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Let $X$ be a spectral topological space, $U\subset X$ be a quasi-compact open subspace. Is there necessarily some scheme structure on $X$ (we do not require it to be affine) such that $U$ endowed with the restriction of the structure sheaf is an affine scheme?

I think I can answer this for $U=X$ (a spectral space is the underlying space of an affine scheme; depending on your definitions, that is either trivial or a theorem of Hochster).

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    $\begingroup$ I think that this question is equivalent then to asking if a qc open subspace of a spectral space is quasiseparated by Hochster's theorem. I think it's true, but I'd need to work it out. $\endgroup$ – Harry Gindi May 17 at 13:00
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    $\begingroup$ Yes, it is actually separated and qc, so this follows by Hochster's characterization as the qcqs sober spaces with topology generated by qc opens. Separation follows from: stacks.math.columbia.edu/tag/01P5 $\endgroup$ – Harry Gindi May 17 at 17:08
  • $\begingroup$ @HarryGindi I do not understand this. Sure, you can put an affine scheme structure on $U$. Why would it extend to $X$? If I am missing something obvious, well, sorry, I am bad with this stuff. $\endgroup$ – user138661 May 17 at 17:26
  • $\begingroup$ Oh, I misread the question. No, it's not true then. You can equip U with the restriction of the structure sheaf to be non-affine, then take the ring of global sections and U will be homeomorphic to the spec of the ring of global sections. That is, $U\cong \operatorname{Spec}(\mathcal{O}_X(U))$ as topological spaces (not as schemes!!!) This is because the underlying space of any qcqs scheme is homeo to spec of the ring of global sections. $\endgroup$ – Harry Gindi May 17 at 21:50
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    $\begingroup$ @HarryGindi what is not true then? Could you kindly give an explicit example where there is no scheme structure on $X$ (affine or non-affine), whose restriction to $U$ would define an affine scheme structure? $\endgroup$ – user138661 May 18 at 5:18

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