# Closure of an open set in the Zariski topology

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.
Is there any description for the closure of an open set $D(E)$ in the Zariski topology?

1) An arbitrary subset $Y\subset X=\operatorname {Spec}(R)$ has as closure $$\overline Y=V(\bigcap_{y\in Y}\mathfrak p_y)=\{\mathfrak p\in X\vert \mathfrak p\supset \bigcap_{y\in Y}\mathfrak p_y\}$$ For example if $R=\mathbb C[T]$ the closure in $X=\mathbb A^1_\mathbb C=\operatorname {Spec}(\mathbb C[T])$ of $Y=\{\mathfrak p _z=\langle T-z\rangle \vert z\in \mathbb Z\}$ is $\overline Y=X$, because $\bigcap_{z\in Y}\{\mathfrak p_z\vert z\in \mathbb Z\}=\langle 0\rangle$.
2) If by chance $X$ is irreducible, then every non-empty open subset $Y\subset X$ is dense so that $\overline Y=X.$ This applies to $D(E)$ as soon as $E\subsetneq \operatorname {Nil}(R)$ (i.e as soon as $E$ doesn't consist exclusively of nilpotent elements).
• Thanks you very mach your help Georges Elencwajg. But I want to know two facts \\a) What is the meaning of $\frak{p}_y$ b) Is there any verification for your claim? – K.Z Dec 18 '16 at 14:40
• Proof: Bosch page 206. The notation $\mathfrak p_y\subset R$ means the prime ideal corresponding to $y\in X$. – Georges Elencwajg Dec 18 '16 at 15:15