Let $k$ be a field. Given a (not necessarily commutative) $k$ graded ring $A$, M. Artin and J.J. Zhang introduced a notion of "noncommutative projective scheme" $Proj(A)$ in this paper. It is defined as a $k$-linear abelian category $$ Tails(A)=Gr(A)/Tor(A) $$ where $Gr(A)$ is the abelian category of right graded $A$-modules and $Tor(A)$ is its full subcategory consisting of torsion modules. (Their definition also contains "structure sheaf" and "shift functor", but we ignore them). When $A$ is commutative, $Tails(A)$ is equivalent to $Coh(Proj(A))$ by Serre's theorem.

Assume $A$ is very simple, say $$ A=k[x_{1},\dots,x_{n+1}]/(x_{i}x_{j}=q_{i,j}x_{j}x_{i})_{i< j}. $$ Since $A$ is AS regular algebra, it is known that $\mathrm{gl.dim}(Tails(A))=n$. I now would like to explicitly show this result;

Given a graded right $A$-module module, take a minimal graded injective (not projective, sorry) resolution of $M$ and show that the $k$th-term is torsion for $k>n$.

Is it possible? Thanks in advance.

  • $\begingroup$ The agebra $A$ is of global dimension $n+1$, and the base field, with its standard module structure, is a module of projective dimension $n+1$. Its minimal resolution is graded projective, and its $(n+1)$th term is not torsion. $\endgroup$ Commented Aug 7, 2012 at 3:20
  • $\begingroup$ When you say "right $A$-graded module", do you mean "graded right $A$-module"? The former indicates to me that the module is graded by elements of $A$, while the latter indicates that the module is graded by $\mathbb{Z}$. Which do you mean? $\endgroup$
    – MTS
    Commented Aug 7, 2012 at 3:22
  • $\begingroup$ @Mariano You are right. I should have mentioned this, but $k=A/A_{+}$ is torsion and it is 0 when one pass it to $Tails(A)$. I don't know how to formulate my question, but I want to know $\mathrm{gl.dim}(Tails(A))$, so please ignore your case. $\endgroup$
    – Michel
    Commented Aug 7, 2012 at 4:14
  • $\begingroup$ There is a notion of «smoothness» for non-commutative "projective schemes" which should make your idea precise. You can find it explained in a recent paper by Michel van den Bergh and Paul Smith on non-commutative quadrics, for example. $\endgroup$ Commented Aug 7, 2012 at 4:24
  • $\begingroup$ (And I am sure the projective spaces in your question are smooth, but I would have to track down references to find who proved it0 $\endgroup$ Commented Aug 7, 2012 at 4:30

1 Answer 1


The usual strategy is to take first the resolution of the diagonal --- that is a projective resolution of $A$ as $A\otimes A$-module up to $Tor_2(A\otimes A)$, where $Tor_2(A\otimes A)$ is the subcategory of bigraded $A\otimes A$-modules $M$ such that $M_{i,j} = 0$ for all $i,j \ge N$ for some $N$. Then tensoring this resolution with any graded $A$-module $M$ one obtains its projective resolution.

The required resolution of the diagonal can be constructed in the same way as for the usual polynomial algebra, by using Koszul complexes. You can find an explicit formula in section 4.4 of the Kapustin-Kuznetsov-Orlov paper.

  • $\begingroup$ Thanks for the response. The paper you referred explain things very explicitly and seems very nice. I would appreciate it if you could give me a reference for the "usual strategy". $\endgroup$
    – Michel
    Commented Aug 7, 2012 at 8:48
  • $\begingroup$ I also would like to extend my question to non-Koszul algebras. Are there any good bimodule resolution of, for example, a Fermat hypersurface of our algebra? $\endgroup$
    – Michel
    Commented Aug 7, 2012 at 8:54
  • $\begingroup$ The link to springerlink.com is broken. Possibly it is meant to point to the following paper: Kapustin, Anton; Kuznetsov, Alexander; Orlov, Dmitri, Noncommutative instantons and twistor transform, Commun. Math. Phys. 221, No. 2, 385–432 (2001). Zbl 0989.81127. $\endgroup$ Commented Apr 30, 2023 at 23:40

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