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Let $A_1$ and $A_2$ be two ideals, and $P_1$ and $P_2$ be two prime ideals in a commutative ring $R$. Assume that $A_1 ∩ A_2 ⊆ P_1 ∩ P_2$. Is there at least an $i$ and $j$ such that $A_i ⊆ P_j$ is true?

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closed as off-topic by Davide Giraudo, Magdiragdag, M Turgeon, colormegone, apnorton May 1 '14 at 0:17

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  • $\begingroup$ I wonder what is unclear with this post that makes it to be put on hold as off-topic. I'm just trying to be as concrete an concise as possible. $\endgroup$ – Joakim von Anka May 1 '14 at 8:12
  • $\begingroup$ Dear @JoakimvonAnka : Welcome to math.SE! Along with a clearly stated question, it is usually expected that the poster includes things that they have tried so far, or thoughts they've had about how to begin. Using this information, users can better help you find a solution that is appropriate for you. And also, this helps convince readers that they are not doing all the work for you. In a nutshell, people enjoy helping, not serving :) Writing in your partial work helps them feel like they are helping. Regards $\endgroup$ – rschwieb May 1 '14 at 13:47
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    $\begingroup$ You can even say that $A_1\subseteq P_1$ or $A_2\subseteq P_1$. Since $P_1\cap P_2\subseteq P_1$, your hypothesis implies $A_1\cap A_2\subseteq P_1$ and, in turn, this implies $A_1A_2\subseteq P_1$. Now, if $P$ is a prime ideal, $I$ and $J$ are ideals and $IJ\subseteq P$, then …. In other words, $P_2$ is just a distraction. $\endgroup$ – egreg May 10 '14 at 8:41
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Yes.

Suppose $A_1 \not\subseteq P_1$ and $A_2 \not\subseteq P_1$, then $\exists x_i \in A_i$ such that $x_i \not\in P_1$. So $x_1x_2 \not\in P_1$, but $x_1x_2 \in A_1A_2 \subseteq A_1 \cap A_2$. Contradiction.

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