Timeline for Does hereditary and connected imply that the underlying ring $k$ of a $k$-algebra is a field?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 16 at 8:51 | vote | accept | kevkev1695 | ||
| Feb 16 at 8:48 | answer | added | Jeremy Rickard | timeline score: 6 | |
| Feb 16 at 0:57 | comment | added | kevkev1695 | Yes, it means that the only central idempotents are 1 and 0. | |
| Feb 15 at 20:41 | comment | added | Manny Reyes | What does connected mean here? Is it that $R$ does not decompose as a direct product in a nontrivial way (i.e., trivial central idempotents), or something else? | |
| Feb 15 at 16:06 | comment | added | Benjamin Steinberg | I need to think more. I thought I had an argument to show that $J^2R=JR$ if $JR$ is projective and then I could use Nakayama. But I need to be more careful. | |
| Feb 15 at 15:58 | history | edited | kevkev1695 | CC BY-SA 4.0 |
added 129 characters in body
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| Feb 15 at 15:56 | comment | added | kevkev1695 | Sorry, I forgot to mention that $k$ should be faithfully embedded in the center of $R$, I will edit the question. How does it follow that $JR = 0$? | |
| Feb 15 at 15:48 | comment | added | Benjamin Steinberg | If R is finitely generated as a right k-module and J is the radical of k, then if R is hereditary, then I claim JR=0 and so it must really be an R/J-algebra. | |
| Feb 15 at 15:40 | comment | added | Benjamin Steinberg | Do you want to assume that k is faithfully embedded in the center of R? Otherwise, as @YCor points out you can just take any connected hereditary algebra R over k/m with m a maximal ideal. | |
| Feb 15 at 15:37 | comment | added | YCor | For me the question is strange: if $k$ is say, local and not a field, e.g., $k=L[x]/x^2$ with $L$ a field, and $R=L$ is the residual field, then $R$ is connected and hereditary (it is a field!) but $k$ is not a field. | |
| Feb 15 at 15:35 | comment | added | YCor | ($R$ hereditary means that submodules of projective $R$-modules are projective.) | |
| Feb 15 at 15:29 | comment | added | Benjamin Steinberg | Is $R$ also finitely generated as a $k$-module or can it be infinitely generated? | |
| Feb 15 at 15:09 | history | asked | kevkev1695 | CC BY-SA 4.0 |