## Galois Extensions, Galois Groups

### Galois Groups

Let F/K be a field extension.
The galois group of F over K is the set of automorphisms of F that fix K.
Since automorphisms can be composed and reversed,
this does indeed form a group.
Let u be algebraic over K, the root of an irreducible polynomial p(x).
Let c() be an automorphism taken from the galois group of F over K,
hence it fixes K.
Apply c to the equation p(u) = 0,
and the coefficients of p are fixed, hence c(u) is another root of p(x).
There are at most n roots, if p has degree n,
so we only have a few choices.

If F/K is normal,
every root of p(x)
is a possible destination for c(u).
We proved this for the
splitting field of p,
and then extended it to a larger,
normal extension.
Using the terminology of group theory,
the galois group
acts transitively on the roots of p(x).

Just to review,
If c() is a field endomorphism on F/K, and F is algebraic, c is a field automorphism.
Suppose there is an element u that is not in the image of c(F), and let u be a root of p(x).
Let E be the intermediate field extension K(u).
This includes u, and some of its conjugates.
Now c is a monomorphism that carries roots to roots.
In other words, it maps a finite set of roots into itself,
hence the map is onto.
There is some root v such that c(v) = u,
and that means c is an automorphism on F.
When F/K is algebraic, all endomorphisms are automorphisms.
They all belong to the galois group.

If K(u) has dimension n over K, and u is a root of p(x),
the automorphisms of K(u) carry u to a root of p(x), and nowhere else.
There are at most n such roots.
Furthermore, the entire automorphism is determined by the image of u.
Hence there are at most n automorphisms in the galois group.
Of course n is an upper bound,
as shown by adjoining the real cube root of 2 to the rationals.
Everything in the extension is real, and the root cannot be mapped onto a complex cube root of 2,
so the galois group is trivial,
even though the dimension of the extension is 3.

Let E be an intermediate extension between F and K.
Every automorphism on F/E also fixes K, hence the galois group of F over E is a subgroup
of the galois group of F over K.

Let F include a transcendental element u.
For every a in K,
the automorphism fixing K and mapping u to a*u is a K automorphism,
and the galois group has the multiplicative group of K as subgroup.
Replace u with u+a to show the galois group has the additive group of K as subgroup.