Show that $x^p - m$ is irreducible for prime $p$ and $m \in \mathbb{Q^{\times}}\setminus \left(\mathbb{Q^{\times}}\right)^p$

Question

I'm stuck if $m$ is not a prime or has a single prime divider (Then using Eisenstein's criterion), e.g, $m=4$ and $p=5$.

Any suggestions?

Answer 1

If $x^p-m=A(x)B(x)$ with $\deg(A)=a$ and $\deg(B)=b$, then $|A(0)|=|m|^{a/m}$ and $|B(0)|=|m|^{b/m}$ (using roots-coefficients relations). This should lead to a contradiction when you decompose $m$ in prime factors.