Timeline for What kind of subset is Spec(R_P) in Spec(R)?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 11, 2013 at 17:50 | vote | accept | Georg S. | ||
| Feb 11, 2013 at 8:52 | answer | added | Qing Liu | timeline score: 6 | |
| Feb 9, 2013 at 21:14 | answer | added | Will Sawin | timeline score: 2 | |
| Feb 9, 2013 at 18:17 | comment | added | Martin Brandenburg | $\mathrm{Spec}(R_f)$ is the complement of $V(f)$ (for $f \in R$), but $\mathrm{Spec}(R_P)$ is not the complement of $V(P)$ (for $P \in \mathrm{Spec}(R)$). | |
| Feb 9, 2013 at 14:41 | answer | added | Karl Schwede | timeline score: 2 | |
| Feb 9, 2013 at 14:39 | answer | added | Martin Brandenburg | timeline score: 3 | |
| Feb 9, 2013 at 13:14 | comment | added | Fred Rohrer | @Georg: You are of course right, since spectra are not necessarily totally ordered by inclusion. | |
| Feb 9, 2013 at 13:13 | comment | added | Georg S. | Well, V(P) contains all $Q \supset P$, so its complement is the set of all $Q \not\supset P$. But $Spec(R_P)$ consists of all $Q$ with $Q \cap (R \setminus P) = \emptyset$, so $Q \subset P$. This is not the complement of $V(P)$. Am I wrong? This would be great. :-) | |
| Feb 9, 2013 at 13:12 | answer | added | Fred Rohrer | timeline score: 4 | |
| Feb 9, 2013 at 13:08 | comment | added | Joe Silverman | Isn't Spec$(R_P)$ (considered as a subset of Spec$(R)$) the complement of $V(P)$? That would make it open. (This question seems a bit elementary for MO.) | |
| Feb 9, 2013 at 12:54 | history | asked | Georg S. | CC BY-SA 3.0 |