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<record version="7" id="5801">
 <title>Bezout domain</title>
 <name>BezoutDomain</name>
 <created>2004-04-23 18:50:40</created>
 <modified>2005-01-28 00:28:02</modified>
 <type>Definition</type>
 <creator id="3771" name="CWoo"/>
 <author id="3771" name="CWoo"/>
 <classification>
	<category scheme="msc" code="13G05"/>
 </classification>
 <defines>
	<concept>Bezout identity</concept>
 </defines>
 <synonyms>
	<synonym concept="Bezout domain" alias="B\'ezout domain"/>
 </synonyms>
 <related>
	<object name="GcdDomain"/>
	<object name="DivisibilityByProduct"/>
 </related>
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 <content>A \emph{Bezout domain} $D$ is an integral domain such that every finitely generated ideal of $D$ is \PMlinkname{principal}{PID}.

\textbf{Remarks}.  
\begin{itemize}
\item A PID is obviously a Bezout domain.  
\item Furthermore, a Bezout domain is a gcd domain.  To see this, suppose $D$ is a Bezout domain with $a,b\in D$.  By definition, there is a $d\in D$ such that $(d)=(a,b)$, the ideal generated by $a$ and $b$.  So $a\in (d)$ and $b\in (d)$ and therefore, $d\mid a$ and $d\mid b$.  Next, suppose $c\in D$ and that $c\mid a$ and $c\mid b$.  Then both $a,b\in (c)$ and so $d\in (c)$.  This means that $c\mid d$ and we are done.
\item From the discussion above, we see in a Bezout domain $D$, a greatest common divisor exists for every pair of elements.  Furthermore, if $\operatorname{gcd}(a,b)$ denotes one such greatest common divisor between $a,b\in D$, then for some $r,s\in D$:
$$\operatorname{gcd}(a,b)=ra+sb.$$
The above equation is known as the \emph{Bezout identity}, or Bezout's Lemma.
\end{itemize}</content>
</record>
