# An explicit negative solution to the Lüroth problem for non-algebraically closed fields

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.

• @PaceNielsen yes I wanted the field fixed. I wil edit the question. Nice argument though! – jg1896 Jul 3 at 17:42
• In positive characteristic if you allow finite group schemes, there are such examples with field of invariants general type. – Mohan Jul 3 at 19:17
• @Mohan this is very interesting. Do you have an easy example/reference for such an example? – jg1896 Jul 3 at 19:22
• I think they are called Zariski surfaces. May be you could look them up. – Mohan Jul 3 at 19:25
• This version of the Luroth problem for the fields of invariants is sometimes called the Noether problem: encyclopediaofmath.org/wiki/…. – Evgeny Shinder Jul 7 at 21:37

According to the first paragraph in Shafarevich's paper "On Luroth's problem" (found here http://www.math.ens.fr/~benoist/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.
• This gives an example of a non-rational subfield of $k(x,y)$, but does it arise as the fixed field under some group $G$ of automorphisms of $k(x,y)$? (I don't think that is automatic, is it?) – RP_ Jul 3 at 18:45
• I second that. It does not seem clear to me how this field is of the form $\mathbb{R}(x,y)^G$ for an appropriate finite $G$. But I might be missing something. – jg1896 Jul 3 at 19:08
• Just to amplify: of course it is not automatic for a subfield $L$ of $k(x,y)$ (with the degree of the extension assumed finite) to be the field of invariants under some group $G$. If it were, then it would have to be Galois, but e.g. $k(x^3,y) \subset k(x,y)$ is not a Galois extension if $k$ does not contain a primitive 3rd root of unity. – RP_ Jul 3 at 19:21
• Geometrically, the question translates into degree of unirationality of this surface. Namely, if there is a degree two dominant rational map from $\mathbf{P}^2$ to this cubic, then the corresponding field extension would automatically be Galois. Typically, degree two parametrization of cubic hypersurfaces comes from a line defined over a ground field, as in math.uni-bonn.de/~huybrech/Notes.pdf, Corollary 1.18. However in this case the only line I see (at infinity) is passing through singular points, and the construction doesn't work. – Evgeny Shinder Jul 7 at 21:34