Timeline for an easy example of valuation ring which is not noetherian? [duplicate]
Current License: CC BY-SA 3.0
14 events
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S Oct 23, 2017 at 18:45 | history | suggested | user 1 |
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Oct 23, 2017 at 17:08 | review | Suggested edits | |||
S Oct 23, 2017 at 18:45 | |||||
Nov 18, 2014 at 2:45 | history | closed |
YCor S. Carnahan♦ |
Duplicate of non discrete valuation ring [closed] | |
Nov 18, 2014 at 0:14 | review | Close votes | |||
Nov 18, 2014 at 2:45 | |||||
Nov 29, 2011 at 9:26 | answer | added | Hagen | timeline score: 2 | |
Nov 28, 2011 at 22:17 | answer | added | Leonardo | timeline score: 2 | |
Nov 28, 2011 at 19:01 | answer | added | Tom Goodwillie | timeline score: 3 | |
Nov 28, 2011 at 16:28 | answer | added | Karl Schwede | timeline score: 10 | |
Nov 28, 2011 at 9:52 | comment | added | Damian Rössler | A valuation ring is a discrete valuation ring if and only if it is noetherian. See Matsumura, "Commutative rings", Th. 11.1, p. 78. So that provides you with a slew of examples. | |
Nov 28, 2011 at 9:34 | answer | added | Martin Brandenburg | timeline score: 11 | |
Nov 28, 2011 at 7:46 | comment | added | Sándor Kovács | @pm: if you quote a theorem, it is a reasonable expectation that you quote it correctly. Someone might think that Serre made a mistake! The criterion you describe holds for any noetherian local ring. Serre actually says (correctly) that the maximal ideal is generated by a single non-nilpotent element (in other words its a regular local noetherian ring of dimension $1$). | |
Nov 28, 2011 at 7:19 | answer | added | mnr | timeline score: 2 | |
Nov 28, 2011 at 7:07 | comment | added | Marc Palm | Proposition 2, page 7 in Serre: Local fields says that a commutative ring is a discrete valuation ring iff it is local and noetherian and that its maximal ideal is finitely generated. Hence you noetherian ring must be nondiscrete. | |
Nov 28, 2011 at 6:49 | history | asked | iff | CC BY-SA 3.0 |