## 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.