Let $\bar{K}$ be a perfect field and let $f \in \bar{K}(C)$ be a nonzero function in the function field of $C$, a smooth curve (projective variety of dimension 1). I'm trying to understand why (even though it feels obvious) any nonzero function $f$ has only finitely many zeros and poles. I only have a really minor knowledge on varieties - because I'm working through Silverman's The Arithmetic of Elliptic Curves. I've been trying to think in order to find a contradiction that occurs when $ord_P(f) > 0$ for all $P \in C$ to no avail. Is it that the set of all local rings, $\bar{K}[C]_P$ where $f \notin \bar{K}[C]_P$ is finite?
Some relevant definitions, $f$ has a zero at $P$ if $ord_P(f) > 0$ and a pole if $ord_P(f) < 0$.
This is a statement purely about being of dimension $1$, proper over $k$. Thus, an open neighborhood $U \subset C$ is of the form $U = Spec(R)$ for a ring of finite type over $k$ and of dimension $1$. It suffices to show that for $f \in R$, $V(f) \subset Spec(R)$ is finite.
By Noether normalization, there exists a finite extension $k[t] \hookrightarrow R$. Thus, we in particular have a surjective closed map with finite fibers $\varphi : Spec(R) \rightarrow Spec(k[t]) = \mathbb{A}^1_k$, where in $\mathbb{A}_k^1$ closed sets are finite.
Thus, $\varphi(V(f))$ is closed and hence finite. By finiteness of fibers, also $V(f)$ is finite.