TITLE: Determining subfields of $\mathbb{Q}(\zeta_3,\sqrt[3]{2})$ without Galois theory?
QUESTION [7 upvotes]: Somehow I had convinced myself recently that one could determine the subfields of $\mathbb{Q}(\zeta_3,\sqrt[3]{2})$ (here $\zeta_3 = \frac{-1+\sqrt{3}i}{2}$ is a primtive $3$rd root of $1$) without using Galois theory, but I'm not so sure any more. 
The subfields should be $\mathbb{Q}$, $\mathbb{Q}(\zeta_3)$, $\mathbb{Q}(\sqrt[3]{2})$, $\mathbb{Q}(\zeta_3\sqrt[3]{2})$, $\mathbb{Q}(\zeta_3^2\sqrt[3]{2})$, $\mathbb{Q}(\zeta_3,\sqrt[3]{2})$  (corresponding to the subgroups of $\mathrm{Aut}(\mathbb{Q}(\zeta_3,\sqrt[3]{2})/\mathbb{Q})\simeq S_3$. 
Is there a way to see this without Galois theory or extremely disgusting calculations?

REPLY [6 votes]: Let $K$ be a non-trivial subfield of $\Bbb Q(\omega, \root 3 \of 2)$.
Since $[\Bbb Q(\omega, \root 3 \of 2):\Bbb Q\textbf{]}=6$, we have $[K:\Bbb Q\textbf{]}\in\{2,3\}$.
If $[K:\Bbb Q\textbf{]}=2$, then $[\Bbb Q(\omega, \root 3 \of 2):K\textbf{]}=3$ and $\root 3 \of 2 \not \in K$.
Suppose $\omega \not \in K$. It follows that $[K(\omega):\Bbb Q\textbf{]}=4$, which is absurd because $[K(\omega):\Bbb Q\textbf{]}=4$ must divide $[\Bbb Q(\omega, \root 3 \of 2):\Bbb Q\textbf{]}=6$.
Therefore $\omega \in K$. It's easy to conclude that $K=\Bbb Q(\omega )$.
I'm not positive because I didn't check, but I believe the other cases are similar.
Edit 1: I don't see how it is similar, so I propose the following argument to conclude:
If $[K:\Bbb Q\textbf{]}=3$, since $[\Bbb Q(\root 3 \of 2):\Bbb Q\textbf{]}=3$, there exists a $\Bbb Q$-isomorphism $\varphi:\Bbb Q(\root 3 \of 2)\to K$.
It is true that $\varphi (\root 3\of 2)$ and $\root 3 \of 2$ share the same minimal polynomial over $\Bbb Q$, because $\varphi$ is a $\Bbb Q$-morphism.
Therefore, since $t^3-2$ is the minimal polynomial of $\root 3\of 2$ over $\Bbb Q$, it comes $\varphi (\root 3 \of 2)\in \{\root 3\of 2,\omega  \root 3\of 2, \omega ^2 \root 3\of 2\}$. 
Recall that $\varphi$ ranges in $K$. It follows that $K\in \{\Bbb Q(\root 3\of 2), \Bbb Q(\omega \root 3 \of 2), \Bbb Q(\omega ^2\root 3 \of 2)\}$
Edit 2: As suggested by the OP, the case $[K:\Bbb Q\textbf{]}=3$ is in fact similar:

If $K$ contains a root of $t^3-2$ then $K$ must be one of the fields we expect it to be.
If it doesn't, then $t^3-2$ will be irreducible over $K$ and it's easy to derive a contradiction.