An extension F/K is normal if, for any irreducible polynomial p(x) in K with a root in F, p(x) splits in F. If F has one root, it has them all.
Note that the intersection of arbitrarily many normal extensions is normal.
A purely transcendental extension is normal, by the above definition, since it has no roots. However, I will deviate from most textbooks and define a normal extension as algebraic; that's usually what we mean by normal anyways.
Let's develop some equivalent criteria for a normal extension, starting with the definition given above.
Suppose F is not normal over K, and let v be any root of p(x) outside of F. An isomorphism takes K(u) to K(v). Extend this isomorphism to the generators of F, using the process described in the previous page. Thus we have an isomorphism from F into E, and u maps to v, outside of F. The map is not an automorphism, and that is a contradiction. Therefore F/K is normal.
When proving normality, the second of these three criteria is the most common. Start with a field K and adjoin all the roots of p(x). In fact, adjoin all the roots of all the polynomials in a set, even an infinite set. These adjoined roots act as generators. The conjugates of each generator are present - that's the way we built the extension - hence the extension is normal.
Every algebraic extension of a finite field is normal. Each finite field consists of the nth roots of 1, for some n. Adjoin the base root and you get them all. In other words, successive powers of the base root produce all the roots of 1. The generator implies its conjugates, and the extension is normal.
The normal closure of an algebraic extension E/K is the smallest extension F/K that is normal. Let S be the set of polynomials having roots in E, and build the splitting field for S.