Does anyone know anything about the normal regular sequences in the quantum plane? 


Here are the definitions: 


Normal regular sequence: Let $R$ be a ring (not necessarily commutative). A sequence $x\_1, \ldots, x\_n$ is a normal regular sequence in $R$ if


1) $x\_1$ is a normal, regular element in $R$, and 


2) $x\_i$ is normal, regular element in $R/(x\_1, \ldots, x\_{i-1})$ for all $2 \leq i \leq n$. 


Quantum plane: A $k$-algebra generated by $x, y$ subject to the relation $yx = qxy$, denoted by $k\_q[x, y]$, that is, 
$$k\_q[x, y] = \frac{k\langle x, y\rangle}{(yx - qxy)}. $$
Here $q\in k$ is a primitive root of 1. 


The length of any regular sequence in $k\_q[x, y]$ is no more than 2, and I know there is some result about such normal regular sequence when $x\_1, x\_2$ are homogeneous elements. But anything else? 


Thanks, 



Normal regular sequence in noncommutative algebras
Let $g \in k\_q[x,y]$ be normal with $q$ a primitive $\ell$th root of unity, $q \neq 1$. Then $g = \sum c\_i y^{m\_i} x^{n\_i}$ where $c\_i \in k^\times$ for all $i$ and $m\_i \cong m\_j\mod \ell$, $n\_i \cong n\_j\mod \ell$ for all $i,j$.


It's not hard to see that such elements are normal. Checking that these are all the normal elements requires a bit of computation, and it helps to know that $Aut(k\_q[x,y])=(k^\times)^2$ unless $\ell=2$, in which case there is an additional automorphism switching the generators.


I'm not sure what this means in terms of finding a regularizing sequence, but of course your second element must now be in $k\_q[x,y]/gk\_q[x,y]$.


