Lüroth's theorem for complex trigonometric polynomials Is it true that a subfield $K$ of $C_t(s)$ (the quotient field of the ring of trigonometric polynomials with complex coefficients) containing a non-constant trigonometric polynomial satisfies that $K=\mathbb C(r)$ for some trigonometric polynomial $r$? (This is Lüroth's theorem for complex trigonometric polynomials.)
 A: The answer is: yes, it is true.     
The field you are studying is $K:=\mathbb C(x,y)$ with $x,y$ transcendental over $\mathbb C$ satisfying $x^2+y^2=1$.   In the analytic context we have $x=\cos \theta,y=\sin  \theta$.
The well known parametrization  $x=\frac{1-t^2}{1+t^2}, y=\frac{2t}{1+t^2}$ with $t=\frac{y}{1+x}$ shows that $K=\mathbb C(t)$, the rational function field in one indeterminate.
We can thus apply the usual Lüroth theorem stating that any subextension  $\mathbb C\subset F\subset K$ is of the form $F=\mathbb C(\phi(t))=\mathbb C(\phi(\frac{y}{1+x}))=\mathbb C(\phi(\frac{\sin \theta}{1+\cos \theta}))$ for some rational function $\phi\in \mathbb C(t)$.
This shows that Lüroth's result also applies to complex trigonometric polynomials.
Edit
Consider the isomorphism invoked above $u:K=\mathbb C(x,y)\stackrel{\cong}{\to} \mathbb C(t):x\mapsto \frac{1-t^2}{1+t^2},y\mapsto \frac{2t}{1+t^2}$ and its restriction $u_\mathbb R:\mathbb R(x,y)\stackrel{\cong}{\to} \mathbb R(t).$
It is proved in the following paper by Cima, Gasull and Mañosas, Lemma 
3 
(reference friendly provided by @user26857)  that the image of $\mathbb 
R[x,y]$ under $u_\mathbb R$ is the set of rational functions $\frac 
{f(t)}{(1+t^2)^n}$ with $n\geq 0$ and $f(t)\in \mathbb R[t]$ a polynomial of degree $\leq 2n$.
This allows one to prove that not every subfield of $\mathbb R(x,y)$ is of the form $\mathbb R(P(x,y))$ with $P(x,y)\in \mathbb R[x,y]$ a polynomial rather than an arbitrary rational function.
In fact even $\mathbb R(x,y)$ itself cannot be written in this form!
Indeed, let's  transport the problem in terms of $t$:
The only generators over $\mathbb R$ of $\mathbb R(t)$ are fractions of the form $\frac {at+b}{ct+d}$ with $a,b,c,d\in \mathbb R, ad-bc\neq 0$ and it is clear that none of these fractions is of the form $\frac {f(t)}{(1+t^2)^n}$ .
Hence transporting back we see that we cannot write $\mathbb R(x,y)=\mathbb R(P(x,y))$  
Although the analogue of the  Proposition  by  Cima et al. is still valid over $\mathbb C$ (with the same proof) the  result I just showed is no longer valid over $\mathbb C$ because $1+t^2$ not irreducible over $\mathbb C$.
Indeed $\mathbb C(x,y)$ is generated by the single polynomial $ x+iy$ over $\mathbb C$. Explicitly:  $$\mathbb C(x,y)=\mathbb C(x+iy)=\mathbb C(\frac {1+it}{1-it})$$ and  $\frac {1+it}{1-it}$ is of the required form $\frac {at+b}{ct+d}$ if you allow complex coefficients $a,b,c,d$.
However I don't know whether an arbitrary subextension $ \mathbb C\subset \mathbb C(x,y)$ can also be written as  $\mathbb C(P(x,y))$ with $P(x,y)\in \mathbb C[x,y]$.
A: An attempt to prove that the answer to Claudia X's question is Yes - but with a gap:
Under the link
https://commons.wikimedia.org/wiki/File:Another_elementary_proof_of_Luroth%27s_theorem-06.2004.pdf
one can find a proof for Lüroth's theorem, that tells us something about the generator of the intermediate field (I didn't check the proof!). So let $M$ be subfield of the rational function field $k(X)$, $M\neq k$, $M\neq k(X)$. Let $p$ be the minimal polynomial of $X$ over $M$. Then in the article mentioned above it is shown that $M$ is generated by any of the coefficients of $p$, that does not lie in $k$.
In our situation $k=\mathbb{C}$. Moreover the ring of trigonometric functions with complex coefficients equals $\mathbb{C}[x,y]$, where $x^2+y^2=1$ as Georges has already explained. The algebraic curve over $\mathbb{C}$ given by this equation has no singularities, therefore the ring $\mathbb{C}[x,y]$ is integrally closed.
Due to the equation $(x+iy)(x-iy)=1$, the fraction field $\mathbb{C}(x,y)$ can be generated by the trigonometric polynomial $x+iy$.
Now let $M\neq\mathbb{C}$ be a proper subfield of $\mathbb{C}(x,y)$, then $\mathbb{C}(x,y)|M$ is a finite extension and the ring $R:=M\cap\mathbb{C}[x,y]$ is integrally closed in $M$.
If we would know that $\mathbb{C}[x,y]$ is integral over $R$, then we are done: the generator $x+iy$ is integral over $R$, hence every coefficient lies in $R$, which proves the assertion.
A: Strictly speaking the answer to the question is no, since $K$ has countable subfields: $\mathbf Q(\sin z)$.
A: The answer to Claudia X's question is Yes.
Let $K$ be an algebraically closed field of characteristic $\neq 2$. Consider the function field $F=K(x,y)$ with $x^2+y^2=1$ and the ring $T:=K[x,y]$. Let $z:=x+iy$, $i^2=-1$. Then $F=K(z)$ and $T=K[z,z^{-1}]$; in particular every $t\in T$ has the form $\frac{f}{z^m}$ for some $f\in K[z]$ not divisible by $z$.
Claim 1: A subfield $E\neq K$ of $F$ is generated by an element $t\in T\setminus K$ if and only if it is generated by an element of the form
$\frac{af+bz^m}{cf+dz^m},\; f\in K[z],\; a,b,c,d\in K, ad-bc\neq 0$.
Proof: $\Rightarrow$: let $E=K(t)$, $t=\frac{f}{z^m}$, $f\in K[z]$ not divisible by $z$. Every generator of $E$ then has the form
$
\frac{at+b}{ct+d}=\frac{a\frac{f}{z^m}+b}{c\frac{f}{z^m}+d}=\frac{af+bz^m}{cf+dz^m}
$
with $a,b,c,d\in K$, $ad-bc\neq 0$.
$\Leftarrow$: let
$
g:=\frac{af+bz^m}{cf+dz^m}
$
be a generator of $E$. Then
$
\frac{1}{ad-bc}\frac{dg-b}{-cg+a}=\frac{f}{z^m}\in T
$
is a generator of $E$.
Claim 2: A subfield $E\neq K$ of $F$ is generated by an element $t\in T\setminus K$ if and only if $T\cap E\neq K$.
Proof: let $t=\frac{f}{z^m}\in E\setminus K$, $f\in K[z]$ not divisible by $z$. Let $E=K(\frac{p}{q})$ with coprime polynomials $p,q\in K[z]$. Then there exist coprime polynomials $G,H\in K[X]$ such that
$
\frac{f}{z^m}=\frac{G(\frac{p}{q})}{H(\frac{p}{q})}.
$
Since both polynomials $G$ and $H$ split into linear factors $X-\gamma$, $\gamma\in K$, and since
$
\frac{p}{q}-\gamma=\frac{p-\gamma q}{q}
$
one gets 
$
\frac{f}{z^m}=q^n \frac{\prod\limits_{i=1}^r(p-\alpha_i q)^{e_i}}{\prod\limits_{j=1}^s(p-\beta_j q)^{f_j}},\; n\in\mathbb{Z},\quad (1)
$
where the elements $\alpha_i$ and $\beta_j$ are pairwise distinct.
The following properties are present:


*

*By assumption about $p$ and $q$, the polynomials $p-\alpha_i q$ und $p-\beta_j q$ possess no zero in common with $q$.

*Two polynomials $p-\alpha q$ und $p-\beta q$ with $\alpha\neq\beta$ possess no common zero.


Due to these properties in equation (1) no linear factors can cancel out. Since the denominator of the left-hand-side has $z$ as its only linear factor, only the following cases are possible:
Case 1: $m=0$.
Case 2: $s=1$.
Case 3: $H$ is constant.
In the first case the polynomial $H$ must be constant, thus one actually is in the third case.
In the second case case $p-\beta_1 q=z^\ell$ and therefore
$
\frac{p}{q}=\frac{\beta_1 q +z^\ell}{q}.
$
The first claim then yields that $\frac{q}{z^\ell}\in T$ is a generator of $E$.
In the third case $n<0$ hence $q=z^\ell$ must hold, which gives $\frac{p}{q}\in T$.
