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!