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It is a well-known theorem in Field Theory that if $F$ is a field:

  1. Which contains the $n$-th roots of unity for some $n\ge 1$.

  2. Is of characteristic not dividing $n$.

    And if $0\ne a\in F$ is some element, then the primitive radical extension $F(\sqrt[n]{a})/F$ is a Galois extension with $\textbf{Gal}(F(\sqrt[n]{a})/F)\cong \mathbb{Z}_d$ such that $d|n$.

I am looking for examples (preferably over characteristic 0, is such examples exist) for such extensions where $d<n$.

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    $\begingroup$ Sure: $a=4$, $n=2$, where $\Bbb Q(\sqrt4\,)=\Bbb Q$. $\endgroup$
    – Lubin
    Jul 4, 2018 at 21:36
  • $\begingroup$ I actually thought about this case and now I'm not sure why I decided it's not a satisfying counterexample. $\endgroup$
    – Gadi A
    Jul 4, 2018 at 21:42
  • $\begingroup$ Probably because $\Bbb Q$ doesn’t contain the fourth roots of unity. I should have taken the base field to be the Gaussian numbers, $\Bbb Q(i)$. Otherwise, same example. $\endgroup$
    – Lubin
    Jul 5, 2018 at 1:59

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Let $F$ be the extension of the rationals got by joining a primitive 15th root of unity. Take $a=8$. Then the extension $F[ 8^{1/15}]$ of $F$ will have Galois group cyclic of order 5. (Because 8 already has cubic roots in $F$).

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  • $\begingroup$ So basically I can only expect $d < n$ when $a$ is already a $n/d$-power of an element of F? $\endgroup$
    – Gadi A
    Jul 4, 2018 at 21:43
  • $\begingroup$ Yes. Because in this examole qe srw adjoining actually the fifth root of 5. This is already implicit in the comment by Lubin. $\endgroup$ Jul 5, 2018 at 0:35

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