# Making a quasi-compact open into an affine open

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).

• 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. – Harry Gindi May 17 at 13:00
• 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 – Harry Gindi May 17 at 17:08
• @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. – user138661 May 17 at 17:26
• 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. – Harry Gindi May 17 at 21:50
• @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? – user138661 May 18 at 5:18