Let $p$ be a prime number and $F$ be a field such that $\textrm{char}(F)\neq p$. Assume that $X^p-1$ splits over $F$ and let: $$\mu_p:=\{x\in F\textrm{ s.t. }x^p=1\}.$$
Proposition 1. One has that:
(i) $X^p-1$ is a separable polynomial over $F$.
(ii) The cardinality of $\mu_p$ is $p$.
Proof. Since $\textrm{char}(F)\neq p$ and is either zero or prime, one has: $$p\not\in\textrm{char}(F)\mathbb{Z}.$$ Therefore, $p\in F^\times$ and one gets: $$X^{p}-1=(pX^{p-1})(p^{-1}X)-1.$$ Hence, $X^p-1$ and $pX^{-1}$ are coprime and since the formal derivative of $X^p-1$ is $pX^{p-1}$, $X^p-1$ is separable over $F$. Therefore, $X^p-1$ has only simple roots in $F$. Besides, since $X^p-1$ splits over $F$, $X^p-1$ has all his roots in $F$ and one gets $|\mu_p|=p$. $\Box$
Let $\zeta\in\mu_p\setminus\{1\}$, using proposition $1.$ (i) along with Lagrange's theorem, since $p$ is prime, one has: $$\mu_p=\langle\{\zeta\}\rangle.$$
Let $F'/F$ be a galois extension such that $\textrm{Gal}(F'/F)$ is cyclic of order $p$ and let $\sigma\in\textrm{Gal}(F'/F)\setminus\{\textrm{id}\}$.
Proposition 2. There exists $a\in (F')^*$ such that: $$a^{-1}\sigma(a)=\zeta.$$
Proof. One has: $$\textrm{Gal}(F'/F)=\langle\{\sigma\}\rangle.$$ Besides, since $F'/F$ is galois, one gets $(F')^{\textrm{Gal}(F'/F)}=F$ and therefore: $$N_{F'/F}(\zeta)=1.$$ Using Hilbert's theorem 90, there exists such a $a\in(F')^*$. $\Box$.
One gets $\sigma(a)=a\zeta$ and therefore: $$\sigma(a^p)=\sigma(a)^p=(a\zeta)^p=a^p.$$ Hence, since $\textrm{Gal}(F'/F)=\langle\{\sigma\}\rangle$, $a^p\in(F')^{\textrm{Gal}(F'/F)}=F$.
I have hard time proving the following statement:
Proposition 3. $F'$ is a splitting field of $X^p-a^p$ over $F$.
Proof. The set of the roots of $X^p-a^p$ is $a\mu_p\subset F'$, therefore $X^p-a^p$ splits over $F'$ and one has to show: $$F'=F(a\mu_p).$$ Since $F'/F$ is galois, using fixed subfield formula, one has: $$\left[F':F\right]=p.$$ If $a\mu_p$ is linearly independent over $F$, I'm done, but I cannot figure why it is true. Any help will be appreciated.
