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).
Let be a Galois extension of number fields, let and be their respective class numbers and let be a prime number such that does not divide , the degree of the extension. Then, if divides , the class number of , , is also divisible by .
Let , and be as in the statement of the theorem. Assume that . Thus, by the corollary above, there exists an unramified Galois extension field of of degree . Notice that the fact that does not divide the degree of the extension implies that . In particular, the compositum is a Galois extension of and
Thus, the extension is of degree , Galois, and therefore abelian. By the corollary above, in order to prove the theorem it suffices to show that the extension is unramified. Suppose for a contradiction that is a prime ideal which ramifies in the extension . Let be a prime lying above and let be a prime of such that lies above it. Similarly, let be a prime of lying above and such that the prime lies above . For an arbitrary extension , the ramification index of a prime is denoted by . Then, by the multiplicativity of the ramification index in towers, we have:
Since we assumed that is ramified in , and the degree of the extension is , we must have . Therefore, by the equality above, divides . Notice that the extension is everywhere unramified, therefore . Also, which, by hypothesis, is relatively prime to . Thus is also relatively prime to , and so, is not a divisor of , 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|
|Date of creation||2013-03-22 15:04:17|
|Last modified on||2013-03-22 15:04:17|
|Last modified by||alozano (2414)|