5
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ The notation $R^*$ for the integral closure isn't the best one. More often it's denoted by $\overline{R}$ and this is consistent with the notation of algebraic closure of a field. $\endgroup$
    – user26857
    Jan 16, 2017 at 21:38

1 Answer 1

0
$\begingroup$

First note that the localization of $R^*$ at a maximal ideal is indeed a valuation ring of $L$ extending the $R$. As you said there are only finitely many of those.

Now, let $(\mathcal{O},\mathcal{M})$ be a valuation ring of $L$ extending $R$. Then $R\subset \mathcal{O}$ and $\mathcal{O}$ is integrally closed so $R^*\subset \mathcal{O}$. Consider $\mathcal{P}=\mathcal{M}\cap R^*$. Since $$\mathcal{P}\cap K=\mathcal{M}\cap R^*\cap K=\mathcal{M}\cap R=\mathfrak{m},$$ the ideal $\mathcal{P}$ lies over $\mathfrak{m}$, so it is a maximal ideal of $R^*$. It follows that $R^*_\mathcal{P}\subset \mathcal{O}$ and, as noted above, $R^*_\mathcal{P}$ is a valuation ring. Note also, that $\mathcal{M}\cap R^*_\mathcal{P}=\mathcal{P}R^*_\mathcal{P}$ the maximal ideal of $R^*_\mathcal{P}$. In other words, $(\mathcal{O},\mathcal{M})$ dominates $(R^*_\mathcal{P},\mathcal{P}R^*_\mathcal{P})$. Finally recall that valuation rings are maximal with respect to the domination relation of local rings. Hence $(\mathcal{O},\mathcal{M})=(R^*_\mathcal{P},\mathcal{P}R^*_\mathcal{P})$ and all extensions of $R$ to $L$ are localizations of $R^*$ at maximal ideal.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .