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
= 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.