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

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 $F$?

Solution 1:

Here is a classic result:

THEOREM $\ $ Suppose $\rm\:c\in F\:$ a field, and $\rm\:0 &lt n\in\mathbb Z\:.$

$\rm\quad x^n - c\ $ is irreducible over $\rm\:F \iff c \not\in F^p\:$ for all primes $\rm\:p\: |\: n\:$ and $\rm\ c\not\in -4\:F^4\:$ when $\rm\: 4\ |\ n\:. $

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

Solution 2:

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.