unramified extensions and class number divisibilityUnramifiedExtensionsAndClassNumberDivisibility2005-02-17 17:28:142005-02-17 17:30:07Corollaryexistence of Hilbert class fieldclass number divisibility% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\Rats}{\mathbb{Q}}The following is a corollary of the existence of the Hilbert class field.
\begin{cor}
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 $p$.
\end{cor}
\begin{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.
\end{proof}