Timeline for Quartic form which is irreducible but not geometrically irreducible
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| Sep 2, 2010 at 16:01 | answer | added | inkspot | timeline score: 3 | |
| Sep 2, 2010 at 14:01 | history | edited | Wanderer | CC BY-SA 2.5 |
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| Sep 2, 2010 at 13:52 | answer | added | Felipe Voloch | timeline score: 2 | |
| Sep 2, 2010 at 13:20 | history | edited | Wanderer | CC BY-SA 2.5 |
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| Sep 2, 2010 at 12:46 | comment | added | Daniel Loughran | Sorry I should have made clear in my post that I was refering to the case n=2, which is essentially the same as that of inhomogeneous polynomials in one variable. | |
| Sep 2, 2010 at 12:27 | comment | added | Wanderer | These are multivariable polynomials! So many polynomials will not factor at all. | |
| Sep 2, 2010 at 11:48 | comment | added | Daniel Loughran | Any polynomial of degree d will factor over its splitting field, which in general can be an extension of degree as large as d factorial. Perhaps it might be smaller in this specific finite field case though. | |
| Sep 2, 2010 at 11:33 | history | edited | Wanderer | CC BY-SA 2.5 |
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| Sep 2, 2010 at 11:04 | comment | added | Daniel Loughran | The galois group of a finite extension of finite fields is always cyclic, generated by the frobenious. | |
| Sep 2, 2010 at 10:20 | comment | added | Wanderer | Why a cyclic Galois group? | |
| Sep 2, 2010 at 10:04 | comment | added | Martin Bright | I don't know what sort of answer you're after, but it seems to me that what you have is some sort of combinatorial object (a configuration of 4 points/lines/planes) with an action of a (cyclic) Galois group. That might be a good point of view to start from. | |
| Sep 2, 2010 at 8:46 | history | asked | Wanderer | CC BY-SA 2.5 |