$\mathbb{Q}(\sqrt[3]{2}, \zeta_{9})$ Galois group How do I calculate the degree of $\mathbb{Q}(\sqrt[3]{2}, \zeta_{9})$ over $\mathbb{Q}$. Should it be 18, as $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3$, and $[\mathbb{Q}(\zeta_{9}):\mathbb{Q}] = 6$? 
However $(\sqrt[3]{2})^{3} \in \mathbb{Q}(\zeta_{9})$, how this affect the calculation?
Thanks

If $X^3-2$ is not irreducible over $\mathbb{Q}(\zeta_9)$, it must have a root $\alpha \in \mathbb{Q}(\zeta_9)$ (this holds only because the degree of the polynomial is $3$). Now $\mathbb{Q}(\zeta_9)/\mathbb{Q}$ has a subextension $\mathbb{Q}(\alpha)/\mathbb{Q}$ which is not normal, since we know that the splitting field of $X^3-2$ over $\mathbb{Q}$ has degree $6$. This is not possible, because the Galois group of $\mathbb{Q}(\zeta_9)/\mathbb{Q}$ is abelian (generally, $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \simeq (\mathbb{Z}/n\mathbb{Z})^{\times}$), therefore every subgroup is normal, and hence (using the Galois correspondence) every subextension of $\mathbb{Q}(\zeta_9)/\mathbb{Q}$ is normal.
