互联网造车量产前夜：智能汽车冲击功能汽车？
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| 21 hours ago | history | became hot network question | |||
| 22 hours ago | comment | added | Kevin Casto | @Z.M I think this should be understood in reverse, i.e. that the interesting examples are non-Noetherian. Think about, say, the ring of all algebraic integers. | |
| yesterday | answer | added | rschwieb | timeline score: 4 | |
| yesterday | comment | added | rschwieb | @Z.M Huzzah, that's what I was aiming for. I've gotten a lot of mileage out of the JOK construction over the years. I guess I should formulate a solution... | |
| yesterday | comment | added | Z. M | @rschwieb Yes, as long as the value group is $2$-divisible, it works, since ideals of valuation rings one-to-one corresponds to ideals of the value group, [SP, Tag 00IH]. And these valuation rings are not von Neumann regular. | |
| yesterday | comment | added | rschwieb | I might not even be remembering correctly if both those things are totally ordered groups :'( | |
| yesterday | comment | added | rschwieb | I'm not sure if I'm remembering how the ideal correspondence works, but I think if you use the Jaffard-Ohm-Kaplansky construction with some appropriate group, you might be able to guarantee a valuation domain with this property. Maybe $(\mathbb R,+)$? or $(\mathbb R^{>0},\cdot)$? It just feels like being able to halve/sqrt the values in an ideal of the value group means that you should be able to split the corresponding ideal in the valuation domain. But I never have been very practiced with valuation domains so maybe it doesn't work. | |
| yesterday | comment | added | Benjamin Steinberg | The op might mean von Neumann regular rings, in which case all ideals are idempotent. | |
| yesterday | comment | added | Z. M | Suppose that $R$ is a Noetherian ring with such property. Then for every prime ideal $\mathfrak p$, you pick an ideal $\mathfrak q$ such that $\mathfrak p=\mathfrak q^2$, which forces $\mathfrak q=\mathfrak p$, and in particular, $\mathfrak p=\mathfrak p^2$, and thus by Nakayama's lemma, $\mathfrak p$ vanishes in the localization $R_{\mathfrak p}$. In other words, every local ring of $R$ is necessarily a field, which forces that $R$ is a reduced Artinian ring, i.e. a finite product of fields. | |
| yesterday | comment | added | Brian Shin | I don't think this holds for regular rings. For example, it fails for any discrete valuation ring. | |
| yesterday | history | asked | Ya MA e. r | CC BY-SA 4.0 |