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Dec 24, 2012 at 6:30 vote accept Leonard
Dec 20, 2012 at 1:16 comment added Harry Altman To expand on what Angelo says -- we only define "regular ring" the way we do because of this property of regular local rings. Or, in other words, the fact that we have the term "regular ring" at all encapsulates this fact.
Dec 19, 2012 at 20:27 answer added Yemon Choi timeline score: 5
Dec 19, 2012 at 20:05 comment added Yemon Choi I think the "regular" example is poor, as Angelo says, but I think there may be merit in other examples - e.g. analogies between results for groups and results for von Neumann algebras, some of which are even true rather than just "advertising"...
Dec 19, 2012 at 10:36 answer added Peter Michor timeline score: 3
Dec 18, 2012 at 19:09 answer added Peter Michor timeline score: 11
Dec 18, 2012 at 5:55 comment added Angelo The fact that a localization of a regular local is regular is, in fact, a deep result. The fact that this can be stated as "a regular local ring is regular" is mildly amusing; but this is just word play, and, in my opinion, of no interest whatsoever. I voted to close as "not a real question".
Dec 18, 2012 at 5:18 history made wiki Post Made Community Wiki by S. Carnahan
Dec 17, 2012 at 23:57 history edited Leonard
Retag.
Dec 17, 2012 at 22:24 comment added Qfwfq An anti-example could be: "A function that coincides almost everywhere with a continuous function is not necessarily almost everywhere continuous" ...Or are you looking for something considerably less trivial?
Dec 17, 2012 at 22:13 answer added Alexandre Eremenko timeline score: 9
Dec 17, 2012 at 21:43 history asked Leonard CC BY-SA 3.0