This is about from Forster, Lectures on Riemann Surfaces, Exercise 8.1.
The exercise is the following: ($\mathcal{M}(X)$ represents the field of meromorphic functions on $X$, and likewise for $Y$.)
Suppose $X$ and $Y$ are compact Riemann surfaces such that $\mathcal{M}(X)$ and $\mathcal{M}(Y)$ are isomorphic as $\mathbb{C}$ algebras. Prove that $X$ and $Y$ are isomorphic.
There's a hint which suggests representing $X$ and $Y$ as the Riemann surfaces of algebraic functions of the same polynomial $P$ from $\mathcal{M}(\mathbb{P}^1)[T]$ and the fact (which is proved later) that the meromorphic functions separate points, over a compact Riemann surface.
To get started, it seems like I need a branched holomorphic covering map $\pi:X\to \mathbb{P}^1$. One candidate seems like using an arbitrary non constant element of $\mathcal{M}(X)$ and consider that as a surjective map into $\mathbb{P}^1$. However I'm not quite sure how to proceed. To that end I would be very grateful if someone could give me a hint where to start.
(I am aware that something along these lines has already been asked here If two meromorphic function fields on compact riemann surfaces isomorphic, then they are isomorphic, but it doesn't shed much light on how I might use the hint with whatever Forster has developed so far.)
Best Answer
Let $K:= \mathcal{M}(\mathbb{P}^1)$ and $L_X := \mathcal{M}(X)$ and $L_Y :=\mathcal{M}(Y)$. Let $\Phi: L_X \to L_Y$ some $\mathbb{C}$-algebra isomorphism. As you said, take some non-constant function $f\in L_X$ (existence by the hint) and let $g:=\Phi(f)$. Then $f: X \to \mathbb{P}^1$ and $g: Y \to \mathbb{P}^1$ are branched holomorphic coverings, finitely sheeted, because $X$ and $Y$ are compact.
Now, $f^*:K\to L_X$ is a finite algebraic field extension of degree $\leq n$, if $f$ is $n$-sheeted. By the primitive element theorem there is some $F_X \in L_X$ with $f^*K[F_X] = L_X$. Let $P \in K[T]$ be the minimal polynomial of $F_X$, i.e. $$\rho_X:K[T]/(P) \to L_X, q(T) \mapsto (f^*q)(F_X)$$ is an isomorphism, in particular $(f^*P)(F_X) = 0$. But it is also $g^*:K\to L_Y$ a finite algebraic field extension and $g^* = \Phi\circ f^*$, since per definition $g = \Phi(f)$. Set $F_Y := \Phi(F_X)$. Then $$\rho_Y:K[T]/(P) \to L_Y, q(T) \mapsto (g^*q)(F_Y)$$ is an isomorphism, because $\rho_Y = \Phi\circ\rho_X$, since $\Phi(\rho_X([q])) = \Phi((f^*q)(F_X)) = \Phi(f^*q)(\Phi(F_X)) = (g^*q)(F_Y) = \rho_Y([q])$. In particular $(g^*P)(F_Y) = 0$. This means, per definition, that $(X,f,F_X)$ and $(Y,g,F_Y)$ are algebraic functions defined by the irreducible polynomial $P\in K[T]$.
By uniqueness of the triples there is exactly one fiber-preserving biholomorphic mapping $\varphi: X \to Y$. This proves the claim.