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!

• I think you are confusing $\sqrt3$ and $\sqrt[3]3$. – Angina Seng Jun 4 '17 at 10:04
• @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 – jcm Jun 4 '17 at 10:08
• Would it be easier to work in terms of $\sqrt[3] 3$ and $\zeta_3$, instead of $\alpha_1, \alpha_2, \alpha_3$? – eatfood Nov 5 '19 at 5:49

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.

• $\tau$ 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. – Claudius Jun 6 '17 at 8:30
• Sorry for the messy formulas. I edit them. – nguyen quang do Jun 6 '17 at 15:58

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 $\sqrt3$.