unramified extensions and class number divisibility

The following is a corollary of the existence of the Hilbert class field.

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:

$$|\mathrm{Gal}(H/K)|=[H:K]=h_K.$$

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 $\mathbb{Z}/p\mathbb{Z}$). Since $H$ is the maximal abelian unramified extension of $K$ the following inclusions occur:

$$K⊊F\subseteq H$$

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 $h_K=|\mathrm{Gal}(H/K)|$. Since $G=\mathrm{Gal}(H/K)$ 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.∎