# Discrete valuation ring associated with a prime ideal of a Dedekind domain

Let $A$ be a Dedekind domain. Let $K$ be the field of fractions of $A$. Let $P$ be a non-zero prime ideal of $A$. Let $v_P$ be the valuation of $K$ with respect to $P$. Then the localization $A_P$ of $A$ at $P$ is the valuation ring of $v_P$. How would you prove this?

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What's the reason for the downvote? Unless you make it clear, I cannot improve my question. –  Makoto Kato Aug 17 '12 at 6:07
There are some people who don't like the OP answering the question. I beg those people to consider the following points. (1) A good question is a valuable asset of this site, not a liability. This is true regardless whether the OP knows the answer or not. The audience can learn reading a question and its answers. (2) There are almost always several different answers to a question. So even if the OP posts a correct answer, this does not necessarily mean that the thread is over. –  Makoto Kato Aug 17 '12 at 10:18
(3) The OP's answer may not be perfect, may be wrong. (4) Answering one's own question is not only perfectly legitimate, but also encouraged in ths site. –  Makoto Kato Aug 17 '12 at 10:19
I noticed that someone serially upvoted for my questions. While I appreciate them, I would like to point out that serial upvotes are automatically reversed by the system. –  Makoto Kato Nov 27 '13 at 7:05

The ring $A_P$ is a discrete valuation ring because it is a local Dedekind domain, and one can show that, in general, a Dedekind domain with finitely many primes is a principal ideal domain (the argument uses the CRT). It follows that any element $x\in K^\times$ can be uniquely written as $\pi^nu$ where $\pi\in A_P$ is a uniformizer and $u\in A_P^\times$. By definition, $v_P(x)=n$, so $v_P(x)\geq 0$ if and only if $n\geq 0$, if and only if $x\in A_P$.

Alternatively, the valuation ring of $v_P$ is clearly a local extension of $A_P$, and since $A_P$ is a discrete valuation ring itself, the two must coincide, since a valuation ring is maximal for the relation of domination on local subrings of $K$ with fraction field $K$.

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Could you explain why $v_P(x) = n$? Regards, –  Makoto Kato Aug 17 '12 at 3:43
Let $\pi\in A$ be any element with $P$ exactly dividing $(\pi)$, i.e. $\pi\in P\setminus P^2$. Then $\pi$ will be a uniformizer in $A_P$, and $u$ is of the form $a/s$ with $a,s\in A\setminus P$. So $x=\pi^na/s$, and the definition of $v_P$ then gives $v_P(x)=nv_P(\pi)+v_P(a)-v_P(s)=n$. –  Keenan Kidwell Aug 17 '12 at 3:52
I'm sorry but I need to ask you again. Could you explain why $\pi$ is a uniformizer in $A_P$? Regards, –  Makoto Kato Aug 17 '12 at 4:20
Because we can write $(\pi)=PI$ where $I$ is an ideal prime to $P$. Then $\pi A_P=(PA_P)(IA_P)=(PA_P)A_P=PA_P$ is the maximal ideal of $A_P$. The reason $IA_P=A_P$ is that there is some $r\in I$ with $r\notin P$, and thus $r\in A_P^\times$, so $A_P=rA_P\subseteq IA_P$. –  Keenan Kidwell Aug 17 '12 at 4:23

Let $\pi \in P - P^2$. Let $a \in A -$ {$0$}. Suppose $v_P(a) = n$. We claim that there exist $t, s \in A$ such that $a = \pi^n t/s$ and both $t$ and $s$ are not divisible by $P$.

$aA = P^n I$, where $I$ is not divisible by $P$. $\pi^n A = P^n J$, where $J$ is not divisible by $P$.

By Chinese remainder theorem, there exists $s \in A$ such that

$s \equiv 0$ (mod $J$)

$s \equiv 1$ (mod $P$)

Then $sA = JM$, where $M$ is not divisible by $P$.

Since $(a/\pi^n)A = I/J$, $(sa/\pi^n)A = sI/J = JMI/J = MI$. Hence $saA = \pi^n MI$. Hence $sa = \pi^n t$, where $t \in A$ is not divisible by $P$. Then $a = \pi^n t/s$.

Let $\alpha \in K -$ {$0$}, $\alpha = a/b$ where $a, b \in A$. Suppose $v_P(\alpha) = n$. By the above result, $\alpha = \pi^n t/s$, where $t$ and $s$ are not divisible by $P$. Hence $v_P(\alpha) \geq 0$ if and only if $\alpha \in A_P$. This completes the proof.

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