When an intersection is contained in a minimal prime ideal For a commutative ring $R$ with identity, it is well known that if a finite intersection of ideals is contained in a prime ideal $\frak{p}$, then one of them is contained in $\frak{p}$. I am looking for an equivalent condition on $R$ under which if an arbitrary intersection of ideal is contained in a minimal prime ideal $\frak{p}$, then one of them is contained in $\frak{p}$. Or is there any research paper related to that?
 A: I would extend Jason's comment to say that this condition trivially holds in any Artinian ring since in those rings every intersection of ideals is a finite intersection. It seems that indeed this is the only case. 
Let $R$ be an arbitrary noetherian ring and $\mathfrak m$ an arbitrary maximal ideal in $R$. If $R$ is not an integral domain, let $\mathfrak p\subset R$ be a minimal prime ideal contained in $\mathfrak m$ and let $A:=R/\mathfrak p$.
Clearly $A$ is an integral domain and $\overline{\mathfrak m}:=\mathfrak m/\mathfrak p$ is a maximal ideal in $A$. By the Krull intersection theorem $\cap_n \overline{\mathfrak m}^n=0$. 
It follows that then $\cap_n\mathfrak m^n\subseteq \mathfrak p$ in $R$. If $R$ satisfied the desired condition, then this would imply that  $\mathfrak m^n\subseteq \mathfrak p$ for some fixed $n\in \mathbb N$. But then $\mathfrak m= \mathfrak p$ is both maximal and minimal. This is true for every maximal ideal, so $\dim R=0$ and hence $R$ is Artinian. 
I leave it for you to decide about the non-noetherian case. 
A: Let us say that an ideal $I\leq R$ has property $(\ast)$, by way of definition, if whenever an intersection of ideals is contained in $I$ then one of the ideals in the intersection is contained in $I$.
Lemma:  Let $R$ be a commutative ring with $1.$  Let $P\leq R$ be a prime ideal.  If $P$ satisfies $(\ast)$ then $P$ is a maximal ideal.
Proof.  Assume $P$ satisfies $(\ast)$.  After replacing $R$ by $R/P,$ we may as well assume $P=0.$
Let $L$ be the intersection of all the non-zero ideals of $R.$  Since $(0)$ satisfies $(\ast),$ we see that $L\neq (0).$  Such an ideal is called a little ideal.  It exists if and only if a ring is subdirectly irreducible.  In particular, since $R$ is a domain it must be a field (by Proposition 12.4 in Lam's "A First Course in  Noncommutative Rings").  QED
So, a ring satisfying your condition must always have Krull dimension $0$.  I'm sure your condition has been studied in the literature, but I'm not sufficiently familiar with the commutative algebra literature to find it quickly.
