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 (13.2.1) 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 (13.2.1) must also be divisibly by in order to cancel the in the numerator. Thus for any we have

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 . Since is a lattice in , we have

which completes the proof.

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