# class number divisibility in extensions

###### Theorem 1.

Let $F/K$ be a Galois extension of number fields, let $h_{F}$ and $h_{K}$ be their respective class numbers and let $p$ be a prime number  such that $p$ does not divide $[F:K]$, the degree of the extension. Then, if $p$ divides $h_{K}$, the class number of $F$, $h_{F}$, is also divisible by $p$.

###### Proof.

Let $F$, $K$ and $p$ be as in the statement of the theorem. Assume that $p|h_{K}$. Thus, by the corollary above, there exists an unramified Galois extension field $E$ of $K$ of degree $p$. Notice that the fact that $p$ does not divide the degree of the extension $F/K$ implies that $F\cap E=K$. In particular, the compositum $FE$ is a Galois extension of $F$ and

 $\operatorname{Gal}(FE/F)\cong\operatorname{Gal}(E/E\cap F)\cong\operatorname{% Gal}(E/K).$

Thus, the extension $FE/F$ is of degree $p$, Galois, and therefore abelian. By the corollary above, in order to prove the theorem it suffices to show that the extension $FE/F$ is unramified. Suppose for a contradiction   that $\mathcal{Q}_{F}$ is a prime ideal  which ramifies in the extension $FE/F$. Let $\mathcal{Q}_{EF}$ be a prime lying above $\mathcal{Q}_{F}$ and let $\mathcal{Q}_{K}$ be a prime of $K$ such that $\mathcal{Q}_{F}$ lies above it. Similarly, let $\mathcal{Q}_{E}$ be a prime of $E$ lying above $\mathcal{Q}_{K}$ and such that the prime $\mathcal{Q}_{EF}$ lies above $\mathcal{Q}_{E}$. For an arbitrary extension $A/B$, the ramification index of a prime $\mathcal{Q}_{A}$ is denoted by $e(\mathcal{Q}_{B}|\mathcal{Q}_{A})$. Then, by the multiplicativity of the ramification index in towers, we have:

 $e(\mathcal{Q}_{EF}|\mathcal{Q}_{K})=e(\mathcal{Q}_{EF}|\mathcal{Q}_{F})\cdot e% (\mathcal{Q}_{F}|\mathcal{Q}_{K})=e(\mathcal{Q}_{EF}|\mathcal{Q}_{E})\cdot e(% \mathcal{Q}_{E}|\mathcal{Q}_{K})$

Since we assumed that $\mathcal{Q}_{F}$ is ramified in $EF/F$, and the degree of the extension is $p$, we must have $e(\mathcal{Q}_{EF}|\mathcal{Q}_{F})=p$. Therefore, by the equality above, $p$ divides $e(\mathcal{Q}_{EF}|\mathcal{Q}_{E})\cdot e(\mathcal{Q}_{E}|\mathcal{Q}_{K})$. Notice that the extension $E/K$ is everywhere unramified, therefore $e(\mathcal{Q}_{E}|\mathcal{Q}_{K})=1$. Also, $[EF:E]=[F:K]$ which, by hypothesis   , is relatively prime to $p$. Thus $e(\mathcal{Q}_{EF}|\mathcal{Q}_{E})$ is also relatively prime to $p$, and so, $p$ is not a divisor  of $e(\mathcal{Q}_{EF}|\mathcal{Q}_{E})\cdot e(\mathcal{Q}_{E}|\mathcal{Q}_{K})$, which leads to the desired contradiction, finishing the proof of the theorem. ∎

Also, read the entry extensions without unramified subextensions and class number divisibility for a similar and more general result.

 Title class number divisibility in extensions Canonical name ClassNumberDivisibilityInExtensions Date of creation 2013-03-22 15:04:17 Last modified on 2013-03-22 15:04:17 Owner alozano (2414) Last modified by alozano (2414) Numerical id 9 Author alozano (2414) Entry type Theorem Classification msc 11R37 Classification msc 11R32 Classification msc 11R29 Related topic IdealClass Related topic ExistenceOfHilbertClassField Related topic CompositumOfAGaloisExtensionAndAnotherExtensionIsGalois Related topic DecompositionGroup Related topic ExtensionsWithoutUnramifiedSubextensionsAndClassNumberDivisibility Related topic ClassNumbersAndDiscriminantsTopicsOnClassGroups