Timeline for Existence of an irreducible polynomial that does not divide $x^n + a$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 10, 2021 at 20:59 | comment | added | Paul Levy | @Wojowu Ok, the absolute Galois group of the field obtained from $K$ by adjoining all roots of unity should be non-abelian... | |
| Oct 10, 2021 at 20:37 | comment | added | Wojowu | @PaulLevy No, that's not the case - $x^4-2$ has nonabelian Galois group. It is, however, solvable, so unsolvability of the absolute Galois group is a sufficient condition. | |
| Oct 10, 2021 at 19:14 | comment | added | Paul Levy | I think the Galois group of a splitting field for $x^n+a$ is abelian (right?) so I would guess that a sufficient condition is that the absolute Galois group of $K$ be non-abelian. | |
| Oct 10, 2021 at 16:38 | comment | added | Wojowu | Separably closed fields are also examples: if $L/K$ is purely inseparable, then for any $a\in L$, you have $a^{p^n}\in K$ for some $n$. | |
| S Oct 10, 2021 at 16:16 | review | First questions | |||
| Oct 10, 2021 at 17:23 | |||||
| S Oct 10, 2021 at 16:16 | history | asked | L. Prasad | CC BY-SA 4.0 |