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Let $\mathbb{Q}(x,y)$ be the field of rational functions in the variables $x, y$ with rational coefficients. Choose $p, q, r \in \mathbb{Q}(x,y)$ and consider the subfield $K=\mathbb{Q}(p,q,r)$. Is it generally true that there exist $s, t \in K$ such that $K=\mathbb{Q}(s,t)$?

Any help is welcome.

NOTE. This question is a particular, but very interesting case of question (II) of my previous post Subextensions of Finitely Generated Fields. In his answer to my other post Can $\mathbb{Q}(x^3,y^3,x+y)$ be generated by only two elements? professor René Schoof proved that the answer is positive for the particular choice $p=x^3, q=y^3, r=x+y$, which I had thought to give rise to a counterexample.

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    $\begingroup$ This is a much more complicated question than I expected. If I'm reading everything right, the answer is known to be yes if $\mathbb{Q}$ is replaced by $\mathbb{C}$ and no if $\mathbb{Q}$ is replaced by a field of characteristic $p$. See: en.wikipedia.org/wiki/Rational_variety $\endgroup$ Nov 23, 2020 at 20:21
  • $\begingroup$ @QiaochuYuan Dear Qiaochu, thank you very very ... much for having linked this Wikipedia page: it seems full of wonders! Unfortunately, it is almost completely unintelligible for me, since I have no knowledge at all of algebraic geometry. Well, I hope in any case to study it in the future, since it is fore sure one of the most fascinating fields of all mathematics, and also because algebraic geometry has been a favorite subject of the Italian mathematical school for a long period. As for this specific problem, I simply give up, because in any case its solution is beyond my present skills. $\endgroup$ Nov 24, 2020 at 9:59
  • $\begingroup$ @QiaochuYuan Thanks. By the way Maurizio maybe you don't know who is Schoof (who answered your previous question) ? $\endgroup$
    – reuns
    Nov 24, 2020 at 16:44
  • $\begingroup$ @reuns Dear reuns, I humbly admit my ignorance, I really never heard of professor Schoof before! I have just googled his name and .... Woooow! He is really a remarkable mathematician! What an incredibly undeserved honour that he kindly answered my question!!! $\endgroup$ Nov 25, 2020 at 10:46
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    $\begingroup$ For a tower $k(t)/F/k$ with $k(t)/F$ algebraic I found a proof that $F = k(f(t))$ similar to thisone except that for $k(t)/F$ non-normal the conjugates of $t$ won't be Möbius transformations so I'll need to count their number of poles on $M$ the smooth curve such that $k(M)$ is the normal closure of $k(t)/F$ (maybe assume first that $k$ is algebraically closed then extend to arbitrary $k$). Is it the standard way ? @QiaochuYuan $\endgroup$
    – reuns
    Nov 30, 2020 at 15:19

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