extensions without unramified subextensions and class number divisibility

Theorem 1.

Let $F\mathrm{/}K$ be an extension of number fields such that for any intermediate Galois extension $L\mathrm{/}K$, with $K\mathrm{⊊}L\mathrm{⊊}F$, there is at least one finite place or infinite place which ramifies in the extension $L\mathrm{/}K$. Then, $h_K$, the class number of $K$, divides the class number of $F$, $h_F$.

First, we deduce some immediate corollaries.

Corollary 1.

Let $F\mathrm{/}K$ be an extension of number fields which is totally ramified at some prime (or at an archimedean place). Then $h_K$ divides $h_F$.

Proof.

The proof is clear since there cannot be unramified subextensions. The theorem applies.∎

Corollary 2.

Let $F\mathrm{/}K$ be a Galois extension of number fields such that $\mathrm{Gal}\mathit{}\mathrm{(}F\mathrm{/}K\mathrm{)}$ is a non-abelian simple group. Then $h_K$ divides $h_F$.

Proof.

In this case, there cannot be subextensions with abelian Galois group and the theorem applies.∎

Proof of the Theorem.

Let $H$ be the Hilbert class field of $K$. By definition, $H$ is the maximal unramified abelian extension of $K$, $\mathrm{Gal}(H/K)$ is isomorphic to $\mathrm{Cl}(K)$, 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 $\mathrm{Gal}(FH/F)\cong \mathrm{Gal}(H/K)$). 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 $h_F$.∎