Let $S$ be a regular surface over a field $k$ and $C$ an irreducible curve on it. if $p \in C$ is a (closed) point, then I know that $C$ is cut out locally (in $\mathcal O_{S,p}$) by one function.
The proof (sketch) I came up with uses some heavy tools, namely that $\mathcal O_{S,p}$ is a regular local ring and hence factorial. Then, if $f \in \mathcal O_{S,p}$ is such that $V(f) \supset C$, we can find a factor $g|f$ such that $V(g) = C$.
This is because $V(f)$ is a union of irreducible curves and since $C$ is irreducible, one of these curves has to in fact be $C$. On the other hand, since $\mathcal O_{S,p}$ is a UFD, irreducible elements correspond to prime ideals and we are done.
Is there a simpler proof of this fact?
Edit: Ideally, I would like to completely avoid the result "regular local ring $\implies$ UFD" but this might be too much to ask for. However, perhaps there is an easier proof of this result in the case of surfaces (local ring is dimension 2)? There is a reasonable easy proof in the case of dimension one at least...