Timeline for When an element of a ring that is divisible by a finite set of elements is necessarily divisible by their product?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 4, 2022 at 10:48 | vote | accept | Mikhail Bondarko | ||
| Dec 4, 2022 at 10:48 | vote | accept | Mikhail Bondarko | ||
| Dec 4, 2022 at 10:48 | |||||
| Dec 4, 2022 at 0:03 | history | became hot network question | |||
| Dec 3, 2022 at 21:21 | comment | added | Mohan | I have added some explanations as an answer below. | |
| Dec 3, 2022 at 20:47 | answer | added | Mohan | timeline score: 2 | |
| Dec 3, 2022 at 20:36 | history | edited | Mikhail Bondarko | CC BY-SA 4.0 |
added 12 characters in body
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| Dec 3, 2022 at 16:44 | answer | added | Martin Brandenburg | timeline score: 4 | |
| Dec 3, 2022 at 16:35 | comment | added | Mikhail Bondarko | Sorry; I want a criterion when the $r_i$ are fixed. This can be a generalization of the Chinese remainder theorem. | |
| Dec 3, 2022 at 16:34 | history | edited | Mikhail Bondarko | CC BY-SA 4.0 |
added 23 characters in body
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| Dec 3, 2022 at 16:17 | comment | added | Laurent Moret-Bailly | Taking $r_1=r_2=r$ you get $r^2\mid r$. This is equivalent to $R$ being absolutely flat, i.e. reduced and $0$-dimensional. | |
| Dec 3, 2022 at 16:03 | history | asked | Mikhail Bondarko | CC BY-SA 4.0 |