Assume $L$ is a finite Galois extension of $K$, and $R$ is a valuation ring on $K$ with maximal ideal $\mathfrak{m}$. Is it true that there are only finitely many valuation rings $(O,\mathcal{M})$ on $L$ such that $R\subseteq O$ and $\mathfrak{m}=\mathcal{M}\cap R$?
I know the usual technique is that for every valuation $O$ ring extending $R$, the intersection with the maximal ideal $\mathcal{M}$ of $O$ with the integral closure $R^*$ of $R$ is a maximal ideal of $R^*$ that lies above $\mathfrak{m}$. I also know that given two ideals of $R^*$ lying above $\mathfrak{m}$, there is an automorphism of $Gal(L/K)$ that sends one into the other. Thus, there are at most many maximal ideals of $R^*$ lying above $\mathfrak{m}$ as automorphisms in $Gal(L/K)$, thus finitely many.
I also know that $R^*$ is the intersection of all valuations rings in $L$ extending $R$.
However, I still don't know if there might be infinitely many valuation rings whose maximal ideals intersected with $R^*$ will be equal.
On the other hand, I need the result to conclude that for every valuation ring $(O,\mathcal{M})$ extending $R$, $O$ is equal to the localization of $R^*$ by the maximal ideal $R^*\cap \mathcal{M}$, so I cannot use this last statement as part of my proof that there are finitely many valuation rings extending $R$.
The literature that I have read so far seem to always prove first several technical results (assuming $L$ is algebraic extension of $K$) and conclude as a corollary that there are only finitely many valuations when $[L:K]<\infty$. I was trying to find a shortcut in the case when we assume finite index from the beginning.
Thank you in advance for any suggestions of ideas or bibliography.