Is the integral closure of an integrally closed Noetherian domain in a finite extension field Noetherian? Just as the title says.  

Let $R$ be a Noetherian integral domain, let $K$ be its field of fractions, let $L$ be a finite extension of $K$, and let $S$ be the integral closure of $R$ in $L$.  Must $S$ be Noetherian, or do I need some additional assumptions on $R$?

EDIT: I meant to assume that $R$ itself is integrally closed in $K$ to start with.  Does that change things?
 A: This is true if $\dim R = 1$, and is known as the Krull-Akizuki theorem. In fact, it is commonly stated with the stronger conclusion that any subring $S \subseteq L$ containing $R$ is Noetherian. If $\dim R = 2$, it is still true that $\overline{R}^L$ is Noetherian, although there may be subrings of $L$ that are not. In dimension $3$ though, Nagata has given examples of $3$-dimensional Noetherian domains whose integral closures are not Noetherian.
If $R$ is a Nagata ring (e.g. an excellent ring, and thus virtually any geometric ring), then the desired conclusion holds.
A: If $L/K$ is a finite separable extension, then $S$ is Noetherian.
Lemma. $S$ is a finitely generated $R$-module.
Theorem. $S$ is a Noetherian ring.
Proof. Suppose $I \subset S$ is an ideal. By the lemma above, $S$ is a finitely generated $R$-module. Since $R$ is Noetherian, $I$ is a finitely generated $R$-module. Of course $I$ is also a finitely generated $S$-module, i.e., a finitely generated ideal of $S$, hence $S$ is Noetherian. 
