 Solution

For the zero ideal, choose $d=0$. Otherwise choose a nonzero $d\in I$ for which the Euclidean function $\delta (d)$ is minimal; the well-ordering principle guarantees such a choice. For any $a\in I$, perform Euclidean division:Because $I$ is an ideal, $r=a-qd\in I$. A nonzero remainder would contradict the minimal choice of $d$, so $r=0$ and $a\in Rd$. Thus $I\subseteq Rd$. Conversely, $d\in I$ implies $Rd\subseteq I$, provingEvery ideal is therefore a principal ideal; in particular, a Euclidean domain is a principal ideal domain.