valuation ring of a field
In this article, $K$ is a field with a nontrivial nonarchimedean absolute value^{} (valuation) $\cdot $ and ${K}^{*}$ its multiplicative group^{} of units (nonzero elements).
Proposition 1.

1.
$A{=}_{df}\{x\in K\mid \leftx\right\le 1\}$ is a ring, called the valuation ring^{} of $(K,\cdot )$,

2.
$$ is the unique maximal ideal^{} of $A$, and ${A}^{*}=\{x\in K\mid \leftx\right=1\}$,

3.
$K$ is the fraction field of $A$.
Proof.
For (1), note that $1\in A$, that $x,y\in A\Rightarrow \leftx\right\le 1,\lefty\right\le 1\Rightarrow \leftxy\right\le 1\Rightarrow xy\in A$, and that $x,y\in A\Rightarrow \leftxy\right\le \mathrm{max}(\leftx\right,\lefty\right)\le 1\Rightarrow xy\in A$.
For (2), it is obvious that $\U0001d52a+\U0001d52a\subset \U0001d52a$ and that $\U0001d52aA\subset \U0001d52a$ so that $\U0001d52a$ is an ideal. Clearly $A\U0001d52a=\{x\in K\mid \leftx\right=1\}$ which is obviously ${A}^{*}$ and the result follows from general considerations regarding units in a local ring^{}.
Finally, to prove (3), choose some $x\in K$ with $$ (to do this, choose any $z$ whose valuation is not $1$; then either $z$ or ${z}^{1}$ will suffice). Given $y\in {K}^{*}$, there is some $n$ such that $$, so that $y{x}^{n}\in A$ and thus
$$\frac{y{x}^{n}}{{x}^{n}}=y$$ 
is in the fraction field of $A$. ∎
We say that the absolute value $\cdot $ is discrete if $\left{K}^{*}\right$ is a discrete subgroup of ${\mathbb{R}}_{>0}$. Note that ${\mathbb{R}}_{>0}\cong (\mathbb{R},+)$ via $\mathrm{log}$, so discrete subgroups are isomorphic^{} to $\mathbb{Z}$ (are a lattice in $\mathbb{R}$), and thus a discrete absolute value is of the form $\left{K}^{*}\right={\alpha}^{\mathbb{Z}}$ for some $\alpha \ge 1$, and $\alpha =1$ corresponds to the trivial absolute value.
Proposition 2.
In the notation of the preceding theorem, TFAE:

1.
$A$ is principal

2.
$\cdot $ is discrete

3.
$A$ is Noetherian^{}
If any of these hold, $A$ is a discrete valuation ring (DVR).
Proof.
($1\Rightarrow 2$): If $A$ is principal, then $\U0001d52a=(\pi )$ with $$. Since $A$ is a UFD, any element $x\in A\{0\}$ can be written uniquely as $x=u{\pi}^{n}$ for $u\in {A}^{*},n\ge 0$, and then $\leftx\right=\leftu\right\cdot {\left\pi \right}^{n}={\left\pi \right}^{n}$. Thus $\leftA\{0\}\right={\left\pi \right}^{\mathbb{N}}$ and $\left{K}^{*}\right={\left\pi \right}^{\mathbb{Z}}$ so that $\cdot $ is discrete.
($2\Rightarrow 1$): If the absolute value is discrete, we may choose $\pi \in {K}^{*}$ with $$ but with the largest possible absolute value strictly less than $1$. Then for $x\in \U0001d52a$, we have $$, so $\leftx\right\le \left\pi \right$ and thus $\left{\displaystyle \frac{x}{\pi}}\right\le 1$ so that $\frac{x}{\pi}}\in A$. It follows that $x\in \pi A=(\pi )$, so $A$ is principal.
Clearly principal implies Noetherian, so it suffices to prove that $3\Rightarrow 2$: if $\cdot $ is not discrete, then $A$ is not Noetherian. But if the absolute value is not discrete, we can choose a convergent sequence of absolute values and, using the fact that the valuations form an additive subgroup^{} of $\mathbb{R}$, we can find a convergent sequence $({r}_{n})$ with ${r}_{n+1}>{r}_{n}$, $lim{r}_{n}=1$, and a sequence of elements of $A$ with $\left{x}_{n}\right={r}_{n}$. Now consider ${I}_{n}=\{x\in A,\leftx\right\le {r}_{n}\}$. Then
$${I}_{1}\subset \mathrm{\cdots}\subset {I}_{n}\subset {I}_{n+1}\subset \mathrm{\cdots}$$ 
and ${x}_{n+1}\in {I}_{n+1}\backslash {I}_{n}$, so that $A$ is not Noetherian.
The fact that $A$ is a DVR follows trivially if any of these conditions holds. ∎
Title  valuation ring of a field 
Canonical name  ValuationRingOfAField 
Date of creation  20130322 19:03:25 
Last modified on  20130322 19:03:25 
Owner  rm50 (10146) 
Last modified by  rm50 (10146) 
Numerical id  4 
Author  rm50 (10146) 
Entry type  Theorem 
Classification  msc 11R99 
Classification  msc 12J20 
Classification  msc 13A18 
Classification  msc 13F30 
Related topic  HenselianField 
Related topic  RingOfExponent 
Defines  valuation ring 
Defines  discrete valuation 