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Trinomials with interesting Galois groupsTrinomials axn+bx+cwith interesting Galois groupsIn 1969 Trinks discovered that the irreducible trinomialx7 - 7 x + 3,factored modulo small primes (other than the primes 3 and 7dividing its discriminant 218), always yieldedpolynomials of degrees 7, 4+2+1, 3+3+1, or 2+2+1+1+1.[The pattern 1+1+1+1+1+1+1, for complete factorizationinto linear polynomials, occurs too, but not until we reach1879, the 289th prime.] This suggested thatthe trinomial had Galois group G168,the simple group of order 168 consisting of the invertible3-by-3 matrices mod 2, acting on the 7 nonzero vectors in(Z/2Z)3(that is, on the point of the finite projective planeof order 2). Matzat then proved that the Galois groupwas in fact G168. Of course every trinomialequivalent to x7-7x+3(i.e., proportional to (mx)7-7mx+3for some nonzero m) has the same Galois group.Ten years later, Erbach, Fischer and McKay [EFM] published the trinomialx7 - 154 x + 99,not equivalent with the Trinks-Matzat trinomial, and showedthat it too has Galois group G168. SeeIn 1999 I analyzed the general problem of trinomialsax7+bx+cwith Galois group contained in G168.The analysis led me to the new examples372 x7 - 28 x + 9, 4992 x7 - 23956 x + 34 113,and to the conjecture that every G168-trinomialis equivalent to either Trinks-Matzat, Erbach-Fischer-McKay,or one of the two new trinomials. This was proved in 2001by Nils Bruin. (A copy of the input files of his computationsis here.)In fact, I showed:Theorem (NDE 1999): Trinomialsax7 + bx + cover a field K of characteristic zerowhose Galois group is contained in G168are parametrized by the curveC168:y2 = x(81 x5 + 396 x4+ 738 x3 + 660 x2+ 269 x + 48).A search for rational points on this curve finds seven,with x=0, -3, 1/9, and Infinity.Of these, the two points above x=-3,and one of the points with x=1/9,yield degenerate trinomials.The Trinks-Matzat trinomial comes from the Weierstrass point x=0;one of the two points at infinityyields the Erbach-Fischer-McKay trinomial.The two new examples come from the other point at infinityand the nondegenerate point with x=1/9.Now C168 is a curve of genus 2.As is well known, Mordell conjectured, and Faltings proved [F1,F2],that every algebraic curve of genus at least 2has only finitely many points over any number field.In our case, that means that there areonly finitely many equivalence classes ofG168-trinomials over any number field.But it can be much harder to list all casesthan to prove that the list is finite.None of the known proofs of the Mordell conjectureprovides an effective procedure for finding all the rational pointsand proving that the list is complete, even in the first caseof a genus-2 curve over Q. In recent years,much effort has gone into solving this problem in practice,and ever more curves are yielding to the combination oftheoretical insight and computational power; see for instance [B].Applying these tools to C168,Nils Bruin proved:Theorem (N.Bruin, 2001):There are no more rational points on this curve.Therefore, every trinomialax7 + bx + cover Q whose Galois group is contained inG168 is equivalent to one of the four trinomialsdisplayed above.Similar analysis applies to trinomialsaxn+bxk+c(n,k coprime) of arbitrary degree nwhose Galois group is contained in some given subgroup Gof the symmetric group on n letters.Several cases of n<7 have appeared in the literature,all with k=1. For instance, take n=5,and let G be the 20-element subgroup of S5(a.k.a. the semidirect product of Z/4Zwith Z/5Z,or the ax+b group mod 5).Then the general trinomial with Galois group contained in Gis equivalent to(4u2+16) x5+ (5u2-5) x+ (4u2+10u+6)(attributed by Matzat [M2, pages 90-91 (Satz 3)] to Weber[W, section 189]). Moreover, the Galois group is containedin the 10-element dihedral group if and only ifu=t-1/t, when the quinticis equivalent to(2t2+2)2 x5+ 5(t4-3t2+1) x+ (4t4+10t3-2t2-10t+4);this is attributed by Matzat [M2, p.93 (Satz 4)] to [JRYZ].Likewise, if n=6, we can ask that G bethe transitive 120- or 60-element subgroup of S6.(These are the images of the obvious subgroups S5and A5 under an outer automorphism ofS6; they can also be obtained asPGL2(Z/5Z) andPSL2(Z/5Z)acting on the six points of the projective line overZ/5Z.)Here the general such sextic is(125-u) x6+ 12u(u+3)2 x+ u(u-5)(u+3)2for the 120-element group; for the 60-element group,take u=t2. Matzat reportsthis on page 94 of [M2], and attributes it, together withthe formulas for sextic trinomials for several other(imprimitive but) transitive subgroups of S6,to Malle [M1]. (Matzat also mentions the genus-2 curve we callC168 on page 95, but does not exhibit equationsor points on this curve, and gives its genus incorrectly as 3.)The first possibility with k>1 is n=5 and k=2. We found(but seeaddendum below) :Theorem (NDE 1999): Trinomialsax5 + bx2 + cover a field K of characteristic zerowhose Galois group is contained in the 20-element subgroupof S5 are parametrized by the elliptic curveC20:y2 + xy + y = x3 + x2 + 35 x - 28.This is an elliptic curve of conductor 15,labelled 15F in Tingley's ``Antwerp'' tables and 15-A4 by Cremona.Over Q, this curve has rank zero,so that there are only finitely many equivalence classesof trinomials ax5+bx2+cwith solvable Galois groups. More specifically,C20(Q) is cyclic of order 8;three of its points (the origin and the generators (2,6) and (32,171))yield degenerate quintics, three produce trinomials with dihedralGalois group, and the remaining two(corresponding to the other two points not divisible by 2in C20(Q))produce quintics with 20-element Galois group:5 x5 - 10 x2 - 1, x5 - 100 x2 - 1000.The three dihedral quintics arex5 - 5 x2 - 3, x5 - 5 x2 + 15, x5 - 25 x2 - 300;curiously the first two of these generate the same field! Added 8/2002: Blair K. Spearman and Kenneth S. Williams draw my attention to their paper [SW], which obtains all the solvable quintic trinomials using the sextic resolvent of a quintic. For trinomials X5+aX+b, their work is anticipated by [W] and [JRYZ]. For X5+aX2+b, they in turn anticipated my 1999 computation. There are a few minor differences: besides the alternative approach to determining the curve C20, Spearman and Williams compute C20(Q) themselves by carrying out a 2-descent rather than citing previous work on the arithmetic of that curve; and they give explicit solutions by radicals of each of the five quintics, but do not apparently notice that two of them generate the same field. For our final example, take n=8 and k=1,and let G be the group G1344of affine linear transformations of(Z/2Z)3.[This is the semidirect product of G168 with(Z/2Z)3;equivalently, G1344 is the group of invertible4-by-4 matrices over Z/2Zwhose bottom row is 0001, acting on the 8 column vectorsof height 4 with last coordinate 1.]We then have:Theorem (NDE 1999): Trinomialsax8 + bx + cover a field K of characteristic zerowhose Galois group is contained in G1344are parametrized by the curveC1344:y2 = 2 x6 + 4 x5+ 36 x4 + 16 x3- 45 x2 + 190 x + 1241.Current ``technology'' is not yet up to provably listingall the Q-points of C1344.Still, we can search for points of small height, i.e., whosex-coordinates have small numerator and denominator;the larger the height, the harder it is for2x6+4x5+36x4+16x3-45x2+190x+1241to be a perfect square. We found a total of eight pointswith numerator and denominator of at most three digits:the four pairs with x=2, 1, -1, and -5.Eliminating three degenerate polynomials,we recovered the polynomialsx8 + 16 x + 28, x8 + 576 x + 1008, 194 53 x8 + 19 x + 2,each arising once, and the polynomialx8 + 324 x + 567,remarkably arising twice! The explanation for this is thateach of the first three trinomials has Galois group G1344 as expected, but the last onehas a smaller Galois group, which embeds as a transitive subgroupinto G1344 in two different ways.This Galois group is isomorphic with G168,acting on 8 letters via the other guise of that group, asPSL2(Z/7Z)(the 8 letters being the points of the projective lineover Z/7Z).All of these degree-8 trinomials seem to be new.Michael Stoll extended the search for rational pointson C1344 from three to five digits,and found none besides the four pairs already known.We naturally conjecture that there are no further rational pointson C1344, and thus that every trinomialax8+bx+cwith Galois group contained in C1344is equivalent to one of the four displayed above.REFERENCES[B] N. Bruin: Leiden Technical Reports W99-14 and W99-15 (``Chabauty methods using elliptic curves'' and ``Chabauty methods using covers on elliptic curves of genus 2''), 1999.[EFM] D.W. Erbach, J. Fischer, and J. McKay: Polynomials with Galois group PSL(2,7), J. Number Theory 11 (1979), 69-75.[M1] G. Malle: Polynomials for primitive nonsolvable permutation groups of degree d<=15. J. Symbolic Computation 4 (1987) #1, 83-92.[M2] B.H. Matzat: Konstruktive Galoistheorie, LNM 1284 (1987).[F1] G. Faltings:Endlichkeitssätze für abelsche Varietätenüber Zahlkörpern, Invent. Math. 73(1983), 349-366.[F2] G. Faltings: Diophantine approximation on Abelian varieties, Annals of Math.(2) 133 (1991) #3, 549-576.[JRYZ] G. Roland, N. Yui, D. Zagier: A parametric family of quintic polynomials with Galois group D5, J. Number Theory 15 (1982), 137-142; C.U. Jensen, N. Yui: Polynomials with Dp as Galois group, J. Number Theory 15 (1982), 374-375. [SW] B.K. Spearman, K.S. Williams: On solvable quintics X5+aX+b and X5+aX2+b, Rocky Mountain J. Math. 26 #2 (1996), 753-772. [W] H. Weber: Lehrbuch der Algebra I. Braunschweig: Vieweg, 1908. |
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