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


1 Answer 1


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, $[\mathcal{O} / \mathfrak{p} : A / \mathfrak{p} ] = [\overline{\Bbb{Q}} : \Bbb{Q}] = \infty$.

  • $\begingroup$ Is there some condition we can impose on $A$ to obtain a positive result? $\endgroup$ Mar 22, 2016 at 22:06
  • 1
    $\begingroup$ I don't think so. This problem sounds like asking: "suppose we have an algebraic extension of fields: under which assumptions it is a finite extension?". This question has no meaningful answers in my opinion. $\endgroup$
    – Crostul
    Mar 22, 2016 at 22:11

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