# 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.

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I recommend T.Y. Lam's Lectures on Modules and Rings. In particular, Section 5B gives methods of computing the right global dimension of $R/I$ when $I$ is nice. I don't know much about the application you have in mind, but it might fit into Lam's framework. Are you aware of the ways to bound right global dimension via regular sequences? That might help, since you know how the ideal $I$ is defined. Lam's book is really a master-piece; I can't recommend it highly enough! – David White Aug 1 '12 at 2:50
Thanks for the reference. Actually it doesn't matter whether left or right as long as R is Noetherian, so this seems helpful. I don't know anything about regular sequence method in non-commutative case. Many thanks again, David! – user2013 Aug 1 '12 at 3:32

## 2 Answers

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 :-)

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Thanks for the detailed answer, Mariano! Embarassingly I was not aware of the inequality $gldim(A)≤pdim_{A^e}(A)$, which turns out to be very useful! I was sure that my example was well-known and could be worked out in detail (because it is the simplest non-commutative ring) but I didn't know any references. Many thanks! – user2013 Aug 1 '12 at 8:03
I am bit confused with Wembst's paper. Actually I am only aware of bar complex as a bimodule resolution and I have never seen any finite bimodule resolution of length. Could you kindly give me an example (even commutative case)? – user2013 Aug 1 '12 at 9:45
Let $A=k[x_1,\dots,x_n]$ and $B=A\otimes A$. The elements $r_i=x_i\otimes1-1\otimes x_i$, taken in any order, form a regular sequence in $B$, so there is a (commuative!) Koszul complex $K(B;r_1,\dots,r_n)$ which is a resolution of $B/(r_1,\dots,r_n)=A$ as a $B$-module which is of finite length. This is the same thing as an $A$-bimodule resolution of $A$. The complex Wambst gives is the generalization of this construction to quantum affine spaces. – Mariano Suárez-Alvarez Aug 1 '12 at 16:13
This makes perfect sense. Thank, Mariano! – user2013 Aug 1 '12 at 17:24

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 $$\mathrm{l.gl.dim} R_i = \mathrm{r.gl.dim} R_i = i.$$

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(Start with $k$!) – Mariano Suárez-Alvarez Aug 1 '12 at 21:02
Sure, starting with $k$ would be even simpler. Good point, Mariano. – MTS Aug 1 '12 at 21:11
Thanks, MTS! Is it true that every skew polynomial ring is a quantum projective space? By this I mean that it has finite global dimension n (this is OK) and Hilbert series 1/(1-t)^n, and Gorenstein (this is non-trivial). – user2013 Aug 1 '12 at 22:00
The Hilbert series is what you say it is. The fact that the monomials $x_1^{e_1} \dots x_n^{e_n}$ are linearly independent follows from Bergman's Diamond Lemma. There is only one type of ambiguity to resolve, and it just works out. These monomials clearly span, so they are a basis. This gives the Hilbert series. About the Gorenstein condition: do you mean AS-Gorenstein, which is I think more common for this type of algebra? i.e. $\mathrm{Ext}^i_R(k,R) = k$ for $i = n$ and is 0 for all other $i$? If so, that follows from an explicit resolution of $k$ that can be constructed by hand... – MTS Aug 2 '12 at 5:47
[continued] The first term of the resolution is $R$ itself, with the augmentation map. The next term is $n$ copies of $R$. For more detail, I recommend Uli Krahmer's Notes on Koszul Algebras, available at his homepage: www.maths.gla.ac.uk/~ukraehmer/connected.pdf and the references therein. – MTS Aug 2 '12 at 5:51