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Galois Extension of Q isomorphic to Z/3Z

  1. Apr 27, 2010 #1
    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.
     
  2. jcsd
  3. Apr 27, 2010 #2
    There is a very good discussion of this on the http://en.wikipedia.org/wiki/Inverse_Galois_problem" [Broken]
     
    Last edited by a moderator: May 4, 2017
  4. Apr 27, 2010 #3
    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.
     
  5. Apr 28, 2010 #4
    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!
     
  6. Apr 28, 2010 #5
    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.
     
  7. May 2, 2010 #6

    TMM

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    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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