Timeline for A geometric proof of Krull's Principal ideal theorem
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 3, 2020 at 22:06 | vote | accept | Asvin | ||
| Dec 3, 2020 at 22:07 | |||||
| Jul 18, 2018 at 3:59 | answer | added | Fan Zheng | timeline score: 11 | |
| Jul 15, 2018 at 12:12 | answer | added | anon | timeline score: 5 | |
| Jul 14, 2018 at 17:44 | comment | added | Samir Canning | @R.vanDobbendeBruyn I think of that result as "anti-geometry" in the sense that it tells you Noether normalization can't be generalized too far from the usual geometric thing about finiteness of generic projections. | |
| Jul 13, 2018 at 22:49 | comment | added | R. van Dobben de Bruyn | Moreover, if the Noetherian hypothesis is dropped, a minimal prime over $(f)$ can actually have infinite Krull dimension! I'm not sure what this says about the geometry behind the proof... | |
| Jul 13, 2018 at 22:48 | comment | added | R. van Dobben de Bruyn | Commutative algebra is to algebraic geometry as calculus to differential geometry. Some 'geometric' statements in AG rely on CA, just like some 'geometric' statements in DG rely on calculus. | |
| Jul 13, 2018 at 20:03 | comment | added | Samir Canning | The proof in the Stacks Project here stacks.math.columbia.edu/tag/00KD uses the interplay between the notion of ideal of definition, Hilbert polynomial, and Krull dimension. I think using the Hilbert polynomial is pretty geometric. But of course it isn't a completely geometric proof. | |
| Jul 13, 2018 at 18:34 | history | asked | Asvin | CC BY-SA 4.0 |