Let $(R,\mathfrak{p})$ be a local integral domain, $K=Quot(R)$ its field of quotients and $k:=R/\mathfrak{p}$ is residue field. Assume for simplicity that $R$ is integrally closed in $K$. It is well known that there exists valuation rings $R\subseteq\mathcal{O}\subseteq K$ that dominate $R$, i.e. $\mathfrak{P}:=J(\mathcal{O})$ satisfies $\mathfrak{P}\cap R=\mathfrak{p}$. Therefore $k \to \mathcal{O}/\mathfrak{P}$ is a field extension.
My question is simply this:
What field extensions of $k$ can occur this way? How small can it be? Does there exist a $\mathcal{O}$ with $\mathcal{O}/\mathfrak{P} = k$ ?
The usual proof of existence constructs $\mathcal{O}$ via Zorn's lemma as a maximal element in a set of rings with certain properties. Thus $\mathcal{O}/\mathfrak{P}$ is probably quite large if $\mathcal{O}$ is constructed in this way.
On the other hand, there are reasonable conditions one could impose in order for small valuation rings to exist. For example for Dedekind rings, every localisation is already a DVR so that the usual rings in algebraic number theory work fine. So there can also be very small $\mathcal{O}$ in certain circumstances. Also the set of all valuation rings between $R$ and $K$ is closed under intersections of chains so that there also exist minimal valuation rings $\mathcal{O}_0$ with $R\cap J(\mathcal{O}_0)=\mathfrak{p}$ by Zorn's lemma. However there does not seem to be any easy way to determine $\mathcal{O}_0/J(\mathcal{O}_0)$.
EDIT: I have found out that the usual existence proof of valuation rings can be modified to show that there exists at least one valuation ring such that $k\to \mathcal{O}/\mathfrak{P}$ is algebraic. There does not seem to be any way to make the degree or any other structural property of the extension $k\to \mathcal{O}/\mathfrak{P}$ visible though.