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Every field $\mathbb{Q}({\zeta }_{59^r})$ contains $\mathbb{Q}(\sqrt{-59})$ and is totally ramified over it at the prime above $59$. The class number divisibility under total ramification therefore shows that its class number is divisible by three. Its ring of integers is not a principal ideal domain. The degrees $58\cdot 59^{r-1}$ strictly increase, giving infinitely many distinct examples.

Use the Hilbert class field theorem stated in part (a), and let $H_L$ be the Hilbert class field of $L$. Put $F=M\cap H_L$ inside a common algebraic closure.

Any intermediate extension of $M/L$ is totally ramified at the given prime. To verify this even when $M/L$ is not Galois, there is a unique prime of $F$ above $\mathfrak{p}$, its residue field embeds into the residue field at the unique prime of $M$ and therefore equals $k_{\mathfrak{p}}$, and the fundamental identity for prime decomposition gives ramification index $[F:L]$. On the other hand $F/L$ is a subextension of $H_L/L$, so it is unramified there. Its ramification index is consequently both $[F:L]$ and $1$, forcingBecause $H_L/L$ is Galois, this intersection identity givesThus $M{H}_L/M$ is abelian. We use the standard local fact that unramified extensions remain unramified under finite base extension: after base change each factor is obtained by lifting an extension of the residue field. Since $H_L/L$ is unramified at every finite prime, $M{H}_L/M$ is also unramified at every finite prime. There are no real places of $M$, since every embedding of $M$ restricts to an embedding of the totally imaginary field $L$. Hence there is no infinite-place obstruction.

The maximality of $H_M$ now implies $M{H}_L\subseteq H_M$. Taking degrees over $M$ proves the class number divisibility under total ramification:The argument uses the degree of an unramified class-field extension, and does not assume that extension of ideals induces an injection of ideal class groups.