Irreducibility of $X^n-a$

Question

Let ${\mathbb K}$ be a subfield of ${\mathbb C}$. Let $a\in{\mathbb K}$ such that $X^d-a$ has no root in ${\mathbb K}$, for any divisor $d>1$ of $n$. Does it follow that $X^n-a$ is irreducible over $\mathbb K$ ?

Answer 1

No, it doesn't. Take $\mathbb{K} = \mathbb{R}$, $a=-1$, $n = 4$. Observe that $X^2 + 1$ and $X^4 + 1$ have no roots in $\mathbb{R}$, but $$X^4+1 = (X^2 + \sqrt{2}X + 1)(X^2 - \sqrt{2}X + 1)$$

Answer 2

Here is the general theorem on such

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

$\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.