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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.

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    $\begingroup$ @PaceNielsen yes I wanted the field fixed. I wil edit the question. Nice argument though! $\endgroup$
    – jg1896
    Commented Jul 3, 2020 at 17:42
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    $\begingroup$ In positive characteristic if you allow finite group schemes, there are such examples with field of invariants general type. $\endgroup$
    – Mohan
    Commented Jul 3, 2020 at 19:17
  • $\begingroup$ @Mohan this is very interesting. Do you have an easy example/reference for such an example? $\endgroup$
    – jg1896
    Commented Jul 3, 2020 at 19:22
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    $\begingroup$ I think they are called Zariski surfaces. May be you could look them up. $\endgroup$
    – Mohan
    Commented Jul 3, 2020 at 19:25
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    $\begingroup$ This version of the Luroth problem for the fields of invariants is sometimes called the Noether problem: encyclopediaofmath.org/wiki/…. $\endgroup$
    – Evgeny Shinder
    Commented Jul 7, 2020 at 21:37

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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.

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    $\begingroup$ 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?) $\endgroup$
    – R.P.
    Commented Jul 3, 2020 at 18:45
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    $\begingroup$ 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. $\endgroup$
    – jg1896
    Commented Jul 3, 2020 at 19:08
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    $\begingroup$ 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. $\endgroup$
    – R.P.
    Commented Jul 3, 2020 at 19:21
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    $\begingroup$ 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. $\endgroup$
    – Evgeny Shinder
    Commented Jul 7, 2020 at 21:34
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    $\begingroup$ Higher degree unirational parametrizations of cubics with a rational point are explained in arxiv.org/pdf/math/0005146.pdf. $\endgroup$
    – Evgeny Shinder
    Commented Jul 7, 2020 at 21:36

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