## Fields, Normal Extension

### Normal Extension

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.

1. Let F/K be a normal extension, as defined above, and let G be any set of generators for that extension. In other words, F = K(G). Each generator is a root of some irreducible polynomial, and since F is normal, all those polynomials split in F. Each generator brings in its conjugates.

2. Assume F includes all the conjugates of its generators. In other words, F is a splitting field for the polynomials associated with the generators of F. Let E be any extension of F, and let c be a monomorphism from F into E that fixes K. Recall that c is determined by its action on the generators. Each generator is a root of some p(x), and is mapped to another root of p(x), which is contained in F. In fact, repeated invokations of c step through the roots of p(x), until you return to the original generator. Thus c is invertible on each generator in G. This is not the reciprocal, it is the inverse function of c. The inverse function is defined on all the generators, hence on all of F. Therefore c is an automorphism on F. Every embedding of F into a larger extension remains an automorphism.

3. To complete the circle, let E be any extension of F, where F/K is algebraic, and let c be any monomorphism of F into E that fixes K, and assume c is always an automorphism on F. Consider any u in F with an irreducible polynomial p(x). If E is not a splitting field for p(x), adjoin the remaining roots of p to E. Do the same for all the generators of F. Adjoin the conjugates of each generator to E, making E that much larger. If the generators of F aren't specified, or are uncountable, take the algebraic closure of E. Now we're ready to build an isomorphism from F into E.

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.