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I have an inclusion of closed subsets $V(J) \subset V(I)$ in an affine scheme $Spec(R)$ with the property that $V(I) = V(J) \cup \partial V(I)$. I would like to conclude that $V(J)=V(I)$. (Here $\partial$ denotes the boundary of a set.) Note that the interior of $V(I)$ is contained in $V(J)$ and hence the closure of the interior of $V(I)$ is contained in $V(J)$. But recall from basic topology that the closure of the interior of a closed set need not be the closed set. (The closed set could have empty interior for example.)

The most important special case for me is when $V(J) = \emptyset$. In this case I have a closed subset $V(I)$ of an affine scheme with the property that $V(I) = \partial V(I)$ and I would like to claim that $V(I) = \emptyset$.

Is there any hope for me? I'm willing to assume that the ring $R$ is Noetherian if that will help.

Does an affine scheme always have non-empty interior? Is $V(I) = \partial V(I)$ only possible if $V(I)$ is empty?

I may be grasping at phantoms but if there is something about the topology of affine schemes that could make this work I would be elated. Thanks for your attention.

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Your question "Does an affine scheme always have non-empty interior?" has a trivial answer: every non-empty topological space $T$ has non-empty interior, namely $T$ ! –  Georges Elencwajg Jan 15 '13 at 10:21

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up vote 4 down vote accepted

You should beware that the intuition for topology obtained from calculus, say, is completely inadequate for the Zariski topology.
For example if $R$ is a domain (noetherian or not) the affine scheme $X=Spec(R)$ is irreducible and any non-empty open subset $U\subset X$ is dense.
This implies that any closed subset $V(I)\subsetneq X$ has empty interior and thus that $\partial V(I)=V(I)$, which is exactly the opposite of what you thought, namely that $\partial V(I)=V(I)$ would imply that $V(I)=\emptyset$.
To put it bluntly, the concept of the boundary of a subset is essentially useless for a scheme endowed with its Zariski toplogy.

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