(1) Use induction on n. If n=1, then the splitting field of f(x) is F itself, hence the statement is obviouly true.
Suppose the statement is true for n ≦ k. Let deg f(x) = k+1.
Let α be a root of f(x) in some extension E of F. We have [F(α):F] ≦deg f = k+1.
In the field F(α)
f(x) = (x-α)g(x), where g(x) is a polynomial of degree k in F(α)[x].
By induction hypothesis, the splitting K field of g(x) exists, and [K:F(α)]≦ k!
Note that f(x) splits in K, hence the splitting field of f(x) exists, and the degree is
[K:F] = [K:F(α)][F(α):F] ≦(k+1)!
(2) Suppose α1 and α2 have the same minimal polynomial f(x). Then the map
σ: F(α1) → F(α2) defined by
σ|F = identity, σ(α1) = α2 is a well defined isomorphism. (it is straight forward to verify this claim, try it yourself.)
Extent σ to an isomorphism of K, we see that α1 and α2 are conjugate.
(3) It suffice to assume f(x) is irreducible over F1. Let α be a root of f(x) in K1. Then f(x) is the minimal polynomial of α. Let β be a root of σ(f(x)) in K2. If g(x) is the minimal polynomial of β over F2, then g(x) divides σ(f(x)), and hence σ-1(g(x)) divides f(x). Since f(x) is irreducible, f(x) = σ-1(g(x)) , and g(x) = σ(f(x)). Therefore, we can extent σ to an isomorphism F1(α) → F2(β) by defining
If K1=F1(α), then σ(f(x)) splits in F2(β), hence K2=F2(β), and is K1 isomorphic to K2. Otherwise, use induction on n=[K1:F1].
(4) Suppose α is an element of K. Then f(σ(α)) = σ(f(α)) = 0. Hence σ(α) is a root of f(x). Since K is the splitting field of f(x), σ(α) must be in K. Therefore σmaps K into K.
Repeat the same argument for the isomorphism σ-1. We see that σ maps K onto K.