It is well known that in a PID an ideal is maximal iff it is generated by an irreducible element. My query is 'is the result true only in a PID?'. My question is what will happen if the PID condition is removed. That is I want to know if in an ID an ideal is maximal then is it necessary that it is not generated by an irreducible element? i.e if an ID is given then will we assume that no Maximal ideal can be generated by an irreducible element? And what about the converse ? i.e if an ideal is generated by an irreducible element in an ID,then will we assume that it is not maximal? I am thinking that in an ID the results are not necessarily false,and my reason behind it is that 'maximal ideal $\Rightarrow$ it is generated by an irreducible element and the converse of it' doesn't characterize an PID. I am not confident as I am unable to find any such example. Sorry for the lengthy question. Hope to get cleared. Thank you

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    $\begingroup$ "Assume the ideal is not maximal" and "Not assume the ideal is maximal" are two very different things. $\endgroup$ – Arthur May 4 '17 at 5:25
  • $\begingroup$ Sorry sir I didn't get you.Will please explain.? $\endgroup$ – hiren_garai May 4 '17 at 5:26
  • $\begingroup$ How about being more objective in your question? $\endgroup$ – M. Wolf May 4 '17 at 5:49
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    $\begingroup$ In the noetherian case, if any maximal ideal of an integral domain is generated by one element (we dont even need to assume that it is irreducible), then it is a PID. In the non-noetherian case, there are counter examples, since principal ideals can have height $2$ in that scenario, see here: math.stackexchange.com/questions/183199/… $\endgroup$ – MooS May 4 '17 at 6:05

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