Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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.

share|improve this answer

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 ${x: x \ \mathrm{has \ valuation \ at \ least \ } n}.$)

share|improve this answer
2  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.