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Given a rational number $a>0$ and an $n\in\mathbb{N}$ such that $x^n - a$ is irreducible over $\mathbb{Q}$, it is known that every subfield of $\mathbb{Q}\bigl(\sqrt[n]{a}\bigr)$ has the form $\mathbb{Q}\bigl(\sqrt[d]{a}\bigr)$ for some $d\mid n$.

Is there an elementary proof of this result? All of the references I could find use cogalois theory. (See this paper, for example, immediately after the proof of Theorem 1.6.)

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  • $\begingroup$ What you claim "is known" is false (even if you impose the intended hypothesis that $X^n-a$ is irreducible over $\mathbf{Q}$). (No idea what "cogalois" theory is.) For $e \ge 3$ and a nonzero rational $h$, $X^{2^e}+h^2$ is irreducible yet the extension generated by one root contains $\mathbf{Q}(i, \sqrt{2h})$. In general, if $X^n-a$ is irreducible over $\mathbf{Q}$ with $4|n$ and $a=-h^2$ then "extra" subfields always exist; the situation is especially thorny when $8|n$. It is possible to determine all counterexamples. I have discovered a proof which this comment box is too small to contain. $\endgroup$
    – grghxy
    Commented Jun 9, 2015 at 16:36
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    $\begingroup$ @grghxy Note that $a>0$ in the question. $\endgroup$
    – Jim Belk
    Commented Jun 9, 2015 at 16:37
  • $\begingroup$ Ah, sorry for misreading. $\endgroup$
    – grghxy
    Commented Jun 9, 2015 at 16:53
  • $\begingroup$ @grghxy Good catch on the implied irreducibility hypothesis, though. I've added it to the question. $\endgroup$
    – Jim Belk
    Commented Jun 9, 2015 at 17:11
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    $\begingroup$ Fix a subextension $F$, it must have degree $d$ dividing $n$. Then try to calculate the norm of $\sqrt[n]{a}$ to $F$. This element certainly lives in $F$ and it can be shown to be equal to $\pm \sqrt[d]{a}$ (the norm is the product over some galois conjugates, and you know them). Therefore you must have equality. You should fill the details, but I hope it is clear! $\endgroup$
    – Daniel Kohen
    Commented Jun 9, 2015 at 23:17

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Regular Galois theory is more than enough for this, see the answers to this question.

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