Corollary 1   Let $K$ be a number field, $h_K$ is its class number and let $p$ be a prime. Then $K$ has an everywhere unramified Galois extension of degree $p$ if and only if $h_K$ is divisible by $p$ .

Proof. Let $K$ be a number field and let $H$ be the Hilbert class field of $K$ . Then: $$|\operatorname{Gal}(H/K)|=[H:K]=h_K.$$ Let $p$ be a prime number. Suppose that there exists a Galois extension $F/K$ , such that $[F:K]=p$ and $F/K$ is everywhere unramified. Notice that any Galois extension of prime degree is abelian (because any group of prime degree $p$ is abelian, isomorphic to $\Ints/p\Ints$ ). Since $H$ is the maximal abelian unramified extension of $K$ the following inclusions occur: $$K \subsetneq F\subseteq H$$ Moreover, $$h_K=[H:K]=[H:F]\cdot[F:K]=[H:F]\cdot p.$$ Therefore $p$ divides $h_K$ .

Next we prove the remaining direction. Suppose that $p$ divides $h_K=|\operatorname{Gal}(H/K)|$ . Since $G=\operatorname{Gal}(H/K)$ is an abelian group (isomorphic to the class group of $K$ ) there exists a normal subgroup $J$ of $G$ such that $|G/J|=p$ . Let $F=H^J$ be the fixed field by the subgroup $J$ , which is, by the main theorem of Galois theory, a Galois extension of $K$ . This field satisfies $[F:K]=p$ and, since $F$ is included in $H$ , the extension $F/K$ is abelian and everywhere unramified, as claimed.