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

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.

Answer 1

Hint: If $I_k\not\subset P$ for all $k\in\{1,\ldots,n\}$, then there is at least one $r_k\in I_k\setminus P$ for all $k\in\{1,\ldots,n\}$. What can you say about the product $r_1\cdots r_n$?

Answer 2

The definition of an ideal P being prime is, given any ideals A and B such that AB $\subset P$, either A $\subset P$ or B $\subset P$. Thus, if $\bigcap_1^{n+1} I_i \subset P$, then $(\bigcap_1^n I_i)I_{n+1} \subset \bigcap_1^{n+1} I_i \subset P$ so that either $\bigcap_1^n I_i \subset P$ or $I_{n+1} \subset P$. If the former, use induction hypothesis. If the latter, done.

NOTE: For commutative rings, the definition for prime ideals given above is equivalent to the usual one: for any two elements $a,b \in R$, $ab \in P \Rightarrow a \in P$ or $b \in P$ iff, for any two ideals $A$ and $B$, $AB \subset P \Rightarrow A \subset P$ or $B \subset P$.

Answer 3

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 $P$.