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I'm looking at the Galois group for $F = \mathbb{Q}(\sqrt[3]{2}, \omega)$ where $\omega$ is the first primitive third root of unity. I know this is the splitting field for $x^3 - 2$ and, I'm told, that $G = Gal(F/\mathbb{Q}) \cong S_3$ (which, for now, I believe). I can also directly see how most of the subgroups hold their corresponding fixed fields fixed. The only one I can't see directly is how the subgroup of $G$ isomorphic $\\{1, (123), (132)\\}$ holds $\mathbb{Q}(\omega)$ constant. Since it must be generated by some $\sigma$ I tried

$$\sigma(\omega) = \sigma(\omega \sqrt[3]{2}/\sqrt[3]{2})$$

but I don't have anywhere to go with that. Can someone help me see what $\sigma$ looks like and how I'd get that from the subgroup/group element of $S_3$? Thanks!

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Note that $\sigma$ is a field automorphism. It respects addition, subtraction, multiplication, and division.

The automorphism corresponding to $\sigma$ fixes every rational, and sends $\sqrt[3]{2}$ to $\omega\sqrt[3]{2}$, sends $\omega\sqrt[3]{2}$ to $\omega^2\sqrt[3]{2}$, and sends $\omega^2\sqrt[3]{2}$ to $\sqrt[3]{2}$. Therefore, it sends $\omega$ to $$\begin{align*} \sigma(\omega) &= \sigma\left(\frac{\omega\sqrt[3]{2}}{\sqrt[3]{2}}\right)\\ &= \frac{\sigma(\omega\sqrt[3]{2})}{\sigma(\sqrt[3]{2})}\\ &= \frac{\omega^2\sqrt[3]{2}}{\omega\sqrt[3]{2}}\\ &=\omega. \end{align*}$$ So $\sigma$ fixes $\mathbb{Q}(\omega)$.

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