Graded global dimension of a graded algebra Let $k$ be an algebraically closed field of characteristic $0$.
Let $A := k \langle x,x^{-1},y \rangle /(xy-qyx, x^{d_1}-ay^{d_2})$, where deg$(x)>0$, deg$(y)>0$, $q,a \in k^*$ and $d_1\text{deg}(x)=d_2\text{deg}(y)$.
From this, we think of $A$ as a $\mathbb{Z}$-graded algebra over $k$.
We also assume $q^{d_1}=q^{d_2}=1$.
Let $\text{Mod}(A)$ be the category of right $A$-modules and $\text{Gr}(A)$ be the category of graded right $A$-modules.
We define the graded global dimension $\text{gr.gl.dim}(A) := \text{sup}_{M \in \text{Gr}(A)} \text{ gr.pd}(M)$ and the global dimension $\text{gl.dim}(A):= \text{sup}_{M \in \text{Mod(A)}} \text{ pd}(M)$, where $\text{ gr.pd}(M)$ is the minimum length of graded projective resolution of $M$ and $\text{pd}(M)$ is the minimum length of projective resolutions of $M$.
Question
$\text{gr.gl.dim}(A) = 0$ ?
It is known that $\text{gr.gl.dim}(A) \leq \text{gl.dim}(A)$ and $\text{gr.pd}(M) = \text{pd}(M)$ for any $M \in \text{Gr}(A)$.
Thank you.
 A: Yes, $\text{gr.gl.dim}(A) = 0$.
Assume $\gcd(\deg x, \deg y)=1$. (If $\gcd(\deg x, \deg y)=s >1$, then regrade the algebra by dividing all degrees by $s$.)
Assume also that $A$ is not a trival ring, so $1 \neq 0$.
For all $b \in \mathbb Z$, $c \geq 0$, the monomial $x^b y^c \in A$ has a right inverse: $$(x^b y^c)(\tfrac 1 {a^l} y^{ld_2-c}x^{-ld_1-b})=1,$$ where $l$ is chosen such that $ld_2 \geq c$. In particular $1 \in A_n A_{-n}$ for all $n \in \mathbb Z$, and $A$ is is a strongly $\mathbb{Z}$-graded ring.
According to
Dade, Everett C., Group-graded rings and modules, Math. Z. 174, 241-262 (1980). ZBL0424.16001,
there is an equivalence of categories $\operatorname{Gr}(A) \to \operatorname{Mod}(A_0)$.
Let $$z=x^{-\deg y}y^{\deg x}.$$ Since $d_1 \deg x$ is a common multiple of $\deg x$ and $\deg y$, and $\gcd(\deg x, \deg y)=1$, we get that $h= \frac {d_1} {\deg y}= \frac {d_2} {\deg x}$ is a positive integer. Using the relation $y x^{-1}=q x^{-1}y$, we get that there is an integer $r$ such that $$z^h=q^r x^{-d_1}y^{d_2}=\frac {q^r} a=v,$$ a non-zero element in $k$.
So a $k$-basis for $A_0$ is $\{1, z, \dots, z^{h-1} \}$, and $$A_0 \cong k[z]/(z^h-v).$$ Since the polynomial $z^h-v$ and its formal derivative have no common roots, the algebra $A_0$ is semi-simple. By Dade's result, $$\text{gr.gl.dim}(A) = \text{gl.dim}(A_0) =0.$$
