If $p$ and $q$ are prime numbers and $m\gt n$ show that $\sqrt[m]{p}\notin \mathbb Q(\sqrt[n]{q})$

Question

If $p$ and $q$ are prime numbers and $m\gt n$ show that $\sqrt[m]{p}\notin \mathbb Q(\sqrt[n]{q})$

I really have no idea how to prove this problem. I started to consider:

Assume $\sqrt[m]{p}\in \mathbb Q(\sqrt[n]{q})$. Then I wanted to try and show that the dimension of $\mathbb Q(\sqrt[n]{q})$ is $[\mathbb Q(\sqrt[n]{q}):Q]=n-1$ because the basis would be $\{1,\sqrt[n]{q},\sqrt[n]{q}^2 \sqrt[n]{q}^3...\sqrt[n]{q}^{n-1}\}$. I thought this could help me see that $\sqrt[m]{p}\in \mathbb Q(\sqrt[n]{q})$ leads to a contradiction but I'm not connecting any dots and what I have already may just be nonsense as well.

Answer 1

Fact: If $L,L'$ are field extensions of a field $K$ with $L \subseteq L'$, then $[L:K]$ divides $[L':K]$. In fact, we have $[L':K]=[L':L][L:K]$.

If $\mathbb{Q}(\sqrt[m]{p}) \subseteq \mathbb{Q}(\sqrt[n]{q})$, we would get that $m | n$.

Answer 2

All you really need is to note that the $m$ "vectors" $p^{1/m},p^{2/m},\ldots, p^{m/m}$ are linearly independent over $\mathbb{Q}$, but if they were each expressible as a linear combination (over $\mathbb{Q}$) of the $n$ "vectors" $q^{1/n},q^{2/n},\ldots,q^{n/n}$ with $n\lt m$, then they would not be linearly independent.