INVESTIGATION INTO SOLVABLE QUINTICS

Diploma

ABSTRACT

Solving quintics has fascinated and challenged mathematicians for centuries. David Dummit in Solving Solvable Quintics [3] gives a powerful method that permits one to determine when a quintic is solvable and to solve for its roots. G.N. Watson in 1948, and Kobayashi and Nakagawa have come up with similar methods.

This paper focuses on two families of quintics that pose different challenges for solving them. The first family is a famous group of quintics that are called Emma Lehmer’s quintics. These quintics are known to have Z5 as their Galois group and one might hope that expressing the roots in terms of radicals would give simple expressions from which Emma Lehmer’s polynomials could be recovered. However, we show that the expressions of the roots in terms of radicals is much more complicated than expected.

We also consider the simple equation f(x) = x5 + ax + p and show that, for a fixed nonzero integer p, the polynomial f is solvable by radicals for only finitely

many a Z.

Chapter 1

Summary of Dummit’s method

In this section, Dummit’s method is summarized. The main steps are:

(1)  a sextic resolvent is constructed which has a rational root if and only if the general reduced quintic f(x) = x5 + px3 + qx2 + rx + s Q[x] is solvable;

(2)  the Lagrange resolvents ri of the roots of f are defined;

(3)  the fifth power of the resolvents are expressed as linear combinations of roots of

unity.

Let x1,x2,x3,x4 and x5 be the roots of the general quintic polynomial x5 s1x4 + s2x3 s3x2 + s4x s5 where the si are the elementary symmetric functions in the roots. We assume that s1,s2,s3,s4,s5 Q. Let

The stabilizer of θ in S5 is precisely F20, the Frobenius group of order 20, with generators (12345) and (2354). Since S3, generated by (12) and (123), is a complement of F20 in S5, (that is, every element of S5 can be written uniquely as an element of S3 times an element of F20), it follows that θ and its conjugates satisfy a polynomial g(x) of degree 6 over Q.

By making a translation, we may assume our quintic is

                                                                 f(x) = x5 + px3 + qx2 + rx + s                                         (1.1)

Dummit proves the following theorem:

Theorem 1 The irreducible quintic f(x) = x5+px3+qx2+rx+s Q[x] is solvable by radicals if and only if the polynomial g(x) has a rational root. If this is the case, the sextic g(x) factors into the product of a linear polynomial and an irreducible

quintic.

Proof: The polynomial f(x) is solvable if and only if the Galois group of f(x), considered as a permutation group on the roots, is contained in a solvable subgroup of S5. It can be shown that all solvable subgroups of S5 are contained in the conjugates of F20. It follows that f(x) is solvable by radicals if and only if θ or one of its conjugates is rational. This proves the first assertion. We may assume θ is rational so the Galois group of f is contained in the specific group F20 above.

If g(x) has a rational root, then it factors as a linear times a quintic. It can be shown that this quintic is irreducible. See [3] for details.

Henceforth, we assume that the Galois group of f(x) is solvable, hence is a subgroup of F20. Let ζ be a fixed primitive 5th root of unity. Dummit defines the usual Lagrange resolvents of x1:

(x1,1) = x1 + x2 + x3 + x4 + x5 r1 = (x1) = x1 + x2ζ + x3ζ2 + x4ζ3 + x5ζ4 r2 = (x12) = x1 + x2ζ2 + x3ζ4 + x4ζ + x5ζ3 r3 = (x13) = x1 + x2ζ3 + x3ζ + x4ζ4 + x5ζ2 r4 = (x14) = x1 + x2ζ4 + x3ζ3 + x4ζ2 + x5ζ so that:

x1 = (r1 + r2 + r3 + r4)/5 x2 = (ζ4r1 + ζ3r2 + ζ2r3 + ζr4)/5

x3 = (ζ3r1 + ζr2 + ζ4r3 + ζ2r4)/5 x4 = (ζ2r1 + ζ4r2 + ζr3 + ζ3r4)/5 x5 = (ζr1 + ζ2r2 + ζ3r3 + ζ4r4)/5

Let


where l0 is the sum of terms involving powers ζj of ζ with j divisible by 5, l1 involves powers of ζ where j ≡ 1 (mod 5), etc. Explicitly,