Finding the fixed subfield of an automorphism subgroup of $k(x)$. I am dealing with the following field theory exercise:
Let $k$ be a field and $k(x)$ the field of rational functions over $k$.
Let
$$
\sigma_{0}(x)=x \quad \sigma_{1}(x)=\dfrac{1}{1-x} \quad \sigma_{2} (x) = \dfrac{x-1}{x}
$$
be three automorphisms. I have proved and I already know that the automorphisms on $k(x)$ are fixed by it's image on $x$, moreover they are exactly
$$
\left\{ \dfrac{a+bx}{c+dx} \Big| a,b,c,d \in k, ad-bc \neq 0 \right\}
$$
I already proved that those three automorphisms form a group, which is of course cyclic of order $3$ and it's generated by either non identity element. Let's call that group $G$.
My problem is that I cannot use the Fundamental Theorem of Galois theory as the extension $k(x)$ is not finite. But, in virtue of Luroth theorem I know that for every field $K$ such that $k \subset K \subset k(x)$, there exists $\alpha \in k(x)$ such that $K=k(\alpha)$. So, I would like to find some $\alpha \in k(x)$ such that the fixed field of $G$, $k(x)^{G}$ is equal to $k(\alpha)$. I guess that as $G$ is of order $3$, we have $k(x)/k(x)^{G}$ is an extension of order $3$.
Well, my only idea to start looking for a generator is the following, so I guessed that if $s(u,v,w)$ is a symmetric function on $3$ variables, then $s(x,\frac{1}{x-1},\frac{x-1}{x})$ is an element of $K:=k(x)^{G}$ as every automorphism on $G$ simply permutates those $3$. I know that we only need three rational functions to generate this.
That is
$$
\begin{array}{l}
u+v+w \\
\\
uv+uw+vw \\
\\
uvw
\end{array}
$$
The third one goes to $-1$ when I evaluate on the trio of rational funtions. And constants are obviously fixed. The other two go to two rational functions which have a constant as difference
https://es.symbolab.com/solver/rational-expression-calculator/t%2B%5Cfrac%7B1%7D%7B1-t%7D%2B%5Cfrac%7B%5Cleft(t-1%5Cright)%7D%7Bt%7D
https://es.symbolab.com/solver/rational-expression-calculator/t%2F%5Cleft(1-t%5Cright)%2Bt%5Ccdot%5Cleft(t-1%5Cright)%2Ft%2B1%2F%5Cleft(1-t%5Cright)%5Ccdot%5Cleft(t-1%5Cright)%2Ft
I used this site as I wanted to be sure that I had do the computations right.
I picked
$$
t+ \dfrac{1}{1-t}-\dfrac{1}{t} = \dfrac{t^{3}-t^{2}-2t+1}{t(t-1)}
$$
as I guessed it was somehow more simple than $t+ \dfrac{1}{1-t}-\dfrac{1}{t} +1 =t+ \dfrac{1}{1-t}+\dfrac{t-1}{t} $.
Well. So I gussed that perhaps that element generates the field. If we write it on reduced form as $f(x)/g(x)$, by examining Luroth's theorem proof I know that $x$'s minimal polynomial on $k( f(x)/g(x) )[T]$ is
$$
p(T)=f(T)-\dfrac{f(x)}{g(x)}g(T)
$$
And $deg_{T} P = \max (deg_{x} f, deg_{x} g)= 3$.
But I have the following problems. First, I'm not sure if I'm on the right track. Second, I don't know how might I be able to prove that THIS is my fixed field, and it's not larger. If it were a finite extension, I guess that this would be a work for Galois fundamental Theorem. Perhaps I should examine the proof and see what stills holds on the infinite dimensional case, but as I'm not sure if I'm on the right track I guessed that after this much time it would be wiser to at least ask on internet if this makes anysense before continuing.
Any hints?
Edit: I am looking at David Dummit's Abstract Algebra's book, and corollary 11 of chapter 14 states:
Let $G$ be a finite subgroup of automorphisms of a field $K$, and let $F$ be the fixed field. Then every automorphism of K fixing F is contained in G, i.e., $Aut(K/F)=G$, so that $K/F$ is Galois, with Galois group $G$.
So if $k(x)$ plays the role of $K$ and $K=k(x)^{G}$ the role of $F$, we should have that $|Aut(K/F)|=[K:F]$ and as I have proved that $|Aut(k(x)/K)|=3=[k(x)/k(f(x)/g(x))]$ and by construction $k(f(x)/g(x)) \subset K$, then
$$
3=[k(x):k(f(x)/g(x))] = [k(x)/K] \cdot [K/k(f(x)/g(x))] = 3 \cdot \cdot [K/k(f(x)/g(x))]
$$
So $[K/k(f(x)/g(x))]$, therefore $K = k(f(x)/g(x))$.
I am right? I am quite confused but I guess some of the confusion comes from a overdose of zoom and lack of interaction with a blackboard which helps to clear my mind.
Edit 2:
Let's see if I can write a complete solution in a somehow ordered and concise fashion.
Let $G= \left\langle \sigma_{1} \right\rangle$. We know that $k(x) : K$ where $K=k(x)^{G}$ is an extension of degree three as it's by definition a Galois extension. So, by the fundamental theorem of Galois theory, we have that $| k(x) : K| = |G|=3$. Let's be $L=k(\alpha)$ where $\alpha = \sigma_{0} (t) + \sigma_{1} (t) + \sigma_{2} (t)$. Clearly $\sigma_{1} (\alpha )  = \alpha$ and as $G = \left\langle \sigma_{1} \right\rangle$, $\alpha$ is fixed by $G$, that is $\alpha \in K$. 
Let $L=k(\alpha)$. We want to see that $L=K$.
$$
\alpha=
\dfrac{x^{3}-3x+1}{x^2-x}
$$
Observe that this is a irreducible fraction as $x=1,x=0$ are the roots of the polynomial of the denominator (which can be splitted into linear terms) and neither are roots of $x^{3}-3x+1$.
By studying a proof of Luroth theorem we know that the minimal polynomial of $x$ over $k(\alpha)$ is
$$
T^{3}-3T+1 - \alpha (T^{2}-T)
$$
which has clearly degree $3$ on $k(\alpha)[T]$. So $[k(x):k(\alpha)]=3$.
Then
$$
[K:L]=\dfrac{[k(x):K]}{[k(x):L]} = \dfrac{3}{3}=1
$$
So $L=K$, that is $k(x)^{G}=k(\alpha)$.
 A: Let $L=\mathbf{Q}(X)$ and $G$ be the cyclic subgroup of order $3$ generated by $\sigma:X\mapsto \frac{1}{1-X}$. You want to find a primitive element for the fixed field $L^G$.
You are on the right track, you just have to recall the following definition of finite Galois extensions (which is equivalent to the "normal and separable" definition.)
Definition. A field extension $K\subset E$ is finite Galois if there exists a finite subgroup $G<\operatorname{Aut} L$ such that $E^G=K$.
In this case, we have $K=L^G$ and $E=\mathbf{Q}(X)$. (For this problem, you don't have to worry about the extension $\mathbf{Q}(X)\supset \mathbf{Q}$. By the way, this extension is transcendental, so we can't even speak about the "degree" as we would for an algebraic extension.)
By the Galois correspondence, $[L:L^G]=|G|=3$. The rational function
$$X+\frac{1}{1-X}+\frac{X-1}{X}=\frac{X^3-3X+1}{X(X-1)}=:f(X)$$
is clearly invariant under $G$, so $\mathbf{Q}(f)\subset L^G$.
Furthermore, $X$ is a root of $T^3-3T+1-f(X)T(T-1)\in \mathbf{Q}(f)[T]$, so the minimal polynomial of $X$ over $\mathbf{Q}(f)$ is of degree $\leqslant 3$, i.e. $[L:\mathbf{Q}(f)]\leqslant 3$.
This forces $L^G=\mathbf{Q}(f)$, as wished.
