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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?

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    $\begingroup$ For the types of commutative rings that arise most often in algebraic geometry, this will almost never be true. I guess that it is trivially true for an Artinian local ring, but that is the only "natural" example that I can see. $\endgroup$ – Jason Starr Jan 3 '17 at 14:30
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    $\begingroup$ Just some food for thought, imagine I'm working on ${\mathbb A}^n_{\mathbb C}$ and I choose some curve. Consider the maximal ideals $\mathfrak{m}$ corresponding to the closed points of that curve. Their intersection is the prime ideal $P$ corresponding to the curve itself, but obviously no $\mathfrak{m}$ is contained in $P$. $\endgroup$ – Karl Schwede Jan 4 '17 at 17:28
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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.

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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.

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