Assume $F$ is a subfield of the complex numbers containing the $p$-th roots of unity. If $\alpha$ is a root of $x^p-a$ for some $a\in F$, and $\alpha \not\in F$ then $x^p-a$ is irreducible.
To me it seems obvious that the polynomial $f$ of smallest degree in $F[x]$ such $f(\alpha)=0$ is $x^p-a$. Proving that however has been a challenge. Given another polynomial $g(x)\in F[x]$ with $deg(g)<p$ and $g(\alpha)=0$ what contradiction could I arrive at?
If $a\in F$, it's known that for $p$ a prime, $X^p-a$ has no root in $F$ iff $X^p-a$ is irreducible over $F$. A proof on this site can be found here, and in other linked duplicates.
The roots of $X^p-a$ in a splitting field are of form $\zeta^k\alpha$ for $0\leq k\leq p-1$, where $\zeta$ is a primitive $p$-th root of unity. If $F$ contains any of these roots, then since $F$ contains $\zeta$, it must also contain $\alpha$, a contradiction. So $X^p-a$ has no root in $F$, so by the linked result, it must be irreducible.
