proof that a Euclidean domain is a PID

Let $D$ be a Euclidean domain, and let $\mathfrak{a}\subseteq D$ be a nonzero ideal. We show that $\mathfrak{a}$ is principal. Let

A=\{\nu(x):x\in\mathfrak{a},x\neq 0\}

be the set of Euclidean valuations of the non-zero elements of $\mathfrak{a}$ . Since $A$ is a non-empty set of non-negative integers, it has a minimum $m$ . Choose $d\in\mathfrak{a}$ such that $\nu(d)=m$ . Claim that $\mathfrak{a}=(d)$ . Clearly $(d)\subseteq\mathfrak{a}$ . To see the reverse inclusion, choose $x\in\mathfrak{a}$ . Since $D$ is a Euclidean domain, there exist elements $y,r\in D$ such that

x=yd+r

with $\nu(r)<\nu(d)$ or $r=0$ . Since $r\in\mathfrak{a}$ and $\nu(d)$ is minimal in $A$ , we must have $r=0$ . Thus $d\lvert x$ and $x\in(d)$ .

Title	proof that a Euclidean domain is a PID
Canonical name	ProofThatAEuclideanDomainIsAPID
Date of creation	2013-03-22 12:43:11
Last modified on	2013-03-22 12:43:11
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	7
Author	rm50 (10146)
Entry type	Result
Classification	msc 13F07
Related topic	PID
Related topic	UFD
Related topic	IntegralDomain
Related topic	EuclideanValuation