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Galois Extension of Q isomorphic to Z/3Z 
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#1
Apr2710, 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. 


#2
Apr2710, 02:45 PM

P: 97

There is a very good discussion of this on the wikipedia page on the inverse Galois problem.



#3
Apr2710, 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.



#4
Apr2810, 01:24 AM

P: 97

Galois Extension of Q isomorphic to Z/3Z



#5
Apr2810, 07:30 AM

P: 23




#6
May210, 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 2cycles 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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