Is an intersection of prime ideals equal to their product? Suppose $R$ is a commutative ring with identity, and let $P_1 , P_2, ..., P_n$ be distinct prime ideals in $R$.
Is it then true that
$$ P_1 P_2 \cdots P_n = P_1 \cap P_2 \cap \cdots \cap P_n?$$
My hunch is that this is not true.
The inclusion $P_1P_2\cdots P_n\subset P_1\cap P_2\cap\cdots \cap P_n$ certainly holds (as this holds for any ideals in $R$). Also, I know that the equality holds if the ideals are maximal. So, in pursuit of a counterexample, I've been looking at rings in which prime ideals are not maximal; and so, PIDs and finite rings are out. My first thought was, of course, $\mathbb{Z}$; but the equality seems to hold in that case.
Am I on a wild goose chase here?
 A: A counterexample is in the ring $R=\mathbb{R}[X,Y]$. Take $P_1=XR$ and $P_2 = XR+YR$. Then $P_1\cap P_2 = P_1$. And $X\in P_1$ is not an element of $P_1 P_2$.
Now let me prove something that does work: $\sqrt{P_1\dots P_n}=P_1\cap\dots \cap P_n$. And for that, the prime $P_i$'s do not need to be distinct. For the inclusion $\subset$, let $x\in R$ such as $x^k\in P_1\dots P_n$. As you said yourself, we have $x^k\in P_1\cap \dots\cap P_n$. And then $x\in P_1\cap \dots\cap P_n$ because each $P_i$ is prime. For the inclusion $\supset$, let $x\in P_1\cap\dots\cap P_n$. Now suppose for the sake of contradiction that for all $k\in\mathbb{N}, x^k\notin P_1\dots P_n$. By Krull's theorem, there exists a prime ideal $Q$ such as $P_1\dots P_n\subset Q$ and $x\notin Q$. Because $Q$ is prime we get $P_i\subset Q$ for a certain  $i\leq n$. That contradicts $x\in P_i$.
A: In this case $P_1 \cap P_2 = (x^2 + xy) \neq P_1$. It is always the case that for two ideals, $I$ and $J$ in a commutative ring $R$, $IJ \subseteq I \cap J$. You can see this as $IJ \subseteq I$ and $IJ \subseteq J$.
