###### INVESTIGATION INTO SOLVABLE QUINTICS

###### Postgraduate

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 **Z**_{5
}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*) = *x*^{5
}+ *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*) = *x*^{5
}+ *px*^{3 }+ *qx*^{2 }+ *rx *+ *s *∈ **Q**[*x*]
is solvable;

(2) the
Lagrange resolvents *r _{i }*of
the roots of

*f*are defined;

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

unity.

Let
*x*_{1}*,x*_{2}*,x*_{3}*,x*_{4 }and *x*_{5 }be the roots of the general quintic polynomial *x*^{5 }− *s*_{1}*x*^{4 }+
*s*_{2}*x*^{3 }− *s*_{3}*x*^{2 }+ *s*_{4}*x *− *s*_{5 }where the *s _{i }*are the elementary
symmetric functions in the roots. We assume that

*s*

_{1}

*,s*

_{2}

*,s*

_{3}

*,s*

_{4}

*,s*

_{5 }∈

**Q**. Let

The stabilizer of *θ *in *S*_{5
}is precisely *F*_{20},
the Frobenius group of order 20, with generators (12345) and (2354). Since *S*_{3}, generated by (12) and
(123), is a complement of *F*_{20 }in
*S*_{5}, (that is, every
element of *S*_{5 }can be
written uniquely as an element of *S*_{3
}times an element of *F*_{20}),
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*)
= *x*^{5 }+ *px*^{3 }+ *qx*^{2 }+ *rx *+ *s *(1.1)

Dummit proves the following theorem:

**Theorem
1 ***The irreducible quintic f*(*x*) = *x*^{5}+*px*^{3}+*qx*^{2}+*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 S_{5}. It can be shown that all solvable subgroups of S_{5 }are contained in the
conjugates of F_{20}. 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 F_{20
}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 *F*_{20}. Let *ζ *be a fixed primitive 5th root of unity. Dummit defines the usual
Lagrange resolvents of *x*_{1}:

(*x*_{1}*,*1) = *x*_{1 }+ *x*_{2 }+
*x*_{3 }+ *x*_{4 }+ *x*_{5 }*r*_{1 }= (*x*_{1}*,ζ*) = *x*_{1 }+ *x*_{2}*ζ *+ *x*_{3}*ζ*^{2 }+ *x*_{4}*ζ*^{3 }+ *x*_{5}*ζ*^{4 }*r*2 = (*x*1*,ζ*2) = *x*1 + *x*2*ζ*2 + *x*3*ζ*4 + *x*4*ζ *+ *x*5*ζ*3 *r*3 = (*x*1*,ζ*3) = *x*1 + *x*2*ζ*3 + *x*3*ζ *+ *x*4*ζ*4 + *x*5*ζ*2 *r*4 = (*x*1*,ζ*4) = *x*1 + *x*2*ζ*4 + *x*3*ζ*3 + *x*4*ζ*2 + *x*5*ζ *so that:

*x*_{1 }= (*r*_{1 }+
*r*_{2 }+ *r*_{3 }+ *r*_{4})*/*5 *x*_{2
}= (*ζ*^{4}*r*_{1 }+ *ζ*^{3}*r*_{2 }+
*ζ*^{2}*r*_{3 }+ *ζr*_{4})*/*5

Let |

where *l*_{0 }is the sum of terms
involving powers *ζ ^{j }*of

*ζ*with

*j*divisible by 5,

*l*

_{1 }involves powers of

*ζ*where

*j*≡ 1 (mod 5), etc. Explicitly,

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