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I have the following question after reading Chapter V, prop. 4.3 of Miranda's book Algebraic Curves and Riemann Surfaces.

The setting is as follows: we have a Riemann surface $X$ and a holomorphic map $\phi: X\to \mathbb P^n$. We let $(x_0:\cdots : x_n)$ be the homogeneous coordinates on $\mathbb P^n$. The goal is to prove that $\phi$ can be defined by an $(n+1)$-tuple of meromorphic functions on $X$.

He proceeds as follows: assume that $x_0$ is not identically zero on $\phi(X)$ and define $f_i$ on $X$ to be the composition of $\phi$ with the function $x_i/x_0$. Next, he wants to prove that every such $f_i$ is meromorphic on $X$: locally around a point $p\in X$ the map $\phi$ is given by $\phi(z) = (g_0(z):\cdots :g_n(z))$ where, after fixing some local coordinate $z$ at $p$, each $g_i$ is a holomorphic function of $z$.

Then he says: note that $g_0$ is not identically zero near $p$. This should apparently contradict the fact that $x_0$ is not identically zero on $\phi(X)$.

What is the easiest way to make this completely rigorous? Thanks in advance!

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Every point has a connected coordinate neighbourhood $U$ such that on $U$ the map $\phi$ is given by $n+1$ holomorphic functions $g_0^U,\dotsc,g_n^U$.

Let $A$ be the union of all such $U$ where $g_0^U \equiv 0$, and $B$ the union of such $U$ where $g_0^U \not\equiv 0$.

If $U,V$ are such coordinate neighbourhoods with $U\cap V \neq \varnothing$, then there is a $\lambda\neq 0$ such that $g_k^U \equiv \lambda g_k^V$ on $U\cap V$, so if $g_0^U \equiv 0$, then also $g_0^V \equiv 0$ by the identity theorem, and vice versa. So for any two overlapping [connected] coordinate neighbourhoods $U,V$, either both of $g_0^U,g_0^V$ vanish identically or neither. Hence $A\cap B = \varnothing$. But evidently $X = A \cup B$, and both, $A$ and $B$ are open. Since $X$ is connected, that means $A = \varnothing$ or $B = \varnothing$. If we had $B = \varnothing$, that would mean that $x_0 \equiv 0$ on $\phi(X)$, contradicting the assumption.

Hence none of the $g_0^U$ vanishes identically.

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