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Let $A$ be a DVR in its field of fractions $F$, and $F'\subset F$ a subfield. Then is it true that $A\cap F'$ is a DVR in $F'$? I can see that it is a valuation ring of $F'$, but how do I I show, for example, that $A\cap F'$ is Noetherian?

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up vote 2 down vote accepted

Restrict the valuation corresponding to A to $F'$. Then the restriction is also a valuation with discreate value group( a subgroup of ring of integers) if it is non trivial. But if the value group is trivial, then it is field and therefore not a DVR.

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This is not necessarily true. For instance, you could have $A = k[[t]]$ for a field $k$ and $F' \subset k((t))$ equal to $k$. Then $A \cap F' = F'$ and a field is often not considered a DVR.

But this is the only thing that can go wrong. In general, if a domain has a nontrivial valuation taking values in the nonnegative integers, such that the elements of valuation zero are units, then it is a DVR (and noetherianness follows --- for instance, one checks that the only ideals occur as ${x: x \ \mathrm{has \ valuation \ at \ least \ } n}.$)

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it is a matter of convention whether a field is a DVR. I would actually say it is, albeit a trivial case of one. – Pete L. Clark Feb 27 '11 at 20:56
@Pete: Dear Pete, OK. I am simply following the conventions on Wikipedia. I've made a note in the answer, though. – Akhil Mathew Feb 27 '11 at 20:59

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