Let $F$ be a field containing a primitive $n$-th root of unity (for $n \geq 2$) and let $E = F(\alpha)$ where $\alpha \in E$ is an element whose $n$-th power (but no smaller power) is in $F$. Let $\alpha^n := a$.
Question: Assuming $p$ is an arbitrary prime dividing $n$ and $a$ is not a $p$-th power, why is the polynomial $X^n - a$ irreducible in $F[X]$?
What I did by myself?
I know that the roots of $X^n - a$ must be $\zeta^i \alpha$ for $i=0,\dots,n-1$ and $\zeta \in F$ being the primitive $n$-th root of unity from the assumption.
I also know that $E/F$ is a cyclic Galois of order $n$. A generator is $\sigma: \alpha \mapsto \zeta \alpha$.
Is there a way to use one of these facts to prove the statement? If not, what can I do differently?
Thanks in advance!
I don't see why you need the assumption of $\alpha$ not being a $p$ th power. Maybe I'm missing something, here's a (hopefully correct) answer:
Suppose that $X^n-a$ is reducible, so that it has a proper divisor $g$. Its roots are some of the $\zeta^i\alpha$, and their product is $g$'s last term.
There exists then $J \subsetneq \{1, \ldots, n\}$ such that $\prod_{j \in J}\zeta^j \cdot \alpha \in F$. Since $\zeta \in F$, we get $\alpha^{|J|} \in F$. But $|J| < n$, which is absurd.
Edit: if you already know that $E/F$ has degree $n$, then no calculation is needed. Since $\alpha$ is a root of $X^n-a$, we have that $m(\alpha,F) \mid X^n-a$. Equality follows from the fact that
$$\deg m(\alpha,F) = [E:F] = n$$
and so $X^n-a$ is irreducible (since it is a minimal polynomial).