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!

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

  • $\begingroup$ $\tau$ 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

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 $\sqrt3$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.