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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!

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up vote 4 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.

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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
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

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