Timeline for A $0$-dimensional ring that is not noetherian
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 9, 2018 at 8:41 | answer | added | user124266 | timeline score: -3 | |
| Apr 6, 2012 at 9:32 | comment | added | Martin Brandenburg | Now, I have also answered Q2 in my answer. | |
| Apr 6, 2012 at 9:01 | comment | added | jmc | Ok, it appears that my flaw of reasoning was exactly that spectra of $0$-dimensional rings need not be discrete. | |
| Apr 6, 2012 at 9:00 | vote | accept | jmc | ||
| Apr 6, 2012 at 8:58 | history | edited | jmc | CC BY-SA 3.0 |
Reformulated q.2 (which was not equivalent to q.1)
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| Apr 6, 2012 at 8:50 | comment | added | jmc | Martin, ok. Then I made a mistake somewhere. Actually I am very interested in an example of a non-noetherian ring $A$ for which $\Spec A$ is discrete. | |
| Apr 6, 2012 at 8:43 | answer | added | Martin Brandenburg | timeline score: 30 | |
| Apr 6, 2012 at 8:40 | comment | added | Martin Brandenburg | $dim(A)=0$ does not imply that $\mathrm{Spec}(A)$ is discrete. Instead, it implies that $\mathrm{Spec}(A)$ is Hausdorff. | |
| Apr 6, 2012 at 8:36 | answer | added | Filippo Alberto Edoardo | timeline score: 6 | |
| Apr 6, 2012 at 8:23 | answer | added | Mariano Suárez-Álvarez | timeline score: 20 | |
| Apr 6, 2012 at 8:21 | history | asked | jmc | CC BY-SA 3.0 |