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

Something I've been thinking about lately has led me to wonder about the following. Consider the quantum polynomial ring $ Q= \mathbb{C}_{-1}[x_1,...x_n]$ generated as a graded ring in degree 1 with relations $x_i*x_j=-x_j*x_i$ for $i\neq j$. Set $u_i=x_i^2$ consider central homogeneous functions of the form $f(u_1,...,u_n)$ and the quotient ring R. The question is when, if ever, does Proj(R) have finite homological dimension. By Proj(R) I mean the quotient abelian category Gr(R)/Tor(R), where Gr(R) consists of finitely generated left modules and Tor(R) is the subcategory of torsion modules? Based upon some calculations, I believe the answer should be whenever $C[u_1,...u_n]/(u_idf/du_i)$ is finite dimensional.This is a condition about how the zero locus of f intersecting the coordinate axes $u_i$, which I believe should have a geometric interpretation in terms of the map $Z(Q)=C[u_1,...u_n] \mapsto Q$ when the $u_i$ are not zero, the fibers are matrix algebras which degenerate at the locus when some of the $u_i=0$. By analogy with the ordinary projective space, I'd like to think of this as a smooth hypersurface in the quantum projective space $P_{-1}^{n-1}$ but I can't seem to find this type of stuff analyzed anywhere. Does this ring a bell for anyone? If not, does anyone understand the map on primes $Spec(Q)\mapsto Spec Z(Q)$?

share|cite|improve this question
What doyou mean by "as a graded ring in degree $1$"? – Mariano Suárez-Alvarez Nov 21 '10 at 3:46
generated by variables of degree 1. So each x_i has degree 1. – Daniel Pomerleano Nov 21 '10 at 9:10
How is the ring $R$ defined? – Jonas Hartwig Nov 21 '10 at 20:22
@Jonas Take the two sided homogeneous ideal generated by f, I and form the graded quotient ring. As f is central, this is the same as considering the left or right ideal generated by f. – Daniel Pomerleano Nov 21 '10 at 20:53

Your Answer


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

Browse other questions tagged or ask your own question.