Past exam of the introduction to schemes course of the University of Oxford / 2023 / 3 / b / Solution

Suppose, to the contrary, that $H^1(X,\mathcal{F})=0$ for every quasi-coherent sheaf $\mathcal{F}$. Since $X$ is a quasi-compact scheme, part (a) makes it an affine scheme.

Every affine scheme $X=\mathrm{Spec}A$ is a separated scheme: its diagonal morphism over $\mathrm{Spec}\mathbb{Z}$ is induced by the surjective ring homomorphism $A{\otimes }_{\mathbb{Z}}A\to A$, $a\otimes b\mapsto ab$, so it is a closed immersion. This contradicts the assumption that $X$ is not a separated scheme. ThereforeIn fact, the version of the Serre criterion for affineness using ideals shows that $\mathcal{F}$ may be chosen to be a quasi-coherent sheaf of ideals. This argument does not use a comparison with Čech cohomology on the nonseparated scheme.