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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

$a_0+a_1\sqrt[n]{q}+...+a_{n−1}\sqrt[n]{q}^{n-1}$

with $a_0,...,a_{n−1}∈ℚ$

and then if we arrive at a contradiction by

$(a_0+a_1\sqrt[n]{q}+...+a_{n−1}\sqrt[n]{q}^{n-1})^m=r$

the proof would be complete. I tried to show that the only way

$(a_0+a_1\sqrt[n]{q}+...+a_{n−1}\sqrt[n]{q}^{n-1})(b_0+a_1\sqrt[n]{q}+...+b_{n−1}\sqrt[n]{q}^{n-1})=k$

for a $k∈ℚ$ 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 cases 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.

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  • $\begingroup$ The only roots in the field $\mathbb{Q}(\sqrt[n]{q})$ are the ones commensurable with some power of $\sqrt[n]{q}$. $\endgroup$ – Orest Bucicovschi Sep 20 '17 at 11:45
  • $\begingroup$ 1) We may assume that $q,r$ are integers (since $b (a/b)^{1/n} = (ab^{n-1})^{1/n}$ generates the same field as $(a/b)^{1/n}$). 2) If $r$ is square-free and $|r|>1$, then $X^m - r$ is irreducible (by Eisenstein), so $\Bbb Q(\sqrt[m]{r})$ has degree $m$. Thus it can't be contained in $\Bbb Q(\sqrt[n]{q})$ if $n$ and $m$ are coprime, and $q$ is square-free and $|q| > 1$. $\endgroup$ – Watson Nov 23 '18 at 16:27
  • $\begingroup$ Possible duplicate of Show the two fields are not isomorphic $\endgroup$ – Watson Nov 24 '18 at 13:18

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