Let $K = \mathbb{Q}_3$ and $L = \mathbb{Q}_3( \sqrt[4]{-3}, \sqrt[3]{2}, \xi_4)$ where $\xi_4$ denotes a $4$-th root of unity.
One can show that $\operatorname{Gal}(L/K) \simeq D_4 \times C_3$. Denote $D_4 = \langle r,s \rangle$. Now I would like to compute the extension $M/K$ such that $\langle (r,1) \rangle \simeq \operatorname{Gal}(L/M)$.
I know $(r,1)$ corresponds to the element in $\operatorname{Gal}(L/K)$ with $\sqrt[4]{-3} \mapsto \xi_4 \sqrt[4]{-3}$, $\xi_4 \mapsto \xi_4$ and $\sqrt[3]{2} \mapsto \sqrt[3]{2}$, so $\xi_4$ and $\sqrt[3]{2}$ are obviously elements of $M = L^{\langle (r,1) \rangle}$.
I thought it might be $M = \mathbb{Q}_3(\xi_4, \sqrt[3]{2})$ but this seems to have no third root of unity which is required since $M/K$ must be Galois (which again follows from the fact that $\langle (r,1) \rangle$ is normal since it is cyclic). Then again, $M/K$ has degree $6$, i.e. it matches the cardinality of $\operatorname{Gal}(L/K)/\langle (r,1) \rangle$.
Could you please tell me where my mistake is resp. if I forgot to include another element?