F-automorphism group of the field of rational functions

[Barre]

I've been doing some exercises in introductory Galois theory (self-study hence PF is the only avaliable validator :) ) and a side-result of some of them is surprising to me, hence I would like you to set me straight on this one if I'm wrong.

Homework Statement

Let K(x) be the field of rational functions with coefficients in the field K of characteristic zero. Then K(x) can be seen as an extension of K by a transcendent indeterminate x, and K(x)/K is actually Galois. Let G = Gal(K(x)/K), which is infinite (for example \phi(f(x)) = f(x+1) generates a cyclic subgroup of infinite order) and let H be any infinite subgroup of G. Then the field fixed by H is K.

Homework Equations

Here are a couple of results from previous exercises that I've used.

1. If E \neq K is an intermediate field of the extension K \subset K(x), then [K(x):E] has finite degree.

2. Let H' denote the intermediate field of the extension K(x)/K fixed by automorphisms in the subgroup H. Equivalently, let E' denote the subgroup of G such that all automorphisms in this group fix E. If H' = E, then H \subset E'.

The Attempt at a Solution

Let H be any subgroup of G = Gal(K(x)/K) of infinite order and let H' = E, and assume that E is not K, hence it is a proper extension of K. Because K(x) is finite-dimensional over E and an E-automorphism of K(x) is determined by its action on the basis of K(x) over E, and basis is finite, then there can only be a finite amount of E-automorphisms of K(x). Hence |E'| is finite. By 2, H \subset E', but the first one was infinite which is a contradiction. Then the assumption that E is not K was wrong, and it would follow that H' = E = K.Could this be correct? The follow-up exercise asked to prove that no infinite subgroup H has the property that H = H'' = (H')' but I've arrived at a slightly stronger result, hence my question.

 

[Barre]

Hello, I would like to extend my question to this exercise as well. I used some of my work on this to prove assertion 1) in previous post.

Let \phi \in K(x) but \phi \not\in K. Then \phi = f/g is a rational function and we assume f,g are coprime. Prove[K(x):K(\phi)] is finite and has degree max(deg(f), deg(g)).

I can see how it is finite, as x (the indeterminate in K(x)) is actually a root of a polynomial in K(\phi), namely (\phi)g(y) - f(y). Hence for sure there exists a minimal polynomial, and by the exercise, I know that it should have degree max(deg(f), deg(g)). But my polynomial has the property that it's degree is the maximum of degrees of f and g, hence it is the minimal polynomial I am looking for. To prove this, I need to prove that it is irreducible.

I will be honest that I have no idea how that can be done. A hint was given to consider reducibility in K(\phi)[x], the ring of polynomials and that the linearity of the equation (in \phi = f/g) together with the fact that f,g are coprime is the way to go. But I barely understand the element \phi here.

 

[Barre]

One last bump :)