1998/99 GCW
Everybody should know the formula for the solutions to the quadratic equation
namely
They may perhaps also know Fontana's solution of
which is
from which a formula for the roots of a general cubic can be derived.
Girolamo Cardano's Ars Magna, published in 1545, contains a method, due to Ludovico Ferrari, of solving quartic equations. All these formulae are built up from the coefficients by repeated addition, subtraction, multiplication, division and taking roots. Expressions of this kind are called radical expressions.
In 1824 Abel showed that there is no formula for the roots of a general quintic in terms of radical expressions. So, if we have a polynomial equation, when can its solutions be expressed as radical expressions in the coefficients? Abel died in 1829 working on this problem. In 1832 E.Galois was killed in a duel at the age of 21. In 1843 J.Liouville discovered among his papers, which had all been rejected by the Academy of Sciences in Paris, a complete solution to this problem. Galois' work also provides answers to many ancient problems - why is it impossible for there to be a ruler and compass construction to trisect a general angle? Which regular polygons can be constructed by ruler and compass?
Rather than look at individual numbers and equations, the approach of modern algebra is to look at all the numbers that can be obtained from some given initial collection by using addition, subtraction, multiplication and division. The resulting collection is called a field. More technically, a field is a nontrivial commutative ring in which all the elements, apart from zero, have an inverse. Galois theory is essentially the study of fields.
Let 
be a ring and let 
denote its unit element and 
denote its zero element. If 
is a positive integer and 
is an element of , 
will denote the sum of copies of . We extend this notation to negative integers by defining 
to be 
and of course we define 
to be 
.
Let us recall the concept of the characteristic of a ring. There is a unique homomorphism of rings (function preserving addition, subtraction, multiplication, zero and unit)
given by 
. The kernel of this homomorphism is an ideal of 
and therefore has the form 
, the set of all multiples of some non-negative integer 
. We call this the characteristic of . If the characteristic of is 0 then 
is injective, and so 
contains an isomorphic copy of 
as a subring - all the elements you can get by adding and subtracting 
's.
If 
then any sum of copies of an element of must be zero, and is minimal with this property. In this case must contain a subring, namely the image of 
, which is isomorphic to 
, the ring of integers modulo .
If the ring is a field, either its characteristic is 0, in which case it must contain a subfield isomorphic to 
, the field of rational numbers ( namely all the elements of the form 
or its characteristic must be a prime number. This is because,
and if 
then, in a field, either 
or 
. So if we had 
we would have a contradiction to the minimality of 
.
If 
is a field, and is any nontrivial ring, then any homomorphism of rings 
must be injective. For suppose that 
are elements of such that 
. Then 
. Now if 
were nonzero it would have an inverse 
. From the equation
we get, by applying 
which contradicts the assumption that is nontrivial. So we must have 
. So is injective. This means that the subring of given by the image of is isomorphic to .
By an extension of fields we mean a ring homomorphism from one field to another. If 
is an extension of fields we have that the image of is isomorphic to 
. In circumstances where the abuse does not lead to ambiguity we write 
and we say that is a subfield of 
and that is a superfield or an extension of .
Given a field we have the notion of a vector space over , i.e. vector spaces where scalars are elements of . In particular, given a homomorphism of rings 
we can think of as a vector space over 
. So vector addition is just addition in and scalar multiplication of 
by 
gives the vector 
. The degree of is defined to be the dimension of when it is regarded as a vector space over in this way. In particular, if a field contains a subfield we write 
for the degree of the extension .
The Multiplication Theorem: Suppose that we have extensions
Then the degree of the composite extension 
is the product of the degrees of its factors, i.e.
Proof: We give this in case both extensions are finite, which is the case of most importance in the course. Suppose that 
. Choose a basis 
of over and a basis 
of 
over 
. We will show that the 
-ple of elements 
is a basis for 
over . First we must show that these elements span 
over .
Every element of 
can be written as a linear combination 
for some elements 
of 
. Each 
can be written as a linear combination 
for some 
-ple of elements 
of . Hence every element of 
can be written in the form
Now we must show that the 
elements 
are linearly independent over . Suppose we had
From the linear independence of the 
's over it follows that for each 
we have
From the linear independence of the 
's over we have 
.
If we have an extension of fields 
an intermediate extension of it is a field and extensions 
so that the composite is the given extension 
. It follows from the multiplication theorem that 
has to be a factor of 
. In particular, if 
is a prime number there cannot be any proper nontrivial intermediate extensions.