3Solutions to polynomial equations

II Galois Theory

3 Solutions to polynomial equations

We have now proved the fundamental theorem of Galois theory, and this gives a

one-to-one correspondence between (intermediate) field extensions and subgroups

of the Galois group. That is our first goal achieved. Our next big goal is to use

this Galois correspondence to show that, in general, polynomials of degree 5 or

more cannot be solved by radicals.

First of all, we want to make this notion of “solving by radicals” precise.

We all know what this means if we are working over

Q

, but we need to be very

precise when working with arbitrary fields.

For example, we know that the polynomial

f

=

t

3

−

5

∈ Q

[

t

] can be “solved

by radicals”. In this case, we have

Root

f

(C) = {

3

√

5, µ

3

√

5, µ

2

3

√

5},

where

µ

3

= 1

, µ 6

= 1. In general fields, we want to properly define the analogues

of µ and

3

√

5.

These will correspond to two different concepts. The first is cyclotomic

extensions, where the extension adds the analogues of

µ

, and the second is

Kummer extensions, where we add things like

3

√

5.

Then, we would say a polynomial is soluble by radicals if the splitting field

of the polynomial can be obtained by repeatedly taking cyclotomic and Kummer

extensions.