How to show that $x^9-3$ is irreducible over $\mathbb Q\Big(\exp\left(\frac{2\pi i}5\right)\Big)$?

Question

Show that $x^9-3$ is irreducible over $\mathbb Q\Big(\exp\left(\frac{2\pi i}5\right)\Big)$. Is it possible to solve it by the roots?

Answer 1

Let $\alpha:=\sqrt[9]{3}$ and $\beta:=\exp\left(\dfrac{2\pi\text{i}}{5}\right)$. Denote by $\mathbb{K}$ the field $\mathbb{Q}(\alpha,\beta)$. Observe that $$[\mathbb{Q}(\alpha):\mathbb{Q}]=9$$ because $\alpha$ is a root of the polynomial $x^9-3$, which is an irreducible polynomial in $\mathbb{Q}[x]$ by Eisenstein's Criterion. Furthermore, $$[\mathbb{Q}(\beta):\mathbb{Q}]=4$$ because $\beta$ is a root of the polynomial $x^4+x^3+x^2+x+1$, which is an irreducible polynomial in $\mathbb{Q}[x]$. (Cyclotomic polynomials $\Phi_n(x)$ are irreducible over $\mathbb{Q}$; in particular, Eisenstein's Criterion can be used to prove irreducibility of $\Phi_p(x)$ for each prime natural number $p$.) By the field version of Langrange's Theorem, $4$ and $9$ both divide $[\mathbb{K}:\mathbb{Q}]$. Prove that $[\mathbb{K}:\mathbb{Q}]=4\cdot 9=36$, and use this to show that $$\big[\mathbb{K}:\mathbb{Q}(\beta)\big]=9\,.$$ What can you say about reducibility/irreducibility of $x^9-3$ of over $\mathbb{Q}(\beta)$?