Timeline for Condition for a local ring whose maximal ideal is principal to be Noetherian
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 25, 2014 at 17:35 | vote | accept | Alexander Gruber | ||
| Jan 25, 2014 at 15:01 | comment | added | Emil Jeřábek | And in order not to dump all confusion on Archimedes, $\mathbb R$ has rank $1$ as an ordered abelian group, but rank $2^\omega$ as an abelian group. | |
| Jan 25, 2014 at 14:24 | comment | added | Emil Jeřábek | Right. And conversely, any nonarchimedean ordered field of cardinality at most $2^\omega$ carries an archimedean absolute value (induced by a field embedding into $\mathbb C$). | |
| Jan 25, 2014 at 14:14 | comment | added | Torsten Schoeneberg | Ah, of course, thanks. (In this setting the terminology confused me: So a DVR has a non-archimedean valuation with archimedean value group ...) | |
| Jan 25, 2014 at 14:03 | comment | added | Emil Jeřábek | @TorstenSchoeneberg: A totally ordered group $G$ is archimedean if for every positive $a,b\in G$, there exists an integer $n$ such that $a\le nb$. This is equivalent to having rank $1$. | |
| Jan 25, 2014 at 13:42 | comment | added | Torsten Schoeneberg | @EmilJeřábek: Is "discretely ordered but nonarchimedean value group" what you really want to say? I would have expected "value group that is discrete but not of rank 1". -- Btw, your answer is amazingly elegant. | |
| Jan 25, 2014 at 12:09 | comment | added | Jesse Elliott | Also related: mathoverflow.net/questions/36611/… | |
| Jan 25, 2014 at 0:09 | comment | added | Torsten Schoeneberg | Related: mathoverflow.net/questions/30715 , mathoverflow.net/questions/39866 | |
| Jan 24, 2014 at 22:36 | answer | added | Emil Jeřábek | timeline score: 10 | |
| Jan 24, 2014 at 21:09 | history | edited | Alexander Gruber | CC BY-SA 3.0 |
added 8 characters in body
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| Jan 24, 2014 at 20:21 | answer | added | Fred Rohrer | timeline score: 4 | |
| Jan 24, 2014 at 19:58 | comment | added | Emil Jeřábek | Any valuation ring with a discretely ordered but nonarchimedean value group is a counterexample to the statement you claim to be true in 1. And for any field $K$, $K[x]/(x^2)$ is an example for the second part of 1. | |
| Jan 24, 2014 at 19:46 | history | asked | Alexander Gruber | CC BY-SA 3.0 |