Let $R$ be a commutative ring with $1 \neq 0$, $I$ and $P$ are ideals of $R$. If $P$ is prime and $I \cap P \neq 0$, does it follows that either $I \subseteq P$ or $I$ is also a prime ideal incomparable to $P$? Does it also extend over the case where $P$ is maximal?

  • $\begingroup$ If $P$ is a prime but not maximal and $M$ is a maximal ideal which contains $P$, then $M\cap P\neq 0$ but both conditions does not hold. $\endgroup$ – Hanul Jeon Nov 9 '14 at 3:35
  • $\begingroup$ @tetori what about the case where $M$ is not a maximal ideal? $\endgroup$ – Narzie Nov 9 '14 at 3:46

The ring $\mathbb{Z}$ already has counterexamples - for instance consider $P=(2)$ and $I=(15)$, we have $I\cap P=(30)$ and $P$ is maximal, but $I\not\subseteq P$ and $I$ is not a prime ideal.

  • $\begingroup$ what's the part that makes it a counterexample? $\endgroup$ – Narzie Nov 9 '14 at 3:43
  • $\begingroup$ I made a typo, I meant to say $I\not\subseteq P$. Therefore it satisfies neither of the two possible conditions you specified. $\endgroup$ – curious Nov 9 '14 at 3:44
  • $\begingroup$ awwwww.. thanks a lot @curious! $\endgroup$ – Narzie Nov 9 '14 at 3:51

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