# Global dimension of quantum $\mathbb{P}^{n}$

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.

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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. –  Mariano Suárez-Alvarez Aug 7 '12 at 3:20
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? –  MTS Aug 7 '12 at 3:22
@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. –  Michel Aug 7 '12 at 4:14
Thank you for indicating the typo. I fixed it. –  Michel Aug 7 '12 at 4:16
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. –  Mariano Suárez-Alvarez Aug 7 '12 at 4:24

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.