Timeline for prime ideals in C([0,1])
Current License: CC BY-SA 3.0
19 events
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| May 29, 2017 at 5:58 | comment | added | M10687 | Since I was recently thinking about your last question and I don't see this argument in the answers below, I thought I'd add it: Suppose every prime ideal in this ring is maximal. Then $C([0,1])/N$ is absolutely flat, where $N$ is the nilradical, which is trivial, so $C([0,1])$ is absolutely flat. Then for every $f$ there is a $g$ such that $f^2g=f$. But this is clearly impossible, take $f \doteq x$. There is no continuous $g$ which satisfies this. | |
| Feb 1, 2017 at 17:12 | comment | added | Gro-Tsen | Since this question came up to the front page, and although this is marginally off-topic, I think it's worth mentioning that the excellent book Rings of Continuous Functions by Gillman & Jerison, already mentioned in Yemon Choi's comment, has sequel of sorts, Rings of Quotients of Rings of Functions by Fine, Gillman & Lambek, which is also very beautifully written. | |
| Feb 1, 2017 at 15:52 | answer | added | Włodzimierz Holsztyński | timeline score: 5 | |
| Feb 1, 2017 at 15:06 | history | edited | Denis Serre | CC BY-SA 3.0 |
added 23 characters in body
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| Feb 1, 2017 at 8:23 | answer | added | Tensor_Product | timeline score: 5 | |
| S Jan 12, 2017 at 9:47 | history | suggested | Ali Taghavi |
I add a tag.
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| Jan 12, 2017 at 9:38 | review | Suggested edits | |||
| S Jan 12, 2017 at 9:47 | |||||
| Aug 27, 2010 at 21:38 | history | edited | Nikita Kalinin | CC BY-SA 2.5 |
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| Aug 21, 2010 at 8:05 | vote | accept | Nikita Kalinin | ||
| Aug 17, 2010 at 3:42 | answer | added | Keivan Karai | timeline score: 27 | |
| Aug 16, 2010 at 23:20 | answer | added | Somnath Basu | timeline score: 2 | |
| Aug 16, 2010 at 21:40 | answer | added | Nate Eldredge | timeline score: 5 | |
| Aug 16, 2010 at 21:35 | answer | added | Bill Johnson | timeline score: 23 | |
| Aug 16, 2010 at 21:24 | comment | added | Akhil Mathew | Any filter of closed subsets of the interval gives a proper ideal in $C(0,1)$ consisting of functions whose zero set lies in this filter. (Indeed, the zero set of $f+g$ contains the intersection of the zero sets of $f$ and $g$, so this is actually an ideal.) This handles the example you gave. Gilman and Jerison call ideals defined in this manner $z$-ideals. I suspect that there are conditions where every ideal is a $z$-ideal, but I'm not aware of them. | |
| Aug 16, 2010 at 21:05 | comment | added | Yemon Choi | @Mikola: take a dense proper subset S of [0,1]. Any continuous function which vanishes on that set is identically zero, and so there is no ideal in C[0,1]) - closed or otherwise - whose zero set is S. | |
| Aug 16, 2010 at 20:53 | comment | added | Mikola | Maybe I am just naive, but why wouldn't the space of ideals just be the set of all subsets of [0,1] ? I don't see how having the functions be continuous would get you out of it, since you could always take a limit of continuous functions until it converges to whichever subset you wanted. (This is not a rigorous argument and I am not trying to fully justify this assertion, rather it is a related subquestion.) | |
| Aug 16, 2010 at 20:44 | comment | added | Yemon Choi | Former colleagues, more knowledgeable than I am about these things, usually recommend looking in Gillman and Jerison's book "Rings of continuous functions". (Try Googling the two names in conjunction with "prime ideal".) On the other hand, since you are dealing with functions on [0,1] and not some more general Hausdorff space, it should be possible to resolve your questions more directly. | |
| Aug 16, 2010 at 20:39 | history | edited | Greg Kuperberg |
edited tags
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| Aug 16, 2010 at 20:37 | history | asked | Nikita Kalinin | CC BY-SA 2.5 |