Galois correspondence for the splitting field of $x^3-3$ over $\mathbb{Q}$ I am struggling to find the fixed fields, am I missing a trick here or is it really this nitty gritty?
By the tower law $[\mathbb{Q}(\sqrt[3]{3}, \zeta_3):\mathbb{Q}] = [(\mathbb{Q}(\sqrt[3]{3}, \zeta_3): \mathbb{Q}(\sqrt[3]{3})][\mathbb{Q}(\sqrt[3]{3}):\mathbb{Q}] = 2\times3$ = 6, since the minimal polynomial for each respective extension is $\Phi_3$, $x^3-3$.
We also know that $E/F:=\mathbb{Q}(\sqrt[3]{3},\zeta_3)/\mathbb{Q}$ is Galois since each element $\alpha$ has a separable polynomial in $\mathbb{Q}[x]$ which splits completely in $\mathbb{Q}(\sqrt[3]{3},\zeta_3)[x]$, namely $x^3-3$, $\Phi_3$, respectively.
So invoking the fundamental theorem we can write $\operatorname{Gal}(E/F) =[\mathbb{Q}(\sqrt[3]{3}, \zeta_3):\mathbb{Q}] = 6$.
Furthermore, we have the injection $\operatorname{Gal}(E/F) \to S_3$ since $E$ is the splitting field for $x^3-3$.
Hence, $\operatorname{Gal}(E/F) \cong S_3$. Let $\{\alpha_1,\alpha_2,\alpha_3\}:= \{\sqrt[3]{3},\sqrt[3]{3}\zeta_3,\sqrt[3]{3}\zeta_3^2\}$ be an enumeration of the roots
Using the group structure of $S_3$ we know that the following maps are automorphisms of $E$ fixing the base field
\begin{align*}
\sigma_1 = ID,\hspace{0.2cm}
&\sigma_2: \begin{cases}
\alpha_1 \mapsto \alpha_2\\
\alpha_2 \mapsto \alpha_3\\
\alpha_3 \mapsto \alpha_1
\end{cases}
\sigma_3: \begin{cases}
\alpha_1 \mapsto \alpha_3\\
\alpha_2 \mapsto \alpha_1\\
\alpha_3 \mapsto \alpha_2
\end{cases}
\sigma_4: \begin{cases}
\alpha_1 \mapsto \alpha_2\\
\alpha_2 \mapsto \alpha_1\\
\alpha_3 \mapsto \alpha_3
\end{cases}\\
&\sigma_5: \begin{cases}
\alpha_1 \mapsto \alpha_3\\
\alpha_2 \mapsto \alpha_2\\
\alpha_3 \mapsto \alpha_1
\end{cases}
\sigma_6: \begin{cases}
\alpha_1 \mapsto \alpha_1\\
\alpha_2 \mapsto \alpha_3\\
\alpha_3 \mapsto \alpha_2
\end{cases}
\end{align*}
The subgroups of $S_3$ are 3 copies of $S_2$ and $A_3$. These are given by
\begin{align*}
H_1 = \langle \sigma_4 \rangle, \hspace{0.2cm}
H_2 &= \langle \sigma_5 \rangle, \hspace{0.2cm}
H_3 = \langle \sigma_6 \rangle\\
H_4 &= \langle \sigma_2, \sigma_3 \rangle
\end{align*}
The fixed fields are given by
\begin{align*}
E^{\langle \sigma_4 \rangle} &= \mathbb{Q}(\alpha_3,\alpha_1+\alpha_2)\\
E^{\langle \sigma_5 \rangle} &=\mathbb{Q}(\alpha_2,\alpha_1+\alpha_3)\\
E^{\langle \sigma_6 \rangle} &=\mathbb{Q}(\alpha_1,\alpha_2+\alpha_3) \\\
E^{\langle \sigma_2, \sigma_3 \rangle} &=  \mathbb{Q}(\alpha_1 + \alpha_2 + \alpha_3) = \mathbb{Q}
\end{align*}
$\Phi_3 = x^2 + x + 1$, so by the Quadratic formula $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$, it has roots $\frac{-1 \pm i\sqrt{3}}{2}$. The roots are exactly the roots of unity, so
\begin{align*}
\frac{-1 + i\sqrt{3}}{2} &= \zeta_3\\
\frac{-1 - i\sqrt{3}}{2} &= \zeta_3^2
\end{align*}
The roots of $x^3-3$ are therefore
\begin{align*}
\alpha_1 &=\sqrt{3}\\
\alpha_2 &=\frac{-\sqrt{3} + i3}{2} = \sqrt{3}\zeta_3\\
\alpha_3 &=\frac{-\sqrt{3} - i3}{2} = \sqrt{3}\zeta_3^2
\end{align*}
From this we can easily find the sums
\begin{align*}
\alpha_1 + \alpha_2 &= \sqrt{3} + \frac{-\sqrt{3} + i3}{2}\\
&= \frac{\sqrt{3} + i3}{2}\\
\alpha_1+\alpha_3 &= \sqrt{3} + \frac{-\sqrt{3} - i3}{2}\\
&=\frac{\sqrt{3} - i3}{2}\\
\alpha_2+\alpha_3 &= \frac{-\sqrt{3} + i3}{2} +\frac{-\sqrt{3} - i3}{2}\\
&= -\sqrt{3}
\end{align*}
So the fixed fields are
\begin{align*}
E^{\langle \sigma_4 \rangle} &= \mathbb{Q}(i,\sqrt{3})\\
E^{\langle \sigma_5 \rangle} &=\mathbb{Q}(i,\sqrt{3})\\
E^{\langle \sigma_6 \rangle} &=\mathbb{Q}(\sqrt{3}) \\\
E^{\langle \sigma_2, \sigma_3 \rangle} &=  \mathbb{Q}(\alpha_1 + \alpha_2 + \alpha_3) = \mathbb{Q}
\end{align*}
Is my process for finding the fixed fields correct? Thanks for your time!
 A: I think you can avoid (most of the) "nitty gritty"calculations by applying the full force of Galois theory. Your field $E=\mathbf Q (\sqrt [3] 3, \zeta_3)$ is the splitting field of the polynomial $f=X^3 - 3$, hence its Galois group $G=\operatorname{Gal}(E/\mathbf Q)$ is a subgroup of the symmetric group $S_3$ because $G$ permutes the 3 roots of $f$. Since $[E:\mathbf Q]$ is obviously $>2$, necessarily $G = S_3$. To determine  the subextensions of $E/\mathbf Q$ amounts then to determine the subgroups of $G$. But $S_3$ is generated by a 3-cycle $\sigma$ and a transposition $\tau$, which can here be chosen as :
EDIT : $\tau$ generates $\operatorname{Gal}(E/\mathbf Q(\sqrt [3]3))$, say $\tau(\sqrt [3]3))=\sqrt [3]3, \tau(\zeta_3)= \zeta_3^{2}$, and $\sigma$ generates $\operatorname{Gal}(E/\mathbf Q (\zeta_3))$, say $\sigma (\zeta_3) = \zeta_3, \sigma (\sqrt [3]3) = \zeta_3 \sqrt [3]3$. The remaining calculations are just a matter of book keeping.
A: How can $\sigma_4$ and $\sigma_5$ have the same fixed field?
The fixed field of $\sigma_4$ is $\Bbb Q(\alpha_3)=\Bbb Q(\zeta_3^2
\sqrt[3]3)$.
The fixed field of $\sigma_5$ is $\Bbb Q(\alpha_2)=\Bbb Q(\zeta_3
\sqrt[3]3)$.
The fixed field of $\sigma_6$ is $\Bbb Q(\alpha_1)=\Bbb Q(\sqrt[3]3)$.
The fixed field of $\sigma_2$ is $\Bbb Q(\sqrt{-3})=\Bbb Q(i\sqrt3)$.
$E$ contains neither $i$ nor $\sqrt3$.
