# An ideal and its J-radical

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 $B$?

• 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 '18 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 $$N$$.