Timeline for Is irreducibility of an affine $k$-scheme, an open condition?
Current License: CC BY-SA 3.0
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| Jan 18, 2018 at 18:03 | 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. | |
| Dec 19, 2017 at 18:02 | 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. | |
| Nov 19, 2017 at 17:40 | 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 20, 2017 at 17:31 | 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 20, 2017 at 16:37 | 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. | |
| Aug 21, 2017 at 16:18 | 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. | |
| Jul 22, 2017 at 16:03 | answer | added | dromedaris | timeline score: 1 | |
| Jul 21, 2017 at 15:49 | comment | added | Jinhyun Park | @nfdc23 Thank you very much! This is enormously helpful! | |
| Jul 21, 2017 at 12:48 | comment | added | nfdc23 | In the presence of properness (which for a single polynomial one could pass to via homogenization) and subject to the more "natural" condition of geometric integrality, an affirmative answer is provided by EGA ${\rm{IV}}_3$ 12.2.1(x). | |
| Jul 21, 2017 at 12:45 | comment | added | Damian Rössler | Geometric irreducibility is a constructible condition but irreducibility is not. | |
| Jul 21, 2017 at 12:38 | history | edited | Jinhyun Park | CC BY-SA 3.0 |
Rewrote the question to improve readability.
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| Jul 21, 2017 at 12:12 | history | edited | Jinhyun Park | CC BY-SA 3.0 |
correction of the question reflecting comments.
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| Jul 21, 2017 at 12:08 | comment | added | Jinhyun Park | @JasonStarr Oh, thank you for an example. This suggest that I should modify the question a bit further to be more useful. If I begin with the assumption that the given scheme is integral, then would "small deformations" of it be irreducible at least (but not necessarily reduced)? | |
| Jul 21, 2017 at 11:59 | comment | added | Jason Starr | In that case, the answer to your question is "no, irreducibility is not preserved by small deformations." For instance, in $\mathbb{A}^1 = \text{Spec}\ k[y]$, the zero scheme of $(1+y)^2 = 1 + 2y+1y^2$ is irreducible. However, for a generic choice of $(a_0,a_1,a_2)\in k\times k\times k$, the zero scheme of $a_0+a_1y+a_2y^2$ is reducible. | |
| Jul 21, 2017 at 11:50 | comment | added | Jinhyun Park | @JasonStarr Sorry for the confusion. I am asking about irreducibility of the affine scheme given by it. I just edited the text. Thank you for asking for clarification. | |
| Jul 21, 2017 at 11:48 | history | edited | Jinhyun Park | CC BY-SA 3.0 |
improved writing
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| Jul 21, 2017 at 11:48 | comment | added | Jason Starr | Are you asking about irreducibility of a polynomial in a polynomial ring $k[y_1,y_2,y_3]$, or are you asking about irreducibility of a Zariski closed subset of affine space? These are not the same thing. | |
| Jul 21, 2017 at 11:39 | history | asked | Jinhyun Park | CC BY-SA 3.0 |