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Do we have some good examples of local subrings of number fields which are not valuation rings?

Do we have an easy criterion for determining whether a local subring of a number field is a valuation ring? There are certainly many such criteria on Wikipedia and Atiyah-Macdonald. What I want to ask for is a criterion which use in an essential way the condition of being a subring of a number ring.

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    $\begingroup$ There are lots of examples. For instance, let $A={\mathbb Z}[\sqrt{-5}]$, $P = (2, 1+\sqrt{-5})$, $R=A_P$, and ${\mathfrak m} = P A_P$. Then $\mathfrak m$ is finitely generated but isn't principal, so $R$ isn't a valuation ring. As for a criterion, I believe that a local subring of a number field is a valuation ring iff its maximal ideal is principal. $\endgroup$
    – Neil Epstein
    Aug 20, 2014 at 2:05
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    $\begingroup$ Bad example : \mathfrak{m} is in fact generated by $1+\sqrt{-5}$ (note that $3$ is invertible in $R$, and $2.3=(1+\sqrt{-5})(1-\sqrt{-5}))$. Replace $\sqrt{-5}$ by $\sqrt{5}$. The point is that $\frac{1}{2}(1+\sqrt{5}) $ is integral over $R$ but does not belong to $R$: a local integral domain of Krull dimension 1 is a DVR iff it is integrally closed. $\endgroup$
    – abx
    Aug 20, 2014 at 5:29
  • $\begingroup$ The problem with the above criterion is that it is probably quite a good criterion to show that something is not a valuation ring, but not very good at showing something is a valuation ring (which is what I need). $\endgroup$
    – abcdxyz
    Aug 20, 2014 at 6:30
  • $\begingroup$ @abx Sorry for my typo (brain-o?), and thanks for catching it. $\endgroup$
    – Neil Epstein
    Aug 22, 2014 at 3:50

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A localization of a Dedekind ring at a (non-zero) prime is a Dedekind ring with a unique non-zero prime, hence finitely-many, hence is a principal ideal domain with a unique non-zero prime...

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  • $\begingroup$ Thanks for answering but this is not what I need. My problem essentially involved proving the localization of a non Dedekind ring is a valuation ring. I managed to prove it but the proof is too long, so i wonder if there is some easy usable result somewhere. $\endgroup$
    – abcdxyz
    Aug 20, 2014 at 16:32
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    $\begingroup$ How is your non-Dedekind ring described, then? Is it an "order" of a ring of algebraic integers, for example? If so, then localizing at any prime not dividing the "conductor" will make the localization be the same as the localization of the full ring of algebraic integers... Or what context do you have? $\endgroup$
    – paul garrett
    Aug 20, 2014 at 16:51
  • $\begingroup$ Yes, the ring I am concerning about is an order. Thanks for the answer. $\endgroup$
    – abcdxyz
    Aug 20, 2014 at 19:41

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