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

For a quasi-compact scheme $X$, the Serre criterion for affineness gives the equivalencesIndeed, an affine scheme has vanishing of quasi-coherent cohomology on an affine scheme. Conversely, vanishing of the first sheaf cohomology for every quasi-coherent sheaf includes every quasi-coherent sheaf of ideals, which suffices for the criterion. No separated scheme hypothesis is needed in this formulation. Nor may one test only the structure sheaf.^[ref]

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.

Let $U=\mathrm{Spec}A$ and $V=\mathrm{Spec}B$ be affine open subschemes of the separated scheme $X$. Its diagonal morphism is a closed immersionRestricting this closed immersion to the open subscheme $U{\times }_{\mathrm{Spec}\mathbb{Z}}V$ identifies its inverse image with $U\cap V$. Henceis a closed immersion. A closed subscheme of an affine scheme is an affine scheme, soThe argument iterates to every finite intersection of affine open subschemes. The empty intersection of two disjoint charts causes no exception: the empty scheme is $\mathrm{Spec}(0)$.

Suppose $X=U_1\cup \cdots \cup U_n$ with each $U_i$ an affine open subscheme. Every affine scheme is a quasi-compact topological space: a cover by principal open subsets $D(f_{\lambda })$ means that the $f_{\lambda }$ generate the unit ideal; an expression $1={\sum }_{j=1}^r{a}_j{f}_{{\lambda }_j}$ gives a finite subcover. Every open cover has a refinement by principal open subsets, so this proves the general covering property.

Restrict an arbitrary open cover of $X$ to each $U_i$ and choose a finite subcover there. The union of these finitely many finite families covers $X$. Thus $X$ is a quasi-compact scheme.

Now assume $X$ is a separated scheme. By (c), every finite intersection of the $U_i$ is an affine scheme, and hence has vanishing positive sheaf cohomology for the structure sheaf. This affine open cover therefore computes sheaf cohomology using its Čech cochain complex. Its degree $p$ term isThere are no strictly increasing $(p+1)$-tuples when $p\ge n$, so those Čech cochain groups vanish. ConsequentlyThe same argument works for every quasi-coherent sheaf. The relevant comparison requires the acyclicity of the finite intersections; the bare assertion that arbitrary Čech cohomology equals sheaf cohomology would be too broad.

Interpretation of the removed locus. Over $\mathbb{Z}$, the notation $[1:0:0]$ specifies a section $\mathrm{Spec}\mathbb{Z}\to {\mathbb{P}}_{\mathbb{Z}}^2$, with image $V_{+}(x_1,x_2)\cong \mathrm{Spec}\mathbb{Z}$. We remove this entire closed subscheme. This is the interpretation forced by the asserted open cover $X=U_1\cup U_2$. Removing one closed point in a particular fiber would not give that open cover. The native PDF uses the same notation as the transcription.

Put $a=x_0/x_1$, $b=x_2/x_1$ on $U_1$. On $U_2$ put $c=x_0/x_2$, $d=x_1/x_2$. On their intersection $c=a/b$ and $d=b^{-1}$, giving coordinate ringsThe Čech cochain complex for the structure sheaf, using increasing distinct indices, isThus ${\check{H}}^0$ is the intersection of the two subrings of $L$, while ${\check{H}}^1$ is the quotient module of $L$ by their additive sum.

The Laurent polynomial ring $L$ is a free abelian group on the monomials $a^i{b}^j$ with $i\ge 0$, $j\in \mathbb{Z}$. The first subring has the basis $j\ge 0$. The second has the basisso its allowed exponents satisfy $j\le -i$. A monomial belongs to both precisely when $i=j=0$. The monomials belonging to neither satisfy $i\ge 2$ and $-i<j<0$. Since the subgroups are spanned by subsets of the same basis, their sum introduces no other relations. ThereforeHere the square brackets denote cohomology classes, and the direct sums are as abelian groups. In particular, $[a^2{b}^{-1}]=[x_0^2/(x_1{x}_2)]$ is a nonzero cohomology class. This scheme is a separated scheme, being an open subscheme of projective space over $\mathbb{Z}$, and is a quasi-compact scheme by (d). Both charts and their intersection are affine schemes, so this Čech cohomology equals sheaf cohomology. The nonzero first sheaf cohomology of the quasi-coherent structure sheaf rules out an affine scheme:

A Noetherian scheme is a quasi-compact scheme, so choose a finite affine open cover $X={\bigcup }_{i=1}^s\mathrm{Spec}{A}_i$ with all $A_i$ Noetherian rings. The nilradical sheaf restricts on this chart to the quasi-coherent sheaf associated to $\mathrm{Nil}(A_i)$: the nilradical commutes with localization.

By the allowed fact, choose $d_i\ge 1$ such that $\mathrm{Nil}(A_i{)}^{d_i}=0$ and put $m={\max }_i{d}_i$. On every chart, the ideal sheaf ${\mathcal{N}}^m$ is zero, and hence it is zero globally. Higher powers are also zero. ThusIf $X$ is empty, take $m=1$. The finite affine open cover is essential for choosing one bound valid everywhere.

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.

Let $\mathcal{F}$ be any quasi-coherent sheaf on $X$, and choose $m$ from (i). Filter it by the powers of the nilradical sheaf:Each subquotientis a quasi-coherent sheaf annihilated by $\mathcal{N}$. Apply (ii) to the quasi-coherent sheaf ${\mathcal{N}}^j\mathcal{F}$. Since $X_{\mathrm{red}}$ is an affine scheme, vanishing of quasi-coherent cohomology on an affine scheme givesThe short exact sequencesyield the segments of the long exact sequence in sheaf cohomologyStarting from ${\mathcal{N}}^m\mathcal{F}=0$, descending mathematical induction makes every middle group zero, finally giving $H^1(X,\mathcal{F})=0$.

This holds for every quasi-coherent sheaf $\mathcal{F}$. Since a Noetherian scheme is a quasi-compact scheme, the Serre criterion for affineness in (a) applies and provesThis is an instance of the general fact that affineness is invariant under nilpotent thickening.