Proposition 1    

1. $A=_{df}\{x\in K\suchthat \Abs x\leq 1\}$ is a ring, called the valuation ring of $(K,\Abs{\cdot})$ ,
2. $\smm =_{df}\{x\in K\suchthat \Abs x < 1\}$ is the unique maximal ideal of $A$ , and $\U{A}=\{x\in K\suchthat \Abs x =1\}$ ,
3. $K$ is the fraction field of $A$ .

Proof. For (1), note that $1\in A$ , that $x,y\in A\Rightarrow \Abs x\leq 1, \Abs y \leq 1\Rightarrow\Abs{xy} \leq 1\Rightarrow xy\in A$ , and that $x,y\in A\Rightarrow \Abs{x-y}\leq\max(\Abs x,\Abs {-y})\leq 1\Rightarrow x-y\in A$ .

For (2), it is obvious that $\smm+\smm\subset\smm$ and that $\smm A\subset\smm$ so that $\smm$ is an ideal. Clearly $A-\smm = \{x\in K\suchthat \Abs x = 1\}$ which is obviously $\U{A}$ and the result follows from general considerations regarding units in a local ring.

Finally, to prove (3), choose some $x\in K$ with $\Abs x < 1$ (to do this, choose any $z$ whose valuation is not $1$ ; then either $z$ or $z^{-1}$ will suffice). Given $y\in\U{K}$ , there is some $n$ such that $\Abs y\cdot\Abs{x}^n<1$ , so that $yx^n\in A$ and thus $$ \frac{yx^n}{x^n}= $$ is in the fraction field of $A$ .

Proposition 2   In the notation of the preceding theorem, TFAE:

1. $A$ is principal
2. $\Abs\cdot$ 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 $\smm=(\pi)$ with $\Abs{\pi} < 1$ . Since $A$ is a UFD, any element $x\in A-\{0\}$ can be written uniquely as $x=u\pi^n$ for $u\in\U{A}, n\geq 0$ , and then $\Abs x = \Abs u \cdot \Abs{\pi}^n = \Abs{\pi}^n$ . Thus $\Abs{A-\{0\}} = \Abs{\pi}^{\BN}$ and $\Abs{\U{K}} = \Abs{\pi}^{\BZ}$ so that $\Abs \cdot$ is discrete.

($2 \Rightarrow 1$ ): If the absolute value is discrete, we may choose $\pi\in \U{K}$ with $\Abs{\pi}<1$ but with the largest possible absolute value strictly less than $1$ . Then for $x\in\smm$ , we have $\Abs x < 1$ , so $\Abs x\leq \Abs{\pi}$ and thus $\displaystyle \Abs{\frac{x}{\pi}}\leq 1$ so that $\displaystyle \frac{x}{\pi}\in A$ . 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 $\Abs\cdot$ 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 $\BR$ , 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 $\Abs{x_n} = r_n$ . Now consider $I_n=\{x\in A, \Abs x\leq r_n\}$ . Then $$ I_1\subset\cdots\subset I_n\subset I_{n+1}\subset\cdot $$ 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.