Let $\mathsf{k}$ be a field of characteristic $0$, and consider $\mathsf{k}(x,y)$.


If $\mathsf{k}$ is algebraically closed, then every field $L$ such that the inclusion $\mathsf{k} \subset L \subset \mathsf{K}(x,y)$ holds is a purely transcendental extension of the base field (i.e., Castelnuovo's Theorem implies a positive solution to the Lüroth problem in two dimensions).


Now suppose that $\mathsf{k}$ is **not algebraically closed**.


Question: can we have a finite group $G$ of field automorphisms of $\mathsf{k}(x,y)$, fixing $\mathsf{k}$, such that $\mathsf{k}(x,y)^G$ is not a purely transcendental extension of $\mathsf{k}$?


I am looking for an explicit example of $G$ such that $\mathsf{k}(x,y)^G$ is not rational.



An explicit negative solution to the Lüroth problem for non-algebraically closed fields
According to the first paragraph in Shafarevich's paper "On Luroth's problem" (found here [http://www.math.ens.fr/~benoist/refs/Shafarevich.pdf](http://www.math.ens.fr/%7Ebenoist/refs/Shafarevich.pdf)) the field of rational functions on the surface $z^2+y^2=x^3-x$ over $\mathbb{R}$ is an example of a non-rational field $F$, containing $\mathbb{R}$ of transcendence degree $2$, that embeds in $\mathbb{R}(u,v)$ (fixing $\mathbb{R}$). I don't know whether or not this embedding can be chosen so that $\mathbb{R}(u,v)/F$ is Galois.


