Timeline for Motivation for Henselian rings in algebraic geometry
Current License: CC BY-SA 4.0
15 events
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Sep 21, 2023 at 13:10 | vote | accept | user267839 | ||
S Jun 18, 2021 at 22:11 | history | suggested | Daniel Donnelly | CC BY-SA 4.0 |
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Jun 18, 2021 at 21:52 | review | Suggested edits | |||
S Jun 18, 2021 at 22:11 | |||||
S Mar 30, 2021 at 22:56 | history | bounty ended | user267839 | ||
S Mar 30, 2021 at 22:56 | history | notice removed | user267839 | ||
Mar 29, 2021 at 10:20 | answer | added | ali | timeline score: 6 | |
S Mar 29, 2021 at 0:06 | history | bounty started | user267839 | ||
S Mar 29, 2021 at 0:06 | history | notice added | user267839 | Canonical answer required | |
Mar 27, 2021 at 2:08 | history | edited | user267839 | CC BY-SA 4.0 |
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Mar 26, 2021 at 4:06 | comment | added | Will Sawin | I'm not an expert on formal schemes. I would say that for $X \to S$ formally smooth, $S$-points of a complete local ring lift to $X$ if they lift modulo the maximal ideal, while if $X \to S$ is smooth, $S$-points of a henselian local ring lift to $X$ if they lift modulo the maximal ideal. So, at least for lifting points, henselian is as good as complete if your morphisms are really smooth and not just formally smooth. | |
Mar 26, 2021 at 3:56 | comment | added | user267839 | so here over henselian rings works a lot already in the category of schemes what in general works only for formal schemes? | |
Mar 26, 2021 at 3:52 | history | edited | user267839 | CC BY-SA 4.0 |
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Mar 26, 2021 at 3:51 | comment | added | Will Sawin | Lifting to a complete local ring with maximal ideal $m$ is equivalent, by the definition of complete, to lifting to the ring mod $m^n$ for all $n$, and on those quotient rings, $m$ is nilpotent. | |
Mar 26, 2021 at 3:46 | history | edited | user267839 | CC BY-SA 4.0 |
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Mar 26, 2021 at 3:40 | history | asked | user267839 | CC BY-SA 4.0 |