When $X^n-a$ is irreducible over F?

Question

Let $F$ be a field, let $\omega$ be a primitive $n$th root of unity in an algebraic closure of $F$. If $a \in F$ is not an $m$th power in $F(\omega)$ for any $m\gt 1$ that divides $n$, how to show that $x^n -a$ is irreducible over $F$?

Answer 1

I will assume "$m \geq 1$", since otherwise $a \in F(\omega)$, but $F(\omega)$ is $(n-1)$th extension and not $n$th extension, so $x^n-a$ must have been reducible.

Let $b^n=a$ (from the algebraic closure of $F$).

$x^n-a$ is irreducible even over $F(\omega)$. Otherwise $$f= \prod_{k=0}^n (x-\omega^k b) = (x^p + \cdots + \omega^o b^p)(x^{n-p} + \cdots + \omega^ó b^{n-p}),$$ so $b^p$ and $b^{n-p}$ are in $F(\omega)$. Consequenty $b^{\gcd(p,n-p)}$ is in $F(\omega)$, but $\gcd(p,n-p)$ divides $n$, so $(b^{\gcd})^\frac{n}{\gcd} = a$, a contradiction.

Answer 2

Below is a classical result:

Theorem $\ $ Suppose $\,c\in F\,$ a field, and $\,0 < n\in\mathbb Z$.

$\ \ \ x^n - c\ $ is irreducible over $F\! \iff c \not\in F^p\,$ for all primes $\,p\mid n,\,$ and $\ c\not\in -4F^4$ when $\, 4\mid n$

A proof is in many Field Theory textbooks, e.g. Karpilovsky, Topics in Field Theory, Theorem 8.1.6.