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

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For any $a,b\in D$ with $b\ne {0}_{D}$, there exist $q,r\in D$ such that $a=bq+r$ with $$ or $r={0}_{D}$.

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For any $a,b\in D\setminus \{{0}_{D}\}$, we have $\nu (a)\le \nu (ab)$.
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, $\nu ({1}_{D})\le \nu (a)$ 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
$$\nu ({1}_{D})\le \nu ({1}_{D}\cdot a)=\nu (a).$$ 
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If $u\in D$ is a unit, then let $v\in D$ be its inverse^{} (http://planetmath.org/MultiplicativeInverse). Thus,
$$\nu ({1}_{D})\le \nu (u)\le \nu (uv)=\nu ({1}_{D}).$$ Conversely, if $\nu (u)=\nu ({1}_{D})$, then there exist $q,r\in D$ with $$ or $r={0}_{D}$ such that
$${1}_{D}=qu+r.$$ Since $$ 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
$$\nu (a)\le \nu (au)\le \nu (auv)=\nu (a).$$
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 ${a}_{0},\mathrm{\dots},{a}_{n}\in F$ with ${a}_{n}\ne {0}_{F}$, 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  20130322 12:40:45 
Last modified on  20130322 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 