Suppose that $R$ is a non-commutative Noetherian ring. Recall that an ideal $P$ of $R$ is prime if for all ideals $I,J$ of $R$, $IJ\subseteq P$ implies that $I\subseteq P$ or $J\subseteq P$. Also, $P$ is completely prime if for all $a,b\in R$, $ab\in R$ implies that $a\in R$ or $b\in R$.
Now, take $\{P_i:i\in I\}$ to be a set of prime ideals of $R$, and suppose that the intersection $J:=\cap_{i\in I}{P_i}$ is a completely prime ideal of $R$.
Is it true that $J=P_i$ for some $i\in I$?
This holds if $I$ is a finite set, because if $\{P_1,...,P_r\}$ is a finite set of ideals and $J=P_1\cap...\cap P_r$ is completely prime, then if $J\neq P_1$, taking $a\in P_1/J$, $b\in P_2\cap...\cap P_r$, $ab\in J$, so $b\in J$, and hence $J=P_2\cap...\cap P_r$. So applying induction gives the result, and in fact, this argument doesn't require that the ideals $P_i$ are prime.
Can anyone think of an argument showing that this works if $I$ is infinite, or a counterexample otherwise?
