Let $K = \mathbb{Q}_2(\zeta_3)$ where $\zeta_3$ is a primitive third root of unity, and $F = \mathbb{Q}_2(\zeta_3,\sqrt[3]{2})$. Furthermore, let $L = \mathbb{Q}(\zeta_3,\beta)$ where $\beta$ is defined by $\beta^3 = \zeta_3 \sqrt[3]{2}$.
My conclusion: Since $\beta^3/\zeta_3 = \sqrt[3]{2}$, the field $F$ is contained in $L$. Therefore, the compositum $FL$ is $L$ and therefore, $FL/L$ has degree $1$.
Problem: My professor told me briefly that $F$ can not be contained in $L$ because $(\frac{\beta}{\sqrt[3]{2}})^3 = \zeta_3$ which means that $FL$ has a primitive ninth root of unity which neither exists in $F$ not in $L$.
Could you please explain me why neither $F$ nor $L$ have a primitive ninth root of unity? And why was my argument wrong? I thought in order to show $F\subset L$ it enough to show that the elements generating $F$ are in $L$ too, so $\zeta_3 \in L$ (which is obvious) and $\sqrt[3]{2} \in L$ (which follows from my computation above).
Any help is really appreciated!