If $\mathbb Q_p(\sqrt[3]{2})$ is galois over $\mathbb Q_p$ then does $\mathbb Q_p$ necessarily contain a cube root of unity?

Question

Denote by $\omega$ any non-trivial cube root of unity. If $\mathbb Q_p(\sqrt[3]{2})$ was galois then it contains all cube roots of $2$ i.e. $\sqrt[3]{2}$, $\omega\sqrt[3]{2}$ and $\omega^2\sqrt[3]{2}$ hence it contains $\omega$. But can $\omega \in \mathbb Q_p(\sqrt[3]{2}) - \mathbb Q_p$? If this was true then $\omega$ generates a degree $2$ sub-extension of a degree $3$ extension? This is not possible, is it?

Answer 1

If $\mathbb{Q}_p(\sqrt[3]{2})$ is a degree 3 Galois extension, then $\mathbb{Q}_p$ must contain $\omega$ for this reason.

But it could still happen that $\sqrt[3]{2} \in\mathbb{Q}_p$ and $\omega \notin \mathbb{Q}_p$; and that does happen for $p=5$: The polynomial $X^3-2$ has a root by Hensel's Lemma, but $X^2+X+1$ has no root in the residue field $\mathbb{F}_5$, hence no root in $\mathbb{Z}_5$ and because roots of unity are integral, also not in $\mathbb{Q}_5$.