Timeline for When is $f(x^d)$ irreducible?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 5, 2015 at 13:18 | history | edited | pinaki | CC BY-SA 3.0 |
added 344 characters in body
|
| Jun 5, 2015 at 12:50 | comment | added | LSpice | As @VladimirDotsenko's comment points out, it is important not to confuse "no rational or repeated roots" with "irreducible" (which is much stronger). The real problem is with non-transitivity of the Galois action on roots; in the given example, the Galois group has the two orbits $\{1 \pm 2\sqrt2\}$ and $\{2(1 \pm \sqrt2)\}$, leading to the two factors indicated. | |
| Jun 5, 2015 at 12:16 | history | edited | pinaki | CC BY-SA 3.0 |
added 75 characters in body
|
| Jun 5, 2015 at 11:58 | comment | added | Vladimir Dotsenko | $x^2+1$ is irreducible in $\mathbb{F}_3$, but $x^4+1=(x^2+x-1)(x^2-x-1)$ | |
| Jun 5, 2015 at 11:46 | history | answered | pinaki | CC BY-SA 3.0 |