From: lazard@posso.ibp.fr (Daniel LAZARD) Newsgroups: sci.math.numberthy Subject: Re: Discriminant formula Date: 2 Oct 95 12:59:05 GMT Many thanks for all (12) people who answered to my questions. I have not yet checked all the references I have received. ************************************** As it seems that many people seem interested by the problem of finding subfields, let me set a challenging problem which is a strong test for factorizers over algebraic extensions: Let P40 the following polynomial proposed to me by G. Bjork: p40:= 7057801275457063385893 - 10629520803193709903880*s + 78450078315486832226357*s^2 - 187966742518186234660038*s^3 + 100184208695576521111933*s^4 + 138524234196323700539396*s^5 - 173138564231129651232857*s^6 - 2712968414028461405418*s^7 + 91700548881113811658426*s^8 - 32675347429924388587607*s^9 - 22653473046721498313233*s^10 + 16999989172236574721320*s^11 + 1440918665088174301840*s^12 - 4418058564958867582535*s^13 + 769494645028196970860*s^14 + 646725310260982947946*s^15 - 265283934931055392917*s^16 - 41867491513233593254*s^17 + 42102543263365146783*s^18 - 2357344486297791023*s^19 - 3856005338134460577*s^20 + 745141821965932237*s^21 + 200736719645454859*s^22 - 71752452562944311*s^23 - 4414713412013052*s^24 + 3954308844337770*s^25 - 122973006802033*s^26 - 136057114420269*s^27 + 12408126952305*s^28 + 2860211279898*s^29 - 412637940242*s^30 - 33340410389*s^31 + 7272303005*s^32 + 158288849*s^33 - 70822744*s^34 + 319263*s^35 + 368411*s^36 - 4438*s^37 - 972*s^38 + 8*s^39 + s^40; The field generated by a root of p40 has 2 subfields; one is generated by 2cos(2Pi/11) = RootOf(x^5+x^4-4x^3-3*x^2+3*x+1); the second subfield is of degree 8, has a symmetrical Galois group which becomes alternate over Q(sqrt(-11)). Despite all this information, I have not yet been able to compute an equation for this field of degree 8. For getting such an equation, the simplest way (for me) would be to find an automorphism of order 5 of the field of degree 40 and to compute the minimal polynomial of an invariant under this automorphism. One way for doing this would be to factorize p40 over the field defined by itself; this would give 5 linear factors and five factors of degree 7, the former giving the 5 automorphism of the field. An easyer way for finding the automorphism would be to do the following (in Maple syntax) alias (ru=RootOf(x^5+x^4-4x^3-3*x^2+3*x+1)): p8:=23*x^8-14*ru^4*x^7+27*ru^8*x^7-14*ru^3*x^7-35*ru^7*x^7-73*ru^6*x^7 -64*ru^5*x^7+2*ru^9*x^7-30*x^7-15*ru^2*x^7+10*ru*x^7-8*ru*x^6-94*x^6 +126*ru^9*x^6+35*ru^7*x^6-110*ru^6*x^6+ru^2*x^6-ru^5*x^6+88*ru^3*x^6 -54*ru^8*x^6+67*ru^4*x^6+134*ru^4*x^5+146*ru^9*x^5+5*ru^2*x^5+4*ru*x^5 -25*x^5-99*ru^8*x^5+57*ru^7*x^5+38*ru^3*x^5+23*ru^5*x^5+24*ru^6*x^5-46*x^4 -63*ru^3*x^4+19*ru^5*x^4+75*ru^9*x^4-45*ru^2*x^4+2*ru^4*x^4-8*ru*x^4 +110*ru^7*x^4-4*ru^6*x^4-14*ru^8*x^4-131*x^3-247*ru^6*x^3-58*ru^4*x^3 -132*ru^2*x^3-131*ru^3*x^3-155*ru^8*x^3-40*ru^7*x^3-52*ru^9*x^3-63*ru^5*x^3 -81*ru*x^3+13*ru^5*x^2-73*ru*x^2+18*ru^4*x^2-138*ru^6*x^2-201*ru^8*x^2 -52*ru^7*x^2-105*ru^2*x^2-14*x^2-124*ru^3*x^2+37*ru^5*x-21*ru^3*x-18*x+ru^2*x -10*ru*x-32*ru^7*x+34*ru^4*x-46*ru^8*x+47*ru^6*x-22*ru^9*x-3+7*ru^5+3*ru^6 +4*ru^7+3*ru^8-5*ru^4-2*ru-ru^9+8*ru^2+20*ru^3: ### p8 is one of the factors of factor(p40,ru) ### factor(p40,a) needs more than one CPU hour on a DEC alpha station ### running Maple p8b:= evala(subs(ru=ru^2-2,p8)): ### p8b is another of the factors of factor(p40,ru) alias(r1=RootOf(p8)): factor(p8b,p8) This last factorization should lead to a linear factor and a factor of degree 7; only the linear factor is of interest. The challenge consists in computing this linear factor. In other words, are the available factorizer able to compute this factorization (or this linear factor) in a few days of CPU? Another, more mathematical question is : Is there a better way for finding this polynomial of degree 8 over the integers? Sincerely -- Daniel Lazard ---------------------------------------------------------------------------- LITP/IBP, Universite' Paris VI, case 168, 4 pl. Jussieu, 75252 Paris Cedex 05 Tel:(33)1-44 27 62 40 Fax:(33)1-44 27 40 42 E-mail: lazard@posso.ibp.fr ---------------------------------------------------------------------------- ============================================================================== From: victor@ccr-p.ida.org (Victor S. Miller) Newsgroups: sci.math.numberthy Subject: Lazard's subfield problem Date: 3 Oct 95 19:07:18 GMT Prof. Lazard's query generated a number of responses. Here they are: -------------- Date: Tue, 3 Oct 1995 04:02:20 +0100 (MET) From: Mario Daberkow Subject: Re: Discriminant formula In-Reply-To: <199509280933.KAA18635@sysal.ibp.fr> Message-Id: Dear Mr Lazard, I tried to compute the subfield of degree 8 of the number field given by the polynomial p40 you mentioned by using Kant V4. I could finish the computation within 45 minutes on a HP 735 workstation. KANT is public domain and you can get a precompiled binary from our ftp-server ftp.math.tu-berlin.de:/pub/algebra/Kant/Kash/Binary. A documentation is available under /pub/algebra/Kant/Kash/Doc. The following Kash (KAnt -- SHell) session will give you the result: kash> s := Poly (Zx,[1,0]); kash> p40:= 7057801275457063385893 - 10629520803193709903880*s + 70822744*s^34 + 158288849*s^33 + 7272303005*s^32 - 33340410389*s^31 - 412637940242*s^30 + 2860211279898*s^29 + 12408126952305*s^28 - 136057114420269*s^27 - 122973006802033*s^26 + 3954308844337770*s^25 - 4414713412013052*s^24 -71752452562944311*s^23 + 200736719645454859*s^22 + 745141821965932237*s^21 - 3856005338134460577*s^20 - 2357344486297791023*s^19 + 42102543263365146783*s^18 - 41867491513233593254*s^17 - 265283934931055392917*s^16 + 646725310260982947946*s^15 + 769494645028196970860*s^14 - 4418058564958867582535*s^13 + 1440918665088174301840*s^12 + 16999989172236574721320*s^11 - 22653473046721498313233*s^10 - 32675347429924388587607*s^9 + 91700548881113811658426*s^8 - 2712968414028461405418*s^7 - 173138564231129651232857*s^6 + 138524234196323700539396*s^5 + 100184208695576521111933*s^4 - 187966742518186234660038*s^3 + 78450078315486832226357*s^2 - 10629520803193709903880*s + 7057801275457063385893; kash> k := Order (p40); kash> L := OrderSubfield (k, 8); [ Generating polynomial: x^8 - 36688*x^7 - 353574160*x^6 - 189512351885*x^5 +\ 945725045119287*x^4 - 732150450978572042*x^3 + 182289427558444750649*x^2 - 92\ 4802424343333791126*x + 7057801275457063385893 ] A representation of a root of this polynomial in terms of a root of the polynomial p40 can be obtained by kash> a := EltMove (OrderBasis (L[1])[2], k); I omit this output, since it would occupy several pages. Anyway, the maximal order of the field is given by kash> M := OrderMaximal (L[1]); F[1] | F[2] / / Q F [ 1] Given by transformation matrix F [ 2] x^8 - 36688*x^7 - 353574160*x^6 - 189512351885*x^5 + 9457250451192\ 87*x^4 - 732150450978572042*x^3 + 182289427558444750649*x^2 - 9248024243433337\ 91126*x + 7057801275457063385893 Generating polynomial: x^8 - 36688*x^7 - 353574160*x^6 - 189512351885*x^5 + 9\ 45725045119287*x^4 - 732150450978572042*x^3 + 182289427558444750649*x^2 - 9248\ 02424343333791126*x + 7057801275457063385893 Discriminant: 1589910113 kash> OrderBasis (M); [ 1, [0, 1, 0, 0, 0, 0, 0, 0], [1, 2, 1, 0, 0, 0, 0, 0] / 11, [9, 8, 0, 1, 0, 0, 0, 0] / 11, [1, 4, 6, 4, 1, 0, 0, 0] / 121, [117, 117, 2, 1, 0, 1, 0, 0] / 121, [1200, 1172, 70, 97, 4, 6, 1, 0] / 1331, [261270555020621880777539667499569820495100459315571234460148093733006489068\ 2551370346, 321790256623151395644796371963756028822282612247510042654103493770\ 8701696642662081906, 594653456595084465632249014965176405084143751863705867871\ 00329292370139264434285256, 15197275016128858958169308714853566885447740948183\ 081208634744918219359330229053654, 1646386937897601531012074667123873905824269\ 9139026137710285553673914580916588337496, 787608055808709191566034821094955521\ 6218320851450750212369702612811481307325714788, 117390701918072930159144726246\ 2276274480390150031939121477865833648969071810106139, 1] / 3382114555309643449\ 231915245064656064883895944018657487675927266669814937727505070105 ] I hope this can help you. Best wishes, Mario Daberkow ------------------------------------------------------- Mario Daberkow daberkow@math.tu-berlin.de Technische Universit"at Berlin Fachbereich 3 Mathematik Sekretariat Ma 8-1 Stra"se des 17. Juni 136 D-10623 Berlin ---------------------------------------------------------------------------- Subject: Subfields of number field of degree 40 Date: Tue, 03 Oct 1995 12:59:03 -0400 From: Juergen Klueners Message-ID: <9510031659.aa28271@manitou.cs.concordia.ca> My name is Juergen Klueners and I am a student of Prof. Pohst at the Technical University of Berlin. At the moment I am a guest at Concordia University in Montreal. In my "Diplomarbeit" I developed an algorithm for computing subfields. I generalized and improved the methods of Dixons algorithm. Prof. McKay forwarded me the following email (a question of D. Lazard): [omitted: Ed.] With my algorithm I calculated 4 subfields. (Degree 4,5,8,20). Sincerely Juergen Klueners (klueners@cs.concordia.ca) (klueners@math.tu-berlin.de) ================================================================================= ================================================================================= The field of degree 40 is created by f in Z[x]. (polynomial p40 in your notation) The subfield of degree 8 is generated by g(x) = x^8 - 36688*x^7 - 353574160*x^6 - 189512351885*x^5 + 945725045119287*x^4 - 732150450978572042*x^3 + 182289427558444750649*x^2 - 924802424343333791126*x + 7057801275457063385893 The embedding polynomial h in Q[x] has the form h(x) = 1/den * sum_{i=0}^39 a_i x^i and the a_i and den are defined by [a_0,a_1,a_2,...,a39] / den Embedding polynomial means: h(alpha) = beta with alpha zero of f and beta zero of g. Or you can say: f | g(h) . The maximal order of the subfield of degree 8 has discriminant 1589910113. All calculations (for subfield of degree 8) are done within one hour. (1 min for the calculation of g, 49 min for h and 10 min for the maximal order of the subfield.) There are two more subfields: x^4 + 122111359*x^3 + 441284550591715*x^2 - 2868510407839796790*x + 7057801275457063385893 and x^20 + 316*x^19 + 20497*x^18 - 1220315*x^17 - 120719738*x^16 - 858476459*x^15 + 107284763042*x^14 + 2364093617871*x^13 - 11436266583320*x^12 - 804111496163231*x^11 - 5293728100770241*x^10 + 80084901023882949*x^9 + 1181007213585185691*x^8 + 697727152722365805*x^7 - 69617731616030332373*x^6 - 389128311209373617964*x^5 + 300480503268741001248*x^4 + 8522840184286372964897*x^3 + 24921107246172262623162*x^2 + 22585672849369149382110*x + 7057801275457063385893 (The embeddings are calculated, too.) Now the embedding of the degree 8 polynomial: [-68970272660963083149179981072698547819102576517027555825736346087875625815253826492417723372131412811876378384548927324957675707617835863769300844151085643406724625106008051309493392330451610880517762942153178159272716466944854991344221682099042342804329083854652259037980263594955204505696721211426802530781436582370476697265402733229, 73359231451023968696130133127625125865572726676628933813551707079935591661500909856214350238451981252997697321926250007202555761445904469845590471840030721443484997794592811705954306109915484549187226837556246199905793573627765456376223507290924133951484606124278103536897196362647006861002706992733834356299264176196700816577760062211, 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------------------------------------------------------------------------------- Date: Tue, 3 Oct 1995 12:36:16 CDT From: "A.O.L.Atkin 312-413-2155" Subject: A problem of Lazard This relates to the problem posed by Lazard and (indirectly)Bjork. A polynomial p40 of degree 40 is given, & it is desired to find some defining equation for a subfield of degree 8. It appears to me that 0= x^8+ -x^7 * 36688 - x^6 * 3535 74160 - x^5* 18 95123 51885 + x^4* 94572 50451 19287 - x^3 * 732 15045 09785 72042 +x^2 * 1 82289 42755 84447 50649 -x * 9 24802 42434 33337 91126 + 70 57801 27545 70633 85893 is such an equation, though I do not claim to have formally proved this. There are some inconsistencies between my calculations and the data provided by Professor Lazard. The Galois group of the equation is not in A8,agreed; but I need to adjoin sqrt(113) and not sqrt(-11) to bring it in. Since the discriminant of the original p40 is 113*square, this seems reasonable. Another point is that Professor Lazard's p8 (" one CPU hour on a DEC alpha running Maple") contains powers of ru between the 5th and 9th, whereas ru is a root of a quintic equation. The analogous equation that I obtain with powers of ru running from 0 to 4 is monic, and only takes a few seconds on an IBM3090. ============================================================================== From: lazard@posso.ibp.fr (Daniel LAZARD) Newsgroups: sci.math.numberthy Subject: Re: Lazard's subfield problem Date: 5 Oct 95 17:26:57 GMT I was probably very tired when I sent my subfield problem : I choose twice a wrong polynomial for copying in the mail: In fact, G. Bjork submitted to me two polynomials of degree 40. The comments (Galois structure) were related to the second one and I have included the first one, for which the subfield of degree 8 has a Galois group of order 384 and for which I already knew generators for the subfields. Nevertheless, I am very impressed by KANT which finds (much easier than I did) an equation for the subfield of degree 8. After simplification using function POLRED of PARI, this first field is generated by a=RootOf(y^4 - y^3 - 2*y^2 - 3*y - 2), b=RootOf(x^2+x-(a+1)(a^2+a+1)) and c=RootOf(z^5+z^4-4*z^3-3*z^2+3*z+1), which makes the Galois structure almost evident and shows that subfields computation is a fundamental tool for simplification of algebraic extensions. I include below the second polynomial as well as the correct (I guess) factor of the factorization over the field of degree 5, because my first question remains open : which is the best available factorizer for polynomials over algebraic extensions? This second polynomial of degree 40 is not monic, but it becomes monic after the substitution s->s/23, which increases the size of the coefficients. KANT again gives a polynomial of degree 8 after 2h30mn over a Sparc 10 station; after the substitution x->23^2*x, this polynomial becomes x^8 - 23469647*x^7 + 125475322630851*x^6 + 125545369937650899017*x^5 + 267869229964383551150375216*x^4 - 19913826023084512839062976417172*x^3 + 46093022375101489868037396982008857*x^2 + 9332479171418993037251282814576955874311*x + 60557702471047014568262217289549013406860303 The simplification with polred is yet running. The second polynomial is polforty2:= 811597135529898169 + 7152980982346226040*s + 18171476841490003710*s^2 - 11355446119881189384*s^3 - 117712533422275973284*s^4 - 112015867904431532196*s^5 + 230165980213575319883*s^6 + 459451216519714656526*s^7 - 83270519500276293878*s^8 - 734169416313703303879*s^9 - 309701724291250465911*s^10 + 578377903845523688438*s^11 + 514466502292905094369*s^12 - 189737074857945494650*s^13 - 366983017878149748835*s^14 - 36220213376169001613*s^15 + 135619533598051236796*s^16 + 55227774448506262996*s^17 - 20250691917531161954*s^18 - 19453432571061340687*s^19 - 3044251267070794660*s^20 + 2300821073721698022*s^21 + 1781903363906052715*s^22 + 392882585322815188*s^23 - 254426897796933790*s^24 - 173336635271317655*s^25 - 2093649224751164*s^26 + 25106824391937532*s^27 + 4914461478555900*s^28 - 1793570319828018*s^29 - 664273174842926*s^30 + 58086065686110*s^31 + 47966892655022*s^32 + 581668312493*s^33 - 2146765884442*s^34 - 114609011287*s^35 + 59330876659*s^36 + 3408895573*s^37 - 905589810*s^38 - 34700284*s^39 + 6436343*s^40; One factor of its factorization over the field generated by c=RootOf(z^5+z^4-4*z^3-3*z^2+3*z+1) if (if I am not wrong this time) p8:=529*s^8+69*s^7*c^4-3381*s^7*c^3-1633*s^7*c^2+9614*s^7*c+1242*s^7 -6134*s^6*c^4-6841*s^6*c^3+22027*s^6*c^2+19017*s^6*c-9810*s^6 -22140*s^5*c^4+34980*s^5*c^3+88081*s^5*c^2-121202*s^5*c-27047*s^5 +18243*s^4*c^4+86586*s^4*c^3-62127*s^4*c^2-262406*s^4*c+65645*s^4 +125329*s^3*c^4-54434*s^3*c^3-476976*s^3*c^2+263201*s^3*c+242295*s^3 +72000*s^2*c^4-182194*s^2*c^3-283439*s^2*c^2+628581*s^2*c+72607*s^2 -92936*s*c^4+11053*s*c^3+369166*s*c^2-89313*s*c-270055*s -67946*c^4+65720*c^3+270438*c^2-239155*c-153229 My factorization problem is to find the (existing and unique, if the factor is correctly copied) linear factor of subs(c=c^2-2,p8) over the field generated by c and RootOf(p8). Many thanks again for these numerous answers -- Daniel Lazard ---------------------------------------------------------------------------- LITP/IBP, Universite' Paris VI, case 168, 4 pl. Jussieu, 75252 Paris Cedex 05 Tel:(33)1-44 27 62 40 Fax:(33)1-44 27 40 42 E-mail: lazard@posso.ibp.fr ----------------------------------------------------------------------------