How to find the intermediate subfield between the splitting field $K$ of the polynomial $f(x)=x^{29}-1$ and the field $\mathbb{Q}$?

$K$ has all 29th roots of unity. So $K=Q(\omega_{29})$ ($\omega_{29}^{29}-1=0$). Right? I cannot see there is an intermediate subfield, since $Gal(K)$ isomorphic to $U(29)=\{x:\gcd(x,29=1\}$ and this is a cyclic group all its elements are generator.

Where am I wrong ?

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    $\begingroup$ There are quite a few different intermediate fields between $\;\Bbb Q\;$ and $\;K\;$ ...in fact, exactly one for every divisor of $\;\phi(29)=28\;$ . $\endgroup$ – DonAntonio Jul 31 '16 at 20:01
  • $\begingroup$ $Gal(K)$ is the cyclic group of order $28$. It has a whole bunch of subgroups, which correspond to a whole bunch of intermediate fields. $\endgroup$ – Anon Jul 31 '16 at 20:04
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    $\begingroup$ Let $e^{i\theta}$ be a $29$-th root of unity other than $1$. Then the field generated by $\cos\theta$ is intermediate. $\endgroup$ – André Nicolas Jul 31 '16 at 20:05
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    $\begingroup$ Oh. I failed by my quick calculation the group has order 28 not 29. $\endgroup$ – Leonardo Jul 31 '16 at 20:19
  • $\begingroup$ A challenge is to spot that $\Bbb{Q}(\sqrt{29})$ is one of the intermediate fields. Look up Gauss sums for an explanation. $\endgroup$ – Jyrki Lahtonen Jul 31 '16 at 20:31

$G = Gal(K/Q)$ is the set of autmorphisms defined by

$$\zeta_{29} \mapsto \zeta_{29}^a$$

where $(a, 29) = 1$. The structure is identical to $(\mathbb{Z}/29\mathbb{Z})^*$, which is cyclic of order $28$.

Where you go wrong is where you say that any element is a generator. In particular, there are elements of order $1, 2, 4, 7, $ and $14$ in this group which, by the fundamental theorem of Galois theory, determine the intermediate fields.

For example, the element $\sigma : \zeta_{29} \mapsto \zeta_{29}^{-1}$ has order $2$. The corresponding fixed field is $\mathbb{Q}(\zeta_{29} + \zeta_{29}^{-1})$.

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