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Let $\zeta_n:=$ primitive $n$-th root of $1$ over $\mathbb Q$. And let $\alpha \in \mathbb Q[\zeta_n]$ satisfy $\alpha^m=2$ for some positive integer $m$. Then I have to show that $m=1$ or $2$.

We have the tower of fields $\mathbb Q \subset \mathbb Q(\alpha) \subset \mathbb Q(\zeta_n)$ with $|\mathbb Q(\zeta_n):\mathbb Q|=\phi(n)$ and $|\mathbb Q(\alpha):\mathbb Q|=m$. Since $Gal(\mathbb Q(\zeta_n)/\mathbb Q)$ is abelian $\mathbb Q(\alpha)/\mathbb Q$ is also Galois extension. Also clearly $m \mid \phi(n)$. How can I conclude that $m=1$ or $2$. I need some help. Thanks.

user371231
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  • Can figure out why it suffices to handle the case where $\alpha$ is real? And then show that you don't have a splitting field of $x^m-2$? – Jyrki Lahtonen Jun 29 '21 at 07:58
  • @Jyrki Lahtonen Something else is in my mind. Since $\mathbb Q(\alpha)/\mathbb Q$ is Galois, $\mathbb Q(\alpha)$ must be the splitting field of $x^m-2$ over $\mathbb Q$. Is it the case that the splitting field of $X^n-2$ over $\mathbb Q$ always has degree $n\phi(n)$? Then $m\phi(m)=m$ implies that $\phi(m)=1$ and hence $m \leq 2$. – user371231 Jun 29 '21 at 08:33
  • @JyrkiLahtonen Can you kindly check the arguments? – user371231 Jun 29 '21 at 08:36
  • That's one way. You do need to justify the claim about the degree of the splitting field. I had in mind a step that if $\alpha$ is real then $\Bbb{Q}(\alpha)$ can only contain the other zeros of $x^n-2$ when all of them are real... – Jyrki Lahtonen Jun 29 '21 at 09:42
  • Now $\alpha \in \mathbb R \iff \alpha$ is fixed by conjugation. But I can't see it how this will hold. – user371231 Jun 29 '21 at 09:58
  • You already deduced in the question that $\Bbb{Q}(\alpha)/\Bbb{Q}$ is Galois, hence normal... – Jyrki Lahtonen Jun 29 '21 at 10:01
  • You may benefit from studying {this thread](https://math.stackexchange.com/q/460397/11619)? – Jyrki Lahtonen Jun 29 '21 at 10:07
  • As you said if $\alpha \in \mathbb R$ then we are done, since $m$-th primitive root of $1$ is real iff $m\leq 2$. Thus enough to show that $\alpha \in \mathbb R$. We already know that $\mathbb Q(\alpha)/\mathbb Q$ is normal. Thus conjugation will give an automorphism of $\mathbb Q(\alpha)$. Then I am lost. – user371231 Jun 29 '21 at 10:17
  • For every choice of $\alpha$ we have $\Bbb{Q}(\alpha)\simeq\Bbb{Q}(\root n\of2)$. If one is normal over the rationals, so is the other. – Jyrki Lahtonen Jun 29 '21 at 10:27
  • Yes Sir, since $\mathbb Q(\sqrt[m]{2})\subset \mathbb Q(\alpha)$ and the prior also has degree $m$ over $\mathbb Q$, so must be $\mathbb Q(\sqrt[m]{2})= \mathbb Q(\alpha)$. Thus $\alpha \in \mathbb Q(\sqrt[m]{2})$, hence $\alpha \in \mathbb R$. – user371231 Jun 29 '21 at 10:32
  • There. If so inclined, feel free to compile the entire argument in an answer. That way you get one more round of checks, and the unclear steps come to light. I think I have used the idea enough many times on the site already, so I won't post an answer. – Jyrki Lahtonen Jun 29 '21 at 10:35
  • Ok Sir, I will write it down as an answer here. – user371231 Jun 29 '21 at 10:37

1 Answers1

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Based on discussion with Jyrki Lahtonen in comment I am writing down an answer.

Since $Gal(\mathbb Q(\zeta_n)/\mathbb Q) \cong (\mathbb Z/n\mathbb Z)^{*}$, $Gal(\mathbb Q(\zeta_n)/\mathbb Q)$ is abelian. Hence every intermediate extension of $\mathbb Q(\zeta_n)/\mathbb Q$ is Galois. As $\alpha \in \mathbb Q(\zeta_n)$, $\mathbb Q(\alpha)/\mathbb Q$ is Galois, and hence normal extension. Thus $\mathbb Q(\alpha)$ is the splitting field of $x^m-2$ over $\mathbb Q$. In particular, $\mathbb Q \subset \mathbb Q(\sqrt[m]{2})\subset Q(\alpha)$, where $|\mathbb Q(\alpha):\mathbb Q|=|\mathbb Q(\sqrt[m]{2}):\mathbb Q|=m$. Thus we have $\mathbb Q(\alpha)=\mathbb Q(\sqrt[m]{2})$. Hence $\alpha \in \mathbb R$. Now for a natural number $n$, all roots of $x^n-1$ are real if and only if $n\leq 2$. Since $\mathbb Q(\alpha)$ is a splitting field of $x^m-2$ over $\mathbb Q$, $m=1,2$ are the only possibilities.

user371231
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