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Let $F$ be a field and $n$ be any integer satisfying $(n,\text{char}(F))=1$ when $\text{char}(F)\ne 0$.

Let $\xi$ be a primitive $n$-th root of unity. I know that $\text{Gal}(F(\xi)/F)$ is isomorphic to a subgroup of $(\mathbb{Z}_n)^{\times}$ and it is isomorphic when $F=\mathbb{Q}$. (In this case, deg (irr($\xi$,$\mathbb{Q}$))=$\phi(n)$ where $\phi$ is Euler Phi-function.)

The reason (I think) why this does not hold in general is that $F$ may contain $\xi$. So if $\xi\notin F$, can we assure $|F(\xi)/F|=\phi(n)$?

In other words, is the polynomial $\prod_{1\le k \le n, \ (n,k)=1} (t-\xi^k)\in F[t]$ irreducible?

If not, please let me know some counter example.

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  • $\begingroup$ A sneaky one is $\mathbb Q(\zeta_5)$ over $\mathbb Q(\sqrt 5)$. $\endgroup$ – user208649 Apr 23 '16 at 7:28
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No, it is still not sufficient to assume that $\xi\notin F$. For example, if $F=\mathbb{Q}(i)$ and $$\xi=e^{2\pi i / 8}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$$ then $[F(\xi):F]=2$ but $\xi\notin F$. Observe that $\varphi(8)=4$, and $$\underbrace{[F(\xi):\mathbb{Q}]}_{4}=\underbrace{[F(\xi):F]}_{2}\underbrace{[F:\mathbb{Q}]}_{2}$$

The Galois group of the $n$th cyclotomic field is $\mathrm{Gal}(\mathbb{Q}(\xi)/\mathbb{Q}))\cong(\mathbb{Z}/n)^\times$. If (and only if) $\varphi(n)=|(\mathbb{Z}/n)^{\times}|$ is a composite number, there will be a non-trivial subgroup of $(\mathbb{Z}/n)^\times$, and hence (by the Fundamental Theorem of Galois Theory) a field $F$ with $\mathbb{Q}(\xi)\supsetneq F\supsetneq \mathbb{Q}$, which will have the property that $1<[F(\xi):F]<\varphi(n)$.

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  • $\begingroup$ +1 Another small counterexample is with $F=\Bbb{Q}(\sqrt5)$ and $n=5$. The number $\xi+\overline{\xi}=2\cos(2\pi/5)$, and the formula $2\cos(2\pi/5)=(1+\sqrt5)/2$ shows that $\xi$ is of degree $2$ over $F$. $\endgroup$ – Jyrki Lahtonen Apr 23 '16 at 7:25
  • $\begingroup$ @Zev Chonoles, Thank you for letting me know this good example. It helped me very much! $\endgroup$ – user29422 Apr 23 '16 at 8:51
  • $\begingroup$ @Jyrki Lahtonen, How did you get the formula $2\cos(\frac{2\pi }{5})=\frac{1+\sqrt{5}}{2}$? $\endgroup$ – user29422 Apr 23 '16 at 9:01
  • $\begingroup$ @Jyrki Lahtonen, Oh, I see. Thank you!^^ $\endgroup$ – user29422 Apr 23 '16 at 9:08

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