From: Igor Schein
Subject: degree-8 polynomial
Date: Fri, 12 Feb 1999 00:18:03 GMT
Newsgroups: sci.math.symbolic
Keywords: Solving a solvable octic polynomial
[ This is a repost of the following article: ]
[ From: Igor Schein ]
[ Subject: degree-8 polynomial ]
[ Newsgroups: sci.math ]
[ Message-ID: ]
Hi,
I'm trying to figure out how to solve polynomial
P8=x^8 - x^7 + 29*x^2 + 29 symbolically. Galois
group of P8 is E(8):7, which is a solvable group
of order 56. Number field defined by a root of P8
doesn't have any subfield of order 4 or 2 ( otherwise
the solution would follow ).
Any ideas?
Thanks
Igor
==============================================================================
From: differential.equation@complex.net (Euclid)
Subject: Re: degree-8 polynomial
Date: Fri, 12 Feb 1999 01:21:33 GMT
Newsgroups: sci.math.symbolic
In some ancient text books on the theory of equations, there is detail
on a method by LaGrange for solving polynomials symbolically. I think
that the author's name was Uspensky but I am not sure.
A group of order 56 is solvable. The lowest order simple group that is
NOT solvable is A5 which has order 60.
Hope this helps.
On Fri, 12 Feb 1999 00:18:03 GMT, Igor Schein
wrote:
[as above -- djr]
==============================================================================
From: mckay@cs.concordia.ca (MCKAY john)
Subject: Roots of solvable x^8-x^7+29*x^2+29, Gal/Q = 2^3:7
Date: 12 Feb 1999 18:33:10 GMT
Newsgroups: sci.math.symbolic
Igor Schein asked about this.
H. Anai et al have a paper on solving solvables.
Use Lagrange resolvents for your degree 8.
There is a description in Knopfmacher's book.
Also there may be something in Klein: Icosahedron.
The main problem is the output size. I suggest a sequence
of transformations is the way to go.
Where does the polynomials arise?
John McKay
--
But leave the wise to wrangle, and with me
the quarrel of the universe let be;
and, in some corner of the hubbub couched,
make game of that which makes as much of thee.
==============================================================================
From: Robin Chapman
Subject: Re: degree-8 polynomial
Date: Fri, 12 Feb 1999 13:44:37 +1100
Newsgroups: sci.math
Igor Schein wrote:
>
> Hi,
>
> I'm trying to figure out how to solve polynomial
> P8=x^8 - x^7 + 29*x^2 + 29 symbolically. Galois
> group of P8 is E(8):7, which is a solvable group
> of order 56. Number field defined by a root of P8
> doesn't have any subfield of order 4 or 2 ( otherwise
> the solution would follow ).
>
For this one first has to solve a seventh-degree equation with cyclic
Galois group, then a bunch of quadratics. Suppose the roots of the
original polynomial are a,...,h. Form a new polynomial with roots
all expressions like (a+b+c+d)(e+f+g+h). This has degree 35, and it willl
factor into a degree 28 and a degree 7 factor. This degree 7 factor
will be an irreducible equation with cyclic Galois group. Now one
has to solve this. One will need to adjoin a 7th root of unity for this,
but, er, the only ways I can see right now to find a solution are a bit
impractical. The thing is there's a Galois resolvent t_0 + t_1 z + .. + t_6 z^6
where t_j are the roots of this septic (no sniggering) and z^7 = 1, whose
7th power lies in Q(z) but there are a lot of ways that the t_j can be
ordered....
Anyway when that`s done we have got 7 values like (a+b+c+d)(e+f+g+h).
Which seven do we get? We can label the roots with the points of
affine 3-space over F_2, and the seven expression we get are the product
of sums of points in complementary afiine planes in this affine space.
As we know a+b+...+h we get a+b+c+d etc. by solving some quadratics.
(Again we need to get the labelling right: the action under the order
7 Galois group will help us, and there are only a finit number of
possibilities.)
Once we have done this twe can easily find a, b, c etc.
--
Robin Chapman + "Going to the chemist in
Department of Mathematics, DICS - Australia can be more
Macquarie University + exciting than going to
NSW 2109, Australia - a nightclub in Wales."
rchapman@mpce.mq.edu.au + Howard Jacobson,
http://www.maths.ex.ac.uk/~rjc/rjc.html - In the Land of Oz
==============================================================================
From: pmontgom@cwi.nl (Peter L. Montgomery)
Subject: Re: degree-8 polynomial
Date: Fri, 12 Feb 1999 14:42:43 GMT
Newsgroups: sci.math
In article <36C39595.69177351@mpce.mq.edu.au>
Robin Chapman writes:
[above article was quoted -- djr]
The discriminant of f is 29^6 * 49109^2, suggesting that the
behavior of f may be related to the primes 29 and 49109.
Experiments show that f usually splits into degree-1 and degree-7
factors. Since the Galois group has 48 elements of order 7 we
expect this to happen 48/56 = 6/7 of the time.
Modulo the primes 17, 41, 59, 157, 173, 191, 307, 331, 347, 349, ...
f splits into four quadratic factors. These seem to be
the primes == +-1 and +-12 (mod 29). These remainders
are the fourth roots of unity (mod 29).
Since the Galois group has order 56, we expect the polynomial
to split completely (into eight distinct linear factors)
modulo one prime in 56. The first primes for which this happens are
1409, 1549, 1607, 1913, 1931, 2029, 3323.
All of these are == +-1 or +-12 (mod 29), but I do not
immediately see how to distinguish them from the primes
which give four quadratic factors.
I have seen similar splitting behavior
(that is, for primes == +-1, +-12 mod 29) when looking at
the degree-7 polynomial with roots
w^1 + w^12 + w^17 + w^28 where w^29 = 1 and w <> 1.
I guessed that the degree-7 polynomial Robin mentions might split
in that field, and my guess was confirmed. A Maple program follows.
P8 := x^8 - x^7 + 29*x^2 + 29;
ifactor(discrim(P8, x));
# If a satisfies apoly (below) then fquad divides P8.
# (a is the cubic coefficinet of the monic polynomial
# -- it might be better to express everything in terms
# of a different coefficient of fquad.)
fquad := 313143573077*x^4+313143573077*a*x^3+(8015720721142+74346432167252*a^2+
826218132770*a^6-45020236888*a^10+4998962098*a^13+29324103525227*a^5-\
3873000433088*a^7+36867534395703*a-928066747997*a^9+32493253637*a^12+
56801009658*a^11+64300096907189*a^4+90736781050327*a^3-3302509904316*a^8)*x^2+
(-42270807772190-202448126330*a^11-331775438820633*a^3-922588865*a^13+
6365311195104*a^8-192884630144114*a^4+13821445963767*a^6-22962460312133*a^5+
647002931620*a^9-249511995597*a^10+18802010964188*a^7-219926349737889*a-\
37927395204*a^12-378887476165464*a^2)*x-34788429764788*a+1450296000758*a^9-\
49538470913*a^12-77111346656*a^11-90008388109961*a^4-119296398523197*a^3-\
84756850435029*a^2-2689728047807*a^6-45887298555716*a^5+4848597393802*a^7+
92820954812*a^10-7936621268*a^13+4863305451880*a^8-4994312310868:
frem := factor(rem(P8, fquad, x)):
apoly := primpart(content(frem, x), a);
# This degree-14 polynomial (in a) divides
# all coefficients (of x) in frem.
# That is, frem vanishes if this vanishes.
vpoly := rem(primpart(apoly, a), a^2 + a - v, a);
# Reduce degree to 7, in v
# Given v, this is a quadratic for a.
wpoly := normal((w^29 - 1)/(w - 1)); # 29-th roots of unity
v := -2*(w + w^12 + w^17 + w^28) - (w^2 + w^5 + w^24 + w^27)
+ (w^3 + w^7 + w^22 + w^26) + (w^4 + w^10 + w^19 + w^25)
- (w^8 + w^9 + w^20 + w^21) + 2*(w^11 + w^13 + w^16 + w^18);
rem(vpoly, wpoly, w); # Zero, proving v ia a root of vpoly
# if w is a 29-th root of unity.
# For other roots, replace w by a power of w.
;quit;
--
Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California
Microsoft Research and CWI
If Clinton and Lewinsky merely partied together, why the fuss?
If they went further, where's the baby?
==============================================================================
From: Robin Chapman
Subject: Re: degree-8 polynomial
Date: Mon, 15 Feb 1999 09:49:49 +1100
Newsgroups: sci.math
Peter L. Montgomery wrote:
>
> >Igor Schein wrote:
> >>
> >> Hi,
> >>
> >> I'm trying to figure out how to solve polynomial
> >> P8=x^8 - x^7 + 29*x^2 + 29 symbolically. Galois
> >> group of P8 is E(8):7, which is a solvable group
> >> of order 56. Number field defined by a root of P8
> >> doesn't have any subfield of order 4 or 2 ( otherwise
> >> the solution would follow ).
>
> The discriminant of f is 29^6 * 49109^2, suggesting that the
> behavior of f may be related to the primes 29 and 49109.
This is already enough to identify the septic subfield. By the
Kronecker-Weber theorem this is contained in a cyclotomic field
Q(zeta_n). The prime factors of n must divide the discriminant, but
49108 can't arise as its not = 1 (mod 7). Thus the septic field is
the unique septic subfield of Q(zeta_29).
> Experiments show that f usually splits into degree-1 and degree-7
> factors. Since the Galois group has 48 elements of order 7 we
> expect this to happen 48/56 = 6/7 of the time.
> Modulo the primes 17, 41, 59, 157, 173, 191, 307, 331, 347, 349, ...
> f splits into four quadratic factors. These seem to be
> the primes == +-1 and +-12 (mod 29). These remainders
> are the fourth roots of unity (mod 29).
>
> Since the Galois group has order 56, we expect the polynomial
> to split completely (into eight distinct linear factors)
> modulo one prime in 56. The first primes for which this happens are
> 1409, 1549, 1607, 1913, 1931, 2029, 3323.
> All of these are == +-1 or +-12 (mod 29), but I do not
> immediately see how to distinguish them from the primes
> which give four quadratic factors.
As the Galois group is non-abelian, one wouldn't expect an easy
characterization of these primes.
> I have seen similar splitting behavior
> (that is, for primes == +-1, +-12 mod 29) when looking at
> the degree-7 polynomial with roots
> w^1 + w^12 + w^17 + w^28 where w^29 = 1 and w <> 1.
> I guessed that the degree-7 polynomial Robin mentions might split
> in that field, and my guess was confirmed. A Maple program follows.
--
Robin Chapman + "Going to the chemist in
Department of Mathematics, DICS - Australia can be more
Macquarie University + exciting than going to
NSW 2109, Australia - a nightclub in Wales."
rchapman@mpce.mq.edu.au + Howard Jacobson,
http://www.maths.ex.ac.uk/~rjc/rjc.html - In the Land of Oz