Difference between revisions of "Euclidean domain"

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 $R$ is a Euclidean domain if:

It is an integral domain.
There a function $N:R\setminus\{0\}\to \mathbb Z_{\ge0}$ called a Norm such that for all nonzero $a,b\in R$ there are $q,r\in R$ such that $a = bq+r$ and either $N(r)<N(b)$ or $r=0$ .

Some common examples of Euclidean domains are:

The ring of integers $\mathbb Z$ with norm given by $N(a) = |a|$ .
The ring of Gaussian integers $\mathbb Z[i]$ with norm given by $N(a+bi) = a^2+b^2$ .
The ring of polynomials $F[x]$ over any field $F$ with norm given by $N(p) = \deg p$ .

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

@@ Line 9: / Line 9: @@
 * 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>.
+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==

Page

Toolbox

Search

Difference between revisions of "Euclidean domain"

Latest revision as of 14:28, 22 August 2009

See also