The number of subfield $K$ of $L$ such that $\,\mathbb{Q}\subsetneq K\subsetneq L$ $\omega\neq 1 \in\mathbb{C}$ such that $\omega^3=1$, suppose $L$ be the field generated by $\omega, 2^{1/3}$ over $\mathbb{Q}$ i.e $L=\mathbb{Q}, (\omega,2^{1/3})$, the number of subfields $K$ of $L$ such that $\mathbb{Q}\subsetneq K\subsetneq L$ is which integer?
Could any one tell me how to proceed?
Thank you.
 A: Hints:
$$(1)\;\;\;\;[\,\Bbb Q(w,\sqrt[3]2),\Bbb Q\,]=[\,\Bbb Q(\sqrt[3]2,w):\Bbb Q(w)\,][\,\Bbb 
Q(w):\Bbb Q\,]$$
$$(2)\;\;\;\;x^2+2+1\in\Bbb Q[x]\;\;\text{is an irreducible polynomial }$$
$$(3)\;\;\;\;\text{There are only two groups of order $\,6\,$ up to isomorphism:$\\$ one has $\,3\,$ proper non-trivial subgroups, the other one has $\,4\,$ proper non-trivial subgroups}$$
Added: As the OP answered a comment saying (s)he hasn't studied any Galois Theory, we can try to following approach:
From $\;(1)-(2)\;$ we must be able to gather $\,\dim_{\Bbb Q}\Bbb Q(\sqrt[3]2\,,\,w)=6\,$ , and from basic linear algebra we then know that for any $\,\Bbb Q\lneq K\lneq L \,$ we have to have $\,\dim_{\Bbb Q}K=2\,\vee\,3\,$ .
In order to find all the subfields of $\,L\,$ of  dimensions $\,2\,,\,3\,$ we could also use the fact that $\,\{\,1\,,\,\sqrt[3]2\,,\,\sqrt[3]4\,,\,w\,,\,w\sqrt[3]2\,,\,w\sqrt[3]4\,\}\;$ is a basis of $\,L_{\Bbb Q}\,$ , and then from here try to find minimal polynomials that fit some of these elements.
For example, $\,x^3-4\in\Bbb Q[x]\,$ is irreducible (why? You can't use Eisenstein...) , so that $\,[\,\Bbb Q(\sqrt[3]4\,:\,\Bbb Q\,]=3\,$ and clearly $\,\Bbb Q(\sqrt[3]4)\le L\,$ ...
