An ideal $s$ of an Euclidean ring $R$ is maximal if and only if $s$ is generated by some prime element of $R$.
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
Try using the fact that Euclidean domains are principal ideal domains, and that maximal ideals are prime ideals in any ring. Note further that in a principal ideal domain you cannot have a chain of prime ideals $(0)\subset P\subset P'$ (prove this by contradiction).
HINT $\ $ In a Euclidean domain a nonzero ideal $\rm\:I\:$ is generated by any element $\rm\:i\:$ of minimal Euclidean value (else $\rm\ i\nmid j\in I\ \Rightarrow\ j\ mod\ i\ =\ j-r\ i\:,\:$ some $\rm\:r\in R\:,\:$ is an elt of $\rm\:I\:$ of smaller value). Furthermore, $\rm\:I = (i)\:$ prime $\rm\iff\ i\:$ prime $\rm\iff\ i\:$ irreducible by Euclid's Lemma.