Timeline for How to apply Hilbert's Irreducibilty theorem?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 3, 2013 at 3:45 | vote | accept | P Vanchinathan | ||
| Jan 2, 2013 at 21:36 | answer | added | Peter Mueller | timeline score: 3 | |
| Jan 2, 2013 at 4:46 | comment | added | P Vanchinathan | Successful specialisation is ok. When a specialisation fails what exactly is the meaning? The polynomial fails to be irreducible? Or it remains irreducible but with different Galois group for the splitting field? I was under the impression that if a specilisation is irreducible then Galois group is the same. | |
| Jan 2, 2013 at 2:41 | comment | added | Felipe Voloch | I don't think you are making any mistake. Hilbert irreducibility implies that, for your cubic $f(t,X)$, $f(a,X)$ is irreducible for most values of $a$ and that the Galois group of the splitting field of $f(a,X)$ is $S_3$ for most values of $a$, but the set of $a$ in the first statement is not the same (as you have discovered with $a=4$) as in the second statement. | |
| Jan 2, 2013 at 0:51 | history | edited | Andrés E. Caicedo |
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| Jan 2, 2013 at 0:50 | history | asked | P Vanchinathan | CC BY-SA 3.0 |