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Let $R$ be a commutative ring with identity, and $Spec (R) $ the set of all prime ideals of $R $ with Zariski topology. How can we verify that $int (V (I))= Spec (R)-V (ann (I))$, where $I $ is an ideal of $R $, int= interior, ann= annihilator, and $V (A)=\{p\in Spec (R)|A\subseteq p\}$.

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This isn't true in general. For instance, let $R=k[x,y]/(x^2y)$ and $I=(x)$. Then $\operatorname{ann}(I)=(xy)$ consists of nilpotents, so $V(\operatorname{ann}(I))$ is all of $\operatorname{Spec}(R)$. But $V(I)$ has nonempty interior (it contains $D(y)$).

However, it is true if you assume $R$ is reduced. In that case, suppose $f\in R$ is such that $D(f)\subseteq V(I)$. This means that the image of $I$ in the localization $R_f$ consists of nilpotent elements. That is, for each $i\in I$, there exists an $n$ such that $f^ni^n=0$. Since $R$ is reduced, this means $fi=0$ for each $i\in I$, i.e. $f\in \operatorname{ann}(I)$. It is also easy to see that every step of this is reversible, so we have shown that $f\in\operatorname{ann}(I)$ iff $D(f)\subseteq V(I)$. But $f\in\operatorname{ann}(I)$ iff $D(f)\subseteq\operatorname{Spec}(R)-V(\operatorname{ann}(I))$. We conclude that the $V(I)$ and $\operatorname{Spec}(R)-V(\operatorname{ann}(I))$ have the same interior. Since $\operatorname{Spec}(R)-V(\operatorname{ann}(I))$ is open, this means it is equal to the interior of $V(I)$.

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  • $\begingroup$ Thank you very much for your answer and help Eric Wofsey. $\endgroup$
    – user296989
    Dec 24 '15 at 8:49

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