Timeline for Morita equivalences and centers of some algebras
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
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| Sep 17, 2023 at 16:14 | vote | accept | YkMz | ||
| Sep 17, 2023 at 10:00 | comment | added | YkMz | 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 ? | |
| Sep 16, 2023 at 22:34 | comment | added | Dave Benson | Did you ever sort this out in your own mind? Did you want accept my answer, or are you still confused? | |
| Sep 10, 2023 at 8:48 | comment | added | Dave Benson | I think that may be the source of your confusion about left and right multiplications. | |
| Sep 10, 2023 at 8:38 | comment | added | YkMz | Perhaps, have I misunderstood how $E_0'$ is embedded in $E''$ ? | |
| Sep 10, 2023 at 8:23 | comment | added | Dave Benson | 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 $x_0$. | |
| Sep 10, 2023 at 8:20 | comment | added | YkMz | I apologize for the inconvenience. Is what I wrote in Edit wrong? | |
| Sep 10, 2023 at 8:03 | comment | added | Dave Benson | Read my edited answer, and then play with the matrices. | |
| Sep 10, 2023 at 8:02 | comment | added | YkMz | I am so sorry. I am confused and still don't understand. | |
| Sep 10, 2023 at 6:21 | history | edited | YkMz | CC BY-SA 4.0 |
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| Sep 9, 2023 at 0:53 | history | edited | YkMz | CC BY-SA 4.0 |
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| Sep 8, 2023 at 9:12 | history | edited | YkMz | CC BY-SA 4.0 |
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| Sep 8, 2023 at 5:51 | history | edited | YkMz | CC BY-SA 4.0 |
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| Sep 8, 2023 at 3:55 | history | edited | YkMz | CC BY-SA 4.0 |
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| Sep 8, 2023 at 3:47 | history | edited | YkMz | CC BY-SA 4.0 |
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| Sep 8, 2023 at 0:42 | history | edited | YkMz | CC BY-SA 4.0 |
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| Sep 8, 2023 at 0:18 | history | edited | YkMz | CC BY-SA 4.0 |
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| Sep 7, 2023 at 14:18 | answer | added | Dave Benson | timeline score: 8 | |
| Sep 7, 2023 at 13:59 | history | edited | YkMz | CC BY-SA 4.0 |
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| Sep 7, 2023 at 13:22 | comment | added | YkMz | 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 $E''$ ? | |
| Sep 7, 2023 at 13:06 | comment | added | Dave Benson | 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? | |
| Sep 7, 2023 at 12:40 | history | edited | YkMz | CC BY-SA 4.0 |
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| Sep 7, 2023 at 12:09 | history | asked | YkMz | CC BY-SA 4.0 |