Intermediate subfield in splitting field of the polynomial $x^{29}-1=0$

Question

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 ?

Answer 1

$G = Gal(K/Q)$ 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 $\mathbb{Q}(\zeta_{29} + \zeta_{29}^{-1})$.