Let $R$ be a Noetherian domain with fraction field $K$. Let $\overline{R}$ be the integral closure of $R$ in $K$. Is it possible to prove that $\overline{R}$ is a finitely generated $R$-module?
The condition of the original problem is that $R$ is a finitely generated module of its PID subring $S$.
Eisenbud has a discussion of this question in Section 13.3 of Commutative Algeba with a View Toward Algebraic Geometry, as does Matsumura in Section 33 of Commutative Ring Theory. If $R$ is a domain that is a finitely generated algebra over a field, then the integral closure of $R$ is finitely generated as an $R$-module. If $R$ is a complete Noetherian local domain, then the integral closure of $R$ is finitely generated as an $R$-module.
In 1935/6, Akizuki and Schmidt (separately) constructed examples of one-dimensional local Noetherian domains $R$ whose integral closure is not finitely generated as an $R$-module. There are many more examples known today. For instance, Goodearl and Lenagan gave a nice construction in 1989 of such examples here, though they are still a bit involved.
The Krull-Akizuki Theorem says that in the examples $R$ above (of dimension $1$), the integral closure is a Noetherian ring; this is also true if $R$ is $2$-dimensional. In 1953, at the link given by @mathmath, Nagata constructed a $3$-dimensional local Noetherian domain whose integral closure is not Noetherian.
