$X^5 - 2$ is irreducible over $\mathbb{Q}[\zeta_5]$

Question

How can I prove that $X^5 - 2$ is irreducible over $\mathbb{Q}[\zeta_5]$ please ?

Answer 1

$$\left[\Bbb Q(\zeta_5):\Bbb Q\right]=\phi(5)=4\;,\;\;[\Bbb Q(\sqrt[5]2):\Bbb Q]=5$$

so

$$\,[\Bbb Q(\zeta_5,\sqrt[5]2):\Bbb Q]=\begin{cases}[\Bbb Q(\zeta_5,\sqrt[5]2):\Bbb Q(\zeta_5)][\Bbb Q(\zeta_5):\Bbb Q]\\{}\\ [\Bbb Q(\sqrt[5]2,\zeta_5):\Bbb Q(\sqrt[5]2)][\Bbb Q(\sqrt[5]2:\Bbb Q]\end{cases}\;\;\;\;\;$$

and now go back to Greg's hint...