From: Leslie Wright
Subject: Re: ALgebraic solutions
Date: Mon, 22 Nov 1999 23:04:03 GMT
Newsgroups: sci.math.symbolic
Keywords: Which quintics _can_ be solved by radicals?
Greetings all--
I don't have an answer to Dr. Basti's question, but it is timely, since I am
presently wrestling with understanding the proof of the Abel Impossibility
Theorem (as presented by Heinrich Dorrie in 100 Great Problems of Elementary
Mathematics, trans. David Antin, Dover Pubs, 1965, at p. 116). I, too, have
discovered that Maple will not provide a general solution for equations
above degree four--precisely normal behavior, given what we know about
algebra. I would like to know as well if one can use Maple to determine if a
specific quintic (or higher degree) equation can be solved algebraically,
even if the general case cannot. For some reason, I believe that Sturm's
"number of roots" theorem has something to do with it, since the above
mentioned work uses this to conclude that the quintic x^5 - a*x - b = 0 is
algebraically insoluble when (4^4)*(a^5) > (5^5)*(b^4). Similarly, the
seventh degree eqn. x^7-a*x-b = 0 is algebraically insoluble when
(6^6)*(a^7) > (7^7)*(b^6).
If any of this makes sense to algebra experts out there, I would love to
know, since I find this whole problem of algebraic solubility fascinating,
even though I don't completely comprehend it.
Les
Hoskins+ wrote:
> Dear Sir/Madam:
>
> I have a question which I hope the Maple group can answer.
>
> We know that the polynomial equations with degree greater than 4, in
> general do not have algebraic solutions.(i.e. solution can not be
> extracted by radicals or four operations (+,-,* and division).
>
> I believe that there is a procedure with groups that maple can calculate
> such that, we know whether the solution is algebraic or not (although we
> may not be able to solve it ).
==============================================================================
From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: ALgebraic solutions
Date: 23 Nov 1999 06:05:20 GMT
Newsgroups: sci.math.symbolic
>Hoskins+ wrote:
...nothing. There was a message posted by someone else named Basti but with
this person's name in the "From:" field. The poster asked,
> We know that the polynomial equations with degree greater than 4, in
> general do not have algebraic solutions.(i.e. solution can not be
> extracted by radicals or four operations (+,-,* and division).
It's a little incorrect to use "algebraic solution" to mean "solution
by radicals", since e.g. the roots of x^5+x+3=0 are algebraic numbers,
even if they cannot be expressed using only radicals and rational numbers.
> I believe that there is a procedure with groups that maple can calculate
> such that, we know whether the solution is algebraic or not (although we
> may not be able to solve it ).
Well, Maple has enough information to do this for polynomials up to
degree 7, I believe. What you need is to know whether or not the
Galois group of a polynomial is a solvable group. AFAIK there is no way
to do this short of actually determining the Galois group. (Well, one
may stop the computation when the group is shown to contain an nonsolvable
subgroup or to be contained in a solvable group, I suppose.)
You need only type in, e.g. galois(x^5+x+3); if you are willing to
trust to a black box.
In article <3839CB95.39B12D48@yahoo.com>,
Leslie Wright wrote:
>I don't have an answer to Dr. Basti's question, but it is timely, since I am
>presently wrestling with understanding the proof of the Abel Impossibility
>Theorem (as presented by Heinrich Dorrie in 100 Great Problems of Elementary
>Mathematics, trans. David Antin, Dover Pubs, 1965, at p. 116).
>[...] I believe that Sturm's
>"number of roots" theorem has something to do with it, since the above
>mentioned work uses this to conclude that the quintic x^5 - a*x - b = 0 is
>algebraically insoluble when (4^4)*(a^5) > (5^5)*(b^4).
I hadn't seen this before. I don't think Sturm's theorem is really the
right answer, although it may be sufficient to proving nonsolvability in
certain special cases.
The "right" answer is in Dummit's article, "Solving Solvable Quintics",
Math Comp 57 (1991) 387-401. He presents an explicit sextic polynomial whose
coefficients may be derived from the coefficients of the original
quintic, with the feature that the quintic is solvable in radicals iff
the sextic has a rational root. In the special case of a quintic of the
form x^5 + a x + b the sextic may be written easily: it's
6 5 2 4 3 3 4 2 5 4
x + 8 a x + 40 a x + 160 a x + 400 a x + (512 a - 3125 b ) x
5 4
+ (256 a - 9375 b ) a
Defining epsilon := (5^5*b^4)/(4^4*a^5), we see the sextic is expressible
in terms of x, a, and epsilon (for non-zero a). We may scale this
polynomial (i.e. set x = a y) and see that solvability depends only
on whether or not the sextic (y^2+16)*(y+2)^4-256*epsilon*(3+y)
has a rational root.
Well, the rational number y = y0 solves this equation iff
2 4
(y0 + 16) (y0 + 2)
epsilon = 1/256 ------------------
3 + y0
With a little calculus you can show the right-hand side of this equation
never attains any value between -1 and 0, so if epsilon is in this
range, then there can be no rational root y0 to the sextic, and hence
the quintic is not solvable by radicals. This is equivalent to the
proposition Wright quotes.
On the other hand, Sturm-like techniques are used to decide how many
_real_ roots there are in an interval, and these answers will describe
open subsets of the coefficient space, while clearly the sextic
above does not have _rational_ roots on any open set of epsilon's.
So that's not really the way to go.
For those who are keeping score: it follows from all this that we can
explicitly describe all solvable quintics of the form x^5 + a x + b:
After a stream of invertible substitutions we find that, up to a
scaling of x and multiplication by an overall constant, all such
quintics are of the form
2 5
(d + 3) (d + 2) + 5 (d + 3) x + (16 + d ) x
for some rational d, and all these polynomials (when irreducible) have
Galois group contained in the Frobenius group of order 20, and hence
are solvable.
For Galois theory see e.g. http://www.math-atlas.org/index/12FXX.html
dave
==============================================================================
From: Leslie Wright
Subject: Re: ALgebraic solutions
Date: Tue, 23 Nov 1999 12:47:43 GMT
Newsgroups: sci.math.symbolic
Dave Rusin wrote:
> In article <3839CB95.39B12D48@yahoo.com>,
> Leslie Wright wrote:
>
> >I don't have an answer to Dr. Basti's question, but it is timely, since I am
> >presently wrestling with understanding the proof of the Abel Impossibility
> >Theorem (as presented by Heinrich Dorrie in 100 Great Problems of Elementary
> >Mathematics, trans. David Antin, Dover Pubs, 1965, at p. 116).
> >[...] I believe that Sturm's
> >"number of roots" theorem has something to do with it, since the above
> >mentioned work uses this to conclude that the quintic x^5 - a*x - b = 0 is
> >algebraically insoluble when (4^4)*(a^5) > (5^5)*(b^4).
>
> I hadn't seen this before. I don't think Sturm's theorem is really the
> right answer, although it may be sufficient to proving nonsolvability in
> certain special cases.
I looked at my reference (referred to above) more closely. The above statement I
made is incomplete. I should also have mentioned that a and b need to be integers
and that if a and b are divisible by some prime p b is not divisible by p^2. One
needs to combine Sturm's Theorem (to determine the # or real roots), Kroenecker's
Theorem (stating that an algebraically soluble equation of odd prime degree has
one real roots or all real roots), and Schoenemann's Theorem to determine if an
equation with integral coefficients is irreducible over the so-called natural
rationality domain (it apparently is if a and b are divisible by some prime p
while b is not divisible by p^2).
Still wrestling with the issue--including the distinction between algebraic
solubility and irreducibility over a particular number field. They evidently are
not the same.
Les
==============================================================================
[Addendum
As noted above, all solvable quintics of the form x^5 + a x + b
may be rewritten, after a scaling of x and multiplication by an
overall constant, in the form
2 5
(d + 3) (d + 2) + 5 (d + 3) x + (16 + d ) x
for some rational d. For generic values of d, the Galois group is
the Frobenius group of order 20.
Computing discriminants, we find that the Galois group is contained
in the dihedral group of order 10 iff 5(d^2+16) is a square. For
almost all d this is not the case, and we may parameterize those
d for which it _is_ true as d = 2(m^2-10m+5)/(m^2-5). Thus a
parameterization of all the polynomials of the form x^5 + a x + b
with Galois group in D_10 is
2 2 2
4 m (m - 5) (m - 4 m - 1) + 5 (m - 4 m - 1) (m - 5) x
2 2 5
+ 4 (m - 2 m + 5) x
That's g:=4*m*(m-5)*(m^2-4*m-1)+5*(m^2-4*m-1)*x*(m^2-5)+4*(m^2-2*m+5)^2*x^5;
I believe the group of these trinomials can never be cyclic, that is, the
Galois group is precisely D_10 if g is irreducible.
--djr]
==============================================================================
Here is a table of irreducible, monic, integral, depressed [0 = coeff of x^4 ],
small [max |coeff| <= 10] quintics with galois group of order smaller than
Sym(5): p(x) = x^5 + add(a[i] * x^(4-i), i=1..4) where a is on this list,
or else -p(-x) for one of these p .
[Missing from these lists is an example with 5 real roots and group F20.
Are there any such polynomials?]
The following have all roots real:
[-10, 5, 10, -3], # Alt 5
[-10, 5, 10, 1], # cyclic
[-9, 4, 10, -4], # dihedral
[-8, 3, 10, -4] # dihedral
No example with group F20 is found.
The following have three real roots:
None found. I guess none exist, since in such a case complex conjugation
is a 2-cycle in the permutation group, and a transitive subgroup of
S5 containing a 2-cycle must be all of S5, I think.
The following have one real root:
1. Galois group is Alt(5):
[-5, 1, 9, -7], [-4, 1, -2, 9], [-3, 1, 1, 3], [-3, 3, 9, 9],
[-3, 4, 6, -3], [-3, 7, -5, 3], [-3, 7, -3, 3], [-3, 7, 6, 3],
[-2, 2, 3, -2], [-2, 2, 5, 6], [-2, 2, 7, -2], [-2, 4, -6, 4],
[-2, 4, 6, 4], [-2, 4, 10, 8], [-2, 6, -1, 2], [-2, 8, 2, 4],
[-2, 9, 10, 5], [-1, 7, 8, 5], [0, 1, -2, 3], [0, 4, 1, -4],
[1, 1, 6, 8], [1, 3, 0, -3], [1, 5, -4, 7], [2, 2, -1, -2],
[2, 4, -3, 4], [2, 4, -2, -4], [2, 4, 6, 4], [2, 5, 2, 1],
[2, 5, 5, -5], [2, 7, 10, -3], [2, 8, -10, 4], [3, 0, 5, -10],
[3, 0, 5, 10], [3, 3, -9, -9], [3, 4, -6, 9], [3, 4, 6, 9],
[3, 5, 3, 3], [3, 5, 6, -9], [3, 5, 10, -5], [3, 6, 0, 9],
[3, 6, 2, 1], [3, 7, -9, 3], [3, 7, 7, -7], [5, 10, 0, -9],
[5, 10, 0, -5], [5, 10, 0, 1], [6, 0, -7, -8], [6, 0, -7, 8],
[6, 6, -4, 8], [6, 6, -1, -2], [6, 7, 6, 3], [6, 10, 7, 2],
[7, 1, -9, -7], [7, 1, 6, 8], [7, 4, -3, 4], [7, 5, 0, 5],
[8, 4, -5, 8], [8, 5, -4, 1], [10, 0, -10, -4],[10, 0, -10, 4],
[10, 5, 0, 1], [10, 6, -9, -6], [10, 6, -1, -2]
2. Galois group is F20:
[-1, 4, 7, -2], [0, 5, -5, 8], [0, 10, -10, 6],[0, 10, 10, -2],
[2, 4, -1, 4], [4, 4, 8, 8], [5, 1, 9, 5], [5, 10, 5, 4],
[6, 4, -9, 4], [8, 2, 2, 2], [10, 0, 0, -5], [10, 0, 0, 5],
[10, 10, 10, 3] and the families
[-5, 0, 5, k] (|k| = 3 .. 10; when |k|=0,1,2 then p is reducible)
[5, 0, 5, k] (|k| = 1 .. 10; when |k|=0 then p is reducible)
[0, 0, 0, k] (|k| = 2 .. 10; when |k|=0,1 then p is reducible)
3. Galois group is D10:
[-10, 10, 5, 2], [-9, 9, 4, 3], [-8, 8, 3, 4], [-7, 7, 2, 5],
[-7, 10, -5, 2], [-6, 1, 10, -4], [-6, 6, 1, 6], [-5, 0, 10, -9],
[-5, 0, 10, -4], [-5, 0, 10, 4], [-5, 0, 10, 9], [-5, 5, 0, 7],
[-5, 7, 8, 5], [-5, 10, -10, 8],[-4, 1, 10, 4], [-4, 4, -1, 8],
[-3, 1, 7, 5], [-3, 2, 6, 3], [-3, 2, 10, 4], [-3, 3, -2, 9],
[-3, 4, 8, -1], [-2, 2, -3, 10], [-2, 3, 10, 4], [-2, 9, -2, 3],
[-1, 2, -2, 1], [-1, 2, 3, -4], [-1, 4, 4, 1], [-1, 4, 10, 4],
[-1, 7, -1, 3], [-1, 10, 8, -9], [0, 3, 2, -1], [0, 5, -4, 1],
[0, 5, 0, 3], [0, 5, 10, 4], [1, 3, 1, 3], [1, 6, 4, 3],
[1, 6, 10, 4], [2, 1, 2, 3], [2, 5, -4, 4], [2, 7, 10, 4],
[2, 8, 9, 8], [3, 1, 3, -3], [3, 6, -1, 4], [3, 8, 6, 9],
[3, 8, 10, 4], [4, 3, 4, -3], [4, 9, 10, 1], [4, 9, 10, 4],
[5, 5, 5, -3], [5, 10, 0, 5], [5, 10, 10, 4], [6, 7, 6, -3],
[7, 3, 5, 5], [7, 5, 5, 1], [7, 5, 9, 5], [7, 9, 7, -3],
[7, 10, -6, 5], [8, 5, 6, 7], [9, 5, 3, 9]
4. Galois group Z5: None exist. (Complex conjugation is an auto.!)
--djr]