§ Galois extension
- Let be a finite extension of . Let . Then is said to be Galois iff:
- is normal and separable (over ).
- . We will show that . So is "symmetric as possible" --- have the largest possible galois group
- [The fixed poits of under ]. This is useful for examples.
- is the splitting field of a separable polynomial over . Recall that a polynomial is separable over if it has distinct roots in the algebraic closure of . Thus, the number of roots is equal to the degree.
- and : There is a 1-1 correspondece between [NOT ! ], and the other way round, to go from to . This is a 1-1 correspondence. is in the denominator because we want to fix when we go back.
- We'll show (1) implies (2) implies (3) implies (4) implies (1)
§ (4) implies (1)
- We've shown that splitting fields of sets of polynomials are normal, so this case is trivial.
- Just to recall the argument, let be the splitting field of some separable polynomial over . We need to show that is normal and separable.
- It's separable because it only adds elements to new elements to which are the roots of , a separable polynomial. Thus, the minimal polynomial of new elements will also be separable, and the base field is trivially separable.
- We must now show that is normal. We proceed by induction on degree. Normality is trivial for linear polynomials, if contains one root it contains all of the roots (the only one).
- Let have a root . If , then divide by and use induction. So suppose .
- Then is some element that is generated by the roots