Let $A$ be a commutative ring with identity. Let $p_{i}, i\in I$ and $p$ be prime ideals in $A$, where the index set $I$ is infinite. If we have $$ p\supset \bigcap_{I}p_{i} $$ Do we still have $p\supset p_{i}$ for some $i\in I$ like the finite index case? In the finite case, assume contrary we pick up elements $x_{i}\in p_{i}\cap p^{c}$ we would have $\prod_{i\in I'}p_{i}\in \bigcap_{I'}p_{i}\in p$, and the proof is immediate. But it is not clear to me what happens in the above case. I now suspect it would not hold in general, but I do not have a quick 5-line proof of this or an enlightening counter-example to myself. Since this is simple, I do not know if this is really up to research level. I did googled around and checked some references, but it seems it is not covered anywhere. The obvious case is when $\bigcap p_{i}=0$ (as pointed out by Qiaochu). I am wondering if there is any other non-trivial cases.
It is not very clear to me how to show if $\bigcap p_{i}=0$, then the closure of the set $p_{i}$ is the whole space $Spec A$. So even in this case I have some trouble.
Motivation: This showed up in a step when I attempt to show $V(I(S))\subset \overline{S}$ when $S$ has infinite cardinality. Perhaps my approach was not right and this is simply wrong in general.
No. Take, for example, $A = \mathbb{Z}, p_i$ the prime ideals generated by the primes greater than $3$, and $p = (2)$.