Let $p,q$ be distinct prime numbers and assume that $p<q$.
Prove that if a subfield $K \subset \mathbb{Q}(\sqrt[p]{2},\sqrt[q]{2})$ satisfies $[K : \mathbb{Q}]=p$, then $K= \mathbb{Q}(\sqrt[p]{2})$.
I have shown that $\mathbb{Q}(\sqrt[p]{2}) \cap \mathbb{Q}(\sqrt[q]{2}) = \mathbb{Q}$ and $[\mathbb{Q}(\sqrt[p]{2},\sqrt[q]{2}) : \mathbb{Q}(\sqrt[p]{2})]=q$. Even a little more that $K \cap \mathbb{Q}(\sqrt[q]{2}) = \mathbb{Q}$ and $[\mathbb{Q}(\sqrt[p]{2},\sqrt[q]{2}) : K]=q$. However I couldn't reach the solution.
Any help would be much appreciated.
EDIT: I think I have found a solution for the rest. Anyway you are always welcome to put comments/feedbacks
Set $H=\mathbb{Q}(\sqrt[q]{2})$. Denote by $KH$ the composite field of $K$ and $H$. Since $[K:\mathbb{Q}]=p$ and $[H:\mathbb{Q}]=q$ and gcd$(p,q)=1$, we have $$[KH: \mathbb{Q}] =[K:\mathbb{Q}][H:\mathbb{Q}] \implies [KH: \mathbb{Q}] =pq.$$ It is easily seen that $KH \subseteq \mathbb{Q}(\sqrt[p]{2},\sqrt[q]{2})$ thus $KH =\mathbb{Q}(\sqrt[p]{2},\sqrt[q]{2})$. Therefore, $\sqrt[p]{2} \in KH$ hence $\sqrt[p]{2} \in K$ since $\sqrt[p]{2} \notin H$. It follows that, $\mathbb{Q}(\sqrt[p]{2}) \subseteq K$. By multiplicity of degrees, we can easily find that $K= \mathbb{Q}(\sqrt[p]{2})$.
We have $[\Bbb Q(\sqrt[p]2,\sqrt[q]2):K]=q$, hence $[K(\sqrt[p]2):K]$ is either $1$ or $q$, but it is certainly $\le p$.