TITLE: Global dimensions of non-commutative rings
QUESTION [9 upvotes]: 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.

REPLY [7 votes]: 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 
$$
\mathrm{l.gl.dim} R_i = \mathrm{r.gl.dim} R_i = i.
$$