Talk:Regular local ring

I think the non-example of the dual numbers is outright false. One can easily check the krull dimension to be 1 and the vector space m/m_2 obviously also has. I recommend its removal.— Preceding unsigned comment added by 2A01:CB04:4FF:7600:75C9:DD45:697:4EBF (talk) 19:27, 14 November 2019 (UTC)Reply

You are wrong. ${\displaystyle k[x]/(x^{2})}$  is Artinian, hence has Krull dimension zero. Rschwieb (talk) 20:49, 14 November 2019 (UTC)Reply

Maybe ${\displaystyle k[\![x]\!]/(x^{2})}$  (still the same ring but a different presentation) is a better example?, which is more clearly a local ring. —- Taku (talk)

I disagree that it is "more clearly local." I think ${\displaystyle k[x]}$  is more accessible and therefore preferable. Rschwieb (talk) 21:23, 14 November 2019

In the definition of regular, I think a stronger condition than just m=(a_1, \ldots, a_n) is needed. Perhaps that a_1, \ldots a_n is a minimal generating set for m would do.

Hold up,

"Another property suggested by geometric intuition is that the localization of a regular local ring should again be regular. Geometrically, this corresponds to the intuition that if a surface contains a curve, and that curve is smooth, then the surface is smooth near the curve."

What about the example of say the surface of two cones joined at their pointy ends. A straight line that runs from one to the other is smooth, but the surface is not locally smooth around the point of intersection?!?129.215.104.179 (talk) 15:03, 29 January 2010 (UTC)Reply

There needs to be a consensus throughout the entry on whether a regular local ring needs to be a domain. If not, then certainly it needn't be a UFD (as claimed twice). If yes, then that should be added to the definition in the introduction. 99.191.229.225 (talk) 15:28, 25 July 2010 (UTC)Reply

Oops...I see that it is a consequence of the definition. 99.191.229.225 (talk) 15:31, 25 July 2010 (UTC)Reply

In the example

"If A is a regular ring, then it follows that the polynomial ring A[x] is regular."

the ring A[x] is not a lokal ring. We need here the definition

A Noetherian Ring R ist a regular ring, if every localization with a prime ideal a regular lokal ring is (see the last two sentences of the article). —Preceding unsigned comment added by 84.146.63.63 (talk) 19:13, 21 March 2011 (UTC)Reply

Why are all the examples of regular rings in the article completions, rings of formal power series like kx? Is not the localization of k[x] at say (x) a regular local ring, a simpler example? Is there some reason why the only examples given in the article are completions? -lethe ^{talk +} 22:10, 18 January 2018 (UTC)Reply

For the sake of accuracy and efficiency, I think it's better to have just one article on regular rings and regular local rings (as they are essentially the same concept). Since "regular local ring" is a more common term, that means to merge regular ring into this article. -- Taku (talk) 01:23, 15 November 2019 (UTC)Reply

    Y Merger complete. Klbrain (talk) 10:56, 14 September 2020 (UTC)Reply

There are June template to merge Geometrically regular ring to here, but given that Geometrically regular ring isn't mentioned on the page (except as a 'see also'), the case is far from clear. Hence oppose. Klbrain (talk) 21:13, 3 August 2024 (UTC)Reply

That’s not necessarily an argument against the merger, though. By means of merger, the article will acquire a discussion of geometrically regular ring (which may be a good thing from the point of view of this article). The article geometrically regular ring has some examples relevant to this article so that also argues for the merger. But anyway we need a tiebreaker. —- Taku (talk) 07:51, 4 August 2024 (UTC)Reply

My point was that there was not case made when proposing the merge, and the case still isn't clear. The definition of Geometrically regular ring refers to them being properties of Noetherian rings, not of Local rings (a subset), so it's not clear that Regular local ring is the correct target. Can there be ring that aren't local but that are regular? Note that I'm not a subject expert, but am trying to follow the argument based on what I read on the pages. Klbrain (talk) 13:23, 28 October 2024 (UTC)Reply