Suppose that $p>2$ is prime and that $a$ is a rational number for which $\sqrt[p]{a}$ is in $\mathbb C\backslash\mathbb Q$.
The cyclotomic polynomial $\Phi_p$ is well-known to be irreducible over $\mathbb Q$. Is it also irreducible over $\mathbb Q(\sqrt[p]{a})$?
I believe this is true.
I was able to argue it partially by contradiction, after assuming that $\Phi_p(x)=f(x)g(x)$ is reducible over $\mathbb Q(\sqrt[p]{a})$. I say "partially" because my particular argument only works when I assume that the roots of $f$ do not consist of pairs of multiplicative inverses in the group of $p$-th roots of unity. (So for example, my argument works when the degrees of $f$ and $g$ are odd, but may fail when the degrees are even.)
Any help would be appreciated.
Note: I do not wish to use the order of the Galois group as part of the argument.