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Let $K=\Bbb Q(\sqrt[n]{a})$, where $a \in \Bbb Q^+$ and suppose $[K:\Bbb Q]=n$ (i.e., $x^n-a$ is irreducible over $\mathbb Q$). Let $E$ be any subfield of $K$ and let $[E:\Bbb Q]=d$. Prove that $E=\Bbb Q(\sqrt[d]{a})$.

Now $N_{K/F}(\sqrt[n]{a})=a\in \Bbb Q$. I also know that $N_{K/E}(\sqrt[n]{a})\in E$. What I have observed is $K/\Bbb Q$ need not be cyclic as the $n$th root of unity is not in $\Bbb Q$. So what to do next? I am clueless here.

Edit. In the comment I got some help considering the Galois closure of $K$. The Galois closure will be $\Bbb Q(\zeta_n,\sqrt[n]{a})$ which has cyclic Galois group $\Bbb Z_n$ over $\Bbb Q(\zeta_n)$. Now if $Gal(K(\zeta_n)/\Bbb Q)=\left<f\right> \cong \Bbb Z_n$ then $Gal(K(\zeta_n)/E(\zeta_n))=\left<f^d\right>$. So the element that will be fixed by $f^d$ is $\sqrt[n]{a}+f^d(\sqrt[n]{a})+\cdots + f^{(n/d)-1}(\sqrt[n]{a})=\sqrt[n]{a}+\zeta_n^d\sqrt[n]{a}+\cdots + \zeta_n^{(n/d)-1}\sqrt[n]{a}$. But is it $\sqrt[d]{a}$?

user26857
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Ri-Li
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  • Think about the Galois closure of $K$? – Angina Seng May 01 '19 at 02:41
  • The Galois closure will be $\Bbb Q(\zeta_n,\sqrt[n]{a})$. – Ri-Li May 01 '19 at 02:43
  • Do you know the Galois group of the Galois closure as an extension of the rationals? If so, you can use Galois Theory to find all the fields between the Galois closure and the rationals, so, in particular, all the fields between $K$ and the rationals. – Gerry Myerson May 01 '19 at 03:14
  • @LordSharktheUnknown I need more clue – Ri-Li May 01 '19 at 03:32
  • I'm not sure I understand. Say $n=6$, $d=3$. I don't see why c=$b+f(b)+f^2(b)$ [I'm abbreviating $\root n\of a$ to $b$] should be fixed by $f^2$. Applying $f^2$ gives $f^2(b)+f^3(b)+f^4(b)\ne c$. And I don't see how one element can qualify as an answer to the question. – Gerry Myerson May 01 '19 at 03:51
  • @GerryMyerson will it $\sqrt[n]{a}+f^d(\sqrt[n]{a})+\cdots +f^{(n/d-1)d}(\sqrt[n]{a})$ as this will be fixed by $<f^{d}>$, where $Gal(K(\zeta_n)/\Bbb Q)=$?. I think i am baffled after doing a lot of Galois problems in a day. – Ri-Li May 01 '19 at 04:08
  • @GerryMyerson for $n=6,d=3$ $c=b+f^3(b)$ then $f^3(c)=c$ now I have edited my question and there why would it be $\sqrt[d]a$ – Ri-Li May 01 '19 at 04:24
  • @LordSharktheUnknown I have progressed whatever I can do, now please help me. – Ri-Li May 01 '19 at 04:28
  • Does the answer to the duplicate target help you enough? If there is a difference I missed, or you want to ask about a specific step, do edit! Reverting the duplicate closure is as easy as the closure. – Jyrki Lahtonen May 01 '19 at 04:33
  • +1 for showing how you might go about it. The Galois group $Gal(K(\zeta_n)/\Bbb{Q})$ is more complicated than what you described. You get $Gal(K(\zeta_n),\Bbb{Q}(\zeta_n))\simeq\Bbb{Z}_n$ instead. What may also happen is that $Gal(K(\zeta_n)/K)$ is isomorphic to $\Bbb{Z}_n^*$, but there are corner cases. – Jyrki Lahtonen May 01 '19 at 04:38
  • See here for a full account of the bigger Galois group. – Jyrki Lahtonen May 01 '19 at 04:40
  • And here for another incarnation of this question. A case can be made for selecting this as the duplicate target instead. – Jyrki Lahtonen May 01 '19 at 04:40
  • I should keep this problem as unknowingly I was stuck at a good problem. Thanks for all the help – Ri-Li May 01 '19 at 04:48
  • Don't worry about it too much. Searching on this site is not easy unless you know a few tools. Approach0 helped. – Jyrki Lahtonen May 01 '19 at 05:02
  • @JyrkiLahtonen can you also help me here "https://math.stackexchange.com/questions/3209005/which-subfields-of-the-galois-group-of-x48x28x4-are-galois-and-find-the-s"? – Ri-Li May 01 '19 at 05:14

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