From: tim@maths.tcd.ie (Timothy Murphy)
Subject: Re: [Q] Galois Theory?
Date: 27 Sep 1999 17:48:19 +0100
Newsgroups: sci.math
Keywords: Example of the Galois correspondence (infinite groups)
writes:
>We know that if F is the extension field of K. then for Galois group G(F/K)
>and it's subgroup H, J satisfying H < J.
>if [J : H] is finite, [H' : J'] <= [J : H] holds in general (ref.
>Hungerford's Algebra)
>H' is the field fixed by all automorphism of H.
>Then, when inequality(not equality!) holds?
>Please tell me some example, and give me a sketch of proof!
I'm not sure if this is what you are asking for,
but consider the extension Q(x)/Q (where Q is the rationals).
I think the automorphisms of Q(x) are the maps
x -> (ax + b)/(cx + d) with ad - bc <> 0.
In other words, G = G(Q(x)/Q) = PLG(2,Q).
On the other hand, if we take the subgroup S of maps x -> x + a
(isomorphic to the additive group of Q)
then the invariant space of S is Q,
if the rational function f(x) satisfies f(x+a) = f(x) for all a in Q
then f(x) is constant.
Now any group between S and G will have invariant space Q.
--
Timothy Murphy
e-mail: tim@maths.tcd.ie
tel: +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland