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