Notation:
If $X$ is a Riemann surface, $\mathscr O(X)$ is the ring of holomorphic functions on $X$ and $\mathscr M(X)$ is the field of meromorphic functions on $X$.
If $X$ and $Y$ are two isomorphic Riemann surfaces, then is not true that $\mathscr M(X)\cong \mathscr M(Y)$ in fact for example $\mathscr M(\,\overline{\mathbb C}\,)=\mathbb C(z)$ while
$$\mathscr M(\mathbb P^1)=\left\{\frac{p}{q}\in\mathbb C(z,w)\,:\,\textrm{$p$ and $q$ are homogeneous with the same degree}\right\}$$
For holomorphic functions, is it true the following proposition?
$X$ and $Y$ are isomorphic as Riemann surfaces if and only if $\mathscr O(X)\cong \mathscr O(Y)$
Where can I find the proof?
Edit: The above statement about meromorphic functions is wrong. I apologize.
Best Answer
For isomorphic Riemann surfaces $X$ and $Y$, every biholomorphic $\varphi \colon Y \to X$ induces isomorphisms $\varphi^\ast \colon \mathscr{O}(X) \to \mathscr{O}(Y)$ and $\overline{\varphi}\colon \mathscr{M}(X) \to \mathscr{M}(Y)$ by composition with $\varphi$, we have $\overline{\varphi}(f) = f \circ \varphi$, and $\varphi^\ast$ is the restriction of $\overline{\varphi}$ to $\mathscr{O}(X)$.
However, for every compact Riemann surface $C$, we have $\mathscr{O}(C) \cong \mathbb{C}$, and since there are non-isomorphic compact Riemann surfaces, the isomorphism of the rings of holomorphic functions doesn't imply an isomorphism of the surfaces.
A $\mathbb{C}$-algebra isomorphism between the fields of meromorphic functions induces a biholomorphism between the surfaces, and for open surfaces also a $\mathbb{C}$-algebra isomorphism between the rings of holomorphic functions induces a biholomorphism.
Farkas/Kra, Riemann Surfaces, $2^{\text{nd}}$ edition, GTM 71, Springer-Verlag, prove that for the fields of meromorphic functions on compact Riemann surfaces in $\mathrm{IV}.11.16$ and following, and remark the analogue(s) for the open surfaces thereafter.