Timeline for When a finitely generated ideal is contained in a union of maximal ideals
Current License: CC BY-SA 3.0
11 events
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| Nov 11, 2016 at 14:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Oct 12, 2016 at 13:46 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 12, 2016 at 12:53 | history | edited | Steven Landsburg | CC BY-SA 3.0 |
added 1 character in body
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| Sep 12, 2016 at 12:48 | comment | added | Jason Starr | @Sasha. Your example is definitely simpler than mine. | |
| Sep 12, 2016 at 12:46 | answer | added | Jason Starr | timeline score: 2 | |
| Sep 12, 2016 at 12:37 | comment | added | SashaP | Take $R=k[x,y],I=(x,y)$ for an algebraically closed $k$ and the set of all maximal ideals of the form $(x-a,y-b)$ for $(a,b)\neq (0,0)$. Then their union is $R\setminus k^{\times}$ so it contains $I$ though $I$ is not contained in any of these maximal ideals. | |
| Sep 12, 2016 at 11:49 | comment | added | Jason Starr | Oops, my ideal is not finitely generated. | |
| Sep 12, 2016 at 11:23 | review | Close votes | |||
| Sep 12, 2016 at 13:52 | |||||
| Sep 12, 2016 at 11:17 | comment | added | Jason Starr | That is certainly not always true. Let $R$ be the commutative ring with $1$ consisting of differentiable functions on the circle $S$. Let $s\in S$ be a specified point, and let $I_s$ denote the ideal of differentiable functions that are zero on some open neighborhood of $s$. For every point $t\in S\setminus\{s\}$, let $\mathfrak{m}_t$ be the maximal ideal of differentiable functions that are zero at $t$. Then $I_s$ is contained in the union of the ideals $\mathfrak{m}_t$, yet $I_s$ is contained in no single maximal ideal $\mathfrak{m}_t$. | |
| Sep 12, 2016 at 11:08 | review | First posts | |||
| Sep 12, 2016 at 12:07 | |||||
| Sep 12, 2016 at 11:04 | history | asked | Arena | CC BY-SA 3.0 |