# 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 ProofThatAEuclideanDomainIsAPID 2013-03-22 12:43:11 2013-03-22 12:43:11 rm50 (10146) rm50 (10146) 7 rm50 (10146) Result msc 13F07 PID UFD IntegralDomain EuclideanValuation