Finding proper subfields Let $\omega$ denote the cube root of unity such that $\omega\neq 1$. I want to find the subfields properly contained in $\mathbb Q(\sqrt[3]{2},\omega)$ and containing $\mathbb Q$ properly. 
Two of then certainly $\mathbb Q(\sqrt[3]{2})$ and $\mathbb Q(\omega)$. I have been told that the correct answer is 4. But what are the other two subfields? Are they $\mathbb Q(\omega+\sqrt[3]{2})$ and $\mathbb Q(\omega\sqrt[3]{2})$? I have no idea. Please help me to understand this.
 A: *

*$\mathbb Q(\omega + \sqrt[3]2)$ is actually equal to $\mathbb Q(\omega,\sqrt[3]2)$. (I can't say this with 100% certainty since I didn't work it out by hand, but it should be.)

*$\mathbb Q(\omega\sqrt[3]2)$ is a third subfield.

*$\mathbb Q(\omega^2\sqrt[3]2)$ is the fourth subfield.


You're probably wondering where $\mathbb Q(\omega^2\sqrt[3]2)$ comes from. This field, along with $\mathbb Q(\sqrt[3]2)$ and $\mathbb(\omega\sqrt[3]2)$, is generated by a root of the polynomial $x^3-2$, for which $\mathbb Q(\omega,\sqrt[3]2)$ is a splitting field. In field theory, a good strategy is to take some special element $\alpha$ and look at the other roots of the minimal polynomial of $\alpha$ over the base field, called the conjugates of $\alpha$.
By the way, what I just said isn't a proof that these are the only nontrivial subfields, or that these subfields are distinct. You should be able to prove distinctness without too much trouble, but I'm not sure how to prove that we didn't miss any other subfields without using tools from Galois theory.
