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

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

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

Proof.

For (1), note that $1\in A$, that $x,y\in A\Rightarrow \left|x\right|\le 1,\left|y\right|\le 1\Rightarrow \left|xy\right|\le 1\Rightarrow xy\in A$, and that $x,y\in A\Rightarrow \left|x-y\right|\le \max (\left|x\right|,\left|-y\right|)\le 1\Rightarrow x-y\in A$.

For (2), it is obvious that $𝔪+𝔪\subset 𝔪$ and that $𝔪A\subset 𝔪$ so that $𝔪$ is an ideal. Clearly $A-𝔪=\{x\in K\mid \left|x\right|=1\}$ 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 ${ℝ}_{>0}$. Note that ${ℝ}_{>0}\cong (ℝ,+)$ via $\log$, so discrete subgroups are isomorphic to $\mathbb{Z}$ (are a lattice in $ℝ$), 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 $𝔪=(\pi )$ 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 $\left|x\right|=\left|u\right|\cdot {\left|\pi \right|}^n={\left|\pi \right|}^n$. Thus $\left|A-\{0\}\right|={\left|\pi \right|}^{\mathbb{N}}$ 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 $x\in 𝔪$, we have , so $\left|x\right|\le \left|\pi \right|$ and thus $\left|{\displaystyle \frac{x}{\pi }}\right|\le 1$ 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 $|\cdot |$ 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 $ℝ$, 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 $I_n=\{x\in A,\left|x\right|\le r_n\}$. 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	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