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An ideal $s$ of an Euclidean ring $R$ is maximal if and only if $s$ is generated by some prime element of $R$.

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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).

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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.

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