Let (R,m) be a non-field local domain with fraction field Q(R) . Let kR:=R/m.

We know that there is a Valuation ring (V,mV) such that R⊆V⊊ 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.

**Answer**

Yes, an algebraic extension is always possible. If one applies Zorn’s lemma not to

\{R\subseteq V\subseteq Q(R) \mid V\,\text{local}\}

but to

\{R\subseteq V\subseteq Q(R) \mid V\,\text{local} \wedge R/\mathfrak{m}_R \subseteq V/\mathfrak{m}_V \,\text{algebraic}\}

instead, one can prove the existence of maximal elements and show that those are valuation rings as well using the usual proof as a blue print.

If it is possible to obtain finite extensions, I don’t know. In fact I asked this a while back on MSE but got no satisfactory answer.

**Attribution***Source : Link , Question Author : Community , Answer Author : Johannes Hahn*