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Aug 9, 2014 at 13:10 vote accept Mikhail Bondarko
Jul 29, 2014 at 16:30 history edited Mikhail Bondarko CC BY-SA 3.0
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Jul 29, 2014 at 9:47 answer added Cantlog timeline score: 4
Jul 26, 2014 at 6:35 comment added Mikhail Bondarko This is probably true. Yet do you now any references for this (where some terms are introduced)?
Jul 25, 2014 at 21:31 comment added Will Sawin I believe one wants to consider models of $F$ of finite type only. Then one needs $A$ to be a localization of a finitely generated subring of itself - this is clearly a necessary and sufficient condition. I don't know a better one.
Jul 25, 2014 at 20:01 comment added Mikhail Bondarko Possibly I am getting something wrong; yet in the 'geometrical' case there are 'bad' valuations; see mathoverflow.net/questions/135544/…
Jul 25, 2014 at 15:53 comment added Daniel Loughran Perhaps I am misunderstanding your question, but don't all discrete valuations on $F$ arise this way? Namely, given a discrete valuation $v$ on $F$, one can consider the associated discrete valuation ring $A \subset F$. Then $\mbox{Spec } A$ is a model for $F$, whose point of codimension $1$ gives rise to the required discrete valuation $v$.
Jul 25, 2014 at 15:48 comment added abx Any closed irreducible subvariety gives rise to a discrete valuation, because you can blow it up and get a divisor.
Jul 25, 2014 at 13:07 history asked Mikhail Bondarko CC BY-SA 3.0