Let $k $ is an algebraically closed field of $\text{ch}(k)=0$.

Let $$E := k \langle x_0, x_1, x_2 ,x_3 \rangle/(x_ix_j-q_{ij}x_jx_i )_{0 \leq i,j \leq 3},$$ where $$(\text{deg}(x_0), \text{deg}(x_1), \text{deg}(x_2), \text{deg}(x_3)) = (1,1,2,2)$$ and $$ (q_{ij}) = \begin{pmatrix} 1 & 1& 1 & \omega^2 \\ 1 & 1& \omega^2 &1 \\ 1 & \omega & 1 &1 \\ \omega & 1 &1 &1 \end{pmatrix}. $$ $\omega$ is a primitive 3-th root of unity.

We consider the localization $E' := E[x_0^{-1}]$ of $E$ by $x_0$.

Then, we have an equivalence $$\text{grmod}(E') \simeq \text{grmod}(\text{End}(E'\oplus E'(1)))$$ by graded Morita equivalence. Here, we denote the category of finitely generated right graded $R$-modules by $\text{grmod}(R)$ for any noetherian graded algebra $R$.

Since $\operatorname{deg}(x_0)=1$, we have $E'$ and $\text{End}(E'\oplus E'(1))$ are strongly graded. Moreover, we have $\operatorname{mod}(E'_{0}) \simeq \operatorname{mod}(E'')$, where $E''$ is $$ E'' := \text{End}(E'\oplus E'(1))_0 \simeq \begin{pmatrix} E'_{0} & E'_{1} \\ E'_{-1} & E'_{0} \end{pmatrix}. $$ $E’_{i}$ is the degree $i$ part of $E’$.

On the other hand, since $\operatorname{mod}(E'_0) \simeq \operatorname{mod}(E'')$, we also have an isomorphism of the centers of algebras $Z(E'') \simeq Z(E'_{0})$. (cf. Theorem 4.4. in this paper. Here we use $E'_0, E''$ are finite over the centers.)

And, $$E'_0 \simeq k \langle X_1,X_2,X_3 \rangle /(X_iX_j-Q_{ij}X_jX_i),$$ where $$X_1 = x_1x_0^{-1},X_2=x_2x_0^{-2},X_3=x_3x_0^{-2}$$ and $$ (Q_{ij})= \begin{pmatrix} 1& \omega^2 & \omega \\ \omega& 1 &\omega^2 \\ \omega^2 & \omega &1 \end{pmatrix}. $$

So, we have $Z(E'_0)$ is generated by $X_1^3,X_2^3,X_3^3,X_1X_2X_3$. Although $Z(E'') \subset Z(E'_0) \subset E’_0$, $Z(E'') \not\simeq Z(E'_0)$. This is because $X_1X_2X_3$ is not in $Z(E'')$. In detail, $$ (X_1X_2X_3) \begin{pmatrix} 0 & x_0 \\ 0 & 0 \end{pmatrix} \neq \begin{pmatrix} 0 & x_0 \\ 0 & 0 \end{pmatrix} (X_1X_2X_3) $$ in $E''$. Here, we identify $$ X_1X_2X_3 = \begin{pmatrix} X_1X_2X_3 & 0 \\ 0 & X_1X_2X_3 \\ \end{pmatrix}. $$ By this identification, we think of $X_1X_2X_3$ as an element of $E''$.

However, this contradicts $Z(E'') \simeq Z(E'_{0})$.

**Question**
Where do we have a mistake in the above discussion?

Any comments welcome. Thank you.

**Edit(9/10)** : (What I wrote here was not correct. So, I will modify for now. The conclusion may not change ...)

First, note that all graded modules which appear are *right* graded modules from the above definition.
(You can find it below the definition of $E'$.)

$\begin{pmatrix} X_1X_2X_3 & 0 \\ 0 & 0 \end{pmatrix}$ represents the homomorphism $ E'(1) \rightarrow E'(1)$ which maps $e$ to $(X_1X_2X_3)e$ (left multiplication by $X_1X_2X_3$.) .

$\begin{pmatrix} 0 & 0 \\ 0 & X_1X_2X_3 \end{pmatrix}$ represents the homomorphism $ E' \rightarrow E'$ which maps $e$ to $(X_1X_2X_3)e$ (left multiplication by $X_1X_2X_3$ ).

$\begin{pmatrix} 0 & x_0 \\ 0 & 0 \end{pmatrix}$ represents the homomorphism $E' \rightarrow E'(1)$ which maps $e$ to $x_0 e$ (left multiplication by $x_0$).

These 3 maps are homomorphisms of right graded modules.

rightmultiplication by this element, and so it commutes withleftmultiplication by $X_1X_2X_3$. Could this be the explanation? $\endgroup$5more comments