# All intermediate subfields

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.

• Hi and welcome to MSE! What have you tried? Have you found the degree of these splitting fields? Commented May 2, 2020 at 19:35
• Have you tried using the factorization $a^3 + b^3 = (a+b)(a^2-ab+b^2)$? Commented May 2, 2020 at 19:43
• I found the degree of L\Q is 6.. Commented May 2, 2020 at 20:11
• I need help in finding all of the subfields of L Commented May 2, 2020 at 20:19

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 $$L$$: $$$$\rho:L\rightarrow L, \begin{cases} \rho(\alpha) = \alpha\omega\\ \rho(\omega) = \omega \end{cases} \qquad \sigma:L\rightarrow L, \begin{cases} \sigma(\alpha) = \alpha\\ \sigma(\omega) = \omega^2 \end{cases}$$$$ They are well defined automorphisms of $$\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:

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

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 $$L$$.

• Thank you very much!.. Can you help me on finding all intermediate subfields of L? Commented May 2, 2020 at 21:52
• @1Math12 I edited the answer, sorry for the misunderstanding :) Commented May 2, 2020 at 22:15
• @ Menezio I want to ask...Is there any difference between the two questions "Find all the subfields of L " and "Find all intermediate fields of L"...?? Commented May 3, 2020 at 19:44
• When I computed $L^{\langle \sigma\rho \rangle}$ I've got $\mathbb Q(\alpha^{2}\omega)$...is it ok or not!! Commented May 3, 2020 at 20:28
• @1Math12 Well, by "intermediate" conventionally we intend all the fields except the base field and the splitting field. For the last comment: $\sigma\rho(\alpha \omega) = \sigma(\alpha\omega^2) = \alpha\omega^4 = \alpha\omega$ so $\mathbb Q(\alpha\omega)$ is fixed by $\langle \sigma\rho\rangle$. We also have $\sigma\rho(\alpha^2 \omega) = \sigma(\alpha^2\omega^3) = \sigma(\alpha^2) = \alpha^2$, so it seems that $\alpha^2\omega$ is not fixed. Commented May 3, 2020 at 21:04