Galois Extension of Q isomorphic to Z/3Z


by AlbertEinstein
Tags: extension, galois, isomorphic, z or 3z
AlbertEinstein
AlbertEinstein is offline
#1
Apr27-10, 02:35 AM
P: 113
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.
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mrbohn1
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#2
Apr27-10, 02:45 PM
P: 97
There is a very good discussion of this on the wikipedia page on the inverse Galois problem.
ABarrios
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#3
Apr27-10, 05:20 PM
P: 23
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
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#4
Apr28-10, 01:24 AM
P: 97

Galois Extension of Q isomorphic to Z/3Z


Quote Quote by ABarrios View Post
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
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#5
Apr28-10, 07:30 AM
P: 23
Quote Quote by mrbohn1 View Post
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
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#6
May2-10, 09:11 PM
P: 96
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.


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