# 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.

• 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. Dec 1, 2022 at 9:30
• It’s just as you said. I added an additional condition. Dec 1, 2022 at 11:20
• In Jeremy's example, $0=x^{2d_1} x^{-2d_1}=1$, so $A$ is trivial. Dec 7, 2022 at 0:55

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

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