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Let $X$ be a scheme which contains a closed point and also assume that for every two closed subsets $Y_1$ and $Y_2$ of $X$, we have either $Y_1 \subseteq Y_2$ or $Y_2 \subseteq Y_1$. Also assume that $X$ has finite Krull dimension. Then, is it true that $X$ is affine ?

My thoughts: Since the closed subsets of $X$ are comparable, so $X$ has exactly one closed point, say $x \in X$ and also every non-empty closed subset contains $x$ . Hence for any abelian sheaf $\mathcal F$ on $X$, we have $\mathcal F_x =\mathcal F(X)$. Since taking stalks of sheaves is exact, we get $H^j(X, \mathcal F)=0, \forall j>0$ for any abelian sheaf $\mathcal F$ on $X$. However , since I'm not assuming the scheme to be quasi-compact , I cannot quite apply Serre's criteria now.

Please help.

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The assumption of finite Krull dimension is unnecessary. Let $x\in X$ be a closed point and note that for any $y\in X$, we must have $\{x\}\subseteq \overline{\{y\}}$. This means that any open set containing $x$ must contain $y$, so the only open set containing $x$ is all of $X$. Since there is some affine open set containing $x$, this means $X$ itself is affine.

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  • $\begingroup$ This is super slick. Beautiful! $\endgroup$ – Alex Wertheim Jul 14 '19 at 21:14

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