# Normal regular sequence in noncommutative algebras

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,

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If you explained what is that «some result» in the homogeneous case, it might be easier to see what you are after in the general case. – Mariano Suárez-Alvarez May 12 '13 at 5:14
The result I referred to is as follows: Let $A$ be locally finite graded $k$-algebra. Let $x_1, \ldots, x_n$ be a normal sequence of homogeneous elements with deg$(x_i) = d_i$. Then $\{ x_1, \ldots, x_n\}$ is a regular sequence if and only if the Hilbert polynomials of $A$ and $A/(x_1, \ldots, x_n)$ satisfy $$H_{A/(x_1, \ldots, x_n)} (t) = \Pi_{i=1}^n (1 - t^{d_i}) H_A(t).$$ – Leslie Wu May 13 '13 at 2:29
I am more interested in this question: given $\{ a, b\}$ in $k_q[x, y]$ where $q^n = 1$, what are the characteristics of such pair $\{ a, b\}$?. This question is much easier in the commutative case, that is, the polynomial ring with two variables, where we can say two elements $\{ f, g\}$ in $k[x, y]$ are regular sequence if and only if gcd$(f, g) = 1$ if and only if $\dim_k k[x, y]/(f, g) < \infty$. I am expecting an analogue in the noncommutative case, more specifically, in the quantum plane $k_q[x, y]$ where $q$ is not generic. – Leslie Wu May 13 '13 at 2:35

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 $k_q[x,y]/gk_q[x,y]$.
This should probably be a comment, since it is too basic an observation. Suppose that the elements $f,g\in k_q[x,y]$ are both normal in $k_q[x,y]$ (this, at least a priori is more restrictive than you want). Then, as in the answer of J.Gaddis here, there exist $a,b,c,d$ such that all monomials $x^iy^j$ appearing in $f$ have $i\equiv a\pmod{\ell}$ and $j\equiv b\pmod{\ell}$, and all monomials $x^iy^j$ appearing in $g$ have $i\equiv c\pmod{\ell}$ and $j\equiv d\pmod{\ell}$. Therefore, $gf=q^{ad-bc}fg$. This suggests that the usual Koszul complex criterion would be modifiable for this case: define $$B=k\langle X,Y\mid X^2=Y^2=YX+q^{ad-bc}XY=0\rangle,$$ and define a structure of a chain complex on $k_q[x,y]\otimes B$ as $$d(h(x,y))=0, d(h(x,y)X)=h(x,y)f(x,y), d(h(x,y)Y)=h(x,y)g(x,y),$$ and $$d(h(x,y)YX)=h(x,y)g(x,y)X-q^{ad-bc}h(x,y)f(x,y)Y.$$ In this case, it seems to be fairly straightforward to prove that this complex is acyclic in positive homological degrees (we of course put the homological degrees of $X$ and $Y$ to be equal to $1$; homology in degree zero is $k_q[x,y]/(f,g)$) if and only if $f$ and $g$ form a regular sequence. Maybe that helps.