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.
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