Let A be a DVR in its field of fractions F, and F'\subset F a subfield. Then is it true that P

Mary Reyes

Mary Reyes

Answered question

2022-01-12

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

Answer & Explanation

soanooooo40

soanooooo40

Beginner2022-01-13Added 35 answers

Step 1
This is not necessarily true. For instance, you could have A=k[[t]] for a field k and Fk((t)) equal to k. Then AF=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 has valuation at least n.)

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