Euclidean valuation

Let $D$ be an integral domain. A Euclidean valuation is a function from the nonzero elements of $D$ to the nonnegative integers $\nu\colon D\setminus\{0_{D}\}\to\{x\in\mathbb{Z}:x\geq 0\}$ such that the following hold:

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For any $a,b\in D$ with $b\neq 0_{D}$ , there exist $q,r\in D$ such that $a=bq+r$ with $\nu(r)<\nu(b)$ or $r=0_{D}$ .
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For any $a,b\in D\setminus\{0_{D}\}$ , we have $\nu(a)\leq\nu(ab)$ .

Euclidean valuations are important because they let us define greatest common divisors and use Euclid’s algorithm. Some facts about Euclidean valuations include:

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The minimal (http://planetmath.org/MinimalElement) value of $\nu$ is $\nu(1_{D})$ . That is, $\nu(1_{D})\leq\nu(a)$ for any $a\in D\setminus\{0_{D}\}$ .
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$u\in D$ is a unit if and only if $\nu(u)=\nu(1_{D})$ .
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For any $a\in D\setminus\{0_{D}\}$ and any unit $u$ of $D$ , we have $\nu(a)=\nu(au)$ .

These facts can be proven as follows:

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If $a\in D\setminus\{0_{D}\}$ , then

$\nu(1_{D})\leq\nu(1_{D}\cdot a)=\nu(a).$
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If $u\in D$ is a unit, then let $v\in D$ be its inverse (http://planetmath.org/MultiplicativeInverse). Thus,

$\nu(1_{D})\leq\nu(u)\leq\nu(uv)=\nu(1_{D}).$

Conversely, if $\nu(u)=\nu(1_{D})$ , then there exist $q,r\in D$ with $\nu(r)<\nu(u)=\nu(1_{D})$ or $r=0_{D}$ such that

$1_{D}=qu+r.$

Since $\nu(r)<\nu(1_{D})$ is impossible, we must have $r=0_{D}$ . Hence, $q$ is the inverse of $u$ .
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Let $v\in D$ be the inverse of $u$ . Then

$\nu(a)\leq\nu(au)\leq\nu(auv)=\nu(a).$

Note that an integral domain is a Euclidean domain if and only if it has a Euclidean valuation.

Below are some examples of Euclidean domains and their Euclidean valuations:

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Any field $F$ is a Euclidean domain under the Euclidean valuation $\nu(a)=0$ for all $a\in F\setminus\{0_{F}\}$ .
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$\mathbb{Z}$ is a Euclidean domain with absolute value acting as its Euclidean valuation.
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If $F$ is a field, then $F[x]$ , the ring of polynomials over $F$ , is a Euclidean domain with degree acting as its Euclidean valuation: If $n$ is a nonnegative integer and $a_{0},\dots,a_{n}\in F$ with $a_{n}\neq 0_{F}$ , then

$\nu\left(\sum_{j=0}^{n}a_{j}x^{j}\right)=n.$

Due to the fact that the ring of polynomials over any field is always a Euclidean domain with degree acting as its Euclidean valuation, some refer to a Euclidean valuation as a degree function. This is done, for example, in Joseph J. Rotman’s .

Title	Euclidean valuation
Canonical name	EuclideanValuation
Date of creation	2013-03-22 12:40:45
Last modified on	2013-03-22 12:40:45
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	15
Author	Wkbj79 (1863)
Entry type	Definition
Classification	msc 13F07
Synonym	degree function
Related topic	PID
Related topic	UFD
Related topic	Ring
Related topic	IntegralDomain
Related topic	EuclideanRing
Related topic	ProofThatAnEuclideanDomainIsAPID
Related topic	DedekindHasseValuation
Related topic	EuclideanNumberField