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Let $n$ be a positive integer, $K$ be a field with $charK$ does not divide $n$, and $F$ be the cyclotomic extension of $K$ of order $n$.

Theorem says that $Gal_{K}{F}$ is isomorphic to a subgroup of $\mathbb Z_n^*$, i.e., the multiplicative group of units of $\mathbb Z_n$.

I don't know why $Gal_{K}{F}$ only isomorphic to a subgroup of $\mathbb Z_n^*$, but not necessarily the whole group. I know that for $K=\mathbb Q$, $Gal_{\mathbb Q}{F} \cong \mathbb Z_n^*$, but is there any field $K$ such that $Gal_{K}{F}$ is only a proper subgroup of $\mathbb Z_n^*$?

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Yes. $K$ may intersect non-trivially with a cyclotomic extension of $\Bbb{Q}$. The first example that comes to mind is $n=5$, $K=\Bbb{Q}(\sqrt{5})$. Here $K\subseteq\Bbb{Q}(\zeta_5)$. Or more precisely, $K=\Bbb{Q}(\zeta_5)\cap \Bbb{R}$. So $F=K(\zeta_5)$ is only a degree two extension of $K$, and the Galois group is cyclic of order two. The complex conjugation in $F$ is its only non-trivial automorphism, so $Gal(K/F)=\{\overline{1},\overline{-1}\}\le\Bbb{Z}_5^*$.

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  • $\begingroup$ Thanks for your anwser, but may I ask how could I show that $\sqrt 5 \in \mathbb Q (\zeta_5)$, I can see $cos(\frac{2\pi}{5})$ and $cos(\frac{4\pi}{5})$ are in $\mathbb Q (\zeta_5)$ $\endgroup$ – AG learner Aug 2 '15 at 21:46
  • $\begingroup$ Good. Have you seen a proof of $$\cos\frac{2\pi}5=\frac{-1+\sqrt{5}}4?$$ A more general result is that for a prime $p>2$ the unique quadratic subfield of $\Bbb{Q}(\zeta_p)$ is $\Bbb{Q}(\sqrt{p^*})$, where $p^*=p$, if $p\equiv1\pmod4$ and $p^*=-p$, if $p\equiv -1\pmod4$. Here is my proof for the fact that $\sqrt{-7}\in\Bbb{Q}(\zeta_7)$. One uses Gauss' sums to prove this for a general prime $p$. $\endgroup$ – Jyrki Lahtonen Aug 2 '15 at 21:53
  • $\begingroup$ I see it, that's awesome. Thanks a lot! $\endgroup$ – AG learner Aug 2 '15 at 22:19
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By "cyclotomic extension of order $n$," you mean $K(\zeta)$ for a primitive $n$th root of unity. What happens if $K$ already has some $d$th roots of unity for divisors nontrivial $d\mid n$? Indeed, in the most extreme case, if $K$ already has all of the $n$th roots of unity, then $K(\zeta)/K$ is trivial so ${\rm Gal}$ is the trivial subgroup of $U(n)=(\Bbb Z/n\Bbb Z)^\times$.

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