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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.

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I'm not familiar with the definition of $SU_q(2)$, but there is some chance that the fact you ask about is related to the corresponding fact for the universal enveloping algebra $U\mathfrak{su}(2)$. For this algebra, the associated graded of the filtration by degree is a polynomial ring, which hence has no zero divisors. It is easy to see that this implies that $U\mathfrak{su}(2)$ itself has no zero divisors. – Matt E Sep 23 '10 at 21:58
up vote 2 down vote accepted

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.

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I am curious, what is SU_q(2)? – BBischof Sep 25 '10 at 2:59
@BBishof, $SU_q(2)$ is a «quantum group» which can be seen as a deformation of the usual Lie group $SU(2)$. It is in fact a Hopf algebra, similar in form to the Hopf algebra of regular functions on the Lie group, but with both its multiplication and its comultiplication «deformed». The best way to see what all this means is to read the first few chapters of Christian Kassel's Quantum Groups book which extremely lucidly and uncomplicatedly introduce $SL_q(2)$, the corresponding object for the Lie group $SL(2)$. – Mariano Suárez-Alvarez Sep 25 '10 at 19:48

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