# Difference between revisions of "Euclidean domain"

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* The ring of [[Gaussian integers]] <math>\mathbb Z[i]</math> with norm given by <math>N(a+bi) = a^2+b^2</math>. | * The ring of [[Gaussian integers]] <math>\mathbb Z[i]</math> with norm given by <math>N(a+bi) = a^2+b^2</math>. | ||

* The [[polynomial ring|ring of polynomials]] <math>F[x]</math> over any [[field]] <math>F</math> with norm given by <math>N(p) = \deg p</math>. | * The [[polynomial ring|ring of polynomials]] <math>F[x]</math> over any [[field]] <math>F</math> with norm given by <math>N(p) = \deg p</math>. | ||

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+ | It can be easily shown through [[infinite descent]] that any Euclidian domain is also a [[principal ideal domain]]. Indeed, let <math>I</math> be any [[ideal]] of a Euclidean domain <math>R</math> and take some <math>a\in I</math> with minimal norm. We claim that <math>I=(a)</math>. Clearly <math>(a)\subseteq I</math>, because <math>I</math> is an ideal. Now assume <math>(a)\ne I</math> and consider any <math>b\in I\setminus (a)</math>. Applying the division algorithm we get that there are <math>q,r\in R</math> such that <math>b=aq+r</math> with <math>N(r)<N(a)</math> (we cannot have <math>r=0</math> as <math>b\not\in (a)</math>). But now as <math>I</math> is an ideal, and <math>a,b\in I</math>, we must have <math>r=b-aq\in I</math>, contradicting the minimality of <math>a</math>. Hence <math>I=(a)</math> and <math>R</math> is indeed a principle ideal domain. | ||

==See also== | ==See also== |

## Latest revision as of 14:28, 22 August 2009

A **Euclidean domain** (or **Euclidean ring**) is a type of ring in which the Euclidean algorithm can be used.

Formally we say that a ring is a Euclidean domain if:

- It is an integral domain.
- There a function called a
**Norm**such that for all nonzero there are such that and either or .

Some common examples of Euclidean domains are:

- The ring of integers with norm given by .
- The ring of Gaussian integers with norm given by .
- The ring of polynomials over any field with norm given by .

It can be easily shown through infinite descent that any Euclidian domain is also a principal ideal domain. Indeed, let be any ideal of a Euclidean domain and take some with minimal norm. We claim that . Clearly , because is an ideal. Now assume and consider any . Applying the division algorithm we get that there are such that with (we cannot have as ). But now as is an ideal, and , we must have , contradicting the minimality of . Hence and is indeed a principle ideal domain.

## See also

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