From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Computing Galois groups Date: 20 Jul 1998 16:36:12 GMT MCKAY john wrote: >A hard problem is to find polynomials with >galois groups /Q from a given group. It is a major open problem as to >which groups are realisable as galois groups /Q. Well, that's perhaps not the best way to phrase it. I suspect most believe the answer to the question, "Which (finite) groups which are Galois groups of number fields?" is "All of them". The problem is not so much to intuit the right answer as to prove the answer correct. All the solvable groups are known to be Galois groups and quite a lot of work has been done with the simple groups, but no proof for the general case is known, and any proof along the current lines of thought would be quite intricate and not a little tedious. dave ============================================================================== Newsgroups: sci.math From: pmontgom@cwi.nl (Peter L. Montgomery) Subject: Re: Computing Galois groups Date: Tue, 21 Jul 1998 07:52:54 GMT In article <01bdb434$965b3220$0100a8c0@mgreen> "Martin Green" writes: >All groups are Galois groups of number fields??? Does this mean that, for >example, the cyclic group on 3 elements is the Galois group of some >polynomial over the rationals? If so, could someone post an example? One such polynomial is x^3 - 3*x + 1. Consider the trigonometric equation cos(3*theta) = -1/2. Solutions in (0, 2*pi) are theta = 2*pi/9, 4*pi/9, 8*pi/9, 10*pi/9, 14*pi/9, 16*pi/9. Denote x = 2*cos(theta). Then 2*cos(2*theta) = 4*cos(theta)^2 - 2 = x^2 - 2 and 2*cos(3*theta) = 4*cos(theta)*cos(2*theta) - 2*cos(theta) = x*(x^2 - 2) - x = x^3 - 3*x. Since we specified cos(3*theta) = -1/2, our equation is (*) x^3 - 3*x + 1 = 0 It is easy to check that (*) has no rational roots, so the cubic must be irreducible. Looking back at our values of theta, however, we observe that 2*theta is a solution whenever theta is a solution. This is easy to check; if cos(3*theta) = -1/2, then cos(6*theta) = 2*cos(3*theta)^2 - 1 = 2*(-1/2)^2 - 1 = -1/2. As noted earlier, x = 2*cos(theta) implies 2*cos(2*theta) = x^2 - 2. If x is a root of (*), then x^2 - 2 is another root. Formally (x^2 - 2)^3 - 3*(x^2 - 2) + 1 = (x^6 - 6*x^4 + 12*x^2 - 8) - (3*x^2 - 6) + 1 = x^6 - 6*x^4 + 9*x^2 - 1 = (x^3 - 3*x)^2 - 1 = (-1)^2 - 1 = 0 if x satisfies (*). The three roots of (*) are {x1, x2, x3} where x2 = x1^2 - 2 and x3 = x2^2 - 2 and x1 = x3^2 - 2. If tau is a field automorphism, then tau(x2) = tau(x1^2 - 2) = tau(x1)^2 - 2 is uniquely determined once tau(x1) is specified, and tau(x3) is uniquely determined too. The image tau(x1) can be x1, x2, or x3, so there are exactly three field automorphisms. -- Peter-Lawrence.Montgomery@cwi.nl San Rafael, California ============================================================================== From: bobs@rsa.com Newsgroups: sci.math Subject: Re: Computing Galois groups Date: Tue, 21 Jul 1998 16:44:22 GMT In article , pmontgom@cwi.nl (Peter L. Montgomery) wrote: > In article <01bdb434$965b3220$0100a8c0@mgreen> > "Martin Green" writes: > >All groups are Galois groups of number fields??? Does this mean that, for > >example, the cyclic group on 3 elements is the Galois group of some > >polynomial over the rationals? If so, could someone post an example? > > One such polynomial is x^3 - 3*x + 1. Yep. The only transitive subgroups of S(3) are either C(3) [cyclic] or S(3) ~ D(3) itself. If the discriminant is a square then we have C(3), else S(3). -----== Posted via Deja News, The Leader in Internet Discussion ==----- http://www.dejanews.com/rg_mkgrp.xp Create Your Own Free Member Forum