I have come across the following result in Hartshorne, Algebraic Geometry, I.6.5 for those who have the book.
The result says that if $K$ is a finitely generated extension of some base (algebraically closed) field $k$ of transcendence degree $1$, and if $\mathcal{C}_{K}$ is the set of discrete valuation rings of the extension $K/k$, then we have that the set $\left\lbrace R \in \mathcal{C}_{K} \mid x \notin R \right\rbrace $ is finite.
So basically, if we take a rational function $x$ on a smooth projective curve, then the number of discrete valuations that $x$ is an element of is finite. Would someone be able to give a geometric interpretation of this result? I think that would vastly improve my understanding of the underlying commutative algebra. Any insight would be greatly appreciated.
Thanks
The intuition is that if you write $K=\text{Quot}(R)$ for some discrete valuation ring $R$ with prime element $\pi$, then this presentation corresponds to fixing a point on the curve, and writing an element ('meromorphic function') of $K$ in the form $\pi^k\varepsilon$ with $k\in{\mathbb Z}$ and $\varepsilon\in R^{\times}$ determines whether it has zero of order $k$ (for $k\geq 0$) or a pole of order $-k$ (for $k\leq 0$) at that point. Therefore, fixing $x\in K$ and looking at $R$ such that $x\notin R$ means looking at the set of points at which $x$ has a pole, of which there should be only finitely many.