Determining subfields of $\mathbb{Q}(\zeta_3,\sqrt[3]{2})$ without Galois theory?

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?

-

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:

1. If $K$ contains a root of $t^3-2$ then $K$ must be one of the fields we expect it to be.
2. If it doesn't, then $t^3-2$ will be irreducible over $K$ and it's easy to derive a contradiction.
-
I don't quite see why your Edit is correct: is it really the case that every extension of degree $3$ is isomorphic to $\mathbb{Q}(\sqrt[3]{2})$? This can't be. –  vgty6h7uij Mar 4 '13 at 0:33
Anyway, the case of $[K:\mathbb{Q}]=3$ goes similarly; if $K$ contains a root of $X^3-2$ then we know what $K$ is, but if not then $[K(\sqrt[3]{3}):K]=3$, which is impossible. –  vgty6h7uij Mar 4 '13 at 0:40
@vgty6h7uij Regarding your first comment see this. It should be on any Galois Theory book too. About your second comment I supppose you meant to consider $K(\root \large 3 \of 2)$, but I don't follow why it should be the case that $[K(\root \large 3 \of 2):K\textbf{]}=3$. –  Git Gud Mar 4 '13 at 7:22
Without using Galois theory I don't see any reason a priori why $K$ should be isomorphic to $\mathbb Q(\sqrt[3]{2})$. I think vgty6h7uij's argument is correct though. The main point being that if $K$ does not contain a root of $X^3-2$ then $X^3-2$ is irreducible over $K$. –  JSchlather Mar 4 '13 at 7:35
I think it's a matter of when does Field Theory become Galois Theory. I wouldn't consider this as using Galois Theory. And of course his reasoning works, I was blind. –  Git Gud Mar 4 '13 at 8:30

There is a way to do it, but it does require a tiny bit of Galois theory.

1. Find a primitive element $\alpha$ for the whole extension.
2. Find its minimal polyomial $f$ over $\mathbb{Q}$ and all of its roots.
3. Find all factors of $f$ that are multiples of $x - \alpha$.
4. Now all subfields are obtained by extending $\mathbb{Q}$ with the coefficients of one of the polynomials in the previous step.

The rationale is that if $F$ is an intermediate field, and $g = a_{0} + a_{1} x + \dots + x^k$ is the minimal polynomial of $\alpha$ over $F$, then $g$ divides $f$, and $F = \mathbb{Q}(a_0, a_{1}, \dots)$.

The tiny bit is in step 2, because you will be (implicitly, perhaps) using elements of the Galois group to get all the roots.

PS By the way, the above is straight from the proof that a simple extension has a finite number of intermediate fields.

-