This is exercise II.4.7 in Shafarevich's BAG. The task is to show that if $X$ is a smooth projective curve and $O$ is a ring satisfying:
- $k\subset O \subset k(X),$
- $O$ is local with a principal maximal ideal $(m)$,
- $k(X)$ is the field of fractions of $O$,
then for any $u\in k(X)$ either $u\in O$ or $u^{-1}\in O$. This is the first step in showing that all such rings are actually exactly the local rings $O_x$.
I start by noting that any $u$ may be written as $\frac{p}{q}$ for $p,q\in O$. Assume that $u\notin O$. Then $q\in (m)$, otherwise it would be invertible and we would have $u=pq^{-1}\in O$. If $p\in(u)$, we get $u^{-1}=qp^{-1}\in O$ and we are done. Otherwise, we can write $u=\frac{tm}{sm}=\frac{t}{s}.$ Then we may repeat this reasoning again. This yields either $u\in O$ or $u^{-1}\in O$ in a finite number of steps, unless we have $p,q\in\bigcap\limits_{n=1}^{\infty} (m^n)$.
If we additionally knew that $O$ is Noetherian, we would immediately eliminate this possibility. Moreover, if the exercise is correct as is, after a little bit more work we would show that $O=O_x$ for some $x\in X$, and the local rings of nonsingular points are Noetherian. That means that we must already be capable of showing that $O$ is Noetherian. So my question is: are the properties (1-3) really enough to show that $O$ is Noetherian?