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Let $A$ be some ring. It is well known that if $A$ is an integral domain, then an element $x$ is irreducible iff the principal ideal $xA$ is maximal in the set of principal ideals of $A$.

Questions: is it true that

1) if $A$ is a factorial ring and $I$ an ideal of $A$, if it is maximal then it is generated by irreducible elements,

2) if $A$ is Noetherian (and an integral domain?), and $I$ an ideal of $A$, if it is maximal then it is generated by a finite number of irreducible elements.

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More generally, if $A\,$ is a ring whose nonunits $\neq 0\,$ factor into irreducibles then every prime ideal $\,P\ne 0\,$ in $A\,$ can be generated by irreducibles. If not, by Zorn the set of ideals $\,\subseteq P\,$ generable by irreducibles contains some $I$ maximal (in subset order). Since $\,I\neq P\,$ there is some $\,a\in P\backslash I.\,$ By hypothesis $\, a = p_1\cdots p_n\,$ for irreducible $\,p_i,\,$ so $\,a\in P\,$ prime $\,\Rightarrow\,$ some $\,p_i\in P.\,$ Further $\,p_i\not\in I\,$ (else $\,a\in I),\,$ therefore $\,I\subsetneq I+(p_i)\subseteq P,\,$ contra maximality of $\,I.$

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EDIT: This answer was to the original question, which has since changed.

Consider $\mathbb{Z}[x]$ which is a Noetherian UFD. The ideal $(x)$ is generated by an irreducible element but is not maximal.

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  • $\begingroup$ Ah I'm dumb, this is not what I wanted to say. I wanted to say "If, then", certainly not iff. Thanks for the counter example tho $\endgroup$ – sure Mar 20 '15 at 18:40

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