Is there a standard way to find the subfields of $\Bbb{Q}(\zeta_n)$ when $n$ is not prime?

Solution 1:

Actually the theory of Gaussian periods (which Gauss developped in his study of the construction of regular $n$-gons by ruler and compass) allows to describe explicitly the subfields of the cyclotomic field $K_n=\mathbf Q (\zeta_n)$ when $n$ is square free. If $n$ has a factor $p^2$, there is a conjectural characterization (see precisely below). Here is brief summary:

Fix $n$ and drop the index $n$ in the notations. Write $\mathcal G =\mathrm{Gal}(K/\mathbf Q) \cong (\mathbf Z/n\mathbf Z)^*$. For any subgroup $G \le \mathcal G$, for any $a \in (\mathbf Z/n\mathbf Z)^* $, define the Gaussian period $S(G, a):=\sum g({\zeta}^{a})$ for $g\in G$ (so it's a special trace value, and in particular $S(G,1)\in K^{G}$). The Gaussian periods behave well under Galois action, in particular, if $H \le G$, then $S(G, a)= \sum s(S(H,a))$ for $s$ running through $G/H$.

Theorem: The periods $S(G,b)$ for all $b\in \mathcal G/G$ form an integral basis of the ring of integers of $K^{G}$ iff $n$ is square-free. A lemma in the course of the proof states that if $n$ is square-free, then $K^{G}=\mathbf Q(S(G,1))$, which answers your question in this case.

Conjecture (based on numerical examples): $K^{G}=\mathbf Q(S(G,1))$ iff $G$ contains no subgroup of the form $\left<1+ \frac n p\right>$ for any prime $p$ s.t. $p^2\mid n$.