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Define n-quantum vector space to be the algebra $$ {\mathbb C}_q^n := \mathbb{C}\left< x_i \mid i =1, \ldots, N\right>/\left<x_i x_j = q x_j x_i \mid i<j\right>. $$ For $q=1$, we get the usual polynomial ring in $n$-variables, and so, Serre's conjecture (Quillen–Suslin theorem) tells that every finitely generated projective module over ${\mathbb C}_1^n$ is free. How does this work for $q \neq 1$? Is there a $q$-deformed Quillen–Suslin theorem? The not a root of unity case is the most interesting to me.

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    $\begingroup$ Have you checked out Lam's book "Serre's problem on projective modules" (the 2006 edition)? Chapter VIII (New developments since 1977) contains subsections on non-commutative and quantum versions. I do not have my copy handy, so I can not check right now if your question is answered there... $\endgroup$ – Matthias Wendt Sep 22 '14 at 20:54
  • $\begingroup$ Thanks, but I checked the book and it gives a result only when $N=2$. $\endgroup$ – Juan Corrida Sep 22 '14 at 21:13
  • $\begingroup$ See also this remark on PhysicsOverflow to the same questions: physicsoverflow.org/24083/… $\endgroup$ – Dilaton Sep 30 '14 at 6:39
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See extensive 1998 survey

Uspekhi Mat. Nauk, 1998, Volume 53, Issue 4(322),
Serre's quantum problem V. A. Artamonov

Russian pdf is available for free from that page. See also other papers by the author.

At the bottom of second page he writes (translated from Russian):

F.g. projective modules of sufficiently HIGH RANK are free, at least at two cases: (1) $q_{ij}$ are roots of unity (2) $q_{ij}$ are at general position.

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  • $\begingroup$ This result is also mentioned in Lam's book on Serre's problem, in section VIII.11 (it came up in the comments to the question) $\endgroup$ – Matthias Wendt Oct 7 '14 at 19:27

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