extensions without unramified subextensions and class number divisibilityExtensionsWithoutUnramifiedSubextensionsAndClassNumberDivisibility2005-03-10 10:53:242005-03-10 10:53:24Theoremunramified extensions and class number divisibilitydivisibilityclass numbertower of number fields% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\Cl}{\operatorname{Cl}}\begin{thm}
Let $F/K$ be an extension of number fields such that for any intermediate Galois extension $L/K$, with $K\subsetneq L \subsetneq F$, there is at least one finite place or infinite place which ramifies in the extension $L/K$. Then, $h_K$, the class number of $K$, divides the class number of $F$, $h_F$.
\end{thm}
First, we deduce some immediate corollaries.
\begin{cor}
Let $F/K$ be an extension of number fields which is totally ramified at some prime (or at an archimedean place). Then $h_K$ divides $h_F$.
\end{cor}
\begin{proof}
The proof is clear since there cannot be unramified subextensions. The theorem applies.
\end{proof}
\begin{cor}
Let $F/K$ be a Galois extension of number fields such that $\Gal(F/K)$ is a non-abelian simple group. Then $h_K$ divides $h_F$.
\end{cor}
\begin{proof}
In this case, there cannot be subextensions with abelian Galois group and the theorem applies.
\end{proof}
\begin{proof}[Proof of the Theorem]
Let $H$ be the Hilbert class field of $K$. By definition, $H$ is the maximal unramified abelian extension of $K$, $\Gal(H/K)$ is isomorphic to $\Cl(K)$, the ideal class group of $K$ and $[H:K]=h_K$. Since there are no nontrivial unramified abelian subextensions of $F/K$, we have $F\cap H=K$ and so $[FH:F]=[H:K]=h_K$. One can show that the extension $FH/F$ is unramified and abelian (in fact $\Gal(FH/F)\cong \Gal(H/K)$). Therefore $FH$ is contained in $L$, the Hilbert class field of $F$. Hence:
$$h_F=[L:F]=[L:FH]\cdot[FH:F]=[L:FH]\cdot [H:K]=[L:FH]\cdot h_K$$
and so, $h_K$ divides $h_F$.
\end{proof}