']$?
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I fully expect this is known but after (what I think is) fairly comprehensive literature searching, including a large proportion of the best-known quantum groups texts, I have been unable to find an answer. A reference would be appreciated just as much as a proof.


Some comments:


* I know the (algebra) automorphism group: by work of Alev-Chamarie this is $(k^{\*})^2$ unless $q=-1$ (when it is a semi-direct product of the torus with the group of order two generated by the map that interchanges the two variables). Hence I don't need to worry about $q=r$.
* I want algebra isomorphisms but information on Hopf algebra maps would be nice too (NB. the Hopf automorphisms for the usual Hopf structure are also those just described)
* if $q$ has finite order $N$ in $k^{\*}$ and $r$ is of infinite order then the corresponding quantum planes are not isomorphic, as in the first case the centre is non-trivial (generated by $x^{N}$ and $y^{N}$) but in the second the centre is just $k$
* if $q$ has order $M$ and $r$ has order $N\neq M$, then the quotients by the centres are both finite-dimensional but of different dimension, hence the quantum planes are not isomorphic
* I would be happy to know the answer just for $k=\mathbb{C}$



Isomorphisms of quantum planes
This answer feels so glib I'm quite worried it's wrong, but anyway:


Write $D\_q$ for the full ring of fractions of $k\_q[x,y]$. By Alev-Dumas, "Sur le corps des fractions de certaines algebres quantiques", Corollary 3.11c, we know that for $q$, $r$ non-roots of unity, $D\_q \cong D\_r$ iff $r = q^{\pm1}$.


It's clear that $k\_q[x,y] \cong k\_r[x',y'] \Rightarrow D\_q \cong D\_r$, so we have
\[k\_q[x,y] \cong k\_r[x',y'] \Rightarrow r = q^{\pm1},\]
 and the converse should also be clear.


