Sextic resolvent has no rational root

Question

An irreducible quintic $f(x)\in\mathbb{Q}[x]$, is solvable by radicals if and only if its sextic resolvent $\theta_f (y)=(y^3+py^2+qy+r)^2-2^{10}\Delta(f)y$ has a rational root ($\Delta$ is the discriminant of $f$, and $a,b,c$ are some defined rationals, see the Cox reference).

In many cases, this associated sextic is not pretty. More troublesome is that in my work I am often coming across quintics with the coefficients in terms of some parameter, where in general we cannot use the Rational Root Test on the sextic.

Any advice on proving the (non)existence of a rational root of the sextic resolvent of an irreducible rational quintic (especially sufficient criteria for non-existence)? Examples or references to special cases are fine. Is the usual technique just ad-hoc tricks?

References:

1. Cox, Galois Theory, Theorem 13.2.6, p372

2. Dummit & Foote, Abstract Algebra, Problem 14.7.21, p639

Answer 1

If I understand you correctly, you are interested in the case when your quintic polynomial has coefficients depending on a parameter $t$. If that is so, then the resolvent sextic defines a curve, and you are asking for a method to prove that there are no rational points on this curve. In general, this is a difficult problem. I won't try to survey all the methods for doing this, but point you instead to Bjorn Poonen's survey (available here). Perhaps the easiest and most general method is to try to prove there are no rational points in $\mathbb{Q}_{p}$ for some prime $p$ (although this isn't always true - your curve could be a counterexample to the local-to-global principle).

In some cases, the genus of the curve you obtain might be zero or one, and in this case you're likely to be able to get a computer to decide if the curve has rational points or not. For example, Spearman and Williams (see here) use the resolvent sextic to prove that there are only 5 irreducible solvable quintics of the form $x^{5} + ax^{2} + b$. (This problem was also solved by Elkies - see this page). In this case, the resulting curve had genus $1$ and was an elliptic curve with rank zero.

Answer 2

While this does not deal with coefficients in terms of parameters, GAP can compute Galois groups of polynomials of degree up to 15. -- For example:

gap> TransitiveGroup(5,GaloisType(x^5-x-1));
S5
gap> TransitiveGroup(5,GaloisType(x^5+2));
F(5) = 5:4
gap> TransitiveGroup(6,GaloisType(x^6+2*x^4+2*x^3+x^2+2*x+2));
F_36(6):2 = [S(3)^2]2 = S(3) wr 2
gap> TransitiveGroup(7,GaloisType(x^7+2));
F_42(7) = 7:6
gap> TransitiveGroup(8,GaloisType(x^8-4*x^4-1));
[1/4.cD(4)^2]2
gap> TransitiveGroup(12,GaloisType(x^12-x-1));
S12
gap> TransitiveGroup(12,GaloisType(x^12-3*x^6-1));
D(4)[x]S(3)