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?
closed as off-topic by Davide Giraudo, Magdiragdag, M Turgeon, colormegone, apnorton May 1 '14 at 0:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
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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.