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Let $R$ be a commutative ring and let $S \subset \mathrm{Spec}(R)$ be a subset. Suppose that for each $P \in S$ there exists a Zariski open neighborhood $U$ of $P_P \in \mathrm{Spec}(R_P)$ such that $^aj_{P}(U) \subset S$, where $j_{P}:R \rightarrow R_{P}$ is the canonical morphism; so $S$ can be covered by images of Zariski open subsets under flat morphisms. Is it true that $S$ is open in the flat topology on $\mathrm{Spec}(R)$? If not, is there any useful topology on $\mathrm{Spec}(R)$ where $S$ is open?

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The flat "topology" is not a topology in the classical sense... – Zhen Lin Feb 11 '13 at 18:13
A delicate aspect. I know that as a Grothendieck topology I would have to consider jointly surjective flat morphisms (+ some properties perhaps). But in "Going-down implies generalized going-down" by Dobbs--Hetzel (Lemma 2.1) there is a topology (in the "classical" sense) on Spec(R) defined (but this should have been considered before), which I thought is also referred to as the "flat topology". I clearly don't know how this is connected to the flat topology in the Grothendieck sense. – Georg S. Feb 11 '13 at 19:47
The only Zariski neighborhood of the closed point in a local scheme $X$ is $X$. So, $U=\mathrm{Spec\,}(R_P)$ and your condition just means that $S$ is stable under generalization. – Laurent Moret-Bailly Feb 11 '13 at 20:25
@Laurent: Oh, that's a good point. Thank you! – Georg S. Feb 11 '13 at 20:29
I should probably stop before I upset some people, but seriously, the term is "generization." Look in Hartshorne or any textbook that talks about Zariski spaces or sober spaces or even PlanetMath: – Matt Feb 12 '13 at 3:31

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