# class number divisibility in extensions

Throughout this entry we will use the following corollary to the existence of the Hilbert class field^{} (see the parent entry, unramified extensions and class number divisibility for details of the proof).

###### 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 $p$.

In this entry we are concerned about the divisibility properties of class numbers of number fields in extensions^{}.

###### Theorem 1.

Let $F\mathrm{/}K$ 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 $\mathrm{[}F\mathrm{:}K\mathrm{]}$, 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

$$\mathrm{Gal}(FE/F)\cong \mathrm{Gal}(E/E\cap F)\cong \mathrm{Gal}(E/K).$$ |

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.
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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 |