Timeline for Morita equivalences and centers of some algebras
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 17, 2023 at 16:14 | vote | accept | YkMz | ||
| Sep 10, 2023 at 8:29 | history | edited | Dave Benson | CC BY-SA 4.0 |
edited body
|
| Sep 10, 2023 at 8:01 | history | edited | Dave Benson | CC BY-SA 4.0 |
Edited to make the answer for right modules rather than left modules.
|
| Sep 8, 2023 at 11:55 | comment | added | Dave Benson | I've edited the answer to explain this better. | |
| Sep 8, 2023 at 11:53 | history | edited | Dave Benson | CC BY-SA 4.0 |
added 231 characters in body
|
| Sep 8, 2023 at 10:13 | comment | added | Dave Benson | 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. | |
| Sep 8, 2023 at 9:47 | comment | added | Dave Benson | 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. | |
| Sep 8, 2023 at 9:42 | comment | added | YkMz | Sorry for the repeated trouble, I wanted to ask you about what I wrote in Edit. In this case, I also had the same misunderstanding again ? | |
| Sep 8, 2023 at 7:56 | comment | added | Dave Benson | 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. | |
| Sep 8, 2023 at 1:36 | comment | added | YkMz | From the comments above I think multiplication from either side would make sense. Is that not the case? | |
| Sep 8, 2023 at 1:25 | comment | added | YkMz | In the question, I identify $X_1X_2X_3 = (\begin{smallmatrix} X_1X_2X_3 & 0 \\ 0 & X_1X_2X_3 \\ \end{smallmatrix}). $ So, $(X_1X_2X_3) (\begin{smallmatrix} 0 & x_0 \\ 0 & 0 \end{smallmatrix})$ and $ (\begin{smallmatrix} 0 & x_0 \\ 0 & 0 \end{smallmatrix}) (X_1X_2X_3) $ mean $(\begin{smallmatrix} X_1X_2X_3 & 0 \\ 0 & X_1X_2X_3 \\ \end{smallmatrix}) (\begin{smallmatrix} 0 & x_0 \\ 0 & 0 \end{smallmatrix} )$ and $ (\begin{smallmatrix} 0 & x_0 \\ 0 & 0 \end{smallmatrix} ) (\begin{smallmatrix} X_1X_2X_3 & 0 \\ 0 & X_1X_2X_3 \\ \end{smallmatrix} )$ respectively. | |
| Sep 7, 2023 at 23:59 | comment | added | Dave Benson | 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: $(xa)y=x(ay)$. | |
| Sep 7, 2023 at 23:38 | comment | added | YkMz | 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. | |
| Sep 7, 2023 at 14:18 | history | answered | Dave Benson | CC BY-SA 4.0 |