This is the problem 1.1 from the book, A Gentle Course in Local Class Field Theory.
Let $p$ be a prime, let $F$ be a field of characteristic $\neq p$, and let $a \in F^{\times} \backslash F^{\times p}$. We make no assumption on the presence of roots of unity in $F$, and we wish to prove that $[F(\alpha):F]=p$, where $\alpha$ is a root of $X^p-a$. For this, we assume $[F(\alpha):F] \neq p$, and work toward a contradiction:
- Consider the composite field $F(\mu_p, \alpha)$, and show that $F(\alpha) \subset F(\mu_p)$.
2. Deduce that $F(\alpha)/F$ is Galois, and then that $F(\alpha)=F(\mu_p)$
- From the fact that $[F(\alpha):F]$ does not depend on the choice of root $\alpha$, conclude.
Note that $\mu_p$ stands for a pth primitive root of unity in the algebraic closure of $F$, aka $F^{alg}$. Also, $F^{\times p} = \{f^n|f \in F\}$.
I've had a hard time proving that $F(\alpha) \subset F(\mu_p)$ and $F(\alpha)/F$ is Galois, and didn't get how the assumption for absurdness would fit in the proof. Could anyone please give some help on this?
I have been aware that the problem is kind of equivalent to the proposition that $x^p-a$ is irreducible iff $x$ has no root in $F$ (there is an answer regrading this). However, I am trying to understand the way from the book.