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

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 ?

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