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