class number divisibility in extensions

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

Corollary 1.

Let K be a number fieldMathworldPlanetmath, hK is its class numberMathworldPlanetmathPlanetmath and let p be a prime. Then K has an everywhere unramified Galois extensionMathworldPlanetmath of degree p if and only if hK is divisible by p.

In this entry we are concerned about the divisibility properties of class numbers of number fields in extensionsPlanetmathPlanetmathPlanetmath.

Theorem 1.

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


Let F, K and p be as in the statement of the theorem. Assume that p|hK. 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 FE=K. In particular, the compositum FE is a Galois extension of F and


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 contradictionMathworldPlanetmathPlanetmath that 𝒬F is a prime idealPlanetmathPlanetmath which ramifies in the extension FE/F. Let 𝒬EF be a prime lying above 𝒬F and let 𝒬K be a prime of K such that 𝒬F lies above it. Similarly, let 𝒬E be a prime of E lying above 𝒬K and such that the prime 𝒬EF lies above 𝒬E. For an arbitrary extension A/B, the ramification index of a prime 𝒬A is denoted by e(𝒬B|𝒬A). Then, by the multiplicativity of the ramification index in towers, we have:


Since we assumed that 𝒬F is ramified in EF/F, and the degree of the extension is p, we must have e(𝒬EF|𝒬F)=p. Therefore, by the equality above, p divides e(𝒬EF|𝒬E)e(𝒬E|𝒬K). Notice that the extension E/K is everywhere unramified, therefore e(𝒬E|𝒬K)=1. Also, [EF:E]=[F:K] which, by hypothesisMathworldPlanetmathPlanetmath, is relatively prime to p. Thus e(𝒬EF|𝒬E) is also relatively prime to p, and so, p is not a divisorMathworldPlanetmath of e(𝒬EF|𝒬E)e(𝒬E|𝒬K), 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