Timeline for Can every idempotent ideal be generated by an idempotent?
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Jul 20, 2023 at 0:28 | vote | accept | GuoJi | ||
| Jul 19, 2023 at 14:59 | comment | added | YCor | Just to repharase my example, it is the ring $\bigcup_n k[t^{1/n!}]$ with the ideal $\bigcup_n t^{1/n!}k[t^{1/n!}]$. Of course it has plenty of variants. | |
| Jul 19, 2023 at 14:55 | comment | added | Alex M. | @GuoJi: The title speaks about an idempotent, but the body of the question does not say anything about the element $e$. Do you want $e^2 = e$? | |
| Jul 19, 2023 at 13:46 | answer | added | Jared White | timeline score: 4 | |
| Jul 19, 2023 at 13:44 | comment | added | YCor | The title has been edited by OP to reflect the intended question (which I answered negatively in the first comment). | |
| Jul 19, 2023 at 13:36 | comment | added | Jeremy Rickard | @HenrikRüping Does anybody use “infinitely generated” in that way? Are there really people who would say that $\mathbb{Z}$ is an infinitely generated abelian group? | |
| Jul 19, 2023 at 13:19 | history | edited | GuoJi | CC BY-SA 4.0 |
edited title
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| Jul 19, 2023 at 13:18 | review | Close votes | |||
| Jul 25, 2023 at 3:06 | |||||
| Jul 19, 2023 at 13:18 | comment | added | HenrikRüping | Well that assumes that 'infinitely generated' really means 'not finitely generated'. Well one could also interpret 'infinitely generated' as generated by an infinite set. Then the answer to the question in the title is trivially 'yes'. | |
| Jul 19, 2023 at 13:16 | comment | added | GuoJi | @JeremyRickard I just want to know idempotent ideal $I$ in the ring $R$ (Not necessarily finite generated), can there still be an $e\in R,e^2=e$, such that $I=Re$? | |
| Jul 19, 2023 at 13:05 | comment | added | Jeremy Rickard | Is the question you want to ask “Is there an infinitely generated idempotent ideal?”. Because the answer to the question in the title is trivially “no”. | |
| Jul 19, 2023 at 12:45 | history | edited | GuoJi | CC BY-SA 4.0 |
added 160 characters in body
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| Jul 19, 2023 at 12:42 | comment | added | YCor | For a semigroup $(S,+)$, $K[S]$ is the vector space with basis $(e_s)_{s\in S}$ and this becomes a $K$-algebra with product $e_se_t=e_{s+t}$ (this is called "semigroup ring" or "semigroup algebra"). For instance $K[\mathbf{N}]$ is just the polynomial algebra $K[t]$ (assuming the convention $0\in\mathbf{N}$). | |
| Jul 19, 2023 at 12:17 | comment | added | GuoJi | oh, that would be ideal generated by idempotent. What does the $K[\mathbb{Q}_{\geq 0}]$ above mean? | |
| Jul 19, 2023 at 12:04 | comment | added | YCor | GuoJi: no, this is certainly not what I wrote. | |
| Jul 19, 2023 at 11:50 | comment | added | YCor | Of course the question is strangely posed. If the ideal is not f.g., obviously it cannot be generated by an idempotent. The question is rather whether "finitely generated" can be removed from the assumption (and the answer is no, i.e. there exists an infinitely generated example). | |
| Jul 19, 2023 at 11:48 | comment | added | YCor | What about the ideal $K[\mathbf{Q}_{>0}]$ in $K[\mathbf{Q}_{\ge 0}]$? | |
| Jul 19, 2023 at 11:31 | history | asked | GuoJi | CC BY-SA 4.0 |