If the intersections of ideals equal to a prime ideal, then at least one of the ideal equals to the prime ideal

Question

Let A be a commutative ring, and Let $I_{1}$, $I_{2}$, ... , $I_{n}$ be ideals in A.

How to show that if $I_{1}$ $\cap$ $I_{2}$ $\cap$ ... $\cap$ $I_{n}$ = $P$ for some prime ideal $P$, then $I_{i}$ = $P$ for at least one of the $I_{i}$ ?

Answer 1

This easily follows from the $n=2$ case.

If $I\cap J=P$ but $I\ne P$, $J\ne P$, consider $a\in I\setminus P$, $b\in J\setminus P$. Prove that $ab\in P$ and also $ab\notin P$.

Answer 2

Use the fact that for two ideals $I$ and $J$, we have $IJ \subset I \cap J$.

Suppose $P = I \cap J$ and that $P$ is strictly contained in $I$ or $J$. Then we have non zero elements $\bar{a} \in I/P$ and $\bar{b}\in J/P$, and we can see that know $ab \in IJ \subseteq P$.

But then $a \in P$ or $b \in P$, so this contradicts that our elements $\bar{a}$ and $\bar{b}$ were nonzero. This generalizes obviously for an $n$-fold intersection.