Past exam of the rings and modules course of the University of Oxford / 2017 / 2 / a

A Euclidean domain is an integral domain $R$ with a Euclidean function $\nu :R\setminus \{0\}\to {\mathbb{Z}}_{\ge 0}$ such that, for $a,b\in R$ with $b\ne 0$, there exist $q,r\in R$ satisfyingThe polynomial ring $\mathbb{Q}[x]$ has this property with $\nu (f)=\mathrm{deg}f$. To perform polynomial division, subtract an appropriate multiple of the divisor to cancel the leading term of the current remainder. Its polynomial degree decreases strictly at each step, so the process terminates below the divisor's degree. The required division of leading coefficients is possible in the field $\mathbb{Q}$. Thus a polynomial ring over a field is Euclidean, with the same argument applying to every field.

Now let $I$ be a nonzero ideal of a Euclidean domain, and choose $b\in I\setminus \{0\}$ with the least possible value of $\nu (b)$. For every $a\in I$, divide $a=qb+r$. The remainder belongs to $I$, since $r=a-qb$. If it were nonzero, its smaller Euclidean value would contradict the choice of $b$. Thus $r=0$ and every $a\in I$ lies in $(b)$. Conversely $(b)\subseteq I$ because $I$ is an ideal. Hence $I=(b)$. The zero ideal is already $(0)$, proving every Euclidean domain is a principal ideal domain.