Let $ℚ(\sqrt[n]{q})$ be the field of rational numbers with $\sqrt[n]{q}$ adjoined; $n,q∈ℚ$
I have been trying to prove that $\sqrt[m]{r}∉ℚ(\sqrt[n]{q})$, for different $m,r∈ℚ$.
My approach:
We write $ℚ(\sqrt[n]{q})$ as all the elements of the form $$a_0+a_1\sqrt[n]{q}+\dots+a_{n−1}\sqrt[n]{q}^{n-1}$$ with $a_0,...,a_{n−1}∈ℚ$, and then if we arrive at a contradiction by $$(a_0+a_1\sqrt[n]{q}+\dots+a_{n−1}\sqrt[n]{q}^{n-1})^m=r$$ the proof would be complete.
I tried to show that the only way $$(a_0+a_1\sqrt[n]{q}+\dots+a_{n−1}\sqrt[n]{q}^{n-1})(b_0+b_1\sqrt[n]{q}+\dots+b_{n−1}\sqrt[n]{q}^{n-1})=k$$ for $k∈ℚ$ is for all the $a_i$ and $b_i$ to be equal to $0$ (except for $a_0$ and $b_0$). From here one could use induction to arrive at a contradiction in the case where $m$ is an integer. But it is still far too complicated to go that way, since -I think- one would have to prove that certain combinations of some sums of the $a_ib_j$ have to equal $0$. This approach seems rather messy, long and complicated.
I was wondering if there was a more intuitive or at least more direct way of proving the statement above. I also know how to solve this for the cases where we are talking about square roots, so if someone wants to mark me as a duplicate at least don't let it be of one of the examples where only the cases $m=2$ or $m=3$ are proven.
I would truly appreciate any help/thoughts!
Edit. There are exceptions to this rule, for instance, whenever $\sqrt[m]{r}$ is already a rational. The cases where $m=kn$, $r=qk$ are also excluded. I'm unsure if there are other exceptions, in that case please tell so that I can further edit the question.
