Why is $F(\zeta_n)/F$ is Galois.
If $F$ is arbitrary field, and $\zeta_n$ a primitive n-th root of unity then why it is Galois, I know all the roots of $X^n-1$ lie in the extension field but what does it mean for the normality. I mean if $F$ were finite then it would be a splitting field of $X^n-1$, so normal (by a theorem) but here $F$ can be also infinite and separability is also a problem
EDIT: OK Separability issue has been solved, but normality not yet
Asumme that $(n, char(F))=1$. We have that $x^n-1$ splits in $F(\zeta_n)$, so by a theorem (see chapter 9, section 9.2 of Stewart's book "Galois Theory") $F(\zeta_n)/F$ is normal. Now, $(x^n-1)'=nx^{n-1}$ and then $(x^n-1, nx^{n-1})=1$. So $F(\zeta_n)/F$ is separable and hence galoisian.