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](http://www.sciencedirect.com/science/article/pii/S0001870884710875). 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. 



Global dimension of quantum $\mathbb{P}^{n}$
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](http://www.springerlink.com/content/vh2mfmbyv3g2f0el/).


