Let $C$ be a reduced, irreducible projective curve (=1D proper $k$-scheme) and $f \in k(C)$ a nonconstant rational function. The number of singular points of $C$ is finite, Assume that $f$ is contained in every local ring at each singular point, i.e. $f \in O_{C,c}$ for every singular $c \in C$.
Let $C_0:= C \backslash (\text{poles of }f)$. Define $A=H^0(C_0, O_{C_0})$.
Q: Why $A$ is a finite $k[f]$-module?
We can (at least I think that it's equivalent) rephrase the question to: Forget the curve stuff. Let $k$-algebra $A$ be an $1$-dimensional, Noetherian integral domain and $f \in A \backslash k$. Is $A$ is a finite $k[f]$-module?
The question is motivated by my MO question and the claim was found in proof of Lemma 1.20.5 from Janos Kolloar's Lecture on Resolution of Singularities (page 19) and is related to finiteness of nonconstant maps $C \to \mathbb{P}^1$ as such maps are defined by rational functions.
The geometrical argument I'm familar with uses proper + quasi-finite => finite. Now I want to know if it possible to conclude that $A$ is a finite $k[f]$-module with pure methods from commutative algebra.