From: Robin Chapman <rjc@maths.ex.ac.uk>
Newsgroups: sci.math
Subject: Re: Third degree polynomials.
Date: Fri, 27 Feb 1998 08:58:10 -0600

In article <6d3n7g$ncp$2@news.ox.ac.uk>,
  mert0236@sable.ox.ac.uk (Thomas Womack) wrote:
>
> Mike McCarty (jmccarty@sun1307.spd.dsccc.com) wrote:
> : In article <6d063e$25g$1@news.usf.edu>,
> : Jorell Hernandez <johernan@suntan.eng.usf.edu> wrote:
> : )Is there a formula similar to the quadratic equation for third degree
> : )polynomials?
>
> : Yes, and another for biquadratic equations. For fifth and higher degree
> : polynomial equations there is, in general, no solution formula
> : involving combinations of addition, subtraction, multiplication,
> : division, and extraction of roots.
>
> Let theta = 2pi / 11.
>
> Consider the polynomial (x-2 cos theta) (x-2 cos 2theta) (x-2 cos 3theta)
> (x-2 cos 4theta) (x-2 cos 5theta).
>
> According to Maple, this has Galois group of order 5. Does this mean that
> I can write the roots by using a single fifth-root operation on some
> appropriate member of Q?
>
> [I realise that the five roots are all polynomials in 2 cos theta; that's
> how I went about showing that the Galois group was Z5; but can I write
> 2 cos theta in radicals?]

Let alpha_j = 2 cos(j pi/11). A generator of the Galois group acts as follows
on the alphas:
	(alpha_1 alpha_2 alpha_4 alpha_3 alpha_5).
Let K be the field Q(zeta) where zeta is a fifth root of unity. One
can express zeta in terms of radicals, this is essentially in Euclid!
Let L = K(alpha_1). Then L/K is cyclic of degree 5 also. Form Galois
resolvents: let
sigma_j = alpha_1 + zeta^j alpha_2 + zeta^{2j} alpha_4
		+ zeta^{3j} alpha_3 + zeta^{4j} alpha_5
for j = 0,1,2,3,4. Now the generator of the Galois group of L/K multiplies
sigma_j by zeta^j sigma_j and so sigma_j^5 lies in K. (Indeed sigma_0 already
lies in Q.) Thus we can write the sigma_j s as 5-th roots of elements
of K, and so we can get the alpha_j s in terms of radicals.

This method works for any cyclic extension.

For a more interesting challenge solve X^5 - 5X + 12 = 0 in radicals.

Robin Chapman                           "256 256 256.
Department of Mathematics                O hel, ol rite; 256; whot's
University of Exeter, EX4 4QE, UK        12 tyms 256? Bugird if I no.
rjc@maths.exeter.ac.uk                   2 dificult 2 work out."
http://www.maths.ex.ac.uk/~rjc/rjc.html  Iain M. Banks - Feersum Endjinn

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