How to show an invariant subfield of rational function field $\mathbb{Q}(x)$ under a certain group action is actually a simple extension? Let $K=\mathbb{Q}(x)$ be the rational functions in one variable $x$ and let the automorphisms $\phi,\psi$ of $K$ be defined as $\phi(x)=-\frac{1}{x+1}$ and $\psi(x)=\frac{1}{x}$.
Let $G$ be the group generated by $\phi,\psi$, then $G=\langle \phi,\psi|\phi^3=\psi^2=1,\phi\psi=\psi\phi^2\rangle  $.
To be specific, $ G=\{1,\phi,\phi^2,\psi,\phi\psi,\psi\phi\} $ and
$$ \phi(x)=-\frac{1}{x+1}\\ \phi^2(x)=-\frac{x+1}{x}\\ \psi(x)=\frac{1}{x}\\ \phi\psi(x)=-x-1\\ \psi\phi(x)=-\frac{x}{x+1} $$
let $K_0$ be the invariant subfield of $K$ under the $G$-action.
The point is to show that $K_0$ is a simple extension of $\mathbb{Q}$.
The threads I hold: Regard $K$ as a $G$-module then $K_0$ is nothing but the 0-th cohomology of $G$ with coefficient $K$ and $\text{Gal}(K/K_0)=G$. On the other hand set $N=\sum\limits_{g\in G}g$ to be the norm element of $G$, then there is a cannonical map $\alpha$ form $K$ to $K_0$ sending every rational function $f\in K$ to $Nf$. And I guess $\alpha$ is a surjection which I'm not sure. Notice that $Nx=-3$ and $N(x^2)=(t+1)^2+(\frac{1}{t+1})^2+t^2+\frac{1}{t^2}+(\frac{t+1}{t})^2+(\frac{t}{t+1})^2$ and I have a vague sense that $N(x^i)$ could be expressed in $N(x^2)$ for any $i\in\mathbb{Z}$. So I guess $K_0=\mathbb{Q}(N(x^2))$. Again, I'm not sure about this.
 A: As hinted in a comment, this is a special case of Lüroth's theorem, but it's not hard to find an explicit generator for $K_0$.
Note that $$x+\phi(x)+\phi^2(x) = \frac{x^3-3x-1}{x(x+1)}$$ is invariant under $\phi$, while $\psi$ interchanges it with $$\frac{1-3x^2-x^3}{x(x+1)}.$$ The sum of the two, as you've already noted, is a constant; but the product of the two or, for convenience, minus the product of the two is a non-constant element $$y=\frac{(x^3-3x-1)(x^3+3x^2-1)}{x^2(x+1)^2}$$
of $K$ and invariant under the whole group, thus in $K_0$. Rearranging this last displayed equation gives a monic equation for $x$ of degree 6 over $\mathbb{Q}(y)$, thus $\mathbb{Q}(y)$ is already all of $K_0$.
(One moral is: When life gives you a tower of field extensions, it's usually helpful to exploit this.)
A: The answer as to the surjectivity of $\alpha$ is no. As in algebraic number theory, the simplest way to prove that an element is not a norm is by local considerations. Let us consider
$$
y=\frac{(x^3-3x-1)(x^3+3x^2-1)}{x^2(x+1)^2},
$$
and ask whether $y$ is a norm of $\mathbb{Q}(x)$. Let us suppose that it is, and derive a contradiction. If $y$ is a norm, we would have
$$
y = \prod_{\sigma \in G} f(\sigma(x)) = f(x)f(\phi(x))f(\phi^2(x))f(\psi(x))f(\psi(\phi(x)))f(\psi(\phi^2(x))),
$$
for some rational function $f \in \mathbb{Q}(x)$, where $G \subset \operatorname{Aut}(\mathbb{Q}(x))$ and $\phi,\psi \in \operatorname{Aut}(\mathbb{Q}(x))$ are as you defined them.
Now, since $g=x^3-3x-1$ is an irreducible factor of the numerator of $y$, it must appear as a numerator in at least one of the six factors in the above product representation of $y$ as well (when written "in lowest terms", i.e. after cancelling any common factors of numerator and denominator). As can be easily checked, $\phi$ permutes the roots of $g$, which means $g$ must appear in either three or all six of the factors. Letting $\xi$ be one of the zeros of $g$, this would imply that the order of the zero of the function $y$ (say considered as a meromorphic function in the variable $x$) at $x=\xi$ is a multiple of three. However from the formula for $y$ it is clear that the order of the zero at $x=\xi$ equals $1$, contradiction.
