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

Let $i:X_{\mathrm{red}}\hookrightarrow X$ be the reduced induced subscheme, with structure sheaf ${\mathcal{O}}_{X_{\mathrm{red}}}={\mathcal{O}}_X/\mathcal{N}$. Its underlying map of topological spaces is a homeomorphism, since every prime ideal contains the nilradical.

Put $\mathcal{G}=\mathcal{F}/\mathcal{N}\mathcal{F}$. Multiplication by $\mathcal{N}$ annihilates $\mathcal{G}$, so its action by ${\mathcal{O}}_X$ factors through ${\mathcal{O}}_X/\mathcal{N}$. Thus $\mathcal{G}$ is naturally a sheaf of modules over ${\mathcal{O}}_{X_{\mathrm{red}}}$. On an affine open subscheme $\mathrm{Spec}A$, with $\mathcal{F}=\tilde{M}$, it corresponds to $M/\mathrm{Nil}(A)M$ over $A/\mathrm{Nil}(A)$, so it remains a quasi-coherent sheaf.

The sheaf cohomology of a sheaf of modules is the sheaf cohomology of its underlying sheaf of abelian groups. Under the above homeomorphism, that underlying sheaf is identical on the two spaces: the global sections functor and its flasque resolutions are identical. Equivalently, use sheaf cohomology under a closed inclusion with $i_{\ast }\mathcal{G}$. Therefore there are canonical isomorphisms, in every degree,In particular, this gives the requested equality in degree one. Equality of the underlying topological spaces alone would not compare arbitrary coefficient sheaves; the factorization of the action explains why the indicated coefficients are the same.