Global dimensions of non-commutative rings This is related to my previous question: When is a quantum affine space $\mathbb{A}^{n}$ Calabi-Yau? I now would like to know the global dimension of the ring $R=\mathbb{C}\langle x_1,\dots,x_n\rangle/I$, where I is the two-sided ideal generated by $x_ix_j=a_{ij}x_jx_i$ for some $a_{ij}=a_{ji}^{-1}\in \mathbb{C}$ for $1≤i,j≤n$. Recall that the global dimension $gl\dim(S)$ of a ring $S$ is defined to be the supremum of the set of projective dimensions of all left $S$-modules.  
In case your ring $R$ is defined by a quiver with relations, there seem some techniques. Are there another approach to compute $gl\dim(R)$? Moreover, are there any standard way to compute $gl\dim(S)$ when $S$ is non-commutative?  
There are many characterization of $gl\dim(R)$ when $R$ is commutative and I am aware of similar questions, such as Commutative Ring of Finite Global Dimension.
 A: There is another way to see this than constructing an explicit resolution.  This involves viewing $R$ as an iterated skew polynomial ring.
I am assuming that you want $a_{ii} = 1$; otherwise $x_i^2 = 0$ for all $i$.
Also, since this works over any field, I am just going to denote the base field by $k$.
Start with $R_1 = k[x_1]$.  Then let $\sigma_1$ be the $k$-algebra automorphism of $R_1$ defined by $\sigma_1(x_1) = a_{21} x_1$, and let $R_2$ be the skew-polynomial ring
$$
R_2 = R_1[x_2; \sigma_1].
$$
Thus $R_2$ is generated by $x_1$ and $x_2$ with the relation
$$
x_2 x_1 = \sigma_1(x_1)x_2 = a_{21} x_1 x_2.
$$
Then we continue this game.  Having constructed $R_i$, define
$$
R_{i+1} = R_i[x_{i+1}, \sigma_i],
$$
where $\sigma_i \in \mathrm{Aut}(R_i)$ is defined by $\sigma_i(x_j) = a_{i+1,j} x_j$ for $1 \le j \le i$.  Then for $j < i$ we have the relations
$$
x_i x_j = \sigma_{i-1}(x_j) x_i = a_{ij} x_j x_i,
$$
and hence your ring $R$ coincides with $R_n$.
This gives us two things.  First, there is an analogue of the Hilbert Basis Theorem for skew polynomial rings; if $A$ is left Noetherian then the skew polynomial ring $A[x;\sigma]$ is left Noetherian for any automorphism $\sigma$ of $A$.  You can find this in Section 1.2.9 of McConnell-Robson or Theorem 1.14 of Goodearl-Warfield.
The other fact is that there is an analogue of the (generalized) Hilbert Syzygy Theorem for skew polynomial rings over Noetherian rings.  This is in Section 7.9.10 of McConnell-Robson.  Explicitly, it says the following: if $A$ is left Noetherian with $\mathrm{l.gl.dim} \, A = n < \infty$, then $\mathrm{l.gl.dim} \, A[x;\sigma] = n+1$ for any automorphism $\sigma$ of $A$.
Starting with $\mathrm{l.gl.dim}k[x] = 1$ and iterating shows that each $R_i$ is both left and right Noetherian and has 
$$
\mathrm{l.gl.dim} R_i = \mathrm{r.gl.dim} R_i = i.
$$
A: Marc Wambst constructed a resolution for your algebra $A$, which he calls naturally enough a quantum Koszul complex. This is a projective resolution of $A$ as an $A$-bimodule and it looks like the usual bimodule Koszul resolution of polynomial rings sprinkled with $q$'s all over the place.
His quantum Koszul complex has length $n$ showing that $\operatorname{pdim}\_{A^e}A\leq n$. Now, a little homological algebra shows that $\operatorname{gldim}A\leq\operatorname{pdim}\_{A^e}A$ (this is explained in Cartan-Eilenberg, in the chapter on augmented rings), so now we know that the global dimension of $A$ is at most $n$. Computing $\operatorname{Tor}_A^n(k,k)$ is easy using his complex, and it is non-zero: this shows that we in fact have an equality, and
$$\operatorname{gldim}A=n.$$
As for your general question, there is no general method of computing the global dimension of rings. You will find several useful techniques which apply to interesting classes of algebras in the books by McConnell and Robson, and by Lam, among several other places —in general, it is somewhat of an art.
The paper I referred to above is [M. Wambst, Complexes de Koszul quantiques. Ann. Inst. Fourier (Grenoble) 43, 3 (1993), 1089–1159]. Every single time I have needed the explicit form of this resolution, though, I have found it easier to consruct it by hand :-)
