# Local subring of a DVR and finite residue field extension

Let $\mathcal O$ be a complete DVR with fraction field $K$, maximal ideal $\mathfrak p$ and residue field $\widetilde K=\mathcal O/\mathfrak p$. Now consider a subring $A\subset \mathcal O$ with the following properties:

1. $A$ is a local ring with maximal ideal $\mathfrak m$ and residue field $L=A/\mathfrak m$
2. $\mathcal O$ is the integral closure of $A$ in $K$.

Then, is it true or not the following claim?

$\widetilde K|L$ is a finite field extension

This is not true. Consider the following counterexample: denote by $\overline{\Bbb{Q}}$ the algebraic closure of $\Bbb{Q}$.
Call $\mathcal{O} = \overline{\Bbb{Q}} [[X]]$ (it is well known that this is a complete DVR), $\mathfrak{p}$ its maximal ideal, and $A= \Bbb{Q} + \mathfrak{p}$.
Then $A$ is a local subring of $\mathcal{O}$ with maximal ideal $\mathfrak{p}$, and since $\mathcal{O}$ is an integral extension of $A$, it is its integral closure.
But now, $[\mathcal{O} / \mathfrak{p} : A / \mathfrak{p} ] = [\overline{\Bbb{Q}} : \Bbb{Q}] = \infty$.
• Is there some condition we can impose on $A$ to obtain a positive result? – ginevracoal Mar 22 '16 at 22:06