For example, let . We know that . Thus , so for we have and . To factor , we note that working modulo we have

Next we consider what happens in the Galois closure of . Since the three embeddings of in are , , and , we have

Let's figure out , , and for the prime relative to the degree six Galois field by using Theorem 13.2.2 and what we can easily determine about and . First, we know that . We have , so , and the prime factors of are disjoint from the prime factors of . Thus is even and also . The only possibility for satisfying these two conditions is , , , so we conclude that without doing any further work, and without actually knowing the explicitly.

Here's another interesting deduction that we can make ``by hand''. Suppose for the moment that (this will turn out to be false). Then the factorization of in would be exactly reflected by the factorization of in . Modulo we have , which would imply that for some prime of , i.e., that and , which is incorrect. Thus . Indeed, this conclusion agrees with the following computation, which asserts that :

> R<x> := PolynomialRing(RationalField()); > K := NumberField(x^3-2); > L := NumberField(x^2+3); > M := CompositeFields(K,L)[1]; > O_M := MaximalOrder(M); > a := M!K.1; > b := M!L.1; > O := Order([a,b]); > Index(O_M,O); 24

William Stein 2004-05-06