Field of algebraic functions We assume $K$ as a field of characteristic zero. By a field of algebraic functions of one variable over $K$ we mean a field $R$ satisfying $R=K(x,y)$ with $x$ being transcendental over $K$, and $R$ is algebraic over $K(x)$.
My question is: Whether there exist a subfield $F \subset R$ and $a \in R$ such that $R=F(a)$ with $a$ being transcendental over $F$. (In the case of rational function fields, this result follows easily.)
 A: I believe the answer is no, not in general. I construct a purported example below - I would scrutinize this carefully before believing it.
Assume that $K$ is algebraically closed.
From $R = F(a)$, with $a$ trascendental over $F$, we have that $\text{trdeg}_F(R) = 1$.
We also have, by assumption, that $\text{trdeg}_K(R) = 1$. By the fact that transcendence degree is additive in towers  (Lemma 9.26.5 in the stacks project), we have that $\text{trdeg}_K(R) = \text{trdeg}_K(F) +  \text{trdeg}_F(R)$, from which $\text{trdeg}_K(F) = 0$ follows.
Since an extension of transcendence degree $0$ is algebraic, this implies that $F$ is an algebraic extension of $K$. Since $K$ is algebraically closed, we have $K = F$.
Thus, $R = F(a) = K(a)$.
We recall that if $K\subset R$ is any field extension, and $a \in R$ is transcendental over $K$, then $K(a) \cong K(x)$, where $K(x)$ denotes the field of rational functions in formal variable $x$.
Now, if $R$ is the function field of any plane curve over $K$ that is not birational to $\mathbb{P}^1_k$, we have that $R = K(x,y)$, with $x$ transcendental over $K$ and $y$ algebraic over $K(x)$, but cannot have $R \cong K(x)$. An elliptic curve provides an example, for instance see this question.
Thus, we conclude that for such $R$, one cannot find a subfield $F$ with $R = F(a)$ and $a$ transcendental over $F$.
