Let $A$ be a ring and $I$ an ideal of $A$. Consider the set $\text{Min}(I)$ of minimal prime ideals over $I$. This set may be infinite.
Is it possible to have a prime $\mathfrak{p}\in \text{Min}(I)$ such that $\displaystyle \bigcap_{\mathfrak{q} \in \text{Min}(I) \\ \ \ \ \ \mathfrak{q}\neq \mathfrak{p}} \mathfrak{q} \subseteq \mathfrak{p}$ ?
Certainly, if the set $\text{Min}(I)$ is finite (for instance when $A$ is Noetherian) this is not true, since otherwise we can consider the product $\displaystyle \prod_{\mathfrak{q} \in \text{Min}(I) \\ \ \ \ \ \mathfrak{q}\neq \mathfrak{p}} \mathfrak{q} \subseteq \bigcap_{\mathfrak{q} \in \text{Min}(I) \\ \ \ \ \mathfrak{q}\neq \mathfrak{p}} \mathfrak{q} \subseteq \mathfrak{p}$ and hence there has to be some minimal prime $\mathfrak{q}$ such that $\mathfrak{q} \subseteq \mathfrak{p}$, which is a contradiction.
What can we say in the case when the set of minimal primes over $I$ is infinite?