Timeline for Can a quotient ring R/J ever be flat over R?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 19, 2022 at 14:33 | history | edited | darij grinberg | CC BY-SA 4.0 |
latex
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| Oct 10, 2009 at 19:44 | history | edited | David Rydh | CC BY-SA 2.5 |
Added example of connected scheme.
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| Oct 9, 2009 at 19:20 | comment | added | Anton Geraschenko | It looks like you can show that a ring is absolutely flat if every ideal is generated by idempotents (the failed proof in my question contains the main idea, I think). But in the infinite product of fields, every element is (up to a unit) idempotent, so every ideal is generated by idempotents, so the ring is absolutely flat. | |
| Oct 9, 2009 at 16:53 | comment | added | David Rydh | I know this for a fact... Also, note that the infinite product has more points then the factors (which are open and closed). There are also more enlightening examples, such as the absolutely flat ring associated to any ring (this is a bijection on the spectrum) introduced by Olivier. | |
| Oct 9, 2009 at 16:50 | history | edited | David Rydh | CC BY-SA 2.5 |
Fixed proof of 1).
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| Oct 9, 2009 at 16:41 | comment | added | Anton Geraschenko | How do you know that an infinite product of fields is absolutely flat (which I assume means that every module over it is flat)? | |
| Oct 9, 2009 at 16:23 | comment | added | Ben Webster♦ | Can you give a quick indication of why this is so? | |
| Oct 9, 2009 at 15:59 | vote | accept | Anton Geraschenko | ||
| Oct 9, 2009 at 15:42 | history | answered | David Rydh | CC BY-SA 2.5 |