= Solution
A <Euclidean domain> is an <integral domain> $R$ with a <Euclidean function> $\nu:R\setminus\{0\}\to\mathbb Z_{\geq0}$ such that, for $a,b\in R$ with $b\ne0$, there exist $q,r\in R$ satisfying
$$
a=qb+r,\qquad r=0\ \text{or}\ \nu(r)<\nu(b).
$$
The <polynomial ring> $\mathbb Q[x]$ has this property with $\nu(f)=\deg f$. 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 \b[<every Euclidean domain is a principal ideal domain>].
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