What are the intermediate fields of $\mathbb Q(\sqrt[3]2,e^{\frac{2i\pi}{3}})$ (Galois group)

Question

The elements of Galois group are

\begin{align*} \sigma _1:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\ \sqrt[3]{2}&\longmapsto \sqrt[3]{2},\\ e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{2i\pi}{3}}, \end{align*} \begin{align*} \sigma _2:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\ \sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{2i\pi}{3}},\\ e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{2i\pi}{3}}, \end{align*} \begin{align*} \sigma _3:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\ \sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{4i\pi}{3}},\\ e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{2i\pi}{3}}, \end{align*} \begin{align*} \sigma _4:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\ \sqrt[3]{2}&\longmapsto \sqrt[3]{2},\\ e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{4i\pi}{3}}, \end{align*} \begin{align*} \sigma _5:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\ \sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{2i\pi}{3}},\\ e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{4i\pi}{3}}, \end{align*} \begin{align*} \sigma _6:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\ \sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{4i\pi}{3}},\\ e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{4i\pi}{3}}. \end{align*}

We remark that

\begin{align*} \sigma _2^3=1\\ \sigma _3^3=1\\ \sigma _4^2=1\\ \sigma _5^2=1\\ \sigma_6^2=1 \end{align*} But $$\sigma _2\sigma _4(\sqrt[3]2)=\sigma _2(\sqrt[3]2)=\sqrt[3]2e^{\frac{2i\pi}{3}}$$ and $$\sigma _4\sigma _2(\sqrt[3]2)=\sigma _4(\sqrt[3]2e^{\frac{2i\pi}{3}})=\sqrt[3]2e^{\frac{4i\pi}{3}},$$ therefore $\{\sigma _i\}_{i=1}^6$ is not a commutatif group and thus $$\text{Gal}(E/\mathbb Q)=\{\sigma _i\}_{i=1}^6\cong \mathfrak S_3.$$

My problem is that it should have a $\sigma _i$ such that $\sigma _i^6=1$ and $\sigma _i^n\neq 1$ for $i\in\{1,2,3,4,5\}$ because one of my intermediate field must be $\mathbb Q$. And if not, is there a mistake in my applications $\sigma _i$?

Answer 1

The subfields are the fields fixed by a subgroup of the Galois group. Not every subgroup is generated by a single element, whence it is no problem that there is no element of order $6$.

Answer 2

As you have found $Aut_{\mathbb{Q}}L \simeq S_3$ then the intermediate fields of $Gal (X^3-2, \mathbb{Q})$ are in correspondence to the subgroups of $Aut_{\mathbb{Q}}L$, according to the Fundamental Theorem of Galois. Now,

$$S_3 = \{Id, \sigma, \sigma^2, \tau, \sigma\tau, \sigma^2\tau\}$$

Notice that there is a element $\sigma$ of order $3$ generating a cyclic subgroup of order $3$, i.e., $\langle\sigma\rangle$ , and 3 elements of order $2$. What is missing is to identify such elements $\sigma$ and $\tau$ among your automorphisms $\sigma_i$ in order to find exactly the intermediate fields of $L = Gal (X^3-2,\mathbb{Q})$ through Galois correspondence.

Hint: $\sigma = \sigma_2$ (need to check)