Newsgroups: sci.math From: pmontgom@cwi.nl (Peter L. Montgomery) Subject: Re: Galois theory question Date: Sat, 14 Nov 1998 12:34:23 GMT In article <364CE520.CB9DD864@cam.ac.uk> Kevin Costello writes: >Let f(x)=x^3+2x+6. Let it's roots be a,b and b' (the complex conjugate >of b). It's Galois group is S3. But what is the fixed field of A3 in >terms of the roots of the polynomial? > If the Galois group is S3, the splitting field has order 6. The normal subgroup A3 of S3 has order 3 and index 2, so the corresponding fixed field is quadratic. A quadratic extension of Q has the form Q(sqrt(d)) where d is a non-square. The group A3 should map sqrt(d) to itself while the rest of S3 (i.e., the transpositions) maps sqrt(d) to -sqrt(d). Find a formula involving a, b, b' which is negated whenever you interchange two of these quantities, and which is not identically zero. -- Peter-Lawrence.Montgomery@cwi.nl San Rafael, California The bridge to the 21st century is being designed for the private autombile. The bridge to the 22nd century will forbid private automobiles.