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.

  • 2
    $\begingroup$ I'm not sure, but your endomorphism $\left(\begin{smallmatrix} 0 & x_0 \\ 0 & 0\end{smallmatrix}\right)$ is probably right multiplication by this element, and so it commutes with left multiplication by $X_1X_2X_3$. Could this be the explanation? $\endgroup$ Commented Sep 7, 2023 at 13:06
  • $\begingroup$ Thank you very much for the comment. You mean that $ (X_1X_2X_3) \begin{pmatrix} 0 & x_0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & x_0 \\ 0 & 0 \end{pmatrix} (X_1X_2X_3) $ in $E''$ ? $\endgroup$ Commented Sep 7, 2023 at 13:22
  • $\begingroup$ I am so sorry. I am confused and still don't understand. $\endgroup$ Commented Sep 10, 2023 at 8:02
  • 1
    $\begingroup$ Here's another way of putting it. Although left multiplication by $X_1X_2X_3$ is an endomorphism of $E'(1)$, it's not the endomorphism you need. What you need is the endomorphism $X_1X_2X_3$ of $E'$, conjugated by the isomorphism between $E'$ and $E'(1)$ given by multiplication by $x_0$. $\endgroup$ Commented Sep 10, 2023 at 8:23
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    $\begingroup$ I think I understand now. The reason why I misunderstood is that I thought $X_1X_2X_3$ as an element in $E''$ is given by a $\textit{left}$ multiplication of $X_1X_2X_3$ in $E''$. In order for $X_1X_2X_3$ to be a center in $E''$, I should have thought $X_1X_2X_3$ as an element in $E''$ is given by a $\textit{right} $ multiplication of $X_1X_2X_3$ in $E''$. This is what you wanted to tell me, isn't it ? $\endgroup$ Commented Sep 17, 2023 at 10:00

1 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.

  • $\begingroup$ I'm sorry, I still don't understand, but the matrices $\begin{pmatrix} 0 & x_1\\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0\\ x_0^{-1} & 0 \end{pmatrix}$ also commute with multiplication by $X_1X_2X_3$ in $E''$ ? I need to verify that to show $X_1X_2X_3$ is a central element. $\endgroup$ Commented Sep 7, 2023 at 23:38
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    $\begingroup$ Your matrices representing endomorphisms are presumably all right multiplications, because otherwise they're not homomorphisms of modules with respect to left multiplication. Associativity says that right multiplications commute with left multiplications: $(xa)y=x(ay)$. $\endgroup$ Commented Sep 7, 2023 at 23:59
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    $\begingroup$ No, that is not the case. Left multiplication by $x_0$ is not a homomorphism of left modules, so you have to use right multiplication. And left multiplication by a central element is what you need for $X_1X_2X_3$. You can't just choose. $\endgroup$ Commented Sep 8, 2023 at 7:56
  • 3
    $\begingroup$ Yes, you still have the same misunderstanding. The point is that for a ring $A$, the endomorphisms of $A$ as a left module are isomorphic to $A^{\mathop{\rm op}}$. You insist on confusing $A$ with $A^{\mathop{\rm op}}$, with the obvious consequence that things don't multiply the way you think they do. This is the last time I'm going to say this. $\endgroup$ Commented Sep 8, 2023 at 9:47
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    $\begingroup$ I will add one further comment, then I'm done. Left multiplication by $X_1X_2X_3$ is the endomorphism that commutes with right multiplication by $\left(\begin{smallmatrix}0&x_0\\0&0\end{smallmatrix}\right)$. It is equivalent to right multiplication by the diagonal matrix with entries $X_1X_2X_3$ and $\omega X_1X_2X_3$ (or is it $\omega^2$? I can't be bothered to check). Whichever it is, these endomorphisms commute. $\endgroup$ Commented Sep 8, 2023 at 10:13

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