Timeline for When do we genuinely need Noetherian conditions?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 9, 2016 at 20:22 | comment | added | Georges Elencwajg | Thanks @R. van Dobben de Bruyn: you nailed it, and I'm happy that Fred and I actually agree on the mathematics. | |
| Dec 9, 2016 at 19:18 | comment | added | R. van Dobben de Bruyn | @FredRohrer: the confusion about your first comment comes from the word 'except', which you use to mean in all examples except (1). I (and probably other people too) read it as stating your objection: everything you say is correct, except that in (1), .... | |
| Dec 9, 2016 at 19:02 | comment | added | Ben Morley | Dear @Georges, thank you for the many examples. I agree with Fred though that while the first is an example of the necessity noetherian conditions, the later examples 2,3,6 showcase more why we need conditions such as quasicompactness or that the underlying topological space is noetherian (and I certainly agree that the resulting noetherian induction arguments are powerful and interesting). Apologies if this sounds like I am quibbling over details - all your examples are certainly very useful for understanding what can go wrong in the non-noetherian case. | |
| Dec 9, 2016 at 12:07 | comment | added | Georges Elencwajg | Dear @Fred, oh yes I agree. Often quasi-compactness suffices, but maybe the philosophy might be to tread carefully when applying "well-known theorems" to non-noetherian schemes... . | |
| Dec 9, 2016 at 12:02 | comment | added | Fred Rohrer | Dear Georges, in your example 1) noetherianness is indeed necessary. But this is not true for the other examples. | |
| Dec 9, 2016 at 11:58 | comment | added | Georges Elencwajg | @Heinrich: yes, I know, but I didn't want to become polemical... :-) | |
| Dec 9, 2016 at 11:55 | comment | added | Georges Elencwajg | @Fred: Sorry, I don't understand your objection. My claim is that the implication "a scheme $X$ has only one point $\implies$ $ (X=Spec A)$ & ($A$ is local artinian)" is true for every noetherian $X$ and false for every non-noetherian $X$. So, yes, I maintain that noetherianness of $X$ is necessary for the implication to hold. | |
| Dec 9, 2016 at 9:54 | comment | added | HeinrichD | Regarding 5), many people do not even know the definition of "coherent" in the non-noetherian case. Thanks to a certain popular textbook in algebraic geometry. | |
| Dec 9, 2016 at 9:43 | comment | added | Fred Rohrer | Except in 1), noetherianness is not necessary (but, of course, cannot just be omitted). | |
| Dec 9, 2016 at 8:40 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
added 772 characters in body
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| Dec 9, 2016 at 0:08 | history | answered | Georges Elencwajg | CC BY-SA 3.0 |