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Let $R$ be a ring which is not a PID. The set $S$ of all ideals of $R$ which are not principal has a maximal element with respect to the inclusion (Zorn's Lemma), I already proved this! Now, I need to show that if $P$ is a maximal element of $S$, then $P$ is prime.

I tried by contradiction: suppose $a,b \not\in P$, but $ab \in P$. So $P+(a)$ and $P+(b)$ are not in $S$, and therefore they are principal ideals, say $P+(a) = (x)$ and $P+(b) = (y)$. Now, I just need to show that $P$ is principal. Any hint will be helpful.

Thus, I will conclude that for any ring, if all prime ideals are principal, then any ideal is principal.

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    $\begingroup$ I believe $S$ is the set of ideals which are NOT principal. $\endgroup$
    – Tim Raczkowski
    Commented Jan 23, 2015 at 2:39
  • $\begingroup$ Yes! $S$ is the set of ideals which are NOT principal. $\endgroup$
    – Gauss
    Commented Jan 23, 2015 at 2:41
  • $\begingroup$ Showing $P$ is principal would give a contradiction. Another approach: If you can show $ab+P$ is not principal it violates the maximality of $P$. $\endgroup$
    – Tim Raczkowski
    Commented Jan 23, 2015 at 2:42
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    $\begingroup$ There seems to be a confusion here, since what you want to see is that if a ring has nonprincipal ideals it has nonprincipal prime ideals, so you want to show your maximal element, which I'll denote by $\mathfrak p$, is prime. It is nonprincipal by definition, since it is a maximal element of a collection of nonprincipal ideals. $\endgroup$
    – Pedro ♦
    Commented Jan 23, 2015 at 2:42
  • $\begingroup$ I think that $ab + P = P$, since $ab \in P$. $\endgroup$
    – Gauss
    Commented Jan 23, 2015 at 2:44

1 Answer 1

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See this link (includes proofs). Two steps:

  1. The family of principal ideals of a ring is an Oka family.

  2. If $\cal F$ is an Oka family of ideals, then any maximal element of the complement of $\cal F$ is prime.

In ring $R$, a set of ideals $\cal F$ is an Oka family if $R\in \cal F$ and whenever $I$ is an ideal such that $(I:a)\in \cal F$ and $(I,a)\in \cal F$ for some $a\in R$, then $I \in \cal F$.

Pedro Tamaroff-suggested other good link is here. Also the ones pointed out by Bill Dubuque in an MSE answer (for another standard Oka family, ideals that do not intersect a given multiplicative subset in the ring) here.

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    $\begingroup$ A good source is this one, too. $\endgroup$
    – Pedro ♦
    Commented Jan 23, 2015 at 2:51
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    $\begingroup$ See also the links here. $\endgroup$
    – Bill Dubuque
    Commented Jan 23, 2015 at 3:49
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    $\begingroup$ @ir7 This solved my problem completely. Thanks! $\endgroup$
    – Gauss
    Commented Jan 23, 2015 at 13:31
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    $\begingroup$ For those who are wondering (like myself) $(I:a) = \{ r\in R : ra\in I \}$ $\endgroup$
    – Myself
    Commented Jan 24, 2015 at 14:34

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