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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2$\begingroup$ 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 WendtCommented Sep 22, 2014 at 20:54
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$\begingroup$ Thanks, but I checked the book and it gives a result only when $N=2$. $\endgroup$– Juan CorridaCommented Sep 22, 2014 at 21:13
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$\begingroup$ See also this remark on PhysicsOverflow to the same questions: physicsoverflow.org/24083/… $\endgroup$– DilatonCommented Sep 30, 2014 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.
answered Oct 7, 2014 at 19:14
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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) $\endgroup$ Commented Oct 7, 2014 at 19:27