Timeline for Algebraic characterization of commutative rings of Krull dimension 1,2, or 3
Current License: CC BY-SA 3.0
12 events
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| Jun 14, 2018 at 14:43 | comment | added | YCor | A question is whether, say, having Krull dimension $\le d$ can be characterized by a 1-st order sentence, or by a family thereof. The answer is, I think, negative for $d\ge 1$, because any ultrapower of $\mathbf{Z}$ has infinite Krull dimension; yet this make sense to ask if it is in a more restricted class (among noetherian rings? among finitely generated algebras over PIDs?). Note that the sentence in Matthé's answer is not 1st order in the language of rings, because of the quantifier $\exists n,m$. | |
| Jun 14, 2018 at 14:23 | answer | added | Matthé van der Lee | timeline score: 3 | |
| Feb 16, 2015 at 22:07 | answer | added | Neil Strickland | timeline score: 2 | |
| Feb 16, 2015 at 4:44 | answer | added | Karl Schwede | timeline score: 5 | |
| Feb 6, 2015 at 18:04 | review | Close votes | |||
| Feb 7, 2015 at 9:04 | |||||
| S Feb 6, 2015 at 17:56 | history | suggested | user 1 |
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| Feb 6, 2015 at 17:55 | review | Suggested edits | |||
| S Feb 6, 2015 at 17:56 | |||||
| Feb 6, 2015 at 17:48 | comment | added | abx | A noetherian, integrally closed domain of Krull dimension 1 is called a Dedekind ring, and has many nice properties. If you drop the "integrally closed" assumption, you'll get all sorts of singularities, so there is no hope for a structure theorem. This becomes much worse of course in higher dimension. | |
| Feb 6, 2015 at 17:44 | comment | added | Andrew Chiriac | Yes, I asked the same question on mathstackexchange. I hope there is a better chance of getting an answer here. | |
| Feb 6, 2015 at 17:40 | comment | added | user26857 | Cross-posted: math.stackexchange.com/questions/1134804/… | |
| Feb 6, 2015 at 17:40 | review | First posts | |||
| Feb 6, 2015 at 17:41 | |||||
| Feb 6, 2015 at 17:35 | history | asked | Andrew Chiriac | CC BY-SA 3.0 |