Valuation ring, containing a Noetherian local domain, whose fraction field is a finite extension of that of the Noetherian domain

Question

Let $R$ be a Noetherian local domain with fraction field $K$. Let $L$ be a finite extension field of $K$. Let $V$, containing $R$, be a valuation ring with fraction field $L$.

My question is: Is $V$ necessarily a Noetherian ring (i.e. is $V$ a DVR) ? If this is not true in general, what if we also assume $L=K$ ?.

Answer 1

Here's a helpful counterexample to keep in mind, which I'm stealing from Karl Schwede's MO answer.

Let $k$ be a field, and consider the chain of rings $k[x,y] \subseteq R = k[x,y,x/y,x/y^2, x/y^3, \ldots] \subseteq k(x,y)$.

Note $\mathfrak{m} = (x,y, x/y, x/y^2, x/y^3)R$ is a maximal ideal of $R$, which contracts to $\mathfrak{m}' = \mathfrak{m} \cap k[x,y] = (x,y)k[x,y]$.

Localizing we see that $R_\mathfrak{m}$ is an overring of the local Noetherian ring $k[x,y]_{\mathfrak{m}'}$. Moreover it can be shown that $R_\mathfrak{m}$ is a valuation domain of Krull dimension $2$.