First of all you should read about the
Galois group. Galois theory,
was first formulated by
Evariste Galois in order to study
solutions of
polynomial equations in one variable. It
is one of the most beautiful moments in
mathematics.
Given the technology of the time it was a truly remarkable achievement.
It allows us to translate problems about
fields into problems about
groups.
We need some notation. We write K <= M to mean that K
and M are fields and that M is a field extension of K.
Associated to such an extension we have the Galois group Gal(M/K)).
This raises the question if we have an intermediate field
K <= L <= M what can we say about Gal(M/L) and
Gal(L/K)? Galois theory gives us a good answer to this question.
First notice that Gal(M/L) is a subgroup of Gal(M/K)
which we shall write as Gal(M/L) <= Gal(M/K), for short.
Proof: To see this recall that, by definition, an element of Gal(M/L)
is an Lautomorphism of M. Since L contains
K any such automorphism must fix the elements of K.
Thus we can see that Gal(M/L) is a subset of Gal(M/K).
But it is clear that if we compose two Lautomorphisms
or invert an Lautomorphism then we obtain another one.
Thus it is a subgroup, as required.
This result gives us a way to turn a picture involving fields into
a picture involving groups. We just apply Gal(M/).
But does this process retain information or destroy it? It turns out
that to retain information we must insist that the field extension
K <= M has some particular properties.
Defintion
We say that a field extension K <= M
is Galois if it is finitedimensional, normal
and separable.
As an example, if M is a splitting field over K
of a separable polynomial then it is Galois.
We need some notation. Write
[M:K] for the dimension of M considered
as a Kvector space. If G is a subgroup
of Gal(M/K) we write M^{G} for the subset
{m in M : g(m)=m for all g in G}.
This is called the fixed field of G and, as its name
suggests, we have K <= M^{G} <= M.
Here's the first main theorem.
Theorem Let K <= M be a Galois field extension.
Then
 Gal (M/K)=[L:K]

There is an bijection between fields L
with K <= L < M and subgroups of Gal(M/K).
The bijection works like this. If L is such a field the
corresponding subgroup is Gal(M/L). If G is such
a group the corresponding field is M^{G}
Note that these bijections turn a <= into a >=
because as the field L gets bigger Gal(M/L) gets
smaller and as the group G gets bigger the field
M^{G} gets smaller.
Let's consider an example. Let a be the real cube root
of 2 and let w=e^{2pii/3} be a primitive complex
root of unity. The field extension K=Q <= Q(a,w)=L
is Galois (L is the splitting field of x^{3}2
over Q) so we can relate the subfields of L to the
subgroups of Gal(L/K). Firstly then we should compute
the Galois group.
Let f be an element of Gal(L/K). Then f(a)
must be a root of the minimal polynomial of a over
Q, which is x^{3}2. Thus
f(a) is one of a,wa,w^{2}a. Likewise
f(w) must be one of w and w^{2}. Since
a Qautomorphism is determined by its value on a and
w then this means that there are at most six elements in the group.
By the theorem (since [L:K]=6) we know that there are
6 automorphisms in the group. So these 6 possibilities all occur.
Let g be the Qautomorphism with g(a)=aw and g(w)=w
and let h be the Qautomorphism with
h(a)=a and h(w)=w^{2}. Then the
six elements of the Galois group are 1,g,g^{2},h,gh,g^{2}h.
Note that the group is not abelian because gh(a)=g(h(a))=g(a)=aw
and hg(a)=h(g(a))=h(aw)=h(a)h(w)=aw^{2}. Thus gh is
not equal to hg. But there is only one group of order 6 that is not
abelian the symmetric group S_{3} of permutations of three
objects. Can we see three natural things that the Galois group permutes?
The answer is yes: a,aw,aw^{2}.
What subfields does Q(a,w) have? Well it's not too hard to
compute them, they are Q(a), Q(aw),
Q(aw^{2}), Q(w),
Q(w^{2}) and Q.
See if you can figure out the corresponding subgroups
of the Galois group. For example, the first subfield
I listed corresponds to the subgroup <h>.
The next main theorem tells us about normal subgroups of the Galois
group which are the most interesting kind of subgroups.
Theorem Let K < L <= M with M a Galois
extension of K.
TFAE
 K <= L is Galois
 [L:K]=Gal(L/K)
 Gal(M/L) is a normal subgroup of Gal(M/K)
When these equivalent conditions hold true we have that
Gal(
M/K)/Gal(
M/L) is isomorphic to Gal(
L/K)
These are not
the only theorems in Galois theory but they are the most basic ones.