All intermediate subfields

Question

I have this multi-part problem in Galois Theory.

Find all the subfields of the splitting field $L$ of the polynomial $X^3+3 \in \mathbb{Q}[x]$. (Exhibit explicit generators for these).

Prove or disprove: $X^6 + 3$ splits over $L$.

I know that $L$ is Galois extension over $\mathbb{Q}$.

Thanks.

Answer 1

As you commented $[L:\mathbb Q]=6$. This is because the polynomial $X^3+3$ is irreducible (thanks to Eisenstein criterion, $p=3$) so $\sqrt[3]{-3} = \alpha$ has degree $3$ over $\mathbb Q$.

The splitting field $L$ is $\mathbb Q(\alpha, \omega)$ where $\omega$ is the third primitive root of unity. Since $\omega$ has degree $2$ (it's minimal polynomial is $x^2+x+1$), then $[L:\mathbb Q]=6$ (because $2$ and $3$ are coprime).

Also the extension $L/\mathbb Q$ is normal and separable, hence it is a Galois extension.

Define the following automorphisms of $L$: \begin{equation} \rho:L\rightarrow L, \begin{cases} \rho(\alpha) = \alpha\omega\\ \rho(\omega) = \omega \end{cases} \qquad \sigma:L\rightarrow L, \begin{cases} \sigma(\alpha) = \alpha\\ \sigma(\omega) = \omega^2 \end{cases} \end{equation} They are well defined automorphisms of $\text{Gal}(L/\mathbb Q))$. Now $\rho$ has order $3$, $\sigma$ has order $2$, $\langle \rho\rangle \cap \langle \sigma \rangle = \emptyset$ and $\sigma\rho\sigma^{-1} = \rho^{-1}$. This easily implies (cardinality) that: $$ \text{Gal}(L/\mathbb Q)) = \langle \rho, \sigma \ | \ \rho^3=\sigma^2=id, \sigma\rho\sigma^{-1} = \rho^{-1}\rangle \cong S_3 $$

Thanks to the Galois corrispondance theorem we have that the intermediate subfields are in bijection with the subgroups of $\text{Gal}(L/\mathbb Q)=S_3$.

We have only $6$ subgroups, so we have only $6$ intermediate subfields that are:

\begin{gather} S_3 \quad \longleftrightarrow L^{S_3} = \mathbb Q\\ \langle \rho \rangle \quad \longleftrightarrow L^{\langle \rho \rangle} = \mathbb Q(\omega)\\ \langle \sigma \rangle \quad \longleftrightarrow L^{\langle \sigma \rangle} = \mathbb Q(\alpha)\\ \langle \sigma\rho \rangle \quad \longleftrightarrow L^{\langle \sigma\rho \rangle} = \mathbb Q(\alpha\omega)\\ \langle \sigma\rho^2 \rangle \quad \longleftrightarrow L^{\langle \sigma\rho^2 \rangle} = \mathbb Q(\alpha\omega^2)\\ \{id\} \quad \longleftrightarrow L^{\{id\}} = L\\ \end{gather}

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Let's denote with $K$ the splitting field of $X^6+3$ over $\mathbb Q$. $K = \mathbb Q(\sqrt[6]{-3}, \omega_6) = \mathbb (\sqrt[6]{-3},\omega)$ where $\omega_6$ is the sixth primitive root of unity.

Observe that $\left(\sqrt[6]{-3}\right)^2 = \sqrt[3]{-3}$; hence $L\subset K$.

Now if we show that the degree $[K:\mathbb Q]=6$, then we have $L=K$ (a containment and equal degree).

Observe that $\left(\sqrt[6]{-3}\right)^3 = i\sqrt{3}$, and $\mathbb Q(\omega) = \mathbb Q(i\sqrt 3)$. Hence $K=\mathbb Q(\sqrt[6]{-3},\omega) = \mathbb Q(\sqrt[6]{-3})$ and this field has degree $6$ over $\mathbb Q$ because $X^6+3$ is irreducible by Eisenstein.

So $L=K$ and $X^6+3$ splits over $L$.