# valuation ring of a field

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

###### Proposition 1.

1. 1.

$A=_{df}\{x\in K\ \mid\ \left\lvert x\right\rvert\leq 1\}$ is a ring, called the valuation ring of $(K,\left\lvert\cdot\right\rvert)$,

2. 2.

$\mathfrak{m}=_{df}\{x\in K\ \mid\ \left\lvert x\right\rvert<1\}$ is the unique maximal ideal of $A$, and ${A}^{*}=\{x\in K\ \mid\ \left\lvert x\right\rvert=1\}$,

3. 3.

$K$ is the fraction field of $A$.

###### Proof.

For (1), note that $1\in A$, that $x,y\in A\Rightarrow\left\lvert x\right\rvert\leq 1,\left\lvert y\right\rvert% \leq 1\Rightarrow\left\lvert xy\right\rvert\leq 1\Rightarrow xy\in A$, and that $x,y\in A\Rightarrow\left\lvert x-y\right\rvert\leq\max(\left\lvert x\right% \rvert,\left\lvert-y\right\rvert)\leq 1\Rightarrow x-y\in A$.

For (2), it is obvious that $\mathfrak{m}+\mathfrak{m}\subset\mathfrak{m}$ and that $\mathfrak{m}A\subset\mathfrak{m}$ so that $\mathfrak{m}$ is an ideal. Clearly $A-\mathfrak{m}=\{x\in K\ \mid\ \left\lvert x\right\rvert=1\}$ 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 $\left\lvert x\right\rvert<1$ (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 $\left\lvert y\right\rvert\cdot\left\lvert x\right\rvert^{n}<1$, so that $yx^{n}\in A$ and thus

 $\frac{yx^{n}}{x^{n}}=y$

is in the fraction field of $A$. ∎

We say that the absolute value $\left\lvert\cdot\right\rvert$ is discrete if $\left\lvert{K}^{*}\right\rvert$ is a discrete subgroup of $\mathbb{R}_{>0}$. Note that $\mathbb{R}_{>0}\cong(\mathbb{R},+)$ via $\log$, so discrete subgroups are isomorphic to $\mathbb{Z}$ (are a lattice in $\mathbb{R}$), and thus a discrete absolute value is of the form $\left\lvert{K}^{*}\right\rvert=\alpha^{\mathbb{Z}}$ for some $\alpha\geq 1$, and $\alpha=1$ corresponds to the trivial absolute value.

###### Proposition 2.

In the notation of the preceding theorem, TFAE:

1. 1.

$A$ is principal

2. 2.

$\left\lvert\cdot\right\rvert$ is discrete

3. 3.

$A$

If any of these hold, $A$ is a discrete valuation ring (DVR).

###### Proof.

($1\Rightarrow 2$): If $A$ is principal, then $\mathfrak{m}=(\pi)$ with $\left\lvert\pi\right\rvert<1$. Since $A$ is a UFD, any element $x\in A-\{0\}$ can be written uniquely as $x=u\pi^{n}$ for $u\in{A}^{*},n\geq 0$, and then $\left\lvert x\right\rvert=\left\lvert u\right\rvert\cdot\left\lvert\pi\right% \rvert^{n}=\left\lvert\pi\right\rvert^{n}$. Thus $\left\lvert A-\{0\}\right\rvert=\left\lvert\pi\right\rvert^{\mathbb{N}}$ and $\left\lvert{K}^{*}\right\rvert=\left\lvert\pi\right\rvert^{\mathbb{Z}}$ so that $\left\lvert\cdot\right\rvert$ is discrete.

($2\Rightarrow 1$): If the absolute value is discrete, we may choose $\pi\in{K}^{*}$ with $\left\lvert\pi\right\rvert<1$ but with the largest possible absolute value strictly less than $1$. Then for $x\in\mathfrak{m}$, we have $\left\lvert x\right\rvert<1$, so $\left\lvert x\right\rvert\leq\left\lvert\pi\right\rvert$ and thus $\displaystyle\left\lvert\frac{x}{\pi}\right\rvert\leq 1$ 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 $\left\lvert\cdot\right\rvert$ 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 $\left\lvert x_{n}\right\rvert=r_{n}$. Now consider $I_{n}=\{x\in A,\left\lvert x\right\rvert\leq r_{n}\}$. Then

 $I_{1}\subset\cdots\subset I_{n}\subset I_{n+1}\subset\cdots$

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 2013-03-22 19:03:25 Last modified on 2013-03-22 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