Morita equivalences and centers of some algebras

Question

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.

Answer 1

The answer is that in your matrix $\left(\begin{smallmatrix} 0&x_0\\0&0\end{smallmatrix}\right)$, the $x_0$ denotes the isomorphism of modules given by left multiplication by $x_0$, so it commutes with right multiplication by $X_1X_2X_3$ in $E''$, which is the same as left multiplication by $\left(\begin{smallmatrix} X_1X_2X_3&0\\0&x_0^{-1}X_1X_2X_3x_0\end{smallmatrix}\right)$. So this is your central element of the endomorphism ring, and not left multiplication by $X_1X_2X_3$.

Edited to make the answer for right modules rather than left modules.