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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