Let $(R, \mathfrak m)$ be a non-field local domain with fraction field $Q(R)$ . Let $k_{R}:=R/m$.

We know that there is a Valuation ring $(V,\mathfrak m_V)$ such that $R \subseteq V \subsetneq Q(R)$ such that $\mathfrak m_V \cap R=\mathfrak m$ (in fact any local ring between $R$ and $K$ maximal w.r.t. "dominance" is a valuation ring). Then $k_V:=V/\mathfrak m_V$ is an extension field of $k_R$.

My questions is: Let $(R, \mathfrak m)$ be a non-field local domain with fraction field $Q(R)$ and let $k_{R}:=R/m$ ;

Does there exist a Valuation ring $(V,\mathfrak m_V)$ such that $R \subseteq V \subsetneq Q(R)$ and $\mathfrak m_V \cap R=\mathfrak m$ and $V/\mathfrak m_V$ is a finite extension field of $R/m$ ? If this is not always possible, then can we at least have a valuation ring as above with the extension $V/\mathfrak m_V$ over $R/\mathfrak m$ being algebraic ?

UPDATE: As an answer by Johannes Hahn shows ... the algebraic extension claim is true .

The claim of existence of the dominating valuation ring with finite extension of residue field remains unsettled.

One might ask whether this finite extension claim can be answered when $R$ is integrally closed in $Q(R)$ or say if $R$ has small Krull-dimension.

divisorial valuationassociated with the prime divisor $V(x_1) \subseteq \operatorname{Spec} R$. It's just the classical order of vanishing of a function along a codimension $1$ subvariety. $\endgroup$ – R. van Dobben de Bruyn May 6 '18 at 17:463more comments