Galois Extension of Q isomorphic to Z/3Z

[AlbertEinstein]

Hi...

How do I construct a Galois extension E of Q(set of rational numbers) such that Gal[E,Q] is isomorphic to Z/3Z.

Thanks.

 

[mrbohn1]

There is a very good discussion of this on the http://en.wikipedia.org/wiki/Inverse_Galois_problem"

 

[ABarrios]

Let d=cubic root of some number x s.t. d is not in Q. Then [Q(d):Q]=3 => o(Gal(Q(d)/Q))=3 and only group of order 3 is Z/3Z so you have your desired group.

 

[mrbohn1]

Let d=cubic root of some number x s.t. d is not in Q. Then [Q(d):Q]=3 => o(Gal(Q(d)/Q))=3 and only group of order 3 is Z/3Z so you have your desired group.

This is not necessarily true. For example, adjoining a primitive cube root of unity to Q generates a field extension of order 2. It is a bit more complicated than that, unfortunately!

 

[ABarrios]

This is not necessarily true. For example, adjoining a primitive cube root of unity to Q generates a field extension of order 2. It is a bit more complicated than that, unfortunately!

When I wrote the above I was thinking of x as a natural number or integer, in which case it should work as an example of a Galois group isomorphic to Z/3Z. But as for a complete generalization, I do not know of how to give it.

 

[TMM]

Consider the splitting field of any cubic polynomial with a perfect square discriminant. The root of the discriminant is rational, so it is fixed by all the Galois automorphisms. This means that there are no 2-cycles in the Galois group. If the polynomial is irreducible, then its Galois group is a nontrivial subgroup of S_3, hence is Z/3Z by eliminating all other possibilities.