I am considering the Galois extension $L/\mathbb{Q}$, where $L = \mathbb{Q}(\sqrt[3]{2}, \zeta)$, and $\zeta$ is a primitive cube root of unity.
I've found that $Aut(L/\mathbb{Q})$ can be viewed as being the same as $S_3$. Also, I've noted that $L$ is the splitting field of $x^3 -2$, and have now ordered the roots as $\alpha_1 = \sqrt[3]{2}$, $\alpha_2 = \zeta \sqrt[3]{2}$ and $\alpha_3 = \zeta^2 \sqrt[3]{2}$.
I want to calculate $\mathbb{Q}(\zeta)^*$; i.e.: $Aut(L/\mathbb{Q}(\zeta))$ as a subgroup of $S_3$.
I know can write $\zeta = \frac{\alpha_3}{\alpha_2} = \frac{\alpha_2}{\alpha_1}$. I want to find all $\sigma \in S_3$ s.t. $\sigma(\zeta) = \zeta$.
Of course, the identity will be one such $\sigma$. My first thoughts for others would be take the permutation $(1, 3, 2)$, as this means $\sigma(\frac{\alpha_3}{\alpha_2}) = \sigma(\frac{\alpha_2}{\alpha_1})$. However, this leaves us with something not nice for $\sigma(\frac{\alpha_2}{\alpha_1})$, so I'm thinking that's wrong.
I don't think any permutations swapping two elements will work, ruling out $(1,2)$, $(1,3)$ and $(2,3)$, and $(1, 2, 3)$ has similar problems to above.
So, am I right in saying $\mathbb{Q}(\zeta)^* = \{1\}$ ?
Thanks in advance!