## The Cube Roots of Two

Suppose is not Galois. Then , , and  are defined for each prime , but we need not have or . We do still have that , by the Chinese Remainder Theorem.

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

so

Thus , , , and . Next, working modulo we have

and the quadratic factor is irreducible. Thus

Thus here , , , and .

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

where , since is a primitive cube root of unity. The notation means the compositum of and '', which is the smallest field generated by and .

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