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[Mathqa]Re: Galois group of the rationals
Posted:
Mar 10, 2001 4:50 PM


On Sat, 10 Mar 2001, Nick Halloway wrote:
> On Sat, 10 Mar 2001, Charles Matthews wrote: > > The Galois group you get by adding square roots of all the > positive integers to Q is a vector space of dimension 2^countable over Z/2Z. > You could find a basis for that, starting with the countable basis for the > vector space which has all but finitely many components = . Call that > basis B. Then you can fill that out to a basis B' for the whole vector space. > > A basis for B is the automorphisms that send a square root of one prime > to its negative, and leave everything else as it is. You could define a > character on the Galois group to be 1 for elements of B, and 0 for all > other elements of B'. The kernel of this character is a subgroup H of > index 2. > > Then suppose you have a subgroup of index 2 which is the fixed field of > Q(sqrt(n)), some positive n. The character which has this subgroup as its > kernel is 0 on some automorphism which sends the square root of some > prime not a factor of n to its negative. So, H isn't such a subgroup.
The preimage of H in the Galois group of the algebraic closure of the rationals over the rationals would be a subgroup of index 2 in this Galois group which isn't the fixed group for a field.
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