Let $X$ be a scheme. Is it true that the morphism $X\rightarrow \mathrm{Spec}\,\mathbb{Z}$ is quasi-separated iff the intersection of two quasi-compact open subspaces of the underlying space of $X$ is quasi-compact (so in particular, quasi-separatedness is a purely topological condition)? If the answer is positive, what is a published reference where this is proved in detail?
2 Answers
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Yes. A reference is EGA IV, première partie, Proposition (1.2.7), b'), page 228.
answered May 16, 2019 at 10:56
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$\begingroup$ It seems to be an error in the statement of (1.2.7), c), see this. $\endgroup$ Commented Jun 15, 2023 at 7:59
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$\begingroup$ Thanks for the info. I guess that it has remained unnoticed because it works for an affine cover which is used most of the time. But it is good to have it sharp. $\endgroup$ Commented Jun 15, 2023 at 8:07
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$\begingroup$ it seems that the condition of quasi-separatedness is only ever formulated for schemes, even though it's a purely topological condition ... or maybe there's another word for it in point-set topology? $\endgroup$ Commented Mar 1 at 22:54
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$\begingroup$ See Kempf: "Some elementary proofs of basic theorems in the cohomology of quasi-coherent sheaves" Rocky Mountain J. Math. Volume 10, Number 3 (1980), 637-646. (projecteuclid.org/euclid.rmjm/1250128841). In this paper the underlying spaces of quasi-compact and quasi-separated schemes are called quasi-noetherian. A space is quasi-noetherian if it possesses a basis (closed under finite intersection) of quasi-compact open subsets and is itself quasi-compact. I much prefer the terminology concentrated. $\endgroup$ Commented Mar 4 at 12:01
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Another reference is Tag 01KO in the Stacks project (note that when $S$ is affine the hypothesis that the two opens map into a common affine is empty).
Of course, the condition that the intersection of any two affine is quasi-compact, is equivalent to the condition that the intersection of any two quasi-compact open subsets is quasi-compact.
answered May 16, 2019 at 11:20