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.

Another result I would really appreciate some help with:

Suppose $R$ is a DVR and let $K$ be its field of fractions. Let $L$ be a finite extension of $L$. Prove that any valuation domain inside of $L$ containing $R$ must also be a DVR.

This is exercise 11.2 from Matsumura's Commutative Ring Theory. I suppose it uses the Krul-Akizuki theorem but I don't see it. Thank you!

share|improve this question

1 Answer 1

up vote 3 down vote accepted

A valuation ring is noetherian if and only if it is a DVR. Krull Akizuki now concludes.

My claim is a rather classical result. Indeed, a DVR is noetherian. If $A$ is a noetherian valuation ring, let $\mathfrak m$ be its maximal ideal. Since $A$ is noetherian, it is generated by $n$ elements, $a_1,\ldots,a_n$. Choose the one with minimal valuation, let's say that it is $a_1$. Then $\frac{a_i}{a_1} \in A$, i.e. $\mathfrak m = (a_1)$. Hence $\mathfrak m$ is principal, and so $A$ is a DVR.

share|improve this answer
    
Indeed, but the two things are perfectly equivalent: if $A$ is a valuation ring, without any reference to a valuation, then simply define $\Gamma := K^* / U$, where $K$ is the fraction field of $A$ and $U = A^\times$ is the group of units of $A$. Then using the property you are saying, you can define an order over $\Gamma$ and the natural projection $K^* \to \Gamma$ becomes the valuation defining $A$. This is exercise 5.30 in Atiyah-Macdonald, for example. –  Mauro Porta Dec 4 '12 at 22:32
1  
Since mathematicians usually choose names for things with intelligence, I totally agree. I just wanted to sketch the idea of the construction, which is not totally obvious, imho. I didn't mean to attack you. –  Mauro Porta Dec 4 '12 at 22:40
    
Thank you! Nice lemma –  Dquik Dec 23 '12 at 15:55

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.