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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 $Gal(E/F) =[\mathbb{Q}(\sqrt[3]{3}, \zeta_3):\mathbb{Q}] = 6$.

Furthermore, we have the injection $Gal(E/F) \to S_3$ since $E$ is the splitting field for $x^3-3$.

Hence, $Gal(E/F) \cong S_3$. 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!

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  • $\begingroup$ I think you are confusing $\sqrt3$ and $\sqrt[3]3$. $\endgroup$ – Lord Shark the Unknown Jun 4 '17 at 10:04
  • $\begingroup$ @LordSharktheUnknown Yep, I just noticed that when finding the roots of $x^3-3$ I multiplied by square roots instead of the cube root. Putting that down to error, are my methods of computation correct? I will try with the cube root and get those fixed fields you mentioned in your answer. Cheers $\endgroup$ – jamesmartini Jun 4 '17 at 10:08
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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 $G=Gal(E/\mathbf Q)$ 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 $Gal(E/\mathbf Q(\sqrt [3]3))$, say $\tau(\sqrt [3]3))=\sqrt [3]3, \tau(\zeta_3)= \zeta_3^{2}$, and $\sigma$ generates $Gal(E/\mathbf Q (\zeta_3))$, 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.

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  • $\begingroup$ $\tau$ cannot map $\sqrt[3]3$ to $-\sqrt[3]3$, since $-\sqrt[3]3$ is not a root of $X^3-3$. Also, your $\sigma$ has order 2. $\endgroup$ – Claudius Jun 6 '17 at 8:30
  • $\begingroup$ Sorry for the messy formulas. I edit them. $\endgroup$ – nguyen quang do Jun 6 '17 at 15:58
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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 $\sqrt3$.

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