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 $m=2$).