# F-automorphism group of the field of rational functions

1. Jun 4, 2012

### 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.

1. The problem statement, all variables and given/known data
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 $K$.

2. Relevant 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 $H \subset E'$.

3. 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.

Last edited: Jun 4, 2012
2. Jun 5, 2012

### 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.

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.

3. Jun 12, 2012

### Barre

One last bump :)