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).
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