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

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?

• I think we have a perfect duplicate somewhere. This is the best match I've found so far. – Jyrki Lahtonen Apr 6 '19 at 15:38

## 1 Answer

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.