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 $M$.


$\text{gr.gl.dim}(A) = 0$ ?

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.

  • $\begingroup$ Are you missing conditions on $q$? The first relation forces $x^{d_1}y^{d_2}=q^{d_1d_2}y^{d_2}x^{d_1}$, but then, unless $q^{d_1d_2}=1$, the second relation forces $x^{2d_1}=y^{2d_2}=0$ and $A$ is a finite-dimensional local algebra that definitely doesn't have graded global dimension zero. $\endgroup$ Dec 1, 2022 at 9:30
  • $\begingroup$ It’s just as you said. I added an additional condition. $\endgroup$ Dec 1, 2022 at 11:20
  • $\begingroup$ In Jeremy's example, $0=x^{2d_1} x^{-2d_1}=1$, so $A$ is trivial. $\endgroup$ Dec 7, 2022 at 0:55

1 Answer 1


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, $$\text{gr.gl.dim}(A) = \text{gl.dim}(A_0) =0.$$

  • 1
    $\begingroup$ Thank you very much for the answer ! In the above example, I think $x^{-3}y^2 = 1$ from the definition of $A$. $\endgroup$ Dec 1, 2022 at 6:59
  • 1
    $\begingroup$ Yes, you are right. $\endgroup$ Dec 1, 2022 at 7:17
  • 1
    $\begingroup$ I think I got it right now. $\endgroup$ Dec 6, 2022 at 18:08
  • $\begingroup$ Thank you for your answer ! Now I understand. $\endgroup$ Dec 7, 2022 at 2:09

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