Timeline for A cubic polynomial which contains a linear factor with irreducible residual quadratic form
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 5, 2012 at 15:15 | comment | added | user2013 | @Peter, I misunderstood your answer. You are totally right. | |
| Sep 5, 2012 at 7:29 | comment | added | Peter Mueller | @Muon: Not sure what the problem is: By $g^\sigma$ I mean the field automorphism $\sigma$ applied to the coefficients of $g$. So if one coefficient of $g$ is $1$, then applying $\sigma$ doesn't change this coefficient. So if $g^\sigma$ is a scalar multiple of $g$, we must have $g^\sigma=g$. | |
| Sep 4, 2012 at 19:16 | comment | added | user2013 | How do you conclude that $\mu=1$ for any $\sigma \in \mathrm{Aut}(\mathbb{C})$ | |
| Sep 4, 2012 at 9:26 | history | answered | Peter Mueller | CC BY-SA 3.0 |