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