Timeline for Polynomials with rational coefficients
Current License: CC BY-SA 2.5
18 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Dec 9, 2010 at 4:37 | vote | accept | Wadim Zudilin | ||
| Jun 8, 2010 at 8:01 | history | edited | Wadim Zudilin | CC BY-SA 2.5 |
further edit
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| Jun 6, 2010 at 21:53 | comment | added | Wadim Zudilin | Harry and Will, thank you for these comments. So, the problem is reduced to finding just one counter example for some $n>1$. (On the other hand, I guess that your comments have resulted in somebody's downvote.) | |
| Jun 6, 2010 at 18:55 | comment | added | Will Jagy | @Harry, on the other hand, if there is an injective example in $ n \geq 3$ variables, by setting $n-2$ of them to $0$ we get an injective example in dimension $2.$ So you have shown that there is an injective polynomial in dimension 2 if and only if there is an example for every $n \geq 2.$ | |
| Jun 6, 2010 at 17:58 | comment | added | Harry Altman | As regards the new question, if there's a counterexample $f$ for $n=2$, there's a counterexample for any $n$, as you can just take $f(f(x,y),z)$ when $n=3$, etc. So if we expect there is a counterexample for $n=2$ then we shouldn't be able to prove this at all; I guess considering $n>2$ might still be helpful if that makes finding counterexamples easier? | |
| Jun 6, 2010 at 11:17 | history | edited | Wadim Zudilin | CC BY-SA 2.5 |
once against sharpened
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| Jun 6, 2010 at 10:54 | history | edited | Wadim Zudilin | CC BY-SA 2.5 |
sharpened the question
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| Jun 6, 2010 at 10:26 | comment | added | Harry Altman | Though the fact that it doesn't work for general $\mathbb{Q}_p$ should make us suspicious of any pure continuity argument for $\mathbb{R}$. | |
| Jun 6, 2010 at 10:14 | comment | added | Wadim Zudilin | Don't be so self-critical: it's hard to trust our "continuous" intuition! | |
| Jun 6, 2010 at 10:02 | comment | added | pinaki | I see - that was pretty stupid of me! | |
| Jun 6, 2010 at 9:53 | comment | added | Wadim Zudilin | @auniket: Please check the solution below ($n=1$). | |
| Jun 6, 2010 at 9:51 | comment | added | pinaki | For n = 1: doesn't the continuity of the graph of f imply that the answer is "yes"? Because in this case the condition is equivalent to saying that for some c in R, the horizontal line (call it L) y = c cuts the graph of f in at least two points. But then this property (of cutting the graph of f in at least two points) will not be destroyed if either we move L a bit up or a bit down. So certainly we can ensure c to be rational. | |
| Jun 6, 2010 at 9:39 | answer | added | Hailong Dao | timeline score: 33 | |
| Jun 6, 2010 at 9:25 | comment | added | Wadim Zudilin | Oh yes, I know this from the cited above answer. But I don't see why no rational pair $a',b'$ with this property exists. | |
| Jun 6, 2010 at 9:18 | comment | added | Guy Katriel | For any polynomial in two variables there exist distinct $a,b$ so that $f(a)=f(b)$, so the condition you put on $f$ is always fulfilled. | |
| Jun 6, 2010 at 9:02 | comment | added | Harry Altman | Well, it's clear that this won't work if you replace $\mathbb{R}$ with $\mathbb{C}$ from considering $x^n$ when $n$ is odd. That also shows it won't work if you replace it with general $\mathbb{Q}_p$, either... | |
| Jun 6, 2010 at 8:47 | history | asked | Wadim Zudilin | CC BY-SA 2.5 |