Timeline for Can a non-surjective polynomial map from an infinite field to itself miss only finitely many points?
Current License: CC BY-SA 2.5
25 events
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| Feb 8, 2021 at 20:42 | comment | added | Erik Walsberg | Maybe also worth pointing out that the Podewski conjecture is a notorious open problem, so Lampe's question is deep. | |
| Oct 16, 2020 at 16:31 | comment | added | Erik Walsberg | This is a very nice question, so it seems worth pointing out that a similar question has already been asked. Reineke conjectured that if K is a field and the image of every non-constant polynomial in K[t] is cofinite then K is algebraically closed. I saw this in Frank Wagner's paper "Minimal Fields" (MR1812183). Reineke's conjecture implies Podewski's conjecture that a minimal field is algebraically closed. (a first order structure M is minimal if every definable subset of M is either finite or cofinite, algebraically closed fields are minimal by quantifier elimination). | |
| Nov 18, 2019 at 22:43 | comment | added | cnpJj2dwc | I posted what I believe to be a proof of this result here on MSE. | |
| Jul 1, 2018 at 9:51 | comment | added | Watson | Related question on MSE: math.stackexchange.com/questions/1792464 | |
| Apr 17, 2014 at 12:22 | answer | added | Michiel Kosters | timeline score: 8 | |
| Nov 6, 2013 at 21:32 | history | edited | user9072 |
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| Nov 6, 2013 at 13:25 | answer | added | Michiel Kosters | timeline score: 14 | |
| Dec 4, 2009 at 10:39 | comment | added | Philipp Lampe | I am deeply impressed with the strong interest MOers show in the problem. So far we have 10 answers, 57 comments, 28 upvotes, and 11 favorites. We have no final answer but I guess I have to accept an answer because the bounty is going to end. I'll choose to accept the answer with the highest number of upvotes, albeit that answer was given before the bounty was released. Accepting the answer doesn't mean that I want to discourage people from further thinking about the problem. I just don't want to waste the bounty. The case of Hilbertian fields seems to be the current state of knowledge. | |
| Dec 4, 2009 at 10:38 | vote | accept | Philipp Lampe | ||
| Dec 4, 2009 at 10:38 | history | bounty ended | Philipp Lampe | ||
| Nov 30, 2009 at 14:06 | answer | added | user1884 | timeline score: 2 | |
| Nov 30, 2009 at 6:35 | answer | added | Greg Kuperberg | timeline score: 7 | |
| Nov 30, 2009 at 3:01 | answer | added | Jason DeVito - on hiatus | timeline score: 3 | |
| Nov 27, 2009 at 21:20 | answer | added | AFK | timeline score: -3 | |
| Nov 27, 2009 at 15:20 | history | bounty started | Philipp Lampe | ||
| Nov 26, 2009 at 18:45 | answer | added | Guillermo Mantilla | timeline score: 4 | |
| Nov 26, 2009 at 5:44 | history | edited | Philipp Lampe | CC BY-SA 2.5 |
Typo corrected.
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| Nov 25, 2009 at 21:24 | answer | added | Lior Bary-Soroker | timeline score: 30 | |
| Nov 25, 2009 at 20:30 | answer | added | D. Savitt | timeline score: 4 | |
| Nov 25, 2009 at 19:39 | answer | added | Pete L. Clark | timeline score: 33 | |
| Nov 25, 2009 at 16:10 | comment | added | Harrison Brown | Why? Because I was being stupid and misread the question :D. I don't really see why that construction has any better chance than any other, but I concede it might work. | |
| Nov 25, 2009 at 16:07 | comment | added | Dror Speiser | Why? I'm thinking of some kind of inverse limit of extensions of function fields over algebraic closure of finite field might occidentally work. | |
| Nov 25, 2009 at 15:45 | comment | added | Harrison Brown | Correct me if I'm wrong, but can't you even say that if you have an embedding of an algebraically closed field into $k$, no such polynomial exists? | |
| Nov 25, 2009 at 15:35 | answer | added | Qiaochu Yuan | timeline score: 7 | |
| Nov 25, 2009 at 15:11 | history | asked | Philipp Lampe | CC BY-SA 2.5 |