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 <kjc23@cam.ac.uk> 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.