Timeline for Rings whose finitely-generated modules are co-hopfian
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 10, 2020 at 15:39 | comment | added | rschwieb | @TimCampion I’m open to adding new conditions, especially when you can help provide interconnections between conditions and good papers for sources. Register and use a submission form and we can talk about it :) | |
| May 10, 2020 at 13:17 | comment | added | Tim Campion | @rschwieb Wow, DaRT looks awesome! I wish it had "pure semisimple" and "finite representation type", but what it has so far is quite impressive! | |
| May 9, 2020 at 17:29 | history | edited | Tim Campion | CC BY-SA 4.0 |
added 36 characters in body
|
| May 8, 2020 at 19:39 | comment | added | rschwieb | Since "strong von Neumann regularity" isn't Morita invariant, it would not be possible to use it as a replacement as an example of "very strong $\pi$-regularity" | |
| May 8, 2020 at 19:33 | comment | added | rschwieb | Just would like to +1 Benjamin's comment: von Neumann regular does not imply strongly $\pi$-regular. A strongly (von Neumann) regular ring is strongly $\pi$-regular, and a von Neumann regular ring is $\pi$-regular. Here are a few other examples of VNR rings that aren't strongly $\pi$-regular. | |
| May 8, 2020 at 18:51 | comment | added | Benjamin Steinberg | In general the path algebra of a finite connected acyclic quiver with at least one edge over K has center K and is not von Neumann regular but is strongly $\pi$-regular. | |
| May 8, 2020 at 18:48 | comment | added | Benjamin Steinberg | The algebra suggested by @YCor is isomorphic to 2x2 upper triangular matrices over K which obviously has the desired properties since it has a non-trivial Jacobson radical the strictly upper triangular matrices | |
| May 8, 2020 at 16:29 | comment | added | YCor | (2) Let $M$ be the monoid $\{1,x,y\}$ with $xy=xx=x$, $yx=yy=y$ and unit $1$. Let $K$ a field, and $A=KM$. Then the center of $A$ is reduced to $K$ (so is von Neumann regular). $A$ itself is not von Neumann regular because $x-y\notin (x-y)A(x-y)=0$. But since $A$ is a finite-dimensional $K$-algebra, clearly every finitely generated left $A$-module is cohopfian. | |
| May 8, 2020 at 16:05 | history | edited | Tim Campion | CC BY-SA 4.0 |
deleted 7 characters in body
|
| May 8, 2020 at 15:59 | history | asked | Tim Campion | CC BY-SA 4.0 |