Timeline for Variants of Eisenstein irreducibility
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Mar 15, 2020 at 13:37 | history | edited | YCor |
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| Jul 12, 2010 at 3:11 | answer | added | Bill Dubuque | timeline score: 2 | |
| Feb 18, 2010 at 10:20 | vote | accept | Franz Lemmermeyer | ||
| Feb 17, 2010 at 14:20 | answer | added | Franz Lemmermeyer | timeline score: 2 | |
| Feb 17, 2010 at 10:53 | answer | added | Vladimir Dotsenko | timeline score: 7 | |
| Feb 17, 2010 at 9:37 | answer | added | Gjergji Zaimi | timeline score: 11 | |
| Feb 16, 2010 at 15:29 | comment | added | Kevin Buzzard | On the other hand if p exactly divides a_j for j>0 but the a_i for 0<=i<j are allowed to be divisible by arbitrary powers of p then you can't possibly deduce irreducibility: consider (x^(m-j)-p)*(x^j-p^1000). | |
| Feb 16, 2010 at 15:28 | comment | added | Kevin Buzzard | If you have a kink in the NP (i.e. if it's not just one line) then surely an arbitrarily small perturbation of your polynomial will factor over Q? Here's an example of something that is true: if f is monic of odd degree m, and p^2 divides all the a_i for 0<=i<=m-1, but p^3 doesn't divide a_0, then f will be irreducible, because the local extension will contain things of valuation 2/m and hence things of valuation 1/m. | |
| Feb 16, 2010 at 14:51 | history | asked | Franz Lemmermeyer | CC BY-SA 2.5 |