Let $R$ be a commutative ring with $1$ and $I$ be an ideal of $R$. Now let $J=\cap_{I\subseteq m\in Max(R)}m$. Set $A:=\{p\in Spec(R): I\subseteq p\}$ and $B:=\{p\in Spec(R): J\subseteq p\}$, where $Spec(R)$ and $Max(R)$ are the set of all prime and maximal ideals of $R$ respectively. When considering $Spec(R)$ with the Zariski topology, and $B$ and $A$ as two subspaces of $Spec(R)$.I want if $B$ is dense in $A$? Or in general, what kinds of topological properties hold between $A$ and $B$?

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    With this degree of generality, essentially nothing: take for $R$ a local ring with maximal ideal $\mathfrak{m}$, and $I=0$. Then $A=\operatorname{Spec}(R) $ and $B=\{\mathfrak{m}\} $. – abx Sep 13 at 17:39

This is true if and only if $R/I$ is a Jacobson ring (i.e., the intersection $J=J(R/I)$ of maximal ideals equals the intersection $N=N(R/I)$ of prime ideals).

We can suppose that $I=0$. So you asking when $V(J)=\mathrm{Spec}(R/J)$, viewed as the set of prime ideals of $R$ containing $J$, is dense in $\mathrm{Spec}(R)$. Since $V(J)$ is closed, this just means to ask whether $V(J)=\mathrm{Spec}(R)$. If it is the case, $J$ is contained in every prime ideal, and hence $J=N$. Conversely, if $V(J)\neq\mathrm{Spec}(R)$, there exists a prime ideal $P$ such that $J$ is not contained in $P$; hence $J$ is not contained in $N$.

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