 Solution

We use the following standard facts about cyclotomic fields. For an odd prime $\ell$, the unique quadratic subfield of $\mathbb{Q}({\zeta }_{\ell })$ isAlso $\mathbb{Q}({\zeta }_{{\ell }^r})$ has degree $\phi ({\ell }^r)={\ell }^{r-1}(\ell -1)$, and $\ell$ is totally ramified, with a unique prime and residue field ${\mathbb{F}}_{\ell }$. These are the quadratic subfield of a prime cyclotomic field and total ramification in a prime-power cyclotomic field results.

TakeSince $59\equiv 3(\mathrm{mod}4)$, every $E_r$ contains $K=\mathbb{Q}(\sqrt{-59})$, through its subfield $E_1$. The unique prime above $59$ is totally ramified in $E_r/K$: the absolute ramification index is $58\cdot 59^{r-1}$, the index in $K/\mathbb{Q}$ is $2$, and the relative index is thereforeThe field $K$ has no real places. By part (b), $h_K\mid h_{E_r}$, and part (a) showed $3\mid h_K$. ThusThe ring of integers of a number field is a Dedekind domain, and is a principal ideal domain exactly when its ideal class group is trivial. Hence none of these rings of integers is principal. Finally the degrees $58\cdot 59^{r-1}$ strictly increase, so the fields are pairwise distinct. The explicit infinite family of nonprincipal cyclotomic fields of conductor a prime power is