Galois Theory?

Question

This is an exercise of Morandi, that i couldn't solve. I have done the rest of my homework, so please, give me some help!

Let K a field and suposse that s is an automorphism of K, and s has infinite order. Let F be the fixed field of s. If the extension K / F is algebraic, show that K is normal over F.

Answer 1

let a in K. since K/F is algebraic, consider the minimal polynomial f of a over F. we know that s must permute the roots of f in K. therefore s(a), s^2(a), ... are all roots of f. since f has only finitely many roots, there is some n such that s^n(a)=a. in that case, a, s(a), s^2(a), ..., s^(n-1)(a) are the distinct roots of f in K. so the polynomial
g(X)=(X-a)(X-s(a))...(X-s^(n-1)(a))
divides f. moreover, the coefficients of g are the elementary symmetric polynomials in the s^i(a), so g is fixed by s. this means that g is in F[X]. hence g=f. so f splits over K. so any polynomial over F with a root in K splits over K.

Answer 2

I will think about it
edit lkjh:
" in that case, a, s(a), s^2(a), ..., s^(n-1)(a) are the distinct roots of f in K."
better to say:
in that case, a, s(a), s^2(a), ..., s^(n-1)(a) are distinct roots of f in K.
You don't know in that moment that they are all the roots of f.
Moreover they are "distinct" if we consider with their order.

I give you a thumbs up. I was thinking about the problem but I didn't see the essential thing that g is fixed by s.