Zero divisors in $SU_q(2)$

Question

I'm looking at the quantum group $SU_q(2)$ (over ${\mathbb C}$) and can't see why it has no zero divisors. It's clear that $M_q(2)$, the quantum $2 \times 2$ matrices have no zero divisors, but I can't seem to see why/if this extends to the quotient over $det_q - 1$. I think one can argue it from the grading on $M_q(2)$, but that's just a guess.

Answer 1

The algebra can be constructed as an iterated Ore extension starting from the base field. By standard properties of such extensions, it is then a domain.