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 L-automorphism 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 L-automorphisms or invert an L-automorphism 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 finite-dimensional, 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 K-vector space. If G is a subgroup of Gal(M/K) we write MG 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 <= MG <= M.

Here's the first main theorem.

Theorem Let K <= M be a Galois field extension. Then

  1. |Gal (M/K)|=[L:K]
  2. 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 MG

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 MG gets smaller.

Let's consider an example. Let a be the real cube root of 2 and let w=e2pii/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 x3-2 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 x3-2. Thus f(a) is one of a,wa,w2a. Likewise f(w) must be one of w and w2. Since a Q-automorphism 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 Q-automorphism with g(a)=aw and g(w)=w and let h be the Q-automorphism with h(a)=a and h(w)=w2. Then the six elements of the Galois group are 1,g,g2,h,gh,g2h. 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)=aw2. Thus gh is not equal to hg. But there is only one group of order 6 that is not abelian the symmetric group S3 of permutations of three objects. Can we see three natural things that the Galois group permutes? The answer is yes: a,aw,aw2.

What subfields does Q(a,w) have? Well it's not too hard to compute them, they are Q(a), Q(aw), Q(aw2), Q(w), Q(w2) 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.

  1. K <= L is Galois
  2. [L:K]=|Gal(L/K)|
  3. |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.

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