Let $k$ be an arbitrary field (I do not mind to take $k=\mathbb{C}$, if things are easier in this case).

A more general version of Lüroth theorem says that a field $L$, $k \subset L \subset k(x,y)$, of transcendence degree one over $k$ is equal to $k(h)$ for some $h \in k(x,y)$. Moreover, Schinzel in "Selected Topics on Polynomials" (Theorem 4, page 10) showed that if $L$ contains a nonconstant polynomial, then $h \in k[x,y]$ suffices (see this question).

Let $k \subset L \subset k(x_1,\ldots,x_n)$, $n \geq 2$, $L$ is a field.

If $L$ is of transcendence degree $m$ over $k$, $1 \leq m \leq n$, is it true that $L=k(h_1,\ldots,h_m)$ for some $h_1,\ldots,h_m \in k(x_1,\ldots,x_n)$?

I guess that the answer is negative, by what Georges Elencwajg has written in his answer to this question: "The analogue of Lüroth is in general false for the rational function fields $k(x_1,...,x_n) \; (n \gt 1)$ : its subfields are not all purely transcendental extensions of $k$."

However, perhaps when $m \in \{1,n\}$ the answer is positive? Am I right?

In cases that have a positive answer (if there exist such cases, except the above mentioned one), when it suffices to take $h_1,\ldots,h_m \in k[x_1,\ldots,x_n]$?

I hope that my questions are not trivial. Any comments are welcome!

Edit: After receiving a few helpful comments, I have asked this question (asking for a pure algebraic proof for the special case $m=2$).

  • $\begingroup$ The answer is negative for $m\geq 3$ for all fields $k$ (and for $m\geq 2$ for many fields). The value of $n$ is irrelevant. This is completely classical by now, just google "Lüroth problem". $\endgroup$ – abx Nov 26 '17 at 13:18
  • $\begingroup$ Both of your questions have a negative answer for every n at least 3, even over the complex numbers, because there are unirational varieties which are not rational. $\endgroup$ – Tippi Nov 26 '17 at 13:18
  • $\begingroup$ @abx, thank you! Please can you elaborate your comment? For which fields $m=2$ has a positive answer? Any recommended references? $\endgroup$ – user237522 Nov 26 '17 at 13:22
  • $\begingroup$ @Tippi, did you mean that for $k(x,y)$ and $m=2$, there is a positive answer? $\endgroup$ – user237522 Nov 26 '17 at 13:35
  • $\begingroup$ Is it true that the case $m=1$ has a positive answer for all $k(x_1,\ldots,x_n)$, $n \geq 2$? (by a similar proof to the proof for $k(x,y)$?). $\endgroup$ – user237522 Nov 26 '17 at 13:37

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