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?
 A: 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 $K$.
A: 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.
