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Background

The following is taken from Hungerford's Abstract algebra an Introduction and F Jarvis' Algebraic Number Theory

Exercise 1 Prove that an ideal $(p)$ in a PID (principal ideal domain) is maximal if and only if $p$ is irreducible.

Lemma Let $R$ be an integral domain, and let $p\in R$. if $(p)$ is maximal, then $p$ is irreducible.

Exercise 2 If $R$ is a principal ideal domain, show that the converse of the above lemma is also true.

Question

I often see an exercise showing that an ideal is mxiamal iff it is irreducible in a principal ideal domain in the form expressed in Exercise 1. I found in an algebraic number theory text, that the statement of Exercise 1, in one direction, it seems that in one direction of the statement, it holds true for a ring to be an integral domain, while in another direction, it need the stronger requirement to be a principal ideal domain. What I want to know is why there is this difference.

Thank you in advance

Seth
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  • An element $p$ is irreducible if and only if $(p)$ is maximal among principal ideals. – Arturo Magidin May 30 '24 at 17:43
  • Simply because $(p)$ is max among all principal ideals is not equivalent to $(p)$ is max among all ideals, i.e. direction $(\Leftarrow)$ in the first $(!!\iff!!)$ in the standard proof in the dupe may fail in a non-PID. Generally you can find counterexamples by going through the proof line-by-line to find which step first fails. – Bill Dubuque May 30 '24 at 17:43
  • @BillDubuque so is it better to express the statement ih both directions assuming the ring is a principal ideal domain then? – Seth May 30 '24 at 17:47
  • Doing so allows us to see how the proof follows immediately from the fundamental (bidrectional) equivalence that "contains = divides" for principal ideals. – Bill Dubuque May 30 '24 at 17:54
  • @BillDubuque sorry I don't understand what you mean by "immediately from the fundamental (bidrectional) equivalence that "contains = divides" for principal ideals" – Seth May 30 '24 at 18:11
  • I mean that it makes obvious the linked standard proof. – Bill Dubuque May 30 '24 at 18:17

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