# Decomposition of Primes:

If is any ideal in the ring of integers of a Galois extension of and , then

is also an ideal of .

Fix a prime and write , so .

Definition 9.2.1 (Residue class degree)   Suppose is a prime of lying over . Then the residue class degree of is

i.e., the degree of the extension of residue class fields.

If is a tower of field extensions and is a prime of over , then

so the residue class degree is multiplicative in towers.

Note that if and , then induces an isomorphism of finite fields that fixes the common subfield . Thus the residue class degrees of and are the same. In fact, much more is true.

Theorem 9.2.2   Suppose is a Galois extension of number fields, and let be a prime of . Write , and let . Then acts transitively on the set of primes , and

Morever, if we let be the common value of the , the common value of the , and , then

Proof. For simplicity, we will give the proof only in the case , but the proof works in general. Suppose and , and . We will first prove that acts transitively on . Let for some . Recall that we proved long ago, using the Chinese Remainder Theorem (Theorem 5.1.4) that there exists such that is an integral ideal that is coprime to . The product

 (9.1)

is a nonzero integral ideal since it is a product of nonzero integral ideals. Since we have that . Thus the numerator of the rightmost expression in (9.2.1) is divisible by . Also, because is coprime to , each is coprime to as well. Thus is coprime to . Thus the denominator of the rightmost expression in (9.2.1) must also be divisibly by in order to cancel the in the numerator. Thus we have shown that for any ,

By unique factorization, since every appears in the left hand side, we must have that for each  there is a  with .

Choose some and suppose that is another index. Because acts transitively, there exists such that . Applying to the factorization , we see that

Taking on both sides and using unique factorization, we get . Thus .

As was mentioned right before the statement of the theorem, for any we have , so by transitivity . We have, upon apply CRT and that , that

which completes the proof.

The rest of this section illustrates the theorem for quadratic fields and a cubic field and its Galois closure.

Subsections
William Stein 2012-09-24