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If $I$ is an ideal which is maximal among the ones that are not principal, then $I$ is prime.

This would mean that for all $f \in R$, $(f) \subset I$. Could I then use colon ideals? I was thinking maybe that for $P$ prime, $(P: (f) )= R$ if $f \in P$ or $(P: (f) )= P$ if $f \not\in P$.

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  • $\begingroup$ No, it wouldn't mean what you say it would...maximal doesn't mean that. $\endgroup$ – DonAntonio Mar 23 '14 at 12:36
  • $\begingroup$ All the possible principal ideals be (f) where f is any element of R, correct? Then would it be that if I is maximal, then $I \subset (f)$ implies $(f)=R$? $\endgroup$ – math1234567 Mar 23 '14 at 18:05
  • $\begingroup$ Yes @violin.lover, that last you wrote is true...if you actually meant $\;I\subset (f)\;,\;\;I\neq (f)\;$ . Still, my first comment remains... $\endgroup$ – DonAntonio Mar 23 '14 at 19:21
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Pretty much the same answer I gave to your other question (Maximal Ideal Must be Prime).

This time, you'll need to prove that $J = I + (f)$ is not principal if $J$ is not principal. The method of proof is pretty much the same.

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    $\begingroup$ "The method of proof is pretty much the same." If you embrace a large view, then this is true, but the proof in this case is a little more tricky. (-1) $\endgroup$ – user26857 Dec 24 '15 at 11:26
  • $\begingroup$ @chriseur What do you mean by J=I+(f) is not principal if J is not Principal $\endgroup$ – Jhon Doe Mar 2 '18 at 17:14

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