Fix a prime and write , so .

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.

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 ,

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

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.