= Solution
A <Euclidean domain> is an <integral domain> $R$ with a <function> $d:R\setminus\{0\}\to\mathbb N_0$ such that, for $a\in R$ and nonzero $b\in R$, there exist $q,r\in R$ with $a=qb+r$ and either $r=0$ or $d(r)<d(b)$. This is the Euclidean-ring convention used here.
If an <ideal> $I$ is zero, it is generated by zero. Otherwise choose nonzero $a\in I$ with $d(a)$ minimal. For any $b\in I$, Euclidean division gives $b=qa+r$, with $r\in I$. A nonzero remainder would have $d(r)<d(a)$, contradicting the choice. Hence $r=0$ and $b\in aR$. The reverse inclusion follows from $a\in I$, proving
$$
\boxed{I=aR.}
$$
Now suppose $a$ is an <irreducible element>. Its <principal ideal> is proper because $a$ is not a unit. If $aR\subseteq J\subseteq R$, write $J=bR$ by the result just proved. Then $a=bc$ for some $c\in R$. Irreducibility gives either $b$ a unit, so $J=R$, or $c$ a unit, so $bR=aR$. There is no intermediate proper <ideal>, and therefore \b[$aR$ is a <maximal ideal>].
For a <field> $F$, the <polynomial ring> $F[X]$ is an <integral domain> and the <polynomial division algorithm> gives $A=QB+R$ with $R=0$ or $\deg R<\deg B$. Thus the <polynomial degree> is a Euclidean function, proving \b[$F[X]$ is a <Euclidean domain>].
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