Timeline for Isomorphism concerning $Soc(M_n(R))$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 22, 2017 at 16:57 | comment | added | Ehud Meir | Take then $R=\mathbb{Z}/4$. The quotient by the socle is $\mathbb{Z}/2$. | |
| Jan 22, 2017 at 16:56 | comment | added | karparvar | Yes, I do. But, the ring $R=\mathbb Z/2$ is a field, so both $R/Soc(R)$ and the full matrix of it are $0$, which "are" Boolean rings! | |
| Jan 22, 2017 at 16:49 | comment | added | Ehud Meir | by Boolean you mean that all elements are idempotents? because in this case the answer is no, take for example the ring $\mathbb{Z}/2$. | |
| Jan 22, 2017 at 16:48 | comment | added | karparvar | Meier: By Morita equivalence, how is it deduced that the ring $M_n(R/Soc(R_R))$ is Boolean if the ring $R/Soc(R_R)$ is Boolean, or I am wrong...? | |
| Jan 22, 2017 at 15:37 | vote | accept | karparvar | ||
| Jan 22, 2017 at 15:30 | comment | added | Ehud Meir | Well, for any $R$-submodule $A\subseteq R$ you have an isomorphism of $M_n(R)$ modules: $M_n(R/A)\cong M_n(R)/M_n(A)$ given in the obvious way. | |
| Jan 22, 2017 at 15:09 | comment | added | karparvar | Thanks for the answer! Now, how do we conclude the isomorphism $M_n(R/Soc(R_R))\simeq M_n(R)/M_n(Soc(R_R))$? | |
| Jan 21, 2017 at 19:38 | history | answered | Ehud Meir | CC BY-SA 3.0 |