MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the usual (fomal) construction of the quantum general linear group $GL_q(N)$, an Ore extension is used. See for example Kassel. Why is this necessary? Surely one can just augment the set of generators with an element det$^{-1}$ and the set of relations with the relations det.det$^{-1} = 1$ and det$^{-1}$.det$=1$ and get the same thing without going to all the trouble of considering Ore extensions?

share|cite|improve this question
up vote 4 down vote accepted

I have hard time understanding what is your notion of usual; most references indeed define the quantum linear group $M_q(N)$ by generators and relations (or by the universal property as defined by Manin) and then localize at $det_q$. So far as the construction goes. But it is of course very useful to know that one can start with the polynomial ring and then do the iterated Ore extension $N^2-1$ times to obtain $M_q(N)$. This is useful to infer many useful ring-theoretic properties. For example, it follows that the matrix bialgebra $M_q(N)$ and its Hopf envelope $GL_q(N)$ are Ore domains (the same O. Ore!), hence they are contained in the Ore quotient ring in which we can do many useful computations. Various intermediate Ore localizations are also useful.

share|cite|improve this answer
Well said. Another example of one of those useful ring-theoretic properties is that every prime ideal is completely prime (for $q$ not a root of unity). This would seem to go without saying, but since the question was asked--the study of any mathematical object is generally advanced by being able to view the object as a special case of something with a well-developed theory. – Greg Marks Mar 3 '11 at 20:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.