From: Brian Stewart
Newsgroups: sci.math
Subject: Re: Help with Gauss's polynomial proof
Date: Wed, 23 Dec 1998 09:23:31 +0000
Martin Green wrote:
> >
> > I didn't see the start of this thread and only barely understood the
> > flow of your logic, but the pattern is this: in order to solve a cubic
> > you need to solve a certain "resolvent" equation first, formed from the
> > symmetries of the original cubic. This resolvent is quadratic.
>
> I found a quadratic equation. Its roots are
>
> a + wb + w^2c
> a + w^2b + wc
>
> where a,b,c are the roots of the cubic. You can tell this is a quadratic
> because its sum and product are
> symmetrical polynomials in a,b,and c.
>
> Once I solve this quadratic, the cubic falls out.
>
> > In order to solve a quartic, you likewise need to solve the resolvent;
> it's
> > a cubic.
>
> I found a cubic equation. Its roots are
>
> ab + cd
> ac + bd
> ad + bc
>
> where a,b,c, and d are the roots of the cubic. You can tell this is a cubic
> because its (generalized) sums and products are symmetrical polynomials in
> a,b,c, and d.
>
> Once I solve this cubic, the quartic falls out.
>
> > You can set things up likewise for a quintic, but the resolvent turns out
> > to have _degree 6_, which is to say it just makes the problem harder.
> >
>
> No idea how to do this. I have seen this statement before and don't know
> what it means.
The key words are ``Lagrange resolvents''.
There is a moderately accessible account of what Lagrange was up to in Kline's
Mathematical Thought from Ancient to Modern Times; chapter on Algebra in the
Eighteenth Century, section on the Theory of Equations, esp pages 604, 605.
B.
==============================================================================
From: Robin Chapman
Newsgroups: sci.math
Subject: Re: Roots of the fifth degree
Date: Wed, 06 Jan 1999 09:41:57 GMT
In article <01be38f7$734a9da0$0100a8c0@mgreen>,
"Martin Green" wrote:
> Suppose the roots of a third degree equation are a,b, and c.
>
> I can find combinations of these terms that are roots of a second-degree
> equation: they are
>
> a^2b + b^2c + c^2a , ab^2 + bc^2 + ca^2
>
> This helps me solve the third degree equation.
>
> Suppose the roots of a fourth degree equation are a,b,c, and d.
>
> I can find combinations of these terms that are roots of a third-degree
> equation: they are
>
> ab+cd, ac+bd, ad+bc
>
> This helps me solve the fourth-degree equation.
>
> Suppose the roots of a fifth degree equation are a,b,c,d, and e.
>
> I have looked for combinations of these roots that are solutions of a
> sixth-degree equation, because I know that the "resolvent" is of sixth
> degree. I can't find any such combinations. Does anyone know what they are?
>
One way to attack such a problem is to use a little group theory. There's
a sixth degree resolvent, since the symmetric group of all permutations of
{a,b,c,d,e} has a subgroup H of index 6. Since this symmetric group has order
5! = 120, H has order 120/6 = 20. This group is generated by the permutations
(a b c d e) and (b c e d). (Relabelling the variables as a_1, ..., a_5 the
first of these adds one to the subscript modulo 5, while the second doubles
the subscript modulo 5).
Now one takes the term a^4 b^3 c^2 d which has the property that any
non-trivial permutation alters it, and operates on it with the elements of H
and take the sum of the 20 terms that arise. One gets (I think)
a^4 b^3 c^2 d + b^4 c^3 d^2 e + c^4 d^3 e^2 a + d^4 e^3 a^2 b + e^4 a^3 b^2 c
+ a^4 c^3 e^2 b + b^4 d^3 a^2 c + c^4 e^3 b^2 d + d^4 a^3 c^2 e + e^4 b^3 d^2
a + a^4 e^3 d^2 c + b^4 a^3 e^2 d + c^4 b^3 a^2 e + d^4 c^3 b^2 a + e^4 d^3
c^2 b + a^4 d^3 b^2 e + b^4 e^3 c^2 a + c^4 a^3 d^2 b + d^4 b^3 e^2 c + e^4
c^3 a^2 d.
Starting with the other terms of this form one gets a total of 6 of these
resolvents.
However this is a fairly mechanical procedure, which always works, but
doesn't usually give the most tractable form.
In this case an alternative resolvent is
a^2 b e + b^2 c a + c^2 d b + d^2 e c + e^2 a d + a^2 c d + b^2 d e + c^2 e a
+ d^2 a b + e^2 b c
(together with 5 similar).
Dummit (Solving soluble quintics, Mathematics of Computation vol 57 pp 387-401
1991) uses this resolvent to calculate explicitly the sixth degree equation
this satisfies. He proves that an irreducible quintic is soluble in radicals
if and only if this resolvent equation has a rational root. He also gives
an algorithm to solve the quintic in radicals whenever this is possible.
Robin Chapman + "They did not have proper
SCHOOL OF MATHEMATICal Sciences - palms at home in Exeter."
University of Exeter, EX4 4QE, UK +
rjc@maths.exeter.ac.uk - Peter Carey,
http://www.maths.ex.ac.uk/~rjc/rjc.html + Oscar and Lucinda, chapter 20
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