Ten years later, Erbach, Fischer and McKay [EFM] published the trinomial
In fact, I showed:
Theorem (NDE 1999): Trinomials
ax7 + bx + c
over a field K of characteristic zero
whose Galois group is contained in G168
are parametrized by the curve
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 infinity yields the Erbach-Fischer-McKay trinomial. The two new examples come from the other point at infinity and 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 2 has only finitely many points over any number field. In our case, that means that there are only finitely many equivalence classes of G168-trinomials over any number field.
But it can be much harder to list all cases than to prove that the list is finite. None of the known proofs of the Mordell conjecture provides an effective procedure for finding all the rational points and proving that the list is complete, even in the first case of 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 of theoretical 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 trinomial ax7 + bx + c over Q whose Galois group is contained in G168 is equivalent to one of the four trinomials displayed above.
Similar analysis applies to trinomials axn+bxk+c (n,k coprime) of arbitrary degree n whose Galois group is contained in some given subgroup G of 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/4Z with Z/5Z, or the ax+b group mod 5). Then the general trinomial with Galois group contained in G is equivalent to
(attributed by Matzat [M2, pages 90-91 (Satz 3)] to Weber [W, section 189]). Moreover, the Galois group is contained in the 10-element dihedral group if and only if u=t-1/t, when the quintic is equivalent to
this is attributed by Matzat [M2, p.93 (Satz 4)] to [JRYZ]. Likewise, if n=6, we can ask that G be the transitive 120- or 60-element subgroup of S6. (These are the images of the obvious subgroups S5 and A5 under an outer automorphism of S6; they can also be obtained as PGL2(Z/5Z) and PSL2(Z/5Z) acting on the six points of the projective line over Z/5Z.) Here the general such sextic is
for the 120-element group; for the 60-element group, take u=t2. Matzat reports this on page 94 of [M2], and attributes it, together with the formulas for sextic trinomials for several other (imprimitive but) transitive subgroups of S6, to Malle [M1]. (Matzat also mentions the genus-2 curve we call C168 on page 95, but does not exhibit equations or 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 see addendum below) :
Theorem (NDE 1999): Trinomials
ax5 + bx2 + c
over a field K of characteristic zero
whose Galois group is contained in the 20-element subgroup
of S5 are parametrized by the elliptic curve
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 classes of trinomials ax5+bx2+c with 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 dihedral Galois group, and the remaining two (corresponding to the other two points not divisible by 2 in C20(Q)) produce quintics with 20-element Galois group:
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 G1344 of affine linear transformations of (Z/2Z)3. [This is the semidirect product of G168 with (Z/2Z)3; equivalently, G1344 is the group of invertible 4-by-4 matrices over Z/2Z whose bottom row is 0001, acting on the 8 column vectors of height 4 with last coordinate 1.] We then have:
Theorem (NDE 1999): Trinomials
ax8 + bx + c
over a field K of characteristic zero
whose Galois group is contained in G1344
are parametrized by the curve
Current ``technology'' is not yet up to provably listing all the Q-points of C1344. Still, we can search for points of small height, i.e., whose x-coordinates have small numerator and denominator; the larger the height, the harder it is for 2x6+4x5+36x4+16x3-45x2+190x+1241 to be a perfect square. We found a total of eight points with 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 polynomials
All of these degree-8 trinomials seem to be new. Michael Stoll extended the search for rational points on C1344 from three to five digits, and found none besides the four pairs already known. We naturally conjecture that there are no further rational points on C1344, and thus that every trinomial ax8+bx+c with Galois group contained in C1344 is equivalent to one of the four displayed above.