# ac.commutative algebra – 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 $$G=langle phi,psi|phi^3=psi^2=1,phipsi=psiphi^2rangle$$.

To be specific, $$G={1,phi,phi^2,psi,phipsi,psiphi}$$ and
$$phi(x)=-frac{1}{x+1}\ phi^2(x)=-frac{x+1}{x}\ psi(x)=frac{1}{x}\ phipsi(x)=-x-1\ psiphi(x)=-frac{x}{x+1}$$
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 $$N=sumlimits_{gin G}g$$ to be the norm element of $$G$$, then there is a cannonical map $$alpha$$ form $$K$$ to $$K_0$$ sending every rational function $$fin K$$ 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 $$iinmathbb{Z}$$. So I guess $$K_0=mathbb{Q}(N(x^2))$$. Again, I’m not sure about this.