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For a commutative ring $R$ with 1, if $\mathrm{Spec} (R)$ is the set of all prime ideals of $R$, we can define the Zariski Topology on $\mathrm{Spec} (R)$ as follows:
1) The sets of the form $$V(E)=\{P \in \mathrm{Spec}(R)|E \subseteq P\}$$ are closed subsets.

2) The sets of the form $$D(E)=\{P \in \mathrm{Spec}(R) | E \not\subseteq P \},$$ are open subsets.

Let $rad(I)$ denote the radical of an ideal $I$ of $R$; if $I$ and $J$ are two ideals of $R$, the relation $V(I) \subseteq V(J)$ is equivalent to $rad(J)\subseteq rad(I)$. Is there any equivalence conditions for $V(I) \subseteq D(J)$? \ And is there any equivalence conditions for $D(I) \subseteq V(J)$?

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    $\begingroup$ $V(I)\subseteq D(J)\iff V(I)\cap V(J)=\emptyset\iff I+J=R$. $\endgroup$ – user26857 Dec 27 '16 at 8:24

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