TITLE: If the intersection of ideals $I_{1},\ldots,I_{n}$ is contained in a prime ideal $P$, then one of them is contained in $P$
QUESTION [8 upvotes]: Let $A$ be a commutative ring and $I_{1},\ldots, I_{n}$ and $P$ ideals in $A$ with $P$ prime so that $\cap_{i=1} ^{n} I_{i} \subset P $. Show that there's an $i_0 \in \{1,...,n \}$ so that $I_{i_0} \subset P$.

My first idea was to try to do induction on $n$. The case $n=1$ is trivial of course, but then I couldn't go farther since removing one of the ideals to consider the intersection of $n-1$ ideals means I can't guarantee that the intersection is inside $P$.
I'm trying to prove the straightforward way now but can't really accomplish much. I tried assuming that every ideal has an element that's not in the rest and trying to work from there but again, seems to not get me anywhere. I'm just not sure how to approach the problem.
Any help would be greatly appreciated.

REPLY [4 votes]: Suppose that $I_i \not\subseteq P$ for all $i$. Then there exist $x_i \in I_i$; $ x_i \notin  P$ $(1 \leq i \leq n)$, and
therefore $\prod x_i \in \prod I_i\subset \bigcap I_i$; but $\prod x_i\notin P$ ($P$ is prime). Hence $\bigcap I_i$ is not a subset of $P$.