Let $R$ be normal, local ring of dimension at least $2$. Let $M$ be a reflexive $R$-module and let $A=Hom_R(M,M)$. Suppose $A$ has finite global dimension. Then one can view $A$ as a weak non-commutative desingularization of $R$ (note that, a) there is a natural map $R\to A$ and b) in the commutative case, finite global dimension implies regularity, hence the name).

This concept imitates Van den Bergh's definition of non-commutative crepant resolution (NCCR). His definition arises from a proof of dimension $3$ case of Bondal-Orlov conjecture. This is a long story, but an excellent account of the reasons behind the definition can be found in Section 4 of this paper. For existence of NCCR in some high dimensions case, check out this.

Now, non-commutative crepant resolution does not always exist (the above papers proves the equivalence, in some cases, with existence of projective crepant resolutions, which exists rarely in high dimensions). What we do know from Hironaka, in characteristic $0$ at least, is that resolution of singularity exist. So:

Question: Does weak non-commutative desingularizations of $R$ (as specified in the first paragraph) always exist?

Some discussions (I am a beginner in this, so feel free to correct me):

  1. Why weak? By Morita equivalence, to ensures the desingularization is an isomorphism on the regular locus one needs $M$ to be free on that locus of $R$.

  2. If one requires extra conditions (like $M$ being an generator-cogenerator) then there are examples when such desingularization does not exist (in some paper by Iyama which I forgot the name). I have not seen an example in the generality above.

  3. There are positive results in diemension $0,1$, but I care about normal rings, so we start in dimension $2$.

  4. Some people like Van den Bergh or Lieven le Bruyn probably know. May be they are even on MO!

I would appreciate even heuristic reasons for one way or another.

EDIT: The question is now resolved, by Lieven's answer below (as expected (:). I will provide a little bit more details in case someone is interested: Van den Bergh and Stafford proved that, in characteristic $0$, if $A$ is a non-commutative crepant resolution, then $R$ has rational singularity. The definition of NCCR is stronger, but if, for example, $R$ is Gorenstein of dimension $2$, it coincides with my version. So a counterexample is something like $R=k[x,y,z]/(x^3+y^3+z^3)$, which is a non-rational hypersurface.

  • $\begingroup$ I know Lieven is on MO (as mathoverflow.net/users/2275/lieven-lebruyn) but haven't seen Michel around, and there seems not to be a way to search for users? $\endgroup$ Feb 6, 2010 at 14:00
  • $\begingroup$ You can search for users by click on Users and type the name. I don't think Michel is here though. $\endgroup$ Feb 6, 2010 at 15:26

3 Answers 3


There are already counter-examples in dimension 2. If you take a 2-dml non-rational singularity, then there cannot exist a non-commutative resolution in your sense. In fact, any 2-dml nc-resolution your sense is also an nc-resolution in Michel's sense and so must have rational singularities by a result of Toby Stafford and Michel.

In dimension two you can resort to ancient stuff such as the book 'Graded orders' by Fred, Michel and me, an online scanned version can be found here (Wayback Machine). Lemma IV.2.3 says that a tame order (such as your End(M)) of global dim 2 is 'moderated regular' (defn IV.1.4) and hence a tame order of finite representation type which are classified further in the book to yield rational central singularities. That lemma (and dfn IV.1.1 and thm IV.1.2) also shows that your and Michel's dfns coincide in dim 2.

Further, I didn't understand your remark about an Iyama counterexample in the case M is projective? In that case End(M) is Azumaya, so if it has finite gldim, then the center has to be regular too.

  • 1
    $\begingroup$ As for my remark about counter example: it was in one of Iyama paper, but not by his, I think it is well-known. It says that if we ask whether the Auslander representation dimension is always finite, then the answer is no. $\endgroup$ Feb 6, 2010 at 14:49
  • 1
    $\begingroup$ For interested passer-by: definition of Auslander rep dim was in Brian first answer, and informally in my comment 2) of the original question. If $R$ is Gorenstein, it says $M$ has a free summand (but M is not necessarily free). Of course, if $M$ is free, then $R$ has to be regular, see Lieven answer and my comment 1) of original question. $\endgroup$ Feb 6, 2010 at 15:02

If you haven't already, definitely check out the paper by Burban Iyama Keller and Reiten on Cluster Tilting for One-Dimensional Hypersurface Singularities.

Also, you may google "Auslander representation dimension." It is defined by rep.dim(R) = inf{gl.dim(End_R(M)) | M is a generator-cogenerator}.

In the ADE case the preprojective algebra of the affine dynkin quiver arises this way, and it has finite global dimension.

But you are dropping the condition that M be a generator-cogenerator.

Here is a very general result of Van den Bergh: Assume X is separated. Then there exists a perfect complex E such that D(Qcoh X) is equivalent to D(A) where A is the DG-algebra RHom_X(E,E).

Here is another article which is highly related: D. Orlov, Remarks on generators and dimensions of triangulated categories.

  • $\begingroup$ are you sure that assume X is separated. Then there exists a perfect complex E such that D(Qcoh X) is equivalent to D(A) where A is the DG-algebra RHom_X(E,E) is due to Van den Berg? $\endgroup$ Feb 6, 2010 at 5:19
  • $\begingroup$ Dear Brian, thanks! I know the BIKR paper. Most current results are over Gorenstein rings, and the module $M$ often has free summands, so it is a generator-cogenerator. One of the reasons I asked this question is I don't see a philosophical reason for that condition. I don't see how VdB result helps, can you explain more? $\endgroup$ Feb 6, 2010 at 5:23
  • $\begingroup$ The theorem, as I know it (Remark 2.22 of Katzarkov-Kontsevich-Pantev), requires X to be quasi-compact, quasi-separated, and finite type. Also I think it's due to van den Bergh and Bondal. $\endgroup$ Feb 6, 2010 at 7:47

(sorry i cannot reply bc i don't have enough points) The VdB result I'm quoting is from his survey paper on FM transforms; he references both himself and Rouquier for proofs. This helps because maybe you don't have an algebra with finite global dimension, but you do have a dga. Philosophically, in my humble opinion, D(A) = Perf Y, where Y is some would-be smooth variety.

As for the Gorenstein condition, maybe someone who knows about the minimal model program could comment on this. All I know is there is a paper by Yoshino, "Graded CM modules over graded CM domains" where he relates the stable category of graded CM modules of a non-Gorenstein ring to one where it is Gorenstein. basically the idea there is that Proj R (in the stacky sense) is equivalent to Proj R' (stacky sense). The example he gives is easy to see in terms of Q-divisors on P1. but anyway, it's going off on a tangent..

  • $\begingroup$ A reason for Gorenstein is (I think) in the dimension 3 case, two things related by a flop contracted down to a Gorenstein terminal singularity. Interesting references, I will upvote your first answer to keep them in order. $\endgroup$ Feb 6, 2010 at 7:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .